NASA Astronautic Structural Manual Volume 1

S

NASA TECHNICAL

MEMORANDUM

NASA TM X- 73305

ASTRONAUTIC STRUCTURESMANUAL

VOLUME I

(NASA-T_-X-733C5) AS_EONAUTIC STRUCTURES

MANUAL, VOLUME I (NASA) 8_6 p

Structures and Propulsion Laboratory

N76-76166

Unclas

_/98 _a_05

August 197 5

NASA

George C.

Marshall

MSFC - For_" JlgO (l_ev June 1971)

Marshall 5pace Flight Center

Space Flight Center, Alabama

APPROVAL

ASTRONAUTIC STRUCTURES MANUAL

VOLUME I

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NASA TM X-73305 I4 TITLE AND SUBTITLE

ASTRONAUTIC STRUCTURES MANUAL

VOLUME I

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George C. Marshall Space Flight Center

Marshall Space Flight Center, Alabama 35812

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Prepared by Structures and Propulsion Laboratory, Science and Engineering

_ IG, ABSTRACT

This document (Volumes I, II, and III) presents a compilation of industry-wide methods in

aerospace strength analysis that can be carried out by hand, that are general enough in scope to

cover most structures encountered, and that are sophisticated enough to give accurate estimates

of the actual strength expected. It provides analysis techniques for the elastic and inelastic stres_

ranges. It serves not only as a catalog of methods not usually available, but also as a reference

source for the background of the methods themselves.

An overview of the manual is as follows: Section A is a general introduction of methods

used and includes sections on loads, combined stresses, and interaction curves; Section B is

devoted to methods of strength analysis; Section C is devoted to the topic of structural stability;

Section D is on thermal stresses; Section E is on fatigue and fracture mechanics; Section F is

on composites; Section G is on rotating machinery; and Section H is on statistics.

These three volumes supersede NASA TM X-60041 and NASA TM X-60042o

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STRUCTURES MANUAL

FOREWORD

f-This manual is issued to the personnel of the Strength Analysis

Branch to provide uniform methods of structural analysis and to pro-

vide a ready reference for data. Generally, the information contained

in this manual is a condensation of material published by universities,

scientific journals, missile and aircraft industries, text book pub-

lishers, and government agencies.

Illustrative problems to clarify either the method of analysis or

the use of the curves and tables are included wherever they are con-

sidered necessary. Limitations of the procedures and the range of

applicability of the data are indicated wherever possible.

It is recognized that all subjects in the Table of Contents are not

present in the body of the manual; some sections remain to be devel-

oped in the future. However, an alphabetical index of content material

is provided and is updated as new material is added. New topics not

listed in the Table of Contents will be treated as the demand arises.

This arrangement has been utilized to make a completed section avail-

able as soon as possible. In addition, revisions and supplements are

to be incorporated as they become necessary.

Many of the methods included have been adapted for computerized

utilization. These programs are written in Fortran Language for utili-

zation on the MSFC Executive VIII, Univac 1108, or IBM 7094 and are

cataloged with example problems in the Structural Analysis Computer

Utilization Manual.

It is requested that any comments concerning this manual be

directed to:

Chief, Structural Requirements Section

Strength Analysis Branch

Analytical Mechanics Division

Astronautics Laboratory

National Aeronautics and Space Administration

Marshall Space Flight Center, Alabama 35812

ii

August 15, 1970

SECTIONAI

STRESSAND STRAIN

f

At.0.0

TABLE OF CONTENTS

Page

Stress and Strain ................................. 1

i.I.0

I.I.

1.1.

I.I.

1.1.

i.i.

1.2.0

1.3.0

1.3.

1.3.

1.3.

1.3.

1.3.

1.3.

1.3.

1.4.0

1.4.

1.4.

Mechanical Properties of Materials ................... 1

1 Stress-Strain Diagram ......................... i

2 Other Material Properties ....................... 3

3 Strain-Time Diagram .......................... 5

4 Temperature Effects ........................... 7

5 Hardness Conversion Tables ..................... 12

Elementary Theory of the Mechanics of Materials ......... 17

Elementary Applications of the Theory of Elasticity ........ 18

1 Notations for Forces and Stresses .................. i8

2 Specification of Stress at a Point ................... 19

3 Equations of Equilibrium ........................ 21

4 Distribution of Strains in a Body ................... 23

5 Conditions of Compatibility ...................... 25

6 Stress Functions ............................. 27

7 Use of Equations from the Theory of Elasticity ......... 28

Theories o¢ Failure ............................. 34

i Elastic Failure .............................. 35

2 Interaction Curves ............................ 36

A1 iii

Section A I

March i, 1965

Page I

AI. 0.0 Stress and Strain

The relationship between stress and strain and other material properties,

which are used throughout this manual, are presented in this section. A brief

introduction to the theory of elasticity for elementary applications is also pre-sented in this section.

AI. I.0 Mechanical Properties of Materials

A brief account of the important mechanical properties of materials is given

in this subsection; a more detailed discussion may be found in any one of a num-

ber of well known texts on the subject. The numerical values of the various

mechanical properties of most aerospace materials are given in MIL-HDBK-5

(reference 1). Many of these values are obtained from a plotted set of test re-

sults of one type or another. One of the most common sets of these plotted sets

is the stress-strain diagram. A typical stress-strain diagram is discussed in

the next subsection.

AI. I.i Stress-Strain Diagram

Some of the more useful properties of materials are obtained from a stress-

strain diagram. A typical stress-strain curve for aerospace metals is shown in

Figure AI. I.i-I.

The curve in Figure A1.1.1-1 is composed of two regions; the straight line

portion up to the proportional limit where the stress varies linearly with strain,

and the remaining part where the stress is not proportional to strain. In this

manual, stresses below the ultimate tensile stress (Ftu) are considered to be

elastic. However, a correction (or plasticity reduction) factor is sometimes

employed in certain types of analysis for stresses above the proportional limitstress.

Commonly used properties shown on a stress-strain curve are described

briefly in the following paragraphs:

E Modulus of elasticity; average .ratio of stress to strain for

stresses below the proportional limit. In Figure A1.1.1-1E = tan 0

A1. I.1 Stress-Strain Diagram (Cont'd)

Section A 1

March 1, i965

Page 2

_--Elastic, _ Plastic, ep

|eeFtu

(pfsi) _'_ _/ _eld Point I I

e u e Fracture-4e (inches/inch)

Figure AI. 1.1-1 A Typical Stress-Strain Diagram

ES

E t

Secant modulus; ratio of stress to strain above

the proportional limit; reduces to E in the pro-

portional range. In Figure AI. t. 1-1 E s =

tan 01

Tangent modulus; slope of the stress-strain curve

at any point; reduces to E in the proportionaldf

range. In FigureAl. l.i-i E t - de - tan 02

-f..Section A1

March 1, 1965

Page 3

AI. i.I Stress-Strain Diagram (Cont'd)

Fry or Fcy Tensile or compressive yield stress; since many

materials do not exhibit a definite yield point,

the yield stress is determined by the . 2% offsetmethod. This entails the construction of a

straight line with a slope E passing through a

point of zero stress and a strain of. 002 in./in.

The intersection of the stress-strain curve and

the constructed straight line defines the magni-

tude of the yield stress.

Ftp or Fcp Proportional limit stress in tension or compres-

sion; the stress at which the stress ceases to

vary linearly with strain.

Ftu Ultimate tensile stress; the maximum stress

reached in tensile tests of standard specimens.

FCU Ultimate compressive stress; taken as Ftu un-

less governed by instability.

EU The strain corresponding to Ftu.

Ee

Elastic strain; see Figure AI. I.I-i.

E

Pplastic strain; see Figure Ai. I.i-i.

efracture (% elongation) Fracture strain; percent elongation in a pre-

determined gage length associated with tensile

failures, and is a relative indication of ductility

of the material.

Ai. i.2 Other Material Properties

The definition of various other material properties and terminology used

in stress analysis work is given in this subsection.

SectionAiMarch i, 1965Page 4

A1.1.2 Other Material Properties (Cont'd)

Fbry' FbruYield and ultimate bearing stress; determined

in a manner similar to those for tension and

compression. A load-deformation curve is

plotted where the deformation is the change in

the hole diameter. Bearing yield (Fbry) is de-fined by an offset of 2% of the hole diameter;

bearing ultimate (Fbru} is the actual failingstress divided by 1. t5.

FSU

Fsp

Ultimate shear stress.

Proportional limit in shear; usually taken equal

to 0. 577 times the proportional limit in tensionfor ductile materials.

Poisson's ratio; the ratio of transverse strain

to axial strain in a tension or compression test.

For materials stressed in the elastic range, v

may be taken as a constant but for inelasticstrains v becomes a function of axial strain.

VP

EG-

2(I + v)

Isotropic

Anisotropic

Orthotropic

Plastic Poisson's ratio; unless otherwise stated,

Vp may be taken as 0.5.

Modulus of rigidity or shearing modulus of

elasticity for pure shear in isotropic materials.

Elastic properties are the same in all directions.

Elastic properties differ in different directions.

Distinct material properties in mutually per-

pendicular planes.

Section AiMarch I, 1965Page5

A1.1.3 Strain-Time Diagram

The behavior of a structural material is dependent on the duration of loading.

This behavior is exhibited with the aid of a strain-time diagram such as that

shown in Figure A1. i. 3-1. This diagram consists of regions that are dependent

Strain

Elastic fStrain

Constant

Loading

Loading

/,/Fracture///_J_ Creep Limit (no fracture) Curve

I I Elastic Recovery = Elastic Strain

P_rmanent Setv _ Strain Recovery _ Time

Unloading

Figure A1.1.3-t Strain-Time Diagram

upon the four loading conditions as indicated on the time coordinate. These

loading conditions are as follows:

1. Loading

2. Constant loading

SectionA1

March 1, 1965

Page 6

AI. 1.3 Strain-Time Diagram (Cont'd)

3. Unloading

4. Recovery (no load)

The interval of time when the load is held constant is usually measured in

weeks or months. Whereas the time involved in loading and unloading is rela-

tively short (usually seconds or minutes) such that the corresponding strain-

time curve can be represented by a straight vertical line.

The following discussion of the diagram will be confined to generalities due

to the complexity of the phenomena of creep and fracture. A more detailed dis-

cussion on this subject is presented in reference 5.

The condition referred to as "loading" represents the strain due to a load

which is applied over a short interval of time. This strain may vary from zero

to the strain at fracture (_fracture - See Figure AI. I.l-l) depending upon the

material and loading.

During the second loading condition, where the load is held constant, the

strain-time curve depends on the initial strain for a particular material. The

possible strain-time curves (Figure A1.1.3-1) that could result are discussedbelow.

a. In curve t, the initial strain is elastic and no additional strain is

experienced for the entire time interval. This curve typifies elastic action.

b. In curve 2, the initial strain increases for a short period after the

load becomes constant and then remains constant for the remainder of the period.

This action is indicative of slip which is characterized by a permanent set re-

sulting from the shifting (slip) of adjacent crystalline structures along planes

most favorably oriented with respect to the direction of the principal shearing

stress.

c. In curve 3, there is a continuous increase in strain after the initial

slip until a steady state condition is attained. This curve is indicative of creep

which is generally the result of a combined effect of the predominantly viscous

inelastic deformation within the unordered intercrystalline boundaries and the

complex deformations by slip and fragmentation of the ordered crystalline domains.

Section A 1

March I, 1965

Page 7

Ai. i.3 Strain-Time Diagram (Cont'd)

d. Curve 4 is also a combination of slip and creep. The only dif-

ference from curve 3 is that the creep action continues until the material fails

in fracture. This fracture may take place at any time during the constant load

period and is indicated by the upper shaded area in Figure A 1.1.3-1.

During unloading, the reduction in strain of curves l, 2 and 3 is equal to

the elastic strain incurred during loading. This reduction is referred to as the

"elastic recovery. " It can be seen in Figure Ai. I. 3-I that in the case of curve

I the structural member will return to its initial configuration immediately after

unloading. This is not the case for curves 2 and 3 as there will be some residual

strain.

The last condition to be discussed on the strain-time diagram concerns the

recovery period. In this period, some of the strain indicated as inelastic strain

is recoverable. This is true particularly for many viscoelastic materials (suchI

as flexible plastics) that do not show real creep, only delayed recoverable strains.

_f The height of the lower shaded area in Figure A 1.1.3-1 is called the elastic

after effect. The upper bound is the maximum possible permanent set and is in-

dicated by the solid horizontal line. The lower bound could be any one of the

family of possible strain-time curves confined within the lower shaded area.

The limiting curve of the lower bound would approach the permanent set curve

due to slip as indicated by the horizontal dashed line. If slip action is negligible,

this limiting curve would be represented by a line that approaches zero asymp-

totically with increasing time.

A1.1.4 Temperature Effects

The mechanical properties of a material are usually affected by its tem-

perature. This effect will be discussed in general terms in this section. For

specific information, see the applicable chapter in reference 1.

In general, temperatures below room temperature increase the strength

properties of metals. Ductility is usually decreased and the notch sensitivity

of the metal may become of primary importance. The opposite is generally

true for temperatures above room temperature.

A representative example for the effect of temperature on the mechanical

properties of aluminum alloys is given in Figures A1.1.4-1 through 4. Most

steels behave in a similar manner but generally are less sensitive to tempera-

ture magnitudes.

A1.1.4 Temperature Effects (Cont'd)

Section A I

March i, 1965

Page 8

Q)

2

cD

QJ

OO

¢J

;h

120

100

8O

60.

40 _

2O

0

\\\\\

\',,

/_ hr

/100 hr_10,000 hr

-400 -200

Figure AI. I.4-I

0 200 400 600 800

Temperature, °F

Effects of Temperature on the Ultimate Tensile

Strength (Ftu) of 7079 Aluminum Alloy (from

Ref. I)

f_

8

AI. 1.4

140

120

100

8O

>_ 60

,t0

20

Temperature Effects (Cont'd)

Section AI

March i, 1965

Page 9

\\

-_ hr

I 100 hr

_.10,000 hr

0-400 -200

Figure A1.1.4-2

0 200 400 600

Temperature, ° F

Effects of Temperature on the Tensile Yield

Strength (Fty) of 7079 Aluminum Alloy (fromRef. 1)

800

OOa_

At.i.4

140

Section A 1

March 1, 1965

Page t0

Temperature Effects (Cont'd)

120

100

80

I

\

60

2O

0-400 -200

Figure AI. I. 4-3

0 200 400 600 800

Temperature, ° F

Effect of Temperature on the Tensile and Compressive

Modulus (E and Ec} of 7079 Aluminum Alloy (fromRef. i}

Section A 1

March 1, 1965

Page 11

AI. 1.4 Temperature Effects (Cont'd)

100

8O

6O

O

O

40

2O

,

0

/

I00 200 300 400 5C 600

Figure AI. i. 4-4

Temperature, ° F

Effect of Temperature on the Elongation of

7079-T6 Aluminum Alloy (from Ref. 1)

Section A I

March 1, 1965

Page 12

Al. I. 5 Hardness Conversion Table

A table for converting hardness numbers to ultimate tensile strength values

is presented in this section. In this table, the ultimate strength values are in

the range, 50 to 304 ksi. The corresponding hardness number is given for each

of three hardness machines; namely, the Vtckers, Brinell and the applicable

scale(s) of the Rockwell machine.

This table is given In the remainder of this section. The appropriate

materials-property haaktbook should be consulted for additional information

whenever necessary.

Tensile

Strength

50

52

54

56

58

60

62

64

Vickers-

Firth

ksi

Diamond

Hardness

Number

104

108

112

116

120

125

129

135

Brinell

3000 kg10ram Stl

Ball

Hardness

Number

92

96

I00

104

108

I13

ii7

122

A Scale

6O kg

120 degDiamond

Cone

,&l

Rockwell

B Scale

100 kg

1/16 in.Dia Stl

Ball

58

61

64

66

68

70

72

74

C Scale

150 kg

120 deg

Diamond

Cone

mm

1B

_W

_m

Table AI_'I.5-1 Hardness Conversion Table

AI. I. 5 Hardness Conversion Table (Cont'd)

Section A I

March i, 1965

Page 13

Tensile

Strength

ksi

66

Vickers-

Firth

D Jam ond

Hardness

Num be r

139

Brinell

3000 kg

I0m m Stl

Ball

Hardness

Number

127

Rockwell

A Scale B Scale C Scale

60 kg

120 deg

D Jam ond

Cone

I00 kg

1/16 in.

Dia Stl

Ball

76

i50 kg

120 degDmmond

Cone

68

70

143

i49

i31

136

77.5

79

72

74

76

78

153

157

162

167

140

145

150

154 51

80.5

82

83

84.5

8O

82

83

85

87

89

171

177

179

186

189

196

158

162

165

171

174

180

52

53

53.5

54

55

56

85.5

87

87.5

89

90

91

Table AI. I.5-i Hardness Conversion Table (Cont'd)

AI. 1.5 Hardness Conversion Table (Cont'd)

Section At

March 1, 1965

Page 14

Tensile

Strength

ksi

Vickers-

Firth

9t

93

95

97

99

102

104

107

110

t12

i15

118

120

D Jam ond

Hardness

Number

Brinell

3000 kg10m m Stl

Ball

Hardness

Number

Rockwell

[


60 kg

120 deg

Diamond

Cone

100 kg

1/16 in.Dia Stl

Ball

150 kg

120 deg

D iam ond

Cone

203 186

207 190

211 193

215 t97

219 201

227 210

235 220

240 225

245 230

250 235

255 241

261 247

267 253

Table A 1.1.5-1

56.5

57

57

57. 5

57.5

59

60

60.5

61

61.5

62

62.5

63

92. 5

93.5

94

95

95.5

97

98

99

99, 5

100

101

i01.5

102

w--

--m

19

2O

21

22

23

24

25

Hardness Conversion Table (Cont'd)

Section A1

March 1, 1965

Page 15


Tensile

Strength

ksi

123

126

129

132

136

139

142

147

150

155

160

165

170

176

Vickers-

F irth

D iam ond

Hardness

Num be r

Brinell

3000 kg

10ram Stl

Ball

Hardness

Number

Rockwell


60 kg

120 deg"

Diamond

100 kg

1/16 in.

Dia Stl

150 kg

120 deg

Diamond

274

281

288

296

304

312

321

330

339

348

357

367

376

386

259

265

272

279

286

294

301

309

318

327

337

347

357

367

C one

63.5

64

64.5

65

65.5

66

66.5

67

67.5

68

68.5

69

69.5

7O

Ball

103

Cone

26

27

28

29

30

31

32

33

34

35

36

37

38

39

Table A I. i.5-i Hardness Conversion Table (Cont'd)

Ai. i.5 Hardness Conversion Table (Cont'd)

Section A i

March l, i965

Page 16

Tensile

Strength

ksi

Vickers-

Firth

Diamond

Hardness

Number

Brinell

3000 kg10ram Stl

Ball

Hardness

Number

Rockwell


60 kg

i20 degDiamond

Cone

i00 kg

I/t6 in,

Dia Stl

Ball

150 kg

120 deg

Diamond

Cone

181 396

188 406

194 417

201 428

208 440

215 452

221 465

231 479

237 493

246 508

256 523

264 539

273 556

283 573

377

387

398

408

419

430

442

453

464

476

488

5OO

512

524

70.5

71

71.5

72

72.5

73

73.5

74

75

75.5

76

76.5

77

77.5

40

41

42

43

44

,45

46

47

48

49

5O

51

52

53

Table Ai. i.5-i Hardness Conversion Table (Cont'd)


Section A 1

March 1, 1965

Page 17

Tensile

Strength

ksi

294

304

Vickers-

Firth

D ia m ond

Hardness

Num be r

592

611

Brinell

3OOO kg

10mm Stl

Ball

Hardness

Num be r

536

548

Rockwell.m


6O kg

120 degDiamond

Cone

78

78.5

100 kg

1/16 in.

Dia Stl

Ball

150 kg

120 degDiamond

Cone

54

55

Table A1.1.5-1 Hardness Conversion Table {Concluded)

A1.2.0 Elementary Theory of the Mechanics of Materials

In the elementary theory of mechanics of materials, a uni-axial state of

strain is generally assumed. This state of strain is characterized by the simpli-

fied form of Hooke's law; namely f = E _, where • is the unit strain in the direc-

tion of the unit stress f, and E is the Modulus of Elasticity. The strains in the

perpendicular directions { Poisson's ratio effect) are neglected. This is generally

justified in most elementary and practical applications considered in the theory

of mechanics of materials. In these applications, the structural members are

generally subjected to a uni-axial state of stress and/or the strains and dis-

placements are of secondary importance. Also, in these applications, the

magnitude of each of a set of bi-axial stresses (when this occurs) is generally

independent of the Poisson's ratio effect.

Frequently in design, there are applications in which the magnitude of each

of a set of bi-axial (or tri-axial) stresses are dependent upon the Poisson's

ratio effect; and/or the magnitude of the strains and displacements are of pri-

mary importance. This type of application must be generally analyzed by the

theory of elasticity. A brief account on the use of the theory of elasticity for

elementary applications is given in the next subsection.

Section A1

March 1, 1965

Page 18

AI. 3.0 Elementary Applications of the Theory of Elasticity

The difference between the method of ordinary mechanics and the theory of

elasticity is that rio simplifying assumption is made concerning the strains in the

latter. Because of this, it becomes necessary to take into account the complete

distribution of the strains in the body and to assume a more general statement

of Hooke's law in expressing the relation between stresses and strains. It is

noted that the stresses calculated by both methods are only approximate since

the material in the physical body deviates from the ideal material assumed by

both methods.

Some of the following subsections are written for a three dimensional stress

field but are applicable to problems in two dimension simply by neglecting all

terms containing the third dimension.

A1.3. l Notation for Forces and Stresses

The stresses acting on the side of a cubic element can be described by six

components of stress, namely the three normal stresses fll, f22, f33, and the

three shearing stresses fl2 = f21, f13,= f3t, f23 = f32.

In Figure AI. 3. l-i shearing stresses are resolved into two components

parallel to the coordinate axis. Two subscript numbers are used, the first

indicating the direction normal to the plane under consideration and the second

indicating the direction of the component of the stress. Normal stresses have

like subscripts and positive directions are as shown in the figure. An analogous

notation for the x-y coordinate system is:

fll =fX

f22 = fy

f12 = fS

xa

f22

!

f"'- xjfs3

I"

Figure AL. 3. 1-1 Representation of Stresses on

an Element of a Body J

F _

f

Section A 1

March 1, 1965

Page 19

A1.3. 1 Notation for Forces and Stresses (Cont'd)

Surface forces

Forces distributed over the surface of the body, such as pressure of one

body on another, or hydrostatic pressure, are called surface forces.

Body forces

Body forces are forces that are distributed over the volume of a body, such

as gravitational forces, magnetic forces, or inertia forces in the case of a body

in motion.

A1.3.2 Specification of Stress at a Point

If the components of stress in Figure A1.3. 1-2 are known for any given

point, the stress acting on any inclined plane through this point can be calculated

from the equations of statics. Body forces, such as weight of the element, can

generally be neglected since they are of higher order than surface forces.

x_

Figure AI. 3. I-2

X2

C

N

x_

An Element Used in Specifying Stress at a Point

SectionA IMarch I, 1965

Page 20

Ai. 3.2 Specification of Stress at a Point (Cont'd)

If A denotes the area of the inclined face BCD of the tetrahedron in Figure

AI, 3. t-2, then the areas of the three faces are obtained by projectin_A on thethree coordinate planes. Letting N be the stress normal to the plane BCD, the

three components of stress acting parallel to the coordinate axes, are denoted

by NI, N 2, and N 3. The components of force acting in the direction of the co-

ordinates X|, Xz, X 3 are AN_, AN2, and AN 3 respectively. Another useful

relationship can be written as:

cos (NI) = k, cos (N2) = m, cos (N3) = n (1)

and the areas of the other faces are Ak, Am, An.

The equations of equilibrium of the tetrahedron can then be written as:

NI = fil k + f12 m + f13 n

N2 = fi2 k + f22 m + f32 n

Na -_ fl3 k + f23 m + f33 n

(2)

The principal stresses for a given set of stress components can be deter-

mined by the solution of the following cubic equation:

fp3 _ (fli+ f22+ f33)fp2 + (fllf22÷ f22f33+ fllf33- f232

- f132 - f122) fp - (fli f22 f33 + 2f23 f13 f12 - fll f232 - f22 f132 - f33 f122) = 0

(3)

The three roots of this equation give the values of the three principal stresses.

The three corresponding sets of direction cosines for the three principal plan_scan be obtained by substituting each of these stresses (one set for each principal

stress) into Equations 3 and using the relation k 2 + m 2 + n 2 = i.

A1.3.2 Specification of Stress at a Point (Cont'd)

Section A 1

March 1, 1965

Page 21

(fp - fit) k - fl2 m - ft3 n = 0

f12 k + (f - f22) m - f23 n = 0

f13k- f23m + (fp- f33) n= 0

(4)

The shearing stresses associated with the three principal stresses can be

obtained by:

t , fl3 !fl2 = + _-(fp! _ fp2) = + 2-(fpl - fp3),

!f23 = + _- (fp2 - fp3)

(5)

where the superscript notation is used to distinguish between the applied shearing

stresses and the stresses associated with the principal normal stresses fpl,

fp2' and fP3"

The maximum shearing stress acts on the plane bisecting the angle between

the largest and the smallest principal stresses and is equal to half the difference

between these two principal stresses.

AI. 3.3 Equations of Equilibrium

Since no simplifying assumption is permitted as to the distribution of strain

in the theory of elasticity, the equilibrium and the continuity of each element

within the body must be considered. These considerations are discussed in this

and the subsequent subsections.

Let the components of the specific body force be denoted by X1, X2, X3,

then the equation of equilibrium in a given direction is obtained by summing all

the forces in that direction and proceeding to the limit. The resulting differen-

tial equations of equilibrium for three dimensions are:

AI. 3.3 Equations of Equilibrium (Cont'd)

Section A1

March i, 1965

Page 22

afl2 8f13afli +_ + +Xi = 08x i 8x2

_+ + +X2= 08x 2 axl 8x 3

(6)

8f33 afi3 8f23--+ +--+X3= 08x3 _ _)x2

These equations must be satisfied at all points throughout the body. The

internal stresses must be in equilibrium with the external forces on the surface

of the body. These conditions of equilibrium at the boundary are obtained by

considering the stresses acting on Figure AI. 3.3-1.

%

----_ _x1

Figure Ai. 3.3-1 An Element Used in Deriving the Equations of Equilibrium

Section A i

March I, 1965

Page 23

AI. 3. 3 Equations of Equilibrium (Cont'd)

By use of Equations 1 and summing forces the boundary equations are:

Xl = fll k + f12 m +f13 n

X2 = f22 m + f23 n + fl2 k

X3 = f33 n+ft3k+f23 m

(7)

in which k, m, n are the direction cosines of the external normal to the surface

of the body at the point under consideration and X1, X2, X 3 are the components

of the surface forces per unit area.

The Equations 6 and 7 in terms of the six components of stress, fll, f22,

f33, f12, f13, f23 are statically indeterminate. Consideration of the elastic deforma-

tions is necessary to complete the description of the stressed body. This is

done by considering the elastic deformations of the body.

A1.3.4 Distribution of Strains in a Body

The relations between the components of stress and the components of strain

have been established experimentally and are known as Hooke's law. For small

deformations where superposition applies, Hooke's law in three dimensions fornormal strain is written as:

1el = _ [fll - v (f22 ÷ f33) ]

1

£2 = E- [f22 - v (fll + f33) ] (8)

1e3 = E- [f33 - v (fil + f22) ]

A1, 3.4 Distribution of Strains in a Body (Cont'd)

and for shearing strain

Section A 1

March 1, i965

Page 24

2(I+ v) +_.,v,• l_ = E fi2 = G

2(t + v) (9)Tts = E ft3 = G

T_-a = 2(i + v) f23E f23 - G

These six components of strains can be expressed in terms of the three

components of displacements. By considering the deformation of a small ele-

ment dxl, dx2, dx 3 of an elastic body with u, v, w as the components of the dis-

placement of the point 0. The displacement in the x 1 - direction of an adjacent

point A on the x 1 axis is

au

u + _xl dxl

due to the increase (au/axl)dx 1 of the function u with increase of the coordinate

x l, It follows that the unit elongation at poiqt 0 in the x 1 direction is au/ax 1.

In the same manner it can be shown that the unit elongations in the x 2 - and x 3 -

directions are given by av/ax2 and aw/ax 3 respectively.

The distortion of the angle from AOB to A'O' B' can be seen from Figure

AI. 3.4-i to be av/ax 1 + au/ax 2. This is the shearing strain between the planes

x I x 3 and x 2 x 3. The shearing strains between the other two planes are obtained

similarly.

The six components of strains in terms of the three displacements are:

au av aw

el - ax 1 , _2 - ax2 ' _s =_..

au av au Dw av aw

Ylg- = 2ax-- + ax t "/i3 ax 3 + ax I _23 ax 3 + ax 2

(10)

At. 3.4 Distribution of Strains in a Body (Cont'd)

Section A l

March I, 1965

Page 25

T-dx 2

1

X_ _ u +iL_. dx 28x2

i I

At

0 + a_v dxtOx!v

0 _ . _ Xl

4- J_x 1 dxt

Figure A1.3.4-1 Distortions Due to Normal and Shearing Stresses Used

to Define Strains in Terms of Displacements

A1.3.5 Conditions of Compatibility

The conditions of compatibility, that assure continuity of the structure,

can be satisfied by obtaining the relationship between the strains in Equations 10.

The relationship can be obtained by purely mathematical manipulation as follows:

Differentiating Q twice with respect to. x2; e2 twice with respect to xt; and

Ti2 once with respect to x t and once with respect x 2. The sum of the derivatives

of (1 and e2 is found to be identical to the derivative of Tt2. Therefore,

ax] + ax_ = axtax2

Section A1March 1, i965Page 26

AI. 3.5 Conditions of .Compatibility(Cont'd)

Two more relationships of the same kind can be obtained by cyclic interchange

of the subscripts 1, 2, 3.

Another set of equations can be found by further mathematical manipulationas follows:

Differentiate e 1 once with respect to x I and once with respect to xs; _/12

once with respect to x t and once with respect to x3; _/13 once with respect to x l

and once with respect to x2; and _'23 twice with respect to x I. It then follows that

8x_0x 3 axl0x3 Oxiax2 0x l"

Two additional relationships can be found by the cyclic interchange of sub-

scripts as before.

The six differential relations between the components of strain are called

the equations of compatibility and are given below.

8x_ = axlax2 ' 8x28x3 8x i\ ax3 ax2 8x i j'

8x{ ax{ ox2ax3 ' axlax3(II)

These equations of compatibility may be stated in terms of the stresses if

the strains in Equations 11 are expressed in terms of the stresses by Hooke's

law (Equations 8 and 9). Differentiating each of Equations 8 and 9 as required

for substitution, we have

A1.3.5 Conditions of Compatibility (Cont'd)

Section A 1

March i, 1965

Page 27

a20

(i +,) _72fil + _x I = 0( I + p) V 2 f23 +

a20

Dx20x 3- 0

_20

(1 + v) V 2 f22 + 0x--_ = 0

020

, (1 + v) V 2 fl3 + 0xlDx3 - 0 (t2)

020

(1 + v) V 2 f33 + --Ox2 -- 0(1 + v) V 2 fl2 + --

020

OxlOx 2- 0

where:

and

V 2 _)2 D2 _2

0 = fll + f22 + f33

For most cases where strains are linear and superposition applies, the

system of Equations 6, 7, and 11 or 12 are sufficient to determine the stress

components without ambiguity. The use of stress functions to aid in the solution

of these equations are discussed below.

A1.3.6 Stress Functions

It has been shown in the previous sections that the differential equations

of equilibrium (Equations 6) ensure a distribution of stress in a body that pre-

serves the equilibrium of every element in the body. The fact that these are

satisfied does not necessarily mean that the distribution of stresses are correct

since the boundary stresses must also be satisfied. The compatibility equations

(Equations 1t) must also be satisfied to ensure the proper strain distribution

throughout the body. The problem is then to find an expression that satisfies all

Section A1,

March 1, 1965

Page 28

A1.3.6 Stress Functions (Conttdl

these conditions. The usual procedure is to introduce a function called a stress

function that meets this requirement. For the sake of simplicity, tliis section

will deal only with problems in two dimensions. The stresses due to the weight

of the body will also be neglected.

In 1862,G. B. Airy introduced a stress function (_b (xl, x2) ) which is an ex-

pression that satisfies both Equations 6 and II (in two dimension) when the

stresses are described by:

fll - , f22 = ax I , f12 = - _xl_x2

By operating on Equations 13 and substitutinginto Equations il, we find that the

stress function _b must satisfy the equation

+ 2 + = V4qb = 0 (I4)

Thus the solution of a two-dimensional problem reduces to finding a solution

Of the biharmonic equation (Equation t4) which satisfies the boundary conditions

(7) of the problem.

At. 3.7 Use of Eqtmtions from the Theory of Elasticity

Proficiency in the use of stress functions is gained mainly by experience.

It is not unusual to find an expression that satisfies Equation i4 first and then try

to determine what problem it solves.

The following problem is presented to illustrate the basic procedure in the

use of stress functions.

Section AiMarch 1, 1965Page 29

At. 3.7 Use of Equations from the Theory of Elasticity (Cont'd)

Statement of the problem:

Determine the stress function that corresponds to the boundary conditions

for a cantilever beam of rectangular cross section of unit width and loaded as

shown in Figure A1, 3.7-1. From this stress function determine the stresses

and compare with the maximum flexure stresses as obtained by the method of

mechanics.

p/unit length

V-

X2

_ _ ii _ _ _ _

L-

V ° = -p L

.__._._.p, Xl

Mo=- 2

Figure At. 3, 7-i Sample Problem

Solution:

Assume that the stress function is

_b = ax2s + bx23xt 2 + cx23 + dx2x 2 + ex 2

Operate on _ to satisfy Equation 14

V4_b = (5-4"3.2) ax 2 + 2( 3.2.2 bx2) = 0

24x 2 (5a+b) = 0

from which a = - b/5 (a)

Section A1

March i, 1965

Page 30

A1.3.7 Use of Equations from the Theory of Elasticity (Cont'd)

Since Equation 14 can now be satisfied by letting a = - b/5, the only other

condition to satisfy is the boundary conditions.

From Figure Ai. 3.7-1 the boundary conditions are as follows:

i. f22 = -P at x 2 = - h/2

2. f22 = 0 at x 2 = h/2

h/2

3. f fl2dx2 = -pL at x 1 = L from ZF=0-h/2

h/2

4. f fltx2dx2 = -pL2/2 at x 1 = L from-h/2

ZM = 0

5. fl2 = 0 at x 2 = h/2

From Equation i3

fli = = 20ax3 + 6bxl2x2 + 6cx2

f22 - = 2bx_ + 2hx 2 + 2e

6bx, , 2hx,f12 = - OxiOx2

(b)

Using boundary condition I

2bh 3 2dhf22 = -P = - 8 2 + 2e (c)

-f-Section A 1

March 1, 1965

Page 31

AI. 3.7 _Use of Equations fro m the Theory of Elasticity (Cont'd)

from boundary condition 2

2bh 3 2dh

f22 = 0 - 8 + 2 + 2e (d)

adding (c) and (d)

4e = -p or e = -p/4 (e)


h/2 h/2

-h/2 f12 dx2 = _hf/2 [- 6bx22xl - 2hxl} dx2

[ 6 bLx23-2hLx2]_/2=2 -_

orbh3

+ 2dh -_p2

= -pL (f)


h/2

hf/ [20ax24 + 6bx}x22 + 6cx22] dx 2- 2

=2[ _ 6b x3+ 6 31 h/2

2___ax_ + _- xl2 5 cx0

ah 5

4bL2h 3 ch 3 _ _ pL2/2

+_+ 2

SectionA 1

March f, 1965

Page 32

Al. 3.7 Use of Ec_uations from the Theory of Elasticity (Cont'd)

substituting Equation a and solving for c

c = -pL2 - b ( L2h a - hs/iO)h a (g)


fi2 = _ bh_xi - 2dxi = 0

3=-x i ( _ bh 2 # 2d )

or

(h)

Solving Equations f and h simultaneously we get

d = 3p and b = -p/h 3 (i)

Substituting b = - p/h 3 intoEquation g

(J)

Section A1

March 1, 1965

Page 33

Ai. 3.7 Use of Equations from the Theory of Elasticity (Cont'd)

The stress function can now be written as

¢) = -px 2 (x_/h 3- 3x2/4h + 1/4)

+(ph2/5) (x25/h 5 - x23/2h 3)

(k)

and the stresses as (see Equations b)

P X 2 X 2 +fll = - 2I ( h2 xz/10 - 2x_/3) (i)

-P---(x23/3 - h2xy/4 + h3/12)f22 = - 2I

-P-- (x22x! - h2xl/4)fi2 = 21

(m)

(n)

where I = h3/12

Comparison of maximum flexure stresses from Equation 1 with x 1 = L,

x 2 = - h/2

felasticity ph /L 2 h_l11 = 4I -

from elementary mechanics

fmiechanics Me _ pL 2 hI 4I

(o)

(P)

The difference is then

felasticity f_i echanics ph 311 - = - 60Ip5 (q)

SectionA 1

March 1, 1965

Page 34

A1.4.0 Theories of Failure

Several theories have been advanced to aid in the prediction of the critical

load combination on a structural member. Each theory is based on the assump-

tion that a specific combination of stresses or strains constitutes the limiting

condition. The margin of safety of a member is then predicted by comparing

the stress, the strain, or combination of stress and strain with the correspond-

ing factors as determined from tests on the material.

Three of the more useful theories are stated in this subsection. A more

detailed discussion on these and other theories of failure can be found in most

elementary strength analysis text books such as references 2 and 3.

The Maximum Normal Stress Theory

The maximum normal stress theory of failure states that inelastic action at

any point in a material begins only when the maximum principal stress at the

point reaches a value equal to the tensile (or compressive) yield strength of the

material as found in a simple tension (or compression) test. The normal or

shearing stresses that occur on other planes through the point are neglected.

The Maximum Shearing Stress Theory

The maximum shearing stress theory is based on the assumption that yield-

ing begins when the maximum shear stress in the material becomes equal to the

maximum shear stress at the yield point in a simple tension specimen. To apply

it, the principal stresses are first determined, then, according to Equation 5,

I f _ fpj)fimJax = 2( pi

where i and j are associated with the maximum and minimum principal stresses

respectively.

The Maximum Energy of Distortion Theory

The maximum energy of distortion theory states that inelastic action at any

point in a body under any combination of stresses begins only when the strain

energy of distortion per unit volume absorbed at the point is equal to the strain

/

Section A 1

March 1, i965

Page 35

A1.4. 0 Theories of Failure (Cont'd)

energy of distortion absorbed per unit volume at any point in a bar stressed tothe elastic limit under a state of uniaxial stress as occurs in a simple tension

(or compression) test. The value of this maximum strain energy of distortionas determined from the uniaxial test is

l+v F 2wl - 3E YP

and the strain energy of distortion in the general case is

W --

.6E [(fpl - fp2 )2 + (fp2 - fP3)2+ (fpl - fp3)2]

where fpl, fP2' fp3 are the principal stresses and Fyp

(For th_ case of a biaxial state of stress, fP3 = 0.)

The condition for yielding is then, w = w 1 or

is the yield point stress.

(fPl- fp2)2+ (fp2- fp3)2+ (fpi- fp3)2 = 2 Fy/

AI. 4. I Elastic Failure

The choice of the proper theory of failure is dependent on the behavior of

the material. It is suggested that the maximum principal stress theory be used

for brittle materials and either the maximum energy of distortion theory or the

maximum-shearing-stress theory for ductile materials.

The choice between the two methods for ductile materials may be made by

considering the particular application. When failure of the component leads to

catastrophic results, the maximum-shearing-stress theory should be used

since the resuits are on the safe side.

Section A J.

March 1, 1965

Page 36

Ai. 4° 2 Interaction Curves

No general theory exists whichapplies in all cases for combined loading

conditions in which failure is caused by instability. Interaction curyes for the

instability case or other critical load conditions are usually determined from

or substantiated by structural tests. The analysis of various loading combina-tions are discussed in Section A3.

AI. 0.0 Stress and Strain

Section A 1

March 1, 1965

Page 37

REFERENCES

1. MIL-HDBK-5, "Metallic Materials and Elements for Flight Vehicle

Structures," Department of Defense, Washington, D. C., August, 1962.

2. Murphy, Glenn, Advanced Mechanics of Materials, McGraw-Hill Book

Company, Inc., New York, 1946.

3. Seely, Fred B. and James O. Smith, Advanced Mechanics of Materials,

Second Edition, John Wiley and Sons, Inc., New York, 1957.

4. Timoshenko, S. and J. N. Goodier, Theory of Elasticity, Second Edition,

McGraw-Hill Book Company, Inc., New York, 1951.

5. Freudenthal, Alfred M., The Inelastic Behavior of Engineering Materials

and Structures, John Wiley and Sons, Inc., New York, 1950.

SECTIONA2

LOADS

A2.0.0

TABLE OF CONTENTS

Page

Space Vehicle Loads .............................. 1

2.1.0 General ................................... 1

2.2.0 Loading Curves ............................. 3

2.3.0 Flight Loads ............................... 42.3.1 General ................................ 4

2.3.2 Dynamic and Acoustic Loads .................. 5

2.3.3 Other Flight Loads ........................ 5

2.4.0 Launch Pad Loads ............................ 6

2.5.0 Static Test Loads .................... ........ 7

2.6.0 Transportation and Handling Loads ................ 7

2.7.0 Recovery Loads ............................. 7

A2-iii

A2

A.2.1

SPACE VEHICLE LOADS.

COORDINATE SYSTEMS.

Section A2

April 15, 1973

Page

The standard coordinate axes which have been used for rockets, missiles,

and launch vehicles are shown in Figure A2.1-1.

taken as positive in the flight direction. The Y and Z

appropriate directions to form a right-handed system.

as determined by the right-hand rule.

The longitudinal X axis is

axes are taken in

Moments are positive

For aircraft analysis the sign conventions used are shown in

Figure A2.1-2. In this figure externally applied loads acting at the airplane

center of gravity are defined as positive when directed aft in the X direction,

outboard to the left in the Y direction, and upward in the Z direction.

Externally applied moments about the airplane center of gravity are defined as

positive when acting as shown in Figure A2.1-2 (left-hand rule). At ,any section

under positive shear the rear, left outboard, or upper part tends to move aft,

left, or up; the right outboard, or upper part, tends to move aft, right, or up.

Any section under positive torsion tends to rotate clockwise when viewed from

the rear, left, or above. Positive bending moments produce compression in

the rear, left, and upper fibers. Positive axial load produces tension across

any section.

The external loads which may act on a space vehicle are categorized

as follows.

1.

2.

Flight Loads

Launch Pad Loads

+X (L, d>, p, u)

Section A2

April 15, 1973

Page Z. 0

IN. _, r, w)

+Z

+Y

(M, (_, q. v)

-Z

FORCESYMBOL

MOMENT

SYMBOL

i ii i,

LINEAR

VELOCITY

LONGITUDINAL X L u

LATERAL Y M v

YAW Z W

SYMBOL ANGULAR

VELOCITY

POSITIVE

DIRECTION

ANGLE

r

PITCH

YAWi i

ROLL ¢ Y to Z

e ZtoX q

Xto Y!

NOTE: Sign convention follows right-hand rule.

4

Figure A2.1-1. Coordinate axes and symbols for a space vehicle. --:

f-

Section A2

April 15, 1973

Page 2. 1

f

F

,,k

\

0_:::u ,-.4

_'_

I-d

0

bD _I

"_ o

._ _._o

°_

NASA--MSFC

Section A2

April 15, 1973

Page 2. 2

3. Transportation and Handling Loads

4. Static Test Loads

5. Recovery Loads.

Since it is universal practice in the airframe industry for the stress

analyst to obtain the magnitudes of external loads for the space vehicle from

the cognizant "Loads Group" in his organization, the methods of calculating

these quantities will not be presented in this manual. Rather, it will be

assumed that these loads are furnished to the stress analyst so that only their

qualitative description is required. These loads are generally resolved along

the coordinate axes for stress and aerodynamic analysis.

MS FC_I_A, A_

w/

Section A2

March 1, 1965

Page 2._ 3

A2.2.0 Loading Curves

The loads are usually presented in the form of load versus vehicle

station curves, where locations along the longitudinal coordinate are referred

to as vehicle stations. These curves are plotted for various times during the

flight of the vehicle. At each of these times, the longitudinal force, the shear

and the bending moment are plotted as a function of the vehicle station. Typical

curves showing the bending moment and longitudinal force distribution along

a vehicle can be seen in Figure A2.2.0-I.

.2I

0

.2

l

Bending Moment

Longitudinal Force(

i L - S -_ __ ! I I

2800 2400 2000 1200 1600 800 400

Vehicle Station ~ Inches

Fig. A2.2.0-1 Typical Bending Moment and Longitudinal ForceDistribution Curves.

Section A2March 1, 1965Page 2._

A2.2.0 Loading Curves (Cont'd)

It is necessary to know the circumferential pressure distribution

along the vehicle at times of critical loading. This circumferential pressure

is applied to the structure along with the critical loads during strength analysis

of the vehicle. Typical distribution of this circumferential pressure at a partic-

ular vehicle station may appear as in Figure A2.2.0-2.

ax P

Figure A2.2.0-2 Typical Circumferential Pressure DistributionCurves at a Vehicle Station

A2.3.0 Flight Loads

A2.3. I General

A space vehicle is subjected to flight loads of varying magnitudes during

its flight. These flight loads must be investigated to determine the critical

loads on the vehicle. Although it is not possible to know when these critical

loads will occur without considering the entire flight history, there are certain

times during the flight where conditions exist which are favorable for the build-

up of critical loads. These times and the loads which occur may be summarizedas follows:

r

Section A2

March 1, 1965

Page 5

1. Liftoff - As the vehicle lifts off the launch pad there is a sudden

application and redistribution of loads on the vehicle. This causes dynamic loads

which may be critical.

2. Maximum Dynamic Pressure (Maximum q) - At this time the combi-

nation of vehicle velocity and air density is such that the maximum airloadsresult.

3. Maximum qo_ - At this time the combination of vehicle velocity, air

density and vehicle angle of attack is such that high bending moments due to air-

loads and vehicle acceleration result.

4. Engine Cutoff - Engine thrust and longitudinal inertia loads are maxi-

mum just before cutoff. During cutoff, high dynamic loads may result becauseof the redistribution of these loads.

A2.3.2 Dynamic and Acoustic Loads

Dynamic loads are loads which are characterized by an intensity that

varies with time. These loads may be analyzed by one of two methods. One

method is to replace the dynamic load by an equivalent static load, and it is the

preferred method for most cases. The other method is a fatigue ,analysis and it

is justified only in those cases where the confidence in the load time-history is

good and the design is felt to be marginal.

Acoustic loads are loads induced by pressure fluctuations resulting

from extraneous disturbances such as engine noise. The effects of these loads

are determined by using an equivalent static pressure load. This equivalent

static pressure acts in both the positive and negative directions, since the pres-

sure fluctuates about a zero mean value. This pressure should be combined with

the design inflight pressure to obtain the total pressure, and should be considered

only in shell or panel stress analysis, not in the analysis of primary or supporting

structure.

A2.3.3 Other Flight Loads

Other flight loads, which are caused by pressure and temperature dif-

ferentials, must be considered in the stress analysis. In addition to the

Section A2March 1, 1965Page6

A2. 3. 3 Other Flight Loads (Cont'd)

longitudinal loads presented in the loading diagrams in Section A2.1.1, there is

a longitudinal load resulting from the difference between the ambient external

pressure and the vehicle internal pressure at any time during flight. The ambient

external pressure is a function of the vehicle's altitude only, while the vehicle

internal pressure depends on vehicle trajectory and venting effects. These pres-

sures in combination usually produce positive net internal pressures which either

increases the tensile or decreases the compressive longitudinal load in thevehicle.

In order to determine the hoop loads at a particular vehicle station, the

difference between the local external pressure and the vehicle internal pressure

must be known at the desired time. The local external pressure is a function of

the angle of attack, dynamic pressure and ambient pressure. The pressure

difference may be positive or negative depending on the circumferential and longi-

tudinal location of the point in question and on the range of values used in the

aerodynamic analysis. This range of values results in a maximum and a mini-

mum design curve.

Temperature magnitudes and temperature differentials caused by aero-

dynamic heating, retro or ullage rocket heating and cryogenic propellants resultin additional vehicle loads which must be considered. The effects of these

temperatures on material properties must also be investigated.

A2.4.0 Launch Pad Loads

The vehicle may be subjected to various loads while it is on the launch

pad. These loads are referred to as launch pad loads and are generally

categorized as follows:

1. Holddown Loads - The vehicle is usually held onto the launch pad by

a holddown mechanism during engine ignition. The loads on the vehicle duringthis time are referred to as the holddown loads.

2. Rebound Loads - During engine ignition it may be necessary to shut

down the engines due to some malfunction. The loads on the vehicle as it

settles back onto the launch pad are referred to as rebound loads.

/

Section A2

March l, 1965

Page 7

A2.4.0 Launch Pad Loads (Cont'd)

3. Surface Wind Loads - While the vehicle is freestanding on the launch

pad, i.c. , unsupported except for the holddown mechanism, it is exposed tosurface wind loads. The magnitude of these loads will depend oil the geograph-

ical location and should be specified in the design specifications.

4. Air-blast Loads - The vehicle may be subjected to an air-blast loa, i

from an accidental explosion at an adjacent vehicle launch site. The potentialeffect of this air-blast on the vehicle must be determined.

A2.5.0 Static Test Loads

The statictest loads are the loads on the vehicle during statictesting

of the vehicle. These loads are summarized as follows:

1. Engine gimbaling loads

2. Longitudinal loads due to various propellant loadings during theholddown and rebound conditions

3. Wind loads

The dynamic and acoustic loads for static firing tests should also be

investigated since they are higher during static test than in flight, in many cases.

A2.6.0 Transportation and Handling Loads

The transportation ,and handling loads arc the loads which occur during

transportation and handling of the space vehicle. In the dcsigm of the vehicle,

these loads are required primarily for the design of ticdown and handling attach-ments.

A2.7.0 Recovery Loads

The recovery loads are the loads which occur during the recovery of a

particular structural component or stage of the vehicle. These recovery loads

also include the loads which may occur during descent and impact.

SECTIONA

GENERAL

GENERAL

SECTION Ai

SECT I ON A2

SECTION A3

SECTION A4

STRENGTH

SECTION BI

SECTION B2

SECTION B3

SECTION B4

SECTION B4.5

SECTION B4.6

SECTION B4.7

SECTION B4.8

SECT ION B5

SECT ION B6

SECTION B7

SECT ION B8

SECTI ON B9

SECTION BlO

STAB IL ITY

SECTION Cl

SECT ION C2

SECT ION C3

SECTION C4

ASTRONAUTICS STRUCTURES MANUAL

SECTION SUBJECT INDEX

STRESS AND STRAIN

L(_DS

COMBINED STRESSES

METRIC SYSTEM

JOINTS AND FASTENERS

LUGS AND SHEAR PINS

SPRINGS

BEAMS

PLASTIC BENDING

BEAMS UNDER AXIAL LOADS

LATERAL BUCKLING OF BEAMS

SHEAR BEAMS

FRAMES

RINGS

THIN SHELLS

TORS ION

PLATES

HOLES AND CUTOUTS

COLUMNS

PLATES

SHELLS

LOCAL INSTABILITY

A-tlJ.

SECTIONSUBJECTINDEX

(CONTINUED)

i

SECTION D THERMAL STRESSES

SECTION E1

SECTION E2

FATIGUE

FRACTURE MECHANICS

SECTION FI

SECTION F2

COMPOSITES CONCEPTS

LAMINATED COMPOSITES

SECTION G ROTATING MACHINERY

SECTION H STAT I ST I CAL METHODS

A -iv

SECTION A3

COMBINED STRESSES

A3.0.0

TABLE OF CONTENTS

Combined Stress and Stress Ratio ....................

Page

i

P

3.1.0 Combined Stresses ............................... 1

3.2.0 Stress Ratios, Interaction Curves, and Factor

of Safety ..................................... 8

3.2.1 A Theoretical Approach to Interaction ....... I0

3.3.0 Interaction for Beam-Columns .................... 12

3.3.1 Interaction for Eccentrically Loaded and

Crooked Columns ........................... 14

3.4.0 General Interaction Relationships ............... 18

3.5.0 Buckling of Rectangular Flat Plates Under Combined

Loading ....................................... 22

3.6.0 Buckling of Circular Cylinders, Elliptical

Cylinders, and Curved Plates Under Combined

Loading ............................ _ .......... 27

3.7.0 Modified Stress-Strain Curves Due to Combined

Loading Effects ................................ 31

A3-iii

---FSection A 3

i0 July 1961

Page 1

A 3.0.0 Combined Stresses and Stress Ratio

A 3.1.0 Combined Stresses

When an element of structure is subjected to combined stresses

such as tension, compression and shear, it is oftentimes necessary to

determine resultant maximum stress values and their respective princi-

pal axes.

The solution may be attained through the use of equations or the

graphical construction of Mohr's circle.

Relative Orientation and Equations of Combined Stresses

fx and fy are appliednormal stresses.

fs is applied shear

stress, fy

fmax and fmin are the

resulting principal

normal stresses.

fSmax is the resulting

principal shear

stress.

0 is the angle of

principal axes.

L

Sign Convention:

Tensile stress is

positive.

Compressive stress is

negative.

Shear stress is positive

as shown.

Positive e is counter-

clockwise as shown.

f8

e

_ fs

fy

Fig. A 3. I. 0-I

45 °

Note:

This convention of signs for shearing stress is adopted for

this work only.

A 3.1.0 Combined Stresses (Cont'd)

Section A 3

I0 July 1961

Page 2

Distributed Stresses on a 45 ° Element

f-"_2

fs =._ j- _ "

fx

-III-------

t t'ytY

/ % fs I

Fig. A 3.1.0-2

Pure TensionFig. A 3.1.0-3

Equal Biaxlal Tensi_',

fx

v

Lv

fx

_em,,.

Ffg. A 3.1.0-4

Equal Tension &

Compression

Fig. A 3.1.0-5Pure Shear

fs

A 3.1.0 Combined Stresses (Cont'd)/

fx + fy ,_ fx

fmax - 2 + VI 2

/fx + fy \// fx

fmin - 2 - V\

2

fy _ + f2s

2

Y _s

Section A3

July 9, 1964

Page 3

............ (1)

............ (2)

TAN 282f s

fx - fy The solution results inl

two angles representing

the principal axes of ....... (3)

fmax and fmin:

fx fy f2f = +Sma x 2 s

(Disregard Sign) ........... (4)

Constructing Mohr's Circle (for the stress condition shown in

Fig. A 3.1.0-6a)

___ + ShearStress fs

f

ht fmin

(a) T handface

fmin

(b)

fx

0

fx + fy

fmax

A

---fn

+ Normal

Stress

(c)

Fig. A 3.1.0-6

Section A 3

I0 July 1961

Page 4


I. Make a sketch of an element for which the normal and shearing

stresses are known and indicate on it the proper sense of these stresses.

2. Set up a rectangular co-ordinate sy_em of axes where the horizontal

axis is the normal stress axis, and the vertical axis is the shearing

stress axis. Directions of positive axes are taken as usual, upward

and to the right.

3. Locate the center of the circle, which is on the horizontal axis at

a distance of (fx + fy)/2 from the origin. Tensile stresses are posi-

tive, compressive stresses are negative.

4. From the right-hand face of the element prepared in step (I), read

off the values for fx and fs and plot the controlling point "A". The

co-ordlnate distances to this point are measured from the origin. The

sign of fx is positive if tensile, negative if compressive; that of fs

is positive if upward, negative if downward.

5. Draw the circle with center found in step (3) through controlling

point "A" found in step (4). The two points of intersection of the

circle with the normal-stress axis give the magnitudes and sign of the

two principal stresses. If an intercept is found to be positive, the

principal stress is tensile, and conversely.

6. To find the direction of the principal stresses, connect point "A"

located in step (4) with the intercepts found in step (5). The princi-

pal stress given by the particular intercept found in step (5) acts

normal to the line connecting this intercept point with the point "A"

found in step (4).

7. The solution of the problem may then be reached by orienting an

element with the sides parallel to the lines found in step (6) and by

indicating the principal stresses on this element.

To determine the maximum or the principal shearing stress and the

associated normal stress:

I. Determine the principal stresses and the planes on which they act

per previous procedure.

2. Prepare a sketch of an element with its corners located on the

principal axes. The diagonals of this element will thus coincide with

the directions of the principal stresses. (See Fig. A 3.1.0-7).

3. The magnitude of the maximum (principal) shearing stresses acting

on mutually perpendicular planes is equal to the radius of the circle.

These shearing stresses act along the faces of the element prepared in

step (2) toward the diagonal, which coincides with the direction of the

algebraically greater normal stress.

"7

F

Section A 3

i0 July 1961

Page 5


4. The normal stresses acting on all faces of the element are equal to

the average of the principal stresses, considered algebraically. The

magnitude and sign of these stresses are also given by the distance from

the origin of the co-ordinate system to the center of Mohr's circle.

\ f' =\ 2

\

\ fmin

fmax + fmin = fx + fy2

fmax " fmin

Fig. A 3.1.0-7

A 3.1.0 Combined Stresses (Cont'd_

Section A 3

I0 July 1961

Page 6

Mohr's Circle for Various Loadin$ Conditions

+ fs fx

fx_ _ fs_ g _-

O_fx_ + fn

Fig. A 3.1.0-8 Simple Tension

f._x + fs

9

_-- fx -_

Fig. A 3.1.0-9 Simple Compression

" fn 0

+ fs

_ fSmax "

+ fn

_y

Fig. A 3.1.0-I0 Biaxlal Tension


Section A 3

I0 July 1961

Page 7

+ fs

0

Mohr's Circle for Various Loadin$ Conditions

Point

fx_fS = 0

fy

Fig. A 3.1.0-II

+ fn

Equal Blaxlal Tension

f

" fn 0

+ fs

fnmin_ x =

" fn

fSma x " fs

+ fn

fs

Fig. A 3.1.0-12

Equal Tension and Compression

Fig. A 3.1.0-13

Pure Shear

Section A3

July 9, 1964

Page 8

A 3.2.0 Stress Ratios, Interaction Curves, and Factor of Safety

A means of predicting structural failure under combined loading

without determining principal stresses is known as the interaction

method.

The basis for this method is as follows:

I°

.

The strength under each simple loading condition (tension,

shear, bending, buckling, etc.) is determined by test or theory.

The combined loading condition is represented by either load or

stress ratios, "R" where

e __

APPLIED LOAD OR STRESS

FAILING LOAD OR STRESS

Failing can mean yield, rupture, buckling, etc.

The effect of one loading R1

represented by an equation or interaction

curve involving R1 and R2. The equation

or curve may have been determined by

theory, by test, or by a combination

of both. 1.0

A schematic interaction curve is

shown in Fig. A 3. Z.0-1. Type of

material or size effects will not

influence it. This curve represents

all the possible combinations of Rl

and Rz thatwill cause failure.

Using the curve:

1. Let the value of R1 and R2

locate point a.

2. Rx and Rz can increase

proportionately until failure

occurs at point b.

on another simultaneous loading Ra is

R1

3. If R1 remains constant,

point c.

4. If Rz remains constant,

at point d.

0

0

/11 \

/ 1o1 \R2 1.0

Fig, A 3.2.0-1

Ra can increase until failure occurs at

R 1 can increase until failure occurs

The factor of safety for (2) is F. S. = (ob+oa)_or(oh+oe),(or og-of)

and the factor of safety for (3) is F.S. = (fc+ fa).

Section A 3

i0 July 1961

Page 9

A 3.2.0 Stress Ratiosj Interaction Curves_ and Factor of Safety (Cont'd)

In general, the formula for the factor of safety stated analyti-

cally for interaction equations where the exponents are only i or 2

(one term may be missing) is as follows:

FoS,

IR+J_2+_21................... (1)

where

R' designates the sum of all first-power ratios.

R'' designates the sum of all second-power ratios.

Section A 3i0 July 1961Page I0

A 3.2.1 A Theoretical Approach to Interaction

For combining normal and shear stresses, the principal stress

equations are convenient to use.

Let F and Fs be defined as the failing stress, such as yielding

or rupture.

Let k=Fs/F; tests of most materials will show this ratio to vary

from 0.50 to 0.75.

Rf = f/F; R s = fs/Fs

Maximum Normal Stress Theory

fmax --7 ++ f2

sRef Eq. (I) Sec. A 3. I. 0

Divide by F; replace fs by RsFs, f/F by Rf and Fs/F by k.

The resulting equation when fmax = F is

Rf _Rf_ 2 2

I:T-+ + (kR s) ................... (I)

A plot of this equation for k = 0.50 and k = 0.70 is shown in

Fig. A 3.2.1-i.

Maximum Shear Stress Theory

fSmax = _(2) 2 + f2s

Ref Eq. (4) Sec. A 3.1.0

Divide by Fs; replace f by RfF,fs/F s by Rs and F/F s by I/k.

The resulting equation when fSmax = Fs is

Rf R21 i-f + s .................................(2)

A plot of this equation for k = 0.50 and k = 0.70 is shown in

Fig. A 3.2.1-1.

f--

Section A 3

i0 July 1961

Page I1

A 3.2.1 A Theoretical Approach to Interaction (Cont'd)

Conclusion

From the foregoing analysis, only Equation (2) with k = 0.5 is

valid for all values of Rf and Rs. It is conservatively safe to use

the resulting Equation (3) for values of k ranging from 0.5 to 0.7,

since all values within curve (_ must also be within the other curves.

The use of other curves of Fig. A 3.2.1-1 may lead to unconservative

results.

2 2Rf + Rs = 1 ....................................... (3)

and the Factor of Safety

1F.S. = ............................. (4)

VR2f +R2s

2 2

For the graphical solution for Factor of Safety, the curve R 1 + R 2 = 1of Fig. A 3.4.0-1 may be used.

Rs

Max. Shear Stress Theory

k -- .5; Rf 2 + Rs 2 = 1

1.6 O

1.=1- \ 4@1.

- L0 ,,,,1 I I

.2 .4 r.O 1.2

k = .7; .5Rf 2 + Rs 2 = 1

Max. Normal Stress Theory

k = .5; Rf +_f2 + Rs 2 = 2

k = .7; Rf +_/Rf2 + (1.4 Rs )2 : 2

@ Valid

@ Partly Valid

@ Invalid

@ Partly Valid

Fig. A 3.2.1-I

Section A 3I0 July 1961Page 12

A 3.3.0 Interaction for Beam-Columns

Fig. A 3.3.O-1

Sinusoidal Moment CurveFig. A 3.3.0-2

Constant Moment Curve

P

P = applied load.

2 E1

Pe = L 2 (Euler load). (Reference Section C 1.0.0).. (I)

_2E I

Po = buckling load = t ......................... (2)L 2

or applicable short column formula. (Reference Section C 1.0.0)

M = maximum applied bending moment as a beam only.

M o = ultimate bending moment as a beam only. (Reference

Section B 4.0.0)

P

Ra = p-_ (column stress ratio) ......................... (3)

Mu

Rb Mo

(beam stress ratio)

f =P+k M-cA I

.......................... (4)

from which the interaction equation is:

R a + kR b = i ............................................ (5)

Po Et

Let _ = Pe E (plasticity coefficient) ............... (6)

For sinusoidal bending moment curves

ik=

I - P/Pe

Rb = (i - Ra) (I - _ Ra) ................................ (7)

A 3.3.0 Interaction for Beam-Columns (cont'd)

Section A 3

i0 July 1961

Page 13

Interaction curves for various va]ues of _ are shown in

Fig. A 3.3.1-5.

For constant bending moment curves

1k=

Interaction curves for various values of _ are shown in Fig.A 3.3. I-6.

Conclusion

Comparison of Figs. A 3.3.1-5 and A 3.3.1-6 show that significant

changes in shape of the primary bending moment diagram do not greatly

influence the interaction curves. Therefore, Figs. A 3.3.1-5 and

A 3.3.1-6 should be adequate for many types of simple beam columns.

A3.3.1

i_e I

Section A 3

I0 July 1961

Page 14

Interaction for Eccentrically Loaded and Crooked Columns

M=Pe

LIIIIIIII

Eccentric Column

Fig. A 3,3.1-1

P e

Crooked Column

Fig. A 3.3.1-2

P

Reference Section A 3.3.0 for beam-column terms

Re = e__ (eccentricity ratio) ........................ (I)eo

M o

=-- (base eccentricity, which is that required

e° Po for Po to induce a moment Mo) ... (2)

For a particular e, M would be a linear function of P as shown in

Fig. A 3.3.1-3. A family of such lines could be drawn which would

represent all eccentric columns.

To obtain Fig. A 3.3.1-4 (a nondimensional one-one diagram of the

same form as the interaction curves of Figs. A 3.3.1-5 and A 3.3.1-6),

P, M, and e of Fig. A 3.3.1-3 may be divided by Po, Mo and eo respectively.

r

M

Fig. A 3.3.1-3

P

Ra =_o

M

Rb = M-_

Fig. A 3.3.1-4

e

e o

Section A 3

i0 July 1961

Page 15

A 3.3.1 Interaction for Eccentrically Loaded and Crooked Columns (Cont'd_

In using Fig. A 3.3.1-6 for eccentric columns and Fig. A 3.3.1-5

for crooked columns the following steps are taken:

I. Determine Po, the buckling load by _ 2Etl/L2 or applicable

short column formula.

2. Calculate Pe = _2EI/L2, the Euler load.

3. Determine Mo, the ultimate bending moment as a beam only

using Section B 4.0.0.

4. Calculate e o = Mo/Po, the base eccentricity.

5. Calculate R e = e/eo.

6. Calculate _ = Po/Pe, the plasticity coefficient.

7. Knowing Re and _ ' Ra = P/Po may be determined from the

appropriate curve. This value of Ra corresponds to a Factor

of Safety of 1.0.

8. The ultimate load is Pu = Pox Ra.

9. The Factor of Safety for an applied load P is

PuF.S.=--

P

A 3.3.1

Section A 3i0 July 1961Page 16

Interaction for Eccentrically Loaded and Crooked Columns (Cont'd_

oOw

mw

II

o_

1.0

0.8

0.6

0.4

0.2

0

0

0.2

/! /

/I

P

Po=0.0 =Y_ = 0.2

=0.4

_=0.6

"q = 0.8

_3 = 1

0.2 0.4 0.6 0.8

Rb = M/M °

Interaction Curves for Straight or Crooked Columns

with Sinusoidal Primary Bending Moment and Compression

1.0

i0.0

Fig. A 3.3.1-5

A3.3.1


Interaction for Eccentrically Loaded and Crooked Columns (Cont'd)

1.0

0.8

0.6O

II

0.4

0.2

R = e/ee o

2

0

1 8

2.0 ,

'i|l*|lll_

0 0.2 0.4 0.6 0.8 1.0

Rb = M/M °

i0.0

Interaction Curves for Columns with Constant Primary

Bending Moment and Axial or Eccentric Compression

Fig. A 3.3. i-6

A3.4.0 General Interaction Relationships

Section A 3

I0 July 1961

Page 18

!

o

r-_

co

0[--i

Om

.0

I

0

o co_._

II

o_

II

o

co co

._ _ oi_1 o r.j

oco

o

0

o _ = _[z_oo o_

co

II

+

I

o

"o

co "o 0.1.,-i I=

._ co o

V 0 _ o _

ml ,_ 00 _ _ ,-_ _ 00

+

oq_

II Jr

c,q _ co

Jr II

I

o

.<

"o

co _0or_

I_ m •

o ,.c_ _J_o0oo

.IJ

(2co .1_ cor_ I_ •

_J

eq

+

_J

! + i+!

o ,._

cN

o

I

,.-4

II II

4J .I..I

+ +

,--4

I

o

..4

! !

o o "_

co

=I= m'_ G 0

cJ -,_ ;_ _ o

!

c3 _ CO cO

0

II

+

c_ _J

I !

oo

"o

co

o

m

_J.J=

4-J

_ =_ ,-__-,_ _ 0

_pf

A 3.4.0 General Interaction Relationships (Cont'd)

Section A 3

I0 July 1961

Page 19

NOTE:

(a)

(b)

Table A 3.4.0-1 (Cont'd)

Care must be exercised in determining whether to check Factor of

Safety for limit or ultimate loads.

For round tubes in compression see Section C.3.0.0.

See Section A 3.2.1 for discussion of range.


Section A 3

i0 July 1961

Page 20 V

1.0

0.9

3+ =I /

0.6 ; :

0_ , - / _1 5 2

R 1" + R2 = 1 /

0.3 I 2 ! /_/ _RI + R = I

0.2 I I _//

R 1 + R 2 = i

0.i

0 "1 I I I

0.1 0.2 0.3 0.4 0.5 0.6 0.7

R1

Interaction Curves

Fig. A 3.4.0-1

\\

0.8 0.9 1.0



30

25

I IAN-IO

Interaction Formula

R _ 2+R t = 1

S

AN-8

20

O,

, 15

mo

• ,AN-6I0 ---_ _

AN-5

5

_-4

\

\\

\

\

\

\\\

\

\\\

\\

\ \

5

\\\

t

NOTE: Curves not

applicable where -

shear nuts are

used. Curves are

based on the

results of com-

bined load tests

of bolts with

nuts finger-

tight.

\i

\

II0 15 20

Shear Load -Klps

Flg. A 3.4.0-2

AN Steel Bolts (125,000 H.T.) Interaction Curves*

\\

\\

\

25

A 3.5.0

Section A 3

19 July 1961

Page 22

Buckling of Rectangular Flat Plates Under Combined Loading

NOTE: See discus=ton in Sec. C 2.1.4

iIIEIHtttttttttt

Combined Loading

Fig. A 3.5.0-1

Table A 3.5.0-I

THEORY

Elastic

Inelastic

LOADING

COMBINATION

Biaxial Com-

pression

Longitudinal

Compression and

Shear

Longitudinal

Compression

and Bending

Bending andShear

Bending, Shear, &Transverse

Compress ion

Longitudinal Com-

pression, Bending

and Transverse

Compression

Longitudinal

Compression

and Shear

FIGURE

A 3.5.0-2

A 3.5.0-2

A 3.4.0-I

A 3.5.0-3

A 3.5.0-2

A 3.4.0-1

INTERACTION

EQUAT ION

For plates that

buckle in sq.

waves,

Rx + Ry = I

For LongPlates.

+ R 2 = IRc s

2 R2R + = Ib s

2R2 +R = Ic s

EQ FOR FACTOR

OF SAFETY

Rx+Ry

Rc +¢R 2 + 4R 2s

_/R 2 + R2s

i

_/ 2 R 2Rc + s

iv

/+--_

A 3.5.0

(Cont 'd)

l.O_

o.__

0.2 \ , 0._o.9\ \

0.2

Section A 3

19 July 1961

Page 23

Bucklin_ of Rectangular Flat Plates Under Combined Loadin$

(a) (b)

\

_o_

o_

Rx= 0

\

¢\

0.6

a/b = O. 8

\\\\

\

\0.8

1 (

0 _

06

R

04

0 2

\

\ o1.0

_--,>,. ,a/b - l. o

\\\\,,\ \i \", _.o._\\\\,\,\\ o_\\ \\\\,

• _\\\\\\....1 _ _\\

0.2 0.4 0.6 0.8

\\1.0

1.0

0.8

0.6

Rb

0.4

0.2

0

(c)

__\_\ \_-o

\\ \\\o_\_,\\\ \ _o._,\,\\_\\ \ o..,\\ \ \\\o::\\\\ \\\....I I

0.2 0.4 0.6 0.8

Ry

alb -- 1.20

\1.0

fY

: a

_IIIIIIIIIIIII'

Interaction Curves for Simply Supported Flat Rectangular

Plates Under Combined Biaxial-Compression and Longitudinal Bending Loadings

Fig. A 3.5.0-2

A 3.5.0(Cont'd)

Section A 3

i0 July 1961

Page 24

Buckling of Rectangular Flat Plates under Combined Loading

(d) a/b = 1.60

1.0

_ --- L.... _

:°o.8"-""'m!..

0.6 -_ r",_z).2o

--_ _ _o ,','Rb _- .

0.4 "-- 0.ii0 _ _

0.80 \ .-;

0.2

0

_ "-0.90_

_,,I\

\\:Ii ' : "

L

\\

'I

l,

O.

O.

RI

3.

3.

(e) a/b = 2.0

1.0 _._ r I

0.8 _

0.6 _'-_

Rb _ __

o.4 _

0.2

oo_ -I1|lit

0 "0.2 0.4 0.6 0._

RY

(f) a/b = 3.0

__- "illlo

-_!_ \ _1 I!1\o_o\ ' I I!1°

• II!o

1.0

(g) a/b = oo

0 I_-- _ R__._..L=C

o8 _ -....,__

_ : -'_ O'_n<_'_I_ I --

Z

) alll

0

%.0.90i\\

\0.2 0.4

\

\0.6

RY

\

1.0

Fig. A 3.5.0-2 (Cont'd)


Buckling of Rectangular Flat Plates under Combined Loading

I1.0

0.8

0.6

Rb

0.4

0.2

0 _nallllln0 0.2

1.0 1

0.4 0.6 0.8 1.0

R s

Interaction Curves for Simply Supported Long Flat Plates

Under Various Combinations of Shear, Bending, and Transverse Compression

Fig. A 3.5.0-3

A 3.5.0

(Cont'd_

Section A 3

I0 July 1961

Page 26

Bucklin$ of Rectansular Flat Plates under Combined Loadin$

1.0

ff

/

0.5 __

___ 10.8

0._f

R s

0.8 0.6 Rc 0.4 0.2

m

l"

0

0.6

Rb

0.4

0.2

0

o o o.''"1" ,.20.4 Rs 0.6

0.2

0.4

Rc

0.6

0.8

1.0

//

//

Rb

Interaction Curves (Cont'd)

Fig. A 3.5.0-3

_J

Sect ion A3

February 15, 1976

Page 27

A3.6.0 Buckling of Circular Cylindersj Elliptical Cylindersj and

Curved Plates Under Combined Loadin$

klE_02_

0

kl

C_O

0

.SZ_z

-4©

,x:l

c_

0

Z

z

I

¢x.1

+O

_>

+

O

II

LJ

+

TO

4

<

g_

r-4

II

I

TO

<

_o

H g_

o_

,.-( +

cy

o ,.Q

_J II

0

4

<

°0'1

C m

(1) _)

II +

+

_ II•,_ II

II II _ c_

_ _ +u_

+

T . T To o oi o

4 $ 41 4

_ ._ _ ,.,_o_ _ _

o o _ o o o o _ _ c _

OLj

Section A 3

i0 July 1961

Page 28

A 3.6.0 Bucklin$ of Circular Cylinders t Elliptical Cylinders t and

Curved Plates under Combined Loadin$ (Cont'd)

=0O%.s

,-4!

O

M

,m

[I3

0

o

r/3

.oo'

ZO z

_0

_CY

Z0

i0CD

I

0

,-4

II

+

+

0

m c _ •= 0.,4 >•,4 .,4 _ 03_ 03 = =:3 03 0 _l

.,4 _ [-_

=eft1,

II

03

+

03

v

+

= =•,4 _1 ¢1'13 03

_ O >•,4 03

• m C m

_ o _,_

"oI I -

.,4 _ >_

,-4

II

¢o#m

¢_ 03

+

¢q 03

+o

= 03 _ • C

•,4 _ 03 _ .,4

o o D _.c o

4-Ieq 03

+¢q

o

+

,m

o

,--4

II

e4 03

+

.m

+

o

_ C C CC o-,_ o

"o 03 C 03D 03 • _

•,4 _ [-__0

o o _ C

oq 03

+

II II

.U¢_1 03 ¢_1 03

O_ O_

+ +

I I

.4 ..4

< <

C i1/ C

C > C O•,4 m _ .,4 .,4-o C _ -O m

o _ _ O

I I

•,4 m-,4 m

_rD-O

f

Section A 3

I0 July 1961

Page 29

A 3.6.0 Bucklin_ of Circular Cylinders_ Elliptical Cylinders_ and

Curved Plates under Combined Loading (Cont'd)

4-J

oo

I

o

,g

,x:l

dV

°_

0 _ 4-J

_ o n_ _

t.J .0_ 0flJ .,-I

¢.1 .,II ,-4 01

o

o

=° _4-1

• _ VlI_ _1--_

o _ _: Ii _ + _ln

,-.-4 121t_

.,-4

N ,._

m

oou_l_ _ Vl

o _J g II II

e,

O_

Section A 3

I0 July 1961

Page 30

A 3.6.0 Buckling of Circular Cylinders_ Elliptical Cylinders a and

Curved Plates under Combined Loading (Cont'd)

1.0

0.8

0.6

R 2

0.4

o2 ....

"| I 11

0 0.2

RI3 - R 2 = 1

2RI• - R 2 = 1

///j/

j'0.4 0.6 0.8 1.0 1.2 1.4

R I

Interaction Curves

Fig. A 3.6.0-1

Section A3July 9, 1964Page 31

A3.7.0 Modified Stress-Strain Curves Due to Combined Loading

Effects

An analysis that uses a uniaxial stress-strain curve or material

properties derived from such a curve (analysis of beams, columns,

thermal effects, plastic bending, elastic and piastic buckling, Elastic-

Plastic Energy Theory of Section B4. 5.7, etc) may require a modified

stress-strain curve or properties derived from a modified curve when

combined loading is involved. Loads or stresses in one plane affect

the loads and stresses in other planes due to the Poisson effect. For

example, a tension member fails when the average stress, (P/A),

reaches the ultimate tensile stress Ftu of the material, but a memberresisting combined loading may fail before the maximum principal

stress reaches Ftu (Reference Section A1). When buckling or other

empirical parameters include combined loading effects, modified

stress-strain curves are not required.

Several methods of modifying uniaxial stress-strain curves have

been developed; the method presented here is derived from the Octa-

hedral Shear Stress Theory.

Assumptions & Conditions:

, fl, f2 and f3, the three principal stresses, are in proportion;

i.e.,

fz = El fl (_)

f3 = K2 fl (2)

K 1 _ K 2

See Fig. A3.7.0-I for direction of principal stresses.

A3.7.0 Modified Stress-Strain Curves Due to Combined

Loadin$ Effect _Cont'd)

Section A3

July 9, 1964

Page 32

3 3

foct

f2- .._

---'-2

Figure A3.7.0-I

Directions of Principal Stresses

2. Prime (') denotes a modified value:

c' = modified strain

V,' = modified modulus of elasticity.

3. In this method, for any principal stress fi' the total strains

and modulus of elasticity are modified to include the effects

of the other principal stresses.

Procedure:

I. Calculate the principal stresses for a given load condition

(Reference Section A3.1.0).

2. Determine the effective uniaxial stress:

= - - (f2 - f3 + (f3 - fl (3)

Section A3July 9, 1964Page 33

A3.7.0 Modified Stress-Strain Curves Due to Combined Loading

Effect (Cont'd)

I

and calculate an effective modulus of elasticity, E l, by:

' (f_ll) El (4)E I =

o Enter the plastic stress-strain diagram for simple tension of-- f

the material, if available, at the value of fl and determine Esp.

(See Figure (A3.7.0_2b) Otherwise, enter the simple tension

stress-strain curve at fl and determine E' (see FigureA3.7.0-2a ) by: sp

fi

E' - (5)sp _i - _I

O

°

1

known, find eI from

P1

[p - E,sp (fl - gp f2 - _p f3 )

E Esi

: _ le ¢I P -F(a) Engineering stress-strain curve

Use this value of E'sp and a value of gp = 0. 5, if not accurately

(6)

__ _- EsPi --

pt. i

I _-- Plastic Secant

I modulus

'¢ip

(b) Plastic stress-strain curve

Figure A3.7.0-2

A3 °7.0 Modified Stress-Strain Curves Due to Combined Loadin 8

Effect (Cont'd_

Section A3

July _, 1,564

Page 34

5. Once E' has been found, c'

of fl by! le

fl

Ie N_

can be determined for any value

(7)

I

6. Determine the total effect strain, el, for each value of fl by:

°

e p (g)

Repeat all steps until sufficient points are obtained to construct

a plot of fl vs cl (see Figure A3.7.0-3 ) which is the modifiedstress-strain curve.

t t t

EE 1 / El s i t

/

/

fl

Any Point I

t

e lO

Ip

¢'1

Figure A3.7.0-3 Modified Stress Strain Diagram Due to Combined

Loading

Section A 3

July 9, 1964

Page 35

Re ferences :

Popov, E. P., Mechanics of Materials, Prentice-Hall, Inc., New York,

1954.

Structures Manual, Convair Division of General Dynamics Corporation,

Fort Worth.

SECTIONA4

METRIC SYSTEM

P/

_F

/

f

TABLE OF CONTENTS

A4.0.0

A4. 1.0

A4.2.0

METRIC SYSTEM .....................

Introduction ......................

The International System of Units (SI) ........

A4.2. 1

A4.3.0

A4.4.0

A4.4. 1

A4.4.2

A4.4. 3

Advantages of SI ...............

Basic SI Units .....................

International Symbols for Units, SI .........

CGS System ...................

Giorgi System ..................

Incoherent Units .................

A4. 5.0 Physical Quantities ..................

A4. 5. 1 Dimensionless Physical Quantities .......

A4.6.0 Other SI Symbols ...................

A4.6. 1 Photometric Units ................

A4.6.2 Rules for Notation ................

A4.7.0 SI Units on Drawings and in Analyses ........

A4.7. 1

A4.7.2

A4.7. 3

A4. 7.4

A4.7. 5

Dual Units ....................

Identification of Units ..............

Tabular Data ...................

Collateral Use, SI and Non-SI Units .......

Temperature Scales ..............

Page

1

1

1

1

2

2

3

3

4

4

4

5

5

5

7

7

8

8

8

9

f

A4-iii

TABLE OF CONTENTS (Continued)

A4.8.0

A4.8. 1

A4.8.2

A4.9.0

A4. 10.0

A4. 10. 1

A4. 10.2

A4. 10.3

A4. 10.4

A4. 10.5

Transitional Indices .................

Mass vs Force .................

Examples of Nomenclature ...........

Measurement of Angles ...............

Preferred Style ...................

Volume .....................

Time ......................

Energy .....................

Tempe rature .................

Prefixes ....................

A4. 11.0

A4. 11. 1

A4. 11.2

A4. 11.3

A4. 12.0

Conversion Factors ................

Basic Linear Unit ...............

Noncritical Conversion ............

Conversion to Other SI Units .........

Conversion Tables .................

Page

9

9

10

11

11

12

12

12

12

14

14

15

15

15

15

A4-iv

metric system and its advantages over the English system.

also presents definitions, symbols, and conversion tables.

Section A4

1 February 1970

Page 1

METRIC SYSTEM

Introduction

The purpose of this section is to acquaint the reader with the

This section

Units of length, mass, and time are basic to both the English

System and to the Metric System. In the English System these are:

length, the foot; mass, the pound; and time, the second. Note that the

second, based on the sexagesimal system, is common to both the English

System and the Metric System.

A4.2.0 The International System of Units (SI)

The International System of Units, or Syst_me Internationale

(SI), is sometimes referred to, in less precise terms, as the Meter-

Kilogram-Second-Ampere (MKSA) system. The SI, therefore, should be

considered as the definitive metric system, although it is much broader

in scope and purpose than any previous system.

A4.2. 1 Advantages of SI

The use of SI has significant advantages in all phases of research

and development work relating to space technology. For instance, the use

of SI will tend to eliminate wasted time and costly errors in computations

now involving varied terms derived from a multiplicity of sources. The

"-- f

Section A4

1 February 1970

Page 2

utilization of a uniform system of measurement such as the SI thus simpli-

fies the exchange of in-house data among NASA centers and installations

and will do so, eventually, among associated contractors and space-ori-

ented organizations throughout the world.

A4.3.0 Basic SI Units

The name, International System of Units, has been recommended

by the ConfErence G_n_rale des Poids et Mesures in 1960 for the following

basic units :

meter

kilogram

second

In addition,

m ampere A

kg degree Kelvin OK

s candela cd

it was also determined that the amount of substance

would be treated as a basic quantity. The recommended basic unit is the

mole, symbol: tool. The mole (tool), a unit of quantity in chemistry, is

defined as the amount of a substance in grams {gram mole; gram molec-

ular weight; or pound mole, or pound molecular weight) which corresponds

to the sum of the atomic weights of all the atoms constituting the molecule.

These atomic weights are based upon Carbon 1Z.

A4.4.0 International Symbols for Units, SI

In order that SI may be in fact an international system, it was

necessary to reach agreement on the symbols, names, and abbreviations.

v _

_F

A4.4. 1 CGS System

In the field of mechanics,

Section A4

1 February 1970

Page 3

the following units of this system

have special names and symbols which have been approved by the General

Conference on Weights and Measures:

1, b, h centimeter cm

t second s

m gram gf, v hertz ( = s- 1) Hz

F dyne ( = g. cm/s 2) dyn

E,U,W,A erg (= g. cm2/s 2) erg

p microbar ( = dyn/cm2) _t bar

poise (= dyn. s/cm 2) p

A4.4.2 Giorgi System

The MKSA system or m-kg-s-A system is a coherent system

of units for mechanics, electricity, and magnetism, based on four basic

quantities : length, mass,

meter

kilogram

second

ampere

time, and electric current intensity.

m

kg

S

A

The system based on these four units was given the name

"Giorgi system" by the International Electrotechnical Committee in 1958.

The mechanical system, which is based on the first three units only, has

the name MKS system.

The MKSA system of units forms a coherent system of units in

the four-dimensional system of equations previously mentioned, and is

most commonly used together with these equations.

A4.4.3 Incoherent Units

Section A4

1 February 1970

Page 4

1 _ngstr_m A

barn ( = 10-Z4 cruZ) b

V liter (= 1 dm 3) 1

t, T minute min

t,T hour h

t, T day d

t, T year a

p atmos pher e atm

p kilowatt- hour kWh

Q . calorie cal

Q kilocalorie kcal

E, Q electronvolt eV

m ton ( = 1000kg) t

Ma, m (unified) atomic mass unit u

p bar (= 10 6 dyn/cm Z)

(= 10 5 N/m z) bar

A4.5.0 Physical Quantities

The symbol for a physical quantity (French:

'physikalische Grosse'; English, sometimes:German:

'grandeur physique';

'phys ical magnitude ')

is equivalent to the product of the numerical value (or the measure, a pure

number) and a unit, i.e., physical quantity = numerical value x unit.

A4.5. 1 Dimensionless Physical Quantities

For dimensionless physical quantities the unit often has no name

or symbol and is not explicitly indicated.

Examples: E = 200 erg nqu = 1.55

F = 27 N v = 3 x 108 s-1

r"

Section A4

1 February 1970

Page 5

A4.6.0 Other SI Symbols

The following units of the MKSA system have special names

and symbols which have been approved by the General Conference on

Weights and Measures:

I ampere A

Q coulomb ( = A. s) C

C farad ( = C/V) F

L henry (= Vs/A) H

E joule ( =kg. mZ/s2) J

m kilogram kg

1, b, h meter m

F newton ( =kg. m/s2) N

R ohm (= V/A)

B tesla (= Wb/m2) T

V volt (= W/A) V

P watt (= J/s) W

$ weber (= V. s) Wb

A4.6. 1 Photometric Units

In the field of photometry an additional basic unit is introduced

corresponding to the basic quantity, luminous intensity. This unit is the

candela, symbol: cd. Special names for units in this field are:

I candela (candle) cd

lumen lm

E lux ( = lm / m2) lx

A4.6.2 Rules for Notation

a. Symbols for units of physical quantities 8hall be printed in

Roman upright type.

b. Symbols for units shall not contain a final full stop (a

period), and shall remain unaltered in the plural, e.g.: 7cm, not 7 cms.

Co

upright type.

shall start with a capital Roman letter,

(weber); Hz (hertz).

Section A4

1 February 1970

Page 6

Symbols for units shall be printed in lower case Roman

However, the symbol for a unit derived from a proper name

e.g.: m (meter); A (ampere); Wb

d. The following prefixes shall be used to indicate decimal

fractions or multiples of a unit.

Prefix Equiv S)rmbol

deci (10 -1) d

centi (10 -2 ) c

milli (10- 3) m

micro (i0-6)

nano (10-9) n

pico (i0 -IZ) p

feint. (10 -15 ) f

atto (10"18) a

deka (I01) da

hecto (10 z) h

kilo (10 3 ) k

mega (106 ) M

giga (109 ) G

tera (1012) T

v

e. The use of double prefixes shall be avoided when single

prefixes are available.

Not: m_ts, but:

Not: kMW, but:

Not: _ F, but:

Section A4

I February 1970

Page 7

ns (nanosecond)

GW (gigawatt)

pF (picofarad)

f.

combination shall be considered as a new symbol,

or cubed without using brackets.

Examples: cm 2, mA 2,

A numerical prefix shall never be used before a unit symbol which

squared, thus, crn2 is never written,

always means (0.01m) 2

go

When a prefix symbol is placed before a unit symbol, the

which can be squared

_s 2

is

and never means, 0.01 (m 2) but

symbols, or prefixes.

following example s :

No periods or hyphens shall be used with SI abbreviations,

Prefixes are joined directly to units, as in the

MN mN

kV kHz

MV mA

GHz cm

A4.7.0 SI Units on DrawinGs and in Analyses

The following paragraphs describe general techniques for using

SI units on drawings and in analyses.

A4.7. 1 Dual Units

When SI units are specified for use on a drawing or in an analysis,

the non-SI units of measure shall be used parenthetically to facilitate

Section A4

1 February 1970

Page 8

comprehension .of the drawing or analysis. Non-SI units shall never be

omitted on the assumption that users are familiar with the SI units.

A4.7. Z Identification of Units

Basic units of measure used frequently on a drawing shall be

identified by a note on the drawing to avoid repetition of unit names through-

out the drawing. For example,

NOTE: ALL DIMENSIONS ARE IN mm (in.).

A4.7. 3 Tabular Data

To provide maximum clarity of presentation, SI and non-SI units

shall be placed in separate columns or in separate tables if the need is

indicated.

A4.7.4 Collateral Use, SI and Non-SI Units

Place the metric units first, followed immediately by the equiv-

alents in parentheses. Intables, other formats may be desirable, such

as one unit in a row or column, followed by the other unit in another row

or column. In some complex tables and drawings it may be desirable to

present the equivalent units in separate tables and drawings. Figure A4-1

shows a drawing with both units given.

"--r_

Section A4

1 February 1970

Page 9

6 MM 4.236 IN)

/ -M' $ 300 MM( ,,.8IN)

FRAME. UPRIGHT (BRASS) 2 REQD

MM 4.236 IN.)DRILL-(2)

OMM C394 IN) SO ROD

FRAME BASE iBRA_,S)- ---I_

t'-'- 5- 0.51 MF (2)

,_.-6.35MM(.251N) DIA ROD

--.. ,SMMC.S ,,N..)--,.i ..j.--300 MMC 11.8 IN,) -,

ROD (ALUMINUM') I REQD

Figure A4-1. Collateral Us_ of Units

A4. 7. 5 Temperature Scales

Either the Kelvin or the Celsius temperature scale may be used

as the SI unit, with the Fahrenheit scale being optional as a parenthetical

non- SI unit.

A4.8.0 Transitional Indices

The following explanations indicate nomenclatures, methods,

and preferred styles which are to be used during the transition from non-

SI systems to SI.

A4.8. 1 Mass vs Force

The term "mass" (and not weight) shall be used to specify the

Section A4

1 February 1970

Page 10

quantity of matter contained in material objects.

The term "weight" shall be defined as the gravitational force

acting on a material object at a specified location. Accordingly, the

statement of the weight of an object should be accompanied by a statement

of the corresponding location of the object or by a statement of the grav-

itational acceleration in m/s2 at the location where the object was weighed

or is assumed to be located.

The pound mass (Ibm), defined as being exactly 0. 453 592 37

kilogram by the U. S. National Bureau of Standards; the pound force (lbf),

defined as being exactly 4.448 221 615 260 5 newtons by the NBS; and the

pound thrust, defined as being exactly 4.448 221 615 260 5 newtons, shall

be abandoned at the earliest practicable date.

During the transition period to SI units, the pound mass shall be

abbreviated Ibm, the pound force shall be abbreviated lbf, and the pound

thrust shall be abbreviated to lbf.

The kilogram (kg), the SI unit of mass, shall not be used as a

unit of force, weight, or thrust.

A4.8.2 Examples of Nomenclature

The dry mass of the S-I (first) stage of the Saturn I launch

vehicle is 48 600 kg (107 139 lbm).

The weight of a man of 70.0 kg (154 Ibm) mass, standing on the

Section A4

1 February 1970

Page il

surface of the moon where the gravitational acceleration is 1.62 m/s 2, is

I13 newtons (25.4 Ibf).

The thrust of the S-I (first) stage of the Saturn I launch vehicle

is 6. 689 MN (i 504 000 Ibf).

The preferred unit of force, weight, and thrust is the newton.

A4.9.0 Measurement of Angles

A circle cannot be divided into a rational number of radians

(tad), there being 2_ radians (approximately 6. 283 tad) in a circle. How-

ever, the radian, arc degree, arc minute, and arc second may all be used

for the measurement of plane angles. Decimal multiples of the degree or

radian are preferred.

The "grad" is a unit of angular measure wherein 100 grads con-

stitute a right angle. This is not an SI unit, but, since it is based on dec-

ades it will be found useful for many purposes.

A4. 10.0 Preferred Style

In order to ensure maximum accuracy, the following style shall

be adhered to wherever practicable in engineering analysis documentation.

a. Spell out a term in full when first used, followed by the

related symbol in parentheses. Thereafter, use the related symbol for

measurement applications.

b. In general, state the measurement in terms of the system

Section A4

1 February 1970

Page 12

of units used, followed by the applicable translated value in parentheses:

for example, 48 000 kg (107 000 lbm), and 25.4 lbf (113 newtons).

c. In using numerical values involving more than three digits,

place a space between each group of three digits. Such spaces shall be

used to the right and left of decimal points. Commas are not used:

126 306. 204 359 60.

A4. 10.1 Volume

The cubic meter (m3) should be used in preference to the liter.

The liter is now defined as exactly 1 dm 3.

A4. 10.2 Time

The preferred unit of time associated with time rates is the

second.

A4.10.3 Energy

The preferred unit of energy (mechanical, electrical, thermal,

and all other forms) is the joule (J). The Btu, calorie, and kilocalorie,

although listed in this document for information, are poorly defined and

should be avoided.

A4. 10.4 Temperature

Either the Thermodynamic Kelvin Temperature Scale, the

International Practical Kelvin Temperature Scale, or the International

Practical Celsius Temperature Scale may be used. Equivalent temper-

atures in degrees Rankine, Fahrenheit, etc. , may be included in

v"

-W"

Section A4

1 February 1970

Page 13

parentheses. Note that temperature differences expressed in Kelvin

degrees (OK) and in Celsius degrees (°Cels) are numerically equal and

that degrees Celsius and degrees Centigrade are identical. See Temper-

ature Nomograph, Figure A4-2, and Table A4-15.

The International Practical Kelvin Temperature Scale of 1960,

and the International Practical Celsius Temperature Scale of 1960 are

defined by a set of interpolation equations based on the reference temper-

atures in Table A4-1.

Table A4-1. Reference Temperatures,

International Practical Temperature Scale

Temperature of o K o C

Oxygen: liquid-gas equilibrium 90. 18 -182.97

Water: solid-liquid equilibrium 273. 15 0.00

Water: solid-liquid-gas equilibrium 273. 16 0.01

Water: liquid-gas equilibrium 373. 15 100.00

Zinc: solid-liquid equilibrium 692. 655 419. 505

Sulphur: liquid-gas equilibrium 717. 75 444.6

Silver: solid-liquid equilibrium 1233.95 960.8

Gold: solid-liquid equilibrium 1336. 15 1063.0

Section A4

1 February 1970

Page 14

A4. 10.5 Prefixes

"Coherent units '_ are units that can be used directly in equations

without the application of numerical coefficients. The exclusive use of

coherent units over the entire range of numerical values of physical quan-

tities is highly desirable. As previously stated, the SI is the only complete

system of coherent units available to meet the needs of all branches of

science and technology. The full range of numerical values of physical

quantities can be represented rationally and conveniently by utilizing SI

units.

To facilitate this, either a power of ten is employed, or an

approved prefix representing a power of ten is placed before an SI unit

(or before any combination of SI units).

Only previously listed prefixes shall be used to indicate decimal

fractions or multiples of an SI unit.

A4. 11.0 Conversion Factors

Measurements in units other than those of the SI are preferably

converted by the application of approved numerical conversion factors.

The number of decimal places should be governed by the purpose

to which the information is to be put and by the degree of accuracy attain-

able with applicable measuring instruments and methods. These conver-

sion factors have been tabulated according to physical quantity.

A4. II. I Basic Linear Unit

The basic unit of measurement is the meter.

Section A4

l February 1970

Page 15

Prefixes shall be

used in accordance with A4.6.2(d). Once a prefix is chosen, no other

prefix shall be used on a drawing or in an analysis. The inch (in.) is

defined as exactly 2.54 cm.

A4. 11.2 Noncritical Conversion

If dimensions are not critical, non-SI data converted to mm

shall be rounded to convenient numbers and the word "nominal" appended

in parentheses.

A4. 11. 3 Conversion to Other SI Units

Conversion to SI units other than mm shall follow the rules as

herein set forth.

A4. 12.0 Conversion Tables

The conversion factors given in the following tables will facil-

itate the conversion of most commonly used units of the English system

to units of the Metric System (or conversion of non-SI units to SI units).

Section A4

1 February 1970

Page 16

Table A4-2. Acceleration

To Convert To Symbol Multiply by

foot/second squared

galileo (gal)

inch/second squared

meter/second squared m/s z

meter/second squared m/s$

meter/second squared m/s z

*3.048 x I0 "l

*I. 000 x I0 "z

*2.54 x I0 -z

Table A4-3. Area


sq foot sq meter m z *9. 290 304 x 10 "z

sq inch sq meter mZ *6. 451 6 x 10 .4

circular rail sq meter m z 5. 067 074 8 x 10 "1_

Table A4-4. Density


gram/cu centimeter kilogram/cu meter kg/m 3

pound mass/cu inch kiiogram/cu meter kg/m 3

pound mass/cu foot kilogram/cu meter kg/m 3

slug/cu foot kilogram/cu meter kg/m 3

*1.00 x I_

2.767 990 5 x I¢

1.601 846 3 x I0l

5.153 79 x I0 z

Table A4-5. Electrical


ampere (Int of 1948) ampere A 9. 998 35 x I0 "I

ampere hour coulomb C = A" s ,'3.60 x I03

coulomb (Int of 1948) coulomb C = A" s 9. 998 35 x i0 "l

faraday (physical) coulomb C = A.s 9. 652 19 x 104

farad (Int of 1948) farad F = A" s/V 9. 995 05 x 10 "l

henry (Int of 1948) henry H = V-s/A 1.000 495

ohm (Int of 1948) ohm _= V/A 1.000 495

gamma tesla T = Wb/m l * 1.00 x 10-9

*Exact, as defined by the National Bureau of Standards.

Section A4

1 February 1970

Page 17

Table A4-5. Electrical (Cont'd)


gauss tesla T = Wb/m z _'1.00 x 10 .4

volt (Int of 1948) volt V = W/A 1.000 330

maxwell weber Wb = V" 8 *1.00 x 10 "a[

Table A4-6. Energy


Btu (mean) joule J=N. m i. 055 87 x I0 _

calorie {mean) joule J=N. m 4. 190 02

calorie (thermochemical) joule J=N. m *4. 184

electron volt joule J=N.m 1,602 10 x 10 "19

erg joule J=N.m *1.00 x 10 "?

foot pound force joule J=N. m 1. 355 817 9

foot poundal joule J=N.m 4.214 011 0 x l0 "z

joule {Int of 1948) joule 5=N. m I. 000 165

kilowatt hour (Int of 1948) joule J=N.m 3. 600 59 x l06

ton {nuclear equiv of TNT) joule J=N. m 4.20 x 109

watt hour joule J=N. m "3, 60 x l0:

Table A4-7. Energy/Area: Time


*$Btu/sq foot.sec

$*Btu/sq foot.rain

*$Btu/sq inch.sec

erg/sq centimeter.sec

watt/sq centimeter

watt/sq meter W/m 2 I. 134 893 1 x 104

watt/aq meter W/m z 1.891 488 5 x I0 z

watt/sq meter W/m 2 1.634 246 2 x 106

watt/sq meter W/m 2 *i.00 x 10 .3

watt/sq meter W/mZ "I. 00 x 10 4


**{the rm ochemical )

Section A4

1 February 1970

Page 18

Table A4-8. Force


dyne

kilogram force (kgf}

pound force (avoirdupois)

ounce force (avoirdupois)

newton N=kg. m/sZ *1.00 x 10 -5

newton N=kg. m/p z *9. 806 65

newton N=kg. m/s z *4. 448 221 615 260 5

newton N=kg.m/s z 2.780 138 5 x 10 "l

Table A4-9. Length


angstrom meter m *1.00 x 10 "1°

astronomical unit meter m 1. 495 x 10 Ix

foot meter m *3. 048 x 10 "i

foot (U. S. survey) meter m '1200/3937

foot (U. S. survey) meter m 3. 048 006 096 x I0 "|

inch meter m "2. 54 x 10 -z

light year meter m 9. 460 55 x 10 ,s

meter wavelengths Kr 86 m *1. 650 763 73x 106

micron meter m *1.00 x 10 -6

rail meter m *2. 54 x 10 -5

mile (U. S. statute) meter m *1.609 344 x 10 3

yard meter m *9. 144 x 10 "l

Table A4-10. Mass


gram

kilogram force, secZ/meter (mass)

kilogram mass

pound mass (avoirdupois)

kilogram kg *I. O0 x I0 "3

kilogram kg *9. 806 65

kilogram kg *I. O0

kilogram kg *4. 535 923 7 x 10 "l


Section _\4

1 February 1970

Page 1 9

Table A4-10. Mass (Cont'd)

To Convert To Symbol Multiply b/

ounce mass {avoirdupois)

ounce mass {troy or apothecary)

pound mass {troy or apothecary)

slug

ton {short, 2000 pound)

kilogram kg

kilogram kg

kilogram kg

kilogram kg

kilogram kg

_;°2.834 952 3i2 5 x lO -z

*3. II0 347 68 x I0 -z

*3.732 417 216 x I0 "j

1.459 390 29 x 10 l

::'9. 071 847 4 x 10 z

Table A4-11. Miscellaneous


degree {angle)

minute {angle)

second {angle)

cu foot/second

cu foot/minute

_:"Btu/pound mass °F

l:_::Kilocalorie/kg °C

,_*Btu/pound mass

Rad (radiation dose absorbed)

roentgen

curie

radian rad 1. 745 329 251 994 3 x [0-

radian rad 2. 908 882 086 66 x 10 -4

radian rad 4. 848 136 811 x 10 -6

cu meter/second m 3 /s :'.:2.831 684 659 2 x 10 -z

cu meter/second m3/s 4. 719 474 4 x 10 -4

joule/kilogram°C J/kg°C *4. 184 x 103

joule/kilogram°C ff/kg°C *4. 184 x 10 3

3joule/kilogram J/kg 2. 324 444 4 x 10

joule/kilogram J/kg ::,1. O0 x 10 -z

coulomb/kilogram A. s/kg *2. 579 76 x 10-4

disintegration/second l/s *3. 70 x 10 l°

Table A4-1Z. Power


**Btu/second

**Btu/minute

**calorie/second

':*calorie/minute

foot pound force/second

watt W= J/s 1. 054 350 264 488 888 x 10 t

watt W= J/s 1.757 250 4 x 10t

watt W= J/s *4. 184

watt W= J/s 6.973 333 3 x 10 -z

watt W= J/s 1.355 817 9

:_'Exact, as defined by the National Bureau of Standards

::_':: (thernlochemical)

Table A4-12. Power (Cont'd)

Section A4

1 February 1970

Page 20


foot pound force/minute watt _'= 5/s 2. 259 696 6 x 10 "z

foot pound force/hour watt W= J/s 3. 766 161 0 x 10 -4

horsepower (550 ft Ib force/sec) watt W= 5/s 7.456 998 7 x l0 z

horsepower (electric) watt W= J/s *7.46 x 10 z

_;_kilocalorie/sec watt W= 5/s *4. 184 x 103

_kilocalorie/min watt W = 3/s 6. 973 333 3 x 101

watt (Int of 1948) watt W= J/s 1. 000 165

Table A4-13. Pressure


atmosphere

centimeter of mercury (0*C)

centimeter of water (4°C)

dyne / sq centimeter

newton/sqmeter N/m z *1.013 25 x 105

newton/sq meter N/m z I. 333 22 x 103

newton/sq meter N/m z 9. 806 38 x I01

newton/sq meter N/m z :',,I.00x I0 =|

foot of water (39.2°F)

inch of mercury (60°F)

inch of water (600F)

kilogram force/eq centimeter

kilogram force/sq meter

newton/sq meter N/m z 2. 988 98 x 103

newton/sq meter N/m z 3. 376 85 x 103

newton/sq meter N/m z 2. 488 4 x I0 z

newton/sq meter, N/m| *9. 806 65 x 104

nev, ton/sq meter N/m z *9. 806 65

pound force/sq inch (psi)

pound force/sq foot

millimeter of mercury (0°c)

torr (O'C)

newton/sq meter N/mZ

newton/sq meter N/m z


newton/sq meter N/m 2

6. 894 757 2 x 103

4. 788 025 8 x 101

1.333 224 x 10 z

1.333 22 x 10 z

*Exact, as defined by the National Bureau of Standards

**( the rmochemical)

Table A4- 14. Speed

Section A4

1 February 1970

Page 21


foot/second meter/second m/s '::3. 048 x 10"i

foot/minute meter/second m/s :_5. 08 x 10 -3

foot/hour meter/second m/s 8.466 666 6 x 10 "s

inch/second meter/second m/s :,_2. 54 x 10 -z

kilometer /hour meter/second m/s 2.777 777 8 x 10 -l

mile/second (U.S. statute) meter/second m/s ;:'1.609 344 x 103

mile/minute (U.S. statute) meter/second m/s '_2.682 24 x 101

mile/hour (U. S. statute) meter/second m/s :,_4. 470 4 x 10 "t

Table A4-15. Temperature

To Convert Symbol To Symbol Computation

*Celsius *Cels. *Centigrade °C *Cels. = *C

*Fahrenheit *F *Centigrade °C *C = 5/9 (*F-32)

°Rankine °R *Centigrade °C °C = 5/9 (*R-491. 69)

*Reaumur *Re °Centigrade *C °C = 5/4 *Re

*Fahrenheit *F *Celsius *Cels. *Cels. = 5/9 (0F-32)

*Fahrenheit °F *Reaumur *Re *Re = 4/9 (*F-32)

*Fahrenheit *F *Rankine °R *R = *F + 459. 69

*Rankine °R *Celsius °Cels. *Cels.= 5/9 (0R-491. 69)

*Rankine *R *Reaumur *Re *Re = 4/9 (*R-491, 69)

*Reaumur *Re *Celsius *Cels. *Cels. = 5/4 *Re

*Centigrade *C *Kelvin *K *K = *C + 273. 16

Table A4-16. Thermal Conductivity


Btu. inch/sq foot. second. *F joule/meter, second-*Kelvin J/m- s. °K 5. 188 731 5 x 10 z


Section A4

1 February 1970

Page 22

ill

m

,i,l,l,l,i,l,l,l,i,

w_

_o

,,,,I,,,,I,,,,I,,, X0

,,,,I,,,,1,,,,1,,,,C

N

,i,l,i,l,,l,I , I,0

0 0 _ _

,i,l,i,l,i,l,i,l,i !

Wli: i.h 0

g_ " g " ° " g SN _ In N I1 InN N .... I'- N

:ii'li'lililiili'lilllllllllllllilllll,liiillilillllll

,,,,I,,,,J,,,,I,,,, ,,,, ,,li Ji,J ,,,ll,,,,I,,,,I,,,I

0

I

, T T T _ ?

lillllililililliliiliilililliliiiililiililliliilililiI

0

I

o o o o o g o g

iT7iiiiiii 'Ill Ill ill ill ill Ill Ill Ill ill ill ill

s !! s s_1' N --

,l,l,l,l,l,l,l,l,i,

,s:l

bg0

0

ii)

l)

!

il

Section A4

1 February 1970

Page 23

Table A4- 17. Time

To Convert To Multiply by

day (mean solar)

day {sidereal)

hour (mean solar)

hour {sidereal)

minute (mean solar)

minute (sidereal)

month {mean calendar)

second (mean solar)

second (sidereal)

tropical year 1900,

Jan, day 0, hour 12

year {calendar)

year {sidereal)

year (tropical)

second (mean solar)

second (mean solar)

second (mean solar)

second (mean solar)

second {mean solar)

second {mean solar)

second {mean solar)

second {ephemeris)

second (mean solar)

second (ephemeris)

second (mean solar)

second (mean solar)

second (mean solar)

*8.64 x 104

8. 616 409 0 x 104

*3. 60 x 103

3. 590 170 4 x 103

*6. 00 x 10'

5.983 617 4x 10

"2. 628 x 106

Use equation of time

9. 972 695 7 x 10 -I

*3. 155 692 597 47 x IO T

*3. 153 6 x IO T

3. 155 815 0 x IO T

?3. 155 692 6 x iO

Table A4-18. Viscosity


8q foot/second

centipoise

pound mass/foot, second

pound force' second/sq foot

poise

pounds1, second/sq foot

slug/foot, second

sq meter/second m'/s

newton, second/sq meter N's/m z

newton, second/sq meter N. s/m z

newton, second/sq meter N. s/mZ

newton, second/sq meter N. s/m z



-2*9. 290 304 x 10

':'1. 00 x 10 "J

1. 488 163 9

4. 788 025 8 x 10*

':'1. 00 x 10"*

1. 488 163 9

4, 788 025 8 x I0 i

:_ Exact, as defined by the National Bureau of Standards.

TableA4-19.Volume

Section A4

1 February 1970

Page 24


fluid ounce (U.S,) cu meter m 3

cu foot cu meter m 3

gallon (U. S. liquid) cu meter m 3

cu inch cu meter m 3

liter cu meter m 3

pint (U, S. liquid) cu meter m 3

quart (U.S. liquid) cu meter m 3

ton (register) cu meter m3

*2. 957 352 956 25 x 10 -s

"2.831 684 659 2 x 10 -z

*3. 785 411 784 x 10 "3

*1.638 706 4 x 10 "s

1,000 000 x 10 -3

.4,731 764 73 x I0 -4

9. 463 529 5 x 10 -4

*2.831 684 659 2

Table A4-20. Alphabetical Listing of Conversion Factors


abampere ampere A *1.00 x 10 l

abcoulomb coulomb C= A- s * 1. 00 x 10 l

abfarad farad F= A. s/V *l. 00 x 10 #

abhenry henry H= V. s/A *1,00 x 10 -9

abmho mho * I. 00 x 109

abohm ohm f_ = V/A *I.00 x I0 -_

abvolt volt V= W/A *I.00 x I0 -s

acre sq meter m z *4. 046 856 422 4 x 10 s

ampere {Int of 1948) ampere A 9. 998 35 x 10 "i

angstrom meter m *1. 00 x 10 -l°

are sq meter m z *1.00 x 10 z

astronomical unit meter m 1.495 98 x 10 It

atmosphere newton/sq meter N/mZ *1.013 25 x 105

bar newton/sq meter N/m z *1.00 x l0 s

barn sq meter m z *I.00 x I0 "z8

* Exact, as defined by the National Bureau of Standards.

f_

Section A4

i February 1970

Page 25

Table A4-Z0. Alphabetical Listing of Conversion Factors (Cont'd)


barye newton/sq meter N/m z ::.'I.00 x I0- L

Btu (Int Steam Table) joule 3= N. m I. 055 04 x 103

3Btu (mean) joule J= N. m 1. 055 87 x I0

Btu (thermochemical) joule J= N. m I. 054 350 264 488 888 x I03

Btu (39°F) joule J= N. m I. 059 67 x 103

3Btu (60°F) joule J= N. m I. 054 68 x 10

bushel (U.S.) cu meter m 3 ,x3.523 907 016 688 x I0 "z

cable meter m "2. 194 56 x I0 z

caliber meter m ,._2. 54 x 10 -4

calorie (Int Steam Table) joule J= N. m 4. 186 8

calorie (mean) joule 3= N. m 4. 190 02

calorie (thermochemical) joule J= N, m _4. 184

calorie (15°C) joule J= N.m 4. 185 80

calorie (20°C) joule J= N. m 4. 181 90

calorie (kilogram, Int Steam Table) joule I= N.m 4. 186 8 x I03

calorie (kilogram, mean) joule J= N.m 4. 190 02 x 103

calorie (kilogram, thermochemical) joule I= N.m _._4. 184 x 103

carat (metric) kilogram kg _2. 00 x 10 -4

*Celsius (temperature) *Kelvin *K *K= °C + 273. 16

centimeter of mercury (0*C) newton/sq meter N/m z 1. 333 22 x 103

centimeter of water (4°C) newton/sq meter N/mZ 9. 806 38 x 101

1chain (surveyor or gunter) meter m _:,2.011 68 x I0

chain (engineer or ramden) meter m _:¢3.048 x i0 l

circular mil sq meter mZ 5. 067 074 8 x I0 -l°

cord cu meter m 3 3.624 556 3

-- *Exact, as defined by the National Bureau of Standards.

Section A4

1 February 1970

Page 26

Table A4-20. Alphabetical Listing of Conversion Factors (Cont'd)


-!

coulomb (Int of 1948) coulomb C= A" s 9- 998 35 x I0

cubit meter m _4.57Z x I0 "l

cup cu meter m _ _2. 365 882 365 x 10 -4

curie disintegration/second I/s *3.70 x 1010

day (mean solar) second (mean solar) *8.64 x 104

day (sidereal) second (mean solar) 8. 616 409 0 x 104

degree (angle) radian rad 1. 745 329 251 994 3 x I0 _z

denier (International) kilogram/meter kg/m *I.00 x I0 "7

dram (avoirdupois) kilogram kg :sl.771 845 195 312 5 x I0 "3

dram (troy or apothecary) kilogram kg _3. 887 934 6 x 10 -3

dram (U.S. fluid) cumeter m3 *3. 696 691 195 312 5 x l0 "_

dyne newton N-- kg.m/s z _I.00 x I0 "5

electron volt joule J= N.m 1.602 I0 x I0 "19

erg joule J= N. m ,',_I. 00 x l0-7

"Fahrenheit (temperature) "Celsius °C °C = 5/9 (°F - 32)

°Fahrenheit (temperature) °Kelvin °K °K = 5/9 (°F + 459, 69)

farad (Int of 1948) farad F= A.s/V 9. 995 05 x I0 "I

faraday (based on carbon 12) coulomb C= A, s 9. 648 70 x 104

faraday (chemical) coulomb C= A. s 9. 649 57 x 104

faraday (physical) coulomb C-- A.s 9. 652 19 x 104

fathom meter m _ I. 828 8

-15¢" rmi meter m _I. 00 x I00

j fluid ounce (U. S. ) cu meter m 3 *2. 957 352 956 25 x I0 -5

I foot meter m '_3.048 x I0 "i

foot (U. S. survey) meter m _ 1200/3937


v

Section A4

l February 1970

Page 27



-!

foot (U.S. survey) meter m 3. 048 006 096 x 10

foot of water (39. Z°F) newton/sq meter N/mZ 2. 988 98 x 10 3

foot-candle lumen/sq meter lm/m z I. 076 391 0 x l0 |

furlong meter m _:_2.011 68 x I0 z

gal meter/second squared m/sZ *I. 00 x I0 -z

gallon (British) cu meter m3 4. 546 087 x 10 -s

gallon (U.S. dry) cu meter mS _4. 404 883 770 86 x 10 -s

gallon (U.S. liquid) cu meter m s _3. 785 411 784 x 10 -s

gamma tesla T= Wb/m z ::_i.00 x i0 "9

gauss tesla T= Wb/rn z *i. 00 x I0 "4

gilbert ampere turn 7. 957 747 2 x i0 -|

gill (British) cu meter m s I. 420 652 x 10 -4

gill (U.S.) cu meter m s I. 182 941 2 x 10 -4

grad degree (angular) 1° ::_9.00 x 10 "!

grad radian rad I. 570 796 3 x i0 "z

grain kilogram kg :,_6. 479 891 x 10 -5

-sgram kilogram kg * 1. 00 x 10

hand meter m '_1. 016 x 10 "!

hectare sq meter mZ ',_ 1.00 x 104

henry (Int of 1948) henry H= V. s/A 1. 000 495

hogshead (U.S.) eu meter mS *2. 384 809 423 92 x 10 1

horsepo_ver (550 foot Ibf/second) watt W = J/s 7. 456 998 7 x 10'

horsepower (boiler) watt W= J/s 9. 809 50 x 10s

horsepower (electric) watt W = J/s *7. 46 x I0 z

horsepower (metric) watt W = .I/s 7. 354 99 x 102

_Exact, as defined by the National Bureau of Standards.

Section A4

1 February 1970

Page 28

Table ,6,4-20. Alphabetical Listing of Conversion Factors (Cont'd)


horsepower (water)

hour (mean solar)

hour (sidereal)

hundredweight (long)

hundredweight {short)

inch



inch of water (39.2°F)

inch of water {60°F)

joule (Int of 1948)

kayser

°Kelvin (temperature)

kilocalorie (Int Steam Table)

kilocalorie (mean)

kilocalorie (thermochemical)

kilogram mass

kilogram force

kilopond force

kip

knot (lnternational)

lambert

lambert

langley

Ibf(pound force, avoirdupois)

watt W= J/s 7. 460 43 x I0 z

second (mean solar) *3.60 x I0z

second (mean solar) 3. 590 170 4 x 103

kilogram kg *5,080 234 544 x I0 |

kilogram kg _4. 535 923 7 x 101

meter m *2.54 x I0 "b

newton/sq meter N/m 2 3. 386 389 x 10 3

newton/sq meter N/m2 3. 376 85 x 10 a

newton/sq meter N/m z 2. 490 82 x l0 z

newton/sq meter N/m z 2. 488 4 x 10b

joule J= N- m I. 000 165

I/meter I/m *I. 00 x 10 b

°Celsius °C °C= °K - 273. 16

joule 3= N. m 4. 186 74 x 103

joule J= N. rn 4. 190 02 x I03

joule J= N. m :_4. 184 x 103

kilogram kg * I. 00

newton N-- kg.m/s b *9. 806 65

newton N= kg.m/s z *9. 806 65

newton N= kg. m/sb *4. 448 221 615 260 5 x 103

meter/second m/s 5. 144 444 444 x I0 "i

candela/sq meter cd/mZ * I/pi x 10 4

candela/sq meter cd/m z 3. 183 098 8 x 103

joule/sq meter J/m2 *4. 184 x 104

newton N= kg.m/sZ _4.448 221 615 260 5

* Exact, as defined by the National Bureau of Standards.

_f.-

Section A4

1 February 1970

Page 29



Ibm (pound mass, avoirdupois)

league (British nautical)

league (Int nautical)

league (statute)

light-year

link (surveyor or gunter)

link (engineer or ramden)

liter

lux

maxwell

mete r

micron

rail

mile (U.S. statute)

mile (British nautical)

mile (Int nautical)

mile (U. S. nautical)

millimeter of mercury (O°C)

millibar

minute (angle)

minute (mean solar)

minute (sidereal)

month (mean calendar)

oersted

ohm (Int of 1948)

kilogram kg _4. 535 923 7 x I0 -L

meter m '_5. 559 552 x 103

meter m "5. 556 x 103

meter m :._4. 828 032 x 103

meter m 9. 460 55 x 10Is

meter m :',_2.011 68 x I0 "i

meter m #3. 048 x I0 °i

cu meter m 3 1. 000 000 x 10 -_

lumen/sq meter lm/m 2 1. 00

weber Wb= V. s *1.00 x 10 "a

wavelengths Kr 86 ::,1. 650 763 73 x l06

meter m _ 1.00 x l0 -6

meter m ;"2. 54 x l0 °5

meter m :_ 1. 609 344 x 10 _

meter m ,*1.853 184 x l03

3

meter m *1.852 x 10

meter m * 1. 852 x 10 _

newton/sq meter N/mZ 1. 333 224 x 10 z

newton/aq meter N/mZ -'*1.00 x l0 z

radian rad Z. 908 882 086 66 x 10 .4

second (mean solar) '_6.00 x i0 1

second (mean solar) 5. 983 617 4 x I01

second (mean solar) ,,_2. 628 x 106

ampere/meter A/m 7. 957 747 2 x 10 i

ohm i'l = V/A I. 000 495

SExact, as defined by the National Bureau of Standards.

Section A4

1 February 1970

Page 30



ounce mass (avoirdupois)

ounce force (avoirdupois)

ounce mass (troy or apothecary)

ounce (U. S. fluid)

pace

parsec

pascal

peck (U. S. )

pennyweight

perch

phot

pica (printers')

pint (U. S. dry)

pint (U. S. liquid)

point (printers')

poise

pole

pound mass (1bin0 avoirdupois)

pound force (lbf, avoirdupois)

pound mass (troy or apothecary)

poundal

quart (U.S. dry)

quart (U. S. liquid)

Rad (radiation dose absorbed)

° Reaumur (temperature)

kilogram kg

newton N= kg. mls z

kilogram kg

cu mete r m3

mete r m

meter m


cu meter m 3

kilogram kg

meter m

lumen/sq meter Im/m z

meter m

cu meter m_

cu meter m3

meter m

newton.second/sqmeter N. slm z

meter m

kilogram kg

newton N =

kilogram kg

newton N =

cu meter m 3

cu meter m 5

joule/kilogram J/kg

*Centigrade * C

*2. 834 952 312 5 x 10 -z

2. 780 138 5 x I0 -i

*3. 110 347 68 x 10 "z

*2. 957 352 956 25 x 10 "5

"7.62 x 10 -t

3. 083 74 x 1016

*I. O0

*8. 809 767 541 72 x I0-'L

*1.555 173 84 x 10 -3

*5. 029 2

1. O0 x 104

'_4. 217 517 6 x lO "3

*5. 505 104 713 575 x lO "4

.4.731 754 73 x lO "4

*3. 514 598 x lO -4

*1.00 x 10 -l

*5. 029 Z

*4. 535 923 7 x 10 "t

kg.m/s z *4. 448 221 615 260 5

*3.732 417 216 x 10 "1

kg.m/sZ .1.382 549 543 76x 10 "t

*1. 101 220 942 715x 10 "a

9. 463 529 5 x 10 .4

* I.00 x I0-z

"C = 5/4 "Re

*Exact, as defined by the National Bureau of Standards. _"


Section A4

1 February 1970

Page 31

]To Convert To Symbol Multiply by

rhe sq meter/newton, second m z /N. s '_1. 00 x 10 l

rod meter m 'x5. 029 2

roentgen coulomb/kilogram C/kg _2.579 76 x 10 -4

second (angle) radian rad 4. 848 136 811 x I0 "6

second (mean solar) second (ephemeris) Use equation of time.

second (sidereal) second (mean solar) 9. 972 695 7 x I0"*

section sq meter mZ ,2. 589 988 II0 336 x 106

scruple (apothecary) kilogram kg ,_I.Z95 978 2 x 10-3

shake second s I.00 x 10 "8

skein meter m ,xl. 097 Z8 x l0 z

slug kilogram kg 1.459 390 Z9 x l0 t

span meter m ,wZ. 286 x 10 "l

statampere ampere A 3. 335 640 x I0 "_

statcoulomb coulomb C = A" s 3. 335 640 x 10 "l°

statfarad farad F= A. s/V I. llZ 650 x I0 "*z

stathenry henry H= V.s/A 8. 987 554 x I0 Li

statmho mho I. 112 650 x i0 "tz

statohm ohm f/ = V/A 8._)87 554 x I011

statvolt volt V = W/A Z. 997 925 x I0 z

stere cu meter m3 *I. 00

stilb candela/sq meter cd/mZ I. 00 x l04

stoke sqmeter/second m z /s '>1.00 x 10 -4

tablespoon cu meter m 3 *I. 478 676 478 125 x l0 "s

teaspoon cu meter m 3 *4. 9Z8 921 593 75 x 10 -6


Section A4

1 February 1970

Page 32



-2

ton (assay) kilogram kg 2. 916 666 6 x I0

ton (short, 2000 pound) kilogram kg *9. 071 847 4x I0z

ton (long) kilogram kg $1. 016 046 908 8 x 103

ton (metric) kilogram kg ,_I. 00 x 103

ton (nuclear equiv, of TNT) joule J= N.m 4. 20 x 109

ton (register) cu meter m3 -*2.831 684 659 2

tort (0°C) newton/sq meter N/m z I. 333 22 x I0 z

township sq meter m z 9. 323 957 2 x I0 v

unit pole weber Wb= V.s 1.256 637 x 10 -7

volt (Int of 1948) volt V= W/A I. 000 330

watt (Int of 1948) watt W= J/s 1.000 165

yard meter m *9. 144 x 10 -l

year (calendar) second (mean solar) *3. 153 6 x l0 T

year {sidereal) second (mean solar) 3. 155 815 0 x 107

year (tropical) second (mean solar) 3. 155 692 6 x 107

year 1900, tropical, Jan, second (ephemeris) s _3. 155 592 597 47 x 107

day 0, hour 12

_Exact, as defined by the National Bureau of Standards. _:

Section A4

1 February 1970

Page 33

Table A4-Zl. Decimal and Metric Equivalents of Fractions of an Inch

Inch Inch Millimeter Centimeter Meter

Decimal Fraction (mm) (cm) (m)

0.015 625 1/64 0.396 87 0. 039 687 0. 000 396 87

0.031 25 1/32 0.793 74 0.079 374 0. 000 793 74

0.046 875 3/64 1. 190 61 O. 119 061 0.001 190 61

O. 062 5 1/16 1. 587 48 O. 158 748 O. 001 587 48

0.078 125 5/64 1.984 35 O. 198 435 0.001 984 35

0.093 75 3/32 2. 381 23 0.238 123 O. 002 381 2-3

O. 109 375 7/64 2-. 778 09 0.277 809 O. 002- 778 09

O. 125 1/8 3. 174 97 O. 317 497 O. 003 174 97

O. 140 625 9/64 3.571 83 O. 357 183 0.003 571 83

O. 156 25 5/32 3.968 71 0.396 871 0.003 968 71

O. 171 875 11/64 4. 365 57 0.436 557 0.004 365 57

O. 187 5 3/16 4.762 45 0.476 245 0.004 762- 45

0.2-03 125 13/64 5. 159 31 O. 515 931 0.005 159 31

0.2-18 75 7/3Z 5. 556 2-0 0. 555 620 0.005 556 20

0.2-34 375 15/64 5. 953 05 0. 595 305 0.005 953 05

0. 25 I/4 6. 349 94 0. 634 994 0. 006 349 94

0. 265 625 17/64 6. 746 79 0. 674 679 0. 006 746 79

0.2-81 2-5 9/32 7. 143 68 0. 714 368 0. 007 143 68

0. 296 875 19/64 7. 540 53 0.754 053 0.007 540 53

0.312 5 5/16 7.937 43 0.793 743 0.007 937 43

0.328 125 21/64 8. 334 27 0.833 42-7 0. 008 334 Z7

0.343 75 11/32- 8.731 17 0.873 117 0.008 731 17

0. 359 375 23/64 9. 128 01 0.912 801 0.009 12-8 01

0.375 3/8 9. 5Z4 91 0.952- 491 0.009 524 91

0.390 625 25/64 9.921 75 0.992 175 0.009 921 75

Section A4

1 F-ebruary 1970

Page 34v

Table A4-21. Decimal and Metric Equivalents of Fractions of an Inch (Cont'd)


Decimal Fraction (ram) (cm) (m)

0.406 25 13/32 10.318 65 1.031 865 0.010 318 65

0.421 875 27/64 10.715 49 1.071 549 0.010 715 49

0.437 5 7/16 11. 112 40 1. 111 240 0.011 112 40

0.453 125 29/64 II.509 23 I. 150 923 0.011 509 23

0.468 75 15/32 11.906 14 1. 190 614 0.011 906 14

0.484 375 31/64 12.302 97 1.230 297 0.012 302 97

0.5 1/2 12. 699 88 1. 269 988 0. 012 699 88

0. 515 625 33/64 13. 096 71 1. 309 671 0. 013 096 71

0. 531 25 17/32 13.493 62 1. 349 362 0. 013 493 62

0. 546 875 35/64 13. 890 45 1. 389 045 0. 013 890 45

0.562 5 9/16 14.287 37 1.428 737 0.014 287 37

0.578 125 37/64 14. 684 19 1.468 419 0.014 684 19

0.593 75 19/32 15.081 11 1.508 111 0.015 081 11

0.609 375 39/64 15.477 93 1.547 793 0.015 477 93

0. 625 5/8 15. 874 85 1. 587 485 0. 015 874 85

0.640 625 41/64 16. 271 67 1.627 167 0.016 271 67

O. 656 25 21/32 16. 668 59 1. 666 859 O. 016 668 59

O. 671 875 43/64 17. 065 41 1. 706 541 O. 017 065 41

0.687 5 I1/16 17.462 34 1.746 234 0.017 462 34

0.703 125 45/64 17. 859 15 1.785 915 0.017 859 15

0.718 75 23/32 18.256 08 1.825 608 0.018 256 08

O. 734 375 47/64 18. 652 89 1.865 289 O. 018 652 89

O. 75 3/4 19. 049 82 I. 904 982 O. 019 049 82

O. 765 625 49/64 19.446 63 1.944 663 O. 019 446 63

0.781 25 25/32 19.843 56 1.984 356 0.019 843 56

SECTION B

STRENGTHANALYS I S

SECTION B l

JOINTS AND FASTENERS

TABLE OF CONTENTS

BI.0.O Joints and Fasteners

Page

1.1.0 Mechanical Joints and Fasteners .................

i.i.i Riveted Joints ..............................

1.1.2 Bolted Joints ...............................

1.1.3 Protruding-Head Rivets and Bolts ............

1.1.4 Flush Rivets ................................

1.1.5 Flush Screws ................................

1.1.6 Blind Rivets ................................

1.1.7 Hollow-End Rivets ...........................

1.1.8 Hi-Shear Rivets .............................

1.1.9 Lockbolts ...................................

i.i.i0 Jo-Bolts ...................................

1.2.0 Welded Joints ...................................

1.2.1 Fusion Welding ..............................

1.2.2 Effect on Adjacent Parent Metal Due to

Fusion Welding ............................

1.2.3 Weld-Metal Allowable Strength ...............

1.2.4 Welded Cluster ..............................

1.2.5 Flash Welding ...............................

1.2.6 Spot Welding ................................

1.2.7 Reduction in Tensile Strength of Parent Metal

Due to Spot Welding .......................

1.3.0 Brazing .........................................

1.3.1 Copper Brazing ..............................

1.3.2 Silver Brazing ..............................

I

i

2

2

19

24

27

39

39

39

41

46

46

46

47

49

49

5O

56

59

59

59

Bl-iii

B 1.0.0 Joints and Fasteners

Section B i

25 September 1961

Page i

B I.i.0 Mechanical Joints and Fasteners

B i.i.i Riveted Joints

Although the actual state of stress in a riveted joint is complex,

it is customary to ignore such considerations as stress concentration

at the edge of rivet holes, unequal division o£ load among fasteners,

and nonuniform distribution of shear stress across the section of the

rivet and of the bearing stress between rivet and plate. Simplifying

assumptions are made, which are summarized as follows:

(l) The applied load is assumed to be transmitted entirely by

the rivets, friction between the connected plates being

ignored.

(2) When the center of cross-sectional area of each of the rivets

is on the line of action of the load, or when the centroid of

the total rivet area is on this line, the rivets of the joint

are assumed to carry equal parts of the load if of the same size;

and to be loaded proportionally to their section areas otherwise.

(3) The shear stress is assumed to be uniformly distributed across

the rivet section.

(4) The bearing stress between plate and rivet is assumed to be

uniformly distributed over an area equal to the rivet diameter

times the plate thickness.

(5) The stress in a tension member is assumed to be uniformly

distributed over the net area.

(6) The stress in a compression member is assumed to be uniformly

distributed over the gross area.

The design of riveted joints on the basis of these assumptions is

the accepted practice, although none of them is strictly correct.

iA

The possibility of secondary failure due to secondary causes, such

as the shearing or tearing out of a plate between rivet and edge of

plate or between adjacent rivets, the bending or insufficient upsetting

of long rivets, or tensile failure along a zigzag line when rivets are

staggered, are guarded against in standard specifications by provisions

summarized as follows:

(1) The distance from a rivet to a sheared edge shall not be less

than 1 3/4 diameters, or to a planed or rolled edge, I 1/2

diameters.

r-

- (2) The minimum rivet spacing shall be 3 diameters.

BI.I.I Riveted Joints (Cont'd_

Section B 1

25 September 1961

Page 2

(3) The maximum rivet pitch in the direction of stress shall be

7 diameters, and at the ends of a compression member it shall

be 4 diameters for a distance equal to 1 1/2 times the widthof the member.

(4) In the case of a diagonal or zigzag chain of holes extending

across a part, the net width of the part shall be obtained

by deducting from the gross width the sum of the diameters of

all the holes in the chain, and adding, for each gauge space

in the chain, the quantity $2/4g, where S = longitudinal

spacing of any two successive holes in the chain and g _ the

spacing transverse to the direction of stress of the same

two holes. The critical net section of the part is obtained

from that chain which gives the least net width.

(5) The shear and bearing stresses shall be calculated on the

basis of the nominal rivet diameter, the tensile stresses on

the hole diameter.

If the rivets of a joint are so arranged that the line of action

of the load does not pass through the centroid of the rivet areas then

the effect of eccentricity must be taken into account.

B 1.1.2 Bolted Joints

Bolted joints that are designed on the basis of shear and bearing

are analyzed in the same way as riveted joints. The simplifying assump-

tions listed in Section B i.i.i are valid for short bolts where bending

of the shank is negligible.

In general when bolts are designed by tension, the Factor of

Safety should be at least 1.5 based on design load to take care of

eccentricities which are impossible to eliminate in practicaldesign.

Avoid the use of aluminum bolts in tension.

Hole-filling fasteners (such as conventional solid rivets) should

not be combined with non-hole-filling fasteners (such as conventional

bolt or screw installation).0

B 1.1.3 Protruding-Head Rivets and Bolts

The load per rivet or bolt, at which the shear or bearing type of

failure occurs, is separately calculated and the lower of the two

governs the design. The ultimate shear and tension stress, and the

ultimate loads for steel AN bolts and pins are given in Table B 1.1.3.1

and B 1.1.3.2. Interaction curves for combined shear and tension load-

ing on AN bolts are given in Fig. B 1.1.3-1. Shear and tension ultimate

loads for MS internal wrenching bolts are specified in Table B 1.1.3.3.

Section B I

25 September 1961

Page 3

B 1.1.3 Protruding-Head Rivets and Bolts (Cont'd)

In computing aluminum rivet shear strength, the correction factors

given in Table B 1.1.3.5 should be used to compensate for the reductions

in rivet shear strength resulting from high bearing stresses on the

rivet at D/t ratios in excess of 3.0 for single-shear joints, and 1.5

for double-shear joints. The basic shear strength for protruding-head

aluminum-alloy rivets is given in Table B 1.1.3.6.

The yield and ultimate bearing stresses for various materials at

room and elevated temperatures are given in the strength properties

stated for each alloy or group of alloys, and are applicable to riveted

or bolted joints where cylindrical holes are used and where D/t < 5.5.

Where D/t _ 5.5, tests to substantiate yield and ultimate bearing

strengths must be performed. These bearing stresses are applicable only

for the design of rigid joints where there is no possibility of relative

motion of the parts joined without deformation of the parts. Yield and

ultimate stresses at low temperatures will be higher than those specified

for room temperature; however, no quantitative data are available.

For convenience, "unit" sheet bearing strength for rivets, based

on a stress of I00 ksi and nominal hole diameters, is given in Table

B 1.1.3.7. Factors representing the ratio of actual sheet bearing

strength to i00 ksi are given in Table B 1.1.3.8. Table B 1.1.3.9

contains unit bearing strength of sheets on bolts. For magnesium-alloy

riveting, it is unnecessary to use the correction factors of Table

B 1.1.3.5, which account for high bearing stresses on the rivet.

B 1.1.3

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25 September

Page 4

1961

BI.I.3 Protruding-Head Rivets and Bolts (Cont'd)

Section B i

25 September

Page 5

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Section B 1

25 September 1961

Page 6

!AN-103O r--_._.

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Shear Load, Kips

Note: Curves not applicable where shear

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with nuts fingertight.

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Section B 1

25 September

Page 8

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Section B 1

25 September 1961

Page 13

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OOOOOOOOOOOOOOOOOOO

Section B i

25 September 1961

Page 14

=.,-4

¢J

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am

=

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ca

BI.I.3

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m

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P_otruding-Head Rivets and Bolts (Cont'd)

Section B I

25 September

Page 15

1961

_-_ • • • • • • • • • ° •

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u_

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Section B i

25 September 1961

Page 16

BIol.3 Protruding-Head Rivets and Bolts (Cont'd)

Section B 1

25 September 1961

Page 17

28

24

20

16

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12

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Bolt or Pin Diai I

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0 .04 .08 .12 .16 .20 .24 .28

Sheet Thk, in

Fig. B 1.1.3-4 Unit: Bearing Strengths of Sheets on Bolts and Pins

Fbr = 100 ksi

BI.I.3

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Protruding-Head Rivets and Bolts (cont'd)

"_ 0 0

O u_O0 O I_.

000OO_O0_

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="_ _ OOOOOO

0 OO0Om_'_ _O_OO_

_ _O_O_OO__ _O_OO_> _ "_ _ O _ _ _ O _

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O OOOOOO OOO__

Section B 1

25 September 1961

Page 18

I

>

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Z

B 1.1.4 Flush Rivets

Section B 1

25 September 1961

Page 19

Table B 1.1.4.1 through B 1.1.4.3 contain ultimate and yield

allowable single-shear strength values for both machine-countersunk

and dimpled flush riveted joints employing solid rivets with a head

angle of i00 °. These strength values are applicable when the edge

distance is equal to or greater than two times the nominal rivet dia-

meter. Other strength values and edge distances may be used if sub-

stantiated by tests.

The allowable ultimate loads were established from test data using

the average failing load divided by a factor of 1.15. The yield loads

were established from test data wherein the yield load was defined as

the average test load at which the following permanent set across the

joint is developed:

(i) 0.005 inch, up to and including 3/16 inch diameter rivets.

(2) 2.5 percent of the rivet diameter for rivet sizes largerthan 3/16 inch diameter.

1.1.4 Flush Rivets (Cont 'd )

Section B 1

25 September 1961

Page 20

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Section B 1

25 September 1961

Page 21

B 1.1.4 Flush Rivets (Cont 'd_

Section B i

25 September 1961

Page 22

4J

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o7oO_

4J

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eq _eq0 0eq ¢q

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I I-4" "_-._

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C',l C4 _1 I:: C'4000 _ 0¢q C_l _'%1 C',l

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.q-

00 ".D

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eq _1 i_

a _ i

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cO -..1"

c'q _ eq0 0o-1 c'.l

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1' 0 i

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¢q ¢q 04

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Section B 1

25 September 1961

Page 23

B i.i.5 Flush Screws

Section B 1

25 September 1961

Page 24

Table B 1.1.5.1 contains ultimate and yield allowable strengthvalues for i00 ° flush-head screws with recessed heads installed in

machlne-countersunk clad 2024 and 7075 sheet. These strength values

are applicable when the edge distance is equal to or greater than two

times the nominal screw diameter. Other strength values and edge

distances may be used if substantiated by tests.

These strength values may be used for the design of dimpled joints.

Higher values may be used for dimpled joints if based on test results.

The allowable ultimate loads were established from test data using

the average failing load divided by a factor of 1.15. The yield loads

were established from test data, wherein the yield load was defined as

the average test load at which the following permanent set across the

joint is developed:

(i) 0.012 inch, up to and including 1/4 inch diameter screws.

(2) 4.0 percent of the screw diameter for screw sizes larger than

1/4 inch diameter.

The test specimens used were made up of two equal-gage sheets lap

jointed and machine countersunk with washers to build up thickness to

minimum grip. All joints had 2D nominal edge distance in the direction

of the load and were either of the three-screws-across or the two-

screws-ln-tandem type. For the latter type, the flush heads were placed

on opposite sides of the joint to assure 2D edge distances.

B 1.1.5Flush Screws (Cont'd}

cq

00°r4O CO

0,--_0'_ r'_

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_m_o_o : :o_ _ o _ o_ _o c_ _- c_ c_ • : :._o o_ e_ _ _ _- oo _ _-_ • • : :

_D _O _ . :

u : : ..... " : "_: :..::::• . .... • : : .: : :._ . • : . : :i !ii.!::.. :

: • • : : : : : • :_ . : ' : : : . : • :

• • : : : .' . :_'_ : : • : . . : •• • _,__ eq O O oh ,-_ O O 6 u_ O O eq m

_ co -.1- u'_ ',O r-- 0o 0", O c-,I ,..o o'_ u'_ ,-_ r--• ,-I O O O O O O O ,-_ ,--_ ,-_ ,-_ ¢q co coi--! . ° . ° ..........4J O O O O O O O C) O O O O O O

J.J

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Section B 1

25 September

Page 25

1961

BI.I.5 Fldsh Screws (Cont' d)

Section B 1

25 September 1961

Page 26

=o

0

Q;

_J

o

I

=

C;

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q_

0

4J

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000000000000 O0

B 1.1.6 Blind Rivets

Section B 1

25 September 1961

Page 27

Tables B 1.1.6.1 through B 1.1.6.6 contain ultimate and yield

allowable single-shear strengths for protruding and flush-head blind

rivets. These strengths are applicable only when the grip lengths and

rivet-hole tolerances are as recommended by the respective manufacturers,

and may be substantially reduced if oversize holes or improper grip

lengths are used.

The strength values were established from test data obtained from

tests of specimens having values of e/D equal to or greater than 2.0.

Where e/D values less than 2.0 are used, tests to substantiate yield

and ultimate strengths must be made. Ultimate strength values of pro-

truding and flush blind rivets were obtained from the average failing

load of test specimens divided by 1.15. Yield strength values were

obtained from average yield load test data wherein the yield load is

defined as the load at which the following permanent set across the

joint is developed:

(i) 0.005 inch, up to and including 3/16 inch diameter rivets.

(2) 2.5 percent of the rivet diameter for rivet sizes larger than

3/16 inch diameter.

For tables B 1.1.6.2 and B 1.1.6.3 the ultimate rivet shear strength

was based on the comparable rivet shear strength of 2117 solid rivets,

as noted in table B 1.1.6.3. Test data on which the strength values of

these tables were based were obtained using standard degreased clad

2024-T4 specimens.

In view of the wide variance in dimpling methods and tolerances

for aluminum and magnesium alloys, no standard or uniform load allow-

ables are recommended. Allowables for ultimate and shear strengths of

blind rivets in double-dimpled or dimpled, machine-countersunk applica-

tion should be established on the basis of specific tests. In the

absence of such data, allowables for blind rivets in machine-countersunk

sheet may be used.

Since blind rivets are primarily shear-type fasteners, they should

not be used in applications where appreciable tensile loads on the

rivets will exist. Reference should be made to the requirements of the

applicable use of blind rivets, such as the limitations of usage on

Drawing MS33522.

B'I,I.6

>

D

0

0=

_0 _JoJ

_J o_oO_C _n

_o o;

J

_ oW

= 0

1.1 ,,M

,r.l

ld

Blind Rivets, (Cont'd_

Section B 1

25 September 1961

Page 28

O eqeO

O

0_0,-4 u_

0_0 _.j ,,,I"

O '_-t

IJ

u'3

0•,_ _ _ 0 _

_J _ _J .,_

W4_

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r.. _o e,I o4 0 o-_ 0.._ e,_c.4 e,I .4..._ u_ oq ,_ _Do.__-,-_._. u,% %o r-. oo o_ oxch

eq ¢,4 u'-, ,t.i _ u-_ _0 • •u-_ 0 ,-I 0o ..J- oo o,,i • °

OO

eq N

o ."eq •

i!::: i

: . :: : :

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! 11 :

: :: :: :: :

: :: :

i!!i::::Be •• , _oo_oooo• : : • __ __ _

: : : :_o__

0 ¢Xl ¢_I _0 ¢xl 0 u'_ O0 ,--I

"_'i : : " :_.••_= : : : • .'ii

A'

._ 000 O0 O00000_ _ _

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B 1.1.6

O>

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U

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q4 4...I

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4J 4,,.Io'3 o0

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0_ -,4

_ mnJ o

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•,-i __ ._

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r_

b-t

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.,44J

0

"0

m

>,

oIJ

0

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ffl4..i

4_1

o

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ill4_

.,4

ffl

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Section B i

25 September

Page 29

1961

O

O

=.,4

I.0

u_

0.,4,4.1

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0

.,-I

0

0 0

4.1

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"tJ ,-'4

0 m

0

_-I 4-1

0 ,"4

m _

0 ,-4,.=:

= O

0

B 1.1.6 Blind Rivets (Cont'd)

Section B 1

25 September 1961

Page 30

,_ ,-.

_-_ _ >_ _ ._

co > _ >i•_ _ •

H od co =_I

_0_00_00_00_

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_O_,

: : : : : :

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: : :

i"_ : : • : : : :. : : :

_00000 O000 _ __,°,....°,°.

0000 O00 O000

co

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Section B 1

25 September 1961

Page 31

O4

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D_.J

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O%0OeqI

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eq, o_ =

,=

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4-1 I _.._ I,.4m O

m _ _ m

.-4

.4

-H.61

_ °

c_ .M

,-_ 0 ;_

m _

(Cont'd)

4_

_J

4J

..44J_=_

: :: :

u'% o_ ¢q %0 00 O0 • •

_o_o : :• :: :

• : : • . . : :

" " " i : : : :: : : : : : :

• : : : : : • :

• i iil• • . . . : :

• •• : • i ": : : • : . : :

• : : : . •: : : . : : : :: : • : : : : :ii::

• : : : : :o iii.:::_

i: : : : . : •• i ": : : : : :

_ . : : • : : • ._.!!:..::_ : " • • • : :

._000000_

_00000000

Section B i

25 September 1961

Page 32

B 1.1.6 Blind Rivets (Cont'dl

Section B I

25 September 196.1

Page 33

o

B 1.1.6

L

4J _O

en c_CO O

_ o m

_ _- -_ .4"

_ 0 _

o _

g _-_ _-_-

o _

CO _

"_ _, ,

_ u3

_JIJ

_D

Blind Rivets (Cont'd)

o,o : :O_oh • •_ : :

!

4J

4=1

Section B I

25 September 1961

Page 34

BI.I.6 Blind Rivets (Cont 'd)

Section B i

25 September 1961

Page 35

_J0303

CO

o_03cOO

c

_J03

>

c

<

_Du_Ou_

C

Ot_

_J

C03

D_

"O

03

C

03_J

E

4J,-q

,-4

03

=-_J

C

_JO'3

_JcO

L)v

_D eq0 c_

U3 "_-

03

0 _

0

03 N,._ 0 ¢'_,

_ 03

_-4 ',0o'3 _ "_

' •_,4

0

,_ i o(3

Lr_

OO

d

v

03

0 _ (D

m > 03 >

• _0 _0_

u)03

>

B 1.1.6

oo "o

> z=

_ooo ,-4

I

<

"o

oO _=

"O oO

o,-4

0 .w

0

.= ,_i

IlJW

.,4

.-.I

11

o2

00=

'O

o

Blind Rivets

_o,-4

(-4o_

u_

oo

o2

eq

oo¢qo

,--I

o2 ,.-4

,-4

0

¢q

I

W ,.-4 _.

m

O¢O ¢O

u,-i ,-4O _"

u'3

(Cont'd)

v

O O2 _ _ O2

_ _J _J o2 _J

.=

bo

w

a.;

o"_ p._ c.q

Ou"_ ,--t

|

o_ i " :,m •

: : : . :• • •

• i

•: : : : :• • : : :

• Q #

ooo

_ ,-4 ¢q

ooo

,-4 u3 0 ¢'_ u% "O_ 0"_ -,T 0 ',.0*r-. oo 0 ¢q C.4 •

,'--_,--_ ,,-_ ,

r'...C'q O_ 0 ,,

,,c,r-ooo_ " :

.!!

: : : : :.

• : : • •

: : • : : •

• . : : :

•• • : : :_: : • : 1_• • : • .

• ,, • O u'_u% O O: : : : _,-_=_: . : :_.o. .

: : _ _ooo_ :

: : : _ooooo: : : _o_,_n

• • ,--II_, ,--Ii"_ u_ WD: : " N__

oo0

u% eq o_,c'q ¢0 c,'_

oo :,-I_DO,-'4

• • |

: I !::i

O0 •

: ::!• . u_o O0 O0 . . : .• • _0 r-,,(%1 0 00 u'% • •

:i i :• _ Ne4eq :

!: . : " :: _ooo_o : i• _o_<o_o_ • : " :• o0o_.._.._ : • " " .

: : : : : .o_oooo : : . • : :_o0000o_o_ : " : . : .

:!'" i": : : : : : • • : : :"i'ii''''": : : .... . :

eo • 4 • • • • • • • • •

o: i i i i ....:... . .m • : • • • • : •o2 • : • • • . • • •

: • : : : : : ....

oo,A_... G,-',,_ooo_,.o•,.-4C',le4 cq .4- u"_ xo r.,.00 0'_0 e,l _.-'_.=: 000000000 ,--4,...-4,.-..44J • • • • • • • • • • • .

0000 O0 O00000

o2

X_u_

Section B i

25 September 1961

Page 36

OL3v

>

cg

_OcO¢q

°_

_0

CO

_J

O

O

CO

tJ

_JJ

+lr-_

tC_

r-q


Section B 1

25 September 1961

Page 37

Ce_

= CO

"_ _ OO

O

,,-I

ffl

ao ._

0

.iJ

_ ,,-p!

_ m.t..1

_0 u--i 0

°r-.I _/3 ,--4

_ j

0 _

r_

¢'_ -- 0

,.--I • °

co ._ : :

"<I"

_Op-q

ce_

cq

u_

oo

,---4

v

0 _ _ qJ _J

c_ ._ ,t_ c_ J .,_

.._ _n,.o,.o r... r-. :!

0 _ ,._ u_ r_

._u_u_. : : : ' " .:

or--00o_ : : : : i "r-.r--cooo • • • : :

m m _ _,-, : : : . : •

_ .'. : : , •

_ooo_c • : : : : :

_m_m : : : : : •

: : • . • . :_o_ . . : : : : .

• .m _ _ • • : • : • •

: : i : i ! ooooo ..: . : • • • o_._-_ _• : : : : :x_ _e4_

: : : oooooum :

• • •" _0 00 0 cq ._ _ "• " *.I_ ,_ ,--I ,--I ,.._ ,--I :

. i

. • u'_ 0 0 0 0 0 0

• :_ _ .

: _o u_o mo : : i

: : : . • , . •

, : • : ...... , •

• • • • e • • • • • • •• i• .... ..

• • • • i i+ • • • • • •

. . • • i , - • • | - -

• • • • I • • • • • + •

I

: : • ; : : • : • : :• . • . . . : . . . . O

• : . • . . . . co. • . : : : : : : :

: : : ' " • . . . oo: : : : : : i : ' :

C 0 0 0 0 0 0 0 0 ,--_ ,-_ ,--_ _n

oooodo_ooSo_

"1D

.,-I

0

,--I

0

{_0 .,-1

0

0

_ "_

i i .I...I

_ m

t.M

0

>

•,..4 .M

t_ ,._


Section B i

25 September 1961

Page 38

Table B 1.1.6.6 Explosive Rivets, DuPont Extended Cavity

Rivet

Size

Ultimate Rivet Load, Lb/Rivet

.025 .032 .064 .072

5/32 320 410 610 610

3/16 -- 495 880 880

Sheet Gauge.'o4o .o51

513 610

620 796

.081

880

B 1.1.7 Hollow-End Rivets

Section B i

25 September 1961

Page 39

If hollow-end rivets with solid cross sections for a portion of

the length (AN 450) are used, the strength of these rivets may be taken

equal to the strength of solid rivets of the same material, provided

that the bottom of the cavity is at least 25 percent of the rivet dia-

meter from the plane of shear, as measured toward the hollow end, and

further provided that they are used in locations where they will not

be subjected to appreciable tensile stresses.

B 1.1.8 Hi-Shear Rivets

The allowable shear load for "Hi-Shear" rivets is the same as

that specified for the standard aircraft bolts heat treated to 125 ksi

and given in table B 1.1.3.2.

B 1.1.9 Lockbolts

Lockbolts and lockbolt stumps shall be installed in conformance

with the lockbolt manufacturer's recommended practices, and shall be

inspected in accordance with procedures recommended by the manufacturer

or by an equivalent method. The ultimate allowable shear and tensile

strengths for protruding and flush-head Huck lockbolts and lockbolt

stumps are contained in table B 1.1.9,1. These strength values were

established from test data and are minimum values guaranteed by the

manufacturer. Shear and tensile yield strengths and ultimate and

yield bearing strengths will be added when available.

For all lockbolts but the BL type (blind) under combined loading

of shear and tension installed in material having a thickness large

enough to make the shear cutoff strength critical for the shear load-

ing, the following interaction equations are applicable:

Steel lockbolts - R t + Rs I0 = 1.0, 7075-T6 lockbolts - R t + Rs 5 = 1.0,

where R t and Rs, are the ratios of applied load to allowable load in

tension and shear, respectively.

B 1.1.9 Lockbolts (Cont' d)

Section B i

25 September 1961

Page 40

.0 _O

W C

•,4 _ O_

C _o c0_ c

,-4

C

0 0 _ I_ m

CO _ r_

u,-i0 cn rn

,-4 0 ,-_ ;>_ ,-_ ",4

t_.r,I

o_

0__

_0000

C_g

:C_gg

"00_0

: oooo

• %0 _D CO u'%

• ,-4

: • : : :

'i'2"

W>

.M

W

m

gm

0

>

C

!

0

U

3m

o

I

_J

0J

0

N0

I

0

O0

0

Ic_

C0

c

0

0

"0_J

m

I

k_

0O

0_.c

0

0_0

A_

A_I

_M

1..1

orD

r

B I.I.i0 Jo-Bolts

Section B I

25 September 1961

Page 41

The ultimate and yield allowable shear strengths for flush-head

steel and aluminum Jo-Bolts in clad aluminum-alloy sheet are given inTables B i.I.i0.i and B 1.1.10.2.

B I.i.i0 Jo-Bolts (Cont'd)

Section B i

25 September

Page 42

1961

_J

,-a

0 _0

I I0

0

_J

0 "_

4-J o_

_J<t

CO

•,.4 &.l

._"_ 0

c_

,-.-I

_J

[--I

0

,,.,-4

2 o

4.1m

00CM

m

r_

,--I

U

,=

[--I!

0

[--I!

-.1"

0

!u'3

0p_

I

"4"

0

[-4!

ue_

0r_

[-.II

0

&J

O0 0 O0 O0 0 O0 0_ _ __ _0 ___0__

O0000 O00 O00

00 O0000000 O0_0__0_ __

00 O0 O0 O0 O000_ _0 _0 _ _ _ __ O_ _ _ _ _

0_0000000000_ 00_0_ _ O_ _

O_ O000000 O00_0__0__ _0 _ _ _ _ _ _

_0,,.,,0t,,,,,,

0•_ 00_ 0 O0 _000_ _

_ __0_ ___0 O0 O000_ _ _ _

_0 O000000000000

FB i.i.i0

o_o

! !Ou'_

0

o9

0"0

c,h

0-,1"

0o'3 cxl

.._0o9

"_ 0-

0

•,-I ¢_ i:_.•IJ ._ 0

Jo-Bolts

_o

r..3 ooq i-_

,.el

_.3 ot.q

n:l i

.i.J0_

o

n:l i

cxl

N

w

00

B

04

O_

0

_9

0.;.-I

Section B i

25 September 1961

Page 43

00000000000

00000000000_0_0_0_

000000000000

___0_

000000000000___0_00_ __ _0 _

000000_00000_0______0__

0000_0000000___0000

0°..,.0,.°,°,

.°,°_°°,,°°°°°

.°0°.,°,°°.°°°,°.°°.°°°°°°°..,,°°°,_°,°°,°°,°.._°.°°.

.............. ::

._,.,,,,°,,,,,,,.,,.,.,..,.,.,

_,°,,.,°,,,,,,._o,,,,,,,,,,,,.0,.°,....,.,°._..°°,.°,...,.

_,....,°....°°,

•_ O0 _ _ 000 _ 000_ ____0__ __0000000___

°,,..°.°.,,..°_00000 O0 O0 O0000

,-..I

>

,--.I

,,.4

.,-4

0

cM

,x:

,....4

0t_

u]0

,-.4

"0

0

t_

BI.I.IO Jo Bolts (Cont'd)

Section B 1

25 September 1961

Page 44

0 _0

! Io u_

0

-,4

0

ODeq= 0OJ ¢q

r-4

_._r._

•,4 o

•,4 _•_ o_ U

m u

,...-4

Ill,--4

Ol

0

<

0

!<

I.I

Ill,=

00

I

<

111

E.-,

U

r._

E.-,I

@q0

E-_I

_c_

0

!

0

OJ

00000 00000

00 O0000 O000

___0_

_z4J

4_

00000000 __0___

_ _ _ _ O_ _ _

.,4

,-4

00000000__ O_ ____0__

°°°,°°,,.,°o,°.o°°°.,,o,,,°.,o°,,,.,,...o°.°,o,o°,°o°°,,.,°°

,oo°l,°l,,,**

._,,°,,.°°°°°°°°.,°°.,o°o°

_,,,°,°.,,o,°_o,,.,°,,,.°°_,o°,°°,,o°°,_,,°,,',°,,°°o

_.°,°°°o,°°°°

•M _ 00_ _ O00 _0 O0_ __ _0_ _ __0000000__

_000000 O00000

B i.I.i0

_J

4"O

¢j

Oc_

t

OOcq

COo4

_0

•_ O

0_ O

_ 0 0,.,_ L) ,--4

[--i

Jo-Bolts (Cont' d)

L) Op..

O_D¢q

4

..D rj 0c',4

2

g4-1m

0

m _

_ r_

00

I<

• ,-4

.,-.4

0

,-.40

.,,.4

Section B 1

25 September 1961

Page 45

0000000000

00_0000000

00000_0000

0000000000

,°°_._°°.°°.°.,°°...°.°,°.,.°0,..°.°.°°°°°°0o.o,

.o°_°.0°°°,°0°°,°°°0.°._..°

_o°,.°,°.°0°,._°°°°,.°,°,°°

°.. .................. °,°.

_.°q.0.°,°,,°_°°,..°°..,°°_°°,°,.o°...°

o._00_000_000___0___ 0000000__

o

>

IJ

0

B 1.2.0 Welded Joints

Section B i

25 September 1961

Page 46

Whenever possible, joints to be welded should be so designed that

the welds will be loaded in shear.

B 1.2.1 Fusion Weldin_ - Arc and Gas

In the design of welded joints, the strength of both the weld

metal and the adjacent parent metal must be considered. The allowable

strength for the adjacent parent metal is given in section B 1.2.2 and

the allowable strength for the weld metal is given in section B 1.2.3.

The weld-metal section will be analyzed on the basis of its loading,

allowables, dimensions, and geometry.

B 1.2.2 Effect on Adjacent Parent Metal Due to Fusion Weldin_

For joints welded after heat treatment, the allowable stresses

near the weld are given in Tables B 1.2.2.1 and B 1.2.2.2.

For materials heat treated after welding, the allowable stresses

in the parent metal near a welded joint may equal the allowable stress

for the material in the heat-treated condition as given in tables of

design mechanical properties of the specific alloys.

Table B 1.2.2.1 Allowable Ultimate Tensile Stresses

Near Fusion Welds in 4130, 4140, 4340, or 8630 Steels a

(Section thickness 1/4 inch or less)

Type of joint Ultimate tensile

bTapered joints of 30 ° or less

All others

stress, ksi

90

8O

a Welded after heat treatment or normalized after weld.

b Gussets or plate inserts considered 0° taper with center line.

Table B 1.2.2.2 Allowable Bending Modulus of Rupture Near

Fusion Welds in 4130, 4140, 4340, or 8630 Steels a

Type of joint Bending modulus

of rupture, ksi

Tapered joints of 30 ° or less b

All others

Fb, figure B 1.2.2-1 for

Ftu = 90 ksi

0.9 of the values of Fb

from figure B 1.2.2-1

for Ftu - 90 ksi

a Welded after heat treatment or normalized after weld.

b Gussets or plate inserts considered 0° taper with center line.

B 1.2.2

Section B i25 September1961Page 47

Effect on Adiacent Parent Metal Due to Fusion Welding (Cont'd)

450

400

350

o_300 _"

Ftu , 260 kgl

250 ,,ok.,

150 _ _ _'_ Flw , IN kjl

100

Ptu' 90 ksl50 ......... J

0 I0 20 30 40 50 60

D/t

Fig. B 1.2.2-1 Bending Modulus of

Rupture for Round

Alloy-Steel Tubing.

B 1.2.3 Weld-Metal Allowable Strength

Allowable weld-metal strengths are shown in Table B 1.2.3.1.

Design allowable stresses for the weld metal are based on 85 percent

of the respective minimum tensile ultimate test values.

B 1.2.3 Weld-Metal Allowable Strength (Cont'd)

•I._ _ u'_ u'_ _ oO 0

o'1 u_ 0

oh

0_D

I

,-4 0 0U 0 0 ¢_1

,.--4 I _ I I

O

0 '_ _ _ _ _ __'_ 0 _o _ co _ _

0 _ 0 0 O

,---4 ",_ _ O_ _ _OO OO

0 I i I I I 0 I c,,I

c/_ I I I I I I I I

•l-I i.-II---I I_ I_ l-.I I-.I

¢)

C_

[-_

O0

._..I

._, 0

0 0_ Z

> >

co o0 _=co O

O_ O_ CY

O

!

I

O_

u'_O0

t_l _ oO

c_

cO

0

=

=O

O

o0 o0

0 0 0,.-i ,--I ,-.I,..-I ,.-.I ,-.I,< < ,<

,.--4

O_

0 O0_ -_'1" <1"

-_ "<1" '<1"

Section B I

25 September 1961

Page 48

Section B 1

25 September 1961

Page 49

B 1.2.4 Welded Cluster

In a welded structure where seven or more members converge, the

allowable stress shall be determined by dividing the normal allowable

stress by a material factor of 1.5, unless the joint is reinforced. A

tube that is continuous through a joint should be assumed as two members.

B 1.2.5 Flash Weldin_

The tensile ultimate allowable stresses and bending allowable

modulus of rupture for flash welds are given in Tables B 1.2.5.1 and

B 1.2.5.2.

Table B 1.2.5.1 Allowable Ultimate Tensile

Stress for Flash Welds in Steel Tubing

Tubing

Normalized tubing - not heat treated

(including normalizing) after

welding.

Heat-treated tubing welded after heat

treatment.

Tubing heat treated (including normal-

izing) after welding. Ftu of

unwelded material in heat-treated

condition:

< i00 ksi

I00 to 150 ksi

>150 ksi

Allowable ultimate tensile

stress of welds

1.0 Ftu (based on Ftu of

normalized tubing)

1.0 Ftu (based on Ftu of

normalized tubing)

0.9 Ftu

0.6 Ftu + 30

0.8 Ftu

B 1.2.5

Section B i25 September1961Page 50

Flash Welding (Cont'd)

Table B 1.2.5.2 Allowable Bending Modulus of

Rupture for Flash Welds in Steel Tubing

Tubing

Normalized tubing-not heat treated

(including normalizing) after

welding.

Heat-treated tubing welded after heattreatment.

Tubing heat treated (including normal-

izing) after welding. Ftu of

unwelded material in heat-treated

condition:

< i00 ksi

I00 to 150 ksi

> 150 ksi

Allowable bending modulus

of rupture of welds (Fb

from Fig. B 1.2.2-1

using values of Ftu listed)

1.0 Ftu for normalized

tubing

1.0 Ftu for normalized

tubing

0.9 Ftu

0.6 Ftu + 30

0.8 Ftu

B 1.2.6 Spot Weldin_

Design shear strength allowables for spot welds in various alloys

are given in Tables B 1.2.6.1, B 1.2.6.2, and B 1.2.6.3; the thickness

ratio of the thickest sheet to the thinnest outer sheet in the combina-

tion should not exceed 4:1. Table B 1.2.6.4 gives the minimum allow-

able edge distance for spot welds, these values may be reduced for non-

structural applications, or for applications not depended upon to

develop full tabulated weld strength. Combinations of aluminum alloys

suitable for spot welding are given in Table B 1.2.6.5.

f

Section B 1

25 September 1961

Page 51

B 1.2.6 Spot Weldin_ (Cont'd_

Table B 1.2.6.1 Spot-Weld Maximum Design Shear Strengths for

Uncoated Steels a and Nickel Alloys

Nominal thickness of

thinner sheet, in.

0.006 .................

0.008 .................0.010 .................

0.012 .................

0.014 .................

0.016 .................

0.018 .................

0.020 .................

0.025 .................

0.030 .................

0.032 .................

0.040 .................

0.042 .................

0.050 .................

0.056 .................

0.060 .................

0.063 .................

0.071 .................

O. 080 .................

O.090 .................

0.095 .................

0.i00 .................

0.112 .................

0.125 .................

Material ultimate tensile

strength, ib

150 ksi and 70 ksi to

150 ksi

Below 70 ksi

70

120

165

220

270

320

390

425

58O

750

835

1,168

1,275

1,700

57

85

127

155

198

235

270

310

425

565

623

85O

920

1,205

°...........

70

92

2,039

2,265

2,479

3,012

3,540

4,100

4,336

4,575

5,088

5,665

above

1,358

1,558

1,685

2,024

2,405

2,810

3,012

3,200

3,633

4,052

120

142

170

198

225

320

403

453

650

712

955

1,166

1,310

1,405

1,656

1,960

2,290

2,476

2,645

3,026

3,440

aRefers to plain carbon steels containing not more than 0.20 percent

carbon and to austenitic steels. The reduction in strength of spot-

welds due to the cumulative effects of time-temperature-stress factors

is not greater than the reduction in strength of the parent metal.

B 1.2.6 Spot Welding (Cont'd)

Table B 1.2.6.2

Section B 1

25 September 1961

Page 52

Spot-Weld Maximum Design Shear Strength

Standards for Bare and Clad Aluminum Alloys a

v

Nominal thickness

of thinner sheet, in.

0.012 ...................

0.016 ...................

0.020 ....................

0.025 ...................

0.032 ...................

0.040 ...................

0.050 ...................

0.063 .......... ,........

0.071 ...................

0.080 ...................

0.090 ...................

0.i00 ...................

0.112 ...................

0.125 ...................

0.160 ...................

Material ultimate tensile strength, ib

Above 56

ksi

28 ksi

to 56 ksi

52

78

106

140

188

248

344

489

578

680

798

933

1,064

1,300

20 ksi to

27.5 ksi

24

56

80

116

168

240

321

442

515

609

695

750

796

840

60

86

112

148

208

276

374

539

662

824

1,002

1,192

1,426

1,698

2,490

19.5 ksJ

and belo_

16

40

62

88

132

180

234

314

358

417

478

536

584

629

aspot welding of aluminum-alloy combinations conforming to QQ-A-277,

QQ-A-355, and QQ-A-255 may be accomplished. The reduction in strength

of spotwelds due to cumulative effects of tlme-temperature-stress

factors is not greater than the reduction in strength of the parent

metal.

BI.2.6 Spot Welding (Cont'd)

Section B I

25 September 1961

Page 53

Table B 1.2.6.3 Spot-Weld Maximum Design Shear Strength

Standards for Magnesium Alloys a

Welding Specification MIL-W-6858


thinner sheet_ in.

0.020 ...................

.022 ...................

.025 ...................

.028 ...................

.032 ...................

.036 ...................

.040 ...................

•045 ...................

.050 ...................

.056 ...................

.063 ...................

.071 ...................

.080 ...................

.090 ...................

.i00 ...................

.112 ...................

.125 ...................

Design shear strength i Ib

72

84

i00

120

140

164

188

220

248

284

324

376

428

496

572

648

720

aMagnesium alloys AZ31B and HK31A may be spot welded in anycombination.

BI.2.6 Spot Weldin_ (Cont'd)

Section B 1

25 September 1961

Page 54

Table B 1.2.6.4 Minimum Edge Distances for Spot-WeldedJoints ab


thinner sheet_ in.

0.016 ...................

0.020 ....................

0.025 ...................

0.032 ...................

0.036 ...................

0.040 ...................

0.045 ...................

0.050 ...................

0.063 ...................

0.071 ...................

0.080 ...................

0.090 ...................

0.i00 ...................

0.125 ...................

0.160 ...................

Edge distance_ E_ in.

3/16

3/16

7/32

1/4

1/4

9/32

5/16

5/16

3/8

3/8

13/32

7/16

7/169/16

5/Salntermediate gages will conform to the requirement for the

next thinner gage shown.

bFor edge distances less than those specified above, appropriate

reductions in the spot-weld allowable loads shall be made.

Fig. B 1.2.6-1 Edge Distances for Spot-Welded Joints.

BI.2.6 Spot Welding (Cont'd)

Section B 1

25 September 1961

Page 55

-'4

"or-q

CO

0

0

O_

0

.rq

4_

.'4

,a

oc_)

o

<I

.'4

<

=.'4

<

CO

<

u_

,-q

CO

(_IO_ o=_)vv _v vv

19z=v-bb

(_IOE P_IO) _-v-bb _-"_= _,"_

(_Og P_ID) _ 9_-V-_)_)

q(gLOt P_ID) tgC-V-bb

(OOII) I9_-v-bb

(ooTt) _-v-bb

(_gO_ P_ID) _9E-V-O0

(_o0_) 6_-v-bb

(_0_ o_) _-v-bb _ _ _ _ _ _

(_0_ _) _-V-bb _' _ _ _' _' _'

(I909) LZ¢-V-bb

(_o_) _[-v-bb

(_o_) _[-v-bb

o.'4

_-_.'4

.'4 u_•"4 0

CO

0000 0 000000• _ __ _ __

,°•°°....o°.°._

,.°°°°°,°.°.,°_

°o°°°°°.°°°°.°_

• ° ............ |

.... °°•°°°°°°°|

.... °•oo.°,°.._

0 ..............•"4 °,.°,,°o.°,.°°_

•.°°°,_°°..°°°_

•"4 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

&&&&&&&&&&&&&&&

B 1.2.7

Section B 1

25 September 1961

Page 56

Reduction in Tensile Strength of Parent Metal Due to Spot

Weldin_

In applications of spot welding where ribs, intercostals, ordoublers are attached to sheet, either at splices or at other points

on the sheet panels, the allowable ultimate strength of the spot-welded

sheet shall be determined by multiplying the ultimate tensile sheet

strength ("_' values where available) by the appropriate efficiency

factor shown on Figures B 1.2.7-1 through B 1.2.7-4. The minimum values

of the basic sheet efficiency in tension should not be considered

applicable to cases of seam welds. Allowable ultimate tensile strengths

for spot-welded sheet gages of less than 0.012 inch for steel and

0.020 inch for aluminum shall be established on the basis of tests.

BI.2.7

Section B 125 September1961Page 57


Weldin_ (Cont'd)

020 in.

= 032 in.U

= • 90 __ '0.040 in.u _ '0.050 in.

ww __ _._ _ 063 in.80 080 in.

O9O in.

"_ 125 in.

_= = 70 sheet

u gage

6o0 0.5 1.0 1.5 2.0 2.5 3.0

Spot Spacing (Center to Center), Inches

Fig. B 1.2.7-1 Efficiency of the Parent Metal

in Tension for Spot-Welded 301-1/2 H

Corrosion-Reslstant Steel

_4auc

O

U

,klII1 .r,I

m _

[.-(U

I00

/ - 032 in.

'°I///,,o o,o,.050 in.

063 in.

!1171 o,o,,.-/ 1090 in.

.125 in.

70_/ sheet

gage

60

50

0 0.5 1.0 1.5 2.0 2.5 3.0

Spot Spacing (Center to Center), inches

Fig. B 1.2.7-2 Efficiency of the Parent Metal

in Tension for Spot-Welded 301-H

Corrosion-Resistant Steel

B 1.2.7

.4

.4

t&4

&J

J= =

U.4

_O

_4JO ;:

.4

• J O

O_4

m

Section B 1

25 September 1961

Page 58


Weldin_ (Cont' d)

I00

9O

80 ___

J

7O

6O

0 0.5 1.0 1.5 2.0 2.5 3.0


Fig. B 1.2.7-3 Efficiency of the Parent

Metal in Tension for Spot-Welded

301-A, 347-A, and 301-1/4 H

Corrosion-Resistant Steel

0. 012 in.

0. 020 in.

O. 032___ in.

0. 040 in.0. 050 in.

O. 063 in.

0. 080 in.O. 090 in.

_0. 125 in.

sheet

gage

I00

[O.02O in.

I_0.032 in.

90 _ID.040 in.

fILL --80

?0

6O

5O

\

0 0.5 I 1.5

O. i00 in: ,

--0.090 in.

O.080 in.

0.071 in;'

O. 063 in.

O. 050 in.

Sheet gage

l l

2 2.5 3


Fig. B 1.2.7-4 Efficiency of the Parent

Metal in Tension for Spot-Welded

Aluminum Alloys

f_

Section B I

25 September 1961

Page 59

B 1.3.0 Brazing

Insofar as discussed herein, brazing is applicable only to steel.

Brazing is defined as a weld wherein coalescence is produced by heating

to suitable temperatures above 800 ° F and by using a nonferrous filler

metal having a melting point below that of the base metal. The filler

metal is distributed through the joint by capillarity.

The effect of the brazing process upon the strength of the parent

or base metal shall be considered in the structural design. Where

copper furnace brazing or silver brazing is employed the calculated

allowable strength of the base metal which is subjected to the tempera-

tures of the brazing process shall be in accordance with the following:

Material Allowable Strength

Heat-treated material

(including normalized)

used in "as-brazed"

condition

Heat-treated material

(including normalized)

reheat-treated during

or after brazing

Mechanical properties of

normalized material

Mechanical properties

corresponding to heat

treatment performed

B 1.3.1 Copper Brazing

The allowable shear stress for design shall be 15 ksi, for all

conditions of heat treatment.

B 1.3.2 Silver Brazin$

The allowable shear stress for design shall be 15 ksi, provided

that clearances or gaps between parts to be brazed do not exceed 0.010

inch. Silver-brazed areas should not be subjected to temperatures

exceeding 900 ° F. Acceptable brazing alloys, with the exception of

Class 3, are listed in Federal Specification QQ-S-561d.

Reference:

Section B 1

25 September 1961

Page 60

(I) MIL-HDBK-5, Strength of Metal Aircraft Elements, Armed Forces

Supply Support Center, Washington 25, D.C., March 1959.

SECTION B2

LUGS AND SHEAR PINS

B2.0.O

TABLE OF CONTENTS

Analysis of Lugs and Shear Pins .....................

Page

1

2.1.0 Analysis of Lugs with Axial Loading ............

2.2.0 Analysis of Lugs with Transverse Loading .......

2.3.0 Analysis of Lugs with Oblique Loading ..........

2

17

23

B2-iii

Section B 2

27 July 1961

Page i

B 2.0.0 Analysis of Lugs and Shear Pins

The method described in this section is semi-empirical and is

applicable to aluminum or steel alloy lugs. The analysis considers

loads in the axial, transverse and oblique directions. Each of these

loads are treated in Sections B 2.1.0, B 2.2.0, and B 2.3.0 respectively.

See Fig. B 2.0.0-1 for description of directions.

Axial load

Oblique load

Transverse load

Fig. B 2.0.0-1

Section B 2

27 July 1961

Page 2

B 2.1.0 Analysis of Lugs with Axial Loadin_

A lug-pin combination under tension load can fail in any of the

following ways, each of which must be investigated by the methods

presented in this section:

i. Tension across the net section. Stress concentration must

be considered.

. Shear tear-out or bearing. These two are closely related

and are covered by a single calculation based on empirical

curves.

3. Shear of the pin. This is analyzed in the usual manner.

4. Bending of the pin. The ultimate strength of the pin is

based on the modulus of rupture.

5. Excessive yielding of bushing (if used).

. Yielding of the lug is considered to be excessive at a

permanent set equal to .02 times pin diameter. This

condition must always be checked as it is frequently

reached at a lower load than would be anticipated from

the ratio of the yield stress, Fty to the ultimate stress,Ftu for the material.

Notes:

a. Hoop tension at tip of lug is not a critical condition,

as the shear-bearing condition precludes a hoop tensionfailure.

Do The lug should be checked for side loads (due to mis-

alignment, etc.) by conventional beam formulas (Fig.

B 2.1.0-1).

Analysis procedure to obtain ultimate axial load.

i. Compute (See Fig. B 2.1.0-1 for nomenclature)

e/D, W/D, D/t; Abr = Dt, At = (W-D)t

. Ultimate load for shear-bearing failure:

Note: In addition to the limitations provided by curves "A"

and "B" of Fig. B 2.1.0-3, Kbr greater than 2.00 shall not

be used for lugs made from .5 inch thick or thicker aluminum

alloy plate, bar or hand forged billet.

F

B2.1.0 Analysis of Lugs with Axial Loading (Cont'd)

Section B 2

27 July 1961

Page 3

Bushing

W

2

P

.10P (Minimum side load assumed

to account for misalignment)

Fig. B 2.1.0-I

,

(a) Enter Fig. B 2.1.0-3 with e/D and D/t to obtain Kbr

(b) The ultimate load for shear bearing failure, P'bru is

P'bru = Kbr Ftux Abr ........................ (I)

where

Ftu x = Ultimate tensile strength of lug material

in transverse direction.

Ultimate load for tension failure:

(a) Enter Fig. B 2.1.0-4 with W/D to obtain K t for proper

material

(b) The ultimate load for tension failure P'tu is

P'tu = Kt Ftu At ........................... (2)

B2.1.O

P/2

Section B2

July 9, 1964

Page 4

Analysis of Lugs with Axial Loading (Cont'd)

where

Ftu= ultimate tensile strength of lug material

4. Load for yielding of the lug

(a) Enter Fig. B 2.1.0-5 with e/D to obtain Kbry.

(b) The yield load, e_ is

P'y = Kbry Abr Fly .......................... (3)

where

Fty= Tensile yield stress of the lug material.

5. Load for yielding of the bushing in bearing (if used):

P_ry = 1.85 Fty Abrb ....................... (4)

where

Fty= Compressive yield stress of bushing material

Abrb = Dpt (Fig. B 2.1.0-1)

6. Pin bending stress.

l'-ITtt

I t2 +(_

(a) P/2-_--

p/2 -_--1

p/2

P/2

JI

t 3

P/2

P/2

7 (t4/2)

(b)

Fig. B 2.1.0-2

B 2.1.0 Analysis of Lugs with Axial Loading (Cont'd)

Section B 2

27 July 1961

Page 5

(a) Obtain moment arm "b" as follows:

(See Fig. B 2.1.0-2a) compute for the inner lug

r = _- - -7 t2 ..................... (5)

Take the smaller of P'bru and P'tu for the inner lug as

(P'u)min and compute (P'u)min/AbrFtux .t ,,

Enter Fig. B 2.1.0-6 with (Pu)min/AbrFtu x and r" to

obtain the reduction factor, "7" which compensates

for the "peaking" of the distributed pin bearing load

near the shear plane. Calculate the moment arm "b"

from

b = +g+7 4 ...................... (6)

Where "g" is the gap between lugs as in Fig. B 2.1.0-2a

and may be zero. Note that the peaking reduction factor

applies only to the inner lugs.

(b) Calculate maximum pin bending moment M, from the

equation

(c) Calculate bending stress resulting from "M", assuming

an My/l distribution.

(d) Obtain ultimate strength of the pin in bending by use

of Section B 4.5.2. If the analysis should show

inadequate pin bending strength, it may be possible to

take advantage of any excess lug strength to show

adequate strength for the pin by continuing the

analysis as follows:

(e) Consider a portion of the lugs to be inactive as

indicated by the _=haded area of Fig. B 2.1.0-2b. The

portion of the thickness to be considered active may

have any desired value sufficient to carry the load

and should be chosen by trial and error to give

approximately equal Factors of Safety for the lugs

and pin.

(f) Recalculate all lug Factors of Safety, with ultimate

loads reduced in the ratio of active thickness to

actual thickness.

(g) Recalculate pin bending moment, M = P (b/2), and Factor

of Safety using _l reduced value of "b" which is obtained

as follows:

Section B 227 July 1961Page 6


Compute for the inner lug, Fig. B 2.1.0-2b

r = 2 2t 4

Take the smaller of P'bru and P'tu for the inner lug,

based upon the active thickness, as (Pu)min and compute

(P_)min/Abr Ftux, where Abr = 2t4 D. Enter Fig. B 2.1.0-

6 with (P_)min/Abr Ftux and "r " to obtain the reduc-

tion factor "7 " for peaking. Then the moment arm is

t3 _ ___4 _b = -_-- + g +7 ..................... (9)

This reduced value of "b" should not be used if the

resulting eccentricity of load on the outer lugs

introduce excessive bending stresses in the adjacent

structure. In such cases the pin must be strong enough

to distribute the load uniformly across the entire lug.

7. Factor of Safety, F.S.

Compute the following Factors of Safety:

(a) Lug

p'bru

Ultimate F.S. in shear-bearingP

...... (lO)

p!

tuUltimate F.S. in tension - p ............. (ii)

p,

--Y-- (12)Yield F.S. = P .........................

(b) Pin

F_U

Ultimate F.S. in shear =fs

F bUltimate F.S. in bending = --

_b

............... (13)

............... (14)

(c) Bushing (if used)

p,

Yield F.S. in bearing pi)rx (15)

An analysis for yielding of the pin and ultimate bearing

failure of the bushing is not required.

v

B2.1.O

cq

O oO

Analysis of Luzs with AxiJl Loading (Cont'd)

4-J

Section B 2

27 July 1961

Page 7

See notc,_; on I_)1]ow[,_;, i_ :,,c,.

B 2.1.0 Analysis of Lugs with Axial Loading (Cont'd_

Section B 2

27 July 1961

Page 8

Curve A is a cutoff to be used for all aluminum alloy hand forged

billet when the long transverse grain direction has the general

direction C in the sketch.

Curve B is a cutoff to be used for all aluminum alloy plate, bar and

hand forged billet when the short transverse grain direction has the

general direction C in the sketch, and for die forging when the lug

contains the parting plane in a direction approximately normal to thedirection C.


Section B 2

27 July 1961

Page 9

o9

°

0

1.0 1.5 2.0 2.5 3.0 3.5 4.0

w/D

Fig. B 2.1.0-4

4.5 5.0

Legend on following pages.

Section B 2

27 July 1961

Page i0


Legend - Figure B 2.1.0-4- L, T, N, indicate grain in direction

F in sketch

L = longitudinal

T = long transverse

N = short transverse (normal)

Curve 1

4130, 4140, 4340 and 8630 steel

2014-T6 and 7075-T6 plate _ 0.5 in (L,T)

7075-T6 bar and extrusion (L)

2014-T6 hand forged billet _ 144 sq. in. (L)

2014-T6 and 7075-T6 die forgings (L)

Curve 2

2014-T6 and 7075-T6 plate > 0.5 in., _ 1 in.

7075-T6 extrusion (T,N)

7075-T6 hand forged billet _ 36 sq.in. (L)

2014-T6 hand forged billet > 144 sq.in. (L)

2014-T6 hand forged billet _ 36 sq.in. (T)

2014-T6 and 7075-T6 die forgings (T)

17-4 PH

17-7 PH-THD

Curve 3

2024-T6 plate (L,T)

2024-T4 and 2024-T42 extrusion (L,T,N)

Curve 4

2024-T4 plate (L,T)

2024-T3 plate (L,T)

2014-T6 and 7075-T6 plate > I in.(L,T)

2024-T4 bar (L,T)

7075-T6 hand forged billet > 36 sq.in. (L)

7075-T6 hand forged billet _ 16 sq.in. (T)

Curve 5

195T6, 220T4, and 356T6 aluminum alloy casting

7075-T6 hand forged billet > 16 sq.in. (T)

2014-T6 hand forged billet > 36 sq.in. (T)

Section B 227 July 1961Page ii


Curve 6

Aluminum alloy plate, bar, hand forged billet, and die forging

(N). Note: for die forgings, N direction exists only at the

parting plane. 7075-T6 bar (T)

Curve 7

18-8 stainless steel, annealed

Curve 8

18-8 stainless steel, full hard, Note: for 1/4, 1/2 and 3/4

hard, interpolate between Curves 7 and 8.

B2.1.O Analysis of Lugs with Axial Loading (Cont'd)

Section B 2

27 July 1961

Page 12

2.0

1.5

1.0

Kbry

.5

0

0

D

11 I!

1.0 2.0 3.0 4.0

e/D

FiB. B 2.1.0-5


I

iIIIN , °i I

IIIIIIl_::I I --

IllI1 -o-IIlll

0

0

(n

0

OOo_iiiii\i\oi\°'-° Ill/IX\t.,-I

oo iii_o,\\-4

oo /_x\U

,-.4 0 C_

0

_4

p_

if3

<t

,--_I c,40

' _ _

---Lfj_-c_-_-=\cO _-- ,_9 u'_ _ o'_ c_I ,-4 0

Section B 2

27 July 1961

Page 13

I0

S

.4

v

0

B2.1.O

Section B 2

27 July 1961

Page 14


Special Applications

I. Irregular lug section - bearing load distributed overentire thickness.

For lugs of irregular section having bearing stress distributed

over the entire thickness, an analysis is made based on an

equivalent lug with rectangular sections having an area equal

to the original section.

1 12 secB T (Sec. B)

(a) (b)

Fig. B 2.1.0-7

Dashed lines show equivalent lug

2. Critical bearing stress

NASA Design Manual Section 3.0.0 lists the values of the

ultimate and yield bearing stress of materials for e/D values

of 2.0 and 1.5, these are valid for values of D/t to 5.5.

The ultimate and yield bearing stress for geometrical

conditions outside of the above range may be determined in

the following manner:

(a) Ultimate bearing stress: For the particular D/t and e/D,

obtain Kbr from Fig. B 2.1.0-3 then

Fbr u = Kbr Ftux

where

Ftu x = Ultimate strength of lug material in transversedirection.

B2.1.O

Section B 2

27 July 1961

Page 15


(b) Yield bearing stress: With the particular e/D obtain

Kbry from Fig. B 2.1.0-5. Then

Fbry = Kbry Ftyx

where

Fty x = Tensile yield stress of lug material in transversedirection.

3. Eccentrically located hole

If the hole is located as in Fig. B 2.1.0-8 (e% less than e2) ,the ultimate and yield lug loads are determined by obtaining

P'bru, P'tu and P'y for the equivalent lug shown and multiply-

ing by the factor

el + e2 + 2Dfactor =

2e 2 + 2D

Actual lug Equivalent lug

Fig. B 2.1.0-8

4. Multiple shear connections

Lug-pin combinations having the geometry shown in Fig.

B 2.1.0-9 are analyzed according to the following criteria:

(a) The load carried by each lug is determined by distributing

the total applied load "P" among the lugs as shown on

Fig. B 2.1.0-9 and the value of "C" is obtained from

Table B 2.1.0.1.

Section B 2

27 July 1961

Page 16

B 2.1.0 Analysis of Lugs with Axial Loadin_ (Cont'd)

P

These lugs of

equal thickness

CPI..9,.---V/I/I/I/

-Two outer lugs of equal thickness

not less than C t' (See Table

B 2.1.0.1)

m

_---mm_ P2

_ P2

P

These lugs of equalthickness t"

Fig. B 2.1.0-9

v

(b)

(c)

The maximum shear load on the pin is given in TableB 2.1.0.1.

The maximum bending moment in the pin is given by the

PIb

formula, M = -=7 where "b" is given in Table B 2.1.0.1.

Table B 2..I.0.1

Total number of

lugs including

both sides

ii

OO

.35

.40

.43

.44

.50

Pin Shear

.50 P1

.53 P1

.54 P1

.54 P1

.50 P1

.28

.33

.37

.39

.50

t' + t"

2

t' + t"

2

t' + t"

2

t' + t"

2

t' + t"

2

Section B 2

27 July 1961

Page 17

B 2.2.0 Analysis of Lugs with Transverse Loading

Shape Parameter

In order to determine the ultimate and yield loads for lugs with

transverse loading, the shape of the lug must be taken into account.

This is accomplished by use of a shape parameter given by

A

Shape parameter =avAbr

where

Abr is the bearing area = Dt

Aav is the weighted average area given by

6

Aav = _+_--_2_ + _--_3_ + _--A4)'--I

AI, A2, A 3 and A 4 are areas of the lug sections indicated in

Fig. B 2.2.0-1.

A 3 least of any

radial section

_w _ radius

all

(a) (b)

Fig. B 2.2.0-1

(i) Obtain the areas AI, A2, A3, and A 4 as follows:

(la) AI, A2, and A 4 are measured on the planes indicated in

Fig. B 2.2.0-ia (perpendicular to the axial center line),

except that in a necked lug, as shown in Fig. B 2.2.0-Ib,

A I and A 4 should be measured perpendicular to the local

line.

Section B2

July 9, 1964

Page 18

B 2.2.0 Analysis of Lugs with Transverse Loading (Cont'd)

(Ib) A3 is the least area on any radial section around thehole.

(ic) Thought should always be given to assure that the areas

AI, A2, A3, and A4 adequately reflect the strength of

the lug. For lugs of unusual shape (e.g. with sudden

changes of cross section), an equivalent lug should be

sketched as shown in Fig. B 2.2.0-2 and used in the

analysis.

Equivalent lug

50

Fig. B 2.2.0-2

Equivalent

lug

' ] 450

(2) Obtain the weighted average

6

Aav = (3/AI) + (I/A2) + (I/A 3) + (I/A4)

(3) Compute Abr = Dt and Aav/ Abr

(4) Ultimate load P'tru for lug failure:

(a) Obtain Ktr u from Fig. B 2.2.0-4

(b) P'tru = Ktru Abr Ftux

(5) Yield load P'y of the lug:

(a) Obtain Ktry from Fig. B 2.2.0-4

(b) P'y = Ktry Abr Fty x

(6) Check bushing yield and pin shear as outlined previously.

Section B 2

27 July 1961

Page 19


(7) Investigate pin bending as for axial load with following

modifications: Take (P'u)min = P' In the equation r =tru"

[e - (D/2)] /t use for the [e - _/2)] term the edgedistance at _ = 90 ° .

Fig. B 2.2.0-3

B2.2.0 Analysis of Lugs with Transverse Loadin_ (Cont'd)

Section B 2

27 July 1961

Page 20

1.6

1.4

1.2

1.0

Ktr u &

Ktry0.8

0.6

0.4

0.2

125,000 HT

I J

Ktry-All alum.

& steel

[_50,000 HT

alloys_

• _'_- _ 8o, O00h_

x

I ! _,_I" I I I Ii i i !

, I I A-Approxlmate cantilever

strength. If Ktr u is below this curve

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

Aav/Abr

Fig. B 2.2.0-4

Legend on following pages

Section B 2

27 July 1961

Page 21


Legend - Fig. B 2.2.0-4

Curve i:

4130, 4140, 4340, and 8630 steels, heat treatment as noted.

Curve 2:

2024-T4 and 2024-T3 plate _0.5 in.

Curve 3:

220-T4 aluminum alloy casting

Curve 4 :

17-7 PH (THD)

Curve 5 :

2014-T6 and 7075-T6 plate_0.5 in.

Curve 6:

2024-T3 and 2024-T4 plate >0.5 in., 2024-T4 bar

Curve 7:

195-T6 and 356-T6 aluminum alloy casting

Curve 8:

2014-T6 and 7075-T6 plate>0.5 in.,_ i in.

7075-T6 extrusion

2014-T6 hand forged billets 36 sq. in.

2014-T6 and 7075-T6 die forgings

Curve 9:

2024-T6 plate

2024-T4 and 2024-T42 extrusion

Curve i0:

2014-T6 and 7075-T6 plate_ 1 in.

7075-T6 hand forged billet _16 sq. in.


Section B 2

27 July 1961

Page 22

Legend Fig. B 2.2.0-4 Cont'd

Curve ii:

7075-T6 hand forged billet >16 sq. in.

2014-T6 hand forged billet>36 sq. in.

All curves are for Ktr u except the one noted as Ktry

Note: The curve for 125,000 HT steel in Fig. B 2.2.0-4 agrees

closely with test data. Curves for all other materials have been

obtained by the best available means of correcting for material

properties and may possibly be very conservative to some places.

In no case should the ultimate transverse load be taken as less

than that which could be carried by cantilever beam action of

the portion of the lug under the load (Fig. B 2.2.0-3). The

load that can be carried by cantilever beam action is indicated

very approximately by curve (A) in Fig. B 2.2.0-4, should Ktr u

be below curve (A), separate calculation as a cantilever beam

is warranted.

\

Section B 2

27 July 1961

Page 23

B 2.3.0 Analysis of Lugs with Oblique Loadin$

Interaction Relation

In analyzing lugs subject to oblique loading it is convenient

to resolve the loading into axial and transverse components (denoted

by subscripts "a" and "tr" respectively), analyze the two cases

separately and utilize the results by means of an interaction equation.

The interaction equation Ra 1'6 + Rtr Io6 = i, where R a and Rtr are

ratios of applied to critical loads in the indicated directions, is to

be used for both ultimate and yield loads for both aluminum and steel

alloys.

where, for ultimate loads

R a = (Axial component of applied load) divided by (smaller of

P'bru and P'tu from Eq. I and Eq. 2.)

Rtr = (Transverse component of applied load) divided by

(P'tru from analysis procedure for _ = 90 deg.)

and for yield load:

R a = (Axial component of applied load) divided by (P'y from

eq. 3.)

Rtr = (Transverse component of applied load) divided by

(P'try from Analysis Procedure for _ = 90 deg.)

Analysis Procedure

(i) Resolve the applied load into axial and transverse components

and obtain the lug ultimate and yield Factor of Safety from

the interaction equation:

F.So

I 1.6 j0.625Ra 1"6 + Rtr

(2) Check pin shear and bushing yield as in Section B 2.1.0.

(3) Investigate pin bending using the procedure for axial load

modified as follows:

P

Take (P'u)mi n =

_Ral'6 + Rtrl.6_ 0"625

'x /

In the equation r = [e - (D/2)] /t use for the [e-(D/2)]

term the edge distance at the value of "_ " corresponding to

the direction of load on the lug.

B 2.3.0 Analysis of Lugs with Oblique Loading (Cont'd)

Section B 2

27 July 1961

Page 24

Reference

Melcone, M. A. and F. M. Hoblit, Developments in the Analysis of

Method for Determining the Strength of Lugs Loaded Obliquely or

Transversely, Product Engineering, June, 1953.

SECTION B3

SPRINGS

i f

TABLE OF CONTENTS

B3.0.O Springs .............................................

3.1.O

3.1.1

3.1.2

3.1.3

3.1.4

3.1.5

3.1.6

3.1.7

3.2.0

3.3.0

Helical Springs .................................

Helical Compression Springs .................

Helical Extension Springs ...................

Helical Springs with Torsional Loading ......

Analysis of Helical Springs by Use of

Nomograph .................................

Maximum Design Stress for Various Spring

Materials .................................

Dynamic or Suddenly Applied Spring Loading...

Working Stress for Springs ..................

Curved Springs ..................................

Belleville Springs or Washers ...................

Page

i

i

i

7

i0

12

15

19

24

25

29

B3-iii

f

Section B 3

19 May 1961

Page 1

B 3.0.0 SPRINGS

B 3.1.0 Helical Springs

B 3.1.1 Helical Compression Springs

Most compression springs are open-coil, helical springs which

offer resistance to loads acting to reduce the length of the spring.

The longitudinal deflection of springs produces shearing stresses in

the spring wire. Where particular load-deflection characteristics

are desired, springs with varying pitch diameters may be used. These

springs may have any number of configurations, including cone, barrel,

and hourglass.

Round-Wire Springs

The relation between the applied load and the shearing stress for

helical springs formed from round wire is

where

8PDf = __

s _d 3

....................................... (1)

fs = shearing stress in pounds per sq. inch.

(not corrected for curvature)

P = axial load in pounds

D = mean diameter of the spring coil (Outside diameter

minus wire diameter or inside diameter plus wire

diameter)

d = diameter of the wire in inches.

Equation (i) is based on the assumption that the magnitude of thc_

stress varies directly with the distance from the center of the wire;

but, actually, the stress is greater on the inside of the cross section

due to the curvature. The stress correction factor (k) used to deter-

mine the maximum shearing stress for static loads is found in

Fig. B 3.1.1-i. This correction factor gives the effect of both

torsion and direct shear. The equation for the maximum stress is

f : k f : k 8PD (2).,..°................,--.,."

max s _d 3

Section B 3

December 3, 1969

Page 2

B 3.1.1 Helical Compression Springs (Cont'd)

where

k = kc + 0.615Cl

Stress concentration factor plus the effectof direct shear

kc = 4C 1 - i Stress concentration factor due to curvature

Cl=,:I

Ratio of mean diameter of helix to the

diameter of the bar or wire

2.1

.

.1.9oAJ

UI 1.

0 1

0

0

= 1.50

°_

_ 1.1.1

U= 1.30

o'1 •

r_ lo

!

X

\

\

\_ r ,

I

0 I 2 3 4 5 6 7 8

DSprJ:n_ index, C l = _[

Fig. B 3.1.l--!

9 i0 Ii 12

4

Section B 3

19 May 1961

Page 3

B 3.1.1Helical Compression Springs (Cont'd)

Stress correction for temperature.

Corrections must be made to account for changes in strength and

in elastic properties of spring materials at elevated temperatures.

This correction is made to the allowable stress of the spring material.

Values for various materials at various temperatures may be obtained

from Section 3.0.0 of the Design Manual.

Deflection

The formula for the relation between deflection and load

when using round wire in helical springs is

8NPD 3_)- . ........................................ (3)

Gd 4

whe r e

= total deflection

N = number of coils

G = modulus of rigidity

The deflection may also be given in terms of the shearing stress

by combining Eqs. (I) and (3). Stress concentration usually does not

affect deflection to an appreciable degree, and no adjustment is needed

in Eq. (I). The expression for the deflection is

Nf _D 2

_ s ......................................... (4)Gd

Sprin$ rate

The spring rate (K) is defined as the amount of force required

to deflect the spring a unit length. By proper substitution of the

previous equations, the spring rate may be shown to be

d4GK =-

8ND 3.......................................... (5)

Bucklin$ of Compression Springs

A compression spring which is long compared to its diameter will

buckle under relatively low loads in the same manner as a column. How-

ever, the problem of buckling is of little consequence if the compres-

sion spring operates inside a cylinder or over a rod.


Section B 3

19 May 1961

Page 4

As the critical buckling load of a colum is dependent upon the

end fixity at the supports, so is the critical buckling load of a

spring dependent upon the fixity of the ends. In general, a compres-

sion spring with ends squared, ground, and compressed between two

parallel surfaces can be considered a fixed-end spring. The following

formula gives the critical buckling load.

Pc = JKL ............................................ (6)

where

J = factor from Fig. B 3.1.1-2 =

L = free length of spring

K = spring rate (See Eq. 5)

Deflection

Free Length

Fig. B 3.1.1-2 (curve i) is for squared and ground springs with

one end on a flat surface and the other on a ball. Curve 2 indicates

buckling for a squared and ground spring both ends of which are com-

pressed against parallel plates. This is the most common conditionwith which the user must contend.

Helical Sprinss of Rectangular Wire.

When rectangular wire is used for helical springs, the value of

the shearing stress can be found by use of the equations for rectangu-

lar shafts. A stress concentration factor is applied in the usual way

to compensate for the effect of curvature and direct shear. For the

springs in Fig. B 3.1.1-3 (a) and (b) the stresses at points A 1 and A2are as follows:

kPR

f = -- for point A 1

s 51bc2

........................ (7)

kPR

f = -- for point A 2s 5 bc 2

2

........................ <,8)

values for51 and52 for various b/c ratios are found in Table B 3.1.1-1.

The stress concentration factor should be applied for point A1 in

Fig. B 3.1.1-3(a) and to point A 2 in Fig. B 3.1.1-3 (b). The stress

concentration factors of Fig. B 3.1.1-1 may be used as an approximate

value for rectangular wire.


Section B 3

19 May 1961

Page 5

1 2

.7

.6

Springs above and to the

right of the curves willbuckle.

_I_

II

.5

.2

.1

0 ,1,0 3

\\\

4 5 6 7

Free Length _ LMean Diameter D

Fig. B 3.1.1-2

\

8 9 I0 II 12

Section B 319 May 1961Page 6

P

P(a)

2RSprin 8 index, C 1 =- C

A I

(b)2R

Spring index, C I =-_-

Fig. B 3.1.I-3

b/c 1.00

_2

.208 .219 .231 .239 .246 .258 .267 .282 .291 .299 .307 .312

.208 .235 .269 .291 .309 .336 .355 .378 .392 .402 .414i.421

.1406 .166 .196 .214 .229 .249 .263 .281 .291 .299 .307 .312

1.20 1.50 1.75 2.00 2.50 3.00 4.00 5.00 6.00 8.00 I0.00 oo

.333

.333

Table B 3.1.I-i

The equation for the relation between the load (P) and the

deflection (5) is

2_pR3N

_Gbc 3

..........................................(9)

where:

is obtained from Table B 3.1.1-1

b and c are as shown in Fig. B 3.1.1-3

R is the mean radius of the spring

N is the number of coils

G is the modulus of rigidity

P is the axial load


B 3.1.2 Helical Extension Springs

Helical extension springs differ from helical compression springs

only in that they are usually closely coiled helices with ends formed

to permit their use in applications requiring resistance to tensile

forces. It is also possible for the spring to be wound so that it is

preloaded, that is, the spring is capable of resisting an initial

tensile load before the coils separate. This initial tensile load

does not affect the spring rate. See Figure B 3.1.2-1 for the load-

deflection relationship of a preloaded helical extension spring.

Load

Initial

tension

Spring with

initial

jfJ

/ i_____Spring without

/ initial tension

J .

Deflection

Fig. B 3.1.2-1

Stresses and Deflection

In helical extension springs, the shape of the hook or end turns

for applying the load must be designed so that the stress concentra-

tion effects caused by the presence of sharp bends are decreased as

much as possible. This problem is covered in the next article.

If the extension spring is designed with initial tension, formulas

(I) through (9) from Section B 3.1.1 are valid, but must be applied

with some understanding of the nature of the forces involved.


B 3.1.2 Helical Extension Springs (toni'd)

Stress concentration in hooks on extension springs.

p P

td

r3

A

r4

r2

(a) (b)

a. Bending Stress

Fig. B 3.1.2-2

32PR 4P

fb =--k +--_d 3 _d 2

................................ (zo)

where k is the correction factor obtained from Fig. B 3.1.2-3, using

the ratio 2rl/d.

rI = radius of center line of maximum curvature.

A simplified equation is:

fb = 32P___RR./rl )_d 3 L_ 3 ................................. (ii)

r3 = inside radius of bend.

The maximum bending stress obtained by this simplified form will

always be on the safe side and, under normal conditions, only slightly

higher than the true stress.

Fig. B 3.1.2-2(a) illustrates the bending moment (PR) due to the

load (P). The bending stress at point A is:


B 3.1.2 Helical Extension Springs (Cont'd)

b. Torsional Stress

At point A', Fig. B 3.1.2-2(b), where the bend joins the helical

portion of the spring, the stress condition is primarily torsion. The

maximum torsional shear stress due to the moment (PR) is

16PR / 4CI - Ifs = _d 3 _ 4C I 4

........................... (12)

2r 2

C1 = d

A simplified form similar to the one for bending is

f =

s _d 3 _4................................ (13)

This will also give safe results.

4

3

k

2

\\k_k

I

IIIIIIIIIIII

i_-

0

0 1.5 2.0 2.5 3.0

R_=D _Dc d h

3.5 4.0 4.5

"- Fig. B 3.1.2-3

Section B 319 May 1961Page I0

B 3.1.3 Helical Springs with Torsional Loading

A helical spring can be loaded by a torque about the axis of the

helix. Such loading, as shown in Fig. B 3.1.3-I(a), is similar to the

torsional loading of a shaft. The torque about the axis of the helix

acts as a bending moment on each section of the wire as shown in

Fig. B 3.1.3-I(b). The stress is then

Mcfb = kn T- .............................. '......... (14)

where the stress concentration factor, Kn, is given as

3CI 2 - C I - 0.8 inner

kl = 3CI(C I - i) edge

3CI 2 + C I - 0.8 outer

k2 = 3CI(C I -I) edge

Rectangular

cross section

wire

2R=-- ; "h" is the depth of section perpendicularC1 h

to the axis.

24C 1 - C I -i inner "

k3 = 4CI(C I - I) edge

24C 1 + C 1 - 1 outer

k4 = 4CI(C I + i) edge

Round

cross section

wire

2RCl = T

Angular deformation

The deformation of the wire in the spring is the same as for a

straight bar of the same length "S'_ The total angular deformation 8

between tangents drawn at the ends of the bar is:

MS

E1........................................ (15)

Angle 8 in some cases may amount to a number of revolutions.

B3.1.3 Helical Springs with Torsional Loadin$ (Cont'd)

Section B 3

19 May 1961

Page iI

7

I

M

h or d

\ x / /k \ f

4-- J

Fig. B 3.1.3-1


B 3.1.4 Analysis of l_elic_Jl S[_rings by Use of Nomograph

The procedure for using the nomographs (Fig. B 3.1.4-1 and

Fig. B 3.1.4-2) for helical springs are as follows:

I. Set the appropriate wire diam. on the "d" scale.

2. Set the appropriate mean diam. on the "D" scale.

3. Connectthe two points and read the curvature correction

factor:

a. For tension and compression springs read the "y" scale

on Fig. B 3.1.4-1.

b. For torsion springs read the "k" scale on Fig. 3.1.4-2.

. Set the correction factor, obtained in step 3, on the

appropriate "y" or "k" scale to the right of Fig. B 3.1.4-1

and Fig. B 3.1.4-2.

5. Set the "calculated" (Eq. I or Eq. 14 with kn = I) stresson the "f" scale.

6. Connect the two point_, from steps 4 and 5, and read the

corrected stress on t le (f') scale.

Section B 3

19 May 1961

Page 13

B 3.1.4 Analysis of Helical Springs by Use of Nomograph (Cont'd)

FIBER STRESS CORRECTION FOR CURVATURE

Helical Extension and Compression Springs

Find Correction Factor Using Correction Factor Found on

From Spring Index Left Half of Chart--Determine

True Fiber Stress

0 OZ-

d

0.031 O_

. OO_

02

i ,ZO

,19 "_

. "17 04) ]

00! ._e c |",', 031

007 -_5 _ |

009 -13 _ 05_

O. "t2 0 0644

_-L, _ °_".I

6 .,-1

J-

°'[ ]0.. 7

157' -3

04.1U

I_I00..4

0

r

:_ 123

> 121,,I-.,:3 L.]9

r.D1.18

116-,I

t i55" '9

1145-.

L125. ,I0

I.II-, II

II0. ,12

109- _13

1085._.14

1.084 15

108' 16

I::

r._0_D

_J

CO

13

6

712,

8

'9.10

01)

°r,4

f'

300,000 -I

2OODOO]

6O/)O

,_ 5o_oo

a 4 0,000.

O-,,-Ir/)

3opoo0

I::-.4

u)

m 2_000O_

4_000

20po0

_000

1,,.=_U(3.,

r._

!

C0

-=,-I

r.o

0E'_

C,,-.4

(,_(t,,t

4,.IrJ3

l-

o,.)

.,-4

"_ Fig. B 3.1.4-1

B 3.1.4 Analysis of

Section B 3

19 May 1961

Page 14

Helical Springs by Use of Nomograph (Cont'd)

FIBER STRESS CORRECTION FOR CURVATURE

Torsion Springs

Find Correction Factor Using Correction Factor Found on

Left Half of Chart--Determlne

True Fiber Stress

d

0021 '_-_D 16£0'0.0 0

"l OO3 t 0000_00_ _U 1.4ei/• _ 1.410,

0,I =o L367,

_-J 1332,

. °

04- 3":

05"06- S-.

0.7£ s-7-

09-:. m-

20-

2-

3-

d

_4Z. StiN"IJ3@t25,I I1_1, #i 115iiio-i lOT.

, 8IN:

liosO-ll 0

1077- II

L072. 12

I.O?o. i3

1066. 14

1064' i5

I,_7, I#

I055" Ill

_9

k

Of" t61_

0¢..)

_1 120-

4.-)

t_:;> IK)-I.

rj

I0_"

r._

c,..)¢J

o,

_J

I

o

in=

=

m(I)

2OOpOO-

00_OO-

701)00-

_ooo-

5QOoo.-

4o/:X_-

30_O00-

zo.ooo-

zoo_ooo.

g

7_oo-

6opoo.

_000 -

4O0OO--

3oooo

zopoo-

Fig. B 3.1.4-2 .__

B3.1.5 Maximum Design Stress for Various Sprin_ Materials

Section B 3

19 May 1961

Page 15

P-4

|

mm

U

&J

-,-4

120

I00

80

60

4O

2O

0

• Ol .02

Phosphor Bronze Spring Wire

Maximum Design Stresses

•04 .I0 .20

Wire Diameter - Inches

•40

_ Fig. B 3.1.5-i

B3.1.5


Maximum Design Stresses for Various Spring Materials (Cont'd)

//

/!

/IS'_ - sse=]S u_3saG mnm3x_]

Fig. B 3.1.5-2

Section B 3

19 May 1961

Page 17

B 3.1.5 Maximum Design Stresses for Various S Materials (Co_

I II I I I I1"0

IS)i - sso=3S u_ls_ mnmlx_i

- Fig. B 3.1.5-3

B 3.1.5

Section B 3

19 May 1961

Page 18

Maximum Design Stresses for Various Spring Materials (Cont'd)

v

15DI - ss_z_S uSTeo(] mnmTxe14

Fig. B 3.1.5-4 _-

Section B 3

19 May 1961

Page 19

B 3.1.6 Dynamic or Suddenly Applied Spring Loading

A freely falling weight, or moving body, that strikes a structure

delivers a dynamic or impact load, or force. Problems involving such

forces may be analyzed on the basis of the following idealizing

assumptions:

1. Materials behave elastically, and no dissipation of energy

takes place at the point of impact or at the supports owing to local

inelastic deformation of materials.

2. The deflection of a system is directly proportional to the

magnitude of the dynamically or statically applied force.

Then, on the basis of the principle of conservation of energy,if

it may be further assumed that at the instant a moving body is stopped,

its kinetic energy is completely transformed into the internal strain

energy of the resisting system, the following formulas will apply:

(a) For very slowly applied loads

b P (16)-- o ..... ° ...... ° ........ ° ° ° ° ........... ° ° °

K

(b) For loads suddenly applied

2P (17)K

(c) For loads dropped from a given height

_2 2P(S + _)G =

K................................. (18)

where:

= Total deflection

K = Spring rate

P = Load on spring

S = Height load is dropped.

The following problems will illustrate such conditions and their

solutions:

Problem I.

Given a spring which compresses one inch for each pound of load,

determine the maximum load and deflection resulting from a 4 lb.

weight.

Section B 3

19 May 1961

Page 20

B 3.1.6 Dynamic or Suddenly Applied Spring Loading (Cont'd)

(a) Case i:

The weight is laid gently on the spring.

P 45 = K i = 4 in. from Eq. 16

since K _ i maximum load = 4 lb.

(b) Case 2:

The weight is suddenly dropped on the spring from zero height.

5 = K2__[Pffi2(4)i = 8 in. from Eq. 17

maximum load = 8 lb.

(c) Case 3:

The weight is dropped on the spring from a height of 12

inches.

52 = 2P(S + 5) = 2(4) (12 + 5) from Eq. 18K 1

or

25 - 85 - 96 = 0

8 + _82 + 4(96)=

2

maximum load 14.6 lb.

= 14.6 in.

From the maximum load produced, Eq. 2 section B 3.1.1 may be

used to calculate the stress produced. This should be within the

limits indicated on Fig. B 3.1.5-1, B 3.1.5-2, B 3.1.5-3 and B 3.1.5-4.

Problem 2.

For many uses it is necessary to know the return speed of a

spring or the speed with which it will return a given weight. A

typical example of this problem could be stated as follows: a spring

made of 5/16 in. by 3/16 in. rectangular steel contains 4 3/8 total

coils, 2 3/8 active coils, on a mean diameter of 1 5/16 in. The

spring compresses 5/32 in. for 200 lb. load. If the spring is com-

pressed and then instantaneously released, how fast will it be moving

at its original free length position of 1 21/32 in.? The solution is

as follows:

Section B 3

19 May 1961

Page 21


Weight per turn = _(i 5/16)(3/16)(5/16)(.283) lb. per cu. in.

= .0685 lb. of steel in each coil

.0685 (2 3/8) active coils (1/3) = .0542 Ib (one-third of the

weight of active spring material involved.)

To this .0542 lb. we add the dead coil at the end plus the moving

weight, if any.

The equivalent total weight (.0542 + .0685) is 0.1227 lb.

The potential energy of a spring is equal to 1/2 the total load

times the distance moved, or 1/2(200)(5/32) = 15.6 in. lb. The

kinetic energy equals 1/2 Mv 2 wherein (M) is the mass and (v) is the

velocity.

Mass = Weight where g is the gravitational acceleration,g

or 32.16 ft/sec./sec.

.1227v 2Therefore 15.6 = or v = 314 in/sec

2(32.16)(12)

Often springs are used to absorb energy of impact. In most

such instances springs must be designed so that they will absorb the

entire energy. In a few cases partial absorption is tolerated. A

typical problem of this type follows.

Problem 3.

A 30 lb. weight has a velocity of 4 ft. per second. How far will

a spring that has a spring rate of i0 ib/in, be compressed?

KINETIC ENERGY

I 30(4)(4) = 7.46 ft lb.K.E. = _ Mv 2 = 2(32.16)

or

7.46(12) = 89.52 in.lb.

i load times deflectionSpring energy =

Load = rate per inch times deflection

Spring energy = _ Kb 2 = 89.52 in. lb.2

Section B 3

19 May 1961

Page 22


! I08 2 = 89.522

6 2 = 17.90

deflection, 6 = 4.23 in.

If springs are used to propel a mass, a parallel attack using

velocity and acceleration applies.

Problem 4.

Let it be required to find the spring load that will propel a

l-lb. ball 15 ft. vertically upward in 1/2 second. It is assumed

that the spring can be compressed a distance of i ft.

In order to travel 15 ft. in 1/2 second the l-lb. load must have

a certain initial velocity. This can be found as follows:

h gtV =--q-t 2

wherein: h = height

g = 32.16 ft. per sec 2

t = time

v = ]--+ 2 = 38.04 ft. per sec.

2

= v 2 _.(__._..04) 2Spring acceleration 2-_ = 2 "T_l) = 723 ft/sec/sec

Spring velocity at

free height.

Force equals mass times acceleration so

723(1)F = 32.16 = 22.5 Ib. avg.

The average spring pressure is 1/2 the total load. Hence the

spring will compress i ft. with 2(22.5) or 45-Ib. of load. Often, it

is desired to know how high the weight would be propelled. This can

be determined by equating the work performed by the spring to the work

of the falling weight; thus work equals force times distance.

In the spring we have 45 (i) ft.2

In the weight we have l-lb.(h)

Hence l(h) = 45 (I) = 22.5 ft., the height to which the weight

would be thrown. 2



If we were to apply the previous formula

_2 = 2P_S + 8)K to the springs,

we must remember (S) is the height the load is dropped.

distance traveled by the weight is (S + _).

The total

Therefore, h = S + g in this case

Substituting i = 2(1)h45

h = 22.5 ft.


B 3.1.7 Working Stress for Springs

If the loading on the spring is continuously fluctuating, due

allowance must be made in the design for fatigue and stress concentra-

tion. A method of determining the allowable or working stress for a

particular spring is dependent on the application as well as the

physical properties of the material.

Section B 3

19 May 1961

Page 25

B 3.2.0 Curved Springs

The analytical expression for determining stresses for curved

springs is

f = +P+-A - M----(I+AR l__y__ZR+y .......................... (19)

in which the quantities have the same meaning defined in section

B 4.3.1.

Displacement of curved springs is determined hy use of

Castigliano's theorem.

M _Mds + EAYoR _p ds

in which

_vTN _N /V _V-_-- _- _d_+p GA _P

M _N ds +/ N _M+ _ _P k_ _ ds

N = normal force

E = Modulus of Elasticity

G = Modulus of RigidityA = Cross-sectional area

R = Radius to centroid

ds = Incremental length

Yo = ZRZ+I

I _ "--Y--dA= - A R+y

y = is measured from the centroid

.............. (20)

P

These expressions for stresses and displacements are quite

cumbersome; therefore, correction factors are used to simplify the

analysis. The correction factors (K) used to determine the stresses

are given in section B 4.3.1. The expression for the stress is

f = K McT ............................................ (21)

See Table B 4.3.1-1 for values of correction factor K.

Deflection formulas for some basic types of curved springs are

given in Table B 3.2.0-1. Complex spring shapes may be analyzed hy

combination of two or more basic types.

B3.2.0 Curved Sprln_s {Cont'd)

Table B 3.2.0-I

Section B 3

19 May 1961

Page 26

Spring types Deflection

A

B

C

D

p 8

PP

B = I_R_3 (m + _)33EI

where (x = _ for finding K

8 - 3El +

where _ = 2_ for finding K

- 3El +


3


* ttl = uR

B 3.2.0 Curved Springs (Cont'd)

Section B 3

19 May 1961

Page 27

0 °

i0.2

r,,-

i,--

0.1

i

0 _1 i I I I!li ]ililllllllJllllllllnllHl[ fill llllllil III:IIIH

0. i 0.2 0.3 0.5 0.7 .9 1.5 2.0

0.15 0,4 0.6 .8 1.0

3 4 5 6 8

7 9

IJ

Ratio R

I0

Fi_. B 3.2.0-2


B 3.2.0 Curved Springs (Cont'd)

For close approximations, the following conditions should be met:

flat springs

spring thickness _ h

radius of curvature R<0.6

round wire

wire diameter d

radius of curvature R<0.6

Figure B 3.2.0-3 is a typical curved spring.

the spring at point A is calculated as follows:

Spring Characteristics _ U 1

h = 2.5" i'_i/_i_< 0.6 u 2 _i

_I = _ RI = R2 = i"

_2 = 2

u I = i" U I

Solution -

The deflection of

"_ u2 _B

4 zh R 2 _//_/

A

P

Fig. B 3.2.0-3

The solution involves two basic types (type B and A of Table B 3.2.0-1).

Type A solution is used for that portion of the spring denoted by sub-

script (2), and type B solution is used for that portion of the spring

denoted by subscript (I).

Correction factor, K (from Fig. B 3.2.0-2)

u2 2.5__Ul= _i = I _ = 180 ° K I = .80 .... 2.5r I I ' i ' r 2 i

_i = 90o

_2 = 90o, K2 = .86

Deflection at point A

2KIPRI3< _i 3 K2PR23 < 3_A = 3El m + _- + 3E_ m + _2

3F.I i + _ 3El 2.5 +

28.4P

E1

f

/

Sc:c-t.o.-_ _ 315 x,-- ' "q7_

Pas_ 29

B 3.3.0 3e!leville Sprln_s or Washers

Belleville type springs are used where space requirements

necessitate high stresses and short range of motion. A c_pleuc

derivation of data that is presented in this section will be found in

"Transactions of Amer. Soc. of M_ch. Engineers", May 1936, Volum_ 58,

No. 4, by Almen and Laszlo.

I

2a =O.D. !_'

i i'D'=2b

Fig. B 3.3.0-I

Symbols

P = Load in pounds

5 = Deflection in inches

t = Thickness of material in inches

h = Free height minus thickness in inchesa = One-half outside diameter in inches

E = Young's modulus

f -- Stress at inside circumference

k = ratio of O.D. _ aI.D. b

v = Poisson's ratio

M, C I and C 2 are constants which can be taken from the chart,

Fig. B 3.3.0-4, or calculated from the formulas given.

The formulas are:

6 (k-l) 2M=

logek k 2

(22)

6

C I " _ logek

3

C 2 - _ logek

]- 1 jlogek

o , o e o, o o, • i o • oe Io • • I * • e,

i

po,o,eoooo,eo Jot,,,, _" _"

(23)

(24)

Section B .315 March, I973

Page 30

B 3.3.0 Belleville Sprin_s orWashers (Cont'd)

The deflection-load formula..usin_._._he_.e constants is

P • (l "- ',Z )M_ZbZ h --_ h - 6 t + t

The stress formula is as follows:

(,. ] ...............: Before using these formulas to calculate a sample problem, there

are some facts which should be considered. In the stress formula it

is possible for the term (h - 5/2) to become negative if (5) is

large. When this occurs, the term inside the brackets should be

changed to read Cl(h - 5/2) - C2t. Such an occurrence means that themaximum stress is tensile.

For a spring life of less than one-half million stress cycles, a

fiber stress of 200,000 p.s.i, can be substituted for f, even though

this might be slightly beyond the elastic limit of the steel. This is

because the stress is calculated at the point of greatest intensity,

which is on an extremely small part of the disc. Iu_edlately surround-

ing this area is a much lo_er-stressed portion which so supports the

hlgher-loaded corner that very little setting results at atmospheric

temperatures. For higher operating temperatures and longer spring

llfe lower stresses must be employed.

Fig B 3.3.0-2 displays the load-deflection characteristics of a.040 in. thick washer for various h/t ratios.

It is noted (from Fig. B 3.3.0-2) that for ratios of (h/t) under

1.41 the load-deflectlon _urve is somewhat similar to that of other

conventional springs. As this ratio approaches the value of 1.41, the

spring rate approaches zero (practically horizontal load-deflectlon

curve) at the flat position. When the (h/t) ratio is 2.83 or over

there I$ a portion of the curve where further deflection produces a

lower load. This is illustrated in the curves for the washers having

(h/t) ratios of 2.83 and 3.50. Such a spring, when deflected to a

certain point, will snap through center and require a negative loadlny

to return it to its original position.

The washers may sometimes be stacked so as to obtain the load-deflectlon characteristics desired. The accepted methods are

illustrated in Fig. B 3.3.0-3.

B 3.3.0 Belleville Sprin_s or Washers (Cont'd)

Section B 3

19 May 1961

Page 31

OO_O

q

Fig. B 3.3.0-2

B 3.3.0 Belleville Springs or Washers (Cont'd)

Section B 3

19 May 1961

Page 32

Series Parallel Parallel - Series

Fig. B 3.3.0-3

As the number of washers used increases, so does the friction in

the stacks. This is not uniform and could result in spring units

which are very erratic in their load-carrying capacities. Belleville

springs, as a class, are one of the most difficult to hold to small

load-llmlt tolerances.

v

B3.3.0 Belleville Springs or Washer (Cont'd)

Section B 3

15 April 1970

Page 33

C1

and

C2

2.3

2.2

2.1

2.0

1.9

1.8

1.7 M j

#.

/1.6 /

/

1.5 /

!14 ! /

/ i1._ / /

F1.2 l ' /

/ ///I.I / /

/I l

lO ]// [1.2 1.6 2.0

//#'

/

/

v

i /f

//

I cl i/

/

/

C2/"f

/

//

//

/"

/

//

/

1.0

0.9

0.8

0.7

0.6

1.0

02.4 2.83.2 3.6 4.04.4 4.8

O.D.

Ratio of , kI.D.

M

Fi_.B 3,3.0-4 Belleville Spring Constants: M, C Iaud C2

Section B 315 April 1970Page 34

B 3.3.0 Belleville Sprlnss or Washers (Cont'd)

Example Problem

Given:

O.D. = 2"

I.D. = 1.25"

Load to deflect .02" = 675 lb.

Required: Fig. B 3.3.0-5

Determine required thickness, t and dimension h.

Solution:

k = O.D. 2.00 = 1.6I.D. - 1.25

The constants M, C1 and C 2 may be taken from the curves in

Fig. B 3.3.0-4 or may be calculated as follows:

M 6 (k-l) 2 6 (1.6 - 1.0) 2= = = 0.57

logek k 2 3.14(0.47) (1.6)2

0 I 0 El01.0ii1123C I " _ logek [l--_gek - i = 3.14(0.47) 0.47

C2 _ logek -- = 3.14(0.47) 2

The Deflection-Stress Formula

(1-v2)M,2

may be written in the form

h fM'2(i"v2) + _ C2" tCIE_ 2 C I

r_

Section B 3

15 April 1970

Page 35

B 3.3.0 Belleville Sprinss or Washers (Cont'd)

Assume that the washer shown in Fig. B 3.3.0-5 is steel

f = 200,000 psimax

E = 30,000,000 psi

v = .3

= 0.02 in.

M = 0.57

C = 1.123i

C 2 = I. 220

a = 1.00 (half outside dia., inches)

try t = .04 and solve for dimension h

t h t _t x_i 2 .57jkl_.32j + .02 1.220 (04)200,000h __ ° _-

(1.123)(.02)(30)106 2 1.123

.120 in.

This value for (h) is then substituted in the deflection-load

formula to obtain the load.

e

(I_E2)Ma21 _h - _>_h - _ _ t + t3 ]

30(106 ) (. 02)

(I-.32 )(.57) (i)2 [('121-'01)('121-'02)('04)+('04)3] = 600 lb.

Since this load is too low, the calculation is repeated using a

stock thickness (t) of .05 in.

Then, solving again for h, the result is

h = .ii0 in.

Therefore, substituting this value of h into the formula used to

calculate the previous value of load (P), the new value of (P) is

P = 665 lb.

This is as close as calculation need be carried. It is not

expected that this or a similarly calculated spring will be deflected

beyond the amount used in the calculation.

REFERENCES

Manuals

I.

,

Section B 3

19 May 1961

Page 36

Chrysler Missile Operation, Design Practices, Sec. 108,

Dtd. I November 1957.

Convair (Fort Worth), Structures Manual, Sec. 10.4.0,

Vol. I.

Handbook

I. Associated Spring Corporation, Mechanical Spring Design,

Bristol, Conn.

Periodicals

I. Klaus, Thomas and Joachim Palm, Product En_ineerlng, Design

Di_est Issue, Mid-September 1960.

Text Books

,

.

Seely, Fred, B. and Smith, J. 0., Advanced Mechanics of

Materials, Second Edition, John Wiley & Sons, Inc.,

New York, 1957.

Spotts, M. F., Design of Machine Elements, Second Edition,

Prentlce-Hall, Inc., Englewood Cliffs, N. J., 1955.

-h

-4

-L

SECTION

BEAMS

BZl.

B4.0.0

4.1.0

4.1.1

4.1.2

4.1.3

4.1.4

4.2.0

4.2.1

4.2.2

4.2.3

4.2.4

4.2.5

4.3.0

4.3.1

4.4.0

4.4.1

4.4.2

TABLE OF CONTENTS

Beams ..... . ......... ......0... .........

Simple Beams .............................

Shear, Moment and Deflection ..................

Stress Analysis ............................

Variable C ross-Section .......................

Symmetrical Beams of Two Different Materials .......Continuous Beams ..........................

Castigliano' s Theorem .......................

Unit Load or Dummy Load Method ................

The Two-Moment Equation .....................

The Three-Moment Equation ...................

Moment Distribution Method ....................

Curved Beams .............................

Correction Factors for Use with the Straight Beam

Formula .................................

Bending-Crippling Failure of Formed Beams ........

Bending Moment Only ........................

Combined Bending Moment and Axial Load ..........

Page

1

1

1

24

25

26

29

29

30

32

34

37

38

38

43

43

46

B4-iii

Section B 4

21 April 196]

Page 1

B 4.0.0 BEAMS

B 4.1.0 Simple Beams

B 4.1.1 Shear_ Moment_ and Deflection

The general equations relating load, shear, bending moment, and

deflection are given in Table B 4.1.1.1. These equations are given in

terms of deflection and bending moments.

Title Y M

Deflection A = y A = 7f_-dxdx

EI

@ = dy/dx 0 = F _-dxSlopeJ EI

Bending Moment M = EI d2y/dx 2 M

Shear V = EI d3y/dx 3 V = dM/dx

Load W = EI d4y/dx 4 W = dv/dx = d2M/dx 2

Table B 4.1.1.1

Sign Convention

a)b)c)

d)

e)

Y

x is positive to the right, lw

• -X __ Xy is positive upward ÷x

M is positive when the compressed i I

fibers are at the top.

W is positive in the direction of

negative y.

V is positive when the part of the beam to the left of the section

tends to move upward under the action of the resultant of the

vertical forces•

The limiting assumptions are:

a) The material follows Hooke's Law.

b) Plane cross sections remain plane.

c) Shear deflections are negligible.

d) The deflections are small.

Section B 4

21 April 1961

Page 2

B4.1.1 Shear_ Moment and Deflection (Cont'd)

The deflection of short, deep beams due to vertical shear may

need to be considered. The differential equation of the deflection

curve including the effects of shearing deformation is:

/M dxdx +fKVY = El _ dx

(K) is the ratio of the maximum shearing stress on the cross section

to the average shearing stress. The value of (K) is given by the

equation:

a

A J b'ydyK I b0

(I) is the moment of inertia of the

cross-section with respect to the

centrodial axis and (a), (b), (b'), and

(y) are the dimensions shown in Fig. B 4.1.1-1.

(A) is the area of the cross-section

IFig. B 4.1.1-1


Section B4

July 9, 1964

Page 3

>

-;-I

r_

O

•_ F)

t_

_ F-.4

O

N '

C _

m _

N cq

+

O II cq

"_ cch I

- I_ _-4

o_

•_ i I |

II II II

0

0 _ >_ I

cn_ tl?

4-1

5"

II

V_!

II

>

©

I

eq

©

II 0 I

o (2)4-1 I I

II

'_ II II

O O O

o < _, m

l----b 0_q _J

ol

4-

I

C',l II

cq ' _

I

c"3

Zod

oq

I

II

;4t_

o

xi\\ _ \'3_' 7o

__ i ...... L

t

t

B 4.1.1 Shear_ Moment and Deflection (Cont'd)

Section B 4

21 April 1961

Page 4

_J

=oO

co

,-4

1_1 q)

, _,,-,4

fu

t_[.-i

,_ >,--,

0_J

oo _._

°,.-4

•l=J *_

0

.._ _le_

° Ii +It X U

0 I __sto

" 11•_ 4J M

=

3-,a cO

.= , +_ Sco c_

,--1 II

-,_ :_I,-_ :_ _ _

°" '< ._1 oom

o= _ , , +4.J

Ii il II li II

r_7co

"= '_ ,_i'

_ +!

m

_ NI I

o ...1-M *

t'M

-I- ,-I

+ c_ +

P_ _.... J t__o eq

c_

r--1

,z3

v I

c-qcq ,_

¢'4c,_ 0

! !

v _ _,

-I- +

+ /;

_ >_ >_ >_ , +

I_l I_ I_l 0 I_ II If

o o o o o :>_ c_

m _ c_

.J e_, " _l

_ _ -- _ --

c_

t_


Section B 4

21 April 1961

Page 5

ou

o"J

o

(lJ

cNI

111

,.Q

>

-r.-I

4-1

¢

o

o

o

o

O

E.I,J

oEo

o,.c:Oq

N

° I_.IJ

++:> II

o :>

o.;.-4

u II

o,0

"E3

o

i_ "_U4 _O _

_ m

_ U u"5

I

II

>

O.I-J _

+

+m co ,-+ I co

_ Jr-

I I

°_

<

t_

I

I!

:>_

N•4-+ ttlm

.-+1 co _1

u_I

X L_

CO

X N b--t _-t

II II II

::m:

>-

,-r,i

.... _.+t..,+l.._

, _ _._II Nc_

> ::E:,-_1 co

r.DI

o.l.J II

::m:v

uEo

II .1..1

cc_

II II

+0 0

II IJ _+J

+

+_ 41

<

+J

t_

.... i

co

r----i

Le_

r --] _._ ,,,.]

I I

cxj

co t_

c_ Xl _r_

I

m

m +r_ co

_ N

,'-- MN _i o

_ X

_ M

N 1'-.-.

o (2),.-4 r---i+ +

_ N

L,"5 __ V

I

N _! I

_-1 cov _

I

N

_ v t_¢M ,...4+ _, +

+ + _

co t _ _

t j ¢N

+_co _I ° +I° _+I I I

II II II

0 0 0

o

<_

+

+

co

__J

+1_ +1_+1°++1+"

, +

U II

E)

N

B4.1.1 Shear, Moment and Deflection <Cont'd)

Section B4

July 9, 1964

Page 6

4J

o_2

,.-4 _0

o _

°r-.I

o

_3

-_ _°o ,--'4I'x:_

_'_1 fl

_i m m+

_ N N

.._, ._ ._ ,-.4

,.-I _ I

N _

°g•,4 , , +

O II II II II

,'¢::j ._

e _ -I_/-.. 'I '

I

'I -_,,,


Section B 4

21 April 1961

Page 7

O

COv

¢o

,-q o_

0r_

°r4

4-1C'4 _

r.4

,.O

0

0

.1_1

0

.,-4

,2r_0A

r_

.r4

.!..J

o.,_ _4..1

t_

.r'4

0,_

0

,<

I

II

0

0 ×+

O H

0

II II

G

% ¢I -%.

+

+

0'0 .-, ¢_1

_" _1_ ,× _

_ _O _

,-.a

0II II II

"0

0 0 0

rn

l i_l

d

I

-- m Ii


Section B 4

21 April 1961

Page 8

=ou

o

e,J >

[--i

o

_ ,.,-4

o

g >

2 o

m

o

_0

o

I

v

II

u

o

/g

II

N

It 4a

_ r._

N

i

r¢3

,t,J

II II

flo o._ 4.1

x

I o1+_-

---_[ t-- [

I

+II

II>

::E:

o.iJ

+

II

>

o.I.J

<

v

+ ,_" o

II

t

0 o

°_

A

u

,,iJ

i

¢q+

+

v cq

II

× +

(D

¢qo_

v

_i_ _ ,_, _ e ='__'] ..1 _ +

I I

II II

0 0

v

I I

II II

+

u +

II

+o

_4

>.

x

Io _

I+_Ira _

O

+ . .i_


Section B4

July 9, 1964

Page 9

oc.)

E

O

O

@,,1 _

• .el

,-.-I u3

4(D

r--4

O

+II

o cD

'4.4 + ._

_ N

c,4 _

_ .o,31

u_

II

,._ _l_

II'_l II .t_

U ;>

° _1_ II oo m

m _II II

o:•,_, _l_ _

¢B

oO

O _ >-

X

_ c-I

v

:xI × c,_

' :x

0

II

0

II

IIO

_ N

v _

_ °ul c._

+ v

IIII

oII 4_1

_l _ "_,× o

+ ,--

_ c,4

II __'_ u

[--IN c_ F---I

¢'J Ocq

+ +

II + + + +

O o i i

_i_ I i i

v v _ !

_ _ _1_

k- --_v-'--_.__ C _ II

I1 II II II I

0 0 0 0

X

o

--Ic_i

--Ic_

B4.1.1 Shear a Moment and Deflection (Cont'd)

Section 4 B

21 April 1961

Page i0

=0_J

m,-4

O

_4!

r-I

m

m

1.1W0

°r.l

I

=

"0m0 _J

o =

O.0

O0

"0

O•rt I_

U 4.4

+ +

£%1.

+ *

°M

>

O

•_ II II

'O

i,=0O

,-4

.@

+¢q

H

N

!I_Im M

IIII m

_ N_ M

_ X m I

, . Ni

• _ o"1 [._.1

II IIo_

I -"

4_

II

._

H

_°r_ _-i

I

II

B 4.1.1 Shear a Moment and Deflection (Cont'd)

Section B 4

21 April 1961

Page II

0

_ ,

.-.-4 O

E

O •

E

¢lau?

f-4 _q

,.o

O

oO

O.M

Q)

I

.,-4

O_:1 II

- >

r._ o11

.Lt °_

o•,_ :_'1_4-1u II

0

0 _J

f_

!

_l,-e

ii

O

°_

I----I

!

_ J

I

II

>

(.3

0

.I-I

v

.IJe_

_ 1<oII

II

e_

r I _ _D

cgI

%1 +

y.'

_1_

0

I> .<

°_

I--I

.d-i r,-_

I f--I _

! .'4

M M rO

+

M _,_ + _"

| i

_, c'j c-,l

,.-1cl

oh _-I.i _, v-_

¢',1 II--t

_:,.o

IIo o.i.I _ o

v

n-

t'-xl

4J / i,-,] I _ "_, '_._.__// <

I II II

,.__._j_ 0 r_ _ II(I)0 0 0

II .I..,I .i.i ,,l,.I

v _ v

t_

,-4

X

7

L


Section B4

July 9, 1964

Page 12

O

v

m

,-4

0r_

I

.4

,1o

[--i

0

>

0,-4C13

•_ 0

r-I II

m

I_ +

._ _ "'_ U _

o_

1.1

_ N

•_ 0 _

t_ II II

0

+ _

_1_I

I

II

0IJ

v

<

+

II

>

U

0

<%J

4-I

° _III

I {'q

_-_ I ._,-, 0

m•_ _ _ +0

+ 0 "-_--"

_ ++ _

<U _

+

0

o_

I

II

II

I

II

4_

,.al_

I

___, _ , %1__I__I_ _ '

II II

II

0 0 CO cb•_ .IJ

X

'.J

-\


Section B4

July 9, 1964

Page 13

-rJ

_J

0

G}E

E

0 &l

1_ o

I

• _2

[--i

0

0

'_0

<

0

4J

E0 _ +

o_

r...)_ mm o

¢1 <1-4 -_(1_ II

o0

_ Ill .i..1

c_.r-I

O

O

[--4 0 e'_

C'-I II

I

c._ _

_ 4-1

I

_-_ _I_I_-Iv t_

• - i II

i N

_III II _ I--4

I_ ,,.-x II

0 0

X

Q_

Q?

4J

+

v

I

II

B4.1.1 Shear a Moment and Deflection (Cont'd)

Section B 4

21 April 1961

Page 14

_z0

o

_ %,

iJ

_ o_ II

.,4

_ II

I ,_ c_

u II

.==

=.,4

O _

_u m

II _1 II

_ aJm

[0 ,_

,,_ II ,._

I

• _

II°_

!

o 14

II II

_ U

O O

I

V

I

i--_

¢q m

_ + +

e,i N N

_ m rn! !

N N

I _,l 0,1,-i ..1- II II

II

0 rO (O

0 0

• .,4

_l,_ ___1_ °

4.1 _

O

e4e4 _

_I _ _ II

_9 -r4

•_ .,4 .Ill

i_ _ I_ _,

_" _" _" _1_

_ _ u"_ .,..i 11-4

/_ II ,l_ II

Ill _II H

l.l

'°t

'_ I%:

i

I

L

v-

B4.1.1 Shear, Moment and Deflection (Cont'd)

Section B4

July 9, 1964

Page 15

ou

r.,.O

o

_) "o

j _

._

_ 4J

,-4 r._

c_

°r...I

"o

o

E

o

>_u

'-0

I ¢,.)

'i o

!

N

-:_ _ ,,c_

I

"_ r...)

II _ O

°_ _

,._ II

C-,I II

I

-_ °_

O

O3

0 Y.l¢_ II

° ".IJ t_

N0£ _.m I o0., _ II

•_ _ >_

O0

_J X,._ II II c,'-)

,,-'-, _ I

_o_1 _"_ "•_ O0 _._ O0 CD

u II_d I_ II

_ _ -_ L-- --

m

+ +

_ v c_v

_I_ IJI _ + >

; oII II ,L)

_4e,4

c_ [--I

,I

II X -__ C'4

,N +

× _ +

I

• - _ <

' [ II_I

II II IIII

v

o o o

C_

- L


Section B 4

2] April 1961

Page 16

s-%

P

0 o_

0

I ,-.4

t_

_ _'j,o

[..i

o

¢q

o ,_1_ot,,l

4.-I IU

4-1

4-)

_, -,-_-_

_ +

-_ _ , ,

o_.,_ _1II I

'1:1

80

%!

IIr_

vo_

_ oI

I

>,

r.II ,,.1-

> _ ._

II '_I N

o

_o -o

v

°_

II

4..I

0r_v

' _i_II

II

<o_

i IJ,o _lo

B4.1.1 Shear, Moment and Deflection (Cont'd)

Section B4

July 9, 1964

Page 17

O0

q-_o

(1,1 C}:

x I

0

I'

B4.1.1 Shear_ Moment and Deflection (Cont'd#

¢,J

I

0J

0J

od

_J

+

or-d IO0

_Io

o3

=o ._.rd

oJ II

=la

II

-.'t

_ II 0

II

¢,)I It

=

°_

+

¢,_ ,4-

o o

I

I

_4

_0¢'4

4.1

_I_II

¢"N°

0

II

oo

oI

II

, _I _

II

X

IIt',,I _

II

°_ I

I

_ v ,

o_

o

I

NJ_ "

I v

II _ II

_!'_ _ _

xl

)

Section B4

July 9, 1964

Page 18


Section B 4

21 April 1961

Page 19

g4J

°Uv

L)o3

,-4 4J

o E

E _J

cq

4J

_J,-_ oo

E_

_J

0P-_ I

It 4.J

_-, oo N

II IO _

O•,--t _1 r,_

UI

E: I t :>_

O _ l::cl

•,-.-" OO_ II 4J

•_ ::E: <

U_J 0

DOII _J _J

_J "_ _-_ :_1oo _ L_

::> II II ,--_II

0 -- P_ ,

_J OII 4J

O0

"El

0 ¢)

0:),.

J,,,,I /__L.

<

<

+_Ic4

cq._ cq II

cq _cq

fl ,._

0

°_

_1 O I,c-4

--, < ' N

+ v

OII ._J +

+

,._cN

+_d II II¢o II

_I 4 v0 0

II _J 4J _

o_¢q

×I

o i

°iI m j

_l. I

II II

_ r6

c; EI I

U II

> >

r_ ,-4

•r4 ._o3 o3o3 o3O O

t_ _3

V A

! !

II II

I I


Section B4

July 9, 1964

Page 20

4-J

0 _.I

rJ I

_ .5!O _

_4

.._ .,4l-J

,.-4 r_

m

b_

P-0

o']

"0

>-,

0-e4_a

¢J

OJ

0 r:_

4-_ 0O 4JrdcJ <

_a A m

_ I .,-4

_°_ _ × _ _1_• m ,-1 m

,._ ,-_ _ II

_¢ ._.c: c'j Cxl

o I_-_

_ II II o'_ C-,I _,1

i

r...) II

0

_Cl n_ N"-" >: Z

_C

_ -

_ m

_1_ ×,

I

_I_

°_

_I_ _I_II II

II _

F

__.i_ _ -_II

_i_

I_I I II

_i;_ _x, _,II

i _ m

I1

×

I--Ic.m

oO

Ii

-\

B4.1.1 Shear t Moment and Deflection (Cont'd)

Section B 4

21 April 1961

Page 21

ouv

o_

.;-4O

I-4

eq >_I ,-q

,-4 O

4J

4-Ir-4

o

rj_

"0

o_

II

o c'N

_ t

;_ %1._ t

?1m

,c +

_ M

o

4_

m

-,-41

_-)I II

r_

-el

o

o _

' [..-4 _ c'e3

i

o

+

+

I

I

II

< 4-

II ",--"f _

>JI i

O _ _

+< o

'o4 _ O

oO II

> -JI II

cq _

_1 0 _

+ u .,_Y_I_ U

+ ,_I_ _<

_ +I _.

I < ._

,.-1 II II m.-4

o o

II ._ 4..i ,_

I

v

Ar6

_),.c:

@I

II

I

v

I

IV

N

_ v

I

IIII

o

N-d N_._ I

N

_ NI

_ M

II II

0 04J ,_I

_ rF

'/m

o

B 4.1.1 Shear_ Moment and Deflection (Cont'd_

Section B4

July 9, 1964

Page 22

O

¢qu_

II

° _I_

,-_ IIu_

I _ r_

m .. II

°

I" ,,--4 ¢qX

• _ _ 0"_ _ _-_

_J I _ I

,_ I_1 II

o= i}!

e_ m

g-I

¢N

I

.4"v

:_ I ,-_ +

! I

II II

I

+

II II

°_

I IM __ M

I IIII

O

O

4J

_= o_-J 4-;

O _J4J m

4J .P1

o _1_+

m

_ o _,II ' II

II II:= >,

o o

v _ _ v


Section B4

July 9, 1964

Page 23

_D eq

coXi

0

_q

+

- _ _-_ cq

o _ _: _ V I

•_ .,-_ A

i 01 U .,_

2 o

i.--4 _c,,I _ _ II II

I_ II

o . +

_ ×

oO

0

0

O. m

_J

0

c_

Section B 4

21 April 1961

Page 24

B4.1.2 Stress Analysis

The maximum bending stress is:

(1)

The limitations are:

a) The loads on the beam must be static loads.

b) The value of f is the result of external forces only.

c) The beam acts as a unit with bending as the dominant action.

d) The initial curvature of the member must be relatively small.

(Radius of curvature at least ten times the dept_)

e) Plane cross sections remain plane.

f) The material follows Hooke's law.

If the calculated stress does exceed the proportional limit, a suitable

reduced modulus must be used.

The maximum shearing stress in a beam in combined bending and

shear is:

Vfs=K -_

• • • ,, • • • .. • • ° • .. • • ..... ...... o o ° ° ° ° o o. o ° o . . , o o ° o o o ° (2)

v

where (K) is the ratio of the maximum shearing stress on the cross

section to the average shearing stress. The maximum shearing stress is

often expressed as:

fs = VO .............................................. (3)It

Where Q = JydAArea

(First moment about the neutral axis of the area

between the neutral axis and the extreme outer

fiber.)

Section B 4

21 April 1961

Page 25

B 4.1.3 Variable Cross Section

The following formulas and figures present a method of analyzing

beams with uniformly tapering cross sections. Figure B 4.1.3-1 shows

a tapered cantilever beam consisting of two concentrated flange areas

joined by a vertical web which resists no bending. The vertical com-

ponents of the loads in the flanges, P tan 5 I, and P tan _ 2, resist

some of the external force V. Letting Vf equal the force resisted by

the flanges and V w the force resisted by the webs, then:

V = Vf + V w .................................................. (i)

Vf = P (tan 51, + tan 5 2) ................................... (2)

hi h2

From Fig. B 4.1.3-1, tan 51 =--, tan 52 , and tan 51c c

hi + h2 h h b

+ tan 52 = = --. From this Vf = P -- and since P = V _,c c c'

then

b (3)Vf = V _ .....................................................

aThe load in the web is V E, so by writing a, b, and c, in terms of

h o and h, we have

ho

V w = V E- .................................................... (4)

(h - ho)

vf = v ---f--- ..............................................(5)

P TANcK I

P TAN _2

Fig. B 4.1.3-I

Section B 4

21 April 1961

Page 26

B 4.1.4 Symmetrical Beams of Two Different Materials

The analysis of symmetrical beams of two or more materials within

the elastic range may be analyzed by transforming the section into

an equivalent beam of one material. The usual elastic flexure formula

then applies.

The transformation is accomplished by changing the dimension

perpendicular to the axis of symmetry of the various materials in the

ratio of their elastic moduli. Examples to illustrate the method for

various conditions follows.

Example i. Consider a beam made of two materials whose cross section

is shown in Fig. B 4.1.4-Ia. Assuming n = Ea/Es, the transformation

in terms of material (S) (Fig. B 4.1.4-ib) is then bI = nb. The maxi-

Mh I

mum stress in member (S) is then f(s) max = -_- and the maximum stress

-Mh 2

in member (A) is f(a) max = n---_ It is noted that the section could

have been transformed in terms of member (A) (Fig. B 4.1.4-ic) giving

the same results for the maximum fiber stresses.

(o) (b) (c)

Fig. B 4.1.4-i

Example 2. Reinforced-Concrete Beams. It is the established practice

in calculating bending stresses in reinforced-concrete beams to assume

that concrete can withstand only compressive stress. The steel or

other reinforcing member then is transformed into an equivalent area

as shown in Fig. B 4.1.4-2b. The distribution of internal forces for

a beam (Fig. B 4.1.4-2a) over any cross section ab is shown in Fig.

B 4.1.4-2c.

v

B 4.1.4

Section B 4

21 April 1961

Page 27

Symmetrical Beams of Two Different Materials (Cont'd)

-,-- b -_

d d i N.A, -J .Rt

(o) (b) (c)

Fig. B 4.1.4-2

To satisfy the equation of equilibrium, the internal moment must equal

the external moment. The mathematical statement is:

kd

b (kd) (-_) = nAs (d-kd) ..............

Concrete Transformed

area arm steel area arm

(i)

From which

kd = nA--'-_Sb_i + nAs2bd ' i_ ................................ (2)

where n =E steel

E concrete

The stress in the concrete fc and the stress in the steel fst is

M (kd) ............................................... (3)fc

I

fst = nM(d-kd)I

.......................................... (4)

where

I = Moment of inertia of the transformed section.

Section B 4

21 April 1961

Page 28

B 4.1.4 Symmetrical Beams of Two Different Materials (Cont'd)

Alternate solution: After kd is determined, instead of computing

I, a procedure evident from Fig. B 4.1.4-2c may be used. The resultant

force developed by the stresses acting in a "hydrostatic" manner on

kdthe compression side of the beam must be located _- below the top of

the beam. Moveover, if b is the width of the beam, this resultant

force Rc = _(kd_ fc max, (average stress times area). The resultant

tensile force Rt acts at the center of the steel and is equal to Asfst,where A s is the cross-sectlonal area of the steel. Then if jd is the

distance between R c and R_, and since Rc = Rt, the applied moment M is

resisted by a couple equa_ to Rcjd or RtJd.

ijd = d - _ kd ................................................ (5)

The stress in the steel and concrete is

M

fs = As jd ................................................... (6)

2Mf ,= .c b(kd) (jd)

........................................... (7)

N

i

B 4.2.0 Continuous Beams

Section B 4

21 April 1961

Page 29

B 4.2.1 Casti_liano's Theorem

Castigliano's Theorem is useful in the solution of problems

involving continuous beams with only one or two redundant supports.

The theorem can be written as

L

_U i / _M8Q - _Q - El M_ dsO

L

BUs I /_ _V

_Q = _-_- = GA JV _ ds

_U

0a =_ -a

o

L

................................ (i)

................................ (2)

M _ ds ................................ (3)E1o a

where

8Q is the deflection at the load Q in the direction of Q.

Q may be a real or fictitious load.

Qa is the slope at the moment Ma in the direction of Ma.

M may be a real or fictitious moment.a

U is the strain energy of the beam.

M is the bending moment due to all loads.

V is the vertical shearing forces due to all loads.

A is the cross-sectional area of the beam.

E is the modulus of elasticity.

I is the moment of inertia.

Section B 4

21 April 1961

Page 30

B 4.2.2 Unit Load or Dummy Load Method

The unit load or dummy load method may be used to determine

deflection at elastic or inelastic members. Deflection of inelastic

members by this method is given in section B 4.5.0. The theorem as

applied to elastic beams is written in integral form as

L

O

(i)

L

e = E-i--dx ..........................................O

(2)

Where (8) is the deflection at the unit load and (8) is the rotation

at the unit moment. The Moment (M) is the bending moment at any

section caused by the actual loads. (m) is the bending moment at any

section of the beam caused by a dummy load of unity acting at the point

whose deflection is to be found and in the direction of the desired

deflection. The bending moment (m') is the bending moment at any

section of the beam caused by a dummy couple of unity applied at the

section where the change in slope is desired. It is noted that although

(m') may be thought of as a bending moment, it is evident from the

expression m' - _M that it is actually dimensionless.a

Section B 4

21 April 1961

Page 31

B 4.2.2 Unit Load or Dummy Load Method (Cont'd)

lllustrative Problem: Find the elastic vertical deflection of the

point A (Fig. B 4.2.2-ia) of the simply supported beam subjected totwo concentrated loads.

R=P R=P

(o) (b)

M

/_ M=PXII---M : P L/4

-_M=PX2

--,--X I X2-. _

m

m= X2/ .-4-

I-.--x, x2-- (el (d)

Fig. B 4.2.2-1

Solution: The actual loading is shown in Fig. B 4.2.2-Ia and the

dummy loading is shown on Fig. B 4.2.2-Ib. The moment for the actual

loading is shown on Fig. B 4.2.2-Ic and the corresponding moment dia-

gram for the dummy loading is shown on Fig. B 4.2.2-Id.

L

L 4

= +J E_,, E1

O O O

3L

The deflection by use of equation (i) noting that Xl starts at

the left and x2 starts at the right isL

4

>_ dxldx = , El \

4

j+ 4-_ dx2 = 48E----I

L

4

Section B 4

21 April 1961

Page 32

B 4.2.3 The Two-Moment Equation

The two-moment equation may be used to determine the bending

moment at one section of the beam when the shear and bending moment at

another section and the loads applied to the beam between the two

sections are known. The expressions for the moment and shear corre-

sponding to Fig. B 4.2.3-1 is

M 2 = M I + Vld + Fx ...................................... (i)

V 2 = V I + F ............................................. (2)

LOADS

Fig. B 4.2.3-1

The two-moment equation is particularly useful in determining the

curve of bending moments and shears in the case of a cantilever beam

subjected to distributed loads, such as shown in Fig. B 4.2.3-2. The

calculations may be done in tabular form as in Table B 4.2.3-i.

,@

'' "I

Fig. B 4.2.3-2

\

B 4.2.3 The Two-Moment Equation (Cont'd)

Section B 4

21 April 1961

Page 33

Station

(inches

from end)

W _-

Load

ib/in

i0

Shear _f_x)V= (WI+W 2 ) -_--

(10+14)10/2 =

0

120

* Bending Momen_ 2h

M=(2WI+W2) _6_VI(Ax)

(2.10+14)102/6 =

i0 14 120

(2.14+18)102/6 =

(14+18)10/2= 160 120"]0 =

20 18 280

22

26

30

40

(18+22)10/2=

(22+26)i0/2=

(26+30)10/2=

50

200

48O

240

720

280

i00030

(2.18+22)I0Z/6 =

280.10 =

(2.22+26)102/6 =

480.10 =

(2.26+30)102/6

720.10 =

0

567

567

767

1200

2534

976

2800

6301

1167

4800

12_268

1367

7200

20_835

Table B 4.2.3-1

*The increment of bending moment between stations may be calculatedfrom the relation:

M = (_V) (Dist. from centroid of trapezoid to inboard station)

+ Vl (_×)

Section B 4

21 April 1961

Page 34

B 4.2.4 The Three-Moment Equation

The three-moment equation is useful in the solution of problems

involving continuous beams with relatively few redundant supports.

equation is:

MaLl 2MbLI 2MbL2 McL2+ + +

I I II 12 12

6E (Ya " Yb) + 6E (Yc - Yb) ...................... (i)KI + K2 + L_ L-_

Where (KI) and (K2) are functions of loading on span (LI) and span

(L2), respectively.

The

Tt__

Lir L21is .......Fig. B 4.2.4-1

One equation must be written for each intermediate support. The

system of simultaneous equations is then solved for the moments at

the intermediate supports.

Values at (K I) and (K2) for various types of loading are tabulatedin Table B 4.2.4-1.

Section B 421 April 1961Pa_e 35

B 4.2.4 The Three-Moment Equation (Cont'd)

Table B 4.2.4-1 K 1 and K 2 Values for Various Load Conditions

Type of Loading Left Bay -K 1 Right Bay - K2

i.

r_ L ID-

°

r '-T

°

.

.

.

b i--a--4

iTI

I-'---------- |

I __

i |

I__ _r

L -I

+WiLl 3

411

+WlC I (3L_-CI 2)

8I 1

+_2L3412

8I 2

KI =+wlEbl2(2Ll2-bl2)-al2(2Ll 2 - a12)]

K 2 = +

411L 1

2 2 2)w2 [b22(2L22-b22)-a2 (2L 2 - a2 ]

412L 2

+2WiLl 3

1511

+7wiLl3

60I 1

+3L_w I

811

+7w2L23

60 12

+2w2L23

1512

+3L22w 2

8I 2

+w2c 2 (3L22- c22)

Section B 4

21 April 1961

Page 36

B 4.2.4 The Three-Moment Equation (Cont'd)

Table 4.2.4-1 K I and K 2 Values for Various Load Conditions

v-

Type of Loading Left Bay - K 1 Right Bay - K 2

o

-FL _1_- t -t

.

I_ L -_

o

10.

+5WiLl3

3211

+wla I _L12_a12 )

IILI

M I / 3al 2,

+5w2L22

3212

+w2b2 _L22_b22_

12L 2

+ M2 (3b2212 \L_2 L2)

jw

qr Ir

E 1 =

12L2M x I dx 1

O

L

6 P

K2 = 12L2 _o M2 x2 dx2

Where M is the banding moment.

Section B 4

21 April 1961

Page 37

B 4.2.5 Moment Distribution Method

The moment distribution method is suggested for the solution of

problems involving continuous beams with many redundant supports. The

discussion and application of the method is given in section B 5.0.0

(Frames).


B 4.3.0 Curved Beams

B 4.3.1 Correction Factors for Use in Straight-Beam Formula

When a curved beam is bent in the plane of initial curvature,

plane sections remain plane, but the strains of the fibers are not

proportional to the distance from the neutral axis because the fibers

are not at equal length. If_denotes a correction factor, the stress

at the extreme fiber of a curved beam is given by

Mcf =K--

I

in which M__M_ [i + c ]K = AR Z(R + c)

Mc

I

where

M is the bending moment

A is the cross sectional area

R is the radius of curvature to the centroidal axis

c is the distance from the centroidal axis to the extreme

outer fiber

I is the moment of inertia

I /_.Z_ dAZ = - -_- R+y

Values of K for different sections are given in Table B 4.3.1.1.


Table B 4.3.1-1

Values of K for Different Sections and Different Radii of Curvature.

.

Section

R

K the same for circle

and ellipse and inde-

pendent of dimensions.

0

e i

K independent ofsection dimensions

7-b

.

I 2b ]

R

C

1.2

1.4

1.6

1.8

2.0

3.0

4.0

6.0

8.0

i0.0

1.2

1.4

1.6

1.8

2.0

3.0

4.0

6.0

8.0

i0.0

1.2

1.4

1.6

1.8

2.0

3.0

4.0

6.0

8.0

I0.0

Factor K

Inside Fiber

3.41

2.40

1.96

1.75

1.62

1.33

1.23

I. 14

1.10

I. 08

2.89

2.13

1.79

i .63

1.52

1.30

1.20

1.12

1.09

1.07

3.01

2.18

i .87

1.69

1.58

1.33

1.23

1.13

I.i0

I.08

Outside Fiber

O. 54

0.60

0.65

0.68

0.71

0.79

0.84

0.89

0.91

0.93

0.57

0.63

0.67

0.70

0.73

0.81

0.85

0.90

0.92

0.94

0.54

0.60

0.65

0.68

0.71

0.80

O.84

O.88

0.91

0.93


Table B 4.3.1-1 (Cont'd)

Values of K for Different Sections and Different Radii of Curvature

Tb

±

o

Section

3b-----_

-m- C -_

I 2b I

-o------- R _

Do

o

/-rI b I

R

C

1.2

1.4

1.6

1.8

2.0

3.0

4.0

6.0

8.0

i0.0

Factor K

Inside Fiber Outside

3.09 0.56

2.25 0.62

1.91 0.66

1.73 0.70

1.61 0.73

1.37 0.81

1.26 0.86

1.17 0.91

1.13 0.94

i.ii 0.95

3.14 0.52

2.29 0.54

1.93 0.62

I. 74 O. 65

i .61 0.68

i.34 0.76

i. 24 0.82

1.15 0.87

1.12 0.91

i.i0 0.93

3.26 O. 44

2.39 0.50

1.99 0.54

1.78 0.57

1.66 0.60

1.37 0.70

1.27 0.75

1.16 0.82

1.12 0.86

1.09 0.88

1.2

1.4

1.6

1.8

2.0

3.0

4.0

6.0

8.0

i0.0

1.2

1.4

1.6

1.8

2.0

3.0

4.0

6.0

8.0

i0.0

Fiber

i

Section B4

July 9, 1964

Page 41


Values of K _or Different Sections and Different Radii of Curvature

o

Section

"- 1.625b "--q"i

A = 1.05b 2

I = O.18b 4

C = 0.70b

3

I I 4t

t ;LJ

o

"------ R _

Rc

1.2

1.4

1.6

1.8

2.0

2.5

3.0

4.0

6.0

8.0

I0.0

1.2

1.4

1.6

1.8

2.0

3.0

4.0

6.0

8.0

I0.0

Inside

3.65

2.50

2.08

1.85

1.69

1.49

1.38

1.27

i. 19

I. 14

1.12

3.63

2.54

2.14

1.89

1.73

i .41

1.29

I. 18

1.13

i. I0

Factor K

Outside

1.2

1.4

1.6

1.8

2.0

3.0

4.0

6.0

8.0

I0.0

3.55

2.48

2.07

1.83

1.69

1.38

1.26

1.15

i.I0

1.08

0

0

0

0

0

Fiber

0.53

0.59

0.63

0.66

0.69

O. 74

.78

.83

.90

.93

.96

0.58

0.63

0.67

0.70

0.72

0.79

0.83

0.88

0.91

0.92

0.67

0.72

0.76

0.78

0.80

0.86

0.89

0.92

0.94

0.95

Fiber

Section B4July 9, 1964Page 42


Values of K for Different Sections and Different Radii of Curvature

i0.

3t±

Section

_t_-T-]

_]

Ii.

m.

12.

4t --1_ p-, F-qr

3t

L_I

_ 2d ----_

2

3

4

6I

8

I0

i

i

1

i

2

3

i 4

6

I 8i0

I

F4t

p'-_t4taJ

!

_C _-

i

±

1

i

1

i

2

3

4

6

8

i0

Rq

c

1.2

1.4

1.6

1.8

.0

.0

.0

.0

.0

.0

.2

.4

.6

.8

.0

.0

.0

.0

.0

.0

.2

.4

.6

.8

.0

.0

.0

.0

.0

.0

Factor K

Inside Fiber Outside

2.52 0.67

1.90 0.71

1.63 0.75

1.50

i.41

1.23

I. 16

i.i0

1.07

i .05

3.28

2.31

1.89

1.70

1.57

1.31

1.21

I. 13

i. I0

1.07

2.63

1.97

1.66

1.51

1.43

1.23

1.15

1.09

1.07

1.06

0.77

0.79

0.86

0.89

0.92

0.94

0.95

0.58

0.64

0.68

0.71

0.73

0.81

0.85

0.90

0.92

0.93

0.68

0.73

0.76

0.78

0.80

0.86

0.89

0.92

0.94

0.95

Fiber

Section B 4February 15,Page43

1976

B 4.4.0 Bending-Crippling Failure of Formed Beams

This section contains methods of analysis applicable to formed

or built-up sections which are critical in the bending-crippling mode of

failure. This method is to be used when plastic bending curves are not

available, otherwise use Section B4.5.

It is noted that a positive margin of safety derived from this

analysis does not preclude failure in another mode.

The analysis procedure is divided into two sections according to

the type of applied loading as follows:

B 4.4.2

Bending moment only.

Combined bending moment and axial load.

Examples are given to illustrate the procedure for each type of

loading.

B 4.4. I Analysis Procedure for Bending Moment Only

/Compression Area

b"-<D0

'J Neutral Axis (n.a.)

Figure I. Bending-Crippling of Formed Shapes

Section B 4

February 15, 1976Page 44

B4.4,1 Analysis Procedure for Bending Moment Only (Cont sd)

(a) Locate the neutral axis (line of zero fiber stress) of the cross-

section assuming a linear stress distribution.

(b) Divide the compression area into elements according to Section C 1,

pages 1 1-16.

(c) Calculate FCCn, m according to Figure C 1.3.1-13 for each element.(d) Calculate the allowable bending-crippling moment by summing

moments about the neutral axis for the compression area and

doubling the result.

= 2 [ Z Fcc n b n t n _n + X FCCm bm tm _'m/2] ........... (1)k __j k .... J

Y

For flange For webmembers members

where:

distance from neutral axis to the resultant force of each

element.

This equation is applicable for all shapes although only a formed

channel is shown in Figure 1.

(e) The margin of safety is given by:

(ult) M.S.

m

M

(FS)ult M............................. (z)

where:

M = applied bending moment at the cross-section in question.

= allowable bending-crippling moment from Eq. (1).

(FS)ult = ultimate factor of safety.

Note: if the section is unsymmetrical, the tension flange should be

analyzed in the conventional manner.

B4.4. l

Section B 4

February 15,

Page 45

Analysis Procedure for Bending Moment Only (Cont'd)

1976

Example 1

Determine the margin of safety in bending-crippling for the cross-

section shown below if the bending moment is 4000 in. -ib and a factor of

safety of I. 4 is desired.

t = .04

3.00 '

_', _

_ 1.75--b

Typ

F{gure 2.

Given:

Mat'l = 6061-T6 Bare Sht.

no a.

Mech. Prop.

Ftu = 42 ksi

F = 35 ksicy

E = 9.9 x 103 ksi

Channels are intermittently rivetedtogether.

Back-to-Back Formed Channels

Analysis

(a) The neutral (centroidal) axis for this case is determined by inspec-

tion.

-1.59_

1.481 4C_ 1. 34

In,a.

r = . 12 + .02 = . 14 in.m

b I = 1.59 + . 535(. 14) = 1.665 in.

b 2 = 1.34 +.535(.14) = 1.415in.

Section B 4

February 15,Page 46

84.4.1 Analysis Procedure for Bending Moment Only (Cont' d)

Example 1 (Cont' d)

J Fcy ff 2...475, Fcc 1 = 275(35)

/bl 35 1.665

(c) E _ = 9.9 x 103 .040 -¥

= 9.6Z ksi

35 1.415

9.9 x 103 .040 - Z.103, Fcc z = .76 (35)

= Z6.6 ksi

(d) M" = Z[ Mflang e + Mweb] Ref. Eq. (1)

1976

(e)

= 2 [(9.62)(1.665)(.04)(1.48) + (26.6)(1.415)(.04)(1.46)/3]

= 3. 36 in-kips (for each channel) See page 278for FCCn, bn, in,

and Yn

Margin o£ safety

FB4.4.2

M 3360

(ult) M.S. = (FS)ul t M/2 - 1 = 1.4(2000) I = +0. Z0

where:

M/Z

(FS)ul t

= 2000 in-lb (per channel)

=1.4Ref. page 278

Analysim Procedure for Combined Bendin_ Moment and AxialLoad

lo Calculate steps (a) through (d) according to the procedure of Section

B 4.6.1. If the neutral axis falls outside the cross-section, con-

sider the section to be stressed as a column and compare with the

maximum applied fiber stress.

Section B4

February 15,

Page 4 7

1976

B4.4.2 Analysis Procedure for Combined Bending Moment and Axial

Load (Cont' d)

Calculate the section modulus of the compression area about the

neutral axis assuming a mirror image.

In. a.

Zcn. a. - Cc (see Example Z) ......................... (3)

3. Compute the equivalent allowable stress

F-

u

M

Zc n .

........................................... (4)

a.

4. The margin of safety is given by

(ult) MS -F

(FS)ul t fc

1 .............................. (5)

where:

p Mcf =-- ±

c A Ic. g.

Note:

(maximum applied compressive stress)

If the normal load (P) is tensile, the tension flange should be

analyzed in the conventional manner.

Example 2

Determine the margin of safety for bending-crippling failure for

the beam column shown in Figure 3.

Section B4

February 15,Page 48

1976

B4.4.2 Analysis Procedure for Combined Bending Moment and AxialLoad (Cont i d)

Example Z (Cont' d)

Given:

Mat'l = 6061-T6 Bare Sht

Mech. Prop.

Ftu = 4Z ksi

F = 35 ksicy

E = 9.9 x 10 3 ksi

Angles are intermittently riveted together.

P = 5000 Ib

M = 6400 in-lb

..(FSJul t = 1.4

.7

3. 000

c.f[. axis

T7c. g.

1

_._oo-.I ._6Ol-I .o_oTyp/ I"1 _[Typ

• 16o'.'_ - r '-_-

Typ-III- _L

_ ___ a

Z. 760

n. a.

Figure 3. Back-to-Back Formed Angle

Section B4

February

Page 49

15, 1976


Load (Cont' d)

Example 2 (Cont' d)

I °

1.26 -_ _i_ Z

.08 _

2.7h

Yc. g.

, ]

a,

ELE

Q

®

®

TOT.

A y

.i008 2.96

.0250 2.89

.2208 1.38

.3466

Ay Ay 2

.298 .883

.072 .209

.305 .420

.675 1.512

Io

• 140

• 140

_A__y .675Yc.g. - EA - .3466

- I. 947 in.

(a)

: LAy Z + EI o - y2 2] A = 1.512 +. 140 - (1.947) 2c.g.

: .338 (for each angle)

P MYn. a.

0 =_-+ Ic.g -

P Ic.g. 5000 (2)(.338)Yn.a. - A M =-2(.3466) 6400 : "'762 in.

(.3466)

f-Section B4

February I5,Page 50

1976

84.4.2 Analysis Procedure for Combined Bendin_ Moment and AxialLoad (Cont' d)

Example 2 (Cont' d)

(b)

_---I. Z6 _ rm

_ _5 b2

= .16 + .04 = .20 in.

= 1. Z6 +. 535(. ZO) = 1. 367. in.

= 1.575 +.535(.20) = 1.682 in.

(c)

bl 35 I. 367

t = 9.9 x 103 .08--= 1.016, Fcc I = .56(35)=19.60 ksi

IE _'- = 9.9x 103 .08- I. 250, Fcc z = 1.1 (35) = 38.5 ksi

(d)

= 2[ Mflang e + Mweb] Ref. Eq. (l)

= 2[ 19.60(I. 367)(. 08)(1.775) + 38.5(1. 682)(. 08)(I. 735)/3]

= 13.60 in-kip (each angle)

Section B4

February 15,

Page 51

1976


-Load (Cont' d)

Example 2 (Cont'd)

2. Section modulus about n. a.

ELE

G

®

®

TOT.

A y Ay

•1008

•0250

.1260

1.775

1.702

.788

• 179

•042

•O99

I = 2[ _Ay 2 + _Io]n.a.

Ay 2 Io

.318 -

•072 -

• 078 .0Z6

.468 .026

1•775 _'

Compression/Area

1. 702

_ 1.815•788 __.__[_II,,II

il

IIii

i

' _--'- Mirrort._o ______2,'

Image

= Z[.468 + .026] = .988 in4 (for each angle)

Zc _ 2(.988) = 1 089 in3n.a. 1.815 "

3. Equivalent allowable stress

_ 2 (M) _ 2{13. 60)

Z c i. 089n,a•

- 25.0 ksi

Section B4

February 15,

Page 52

1976

84.4.2 Analysis Procedure for Combined Bending Moment and AxialLoad (Cont _d)

Example Z (Cont' d)

Applied compressive stress

P Mc

fc=A+ic.g.

5

Z(. 3466) +

6.4(1.053)

Z(. 338)= 17.2 ksi

4. The margin of safety is

(ult) MS =

I

F

(FS)ult fc-1=

25.0

1.4(17.2.)-1= +0.04

where :

(FS)ult = 1.4 Ref. page 281

REFERENCE

Section B4

February 15,

Page 53

1976

B4. 0.0 Beams

4. I. 0 Perry, D.

Co., 1950.

J., PhD. Aircraft Structures, McGraw-Hill Book

Popov, E. P., Mechanics of Materials,Prentice-Hall Inc.,

New York, 1954.

Roark, Raymond J., Formulas for Stress and Strain, Third

Edition, McGraw-Hill Book Company, Inc., New York, 1954.

Seely, Fred B. and Smith,

Materials, Second Edition,

New York, 1957.

J.O., Advanced Mechanics of

John Wiley & Sons, Inc.,

Timoshenko, S., Strength of Materials, Part I, Third

Edition, D. Van Nostrand Company, Inc., New York, 1955.

4. Z. 0 Deyarmond, Albert and Arslan, A., Fundamentals of Stress

Analyses, Aero Publishers, Los Angeles, 1960.

Seely, Fred B. and Smith, J. O., Advanced Mechanics of

Materials, Second Edition, John Wiley & Sons, Inc.,

New York, 1957.

4.3.0

Wilber, John B. and Norris, C. H. , Elementary Structural

Analyses, First Edition, McGraw-Hill Book Company, Inc.,

Roark, Raymond J., Formulas for Stress and Strain, Third


Seely, Fred B. and Smith,

Materials, Second Edition,

New York, 1957.

J. O., Advanced Mechanics of

John Wiley & Sons, Inc.,

4.4.0 Roark, 'Raymond J., Formulas for Stress and Strain, Third


SECTION B4.5

PLASTIC BENDING

4.5.0

4.5.1

4.5.1.1

4.5.1.2

4.5.1.3

4.5.1.4

4.5.1.5

4.5.1.6

4.5.1.7

4.5.2

4.5.2.1

4.5.2.2

4.5.2.3

TABLE OF CONTENTS (Continued)

Plastic Bending ..........................

Analysis Procedure When Tension and Compression

Stress-Strain Curves Coincide ................

Simple Bending about a Principal Axis---

Symmetrical Sections ......................

Simple Bending about a Principal Axis---

Unsymmetrical Sections with an Axis of

Symmetry Perpendicular to the Axis of

Bending ...............................

Complex Bending---Symmetrical Sections; also

Unsymmetrical Sections with One Axis of

Symmetry ..............................

Complex Bending---Unsymmetrical Sections with

No Axis of Symmetry ......................

Shear Flow for Simple Bending about a Principle

Axis---Symmetrical Sections .................

Shear Flow for Simple Bending about a Principle

Axis---Unsymmetrical Section with an Axis of

Symmetry Perpendicular to the Axis of Bending .....

Shear Flow for Complex Bending--AnyCross Section ...........................

Analysis Procedure When Tension and Compression

Stress-Strain Curves Differ Significantly .........

Simple Bending about a Principle Axis---


Simple Bending about a Principle Axis---

Unsymmetrical Sections with an Axis of

Symmetry Perpendicular to the Axis of

Bending ...............................

Complex Bending---Symmetrical Sections; also

Unsymmetrical Sections with One Axis of

Symmetry ....................

Complex Bending---Unsymmetrical "S;ctions w'i;h"

No Axis of Symmetry ......................

Shear Flow for Simple Bending about a Principal

Axis---Symmetrical Sections .................

Page

1

5

5

5

8

8

10

13

17

18

18

22

22

22

23

B4 5-iii

4.5.2.6

4.5.2.7

4.5.3

4.5.4

4.5.4.1

4.5.4.2

4.5.4.3

4.5.5

4.5.5.1

4.5.5.2

4.5.5.3

4.5.5.4

4.5.5.5

4.5.5.6

4.5.6

4.5.6.1

4.5.6.2

4.5.6.3

4.5.6.4

4.5.6.5

4.5.6.6

4.5.7

4.5.7.1

4.5.7.2

4.5.7.3

4.5.7.4

4.5.7.5

4.5.7.6

TABLE OF CONTENTS (Concluded)


Axis---Unsymmetrical Sections with an Axis of

Symmetry Perpendicular to the Axis of Bending .....

Shear Flow for Complex Bending--AnyC ross Section ...........................

The Effect of Transverse Stresses on Plastic

Bending ...............................

Example Problems ........................

Illustrationof Section B4.5.2.1 ...............



Index for Bending Modulus of Rupture Curves for


Stainless Steel-Minimum Properties ............

Low Carbon and Alloy Steels-Minimum Properties...

H eat Resistant Alloys-Minimum Properties .......

Titanium-Minimum Properties ................

Aluminum-Minimum Properties ...............

Magnesium-Minimum Properties ..............

Index for Plastic Bending Curves ..............

Stainless Steel-Minimum Properties ............

Low Carbon and Alloy Steels-Minimum Properties...

Corrosion Resistant Metals-Minimum Properties . . .

Titanium-Minimum Properties ................

Aluminum-Minimum Properties ...............

Magnesium-Minimum Properties ..............

Elastic-Plastic Energy Theory for Bending ........

General ...............................

Discussion of Margin of Safety ................

Assumptions and Conditions ..., ..............D eflnitions .............................

Deflection of StaticallyDeterminate Beams ........

Example Problem .........................

Page

23

23

23

24

24

26

28

32

32

33

33

33

34

34

35

36

36

37

37

37

38

214

214

214

215

215

217

219

B4 5-iv

B4.5.5.6 .Magnesium-Minimum Properties

Section B4.5

February 15,

Page 201

1976

4O

Fb(ks{)

1.0 .5 2.0

Zqck_

Fig. B4.5.5.6-5 Minimum Bending Modulus of Rupture for

Symmetrical Sections ZK60A Magnesium Alloy .

Forgings (Longitudinal}

Section B4,5

February 15,

Page 202

1976

B4.5.6.6 Magnesium-Minimum Properties

o

_.o

°a

%.

,6.4#

°_,_

r

Section A4

1 February 1970

Page 35

Table A4-21. Decimal and Metric Equivalents of Fractions of an Inch (Cont'd)


Decimal Fraction (mm) (cm) (m)r

0.796 875 51/64 20. 240 37 2. 024 037 0.020 240 37

0.812 5 13/16 20.637 31 2.063 731 0.020 637 31

0.828 125 53/64 21.034 Ii 2. ]03 411 0.021 034 i]

0.843 75 27/32 21.431 05 2. 143 105 0.021 431 05

0.859 375 55/64 21.827 85 2. 182 785 0.021 827 85

0.875 7/8 22. 224 79 2.222 479 0.022 224 79

0.890 625 57/64 22.621 59 2.262 159 0.022 621 59

0.906 25 29/32 23.018 53 2.301 853 0.023 018 53

0.921 875 59/64 23.415 33 2.341 533 0.023 415 33

0.937 5 15/16 23.812 28 2.381 228 0.023 812 28

0.953 125 61/64 24. 209 07 2. 420 907 0. 024 209 07

0.968 75 31/32 24. 606 02 2.460 602 0.024 606 02

0.984 375 63/64 25.002 81 2.500 281 0.025 002 81

1.0 I 25.4 2.54 0.025 4

Section B4.5

February 15, 1976

Page l

B4.5.0 Plastic Analysis of Beams

Introduction

The conventional flexure formula f=Mc/I is correct only if the

maximum fiber stress is within the proportional limit. In the plastic

range, the assumption that plane sections remain plane is valid while

the stress corresponds with the stress-strain relationship of the

mater ia I.

This section provides a method of approximating the true stress

which depends upon the shape and the material properties. It is noted

that deflection requirements are investigated when using this method

since large deflections are possible while showing adequate structural

strength.

The method outlined in this section is not applicable if the member

is subjected to high fluctuating loads.

The following glossary is given for convenience:

Simple bending: This condition occurs when the resultant applied

moment vector is parallel to a principal axis.

Complex bending: This condition occurs when the resultant applied

moment vector is not parallel to a principal axis.

Development of the Theory

A rectangular cross section will be used in this development; any

other symmetrical cross section would yield the same results.

Section B4.5

February 15,

Page 2

i 976

B 4.5.0 Plastic Analysis of Beams (Cont, d)

dA fmax

/-

¢2 , fl

--[ ' " -7 f true Stres'

e max max

(a) Section (b) Strain (c) Stress

Figure B4. 5. 0-I

0 el Strain eZ

(d) Stress-Strain Curve

(Tension and Compression)

Since the bending moment of the true stress distribution about

the neutral axis is greater than that of a linear Mc/I distribution as

used in the elastic range, a trapezoidal stress distribution is used to

approximate the true stress distribution, fmax may be defined as ayield stress, a buckling stress, an ultimate stress or any other

stress level above the proportional limit.

£max

Let fo be a fictitious stress at zero strain in the trapezoidal

stress distribution as shown in Figure B4.5.0-Z. The value of fo

may be determined by integrating graphically the moment of (not the

area of) the true stress distribution and equating this to the bending

moment of the trapezoidal stress distribution.

Sect ion B4.5

February 15,Page 3

1976

B 4.5.0 Plastic Analysis of Beams (Cont'd)

fmax

fmaxf

max

(a) True Stress

fmax

fmax

fo

Stress

Strain

(b) Trapezoidal Stress (c) Stress Strain Curve

Figure B4. 5. 0-Z

Mba = Allowable bending moment of the true stress distribution for aparticular cross-section at a prescribed maximum stress level.

6max

F b

Mb c

a

I

Mc

= Fictitious allowable I stress or the bending modulusof rupture for a particular cross-section at a prescribed

maximum stress level.

IMb t = "_-fmax + (ZQ- I_)c fo = The bending moment of a trapezoidal

stress distribution that is equivalent to

Mb a

Where for symmetrical sections,

c

I = ( y2 dA

Q ._

C "

(moment of inertia)

c

0 y dA (static moment of cross-section)

Distance from eentroidal axis to the extreme fiber

Section B4.5

February 15, 1976

Page 4

B 4.5.0 Plastic Analysis of Beams (Cont,d}

Therefore, from F b =

Mb c

twe obtain:

I

Fb = fmax + (k-1)fo (4.5.0-I)

Where,

2Qck --

i (4.5.0-2)

Two types of figures are presented for plastic bending analysis.

One type presents Bending Modulus of Rupture Curves for Symmetri-

cal Sections at yield and ultimate (Reference Section B4.5.4). The

other type presents Plastic Bending Curves {Reference Section B4.5.5)

which are necessary for the following:

1) Limiting stress other than yield or ultimate.

z) Tension and compression stress strain curves which are

significantly different.

3) Unsymmetrical cross-section.

Figure B4.5.0-3 shows values of k for various symmetrical cross-

sections. Generally, validity of the approach has been shown only

for typical aircraft sections which are geometric thin sections.

I Flanges ] Thin Tube

k = 1.0 k'= 1.27 ,

Hourglass Rectangle Hexagon ]Solid Roun_ Diamond

ISHAPE FACTOR

Figure B4.5.0-3

Section B4.5February 15,Page 5

1976

B4. 5. I Analysis Procedure When Tension and Compression Stress-

Strain Curves Coincide.

B4. 5. i. 1 Simple Bending about a Principal Axis---Symmetrical

Sections.

The procedure is as follows:

io Determine k by Equation (4. 5. 0-2) or by the use of Figure B4. 5. 0

-3.

2, For yield or ultimate limiting stress, use the Bending Modulus of

Rupture Curves (F b vs. k - see section B4. 5. 5 for index) to

determine F b.

. For a limiting stress other than yield or ultimate, use the Plastic

Bending Curves (See section B4. 5. 6 for index). Locate the limiting

stress on the stress-strain (or k=l) curve and move directly up to

the appropriate k curve to read F b for the same strain.

, F b from Step 3 and fb, the calculated Mc/l stress, may be used

in determining the bending stress ratio for combined stresses

and the margin of safety for pure bending as follows:

fb

Rb-Fb

M.S.

Where:

(4.5.1.1-I)

1- 1 (4. 5. i. I-2)Rb(S.F.)

S.F. is the appropriate (yield or ult) safety factor

B4.5. 1.2 Simple Bending about a Principal Axis---Unsymmetrical

Sections with an Axis of Symmetry Perpendicular to the

Axis of Bending

This procedure also applies to any section with bending about one

of its principal axes.

Section B4.5February 15,Page6

1976

B4.5.1.2 Simple Bendin G about a Principal Axis---Unsymmetrical


Axis of Bending. (Con t'd)


. Divide the section into two parts, (1) and (Z), on either side of the

principal axis (not the axis of symmetry) similar to that shown in

Figure B4.5. I. Z-l.

, Calculate:

QlCl

k 1 - ii (4.5. I. Z-I)

where I1 is the moment of inertia of part (I) only about the principal

axis of the entire cross-section. This would be identical to k of a

section made up of part (I) and its mirror image. Figure B4.5.0-3

may be used where part (i) and its mirror image form one of the

sections shown.

3. Calculate similarly:

1

Qzcz

k 2 = iZ (4.5. I, Z-Z)

Assuming part (I) is critical in yield (or crippling, ultimate, etc. )

use the P_astic Bending Curves and locate this stress on the stress-

strain ( or k=l) curve. Move directly up to the appropriate k curve

and read for the same strain.Fb 1

5. Read the strain el.

6. Calculate:

ElcZ (4.5.1.2-3)

E2- c1

7. Locate E Z on the strain scale and move directly up to the appro-

priate k curve to read Fb2

8. Calculate Mba by

Fblll Fb212

Mb a = 5_ + E_ (4.5 .l. 2-4)

Section B4.5

February 15, 1976

Pa9e 7

B 4.5.1.2 Simple bending about a principal axis--

u_n_s_ym_letrical sections with an axis of

symmetry perpendicular to the axis of

bending. (Cont,d)

1 M_ mustbe used in determining the moment ratio for bending and

an_dath--emargin of safety for pure bending as follows:

M

R b _ (4. 5. 1.2-5)Mb a

Where:

1M.S. = - 1 (4.5.1.Z-6)

a b (s. F. )

S.F. is the appropriate (yield or ult) safety factor

I!

(i)

rincipal Axis

(z)

B4.5.1.3

Figure B4. 5. i. 2-i

Complex Bending--Symmetrical Sections; Also Unsymmetri-

cal Sections with One Axis of Symmetry.

This condition occurs when the resuItant applied moment vector

is not parallei to a principal axis as shown in Figures B4. 5. 1. 3-1 or

B4. 5. i. 3-2. ,y

Y/"C.G.

M I M

I/;/ __ x

Y

Symmetrical Section

Figure B4. 5. i. 3-I

i X X -X

Y

Unsymmetrical Section

with one axis of symmetry

Figure B4. 5.1. 3-2

Section B4.5

February 15,Page 8

B 4.5. 1.3 Complex Bending-- Symmetrical Sections; also

Unsymmetrical Sections with One Axis of Symmetry.

The procedure is as follows (this procedure is always conservative and

may be very conservative for some cross-sectional shapes):

3. Rb= Rbx + Rby

1.. Obtain M x and My, the components of M with respect to theprincipal axes.

2. Follow the procedure outlined in Section B4.5.1.1 or B4.5.1.2

to determine Rbx and Rby

(4.5.1.3-1)

4. For pure bending,

l

Rb (S.F. )M. S.

Where:

S.F.

B4.5.1.4

1 (4.5.1.3-z)

is the appropriate (yield or ult) safety factor.

Complex Bending--Unsymmetrical Sections with No Axis

of Symmetry

This condition occurs when the resultant applied moment vector is

not parallel to a principal axis as shown in Figure B4.5.1.4-1.

1976

y, y M

\ I

x' (Principal axis)

x(Ref.x - - axis)

X

y' (Principal axis)!

y (Ref. axis)

Figure B4.5.1.4-1

B 4.5.1.4

Axis of Symmetry.


Section B4.5

February 15,

Pa ge 9

Complex Bending-- Unsymmetrical Sections with no

i. Determine the principal axes by the equation

2I

tan 2 0 - xy

Iy - Ix(4.5. I.4-i)

1976

where x and y are centroidal axes and

I x = y dA (Moment of inertia)

o

,

xIy = dA

Ixy=/Xy dA

Obtain M x, and My,,

principal axes.

(Moment of inertia)

(Product of inertia)

the components of M with respect to the

Follow the procedure outlined in Section B4. 5. 1. 2 to determine

Rbx , and Rby ,

4. The stress ratio for complex bending is

So

R b = Rbx , + Rby , (4. 5. 1.4-2)

For pure complex bending,

M. S.

where:

S.F.

I

ab(S. F. )

the margin of safety is

(4. 5. 1.4-3)

is the appropriate (yield or ult) safety factor.

Section B4.5

February 15, 1976

Page I0

B4.5. I. 5 Shear Flow for Simple Bending About a Principal Axis--

Symmetrical Sections

When plastic bending occurs, the classical formula SQ/I is correct

only for cross-sections withk=l.O. For k greater than 1.0, SQ/Iis

unconservative. The derivation of a correction factor ;3 SQ/I is as

follows for an ultimate strength assessment:

P.A.

/ ///J Jr J/i

Cross -Section

__ f __lma-_l •

-t °",%q

Stress Distribution

Figure B4. 5.1.5-1

Let P = load on cross-section between y and c

P= jy tfdy

From the geometry of Figure B4.5. i. 5-1

f = fo + (fmax " fo) y--c

c ff t y dy•".P= " fot dy + (fmax " fo) c-,y

Let A= area of cross-section between y and c

_c

A= / t dy then,

Jy

B 4.5.1.5

P= fo

Shear Flow for Simple Bendin_ about a Principal

Axis-- Symmetrical Section. (Cont'd)

A + (fmax - fo) A__ and since Q=Ay,c

Section B4.5

February 15,

Page 11

1976

P= fmax I--_)4 fo (A Q)c

From Equation (4. 5. 0-i)

Mc

Fb = fmax + fo (k - I) _ I then,

(a)

dfc dM max

I dx dx

df0

dx (k - 1 )

Since, by definition, S = dM/dx and q = dP/dx,

df [ dfc max o (k - 1

-_--(S)- dx 1 + dfmax

] and from (a),

df df dfinax Q o m ax

q = +dx c d f dx

max

Let X = df /dfmax and _ =0

q dfmax= dx

1 +x (c/F- l)l+x (k-l)

then,

But since from (b): df /dx- Sc [max I / 1 +k (k - 1)]

and substituting (d) into (c) we have

q = [1 +k

(Ac/Q - 1)](k - 1)]

Since Q = A-y-',

SQ

q- _-y-

(b)

(c)

(d)

Sect ion B4.5February 15,

Page 12

1976

B 4.5. I. 5 Shear Flow for Simple Bendin_ about a Principal

Axis-- Symmetrical Section. (Cont wd)

The method outlined below shows how to correct SQ/I and calculate

the margin of safety for simple plastic bending about a principal axis

of a symmetrical section.


1. Determine Mc/I

o Determine k by Equation (4.5.0-2) or by the use of FigureB4.5.0-3.

1 Refer to the applicable Plastic Bending Curves and locate

Mc/I on the F b scale. Move across to the appropriate k curveto read the true strain, c.

o By use of the stress.-strain (or k = 1) curve and the fo curve,determine the rate of change of fo _vith respect to f at the true

strain E, which would be expressed as

d'f df /d_O O

)t = df - dr/d, (4.5.1.5-1)

o To determine the shear flow at any point on a cross-section,

determine the distance from the principal axis to the centroid

of the area outside of the point in question as Shown in Figure

B4. 5.1. 5-1. This is defined as _.

1 Determine the shear flow at distance "a" from the principal

axis bySQ

a

qa = _ I (4.5.1.5-2)

where,

= l+X (c/7- l) (4.5.1. 5-3)l+X (k-l)

£Qa = y dA 14.5.1. 5-4)

B 4.5.1.5.

.

,

Sect ion B4.5

February 15,

Page 13

Shear Flow for Simple Bending about a

Principal Axis-- Symmetrical Section. (Cont'd)

Calculate fSm&xf

S l_fl aN

R -s F s

= (r s) = (q/t)a= 0 a= 0

and the stress ratio is

(4.5.1.5-5)

For the margin of safety with pure bending and shear use

1

M.S. = 2- -1 (4. 5. 1. 5-6)+ Rs

1976

,

B4. 5. 1.6

For the margin of safety with axial load, bending and shear

use

1M.S. = 1 (4. 5.1. 5-7)

(S. F. )v_R b + Ra )2 + R2s

where,

fa

Ra- Fa

S.F. is the appropriate (yield or ult) safety factor,

Shear Flow for Simple Bending About a Principal Axis--

Unsymmetrical Section with an Axis of Symmetry Perpen-

dicular to the Axis of Bending

A procedure similar to that shown in Section B4. 5. 1. 5 for Figure

B4. 5.1. 6-1 is as follows:

• Divide the section into two parts, (1) and (2), on either side of

the principal axis (not the axis of symmetry) similar to that

shown in Figure B4. 5. 1. 2-1.

Calculate:

QlCl (4.5. 1.2-1)

k 1 - I1

where I 1 is the moment of inertia of part (1) only about the

principal axis of the entire cross-section. This would be

identical to k of a section made up of part (1) and its mirror

image. Figure B4. 5. 0-3 may be used where part (1) and its

mirror image form, one of the sections shown.

B 4.5.1.6

B

Shear Flow for Simple Bendin_ about a Principal

Axis-- tl_syrnmetrical Section with an Axis of

Symrnetry Perpendicular to the Axis of Bending.

Calculate similarly:

Q2c2

k 2 - i2

Section B4.5

February 15,

Page 14

(4.5.1.2-2)

1976

.

Mc 1 Mc 2Calculate -- and

I I

5. Refer to the applicable Plastic Bending Curves and locate

MCl on the F b scale. Move across to the appropriate kI (use k 1) curve to read the true strain _ 1'

• _1c26. Find E2, similarly, or by _ 2- (4.5. i.2-3)

cI

.

e

.

1 0.

By use of the stress-strain (or k = 1) curve a,_d the fo curve,

determine the rate of change of fo with respect to f for thetrue strainq which would be expressed as

xl = = 7 l (4.5.1.6-i)

Calculate similarly:

(4.5.1.6-2)

To determine the shear flow at any point on a cross-section,

determine the distance from the principal axis to the centroid

of the area above or below the point in question as shown in

Figure B4. 5.1. 6-1. This is defined as _a or _b"

Determine the shear flow in part (1) at distance "a" from the

neutral axis by

t SQa) (4.5.1.6-3)qa = _a -T--

B 4.5.1.6

11.

Sect ion B4.5

February 15,

Page 15


Axis-- Unsymmetrical Section with an Axis of

Symmetry Perpendicular to the Axis of Bending.

where,

c(_ a )i +k I i

/3a = 1 +X1 (kl - 1) (4.5.1.6-4)

c 1

Qa = _a ydA(4. 5.1. 6-5)

Determine the shear flow in part (2) at distance "b" from the

neutral axis by

qb = _3b I

where,

(4. 5.1. 6-6)

)1 +_Z - 1

(4. 5.1. 6-7)i3b = l +%2 (k2 - I)

1976

12.

Qb = _b 2 ydA (4. 5.1. 6-8)

For the shear flow at the principal axis, calculate q using both

parts of the cross-section and use the larger. A rigorous

analysis could be made so that shear flow calculations at the

neutral axis would result in the same value regardless ofwhich side was used in the calculations. This would involve

determining the amount of transverse shear distributed to

each side of the neutral axis. If the bending moment is

developed entirely from external shear, then the shear distri-

buted on each side is proportional to the corresponding mo-

ments and Equations (4. 5.1. 6-3) and (4. 5.1. 6-6) become as

follows:

M 1 Fb 1 11S1Q a

qa = /3a I 1 where S 1 = M1 + M2 S, M 1 = C--1

(See B4. 5. i.2)

Section B4.5

February 15,

Page 16

1976

B 4.5.1.6 Shear Flow for Simple Bendin_ about a Principal


S]rmmetry Per_.endicular to the Axis of Bending.

SZQ b M Z

qb = _b I Z where S Z= M1 + MZ S,

fc z

c1 ta

b

i

I

t z

fm

_i

-i

IzM2= Fb 2 C2

(See B4.5.1.2)

_ Principal axis

Yb

13. Calculate fS

a

ratio s:

1%S a

Figure B4.5.1.6-I

qa= -- and f

,t1 sb

fSa

.=.

F s

fs b

Rs b - FS

qb= -- to obtain the shear

t2

(4.5.1.6-9)

(4.5.1.6-10:)

14. For the margin of safety with pure bending and shear use

1M.S. = - 1 (4.5. i.6-II)

(s.F. )_+ RZs

where:

S.F. is the appropriate (yield or ult) safety factor,

Sect ion B4.5

February 15,

Page 17

B 4.5.1.6Shear Flow for Simple Bendin G about a Principal


Symmetry Perpendicular to the Axis of Bending.

(Cont ' d)

15. For the margin of safety with axial load,

IM.S. = -1

(S. F. )_(R b

where:

Ra )2 R 2

bending and shear use

(4. 5.1. 6-12)

fa

R a - Fa

S.F. is the appropriate (yield or ult) safety factor.

B4. 5. 1. 7 Shear Flow for Complex Bending--Any Cross-Section


l • Determine the principal axes by inspection or, if necessary,

by Equation (4.5.1.4-I),

, Obtain S x, and Sy,, the components of S with respect to theprincipal axes. The principal axes are denoted by x' and y'

as indicated in Figure B4.5.1.4-1o

. Follow the same procedure of Section B4.5.1.6

principal axes to obtain the shear stresses fsx ,at a prescribed point.

about both

and fSy,

4. The shear stress ratios at this point are

So

fSX w

_ (4.5. :.7-l)Rsx' F s

fs ,

Rsy _ Y (4. 5.1.7-2)' F s

For the margin of safety with complex bending and shear use

1M.S. = -1 (4.1. 5.7-3)

S.F b + Rs

1976

Sect ion B4,5February 15_Page 18

1976

B 4.5.1.7. Shear Flow for Complex Bending-- Any Cross-Section. {Cont'd)

where,

S, F.

_R 2 Rsy,2Rs = Sx' + (4.1, 5.7-4)

is the appropriate (yield or ult. ) safety factor.

o For the margin of safety with axial load, complex bending andshear use

B4.5.2

lM.S. = -1 (4.1. 5.7-5)

S.F. R b 4- Ra )2 4- R s

Analysis Procedure When Tension and Compression Stress-

Strain Curves Differ Significantly

This may occur in materials such as the AISI 301 strainless steels

in the longitudinal grain direction where the tensidn stress-strain curve

is higher than the compression curve•

B4.5.2. 1 Simple Bending about a Principal Axis--SymmetricalSections

t-C

P. A.

N.A. _]

A

Cf f

C C

N.A.

f

A. o• C

o

ft

(a) Section (b) Strain (c) True Stress

Figure B4.5.2. I-I

(d)Trapezoi-

dal Stress

B 4.5.2.1

Section B4.5

February 15,

Page 19

Simple Bending about a Principal Axis-- Symmetrical

Sections. (Cont'd)

1976

f

FTension tu

6U

(e) Stress-Strain Curves

Figure B4. 5. 2. I-i (cont.)

The procedure for a symmetrical section such as that in Figure

B4. 5. 2. i-I is as follows:

I , Determine k by Equation (4. 5. 0-2) or by the use of Figure

B4. 5. 0-3.

, For yield or ultimate limiting stress, use the Bending Modulus

of Rupture Curves to determine F b. The correction for shift-

ing the neutral axis away from the principal axis by A has

been taken into account in the development of the curves.

Note Step 8.

. For a limiting stress (or strain) other than yield or ultimate,

use the Plastic Bending Curves. Locate the limiting stress

on the appropriate (tension or compression) stress-strain

(or k=l) curve and call this fl at _ 1 with its trapezoidal inter-

cept fol. Note Step 9.

Section B4.5

February 15,

Page 20

1976

B 4.5. Z. I Simple Bendin_ about a Principal Axis-- Symmetrical

Sections. (Cont'd)

, Locate the neutral axis which would be some distance A from

the principal axis toward the tension side. This may be

difficult to determine, but trial and error type formulae are

provided in Figure B4.5. Z. 1-2 for several symmetrical cross-

sections. See Example B4.5.4. I for typical procedure in

determining A, e Z' fz' fo2' kl' and k 2. Note that k I and k 2

are with respect to the neutral axis.

,

Find Fbl and Fbz for E I and _ Z' using the correct values of

k I and k z which may or may not be equal.

. Calculate I 1 and I Z, the moments of inertia of the elements

with respect to the neutral axis of the entire cross-section.

7. Calculate byMb a

11 Fb IzFbl Z

Mba _ Cl + c2 (4.5.2. I-I)

8. For cases as in Step 2, F b may be used with fb' the calculated

Mc/I stress, in determining the stress ratio for bending and

the margin of safety for pure bending as follows:

IM.S. = -I (4.5.2. I-3)

(S.F.) R b

where:

e

S.F. is the appropriate (yield or ult) safety factor.

For cases as in Step 3, M_. mustbe used in determining thegJa - _

moment ratio for bending and the margin of safety for pure

bending as follows:

M(4.5. Z. 1-4)

R b = Mba

IM.S. = -1 (4.5.2. I-5)

(S.F.)Rb

Section B4.5


B 4. 5.2. ] Simple Bending about a Principal Axis-- Symmetrical

Sections• (Cont,d)

0 _ _ _1

0 _1

o o

_ _ ._ +

?1"_ _ [..-i _2:d) -..D _ll

0 rd _ _ 0!

°_¢_ ,-4 0

_o _<I +

_ 0 _ 0

I,L oI ,

A

r_

o rqi • •

o ¢,a

n u

I ,

÷

<I + u, u _i_

v

_1_ _ +_ cq

u_4 .r-i

, _ m

E , ._ =--.

II

r..)<_I"_ ;_

r_

' £cO _

-.D o0

D.- (3"...D

_ +ed _

E _t_ ..D

÷ _

<ii_ c;

o.1-..D I

• (y.

i _)O

v ,._t

4,_

_ o_

O _

..D

_ _ _

, _ <I

o _I_ ÷

I

0

t'-

Z

+ _m

u _

+ C_

;>

I_'; _ ::::;i I

,, _ _i

+ _m M

mM N

X O

Z

O3

No

<_

O

_d

Section B4.5

February 15, 1976

Page 22

B4.5.2.2 Simple Bending about a Principal Axis--Unsymmetrical


Axis of Bending


. Locate the neutral axis which would be some distanceA from the

principal axis toward the tension side by a method similar to that

outlined in Example B4.5.4.1 for symmetrical sections. This will

require the derivation of an expression relating A, fmaxt, lot,

fmaxc, and f_vc by. use. of. the equilibrium of axial loads due to thebending stress dlstrlbut_on. This expression is likely to contain

higher powers of Aand the solution for _ can best be obtained by

trial and error using the stress-strain (or k = 1) and fo curves.

Refer to Example B4. 5.4.2 for typical procedure.

Z. Now, follow the identical procedure to that of Section B4.5.1.2

with the exception of using the neutral axis instead of the principal

axis. Remember that F b on the compression side is obtainedfrom the compression Plastic Bending Curves.

B4.S.Z.S Complex Bending--Symmetrical Sections; also Unsymmetri-

cal Sections with One Axis of Symmetry

Refer to Section B4.5.1.3 and follow the identical procedure except

for Step 2 which is expressed as follows:

Z. Follow the applicable procedure outlined in Section B4.5.2.1

and/or B4.5.2.2 to determine Rbx and Rby.

B4.5. Z. 4 Complex Bending--Unsymmetrical Sections with No Axis of

Symmetry

Refer to Section B4.5.1.4 an(t follow the identical, procedure except

for Step 3 which is expressed as follows:

3. Follow the procedure outlined in Section B4.5. Z. 2 to deter-

mine Rbx and Rby

i

Section B4.5

February 15,

Page 23

B4. 5. 2. 5 Shear Flow for Simple Bending about a Principal Axis--

Symmetrical Sections

Refer to Section B4. 5. I. 5 and follow the identical procedure with

the following exception:

If shear flow is being determined on the tension side, use the

tension Plastic Bending Curves in the evaluation of 71 for Equation

(4. 5.1. 5-3). The compression side is considered similarly using

the compression Plastic Bending Curves.

B4. 5. Z. 6 Shear Flow for Simple Bending about a Principal Axis--

Unsymmetrical Sections with an Axis of Symmetry Per-

pendicular to the Axis of Bending

Refer to Section If4. 5.1. 6 and follow the identical procedure with

the following exceptions:

If shear flow is being determined on the tension side, use the

tension Plastic Bending Curves in the evaluation of X for Equations

(4. 5.1. 6-i) or (4. 5.1. 6-2). The compression side is considered

similarly using the compression Plastic Bending Curves.

B4. 5.2. 7 Shear Flow for Complex Bending--Any Cross-Section

Refer to Section B4. 5. I. 7 and follow the identical procedure except

for Step 3 which is expressed as follows"

. Follow the same procedure of Section B4. 5. 2. 6 about both

principalaprescribedaxes point,and obtain the shear stresses fsx, and fSy,__at

B4. 5. 3 The Effects of Transverse Stresses on Plastic Bending

The elastic and plastic portions of stress-strain curves for com-

bined loading do not correspond to those of the uniaxial curves due to

the effects of Poisson's ratio. The plastic bending curves depend on

the magnitude and shape of the stesss-strain curve of a given material.

Therefore, under combined loading the plastic bending curves may

have to be modified if the uniaxial stress-strain curve is significantly

affected.

1976


! 976

B 4.5.3 The Effects of Transverse Stresses on Plastic Bending.

When modification of the plastic bending curves is necessary,

procedure is as follows:

,

.

the

Modify the uniaxial stress-strain curve by Section A3.7.0.

Determine and construct the modified plastic bending curves,

F b vs E. Refer to Section B4.5.0 for the theoretical back-

ground; the fo curve must first be plotted and then k curvesconstructed by the use of Equation (4.5.0-1).

Example Problems

A Diamond Cross-Section of AISI 301 Extra Hard (2%

Elongation) Stainless Steel Sheet Is Subjected to PureBending in the Longitudinal Direction at Room Temperature.

Find the Shift in Neutral Axis (A) from the Principa ! Axis

When the Limiting Stress Is (a) Ftu and (b) Fcy. (Fty is

Is Not Used Since Fty > Fqy. )

134.Reference the Minimum Plastic Bending Curves on page 133 and

(a) Ftu = ZOO ksi

c t -= . 020 in. /in.

lot = 146 ksi

/_tenslon

_ide

PlA,

From Figure B4.5. Z. 1-2 use the equation for determining

Awhen k = 2 (for a diamond).

ft + Zfot- 1 [fc +fo (Z + 6A- 3AZ)](I - A 2) c

By trial and error, equality is reached within 0.8% when

is assumed equal to 0.187c as shown below:

From Equation (4.5. i.2-3)

E ic2 E t c c

c 2 - c 1 or _ c - ct

0. 020 x I.187c

c = .813c = 0. 0292 in. /in.

B 4.5.4.1 l_.xample Problem B 4. 5.4. 1 (Cont 'd)

Section B4.5

February 15,

Page 25

1976

(b)

From the stress-strain (k=l) curve page 135 at cc

f = 149 ksiC

f = 107 ksiO C

Therefore,

1 - 0. 1872[ 149 + 107(2 + 6 x 0.187

492-_,488

f = 97 ksicy

c = 0.00575 in./in.c

fOc = 27 ksi

From Figure B4. 5.2.1-2, use the equation for determiningA when k = 2 (for a diamond).

1

- [fc + f°cft + 2fo t (I -A 2)

(2 + 6A - 3A 2) ]

By trial and error, equality is reached within 0. 60% when

A is assumed equal to 0. 024c as shown below:

From Equation (4. 5. I.2-3)

c I c 2 _ cc t

c 2 - Cl or = c t = Cc

0. 00575 x 0.976c

ct = i. 024c = 0. 00548 in. /in.

From the stress-strain (k=l) curve page 132 at

ft = iZl ksi

c t

for= 16.5 ksi

Sect ion B4.,5

February 15,Page 26

1976

B 4.5.4.1 Example Problem B 4.5.4.1. (Cont' d)

B4.5.4.2

Therefore,

121 + 2 x 16.5_ 1 I97 + 27 (2 + 6xO. 0241.0.-0242

154_154.93

Calculate the Ultimate Allowable Bendin_ Moment Aboutthe Principal Axis for the Built-Up Tee Section Shown.

The Flan@e Is in Compression and the Crippling Stress1 Hard StainlessIs 60 ksi. The Material Is AISI 301

Steel Sheet at Room Temperature with Bending in theLongitudinal Grain Direction.

Since the longitudinal tension and compression stress-

strain curves are significantly different, the procedure of

Section B_. 5.2.2 will be followed.

For ultimate design, crippling is the limiting stress

rather than Ftu.

Fcc = 60 ksi /----tension

; 1o.o_ -_ _._.L _.c=O. 1

fOc ----- T

A trial and error equation for shifting the neutral axis from

the principal axis toward the tension side is determined by equat-

ing the load on the tension side to the load on the compression

wide. (See Section B4.5.2.2):

c "2 = Cc 'Fcc + cc

B 4.5.4.2 Example Problem B 4. 5.4. 2 (Cont' d)

Section 84.5

February 15,Page 27

1976

Equality is reached within 0.70% whenAis assumed equal to

0. 065:

c = 0.345 + A = 0.410 in.c

c = 1 - c = 0. 590 in.t c

Using Equation (4. 5.1. 2-3)

E Cc t 0. 0046 x 0. 590

ct - c - 0.410 =c

0. 00662 in. /in.

From page 112 at e t

ft = I13 ksi

fot 38 ksi

Substitution of these values into the above trial and error

equation results in equality within 0. 70%.

Section properties about the neutral axis:

3

2 tc t 2 x 0.05 x 0_ 3

It- 3 - 3= 0. 006845 in.

4

Z _Z

Qt = t c t = 0.05 x 0. 590 = 0. 01740 in.

Qt ct 0. 01740 x 0. 590

kt - I t 0. 006845 = 1 5

(which checks with Figure B4. 5. 0-3 for a rectangle)

Ztc3 2c t, 2x0. 05x0. 4103

Ic - 3 + 2 x 0.450t (c c - -_-) = 3\

+ 2x0. 450x0. 05x0. _-_2 = 0. 008967 in. 4

Section B4.5

February 15,Page 28

1976

B 4.5.4.2 Example Problem B 4.5.4.2 (Cont'd)

Z

Qc = tc c

kC

x 0.05 x 0. 385 = 0. 02572 in.

Qcc c 0. 02572 x 0.410

I 0. 008967C

+2 x 0.450t (Cc _ t,-_-) = 0.05 x--0.4102+ 2x0.450%

3

=1.17

From page li2 using _t

fb = 132 ksit

and k t

Let's check this value by numerically evaluating Equation

(4.5.0-I).

Fb t = ft + (k - I)fOr = 113 + (i.5-I)38 = 132 ksi (check)

From page 114 using _ and kC c

Fbc = 64.5 ksi

Calculate the ultimate allowable moment by Equation (4.5.1.2-4)

It Fb I ct c 13Zx0. 006845

Mbult - ct + cc 0. 590

B4.5.4.3

64.5 x 0.008967 2.94 in. - kips (answer)+ =

0.410

Find the Shear Flow at the Neutral Axis of the Example

Problem B4.5.4.2 If the Transverse Shear, (s), Is

Equal to 5 kips. and the Bending Moment Is Equal to the

Ultimate (2.94 in-kips. )

Shear flow at the neutral axis should be determined by using

the section properties from each side of the neutral axis. The

larger of the two shear flows should be (conservatively) used.

The procedure of Section B4.5.2.6 and consequently Section

B4.5. I.6 will be used.

B4.5.4.3 Example Problem B4.5.4.3 (Cont 'd_

Section B4.5

February 15, 1976

Page 29

Shear Flow From Tension Properties

The required section properties already known from example

B4. 5.4. 2 are:

It

Ic

I

4= 0. 006845 in. (tension side only)

4= 0. 008967 in. (compression side only)

4= I t + I c = O. 015812 in.

ct= 0.590 in.

Qt= 0. 01740 in.

kt= i. 5 (rectangle)

t = 0. 00662 in. /in.

From the stress-strain (k= i} curve on page 112 at _t

/_\[d-v:-]=7 ksi/. 001\co/ 1

(df d-_-_I = 8 ksi/. 001

1

Using Equation (4. 5. I. 6-i)

x i=_d-Y-71=_Ta_ Jl

8

XI- 7 -1.14

Using Equation (4.5.1.6-4)

l+X 1 -

_a = i+_1 (kl - i)

_ c t

Ya- 2 - 0.295 in.

B 4.5.4.3 Example Problem B 4.5.4.3 (Cont' d)

Section B4.5

February 15,

Page 30

1976

-._J

(_ 590 )1 + 1.14 295 1

_a = I + 1.14 (1.5 - I) = 1.36

Shear flow at the NA from Equation (4.5.1. 5-2)

/3 SQa a

qa- I

5 x O. 01740qa = I. 36 O. 015812 = 7.48 kips/in.

SQ[Note that the conventional -- l

]formula is 36% unconserva_ve]L he re. J

Shear Flow From Compression Properties

The required section properties already known from exampleB4.5.4.2 are:

I = O. 015812 in.

c = 0.410 in.C

Q = O. 02572 in.C

4(Reference Page 82 )

k =1.17C

E = 0. 0046 in. /in.C

From the stress-strain (k = 1) curve on page 114 at

dr) = 4. ksi/. 0016

EC

(df ==/I= 5.o ksi/0oi2

B4.5.4.3 Example Problem B 4. 5.4. 3 (Cont'dl

Sect/on B4.5

February 15,

Page 3 11976

Using Equation (4. 5.1. 6-2)

X2 =\dr /2 dr/de z

5.0)t2- 4.6 - 1.09

Using Equation (4.5.1. 6-7)

/Jb1 +x z (k z - t)

Q Qb c

Yb = A - 7b c

A = 2 [0.05 (0.410 + 0.450)Ic

= 0. 086 in.

-- O. 0Z572

Yb - O. 086 _ O. 299 in.

@410 )t + t. 09 299 t• = 1.18

/gb = 1 + 1. 09(1. 17 - 1)

Shear flow at the NA from i'2quat*on (4. 5.1. 6-6).SQ

bqb = /d b I

5 x O. 02572

qb = i.18 O. 015812 = 9. 6 kips/in.

the conventional --]Note that SQ

Iformula is 18% unconservative

here.

The larger shear flow should be used which is

._ q = 9. 6 kips/in.

Section B4.5

February 15,

Page 32

1976

B4. 5.5 Index for Bending Modulus of Rupture Curves for Symmetrical

Sections

These curves provide yield and ultimate modulus of rupture values

for symmetrical sections only. For materials with significantly dif-

ferent tension and compression stress-strain curves, the necessary

corrections for shifting of the neutral axis are already included. In

the case of work hardened stainless steels in longitudinal bending with

all fibers in tension (as in pressurized cylinders), the transverse

Modulus of Rupture Curves are applicable.

It is recommended that MIL-HDBK-5 or other official sources be

used for allowable material properties. Where these values correspond

directly to the values called out on the graphs of this section, the modu-

lus of rupture values are applicable as shown.

Where material allowables vary with thickness, cross-sectional

area, etc., only one or two Modulus of Rupture Curves are presented.

Therefore, for materialproperties slightly higher or lower than those

used in the given curve, the modulus values may be rafioed up or down

(provided the % elongations are practically the same).

B4. 5.5.1 Stainless Steels - Minimum Properties

Page

AISI 301 1/4 Hard Sheet *(RT) ........... 39

AISI 301

AISI 301

ltZ Hard Sheet

3/4 Hard Sheet

(RT) ........... 40

(RT) ........... 41

AISI 301 3/4 Hard Special Sheet (RT) ........... 42

AISI 301 Full Hard Sheet (RT) ........... 43

AISI 301 Extra Hard Sheet (RT) .......... /#4

AISI 321

AM_ 355

_ "_ AM 355

Annealed Sheet (RT) ........... 45

Sheet, Forging, Bar and Tubing (RT) ...... 46

Special Sheet, Forging, Bar and Tubing

(RT) ............................ 47

*(RT) - Room Temperature.

B4.5.5.1

17-4 PH

17-7 PH

17-7 PH

PH 15-7

PH 15-7

19-9 DL

19-9 DL

B4. 5. 5.2

Stainless Steels - Minimum Properties

Bar and Forging (P.T) ..............

Ftu = 180 ksi (P.T) ...............

Section B4.5

February 15,

Page 33

Ftu = 210 ksi (RT) ...............

Mo (THI050) Sheet ...............

Mo (P.H 950) Sheet ...............

(AMS 5526) & 19-9 DX (AMS 5538) .....

(AMS 5527) & 19-9 DX (AMS 5539) .....

Low Carbon and Alloy Steels ::-"- Minimum Properties

Page

48

49

50

51

52

53

54

AISI

AISI Alloy Steel,

AISI Alloy Steel,

AISI Alloy Steel,

AISI Alloy Steel,

AISI Alloy Steel,

1023-1025 (RT) .......................

Normalized, Ftu = 90 ksi (P.T) ....

Normalized, Ftu = 95 ksi (P.T) ....

Heat Treated, Ftu = 125 ksi (P.T)...

Heat Treated, Ftu = 150 ksi(RT)...

Heat Treated, Ftu = 180 ksi(RT)...

AISI Alloy Steel, Heat Treated, Ftu = 200 ksi(P.T)...

B4.5.5. 3 Heat Resistant Alloys - Minimum Properties

Stainless Steels ....................... S.e¢ _Page

A-286 Alloy-Heat Treated (RT) ...............

K-Monel Sheet--Age Hardened (RT) ............

Monel Sheet - Cold Rolled and Annealed (P.T) ......

Inconel-X (P.T) ..........................

B4. 5. 5.4 Titanium - Minimum Properties

Pure Annealed (P.T) .......................

Ti-8 Mn (P.T) ...........................

Ti-6 A1 - 4V (RT) ........................

Ti-4 Mn - 4 A1 (P.T) .......................

98

99

100

lO]

I02

I03

I04

32

I19

120

121

129

130

131

132

1976

*Alloy Steels include AISI 4130, 4140, 4340, 8630, 8735,8740, and 9840.


1976

B4. 5.5.5

2014-T6

2014-T6

2014-T3

Page

6061 -

7075-

7075-

7075-

7075-

7075-

7079-

7079-

7079-

7079-

7079 -

B4. 5. 5. 6

AZ 61A

AZ 61A

HK 31A-0

Aluminum - Minimum Properties

Extrusions (RT) ....................

Forgings (RT) ...................... 140

Sheet and Plate, Heat Treated

Thickness < .250 in. (RT) ............. 14l

2024-T3 & T4 Sheet and Plate, Heat Treated,

Thickness .250_< . 50 in. (RT) ........... 142

2024-T3 Clad Sheet and Plate (RT) ............. ]43

2024-T4 Clad Sheet and Plate (RT) ............. ]h4

2024-T6 Clad Sheet - Heat Treated and Aged (RT) .... 145

2024-T81 Clad Sheet - Heat Treated, Cold

Worked and Aged (RT) ................ 146

T6 Sheet - Heat Treated and Aged (RT) ....... 147

T6 Bare Sheet and Plate (RT) .............. 148

T6 Clad Sheet and Plate (RT) .............. 149

T6 Extrusions (RT) ..................... 150

T6 Die Forgings (RT) ................... 151

T6 Hand Forgings (RT) .................. 152

T6 Die Forgings - Transverse (RT) ......... 153

T6 Die Forgings - Longitudinal (RT) ......... 154

T6 Hand Forgings - Short Transverse (RT) ..... 155

T6 Hand Forgings - Long Trans'_erse (RT) ..... 156

T6 Hand Forgings - Longitudinal (RT) ....... 157

Magnesium-Minimum Properties

Extrusions (RT) ....................

Forgings (RT) ...................... 197

Sheet (RT) ........................ 198

HK 31A-H24 Sheet (RT) .......................

ZK 60A Extrusions (RT): ....................

ZK 60A Forgings (RT) ...................... 201

Section B4.5

February 15,

Page 35

1976

B4. 5. 6 Index for Plastic Bending Curves

These curves represer_t modulus of rupture values relative to

the stress-strain curve for a section on either side of the neutral

axis. These curves are particularly useful for unsymmetrical

sections and when the allowable stress is other than yield or ulti-

mate for symmetrical or unsymmetrical sections.

For materials with significantly different tension and compres-

sion stress-strain curves, Plastic Bending Curves for both are

presented. When the difference is slight, the Plastic Bending

Curves for the lower stress-strain curve only are presented.

It is recommended that MIL-HDBK-5 or other official sources

be used for allowable material properties. Where these values

correspond directly to the wdues called out on the graphs of this

section, the modulus of rupture values are applicable as shown.

Where material allowables vary with thickness, cross-sectional

area, etc., only one or two Plastic Bending Curves are presented.

Therefore, for material properties slightly higher or lower than

those used in the given curve, the modulus values may be ratioed

up or down (provided the % elongations are practically the same).

Stress directions called out in the headings of the Plastic

Bending Curves apply only to stress caused by pure bending. The

bending modulus for compression is higher in the transverse grain

direction (in fact, slightly higher than the tension modulus) than in

the longitudinal grain direction, for the AISI 301 stainless steels.

Separate curves are therefore presented for AISI stainless steels

in the longitudinal compression.

Section B4.5February 15,

Page 36

B4.5.6. 1

AISI 301

AISI 301

AISI 301

AISI 301

AISI 301

AISI 301

AISI 321

AM 355

AM 355

17-4 PH

17-7 PH

17-7 PH

PH 15-7

19-9 DX

B4.5.6.2

Stainless Steels-Minimum Properties

Low

1/4 Hard Sheet ':-'(RT) ..........

1/2 Hard Sheet (RT) ..........

3/4 Hard Sheet (RT) ..........

3/4 Hard Special Sheet (RT) .....

Full Hard Sheet (RT) ............

Extra Hard Sheet (RT) ...........

Annealed Sheet (RT) .............

Sheet, Forging, Bar and Tubing

(RT) .......................

Special Heat Treated Sheett

Forging, Bar and Tubing {RT) ......

Heat Treated Bar and Forging

(RT) .......................

Ftu = 180 ksi (RT) ..............

Ftu = 210 ksi (RT) ..............

Mo , . . . . , . , . . ° . , . . . , • . . . . . ,

19-9DL ...................

Page

55-5A_59 -64

65-70

71-74

75-78

79-82

83 -84

87

88-89

90

91-93

94-97

Carbon and Alloy Steels*-':=-Minimum Properties

AISI 1023-1025 (RT) ......................

AISI Alloy Steel, Normalized, Ftu = 90 ksi

(RT) .......................

AISI Alloy Steel, Normalized, Ftu = 95 ksi

(RT) .......................

AISI Alloy Steel, Heat Treated, Ftu = 125 ksi(RT) .......................

AISI Alloy Steel, Heat Treated, Ftu = 150 ksi

(RT) ...... : ................

I05-I06

I07-108

109-110

111-I13

114

1976

':" (RT) - Room Temperature.

** Alloy Steels include AISI 4130, 4140, 4340, 8630, 8735, 8740 and 9840.

Section B4.5

February 15,

Page 37

B4.5.6.2 Low Carbon and Alloy Steels - Minimum Properties (Cont' d)

B4.


(RT) .........................


(RT) .........................

5.6. 3 Corrosion Resistant Metals=Minimum Properties

A-286 Alloy - Heat Treated (RT) ...............

K-Monel Sheet - Age Hardened (RT) .............

Monel Sheet - Cold Roiled and Annealed (RT) .......

Inconel-X (RT) ...........................

B4. 5. 6.4 Titanium-Minimum Properties

Pure Annealed (RT) ........................

Ti-8 Mn (RT) ............................

Ti-6 AI-4V (RT) ..........................

Ti-4 Mn-4 AI (RT) ........................

B4. 5. 6. 5

2014-T6

2014-T6

2024-T3

Page

115-116

I17-I18

Stainless Steels ...................... S.e.e.P.age __ 36

123

126

2024-T3

Aluminum-Minimum Properties

Extrusions (RT) .................

Forgings (RT) ...................

Sheet and Plate, Heat Treated,

Thickness < .250 in. (RT) ...........

& T4 Sheet and Plate, Heat Treated,

Thickness .250 to . 50 in. (RT) ........

Clad Sheet and Plate (RT) ...........

Clad Sheet and Plate (RT) ...........

Clad Sheet - Heat Treated and

Aged (RT) .....................

Clad Sheet, Heat Treated, Cold

Worked and Aged (RT) .............

Sheet - Heat Treated and Aged (RT) ....

Bare Sheet and Plate (RT) ...........

2024-T3

2024- T4

2024-T6

2024-T81

6061 -T6

7075-T6

127-128

133-134

135-136

138

158

160-161

162-163

165

166-167

168-169

170-171

173

174-175

176-177

1976

B4.5.6.5

7075-

7075-

7075-

7075-

7079-

7079-

7079-

7079-

7079 -

B4. 5. 6.

AZ

AZ

HK

HK

ZK

ZK

Section B4.5

February ]5,

Page 38

Aluminum - Minimum Properties (Cont' d)

Page

T6 Clad Sheet and Plate (RT) ............. 178-I 79

T6 Extrusions (RT) ................... 180-181

T6 Die Forgings (RT) . . . .............. 183__

T6 Hand Forgings (RT) ................ ]85

T6 Die Forgings (Transverse) (RT) ........ ]86

T6 Die Forgings (Longitudinal) (RT) ....... 188-]89

T6 Hand Forgings (Short Transverse) (RT)... 19l

T6 Hand Forgings (long Transverse)(RT) .... 192-193

T6 Hand Forgings (Longitudinal) (RT) ....... ]95

6 ,Magnesium-Minimum Properties

61A Extrusions (RT) ........ •.......... 202

61A Fo rgings (RT) .................... 204-205

31A-0 Sheet (RT) ...................... 206-207

31A-HZ4 Sheet (RT) . . . ...................

60A Extrusions (RT) ...................

60A Forgings (RT) .................... 2]2-2]3

1976

B4.5.5.1

300


Section B4.5

February 15,

Page 39

1976

200

; 75,000_p_;i27 x 10 _ p 3i

Elongation = 25%

T

-!

_+U ltimate

(Transversc)i

100

1.0

Fig. B4.5.5.1-1

1.5 2.0

k= 2QcI

Minimum Benmng Modulus of Rupture Curves for

Symmetrical Sections 1/4 Hard AISI 301 Stainless

Steel Sheet

B4.5.5. I Stainless Steels-Minimum Properties

soo_....

Sect|on B4,5

February 15,Page 40

1976

200_

F b

(ksi]

lO0

5O

1.0 1.5 Z.O

Fig. B4.5.5.1-2Minimum Bending Modulus o£ Rupture Curves for

Symmetrical Sections l/Z Hard AI$1 301 StainlessSteel Sheet

B4.5.5.1 Stainle s s Steels- Minimum Properties

Sect ion B4.5

February 15,Page 41

1976

Fb

. (ksi)

300

,2OO

I00

1.0

Fig. B4.5.5.1-3

Ftu = 175, 000 psi

Fty = 135,000 psiE = 26x 106 psi

Elongation = 12%

1.52Qc

k =I

Minimum Bending Modulus of Rupture Curves for

Symmetrical Sections 3/4 Hard AISI 301 Stainless

Steel Sheet

_.0

Section B4.5

February 15,Page /42

1976

B4.5.5.1 Stainless Steels-Minimum Properties

F b(ksl)

300

ZOO

I00

I.O 1.5 z.o

Fig. B4.5.5.1-4

ZQck=--

I


Symmetrical Sections Special3/4 Hard AISI 301 Stainless Steel Sheet

B4.5.5.1 Stainle s s St e els-Minimum Pr.ope rtie s

Section B4.5

February 15,

Page 43

1976

Fb(ksl)

3OO

Z00

I00

1.0

Ftu = 185,000 psii

Fty = 140, 000 psil

E = 26 x 106 psi

Elongation = 9_o

_44

"ltimate Transverse).

;ii

:ii

i " "

!ft

ttt

LH

1.5 Z.O

Fig. B4.5.5.1-5

k ZQc

Minimum Bending Modulus of "Rupture Curves for

Symmetrical Sections Full Hard AISI 301 StainlessSteel Sheet \

B4.5.5.1 Stainles s Ste els-Minimum Properties

Section B4.5

February 15,Page 44

1976

F

E

300

Fb(ksl)

200

100

1.0

Fig. B4.5.5.1-6

1.5 2.0


Symmetrical Sections Extra Hard AISI 301 StainlessSteel Sheet

_B4.5.5.1

t

Stainle ss Ste els- Minimum Prope rtie s

Section B4.5

February 15,

Pa ge 45

1976

120

• 100

8O

F b(kai)

60

4O

ZO

0

1.0

Fig. B4.5.5.1-7

1.5

ZQcT

t_

_t

_r

11

tl

iit

,If, I

J_

!:: i

2"t,


Symmetrical Sections Annealed AISI 321 Stainless

Steel Sheet

*;lh';l'ii__i .......

;IIHT_IIII

'" H_!

ftI:_ttI_IfI_1;,;11i_1

tlt::!II_!:

_!tf_tt|:tt

!!I12:"--':If:::I!"

i!II_ilIii_:_!!:11!t:t

iitt_!t!i!!

_!;¢!;tt_:

it:,t ftl_:ii:;ii_i:_i

_t:!!:! t_:

.'tt_

_!!!!i!!!il

:'/!_!i!!:i!J

:!!tl;li_Idittpllrl+_4

l,tf,t,_,Jt_tHISH;1

2;.0

-A

Section B4.5February 15, 1976Page 46

Graph to be furnished when available

Section B4.5

February 15, 1976

Page 47



1976

B4.5.5. l Stainless Steels-Minimum Properties

3Z0

Z80

Z40

200

160

IZO

8O

4O

B4. 5. 5.1-10

'.:..:-.-l.i-..'-.-i. ....": ....... I !.-! ...... I...:. I..... _. •...... , :...... i..J. ..............1 _...i..+,.... . ......, :,...I , , _ *-__--_----r .... : ..... -----t---_-_+-_--_ -F-_-.-_L.......t----i_oo__,_,,e-_,,r_ :...t..;-I'-I :': i : i _ !..-:.--.i-'i""i

"" " ' " " m " "" _'" ". .......... ":"1" "; ..... :" "• 1 : : ...... "" " :" I ":'.1)-_-r-"-¢--_-t :-+-Jr-:-'-:--'_".-+-_ _-: '-I-+-_-: -l--_._"-t- , ' I

l ' ' : I ' ' ..... " I : : • :" : ,I--.;-.-_..:-..I- .... | .... , ..... I": 1....... !--.!.. i '! "k ........ :" . "-'-:'-i.--'_-::-l:---_..---I----. :. .1_.:..h-r--.;--.j-.q.-i__.-.-.'.__._.:i..:._L...i_.j_ .; 1__ ..i. '[Ultimate :[ _ | : |' ' ' : ' : I • : / / ' -- _.' ' " ' I!.....:.-,:._!.:_.l _-..,..:!-.' '--.. " . : • -. !.:...: _._._ ' : "! t_,_ _, ,: I: !_ j ,:, I f-_t_l;: ). _I _-t--.,-.-,.....+---_-+--t......_--t--:-;-t--.---i--.---I.: i I :1l--..i...-i-._;---!..-:L.!.-.i...I...._...-i--.:..i ': i.....i..-_..--,:--=-.-...I-.;1 i...,--.;-..-t.-:._:.-i---,.-..i_._" • '" "'-";' " "-''-:'fl"-:-i-'": '--- : I.--- __.__.-- : i ,i; ::: I i: !: !• _._/ _.i_i_tflgiH_g_fl_i_g_i_i_gmfigi_i_flff_m___i.:.:L.)...:..::L.q.._dlllfillllllHlilfllggglggiHlfllgr_MgfilBHlgmLW.";_'_.-! "_!

-i i ;: _.;,,_tl!lillmlilllHMIgglgglgmll,,HH.fl.flllgg_lP.!!":."F. !-_ _"_d"!"'T T-K:])i.,. I..,/,ddlIMIllMIHIIIHIHJHIlgiluuRw,""I !--[..--I' i " i :.:_ ) : !)./_;/i._!iiil!iillilllililii!lllllllit_."._;.__.L.i_..i-._ .-_i _ ]i_._:7!i:7_T ::T.-i

I: ": I1::1 _ : I : 1.L' : ' " -I- .... ... ' '1.- 1.. • :,_: , • I :.L.]¢ • ,. : ,: " "" ', : I ,: l1_.._4__,._..'"" ' : "'l".... ; ......." I I"......................... " "-""'; ........ : --1":''"[ ............ I " l" " :, I"l I ' "

": -T : ,. I " I : F" . "1' . , . . ; . : . :"_. . _ i "-: ..... ' "_" "--' ' II" .... I .:---Jr ".... -_--L--J r ' i i' ._---J

t-:--:.....-,...........-'.......'................I--'---_........_,,-................-, ...............', • , "_,_. -F.-'--.--'.--'.':-'- . ..... 1------!.'-" , .'-T-l"- I : 1

i ..: !" :. .': / • '" " ' • • : : • I • I •tI::[ :-!"-:",;_:1_t:I; I.....i :I .......i-: r-T::_ ........!.":"_'. : ," ,-"i"'---j .... +-:--+-----!'.-----H .... "+'-'---F'--.'--F ..... i-_ , "" _ :

,'--,::-_:::-1--:-:1--iI .. .-i.......i .... i........-,.-:.-' " " !..:. !-i..i.1,: L : _ L:' : _ L , .; __-..,_o.o,,o,,,_. ,"_'.'-"" "--I". .... "_"'_-'-T-- ..... . "";"--'i"'_'_ .t.,I. *' _''" " "-T,"_...", : / : /.: I : / • : • . : _',:.-16s, ooopsi , ....r::!:i,' LI __-tr--:.-r-i.--I-;-.-!--.v-r-_:';--_7.,,,1o_,,_,).--.--z,.--,---.l

'-'..'_', • , : ' " : t ; "d-'---'-I"-'-_'-t-- ..... : -,' - i• I :: I' : I • I , • . . , I t-,zorl@,atl.on = O'_c ! I|' : '............... ; -,"t--'-; - I------ ..I, " ...... '- ...... t ........ : ....... :";',; i :,, t, _ I, :: i , , , ' • ', I:Y 'L,• I . L..;_./ . . :_--..r'-.-. -', i' -I,--"+-_I_....... ..;....)i.... i . --.)....ii. :...I..:.L.i .... _.... _-l-.-n _;..-.:..._-:.: ..... i.... i..--: -- i""- ""I ......... !.-..'.:-1 ....... L..'-.J

,: I_,. _ ,...-r-. !.,., :--1 _ j.._.._i.;; F.._r ;rfl ...... i- i----:..._........v-,..._........i-.--..:..-...-..i............[ , h-r-----.-r---.---!•-_-- _ " !1 _.-4--."--L_-. ........ _ .... "i ..... I " . " "-"" I _ I • I

I • : .... "1 • l • j J • I : • . . i .::• ", "1 •................. ; ' • I : [.. • • I I •• • | , I ................1 !:,., . , . , , , i......... l-!i:.-!..iUI_" _:: I _ .... L _ I J ..... _ ............... : ...... L ........... _ " L . i "--I .--_

1.0 I.Z 1.4 1.6 1.8 Z. 0

ZQek=--

I


Symmetrical Sections ]7-4 PH Stainless Steel -

Heat Treated - Bars and Forgings

B4.5.5.1 Stainle s s Steels-Minimum Prope rtie s

Section B4.5

February 15,

Page 49

1976

Fb

(ksi)

300.. .: : .;: .. :';: • : ". . • ; . T : • .:.._: • .. ,. ; ' . . : : ' . ; ._.i_i../_...:.;"_ _:i:,:_i ',:_:::":::-:!::':'_:: i:i:_....,: :::: I_..:_':"_::l:_i-_-..:--i.-:'._:1.._.!:r.i-.:-I:::.i.::I '_""I:]':'l-:.i ..!"-:4!"'[';"i ..... :.:--_L.-.I. :....:!_ 1Room Temperature-T : I.: ;:_.1".-TT..'-.[:li::!.. I'... ]' .I.: :.. '/. - ,,'_.." l :-:_- '--_.i::i_:-_:.i-:--.:.i,_: .:-..-:.....: I:::_:4-!:-:.':_-I::!:--i-I]-':.:#.':i::-_:::!i:-_:i:::;-i-1-._:: -:-.i::- :-::!::. :-.!--_

_0 :.:::.t--::ti:'#:.4:-:.ii::.-I:.i::.l_.?::.:;l_:;.i_-:il?_i:;::I.:-i',:.-:1:!_41_:1:...__!.l:._i: :l..-_i.._.:..i...L:.:i.}:.!_:i:.l..i::]..I 1. ! I i_,:i-P:.l !i:ii:i.i i i l._.:.l._!:i 1:]. i... i,[!:': ]. :I._i:q:::__.,1::i:.__-_ .q..-T,.,T!-iT _.ltu_,_mat_ __!_[:'::,,'::!i_TiF' 1• :'L":!5" ":'""''iT" "!..... "7F!"v!'Y-'-I"q[F ..... F:L"?:"7"'!: .... !'r"i'V':':'"":'":':" ":"1":...... v" •,40 .......:,._. .::. • • .: . .. . .. ; :_-712i.._i-.. :,:• .- : • • "1--"T--'t::I:I.:" '. -_- .-_.._:_.._..-... :. .:. :...-_._ .: _+, :...-..

•:: :::_ : : : " I :,. :' • I : I • : : '!i ! ii : _ :i -_-- : : _:+-- ++-i-- +-_÷ _:-:- -_:

.l::.._ :.. " .:..::........:: .,... -r _-7 i57.:-,_ •

:::-7..'":-:F!iF" Y!Iz.-i::?77 :::T=[[III._:. i:T:-.i:: ...._-;:': .....::__:: .......... _'1Z0 ,,-['i:--T_,:_ . - !_ ! • i; :'!; ,: ' ' "i, . ': ! ...__

_i!Z iii_ !iiii_i__! Z. _i!._.""i.S:!Z ]:__Z.!_!Z':Zi.:].2:i._.::.]:;.:].i:._.]::Z:.,.i:...i:.:.

[ , " " ' ": :: : ":'":::: :::: ": '::: " "'!: ":'i "

.'1: :I:".: : /:'" .'.|':I :.v".: 1_';:":: I:::.!'.:. I:: '. :: '::;:!_:_ ,r -_,-, "^6 • • _ '. i; :.: .So i_ii_h.....jii.qi_iit.!!il._]:l:ii'l:;]ii_7,?_..-..._li_];ii_iti:!;i._!_:i;!_il]_, = _v x tu ps_ ::TZ-T:V_T_"

:_t:._:iifi:t:ii:_:-l:]_-,-i_q.-.-t-:i:.P.i:F.-÷ti..iit-.=..-t.-.,:-ti_:i:ii_- _.longation = 6% :ti:+[-]_:!":iiiT_]:jlZ[il.] ill: iii!:}[]; : ]L::-:I::]]!LI: ::Ii]ZF :!!!KZZ|F I : _ • I!_i•iiil!::::/l:..4iVzi!ii_i::?ii:ii.li_i_i_]_i:J_iiiiii_it_!:.ih=..l;!::il_!i:_hiqiiili_!]if!t:r_:!_i_i_.._;l..;_.!_[:";-F_.]_:--_

40 i:::_':_'iT:';;::i _:]!;::_ ',::ifi:i':t!ii:i:i_Yiii]:-iti:::_i!!:tii]i}iF:.,'ii!liii:i'iii:::i!_:.]i_:! : i-,!:b:.l :.:; i: I::i::! I:::1.:::::::::::::::::::::::::::::: ::i]ti::l ._]ii-!: ...::::.,:::....',...ui:.,:.,:. :..Iq::_.;::..:;::._.: :!::__ ...._..,.._ :::![ii:':!iii!'_!:l _:iiii[:: ! iili!i !:]:_-'!;!.....:. _...,_ ! :q: ii t I ;_!!-;1::.!'::i'l:::!ii_.t:t!':i!i;:!

'": .....':i.-,-t.. :1:7-' :i_._: ii _!_;,._:ii:,iii!o ,..:..... :,.::. :::,::.: :,,.:., ..... ...... . : i:.ii:il :;ii'!.:i_di]i:::.'_iiii;i]ii!iTi}:i:::;[[ii]i'i:.!i-i['-!..i.i:i_

1.0 l.Z 1.4 1.6 1.8 2.0

2Qck--'_

Ir

Fig. B4.5.5. I-II Minimum Bending Modulus of Rupture Curves for

Symmetrical Sections 17-7 PH Stainless Steel


Section B4.5

February 15,

Page 50

1976

!iilt!_i!i

360 -

iiTii_i!i

:: r",|." |. _.[.. -!..

ii!:'ilii-"

_!iiiiiii!!ii!i,!!ilti_ii!i!i:i_!i,ilil.,iH..:_.r_:_l_!:I;-;; ;:,; ;;I; |.;l ;;., I';; i": li" , ..... I:: I',......... -. ." .-'1.-': .:,... _'_:_-'_-

280 ...... i: ....... '" ....... _: i:,'.'.!FHli:.It:,.':t:I,: :t:: ;_:: "l.: iii!t!::! '_'" :' -_!:j.•":|':!:1::1;1!'1'11"' i';'l "' :;I

_-:4_" -";_.,:::!]: ..... . ':' .

::;r !!l,: "

!]ii .... '::i_ -% .._oo ii]iii!i_,;iii'.ll__. :.._, .,; :,.,..,,:H_ii

.... J.I • .[ ............. I ;: I I ; . : I

"L:I!i:" ii.; d]!|':ll'l_l."i !!: • ' ' : .... ":

:::::::::::::H:H::::::|::::|::::::::, iJ_i:.. i:" ;" ';" ::';.,"_t."TT':,i;,;' :;i .;i- . .... ","r "....... i .... i. h ..... J .,I .... i

i::!i.!:i,i!:i!_i::l, i:l:i ! i.}i;lii!: ...... :H:t! ::

":':: .l::::l;|.:.;:i :.'.l;':'....'.1;:'1 ....

........ ,............ "!t:.ii": [i !::.l; '::

; i I;I;. • ;;I ;. .;'.1 ; I ;;;- I; -; • ;I.

80 :_!i::_i;!HH':i:_:;i_ .._i._÷i]-,.,_ 4:,i_H-_ :.,.:.::.:r.,-,.-:i_ _ ..... ::_ ::i:.:_:;::,::.:i:r.;:_,.i!-_!:l:.i!:i:_ii.-,.:,

: ":1,::',,_ .!, ,: ..,:.., .-a,_:.._l?i i _" i:,"r[:'_]7F."_:i':i_il!!i! tii!li!:. ,.:.h .. ;.:,,:] ;l ;'-:, :.:r:_., .,:J.:l: • !;_"; ::1.i : ':l:l::i4,0

':' ':':'?T|':F.

o i:i:l_iii i ::i_: _- ::':'"i]q: ....."":",,. ::'['i I '_7"IL': :.,':;

";:'"-!:': :.l:_:!: '.:h,:':::i:. ': :;_! ::"!::! .';:, "!:.. :!;;!_!i ..!: :_:,! .7,._!'==H':_ :!.H:]IiL:i:_::=__._il:i:.}.,...i._==-iiiH,l:: "r:itiii!

• El!E'il!':Eiil!:!i::!::i!':':;:iiii:i,:;!:!;i.:i ::::'::":::':;:":" 1;_ . [..; i,.. I'":'.._"T,..'_',.... ;. , ."1"*,"?"[-., ..'"_1"". .. , .-'_-r*-..,':Tl:... .'-','*"_._:::: ::.[: ,', ; 1:'; !: , J;;: I:': ":', I";1', . ]:'," I;'" I ..... i'"|i .........

it_.q!i!;iiiilttti _fit!li; _ii;1;_.I.,..;1_,:,;i_!l;i.: i:: [_i_ii:!li::[ _.":;;[:'.", :;1:_:: : :11:1% : ::./l" .11: ; ::;: ' i.t::." :.:.h: : :."Ji4f., :: :.:..

;:'!i !i: ::': :':',

ii:i ':_:i_ti:ti!'._ :::•::_' _"i1:F'_iEH_'_i=,_-_...._..:-'._ _:_:_.........i_i!!i!i!"'": _i)_ "!ii:!ii;..,._. ii!]t!i!i!'-__

._...,,':';_;:_!1_:' ":i i:i.;ii:lii:!:i:_iiH_i:;]}Ultimate :_::t. :ii!::i :.::i]i!!i

•.: "::':.'.:',::l}.::[•:""l'- :" :'Y."Ei:J""k:.:E_._.i:.Li:_:i;iL_!: :.::iiU::!

: _':;':,i !?:::i::._!:ii:ii::.:i"" :_:r,z: _'"t_! :: ";_-' .' i"

•L".: :_ F:_ i:.:f':!i

• "'1"" " " "'!'' "" ' :::';::_|:':":_.,.:.._ .::_].;.: .-.['_.. _._ :.,..i_ield :!_...i_.J_:.:i:ii::;i•"l::l i: ! !::]:'!i.:::": l': ' "l.: ":_": ""i'::

_:._.:F!l:/a, .. "i:; :ii;

,.ll-.,iiili:;i._i:i!ii!i

_11,/:,ii:ii!i'i i.l:,.ii::.;!I:.-!]

.I : ;.11;.:.i.i' i: :'; .:1--,..;. • , :' • ,. ,-; .,-,.. , ';

: ," _'.: ! • l" " '.: "' . '. :,':" '!. .., . ...!... L.:J.;._

!:7_i:i.

:i:'*F.-:i: , ':'._i'!'. i I ':: i:'_'

l ;; - ;; ;;..;I : : • ;

,':: ;;-.

".':|*.": "'"I" r'-; ':_!_""_'1'.-'-:,':': '_.'.! "_:'_ '_ :, !:.i'_ii_li:,:_',:;:l,.,_..q_:'::_,::_.I:_:-:.:':':ii ;i |-. i , i . ; ; . ; J , - . .::!.! .

-,:.:l_,:_l, ,_1:./:I,, I I:::_.;! _._r- .,_.-..-._,:":',:'-'.,'-"r'-:r'", _-:._i..... :"_:', :::i :!;, _ _ _:;,!_.r :: ': : , ,:,: .., ...., i:i::-i :i:: i: i.'/.i:............:i, .:',_:...,,,,. !ii__..: :.::...: _.,.._-..:., - ., .::. ... i_.:4!i:.r d- t .... :. t. I; .t .... ._. :-'T':.

,.1'!.! i:':., I i._'il ;];:i ':;!"i ";i;.: '1::i

, ,.':{: ':; 1. ' :. '.| .', ..".1, . .I .,/ ":iJ ":: °t':: ,'_."!::_i:i:.::i';!{!!Ftu,.. = ZlO, O00 psi I'! ''::H:i

•,-.I-.i- :_F.,.. = 190, 0OO psi • I ......

,., . _:r,-;S = Z9 x ,.o ps_ -1- ....... -.':--

:.i_l.[V._o,,gatio_, = 6% . ..:il!! ............... ,.,,. , .... ;::.:,..

k.:-'tT:'"':"" i:'" :t?:":1,"i'::' :_!i:'-_

• "!':"! ":"": :'"; "! ';'" : ..... ;! : .... :'l::';_!i It!. ii. L ["ilL! !! !_ i: !'i'l .!:Iri !

:L,.L"...!L:."...:;.:. ;t...:;,:.[ .:.h.:.;i!.,.,ii:,.,. =: I_. ;. , . ; .,. ......

1.0 l.Z 1,4 1.6 1.8 2.0

_c_ -=-'r

Fig. B4, 5.5. l-IZ Minimum Bending Modulus of Rupture Curves forSymmetrical Sections 17-7 PH Stainless Steel

B4.5.5.1 Stainle s s Steels -Minimum Properties

Section B4.5

February 15,

Page 51

1976

350

300

F b

(ksl)

250

2OO

150

1.0

Fig. B4.5.5.1-13

1.5 2.0

k = ZOcI


Symmetrical Sections PH 15-7 Mo (TH 1050).

Stainless Steel Sheet & Strip Thickness 0.0Z0 to0. 185 Inches.


Section B4.5

February 15,Page 52

1976

400

300

Vb(ksl)

200

I00

1.0

Fig. B4.5.5.1-14

1.5 2.0


Symmetrical Sections I=H _15-7 Mo (RH 950) StainlessSteel Sheet & Strip Thickness 0. 020 to 0. 187 Inches


Section B4.5

February 15,

Page 531976

F b

(ksi)

160

140

lZ0

I00

80

60

4O

Z0

1.0

Fig. B4.5.5.1-15

++,

tlt

!!i :_

÷#* tt

ttt _t

}ii :

it:

!t: _t!iT _:

i!i ;_

_tl t-I

t:! !t

t :_

I i"

J_t ,

rtHtVth'1111 It

H_tVfH_HH:

HIHIU

Jill?![T_![![[

[71T:![T

!iti!tf_

"f!ttltr

_!!!!!i!

"!+*t*t

:!Fi_!i

!ii_Fi

_*-+-,-+-e.,_

:!!!i!!!.!;ri!TY

'+lt!lN

:i!!!,h_

tlt _t**.

-_IUttt:

t;:',1+,:tl i-tl_t-

!i1t',tHIH;.t

ttttti_

1.5 2.0

ZQck=--

I


Symmetrical Sections 19-9 DL (AMS 5526) & 19-9

DX (AMS 5538) Stainless Steel

B4.5.5.1 Stainle s s Ste els- Minimum Prope rtie s

Section B4.5

February 15,

Page 54

1976

240

200

160

120rb

(k,i)

8O

4O

0

1.0

Fig. B4.5.5.1-16 Minimum Bending Modulus of Rupture Curves for

Symmetrical Sections 19-9DL (AMS 5527) & 19-9

DX (AMS 5539) Stainless Steel

B4.5.6.1

140

lod

8O

F b(k.i)

6O

40

20

0

Fig. B4.5.6.1-1


0.002 0.004 0.006 0.008 0.010

(inches/inch)

Sect ion B4.5

February 15,Page 55

Minimum Plastic Bending Curves 1/4 Hard AISI 301Stainless Steel Sheet-for Tension or Transverse

Compression and Stress Relieved Material-for

Tension or Compression

k=2.0

k = 1..7

k=l.5

k= 1.25

k=l.O

1976

t

Sect ion B4,5

February 15,

Page 56

1976


u_O f,- _ N O

II II II II II

Q oN

Section B4.5

February 15,

Page 57

1976

B4.5.6.1 Stainless Steels-Minir_am Properties

F b

(ksl)

120

100

8O

6O

4O

2O

(inches/inch)

Fig. B4.5.6.1-3 Minimum Plastic Bending Curves 1/4 Hard AISI 301

Stainless Steel Sheet - for Longitudinal Compression

Section B4.5

February 15,

Page 58

1976

B4.5.6. l StainlessSteels-Minimum Properties

0 _, u_ N O

II II II II II

0 0 0 0_D _ N 0

0 0 0 0 0

O .r_

N 0

_ 0

I

4

Section 84.5

February 15,

Pa 9e 591976

B4.5.6. I Stainless Steels-Mini].num P±'o_erties

F b

(kst)

240

ZOO

160

120

8O

4O

O. 002 O. 004 O, 006 O. 008 O. 010

(inches/inch)

Fig. B4.5.6.1-5 Minimum Plastic Bending Curves 1/2 HardAISI 301

Stainiess Steel Sheet-for Tension or Transverse

Compression and Stress Relieve_Material-for

Tension or Transverse Compression

k=2.0

k=l.7

k=l.5

k= 1.25

k=l.0

Section B4.5

February 15, 1976

Page 60

B4.5.6. 1 Stainless Steels-Minimum Properties

O r-- u'_ N O

II II II II II

a_

'.O

O

Ng-4

c;

Op=4

O

O

O

O

O

A

0,=1

"_o

_ U

_ w

._ _¢II

._ ,_! .0_ 0

I

4

0 0 0

B4.5.6.1 Stainle s s _St:f: ,i.:_i? _ i:,,! _- '2)_! [ [.t"_._[c_:i_F_!:_____!_2..

Section B4.5

February 15,

Page 61

1976

120

100

8O

(kF_)

6O

4O

2O

O. 002

Fig. B4.5.6.1-7

k=2.0

k=l.7

k=1.5

k=1.25

k=l.0

0. 004 0. 006

(inches/inch) I

0. 008 0.010

Minimum Plastic Bending Curves I/2 Hard AISI 301

Stainless Steel Sheet-for Longitudinal Compression

\

Section B4.5February | 5,Page 62

] 976


II II II II II

0 0

N N

0

v

0 0 0 0

00N

O

o4

O

ON

d

,,D

d

N

d

00O

O

<

,.¢

"_ 0

r_

0_U

r_ .,_

0

_ 0

!

4m

&


Section B4.5

February 15,

Page 63

1976

240

200

160

Fb

(kst)

120

80

4O

Room Temperature

Ftu = 150, 000 psi

_cy = 105,000 psi= Z7.0 x106 psi

Elongation = 18%

1i

rt

T:!

EHiii+÷+ttt

:t:

:fl

F

0.002 0.004 0.006 0.008 0.010

(inches/inch)

= 2.'0

=1.7

=1.5

= 1.25

=1.0

Fig. B4.5.6.1-9 Minimum Plastic Bending Curves Stress Relieved

1/2 Hard AISI 301 Stainless Steel Sheet - for

Longitudinal Compression


1976

B4.5.6. l Stainle s s Steels -Minimum Prope rtie s

0

II II II II II

A

0

U

!.I.a

m

u_

0er_

0

e.i

¢)i-q

!

ow.i

0

@

0U

14

_o

o 0 0 0ao _p o _DN N N ,-_

_'_,11

0m

0 Q 0


Section B4.5

February 15,

Page 65

1976

240

2oo

16o

12o

80

40

o

Fig. B4.5.6.1-11

0.002 0.004

:T

Stress-Strain Curve

fo

H-IHH-t-H_t_:

0. 006 0. 008

k=Z.O

k=l.7

k=l.5

k= 1.25

k=l.O

0.010

(inches/inch)

Minimum Plastic Bending Curves 3/4 Hard AI_I301 Stainless Steel Sheet - for Tension or Trans-

verse Compression and Stress Relieved Material -

for Tension or Transverse Compression

B4.5. 6.1 Stainless Steels-Minimum Properties

Sect ion B4.5

February 15,

Page 66

1976

F b(ksi)

360

320

280

Z40

200

160

120

8O

4O

k=2.0

k=l.7

k=l.5

k= 1,25

v

(inches/inch) 'u

Fig. B4.5.6.1-12 Minimum Plastic Bending Curves 314 Hard AISI

301 Stainless Steel Sheet - for Tension or Trans-


for Tension or Transverse Compression



k=2.0

F b

(ksi)

140

120

100

8O

6O

4O

2O

k=l.7

k=l.5

k= l. Z5

k=l.O

0.002 0.004 0.006 0.008 0.010

Fig. B4.5.6.1-13

c (inches/inch)

Minimum Plastic Bending Curves 3/4 Hard AISI

301 Stainless Steel Sheet for Longitudinal Com-

pression

1976

Section B4.5



_0

'I .....! "1 r'_.......r......."......-......, . • s• ' ! , • I , -; : • l, , t ......

I I.... ;-.+ ..... _...L...I ........ i ..... ._

• i : I .

I.._r. '" ..t'':"';i ......." :1 , Ii ....... i

- "'r ...... _ ....... _ ....... t ":-'-_.'---:---!. •j I I I 0

u, ii _ _ I.......- _: -' .... _ -'-"F-- ...... _--_.---I- .......

• • I I • I"_ i . , : .... , .... _......

........... _ "-'-"i ..... ]' " ! C:_[ • "" "i." ....

• I4........ _ ........ " • , i..I-.......... : """r......

i , I • I ' •

.i ..... i ..... I..... l.............. !"_l , ' i _!'-':"i'-! .......... 1...... "

I ' . • c_." ....... i .... ,...... i"! "......_.........1....... "'"'-"i'"':-'i

!i!*.... : ........... ........... _4.---1'" :"'_ :. t ..... i'":. : ...... _0

• i.!.I 'o-i-.-_ ................... "'_ c_I _ I

" _of" ' " I.......I.......i-,.__--i---;--:.......

• "l ......... I, ..:..,.: ......... I, t:_! 1..... ....-..' • I

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: .... :..: ..... :..I..._

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' i _! " ;"1 " i _- _ .-_. I....,4....1_,_ I[_. , i s ,

..,..:' i _ " _

m

7

_44

Section B_°5

February 15,

Page 69

1976


240

200

160

k=2.0

= L7

=1.5

= 1.25

=1.0

F b 1Z0

(ksi)

8O

40

Fig. B4.5.6.1-15

0. 002 0. 004 0:.006 0. 008 0.010

(inche's/inch)

Minimum Plastic Bending Curves Stress Relieved

3/4 Hard AISI 301 Stainless Steel Sheet for Longi-

tudinal Compre s sion

Section B4.5

February 15,

Page 70

1976


O. 02 O. 04 O. 06 O. 08 O. 10 &. 1Z O. 15

e (inches/inch)

Fig. B4.5.6.1-16 Minimum Plastic Bending Curves Stress Relieved3/4"Hard AISI 301 Stainless Steel Sheet for Longi-

tudinal Compression

k=2.0

k=l.?

k=l.5

ik= 1.25

k=l.O


Sect ion B4.5

February 15,

Page 71

1976

240

200

160

Fb

(kst)

120

8O

40

0.002 0.004 0.006 0.008 0.010

(inches/inch)

k:2. O

k=l.7

k:l.5

k: 1.25

I_: 1.0

Fig. B4.5.6.1-17 Minimum Plastic Bending Curves

Special 3/4 Hard AISI 301 StainlessSteel Sheet for Tension or Transverse Com-

pression


Section B4.5

February 15,

Page 72

1976

320

280

Z40

ZOO

160

120

8O

4O

I " : " ' " : ' " "I " ' " : I...... '....

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_ :.+; , ' : _:+ , ._7-_-" "-i"" I ....... t"- .... "" "

........,_., i• ]r[ , ......, i i-::.1171171._, .........

--+;..+i-,.+--i

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.... ix" i.>" --,i.....i'":'---'"----' '

i : .: ::" " = 2¢" ()x LO :>si I I ! , '/ " / : '!' "':"' "" " " I" l'" : " J ..... I ........ ! ........ ::|

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• : • I I ," ' ;..L.J ....:...... -..-.......... :........ I.- .... •...L.._...J

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: I Roonq "['ernl_eratt, re I • _ • ; i _..... !::.....I.....; ' ":........l"""l':!"l'"; "''J

• I • " " I ' ' : ' " " " ".... ;........ i..................... - ....... ,........i.-..-.l---.'.-l.- .... .-I--..+-.-i-!..L :. i,.._..i ....... !. :. I ,....i .......i........ t....:.t....-.:.i• '' ' " " " + _ : " ! i. :."-_-+.-I-.-+..:_.-.:.--I---:--+.--....!........l-.-l---+-.-,-..:._

::;: i i: i i0.02 O. 04 O. 06 O. 08 O. I0 O. IZ

ILl

((Inches/Inch)

Fig. B4.5.6.1-18 Minimum Plastic Bending Curves.

Special 3/4 Hard AISI 301 StainlessSteel Sheet for Tension or Transverse Com-

pression

k=2.0

k=1.7

k=l.5

k= 1.Z5

k=l.O

F b

(kst)

Sect ion B4.5

February 15,

Page 73


150

140

IZO

I00

8O

6O

4O

zO

"'E .... "-- i" "" "R._.'_m

I---:....__.:LL_.....i I

....:"............Y" "!............

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: f

.II : :..... _ ,._o

: I !•" " " i' " "

i

k=2. O

i

k=l.7

k=l.5

...)......I:PP" i

I.........I.........i..........' { '" St:.q:._s-Strair C,,r'.'e : •

....,....i....'.......t....i....-.......!....:.....: i I i

.......i i.]-i!i ..............•,_ : ! i-- i : i

: : _ ..... i......

k= I.Z5

k=l.O

{ (inches/inch)

Fig. B4.5.6, 1-19 Minimum Plastic Bending CurvesSpecial 3/4 Hard AISI 301 Stainless

Steel Sheet for Longitudinal Compression

1976


1976


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Section B4.5

February 15,

Page 75

1976

F b

(ksi)

Z40

200

160

IZ0

80

4O

HIH 7_f

!_!ii

tilii_!!!t T!

Nil! 7i!_]tti :}i

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(inches/inch)

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I

ili++*

iii

tts

II:

t,,

21!

O. 010

k=Z. 0

k=l.7

k=l.5

k= I.Z5

k=l.0

Fig. B4.5.6.1-Zl Minimum Plastic Bending Curves Full Hard AISI

301 Stainless Steel Sheet - for Tension or Trans-


for Tension or Compression

B4.5.6.1 Stainless Steel-Minimum Properties

Section B4.5

February 15,Page 76

1976

3O0

20O

(ksi)

lOO

O. O1 O. 02

Fig. B4.5.6.1-ZZ

O. 05 O. 06 O. 07 O. 08

: (inches/inch)

Minimum Plastic Bending Curves Full Hard AISI 301

Stainless Steel Sheet-for Tension or Transverse

Compression and Stress Relieved Material - for


O. 09

u

B4.5.6.1 Stainles s Steels-Minimum Properties

Section B4.5

February 15,

Page 77

1976

F b(ksl)

200

160

120

80

40

0

O. OOZ 0.004 0.006 0.008

(inches/inch)

k=2.0

k=l.7

k= 1.5

k= 1.25

k=l.0

Fig. B4.5.6. I-Z3 Minimum Plastic Bending Curves Full Hard AISI

301 Stainless Steel Sheet for Longitudinal Com-

pression

B4.5.6. ! Stainless SteeJs-Minimum Propertiee

Sect ion B4.5


,/3

O_ " • ,

tl II II {I II

i :" ;:',

0 0 0 0 0

N ,"q .-_ ,--4

v

0 0 0

00 _"

0

0

uo_

I

_44

Section B4.5

February 15,

Page 79

1976


320

280

240

200

(ksx)

160

120

8O

4O

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ii::__'tu:zoo,ooopsi !;TI_:.:-:_i._T!:_i-Ii_IT7:T!T..-II_......i::)"Ft. = 160,000 psi 'i)lllTI!!ml:TI:_)i":i:]itl)l!)]'::'_):iIT::]:)FT)I;)I:_i!:ZTT:I.-I:F:I_

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0.00_ 0.004 0.006 0.008 0.010 0.012 0.014

(inches/inch) tu

Fig. B4.5.6.1-25 Minimum Plastic Bending Curves Extra Hard AISI 301Stainless Steel Sheet -for Tension or Transverse

Compression and Stress Relieved Material - for


Section B4.5

February 15, 1976

Page 80

B4.5.6.1 Stainless Steels-Minimurn Properties

0

0

0

0

_ _._

_ o

!

4


Sect ion B4,5

February 15,

Page 8 1

1976

Fb

(ksi)

160

IZO

O. OOZ 0.004 0.006 0.008 0.010

' {Inches/inch)

Fig. B_. 5.6. 1-27 Minimum Plastic Bending Curves Extra Hard AISI 301Stainless Steel Sheet for Longitudinal Compression

B4.5.6.1 Stainle ss Steel s -Minimum Prope rtles

Section B4.5

February 15,

Page 82

1976

0 uq

0 _-- u_ e,] CD

II II II II II

i

_q

0 0

0

0

m0

A

I2

w

o0

.'_U_

_'_

. _

N!

,6

4

B4.5.6.1 StMnless Steels-Minimum Properties

Section B4.5

February 15,

Page 83

1976

F' b .(ksi)

60

5O

4O

3O

2O

10

0

O. 001 0. 002 0. 003

(inches/inch)

k= 1.Z5i

0.004 0.005

Fig. B4.5.6.1-30 Minimum Plastic Bending Curves Annealed AISI 3£1Stainless Steel Sheet

Section B4.5

February 15)

Page 84

1976


140

120

100

8O

Fb(ksi)

60

4O

20

Fig. B4.5.6. 1-31

•--i ...................:.-. L..--.._t : i' I

v i ;

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......_........i !i ' • - .. i

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Ik=l.7

k=l.5

i ...... i :k= 1.25

I

k=l.O

i I " • • o"'" i - S-r(:_ -Strain Curve I

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O. 10 O. 20 O. 30 0.40

t (inches/inch) E u

Minimum Plastic Bending Curves Annealed AISI 321

Stainless Steel Sheet

Section B4.5

February 15, 1976

Page 85


Sect ion B4.5



B4.5.6. I Stainle s s Steel s-Minimum Prope rife s

Section B4.5

February 15,

Page 87

1976

!,

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B4.5.6. 1 Stainless Steels-Minlmam Propreties

Section B4.5

February 15,

Page 88

1976

I....

_i .J ,-: .JIf II II If I!

0

c_

m

m

e_

0

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o.._ o

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Sect ion B4.5

February ]5,

Page 89

1976

B4.5. 6. l Stainless Steels-Minimum Properties

F b

(ksi)

320

280

240

ZOO

160

120

8O

4O

k=Z.O

k=l.7

k=l. Z5

k:l.O

B4.5.6. 1 Stalnle s s Steel s- Minimum Propertle s

Section B4.5

February 15,

Page 90

1976

360

320

280

240

200

Fb(k.t)

160

120

80

4O

Fig.

: I • I

k=2.0

k=l.7

k=l.5

k= 1.25

k=l.O

B4.5.6.1-40 Minimum Plastic Bending Curves 17-7 PH StainlessStdel

Section B4.5

February 15,

Page 91

1976

B4.5.6. 1 Stainless Steels-Minimum Properties

k=2.0

350

k=l.7

300k=l.5

"I

250

Fb(ksi)

2OO

k = 1.25

k=l.O

150

100

5O

Fig. B4.5.6.1-43

O.004 O.008 O.012 O.016 O.02

' (inches/inch)u

Minimum Plastic Bending Curves for PH 15-7 Mo

(RI-I 950) Stainless Steel Sheet & Strip Thickness0. 020 to 0. 187 Inches

Section B4.5

February 15, 1976

Page 92


Section B4.5February 15, 1976Page93

Graph to be. furnished whenavailable


Section B4.5

February 15,

Page 94

1976

100

8O

60

F b(ksl)

4O

2O

O. OOZ

Fig. B4.5.6.1-46

k=2.0

k=l.7

k=l.5

k= 1.25

k=l.0

0. 004 0.006

, (inches/inch)

0.008 0.010

Minimum Plastic Bending Curves for 19-gDL (AMS

5526) & 19-9DX (AMS 5538) Stainless Steel

B4.5.6. I Stainless $teels-Minimam Properties

Sect ion B4.5

February 15,

Page 95

1976

0

II

......F_

0 0 0

0 -_D _,_

L_

C" Lr_ _ 0

• • • •

II II II II

A

il!!!!]iiiiiii iiiiii!o"•:_ I '

,i . t • -, - : :

Ifi ..: ": • "" i_ . c_

: ............ .o.. .....

: I ' '_...... •.... !....... _ ......... • 0

iii..i,i.... -.......

i

A

0 0 0

<

<y,_O

I

_ m

o--

t'..-

i

.A_44

B4.5.6. I Stainless Steels-Minirnurn Properties

Section B4.5

February 15,

Page 96

1976

ZOO

160

4O

0.004 0.006 0.008

(inches/inch)

k=2. O

k=l.7

k=l.5

k= 1.25

k=l.O

Fig. B4.5.6. I=48 Minimum Plastic Bending Curves 19-9DL (AMS

55Z7) & 19-9DX (AMS 5539} Stainless Steel

_4.5.6.1 Stainless Steels-Minimum Properties

Section B4.5

February 15,

Pa 9e 97

1976

200

160

F b(ksi)

120

I-i

I

.... o,

!I

i...

' I:i-I :........r---........:i.. i .... : I. I.:..:..I,b._c,rn l'e:X:l)e :.,.tu :.e ' i.

t

I

"! ....

] I :I ..... i' : k=2.0

i

'i ....!i..... • ........ k=l 7; •

I. I "", •.....-.'...........i....i.-

i...... t.........I-W.... iI II I

1 I

k:_ 1.5

k= 1.25

k=l.O

8O

O. 01 O. 02 O. 03 O. 04 O. 054

(inches/inch) Cu

Fig. B4.5,6. 1-49 Minimum Plastic Bending Curves 19 9DL (AMS

5527) & 19-9DX (AMS 5539) Stainless Steel

B4.5. 5.2

Section B4.5

February 15,

Page 98

Low Carbon and Alloy Steels-Minimum Properties

1976

IZO

100

8O

Fb(ksl}

60

40

ZO

Room Temu_

Ultimate

1.0 Z.O

Z_cI


Symmetrical Sections_Carbon Steel AISI 1023-1025

v"

B4.5.5.2 Low Carbon and Alloy Steels-Minimum Properties

Section B4.5

February 15,

Page 99

1976

[._i.:i_ii[ I_i:-;Tti!:ITt:Tih_ :_7.:]i.-i_:_-iT:i:iliii_:i:i!i_ii:iiii"--f-.:.-_:t;::!....;...._o_i_m':,,_,_.r_,___,-H-t_-i ....."i!-.:'"_ ......:-t

16oI_:[i_ ifi:ll:....!/"l::l!:,:!-i--:r--_!-mi:-:.t---I-:--,--::_.....:-_- +d: ....I,_.:_-_:i:-F::t:"d:i_--i-_..i.-i.l::.-l.._.l_i.:d:._:..,?._,.:,51__'d: 'i

...... ,,• : !. : • ' ! ...... i .............. [''I ................. :""I"": ....... I ......... t .... i..,................' _....... .4 f .... -_ • L" .... "... : : " " L I ._' " :" "

.. :! : ]: , !. :. : 1-!: • i ....... :.......... !.... i" ..... ' : ':! [ ')_:._.._!_i.._i_.Z:i__:Z__i..iI..ii!i..:'.... .._/;:I2____..L_......._:::_.0,i:_.........._t._.,- __i.._._t ....:!"_ :i i........ ' i i : : "-"":" !....... _'- _ "':i_7- _:--: --_--

i--.:..-I.-_..:._!:i _it -.:t: t :.1...!-1:.._-.,__!..-i /:_......: .........i...._.' ..,...t.-..:...........`_....:._.__..:_::.:_:._:_:_:._.._:_._*_.___`_._._+_::.._:_:..........,_h :: _. i. _-....i".-':..!.......| i . .(ksi) ....!-" -_- F -!-,......."- :'-:

::::_ _--_-h_: ............60 [!}!iiiii!] "']""t" "_" "'i': .... ""'i. '.i,_""::..... :........I J i..;'"'i...........! .: _ . . , , _ .. :: ;'":':';, .!'_- ...... :- _-'. ...... _-- ....... ,.--4

_----_-_-_-.:1_..!..-1-_.-i__-.i]:...i-_-.I....:.:./..::.:,:...i....:.: .!

o "Z.J."2.Z:jI]_il......d-_i.....i:"12]',::__ddt:i"17:_k::-I! '_jt:.!Z:'j:] d, i 11 ... i I ' ::! i _-_: ......_. ::.::: " : _ -'--t.. -']_ .

: ....... . . :... -.'. i "'..'I:" .;:::.1. 1"_';':1 "1:: : : .: • : • : '. ' ' . .':. ",:: I ......:_:_:-,......_.-_-:_-::.......-.................I............:-....o, :IZI, __:,.........t _,_lr [I Td1.0 l.Z 1.4 1.6 1.8 g.O,

k = _ZQcI

Fig. B4:5.5. Z-Z Minimum Bending Modulus of Rupture Curves for

Symmetrical Sections AISI Alloy Steel, Normalized,> 0. 188 Thick

B4.5. 5.2 Low Carbon and Alloy Steels-Minimum Properties

Section B4.5

February 15,

Page lO0

1976

180 ...... "r- .... _ "'"" ";....... : '"[ " i ' i ......... _ ..... ': : i " i . _ I .. L ..... : .. : ...! ..._,.r.._.

, : ........I "" i "''; • "'" ! • I ' l' •• : : " ' ' • • • ' _ • • . --.._....; .... --4_-..... --............ _ , • • ' , , _ .... i..... i..... _ ....... ! i : ' • i • I _ " Uitimatc_;.-_+----..J_.. ' " ...... _ ...... "1 ............ ' " i160 , ..... .---..---_" -I- ..... .4-":" • . ; T ; _. ! ! ! ;,._ i.: ' _ ' _ ..._.:...'....t..._,i....L,jq__;...!.:._!- IJ'"':"'IRoom 'rempera(ure I • i : : . . _ ! ! • _ --i _A

......÷-----t-:'-r-_"-.............r.....1.......F-_.-i-"----;-9'.1. • _ _ I ..::.._..:..i ...I.. ,.._'-._P-:i

140 I"i'i":._':.']"'I.....i " '_ ._.L.i_..'.........[-_ 'i."-_-."I.................. |._._-..- . . • . [ : , ,1 : _ ' • , _ ., ._"_ .-i....-.'.-_...'-....... r":'"'l : i :" •...... I : _ i _ i: : ! i i i, JC_...__i.u-,-:..-.....-i--_ -_-.- i......:" .-_". ! : l .... -- " " : ' I. .. I , _ ! . j,,,r.__ ........ !... i!'_.....,:, ....... i .... , _-!,___. • . : . . I ;..........__..... ._; ......... -........ |.......'......,_o_--._-_._.......--__;......t........I....'._ .-_I_..L]_-_-I_'-= i'....;I

i "................-<i-..........<;--iI00 ..... _-"I-"'-...... : "_.., .. , , i:: :-:;-F:-;;;-:;;::7;• "'" " • : : [ . • ' I ..:...........-...... I....

F ; : .'. I • . • ' . : : , [ .:.....'...... --4......;......"-•-._. _.-.:_---*-.--.__,;-!;;_-!,_'*, .<-L.LL.,__ ......i

-___.._.._ .... _'.-L.:" """ ----. : _ '_ ....... 1--'--'. .... --I .... I ' " ' • • i80_t-=;_ _!.._;i-;:;<-;_:<::i-!.....!:i-:r-_-.!_I" _. ! ! : I ! • I : .i • " __ ._l ..... J.'.._---L ..... .4: . I" " _.......___ I . I -----t- ..... ?...... :'--': ........ ,.... . t ."'_I : • : I ! l i • i " : : [ • l,..!._.i: i:..!- -..I. _ i:-I"!i: .... I !: ! ..... I.... i: :, :.i _]. _/

_0,...._._,.-+--_---_-_--_-_--;--_-_t-_-:-,i;=S_--_......_--:-..;:_;;i_{.-.:-4-..i..".:.i.:......:':--I........I-:"I :! _'tu: ,_,' , I....: '!:::I

: i I ' I ". '' _L. "l| ...... L---I F*- - 75 0",n ._si "'._-- ....... -r--_...... • ...... --,- ...... s....... _....... 9F "_'| 1 "_" " ' ' ' _' " : : " ' ! " "

i . i ..i i l ! : • ' :....L:' Z9 xli:¢,_s_'.:-I.......!.........[:'':''l'" 'i:: r :' _........ :' ': .......... " " _'_i ' ] " :" '. .!: • _ , .J . i_.L i-......F_.,,,._,'i,__z_..',4....t---+---,

_0 "-r-r_i-:"T. .- _.--T- .... r...... l ." ' i" " _ . . i...... I...: .[..........i..._...!-,._-..I..I..'..!'--:!_!......;......l ....!......F- ....l" : i _ i : _ _ i:_-_Li=-i.--!.._ .....4--=--,-----_-:--4.....--i----e .'T-'N,":-'t-'_} ' ! '_.-"i-i,,"-.:' i : ! i : I l ! • : • _ : _ ". i .....: ...I.......i..J.-:

• . ._ .................... •.... ,, , |..... ] ...... : ..... . . t . : , ,• .. : I , • . , • ._____...;._L---t._-,o _a.._l__-I-_..-,-----b-.-_........----_-.-r • . • • • . ..i-_'.'-[:. : • J . ::1 1 :. ..:.....'.-.:.--.r--.i.---,-._--.-i.--._-:-i:l _I-_--I:i:-:-:.................,-.-,.-I......I- r ,_ ,,::, • . , : • ' • : : ! ": I " -L ..... •...... --L- ..... :'--4"-':'_•;-..._....+--_........ !.....G.---_--.-.T---.-----. !: t I- 1 • :

o i !.....I i: " "l:l .......J _......_'i _ ....! ' ;-.._-;;;-1.0 1.2 1.4 1.6 1.8 2.0

Zack= _

I

Fig. B4.5.5. Z-3 Minimum Bending Modulus of Rupture Curves for

Symmetrical Sections AISI Alloy Steel, Normalized,

< 0. 188 Thick

Section 84.5

February 15,

Page 101

1976

B4.5.5. Z Low Carbon and Alloy Steels-Minimum Properties

Fb(kii)

4O

0


Symmetrical Sections AISI Alloy Steel, Heat

Treated

Section B4.5

February 15,

Page 102

1976

B4.5.5. 2 Low Carbon and Alloy Steels-Minimum properties

280

240

200

F b

160

120

8O

4O

i .: i i : I , [ ...... ": ........ _........T........... 7-[ .......• _; I . ...... _ I i: _ :-1 ........ i :---; i

! • . . . .• ",.-.-..... I,........................"I_ • !I.............] 4 ' .........4.- -':........| ;,..........

• " T" ! '!...." II ...... . _"..... ! !" . .i ....I',' ": "'."I" "'!.....! .:F:--4........... ............,....... ...........,--4: _-:!Room Temperature : .._ ,..i. ; .I. ._ I .... I-':.--.I ...... I - .: ..... i

.... Jr.... I....__...' "/ : / ' : . I . / / : I:;. I. '---[.......1-.....-r:i .......;--r:-t ..... ;:--.-t .............

" 7_ :. i .... I !. "1 ' / ...... :i '". " ! : :'-! ..... :i........."."_-I .......!......-..:1 .....I""q- ......i....... --r"-i .... i-": ..... 4--....-i.: ..--_ I .... I } ...... I ..../.. ,..... I..i ......... _..... I........ :.... | i

; " ; | • , • I I I : . : I ' ,L ..... +-+."i ...... •....... 1-----.---[.-----_........ ' .......+....._ ............. _......... --L.__I ...... !' ! I ! . I I " ' : : : " I ' '

I..........Jl......,, . . ,, .........; . . . . ..4. " t_..... I," ": ......,........ ;.............I . .,1• ---T"7"".'""."-I--'"1 ....... , .... -T',...... {....... +--" ....... 1""'."....... :"Jl-":'"'l" ......_....... '- --I. . -I • _ : ; : ' ' • .! • '•+ I + . : ,.. ,.:+ _: .... :.... I ......I: .... iY,ehl"l-._

--:"I ..... l ...... I' ' ' ' ' ' ,,,,,;,.... _----="-'........j..._..,."_ii_IIIIIIIIIIIIIIIIIII_IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII_IIIIII_IIIIIIIII_iiii".-. :: _: :"i _..J'_iaIIIIIIIIII_IIIIIIIIIIIIIIIIIIIIImIII_IIIIIIIIIIIIIIIIIiIIi._j'"!i i ::........::i ........'FTi • I ': _ •, , ...... "!........ -"i- ---t"':'"T---- .':'"---!". "";--

,_•• i • I i •i• r: , •'•1 ........I " ! •_••_ I .........••1 :•••!•_ ':••:-I•:.........:........".....' 'I-= ......_.... +"-.......+....._-:: .....--:-* .........I - :--q'.........i- , [ _'-I:. I ;' !,I , • .I • I '': : +" .......................... _ : : ............. . ,:'" ....

: i .... +;...... , . • +:+", ""............. !-+r! ......... i......... !---1, •__.;.._1! !. • .r, i !.'"-; ..... :...,..__+_-...I......... i + t:.lq,r,...._...... i ......... ,._ _:-._.: _.....

• I i " i • / ......!-!:-I ..... !.... I...... !" i." ..... !.-.:....., ...... :..i-.-:..... 4- ...._.....-F--i---4..... _---.. ......4.... ' ' • _ ' " Jr-.....__Z.

•" -; ........I......_ "-: ........i--!-"-t ..........._,,.- 13z,::,oop.,,,----!--.... I • , ..I. , ..I. : i ..., ...i..: i.'' z+;<toc, psi _...!---.!.....• ! i : I I i i /.................. _........ : ........... .-g .... -4--- ............ "I " " "0_' -' '

' ' l = " r 1-- E'°nRat"_n-" "8"_:°

" !: ' " ! i ::. ' = +!i: .......:-1........iT: !.................. _.............. ?-....... ;......... :---i--....-...... !-........ !"";--+ ............. t ....... I-----.i--..,.;+ .....

: " : ' : • I ' _ |

• :+ ......... t ': ......... I '1 ! ...... + .... I " i ........ •'. .... _...................... - ..... L...... :..... I....... I __L ........ ,...... J. __.: ........ ! ............ .:1.0 1.2 1.4 1.6 1.8 z.i0

k = __2qcl

Fig. B4.5.5. Z-5 I%4inimum Bending Modulus of Rupture Curves for

Symmetrical Sections AISI Alloy Steel, HeatTreated

Sect ion B4.5

February 15,

Page 103

1976

B4. 5.5.2 Low Carbon and Alloy Steels-Minimum Properties

F b(ksi)

360

320

280

240

200

160

120

80

40


Symmetrical Sections AISI Alloy Steel, Heat

Treated

Section B4.5

February 15, 1976

Page 104

B4.5.5. Z Low Carbon and Alloy steels-Minimum PropertieB

4OO

32O

24O

160

8O

0

i ' Room Temperature ' _' i

tttttttttltItft14,ttlllllltlItltttt111tlltIIIitf!ttttt_ Iit_!!ttiftt_ttltlt[t tlttttlltllttttttltltttlt!_tf ill!i]

_i t_t_tttttttftttl[tlttflttltttl_ttltfI_iliJllftillt _t!tiii_i ittttfttttttl[tt_iIttttttltttt_tttttltltt!IiI1_it!1 iltli!

#i_i_t_ti11t_tt!)iltiitltt111Ittl_ti!lttt_t fltt_!._It_)t_ttfltttt_,....l'[t!ttttlllllltttl!tltI!t!iii!!l!ttl!t!f!_llf_ !;,!II!!" :-i_:"-:......... :...........;..... _.......__:;

_-k-! -_---.___,--." ' , •......_--_" i":....h: _

.r :-i ....:....I....7......l ....... i....... 1.........i .........i .........I....... u.....: ..

I--...i--..i ........ " .............. ; .... t' ....._h_ttltllltlittlllttlilk_.t_ ' '_ F. - zoo, ooo psi ' -_

_tttt lttttlltltlilliIl_li!llfllt_, _**:,,6,ooo_, _,,,_It_tt!tt)ttttt]ttt!tlftt!ll1II!t!/il!lllttlt!_. =_.o_1o6 psi

1.0 1.5 2.0

Fig, B4.5.5.2-7 Minimum Bending Modulus of Rupture Curves for

Symmetrical Sections AISI Alloy Steel, HeatTreated

B4. S.6. Z

70

6O

5O

4O

F b(ksi)

3O

20

I0

0

Fig. B4. 5.6.2-1

Sect ion B4.5

February 15,

Page 105

Low Carbon and Alloy _te_ls-Mtnii-durti i3ro_erttl_i

k=Z.O

k=l.70

k=1.5

k= 1.25

k=l.O

0. 002 0. 004 0. 006 0. 008 0. 010

c (inches/inch)

Minimum Plastic Bending Curves Carbon SteelAISI 1023-1025

i 976

Section B4.5

February 15,

Page I06

1976

B4.5.6. Z Low Carbon and Allo 7 Steels-Minimum Properties

lO0.

60

40.

ZO

44+PP_Room Temperature _ it _,

H I IHIII II ilI II!iN I!HI!III!I!!III!!!!!

f_ t_M_.,TitlIIIP_;_',..'_iLtjlLt-ttttl,, iiiiiiiiiiJi _u!!_l_ ]_;;,tlllll!W_';;iftttllttttittlltft_JtIi !];:ttJlftftttittttt !!!!

II__t-_flt!!!It_iii_itIltt!I!Jl_!Iiil_l_IIIi!!!liIIIiill_'..m,_,,

II_ .-._"'_"'"_'"""'i'"".. •....i........: , i__-'.....!l'..;,Iillli,'.'.:i_.nn.nIli,.ll,-Iiqllll..,,_""t'_.........I , _.",,".......

I, -'IUl.._lllllUllllllllllllll,lllll t " I I I ": ;dl'IjliilliilliililJllllW"" I I ":,,-.:...pp,,!,,,, : _ I I,I'AII !1;_ "' --'" ......... •.... lit I F "" ............... t ....... I ..........

.,,iuilI. , I i =. iI-- 't : I I " :

_'-:mtm ,_m,,,_tlt-_tm-_m,tlttliiiimiiiirt-_ tI !_ _tfiltitltm_-tttJ_'_ttttJitHi_II:::::::_,.=_,ooo_=_tt_lftftt- : IiilIiIJiJlJlilIiI";;";: Ili,Ilt _tv = ao, vuu.psz '_ittlill :r llTltlltltttll?!lttl

71-ii_iiiiii =-= _-<_x_o.,>._i__tii '_lfltttl!liltlttltfiiiiiit I!I:.+._:+_ !iiii'_+_!"!!!!' El°ntati°n = _Z_/oiiiiJiii I _ _ I t ' } ' 1_-+r'..+ ....... ,Hii.i4litlt,tlitHilltiiilfllttHlltltll]i :: II t.i,itil_tlhtitJ'_ tttiJI

tJ....liiihi l,Itfttlitl,,,it-ht,t.'ul_i 11t,it+_ti ttt 'il I _ tit,I!IIt_tttttfllt-tI1tt!ttt;;;;- i I i tfi tHt 'li tflitltl _ tttilllltlifttfttttf.•lttff tt,l,ld,_ddII,,t,,t,t_,_d_fittt'_:::: ,, ,,tili_,tlnltmtui,_iltlJ,tItH7

O. 04 O. 08 O. 12 O. 16 O. ZO

(inches/inch) e u

k=2. O

k=l.7

k 1.5

k=I.25

k=l.0

Fig. B4.5.6.2-2 Minimum Plastic Bending Curves Carbon Steel AISI1023-1025

Sect ion B4.5

February 15,

Page I07

1976

B4.5.6.2 Low Carbon and Alloy Steels-Minlmum Properties

140

120

100

8O

F b(ksl)

60

4O

_0

..... !1 .......

O. 002 O. 004

!iiilililhiill

O. 006 O. 008

¢ (inches/inch)

k_2.0

k=l.7

k=l.5

!

k= 1.25

k=l.0

Fig. B4.5.6.2-3 Minimum Plastic Bending Curves AISI Alloy Steel,

Normalized, > 0. 188 In. Thick


1976

B4.5.6.2 Low Carbon and Alloy Steels-Minlmum Properties

Fb(ksl)

150

100

,- .... "......................................... I.............. i',".... "_--,',"• ; . , . I¢,. --.- _,,_;• , • [ . ' . . : I • ." " ....... I

i...... i"'_ Ro:_m 're:rperaturei . : i ' • i [ i i :"

r.....-i-_ .....;............._ .....I......._......7........t-_ r_-i.........:. ......I.:.--i .... :-_............... ..... _ " ,"_! - 'i " ! " ! 'p....-.i..-.----...--....i.-:..-.,-...-.-........._.._---! ........i......i.........

• ' ! ! • • • • I i .. ! ............I _:_:,_J......._1 " _ ! : , ,-:.i_.,_ ....... i........... lk=,., .i--r" ,-r-:' _-_.- _-T, / , i-,-; :. :.....:._ .,,__ ,.; . ,, .. _ ! • . ......• " : " ! • I i I : • i " i i •

- i_'r-. ! " ! ! _ i ! ........

I /_ .: !.. !. .i.. k-,.zs"" ' " • J I [ I I i :

l" " " '. " I

' ;_,iilIIMIUlIIIIe"_................. :'_i n'. , -.......i...................- .....'......-t.......;........!..............;........; i , I I : , i I " I : • • i. - I t, ,

i ............. . ........ • , • " ..b p, - i . ,l.r ,l I :. i ! I : . S.r.;..,._.Str,"liil Cur-L: -_

'.: I:! ! i 'ii,,,!,,,,,;-""i""7i.... • ; i.---_.".'"'" ' ; ......... : :"l " i _ : . '.

........I_7: i,_::.;....::: ........, i.,:_ !'!:: :;-_i ...........:-i:-!...........---......ir_, .,, _-!

• , I " : : : ; ' • ' , ' iit__ i......'t__ - . .... I •

: ....... : ................ : ' "" I " ' " .... I

_ol_I-i-_ .........! ....... :----t........ .--._-t---" ..................................... •...... : ........... •

.[__-:: ...... i .- , .-- .: -' .... i ! '...... i;...... ":" i " i• i...... _.... I

; .i,. !.. , '" >' i I '" I "'i ........._.... ........,............... "-..... :,...... .4--;...:,- ............I.-

Ii

it ; . i : . i : . • i i : . i • J

it:,, ...,, :.,, :..,,. . ,, ...........................,.........:._,.......:.,.... _..... I ' :: :. i I i I _.. i.....,........,............... ..............,.....................,.......... -:, !i I,, !, I !' -' .....I'_ i !! '!n-"--I ' i .:._J.........,...-....l...:..-.........1.--...-I-.......-........t........................ ..... i-'t- .....!, ; ! : ! i . ', • I " "II--.i.- ! ........ !...... ! ..... I. t i... i ..: ..I ....... _;

." • ' : . : .' I I i ,li...-... , .... ....... : ... ............ ,....... ,.................. i........ ,0O. OZ O.04 O.06 O. 08 O. i0 O. IZ O.14 O. 16

E (inches/inch) 'u

Fig. B4.5.6.2-4 Minimum Plastic Bending Curves AISI Alloy Steel,Normalized, Thickness > 0. 188 In.

Section B4.5

February 15,

Page 109

1976

B4.5.6. _ Low Carbon and Alloy Steels-Minimum Properties

140

IZO

I00

F b

(ksl)

80

60

40

20

Room Temperature

O. 002 0.004

Fig.

(inches/inch)

B4.5.6.2-5 Minimum Plastic Bending Curves AISI Alloy Steel,

Normalized, Thickness_< 0. 188 In.

k=2. O

k=1.7

k= 1.5

k= l. Z5

k=l.O


Sect ion B4.5

February ]5,

Page ] I 0

1976

0 0 0_0 ,_

0

<

u__Vl

0

e_

_z

I

Section B4.5

February 15,

Page Ill

1976


24O

200

16o

120

Fb

(ksl)

8O

4O

O. OOZ O. 004 O. 006

I (Inches/inch)

k=2.0

k= 1.70

k=l.5

k=1.25

k= 1.0.

Fig. B4, 5.6.2-7 Minimum Plastic Bending Curves AISI Alloy Steel,

' Heat Treated

Section B4.5

February 15,

Page 112


!!i!

}iii

iIi:

;!I!

!!!!

i

I

0

0

r,,l

0

• °

II II

,0 0 0 0 0

,,.0 ¢,,.I (13 ',_

o

0

;,:1

o

"i_

I

IM

,d_44m

1976

Sect ion B4.5

February 15,Page I13

1976


B4.5.6.2 Low Carbon and Alloy Steels-Minlmum Properties, J ,,

Sect ion B4.5

February 15,

Page ll4

1976

320

280

240

200

F b

(ksi)

160

IZO

8O

4O

ill R _ _ ; i d_ .... _:_'...._'_ :I._i T_:T: .....-ff:-I-:¢ _ '1 I" I , d_i i_: :_ _:_ :::'_:.::•i'_;ooJ_T_p_r<,,r:...."-I_ " :' _ --_--'_.0"i_.i:_!-.:I.:....i.r__ I::-!-i:!:':'!!'"/'e_"'l .......... !",', -] _ t ?.... " ........ :!lit!it i]i!::ji i _

; il I !i i!!i_.._ k 1 25i

_' S "c_ain Curve l_--4

b

It_ tI_ :'.... . ,..I.:'_ ! ! ! i ....

___Fty =1:! :iii;: ! !_: !:l:: ::, ; :! ;: : 1: IF t = lJ2,O00psiBZ,000psi:i.!_ "I]::: IE = 2-9 x 10 6 psi

7;iir]i!iT_!--_<.,:i7! I::l:::i! :':;!::' :iiii::

O. 02 O. 04 O. 06 O. 08

,'ii i:]7![ iiiT--::i:i,,:...... ........._:::v:: ;Elongation := 18. 5%, ::t-i-: ::

ITT1777T;:: i:T;II71)"

O. 10 O. 12 O. 14

EU(inches/inch)

Fig. B4.5.6.2-10 Minimum Plastic Bending Curves AISI Alloy Steel,Heat Treated

Section B4.5

February 15,

Page II5

1976

B4.5.6.2. Low Carbon and Alloy Steels-Minlmum Properties

280

240

ZOO

160

F b(kst)

120

8O

40

...-:_..:-: , ......... i .................. ! ...._ .... . ...... • : . •

z_x '_i:":" 5-:_--- _'_-- ...._ ........ JF"

i_i: :_ :i:i :i : "i ':':[" t. ! : : ! zy ": ', _ i" i . •"'::-:'::i-÷!!i..... -":....:"'!:"':.......:......:":............::._.::.:i_i:iT:." ::-". :,. ::: ,_::1_.... i i_--_ .:: i

::.:.: I: ........... : IP ::. : . : • :• '" ".. " • ". • • ' ; ' _ D ' --....... _.. ,;

:::i::.:,l -F- T_-.:-lT.r:q-': '! l:! :., l l:- •

:_:_!:i:::[ l::! • ! -i_.:[ -_-'_ ,--'_,J _-I t, ! J Stress-Strain Curve :

_i'i [[::::iif!i !!:I-÷: !" ........ _!i_ii:! .:_:_ !::1: . ./_ :.... ..........l!i!!ii!tli!_.ii_.:_":'_?'""_':_'":-...,:__:,._i_if_!;ii.:_!!_':::....:..,._.....,:,........._......

l:i::::...i!::.:-_/.++_--"--t-. _........ , ...... i .iiFtu : 180,000

F'_!':_t! _:-H----t-/,.........._ _ JFtv: 163,000 psiPSiI'"':-i:,''F" i

......... :" +_-"'= 7" I'----T-" :-'" --T" .. :. " 6 iI:,_:,_A..._pl-r- . -.7...___. =,__,o ,,._,....._:. :i.i:_-,..-:-:[_.-..._,-,_..H-..t---t-_.,oo,_,,o_=,_o,o_.-._--i

i • i': :i : _ :...._[_ : " ' / , "" "...............z ........-::K_:::_:t:: .......__....:: : ":: ...."......:" _..:.J :::: , " L_.............l"i_J0. 00Z 0. 004 0. 006 0. 008 0. 010 0. 012 0. 014

k_Z.

k= 1.70

k= 1.50

k= 1.25

k=l.O

(inches/inch)

Fig. B4.5.6.2-II Minimum Plastic Bending Curves AISI Alloy Steel,

Heat Treated


1976

B4. 5.6. Z Low Carbon and Alloy Steels-Minimum Properties

360

_0 tIII

280 i:=

Z40

200

_b(k.{)

160

120

8O

4O

O. 08 O.I0

(Inches/tnch)

k=Z.O

k=l.7

k=l.5

k= 1.25

k=l,0

Fig. B4.5.6. Z-1Z Minimum Plastic Bending Curves AISI Alloy Steel,Heat Treated

B4.5.6. 2

Section B4.5

February 15,

Page l17

Low Carbon and Alloy Steels-Minimum Propertie_

1976

3ZO

280

240

200

(kF_)

160

120

80

40

0

i.i!"....'.:'::;:.. I I.T7_-:]--F-7-FI "'''I......... Room Temperature

Ftu = ZOO, 000 psi _'-

-'""...,.,........ Fty = 176, 000 psi .=

7_- _ _-,_9.0x 10Gpsi":i .... Elongation = 13. 5% _

".i;....

: : 1 1 } I 1 1 l

_ l?F!l_ ............... ] ....:.!....i_::..:...i._.....• i_

•,..:: i i':i.i'! "! !"

.... ..... i:i:i--:'-1 " !" F .......

:i:i!:_iI :....i- i i::r..... : [ -

i/'li! .:sl .L _..:_:._

7:17i_ _ii.,:i!i.... :;_ TIT T'!"i :i:11:- ....-__'- -i-:--;!-?ii_

.... ._r.-

0. 002 0. 004 0. O. 008 O. 0l{) O. 013

((inches/inch)

Fig. B4.5.6. Z-13 Minimum Plastic Bending Curves AISI Alloy Steel,Heat Treated

B4.5.6.2 Low CRrhon and Alloy Steels-Minlmum Propertlen

Section B4.5

February 15, 1976 _-"

Page 118

2_

Section B4.5

February 15,

Page 119

1976

B4.5.5. 3 Heat Resistant Alloys-Minimum Properties

_00

160

Fb(ksi)

IZO

8O

4O

01.0 1.5 2.0

Fig. B4.5.5.3-I

ZQck =

I


Symmetrical Sections A-286 Alloy, Heat Treated

Sect ion B4.5February 15,Page 120

1976

B4.5.5.3 Heat Resistant Alloys-M[nlmum Properties

F b(ksi)

Z40 :ttit_.._ :_[_: ,_[_::;i_R°°rnl;::-i ...... , i.iii,_:i_i_i_=_iii._Temperature_. .... ; _-_ i "

_;-- " '! .......... i " ! ...... _",'-- ....f-/__-_'- _ I i i . ! ! .':,_,_'I?L!

_*+_ It ........ •" , : • , •....zoo L_{_i _

-trTT_ _ _ ti i-AIJ-14_ $ t i

"_'_ II ....._:i!l_'_"r_l_llil'll!I l!li l..... li i Hliiili!iilIillliIIillllll IllllfIIlltlllI: ,Itll, lliLI2_ll, i ....

, " " '. ; I : ' :i "., • , • • • I :..: ..

lZO .......... :.... : ................. i " " : __.'- -.I ........... !

• " " '. _ I _ .. I

! ................ _..... _....... _................ ;.........i .......' i .... i , • I. • .I - •

', . . ; • i I • !8o_-_ ' _ '_ .... : ::; '., : ' '_

_: i i:!: !! i_i! 6

1.0 1.5 Z.0

'160

ZQck=_

I


Symmetrical Sections Age Hardened, K-Monel

Alloy Sheet

B4.5.5.3

Section B4.5

February 15,

Page 121

Heat Re ststant Alloys -Minimum Prope rtie s

! 976

1ZO

100

8O

6O

4O

ZO

Ultimate:

1.0 1.5

2Qc

I


Symmetrical Sections Monel Alloy-Cold Rolled,Annealed Sheet

2.0

Sect ion B4.5

February 15p 1976

Page 122



1976

B4, 5.6.3 Corrosion Resistant Metals-Minlmum Properties

F b

(ksl)

200

160

120

(inches/inch)

Fig. B4.5.6.3-1 Minimum Plastic Bending Curves A-Z86 Alloy,

Heat Treated


Graph to be furnished whenavailable


Graph to be furnished _en available


197

_4.5.6. 3 Corrosion Resistant Metals-Minimum Properties

240

200

160

IZO

8O

4O

I ..:: .... i ...... '

I ..... I....:.--

O. 01 o. 02 0. 03 0. 04 0. 05 0. 06 0. 07

c (inches/inch) E u

k=2.0

k:l.7

k:l,5

k: l.Z5

k=l.O

Fig. B4.5.6.3-4 Minimum Plastic Bending Curves for Age Hardened

K-Monel Alloy Sheet

Sect ion B4.5

February 15,

Page 127

1976

B4. 5.6. 3 Corrosion Resistant Metals-Minimua'n Properties

(kFs_)

60

50

Room Tempe rature

30

20 ._......

I

I0 i

0

0. 002 0. 004

..... , ....... j .. _".........

|

!

Ftu = 70, 000 psi

Fty = 28, 000 psi

E = 26 x 106 psi

Elongation = 35%

i!iiiii_ii

0. 006 0. 008

E (inches/inch)

k=Z.O

k=l.7

k=l.5

k= 1.25

k=l.O

Fig. B4.5.6.3-5 Minimum Plastic Bending Curves Monel Alloy

Cold Rolled, Annealed Sheet

_4.5.6.3Corrosion Resistant Metals-Minimum Properties

Section B4.5

February 15,Page 128

1976

Q 0 0

B4.5.5.4 Titanium-Minimum Propertle s

Sect ion B4.5

February 15, 1976

Page 129

F b

(k,i)

130

120

110

100

90 ¸

8O

7O

t't_

tt__4+H

ttt_4

t_trtH_H,tt-!

.H

Ji

./i

.ii

..+

_ttH

Itt

LH

iL

ii

H

.+.

.o

U

II1 1 1I : i l " i l• | I

........i.... .! ......._......I........._........! ,..;. _.. i!il]t!t_ _! !,ii!ttfl_[!!!]_], . ]J,tt!j, _ii_tt_h_!Ii!_Jlt

| , , ,

l;tliiti t t_

1.0

_ th,_̧ ,,_,,_IlJi_, iiittIll IIitittI!r.itt!i!!

i!! tli _t f ..,

!_!ii!ti_l'!ili II I !i!i!!;;' tt fiJ

1.5

_-,_ii̧!_,Ii!

. i_; _

2.0

ZQck-

I


Symmetrical Sections Commercially Pure AnnealedTitanium

B4.5.5.4 Tita_[um-Minimum Prope rties

Section B4.5

February 15,

Page 130

1976

Z30

ZlO

190

170

150

130

II0

1.0 1.5

2Qck-

2.0

Fig. B4.5.5.4-Z Minimum Bending Modulus of Rupture Curves for

Symmetrical Sections Ti-8Mn Titanium Alloy

B4.5.5.4 Titanium-Minimum Propertie s

Section B4.5

February 15,

Page 131

1976

F b(ksl)

240

220

200

180

160

140

120

fli!llttlttftttllli_lltli!f!tttt 'il

1.0 1.5 Z°O

2Qck =

I


Symmetrical Sections Ti-6A1-4V Titanium Alloy

B4°5.5.4 Titanium-Minimum Properties

Section B4.5

February 15,

Page 132

1976

Room Temperature

Z50

Ftu = 140,001

Fty = 130,000 psi

E = 15.5X I06.psi

Elongation = I2a/o

F b(ksi)

220

Ultimate

t,--. | I

' • i . "': ...... [........ I'-'" ....I ..........!...........I " I : , ' ' : , • I ; "

: '' ",' , : I" l "" I : I • "' , •t_._..... I. ; . ..;......... L...... I ! ":, : ! '

L..... --....................... •........ ,.. :...'.2 ............. • ........

...._........._, ..... _......__.._oi/ i! " _ i i ! ,_•

_1_=_...._" ................... - ......... i ..... , .. _J .............

1.0 1.5 2.0

20ck=_

I


Symmetrical Sections Ti-4Mn-4A1 Titanium Alloy

B4.5.6.4 TitanlumoMinim_/m Properties

Sect ion B4.5

February 15,

Page 133

1976

F b

(ksi)

160

140

lZO

100

8O

60

4O

ZO

! ,i!' i

! ":',.

: : ]

.... !: ...... _ ....... i

"'--T"--

Fty • •k=2.0

"1

_k=l.7

i

..... !. 2...i k = 1.5

_...__.i; : :k = 1.25

i.---i! } :• : '" ": ' "ik = 1.0

,!I-i-. .,: ..F..i.,. .d ..... , . ,.,

• Stress Strain Curve ......

...... i ...i i.

( (inches/'inch)

• Fig, B4.5.6,4-I Minimum Plastic Bending Curves for Ti-8Mn

Titanium Alloy

Section B4.5

February 15,

Page 136,

1976

B4.5.6. 4 Titanium-Minimum Properties

160

Fb(k.i)

1Z0

8O

4O

0.005

i i

=I.7 'I

, i I

I " " I k- 1.5 I

- --. --F---i ..... I---: i " '.... ! .... i • :" l _ l "[

: I I

[. ..1....... I:..j, k : i. 7-5

• i I I !........ . .................. |

Ila I.m,."--""l ' "1' • ' I [ " I-- ,

II .... l ' " " i- I .l I. " II . _! k : 1 "O' " :..... -_-- S_ress-Strain Curve ....

[ I FI"-T "'-'''''_ fo Curve

"i " i } .I ......... , • •' _ i .... b ...... _ 2 .... J .....

I f I :,. i

• t _ i i...........4 .....................r........b..........." "'_ "--I-'"'"

,.....ll .........i!: " ; ' : } I:} ,i I , ' I I 1 _ ..

..... I .......... _ .... li 1 1 1 1 1 I1 1 _ ......... _ ....

O. 01 0.015 O.02 O. 025 O.030 O.035 O.04

¢ (inches/inch) 'u

Fig. B4.5.6.4-Z Minimum Plastic Bending Curves Ti-8Mn Titanium

Alloy

Sect ion B4.5

February 15, 1976

Page 135

B4.5.6.4 T_tardurn-Mtnirnum Propertiee

b.'_._l'_ [H:[ F I,_L_L_.L_ ,_ i: :_ []_ :; :_ ,:_:I:::::_i:_' • II [ II _ * i ** *[" 'I" [ *_ I: ,.

___ _..... _...........................:.........i...............i.......:.........i.....

i [: ::._f. _._-_, 1 llti _i ii l !ll. I _ " " ll "

' "--41_ _4--_ _ .... " "V--; ...... _ ....... _ _ ..... : l "ll'_--!" "1_ _ l l_ .... _ .... 1........ i! _, i . _, : _, !_, / I i_ i • ! I

il I--_l _. . . _ ' _ "'! .... ,,---,_ ........ _-._--..i... %. • -- l •....... 7 - r ..... :---i _----_--- _ .... _ . • --t

'. . . . • : • ,._'l : : " : : I ; -- . ", : ;. • l • l :,'-,. "_, _i_ . _ • • _ • I' ' ,_. " I . l I • : : " : • I • I

i" :"" ! l/ .... "_ ! _ i :: : _ \ I _ _: : _ : IF " --" : " " "l " .... _ _ I_=_ e_P _ • ........ _--'_I .... ; . "............... ..... • , : "I i _ • .... _ • ..... i_'_ l I• .. . : ..

....... I -- _ _ _ ;" • i : " . • : _, _ _" _ "'_ " "l : I :_'':

_+,_1,° _ _ ,, Lh_,_, i_,,_,_,_i_,_,_, _, _1 _i_l,_,_l

J : ".... tel C_ I_ :_ _ _ , l : , • " ,i

:: i _':_ ._.l....... "= l I

0

'". ::I: :,...........t r.... ............... ".......................... I I I I I" i .... I ...........i !_71.....-ii I L.........: ....i _; _ ........!.....!....ill

o0 0 0

oO NN --_ _ _ _-_

od

oO

>

l

l

o

>

L)

m

i

_Q

&

B4. 5.6.4 Titanium-Minlmum Properties

Section B4.5

February 15,

Page 136

1976

.240

200

F b(kei)

160

120J

8O

4O

O. Ol O. 02 O. 03 O. 04 O. 05 O. 06

t (inches/inch) _u

Fig. B4.5.6.4-4 Minimum Plastic Bending Curves Ti-6A1-4VTitanium Alloy

Section B4.5

February 15, 1976

Page 137

Graph to be furnished when avialable

B4.5.6.4 Titanlum-Minimum Proportioo

Section B4.5

February 15,

Page 138

1976

280_=2.0

Z40

k=l.7

200

k=l.5

160

k= 1.25

k=l.0

120

8O

4OFty = 130, 000 psi15.5 x106 psi

0O. Ol O. 02 O. 03 O. 04

I (inches/inch)

Fig. B4.5.6.4-6 Minimum Plastic Bending Curves for Ti-4MnI_4AI

Titanium Alloy

Section B4.5

February 15, 1976

Page 139


_4.5.5.5 Alumlnum-Minlmum Propertle •

Section B4.5

February 15,

Page 140

1976

120

II0

150

90.

(kst)

80

70

60

50.

1.0

Fig. B4.5.5.5-2

1.5

k = __2QcI

2.0


Symmetrical Sections 2014-T6 Aluminum Alloy

Forgings, (Transverse) Thickness <4 In.

Sect ion B4.5

February 15,

Page 141

1976

B4.5, 5.5 Ahlmintlm-Minimum Propertiill

Fb(kst)

I .....tID_ttNttt!tttt Elongation : Iggo _ _1{

• : i ,

,........:...............,..._....,,,_...,.,....:,...... ./,i

i I80 ......... I ..... i ............_.'"'IT._ ........,.,.;. •,,,..i ...... . i

lffflTfttttltfftlttt,t,l,_ ttttt+t h lf-llfhlTitpltttittfliTi_IflIti!+71ti_ ir rt +t_ _ttftIHHIitt_t titlt:1i tltItlittll!iltltltttltt!llf_411}iI i!" +

i_tttlT: Ii-"i'i'_:"i_'+'+'"tt+ti'_i_t+_iiti<, !i 7t i_tltii!

+,. :,ii+6ol+i+l!lll,,+ll+ll,ll ++_l+,,Isl.tlilttllt+t_ll+ttti+t+lftlll:++tll+!li_tll:rlilltll+tllt+_+++tll+_+ .l- ' I+I llil_+: _! _iil.i + ! I+' :' i , _ - _++i,-+I+I+H++I+II++ll,++II++iI_.¥._,<,l+l++i_+_ _ +Ii+++++++_+++

_+i + ++_++iH+++++,++++i++++l++++i+++!i+!++!++i+ittlE_i H t+i ttl+l+tif_f+tf !t/ftltt..fii-I +Hl+f I+ ++f-i+Til[f:40 I: t! t[t'HriltlltH_+l _llli_llt+!tl-+ittI_ it Htlt-ir

1.0 1.5 2.0

ZQcI

Fig. B4.5.5.5-3 Minimum Bending Modulus of Rupture Curves'for

Symmetrical Sections Z0Z4-T3 Alloy Sheet & Plate-

Heat Treated. Thickness _ 0.250 In.

Section B4.5

February 15, 1976

Page 142

B4.5.5.5 Aluminum-lvtlnimu_n Properties

140

IZ0

100

Fb(ksi)

8O

6O

40,1.0 1.5 2.0

2Gokm'-'-

I

Fig. B4.5.5.5-4 Minimum Bending Modulus 0f Rupture Curves forSymmetrical Sections 2024-T3 _ T4 Aluminum

Alloy Sheet & Plate - Heat Treated. Thickness0.50 In.

B4.5.5.5 Aluminum-Minimum Properttes

Section B4.5

February 15,

Page 143

1976

110

I00

9O

F b(kit)

8O

70

6O

5O

40

1.0 1.5 2.0

2Qck=_

I


• Symmetrical Sections ZOZ4-T3 Aluminum Alloy

Clad Sheet & Plate - 1teat Treated. Thickness

0. 010 to 0.06Z In.

t

B4.5.5.5 Aluminum-Minimum Properties

Section B4.5

February 15,

Page 144

1976

120

1o0

8O

60"

4_

[ "! . i : :

| . . .

[ : : • . .

• i : :

.... |..-

, . . [

•..1... :...1..

i; : ; : Ii............i :.: _.:I:_: :__:___......i::::i !1.,.5

:1

k = ZQ....._cI

Fig. B4.5.5, 5-6 Minimum Bending Modulus of Rupture Curves for

Symmetrical Sections Z0E4-T4 Aluminum Alloy

C_ad Sheet & Plate - Heat Treated. Thickness

0.25 to 0.50 In.

B4.5.5.5 Aluminum-Minlmum Prope rtle s

Section B4.5

February 15,

Page 145

1976

F b(ksl)

II0

I00

9O

8O

70

6O

5O

1.0 1.5 2.0

Fig.

k ZQc=T

B4.5.5.5-7 Minimum Bending Modulus of Rupture Curves for

Symmetrical Sections ZOZ4-T6 Aluminum Alloy

Clad Sheet - Heat Treated & Aged. Thickness< 0. 064 In.

Alumlnum-Minlmum Prol_rtle m

Sect ion B4.5

February 15,

Page ]46

1976

IZO

110

100

9O

80.

7O

80

50

1.0 1.5 2.0

ZOck=-- l-

Fig. B4.5, 5, 5-8 Minimum Bending Modulus of Rupture Curves for

Symmetrical Sections 2DZ4-T81 A1uminim AlloyClad Sheet - Heat Treated, Cold Worked & AgedThickness < 0. 064 In.

B4.5.5.5 Aluminum- Minimum Prope rtie s

Sect ion B4.5

February 15,

Page 147

1976

90

80

70

F b(ksi)

60

5O

4O

Fig.

30

1.0 1.5 2.0

2Qck =

I

B4. 5.5.5-9 Minimum Bending Modulus of Rupture Curves for


Sheet - Heat Treated & Aged. Thickness >_0.020 In.

B4. 5.5.5 Aluminum-Minimum Properties

Sect ion B4.5

February 15,

Page 148

1976

150

140

130

IZO

110

9O

8O,

7o

..... i

i!-_ ............ ..... °

Room Tem, perature-4 ....... 4' : i I: l i °.....t

•!........ :"" :'"]

Fig. B4. 5. 5. 5-10 Minimum Bending Modulus of Rupture Curves forSymmetrical Sections 7075-T6 Aluminum Alloy

Bare Sheet 8_ Plate. Thickness <. 039 In.

F b(kat)

Section B4.5

February 15,

Page 149

B4.5.5. S Aluminum-Mtnimu_n Propertie=

130

1,0 1,.5

k = 2Qc

1"

illIl!!II!!!!!!i!!t,ttff,tt

,°,_,*°,

ttlttttlt

t_IHHII

t"t_it'

!IN!N!I}II

!!!!!!!!_

!!!!!!t!_

!!!!!!!!!!!!!!!!!i

tttlrHlt

Room

:]iiii_ii

_!!!!!!!!

Ill!!!!!

;ii_iiiii

:_!!!!!!!

[![!![!!

Z.O

Fig. B4.5.5.5-t1 Minimum Bending Modulus of Rupture Curves for


Clad Sheet & Plate. Thickness _< .039 In.

1976

Section B4.5

February 15,

Page 150

1976

B4.5. 5.5 Aluminum-Minimum Properties

140

120

I00

Fb

(ksi)

80

6O

Fig.

1.0 1.5 2.0

B4.5.5.5-12

2Qck=--

!

Minimnrn Bending Modulus of Rupture Curves for

Symmetrical Sections 7{)75-T6 Aluminum Alloy

Extrusions. Thickness < 0.25 in.


1976

B4.5.5.5 Aluminum-Minilnum Properties

Fb

(ksi)

140

120

i00

80

6O

1.0

i::: i_ll _:It i; , ii I ':j,,_r'':

_F, ',,_._ Ultimate _ _'., if::

.... _:'tit'. t::t

._ t_t; I::I

', :,, ,It, _ _;t:I:_iI_

i i! :,i ii }iil: l :i!::: _ :;HII: lll_

!: J_tl'.!fli_t'_ It:{

:i;! t1:itt_r;'ltt_'ltt f!ilt]/i!ti_Jt ltt_I I i; ......

i]i: i!t{I',It Room Temperature

;r_ " ; ii _ ;!i:i][ : :!1 _ ':1, t : I '{ ttlt :_ t:l

1.5

ZQck =--

I

;Yt_

2.0

Fig. B4.5.5.5-13 Minimum Bending Modulus of Rupture Curves

for Symmetrical Sections 7075-T6 Aluminum

Alloy Die Forgings. Thickness < 3 in.


1976

]_4. 5.5.5 Aluminum-Minimum PropertiesI i i

Fb

(ksi)

100

8O

6O

'1° 0

Fig. B4.5.5.5-14

1.5 2.0


Symmetrical Sections 7075-T6 Aluminum AlloyHand Forgings Area _ |6 In.

B4.5.5.5

140

Aluminum-Minimum Prope rtie s

Sect ion B4.5

February 15,

Page 153

1976

120

I00

Fb

(ksi)

8O

6O

4O

2O

0

1.0

Fig. B4.5.5.5-15

1.5 2.0

2Ock =--

I



Die Forgings (Transverse). Thickness <6.0 In.

Section B4°5February 15, 1976Page 154

B4.5.5.5 Aluminum-Minimum PrOperties

(ksi)

160

140

120

I00

8O

60

1.0 1.5 Z. 0

ZQck---

I

Fig. B4.5.5.5=16 Minimum Bending Modulus of Rupture Curves for


Die Forgings (Longitudinal) Thickness < 6.0 in.

Section B4.5

February 15,

Page ]55

i 976

B4.5.5. 5 Aluminum-Minimum Properties

F b(ksl)

1.20

110

100

90

8O

7O

6O

Z_cI

Fig. B4.5.5.5-17 Minimum Bending Modulus bf Rupture Curves for


Hand Forgings (Short Transverse) Thickness

<6.0 In.

Section B4.5

February 15,

Page 156

1976


m

(ks_)

130

II0

9O

7O

5O

tu

ty = 58, 000 psi10.3 x106 psi

)erature!

1.0 1.5 Z. 0

ZQck=--

I

Fig. B4.5.5.5-18 Minimum Bending Modulus o£ Rupture Curves for


Hand Forgings-(Long Transverse) Thickness <_6 in.


1976

B4.5.5.5

140

Aluminum - Minimum Properties

120

100

F b

(ksi)

8O

6O

4O

2O

0

1.0

Fig. B4.5.5.5-19

1.5 2.0

2Ock--_

I



Hand Forgings - (Longitudinal}. Thickness < 6 In.

Section B4.5

February 15,

Page 158

1976


100

8O

60

4O

2O

0

fi , ifllHi_ | _" I IIilU WIHHHII

O. OO_ O. 004 O. 006

(inches/inch)

O. 008 Oo 010

k=2.0

k=l.7

k=l.5

k= 1.25

k=l.0

Fig. B4.5.6.5-I Minimum Plastic Bending Curves 2014-T6

Aluminum Alloy Extrusions. Thickness _< .499 in.

Sect ion B4.5February 15, 1976Page 159

Graph te be furni,'_hed _11_enavailable

Section B4.5February 15,

Page 160

1976


100

8O

6O ty = 52, 000 psic 10.5 x 106 psi

4O

2On n nlq" _;!.hhi,WIHl

lltlgmlglll

Fig. B4.5.6.5-3 Minimum Plastic Bending Curves 2014-T6

Aluminum Alloy Die Forgings. Thickness < 4 in.

Section B4.5

February 15,

Page 161

1976


F b(ksl)

120

100

8O

6O

4O

ZO

:_::_ ]-:iil:.!!.,,'lllfllllllilfllllflllllllllilliimw',',,.".',"--"-t_!:_:l_i_:_i:: __!::_:'_]

.. ".; e -- ] : :'_ _ _ '_ '" !"l]: "" " --'.I ..... _.... ' .'. . - '. ' .'. . .-/_,: ::, :::::t:i:: i:_-_l:_::___---

tt:-__iiii! i!..i_7:i_:$;:_!_;ii:_i!St_e_s-_':ra_nCu_i _:• . [: :..; .... I ..... _,__.i.___ ....

I _ _: Ec =10.5X P ;;i!_ i ::!!i:::ill : i!': i: o!i --: ::!:: ! - 106 si !ii. !:,i: :::::1::::!:: ;;: _j]'..i !i :_i [ong&t[on = 10% _i-i-!i-ii:r-.li.:,:::i:_ !li--::_i: ilii: :-I:-ii-I :i_!-: :'t_:_:_: __.. :_:.t":: ' I, .it.i: Ii::.:: I : _:1.: , _:1r!!_::i i:!:[: '::i:i":lill ::1 !. • :i I.:[.! : ".."Ti': ;: _':: ;i;: i: I [ .i t 'i! !:

0. 01 0.02 0. 03 0. 04 0. 05 0. 06 0. 07

• . .: : : . " ., .......: ...:l:i.:J::::::'.i:::ii.i:: .::I.: .: " I .!!:.I :!.., : ..T.: :. :: • r-_-_Tr-[--..-- :

i!i_], Room Temperature,-, ...::':i":l::!:i:i_l::::;i_il:::i!i_:_l:::.:........, ..... :.i.i ::::!:i::ii_!:;i!li:ii• ',..::.:...... ]:::,.::.: .,: .... ::.. ..... ... : -.: :..... .....: :::::. 1,...... _ k= Z.O

-.-__i-:_,:___::_i.!_ :_-:::.:I:::_':.-, ..,_::.:_,_,-_,_:-_,_'_:._:: ::_ii_:-i _-..:,...::,.. ._ . .,::., .... _ ............. .i_.: ...._.: ii

.i.I :_:,:_.l:ii;i::.::_:ii_:::iiii!i:::_-::_i:.i_iii!i!!_!li::::!ili_.i_ l i:i:::_r:ii:! : i • :-_ k _.

k=l.5

k= I.Z5

k=l.0

(tnche s linch) e

Fig. B4.5.6. 5-4 Minimum Plastic Bending Curves 2014-T6

Aluminum Alloy Die Forgings. Thickness _< 4 in.

B4.5.6.5 Alumirlum- Minimum Properties

Sect ion B14.5

February 15,

Page 162

1976

8O

7O

6O

5O

4O

3O

2O

10

Fig. B4.5.6.5-5

0. 002 0, 004 O. 006 O. 008

(inches�inch)

k=2. O

k=l.7

k=l.5

k=1.25

k=l.O

O. 010

Minimum Plastic Bending Curves 2024-T3

Aluminum Alloy Sheet & Plate - Heat Treated.Thickness /_ 0. 250 Inches

Section B4.5

February 15,

Page 163

1976


un

o t,- u'_ N O

II II II II II

A

v

0

o

I

¢,4 4)

o_¢fl I

N

uu_m _

.! _

M

4

Section B4.5

February 15, 1976

Page 164



1976


120

io0

8O

Fb

(ksl)

60

4O

Z0

: : I : I

..... i .... 'i'" i "1 "'"" : [f!: ,,:!:i,-i i :l

-:.: -i...:...:.......'....•.....: i : : I

: :

. , : :

....._i,,;Ifilllilll!l!llllllniUn"_"':I "_ ','I t ' n I

• ,,: ,m !

: I, ........I....i....._......

": I "• , m ,• ,.......I.....i ii • : ......

: i : : I"lllll_--" ,, I

• :

: : .... i..... t": : I I

• :, : _ , , ,- ; .....

: : :

i 'f .... '' "'i" ".... L .......... i .............

i; k=2.0

II .... i

.....i........!• ]

,]k=l.7

.]' k=l.5

•-- e ........ J

i]

i

k= 1.Z5

i. iI 1 :

.... • .... "............... I

-" i : : I

; : :. k=l.0

: > (',,:I'',:' '':

I . . . ;l

: I

i.......

i : i : J ]'t', _"','"'"; !:"_: :• [ ....... i " " ' l.'t.. ;'_. C(;': :,',_ :

................. ,f ." in ." ,[ . ]" " :" I] : :: ill" ."..... .||

. .4.. ....... t .................. : ....... i ....... .. ...... t ...... t ........ } ...................

: I • • n • I : : I ! i . I '

i.... • ,.......I.....i........:.... I ....l......i........:.........0. 02 0. 04 0. 06 0. 08 0. i0 0. 12

E (inches/inch) E u

Fig. B4.5.6.5-8 Minimum Plastic Bending Curves for 2024-T3 & T4

Aluminum Alloy-Heat Treated-Sheet _ Plate.

Thickness < 0.50 Inches

Section B4.5

February 15,

Page 166

1976

]B4. 5.6. 5 Aluminum-Minimum Properties

7O

d_h

6o {ii_i_H_H_ IHil _:':

_0 _

40 !iI_

!Hi

ii!i_o i!ii

_!ii"_i_!_'_ '_'_zo if!!: ,,!!ii :

[! !t:

ii{!i_4_;',

I,_.i!

o :_i!ii___!i_O. 002

Fig. B4.5.6.5-9

k=Z.O

=1.7

k=l.5

k=1.25

k=l.O

!i!_ ir_i:_4 Ir !

0.004 O. 006 0.008 0.010

{inches/inch)


Aluminum Alloy Clad Sheet g_ Plate - Heat Treated.

Thickness 0. 010 to 0.06Z in.


1976

B4. 5.6. 5 Aluminum-Minimum Properties

0 0

I| II

.Q.,_

0

o,1 I_0 _

,..-t

L_ *._ .

U

•_ .• _ _

d

4

Section B4.5

February 15,

Page 168

1976__=

B4. 5. 6. 5 Ahminum-Minirnurn Properties

F b(ksi)

7O

60

5O

4O

30

2O

10

Fig.

k=2.0

k=l.7

k=l.5

k= 1.25

k=l.O

O. 010

Sect ion B4.5

February 15,

Page 169

1976

B4.5.6. 5 Aluminum-Minimum Pr6perties

120 k=2.0

100k=l.7

8O

6O

4O

k=l.5

k= 1.25

=i.0

2O

: • . ," :..

O. OZ O. 04 O. 06 O. 08 O. I0

E (Inches/inch)

Fig. B4. 5.6. 5-1Z Minimum Plastic Bending Curves 2024-T4

Aluminum Alloy Clad Sheet and Plate - Heat

Treated Thickness 0.25 to 0. 50 in.

Section B4,5

February 15,

Page 170

1976


80

70

60

50

%

4O

k-2.0I

• "i :::'": "-!:'"! : : _ ..... |

• " ; " " .,l'tY. =I.7

I':::i "

!._::]:._::i: ..... ii;/; ._ .... , ........: ........"" " ; • " J ' '. , : ::" .i • •

i'i. i''': ...... I ...... ' i"" ..... i/I :: ""....... ! / ; ' i "_: ..... :...:.:.. i :" "|!'"!.i.":':':'_

30 , .:.. " ". ...... • ...... • "_ "' . • • ..i

:.":i: i [/[[i_,, ,/' / :,, i,-_._', i,-:-_,_. : ...i/... • ':' i ' • '/i::!_::!::2]]

_-0 !:................. /p .......... ;, -....... , ........... /___-- .----.----;/ r / ,' /",7,::.'.,'.-- L- :....L

: : i _ I_' :_"-T!i:: :/_ ,: ,:/ '!.....i'-i...........r:-_l

,of..........,/i:....,,: i" =,_; :i:::_:i_i_::i_i_'............':_......': ........ : l ........ : ....... ?....... !' '" : ..... I':'".. "'_': ..... ."

:i/-_ i I: i ..........]!/" :'! • ". / I : _i--__:,l • , . * ; . : . ' . ' _ .: : :" ; ' ":: I.: .:.: :

0 ___ ; /...!: ; " . : l ........ /__]' "":" ; "'. .... I':':'"",':-:::.-:-:

O. 002 O. 004 O. 006 O. 008 O. 010

k=l.5

k=l.O

• (inches/inch)

Fig. B4.5.6.5-13 Minimum Plastic Bending Curves 2024-T6 Aluminum

Alloy Clad Sheet - Heat Treated & Aged. Thickness

< 0. 054 Inches

Section B4.5

February 15,

Page 171

1976

B4. 5.6. 5 Aluminum-Minimum Properties

F b

(ksi)

120

ii0

i00

9O

8O

7O

6O

50

40

30

20

I0

0

V*+++4+

iRoom Temperature

"U!!_

_titt

(tnches/_nch) 'U

k=Z.O

k=l.7

k=l.5

k= 1.25

k=l.0

Fig. B4. 5.6.5-14 Minimum Plastic Bending Curves Z024-T6

Aluminum Alloy Clad Sheet-Heat Treated and

Aged Thickness < 0. 064 in.

Section B4.5

February 15, 1976

Page 172



1976


120

I00

80

Fb

(kst)

60

40

ZO

Ftu = 62, 000 psi= 54, 000 psi

_ty 9. 5 x 106 psi

Elongation = 5%

e (inches/inch) e u

k=2.0

k=l.7

k=l.5

k= 1.28

k=l.0

Fig. B4. 5.6.5-16 Minimum Plastic Bending Curves 2024-T81

Aluminum Alloy Clad Sheet-Heat Treat, Cold

Worked and Aged Thickness < 0. 064 in.

B4.5.6.5 Alurninum-Minimum Properties

Sect ion B4.5

February 15,Page 17.4

1976

60

50

40

30

N

Stress-Strain Curve

k=2.0

k=l. 7

k= 1.5

k= 1.25

k=l.O

fo Curve

10 Room Temperature

0

Fig. B4.5.6.5-17

0.002 0.004 0.006 0.008 0.010

a (inches/inch)

Minimum Plastic Bending Curves 6061-T6 Aluminum

Alloy Sheet Heat Treated & Aged. Thickness >_ 0.0Z0 in.

Section B4.5

February 15,

Page 175

1976

B4.5.6. 5 Aluminum-Minimum Properties

/'.,' i ; i • • .; I ' I

i:.':, i....L.i..:...i, i...! .......i... '... i .J.... :: , . • .

i.........,.:..L._.......i.......-_i--i--.-l:............. t'

8o [.-.-...i...::..........:...', T "...... | ....

L-If ::!;:: :k2. i

60

i ....

5O

F b(ksi)

40 .....

I . . : i .• ..;... l... ].

: i

! i, :;, !

• '| ............. "--I- ....

.... r ...... r" • !I

!

I : I I

I

i

.....ii_ i

. !.i ....

! ""i"

•.! ...... I......... l

..... "......... [ .............. i....... i.. :...I ....

.... i ........ !....... i ;''"

I......; k=2.0

: i......

., .... i

.!

I

I i

! -.

: : I

..... {

..........i.......!.......

t

: i il i' 1....... -_!-_"! ......... t

.. /i'i_,i: ,St res s-Strain Cu¥_'_----_*......'

I i..........!......

i lfo Curve : : :

"-- i _!i_; . : " t-_.............i......i-..-i..._30] .... : " -' ';............. : ............ i.-.q .... :.......... l ..............

• I ! : ! ! , I ' I ; , Ii. ' ' • , ' I ''' _ • "I ...L .... ; ..... :. ": .............t . " ........ ' • I ,t , ; ,,, i.i _, .' , . ' J • t ...i .........i...............................,..................................,.......1 1....l /ii',..1..,,, ! i.........1..'. . : t......... • IIt_ .... /., :] .. I : .i. : .:i:l .'

: :, : : ! i!!'_l.!i i!!i '! i! !I ;:iii' i _:_ :ii :i !: _i! i!i !, ;ii : ii! i::

•i :: L_ E 9 9 xl0 6 si l'::;;:i _! ::!i!::i::,:il:!::ii :i i:i!i:!|. |: :. , I_-F- = • P ! i:ilii ]!'. I [!i 177|ii i!ii _: !I i i ]

' f "!1...... i'" ":" _:: E:],on.gation = [40/0 "'t' • "i ...... I ' 't ......1'" I J... I i ..I....... !lo I..-.t!--:---t.-_-!...!..+u-_--.:-..-+....l...+.........+-,-.!-:-!....,:.....:.1-:......!----:-.-.1--.-I.-!-+........l.--._-i........1

.tl.-.i .. _.._..,...._............I.. .... , ....,..., ....' _.i...'._.'0. 01 0. 0Z 0.03 0. 04 0. 05 0. 06 0. 07 0. 08

. (Inches/Inch) ' u

Fig. B4. 5.6. 5-18 Minimum Plastic Bending Curves 6061-T6

Aluminum Alloy Sheet - Heat Treated _ Aged

Thickness > 0. 020 in.

k= 1.7

k= 1.5

k= 1.25

k= 1.0

Sect ion B4.5February 15,Page ! 76

1976


100

8O

6O

Fb.(kst)

4O

2O

0

0 0.002

Fig. B4.5.6.5-19

0. 004 0.006

(inches/inch)

O. 008 O. 010


Aluminum Alloy Bare Sheet and Plate. Thickness<_. 039 in.

k=2.0

k=l.7

k=l.5

k= l.Z5

k=l.0

Section B4.5

February 15,

Page 177

1976

B4. 5. 6. 5 Aluminum-Minimum Properties

Fb

(ksi)

140

120

100

8O

6O

4O

2O

!i,t;

:t!

lil

4F+

pr.

_r

7

........... 71, i'ii'i; ?i

....... I111 Ii,;;

i ri• ii_ "• _:. i .t

..... Ii:

till!!il_i_i!_ii__iil_ii_'stress-Strain Curve

I ii ii!ii !_ _-:' .2. i ........... ! i!:i

._t_ "'71:7::1:::rf o Curve

' !i1 I till

...... ::2' 2,ii! :::

ii__i_ :ill ii! iii':iiili::_tu = 7_, 000 psi +i_iiii2i..i!2i

_tr = 65,000 psi siii!iiiiiiitii_L.

Elongation = 7% li!ii/! /*:!li]i;;,

i!,ii!1,I i ,t,,ti!,,i1!I1! l!f i!ltt llt,ll=ltItt_lil,i _ Ill ill il_i

0. 03 0. 04 0. 05

Fig. B4. 5.6.5-20

(inche s/inch)


Aluminum Alloy Bare Sheet & Plate Thickness

< . 039 in.

F-UT-7.... _,.

I

i ' i;

,I

±_. -i.....

, .,..,

'.i!.!2

;i:illlI

I [.2-

.l,,,t

'11

!:?:!t:..ill

Tr"

..... ;i

'i_:!llll

IfU

k=2.0

k=l.7

k=l,5

k= 1.25

k= 1.0


1976


I00

8O

6O

Fb

(kst)

4O

2O

0 0.002 0. 004 0. 006

, (inches/inch)

Oo 008 O. 010

k=2.0

k=l.7

k=l.5

k=I.25

k--l.0

Fig. B4. 5. 6. 5-21 Minimum Plastic Bending Curves 7075-T6

Aluminum Alloy Clad Sheet & Plate. Thickness0. 039 in.

Sect ion B4.5

February 15,

Page 179

1976


140

120 ....÷t+÷

iti:

i00 :t"

7i_i

f¢

80 _L_

F b ::; :I(ksi) !_

!rT!!tt*f I

?tltl

:fi:!l

20: _}I

0 _?t

Fig.

l! i ::1!

L2:i L_

!TF ,,;

:2"' 1!U, ,_

i [:

.... L_

_ttt t_

H÷t[2:t

t_ L H1LLL L;HH

E;Xl

:i.... r2

_:LLiL

+4 _4;rn _

_+_

1!!!

.H

:1[?

k.J

TY/:,

'r_f

44_-"tt'}-

0.01

(inches/inch)

E_

B4.5.6.5-ZZ Minimum Plastic Bending Curves 7075-T6 Aluminum

Ailoy Clad Sheet & Plate Thickness < 0.39 in.

k=2.0

k=l. 7

k=l. 5

k= 1.25

k=l. 0

Sect ion B4,5February 15,

Page 180

1976


I00

8O

60

Fb(kst)

4O

2O

.illlllllll

O. 002 0. 004 0.006

, (inches/inch)

fo Curve

O. 008

k=2.0

k=l. 7

k-l. 5

k-l. 25

k=l. 0

0.010


Aluminum Alloy Extrusions. Thickness

<0.25 in.


Section B4.5

February 15,

Page 181

1976

F b

(ksi)

Fig. B4.5.6.5-Z4

(inches/inch)

0.05


Aluminum Alloy Extrusions. Thickness

< 0. Z5 in.

(U

Section B4.5

February 15,

Page 182

1976_J



1976


Fb

I00

(ksi)

! (inches/inch) eu

k=2.0

k:l.7

k=1.5

k= I.Z5

k=l.O


Aluminum Alloy Die Forgings. Thickness

<gin.

Sect ion B4.5


Graph to be furnished T_hen available

Sect ion B4.5

February 15,

Page 185

1976


O I_" _ N O

II II II II II

I

i

O 0 0 0 0 0 0

N

o,-i

I

_ vI

U

m 0

(M

I

,d,44

Section B4.5

Februrary |5, 1976

Page 186


F b(ksl)

tO0

80

6O

;tre

40

2o ti

"II IIII I II" 'IJ .. !! I,.t ,,. I.....I..R..,lll..dl-" I |III "'" |l"il'".... il! , ;.M .II,m,....,....m,NUlil

O. 002 O. 004 O. 006 O. 008 O. 010

. {inchos/inch}

k=2.0

k=l.7

k=l.5

k= 1.25

k=l.O

Fig. B4. 5. 6. 5-29 Minimum Plastic Bending Curves 7079=T6

Aluminum Alloy Die Forgings. {Transverse)Thickness <__6.0 in.

Section B4.5

February 15, 1976

Page 187

Graph to be furnished _en available

Section B4.5

February 15,

Page 188

1976


100

8O

6O

4O

2O

WRoom Temperature

iiuqi. !!qqqlJ,::i. :i immnlUU,Hmnul.,||I,,IIU,I.I. ,lU .11.|11111 |111.11111 |11111111111111

O. 002 O. 004 O. 006

Stre s s- Strain

!if!f!

, (inches/inch)

k= 2.0

k= 1.7

k= 1.5

k= I. 25

k= 1.0


Aluminum Alloy Die Forgings (Longitudinal)

Thickness _< 6.0 in.

Sect ion B4.5

February 15,

Page 189

1976


140

120

I00

8O

Fb(ksi)

60

4O

2O

0

Fig.

;,- 4 '_. li]" :i ' .... :ii . i iif , .it _...... t ttt. ,1.... _ ,._ .... tJ

;z}; ;: tii;t ;;::tltl;|::/l Ill: ::;it:ill ;_i ]ii i ;. ; :_.I

,1 .: ; t i.,.' 't .,*e : 1• -_..... I, _.t I, i ,!+ _ ,+ ', I ,!

t ,1 _ ............ I ......

;2 ; _ ;i; .; [i; ] : ;1' _i_ 4 I '11 ;J ;....... t ..... t .............

-! _ ." 'J _r_r ,. +_. , .: i ......

, i_ J ......... t .... i .....

t_.,t]ii_f i!tlt,t'.)._fl i!tii! ! _ $t !t_!t_i_?tl;;llll;; 'Ll,_ff'_._._l_il ':i!1t},._]1,;!; ,t_t1;_i I,:, _,!i

.... ![ .i"_'" ,_" i .......I!jiL _1 ......... _ .... '. ._1,_ _..... i:':i .!: .... :.... i:.lii...:..i.:,,,_-,.. '_" , ::

;i; ;1 :.<:., .; ,._.... _. i _: . , _ . _ y . ,,I'_ !,,,:: ,:_11! ,: , :!_![ ! ',:!il _ l_j,:i!l,,.,.,'''II I .... _ ..........

i_i I i, +_ _

l;'_- ii_;_[ i ]l ; ,i _ i _ _ [;lIl/_llllill_[il[;: _$;: li;]

_j_ l}Jl;;!ili![:t I_Stress-strain Curve :t:i i _, _l'i " [ ....... I _

:;' ,l_i !_; [I; i ; [ i [ : i J ;_[ J] _ ;;I[ ' I;r.............. _,/, :, [ t .... f..... 1,,

fJ:'lt!_:lt!ti/t1!!/ :flt,!_!l 'ill #'1 ':'1'_ _urve ,l

i1;, , I[ ii ; I : ;;j I: ,_i; ;;]| ; ]ii O -

! 'l ;:].._,_,._-:d___:_:.,!_!!!l_lt;t It!i]44'l_'_l il_: !:::t

J ;; t ,. !i, i ;i: : T : [ ] ! i ] _ 't _N l [ _ S : rr : : ;: ': ;::II:" :I ];: :t I r: ::1 :: ::i

; J ..... ] ..... J]

;:::iil£4tu = 74,000 psi Ill !t i]!l,_iil:::T_._iiil_::!I! , !|!1!_1 t!ttt!:_l_!!t'!r!! :l!tl

,, _ , t ,tl _ , !,d,, Itr i

.... Elongat,on_ "too ,,,, f .... :1,,I,_,, ,_1,I

!t .... t ......... I.... t I _ I .......

_[l: ;],]]I,H':[],I ],', Ill! lli',l !l:ff!t!t]i! lJJJll]f':Ifi:i ,]_,t

0.01 0.02 0.03 0.04

(inches/inch)

B4.5.6.5-3Z Minimum Plastic Bending Curves 7079-T6

Aluminum Alloy Die Forgings (Longitudinal)

Thickness < 6.0 in.

ii:iii.

lti:Itt;

2z_

iiil

I;i:

tF,:

[ii!

0.

t:l;t,t,*,¢t'

!ill

:!i:

:?i;

!'1[

Lalt ,,

1t;

!lli

i , t t

,,1+

:Ii!

i,

,,t

ii:Ii,

.112

05

_w_.k = 2. 0

;:i!

.i? k = I.7

':It

ii;: k = 1. 5

i:11

'_i k =I. 25',;It

:[i;;,4,,

+_k= 1.0

':!t':

i::i I

;_1!!

ll!;1

(U

Sect ion B4.5



Sect ion B4.5February 15,Page 191

1976


!

..... ! "" II I

• I

l :

,. • : :

i -,- : :|: I :

)

: :. 1...:.-1 i , • • :• • I •

.........___._!...._ .....I

i:....... _.-" _- ......i

..... :.... i,.- -_ .... -....'. _. in

I :

I :

: ...... i

; I

i ,. )

" I .... in

__.._................i.......L !......L........i ......_ : L ......

,_'g

I

0 0

_0 rJ)v

_I_ "_

•_ o

ovm

!

U_

4

Sect ion B4.5

February 15,

Page 192

1976


F b

(kei)

100

8O

40

20

0

fo Curve

0. 002 0. 004 0. 006 0. 008 0. 010

(inches/inch)

kin2.0

k=l.7

k=l.5

k= 1.25

k=l.O

Fi s. B4. 5. 6. 5-35 Minimum Plastic Bending Curves 7079-T6

Aluminum Alloy Hand Forgings (Long Transverse)

Thickness < 6 in.


1976


k-2.0

k=l.7

• k= 1.5

Fb(ksi)

8O

6O .._..;.._.L.....:

k= I.Z5

k- 1.0

4O

2O

0.01 0.02 0.03 0.04 0.05

(incLes/inch)

Fig. B4.5.6.5-36

_U ¸


Aluminum Alloy Hand Forgings (Long Transverse)

Thickness < 6 in.

Section B4.5

February 15,

Page 195

1976


40

O. Ol 0.02

k=2.0

0.03 0.04 0.05 0.06 0. 07

(inches/inch) { u

Fig. B4.5.6.5-38 Minimum Plastic Bending Curves 7079-T6

Aluminum Alloy Hand Forgings (Longitudinal)

Thickness < 6.0 in.

Sect ion B4.5

February 15, 1976

Page 196



1976

B4.5.5.6 Magnesium-Minin_urn Properties

60

50

40

F b(kst)

30

20

10

$;;tl

iiii_[l:!

;:St_

t!22!

i[::i,t.,k.,

itli:!i,,

.t,r

!::T

!iil

_1Tr

1.0

ii

:IF

*;4

r!_t

ilt:,:i::i!

_._;

:!ii

'i;t,,t

;:7,

:IL:

t:;:

:ii!

, ,t

1.=l-

;!i:

iiil !!iii::i:i:!iti_: _!! !t!:::I!;!ili_:_i ;_:_!i

i!ij ii_!|ti!!!![:li[ii|iiii"_i_-iJl_oom Temperature

t_. *t " ._ ........... t,, ._1, !, t_f 4 , t

Ult_mate _ ...._I;_._,_ ,,r....

[!i,'t!il; }iii _ i 'Wl:',ii :: _i!i i.ll

_I_'I'_',_I:'_' ; i':'l"i_ r' "lI ;"_

..... I...... _ ......... _[,I,.! .... h.,

:::! .................... 1 t

..,.. ,,-,-d i : :

--_TT *_'YtM,.

.::: it::iili*

l,

::tl t:!l

.,._:it i]]i! f!_

i ......

[t t:!i _; iii

! :

;: i! i!i

! i! I! ;i !_I _t !t

', il

iiiil

_! ![ il

I 11!{i 'i, _1!

!! :t} itt_ ;i I t

[i :ii ii

....... . :!i!i

',t!: ;i:: .......

iI:_ !t}!liill :: t ::::i!i!tiiil :_ :i

.... it,

ii?itliii.............Itl: .... ; .......:. tl;l ii:_ ii, fill

, _,q+Si-_!!!:!:

:;'l!:Ir !ii" ii[iil:

!.i!i::_ Fry : 22, 000 psi::: E = 6. 3 x 106 psi

i i!:;][!i Elongation = 6_o

!iil ;ill i, ::! ::[._ i i i

....._4=_! !:[_,][',J_];II;:.:T :',ii:I11_i ''' , "ill[ I_I I_.H .ii..,'::' [[" I] !iri[:jl!l'l'llliit [:llil

1.5

ZQck =

I

2.0


Symmetrical Sections AZ61A Magnesium Alloy

Forgings (Longitudinal)

Section B4.5

February 15,

Page 198

1976


;: $77i

!!!!i!_

!If iii:

_i; !11;;!t ,..

!r lii

i_=,; ....

Itli:tti!:i!I !1ti

t_t!

H!t

]!I_:,ill;_ Iiii

!!H

*l+ ,++,

!!1 ![!':',hl ;:;i',i! ttfi;ii Iic]

tit Ill',t_: Itll

if:

1i

if

i[t

Z.O

Fi x . B4.5.5.6-Z Minimum Bending Modulus of Rupture for

Symmetrical Sections HK31A-O Magnesium

Alloy Sheet. 0.016 a Thickness <- 0.250 in.

Section B4.5

February 15, 1976

Page 199



Graph to,be furnished whenavailable


1976


5O

4O

30

2O

10

k=l.7

k=l.5

k= 1.25

k=l.O

Fig. B4.5.6.6-3 Minimum Plastic Bending Curves AZ61A

Magnesium Alloy Forgings (Longitudinal)

Sect ion B4.5

February 15,

Page 205

1976


Fb

(ksi)

6O

5o

40

3O

2o

10

:1. • : : ,. : . : ! ,..... .:: ......, :!....J .rI............... :_'.........:::i :, :[......, ::I".:'"'t'-'r-. '.'"_ ................... :---" ............. [ ............... ÷..... 1 ......... _........ ' ......... _....... _i

:" ] . . .I ] Room 'Fe:inperature I , I . • ".I

!: _ ,:1 1: ! , / : !: : _ I _ ..... l : ; : F_.;_l =2. oo[::_ " " ....r!--! .........i-t; .... .---,,_......- t...' -T-I' '::iFtu = '8' ()001'si I ...... i: ::! i ' 'i: ...... 1i .... I _:-I

!....iLrtv ---zz, 000 psi 1 L i I J J L..4 _ • :, .:,: : 6 "/'-_-7 .................. ' i -r--. " '.l"j£,i:..,,.:E. =_>.3xtO i>__. i...... ] ' ' .!..l. " _.::.ik= 1.7

7!!7- :

-, k=l.O

Fig. B4.5.6.6-4 Minimum Plastic Bending Curves AZ61A


Section B4.5

February 15,

Page 206

1976

B4.5.6.6 Masnesium-Minimurn Properties

0 o

v,

A U

T

!

Sect ion B4.5

February 15,

Page 207

1976


i.... /...

0 0 0 00 0 0

v

t.J

0

t_

_oAI

h U

_J

• "4 0

_o

,4D

I

_4

°1-1

Section B4.5

February 15, 1976

Page 208

_J


Section B4.5February 15, 1976

Page 210



1976



1976


:..-.2!-.:._i- -I-::-i--- " '_:i----".........i.. :...: ..L_i..__.:.i....... ,. i....:.:._ .L.._..:.. : .. '.._-." .... -.'..

I_.'. ; !. :.... _" • : :'. : '.'i • . • .. .'%. : . • " ! i .. i ._'_-.----T.--_...._ _ _ _ ....--.-:--'"-4......•......;.....:'_............."'""_" :......_..............'.....

••22 ° i i:i: i ; ; , - : ' :. . , . .._ _•_:: _: ........o o ,-.......: _L.: ; ....-:-.. "....:..! .......i..............-:............... "..-.:..............:

f i : : i " "'_ c' " ' ' ""':_.:.... -_'_...... 1, 1, u - .... " " : [. _ " ' "

• '" :' " :''_ • . .... ; .... ; .......... I • . ...... , •.' "_' _ lid _ ..: " I • " : • : " : " ' : ; " ' " • : : :

_,_ ...i ..... : ....... " .... ..;.L ..... " ............................. " ..... I ...................: : . : . ,. , ,:_." '-" _ _ ....... ..... :: ..... :.: .L.'..: ....... _..... :.: .." .... : : ........... : ...... _..." •

::_:F . .:"- _'- _--' _ ". : _ :.": !- : : [ " :" I; : : i • ' : • .''= .... ""--" _ .................. t"- :'_-r ..... . .: ,.'..._...._.....'..,." ........ 1... : .................... "1

li: ": " ' ..... : : " : ' : " :1........... [ . . ,' .,. ,_ ......... "._ .................................. I ..... t ................ • ...

• :! i : ' I : : "::: : _'': i .. • • : 1 • : , ...: • .,_,'" ......:"1" "'" : ............. :. I..... :................... :...... I.. .. '..-.., .b.. :- "":. ..... :1..... .-...... :............. !:'--:.!.........! .].:.... • "'_.'_'..-'_E_""

l- .! -- .:'---: ..l " . .:. ..... :.- : ...F.:.. ":!:..H....:.. i !':. !...'.! .... :... : ".:....'.. :...;... :.-_i.:. : ..!-.:-- }._l_ .-,-II. i ........... ' ":, : • • ' .... • : • ' : ";'." -' ""_-'11 .-'. I : , ! ] r ' ! ;' I • : " "" , ", • " • r . i . ": , .: ,'" "1 :_ : : , I :: , : - 1 •

o o _ _ ° o o_1_- q_ e4 ,--I

0

0

°

._---pI

4.10pI

0

i_4

4

I

Section B4.5

February 15,

Page 213

1976


F b(kai)

4O

2O

Fig. B4.5.6.6-1Z Minimum Plastic Bending Curves ZK60A


Sect ion B4.5

February ]5,

Page 214

1976

B4.5.7 Elastic-Plastic Energy Theory for Bending

B4. 5.7. I General

The Elastic-Plastic Energy Theory is defined as an extension

of the Elastic Energy Theory into the plastic range of a material.

This section will consider only energy due to bending stresses

which may be in the elastic or plastic range, or both. Plastic

bending curves found in Section B4. 5. 6 will be required. Deflections

of statically determinate structures due to bending with any or all

fiber stresses in the plastic range can be readily determined.

Partially or completely plastic statically indeterminate structures

can also be solved by procedures similar to those used in the

Elastic Energy Theory. Other elastic theories could have been

extended as well to include plastic bending effects but the Elastic

Energy Theory was chosen due to its simplicity and common usage.

Elastic theories are accurate only if no part of a structure is

stressed beyond the proportional limit of a material (no plastic

strain). In structures designed to stress levels beyond the propor-

tional limit, the error of an elastic deflection analysis is depen-

dent on the amount of plastic strain involved. In some cases this

error may be as much as 100% or more. Therefore when deflec-

tion is a limiting factor and plastic strains are involved, an

analysis such as the Elastic-Plastic Energy Theory should be used.

B4. 5.7. Z Discussion of Margin of Safety

In calculating the margin of safety at yield or ultimate, deflec-

tions must be considered as well as loads and/or stress levels, etc.;

e.g., although a positive margin of safety for a structural element

may be shown on the basis of loads, excessive deflections may occur.

If the deflections are then the most critical design condition, the

margin of safety becomes

Permissible LoadM.S. = -1 (4. 5.7.Z-I)

(Safety Factor) (Applied Load)

where the Permissible Load is the calculated load corresponding

to the maximum permissible deflection. Equation (4. 5.7. 5-11)

may be used in obtaining a permissible load level for a maximum

permissible deflection by a trial and error process.

.

4.

Section B4.5

February 15,

Page 215

Assumptions and Conditions

Energy is conserved; i.e., the external work due to a virtual

load moving through a real deflection is equal to the internal

strain energy developed during that deflection.

Plane sections remain plane; i.e., the strain is linearly dis-

tributed across any cross-section.

Poisson's ratio effects are negligible.

The deformations are of a magnitude so small as to not

materially affect the geometric relations of various parts of a

structure to one another.

1976

B4. 5. 7.4 Definitions

dA

c

A

C b

cross-sectional area of an infinitesimal volume, dV.

distance from the neutral axis to the extreme fibers of a

cross-section.

real deformation of an infinitesimal volume, dV, in the

x-direction.

real vertical deflection of a beam at the point of application

of a virtual load.

total (elastic plus plastic) strain of an infinitesimal volume,

dV, in the x-direction.

c - extreme fiber strain of a cross-section.bmax

Fv

m

We

W.

1

Q

fbv

virtual normal force acting on dA.

virtual bending moment in a beam due to the application of a

virtual load.

external work equal to a virtual load moving through a real

deflection.

internal strain energy equal to a summation of internal

virtual forces times their real deflections.

virtual unit load.

virtual bending stress on dA due to m.

Sect ion B4.5February 15,Page2_6

1976

B4.5.?. 5 Deflection of Statically Determinate Beams

Consider the infinitesimal volume dV of Figure B4. 5.7. 5-1(a) and

(b)

fb my (4.5 7_5-I)- I _vV

F v = stress >< area = fb dAv

= -- dA i4. 5.7. 5-2)I

6 = c b dx (4. 5.7. 5-3)

Since plane sections remain plane,

and

Cb = Cbma x c (4. 5.7.5-4)

5 = c Y----dx (4. 5.7.5-5)bma x c

By definition,

W = Q A (4.5. 7.5-6)e

W i =Y F 6v

Since energy is conserved,

(4.5.7.5-7)

W e = W i , (4.5. 7. 5-8)

or

QA = Z F v 6 (4.5. 7.5-9)

Substituting Equations (4.5.7.5-2) and (4.5.7.5-5) into Equation

(4.5.7.5-9) since Q is equal to unity,

Section B4.5

February 15,

Page 217

1976

B4. 5.7.5 Deflection of Statically Determinate Beams (Gont' d)

fAiL (mY_.T._ dA)(( maxb Yc dx)

But,

-/ Soby definition, I =

2 moby dA max

I C

y2 dA,

dx, (4. 5.1. 5-10)

I

L mob/e.

,',A :_ # max dx (4.5.7.5-Ii).So c

Equation (4. 5. 8. 5-ii) can now be solved graphically to findA.

c b can be determined from a plastic bending curve for the applicablemax

material as shown in Figure B4.5._.5-i(c). Enter the F b (modulus of

rupture) scale with Mc/I and move horizontally across to the plastic

bending curve for the specific cross-section; this intersection locates

the corresponding ¢b on the c (strain) scale. For beams with• . max

varylng cross-sectlon, c may be a variable.

(a)

III

(b)

Beam Cross-Section

.I:/ dz Fv

DifferentiaI Volume, dV

_, L "-

_X

(d) Real Load Diagram

(e) Real Moment Diagram

Figure B4.5.7.5-I

Section B4.5

February 15,

Page 218

1976

B4.5.7.5 Deflection of Statically Determinate Beams (Cont' d)

q I lb.!

(f) Virtual Load Diagram

(Real Deflection Shown)

M_.XI

%

l,

• (in./in. ) Cb

(c) Plastic Bending Diagram forBeam Cross-Section

(g) Virtual Moment Diagram

Figure B4.5. _. 5-1 (,Cont 'd)

Section B4.5

February 15,

Page 219

B4.5.-';7.6 Example problem

A rectangular beam, simply supported at the ends, is subjected

to a concentrated vertical load at its center. Find the vertical

deflection of the beam at its center.

Material: 2024-T3 Aluminum Alloy Plate

Beam dimensions: 1/Z in. x 1 in. x 20 in.

Load: P= 400 lb

Procedure:

,

PA I

i 2o -

Apply a virtual unit load, Q, at the beam center and construct

the virtua! moment diagram, m.

2. Construct the real moment diagram, M.

3. Calculate the real bending stress., F b, for each value of x by:

Mc

Fb = T

4. Enter the plastic bending curves on page Z18 (at k=l. 5 for a

rectangular cross-section) with each value of F b from Step 3

and determine c b for each value of x.max

5. Multiply the value of m by the corresponding value of Ebmax

to obtain'a value of m Cbmax for each vaiue of x.

1976

,

Section B4.5

February 15,

Page 220

! 976

Example problem (Gont' d)

Tabulate the results of the previous steps (see Table B4. 5.8.6-i).

Construct a plot of m _bmax vs. x and determine the areaunder the curve. This area represents _'g

(See Figure B4. 5.8. 6-I) J0 m Cbmaxdx.

8. Calculate A.

L mobf_A = J

max

J C0

dx Ref eq (4.5.7.5-11)

c = 0. 250 in.

L

f m c bmax

0

dx = 0.165 by graphicalintegration )

• 0. 165 0. 661 in.• • /% _

0. 250

By an elastic analysis, was found to be 0.6104 in.

which is 7.7% in error. Partially plastic fiber

stresses existed only over the middle 8 inches in

this example. The elastic analysis would be

considerable more in error for higher loadings and

/or beams in which the plastic stresses exist over

greater lengths; e.g., beams of constant moment,

etc.


B4. 5.7.6 Example problem (Cont' d)

P = 400 lb.

x, m M Mc/I, c bmax , m e bmax ,3 -3

in. in. -lb. in. -lb. ksi in. /in. x 10- in. -lb. x lO

1976

0

I

2

3

4

5

6

7

8

9

i0

iI

12

13

14

15

16

17

18

19

2O

0

.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

0

200

400

600

800

i000

1200

1400

1600

1800

2000

0

4.8

9.6

14.4

19.2

24.0

28.8

33.6

38.4

43.2

48.0

0

.4570

.9140

1.3710

1.8280

2. Z850

2.7420

3.14

3.80

4.64

5.83

4. 5

4.0

3.5

3.0

2.5

2.0

1.5

1.0

.5

0

1800

1600

1400

1200

i000

8OO

6OO

43.2 4

38.4 3

33.6 3

28.8 2

24.0 2

19.2 1

14.4 1

400

2OO

0

9.6

4.8

0

.64

.80

.14

.7420

.2850

.8280

.3710

.9140

.4570

0

0

.2285

.9140

2.0565

3.6560

5.7125

8.2260

I0.9900

15.2000

20.8800

29.1500

20.8800

15.2000

I0.9900

8.2260

5.7125

3.656

2.0565

.9140

.2285

0

Table B4. 5. 7. 6-i

B4.5.7.6 Example problem (Cont'd)

Ev

O

x

Xm

I

28

24

2O

_AREA UNDER CURVE EQUALS _

L

"--" J" me b dxm. 0 max

IIIi

k

I _iI

16

12

f

J

/¢

//

f

//

//I

A/I

10

x (inches)

Figure B4.5.7.6-1

\\

\\

Section B4.5

February 15, 1976

Page 222

\\

16

\

SECTION B4. 6

BEAMS UNDERAXIAL LOAD

TABLE OF CONTENTS

Pa ge

B4.6.0 Beams Under Axial Load ................. l

4.6. l Introduction ..................... l

4.6.2 Notation ........................ 3

4.6.3 In-Plane Response ................... 5

4,6,3.1 Elastic Analysis .................. 5

4,6.3,2 Inelastic Analysis ................. 5

4,6.4 Biaxial In-Plane Response ............... 9

4.6.4,1 Elastic Analysis .................. 9

4.6.4.2 Inelastic Analysis ................. 10

4.6.5 Combined Bending and Twisting Response ......... 10

4.6.5.1 Elastic Analysis .................. 12

4.6.5.2 Inelastic Analysis ................. 13

4.6.6 Recommended Practices .................. 14

4.6.6.1 In-Plane Response ................. 16

4.6.6.1,I Elastic Analysis ................ 16

4.6.6.1.2 Inelastic Analysis ............... 21

4.6.6.2 Biaxial In-Plane Response ............. 22

4.6.6.2.1 Elastic Analysis ................ 23


4.6.6.3 Combined Bending and Twisting Response ....... 24

4.6.6.3.1 Elastic Analysis ................ 25


References ........................... 42

B4.6-iii

LIST OF ILLUSTRATIONS

Tables B4.6,1

Tables B4.6.2

Tables B4.6.3

Beam Column Formulas -

Elastic Analysis Tensile Axial Loads .......

Beam Column Formulas -

Elastic Analysis - Compressive Axial Loads ....

Reference Listing for Other Beam Column

Loading Conditions ................

Page

29

32

40

B4.6-iv

f

Section B4.6

February 15, 1976

Page l

B4.6.0 BEAM-COLUMNS

L B4_.6. I Introduction

Beam-columns are structural members which are subjected simultan-

eously to axial and bending loads. The bending may arise from transverse

loads, couples applied at any point on the beam, surface shear loads, or

from end moments resulting from eccentricity of the axial load at one end

(or at both ends) of the member.

There are many problems associated with the analysis of beam-

columns. For example, individual beam and column effects cannot be

superimposed since they are interdependent. Initial curvature, distor-

tion, inelastic effects, and restraint conditions all affect the deform-

ational characteristics and are important factors. Both strength and

overall instability need to be considered. In cases where lateral support

is lacking, lateral instability (torsional instability) must be considered.

If the elements of the section are relatively thin, and the beam-column

is relatively short, local buckling or crippling may occur.

The analyals of shear-web beams is distinctly different from the

analysis of simple beams. As a consequence, the effects of axial loads

on shear-web beams are beyond the scope of this section; they are, how-

ever, considered in the section on shear beams.

It has been shown that beam-columns can have one of three types of

overall response (References I and 2). These are shown in Figure i as

indicated below:

(a) In-Plane Response

(a) IN-PLANE RESPONSE

ZI

Section B4.6

February 15,

Page 2

i 976v

(b) BIAXIAL IN-PLANE RESPONSE

i \ 1

1

(¢) COMBINED BENDING AND TWISTING RESPONSE

FIGURE I BEAM-COLUMN RESPONSES

Section B4.6

February 15, 1976

Pa ge 3

f-.!

/f-

F

(b) Biaxial In-Plane Response

(c) Combined Bending and Twisting Response

The particular response for a given member is dependent mainly

upon five conditions. These are:

(a) Cross Section Shape

(b) Span Length

(c) Amount of Intermediate Lateral (Torsional) Support

(d) Restraint Conditions at Boundaries

(e) Loading

The analysis procedures employed for these responses will be discussed

in the following paragraphs.

In general, the analysis and design procedures for beam-columns

with in-plane response are well developed for both the elastic and in-

elastic stress ranges. However, procedures for blaxial In-plane response

and combined bending and twisting response are limited, especially in the

inelastic stress range.

One of the more powerful techniques employed in the analysis of

beam-columns is referred to as "interaction equations;" these equations

are based on experimental data, and they require that the strength of

the member as a column and the strength as a beam be determined separately.

These strengths are then expressed in terms of stress ratios (applied

load/strength) and incorporated in the interaction equation which ex-

presses the effects of the combined loading. Interaction equations will

be used in the following discussion.

B4.6.2 Notation

A

r

d

E

E t

f

fcb

G

h

I

IY

K

J

L

L'

M

x,M Y

Mac tual

Me,

(Ky)e

(Mx)u,(Sylu

Sect ion B4.6

February 15, 1976

Page 4

Area of Cross Section (in. 2)

Torsion Warping Constant (in. 6)

Eccentricity (in.)

Modulus of Elasticity (Ibs/in. 2)

Tangent Modulus of Elasticity (Ibs/in. 2)

Stress (ibs/in. 2)

Nominal Extreme Fiber Stress at Lateral

Buckling (Ibs/in. 2)

Elastic Shear Modulus (Ibs/in. 2)

Distance between Centers of Flanges (in.)

Moment of Inertia (in. 4)

Moment of Inertia about Minor Axis (in. 4)

UnlformTorsion Constant (In. 4)

El (see Tables I 2, and 3)p

Length of Member (in.)

Effective Length (in.)

Bending Moment (in.-Ib)

Principal Axes Bending Moments (in.-ib)

Bending Moment due to Both Bending and Axial

Loads (in.-Ib)

Bending Moment at the Elastic Stress Limit or

Yield Point (in.-Ib)

Bending Moment at the Inelastic Stress Limit

(in.-Ib)


MI_ M2

Meq

MoS,

P

Pe

(Px)e ' (Py)e

Pu

U

W

w

Y

Zx, Zy

8

Applied Moments at Each End of Beam-Column

Loaded with Unequal End Moments (in.-Ib)

Equivalent Moment for Beam-Column Loaded

with Unequal End Moments (in.-Ib)

Margin of Safety

Applied Axial Load (Ibs)

Critical Column Buckling Load in the Elastic

Stress Range (Euler Load);

Pe = (_2Elmin)/(L')2

Critical Column Buckling Loads in the

Elastic Stress Range for Each Principal

Axis (ibs)

Critical Column Buckling Load in the Inelastic

Stress Range (Ibs)

L/j (see Tables i, 2, and 3)

Transverse Load (Ib)

Transverse Unit Load (Ib per linear in.)

Deflection (in.)

Section Moduli about Principal Axes (in. 3)

Slope of Beam (radians) to Horizontal, Positive

when Upward to the Right

Section B4.6

February 15, 1976

Pa ge 6

B4.6.3 In-Plane Response

In-plane response, as mentioned herein, refers to beam-columns

which, when subjected to combined axial and bending loads acting in one

plane, respond substantially without twisting in the same plane, as shown

in Figure la. This response usually occurs when adequate lateral (tor-

sional) support is provided, or when torsionally stiff sections are used.

Beam-columns which respond in this manner have been investigated exten-

sively for both the elastic stress range and the inelastic stress range;

these are discussed separately below.

B4.6.3.1 Elastic Analysis

The elastic analysis for the strength of beam-columns is based on

the assumption that failure occurs when the computed value of the com-

bined axial and bending stress in the most highly stressed fiber reaches

the yield point or yield strength of the material. This definition, in

a sense, does not consider the danger of buckling; as such, it is not

correct in the limiting case of a pure column. Initial imperfections are

always present, and contribute to the response. Thus, if these imper-

fections are considered, the definition above remains valid. Further

discussion of this concept is given by Massonnet (Ref. 2).

84.6.3.2 Inelastic Analysis

The maximum bending strength of a beam actually is higher than the

hypothetical elastic limit strength. As the applied bending moment in-

creases above the yield moment, yielding penetrates into the cross section

as shown in Figure 2. By comparison of Figures 2b and 2d, it can be seen

f

f

Z

3Q.

a.J

>-

Z

o.

a.J,,i

>-

i,iZm

Z0

I-Z ,,,

Q.J,,i

>-

<,,i

X

.J

rr

wZ

3v

Section B4.6

February 15, 1976

Page 7

"r

_DZwrr

oO

uJ>ro-w

w

rr

e_

uJo-

U-

Z

Ii1ooA

Sect ion B4.6

February 15, 1976

Page 8

that a greater strength i_ obtained by considering the inelastic stress -

distribution. Analysis procedures which utilize this extra, or reserve,

strength are known as inelastic analyses; this extra strength has been

shown to vary, and to depend on three factors (Ref. 2).

(a) The Mmax/P Ratio - The reserve strength is almost zero for

the case of pure buckling (M = 0) and tends, as Mmax/P

increases, toward the value that is associated with pure

bending.

(b) The Shape of the Cross Section - For example, the reserve

strength is smaller for an I-section than it is for a

rectangular section.

(c) The Nature of the Metal - For example, the reserve strength

is much higher for material A than it is for material B in

Figure 3. (The stress-strain diagram for material A does not

have a flat portion, as does the diagram for material B).

Several failure criteria have been used in the inelastic analysis

for the strength of beam-colunms (Refs. 3, 4, 5, and 6). Basically, the

criterion used has been one of the following:

(a) Maximum Stress Criterion - Failure occurs at some prescribed

maximum stress level in the inelastic range.

(b) Maximum Strain Criterion - Failure occurs at some prescribed

maximum strain level in the inelastic range.

(c) Ultimate Load Criterion - Failure occurs at some ultimate, or

collapse, load which utilizes the inelastic behavior of the

material.


Page 9

F ¸

.J

° \

n

w

uJ

UJ

0O

I

m-

LU

I--

Z

DC

I--CO

I.I.I

nr"

I.l.Irr

£9II.I_

Section B4.6

February 15, 1976

Page I0

B4.6.4 Biaxial In-Plane Response

This section considers the strength analysis methods of torsionally

stiff beam-columns which are subjected to applied bending forces that

cause either bending about the major principal axis only or bending

about both principal axes (Fig. ib). The beam-columns are free to deflect

in all directions, without twisting. This type of response can occur as

a result of one of the following two conditions:

(a) Primary Biaxial Bending

(b) Primary Bending in the Strong Direction

Case I is discussed in the following paragraphs.

Little information is available for Case 2. However, it has been

predicted that short members with large bending forces will respond

essentially in the plane of the applied forces and develop as much

strength as if they were restrained from deflecting in the weak direction.

On the other hand, long slender members with small bending forces buckle

in a direction normal to the plane of the bending forces and develop

essentially the strength of the member loaded concentrically, that is,

the bending forces cause no loss of strength. Intermediate length mem-

bers may respond about both principal axes simultaneously, and the

strength will then be less than the strength for responding exclusively

in one direction or the other.


In the elastic range, when the limiting stress failure criterion is

r

f

Section B4.6

February 15, 1976

Page ll

used, the determination of the stresses due to bending about the two

principal axes can be made independently, as there is no coupling of the

flexural actions. Thus, although few test data are available, practical

procedures for the prediction of strength for such cases are based on a

knowledge of in-plane response. Hence, elastic solutions for the case

of In-plane response can be extended to include biaxial in-plane response;

but it has been found (Ref. i) that the limiting stress criterion is even

more conservative for cases of response about both axes than it is for

cases of response about one axis. Austin (Ref. i) extended one form of the

interaction equation for in-plane response to include biaxlal in-plane

response.

B4.6.4.2 Inelastic Analysis

Austin (I) states that there have been no precise theoretical studies

based on the von K_rman theory of the strength of beam-columns which fail

by bending about both principal axes without twisting. From a practical

viewpoint, Austin has extended one form of the interaction equation for

in-plane response to biaxial in-plane response for the inelastic range.

Little test information is available for this phenomenon. In general,

it appears that methods used to predict the inelastic in-plane response,

can be extended to predict the inelastic biaxial in-plane response.

B4.6.5 Combined Bendin_ and Twistln_ Respons _

Structural members of thin-walled open cross section, when subjected

to combined axial and bending forces, may respond by combined bending and

twisting (Figure Ic). This phenomenon is commonly called torsional-

Section B4.6

February 15, 1976

Page 12

flexural buckling or buckling by torsion-bending. The twisting action is

a result of the low torslonal stiffness of members with open cross section

such as I, channel or angle. It should be noted that in the discussion

which follows it Is assumed that the open section members are not subjected

to dlrectly applled torslonal couples, such as arise when the llne of ac-

tion of transverse loading does not pass through the shear center (Figure

4). The shear center Is defined as the point through which the shear

force must pass if the member is to bend without twisting.

CENTROIO SHEAR CENTER

FIGURE 4. SHEAR CENTER

Columns of open cross section wlll respond by combined bending and

twisting under any of the following three conditions:

(a) Axial compression and moments acting to cause bending about

both pTincipal axes;

(b) Axial compression and moments acting to cause bending only


about the major principal axis when the momentof inertia

about the major axis is muchgreater than about the minor

axis; for example, an l-section memberwith axial compression

and end momentsacting to cause bending in the plane of the

web; and

(c) Axial compression and momentsacting to cause bending in a

plane parallel to the plane of either principal axis, whenthe

principal axis does not contain the shear center as well as the

centroid, as may occur for a channel, tee, or an angle section.

Little information is availn_le on the strength and behavior of mem-

bers subjected to the first and third conditions (Ref. i). The greatest amount

of study has been devoted to the second case, primarily for 1-section members

(Refs. 4, 6, 7, 8, 9, 10,and ii). The second case is discussed in the follow-

ing paragraphs.


As mentioned above, most of the elastic analyses for beam-columns

subjected to combinedbending and twisting have been limited to uniform

I-section members. It has been found that the behavior is similar to

the lateral buckling action of an 1-beamsubjected to transverse forces

only. However, exact formulas for critical loads for torsional-flexural

buckling are complex (Refs. 4, 6, and Ii), therefore, interaction equations

have been used. These interaction equations have been shownto agree

closely with available data. An interaction equation has been proposed

by Hill, Hartman, and Clark (Ref. i0) for aluminumbeam-columns; this equation

has been verified by Massonnet (Ref. 2) for steel beamcolumns.

Section B4.6

February 15, 1976

Page 14

In a theoretical study, Salvadorl (Ref. 12) found that the interaction

equation used by Hill, Hartman, and Clark gave safe predictions for the

combinations of axial compression and bending which produce elastic

buckling by torsion-bending. Salvadori considered members whose ends

were free to rotate in the plane of the web, but were elastically restrain-

ed with respect to rotation in the planes of the flanges.

Little analytical work has been done on the elastic response of

tapered members under combined bending and axial load. However, Butler

and Anderson (Ref. 13) have performed tests on tapered steel beam-columns,

and compared the test results to Salvadori's interaction curves.

The outcome of the comparison suggests that Salvadori's curves can be

applied to tapered as well as uniform beam-columns. Also, analytical

studies of "solid" tapered members have been made by Gatewood _Ref. 14),

and the interaction curves obtained are essentially independent of the

degree of taper and are closely approximated by the results of Salvadorl.

B4.6.5.2 Inelastic Analysis

Again, the majority of studies in the inelastic range have been

devoted to 1-sectlon members. Hill and Clark (Ref. 9) have shown that the

interaction equation used in elastic analysis can also be extended for

use in the inelastic range by using the tangent modulus concept.

Massonet has also extended the elastic interaction equation for steel

into the plastic domain. The results of this extension were compared

with results of their tests of steel 1-section columns and found to be

f


in excellent agreement for oblique eccentric loading, momentat one end

only, and equal and opposite end moments(Ref. I). Galambos(Ref. 15) presents

a thorough study of the inelastic lateral buckling of beamswhich maybe

applied in the interaction equation.

B4.6.6 Recommended Practices

Detailed and comprehensive stress analysis shall be performed to

ensure efficiency and integrity of the member or structure; and the design

shall comply with the particular structure or vehicle requirements, e.g.,

reliability. In general, the methods of analysis shall follow those

given herein. In utilizing these methods, minimum weight designs shall

be given prime consideration and the member shall be so designed that

(a) There shall occur no instabilities resulting in collapse of

the member from the application of the design loads; and

(b) Deformations resulting from limit loads shall not be so

large as to impair the function of the member or nearby

components, or so large as to produce undesirable changes in

the loading distribution.

The immediate problem in the design of a beam-column is the choice

of a suitable cross section to withstand the combined axial and bending

loads. Because of the number of variables, direct choice of section is

not usually feasible, except for the selection of a shape, such as round,

rectangular, I-section, etc. Thus, in general, successive trials must

be made to determine the safest and most economical section. The pre-

liminary selection of the cross section at any station along the member


may, in manycases, be based on the elementary formula for Stress

Zx _y '

where the interaction of the transverse and axial loads is neglected.

Naturally, this selection must be improved by a refined analysis.

In choosing an optimum shape, particular attention should be given

to the degree of lateral restraint or end restraint which will or can

be provided. For example, if little or no lateral restraint is provided,

then a torsionally stiff section such as a box or tubular section should

be used. Also, when the type of response is not known, all three con-

ditions

(a) In-plane response

(b) Biaxlal In-plane response

(c) Combined bending and twisting response

should be analyzed to determine the critical response.

If the analyst or designer has the choice of elastic or inelastic

analysis procedures, the following factors should be considered in making

the choice.

(a) Member function

(b) Material

(c) Deflection limitations - Deflections may be excessive in the

inelastic range

(d) Thermal conditions - Little information is available on

thermal effects in the inelastic range


(e) Dynamicconditions - Little information is available for in-

elastic effects due to dynamic loadings

(f) Reliability

(g) Analysis procedures available for the method of analysis and

types of loadings considered.

The recommendedpractices and procedures for the analysis of beam-

columns are discussed in the paragraphs which follow. The analysis and

design for local buckling and crippling should be in accordance with the

methods presented in Section CI. In complying with the references and

recommendationscited above, the designer should keep abreast of current

structural research and development. With this approach, optimummethods

and designs should be achieved.

B4.6.6.1 In-Plane Response

B4.6.6.1.I Elastic Analysis

Tensile Axial Loads - The strength of beam-columns subjected to

combined flexure and tensile axial loads has been investigated for many

commonly encountered cases. In Tables B4.6.1 and B4.6.3 results are

tabulated for some of these cases. In general, adequate solutions and

methods of analysis for other beam-column loading conditions are avail-

able in several references (16, 17, 18, 19 and 20).

Compressive Axial Loads - Many exact solutions have been ob-

tained for the case of beam-columns subjected to combined flexure and

compressive axial loads. Results are given in Tables 2 and 3 for some

of the more commonly encountered conditions. A number of these and other

conditions are discussed below.


Beamcolumns with intermediate supports have been investigated.

Probably the three momentequations (Refs. II, 18, 20,and 21) are the most

powerful method of analysis for this case. A variety of conditions is

covered in this technique; for example:

(a) Any type of transverse loads can be included

(b) Span lengths mayvary

(c) The moment of inertia may vary from span to span

(d) The effects of rigid supports not in a straight llne

(e) The effects of intermediate spring supports (Ref. II p. 23)

(f) The effects of uniformly distributed axial loads (Ref. 21).

Niles and Newell (Ref. 18) present tabulated results for many of the cases

listed above. Another analytical method which can be used for this

problem consists of a solution adapted to matrix form. Saunders (Refs. 22,

23) demonstrates the application of the "transfer matrix" technique to

the analysis of nonuniform beam-columns on multl-supports.

Methods and solutions for beam-columns on an elastic foundation can

be found in several references (6, 17, 24,end 25). Het_nyi (Ref. 17) is a

particularly good source for formulation and solution of the differential

equations for this problem. He considers both axial tensile and axial

compressive loads. See Table B.4.6.3 for illustrations of some of these cases.

For the cases of tapered and stepped members, numerical methods have

been shown to give good results with a savings in time and labor. Newmark's

method (Ref. 26) is particularly applicable to beam-columns of variable

cross section and can be extended to include many commonly encountered


Page 19

loading conditions. Basically, the method is a numerical integration by

a sequence of successive approximations. Salvadori, Baron (Ref. 27) define

finite difference numerical methods which can be used for the analysis

of beam-columns with many loading conditions.

Other conditions not covered in Tables B.4.6.2 and B.4.6.3 can be

found in references (4, Ii, 16, 18_ 19, 26, 28, 29, and 30). The ana-

lytical methods which can be adapted to the solution of problems which

have not been investigated are available in several references (4, ii,

16, 18, 19, 20, 26, and 29).

The abundance of exact solutions and methods of solution for the

elastic analysis of beam-columns which respond in-plane reduces the need

for using interaction equations. However, the interaction equation

can be, and is_used in many instances. The following simple straight

line equation is the basis for several interaction equations,

MP actual-- + = I. (2)

Pe Me

Since Mactual is the moment resulting from both the axial and trans-

verse loads, it can be difficult to obtain for complicated conditions.

However, for members which satisfy all of the following requirements:

(a) Must be simply supported

(b) Must have uniform cross section

(c) May be subjected to any combination of bending forces producing

maximum moment at or near the center of the span

it has been shown (Refs. I, 2, I0, and 31) that a good approximation of the

actual bending moment is given by

Mx

MactuaI - 1 - P/(Px)e

Section B4.6

February 15, 1976

Page 20

(3)

Here (Px)e is the Euler elastic critical load in the plane of the

applied moment and M x is the maximum moment, not considering the moment

due to the axial load interacting with the deflections. For the con-

dition of moment due to interaction of the axial load with deflection,

Eq. (2) becomes

p Mx-- + = 1Pe (Mx) e { 1 - P/(Px) e

and the corresponding margin of safety is given by

(4)

1 ] - 1 (5)

M.S. = M "P

- P/(Px) e }

For eccentrically loaded members, where d is the eccentricity, equal

at both ends, Eq. (2) takes the following form

__P + Pd = 1. (6)

Pe (Mx) e {1 - P/(Px)e[

The margin of safety for this case can be determined analogously to

Eq. (5).

The interaction equation may be written in terms of an equivalent

moment, Meq, for a beam-column subjected to unequal end moments as shown

in Figure 5.


FM I M 2

FIGURE 5. UNEQUAL END MOMENTS

Equation (2) becomes

M

__P + eqPe (Mx) e { I - P/(Px)e }

= 1 (7)

where a good approximation for Meq as given by Austin (I) is

Me q M 2 M= 0.6 + 0.4 -- for 1.0 > 2 > - 0.5

M I M1 M I(8)

and

M M2eq = 0.4 for -0.5 > -- _ -i.0. (9)

MI - M I

Accurate interaction formulas which are simple and general have not

been developed for beam-columns with other than simple supports; for

example, cases where each end can be free, hinged, fixed,or elastically

restrained both with respect to rotation and translation. However, it

is conservative to use Eq. (4) with M x = (Mx)max, where (Mx)ma x is the

maximum moment in the member and is determined by an ordinary structural

analysis without regard to the effects of axial load. In utilizing this

Sect ion B4.6


method, the effective length concept should be employed in determining the

Euler load.

B4.6.6.1.2 Inelastic Analysis

A practical interaction formula for predicting the strength of

metal beam-columns under combined compressive axial loads and bending,

which respond in-plane and in the inelastic region, has been shown (Refs.

i, 2, 7, 10,and 31) to be

pP--_ + (Mx) u { 1 - PICPx) e } = i. Cl0)

where Pu is the strength of the member as a column in the inelastic

range. The value of Pu can he found by the tangent modulus method (Refs. 4

ii), Johnson's modified parabolas (Refs. 29, 31), or by methods presented in

the section on columns. As before, M x _s the moment due to transverse

bending without the axial load, and (Mx) u is the ultimate moment which

the section can inelastically withstand. In determining the ultimate

moment, (Mx)u, for the interaction equation above, one of the three

methods presented in paragraph 4.4.3.2,

(a) Trapezoidal Stress Distribution

(b) Plastic Design

(c) Double Elastic Modull

can be used. The trapezoidal stress distribution method developed by Coz-

zone (Refs. 5, 29) is widely used in the aerospace industry in structural

design. It is especially adaptable to metals which behave like aluminum

alloys. In general, the trapezoidal stress distribution method will be

Section B4.6

February 15, 1976

Page 23

preferable; however, there are certain cases where one of the other meth-

ods may prove superior. For example, the plastic design method (Refs. 3,

16 and 32) has been primarily developed for steels.

Obviously, interaction equation 10 is an extension of the one

presented in the previous section (Eq. 4). As a consequence, it is sub-

ject to t_le same limitations, i.e., simple supports, uniform sections, and

equal end-moments. However, for beam-columns with unequal end-moments

(Fig. I0), the equivalent moment (Meq) as defined by equations 8 and 9

may be substituted for M x in equation i0. Additional investigation is

required to dttermine the limits of applicability of this interaction

equ_tJon for other boundary and loading conditions.

In _mploying the inelastic method, excessive deflections may occur.

In general, except for the e_se of plnstic design, there are few or

no methods for calculating the resulting deflections. Thus, if there is

a prespecified deflection limitation the inelastic method may not be

adequate.

Inelastic analysis procedures for combined tensile axial loads and

bending have not been developed. However, previous elastic analytical

methods can probably be extended to cover inelastic behavior.

B4.6.6.2 Biaxial In-Plane Response

Beam-columns which are torsionally stiff and free to deflect in

all directions may have a biaxial in-plane response condition (Fig. i)

res_.Iting from either of two loading conditions. These two conditions

c_nsist of compressive axial load and

Section B4.6

February 15, 1976

Page 24

(a) Primary Bending in the Strong Direction; or

(b) Biaxial Bending.

For case i, primary bendlng In the strong direction, the member is

always designed so thal it will have only in-plane response in the strong

direction. This is accomplished by checking the buckling value for the

weak direction, and, if necessary providing adequate intermediate supports

or stiffness for that direction. For the condition where a biaxial inplane

response does occur, Austin (Ref. I) and Massonnet (Ref. 2) both provide an

excellent discussion.

It can be seen that only blaxlal in-plane response for case 2 is

of primary practical interest. Therefore, it is discussed in the follow-

ing paragraphs.

B4.6.6.2.1 Elastic Analysis

"The determination of the stresses due to bending about the two

principal axes can be made independently as there is no coupling of the

flexural actions in the elastic range" (Ref. I). Thus, solutions for elastic

In-plane response problems (Paragraph 4.4.6.1.1) can be extended to in-

clude bending about both principal axes by simple superposltlon of results.

For beam-columns which are subjected to axial compression and bend-

Ing moments about both principal axes, Austin has also stated that the

interaction equation for in-plane response, Equation 4, can be modified to

take into account the blaxlal loading. Thus Equation 4 becomes

-- + =

Pe (Mx) e { I - P/(Px) e } + (My) e { I - P/(Py)e } I. (II)


Equation II is subject to the samerestrictions as Equation 4. Also,

the _uivalent momentconcept as discussed in Paragraph 4.4.6.1.1 can be

utilized for unequal end moments.

B4.6.6.2.2 Inelastic Analysis

The recommendedprocedure for inelastic analysis is an extension

of the interaction equation from in-plane response to include blaxial in-

plane response. Equation I0 then is

-- + + =i. (12)Pu (Mx) u { I - P/(Px) e } (M_)u{l - P/(Py_ }

Using this interaction equation, the same restrictions discussed

in Paragraph 4.4.6.1.2 will also apply here.

4.4.6.3 Combined Bending and Twisting Response

The methods summarized in the preceding sections are applicable to

problems of in-plane response, and should be applied only to beam-columns

which are restrained against twisting by adequate bracing or to beam-

columns which possess a high torsional rigidity.

A torsionally weak beam-column of open section such as the wide-

flange, tee, or angle is apt to twist as well as bend during the response.

The various possibilities wherein twist may be involved are summarized as

follows:

(a) If the shear center axis and centroidal axis are not co-

incident, the member may respond by a combination of

twisting and bending, with the tendency toward twist

failure increa_in_ for very thin-walled, torsionally

Sect ion B4.6February 15, 1976Page26

weak, short column sections.

(b) If the shear center axis and the centroldal axis are co-

incident, as in the case of the I- or Z- shapes, buckling

by pure twist may occur without bending.

The recommendedformulae given herein for both elastic and inelas-

tic analysis are in the form of interaction equations and are applicable

only to cases with bending in the strong direction. These equations have

a simple form, are convenient to use, are accurate, and have a wide scope

of application. If a theoretical solution is ueslred, References 4, 6,

and llcontaln analytical investigations of torsional-flexural response for

manycommonsections and loadings. However, most of the work has been

done for axial compression and momentsacting to cause only bending about

the major principal axis when the momentof inertia about the major axis

is muchgreater than about the minor axis (e.g., l-sectlon). It has been

proposed by Austin that the interaction equations can be extended to

include primary bending about both axes. However, there are few data,

experimental or analytical, available to verify this.

B4.6.6.3.1 Elastic Analysis

For the elastic analysis of doubly symmetric 1-section members

subjected to primary bending in the plane of the web the following inter-

action equation is recommended:

p M__ + x . ' = 1. (13)

Pe fcb Zx {I - P/(Px) e}


The value of fcb is the nominal extreme fiber stress at lateral buckling

for a membersubjected to a uniform momentcausing bending in the plane

of the web. The value of fcb is given by

J2El yh KGL

fcb = I+--

2ZxL2 _ 2E r (14)

A complete study of this lateral buckling is given by Clark and Hill in

Reference 8.

When the maximum moment is not at or near the center of the span,

the interaction equation may be excessively conservative. This is par-

ticularly true when end moments are of opposite sign and the maximum end

moment is used for M x. However, the interaction equation cited above can

be used if an equivalent uniform moment is calculated and substituted for

M x. The recommended expression for the determination of the equivalent

uniform moment as given by Massonnet (Ref. 2) is

Meq = _013 (Ml)2 + (M2)2 + 0.4 MIM 2 . (15)

Massonnet also states that the interaction equation is not necess-

arily limited to doubly symmetric 1-shaped sections, but can be used for

all shapes provided that the proper effective lengths, equivalent moments,

and appropriate expressions for fcb are adopted. However, these exten-

sions should be applied with discretion, as little work has been done to

support this.

Relatively few studies have been conducted on the elastic stability

of nonuniform or tapered beam columns which fail by combined bending and

twisting. Butler, Anderson (Ref. 13), and Gatewood (Ref. 14) have investigated


tapered members,and their results indicate that the previous interaction

equatlon=my be used. However, additional tests under a variety of load-

ing conditions will be necessary to establish this possibility conclusive-

ly.

84.6.6.3.2 inelastic Analysis

It has been predicted by various sources (Refs. I, 2, I0 and 31) that

the Interaction equation recommended for elastic analysis will also be

adequate for inelastic analysis provided that the tangent modulus is

used. For this case, Equation 14 becomes

_2Etlyh _' 2fcb: 2Z_L2 1 +

and the associated interaction equation Is

+Pu fcb Zx {I - P/(Px)e}

KSL• (16)

_2E r

=I.(17)

Galambos (Ref. 15) should be consulted for further study of the in-

elastic lateral buckling value to be used In the interaction equation.

0

0

0

Xt_

C

0

C

N O0

m C

C

0

&J

C_-_ 0

U

0 I::

0O4C_DUl

C

C

"0

0

0

14

C

._.

Sect ion B4.6

February 15, 1976

Page 29

0

II

X

C

&J

.CC 0

I !

II II

X

I

!

11oI

C_

,!

X

;II

X

_CC

&J

.3_CU i

I

II

M _

I I

tl II

X X

I

_ ] [i..._

ii i I

---'1r

Section B4.6


uJz

Qo

o<.,) u

ocL

el-

1

oP-O

0

N

I i__ _

-i--__L

,J

_vj'

Section B4.6

February 15, 1976

Page 3 1

,,ioQ

i,i

0

.,,,.4I@I

II 41

+II

Xx J

I¢ r----i

cll

Izl

._ cno _1_U

_l r_ _.._-

0

_ e

I

U

x

_ c'M

,,-3

_leq I

(N

V

II

II

Ig

II

x +

_I _

I

+ _

_ X

l

__ _\L.

j _2

tO

0

,-I0

_0

Io_

Qo_

"7rQ

I-'


0

II

_Jc_

I

_J

I

II

\\

0

II

0O u_

0I O

,-4:

II

/L

&

II

x

l-J

,...1

+

(11

I

I

II

1:N

0

II

X

1.1

I--I

I

t_

+

u j.-_,,'

I

_"'I I _"_ ! oM

+ _

_ ""_I

0 L___I

IIN

X

\\\%\\\\

I

1976

Section B4.6

February 15, 1976

Page 33

0

00

oq!

c_

Q

r_

rOF-

o_

0

.,.44-)U0

q-I

"0

,iN

J

O

N II

_ m

N _

_ "_ _1_if)

mo

o

0

II

N

m

o

4_

!

II

4-1

II

I

I

O

!

I!

_I _ _--\

_ =1_

_I _ 0U

,--I c'_l

, :_1_I1

I

II

X

!

UO

v

II

N

X

O

II

o_O

I "N

+

I

I

I|

Section B4.6

February 15, 1976

Page 34

_JC_0

o°M

°_

o _

_ o

ul

,.Q

¢t

-,-4

o

04.J

0

_l._-

ffl i

•", J

II

,M

.r,i

°_,,_

u

II II

0 _ 04J 4.1

Vo _1

II .t--_ (N u

.,ii

0 o•_4 °r,l

c _ c

V

t_I

A

I

II

II4_

4_ m

r_

I

I

II

II

o

0

II

_4

o._I_J

IT'_J

0

_J_J

_J

I

I_0

II

II

0.u

II

o

II

4..i

II

o

o

_ m

.liiu i

,-4 ii IIkl4

CD C_

I

°'l

Sect ion B4.6

February 15, 1976

Page 35

O

OCD

I

O

×I_O

,--i ._ °_

o"O

E

.r-I

x

E II

O

E II

•,4 O_ 4..1

_ m

°HN _

_ .,.4% °,4 IJ

4-1 _

I

II

°_

IIN

0

I II

.1..I

_1 "'- ,-_lm

• _-I Im

,A +

+ .,-

I

II

(1)

.,--I0

0

¢'+' __ O% .,-I

r_

OII

N

4-1

II

I

I

•r- _

o _lm4-1 -_

.1_1 I0

_j CD

II

4-1 0

o _1_ m

, 0 _I_NO

m.I-I

, Ela"II I

N

-r.t

O

_3

II

>-

Section B4.6

February 15, 1976

Page 36I

CO

o

oL)

I

La_I

C_

p.-

.o

NI_

O m_4 Oco U

+

U co

II

N

4J

'_m II "J-I

I_1_ r----m _1_ _'_

_i m o _le, I 0q o I u

.. I _ ,

' l_lt_l ,

N o 0

iJ II _l ,-Io m m

o Ico io II .IJ ,_1ill ¢:I

0

0

CO

,a "i ,a

_ il

HII II

II

,_le_

II

N

4J

i

x

•-,il(li _ I/>- H , ,

4.,IL____I

I

II

Section B4.6

February 15,

Page 37

1976

,-q

0L_

0

_0

I

¢q

F--

.0

!ibq

li

bZ .

12i

"c

i

i,

i,r:

t:.J ._

!'.

_I-'_I _: _ i_l_' I'_

°r'-_

.,e 1

II

('1

II X

_ _ °

i

!

I

0

o3

+

_l_ I

I_I__I_-i o

_I_] L_l°_

I

l

H

;2

iI

_J°_._

-r 1

_l _II

U

c_

>-

,--4

!

II

II

0

H

N

0

4-1

.IJ

O

x

: I_1

_1 __

Section B4.6

February 15, 1976

Page 38

m

v-4

0

,d4

..Q(II

f--

0

0

I_1_

,_) I= II

I= _ I:1

! °M._-I _

-,.4 _

I_ -_1 II

° _ _1

_I I_ II

°

0

u_0

¢) 0

0

0!

4-I_0 00 00

+!

_I__1"_

g

@

o _

N _

0 oI II

L)I

o xl'_-_._I

(d -PI -_I

Q) L__._.J _-I

I °- (1_)

i!/'_ r

t-I

+

I

_N

I "PIV

+

O0

I

_1" ,._

I

@

._"1

I

0

II

II _:_

.=

,_.-):_ :_I_

I I

II II

0,--__-i -el

._ ._Irj -_I

0

II

4J

4Jv

mtJ

-r-1I

!

I

! !

II II

0

-_I

m.= (_

rj -_I .u

II I

4J

Section B4.6

February 15, 1976

Page 39

ed

0

0

aJ

c_JQ

_0

o

.Q

J0

_J

=-0

4-J0_J

M

0

m

P_

a_ _'_•r-4 _I_ I

o _ :3: _Im

_I_ I

_ II II0

0 _ c_ cd

_D _-_

0 (_c_ _ _-_

0 _J

0

F--1

O

I

_ .r-_

I I

L__

II II

N N

O

o _

u_"m4.1

m _ 1.-1m O _

r._ 12u

,,D _ N

N

+

' _I_ o

_ _ ,

O

II

_ +" M

,._ o0

0

_I_

I0

t__l

!

fl

N

0 ._

4_

N •• .,.-I O

r_ t_-4 N

Sect ion B4.6

February 15,

Page 40

1976

..4

I--

,-i

m

0

O0

0

&J

O

0

o"_

o

","4 o

--.._

E

'1

I

L

-I_L

$

g

Zo

<0

i-4

w-f

0

U _

t_ m

IJ 0

X

- I

-i

I

Section B4.6

February 15,

Page 41

i

E

O_

0 _

U _

,.-_eO

%%

C'I" _

.t( _-]

o

0

0

i

c! ('-_

1976

Section B4.6

February 15, 1976

Page 42

I

o

zo

,-1

0

•,.4 N

- L- I

m !

• =

0 _J

g.\

eL. ,

0

Section B4.6

February 15, 1976

Page 43

_FE_N_S:

I. Austin, W. J.: Strength and Design of Metal Beam-Columns, Proceedings,ASCE, April 1961, page 1802.

2. Massonnet, C.: Stability Considerations in the Design of Steel

Columns, Proceedings, ASCE, ST7, September 1959.

3. Beedle, L. S.: Plastic Design of Steel Frame_, John Wiley and Sons,

Inc., New York, 1958.

4. Blelch, F.: Bucklin_ Strength of Metal Structures, McGraw-Hill Book _

Co., Inc., New York, 1952.

5. Cozzone, J.: Bending in the Plastic Range, Journal Aeronautical

Sciences, May 1943.

6. Seeley, F. and Smith, J.: Advanced Mechanics of Materials, John

Wiley and Sons, Inc., New York, 1957.

7. Clark, J. W.: Eccentrically Loaded Aluminum Columns, Transactions,

ASCE, Vol. 120, 1955, page 1116.

8. Clark, J. W., and Hill, A. N.,: Lateral Buckling of Beams, Proceed-

in___, ASCE, Vol. 86, No. ST7, July, 1960.

9. Hill, H. N. and Clark, J. W.: Lateral Buckling of Eccentrically

Loaded I- and H-Section Columns, Proceedings, First National Congress

of Applied Mechanics, ASME, 1952.

I0. Hill, H. N., Hartmann, E. C., and Clark, J. W.: Design of Aluminum

Alloy Beam-Columns, Transactions, ASCE, Vol. 121, 1956.

Ii. Timoshenko, S. and Gere, J.M.: Theory of Elastic Stability, Second

edition, McGraw-Hill Book Company, Inc., New York, 1961.

12. Salvadori, M. G.: Lateral Buckling of Eccentrically Loaded 1-Columns,

Transactions, ASCE Vol. 121, 1956.

13. Butler, D. J. and Anderson, G. B.: The Elastic Buckling of Tapered

Beam-Columns, Welding Research Journal Supplement, January 1963.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

97.

28.


Gatewood, B. E.: Buckling Loads for Beamsof Variable Cross SectionUnder CombinedLoads, Journal of the Aeronautical Sciences, Vol. 22,1955.

Ga]_.mbos,T. V.: Ine]_etlc L_.ter_1 _ckling_ of Be_.m_: Proceedings,

Bresler, Boris and Lin, T. Y.: Design of Steel Structures, John

Wiley and Sons, Inc., New York, 1960.

Hetenyi, M.: Beams on Elastic Foundation, the University of Michigan

Press, Ann Arbor, Michigan, 1955.

Niles, A. S. and Newell, J. S.: Airplane Structures, John Wiley and

Sons, Inc., Volume II, third edition, New York, 1949.

Roark, R. J.: Formulas for Stress and Strain, Third Edition, McGraw-

Hill Book Company, Inc., New York, 1954.

Timoshenko, S.: Strength of Materials Part II, Advanced Theory and

Problems, 3rd ed., D. Van Nostrand Co., Inc., Princeton, New Jersey,1956.

Laird, W. and Bryson, A.: An Analysis of Continuous Beam-Columns

with Uniformly Distributed Axial Load, Proc. of 3rd U. S. National

Congress of Applied Mechanics, ASME, N.Y., 1958.

Saunders, H.: Beam-Column of Nonuniform Sections by MatrlxMethods,

Journal of the Aerospace Sciences, September 1961.

Saunders, H: Matrix Analysis of a Nonuniform Beam-Column on Multi-

Supports, AIAA Journal, April 1963.

Lee, S., Wang, T. and Kao, J.: Continuous Beam-Columns on Elastic

Foundation, ProceedinKs , ASCE EM2, April 1961.

Sundara, K. and Anantharamu, S.: Finite Beam-Columns on Elastic

Foundation, ProceedlnFLs , ASCE, EM6, December 1963.

Ne_mrk, N. M.: Numerical Procedure for Computing Deflections,

Moments, and Buckling Loads, Transactions, ASCE, Vol. 108, 1943.

Salvadorl, M. G. and Baron, M. L.: Numerical Methods in EnHineering,

Prentlce-Hall, Inc., New Jersey, 1962.

Woodberry, R. F. H.: Analysis of Beam-Columns, Design News, April1959.


29.

30.

31.

32.

Bruhn, E. F.: Analysls and Desig n of Flight Veh_le Structures,

Tri-State Offset Co., Cincinnati, Ohio, 1965.

Griffel, W.: Method for Solving Beam-Column Problems, Product

Engineering, September 14, 1964.

Guide to Desisn Criteria for Metal Compression Members, Column

Research Council, Engrg, Foundation, 1960.

Plastic Design in Steel, American Institute of Steel Construction,

Inc., New York, 1959.

SECTION B4.7

LATERAL BUCKLING OF BEAMS

--.._j

B4.7

4.7.1

4.7.1.1

4.7.2

4.7.2.1

4.7.2.2

4.7.3

4.7.4

4'.7.4.1

4.7.4.2

4.7.4.3

4.7.4.4

4.7.5

TABLE OF CONTENTS

Section B4.7

Lateral Buckling of Beams ......................

Introduction ................................

General Cross Section .........................

Symmetrical Sections .........................

I-Beams ..................................

Iklre Bending ..........................

Cantilever Be,am, Load at End ...............

Simply Supported Beam, Load at Middle ........

Simply Supported Beam, Uniform Load .........

Ie

II.

III.

IV.

Page

1

1

1

5

5

5

6

7

8

Rectm_mlar Beams ........................... 10

Unsymmetrical I - Sections ..................... 13

Special Conditions ............................ 14

Oblique Loads .............................. 14

Nonuniform Cross Section ...................... 14

Special End Conditions ........................ 14

Inelastic Buckling ............................ 15

RE FERENCES .............................. 16

...._ B4.7-iii

Section B4.715 April, 1971

Page 1

B4.7 LATERAL BUCKLING OF BEAMS

4.7. i INTRODUCTION

A beam of general cross section which is bent in the planeof greatestflexural rigidity may buckle in the plane perpendicular to the plane of greatestflexural rigidity at a certain critical value of the load. Concern for lateralbuckling is more significant in the design of beams without lateral support whenthe flexural rigidity of the beam in the plane of bendingis large in comparisonwith the lateral bending rigidity.

Consider the beamwith two planesof symmetry shownin Figure 4.7- i.This beam is assumedto be subjected to arbitrary loads acting perpendicularto the xz plane. By assuming that a small lateral deflection occurs under theaction of these loads, the critical value of load canbe obtained from thedifferential equationsof equilibrium for the deflected beam (Ref. l).

Beams with various cross sections andparticular cases of loading andboundary conditions will be considered in this section.

4.7.1. i General Cross Section

The general expression for the elastic buckling strength of beams can be

expressed by the following equation (Ref. 3).

__ GJ(KL 2f .y w 1 +cr S (KL) 2 2g + C3k+ (C2g+ C3k) 2+ I

c y

(1)where:

f = critical stress for lateral bucklingor

E = modulus of elasticity, lb/in. 2

I = modulus of inertia of beam cross section about the y axis, in.Y

L = distance between Doints of support against lateral bending and

twisting, in.

o

p

m !

P2

Section B4.7

15 April, 1971

Page 2

0

m

Z

-U

Y

|

!

!

FIGURE 4.7-i LATERAL BUCKLING -

/

CW

g ___

G

J __

torsion warping constant, in. 6

Section B4.7

15 April, 1971

P,_e 3

distance from shear center to point of application of transverse

load (positive when load is below shear center and negative

other%vise), in.

shear modulus of elasticity, Ib/in.2

torsion constant, in.4

S = section modulus for stress in compression flange, in. 3C

ki

= c + m rj (x2 _ y2) dA, in.21x A

e = distance from shear center to centroid, _)sitivc ifbct_vcen

centroid and compressi(m flange, in.

CI, C2, C3, K= constants which depend mainly on conditions of loading and support

for the beam (Tsl)Ie 4.7- I).

In the equation above, itis assumed that the lines of action of the loads

pass through the shear center and the ccntroid, and that the loads attach to the

beam in such a mmmcr that their lines of action remain parallel to their initial

directions as the beam deflects. It is also assumed that the shear center lies

on a principal axis through the centroid.

The coefficients Ci, C2, C3, and K are derived in Reference 3. They

depend mainly on the conditions of loading and support for the beam. The values

of Ci, C2, C3, and K given in Table 4.7-I have been obtained from Reference 3.

Table 4.7- I.

Section B4.7

15 April, 1971

Page 4

Values of Coefficients in Formula for Elastic Buckling

Strength of Beams

CASE NO.

!1

12

LOADING RESTRAINT

VALUE OF COEFFICIENTS

K C1 C2 I

M M

1_--4

l

M

+ t

w

t +

w

P

+ +

P

t---+:--!=4

+ I

SIMPLE SUPPORT

FIXED

SIMPLE SUPPORT

FIXED

SIMPLE SUPPORT

FIXED

SIMPLE SUPPORT

FIXED

SIMPLE SUPPORT

FIXED

SIMP.LE SUP PO.R.T

FIXED

SIMPLE SUPPORT

FIXED

SIMPLE SUPPORT

FIXED

SIMPLE SUPPORT

FIXED

SIMPLE SUPPORT

FIXE D

1.0

05

1.0

0.5

I0

0. S

1.0

0.5

1.0

0.S

1.0

0.5

1.0

0.5

1.0

O.S

1.0

0.5

1.0

1.O

1.0

1.31

1.30

1.77

1.78

2.33

229

2.56

2.23

1.13 0.45

0.97 0.29

1.30 1.SS

0.86 0.82

C3

1.0

1.0

1,35 0.55 2.5

1.07 0.42

CANTILEVER BEAMS

WARPING RESTRAINED

AT _,UPPORTED END

WARPING RESTRAINED

AT SUPPORTED END

1.0

10

1.28

1.70 1.42

1.04 0,84

1.04 0.42

2.05

0 64

6.S

4.7.2 SYM ME TRIC AL SE CTIONS

Section B4.7

15 April, 1971

Page 5

For sections that are symmetrical about the horizontal axis or about

a point (channels, zee sections, etc.), the quantity k in equation (1) is equal

to zero. The expression for elastic bucMing strength can then be written

f _ y z w GJ (KL) 2_KL_ 2 2g+ (C2g) + _ 1 +I ; ECor

a y w

Values of C t, C 2, and K can be obtained from Table 4.7-1.

(2)

4.7.2. i I-Beams

Given below are solutions for particular cases of load and boundary

conditions for I-beams. For cases not considered below, equation (2) shouldbe used.

I. Pure Bending

If an I-beam is subjected to couples Mvalue of the moment M is o

o

(M)o -_ Lrr EIy GJ 1 q _--_L---_/cr

at the ends, the critical

(3)

This expression can be represented in the form

EI GJy

(M) = K 1o Lcr

(4)

where

J Ec _2w

K 1 = _ I + GJLT

Values of K 1 are given in "Fable 4.7-2.

Table 4.7-2.

L2GJ-- 0EC

W

K1 oo

L2Gj16

ECW

K l 4.00

Section B4.7

15 April, 1971

Page 6

Values of the Factor K 1 for I-Beams in Pure Bending

0. t i 2 4 6 8

31.4 10.36 7.66 5.85 5.11 4.70

20 24 28 32 36 40

3.83 3.73 3.66 3.59 3.55 3.51

10 12

4.43 4.24

100 oo

3.29

v"

II. Cantilever Beam, Load at End

If a cantilever beam is subjected to a force applied at the centroid

of the end cross section, the critical value of the load P is

P = K 2 L2 (5)cr

where

K 2 =

4.0t3

L2GJFor values of

ECW

4.7-3. For values of

for values of K 2.

greater than 0.1, values of K 2 are given in Table

L2Gjless than 0.1, see Reference 1, page 258,EC

W

Table 4.7-3.

Section B4.7

15 April, 1971

Page 7

Values of the Factor K 2 for Cantilever Beams <)[ I-Section

L2Gj

ECW

0.1

K 2 44.3 15.7 12.2 10.7 9.76 S 69 8.03

L2GJ

ECW

K2

10

7.5_

12

7.20

14 16 24

6.96 6.73 6. 19

,)r)

5.£7

4O

5.64

III. Simply Supported Be.am, L()ad at Middle

If a simply supported I-beam is subjected to a l,m(I P applied at

the eentroid of the middle (.'ross secth)n, the critical value ,ff the load Pis

,/EIyGJ

P = K 3 L2 (6)er

Values of K 3 obtained from Reference I, page 264, are given in Table 4.7-4(a)

Table 4.7-4(a) . Values of K 3 for Simply Supported I-Beams WithConcentrated Load at Middle

L,,md L l (;J I-2("

AptAt_,d w

At (t 4 4 _ 11; 21 2t2 4_

Iplmr

_lallgt. _il 5 2il 1 lti 9 15 4 15 II 14 !1 [4

('{'ntl'OllJ W_; | 31 !) 27, _, 21 h 211 { 1!1 _, I_

I. langr 117 -tq_ o :1_ 2 FILl 3 27 1 25 1 _'_

L,_ad I, 2 (1.1 I-;("

Apph¢_l w

!'q f;4 _11 !l(i ]fi(t 2.1'_ 121_ 41)0

|'lql.r

Fi:m_u' 13 ql 15 It 15 I 15 3 t5 l 15 ti 15._

('lqll[rl,ld 1_4 3 1_. I 17 9 17 5 17 I t7 2 17 2

l'langc 22 I 21 7 21 I Lql u 193 19 i1 I_ 7

Section B4.7

15 April, 1971

Page 8

If lateral support is provided at the middle of the beam, values of K 3 are given

in Table 4.7-4(b).

Table 4.7-4(b). Values of the Factor K 3 for Lateral Support at Middle

L2Gj0.4 4 8

ECw

K 3 466 154 114

16 32

86.4 69.2

96 128

54.5

2OO 4OO

52.4 49.8 47.4

If lateral support is provided at both ends of the beam, values of K 3 are given

in Table 4.7-4(c).

Table 4.7-4(c). Values of the Factor K 3 for Lateral Support at Ends

L2Gj

ECw

K2

0.4 4 8 16 24 32 64 128 200 320

268 88.8 65.5 50.2 43.6 40.2 34. i 30.7 29.4 28.4

IV. Simply Supported Beam, Uniform Load

If a simply supported I-beam is subjected to a uniform load q, the

critical value of this load can be expressed in the form

EIyGJ(ql)cr = K4 L 2 (7)

Values of IQ obtained from Reference l, page 267, are given in Table 4.7-5(a).

F"

Table 4.7-5(a).

Section B4.7

15 April, 1971

Page 9

Values of K 4 for Simply Supported I-Beams with Uniform Load

Load L 2 GJ/ECApplied w

At 0.4 4 8 16 24 32 48

Upper

Flange 92.9 36.3 30.4 27.5 26.6 26. 1 25.9

Centroid 143.0 53.0 42.6 36.3 33.8 32.6 31.5

Lowe r

Flange

Load

Applied

At

223. 77.4 59.6 48.0 43.6 40.5 37.8

L 2 GJ/ECw

64 80 128 200 280 3(;0 400

Upper

Flange 25.9 25.8 26.0 26.4 26.5 26.6 26.7

Centroid 30.5 30.1 29.4 29.0 28.8 28.6 28.6

Lower

Flange 36.4 35. 1 33.3 32. 1 31.3 3 1.0 30.7

If the beam has lateral support at the middle, K 4 is given by Table 4.7-5(b).

Table 4.7-5(b) . Values of K 4 with Lateral Support at Middle

Load

Applied

At 0.4

Upper

Flange 587

Centroid 673

Lower

Flange 774

4 8

194 145

221 164

L 2 GJ/ECW

16 64 96 128 200

112 91.5 73.9 71.6 69.0

126 101. 79.5 76.4 72.8

251 185 142 112 85.7 81.7 76.9

Section B4.7

15 April, 1971

Page i0

If the beam has lateral support at both ends of the beam, K 4 is given by

Table 4.7-5(c).

Table 4.7-5(c). Values of K4 with Lateral Support at Ends

L2Gj

ECw

0.4 4 8 i6

K 4 488 16i i19 91.3

32

73.0

96

58.0

128

55.8

200

53.5

400

5i.2

4.7.2.2 Rectan_lar Beams

For a beam of rectangular section of width b and height h, the warping

rigidity C can be taken as zero; therefore, equation (2) becomesw

f = y GJ(KL) 2cr S (KL) 2 2g + (C2g) 2 + _ EI (8)

e y

Ifthe load is applied at the centroid, g = 0; therefore,

C1 _2_"_yGJf = (9)cr S KL

c

3By taking G=--_- E, J=0.31hb 3, I= hb3, and S = bh-_2c 6

Eb 2f = 1.86 C 1 (iO)cr K Lh

or

Eb 2

f = Kf (ii)cr Lh

where

i. 86 C1Kf = K

Section B4.7

15 April, 1971

Page 11

Values of Kf are given in Figure 4.7-2 and Table 4.7-6 for several load

cases. For cases not available in "[',able 4.7-6 and Figure 4.7-2, refer to

Table 4.7-1 for values of C 1 and K for use in equation 10.

Equation 8 must be used for loads not applied at the eentroid for any

of the given cases.

3.0 _'--

Kf 2.0

L

1

t" L _ m

1.0 ill 111111 1 I 11 110 0.10 0.20 0.30 0.40 0.50

c/L

FIGURE 4.7-2 CONSTANTS FO[_ DETERMINING 'FILE LATERAL STABILITY

OF DEEP RECTANGULAR BEAMS

Table 4.7-6.

Section B4.7

15 April, 1971

Pa_e 12

Constants for Determining the Lateral Stability of

Rectangular Beams

6

7

8

9

I0

11

12

17

Side View Top View

L/2

L/2

1

_v

f_

Section B4.7

15 April, 1971

page 13

4.7.3 UNSYMMETRICAL I-SECTIONS

For I-beams symmetrical about the vertical axis, but unsymmetrical

about the horizontal axis and subjected to uniform bending moment, the following

approximate equation for the elastic buckling stress should be used (Ref. 3).

_ EI I J C ( GJ(KL)2_ 1/f = _ e2 w 1+ ,r2ECw j (8)cr S (KL) 2 e + + Ic y

Section B4.7

15 April, 1971

Page 14

4.7.4 SPECIAL CONDITIONS

4.7.4. i Obliclue Loads

The case of a beam subjected to a uniform bending moment that does

not lie in one of the principal planes of the cross section is discussed in

References 4 and 5. Reference 5 shows that the equation for the critical moment

takes the form of equation lwith C 1= C3= 1.0, C2= 0. The quantityI isY

replaced by the expressicn I:Ix/Ix,y in which - and - denote principal axes andy x

the x axis is the axis normal to the plane of bending.

4.7.4.2 Nonuniform Cross Section

A concise solution for the lateralbuckling strength of a tapered

rectangular beam, subjected to constant bending moment and simply supported

at the ends, is presented in Reference 6. Tapered cantilever I-beams have

been investigated experimentally in Reference 7.

4.7.4.3 Special End Conditions

Solutions have been obtained (Ref. 8) for the buckling strength of

I-beams under a load (either uniform or a concentrated load at the center)

acting perpendicular to the principal plane having maximum bending rigidity and

with various degrees of restraint against rotation of the beam about either

plane. Each type of restraint was considered to vary between zero and complete

fixity. In all cases, the beams were considered to be fixed at the ends against

rotation about a longitudinal axis perpendicular to the plane of the cross section.

Frequently, a cantilever beam is simply the overhanging end of a beam

that extends over two or more supports. In this case, the supported end of the

cantilever beam may not be fixed against lateral bending of the beam flanges

but some restraint is supplied by continuity at the support. In such cases, a

conservative estimate of the buckling strength can be made by considering the

warping constant, Cw, to be zero in the buckling formula.

F

Section B4.7

15 April, 1971

Page 15

If a beam is continuous beyond one or both supports, the end conditions

for any one span are generally between the cases of complete fixity and simple

support covered in Table I. The effect of continuity has been discussed in

References 9 and 10.

4.7.4.4 Inelastic Buckling

It is explained in Reference II that it is possible to obtain a lower limit

to the theoretical buckling stress in the inelastic range by substituting the

tangent modulus, E t, corresponding to the maximum stress in the beam for the

elastic modulus, E, in the elastic buckling formula. Tests on aluminum alloy

beams show that this method gives a close approximation to the experimental

buckling stress when the bending moment is constant along the length (Ref.

i2 and 13). Tests of aluminum alloy beams subjected to unequal end moments,

with the ratio of the moment at one end to the moment at the other end varying

from i.0 to -i. 0, resulted in experimcnt.al critical stresses varying from 8

percent below to 39 percent above the values computed by the tangent moduhls

method.

D

3.

.

.

.

.

e

.

I0.

II.

Section B4.7

15 April, 1971

Page 16

REFERENCES

Timoshenko, S. P., and Gere, J. M: Theory of Elastic Stability.

McGraw-Hill Book Company, Inc., New York, 1961.

Structural Design Manual, Northrop Aircraft, Inc.

Clark, J. W., and Hill, H. N: Lateral Buckling of Beams. Proceedings

of the American Society of Civil Engineers, Journal of the Structural

Division, July, 1960.

Go.diet, J. N: Flexural-Torsional Buckling of Bars of Open Section.

Bulletin No. 28, Cornell University Engineering Experiment Station,1942.

McCalley, R. C., Jr: Discussion of Paper by H. N. Hill, E. C.

Hartmann, and J. W. Clark. Transactions of the ASCE, Vol. 121,

1956, p. i5.

Lee, L. H. N: On the Lateral Buckling of a Tapered Narrow Rectangular

Beam. ASME Journal of Applied Mechanics, September, 1959, p. 457.

Krefeld, W. J., Butler, D. J., and Anderson, G. B: Welding Cantilever

Wedge Beams. The Welding Journal Research Supplement, March, 1959.

Austin, W. J., Yegian, S., and Tung, T. P: Lateral Buckling of

Elastically End-Restrained I-Beams. Separate No. 673, Proceedings

of the ASCE, Vol. 81, 1955.

Salvadori, M. G: Lateral Buckling of I-Beams. Transactions of the

ASCE, Vol. t20, i955, p. 1165.

Nylander, H: Torsion, Bending and Lateral Buckling of I-Beams.

Bulletin No. 22, Division of Building Statics and Structural Engineering,

Royal Institute of Technology, Stockholm, Sweden, 1956.

Bleich, F: Buckling Strength of Metal Structures. McGraw-Hill Book

Company, Inc., 1952, pp. 55 and 165.

f

Section B4.7

15 April, 1971

Page 17

RE FERENCES (Concluded) :

i2. Dumont, C., and Hill, H. N: Lateral Stability of Equal-Flanged

Aluminum Alloy I-Beams Subjected to Pure Bending. T. N. 770, NACA,

t940.

13. Clark, J. W., and Jombock, J. R: Lateral Buckling of I-Beams

Subjected to Unequal End Moments. Paper No. 1291, Journal of the

Engineering Mechanics Division, Proceedings of the ASCE, Vol. 83,

July, 1957.

SECTION B4.8

SHEAR BEAMS

Section B4.8

15 October 1969

TABLE OF CONTENTS

Page

Shear Beams .................................... 1

Plane-Stiffened Shear-Resistant Beams .................. 3

4.8.1.1 4

I. 5

II.

Ill.

IV.

V.

VI.

4.8.1.2

4.8.1.3

I.

II.

III.

4.8.1.4

4.8.1.5

4.8.1.6

4.8.2 Plane

4.8.2.1

4.8.2.2

4.8,2.3

Stability of Web Panel .........................

Transverse Stiffeners-Flexural Rigidity Only .........

Transverse Stiffeners-Effect of Stiffener Thickness and

RIg('! L,,cation .............................. 7

Transverse Stiffeners-Flexural and Torsional Rigidity . . . 8

Transverse and Central Longitudinal Stiffeners ........ t0

Longitudinally Stiffened Web Plates in Longitudinal

Compression ............................... 16

Combined Stresses ........................... 20

Flange Design .............................. 20

Rivet Design ............................... 21

W eb-to-Stiffener ............................. 21

Stiffeners-to-Flange .......................... 22

Web-to-Flange .............................. 22

Design Approach ............................. 22

Stress Analysis Procedure ...................... 23

Other Types of Web Design ...................... 23

Tension Field Beams .......................... 25

General Limitations and Symbols .................. 26

Analysis of Web ............................. 28

Analysis of Stiffeners ......................... 38

B4.8-iii

SectionB4.8i5 October 1969

4.8.2.4

4.8.2.5

4.8.2.6

4.8.2.7

TABLE OF CONTENTS (Concluded)

Analysis of Flange ...........................

Analysis of Rivets ............................

Analysis of End of Beam .......................

Beam Design ...............................

Page

46

47

48

50

B4,:8-iv

t

Section B4.8

15 October 1969

f-- DEFINITIONS OF SYMBOLS

t

tU

thickness of web plate

thickness of attached stiffener I(:g

b

bC

spacing of intermediate stiffeners, or width of unstiffcned web plate

clear web plate distance between stiffeners

dC

clear depth of web plate

o_e

effective ; :_,:t ratio;

D

E

#

I

= b/d for unstiffened w.eb plates, or web plates reinforced by single-sidedc

stiffeners;

= bc/d c for web plates reinforced by double-sided stiffeners

flexural rigidity of unit width of plate = Et3/12( 1-u 2)

Youngts modulus

Poisson's ratio

moment of inertia of stiffener about base of stiffener (next to web)

yL limiting value of T

KS

critical shear-stress coefficient

K L

T}S'_B

Af

limiting value of KS

plasticity coefficients which account for the reduction of modulus of

elasticity for stress above the elastic limit; within the elastic range,

_=1

area of tension or compression flange

B4.8-v

Cr

Dr

fb

fS

FScr

M

P

q

S,V

A

h

B T =

"/T =

C T

B L =

TL -

C L

F L =

Section B4.8

15 October i969

DEFINITIONS OF SYMBOLS (Continued)

rivet factor

p-Dr

P

rivet diameter

applied bending stress

applied web shear stress

critical (or initial) buckling stress in shear

applied bending moment

rivet spacing

applied web shear flow

applied transverse shear on beam

parameter used for type of shear beam selection

height of beam between centroids of flanges

EIT, flexural rigidity of transverse stiffeners

BT/Db, nondimensional flexural rigidity parameter for transverse

stiffeners

torsional rigidity of transverse stifleners

EIL, flexural rigidity of longitudinal stiffeners

B L

Dd ' nondimensional stiffeners ,_arameter for longitudinal stiifenersc

torsional rigidity of longitudinal stiffeners

C L/Ddc

B4.8-vi

f

F T

G

KL H

nl

F BeY

IO

Section B4.8

15 October 1969

DEFINITIONS OF SYMBOLS (Concluded)

C T/D d e

modulus of rigidity

limiting value of K for web reinforced by vertical stiffeners and aS

central horizontal stiffener

distance from edge of plate to longitudinal stiffener

critical (or initial) buckling stress in bending

stiffener moment of inertia for longitudinal stiffener

B4.8-vii

Section B4.8

15 October 1969

Page 1

4.8.0 SttEAR BEAMS

The analysis and design of a metal beam composed of flange members

riveted or welded to web members are eommon problems in aerospace struetural

design.

Shear be_'uns denote a particularly effieient type of beam. In shear beams

the moment resistanee is provided by the flanges, which are eoneentrated near

the extreme fibers, and the shear resistance is provided by the thin web

connecting the tension and compression caps.

The analysis and desit,m of shear beams as structural components are

generally based upon the web response to the applied shear loads. If bueMing

of the web is inhibited within the design ultimate load, the beam is known as a

shear-resistant beam1. If, however, the web is allowed to buckle :ffter some

application of load causing shear to be resisted in part by tension-field action,

the beam is known as a tension-fiehl beam.

The type o1" shear bc'anl most suitabl_' for a particul:_r" dcsiRn apl_lication

may depend on many factors. One of the most common factors is based on

economy of weight, tt. Wagner [ 1 ] offers the following criterion, based upon

the economy of weight:

4Vh

wh ere

V

and

h

= shear load, lb

= depth of web, in.,

SectionB4.815October 1969Page 2

with the recommendation that when A < 7 the tension-field web is best, and

when A > 11 the shear-resistant web is best. When 7 < A < 11, there is little

choice between the two; factors other than weight will then determine the type

of web to be used.

The criterion above should not be adhered to rigidly, however, because new

data and design techniques have become available that have resulted in reduced

weight designs for shear-resistant beams.

Shear-resistant beams and tension-field beams will be discussed in

Paragraphs 4.8.1 and 4.8.2 respectively.

/f-'_.

Section B4.8

15 October 1969

Page 3

4.8.1 PLANE-STIFFENED SHEAR RESISTANT BEAMS

As previously stated, a shear beam whose web is so designed that it does

not buckle under the applied loads is referred to as a shear-resistant beam.

The analysis of this beam is primarily one of stability. That is, with the

exception of the tension flanges, the web, the compression [lange, and the

stiffeners are all designed from a stability standpoint rather than from a

material allowable stress standpoint.

The stability of the shear web can always be increased by increasing its

thickness, but such a desigm will not always be economical with respect to the

weight of the material used. A more economical solution is obtained by keeping

the thickness of the plate as small as possible (just thick enough to fulfill

strength requirements) "rod increasing the stability by introducing stiffeners.

The weight of such stiffeners will usually be less than the additional weight

introduced by an adequate increase in the thickness of the plate.

The design criteria of shear-resistant beams may be stated as follows:

1. Local buckling of the web between the stiffeners under combined

shear, bending, axial, and transverse stresses must not occur.

2. Elements of the tr:,nsverse stiffeners must not buckle locally

under the transverse stresses.

3. Elements of the flange must not buckle locally under the longitudinal

stresses.

If the criteria above are not met, the procedures of analysis that follow are not

applicable.

Design analysis techniques for shear resistant beams are given in the

following paragraphs.

SectionB4.8t5 October i969Page4

4.8.1.1 Stability of Web Panel

The critical buckling stress, F of a web panel of height d width b,S ' C'cr

and thickness t, is given by

Fs K 7r2E

12 ( l-p2 ) for d cS

_b (1. a)

or

Fs K 7r2E 2

Us 12 (l-pz) c, (1. b)

where K is a function of the aspect ratio, d /b, and the edge restraint offeredS e

by the stilfencrs and flanges.

Fil4urc 1 shows how the value of the critical shear stress cocl ficient, Ks,

increases with decreasing values of the aspect ratio, b/d (d _b), for a numberC C

of different edge conditions. This figure indicates quite clearly why vertical

stiffeners are so effective in increasing the buckling stress of rectangular webs.

In vicw of the importance of obtaining the correct desi6m of intermediate

stiffeners, a number of theoretical and experimental investigations have been

made to determine the relationship between the size and spacing of intermediate

stiffeners, and the buckling stress of the stiffened web plate. These investigations

have also included the effects of stiffener torsional rigidity, stiffener thickness

Section B4.8

15 October 1969

Page 5

and rivet location, central longitudinal stiffness, and single-sided or double-sided

stiffness. The procedure for the desi_ma and analysis of these effects will be

given below.

16

FIGURE 1.

14

12

CRITICAL

SHEAR STRESS

COEFFICIENT

K 10$

\

\\

PLATE CLAMPED

ON ALL FOUR EDGES

PLATE SIMPLY ''f

SUPPORTED ON

ALL FOUR EDGESI I

1.5 2.0 2.5

-"v----

4 #%

i.0 3.O 3.5 Go

ASPECT RATIO 10/d c

K VERSUS ASPECT RATIO FOIl DIFFERENT EDGE RESTRAINTSS

I. Transverse Stiffeners -- Flcxural Rigidity Only

Thboretieal work has been performed [ 2-4 ] to ascertain the relationship

EI

between Ks and the nondimensional parameter 3_ (=D--b) for various aspect ratios

and boundary conditions. Figxwe 2 is a typical plot showing the relationship of

these parameters. It will be noted that points of discontinuity occur on the

K /3/ curves. These points of discontinuity denote where changes in the buckleS

Sectioa B4.8

i5 October 1969

Page 6

pattern occur. It can also be seen that when a certain value of _, is reached,

there is no appreciable increase in Ks. This value of _ is called _'L or limiting

value of _, since higher values would result in an inefficient design.

30I

25

2O

15CRITICAL

SHEAR STRESS

COEFFICIENT

Ks 10

7I

ORTHOTROF

I/

V

25 50

.... ......VALUEOF.PANEL SIMPLY

SUPPORTED ONALL FOUR EDGES-

75 100 125 175

¥ = EI/(Db)

FIGURE 2. THE THEORETICAL K/_ RELATIONSHIPS DERIVED BY STEIN

AND FRALICH FOR INFINITELY LONG PLATES SIMPLY SUPPORTED

AT THE EDGES AND REINFORCED BY STIFFENERS 0.5d APART.C

For design purposes, the following relationships should be used. However,

it should be noted that these relationships are valid only for stiffeners whose

thickness is equal to or greater than the thickness of the web plate.

7L = 27.75 (d e)-2 _ 7.5 for double-sided stiffeners, (2. a)

5, L = 21.5 ((_e) -2 _ 7.5 for single-sided stiffeners, (2. b)

Section B4.8

15 October 1969

Page 7

K L = 7.0 + 5.6 (OZe)-2 for both single- and double-sided stiffeners, (3)

where

a = b /d for double-sided stiffeners (b _-d ) ande c c c c

a = b/d for single-sided stiffeners (b->d)e c c

II. Tr:msverse Stiffeners -- Effect of Stiffener Thiclmess and Rivet Location

Investigations have been carried out to investigate fully the various

parameters that affect the behavior of single-sided stiffeners having attached

legs thinner than the web-plate [ 5]. The parameters tu/t and c/t were studied

in these investigations to evaluate their effect on Ks, where t, t u, and c are as

shown in Figure 3. It was shown that the primary influence on K was the values

c and that the tu/t variation had littleeffect on

!

't

ttj

'" I l-[

K . Figure 4 shows the variation of K with theS S

parameter (t /t) (t/c) 1/2. From the figure itU'

can be seen that for the stiffener to provide a

wdue of K equal to KL, it is necessary thatS

(t /t) (t/c) '/2 -> 0.27 . (4)U

FIGUR E 3.

t u ,

VALUES OF t,

AND e

This equation can be used to determine the

position of the rivet for fully effective stiffeners

Section B4.8

15 October i969

Page 8

for various values of tu/t. For values of (tu/t) (t/c) 1/2 less than 0.27 the

value of K should be reduced as shown in Figure 4.S

K

25

2O

15

100

/TI

20 40 60 80 100 120 0.2

Y 0

. ..I I// / KLEC UAT,O.(3>

LINEAR GROWTH

-VALUE MARKED A

0.4 0.6 0.g

(tu/t)(t/c)

FIGURE 4. VARIATION OF K WITH THE STIFFENER THICKNESSS

AND RIVET LOCATION PARAMETER

III. Transverse Stiffeners -- Flexural and Torsional Rigidity

Theoretical results have been obtained [4, 6-8 ] that provide relationships

between K and the flexural rigidity of the stiffeners for various values of thes

ratio of torsional rigidity to flexural rigidity for simply supported or clamped

longitudinal edges. It was assumed that the stiffeners were symmetrically

disposed about the midplane of the web plate as shown in Figure 5.

FIGURE 5. VARIOUS SHAPES OF STIFFENERS

r

Section B4.8

15 October 1969

Page 9

Figures 6 and 7 give K /31 relationships for b = d and b - d/2 with theS

longitudinal edges simply supported. Figures 8 and 9 give K /3/ relationshipss

for b = d ,and b = d/2 with the longitudinal edges clamped. The maximum value

of CT/B T plotted is for a closed circular tube. This value is 0.769.

u cT/B = t0.6 AND 0.769 __

lo .r/.. _-I_ _

LONGITUDINAL EDGESKs 9 SIMPLY SUPPORTED

8 -_ _ c d c

# THEORETICAL P/M

BUCKLING MODES 1/0

7 / 2/I3/1

6 / 4/I

24

FIGURE 6. K VERSUS YT IIELATIONStIIPS FOR b :-dS C

From these figures it is shown that very significant increases in the

buckling resistance, Ks, are obtainable by using closed-section stiffeners in

place of the open-section stiffeners so frequently used. For ex,'unple, by using

a thin-walled circular tube for the stiffeners (CT/B T = 0. 769) the gain in K L

Section B4. S

15 October 1969

Page i0

35-- o._ _-_

0.1--_

0.2--0.4 -._

30 O.OS-

' 0,769

° 0.6

_ LONGITUDINAL EDGES

K I 20 SIMPLY SUPPORTED

,.Eo,,,o UC L.OMO0 S,/0

2/I

S ........... 3/i4/1

24O

FIGURE 7. Ks VERSUS 5,T RELATIONSHIPS FOR b = dc/2

is 25 percent and 60 percent for a = 1 and c_ = 2, respectively, when the

longitudinal edges are simply supported. For the case of clamped edges, the

gains are 13 percent and 43 percent for a = 1 and a = 2, respectively. Thus,

if a minimum weight design is desired, consideration should be given to

closed-section stiffeners.

IV. Transverse and Central Longitudinal Stiffeners

The use of deep beams with webs having a high depth-to-thickness ratio,

may make it desirable to employ both vertical and horizontal stiffening. When

f

St, ction B4.8

15 October 19(;9

Page 1 1

yi Do, K L = 14JJ2

14 A,'f' "_ .... --.,,.-'r- ..... _ .....

12 #r _ _ t ' t _

] / = . LONGITUDINAL. EDGES CLAMPEO

10 _ TH:O:,TIC,L ,UCIk_I/:GMO)E$

2 I 5/'1 ....

3 I .... 1/0 ......41 --.--

9

0 2 4 6 8 10 12 14 16 18 20 22 24 26

Y=

"T BT = 0.769.0.40._

0.20_ \

.or--?0_\" _\ T.... r _--f_-------- -/ /o'os.\' \ x ,_+.+_-.7:'lX--_--7

I iN" I

K s

20

10

0 20 40 60 80 100 120 140 160

¥'r

1 !

),r = c_, KL= 41,55

l

y.r = m, KL. = 29.17

LI i,LONG_TUDtNkL EDGESCLkMPEO

,iTHEORETICAL BUCK k ING MOOE|

P ' M P/M

2 t 4'1 -----

3 ! .... $ I ....

I_ 2@0 2_ 2_ 2JO

FIGUIIE 9. K, "/T RELA'I?IONSIIIPS; ASPECT I{ATIO o_ 2.0

Section B4.8

15 October 1969

Page 12

a web is subjected to shear, the most effective position for a single horizontal

stiffener is at middepth. This combination of vertical and horizontal stiffening

can result in more economical designs than are possible when only vertical

stiffeners are employed. For example, the weight of stiffening required to

achieve a given buckling stress with horizontal and vertical stiffening can be

as little as 50 percent of the weight required when only vertical stiffeners are

used.

If neither of the transverse or central longitudinal stiffeners has torsional

rigidity and the vertical stiffeners have a rigidity equal to or greater than EILv ,

then the value of 7LH necessary to produce the limiting value of KLH is given by

and

_LH = 11.25 (b/de)2 (5)

KLH = 29 +4.5(b/d )-2 . (6)e

Additional weight savings can bc achieved if torsionally strong stif[cners are

used in either the transverse or longitudinal direction. For example, studies

in Reference 6 have shown gains in the value of K L (which is proportional to the

weight of the web) of up to 25 and 60 percent for _ equal to one and two,

respectively, by using closed-section stiffeners in the transverse direction

only. Investigators have given parameter studies on the following in References

4 and 9:

1.

2.

Transverse and longitudinal stiffeners of closed tubular cross section.

Transverse stiffeners of closed tubular cross section; longitudinal

stiffener possessing only flexural rigidity.

Section I34.8

15 October 1969

Page 13

3. Transverse stiffeners possessing only flexural rigidity, the longitudinal

stiffeners being of closed tubular cross section.

Figure 10 enables one to make an assessment of the benefits that result

from using torsionally strong stiffeners when a central longitudinal stiffener is

used in conjunction with a system of equally spaced transverse stiffeners. When

c_ _ 1, it is evident that little increase is obtained in the buckling resistance by

using torsionally strong transverse stiffeners, but that a considerable increase

in buckling resistance will be obtained by using a torsionally strong longitudinal

stiffener. However, as c_ increases (that is, as the transverse stiffeners are

6O

55

5O

45

K s 40

35

3O

25

20

FLANGES ANO LONGITUDINAL ALL EDGES

STIPFEIqER PROVIDE CLAkIPEO_IaC_'/CLAMPE O

SUPPORT. TRANSVERSE _

STIFFENERS SIMPLE SUPPORT •

' _ Y'. I ,CA.CEoR"'o,._ _\ \ "-. I L°"or Tu °l"* "____.___2:':"0.

STIPFENER PROVIDES I

, ALL I!/OGE$ _ P :uOp/DER TCLFA::: GE S AND "_

I51MPLY SUPPORTED LO_IGITUDINAL STIPFENER5;.; _ iUPPORTED

2 1 0

CX: dc/b

FIGURE 10. K VERSUS ASPECT RATIO FO[_ VAIIIOUS EDGE RESTRAINTSS

Section B4.8

15 Oct()ber 1969

Page i4

more closely spaced), the increase in K resulting from the use of torsionally

strong stiffeners becomes more significant; until, when (_ = 1.7, the buckling

resistance obtained with torsionally strong transverse stiffeners and a

longitudinal stiffener without torsional rigidity is equal to that obtained with a

torsionally strong longitudinal stiffener and transverse stiffeners possessing

only flexural rigidity. Thus, it will be readily seen that is is necessary to know

the relationships between K and the stilfener propertics for the cases enumerateds

above. Figures 11, 12, and 13 give typi(':_[ relationships between K_ and YT for

b = d and different positions of the torsionally strong _tiifener. Culves for

other values of the aspect ratio can be obtained from Reference 4. The points

3o

2C

K

10

I

//

20 40

= soo ]lOO._.-----_30 ,,-v i

20

10

80

y.r = EID b

°

b--d

f_ 3I;c"

2/I P/M

........ 3,"2 THEORETICAL

4/il5/I BUCKLINGMODES

100 120 140

FIGURE 11. Ks' YT RELATIONSHIPS; LONGITUDINAL STIFFENER WITH

TORSION (C L :_ 0.769 BL); TRANS\rERSE STIFFENERS WITH ONLY

FLEXURAL RIGIDITY (C T -: 0) ; ASPECT RATIO a _ l. 0

Section B4.815October 1969[>age 15

30

2o

= 500°U

100

A

00 20 40 60 80 100

'y.r= EI'Db

-'--dc

I-------,H

fi I/ rdc¸¸2-E c,f22/'1 } P,'M

3/1 THEORETICAL4:1 BUCKL_NCiS/'1 MODES1 i

120 1,10

FIGURE 12. Ks' %1" I1ELATIONSIIIPS; LONGITUI)INAI_ EDGES SIMI'LY

SUPPORTED: TIIANSVEI1SI'2 STII"FENERS WITII TOIISIONAL RIGIDITY

(C T _ 0.7(19 BT): IAI)NGI'FUI)tNAL STIFFENERS WIT II ()NLY FLEXURAL

RIGIDITY (C _-tl); ASPECT RATI() _' 1.ItL

/'7..---1

3O //:if'//

20_/// YL-5

K 1/ o

1 :

/

o2o

ff

b=ddc/2

ilt 4/I / m_:.km_5/1 / i_li

dc/2

40 60 80 100 120 I140

Y.r = El 'Db

K ' YT RELATIONSHIPS; SIMPLY SUPPOI_TED I,ONGITUDINALS

0. 769 BT'

FIGURE 13.

EDGES; ALL STIFFENERS WITI! TORSIONAL RIGIDITY (CT

C L= 0.769 BL);_ ASPECT RATIO ff 1.0

Section B4.8

i5 October 1969

Page 16

marked A on the curves of Figures 11, 12, and 13 are values of TT which would

give 95 percent of the limiting value of K . This is suggested as a good cutoffs

point for an efficient design. If greater values of _/T are chosen, only a small

increase in the value of Ks would occur. Thus, the extra increase in TT would

not be very beneficial from a weight standpoint.

V. Longitudinally Stiffened Web Plates in Longitudinal Compression

In deep beams, it is often economical to stiffen the web plate by longitudinal

stiffeners in locations where the longitudinal compressive stresses resulting

from bending are high. Two positions of the stiffener will be considered here:

(1) The stiffener located at the longitudinal center line of the web, that is, at

the neutral axis (Fig. 14a) and (2) the stiffener located in the compressive

region at a distance from the edge of the plate (Fig. 14b). In case 1, the

stiffener itself does not carry compressive stresses.

fb STIFFENER fb fb STIFFENER fb

_ _: \ % . \ t_,

dcl2 dc

(o) (b)

FIGURE 14. POSITIONS OF LONGITUDINAL STIFFENERS

Section B4.8

15 October 1969

Page 17

Adding this longitudinal stiffener results in another structural part and more

assembly cost; therefore, such construction is not widely used although it is a

structural arrangement that will save structural weight under certain conditions

of beam depth, span, and external loading.

A. Stiffener at the Centerline

For a stiffener at the eenterline, the largest practical value of the

stiffener moment of inertia is

I= O. 92t3d

_'B c(7)

With this value of I , the critical bending stress F BOcr

is

<)2rib 12(1_g2) for c_ >- 2/3(_)

For values of F B for stiffener moment of inertia less than Io, seeer

Reference 10.

B. Stiffeners Located Between Compression Edge and Neutral Axis

The increase in buckling strength that can be obtained by a stiffener at

the eenterline of the web amounts to only 50 percent of the unstfffened plate in

the inelastic range. Stiffeners at the centerline are therefore not very effective

in improving the stability of web plates in ease of pure bending stresses.

SectionB4.815()ctobcr 1969Pagei8

v

For a stiffener spaced at a distance m -- d /4, the critical buckling stresse'

is given by

F B

cr- 1017r2E (_c) 2_?B 12(1-hz)(_ -> 0.4), (9)

if 3: -> Yo

EIWhere y =

T/BDd c

EI= O

To Dd = (12.6 +506) ot2 - 3.4 c_3 (o_ 1.6) andC

A6 = , a = b/d

rldct c

Comparison of the results obtained above for a stiffener at the centerline

of the plate with the results obtained for a plate stiffened in the compression

region shows that the reinforcement in the latter case is much more effective.

Limited numerical results have been obtained for plates reinforced by a

longitudinal stiffener located at a distance m = dc/5 from the compression edge

of the plate. This information is plotted in Figures 15 and 16. The largest

value of the buckling strength of the plate stiffener system corresponds to

K = 129 and is larger than in the case of a stiffener located at the distance dc/4

from the compression edge.

Section B4.815October i969Page i9

129

120

8O

K

_TJ°/..7

ii//y ' _,,_t b --_---- m

r _',l t4ole dc!

I10 20 30

m _m=

El

DdC

4O

FIGURE 15. K VERSUS y FOIl VARIOUS VALUES OF c_, 5 = 0

K

12_ .....120 _---;I_

AT/

4011-

gj i

I

I

0 10

=5

"-i ]d c

20 30 40

El

Dcl c

5O

FIGURE 16. K VERSUS y FOIl VAIIIOUS VALUES OF c_, 6 = 0.10

SectionB4.815October i969Page20

VI. CombinedStresses

The abovementionedbending, shear, and possibly axial andtransverse

stresses that act uponthe web shouldbe interacted by the following equations.

R2+R _1s cL

R 2+R <1

s c T

R 2 +RB2 _-<1S

RB1.75 +Rc < 1L

(10)

(11)

(12)

(13)

where

f fS e

R = R B = R -s F ' F ' c F Cs B

cr cr er

and the subscript, L, indicates longitudinal and the subscript, T, indicates

transverse. The critical values of F C should be obtained from Section C2.1.1.cr

If the interaction equations above are not satisfied, an iteration of the design

must be performed.

4.8.1.2 Flange Design

The beam flanges are designed for tensile and compressive norma[ forces.

The ultimate allowable stress for the tension flange is equal to Ftu of the

material, reduced by the attachment efficiency factor. For riveted or bolted

connections, the efficiency factor is the ratio of the net area to the gross area

of the cap.

Section B4.8

15 October 1969

Page 2 1

Compression flanges should be designed for column stability. The loads

on the compression flange can be a combination of normal force, longitudinal

shear at the web-flange connection, and transverse forces. Specific analysis

techniques are available in Section B4.4.0.

4.8.1.3 Rivet Design

I. Web-to-Stiffener

Although no exact information is available on the strength required of the

attachment of the stiffeners to the web, the data in Table B4.8-I are recommended.

Table B4.8-I. Rivets: Web-to-Stiffener

Web

Thickness (in.)

0. 025

0. 032

0. 040

0. 051

0. O64

0. 072

0. 081

0. 091

0. 102

0. 125

0. 156

0. 188

Rivet

Size

AD 3

AD 4

AD 4

AD4

AD4

AD 5

AD 5

AD 5

DD 6

DD 6

DD 6

DD 8

Rivet

Spacing (in.)

1.00

1.25

1.10

1. O0

0.90

1.10

1.00

0.90

1.10

1. 00

0.90

1.00

Section B4.8

15 October 1969

Page 22

II. Stiffeners-to-Flange

No information is available on the strength required of the attachment of the

stiffeners to the flange. It is recommended that one rivet the next size larger

than that used in the attachment of the stiffeners to the web or two rivets the

same size be used whenever possible.

III. Web-to-Flange

The rivet size and spacing should lie designed so that the rivet allowable

(bearing or shear) divided by q x p, the applied web shear flow times the rivet

spacing, gives the proper margin of safety. For a good design and to avoid

undue stress concentration, the rivet factor, Cr, should not be less than 0.6.

4.8.1.4 Design Approach

The design of stiffened shear-resistant beams is a trial-and-error method.

Assuming q, b, h, and E are known, the first step is to assume a reasonable

value of t and compute f = q/t. The problem is to find the moment of inertias

of a stiffener required to develop an initial buckling stress, F , in the webs

cr

greater than 1 by the desired margin of safety. The procedure is to choose8

the desired Fs ' and Fs /_s if required. Ks is then found from equation (1)cr cr

as a function of F /7 and d/t or h/t. Then, depending on the type of stiffeningS

cr

arrangement used, the required I of the stiffener is obtained from Paragraph

4.8.1.1. Particular attention should be given to stiffener properties that provide

an efficient design. For a minimum-weight design, consideration should be

given to longitudinal stiffeners and/or torsionally strong stiffeners as discussed

Section B4.8

15 ()ctober 1969

Page 23

in Paragraph 4.8. I.I. Attention should also be given to stiffenerthickness

and rivet location as discussed in Paragraph 4.8.1.1-II.

4.8. I.5 Stress Analysis Procedure

The stress analysis procedure for the web of a stiffenedshear resistant

beam is straightforward and easy to apply. Since q, b, h, E, t, fs' and I are

all tmown, the first step is to obtain K from the appropriate curve in Paragraphs

4.8.1.1, depending upon the aspect ratio, torsional rigidity, stiffener thickness,

etc. Then FS

cr

can be obtained from equation (1). If necessary, values of

77 c,'m be obtained from Section C2.0. Then the margin of safety for the web isS

FS

erM.S. - -1. (14)

fS

4.8.1.6 Other Types of Web Design

The web designs discussed previously require a large number of parts

(stiffeners) to achieve lightness. To keep manufacturing costs low, the

number of parts must be kept to a minimum. The problem is one of weight

trade-off versus manufacturing expense.

Three types of shear-resistant, nonbuckling webs are frequently used in

design to save the expense of stiffeners. Actually, the web in most cases is as

light as, or lighter than, a web with separate stiffeners. There is a general

limitation, however, in that a stiffener must be provided wherever a significant

load is introduced into the beam. The web types are:

I. Web with formed vertical beads at a minimum spacing

II. Web with round lightening holes having 45 degree formed flanges at

various spacing

SectionB4.815October i969Page24

III. Web with round lightening holes having formed beaded flanges and

vertical formed beads between holes.

The webs with holes, II and III, also provide built-in access space for the many

hydraulic and electrical lines that are sometimes required.

Procedures for the design of these beams should be obtained from

Reference 11.

Section B4. S

15 (_tober 1969

Page 25

4.8.2 PLANE TENSION-FIELD BEAMS

If web buckling occurs :ffter some application ¢ff load, tile shear load beyond

buckling is resisted in part by pure tension-field action of the web, and in part

by shear-resistant action of the web (Fig. 17). This action of the web is

defined as an incomplete tension field, or partial tension field.

Th

Ltl-]l

(a) NONBUCKLED ("SHEAR-RESISTANT") WEB

(b) PURE DIAGONAL-TENSION WEB

FIGURE 17. STATE OF STRESS IN A BEAM WEB

Section B4.8

i5 October 1969

Page 26

The theory of pure tension-field beams was published by Wagner [ 1 ] .

In this theory, after initial web buckling, the total shear load is resisted by

pure tension-field action of the web. Such behavior is essentially nonexistent

in practice, and will not be discussed here.

Kuhn, Peterson, and Levin [ 12 ], developed a semiempirical analysis for

partial tension-field beams. Correlation with experimental results indicates

that the analysis is conservative for the beams within the range of beams

tested. The experimental verification for the analysis was restricted to 2024S-T

and 7075S-T aluminum alloy beams. As a consequence, the Kuhn analysis is

limited to beams of these alloys. Extension of the analysis to include other

alloys is not documented, and the designer must exercise considerable caution

in attempting an extension of Kuhn's work in the analysis of beams fabricated

from other alloys.

4.8.2.1 General Limitations and Symbols

The methods of analysis and design given herein are believed to furnish

reasonable assurance of conservative strength predictions, provided that

normal design practices and proportions are used. The most important points

II.

III.

are:

I. The uprights should not be "too thin"; keep

tu/t> 0.6.

The upright spacing should be in the range

0.2 < b/h < 1.0.

The method of analysis presented here is applicable only to beams

with webs in the range 60 < h/t < 1500.

When h/t < 115, the portal frames effect and the effect of unsymmetrical

flanges must be taken into account by using Reference 13.

Section 134.815October 1969Page27

SYMBOLS (in addition to those given in the front matter)

Af area of tension or compression flange

AU

actual area of upright (stiffener)

Aue

effective area of upright

C1,C2,C3

Fco

stress concentration factors

column yield stress (the column stress at m _L_

- O)P

FC

allowable column stress

Fmax

ultimate allowable compressive stress for natural crippling

F0

ultimate allowable compressive stress for forced crippling

FS

ultimate allowable web shear stress

If average moment of inertia of beam flanges

IS

required moment of inertia of upright about its base

Iu

moment of inertia of upright about its base

Msb secondary bending moment in the flange

P applied shear load

PU

upright end load

distance of upright centroid to web

fcent compressive stress at centroidal axis of upright

ff compressive stress in flange because of the distributed vertical

component of the diagonal tension

fU

average lengthwise compressive stress in upright

fsb

fU

max

k

b

h

qr

P

kSS

R d, R h

apD T

N

Q.

hU

Le

o)

4.8.2.2

Section B4.8

15 October 1969

Page 28

secondary bending stress in flange because of the distributed vertical

component of the diagonal tension

maximum compressive stress in upright

diagonal tension factor

spacing of uprights

effective depth of beam centroid of compression flange to centroid

of tension flange

rivet shear load, web-to-flange and web splices

angle of diagonal tension

radius of gyration of upright with respect to its centroidal axis

parallel to web (no portion of web to be included)

critical shear stress coefficient

restraint coefficients

angle of diagonal tension for pure tension field beam

applied transverse load

static moment of cross section

height of stiffener

effective stiffener length

load-per-inch acting normal to end bay stiffener

Analysis of Web

The web shear flow can be closely approximated by

Vq : T (Is)

Section B4.8

15 October 1969

Page 29

From this, it follows that the web shear stress is

f = q/tS

The critical buckling stress of the web is given by

(16)

F

c =kss 12(1-p z) b Hh + 1/2(Rd

and

(17. a)

F

scr kss 12(1-p z) d + 1/2(Rh - Rd) b > d c.

(17. b)

The value of kss is obtained from Figure 18. The values R h and Rd, the

restraint coefficients, are given in Figure 19. Figure 20 provides F for theS

er

case _7¢ 1. When R h is very small, the value of the critical shear stress

calculated from the equation above may be less than the value computed

disregarding the presence of stiffeners. In this case, the stiffeners are

disregarded and the latter value is used.

SectionB4.815October 1969Page30

k$$

10

Ji

//

//

f

r-!

i

f$]

m._r------_ _mm._-------- qlr-----_

r b v

d¢

r±II

b, dc: CLEAR PANEL DIMENSIONS

o 0.2 0.4 0.6

dc 'b

0.8 1.0

FIGURE 18. k VERSUS d /bSS e

The loading ratio, fs/Fs , is used to determine the tension field factor, k.cr

It may be calculated by

k-: tanh(0.5 logl0 fs/Fscr)f > F , (18)

S Scr

Secti_,n B4.8

15 ()ctol)er 1!)6_)

l)age 3 1

Rk, Rd 0.8

/

/°, //

i //

0 0.5

t - WEB THICKNESS

t! - FLANGE THICKNESS

t - STIFFENER THICKNESS

Rh - RESTRAINT COEFF. ALONG STIFFENER

R d - RESTRAINT COEFF. ALONG FLANGE

1.0 1.S 2.0 3.5 3.0

tuT ' t

FIGURE 19. EDGE RESTRAINT COEFFICIENTS FOR \VEB BUCKLING STRESS

SectionB4.815October t969

Page 32

F$cf

4O

3O

20

lO

I

llO

2024-T3

20 30 40 GO 60

F scr" q

FIGURE 20. VALUES OF F WHEN 77 ¢ 1S

cr

or it may read from Figure 21.

unbuckled state.

For values of fS

- F , the web is in the

er

f-

f

Section B4.8

15 October 1969

Page 33

b

.I

2

9

1

• -. ¢D 0 0 0

N

;I

R

2

qr_

i

oo

q' _iO.L_)¥:l NOISNIJ. "IVNOOVI(]

ubl

41

©

<

<©

m

M

N©

r..)<

2:

r_Z

r_

Section B4.8

i5 October i969

Page 34

The angle of the diagonal tension is then obtained from Figure 22, which

shows the variation of tan o_ as a function of k and tb/A . For double stiffeners,ue

A is equal to the cross-sectional area of the stiffeners. For single stiffeners,ue

AU

A = ,,. (19)ue 1 + _e/p) 2

It is recommended that the diagonal tension factor at ultimate load be limited

to a maximum value,

/k = O. 78 - (t - O. 012) 1/2

max, (20)

to avoid excessive wrinkling and permanent set at limit load, thereby inviting

fatigue failure.

The average web shearing stress, fs' may be appreciably smaller than

the maximum web stress. The maximum web stress is given by

f = f (1 +k2C1)(1 +kC2) , (21)S S

max

where C1 ,-rod C2 are empirical coefficients obtained from Figure 23. C 1 is a

correction factor accounting for c_ differing from 45 degrees. C2 is the

stress-concentration factor arising from the flange flexibility.

The web allowable stress, F , is given in Figure 24 as a function of kSail

and cePDT, the angle that the buckles would assume if the web could reach the

Section B4.8

15 October 1969

Page 35

J,.

J

O

I

i/

, jj,i iII/1/,ss

-- z0

Zc) ill

•-'-" i.-..I.<Z

<_ _O-

=77,/

J/j/

/

o_. 0 • • 0

II uol

.<

Zoes_

<Z0q_<

0

Z<

C'I-

;.q

r_

Section B4.8

15 October 1969

Page 36

cl

0.12

0.08

0.04

00.6

a- ANGLE OF DIAGONAL TENSION

0.7 0.8 0.9

tan O[

1.0

c 2, c3

1.2

0.8

0.d

c3

"_ 1/4cob = 0.7 b t

.(I c + It) he .

Ic = MOMENT OF INERTIA OFCOMPRESSION FLANGE

It = MOMENT OF INERTIA OFTENSION FLANGE

J0 1 2 3 4

cob

FIGURE 23. EMPIRICAL COEFFICIENTS FOI_ MAXIMUM SHEAR STRESS

IN WEB AND FOR SECONDARY BENDING MOMENT IN FLANGES

S

]0

25

I0

Section B4.8

15 October 1969

Page 37

t

*p,W

45 4835

30

25

20

O.2 O.4 0.6 0J

k

(e) _)24-T3 ALUMINUM ALLOY. • 62 ksi|tv| t

DASHED LINE IS ALLOWABLE YIELD STREW

1.8

)5

3O

2O

150

II

r

- IJ

0.2 0.4 0.6 0.i 1.0

(b) ALCLADT075-T6 ALUMINUM ALLOY. ttult 72ksi

e_h. DEG4$

4835

30

2S

2O

FIGURE 24. BASIC ALLOWABLE VALUES OF fs

max

Section B4.8

15 October 1969

Page 38

state of pure diagonal tension without rupturing. The values of F have beenSall

established by tests and may be called "basic allowable. "

connections, they are applied as follows:

I.

For different

Bolts, just snug, heavy washers under bolt heads, or web plates

sandwiched between flange angles; use basic allowables.

II. Bolts, just snug, bolt heads bearing directly on sheet; reduce basic

allowables 10 percent.

III. Rivets, assumed to be tight; increase basic allowables 10 percent.

IV. Rivets, assumed to be loosened in service; use basic allowables.

The allowable stresses given are valid if the allowable bearing stresses on the

sheet or rivets are not exceeded. They are not valid for countersunk rivets.

For webs of unusual dimensions arranged unsymmetrically with respect to the

flange, use Figure 25 to obtain FSal I

4.8.2.3 Analysis of Stiffeners

Stiffener loads result from the web diagonal tension and the transverse load

not carried by the web. The stiffener load is given by

I F tb 1

SCI"

P = ktbf tanoL + Nb 1- (22)u s fs (td + Au) '

where N is positive for a transverse compressive load. This load is resisted

by the stiffener and the effective web. The effective width of the web working

with the stiffener ma_ be assumed to be given by

be

- 0.5(l-k) . (23)b

f

Sectiol_ B4.8

15 October 1969

Page 39

....... j

20 _

,!

I0

0

4O

3O

F'eli, kli 20

I0

00

2024 BASED ON F_ : 62ksi

[ ] T0.2 0.4 0.6 0.8 1.0

7075 6ASEDON Ftu

0.2 0.4

FIGURE 25. ALLOWABLE WED STRESSES FOR 2024 AND

7075 AT HOOM TEMPERATURE

Section B4.8

15 October 1969

Page 40

The average stiffener stress is then

Ptl

f = (24)u A + 0.5(l-k) tb

ue

The maximum compressive stress in the stiffener occurs near the neutral axis

of the beam. The ratio of the maximum stiffener stress to the average stiffener

stress, fu /fu' is obtained from Figure 26.max

Stiffeners may fail by column action or by local crippling. Column failure

by true elastic instability is possible only in (symmetrical) double stiffeners.

A single stiffener is an eccentrically loaded compression member whose

failing stress is a function of the web and stiffener properties. To guard against

excessive bowing and column stress, the following must be adhered to:

I. The stress f must not exceed the column yield stress.U

II. The average stress over the column cross section, feent = fu Aue/Au'

must not exceed the allowable stress for a column with the slenderness

ratio hu/2O.

The effective column length for double stiffeners is

hU

L = b < 1.5h (25)

e '41 + ka(3-2 b/hu)

L = h b-> 1.5h (26)O U

J

/

V/

//j//I,

'1I

/

i

/ A ,

Section B4.8

15 October 1969

Page 41

p/r -

fGO

O

II I I" o

9

LqZ_q

Z

_q

E_

_qL_,<

_q><

0E_

;.q

E_D_

©

©

W£Xl

M

Section B4.8

15 October 1969

Page 42

To avoid column failure of double stiffeners, the average stress, f, should be

less than the allowable stress taken from the column curve for solid sections

of the stiffener material, with the slenderness ratio Le/p.

Forced crippling of stiffeners must be considered. In this mode of failure,

the attached leg of the stiffener is deformed by being forced to adapt itself to the

web shear wrinkles.

The allowable forced crippling stress is given by the empirical equation

t 1/s/__\

O \ L/

where C is a constant as follows:

2024-T (Bare)

7075-T (Bare)

Nomographs for Fo

Single Stiffener_

C =26.0

C =32.5

are given in Figure 27. If FO

Double Stiffener

21.0

26.0

exceeds the material

proportional limit, a plasticity factor, 77, equal to Esec/Ec is used in the

equation above.

Torsional stability of single stiffeners is provided by meeting the following

criteria:

(fs- Fs )h t = 0"23E (J-_bh2)I/3e , (28)cr

it'--"

S

1.0

0.9

0.8

0.7

0.6

0.5

0.4

k

0.3

0.2

0.1

F o. ks,

DOUBLE

UPRIGHTS

Y)

.tO

6O

10 -_

8

7

6

3 ---

4O

30

- 20

.- 1.5

- l0

9

--- 8

- 7

-- - 6

5

Fo, kS,

SINGLE

U PRIGH TS

Iol 2024 T3 ALUMINUM ALLOY

Secth_n B4. S

15 ()ctob(,r 19(;!)

Page 43

10

9

8

7

6

-- $

-_: 4

- 3

t u

-- 2 t

-- 1.5

-:: I.o0.9

--. 0.8

0.7

- 0.6

- 0.5

- 0.4

FIGURE 27. NOMOGRAM FOR ALLOWABLE UPRIGHT

STRESS (FORCED CRIPPLING)


10

0,9

08

07

0.6

0.5

04

0.3

O2

0 1

f: k*.lO'

DOUBLE

UPRIGHTS

7r

6fJ

If

_0

20

10

9

8

7

6

o

SINGLE

UPRIGHTS

.

9

I'w

t

2

1.0

0.9

0.11

0.7

0.6

0.5

0.4

qb_ 7075-T6 ALUMINUM ALLOY

FIGURE 27. (Concluded)

where

Section B4.815 October 1969

Page 45

(f -F )hetS Scr

= total web shear load above buckling which can be carried

before the stiffener cripples

J1 = the effective polar moment of inertia of the stiffener

1/3 (developed width} t 3. This applies for formed sheet stiffeners.U

To prevent :_ forced crippling type of failure when the upright resists an

extern_ compressive load in addition to the compressive load resulting from

diagonal tension, an interaction equation such as the following must be used to

evaluate this effect:

m_______,_ + __2

Fo ce/

= 1 , (29)

where f and FU O

In ax

are the maximum upright compressive stress and the

allowable forced crippling stress, respectively, for diagonal tension acting

alone, and f and F are the actual and allowable compressive stress,co ce

respectively, resulting from external compressive stress acting alone.

An effective area of web plus upright may be used in computing f Theee"

allowable crippling stress, Fmax, may be used for Fee.

The effect of the external load should also be investigated with respect to

column failure of the upright. To prevent cohmm failure under combined

SectionB4.8i5 October 1969

Page 46

loading the following criteria should be fulfilled:

and

f +f <Fu ce co

f +f <-Fcent ce c

(30)

(31)

4.8.2.4 Analysis of Flange

The flange stress is the result of the superposition of three individual

stresses: (1) primary bending stresses, (2) axial compression because of

the flange-parallel component of the web diagonal tension, and (3) secondary

bending stresses because of the stiffener-parallel component of the web

diagonal tension.

The primary bending stresses are given by

s If_ c 1- cr

fprim If --_s ( 1- _- )(32)

where

I = moment of inertia of section

and

If = moment of inertia of section (web neglected).

The total axial load because of the flange-parallel component of the web

diagonal tension and applied axial load is

S

Paxial = khtf cotc_ + P 1- crs a IS

(33)

Section B4.815October 1969Page47

Pa

is positive for compressive axial load.

Paxial

faxial = A +A +0.5(l-k)thc t

The axial flange stress is then

(34)

where A _md A are the area of the compression and tension flange.c t

The secondary bending stress is given by

sec seef

wherePb

uM = C 3sec 12

and

PbU

M = C 3sec 24

(over s tiffene r }

(midway between stiffeners)

(35)

C3 is an empirical stress concentration factor, given in Figure 23.

The allowable stress for the compression flange can be found by the methods

of Section C 1.0. The allowable tension stress for a tension flange is given by

F of the material, modified by the attachment efficiency factor.tu

4.8.2.5 Analysis of Rivets

Web-to-Flange: The flange-web shear flow at the line of attachment is

Vq = h--V (I +0.414k) , (36)

where h' = beam depth between attachment line of flange web.


Web-to-Stiffener: The stiffener-web rivets, for double stiffeners, must

develop sufficient longitudinal shear strength to make the two stiffeners act as

a unit until column failure occurs. The shear strength should be

2F Qq = cy

b LS e

(37)

where b = outstanding stiffener flange width.S

The stiffener-web connectors must carry a tension component as follows:

i t

and

N'

--- 0.15t F (double stiffener)tu

(38)

= 0.22t Ftu (single stiffener) (39)

The interaction of shear and tension in the connectors is given in Reference 11.

Stiffener-to-Flange: The stiffener-to-flange connectors are designed with

the empirical relationship

P =f AU U tie

which gives the load in the stiffener.

into the cap.

4.8.2.6 Analysis of End of Beam

The previous discussion has been concerned with the "interior" Oays of a

beam. The vertical stiffeners in these areas are subject, primarily, only to

axial compression loads, as presented. The outer, or "end bay, " is a special

case. Since the diagonal tension effect results in an inward pull on the end

(.i0)

The connection must transfer this load


stiffener, it producesbendingin it, as well as the usual compressive axial

load. Obviously, the end stiffener must be considerably heavier than the others,

or at least supported by additional members to reduce the stresses resulting

from bending.

The component of the running-load-per-inch that produces bending in such

edge members is given by the formulas

w -- kq tml o' (41)

for edge members parallel to the neutral axis (stringers) and

_) = kq cot _ (42)

for members normal to the neutral axis (stiffeners). The longer the unsupported

length of the edge member subjected to w, the greater will be the bending

moment it must carry.

There are, in general, three ways of dealing with the edge member subjected

to bending, the object being to keep the weight down.

I. Simply "beef-up" or strengthen the edge member so it can carry all of

its loads. (This is inefficient for long unsupported lengths. )

II. Increase the thickness of the end bay panel either to make it nonbuckling

or to reduce k, and thereby reduce the running load producing bending

in the edge member. (This is usually inefficient for large panels. )

III. Provide additional members (stiffeners) to support the edge member

and thereby reduce its bending moment because of w. (This requires

additional parts. )

Actually, a combination of these methods might be best.


4.8.2.7 Beam Design

This paragraph presents methods to facilitate efficient preliminary design

of tension field beams.

Allowable Shear Flow: Figure 28 gives the ultimate allowable shear flow,

q, for 7075S-T6 Alclad sheet as a function of the sheet thickness, t, and the

stiffener spacing, b. The dashed line on the left is the approximate boundary

between shear-resistant and tension-field beams when the sheet is loaded to

full strength. The central dashed line is a limitation on b , the m:uximummax

stiffener spacing, in order to minimize the possibility of excessive wrinkling

and permanent set when the web works at full strength. Stiffener spacings

greater than b can be used, if necessary, by using shear flow availablesmax

which conform to the limitation on the value of k given in Paragraph 4.8.2.2.max

The dashed line at the right establishes b' the absolute maximum stiffenerm_iX'

spacing in order to prevent "oil canning. "

One of the assumptions made in the construction of Fig_ure 28 is that the

aspect ratio b/h = 0.5. Varying b/h has only a small effect on the curves, :is

can be seen from the curve for 0. 050 she(-t, where additional curves for

b/h = 0.2 and b/h = 1.0 are plotted. The relationship for other sheet gages is

approximately the same.

Stiffener Area Estimation: Figure 29 presents the stiffener are:_ to web

area ratio plotted as a function of b/h and x/q/h, the square root ,)f ttm

structural index. This index is a measure of the loading intensity on the beam.

These curves are to be used only as a me:ins of roughly approximating the

required area of stiffener for preliminary desig_ and preliminary weight

f

Section B4.8

15 October i969

Page 5 i

Ioooo90001

7000 t

6OO0

._00

4OOO

3OO0

2,$00

20GO

I,$00

quit

lb.

tn.

I000

BOO

70_

6OO

30O

_0

aMAX TO PREVENT EXCESSIVE I

WRINKLING AND PERMANENT SET|

WHEN WEB WORKS AT FULL I

',TR ENGTH J

'I f

, Q

Ii

2. WEB SI/_PLY SUPPORTED

3. THE STRESS CONCENTRATION

FACTOR DUE TO FLANGE

FLEXIBILITY. (2 = 0.2

. 1 iIo 12 h 16 la 20 22 24 26

I--m

28

b. ,_.

NOTE: FOR BARE 7075S-T6, MULTIPLY BY 1.07

FIGURE 28. ULTIMATE ALLOWABLE SHEAR FLOW ALCLAD 7075S-T6 SHEET

i I

Q

Section B4.8

15 October 1969

Page 52

!_- !

i

o

°1-

i

C/3

LC_L'--

D

N

c_c_

_ZZ_

zr_

r_mZ_

M b-

C)

v-

f

Section B4.8

15 October 1969

Page 53

estimation. If the stiffener design is limited to standard sections, the required

area might be larger than that given in Figure 29 since a zero margin of safety

cannot always be obtained. The curves are for 7075S-T single-angle stiffeners.

Curves for double stiffeners or 2024S-T material, similar to Figure 29, can be

found in Reference 12.

Design Method: The preliminary design of 7075S-T web-stiffener can be

arrived at in the following manner:

1. The desi_m shear flow, q, and the depth of beam, h, are usually known.

This fixes q_-_, the square root of the structural index.

2. The stiffener spacing, b, is often determined by considerations not

under the control of the designer. If such is the case, inspection of

Figure 29 shows that a stiffener spacing, b, equal to the beam depth,

h, is desirable for minimum weight design of the web-stiffener system.

However, it is possible that wide stiffener spacing might induce

excessive secondary bending in the flange. In general, b/h ratios from

0.5 to around 0.8 are commonly used for tension-field beams.

3. The required web thickness, t, can be obtained from Figure 28 since

q and b are known. This figure can also be used to check the stiffener

against the maximum allowable spacing, bmax"

4. Estimate the required value Au/bt with the aid of Figure 29.

5. Compute the approximate cross-sectional area of stiffener as follows:

e

Section B4.8

15 October 1969

Page 54

Choose a stiffener with the proper area. Unless the beam is very deep,

or unless there are other design considerations, a single-angle stiffener

is an efficient design. Also, a stocky, equal-legged angle gives greater

resistance against forced crippling, which is usually the dominant mode

of failure.

f

/-

f

Section B4.8

15 October 1969

Page 55

REFERENCES

1. Wagner, H., "Flat Sheet Metal Girders with Very Thin Metal Web, "

Parts I-III, NACA TMS 604-606, 1931.

2. Cook, I. T. , and Rockey, K. C. , "Shear Buckling of Clamped and Simply

Supported Infinitely Long Plates Reinforced by Transverse Stiffeners,"

Aeronautical Quarterly, Vol. XIII, February 1962, p. 41.

3. Cook, I. T., and Rockey, K. C., "Shear Buckling of Rectangular Plates

with Mixed Boundary Conditions," Aeronautical Quarterly, Vol. XIV,

November 1963, pp. 349-356.

4. Rockey, K. C., and Cook, I. T., "Shear Buckling of Orthogonally Stiffened

Infinitely Long Simply Supported Plates -- Stiffeners Having Torsional and

Flexural Rigidities," Aeronautical Quarterly, February 1969, p. 75.

5. Rockey, K. C., "Influence of Stiffener Thickness and Rivet Postition upon

the Effectiveness of Single-Sided Stiffeners on Shear Webs," Aeronautical

Quarterly, February 1964, p. 97.

6. Rockey, K. C. , and Cook, I. T., "Shear Buckling of Clamped and Simply

Supported Infinitely Long Plates Reinforced by Closed-Section Transverse

Stiffeners," Aeronautical Quarterly, Vol. XIII, August 1962, p. 212.

7. Rockey, K. C., and Cook, I. T., "Influence of the Torsional Rigidity of

Transverse Stiffeners upon the Shear Buckling of Stiffened Plates,"

Aeronautical Quarterly, Vol. XV, May 1964, p. 198.

8. Rockey, K. C., and Cook, I. T., "Shear Buckling of Clamped Infinitely

Long Plates -- Influence of Torsional Rigidity of Transverse Stiffeners,"

Aeronautical Quarterly, February 1965, p. 92.

o

Section B4.8

15 October 1969

Page 56

Rockey, K. C., and Cook, I. T., "Shear Buckling of Clamped and Simply

Supported Infinitely Long Plates by Transverse Stiffeners and a Central

Longitudinal Stiffener," Aeronautical Quarterly, Vol. XIII, May, 1962,

pp. 95-114.

10. Bleich, F., Buckling Strength of Metal Structures, McGraw-Hill Book

Company, Inc., New York, 1952.

11. Bruhn, E. F., Analysis and Design of Flight Vehicle Structures, Tri-state

Offset Company, Cincinnati, Ohio, 1965.

12. Kuhn, P., Peterson, J. P., and Levin, L. R., "A Summary of Diagonal

Tension, " Parts I and II, NACA TN266 1 and TN2662, 1952.

13. Levin, L. R., "Strength Analysis of Stiffened Thick Beam Webs with

Ratios of Web Depth to Web Thickness of Approximately 60," NACA TN2930,

May 1953.

BIBLIOGRA PHY:

Rockey, K. C., "The Design of Intermediate Vertical Stiffeners on Web Plates

Subjected to Shear," Aeronautical Quarterly, Vol. VII, November, 1956.

Stein, M., and Fralich, R. W., "Critical Shear Stress of Infinitely Long,

Simply Supported Plate with Transverse Stiffeners," NACA TNi_51,

April 1949.

Structural Design Manual, Northrop Aircraft, Inc.

SECTION B5

FRAMES

B5.0.0

TABLE OF CONTENTS

Frames ................................................

Page

I

5.1.0 Analysis of Statically Indeterminate Frames By

The Method of Moment-Distribution ............... i

5.1.1 Discussion of the Method of Moment-Distribution 2

5.1.2 Sample Problem ................................ 9

5.1.3 Application of Moment-Distribution to Advanced

Problems .................................... ii

5.1.4 Particular Solution of Bents and Semicircular

Arches ...................................... II

5.2.0 Analysis of Arbitrary Ring by Tabular Method ...... 27

5.2.1 Sample Problem ................................ 39

B5-iii

B 5.0.0 FRAMES

Section B5

12 September 1961

Page 1

This section deals with specialized methods of analyzing statically

indeterminate structures. The procedures of analyzing rigid bents and

circular rings are given in detail in Section B 5,1.0 and B 5.2.0

respectively.

A sample problem to illustrate the methods and procedures is

given in each section,

B 5.1.0 Analysis of Statically Indeterminate Frames by the Method of

Moment -d is tr ibut ion,

Moment-distribution is a convenient method of reducing statically

indeterminate structures to a problem in statics. Moment-distribution

does not involve the solution of simultaneous equations, but consists

of a series of converging cycles that may be terminated at the degree

of precision required by the problem.

\,

Section B5

12 September 1961

Page 2

B 5.1.1 Discussion of the Method of Moment-distribution.

The method of moment-distribution requires a knowledge of moment-

area theorems and slope-deflection equations. The five basic factors

involved in the method of moment-distribution are: fixed-end moments,

stiffness factors, distribution factors, distributed moments, and

carry-over moments.

Only structures comprised of prismatic members with no jointtranslation are considered in this article. All members are assumed

to be elastic.

The fixed-end moments are obtained through the use of a general

"end-moment" equation derived in the development of the slope-deflection

method. (See Fig. B 5.1.1-1)

MBA: --f--

where

MAB is the moment acting on the "A" end of member AB.

MBA is the moment acting on the "B" end of member AB.

E is the modulus of elasticity.I is the moment of inertia.

L is the length of member AB.

0 is the rotation of the tangent to the elastic curve at the

end of a member and is positive for clockwise rotation.

is the rotation of the chord joining the ends of the elastic

curve referred to the original direction of the member and

is positive for clockwise rotation.

(_o) A is the static moment about a vertical axis through "A" of the

area under the Mo portion of the bending-moment diagram shown

on Fig. B 5.1.1-1.

(_O)B is the static moment about an axis through "B",

BS.I.I

Section B512 September 1961Page3

Discussion of the Method of Moment-distribution ,(Cont'd_

Any Loading

_A El B [

MAB

MBA

i

AA

Fig. B 5.1. I-I

Section B5

12 September 1961

Page 4

B 5.1.1 Discussion of the Method of Moment-distribution (C0nt'd)

If 9A, eB and _AB are all equal to zero, then both ends of the

member are completely fixed against rotation or translation and the

member is called a fixed-end beam. The last terms of Eqs. (I) and (2)

are therefore equal to the so-called "fixed end moments". Denoting

fixed end moments as FEM and setting CA, _B and _AB equal to zero.

FEMAB = _2 I (_°) A 2 (_°)B 1 ................... (3)

2[ 1_ °°" .... ,°°°,°° .........FEMBA 2(_°) A (_o) BL

Fixed end moments for various type of loading are given in Table B 5.1.1-1

Equations (3) and (4) are summarized by one general equation by

calling the near end of a member "N" and the far end "F". Also let

INFKNF = stiffness factor for member NF = -- .............. (5)

LNF

then the fundamental slope deflection equation is

MNF = 2EKNF (20 N + 9 F - 3_r NF) + FEMNF ............... (6)

The conditions to be met at a joint are: (i) that the angle of

rotation be the same for the ends of all members that are rigidly

connected at that joint and (2) that the algebraic sum of all moments

is zero. The method of moment-distribution renders to zero by iteration

any unbalance in moment at a joint to satisfy the latter condition.

To distribute the unbalanced moment mentioned above, a distribution

factor DF is required. This factor represents the relative portion of

the unbalanced moment which is reacted by a member and for any bar bm,

DFbm is given by

%mDFbm- ........................................... (7)

E Kb

where the summation is meant to include all members meeting at joint b.

The distributed moment in any bar bm is then

Mbm = -DFbmM ....................................... (8)

Equation (8) may be interpreted as follows:

"The distributed moment developed at the 'b' end of member 'bm'

as joint 'b' is unlocked and allowed to rotate under an unbalanced

moment 'M' is equal to the distribution factor DFbm times the unbalan,

moment 'M' with the sign reversed."

-r

Section B5

July 9, 1964

Page 5

B 5.1.1 Discussion of the Method of Moment-dlstribution (Cont'd)

Table B 5.1.1-1 Fixed-end Moments for Beams

Notation: P - load (ib); w = unit load (Ib per linear in.).

M - bending moment (in.-Ib) positive when clockwise.

I. P

^ ti_ L l_

PLM^ - -F..

o

2PL

a B = _ F"

w Ib/In.ltf_! ttttttltttt It flll_,

"V L •

wL 2 wL 2

MA " I-_-- MB = " i-_--

Q

2

MB . . wa_._._ (4aL.3a2)12L 2

!

wL 2 wL 2

MA " T MB " " 3-"0-

2. P

L .Ia 3_ bI L

Pab 2 pa2b

MA "7 MB - e 2

o w lb/in.

a ,m|

L I_

11wL 2

MA" 192

L

2

= _ 5wL 2

MB _

6. _ _ w Ib/In.

a

2

MA . wa (10L2_10aL+3a 2)60--Uwa 3

MB - - -- (5L-3a)60L 2

B

a -i- b __r L -

MA "-_- ( - I), MB- - -_- (3 -I',

10.

A

wL 2 wL 2

MA " 3"-_ MB " " 3"-_

_B

|

A+_..,'mmilt, iiiih,.. B

A _ B-

Section B5July 9, 1964Page6

B 5.1.1 Discussion of the Method of Moment-distribution (Cont'd)

Table B 5.1.1-1 Fixed-end Moments for Beams (Cont'd)

Ii. _P P_.-- a aA B

t- L -,

MA = Pa(1 - _) MB = - MAL

13.

IllIIII" llt11111:BA_'.._,,__..I I.__._ __._

e

MA wa 2= 6---L(3L-2a) M B = - MA

i5. _ _ " - __..t_

A

,_-- LI

MA 30

3_v_

MB- ZOLZ (5L-4a)

B

17. ..w elliptic load

:IItIItItttitt7ttl_'_,,A_

wL 2 wL 2

MA = 13.52 MB = - 15.8"----_

A_/-- e - /

L L L ,

,_B

12.

MB =-M A

15PLMA = 48 MB = - MA

14. !._-_ ___ w Iblin __ a _.

....flIItifIIIl i_A _ L _ '

w (L3_a2L+4a3)MA = 12--"L

16. ------ L

J

A;, ,,

;= L ¸ v _i

wL 2 3wL 2

MA = 3T MB " 160

18.

L. L

MA = l___ x(L_x)2f(x)d x

L2 o

L

_ -i [ x2(L_x)f (x)dxMBL2 ,)

O

_J

B5.1.1

Sect ion B5

12 September 1961

Page 7

Discussion of the Method of Moment-distribution (Cont'd)

_,j--'- r_b _ b Mbm

Fig. B 5.1.i-2

f

The "carry-over" moment is obtained by applying Eq. (6) considering

@m = _bm = o (see Fig. B 5.1.1-2). The carry-over moment is equal to

one-half of its corresponding distributed moment and has the same sign.

Mbm = 4EKbm@ b and Mmb = 2EKbm0 b

Henc e,I

Mmb = _Mbm.......................................... (9)

The sign convention adopted for this work deviates from the usual

convention used in elemen_ beam analysis as found in Sec. B 4.1.1 of

this manual. The positive sense for moments has been adopted from the

convention used in slope-deflection equations; namely, moments acting

clockwise on the ends of a member are positive. (See Fig. B 5.1.1-3)

M

Fig. B 5.1.1-3 Positive sense for bending moments

The following procedure is established for the process of moment-

distribution analysis:

i. Compute the stiffness factor for each member and record

as shown in example problem one.

o

o

o

.

.

J

g

Section B5

12 September 1961

Page 8

Discussion of the Method of Moment-distribution (Cont'd)

Compute the distribution factor for each member at each joint

and record as shown in example problem one.

Compute the fixed-end moments for each loaded span and record

with proper signs as shown in example problem one.

Balance the moments at a Joint by multiplying the unbalanced

moment by the distributor factor, changing sign, and recording

the balancing moment below the fixed-end moment. The unbalanced

moment is the algebraic sum of the fixed-end moments of a joint.

Draw a horizontal line below the balancing moment. The algebraic

sum of all moments at any joint above the horizontal line mustbe zero.

Record the carry-over moment at the opposite ends of the member.

Carry-over moments have the same sign as the corresponding

balancing moments and are one-half their magnitude.

Move to a new joint and repeat the process for the balance and

carry-over of moments for as many cycles as desired to meet the

accuracy required by the problem. The unbalanced moment for

each cycle will be the algebraic sum of the moments at the

joint recorded below the last horizontal line.

Obtain the final moment at the end of each member as the

algebraic sum of all moments tabulated at this point. The

total of the final moments for all members at any joint must

be zero.

Reactions, vertical shear, and bending moments of the member

may be found through statics by utilizing the above mentioned

final moments (step 8).

Section B512 September1961Page9

B 5.1.2 Sample Problem.

PROBLEM #I. Compute the end moments and draw the shear and bending-

moment diagrams for this frame.

w = 2 klp/ft

_A 1=2_ B

_ 25'-_- 35

P1 = 5 kip

i_ 15-,-

1=3.5 C

AB

DF

I04

+19

+I

--- +124 _

×

IBA BC

.08 .I

.44 .56

-104 +18

+38 +48

-4

+2 +2

-4

+2 +2

-62 +62

DC

×0

-3

-6

'D_Z-T-1=2 I

25' P2 = i0 kip

I=5 _

40' _!

ICB CE CD

.i .129 .08

.33 .41 26

-25 +50 0

-8 -i0 -7

+24

-8 -I0 -6

-17 +30 -13

EC

-5O

-5

-5

-60

Fig. B 5.1.2-1

B 5.1.2 Sample Problem (Cont'd)

Section B5

12 September

Page i0

1961

2 kip/ft

I i II

--'--25'----"-

+27.48 kip

[+189 kip-ft

+3.42 kip

-22.52 kip

5 kip

35 I,,,

-1.58

kip

- SHEAR DIAGRAM -

40'

+4.25 kip

-6 klp-ft

i0 kip

----20'---_

'-5.75 kip

-124 kip-ft

+13

kip-fti

|

+7 kip-ft \ +55 kip-ft

-62 kip-ft

- FINAL MDMERT DIAGRAM -

-60 kip-ft

Fig. B 5.1.2-2

f

Section B5

15 February 1970

Page II

B 5.1.3 Application of Moment-dlstribution to Advanced Problems.

In article B 5.1.1, "Discussion of moment-distribution", the basic

principles of moment-distribution were founded. Moment-distribution may

be applied to complex structures involving joint translation, settlement

of supports, non-prismatic members, symmetrical and unsymmetrical bents

and other involved structures. For information on the technique of

solving such structures see the references listed in Section B 5.0.0.

Methods for accelerating the convergence and other short-cuts may also

be obtained in these references.

B 5.1.4 Particular Solution of Bents and Semicircular Arches

In the tables that follow, formulas for computing reactions are

given for several load cases. In all cases constraining moments,

reaction, and applied loads are positive when acting as shown and

K _

K _ m

I2h

IL1

IIS 2

12S 1

for cases i through 18

for cases 19 through 28

B 5.1.4

Section B5,,12 September1961Page 12

Particular Solution of Bents and Semicircular Arches (Cont'd_

Table B 5.1.4.1 Reactions and Constraining Moments in

Two Legged Rectangular Bents

i VERT. CONCENTRATED VA = Qb• LLOAD

,1 !g c ! DFh '9_-"-- L "-"_i

I! _-I I Ii-_

4 HVA VE

2. VERT. CONCENTRATED

LOAD

Q

I '_--'- L _Ii

hi

A E

VA VE

3. HORIZ. CONCE_RATED

C' I/ D

Q B j' I

h 13_a_ I J

L E .__HE

V V

30abH =

2Lh(2K + 3)

FOR SPECIAL CASE:

VA = VE =Q 2

H = 3QL8h(2K + 3)

VE = Q - VA

VA =QbL I I +

La =b= 2

H = 3qab2Lh(K ! 2)

L =" '

l

!

I

@(b - a) VE = Q _ VAL2(6K + i) -I

(b,- a)2L(6K + I)

+ (b -2L(6K + 1)_

V__Qab | l

MA L [ 2(K + 2)

qab 1ME = L 2(K + 2)

FOR SPECIAL CASE:

VA = VE =R2

La = b = -

2 .L,= qL

MA = ME 8(K + 2)

v=q _L HA = Q " HE

HE QaibK(a + h) ]= 2h h2(2K + 3) + I

FOR SPECIAL CASE: b = O,

V . 9-b-L

H E = HA = Q2

a =h

l

-i

/--

f

/

Section B5

12 September 1961

Page 13

B 5.1.4 Particular Solution of Bents and Semicircular Arches (Cont'd)

Table B 5.1.4-1

Two Leg I

4. HORIZ. CONCENTRATED

LOAD

5. VERT. UNIFORM

RUNNING LOAD

E;

ii

Reactions and Constraining Moments in

ed Rectangular Bents (Cont'd)

_ 3(_a2K

V = Lh(6K + i)

HE = Qab h h + b + K (b2h 2 L b h(K + 2)

Qa L b(h + b + bE)MA = 2h h(K + 2)+ h

Qa [ -b(h + b + bE)M E

+ h2h [ h(K + 2)

FOR SPECIAL CASE: b = O. a = h

3qhKV = e(6K + i)

HA = Q HE

a) I

](6K + i)

HA = HE = 2_ _-I "I

MA = ME Qh(3K + I): 2(6K + I) _ _

= wc

wcd

VA :

]Xl + x2 3wcH = _ + _\ = 24Lh(2K + 3)12dL-12d2-c 2

where:

: wc 24-- + 3 + 4c - 2_d 2X 1 24L L L

=__[ d3X2 wc 24_-- "bc2 3_+24L " bT + 2c2 - 48d2 + 24di

w r d4"

BC_'_'_1 )

I-,---L

h I 1 I_

F

VA VF

a bd=L

2 2

FOR SPECIAL CASE:

wLVA VF :T

wL 2H=

4h(2K + 3)

La = O, c = b = L, d =-

2

Ip u

B 5.1.4

Section B512 September1961Page 14

Particular Solution of Bents and Semicircular Arches _Cont'd)

Table B 5.1.4-1 Reactions and Constraining Moments in

Two Legged Rectangular Bents (Cont'd)

6. VERT. UNIFORM

RUNNING LOAD

w _ d_-r

_--- b ----_I

h _------L -.-_

7. VERT. TRIANGULAR

RUNNING LOAD

i

h

.,- I 1

I

td _L

a 2b

3 3

VA wcd + XI " X2= _ L(6K + I)

X1 and X 2 are given in case 5

VF = wc - VA

3(X 1 + X2)

H = 2h(K + 2)

X 1 + X 2

MF = 2(K + 2) +

X 1 + X 2

MA = 2(K + 2)

X I - X2

2(6K + I)

FOR SPECIAL CASE:

wL

VA ffiV F = _-

2wL

H = 4h(K + 2)

X 1 X 2

2(6K + i)

La = O, c = b = L, d = 2

wL 2

MA = MF = 12(K + 2) n_ ,r

wcdVA= 2-i-

wc we + 2cVF= _--VA'2-_ _-

3 [x3+x4 3wc [H ffi_ 2K + _'_= 4LN(2K + 3) de - c 2 d2 ]18

WHERE:

we c 51c 3 c2b . d 2X3 = " _ + _- + _ + 6L

X4 wc d3 c2 51c 3 c2b 2d 2 + dLffi 2L L + _ + 810L 6L

FOR SPECIAL CASE:

wLV s --

6

wL 2H =

8h(2K + 3)

La=o, c=b=L, d =_

.==c_

B 5.1.4

Section B5

12 September 1961

Page 15

Particular Solution of Bents and Semicircular Arches (Cont'd)



8. VERT. TRIANGULAR

RUNNING LOAD

MA_ _ MF

VA V F

a 2bd =L

3 3

9. HORIZ. UNIFORM

RUNNING LOAD

D _---iL ----

__[ Iij h

V V

X3 - X4VA =wcd + X 3 and X4 are

2L L(6K + i) given in case 7

wc 3(X3 + X4)

V F = _- - VA H = 2h(K + 2)

X 3 + X4 X3 - X4

MA = 2(K + 2) 2(6K + I)

X3 + X4 X3 - X4

MF = 2(K + 2) 4 2(6K + I)

FOR SPECIAL CASE:LI

a=o, c=b=L, d = 3

wLI i 1VA = 6-- ] IO(6K + i)

VF = _-- I + 20(6K + I)

wL 2H =

8h(K + 2)

wL 2 [ 5MA--Z-=12vK + 2

wL2 I 5MF=_ K+ 2

i I--+6K+ i

6K+ 1

v =w%_" c272L

w<a 2 - c2) +HF = 4h

FOR SPECIAL CASE:

wh 2V = --

2L

HA = wh - H F

c=o, b=o, a=d=h

4E K 1H F = I + 2(2K + 3)

H4 = w(a - c) - HF

KIw_a2-c2)(2h2-a2.c2) I

8hB(2K + 3)

B5.1.4

Section B512 September 1961Page 16




i0. HORIZ. UNIFORM

RUNNING LOAD

D

HA

V

_---L -----

%-12

JM A _

E

-_MFHF

V

w(a 2 - c2> MA MFV= 2L "_- -L'-

HA = w(a - c) - HF

• 2 2, X= X_(K - I)

nF - 4h - 2-h _ _ _

WHERE:

X5 = l_h2 [d3(4h - 3d) - b3(4h -3b) 1

X6 = w--w---[a3(4h - 3a) - c3(4h -3 c)l12h 2

(3K + I)[ w_a2 -2 c2) - X5 J

MA = 2(6K + i)

2 K +------26K + i + X5

MF =

(3K + l)[w(a2 2" c2) - X51

FOR _PECIAL CASE:

wh2KV =

L(6K + i)

wh(2K + 37HF = 8(K + 2)

2_6K + I)

6K + f

c=o, b=o, a=d=h:

HA = wh - HF

wh 2 F 18K + 5

MF - 24 | 6K + i K+2

w 213°K+7ii HMA = 2-4-- 6K + 1 + K+------2


B 5.1.4 Particular Solution of Bents and Semicircular Arches (Cont'd)



II. HORIZ. TRIANGULAR

RUNNING LOAD

--I

a h

HA _ _ HF

V = _w (a2 + ac - 2c 2) HA = w(a2- c) _ HF

VL KX7

HF = _ + (2K +3)h WHERE:

X7 = w 13(4d5+b5) _ 15h(3d4+b 4) +120h2(d-b)

20h2(2d3+b 3) _ 15bd2(2h_d)2

FOR SPECIAL CASE: b=c=o, a=d=h:

wh

wh2 HA =-- _ HF _

V = 6--L-- 2

)wh 7K

= 1 + 10(mE + 3)

V V

12. HORIZ. TRIANGULAR

RUNNING LOAD

D --LI I

b

V

V w" _ (2a + c)(a - c)

I

f=-i

h

_ HF

1 '

KX

HA w(a c) _ HF HF _ VL + I0= 2 - 2-_ h(2K + 3)

WHERE:

W

Xlo = 120h2(a.c)

FOR SPECIAL CASE:

wh 2V = --

3L

wh

HA =,_- - H F

whI4K + 5 ]HF=_ 2K+3

-30h2c (a2-c 2) + 20h2(a3-c 3)

+ 15c(a4-c 4) - 12(a5-c 5)

b=c=o, a=d=h:

B5.1.4





13. HORIZ. TRIANG_

RUbbING LOAD

--- L

°I,II

i.Jc/

HA ,../MA _._M F FV V

V " wl_a2 _ ac - 2c 2) MA MF6L L L

w(a - c)HA = 2 " HF

u w( a2 + ac - 2c 2) X8 . x9(K-l)

"F = " imh' " _ _

WHERE:

X_ = ._.._w _ [15(h+b)(d4.b4 ) _ 12(d5_b 5)

60h2(d-b) | 3

-20bh(d3-b3)I

Xo = ----_m [lOd2h2(2d.3b) + lObh(4d3+

60hZ(d-b) L

b2h_b 3) _ d4(30h+15b) + 12d 5 + 3b 5 ]J

MA "_"

Ew a2+ac2c2 0 x812(6K + I)

x9 [ I 3K ]+X s+ _- K + 2 + 6K + i

M F

(3K + i)[ w(a2 + ac- 2c2)]. X86

2(6K + I)

xsl 1 3K 1_'- K+ 2 6K+ i

FOR SPECIAL CASE:

wh2K

V = 4L(6K + I)

wh(3K + 4)HF = 40(K + 2)

ii=....i.

blclo , a=d=h

whHA =-_ - HF

wh 2 [ 27K+7HF =6-'_ '2(6K+I)

wh2 I 27K + 7 3K + 7 1MA " _O- 2(6K + I) + K + 2

K+2

B 5.1.4





/

P

f-

14. HORIZ. TRIANGULAR

RUNNING LOAD

H "_-'-'_}B "_---_ I

A _A "_

V

E

F

b.a_MF

V

V =w(2a + c)(a - c_ MA MF6L L L

w(a c)

HA = . HF2

H w(2a2_ac.c 2) _ XII + XI2(K I)F = 12h 2h 2h(K + 2)

where:

x = ----w--w I

r

5hd4-3d 5

II 60h2(d_b)

X = w-----E-- 15 (h+c) (a4-c 4 )

12 60h2(a_c) L

- 20ch(a3-c3)]

x,:lMA = 2(6K + I)

x:2 I 1 + 3K ]+X:I+--2-- K + 2 6K + I

I 3K+I](w(2a2-'ac-c2) ]6 - XII _ X22

MF = 2(6K + i) 2

K+2 6K+l

FOR SPECIAL CASE:

3Kwh 2

V = 4L(6K + I)

4 ]-20hdb3-12b (d+h)

12(a5-c 5)

HA " w__h_h.2

!HF wh(7K + II} wh 2 V97K+22 3 "= 40(K + 2) MA = _-_ L 6K+I + K-_

wh2 I 21K + 6 1 ] JlMF " _- 6K + i K + 2'

B 5.1.4





15. MOMENT ON HORIZ.

SPAN

I

h I1

l tV V

16. MOMENT ON HORIZ.

SPAN

V V

MV _. w

L

H = 3(b - L/2)Mnh(2K + 3)

FOR SPECIAL CASE: a=o, b=L

M

V=-L _-_M

I

3M iH = 2h(2K + 3)'

i

V = 6(ab + L2K)M

L3(6K + i)

H = 3(b - a)M

2Lh(K + 2)

= M -6ab(K+2) - L [a(7K+3) - b(5K-l)lMAb 2L 2 (K+2) (6K+I) J

ME = VL - M - MA

FOR SPECIAL CASE: a=o, b=L

CML<I+ 6_) I{

V _._

3M iH -

2h(K + 2)

(sK- I)MIdA " 2(K + 2)(6K + I)

B5.1.4





f k

17. MOMENT ON SIDE SPAN

C D'6 " I' _b

, h Ii_

b _----- L -P

H____ A _L E

lv1'8. MOMENT ON SIDE SPAN

MV=-- H =

L

FOR SPECIAL CASE:

MV = --

L

3 [K(2ab+a 2) + h_M

2h3(2K + 3)

f-_M

3MH =

2h(2K + 3)

a-o, b--h

q) q)

3bM [2a(K+l) + b ]

b

b

MA_-J

V

y--L

_ ! 12BM:

r I

!A ___._, E- _,_

6bKMV=

hL(6K + I)

-M

MA = 2h2(K+2)(6K+I)

ME =VL-M-M A

H

2h 3 (K + 2)

[4a 2+2ab+b2+K (26a2-5b2)

+ 6aK2(2a-b)]

H

FOR SPECIAL CASE:

6KMV =

L(6K + i)

3MH=

2h(K + 2)

= M(5K- i)MA 2(K + 2)(6K + I)

a=o ; b=h

I

Section B5

15 February 1970

Page 22

B 5.1.4 Particular Solution of Bents and Semicircular Arches _Cont'd)


Triangular Bents (C,:'_'d)


LOAD


LOAD


LOAD

C

VA = Q - VD

ReVD =L

_ Re rb d _a+c) ]

= h Li + 2a2(K + i) J

VA =Q - VD

]D L L - 2a 2 J

, 4, 6La_h(K+l) [L + 2(2L+b)(a+c) + 3ac

M = ^qcd r(a+d)(BK+4) - 2(a+c)]

A 6a_(K+l) L J

qc2d

MD 2a2(K + I)

!H,A=Q -H D

d _h+c) ]2h2(K + i)

B5.1.4




Triangular Bents (Cont'd)


LOAD

C

4 va_ q_- b --_r

HA _ I_ HDe _MD

V V

23. VERTICAL UNIFORM

RUNNING LOAD

' B

1_ ii_j _S 2 I2

24. VERTICAL uNIFORM

RUNNING LOAD

LB 12

S 2

VA VC._ H

___ "dMcL_D-

r- ]

v 11- 1 HAQ Ho= L 2h 2 J ; = -

i

HD = Qc b + d (h+d)(-b[3K+4] - 2L)Lh' 6h2(K+I) [ I]

+ 2(2L+b)(h+e) + 3acl1

qcd , (h+d)(3K+4) - 2(h+c) IMA = 6h2(K + I) [

MD = _cd .6h2(K + i)

(h + 2c + d)

VA = wa I - a_2L]

wa 2

Vc = 2-i-

wa2 14b i ]H = 8-_- l- + 1 +-----f

VA = wa ! I 3a 3wa 2I - 8-_ : VC -- 8L

H __wa

24Lh(K + i)

wa2{3K + 2)

MA = 24(K + I)

2wa

M =c 24(K + i)

b(10 + 9K) + 2L + a ]

B 5.1.4

Section B512 September1961Page24



Triangular Bents (Cont'd)

25. HORIZ. UNIFORM

RUNNING LOAD

W

I1 12

HA _ . _I_

vf L----_v26. HORIZ. UNIFORM

RUNNING LOAD

V V

27. APPLIED MOMENT

AT APEX

!s,iJ

vrL L!

wh 2V :

2L

_A = wh i Hcwh 4b

_c "_- _-+_

3wh 2

8L

_A " wh -M C

whk_

_ 8L(K + I)

wh2(SK+ 2)_A " 24(K + I)

M

L

.I.b 1_'_h'L K+I

K+I

[ b(3K + 4) + a]

wh 2

MC = 24(K + I)

B 5.1.4




Triangular Bents and Semicircular Frames

or Arches (Cont'd)

/

_k

f

28. APPLIED MOMENT

AT APEX

V

S.b_

_- b

cjMc

L -TV

29. S INUSOIDAL

NORMAL PRESSURE

I

b(Ib/in.)

*V b=c sin@ Vt

3MV =

2L

H = 3M(a - bK)2he(K + I)

KM

MA = 2(K + I)

M

MC = 2(K + I)

C_RV -

4

CR

4

M0 - CR24 [(n-28)cos 8 - _ + 3 sin Q]

(Positive moment acts clockwise on

section ahead.)

B5.1.4

Section B512 September 1961Page26


Table B 5.1.4 Reactions and Constraining Moments in

Semicircular Frames or Arches (Cont'd)

30. S INUSOIDAL

NORMAL PRESSURE

b(Zb/in.)

C_RV -

4

13_ 2 - 32 ] = 31974CRH=_L 8 _2

M_ CR2 [_3 -i0_.]_ = .05478CR 24 8 2

b=C sin0

/

rv

31 • UN IFORM NORMAL

PRESSURE

I

in. )

It

F

M 0 = CR 2 |.81974 sin 0

cosO _ }- .s4018 + -7-- (7 - o)

(Positive moment acts clockwise onsection ahead.)

M m 0 at all points since pin points

permit a uniform hoop tension.

T, where :

T m V = bR

H_O

f-

Section B5

July 9, 1968

Page 27

B5.2.0 Analysis of Arbitrary Ring by Tabular Method

The following analysis is a corrected version of work taken

originally from Reference 6.

Rings or frames with ends built in, elastically restrained, or

pinned may be analyzed by the procedure outlined in this section. The

procedure is given in tabular form (Table B5.2.0-1) with added notes to

define the sequence to be followed.

A sample problem is given in See. B5.2.1 to illustrate the pro-

cedure. This sample problem considers the energy due to direct and shear

loads. The effects of shear flow are presented as a supplement to the

sample problem.

Procedure

Procedure to Obtain Initial Geometric and Elastic Data

(i) Set out the neutral axis of the ring between the end points T

and O. (In a complete ring T and O coincide.)

(2) Divide the neutral axis into a _umber of segments which are

conveniently but not necessarily of equal length As. Ten to

twenty segments will usually give sufficient accuracy. Mark

the Joints and central points of the segments. In symmetrical

rings a complete segment should lie each side of the axis of

symmetry.

(3) Calculate the values As/E1, As/EA, As/GA' at the segment

centers.

(4) For a built-in ring, calculate a and b from equation (I) below

which defines the elastic center C and the axes X' and Y' (See

Figures B5.2.0-I and -2.)

For a pinned-end ring, no translation of axes from 0 is necessary.

Use the X and Y axes as they are defined in Figures B5.2.0-3

and -4,

(5) For a built-in ring obtain the co-ordinates x', y', and the angles

_' at the segment centers. For a pinned ring obtaiL_ x, y, and

directly. In syLi_etrica] rings only half the ring need be con-

sidered.

Section B5

July 9,1968

Page 28

B 5.2.0 Analysis of Arbitrary Ring by Tabular Method (Cont!d)

Angles Measured Positive

in Direction of

Arrow 5 4

fo -- 3

9 / 'Neutral_

Fig. B5.2.0-I General Ring with Built-ln Ends

y Note :

i

Neutral 7 6

9 ' 4 X

j2

0 T

Fig. B5.2.0-2 Symmetrical Built-ln Ring

e=O

Section B5

July 9,1968

Page 29

B 5.2.0 Analysis of Arbitrary Ring by Tabular Method (Cont'd)

6 5 4

O

Fig. B 5.2.0-3 General Ring with Pinned End

Y

7 6Neutral Axis

9 4

_S 3

"%\1 71/'O T

Fig. B 5.2.0-4 Symmetrical Pinned-End Ring

(6)

Section B5

July 9, 1968

Page 30

Note that _, 8, and _' are measured as shown in Figure B5.2.0-I.

The angle must originate at the end coming from O, and be measured

counterclockwise to the end going to T.

For the slope of the tangent to the neutral axis of the element,

use the slope at the midspan of the element.

NOTE : As/EA and As/GA' are not usually required, since in

most cases the thrust and shear energies are negligible

compared with the bending energy.

General Notes

I. Limits of Application

The method may be applied to any ring or curved beam in which the

deflections are linear functions of the loads and in which the

elementary formula connecting curvature and bending moment holds.

II. Calculations

(1) The advantage of choosing X, Y, and M c as redundant reactions is

due to their orthogonality. Thus, one can calculate them directly

without having to solve simultaneous equations.

(2) The calculations are shown in tabular form using finite segments.

Graphical integration may also be used and in exceptional cases an

analytical method may be applied.

(3) The referenced table is set out for an arbitrary built-in ring, but

is also applicable when the supports T and 0 "give" elastically

(See General Note III, 6.) In all other cases the size of the table

is reduced but the form is essentially the same. A considerable

reduction in the numerical work is possible when the ring is

symmetrical (See notes IV and VI.)

(4) In practice certain steps of the calculation may be conveniently

done with additional columns, e.g., for x and y one requires x'

cos @, etc.

(5) Numerical Accuracy. Although the data in columns 1-6 and 24-26

will not exceed three-figure accuracy it is necessary to retain

four or five figures in the remaining columns in order that the

final results shall be correct to three figures.


#-

III. Arbitrary Built-In Ring (Fig. B 5.2.0-i)

Equations required and the stage at which they are used are as

follows:

F(i) a .

Z x o AsE1

YO &s

b=

E1

(2) After column 12,

tan 20 =

A_Ks _ As

2Ex'y' E1 >; sin 2_h' _,(

.,) As 2, r_, As Asy,(x 'rz - y' _-- >; cos GA'

,f-

x = x' cos 0 + y' sin 0; y = y' cos 0 - x' sin 0.

(3) M , N , and S are calculated for tile ring built in at T and.0 0 0

tree at O.

(4) After colunu_ 33,

As

-EMoY _ + _3N cos _' As Aso _ _ LSo sin _" _,X = -

Z'Y_ As .-, _, As As

Section B5

July 9,1968

Page 32

B 5.2.0 Analysis of Arbitrary Ring by Tabular Method (Cont'd)

y

As As AsEM x _-__+ ZN sino _I o _- es cos _--o GA'

As As AsZx2 _ + _ sin2 _ E-A + _ c°se _ G--A'

AsZM --

o E1M = -

c As

It. E-"_

(5) After column 39,

(6)

M : M c- Xy + Yx + M °

N = N + Xcos _ + Y sin0

S " S + X sin _ - Ycoso

Effect of Elastic Supports. The elastic characteristics at

the supports T and O are represented by the two sets of three

coefficients, kTm , kTn, krs; kom, kon, koS. Thus kTm is the

moment required to rotate T through one radian, k la is the

force required to move T in a direction defined by $_ through

unit distance, and kTS is the force required to move T in a

direction normal to _ through unit distance. At each

support the rotation and the two displacements are assumed

elastically orthogonal, i.e., when a force is applied to the

support T in the direction @_ , there is no rotation and no

movement normal to _$

For the purposes of the calculations the support flexibilities

may be represented by two additional segments considered con-

centrated at T and O, of which the "elastic weights" are

respectively:

As i As 1 As 1

E1 _m' EA kTn' GA' kTS '

As I As 1 As I

"r

Section B5July 9,1968Page33

r--

B 5.2.0 Analysis of Arbitrary Rin_ by Tabular Method _Cont'd>

! !

Note that _T and _0 need not necessarily be the values of

_' at T and O. The values above and coordinates are entered

in rows T and 0 of the table.

IV. Symmetrical Built-ln Ring (Fig. B 5.2.0-2)

(I) CY is the line of symmetry and columns 4-12 are not required

as X, Y1and _ are calculated directly.

(2) Any loading may be analyzed as a symmetrical and an anti-

symmetrical loading.

(3)

(4)

For Iymmetrical loading Y is statically determinate, being

half the total load in the direction YC and should hence be

included in the external loading. Thus in the table, columns

18, 20, 23, 28, 30, 33, 35, 38, and 39 are not required, and

in the expressions for M, N, S terms involving Y are omitted.

X and M c are determined from the formulas in III (4).

For antisymmetrical loading X and M c are statically determinate

being half the total load in the direction XC and equal and

opposite to half the moment of the external load about C

respectively; they should hence be included in the external

loading. Thus, in the table, columns 19, 21, 22, 27, 29, 31,

32, 34, 36, and 37 are not required, and in the expressions

for M, N, S the terms involving M c and X are omitted. Y is

determined from the formula in III (4).

(5) Note that only half the ring need be considered in the table,

i.e., elements 7-12 and support 0 . For symmetrical loading

M and N are symmetrical and S antisymmetrical. For anti-

symmetrical loading M and N are antisymmetrical and S symmet-

rical. The summations are, of course, only to be taken over

half the ring.

(6) When the supports have symmetrical elastic characteristics the

method of this section still applies by introducing an addi-

tional segment at O, as described in III (6).

V. Arbitrary Pinned Ring (Fig. B 5.2.0-3)

(I) Columns 4, 12, 18, 20, 23, 27, 28, 30, 33, 35, 38, and 39 are

not required.

(2) Y is statically determinate and should hence be included in

the external loadillg.

(3) X is determined from the formula in Ill (4).

B5.2.0


Analysis of Arbitrary Ring by Tabular _Method (Cont'd)

(4) The formulas for M, N, and S are:

(5)

M • M o - Xy; N - N o + X cos _ ; S = So * X sin _.

Effect of Elastic Supports. In this case the procedure of

III (6) should be applied, noting that the terms kTm , komdo not exist.

Vl. Symmetrical Pinned Ring (Fig. B5.2.0-4)

(i) Any loading may be analyzed as a symmetrical and an anti-

symmetrical loading.

(2) For symmetrical loading, the load Y at O is obviously half

the total load in the direction Y, and only half the ring need

be considered in the table, i.e., elements 7-12 and support O.

Otherwise, the procedure is the same as in the general case.

(3) For antisymmetrical loading X is statically determinate, being

half the total load in the direction X. Thus, in this case the

problem is solved purely by statics.

(4) When the supports have symmetrical elastic characteristics the

method of this section still applies. For the symmetrical loading

an additional segment should bu introduced at O, as described in

IV (6) and V (5).


Notation

A : cross-sectional areaA' = effective area of cross section for shear stiffnessC : elastic center : centroid of elastic weights As/E1E = Young's modulusG = shear modulusI = moment of inertia of cross section

k : elastic constants of supports T and O (see note 111,6)

M c - moment at elastic center in fixed ring

Mo, No, S O - bending moment, direct load, and shear load in ring

supported at T, due to external loading and any of

the reactions X, Y, and Mc, which may be statically

determined (e.g., in a pinned ring Y is always

statically determined). M o is positive for com-

pression in outer fibers. N o is positive for com-

pression and S o is positive when acting outwards onthe right-hand side of a section (Subscripts do not

refer to point "O")

M, N, S • total bending moment, direct load, and shear load at

any section. Sign convention as for Mo, No, S O

O = origin, or left-hand e,d point of ring

T s terminus, or right-hand end point of ring or cross-sectional

area of ring

s - length of segment

xo, Yo" co-ordinates referred to arbitrary orthogonal axes

OX', OY' of built-in ring

_' - angle def_nlng slope of neutral axis of built-in ring,

with respect to CX'xp y = co-ordinates referred to geometric and elastic

orthogonal axes CX, CY of built-ln ring; or, co-

ordinates referred to orthogonal axes OX, OY of

pinned ring, where OX passes through T

8 - angle between CX' and CX

- angle defining slope of neutral axis for built-ln

ring, with respect to CX(_ = _' -

X, Y = reactions in directions CX, CY at elastic center for

built-in ring, or in directions OX, OY for pinned

ring

x',y' , coordinates referred to geometric and elastic orthogonal

axes CX' and CY' of built-in ring

!

o

L_

Section B5

July 9,1968

Page 36

[e__]gV sVJ '_Z uT

Section B5

July 9, 1968

Page 37

/f

i

"0

v

!

0

X

o=v

I

0

Section B5

July 9, 1968

Page 38

Section B5

July 9, 1968

Page 39

u

o

!

o

o4

<E-4

¢q

o

O_on

oOon

p-.on

¢0

u_oO

on

Z

soo X

_u._s X

X>*

,aO

÷ cO'.0 I

p_on4- oOt_ ¢,3¢,q ._.

tr3-a-on¢.4 "4-

4,, --..1"

p_,.-.4

,--4

H_

0 XZ_

co

X

>4

bO -,-40

I

Section B5

July 9, 1968

Page 40

B5.2.1 Sample Problem

The frame to be analyzed in this sample problem is the symmetrical

built-ln ring illustrated in Figure 5.2.1-1. The ring is divided into 24

segments for this problem. The section properties at the center of each

segment are used as average values for the entire segment. A typical cross

section of the ring is shown in Figure 5.2.1-2.

The necessary calculations to determine the magnitudes of M, N_and

S at each segment are shown in tabular form on pages 42 through 46. The

results have been plotted on graphs, which are shown on pages 47 through 49.

The problem data, statement, and conditions follow.

Given :

P = I00 Ib R a = 47 in. 6Q : I00 Ib E I0 x I0 psi

M = I000 in.-Ib C = 3.85 x 106 psi

R = 50 in. A = 11 x lo

A' = 5/6 A (See Ref. 7)

Problem:

Determine the bending moment M, direct load N, and shear

load S for the given conditions by use of Sec. B5.2.0.

B5.2. i Sample Problem (Cont'd)

Section B5

July 9,1968

Page 41

>

>

o |O_

/T

0

Imaginary Rigid

Bracket

Figure B5.2.1-1 Frame for Sample Problem

Figure B5.2. 1-2

_-" 1.0 in.

Section A-A

Typical Frame Cross Section

B5.2. i Sample Problem (Cont'd)

Section B5

July 9, 1968

Page 42

I 2 3 4 5 6 7 8.=

Segment

or

Points

T

AS

EI

10-6

AS

EA

AS

GA'X i !

r__q

Operations

10_-6

described in terms of column numbers

1 .11608 .24631

2 .05156 .18591

3 •11608 .24631

.116084 .24631

.18591.051565

6 .I1608 .24631

7 .11608 .24631

8 .05156 .18591

9 .11608

10 •11608

II .05156

12 .11608

13 .11608

¢q

0o

14 .05156

15 .11608

16 .I1608

17 .05156

18 .11608

19 .11608

20 .05156

21 .11608

22 .11608

23 .05156

24 .11608

0

2.26979

24631p"

[.24631

.18591i

.24631

.24631

18591

_24631

1.24631i.18591

.24631

.24631

.18591

• 24631

.24631

.18591

•24631

m this row

10-e

.76774

.57946

.76774

.76774

.57946

.76774

.76774

.57946

.76774

.76774

.57946

.76774

.76774

.57946

.76774

.76774

.57946

.76774

.76774

.57946

.76774

.76774

.57946

.76774

E

Columns 4 through 12 not required

for a symmetrical built-in ring;

Sum of Columns

B5.2.1 Sample Problem (Cont'd)

Section B5

July 9, 1968

Page 43

9

4x 5x I

i0

!

g_

(42 _5z)

×i

iI

W!

(2-3)

x7

12

!

Oo

(2-3)

×8

13

47.

43.

14

Y

15 16

_9

17

_900

See Note IV(i)

277.5 .99144 .i3051

292.5

37. 307.5

28.

- 6.

17.

6. 352.5

7.5

-17.

-28.

322.5

337.5

22.5

37.5

_J

CD

.,-4

0 i

m

o %o2

52.5

070 6. 197

15_5 17. 875

666 28. 902

902 37. 666

875 43. 155

197 47. O7O

197 47. O7(}

875 4:]. 155

9(12 37.(;(i(i

G(i(; 28. D02

155 17. 875

07O 6. 197

070 -6. 197

155 -17. 875

(;66 -28.902

902 -37. 666

875 -43. 155

197 -47. 070

197 -17. 070

875 --13.155

902 -37. (;6(;

(;(it; -2_. 902

155 -17.875

070 I_ 6.1!)7

J

67.5

82.5

97.5

I12.5

127.5

-37.

-43.

.92387 .38267

.79334 .60875

- 60875

-. 38268

.79334

.92387

•99144• 13051

• 13051 .99144

.38267

.60875

.79334

.92387

.99[44

.99144

-43.

-37.

.92387

•79334

.92387

.79334

.60875

.38267

.13051

-•13051

-.38267

-.60875

-28. 142.5 .60875 -.79334

-17. 157.5 •38267 -.92387

- 6. 172.5 .13051 -.99144

.13051 -.99144

262.547.

G.

17. 202.5

2_. 2[7.5

37. 232.5

247 5

187.5

43.

•38268 -•92387

•60875 -.79334

.79334 -.60875

•92387 -.38267

•991.44 .13051

B5. 2. I Sample Problem (Cont'd)

Section B5

July 9, 1968

Page 44

18

t3Zxl

10-6

257.19

96.02

164.69

96,96

16.47

19

%

14 I

10-_

4.46

16.47

96.96

164.69

96.02

4.46 257.19

4.46 257.19

16.47 96.02

2O

u)

16zx2

10 -6

.24212

.15868

.15503

•09128

.02722

•00420

•00420

.02722

.09128

21

0o

172x2

10 -6

.00420

.02722

.09128

.15503

.15868

.24212

,24212

.15868

22

%

162×3

10 ±_

.75466

.49459

•48322

.28451

_,08485•01308

013o8.08485

.28451

23

oo

172×3

10 -6

.01308

.08485

.28451

.48322

.49459

.75466

.75466

.49459

24 25

M o No

See Note II1(3)

11.

•

16.

22.

-1.

-63.0

-211.2

-500.6

26

S o

.15868•0272216.47 96.02

257.19

257.19

4.46

4.46

• 24212.08485

.28451

.48322

1.49459

.75466

.49459

.01308

,24212

•00420

96.96 164.69 .09128 •15503 -1535.4 32,319

16.47 96.02 .02722 .15868 .08485 -1166.6 29.736

.00420 .01308 .75466 25,028

164.69 .09128 .48322

•28451

257.19 4.46 .24212 .00420 -3169.8 -80,198

257,19 4.46 .24212 .00420 .75466 -2527.4 18.909

96.02 16.47 .15868 _02722 .49459 708485 -2244_8 27,523

96.96

.75466

-1920.2

-814.6

-532•4

-337.5

7 .246

9 2.356

2 1.661

7 -1.996

7 -8.366

-16.810

-26•356

-35.790

-43.784

-49.024

-50.361

-46.935

47.798

29.656

12.470

-2.385

-13.896

-21.484

-25. 028

-24.849

.15503

96.96 164.69 .15503 .48322 -880.3 -35,664

164.69 96.96 .15503 •09128 .48322 .28451 -1457.8 -50.058

96,02 t6.47 .15868 .02722 .49459 .08485 -2259.2 -65,318

1.441

•957

-1,251

-5.799

-13.065

-23.116

31.835

19.205

13.224

96.96

164.69 96.96 .15503 .09128 -81.

96.02 16.47 •15868 .02722 .49459 .08485 -41.4 1. i5t

257.19 4.46 .24212 .00420 .75466 .01308 -11.2 0.000

2543.

164.69 •09128 .15503 .28451

.48322

.48322

.28451

-171. 4 -21,647

4 -16.406

-10,278

-4.448

7.890

3,772

16 2543.16 2.7141 2.7141 8.45964 8.45964_'.__

*Calculations on pages 50 through 53

.962

• 320


Section B5

July 9,1968

Page 45

f

27 28 31 32 33

O

24x I

10-6

-1.35

.2O

i .88

2.64

-.09

-7.31

-24.52

-25.81

Lt_ t-4

O

24×13×i

10-6

-63.7

8.7

70.7

76.2

-I .6

-45.3

152.0)

461.4

2953.4

U3 <

U_

,3

03

O

O

OZ

125xt7x2

i0-6

.00791

.16761

.24905

-.39003

29 30

U3 <

_.1 -,q

09

O Oz

24x 14xl 25x 16x2

10-6 10 -6

-8.4 -.06000

3.6 -.40466

54.2 '''t- . 32457

99.4 .29928

-3.7 .59519

-344. t .54037-li54.2 F-.84724-1113.9 -2.5462-3849.0-6.5650-4890.0 -9.5797

-2082.1-8. 6498

-2280.2 -11.462

1818.2 11.672

2068.9 5.094

6442.3 2. 437

6713.2 !- .3576

2595.6 -.9886

4450.9 -.6906

2908.8 .8046

750.9 !1.7679

749.3 3. 2458

273.0 13.2059

38.1 1.7654

8. I i. 0862

13248.01 -9.9626

-1.4369

-4.1050

-6.4362

L0

-r4

03

OO3

26x16x3:

10-6

-.24357

-.51500;

-.87768

-.44727

.27741

.58105

-1. 3091

03

0U

0

26x17x3

10-6

.03206

.2[332

.67347

.58289

-.66972

-4.4140

-9.9447

34 35

x >*

See Note 111(5)

Xxt4 Yxt3

32.403

93.465

151.123

196.948

225.649

246.120

246.120

-1193.318

_-i094.065

-954.908

-732.723

-4,53. [ 67

-157.106

157 .106

.167

-18.563 -89.533 -199.5964567.7-1887.15

-9.45 -355.8 2.4599 -2.267 -151.123 -954.9 08

-2.13 -92.1 .7312 -.6162 -.25523 -93.465 -1094.0 7

-1.30 -61.3 .143 0.0000 0.00000 -32.403 -1193.32

-3.688

-225.649

-196.948

-2.932

-19.89

-453. l 67

-732.723

4.2679

4.2300

-17.40

-4.806

-] .763

-'7.079

-6.1472 -5.1259 -12.375 225.649 453

-102.19 -8.5568 -16.668 -21.722 196.948 732.723

_169.22 6374.0 -7.3507 -30.489 -23.395 151.123 954.908

-116.48 5026.7 -3.5828 -34.968 -14.484 93.465 1094.065

-367.97 17320.4 -1.5088 -61.044 -8.036 32.403 1193.318

-293.39 13810.1 -1.5365 14.393 -1.895 -32.403 1193.318

-115.74 4994.7 -2.1098 14.734 -6.103 -93.465 1094.065

-222.90 8395.8 1.8698 19.390 -14.878 -151.123 954.908

-178.23 5151.2 .46605 15.10.5 -[9.685 -196.948 732.72

-60.14 1075.1 2.3867 6.594 -15.919 -255.649 453.1 67

-94.56 586.0 5.2464 2.508 -19.051 -246.120 157.1 06

-61.80 -382.9 6.1119 -1.924 -14.618 -246.120 -157.1 06


Section B5

July 9, 1968

Page 46

36 37 38 39 40 41

X× 16

-5.184

-4.831-4.148

Oo

X

See Note HI

Xx 17

.682

2.001

3.183

4. 148

.=

>.

(5)Y× 16

25.135

23.422

20.113

15.433

Oo

Yx17

-3. 309

-9.702-15.433

-20.113

M

M c + '24-34 +35

-406.0

-20.113

-352.3

-258.4

N

25+ 37

+ 38

26.063

27.77924.957

17.585

6.167

-8.317

4.831 9.702 -23.422

-.682 5.184 3.309 -25.135 365.2

.682 5.184 -3.309 -25.135 531.2 -24.481

2.001 4.831 -9.702 -23.422 558.4 -40.661

4.148 -15.433 486.9 -55.069

3. !83

3.183

4.1484. 831

-20.113 177.4 -65.954-15.433

2.001 -23.422 -9.702 -427.2 -71.782

5.184 .682 -25.135 -3.309 -1177.5 -71.388

-.682 -25.135 3.309 -470.3 2'i.981

-2.001

-3.183

-23.422

-20.i13-15.433

-9.702

-3. 3093. 309

9.70215.433

20.113

23.422

25.!35

_.184

4.831

4.148-225.8

17.2

9.702

15.433

20.113

23.422

225.6

343.6

25.135 420.0

25.135 388.0

23.422 266.3

20.113 124.2

15.433 -53.8

-210.6

-34o.?

9.702

3. 309

4.233

-10.826

-21.966

-28.429

-29.997

-26.903

42

26 + 36

- 39

-1.555

5.83_12.726

17.887

20.170

18.654

12.752

2.307

-12.368

-30.477

-5O.785

-71.705

20.784

22.652

20.550

15.389

8.315

.575

-6.612

-19.978 -12.199

-10.362 -15.406

.524 -15.809

11. 143

26.005

-13. 382

P

S

_

July 9, 1908

Page 47

"1 ""

[

:.:fill ' :.'>.iii

: i

:!'" _:"_ .... i .................,1.. i .... i • ;

.'I: , • , ",

!..,:1.

: : i

• " i : .....

• | ''':i ...... i........ i........

...'. . •

i": ':

.... | ....... , .....

•,i....i .........._..! ;_-;i[ .....

• .' : ".. 1 ...

, l

; : :

I i

i r li.,,:-,;-:i,; :t

,, ,:'. :,"':'.. I ]

.... : i ...._

Figure B5.2. I-3

! ,i ; i" ':: ..] ..,.....,.........:::::::::::::::::::::::"', ..... .'I: " " ' :""• I "." .:..'::" :'.':: ....

.... ":'l "". .... :': ":;: ::1.::::

• , , : : ,,. ,l , : ": ;',,::':::I

'. ". I ! , ......:: I

Segment (See Fig. B5.2.1-1)!

Bending Moment Diagram of Sample Problem

Section B5July 9, 1968

Page 48

Figure B5.2. t-4 Shear Diagram of Sample Problem v

;'L 'l: :: ; V.':'::':I ::"::': l':..:: " ::':: ":1:':" '"' ":': --": _:'" " !: ' I . :" . : L ": I ::': • . ' : "I.:::".'.:'::". : ". .'"! " • ." ' "s'.:: • : " :" " : " :i

:L i'i'i]lt5 2 I S;irnldC i)roi,l('rn ICont'd) .:: :::.::i..iLF:r : :1' ::i: "} :l } • } . : . .. 'i• - .... . " • .. ....... . .... : .-_-.*'"."%'.'"---.*:-.'-."'--.: ..... ,.... .............. . ............... :... • . : . .

_/...:hl..'i-_7-:..-.,T-.-::.!7.:i_;!:.:iiT_!:.i:.:.L:: l::i:::i i::. : .!. !.. .I:_ !: i :._i: :.:.I ..:': :! :!:.. i i ::_:. i _ i..;;. .F.:: i... i.: .. i : .. :;.... _l:_.i.4 iI'--o-: o-'_. ....... "-'I'--":--.:','-'T"T"" : .... '.':'"I: ......... _,.--:::--"I'." ........ ! ..... :'d': ...... ::,_": ..... : ...... .q........ |_.:::::'":" "': : ....... : ..... :" "':

:.." ! " ." "i.':'i '" ' : "i'" i. ": :'.J. .:'.L ' ..' ': " : " i" . ""._1 : "':" • " ! '" '

I-IL...L-;:.I i :_::i:...:I_.:.!...L:.L.::!....:.:.;....:::::::::::::::::::::::::-:.:;..]..i:.:.::....i:.-l:i..i:::.ii.-!:.:.;l:i:.:.:..!....:...:.. i:.. i .....• |.'" : . • .'.l t • " : ": • " .... , ' " ; ; : • : • • : : :.... • • • J i ........... | .... • • •l:i :...;!i i.:ii_l:::;...!...:..;..: : .. ! ....._.::.;..,:....._..:.. .:::..,::.;.:.. ....!..;.:..!::..;.:, .: ....... ....• ..:..I ..... ' " • " " ' " : " ' ' " "i ...... : .... I : ...... :" _ ;" • ; "I::: .... .: .... : ".:'.I ". I ::.:" '. .. ! "".J : ; : ", • "I";'. I .... :':. I "'" J "' : i'"

:!'_'..::.... !--I;I,'i • i ---: ::;-- i :::. _ i:: ;:::i:-, -.i._ i.-. : • ;!.!i _- '- ; "-_.i .! .- "" .... i......... ":' ""li::.:::...;-... I. ....__. : : _::.i: :_. ! :: i_:::.i.I... ::.i:: .: :.i .:_:_i;...r ..li:._. _ .:.: . i ' i .:ii.lli:Ti :1 ii :i i i:i__ i:: i i, i,::l ::: i:!_::l i :+:;: ; ;:1 : i..... I ...... I' : "l' :" • "1" : :' ":.. ":" " ': ;'.: 1 • . .... :..l ....."" : ," "," • • • i ....... I .... '""'" ..... ' ............ "" [ ,. • I ...... I ....... :.... . ...; :: ,. ... .... : . . . : .. . .,...1;;..{ . -:. "'. ..... : : .... : :..: .'. : :::i..'...: ....... I . .

: " : I I ......... i ....... I ....... :l l , '':" t ": "" .: " .' ........ : • ,':. : "" ..

• " . ._ ..i... i...:..:;_

::.;'" , . , . . ,. , . ." , . . , . _ ":.. . I_ | . _ " | ' " _ .I.

'_! I_ ,_i;'' • i:.. i:! ' :_i:"':i.:. I.., .i .... : :I........ -- ....................................... ' ................ _ "".,l' ....... I ....... | ................ : .... : "" ": : ....... : .... : .....:. -I • : I:i " " " i " ; : :. :" ..":: . :' '...: : " : • ;!:,,::; l Y i; , , , : I, ;• ''. ' • • ": " I_1 : I ; ": :: ' :""I::..,_1...L ,.. .._ I_: _, _-,,I._ ."ri.;.;.......I , ._ :__ .......... :....l_.l--"'! i. .-".:: , " • '-'!" ,', '" . -' T -"; . . ' ;!_".-_-! " I " : ::.... : :_ _! ' I I : % i • i I': .'i''_"'i" '': " "

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Fig_ure B5.2. I-5 Direct Load Diagram of Sample Problem



The values for Mo, No, and So are calculated and shown in the tables

on pages 52 and 53. The necessary formulas for computing the shear flows

are shown below. In these equations, counterclockwise shear flow is con-

sidered positive.

Figure B5.2. t-6 Symbols and Sign Conventions for Computation of

Mo, N o and S o

P -M

qPot = lrR---o sin _ qM_ - 2_R:

= ---Q'-sin (c_ + 90 ° ) - 21rR _qQc_ 7rR ° o qa = qPoL + qQo_ + qMc_

In order to perform the calculations for Mo, No, and So, the shear

force acting on each segment must be known. This shear force is obtained

by multiplying the average shear flow acting on the segment by the length

over which it acts.

Section B5

July 9, 1968

Pa_e 51

B5.2.1 Sample Problem (Concluded)

This average shear flow is determined by using Simpson's Rule.

the typical segment in Figure B5.2.1-7, this average shear flow is:

F_ r

l ÷ tt tit)qavg. = _ {qa 4qa + q_

where:

t

cl_ = Shear flow at left end of the segment.

tt

qc_ = Shear flow at center of the segment.

q_!

= Shear flow at right end of the segment.

f!

q_

Figure B5. 2.I-7 Shear Flow Acting on a Typical Segment

The formulas used for computing the values in the tables on pages 52

and 53 are listed under the cables. The index (n) refers to the column

number and its range is indicated where necessary. The formulas for So,

No, and M o are applicable for all values of n, therefore, the index wasomitted.

B6.2. i Sample Problem (_Cont'dJ_

Section B5

July 9, 1968

Page 52

e_

+

+

+ +

fl H I$

e_ el e_

t!

h

W

II

>

0

II

e_

II

II II fl II

B5.2. I Sample Problem (Cont'd)

Section B5

July 9, 1968

Page 53

o

OZ

O

>

i@4 _ _ _ u_ _ _D _D _ ¢q Q0 x-_ OO Cq OO _ c_D ¢q'O r_- P- ¢q_ _--

_ .........................................

Ill I[I lllliJi

IIIIII

............. ............ %,_°_=_

iilllll

4_'_'_'_'4'_'_'d'_'C'_'_'_'d'_4 "'_'_'i'_ _ _ _ ,I _, _ _ _ _ ,II

°, ., °. °, °. °, °. 0. .. ° .I °1 °i .i .i .i °1 01 °1 ol °1 .i °

" _ " ®" '<>-=_ _,i_ '° " _ " '7I I I I I I

II il

+

0

Ir"

0

I

t._

d

+

I

e-,

q _

Vl I II II II

--;I

I

II

O

O

I

.=

I

II

O

z

c_

+

Oc_

!

O

Section B5

July 9, 1968

Page 54

B5.0.O - FRAMES

References

1 Timoshenko, S., Theory of Structures, McGraw-Hill Book Company, Inc.,

New York, 1945.

, Sutherland, H. and Bowman, H. L., Structural Theory, Fourth Edition,

John Wiley & Sons, Inc., New York, 1954.

, Wilbur, J. B. and Norris, C. H., Elementary Structural Analysis, First

Edition, McGraw-Hill Book Co., Inc., New York, 1948.

. Grinter, L. E., Theory of Modern Steel Structures, Vol. II, The

Macmillan Co., New York, 1949.

o Perry, D. J., Aircraft Structures, McGraw-HiLl Book Co., Inc.,

New York, 1950.

. Argyris, J. H., Dunne, P. C., Tye, W., et al., Structural Principles

and Data, Fourth Edition, The New Era Publishing Co., Ltd., London,

No date.

• Roarke, R. J., Formulas for Stress and Strain, p 1120, Third Edition,

McGraw-Hill Book Co., Inc., New York, 1954.

r

SECTION

RINGS

B6

B6.0.O

TABLE OF CONTENTS

Rings ...............................................

Page

I

6.1.0 Rigid Rings .....................................

6.1.1 In-Plane Load Cases .........................

(Index to In-Plane Cases) ...................

6.1.2 Out of Plane Load Cases .....................

(Index to Out of Plane Load Cases) ...........

6.2.0 Analysis of Frame-Reinforced Cylindrical Shells..

6.2.1 Calculations by Use of Tables ...............

2

7

8

57

58

59

70

..._. B6-iii

.,.r..z

Section B 6

July 9, 1964

Page l

B 6.0.0 Rin_s

Charts and tables are presented to facilitate the analysis of rings

and ring-supported shells. Section B 6.1.0 deals with rings that are

rigid with respect to the resisting medium, i.e., for an out-of-plane

loading, the free body of a ring supported by a thin shell is as follows:

p ".

Only bending is considered in the deflection curves for the in-plane

load cases given on pages 9 through 54. Refer to page 56.1 to include

the effects of shear and normal forces.

Section B 6.2.0 deals with circular cylindrical shells supported

by "flexible" rings. The choice between the use of this section over

section B 6.1.0 for any given problem is left to the analyst. Experience

should be gained on both methods.

--._..

B 6.1.0 Rigid Rings

Section B 6

15 September 1961

Page 2

In general, four basic loadings are required to define all loads

on a ring loaded in the plane of the ring. They are:

I. Loading by a single radial force

2. Loading by a single tangential force

3. Loading by a single moment

4. Loading by a distributed load.

Special cases for out-of-plane loadlngs are considered in SectionB 6.1.2.

The procedure for calculating the Bending Moments of the basic

in-plane loading is briefly reviewed in the following discussion.

Many other loading conditions may be analyzed by using these basic

cases by applying the principal of superposition.

The general equation for the transverse force (shearing force on

a cross-section) is

dm dm

S = ds rd_ ................................ (i)

The general equation for the axial force is derived as follows:

N+dN S+dS I X

P_Rd_ N_-

(a)

(N+dN)

7-

N

de(b) 2

Fig. B 6.1.0-1

B6.1.0 Rigi d Rings (Cont'd)

Section B 6 .

July 9, 1964

Page 3

d_ Rd_7_Fx = 0 = -S + (S + dS) + (N + dN) + N -_-- + P_

= dS + Nd_ + dN d___+ P.Rd_2

Neglecting the second order term, QdN _ _

dSN = - d-_ - P_R .......................................... (2)

The second term (P_R) occurs in the case of a distributed load.

The procedure to obtain an expression for the bending moment is

i11ustrated by the following specific case:

Load

P

(a)

q q

2 2

(h)

Fig. B 6.1.0-2

Because of symmetry Y = 0

The shear flow distribution is obtained from q = SQ/I, where Q

is the static moment of half the ring, S is the shearing force, and I

is the moment of inertia of the section.¢

S f S sin _5 P sin= _R 3 _ Rd95 R cos _ = _R = _R

O

B 6.1.0 Rigid Rin_s (Cont'd)

o P

T

R [1 - cos ( _-e)]

Fig. B 6.1.0-3

Section B 6

15 September 1961

Page 4

dMq = q_ Rde R [1-cos (C-e)]

= P-- sine R2de [1-cos (¢-9)]_R

Mqfl = PRT sin0 [1-cos (¢-0)]d@

_ PR (I _ cos¢_ ¢ sin¢)2

PMp¢ = -_ R sin ¢

M x - -R cos_

R_Y - +I

R sln¢

0

My =-R sin_ M z = +I

Z =+i

Fig. B 6.1.0-4

Section B 6

15 September 1961

Page 5

B 6.1.0 Rigid Rings (Cont'd)

f MiMkd sThe displacement "i" due to load system "k" = 5ik = E1

E1 5xx = / (-R cos _ )2 Rd_

o

E1 5yy = / (-R sin ¢)2 Rd¢

o

E1 5zz -- f (+1) 2 Rd_ = _R

o

_R 3

2

R 3

2

Displacements due to applied forces and reactions

E1 5xo =

o

E1 5zo = /_

o/o

_ xoX;

xx

Deflections due

to unit loads

shown in

Fig. B 6.1.0-4

(Mp + Mq) Mxds

/ < I _cos__ _ _ sin P _ PR (-R c°s_ ) Rd_-2

Because of symmetry MpMx = 0

3PR 3

(Mp + Mq) Mzd

PR_ sin_ + i_2 _ cos_._ psinp_(+I)2_ Rd_ - 3P"2

3P

4_

5zo 3PR

Z - =2_

gz

Redundant obtained by equating

deflections.

By superposition

M = Mp + Mq + XM x + ZM z

_ PR sin_ + < I cos _ _ _sin_ _ PR +< 3P_2 _ - _ 2 n - _ R cos

+ 3PR (+i) = ( _-¢ ) sin ¢ - i2 _ 2 2_

dM [ sin_ + (_-_) cos _ ] PS" R_ = - 2 2--_-

dS [ ] PN = - d-_ = (_" _) sin_ + 3 cos2 2n

Section B 6

15 September 1961

Page 6

B 6.1.0 Rigid Rings (Cont'd)

Sign Convention

Moments which produce tension on the inner fibers are positive.

Transverse forces which act upward to the left of the cut are

positive.

Axial forces which produce tension in the frame are positive.

+S

+M

Fig. B 6.1.0-5 Positive Sign Convention

B6.1.1 In-Plane Load Cases

Section B 6

15 September 1961

Page 7

Coefficients to obtain slope, deflection, bending moment, shear,

and axial force along with equations for these values are given for

some of the frequently occuring load cases.

B 6.1.1

Index

In-Plane Load Cases (Cont'd)

Section B 6

15 September 1961

Page 8

21. P_=PmaxC os

o

13.

_.d M'

18.

P_ =Pmax c os

23.

_a x

P_ =Pmax c os

p

P_ =Pmax co_

.

I0.

i

p IP

20.P_=Pmax cos (2_

ax

25. P_= Pmax (a

+b cos_

+c cos2_)

B6.1.1 In-Plane Load Cases (Cont'd_

Section B 6

15 September 1961

Page 9

k r _f '-f

il . I

f

/

/

\ ./

\i

/,

jl f

/

/

I

>-...

rI

jfJ

/r

09

o o o.• 9

N

\\}

/C

/

i,

/

"ocu

'bo

"o

I_ _- Lf'__

9 o,.q ?.

B 6.1.1 In-Plane Load Cases

//

\

/

\\

/

• F

\

(Cont'd)

Z Ov E /

io

v

/t

/i\!<

//

/d

/

o/f

eo

///

J

/j/

rI

\\

S

/

J

/

_b/

\\kk//

/-

IO4

%\

\

\

/

//

o.

Section B 6

15 September 1961

Page i0

/

/

+

i

IN".\ b

0

Q

0

. J

" /j/ +%% ....

,/ I

" ! °/X o

"/ _ -0

0__ __ _,]

/ / .... _-- ............. i __..... _ - ..........

/ ...... ! g¢_

• r 5J r

/ .o

_,. ----+p-- 0

8O

0

e4

Q

8N

0

0

t ...... O0

t ,t

I °"_ _'_ ___ 0

I f I' I' I"

ON.

(D O4 (Do. o. O

C_

o q - - i_I I I I I

,f

f

B 6.1.1 In-Plane Load Cases (Cont ' d )

Section B 6

February 15,

Page l l

n

b_

4

JJ

f_ /

ff

f

(Xx ..

_<

\

h

I

yl

I,LI f

f

ff

JP

r Z

J-i

1976

B6.1.1 In-Plane Load Cases (Cont 'd

Section B 6

15 September

Page 12

1961

f

\\

J _;oz

_ oz

o-

f/

ii

tv

_" o

J

,if

i

11

z

f/t

\ L

_J

(\

if

Jf

JI'

f

Ii

fl

i

i"

J

i /

j oj 0

I

0

o

%

Y-.

J -.

-...

A

_J

-j1 /

O/t/

/r

• I

_ _ N - oI"

k)

JJ

j-

0

o

0

o

¢q

0

o_0oJ

\.o

)oj °

\ -\

\

7J

i

cO

oJ

o_' o _ co o.I (0 oo o q - -

I I I I I

,/

I_../

B 6.1.1 In-Plane Load Cases (Cont'd)

a.

k

i'

ff

iJ

i\ i

Fs w

\\

J

ii

f4

o.

Jf

/

fJ

-_. q O.

O. q Q

JJ

J

i r

Section B 6

15 September

Page 13

b

J

s vi

h/

I

f

Ji

\\

/

b

b

J

jf

_J

f

\\

k) _>

l

\

/i

o0 0 0 0

I" I' I" I

1961

B6.1.1 In-Plane Load Cases (Cont'd)

Section B 6

15 September

Page 14

1961

Q. f jr !--/ o--_ o .,_" / 7-.... ,o

/ // ,I/, ON

i / ,/ oo

n ii Ii • _

5:oz i b

i 0

'N,_" \ b• ",_,,, _ k'M

0

/ j,- i __

/ / z o

,i /" ," b----7- ....

i" / / -I /,- / ._ o

i , /t / j'/ --

I// / " <'t 0..,, " 0

( f

/ i

-_ °\_ i o

\ o0i; "\ \' "\ _ b

, .,_ _/_ , _.

_oz_ _. _ _ -. o 7 e_ _ _. to.v vv • I' I" I i

,F

f


a.

ro

I!

\\

a.

f

\

\

/(\

\

/

<)=

(Cont'd)

JJ

f

Jf

J

Jf

iI

If

J

jfJ

JJ

f

JJ

JJ

J

_fJ

7

--%.

J

\)

//

Section B 6

15 September

Page 15

o. O

.... O

O

oJ

O

OOro

O

oO

O

oJ

@

O

oJ

0J

O¢qO

OGO

o

O

O

Ox

O

OoJ

O

"\\ 8

O

S

or

jJ \

_/ oO

0 --. 04 rOI I' l"

CO Odo. o -I I I

0

1961

Section B 615 September 1961Page 16


o. o-

0 Zv XII iJ ||

=E 0 Z

0

I

L

I.

Jf

J

J

_J

J

f

/

\

O.

Sf

/f

f

\

_ff

IJ

i r

J

_Jf

J

V

//

--. _o ,

t L

JJ

\j

i\

/

/

//

0I

_Z

I "

c_ toI ! I I

V. N 0

I" I I" I" T

0

0odI¢)

l¢)

o,I

E,

oJ

oJ

od

0

0oO

_p

0

0

0

00

oD

@

0_P

_r

0

0o,I

v


Section B 6

15 September

Page 17

1961

f-

,f

_-'- -_._

a,. _'_1_. I I jI

ff

f

n

f _

lJ

f-

t _

i

I

ffN --Q Q

J

1

_....-

, toO

L

O

O

) oJ O

J _

0

O

0

O

0

ON

) oL.._.._ _

O

O _-- OQ o

i r

O

Jf

J

_ _ s _

0

O

o

O

O

) of J 8

Q

O(10

IJ

)f

f

l_?.

0J

O OO

9.

m

9.


Section B 6

July 9, 1964

Page 18

W

(.I o z_;v v

vi| I| II

//

/

0.

\

I'\,""- _ / l

/

fi

\%

,/

irI

II

\

\

/

/

• A

\

\¢t,

2"(! ,

fJ

//

tI

/"

i

fA

i f

f

/

,I

f

] ,i1/

/

/

i

J

/

f

fi

( I

• i/,, _

>'..._/

/"/

i

(\

ix

/

,¢

..2

/

If

_p.-,t

i /J

/

/

i /i/

//

iJ

f

t/

__ t

I" I I

- - o. O. o o -.I I I

¢0I

I

0

I

O4

-vf

Section B 6

July 9, 1964

Page 19


I

--- - +---- _ _....,.

I - ---

I

'- .,,, i --4 --

I

.... +

i --*-

t-- --I _

1I

/

, _

44 I

- l

i

-1- -K

I

_ . ÷___

_I _ _

_'[_.

i

.... __ ____,

1

I 1

I

i

i

[- ,

.... -_!__

-1 '

, \I, /

\,

/

.... 1 ____+ ....

r

IQ

I I

0 0o. o.

00 0 --o. _. o.I I i

]

b,

I

[I

1

J

\

C

I

I

I.--

I

ro

b

?:>

b

oJ

0

o.!


\. i f"

_L

0 Z _-- ........ "7

Ivv \ I" t liI II II %.

iOZ - I i

'\/I

I "_ %/i \\ k,

_= ................ _ \

\ I_-.. /

.,._ _ _ _t.., 7

_ ............. 7

/

/

1\/"I/'_,

X" \

/ t _

F L ........ _-"/ ....... 7-"

/ /

/ -_/ / /

o. _. _ _. N. o N. _,-I" I'

Oa --. -- 0 _ 0 0 --..• I" I" I

Section B 6

15 September 1961

Page 20

O

O

oi'MrO

O

GO¢J

O

O

N

oo(Xli'M

p _,,s

J oo

0

0

_o

0

o

0o

_. q' T

_ 0I" I"

Section B 6

15 September

Page 21

1961


/

k_

_N

i rI

-\/

11 "+"

/ 11/ /

fl /

O

0

0

0

00

0

O

(\

_'oO4q

LO N cO _1" 0 _ _ oa _ 0

o o o o o o -- --o o 9 o o 9 q• • • ° t' I" " =


,,j

(Cont'd)

/

/

/

\\

\

\\

\\

o.z

II

Z

%vII

(3

IEn

vII

=E

fl"

/ /

// /

/

f

\ I/

\\

!!

!

J

/

\/

tI/

/

0 Z--

f

./ It I

\

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15 September 1961

Page 22

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B 6.1.1

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15 September

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15 September

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Section B 6

15 September 1961

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15 September 1961

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15 September 1961

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Section B 6

15 September 1961

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15 September 1961

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Section B 6

15 September 1961

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15 September 1961

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15 September 1961

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15 September 1961

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15 September 1961

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15 September

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15 September

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15 September 1961

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15 September 1961

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Section B 6

15 September 1961

Page 53

• o _. _ _ o o o o o o• I' I" I" I" I"

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15 September 1961

Page 54

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Section B 6

15 September 1961

Page 55

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15 September

Page 56

I" I" I" I' I" I" I" I' r "r

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Section B 6

July 9, 1964

Page 56. 1

B 6.1. I In-plane Load Cases (Cont'd)

Deflection curves for the three basic load cases due to shear and

normal forces are displayed on the foI!owing pages. A shape factor

(/_) that is to be used with the curves for shear deflection of vario,_s

cross-sections is tabulated below.

Cross-Section Shear Area Shape Factor,

--f-h

_i_

Area of Web

AQ= th

Entire Area

A Q -- b h

Entire Area

AQ= 27rr t13q

Fig. B6.1.1-1

i_= 1.00

_ = 1.20 for b > 0.50h

l_ = 1.00 for b < 0 50 h

N.A. ,---:--e,

p = radius of gyration

with respect to the

neutral axis

Entire Area

AQ = (w) z -

( 2a)(w-2t)

=[1+ 3-(bZ-- aZ) a t__ll ]

[ "1ab3t t-O--tip2 .I

If the flanges are of

nonunifo r m thickne s s,

they may be replaced

by an "equivalent"

section whose flanges

have the same width

and area as '.hos_ of

the actual section.

=2.00

B 6. 1. I In-Plane Load Cases {Co,[',l)

Section B6

August 15, 1972

Page 56.2

\N

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July 9, 1964

Page 56. 3

f

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_oction B 6

July 9, 1064

Pa_e 56. 4

Z

o o o o _ o o o o

B 6.1.2 Out-of-plane Load Cases

Section B 6

March I, 1965

Page 57

Sign Convention

The following sign convention is given to define the positive

directions for out-of-plane loads.

Moments which produce tension on the inner fibers are positive.

Torque "I"' and lateral shear "V" are positive as shown in Fig.

B 6.1.2-1.

+T +V

Fig. B 6.1.2-1

B6.1.Z

Index

Out-of-PlaneLoad Cases (Cont'd)

.

PA

0For all cross sections

Section B6

March 1, 1965

Page 58

0For all cro_s sections

3" rMA V ..

Fo

.cross sections

Fig. B 6. 1.Z-g

v _

B 6. 1.2 Out-of-Plane Load Cases (Conttd)

Section B6

March 1, 1965

Page 58.1

"

B 6. I.Z Out-of-Plane I_,_,iidCas,.s (Contld)

l'uK{_ 5x. 2

B6.l.g Out-of-PlaneLoad Cases (Cont'd)Section B6

March i, i965

Page 58.3

o

B 6. 1.2 Out-of-Plane Load Cases (Contld) M r_'h 1, _t+';5

, m

o i -

---- Z- -7 / ..................... -_

o

X

H

\\,

B 6.1.2 Out-oi-Plane Load Case8 (Cont°d) Secclon B6

March I+ 1965

Page 58.5

_v

. ° ° .

o

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P

Section B 6

15 September 1961

Page 59

B 6.2.0 Analysis of Frame-Reinforced Cylindrical Shells

Tables are presented giving the loads and displacements in a

flexible frame supported by a circular cylindrical shell and subjected

to concentrated radial tangential, and moment loads. Additional

tables give the loads in the shell. The solutions are presented in

terms of two basic parameters, one of which is of second-order

importance. Procedure for modifying the important parameter to

account for certain non-uniform properties of the structureare

presented.

Notation

2.25

A

Lr 2

E Young's modulus - ib /in 2

Ef Young's modulus of unloaded frames - ib/in 2

E o Young's modulus of loaded frame ~ ib/in 2

Esk Young's modulus of skin ~ Ib/in 2

base of natural logarithms

axial force in loaded frame - ib

G shear modulus - ib/in 2

moment of inertia of a typical unloaded frame - in4

moment of inertia of an unloaded frame, distance "_" fromthe loaded frame ~ in4

Io moment of inertia of the loaded frame ~ in4

i I/_o - in3

Kn

n 2 l_Lr_ 2

3 Lc

distance from loaded frame to undistorted shell section - in

Section B 6

15 September 1961

Page 60

B 6.2.0 Analysis of Frame-Reinforced Cylindrical Shells (Cont'd)

Notation (Cont'd)

L c characteristic length (see Glossary) = --[--r [ t'r2 I i/4A/_ T ~ in

L rr %E/_-_-- in

characteristic length (see Glossary) = -_- V G--I-- ~

J_o

M

Mo

Po

P

q

r

S

s

To

t

t'

t e

u

v

w

x

7

7_

frame spacing ~ in

bending moment in loaded frame ~ in-lb

externally applied concentrated moment ~ in.lb

externally applied radial load - Ib

axial load per inch in the shell ~ ib/in

shear flow in shell ~ ib/in

radius of skin line ~ in

transverse shear force in loaded frame ~ ib

transverse shear per inch in shell ~ Ib/in

externally applied tangential load ~ Ib

skin panel thickness N in

effective skin panel thickness for axial loads - in

weighted average of all the bending material (skin and

stiffeners) adjacent to the loaded frame, assumed uniformly

distributed around the perimeter ~ in.

axial displacement of shell ~ in.

tangential displacement of shell - in.

radial displacement of shell - in.

axial co-ordinate of shell - in.

"beef up" parameter Io/2i L c

y for nearby heavy frame

/I

Section B 6

15 September 1961

Page 61


Notations (Cont'd)

rotational displacement ~ radians

polar co-ordinate of frame and shell

Basic Assumptions

In the method of attack with which this section is mainly concerned,

a simplified structural model (Fig. B 6.2.0-1) is used to obtain a

solution for a uniform shell stretching to infinity on both sides of

the loaded frame. Clearly the effects of any frame can be propagated

only a finite distance along the shell. In practice, the perturbations

from the "elementary beam theory" are, at worst, negligible at some

characteristic length "Lc" inches away from the loaded frame. Procedures

for modifying the solution to account for discontinuities and non-

uniform properties are discussed in the following sections. For the

model used, the following assumptions are made:

(i) Concentrated loads are applied to the loaded frame and

are reacted an infinite distance away on either one or

both sides. The shell extends to infinity on both sides.

(2) The loaded frame has in-plane bending flexibility. It

is free to warp out of its plane and to twist. It has

no axial or shearing flexibilities. Its moment of

inertia for circumferential bending is constant.

(3) The effects of the eccentricity of the skin attachment

with respect to the frame neutral axis is ignored for

both the loaded and unloaded frames.

(4) The shell consists of skin, longerons, and frames

similar to the loaded frame, but possibly with different

moments of inertia. The skin and longerons have no

bending stiffness. All properties of the shell are

uniform.

(5) The longerons are "smeared out" over the circumference

giving an equivalent constant thickness, t', (including

effective skin), for axial loads.

(6) The shell frames, but not the loaded frame, are "smeared

out" in the direction of the shell axis, giving an

equivalent moment of inertia per inch, "i", for

circumferential bending loads.

B6.2.0


Analysis of Frame-Reinforced Cylindrical Shells Cont'd_

Basic Assumptions (Cont'd)

The simplified structural model described by the basic assumptions

bear only slight resemblance to practical space vehicleshells, however

the difference is compensated by, modification of certain parameters as

discussed in the following pages.

Glossary of Terminology

Characteristic length - In this section there are two characteristic

lengths, defined as follows: L c is the distance required for the

exponential envelope of the lowest order self-equilibrating stress

system to decay to I/e (e ~ base of natural logarithms) of its value

at x = o, provided that the skin panels are rigid in shear. L r is the

distance required for the envelope of the lowest order self-equili-

brating stress system to decay to I/e of its value at x = o, provided

that the frames are rigid in bending.

Shell with Flexible Frames

H_-.Externally Loaded Frame

to_v

Fig. B 6.2.0-1

W

/u /

fq

Fig. B 6.2.0-2 Load per inch and dis-

placements in the shell (Loaded frame

at x=o)

B 6.2.0

Section B 6

15 September 1961

Page 63

Analysis of Frame-Reinforced Cylindrical Shells (Cont'd)

Evaluation of Parameters Lr, Lo, and 7

Case of uniform shell

In cases where the shell happens to satisfy all the assumptions

listed in the previous pages, and in particular, if the skin thickness,

stringer area, and shell-frame moment of inertia are uniform in both

the axial and circumferential directions, the following formulas may

be used:

Lc=

__E__rV E t'Lr = 2 Gt

7

I o

2 iL c

1/4

....................... (1)

....................... (2)

.................................. (3)

Young's modulus for skin, stiffeners and all frames is assumed

equal. Coefficients are obtained by use of these parameters (Lc, Lr, 7)

in the tables. These coefficients yield the required loads and defor-

mations when substituted into Eqs. 14 thru 21. In non-uniform shells,

use the modified parameters indicated in the following equations:

Case of non-uniform shell

(a) In the case that the shell properties, i, t, and t', vary

over the surface of the shell to a moderate degree, the

following formulas and definitions are appropriate:

Lc r I Esk te r2 ] I/4- . ..................... (4)

_- Ef i

V Esk t'Lr = 2 G t

......................... (5)

=

E o I o

2Ef i L c................................ (6)

The stiffness factors, Gt, Esk, te, and Efi, must be averaged in

the neighborhood of the loaded frame. The factors Gt and Eski shall

be averaged over a length of shell extending approximately one-half of

a characteristic length from the loaded frame in both directions.


B 6.2.0 Analysis of Frame,Reinforced Cylindrical Shells (Cont'd)

Case of non-uniform shell (Cont'd)

(b) When unloaded frames _ave unequal moment of inertia or are

unequally spaced, the following weighting factor is used for

computing Efi:

Efi = (Efi)fw d + (Efi)af t ......................... (7)

i E (WEflf)(Efi)fwd - Lc fwd

................... (8)

I Z (WEflf)(Efi)aft = Lc----aft

................... (9)

Where

X

W = 1 - Lc for x < L c

= o for x > L c

(x is measured forward and aft of loaded frame)

The summations in Eqs. (8) and (9) are to be extended over all

frames except the loaded frame. The method of calculation gives

greater importance to frames closest to the loaded frame and less

importance to those farther away. For the case of a single, particular

heavy, neighboring frame, or for other neighboring discontinuities such

as rigid bulkheads, a free end, or a plane of symmetry, the correction

factors to be discussed is applicable. If those corrections are

applied, the heavy frame or other discontinuity must be ignored in

applying Eqs. (7), (8) and (9). In particular, if the loaded frame

is near the end of the shell, the shell must be continued beyond the

end, fictitiously, in the summations of Eqs. (7), (8), and (9), as

though the shell were symmetric about the loaded frame and extended

for a length greater than L c on both sides of the loaded frame.

The method of calculation indicated in this sub-section exaggerates

the effect of frames which are heavier than average when compared with

the more accurate method of correction given in the next section.

Since L c depends on (Efi) I/4, an initial estimate of Efi is required

in order to calculate the Lc used in Eqs. (7), (8), and (9).

B6.2.0

Section B 6

15 September 1961

Page 65


Corrections to 7 , the "Beef-Up" parameter

The general form of the modified "beef-up" parameter, 7 *, is:

7" = 7 .fa.fb.fc. , etc., ........................ (i0)

where 7 is computed by the methods of the preceding section, and fa,

fb, and fc are factors accounting for effects of nearby heavy framesetc.

Modification for different value of Lr/Lc

The value of Lr/L c used in the graphs are 0.2, 0.4, and 1.0.To account for values of this parameter between 0.2 and 1.0, graphical

interpolation should be used. Otherwise, the following formula may be

Lr h* 2 Lr 2

7 * = 7 + Lc / i + 2 (ii)• 2 / ,, 2

where (Lr/Lc)" is the value of the parameter for the shell, and

(Lr/Lc)* is the value of the parameter closest to (Lr/Lc)", for which

graphs are available.

Modification for finite frame spacing

The modification for finite frame spacing is as follows:

7 * = 7 1 + 2LcK 2 1 + 2 7 K2/

where

z distance from loaded frame to adjacent frames

2

K2 -

Lc

/

Section B 6

15 September 1961

Page 66

B 6.2.0 Analysis of Frame-Reinforced Cylindrical Shells (Cont'd_

Modification for nearby heavy frames and for other similar nearby

discontinuities.

The corrections to "7 " in a previous section are not intended to

account for discontinuities in circumferential bending stiffness. Theform of the correction for these effects is:

7 * : 7 -f(2) ...................................... (13)

Fig. B 6.2.0-3 shows f(2) plotted for nearby heavy frames and for

nearby rigid bulkheads. Fig. B 6.2.0-4 shows f(2) plotted for a

finite length of shell terminated in various ways on one side of the

loaded frame. The validity of the correction is considered doubtful

for f(2) < 0.25, due to the importance of higher order stress systems.

Figures B 6.2.0-3 and B 6.2.0-4 are for Lr/L c = 0.4, but their varia-

tion with Lr/L c is negligible for conventional shell-frame structures

and adequate in other applications for Lr/L c < 0.75. The corrections

for nearby planes of symmetry and antisymmetry can be used to solve

problems where two similar frames are simultaneously loaded. To

illustrate the method the two following examples are given:

Example 1

A frame of moment of inertia 4.0 in4 that is subjected to

concentrated loads is supported in a uniform shell whose characteristic

length, Lc, is 200 inches and moment of inertia per unit length, i,

is 0.i0 in. 3 A heavy frame having a moment of inertia 16.0 in4 is 50

inches to one side of this frame. The loaded frame and shell loads

are required.

The parameters needed are:

7 = 4.0 = 0.i0 by Eq. (3)2(.1)(200)

167 _ = = 0.40

2(.1) 200

_ 50 = 0.25

Lc 200

Using 7_ and _/L c in Fig. B 6.2.0-3 yields f(2) = 0 75

-'- 7" = 0.75 (0.i0) = 0.075 by Eq. (13)

F ¸

f_

Section B 6

15 September 1961

Page 67


Example i (Cont'd)

Use 7 = 0.075 instead of 0.i0 in the curves to account for the

presence of the heavy frame on the stresses in and near the loadedframe.

Example 2

A shell whose characteristic length, Lc, is 250 inches is supported

by a large number of identical frames whose moments of inertia are 2.0

in4, spaced 24 inches apart. A pair of frames 96 inches apart are

subjected to concentrated loads at the same polar angle, _ . The two

radial loads are of equal magnitude but opposite sign, while the

tangential loads are of the same magnitude and sign. The loads in

the loaded frames and shell are to be found.

I 2i = - - .0833

Io 24

Io 2

7 = 2iLc = 2(.0833)(250) = 0.048 by Eq. (3)

f- 48 - 0. 192

Lc 250

For the tangential loads there is a plane of syn_netry midway

between the loaded frames, while for the radial loads a plane of anti-

sy_netry exists at the same place. From Fig. B 6.2.0-4 it is seen

that for the radial load stress system, f(2) = 0.32, while for the

tangential loading f(2) = 1.75. Hence, the values of 7 * to be used

in the graphs are 0.015 and 0.084, respectively.

Eccentricity between skin line and neutral axis of the loaded frame.

In the three types of perturbation just discussed, it is possible

to account for the effects by modifying 7 only, since the "elementary -

beam-theory" part of the solution is always valid. In the case when

the eccentricity between skin line and neutral axis of the loaded

frame exists, the "elementary-beam-theory" solution is also affected.

This particular aspect is discussed in Appendix E of reference i.

B6.2.0


Analysis of Frame-Reinforced Cylindrical Shells (Cont'd_

0

0 0.2

Fig. B 6.2.0-3

I I

7£=0.1

7s= o.5

7_= 1.o

7_ = 2.0

Two Rigid BuIkheads Sy_'_etrically IPlaced about the Loaded Frame --

Lc

A single frame on one side of loaded frame or two

rigid bulkheads symmetrically placed about the loaded

frame curves of f(2) and f(3). Lr/L c = 0.4.

B6.2.0

4.0

Section B 6

15 September 1961

Page 69


I I I i I I 1 I I I I I

,,..---- _ ---__

x Loaded Frame

Free End at x =

Plane of

Symmetry At

Plane of Anti-Symmetry at

x =

Built-ln At x =

_/L c

Fig. B 6.2.0-4 Finite length of shell on one side of loaded frame f(2)

vs _/L c for various boundary conditions at x =£ , Lr/L c - 0.4.

B6.2.1 Calculations by Use of Tables

Section B 6

15 September 1961

Page 70

Eqs. (14) thru (21) are given in this section, by which the effects

of a concentrated load or moment on a shell-supported frame may be

computed by using the tabulatedcoefficients. The method of computing

is indicated in a previous section. These enable the shear flow

and axial load at all points in the shell and the internal loads and

displacements of the loaded frame to be computed.

The following parts of the overall solution are omitted in the

tabulated coefficients:

(i) The "elementary-beam-theory" part of skin shear flow

which is calculated from beam theory.

(2) The "elementary-beam-theory" part of the axial load

intensity in longerons which should be calculated

from beam theory.

(3) The rigid translations and rotation of the loaded

frame.

As a consequence of items (i) and (2), shear flow and axial load

intensity in the shell, as calculated from the tables, can be added

directly to the results of an "engineers bending theory" calculation.

The shear flow and axial load distributions given in the tables are

assumed to be symmetrical with respect to the loaded frame. In a

shell that is unsymmetric about the loaded frame, the shear flows and

axial loads are not symmetric ahout the loaded frame. It is not

possible to derive a simple correction for this effect, but the exact

solutions indicated in reference 2 are applicable.

Distributed load on a frame

The effect of a distributed load on one frame may be obtained by

superimposing the effects of the concentrated loads into which the

distributed load can be resolved. The axial load and shear flow in

the shell can be obtained for loads on several frames by a similar

superposition, since "p" and "q" are tabulated in Ref. 3 NASA TN D402

as a function of x/L c.

Frames adjacent to the loaded frame

At the present time it is not possible by use of tables to compute

the internal forces in frames adjacent to the loaded frame. It is,

however, a simple matter to tabulate the frame-bending moment per

inch, "m", and the other internal forces as a function of x/Lc. The

bending moment in an adjacent frame, due to a force applied at the

loaded frame, is then obtained by multiplying "m" at the frame station

by l_/i (see Appendix D of reference I).

F

Section B 6

15 September 1961

Page 71

B 6.2.1 Calculations by Use of Tables (Cont'd)

Effect of local reinforcement of the loaded frame

It is not practical to attempt to cover, by a set of tables or

charts, the many possible reinforcing patterns that can be used to

locally strengthen frames in the region of applied concentrated loads.

A solution is presented in Appendix A of reference 3, together with a

simple example, to illustrate the numerical procedure. A loaded frame,

whose moment of inertia varies around the circumference in any mannercan be treated as a frame of constant moment of inertia that is rein-

forced to produce the actual inertia variation.

Tables

The loads and displacements of the loaded frame and loads in the

shell are given in terms of the non-dimensional coefficients of the

tables by the formulas below. The tables contained in this section

are for M, S, F, p, and q at x = o.

Coefficients for displacements v, w, and 7 are tabulated in Ref. 3

along with coefficients for "q" and "p" as a function of x/Lc.

Po To Mo

= _+ Cqt + Cqm r--_--q Cqp r r..................... (14)

+ __2__o o (15)p m Cpp r CP t r Cpm 7 r

M = Cmp Por + Cmt Tor + Cmm M o

M___2_o

S = Csp Po + Cst To + Csm r

M

F --Cfp Po + Ctt To + Cfm _._.9__r

........................ (16)

........................ (17)

........................ (18)

2

V = Cvp Po -_ + Cvt To _r3 + Cvm Mo 7 r (19)El o El o El o

W : Cwp po _r3 + To _r3 + Cwm Mo 7 r2El o Cwt El o El o

(20)

8 =, Ce p Po _ + C8 To _r2 + Ce Mo ___ .. ..... (21)El ° t El ° m El °

i •

Section B 6

15 September 1961

Page 72

B 6.2.1 Calculations by use of Tables (Cont'd_

Sign Convention

Loads, moments, and displacements are positive in the loaded

frame as shown in Fig. B 6.2.1-1.

Mo

To

eo Neutral axis

F

M

S

M

f- B 6.2.1 Calculations by Use of Tables (Cont'd_

Section B 615 September 1961Page 73

FrameLoads

Coefficient

Index of Tables

Lr/Lc=.200 Lr/Lc=.400 Lr/Lc=l. O00

AxialLoad,

F

Cmp B 6.2.1-1 B 6.2.1-5 B 6.2.1-9BendingMoment,

M Cmt B 6.2.1-13 B 6.2.1-17 B 6.2.1-21

Cram B 6.2.1-25 B 6.2.1-29 B 6.2.1-33

Csp B 6.2.1-2 B 6.2.1-6 B 6.2.1-10

Shear, Cst B 6.2.1-14 B 6.2.1-18 B 6.2.1-22S

Csm B 6.2.1-26 B 6.2.1-30 B 6.2.1-34

Cfp B 6.2.1-3 B 6.2.1-7 B 6.2.1-11

B 6.2.1-15 B 6.2.1-19 B 6.2.1-23Cft

Cfm

Cqp

Cqt

Cqm

Shear

Flow, q

At Ring

B 6.2.1-27 B 6_2.1-31 B 6.2.1-35

B 6.2.1-4 B 6.2.1-8 B 6.2.1-12

B 6.2.1-16 B 6.2.1-20 B 6.2.1-24

B 6.2.1-28 B 6.2.1-32 B 6.2.1-36

• i r-¸

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Section B 6

15 September 1961

Page 74

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15 September 1961

Page 75

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15 September 1961

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15 September

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Section B 6

15 September 1961

Page 79

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Section B 6

15 September 1961

Page 80

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References

, MacNeal, Richard H., and John A. Bailie, Analysis of Frame-

Reinforced Cylindrical Shells, Part I - Basic Theory. NASA

TN D-400, 1960.

1 MacNeal, Richard H., and John A. Bailie, Analysis of Frame-

Reinforced Cylindrical Shells, Part II - Discontinuities o_

Circumferential-Bending Stiffness in the Axial Directions.

NASA TN D-401, 1960.

, MacNeal, Richard H., and John A. Bailie, Analysis of Frame-

Reinforced Cylindrical Shells, Part III - Applications.

NASA TN D-402, 1960.

I

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NASA Astronautic Structural Manual Volume 1

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Transcript of NASA Astronautic Structural Manual Volume 1