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Dissertations and Theses Dissertations and Theses

1980

The effect of residual stress distribution on the The effect of residual stress distribution on the

ultimate strength of tubular beam-columns ultimate strength of tubular beam-columns

Steven L. Barrett Portland State University

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Recommended Citation Recommended Citation Barrett, Steven L., "The effect of residual stress distribution on the ultimate strength of tubular beam-columns" (1980). Dissertations and Theses. Paper 3050. https://doi.org/10.15760/etd.3045

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I

I _J

AN ABSTRACT OF THE THESIS OF Steven L. Barrett for the

Master of Science in Applied Science presented

February 22, 1980.

Title: The Effect of Residual St~ess Distribution on

the Ultimate Strength of Tubular Beam-Columns.

APPROVED BY MEMBERS OF THE THESIS COMMITTEE:

III, Chairman

H~kErzu

son

- ~~

Using data for the longitudinal residual stress

distribution in welded steel tubes, curves describing

these distributions are selected for study. Each of these

curves are checked for static balance across the tube

cross section. The curves that exhibit an imbalance are

adjusted by a combinatior1 of a simplified model for each

and the use of a computer program that is developed to

calculate the resulting forces and moments on the cross

section. The residual stress in the area of the tube wall

opposite the longitudinal weld is found to be the most

important in the adjustment to obtain exact equilibrium.

The method of adjustment is rational and based on maintain­

ing a smooth curve shape that matches the· raw data the

closest and producing a curve that is balanced within the

accuracy limits required.

The balanced longitudinal residual stress distri­

butions are used to determine M-P-~ relationships,

where M equals moment, P equals axial load, and ~ equals

curvature. A computer program developed by Wagner (13)

is modified to determine these relationships. The program

was originally developed for a bending axis through the

weld and the results are found to be inaccurate for

other orientations. The problem is identified as the

creation of a moment on the cross section from an uneven

distributionof:axial stress caused during the iteration

to determine axial strain from an applied axial force.

The problem is solved by adding another iteration to the

program to redistribute the axial strains and eliminate

the moment.

2

The M-P-~. relationships obtained from this computer

program for three longitudinal residual stress distributions

with five different bending axis orientations are compared

and one residual stress distribution curve is selected

to compare with Wagner's (13) distribution. The beam

column failure load computer program (13) is used with

the M-P-~ relationships generated· and the results are

compared with test data.

The conclusion is reached that residual stresses

have a large influence on the behavior of welded steel

tube beam columns. The difference in effect of each

residual stress distribution is not large. This could be

because all of the curves have the same· general shape with

very large stresses close to the weld and rapidly de­

creasing away from it. With the effect of residual

stresses included, the beam column failure load program

still overestimates actual member strength. Therefore,

other conditions effecting member strength when used as

beam columns require further study and the residual

stress distribution labeled curve number three which is

Chen and Ross's (3,4) curve forced to two maximums from

x/d = 1.0 to x/d =1)-' is recommended for use in further

research.

3

THE EFFECT OF RESIDPAL STRESS

DISTRIBUTION ON THE ULTIMATE STRENGTH

OF

TUBULAR BEAM-COLUMNS

by

STEVEN L. BARRETT

A thesis submitted in partial fulfillment of the requirements for the degree of

MASTER OF SCIENCE

in

APPLIED SCIENCE

Portland State University February 1980

; I. 1

:

TO THE OFFICE OF GRADUATE STUDIES AND RESEARCH:

The members of the Committee approve the thesis

of Steven L. Barrett presented February 22, 19'80.

Wendelin H. Mueller, III, Chairman

APPROVED:

Arnold Wagne~·

Engineering

Stanley E. Rauch, Dean of Graduate Studies and Research

~

TABLE OF CONTENTS

LIST OF FIGURES . . . . . .

CHAPTER

I

II

III

IV

v

VI

INTRODUCTION . . . . . .

REVIEW OF LITERATURE .

COMPUTER PROGRAM FOR CURVE ADJUSTMENT ..

RESIDUAL STRESS DISTRIBUTION CURVES .....

M-P-0 CURVE GENERATION . . . . . . . . .

CONCLUSION . . . . . . .

REFERENCES ..

APPENDIX I.

APPENDIX II

PAGE

iv

1

5

7

25·

39

61

69

71

83

FIGURE

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

LIST OF FIGURES

Longitudinal Residual Stress in a Welded

Steel· Tube . . . ~ . . . . . . . . .

Longitudinal Residual Stress .

Area Under Curve .

Arc Length - Lever Arm

Areas Under Curve ..

Areas Under Curve.

·Flow Chart - Moment, Force Summation

Residual Stress From Tran (11)

Approximate Curve ..

Approximate Curve.

Residual Stress Tran (11)

Simplified Curve

Residual Stress Chen & Ross (3,4)

14. Residual Stress, Baianced Chen & Ross (3,4)

Curve 1. . .

15. Residual Stress First Section Average - Chen·

& Ross (3,4) - Tran (11) Curve 2.

16.

17.

Fluctuation in Curve Shape .

Residual Stress First Portion, Smoothed Curve

· from Tran (11)

PAGE

9

9

11

11

13

14

15

18

21

22

26

27

29

31

32

33

35

FIGURE

18. Residual Stress Chen & Ross (3,4) Forced to

Two Maximums Curve 3. .

19. Flow Diagram for Calculation of M-P-0 Data

(Wagner)

20. Orientation of Axes and Residual Stress

Distribution (Wagner (12,13) . . . . . 21. M-P-.0 Curve 1 (Chen & Ross)· .. . . . .. . . . ..

22. M-P-.0 Curve 2 (First Section Ayerage) . . . . 23. M-P-.0 Curve 3 (Chen & Ross Forced To Two

Maximums). .

24. Moment Imbalance . • .

.

.

.

v

PAGE

37

40

44

46

47

48

53

25. M-P-0 Curves From Improved Generation Program. 56

26. Beam Column Results - Wagner's Distribution. . 59

27. Beam Column Results - Curve 3 (Fig. 18). . . . 60

28. Longitudinal.Residual Stress Distribution

Curves . . .

29. Moment Curvature- Relationships

30. Effect of Residual Stress on Strength Under

31.

32.

Axial Load ...

Ef feet of Residual Stress ..

Example for Testing Curve Program. .

62

63

64

65

81

CHAPTER I

INTRODUCTION

The idealized model that is used to exhibit the

pehavior of steel in most mechanics of materials texts has

well defined characteristics. The idealized material is

stress-strain free prior to application of any load,

there are no thermal strains, the material is homogenous

and it has a bilinear stress strain relationship with con-

stant slope to yield and a constant value beyond. If this

material is used in an idealized member it is possible

to theoretically predict the member's exact behavior under

load. For different cross section shapes it is possible

to determine the effect of slenderness on the member.•· s

behavior if it is used as a compression member, a beam,

or a beam-column.

For a real structural member the prediction of actual

behavior is much more difficult. Actual materials are

not entirely homogenous and do not have bilinear stress

strain curves. A real structural member is not stress

free prior to loading, may have a comple~ cross section

shape which ha~ no simple closed form solution for ·

critical buckling load, may have accidental eccentricity

of load, initial crookedness and any number of stresses

induced during erection. Each of these cond~tions that

can be accurately measured in the laboratory should be

investigated to determine its individual effect on the

strength of the member. Once the relative magnitude of

the individual effect is known, a decision can be made

whether or not to include that effect. in every. member

analysis. To that· end, there has been considerable

research done on certain structural shapes in various

materials. The largest volume of research has been done

on rolled steel wide flange shapes. due to their widespread

application for structures of all types. (l,2,5,8,10)

Because the information available for the steel wide

flange ·shape is. extensive, there has been a tendency t.o

apply it to other cross section shapes where data has not

yet been developed. For any particular case· this practice

may or may not be conservative. It is not possible to

completely discount any of the previously mentioned effects

on a cross section shape other than a wide flange based on

the argument that such an effect did not alter the behavior

of the wide flange as idealized. Nor is it possible to

make a direct comparison without some data being developed

for the particular cross secti6n.

2

The complete description of the effect of all possible

initial conditions on a welded steel tube cross section for

all types of loading is too broad in scope for this study.

Therefore the single consideration of the effect of residual

3

stress was selected to study. Residual stresses are those

stresses which are present in a member after it is fabri­

cated to its finished form. Usually these residual stresses

cause a reduction in strength in fatigue, fracture and

stability· although for some distributions a strengthening

at certain load levels will be exhibited. These stresses

may result from uneven cooling after hot rolling as for

structural shapes and plates, from cold bending during

fabrication, from welding or ·from cutting, drilling and

punching during fabrication. Usually residual stresses

with the largest magpitude result from cooling and welding.

In many cases, residual stresses in the region of a weld

will be greater than the yield stress. It therefore

becomes important to determine the distribution and magni­

tude of residual stress in a member that has extensive

weld~ng during fabrication. Recent research has suggested

some possible distributions of both longitudinal and

circumfere.ntial residual stresses in welded steel tubes.

(3,7,11,12,13)

The primary goal of this project was to analyze and

select one longitudinal residual stress distribution

in welded steel tubes from the literature and test data

available and then determine the effect of this distri­

bution on the ultimate beam-column failure load of the

tube. The circumferential residual stress was not included

since the project was simplified excluding the effects of

local buckling or ovaling from consideration. After deter­

mination of the residual stress distribution, moment­

curvature-axial load curves could be generated by computer.

These describe· the performance of the cross section of the

tube. Then, using these curves, an iterative open form

computer solution calculates the beam column failure ioad~

The solution obtained is c~mpared to the ideal case of no

residual stress and with the results of previous research

using different.distributions. From this comparison it

is possible to determine the relative effect of res{dual

stresses in determining failure load and the effect of

changing the distribution.

4

CHAPTER II

REVIEW OF LITERATURE

The distribution of longitudinal residual stress

described by a linear variation between three peak values

was used by Wagner (12,13) in the absence of actual test

data. This distribution gave a starting point for the

computation of M-P-0 curves, where M equals moment, P

equals axial load and 0 equals curvature with the effect

of some residual stress added. A computer program was

developed using the assumed general shape that could adjust

the value of residual stress on each element of the tube

such that force an~ rotational equilibrium would be

satisfied.

While there is a large volume of information on the

magnitude and distribution of residual stresses in wide

flange sections (4,8), there is a scarcity of information

about the distribution and magnitude.in welded steel tubes

or its effect on the beam column failure load. Some

investigators have suggested possible theoretical distribu­

tions (7,) while others (4) have attempted to use more

realistic patte~ns in correlating experimental results.

They, however, discounted the necessity of static equilibrium

across the cross section considering the imbalance of

6

internal forces to be of little consequence. Further,

no consideration of the uniformity. of the stress field

through the tube wall thickness was made.

Recently however Tran (11) developed test data that

first confirmed a uniform field of longitudinal stress

through the wall of a ·tube and showed a very close

approximation to static equilibrium across the section.

CHAPTER III

COMPUTER PROGRAM FOR CURVE ADJUSTMENT

In order to determine .~he.best curve to describe the

longitudinal residual stress distribution in the welded

tube it was necessary to generate a method of rapidly

checking the balance of forces and moments that each

distribution curve implied. If the residual stress

distribution through the th~ckness of the tube wall was

not uniform, it would be necessary to check the balance

for several layers and use that information in the genera­

tion of moment-curvature axial load curves. However,

Tran (11) has shown that the distribution through the tube

wall is uniform. Therefore, it is·only necessary to

consider one layer equal to the tube wall thickness. To

check the balance of forces and moments accurately and

quickly, it was decided to write a computer program that

could handle any general curve describing the residual

stress pattern.

Consider the general longitudinal residual stress

distribution shown in Figure 1. The ordinates of the

residual stresses are measured along· the X axis with the

weld located on the Y axis. To make the curve easier to

represent, it is convenient to straighten the circumference

of the tube, in effect unrolling the tube so it is flat

(Fig. 2). Now the area under any portion of this curve

will represent a force, since the distance along the

circumference multiplied by the wall thickness is area.

Area multiplied by stress, wh~ch is force per unit area,

8

is force. This force will be located at the center of

gravity of the area under the stress curve. The distance

from any single· point on the tube circumference.to the force,

multiplied by the force will equal moment. Since the tube

is under no external loading and therefore must be in

static equilibrium for internal stresses, the summation .of

forces and the summation of moments over the entire cross

section of the tube must be equal to zero.

In general the problem is to find the area (force),

center of gravity and first moment of area (moment) about

some point for segments under a general curve that the

mathematical descrip~ion for is not known. The method used

will be a linear approximation .of segments of the curve,

in other words, short straight lines between points on

the curve. Sign convention used for this development

were: Tensile stresses positive, plotted above the X

axis (Fig. 2); compressive stresses negative, plotted

below the X axis; positive moment .about any singular

point, clockwi~e; negative moment counter clockwise.

There are several possible conditions that must be

accounted for if.the program is to handle all possible

dL

Figure I. Longitudinal residual stress in o welded steel tube

+

of tu be unrolled

Figure 2. Longitudinal residual stress

9

Refering to Figure 3:

Let Y1

= First ordinate on curve

Y2

= Second ordinate on curve

x1 = First abscissa on curve

x2

= Second abscissa on curve

C.G. 1 =-Distance from Y axis to center of gravity of area1

C.G.2

= Distance from Y axis to center of gravity of area 2

Area1 = (X2 - Xl) (Yl)

C.G. 1 _= (x2 - xl) + ~1·= x2 - xl + 2x1 = xl + x2

2

Area2 = (X2 - Xl) (Y2 - Yl)

2

2

C.G.2

= 2/3 (X 2 x1 ) + x1

= 2X2

2

2~1 + 3x1 = .2x2 + x1 3 3

10

Now since this is for a tube the X distances can be thought

of as arc lengths. (Fig. 4)

C.G.x = Arc Length

~ circumference = 21J-r = 7rr 2

~ (radians) = (C.G.x)

( -n-r ) ( 211) = ( 2)

C.G.x

r

Lever arm = L = r - (COS~)r = (1 ~ COS~)r

Moment = (Force) (Lever Arm)

Force = Area , ·Lever Arm = L

M = (Fl) (L1 ) + (F 2 .L2

)

M = {Area1 ) (r -·COS (C.G. 1 ) + (Area 2 ) (r-COS (C.G. 2 ))

r

M = [cx2 X1 ) (Yl) ][ 1 cos

(Y2 - Y1 )J [ 1-COS (2x2';

(Xl + X2)]

2r

x1 ) J (rl

r

(rl + [<2x2 + x1 J

3

. 1·

Y2

Ya

y

M x2

I

Figure 3. Area under curve

y

crown ~r,,-weld of tube

Figure 4. Arc length - Lever arm

0 lo- >-: ~ E cu i...

..::.J <Cl:

11

x

12

cases.of slope of the curve and axis interception (Fig.

5 & 6). Taking each of these one at a time, we can ex-

press each.area and center of gravity in the same terms

used previously (Fig. 6). The location of the center of

gravity for case 5 and 2 are equivalent (for the X direction)

l· as are case 6 and 1 as well as case 3 and 4. Thinking of

i this in flow chart form, we have the central ·part of the

force-moment program complete (Fig. 7). Starting with

the coordihates:of a known curve, it is possible to sum

the forces and moments for all the segments of the curve

·and arrive at a total that will indicate whether or not the

section is in static equilibrium. The method could be used

for random lengths of curve segments or for ~ constant

increment. Xhe program used for·this project used a

constant increment for the ·abscissa simplifying· the input

required since the ordinates become the only required

input. These ordinates were scaled from each curve

selected for analysis. Since the summations should be

zero about any point, the crown of tube was selected for

convenience.

Assuming that an imbalance of force is discovered,

it is possible to select the largest ordinate, divide by

10.0 and use this value to modify all ordinates until a

change of sign .of the summation of forces is detected.

Reducing the increment applied t~ the ordinates, it is

possible to continue the iteration until a force balance

i

I I I

I I I I i I

I :

y

+

13

0

A1 0

A1 x

oAd J oAI

Case I 2 3 4 5 6

y y

0A1 x 0A1 x

OA1 oA1

Cose 2 8 5 Cose I a 6

Figure ~ Areas under curve

1 · I I I

.y

x

( X2- X 1 )( Y2) A1 =

2

CG _ 2 ( X2 - XI ) _ 2 X 2 + X I - 3 + >t1 - 3

(X2-X1) --Xo= {-y,)

• . Y2 -y,

A ·_ (xoHY1l ' - 2

CG1 = (;0) + X1

X2

.y

A,·=· (x2-x1HY1) 2

14

CG = { X2- x I) +· x - X2 + 2 x I 3 1-3

A _ [x2-(X1 ·tXo~ (y2)

2- , 2

CG _ 2 ~x2-x1)- xoJ . ( ) 2-

3 -t x1+x0

= 2x2 - 2x1- 2x0 + 3x 1 + 3x 0

2x 2+ x1 + xo 3

3

Figure 6. Areas under curve

l> -.. l>

N II

0

)>

.. _ l>

N

3 CJ) c 3 l>

CJ) c 3 ~

. l>

St

(") ()

~0 N-II

0

nn P0 N-

nn (;) Si> ,\)-II

0

nn G> G1 . . N-

nn (;') ~ ~---f--...llCiCf----­N -

"'CD , ~

+

r~ ~

le of zero is obtained. The result of this procedure is

a shift of the X-Axis of the curve. Re-entering the

balancing program, the balance of moment for the new ordin­

ates can be found. Now if the moment is not balanced, the

program becomes how to select the location and amount of

change· required to preserve force balance, the general

curve shape and bring about moment equilibrium.

In order to maintain a smooth curve, the changes

made to balance the moment should vary from ordinate to

ordinate. With a large number of ordinates the changes

possible become large and the number of possible solutions

increases. Because the balanced curve should be the same

general shape as the curve started with, each possible

solution would have to be compared with it and measured

against some criteria to find the best solution. Rather

than design a program to do that it was decided to first

find a solution by simplifying· each curve, approximating

the areas, and doing the balancing calculations by hand.

This way possible curve shapes could be evaluated by

determining the balanced curve ordinates and comparing them

with reasonable· stress values at each element, thereby

eliminating curve-s with impossible or unlikely .distributions.

Then using the more exact summation program previously

described on each selected curve, the imbalance for moment

and force could be determined. Using this information,

changes could be made to the curve and the summation pro-

17

gram run again. This method eliminates some cases and

within four to six runs through the summation program it

is possible to find a solution that satisfies the accuracy

limit required by Wagner's (13) M-P-~ Program.

As one example, the curve suggested by Tran (11) as

shown in Figure 8 might be approximated as shown in

Figure 9. The· summation of forces and moments about the

weld can be done as follows:

d = Diameter ~=Angle from weld to center of gravity

x =Arc length·to C.G. L.A. = Lever arm to center of gravity

r = Radius to center line of element layer = 0.9635 in.

~ = 0.1 ct:.=0.1 radians L.A = (1 - COS~)r = 0.005 in.

~ = 0.3 + 0.35 = 0.65 cf::.= 0.65 radians L.A..= 0.184 in.

~ = 1.00 + 2.14/2 = 2.071 O:.= 2.071 radians L.A. = 1.337

l:F = (Yi) (0.30) (4.7) - (0.7) (4.57) + (1.61) (2.14)

= 0.70 - 3.20 + 3.45 = +0.95

.LM = +0.70 (0.005) 3.20 (0.184) + 3.45 (l.337)

= 0.335 - 0.589 + 3.343 = +4.057

The points Tran's (11) data gives generate a curve

with an imbalance of positive force and positive moment.

The imbalance is perhaps withi~ tolerable experimental

error for the data gathered .( 0. 0 2Py and 0. 0 2 7My where

Py equals axial load at yield, My equals moment at yield)

but the accuracy of balance for use of the residual stress

~ '

'< I 0 0

0 Ut 0 u. (:)

I

'Tldl c '""' (I)

CX> I . . !

::0CD N en 0 .-0.c: g_

en ... """ CD en en N

-4-(JI

~0 3 -I ... Q ~

"'Ut' b .._,

:t

St

19

distr~bution in the M-P-~ program are much more restrictive

(0.00005Py and· 0.001 My). Comparing the results of the

balancing program for Tran's (11) curve and the approximate

solution, close agreement is found confirming the computer

code. A complete check of the program is given in Appendix

I. Using the simple approximation, it can be shown that

changing the portion of the curv~ from x/d = 1.0 to x/d =7r

so that the summation of forces is zero still leaves a

positive unbalanced moment.

What then can be done to develop a curve based on

Tran's (11) data that will satisfy ~F = 0 & ~M = 0? The

first portion (to x/d =·1.0) of Tran (11) & Chen-Ross (4)

curves are of the same shape altho~gh of different amplitude

and therefore, a modification of the portion of the curve

from x/d = 1.0 to x/d =n-- seemed to be a good possibility.

This is the area that previous investigators have regarded

as being of lesser importance probably due to the rapidly

decreasing magnitude of residual stress; although it is

the most important area for accomplishing static

equilibriumL

On·viewing the simplified model, it is quite apparent

that the positive area past x/d = 1.0 is far too large to

be balanced by the negative area close to the summation

axis. Changing the negative area has little effect on the

summation as does changing the even closer positive area.

Changing the area from x/d = 1.0 to x/d ='Tr so that the

force? balance does not balance moments as was previously

shown. Further adjustment is required. Changing.the area

labeled (3) in Figure 9 require~ changing (1) to maintain

20

a force balance. Changing (3) enough to get M = 0 requires

lowering the stress level to 0.016 Fy from x/d = 1.0 to

x/d =~. Then calculating ~he area and moment for the

portion of the curve from x/d = 1.0 to x/d =7'r, the amount

of change to area (1) that is required for equlibrium can

be solved for. With a known value for area (1), the maximum

stress at the weld can be solved for. This calculation

indicates the· stress at the weld would be 2.758 times

the yield stress when the ultimate stress for the tube

as determined by Tran (11) is 1.14 times the yield stress.

A residual stress of 2.758 times yield at the weld is not

reasonable or possible. Nor does the rapid decrease of

residual stress at x/d = 1. 0 to a level of 0. 016 rr/O'y through

the rest of the curve seem logical.

A more likely solution is to change area ( 3) to be

larger positive in the section x/d = 1.0 to x/d = 2.0

and balance that with a smaller negative from x/d = 2.0

to x/d =tr (Fig. 10). This shape begins to look similar

to Chen and Ross's (4) curve (see Figure 13) but the

magnitudes are different. The only viable alternative is

to assume the r.esidual stress pattern is not symmetrical

about the weld axis. There is no available evidence to

supp.ort the assumption that the residual stress distribution

-~

5.0

4.0

3.0

2.0

1.0

Stress w -1.0

-2.0 I

•3.01 -4.0

~~.o

Q)

I

I

~ = 0.1 d

@ '2.50 % 1.5 2..0 2.5 3.0 11'

® 3.20

o< = 5.7295 L.A.= (l - cos o<) r

= (I - cos o<)(0.9035) = 0.005

® ~ = 0.3 + 0.35 = 0.65 L.A.= 0.184

@ ~ = 1.00+ 2 ·!4 = 2.071 L.A.= l. 337

Figure 9... Approximate curve

·21

L

5.0

4.0

3.0

2.0

1.0

Stress

-1.0 I

-2.0~ -3.01 -4.0

-5.0

I

22

@A

3.77

\ 1'( % 1.0 2.0

@a l.26 3.0

® 3.20

Figure I 0. A pproxi mate curve

is not symmetric. Trantookmeasurements from each half

of the tube but he did not select mirror image points.

Chen and Ross (4) also took values around the tube but did

not separate the reported values in thei-r representation

of the data.

23

If the values for the portion of the curve from x/d = 0

to x/d = 1.0 are held constant and only the portion of the

simplified curve from x/d = 1.0 to x/d =7r is changed it

i~ possible to show that for any selected point of change

from positive to negative, there is only one unique set

of areas that will satisfy summation of forces and summation

of moments equal to zero.·

Summarizing then it is possible to use a simplified

model of any residual stress curve to check for an approxi­

mate summation of forces and moments. Then using a more

detailed description of the curve, e.g. 20 to 100 values

aroun~ the tube scaled from a sketch· of the curve, and using

the previously described summation program, a more exact

check for balance can be made. ·rf the result is outside

the required accuracy limit, the ordinates of the curve

can be adjusted and the balance rechecked. By choosing unit

values of change over one part of the curve, an indication

of the effect of changing those ordinates on the total

balance can be obtained. Using that information for

further refinement makes the required iteration procedure

converge on the required accuracy limit very quickly.

Using this technique, it is possible to balance a possible

curve shape in less than ten trial runs with the summation

program. No attempt is made to find a mathematical ex­

pression describing these curves.

24

CHAPTER IV

RESIDUAL STRESS DISTRIBUTION CURVES

Tran (11) has shown a qurve that represents the data

points obtained from his measurements by the hole drilling

technique (Fig. 11). By dividing the curve into sixty-two

segments and scaling the ordinate of the residual stress

at each point and inputing these values into the summation

program, an indication of the degree ·of imbalance.was

obtained. The summation of moments gave an imbalance of

10.9% My and the summation of forces an imbalance of

0.7% Py, bothof which are certainly within experimental

tolerance. The problem is that the M-P-~ Program requires

a much higher degree of accuracy; 0.1% My and 0.005% Py.

Therefore, some adjustment of the curve is required.

Returning to a simplified·model of this curve

(Fig. 12) an approximate balance of forces and moments

can be attempted. The areas labeled 1, 2 and 3 can be

idealized as forces concentrated at the center of gravity

of the areas. Force 1 is positive and close to the crown,

Force 2 is negative and within x/d = 1.0 and Force 3 is

positive at approximately x/d = 2.0. Forces 2 and 3

are close to equal but of opposite sign.

9Z

,,

::u CD (/)

a.. c a

-~

.!. I

9 0 (J1 0 0

I •

t 0 9 0 ~ (>I

~ 0 0

~ ~q I

9 9 0 0 0 9 0 0 N

°' ~ (J1 0 0 0 0 0 0

~ . CJ)

3 'O

:::;; (i. 0.

g .... < CD

LZ

I .1 !

Force lx very small lever arm = very small +M

Force 2x small lever arm = small -M

Force 3x larger lever arm

Summation

= large +M

Large +M

28

When Force 2 and 3 are nearly equal in magnitude, it is not

possible to obtain static equilibrium. Tperefore the curve

Tran (11) suggests cannot.be balanced without some major

changes to the ordinates of the residual stress curve.

Chen and Ross (4) have shown a curve describing

the residual stress distribution (Fig. 13) which they ob­

tained from a combination of their measurements and

theoretical considerations done by Marshall (7). The first

portion of the curve from x/d = 0.0 to x/d = 1.0 is similar

in shape to Tran's but of a different magnitude. This

curve does not match all of their data points, but was

selected based on close agreement. Dividing the curve into

62 segments as was done on the previous curve, scaling the

ordinates and using the summation program gave results

that could be compared with equilibrium and the results

from Tran's curve (Fig. 13). The summation of moments gave

an imbalance of 0.4% My and a Force imbalance of 0.06% Py.

This is again outside the limit required making further

adjustment necessary. Selecting portions of the curve

to change, it was discovered that.very slight changes

in the latter (x/d ~ 1.0) portion of the curve affected the

total moment balance appreciably. The part of the curve

.,, -·

::0 CD.

~~ 0. c 0

g>

::0 0 (It

. (It

6Z

I

0 0. 0 I

with the most difference between the actual data points

and the suggested curve were selected for change. After

five iterations through the summation program, adequate

approximation of equilibrium was obtained. Comparing the

final curve with the starting curve (Fig. 14) it can be

seen that the change required was small. Even though

equilibrium is approximated with this curve, it does not

consistently match the data presented by Chen an·d Ross (4)

and Tran (11) particularly over the portion from x/d = 1.0

to x/d =Tr • Other possibilities needed to be explored.

30

One alternative is to use an average between Chen and

Ross's curve and Tran's.over the range 0 ~ x/d ~ 1.0

with a positive and negative maximum between 1. 0 ~ x/ d s11'".

As has been previously demonstrated,. this is the minimum

number of "fluctuations that can render a possible solution.

After balancing the curve to the required limit, it can

be seen that the· required curve areas imply larger positive

and negative stresses over the latter portion than can be

justified by the available data (Fig. 15).

Examining the curve shape in general, for changes

in the first maximum of compressive stress, the behavior

of the later portions of the curve can be predicted.

Assuming that the curve crosses the axis at approximate

multiples of x/d = 1.0 and beginning with a curve like

Chen and Ross•s· (Curve 1 Fig. 16) it is possible to show

what changes will need to be made to maintain equilibrium

1.00

0.5

0 i \

J

\ ~

\ %

i

y I

'

I -0

.50

-t.0

0

.&-·-----

....

....

Ba

lan

ced

C

urv

e

./

~-~

0.5

/

1.0

1.5

~

3.0

ff

2.

0 2.

5

Init

ial

Cu

rve

Fig

ure

14.

R

esi

du

al

Str

ess

, B

ala

nce

d

[Che

n 8

Ro

ss.(

3,4

) ]-

Cu

rve

I

%

w

.......

1.00

0.5

0

%y

-0.5

0

-1.0

0

--------

Che

n a

Ros

s (3

,4)

Tro

n (

II)

Fir

st

Se

ctio

n

Ave

rao

e-C

urv

e 2

---.;...

__.....

.___ __

_

o D

ata

P

oin

ts (

Che

n 8

Ro

ss) Yc

i

Fig

ure

15

. R

·esi

dual

S

tres

s [F

irst

Sec

tion

Ave

rage

-C

hen

8 R

oss

(3,4

)-T

ran

(l I)]

Cur

ve 2

w

f\

.)

/" /

I I ,_ \ ,, \

'° \

c ' ..... ' CD '

!)) ,, \

\ c

' 0

--tN

c a I

-I N

--- 0 I ::) I

I ::> I

I

0 c ..... < 0 0 I CD c: c I ,

.., .., (J)

< < I n>

" I =r a N I "O oa"t CD \

\ \

\ \ \

'

for a change in magnitude of the first negative maximum.

Curve Number 1 is in equilibrium. If Curve 2 is generated

by adjusting Point 1 to a different magnitude and it is

assumed the curve still crosses the axis at multiples of

34

x/d = 1.0, the general shape for x/d~l.O, assuming a smooth

curve, is determined by consideration of moment and force

equilibrium. Adjusting Point 1 to a larger negative mag­

nitude means Point 2 must also increase in magnitude to

balance the additional negative force from the change in

Point 1. The moments ·are not in balance now since equal

forces have been added but at different lever arms.

Therefore Point 3 must be adjusted to generate more negative

moment to balance the positive moment from adjusting

Point-2. Adding negative force from 3 means point 2 will

have to increase which again results in a moment imbalance,

although one of smaller magnitude. Continuing the iteration

between the portion of the curve at 2 and 3 will result in

a unique solution with larger final magnitudes for Points

2 and 3. Therefore the portion of the curve from x/d = 1.0

·to x/d = rr is very sensitive to Ghanges in magnitude of

Point 1.

An example of this sensitivity is the consideration

of a curve generated by smoothing out Tran's curve from

x/d = 0.0 to x/.d = 1.0 (Fig. 17). After balancing the

curve as previously described, the ordinate of the positive

maximum was three times the yield stress which is consider-

c: .... (1)

:0 (1) (/)

-· 0. c Q

(J)

-.... CD (/) (/)

.... (/)

--0 0 ..... .... o· :J .. (/)

3 0 0 -=r(1) a.

0 c .... < (1)

..... .... 0 3

-t .... Q :J

-

• 0 0 I

ably greater than.the stress at the weld. This means the

single data point Tran found at approximately x/d = 0.8

is inaccurate and should not be considerated·in generating

a curve, since the condition of equilibrium demands un­

realistic stresses using the point.

In general, a smooth curve shape with a maximum

tensile stress at x/d = 0, a maximum compressive stress

at x/d ~ 1.0, and zero stress at approximately x/d = 1.0,

x/d = 2.0 and x/d = 3.0 will exhibit an easily determined

behavior for changes in the magnitude of the first com­

pressive maximum. It has been shown herein that using a

base curve of this shap~, small changes in the compressive

stress maximum at x/d < 1.0 require larger changes in the

remaining tensile and compressive stress regions. It is,

therefore, possible to modify the effect of a reported

data point on the smooth curve by considering the change

required in the rest of the curve. A data point which

causes, by its inclusion i~ a smoothed curve shape, an

unreasonable stress distribution for the other parts of

the curve can be identified using this me~hod.

Several other curves were generated in an attempt

to use the combined data from Chen and Ross (3,4} and Tran

36

(11). The most consistant results were obtained from a

curve using the first part of Chen and Ross's curve from

x/d = 0.0 to x/d = 1.0 and forcing the latter portion of

the curve to one positive maximum and one negative maximum

with the curve approaching zero at x/d =7r (Fig. 18).

1.00

0.5

0

%. y

2.5

1.5

2.0

-0.5

0

-1.0

0

Fig

ure

18

. R

esi

du

al

Str

ess

[C

hen

8 R

oss

(3,4

} fo

rce

d

to

two

ma

xim

um

s]-

Cur

ve 3

%

......

......

......

....

w

....J

38

Three curves of the several curves developed were

selected to generate M-P-~ curves with: a balanced version

of Chen and Ross•s ~urve (Curve l, Fig. 14), section from

I I

x/d = 0.0 to x/d = 1.0 average between Tran and Chen and

Ross 1 s (Curve 2, Fig. 15), and Chen and Ross 1 s curve

I. forced to two maximums from x/d = 1.0 to x/d =7r (Curve 3,

Fig. 18).

CHAPTER V

M-P-~ CURVE GENERATION

Once the description of the residual stress distri­

bution was complete, Wagner's (13) program was used to

generate M-P-~ curves. It is important to understand

the method that·Wagner's program uses to calculate these

curves. There are four major stages: assigning the

appropriate value of residual stress and strain to each

element, application of a percentage of the stub column

yield load to each element, assigning a curvature to the

cross section, and calculating the moment corresponding

to a state of equilibrium. For one value of axial load

at one curvature, the moment is then calculated. By

repeating these steps for several axial loads at different

curvatures, it is possible to generate a family of M-P-~

curves. Several intermmediate steps are involved in each

of the four major stages (Fig. 19).

In the first stage of the program all of the required

data is read; number of layers, number of elements, diameter

of tube, wall thickness, modulus of elasticity, yield

stress, the number of axial load values, and the number

of curvature values. Then the values of axial load and

curvature are read. The program has the·option of using

START

Assign appropriate residual stress and strain to each element.

Apply axial load (P) and calculate the strain (£ = P/AE)

a

STAGE 1

(e ) value r

STAGE 2

Calculate the total strain (ft=£+·£) for each element r a

Using£tand the stress-strain relationship find the total force on the cross section (F) .

~ No,. r Adjust fa

Yes

Assign a value of curvature

Determine the strain on each element due to curvature (£~)

STAGE 3

Calculate the total strain for each element (£t=\+~+e-)

STAGE 4

Usingft andthe stress-strain relationship; determine the total force (F) and the bending moment (M) on the cross section

No

I .. 1STOP

Adjust the location of the neutral axis.

Figure 19 Flow diagram for calculation of M-P-~ data

40

41

a tabulated stress-strain curve or a bilinear stress-strain

curve and this information is entered next. The residual

stress and strain is read in and the applicable value

assigned to each element. For each layer the average

radius, arc length of the elements and the area of the

elements is calculated. From the results of Tran's work

it is apparent that consideration of layers in the tube

wall is riot required. for the description of the residual

stress-strain distribution since these do not vary through

the thickness of the wall. The strain, curvature, bending

moment and axial load at first yield are calculated

next. Then the distance from each element to the centroid

of the cross section, the total cross sectional area,

plastic modulus, flexural stiffness, plastic hinge moment

and shape factor are determined. The description.of the

problem is now complete.

The next stage is to apply the first axial load

(expressedasa percentage of Py) to the cross section and

calculate the axial strain. An iteration loop is performed

next to determine the correct value of axial strain. This

is required since it is possible for the summation of the

residual strain and the axial strain on any particular

element to exceed the yield value. In these cases the

summation of t~e elemental stress available to resist the

axial load is less than predicted by elastic theory. The

residual stress distribution is an initial condition and

cannot be changed, therefore the additional force must

be supplied by the elements that have not yet reached

42

yield. The stress distribution and magnitude is determined

by increasing the strain on all elements by the same amount

and then calculating the resulting stress using the material

stress-strain information (tabulated input of any stress­

strain curve or a bilinear curve as previously described) .

The summation of available elemental stress at each increment

of strain is compared to the total applied force .and when

they are approximately equal the interation is stopped and

the values for each element are stored for further use.

The next stage involves assigning a value of curvature

to the cross section. The neutral axis is assumed initially

to be at the centroid of the cross section and the corres­

ponding strain at each element is calculated. The summation

of strain due to residual stress, axial load and the imposed

curvature is limited to the strain at first yield as pre­

viously described for residual stress plus axial load for

each element. The resulting strains are used with the

material stress-strain information to determine the resulting

stress for each element. These stresses are summed and the

resulting total thrust is compared with the applied axial

load. If the axial load and thrust are not equal the loca­

tion of the neut~al axis is shifted and the strains recal­

culated. The iteration is continued unti~ approximate

equality is reached.

The last stage is to calculate the summation of

moments for the final stress-strain state. The result is a

value of moment at a particular curvature for one value

of axial load. This gives one point for determining one

M-P-0 curve. The process for finding the moment is

repeated for each curvature at one value of axial load.

That gives all the points of the M-P-0 curve for that

particular axial load. The program then selects the next

value of axial load and begins at stage two repeating the

process of solving for moment at each value of curvature

until all values of axial ·load input have been used. The

values of moment, axial load and curvature are then output

in tabular form. This information can then be used in the

failure load program as described by Wagner (13) .

43

In order to completely study the effect of the residual

stress distribution on failure loads it is necessary to

determine the effect orientation of the axis of bending

with respect to the axis of the weid has on the M-P-0

curve. The concern is to find the weakest axis of bending

to be sure that the controlling case has been found. The

axis of bending selected by Wagner (12,13) passed through

the weld (Fig. 20) for which results for the M-P-0 curves

were compatible with theory (l,2,5,8). Considering the

axis orientation with the weld at the top of the pipe and

the bending axis through the horizontal diameter as the

zero or reference po:tnt;, any orientation can be described

as the number of degrees of rotation of the bending axis

from the reference position. Thus the orientation Wagner

---~--~.

.. ~~--·--·-----

· Wag

ner's

A

xis

of

Ben

ding

. --

-c I

T_.

....

We

ld'

28.0

ksi

-12

.3 k

si

1 •

1 R

efe

ren

ce

Ax

is

. .

Fig

ure

20

. O

rie

nta

tio

n

of.

Axe

s an

d ·R

esid

uol

Str

ess

Dis

trib

uti

on

[W

og

ne

r.(1

2,1

3))

~ ~

45

0 (12,13) selected would be at 90 from the reference. When

any other orientation was selected, the results were not

consistant with theoretical considerations (1, 2, 5, s·) •

The moments were-not approaching the value for full plastic

moment at large values of curvature. (tig~. 21~22,23)~

Clearly there was a significant problem with either the

proposed residual stress distribution or with the program.

The first step was to verify the results obtained by

Wagner for the distribution used with the orientation of

90° from the reference. Comparing the results for this

orientation of bending axis showed exact agreement with the

work done by Wagner (12,13).

Inputing zero residual stress and checking several

axis orientations also showed exact agreement with the

theoretically predicted behavior of the M-P-~ curves.

In order to pinpoint the problem Wagner's distribution

(see Figure 20) was used. The axis of bending was assumed

at the reference position and a summation of moments

accomplished for the cross section at the end of each

iteration to determine the correct axial strain. (This is

described in the second stage of the program, see Fig. 19) .

Residual stresses were output to verify that there was no

change occurring. (They are input constants and should

not change.) The total stresses and strains were

also output to verify that they did not exceed yield during

the iteration procedures. All of the output verified the

46

1.40

Zero Residual Stress~----= ~:y ~

1.20

0.40

1.00

0.80

%y 0.60

0.40

0.20

Zero R.S.""""\_

---~----- 0.90 ----------------v

0 1.0 2.0 3.0 4:0

~y

Figure 21.. M-P-' Curve I (Chen 8 Ross)

1.40-1 ~Py 0.00

Zero R.S. ~,....~ /

/ I

~

s.20--{ I

1.00

0.80

%,y·· 0.60

0.40

0.20

0

.,,,.

tl

I I I

/,

Zero .Residual Stre~} ____ _

__ ........... --------,--- - -

1.0 2.0 . 3.0 4.0

~y

0.90

Figure 2.2. ~M-.P-~ Curve 2 (First Section Average)

47

48

%y l.40 - ------ - 0.00

R.S.

1.20

·- 0.40

1,00

0.80

M/My

0.60

0.40

Zero Residual Stress)

----------------· --- 0.90 ,,,,..,,,., 0.20

0 1.0 2.0 3.0 4.0

%y

Figure 23. M-P-~ Curve 3 {Chen 8 Ross f.orced to two maximums)

l -- ---· ~

49

procedure with the exception of the summation of moments

after finding the "correct" axial strain and corresponding

stress. The summation of moments at this point should be

equal to zero since the· residual stress distribution is in

equilibrium to begin with and only axial load has been

added. Because the program does not alter the residual

strain or the· axial strain after this point in the program

any imbalance produced here will be unchanged at the end

of the iterations. The strains due to curvature are added

after this and the total limited to the described material

properties. The strain due to curvature at this value of

axial load is calculated by subtracting strain. This strain

due to curvature is used to find the stress on each element.

Summing moments for these·values the moment for the entire

cross section is obtained. The thrust is calculated and

compared to the.applied force and they are not equal. The

neutral axis is shifted and the moment and thrust are

recalculated. This procedure in no way affects the values

of strain that were solved for due to axial load. Therefore,

if the cross section is not in static equilibrium after the

solution for strains due to axial load, it will still not

be in equilibrium after the "correct 11 location of the neutral

axis has been found and the corresponding moment calculated.

The result of the summation of moments after completing

the iteration to find the ••correct" axial strain was not

equal to zero in any case tested except for those with no

r .. ---- -· residual stress and those with 90° orientation of axis of

bending with respect to the reference. The disparity

between the theoretical plastic moment and the calculated

50

moment for the P/Py ratio was compared to the moment

imbalance created during the solution for axial strain and

found to be equal. In other words the sum of the calculated

moment at some large curvature plus the moment created in

the solution for axial strain equals the theoretically

predicted plastic moment. · It would seem the simplest

solution would be to compute the moment due to curvature

by a summation using the total stress instead·of only that

part of the stress due to curvature. That way the moment

imbalance is incorporated in the final moment. The problem

with this solution is that while the M-P-0 curve has the

correct starting point and is correct for large values of

curvature, the middle portion has its shape changed con­

siderably by the moment imbalance. The solution, then is

to eliminate the moment imbalance created before entering

the iteration loop to find the moment due to curvature.

In this manner a residual stress distribution that is in

static equilibrium will have an axial stress distribution

added to it and the final result will also be in static

equilibrium.. This model is then showing the same behavior

as the actual p,hysical cross section.

To eliminate the moment imbalance, the first step

was to determine how the imbalance is created. When the

"';

r:: r

I· ·~

. \

l

\ '

• \.

i

\ .

\

"•

,-,

51

axial strain is being solved for the first step is to

apply a uniform strain to the cross section computed from

the average axial stress, P/A, and the material properties

of the cross section. The axial strain is added to the

residual strain for ·each element and the stress from this

sum is limited by the described material properties (either

tabulated or bilinear stress strain curve) . If the sum of

axial strain and residual strain is larger than the strain

at first yield the available stress will be less than the

average axial stress. That means the summation of thrust

on the cross section will be less than the applied axial

load. Therefore the strain over all the elements is in-

cremented and the summation recalcul'ated. This procedure

continues until approximate agreement is reached between

the calculated thrust and the axial load. If some of the

elements above the center of gravity·of the cross section

are very close to yield from residual strain alone, the

increase in strain for axial load will put them over yield.

The stress available· for axial load from them is less than

the average, therefore the strain is adjusted. This increase

in strain doe~ not cause much if any increase in stress

available from the elements ·that were at·or over yield

in the first iteration. The other elements now have an

increase in st~ess over the average P/A. As the iteration

process continues the distribution of axial stress may

become even more irregular as other elements yield. When

,-;

f..

•4

!

{' I \

.'

"

the iteration is complete the stress distribution is no

longer uniform and the center of gravity of the resulting

thrust is no longer located at the center of gravity of

the cross section. Therefore a moment has been created

during the iteration due to the yielding of some elements.

(Fig. 24)

To explain why there was no imbalance using this

program for the axis of .bending taken through the weld

(90° to the reference) the -residual stress distribution

used must be examined. The distribution used was assumed

to be symmetrical about the axis of the weld. Therefore

the change in stress due to axial ·load on each element is

also symmetrical about the axis of the weld. Since the

change in stress is symmetrical about this·_ axis, the

resulting thrust is positioned somewhere·along the axis.

Therefore summing moments about this axis yields zero

52

even though summation of moments about any other axis would

not be. The cross section· is not in equilibrium but the

moment is only being taken about one axis in this program so

the imbalance goes undetected for this one special case

(axis of bending through the weld, 90° with respect to

reference chosen.) The M-P-0 curves generated for this

case are then unaffected by the imbalance but for all

other cases th~y would be. The case with the largest

imbalance should be with the· axis of bending at the refer-

ence position (perpendicular to the weld axis) and this was

confirmed by a .test run of the program as written.

; ., I j I !

. ' " \. 1

0 0

I +

-+

+ ~

--Residual Stress Axial Stress

0

Summation

0 0

0

0 ..J

~· ->· I I

53

Not A\lailoble

Summation (Not in Force Equilibrium)

0

Summation (In Force Equilibrium)

0

Not in Moment Equilibrium

Figure· 24. Moment ·Imbalance

\

\ \ ' ~ t ~ ~

54

The problem was identified and several possibilities

for a solution were explored. The. method selected was

to add another iteration after determining the axial

strain·to eliminate the imbalance created. The value of

curvature that would generate ·a moment equal to the

unbalanced moment was calculated assuming the section to

be elastic. The sign of this curvature was reversed there-

\ by changing the sign of the moment. Strains were calculated

\ \

\

based on this curvature and added to the strain determined

fo~ axial load plus residual stress. This in effect would

cancel the unbalanced moment leaving the cross section in

equilibrium. The curvature used is actually the reverse

of the curvature generated by the uneven axial stress

distribution, which means the section ends up with curvature

equal to zero. The procedure actually assumes a curvature

(elastic) and iterates to find the correct curvature

(produced by uneven axial stress). The iteration also must

still limit the stress to that determined from tabulated

or bilinear stress-strain curves. The iteration also

involves checking the summation of thrust on the elements

against the applied axial load for approximate agreement.

The revisions required to the program were extensive and

are shown in Appendix II. Each axis rotation requires a

separate run of the program.

Because various orientations of the bending axis

were desired, the program was also modified to automatically

·1 '

..,

i

' .

,.

"..

,;

\ '\ ~

\

55

rotate the bending axis to any whole element increment

(for example: 0 0 0 0 360 /80 elements = 4.5 /element 45 /4.5 I

element = 10 element rotation) .. Since a large number of

M-P-~ curves were going to be generated, a graphics plot

program was written to automatically plot each set of curves

on a Cal-Com plotter. The p~ots for Wagner's residual

stress distribution and the suggested curve 3 are shown in

Figure 25. The results of these calculations agreed with

that predicted by consideration of the theoretical maximum

possible moment for each axial load ratio.

Another problem was discovered when these M-P-~

curves were used directly in the beam-column failure load

program. None of the iterative solutions supply exact

closed form solutions; they converge on a solution ~ithin

an acceptable degree of error. Since some error is inherent

in each iteration, the errors may accumulate. The degree of

error is still small but it means for some cases of axial

load with the curvature equal to zero, small positive or

negative moments appear when they should be equal to zero.

When the curves with these small negative moments were used·

in the failure load program it caused instability in the

beam-column solution. It was discovered that the curve

interpolation portions of the program could not handle

negative number.s. The solution was to assign all the moments

at zero curvature to exactly zero as they should be. The

M-P-~ curve generation program could be changed to auto-

i l

I

1.4

1.2

1.0

"'I 0.8

~

~ ....... ~

0.6

0.4

0.2

0 0

; i

i

/ /

/

/ /

/.

/ /

,,,,...-·-·-·-·

·" ,,.·"' ,,..

56

P/Pv=O

P/Pv=0.4

-------·- .---·-· - ·-· -·-·-·-·-­_,.

No residual stress

Assumed residual stress (curve 3)

-·-·-·-· Assumed residua I stress [Wagner 0 2,13)- Sherman ( 9)]

P/Pv =0.9

1.0 2.0 3.0 4.0 5.0 '/J l</Jy

Figure 25. M-P-¢ Curves. from improved generation program·

j

j '

I I I

I 1

I

I i-

!

p

~ , I \ . \l

\l

1 '

I !

57

matically assign moment equal to zero for curvature equal

to zero which would eliminate the need to change the output

later. In this case, the program was not changed since the

magnitude of the moment gives an indication of the composite

accuracy of the manipulations of the program.

It should be noted that the M-P-~ curves calculated

for Wagner's (12,13) residual stress distribution exhibit

a flattening effect when compared to the curves calculated

for the suggested residual ·stress distribution (Curve 3,

Figure 18). This means when using Wagner's distribution the

section is weaker than would be predicted using· the suggested

distribution (Curve 3, Figure 18). It should also be noted

that different axis orientations control the flattest (most

critical) curve at different axial load ratios.

It should be noted that the M-P-~ curves generated

using the suggested residual stress· distribution give more

consistent results with respect to the controlling axis

orientation for different slenderness ratios than by··using

the distribution·suggested by Wagner (12,13).

Once the M-P-~ curve generation and interpolation pro-

blems had been solved, the failure load program was used

for the same sections and loading conditions that Wagner

(13) used. M-P-~ curves were generated for five axes

orientations; 00 , I

0 0 0 0 . 45 , 90 , 135 , and 180 with respect to

the reference axis (perpendicular to weld axis), for

Wagner's residual stress distribution and for the suggested

., I

t

58

residual stress distribution. Each of these M-P-0 curves

were used in the beam-column failure load program. This

was done since the M-P-0 curves for Wagner's residual

stress distribution indicated different axes orientations

controlled for different P/Py ratios. The controlling·

results ~ere then graphed as a failure load curve using the

same format as shown by Sherman (9) .

Figure 26 shows the resulting curve for Wagner 1 s

distribution at an orientation of 90° with respect to the

reference axis.

The failure load results for the suggested residual

stress distribution are shown in Figure 27 since the

curves are close enough to those of Figure 26 as to make

representation on the same figure difficult at this scale.

The orientation of the axis for the suggested residual

stress distribution (Curve 3 Figure 18) is 180° with

respect to the reference axis. These curves were compared

to available test data and reasonable agreement was

indicated.

59

1.0 I I I I I 1.0 ,, 1.0 0 0.2 0.4 0.6 0.8

0.8

0.6

a?' ' Q.

0.4

0.2

0

a ----.....:--_.,, - - -----...---- -------~

I 0

'" ..... ..... .... , ' .......... ' ..... ', -- ..... - b

' -LID= 120

' --'..... ---.......... c ----....... --- ------ -~ ............ __ ----._ ..... ,

. ' -------- .... -.....

I I I r 0.2 0.4 0.6 0.8

~MMp

.a Double curvature

b Single end moment c· Single curvature

0.8

0.6

0.4

0.2

~, 0

1.0

Figure 26. Beam column results -Wagner's distribution

i.

i

60

0 0.2 0.4 . 0.6 0.8 1.0 1 .. 0 I ' ' I I I ,I .0

0.8

0.6

~ 0..

a: 0.4

02

0.8

0.6

a ----------- -------- - -----~ - -:- .... - 0.4 , ..... , . ~- ........ _ b L/O= 120 ' -.... . .... , --...... --...._ ..

"'..... ----..... c ..... ___ ~~ ---~ ----------. - ......... __ .- ..... , ---- ..... -..... ' ----- ' --..._ ' -..... ', --..... '

0.2

' ...... \ ...... ' .......... '·

~ 0 I r 0 I I I I l.G 0 0.4 0 .. 6 o.s 0.2

M/Mp

a Double curvature

b Single end moment

c Single curvature

Figure 27. Beam column results - curve 3 {Fig.18).

CHAPTER VI

CONCLUSIONS

Comparing the balanced residual s_tress curve (Curve 3)

with the curves determined by the sectioning method, hole

drilling method and Wagner•s·assumeddistribution, the

balanced curve suggested for use (Curve 3 Figure 18) appears

to be a smooth curve more closely modeling the test data

(Fig. 28). Figure 29 shows a comparison of M-P-~ curves

for no residual stress, Wagnerlsdistribution, that dedu~ed

from member behavior and the balanced residual stress

curve (Curve No. 3). ·sherman (9) presents a curve showing

the effect of residual stress on strength under axial

load which is shown in Figure 30 with the· points determined

usipg the residu~~ stresses from Curve ~ (Figure 18) added

to it. Using the· same data as Sherman used in the beam

column failure load program but with the balanced residual

stress distribution of Curve 3 (Figure 18) instead of

Wagner's curves, new points are determined which are

plotted on Sherman's Figur~ 6 to show the effect of the

balanced distribution .(Fig: 31}.

While there are large differences:·in the shape of the

M-P-~ curves as previously described (Figure 25,29), the

effect on the interaction curve for failure loads is

0.80

~/Cf

v·

., , ..

. '" . ...

......

......

......

......

......

.... -

--.

........

........

...

..........

..........

.... _ .... _

__ ....

........

. .....

..........

. -.....

... -.

..

--

..,_

__

1

Cur

ve

of

sect

ioni

ng m

etho

d

n cu

.rve

of

hole

dri

llin

g m

eth

od

----:-

·--:-

m C

urve

of

assu

med

di

stri

butio

n {S

herm

an)

_..

..,_

_ N

C

urve

of

bala

nced

d

istr

ibu

tion

{C

urve

3)

0.40

-P~~

\.

:'> \ \ ..

~-·--<::_en

\ I

' ~b

!7,/

~

I -r

-0

.00

·t

t' f.

'\ ~/

···

Cu

rve

I2:

-0.4

0

-0.8

0

Fig

ure

28

. Lo

ngitu

dina

l re

sidu

al s

tress

dis

trib

utio

n cu

rves

--

~~·~ -~--~ _

_ ...

_ _

...

....J

°' I\.)

~

~ .......... ~

1.4 I

L2+

0

o.a

0.6

0.4

0.2

0

1.0 2.0 3.0 I I I

P/Py = 0.0

..... ....- .i:;iiiii":;-" ·"",,,

~ .,,,.,, ~ .. ~ b <," /,/ c

_..-· ,. . .,,,,. ...... ...:-~---­,.,,,,.,,,,,, .....

....... -::,,.- ----­_,,,.,.,,,.---·

P/Pu = 0.8

;.::. ·--- - - - = .. / . ...-· ---- ,

/ ~"' . ------------· -/" ,,.- ..--· . / ;',,,. ,/'

;-)">/ •/ / ,/ ri'// .,,/

~.

1.0 2.0 3.0

</J/~y

a. No residual stress b. Assumed residual stress of curve 3 c. Assumed residual stress, Sherman d. Deduced from member behavior

Figure 29.. Moment-curvature relationships

63

4.o· I 1.4

+ 1.2

1.0

0.8

0.6

0:2

4.0

......

......

......

......

.. ""':

....

........

........

........

. -......

. _~~

........

........

......

...-.

......

......

. .....

.. ...

.........

...~~ --

-...

......

.._..

~"--

~" -

-----

--.

0 O.

!:>

I .0

1.

5 2

.0

1.0

le:::::::::

t.'

\ '

p -

' P

y-~2

J.O

. 0.8

~ 0~

6

0:

0.4

0.2.

.

Lin

ear

:. =

, 1-0

.38

5 ~

y

LID

= 1

5 L

IO •

20

F,=

50

kst

·. Par

abol

ic (

SS

RC

)

~ =

l-0

.25

~y

LIO

=3

0

L/O

=4

0

0.8

0.6

0.4

02

LI 0

=5

0

·o 0 ~1

-'--

05

--

---~

~,J"

"~'

-_:_

~~~~

~~~J

F

" 1.

0 11

)..=

2!!

/fi

L

1.5

__

_ IL

JY

o

o a

No r

esi

du

al

stre

ss

c b

Ass

um

ed

're

sidu

al s

tre

ss (

She

rman

)

6.

c D

educ

ed

fro

m m

em

be

r b

eh

avi

or

x d

Ass

umed

re

sid

ua

l st

ress

(cu

rve

3)

Fig

ure.

30

. E

ffe

ct

of

resi

du

al

stre

ss

on

stre

ng

th

unde

r a

xia

l lo

ad

°' .e.

1.0

0.8

0.6

>o,, ·Q.

' 0. .

0.4

0

0.2 I

I

0.2

Fy:: 50 ksi

p ,. ~ l -~

65

0.4 0.6 0.8 1.0 l.O

0.8

0.6

0.4 .

w 0.2

~ I L

o~

• I I ~O 0 0.2 0.4 0.6 0.8 1.0

w L2 -

f6 Mp

---- a No residual stress

------ b Assumed residual stress distribution (Sherman) ______ : c Deduced from member behavior

-·-·--·-- d Assumed residual stress distribution (curve 3)

Figure 31. Effect of residual sfress

66

slight. The maximum difference between predicted failure

loads is in the neighborhood of five percent. The failure

load curve as described from M-P-~ curves using the suggested

residual stress distribution shows an increase in predicted

strength over the fa~lure load curve using the M-P-~ curves

as described by Wagner's residual stress distribution

(Figure 26,27,30,31). This agrees with the prediction

based on the shape of the M-P-~ curves (Figure 25,29).

The interaction curve obtai·ned using the M-P-~ curve

generated based on the suggested residual stress distri­

bution (Curve 3 Figure 18) does not exactly conform to the

test data as plotted but that is to be expected for several

reasons. One is the possible error in the physical test

data, another is the small errors that are inherent in

the process of curve interpolation and iterative solution

for axial load, moment and failure loadi another is the use

of a perfect bilinear stress-strain curve instead of using

tabulated data from physically testing a coupon from the

member tested. The degree of agreement is acceptable in

light of these considerations and .the use of the suggested

residual stress distribution is recommended for future

research into the behavior of welded steel tubes in com­

bined bending and axial load.

It is fur.ther recommended ·that the M-P-~ curve

generation program be changed to facilitate more rapid use

of the program. A data file should be created that would

contain the stresses and strains on each element that

result for each axial load ratio used. These stresses

67

and strains can be obtained using the first portion of the

M-P-~ curve generation program as re-written in this study.

The stresses and strains that result from the application

of residual stress plus axial stress after the moment

imbalance has been resolved are the ones that· should be

saved. By storing these stresses and strains, the program

would not be required to recalculate them· for each axis

rotation thereby reducing the computer time used.

Summarizing, the residual stresses present in a

welded steel tube can be modeled by a smooth curve that

describes stresses such that the cross section is in

equilibrium.

Curve 3 (Fig. 18) models the existing test data wel~,

is in equilibrium and exhibits consistent behavior in the

M-P-~ and beam column failure load program. This curve is

suggested for use in further research on the behavior

of steel tube columns. The effect of these residual

stresses on the predicted failure load of tubular beam

columns is significant and should not be ignored. The

difference in failure load predicted by this family of

curves was five percent at the maximum, indicating that the

exact shape of .all parts of the residual stress curve is not

required to obtain·good results. Because of the large

strength reductions indicated by use of the residual stress

distribution described by Curve 3 (Figure 18) in the beam­

column failure load program, it is evident that residual

stresses can not be ignored when calculating failure loads

in steel tube beam columns.

68

REFERENCES

1. "Plastic Design in Steel, A Guide and Commentary, 11

American Society of Civil Engineers 2nd Edition, 1971.

2. Beedle, L. S., i'Plast~c ·Design of Steel Frames 11,

John Wiley and Sons, Inc., 1958.

3. Chen, W. F. and Ross; D. A., "Residual Stress Measure­ments in Fabricated Tubular Steel Columns," Fritz Laboratories Report, No·. 393.4, July 1975.

4. Chen, W. F. and Ross, D. A., "Tests of Fabricated Tubular Columns," Journal of the· structural Division, Proceedings of ASCE, Vol. 103, No. ST3, March 1977.

5. Disque, R. O. "Applied Plastic Design in Steel" Van Norstrand Reinhold Company, 1971.

6. Dwyer and Galambos, "Plastic Behavior of Tubular Beam-Columns 11

, Journal of the Structural Divis ion, ASCE, Volume 91, ST4, August 1975.

7. Marshall, P. W., "Stability Problems in Offshore Structures", presentation at the Annual Technical Meeting of the Column Research Council, St. Louis, March 25, 1970.

8. Salmon, C. G. and Johnson, J. E., "Steel Structures, Design and Behavior" Intext Educational Publishers 1972.

9. Sherman, D. R., Erzurmlu,. H., Mueller, w., "Behavioral Study of Tubular Beam Columns." Journal of the Structural Division, ASCE, Volume 195, No. ST6, June 1979.

10. Szuladzinski, G., "Moment-Curvature for Elastoplastic Beams," Journal of the Structural Division, Proceedings of ASCE, Vol. 101, No. ST7, July 1975 ..

11. Tran, C. M., "An Investigation of the Hole-drilling Technique for Measuring Residual Stresses in Welded Fabricated Steel Tubes" Masters Thesis, Portland State University, Dec. 1977.

70

12. Wagner, A. L., Mueller, W. H. and Erzurmlu, H., Portland State University, "Design Interaction Curves for Tubular Steel Beam-Columns 11

, .Eighth Annual Offshore Technology Conference, Houston, Texas, May 3-6, 1976.

13. Wagner, A. L., "A Numerical Solution for the Ultimate Strength of Tubular Beam-Columns, 11 Masters Thesis, Portland State University, Nov. 1976.

APPENDIX I

CURVE BALANCING COMPUTER PROGRAM

CURVE BALANCING PROGRAM

DATA INPUT

NOTE: Use of the punch output option gives the user a

deck of cards with the stress-strain information on them in

proper form for direct input to the moment-thrust curvature

program (average over element) . Numbers at left indicate

card columns.

A. Control ·Data; Curve Identity Number.

FORMAT ( 2I3)

1-3 Curve identity number

4-6 Are stress-strain values to be punched on. cards? (- or 0 no, + yes)

B. Control Data.

FORMAT (I3, 5Fl2.4)

1-3 Number of elements in 1/4 tube (Maximum number = 100)

4-15 Outside diameter of tube (in.) "·\

16-27 Tube wall thickness (in.)

28-39 Distance from crown to center of moment (in.)

40-51 Modulus of elasticity (ksi)

52-63 Yield stress (ksi)

C. Res~dual Str~ss Values from Distribution Curve

FORMAT (F9. 6)

1-9 Residual stress (ksi) (Distribution curve ordinates)

FLOW DIAGRAM

RESIDUAL STRESS DISTRIBUTION CURVE BALANCING

START

READ: NC, IP

READ: NELE, QD, T, DC E, FY

CALCULATE: X(N) = Arc Distance, Radius at 4: Wall, Py, My, Accuracy Limits.

READ: RESIDUAL STRESS CURVE

ORDINATES

·e-

y (N)

DO 240 FOR EACH

ELEMENT

+8

73

CALCULATE XO, CGl, CG2 A'l, A2

GO TO 220

CALCULATE CGl, CG2, Al, A2

GO TO 220

0

CALCULATE Al, CGl A2=0,CG2=0

GO TO 220

+

+

CALCULATE CGl, CG2 Al, A2

GO TO 220

74

Al=O, A2=0 CGl=O, CG2=0 GO TO 220

CALCULATE CGl, CG2=0 Al, A2=0

Al=Al*T*FY A2=A2*T*FY

LEVER ARM 1 = 0

+

+

CALCULATE CGl, CG2=0 Al,A2=0 GO TO 220

CALCULATE LEVER ARM 1

75.

LEVER ARM 2 = 0

CALCULATE MOMENT

+

SUM OF FORCES SUM OF MOMENTS

PRINT

CALCULATE LEVER ARM ·2

76

FORCE l, LEVER ARM 1 FORCE 2, LEVER ARM 2 SUM FORCES, SUM MOMENTS

4-

FOR EACH ELEMENT CALCULATE AVERAGE STRESS & STRAIN, FORCE, SUM FORCES ~OMENT, SUM MOMENTS

PRINT RESULTS AND PUNCH .

STRESS & STRAIN

1 2 3 4 0 6 7 e. 9 1l

1 1

1

12

1

3

14

1

5

16

1

7

18

.

19

2L

t 2

1

22

2

3

24

2

5

26

2

7

c (' r ,.. \

, c r. r. c

......

...

......

... ...

_ ...

......

......

......

......

....

-... --~

~·-.

~· -·

• ~·-~

---·-

-JI

-~ .

. .

TH

E PUPFL:S~

OF

T

HIS

P

fWG

RA

M

TS

T

O

:::iu

iv:

FO

HC

fS

AN

O

i'-lM

'lEN

TS

F

OR

I

ON

GI

run

i'N

AL

H

FS

ICU

AL

S

TR

ES

S[S

S

HJ

/), C

IPC

UL

AP

Cl~OSS

SC

CT

IOf\

l.

I =

j·.1U

11 r: ~~ E

R

Of

ST

H [S

S

L 0

C /\

T I

0 f\l

S •

E:

:: Y

OU

f\I G

' S

M

G U

UL

U S

• F

Y=

YE:

I L

. C:

ST

f{ E

f\l G

T ii

G

D

::

'1 UT~ I

iJ E

D

I t\

M [

T [

R

0 F

TU

13 [

T =

TH

I CK

f"W

SS

0

F :tJ

i\ l

L

n F

SE

CT

I 0

l\J •

D

C =

D 1 ~)

T M

J C t::

FR 0

r~

C H

0 \rJ

f ~

T 0

C

EN

T E

H

0 F

!VI

0 !J

I [ j

-.JT

~ NC

=

CURV

E N

UM

RER

J p

-=

0 I\

u c

l\ rrn

s p

u ~Jc

H [

D

:1

ST

RE

SS

+

ST

RA

IN

PU

NC

Hfn

nIMENSIO~

X(l

OO

>,Y

C1

00

) Tf

<E 1

\U:::

2

I l~1

R !

T F

= :-,

REA

D CIRE~0,15)

NC

,IP

1

5

F0

hl0 l/

.\T

(2

13

) P

F t\

D

< I

H [

AU

, 1

0 )

f'J

EL

E ,

0 D

, T

, 0

C ,

E ,

FY

JO

FORM

AT

<I3

,5F

l2.4

>

n= <

un-1

>;;.:

o.50

l=

f-.I

EL

E*

2-t

l D

X=

3.t

41

L9

2G

54

/(J-

l>

X<

l>=

O.u

GO

O

no

25

f\

:::2

,1

25

X<N>=OX*fi~(N~l>

OI=

<2

.0*

R>

-T

PY

=3

.1q

15

93

/4.0

*<

0D

*O

D-O

I*n

I>*

FY

XMY=<3.141593/64.D>*<OO**q•o-DI**4·0>*<~.0/0D>*FY

ST

OP

f'.=

0.00

005*

PY

S

T 0

P r·I

= 0 •

0 0

1 * X

M Y

K

=I-

1

i·I R.

I T

C

< H

v H

! T

E ,

2 9

>

NC

2

9 F

0 R

~1 1 AT

< '

* * *

TH

IS

IS

C

UR

\/ E

i'.J

U 1v1

RE

R ' ,

I 3

>

s u flilf

i.~ =

o • n

st

mA

::: o

. n

\m I

TE

(

nw

.1 T

[ •

3 0

) 3

0

FO

PM

/\T

<'

/\

RE

A

1 L

E\/

F:H

f\

l{M

A

RE

/l,

2 l

su~·

. F

OR

CE

S

SU

M

MO

ME

NT

S

• )

H E

AD

C

I f ~

[ f\

IJ '

4 0

>

( Y

( l\J

> '

N =

1 ' I

)

t~ 0

F

0 R

r~ A

T

( F 9

• 6

}

LCV

ER

AHM

-J

~

---

---

---

---------

-----

-----

-----~

2e

no

2

40

N

=1

.K

?9

IF

<

Y<

NJ)

1

00

,90

,11

0

3 u

9 u

I F

< Y

( !~

+ 1

> )

1

8 0

, 9

5 ,

1 8

0 3

1

95

A

l=O

.n

32

1

\2=

0.0

3

3

CG

l=O

.O

34

C

G2

=0

.0

35

G

O

TO

22

0

36

1

00

IF

(Y(~+l))

12

0t1

90

,17

0

37

1

10

IF

(Y(~+l))

17

0tl

9Q

,13

0

3 j

1 2

0 I

F

< Y

< N

+ 1

> -

Y (

r ,J l

>

1 5

0 •

1 3·!

-J ,

1'+

O

3~

13U

IF

CYC~+ll-Y(NJ>

14

0.1

35

,15

0

4 0

1 3

5 A

1 =

( X ( I

', + l

> -

X (

r .1 )

) * Y

( N

) lf

l C

Gl=

CX

(t-.

J+l)

+X

<r1

> l

/?.0

4~

A2

=0

.0

46

C

G2

=0

.0

t;.4

t.to

46

4

7

48

'+9

5

u

51.

t.-r,

_1

t::

53

5

4

~5

56

5

7

5t~

5':1

60

f,

1

62

6

3

GO

TO

2

20

l~U

CG

l=<

X<

N+

l)+

X(N

))/2

.0

c G

2 =

< x <

r'.l +

1 >

+ 2

• o *

x <

r 1 >

> /

3 • o

I\ 1

= ( x

( t-.

+ 1

) -

x ( ~

·! )

) * y

(.i'J

+ 1

)

A 2

= ( x

{ N +

1 )

-x

( f'J

) } * (

y ( f ·

J ) -

y (

rJ +

1 )

) I

2 •

n

GO

TO

2

?0

1

50

C

Gl=

<X

CN

+l)

+X

(Nl)

/2.0

.

-"'

-

---

----------,--------------·

CG

2=<

2•0*

XC

N+

l>+

X<

N>

>/3

.0

t\ -1

:: <

X C

N + 1

> -

X {

r·J l

> * Y

( ~ !

> A?=<XC~+l)-XCN>>*<Y<~+ll-Y(N))/2.0

en

ro ~20

l 7

0 X

0 =

{ X (

N +

1 )

-X

( ~,

) ) *

( -1

. •

U ) * (

)' ( N )

) I

( Y

{ !'J

+ 1

) -Y

( N

) )

CG

l=C

X0

/3.u

>+

X(N

) CG2=<?·0*X(N+l)+X<N>+~0)/3.0

A l =

( X O

* '( (

1\J ) )

I 2

• 0

f\2

= (

/.. C

l\;+

l >-

( X (

1'1 >

+X

O}

) *

Y (

M+

l) /

2.

0 G

O

TO

22

0

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Compare with program output (next page)

Figure 32. Example for te.sting curve program

81

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APPENDIX II

REVISIONS TO MOMENT-THRUST-CURVATURE PROGRAM

MOMENT-THRUST-CURVATURE PROGRAM

DATA INPUT

NOTE: The last data card must assign the outside diameter

a value of zero to stop the program.

Numbers at left indicate card columns.

A. Control Data; Cross Section and Material Properties.

B.

FORMAT (4I5, 4El5.5)

1-5 Actual stress-strain data used? (+=Yes; -l=No)

6-10 Residual stresses used? (+l=Yes; -l=No)

11-15 Number of layers of elements. (Max. = 5)

16-20· Number of elements in 1/4 circle of one layer. (The product of the last two numbers must not exceed 50.)

21-35 Outside diameter (in.)

36-50 Wall thickness (in.)

51-65 Modulus of elasticity. (ksi)

66-80 Yield stress. (ksi)

Data and Time of Run

FORMAT (4I5)

1-5 Month

6-10 Day

11-15 Year

16-20 Time (001 - 2400)

85

C. Control Data

D.

E.

FORMAT (715)

1-5 Number of P/PY values (Max. = 12)

6-10 Number of PHI/PHIY values (Max. = 25)

11-15 KSKIP (l=l stress-strain data for one half tube) ( =2 stress-strain data for entire tube)

16-20 Rotation of bending axis (Number of elements)

21-25 Curve identification number

26-30 Are curves to be plotted? (- or· 0 = No) (+ = Yes)

31-35 Are M-P-PHI values to be punched on cards? (- or 0 = No) (+ = Yes)

Axial Load Values

FORMAT· (6Fl0. 5)

1-10 P/PY values (Always Positive)

11-20

21-30

31-40

41-50

51-60

Curvature Values .

FORMAT (5Fl0.5,/,5Fl0.5,l,5Fl0.5,l,5Fl0.5,l,5Fl0.5)

1-10 PHI-PHIY values (Always Positive)

11-20

21-30

31-40

41-5_0

NOTE: The data for one problem is now complete if the

actual stress-strain data and residual stresses are not

used. If both options are u~ed, the stress-strain curve

data is read in first.

F. Stress-Strain Curve Data

1. Control Card

FORMAT (IS)

86

1-5 Number of tabulated points on stress-strain ·.curve.

2. For each tabulat~d point

FORMAT (2El5.5)

1-15 Stress value

16-30 Strain value

G~ Residual Stress Data

1. Time of Residual Stress Calculation

FORMAT (4I5)

1-5 Month

6-10 Day

11-15 Year

16-20 Time

2. For each element

FORMA'r (2El5.'5)

1-15 Stress value

16-30 Strain value

FLOW DIAGRAM

CALCULATION OF MOMENT-THRUST-CURVATURE DATA

START

Read: NBS, IRS, NLYR, NELE OD, WT, E, FY

Yes

No

Read IDl, ID2, ID3, ID4

Read NP, NPHI, KSKIP ROTA,· IDCUR, IPLT, IPUN

Read P/Py values ..

Read 0/0y values.

STOP

87

No

used?

Yes

Read stress-strain data.

residual No

Yes

Read residual stress data.

Calculate for each layer:

Average radius Arc length of elements Area of elements.

88

~ .

Calculate the following:

Strain, curvature, bending moment and axial l.oad at first yield.

The distance.from each element to the centroid of the cross section.

The total cross sectional area, plastic modulus, flexural stiffness, pla~tic hinge moment and shape factor.

Calculate Summation ,of forces and moments for residual stress distribution

summations

No

Yes

Print Warning

89

Do 500 for each P/Py value.

Apply an (a new) axial load and calculate the corresponding axial strain (P/AE) .

Is tabular stress-strain

No

No

Axial stress

Go To 600

= P/A

90

e

Go To 300

Yes

Go To 300

Comment: This is the beginning of an iteration to determine the correct axial strain.

used?

Yes

For each element:

Calculate the total strain

No

Go To 21

· (Axial strain + residual strain)

Interpolate the corresponding stress value -

SUBROUTINE INTERP

Calculate the force on the element.

Calculate the total force on

the cross section.

Go To 75

91

For each element:

Calculate the total strain (Axial strain + residual strain)

Determine the corresponding stress value from the bilinear stress-strain relationship.

Calculate the force on the element.

Calculate the total force on the cross section.

Let "DIFF 11 = The total force on the cross section - the applied. axial load.

Yes

No

Adjust the axial strain

Go To 300

92

Calculate the unbalanced moment from residual + axial strain

Calculate curvature from unbalanced moment

Locate neutral axis at the centroid of the cross section

Have iterations balance axial stress been exceeded?

No

Have iterations find neutral axis been exceeded?

No

Is tabulated stress-strain data

used?·

Yes

Yes

Yes

No

Print Stop

Print Stop

Go To 525

93

l For each element

Calculate strain due to bending Calculate the total strain (Axial strain + residual strain + strain due to unbalanced curvature)

Interpolate the corresponding stress value -

SUBROUTINE INTERP

Calculate the force on the element

Calculate total moment for curvature stress and total moment for total stress

--'-~~~~~~~~~~~~~~--.~~~~~~~~~~~~~~~~~.--~~~~ ...

IGo To 5351 I

7 For each element

Calculate strain due to bending

Calculate the total strain

Determine the corresponding stress value from the bilinear stress -strain relationship

Calculate total moment for curvature stress and total moment for total stress

i

94

Calculate total force

Let . "force 11 be the net force on the cross section ·

Is 11 force 11

equal to

No

Yes --Go To 548

Adjust neutral axis location (sign of unbalanced curvature included)

Go To 515

stress+ axial stress·+ unbalanced curvature

less than 0.001 My?

No

Find new unbalanced curvature

95

Write axial load ratio summation of moments after balancing cross section for axial + residual strain.

- ""---------..--------------------'

Is bending axis rotated?

Yes

,,

Rotate axis of bending write; number of degree

rotated

Write; residual stress, residual strain, axial stress (after balancing)

,__ ________ ......, ______________

Do 400 for each ~/~y value.

No

Assume a (a new) curvature value.

Locate the neutral axis at the centroid of the cross section.

~

96

Go To 509

No

For each element

Calculate the strain due to bending

Calculate the total strain (Axial strain (balanced) + residual strain + strain due to bending)

Interpolate the corresponding stress value -

SUBROUTINE INTERP

Calculate the force on the element

Calculate the total force and .bending.moment on the cross section.

Go To 89

Yes

No

Print Stop

Go To 23

97

G For each element:

Calculate the strain due to bending.

Calculate the total strain.

Determine the corresponding stress value from the bilinear stress-strain relationship.

Calculate the total force and bending moment on the cross section.

Let "FORCE" be the net force on the cross section.

No

Adjust the location of the neutral axis.

Go To 450

Yes

Go To 71

98

Save the total moment calculated.

400 - Continue to next ~/~y value.

Return to original residual stress distribution

500 - Continue to next· P/Py value

Print results

Call plot subroutine if specified

Punch results if specified

Go To 999

99

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