2 Axial Loading MKM

1MECHANICS OF MATERIALS

CHAPTER

22 Stress and Strain Axial Loading

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MECHANICS OF MATERIALSContents

Stress & Strain: Axial LoadingNormal StrainStress-Strain TestS S i Di D il M i l

Generalized Hookes LawDilatation: Bulk ModulusShearing StrainE l 2 10Stress-Strain Diagram: Ductile Materials

Stress-Strain Diagram: Brittle Materials Hookes Law: Modulus of ElasticityElastic vs. Plastic BehaviorFatigueDeformations Under Axial LoadingExample 2.01

Example 2.10Relation Among E, , and GSample Problem 2.5Composite MaterialsSaint-Venants PrincipleStress Concentration: HoleStress Concentration: Fillet

2 - 2

Sample Problem 2.1Static IndeterminacyExample 2.04Thermal StressesPoissons Ratio

Example 2.12Elastoplastic MaterialsPlastic DeformationsResidual StressesExample 2.14, 2.15, 2.16


Thomas Young helped to develop the theory of how materials deform elastically.

In particular, he defined an important

Material Properties (Thomas Young, 1810)

material constant, Youngs Modulus.

P P

L

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Let Axial Stress,

Let Axial Strain,

Area.S.CForceNormalInternal=

LengthOriginalLengthinChange=

(in Units of N/m2or Pascals (Pa))

(Dimensionless)

2 - 3

MECHANICS OF MATERIALSStress-Strain Test

2 - 4

3MECHANICS OF MATERIALSStress-Strain Diagram: Ductile Materials

Mild

2 - 5

Youngs ModE

GPa

Mild Steel

210

Aluminium

70

Concrete

18.5

Wood

12.5

Nylon

2.8

Rubber

0.004

MECHANICS OF MATERIALSStress-Strain Diagram: Brittle Materials

2 - 6

4MECHANICS OF MATERIALSHookes Law: Modulus of Elasticity

Below the yield stressE

Elasticity of Modulus or Modulus Youngs=

=E

E

Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of

2 - 7

Elasticity) is not.

MECHANICS OF MATERIALSFatigue

Fatigue properties are shown on S-N diagrams.

When the stress is reduced below

A member may fail due to fatigueat stress levels significantly below the ultimate strength if subjected to many loading cycles.

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the endurance limit, fatigue failures do not occur for any number of cycles.

5MECHANICS OF MATERIALSDeformations Under Axial Loading

AEP

EE ===

From Hookes Law:

From the definition of strain:

L =

Equating and solving for the deformation,

AEPL=

2 - 9

With variations in loading, cross-section or material properties,

=i ii

iiEALP

MECHANICS OF MATERIALSExample 2.01

SOLUTION: Divide the rod into components at

th l d li ti i t

Determine the deformation of the steel rod shown under the

in.618.0 in. 07.1

psi1029 6

===

dD

E

the load application points.

Apply a free-body analysis on each component to determine the internal force

Evaluate the total of the component deflections.

2 - 10

the steel rod shown under the given loads.

6MECHANICS OF MATERIALSSOLUTION:

Divide the rod into three components:

Apply free-body analysis to each component to determine internal forces,

lb1030

lb1015

lb1060

3

32

31

==

P

P

P

lb1030 33 =P

Evaluate total deflection,

( ) ( ) ( )16103012101512106011

333

3

33

2

22

1

11

++=

++==

ALP

ALP

ALP

EEALP

i ii

ii

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221

21

in 9.0

in. 12

====

AA

LL

23

3

in 3.0

in. 16

==

A

L

in.109.75

3.09.09.01029

3

6

=

++=

in. 109.75 3=

MECHANICS OF MATERIALSSample Problem 2.1

SOLUTION:

Apply a free-body analysis to the bar BDE to find the forces exerted by

The rigid bar BDE is supported by two links AB and CD.

Link AB is made of aluminum (E = 70 GPa) and has a cross-sectional area of 500 mm2 Link CD is made of steel (E = 200

links AB and DC. Evaluate the deformation of links AB

and DC or the displacements of Band D.

Work out the geometry to find the deflection at E given the deflections at B and D.

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mm . Link CD is made of steel (E = 200 GPa) and has a cross-sectional area of (600 mm2).

For the 30-kN force shown, determine the deflection a) of B, b) of D, and c) of E.


Displacement of B:

( )( )m3.0N1060 3=

AEPL

BFree body: Bar BDESOLUTION:

Sample Problem 2.1

( )( )( )( )m10514

Pa1070m10500m3.0N1060

6

926-

==

= mm 514.0BDisplacement of D:

=AEPL

D( )

tensionF

F

M

CD

CD

B

kN90

m2.0m6.0kN300

0

+=+=

=

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( )( )( )( )m10300

Pa10200m10600m4.0N1090

6

926-

3

=

=

= mm 300.0D

( )ncompressioF

F

AB

AB

kN60

m2.0m4.0kN300

0MD

==

=

MECHANICS OF MATERIALS

Displacement of D:

=

HDBH

DDBB

Sample Problem 2.1

( )mm7.73

mm 200mm 0.300mm 514.0

=

=x

xx

( )mm7.73400 +=

E

HDHE

DDEE

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= mm 928.1E

( )mm 928.1

mm 7.73mm 300.0=

=

E

E

8MECHANICS OF MATERIALSStatic Indeterminacy

Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate.

A ill b i ll i d i

Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations.

A structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium.

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0=+= RL

Deformations due to actual loads and redundant reactions are determined separately and then added or superposed.


Determine the reactions at A and B for the steel bar and loading shown, assuming a close fit at both supports before the loads are applied.

SOLUTION:

Solve for the displacement at B due to the redundant reaction at B.

SOLUTION:

Consider the reaction at B as redundant, release the bar from that support, and solve for the displacement at B due to the applied loads.

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Solve for the reaction at A due to applied loads and the reaction found at B.

Require that the displacements due to the loads and due to the redundant reaction be compatible, i.e., require that their sum be zero.


SOLUTION: Solve for the displacement at B due to the applied

loads with the redundant constraint released, PPPP 34

3321 N10900N106000 ====

Example 2.04

EEALP

LLLL

AAAA

i ii

ii9

L

4321

2643

2621

10125.1

m 150.0

m10250m10400

==

========

Solve for the displacement at B due to the redundant constraint,

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( ) ====

====

iB

ii

iiR

B

ER

EALP

LL

AA

RPP

3

21

262

261

21

1095.1

m 300.0

m10250m10400


Require that the displacements due to the loads and due to the redundant reaction be compatible,

( )0

39

=+= RL

Example 2.04

( )kN 577N10577

01095.110125.1

3

39

==

==

B

B

R

ER

E

Find the reaction at A due to the loads and the reaction at BkN577kN600kN 3000 +== Ay RF

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kN323=Ay

R

kN577

kN323

==

B

A

R

R

10

MECHANICS OF MATERIALSStatic Indeterminacy

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A Scottish engineer, Rankine made observations about the expansion and contraction of materials due to changes in temperature.

Thermal Strains (William Rankine, 1870)

He noted that these deflections were proportional to

x

y

( )Ty TTo ( )TTT.e.i o =

( )Tx =

the change in temperature the material experienced .

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y

x ( )Tx ( )Ty =

=Coefficient of Linear Expansion (A material property)2 - 22

12

MECHANICS OF MATERIALSThermal Stresses

A temperature change results in a change in length or thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by the supports. t e suppo ts.

( )coef.expansion thermal=

==

AEPLLT PT

Treat the additional support as redundant and apply the principle of superposition.

The thermal deformation and the deformation from th d d t t t b tibl

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( ) 00

=+=+=

AEPLLT

PT

the redundant support must be compatible.

( )( )TE

AP

TAEPPT

===

=+=

0

MECHANICS OF MATERIALSFrom Hookes Law:

y

( )Ty (Due to Forces and Temperature Changes)

Mechanical StrainThermal Strain

x ( )Tx [ ]zyxx E1 =[ ] ( )T

E1

zxyy +=[ ] ( )T

E1

yxzz +=

( )T+

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Coef Th expan

x 10-6/oC

Mild Steel

12

Aluminium

23

Concrete

10.8

Wood

-

Nylon

0.9

Rubber

130-200

2 - 24

13

MECHANICS OF MATERIALSIf a steel bar has a maximum axial stress of 300 MPa, what is the greatest allowable temperature increase?

( )MaxT

E, , , MaxE=200 GPa=0 25 ( ) ?T Max ==0.25=12x10-6 1/oC

( )TEx =( )

=E

T Max,xMax( )( ) ( )69

6

10x1210x20010x300

= C125o=

Compressive stress relates to increase in T.

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+ve (T) i.e. HOT Compression-ve (T) i.e. COLD Tension

Tve + Tve

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14


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Poisson made important observations and theories about lateral deflections of materials.

Material Properties (Simon Poisson, 1825)

When a bar is placed in tension, lateral contractions accompany the extension.

P A real math nut.His only passion has been science: he

p y

Initial Shape

Final Shape

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Pbeen science: he lived and is dead for it.

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15

MECHANICS OF MATERIALSPoissons Ratio

For a slender bar subjected to axial loading:

0=== zyxx E

The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic (no directional dependence),

0= zy

Poissons ratio is defined as

x

z

x

y

===

strain axialstrain lateral

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PoissonsRatio

v

Mild Steel

0.3

Aluminium

0.33

Concrete

0.1-0.2

Wood

-

Nylon

0.4

Rubber

0.45-0.5

There exists a theoretical limit: 0.5 Constant volume

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16

MECHANICS OF MATERIALSShearing Strain

A cubic element subjected to a shear stress will deform into a rhomboid. The corresponding shearstrain is quantified in terms of the change in angle b t th idbetween the sides,( )xyxy f =

A plot of shear stress vs. shear strain is similar the previous plots of normal stress vs. normal strain except that the strength values are approximately half. For small strains,

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zxzxyzyzxyxy GGG ===where G is the modulus of rigidity or shear modulus.


SOLUTION:

Determine the average angular deformation or shearing strain of the block.

A rectangular block of material with modulus of rigidity G = 90 ksi is bonded to two rigid horizontal plates. The lower plate is fixed, while the

l i bj d h i l

Use the definition of shearing stress to find the force P.

Apply Hookes law for shearing stress and strain to find the corresponding shearing stress.

2 - 32

upper plate is subjected to a horizontal force P. Knowing that the upper plate moves through 0.04 in. under the action of the force, determine a) the average shearing strain in the material, and b) the force P exerted on the plate.

17


Determine the average angular deformation or shearing strain of the block.

rad020.0in.2

in.04.0tan == xyxyxy

Apply Hookes law for shearing stress and strain to find the corresponding shearing stress. ( )( ) psi1800rad020.0psi1090 3 === xyxy G

Use the definition of shearing stress to find

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the force P.( )( )( ) lb1036in.5.2in.8psi1800 3=== AP xy

kips0.36=P

MECHANICS OF MATERIALSRelation Among E, , and G

An axially loaded slender bar will elongate in the axial direction and contract in the transverse directions. A i iti ll bi l t i t d i

If the cubic element is oriented as in the bottom figure, it will deform into a rhombus. Axial load also results in a shear strain

An initially cubic element oriented as in top figure will deform into a rectangular parallelepiped. The axial load produces a normal strain.

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( )+= 12GE

Components of normal and shear strain are related,

strain.

18

MECHANICS OF MATERIALSSample Problem 2.5

A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses x = 12 ksi and z = 20 ksi.

For E = 10x106 psi and = 1/3, determine the change in:

a) the length of diameter AB,

b) the length of diameter CD,

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c) the thickness of the plate, and

d) the volume of the plate.

MECHANICS OF MATERIALSComposite Materials

Fiber-reinforced composite materials are formed from lamina of fibers of graphite, glass, or polymers embedded in a resin matrix.

N l t d t i l t d b H k

zz

zy

yy

x

xx EEE

===

Normal stresses and strains are related by Hookes Law but with directionally dependent moduli of elasticity,

Transverse contractions are related by directionally dependent values of Poissons ratio, e.g.,

2 - 36

x

zxz

x

yxy

==

p , g ,

Materials with directionally dependent mechanical properties are anisotropic.

19

MECHANICS OF MATERIALSSaint-Venants Principle

Loads transmitted through rigid plates result in uniform distribution of stress and strain.

d l d l i l

Stress and strain distributions become uniform at a relatively short distance from the load application points.

Concentrated loads result in large stresses in the vicinity of the load application point.

2 - 37

Saint-Venants Principle:Stress distribution may be assumed independent of the mode of load application except in the immediate vicinity of load application points.

p

MECHANICS OF MATERIALSStress Concentration: Hole

2 - 38

Discontinuities of cross section may result in high localized or concentrated stresses. ave

max=K

20

MECHANICS OF MATERIALSStress Concentration: Fillet

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SOLUTION:

Determine the geometric ratios and fi d th t t ti f t

Determine the largest axial load Pthat can be safely supported by a flat steel bar consisting of two portions, both 10 mm thick, and respectively 40 and 60 mm wide, connected by fillets of radius r = 8

find the stress concentration factor from Fig. 2.64b.

Apply the definition of normal stress to

Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor.

2 - 40

ymm. Assume an allowable normal stress of 165 MPa.

pp yfind the allowable load.

21

MECHANICS OF MATERIALS Determine the geometric ratios and

find the stress concentration factor from Fig. 2.64b.

82.1

20.0mm40

mm850.1mm40mm60

=

====

K

dr

dD

Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor.

MPa7.9082.1MPa165max

ave === K

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Apply the definition of normal stress to find the allowable load.

( )( )( )N103.36

MPa7.90mm10mm40

3=== aveAP

kN3.36=P

MECHANICS OF MATERIALSPlastic Deformations

Elastic deformation while maximum stress is less than yield stressK

AAP ave max ==

Maximum stress is equal to the yield stress at the maximum elastic loadingK

AP YY=

At loadings above the maximum elastic load, a region of plastic deformations develop near the hole

2 - 42

As the loading increases, the plastic region expands until the section is at a uniform stress equal to the yield stress

Y

YU

PK

AP

==

22

MECHANICS OF MATERIALSResidual Stresses

When a single structural element is loaded uniformly beyond its yield stress and then unloaded, it is permanently deformed but all stresses disappear. This is not the general resultresult.

Residual stresses will remain in a structure after loading and unloading if

- only part of the structure undergoes plastic deformation

- different parts of the structure undergo different plastic deformations

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Residual stresses also result from the uneven heating or cooling of structures or structural elements

plastic deformations

MECHANICS OF MATERIALSExample 2.14, 2.15, 2.16

A cylindrical rod is placed inside a tube of the same length. The ends of the rod and tube are attached to a rigid support on one side and a rigid plate on theon one side and a rigid plate on the other. The load on the rod-tube assembly is increased from zero to 5.7 kips and decreased back to zero.

a) draw a load-deflection diagram for the rod-tube assembly

b) determine the maximum psi1030

in.075.0

6

2

==

r

r

E

A

psi1015

in.100.0

6

2

==

t

t

E

A

2 - 44

elongation

c) determine the permanent set

d) calculate the residual stresses in the rod and tube.

ksi36, =rY ksi45, =tY

23


a) draw a load-deflection diagram for the rod-tube assembly

( )( )i

kips7.2in075.0ksi36

3

2,, === AP rrYrY

Example 2.14, 2.15, 2.16

in.1036in.30psi1030psi1036 3-

6

3

,

,, =

=== LE

LrY

rYrYY,r

( )( )in.1009in.30

psi1015psi1045

kips5.4in100.0ksi45

3-6

3

,

,,

2,,

====

===

LE

L

AP

tY

tYtYY,t

ttYtY

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tr

tr PPP

==+=

MECHANICS OF MATERIALSb,c) determine the maximum elongation and permanent set

at a load of P = 5.7 kips, the rod has reached the plastic range while the tube is still in the elastic range

( ) kips03kips7275kips7.2,

=====

PPP

PP

t

rYr

Example 2.14, 2.15, 2.16

( )

in.30psi1015psi1030

ksi30in0.1kips0.3

kips0.3kips7.27.5

6

3t

2t

===

===

LE

L

AP

PPP

t

tt

t

t

rt

in.1060 3max== t

the rod-tube assembly unloads along a line parallel to 0Y

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0Yr

( ) in.106.4560in.106.45

in.kips125kips7.5

slopein.kips125in.1036

kips5.4

3maxp

3max

3-

=+=

===

===

m

P

m

in.104.14 3=p

24


calculate the residual stresses in the rod and tube.

calculate the reverse stresses in the rod and tube caused by unloading and add them to the maximum stresses

Example 2.14, 2.15, 2.16

stresses.

( )( )( )( ) ksi822psi101510521

ksi6.45psi10301052.1

in.in.1052.1in.30

in.106.45

63

63

33

======

===

rr

E

E

L

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( )( )( )

( ) ksi2.7ksi8.2230ksi69ksi6.4536

ksi8.22psi10151052.1

,

,

==+===+=

===

tttresidual

rrrresidual

tt

.

E

MECHANICS OF MATERIALSContoh Soal 1

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MECHANICS OF MATERIALSContoh Soal 10 (lanjutan)

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