introduction to axial loading

7/28/2019 Introduction to Axial Loading

1/35

MECHANICS OF

MATERIALS

Fourth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

John T. DeWolf

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

2002 The McGraw-Hill Companies, Inc. All rights reserved.

2Stress and Strain Axial Loading


2/35


MECHANICS OF MATERIALSFourth

Edition

Beer Johnston DeWolf

2 - 2

Contents

Stress & Strain: Axial Loading

Normal Strain

Stress-Strain Test

Stress-Strain Diagram: Ductile Materials

Stress-Strain Diagram: Brittle Materials

Hookes Law: Modulus of Elasticity

Elastic vs. Plastic Behavior

Fatigue

Deformations Under Axial Loading

Example 2.01

Sample Problem 2.1Static Indeterminacy

Example 2.04

Thermal Stresses

Poissons Ratio

Generalized Hookes Law

Dilatation: Bulk Modulus

Shearing Strain

Example 2.10

Relation Among E, n, and G

Sample Problem 2.5

Composite Materials

Saint-Venants Principle

Stress Concentration: Hole

Stress Concentration: Fillet

Example 2.12Elastoplastic Materials

Plastic Deformations

Residual Stresses

Example 2.14, 2.15, 2.16
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Edition


2 - 3

Stress & Strain: Axial Loading

Suitability of a structure or machine may depend on the deformations inthe structure as well as the stresses induced under loading. Statics

analyses alone are not sufficient.

Considering structures as deformable allows determination of member

forces and reactions which are statically indeterminate.

Determination of the stress distribution within a member also requires

consideration of deformations in the member.

Chapter 2 is concerned with deformation of a structural member underaxial loading. Later chapters will deal with torsional and pure bendingloads.

MECHANICS OF MATERIA S


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Edition


2 - 4

Normal Strain

strainnormal

stress

L

A

P

L

A

P

A

P

2

2

LL

A

P

2

2

MECHANICS OF MATERIALSFE


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Edition


2 - 5

Stress-Strain Test



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Edition


2 - 6

Stress-Strain Diagram: Ductile Materials

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7/35 2002 The McGraw-Hill Companies, Inc. All rights reserved.


Edition


2 - 7

Stress-Strain Diagram: Brittle Materials





Edition


2 - 8

Hookes Law: Modulus of Elasticity

Below the yield stress

ElasticityofModulus

orModulusYoungs

E

E

Strength is affected by alloying,

heat treating, and manufacturing

process but stiffness (Modulus ofElasticity) is not.

(i.e. the slope of the linear part of

the stress-strain diagram)

Isotropic Materials: Material properties are independent of direction considered.

Unisotropic Materials: Material properties are direction dependent, e.g. wood.





Edition


2 - 9

Elastic vs. Plastic Behavior

If the strain disappears when the

stress is removed, the material is

said to behave elastically.

When the strain does not return

to zero after the stress is

removed, the material is said tobehaveplastically.

The largest stress for which this

occurs is called the elastic limit.





Edition


2 - 10

Deformations Under Axial Loading

AE

P

EE

From Hookes Law:

From the definition of strain:

L

Equating and solving for the deformation,

AE

PL

With variations in loading, cross-section ormaterial properties,

i ii

ii

EA

LP





Edition


2 - 11

Example 2.01

Determine the deformation of

the steel rod shown under the

given loads.

mm56.15dmm18.27DGPa200E

SOLUTION:

Divide the rod into components at

the load application points.

Apply a free-body analysis on each

component to determine theinternal force

Evaluate the total of the component

deflections.





Edition


2 - 12

SOLUTION:

Divide the rod into three

components:

2

21

21

mm580AA

mm300LL

2

3

3

mm190A

mm.400L

Apply free-body analysis to each

component to determine internal forces,

kN120P

kN60P

kN240P

3

2

1

Evaluate total deflection,

mm.1.73m1073.110190

4.010120

10580

3.01060

10580

3.010240

10200

1

A

LP

A

LP

A

LP

E

1

EA

LP

3

6

3

6

3

6

3

9

3

33

2

22

1

11

i ii

ii

mm.73.1





Edition


2 - 13

Sample Problem 2.1

The rigid barBDEis supported by two linksAB and CD.

LinkAB is made of aluminum (E= 70 GPa)

and has a cross-sectional area of 500 mm2.

LinkCD 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) ofB, b) ofD, and c) ofE.

SOLUTION:

Apply a free-body analysis to the bar

BDEto find the forces exerted by

linksAB andDC.

Evaluate the deformation of linksAB

andDCor the displacements ofBandD.

Work out the geometry to find the

deflection at E given the deflections

at B and D.

MECHANICS OF MATERIALSF

E




Edition


2 - 14

Displacement ofB:

m10514

Pa1070m10500

m3.0N1060

6

926-

3

AEPL

B

mm514.0B

Displacement ofD:

m10300

Pa10200m10600m4.0N1090

6

926-

3

AE

PLD

mm300.0D

Free body: BarBDE

ncompressioF

F

tensionF

F

M

AB

AB

CD

CD

B

kN60

m2.0m4.0kN3000M

kN90

m2.0m6.0kN300

0

D

SOLUTION:

Sample Problem 2.1

MECHANICS OF MATERIALSFo

E
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Edition


2 - 15

Displacement of D:

mm7.73

mm200

mm0.300

mm514.0

x

x

x

HD

BH

DD

BB

mm928.1E

mm928.1mm7.73

mm7.73400

mm300.0

E

E

HD

HE

DD

EE

Sample Problem 2.1


Ed
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Edition


2 - 16

Static Indeterminacy

Structures for which internal forces and reactions

cannot be determined from statics alone are saidto bestatically indeterminate.

0RL

Deformations due to actual loads and redundant

reactions are determined separately and then added

orsuperposed.

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.


Ed



MECHANICS OF MATERIALSourth

Edition


2 - 17

Example 2.04

Determine the reactions atA andB for the steel

bar and loading shown, assuming a close fit at

both supports before the loads are applied.

Solve for the reaction atA due to applied loads

and the reaction found atB.

Require that the displacements due to the loadsand due to the redundant reaction be compatible,

i.e., require that their sum be zero.

Solve for the displacement atB due to the

redundant reaction atB.

SOLUTION:

Consider the reaction atB as redundant, release

the bar from that support, and solve for the

displacement atB due to the applied loads.


Ed




dition


2 - 18

SOLUTION:

Solve for the displacement atB due to the applied

loads with the redundant constraint released,

EEA

LP

LLLL

AAAA

PPPP

i ii

ii9

L

4321

2643

2621

34

3321

10125.1

m150.0

m10250m10400

N10900N106000

Solve for the displacement atB due to the redundant

constraint,

i

B

ii

iiR

B

E

R

EA

LP

LL

AA

RPP

3

21

262

261

21

1095.1

m300.0

m10250m10400

Example 2.04


Ed
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dition


2 - 19

Require that the displacements due to the loads and due to

the redundant reaction be compatible,

kN577N10577

01095.110125.1

0

3

39

B

B

RL

R

E

R

E

Find the reaction atA due to the loads and the reaction atB

kN323

kN577kN600kN3000

A

Ay

R

RF

kN577

kN323

B

A

R

R

Example 2.04


Ed




dition


2 - 20

Thermal Stresses

A temperature change results in a change in length or

thermal strain. There is no stress associated with thethermal strain unless the elongation is restrained by

the supports.

tcoefficienexpansionthermal

AE

PLLT PT

Treat the additional support as redundant and apply

the principle of superposition.

0

0

AE

PLLT

PT

The thermal deformation and the deformation from

the redundant support must be compatible.

TEA

P

TAEPPT

0


Ed




dition


2 - 21

Poissons Ratio

For a slender bar subjected to axial loading:

0 zyx

xE

The elongation in the x-direction is

accompanied by a contraction in the other

directions. Assuming that the material isisotropic (no directional dependence),

0 zy

Poissons ratio is defined as

x

z

x

y

n

strainaxial

strainlateral


Ed




dition


2 - 22

Generalized Hookes Law

For an element subjected to multi-axial loading,

the normal strain components resulting from thestress components may be determined from the

principle of superposition. This requires:

1) strain is linearly related to stress

2) deformations are small

EEE

EEE

EEE

zyxz

zyxy

zyxx

nn

n

n

nn

With these restrictions:


Ed
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dition


Dilatation: Bulk Modulus

Relative to the unstressed state, the change in volume is

e)unit volumperin volume(changedilatation

21

111111

zyx

zyx

zyxzyx

E

e

n

For element subjected to uniform hydrostatic pressure,

modulusbulk

213

213

n

n

Ek

k

p

Epe

Subjected to uniform pressure, dilatation must be

negative, therefore

210 n


Ed

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MECHANICS OF MATERIALSurthdition


2 - 24

Shearing Strain

Consider the effect of the presence of shearing

stresses on the faces of a small element.

In the case of isotropic materials, these stresses come

in pairs of equal and opposite vectors applied to the

opposite sides of the given element and have no

effect on normal strains.

A cubic element subjected to a shear stress will deform into a rhomboid.

The correspondingshearstrain is quantified in terms of the change in angle

between the sides,

xyxy f


Ed

B J h t D W lf


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MECHANICS OF MATERIALSurthition


2 - 25

Shearing Strain

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,

zxzxyzyzxyxy GGG

where G is the modulus of rigidity or shear modulus.

MECHANICS OF MATERIALSFou

Ed

B J h t D W lf


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2 - 26

Example

A rectangular block of material with

modulus of rigidity G = 620 MPa is

bonded to two rigid horizontal plates.

The lower plate is fixed, while the

upper plate is subjected to a horizontal

forceP. Knowing that the upper plate

moves through 1 mm under the action

of the force, determine a) the average

shearing strain in the material, and b)

the forcePexerted on the plate.

SOLUTION:

Determine the average angulardeformation or shearing strain of

the block.

Use the definition of shearing stress to

find the forceP.

Apply Hookes law for shearing stress

and strain to find the corresponding

shearing stress.


Edi

B J h t D W lf


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2 - 27

Determine the average angular deformation

or shearing strain of the block.

rad020.0mm50

mm1tan xyxyxy

Apply Hookes law for shearing stress and

strain to find the corresponding shearing

stress.

MPa6.13rad020.0MPa680G xyxy

Use the definition of shearing stress to find

the forceP.

kN170105.62200MPa6.13 -3 mmmmAP xy

kN170P


Edi



28/35




2 - 28

Relation Among E, n, and G

An axially loaded slender bar will

elongate in the axial direction andcontract in the transverse directions.

n 12G

E

Components of normal and shear strain are

related,

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.


Edi



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MECHANICS OF MATERIALSurthtion Beer Johnston DeWolf

2 - 29

Sample Problem

A circle of diameterd= 225 mm is scribed on

an unstressed aluminum plate of thickness, t,

20 mm. Forces acting in the plane of the plate

later cause normal stresses x = 80 MPa and

z = 140 MPa.

ForE= 70 GPa and n = 1/3, determine thechange in:

a) the length of diameterAB,

b) the length of diameterCD,

c) the thickness of the plate, and

d) the volume of the plate.


Edi



30/35



2 - 30

SOLUTION:

Apply the generalized Hookes Law to

find the three components of normal

strain.

mm/mm10619.1

EEE

mm/mm10047.1

EEE

mm/mm10476.0

MPa1403

10MPa80

MPa000,70

1

EEE

3

zyx

z

3

zyxy

3

zyxx

n

n

n

n

n

n

Evaluate the deformation components.

mm225mm/mm10476.0d 3xAB

mm225mm/mm10619.1d 3zDC

mm20mm/mm10047.1t 3yt

mm11.0AB

mm36.0DC

mm02.0t

Find the change in volume

33

333

mm2038038010048.1

/mmmm10048.1

eVV

ezyx

3mm6.026,3V


Edit Beer Johnston DeWolf
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2 - 31

Saint-Venants Principle

Loads transmitted through rigid

plates result in uniform distribution

of stress and strain.

Saint-Venants Principle:

Stress distribution may be assumed

independent of the mode of load

application except in the immediate

vicinity of load application points.

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.




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2 - 32

Example 2.14, 2.15, 2.16

A cylindrical rod is placed inside a tube

of the same length. The ends of the rodand tube are attached to a rigid support

on 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

elongation

c) determine the permanent set

d) calculate the residual stresses in

the rod and tube.

ksi36

psi1030

in.075.0

,

6

2

rY

r

r

E

A

ksi45

psi1015

in.100.0

,

6

2

tY

t

t

E

A




33/35



2 - 33

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

tube assembly

in.1036in.30psi1030

psi1036

kips7.2in075.0ksi36

3-6

3

,

,,

2,,

LE

L

AP

rY

rYrYY,r

rrYrY

in.1009in.30psi1015

psi1045

kips5.4in100.0ksi45

3-6

3

,

,,

2,,

LE

L

AP

tY

tYtYY,t

ttYtY

tr

tr PPP

Example 2.14, 2.15, 2.16




34/35


MECHANICS OF MATERIALSrthtion Beer Johnston DeWolf

2 - 34

b,c) determine the maximum elongation and permanent set

at a load ofP= 5.7 kips, the rod has reached the

plastic range while the tube is still in the elastic range

in.30psi1015

psi1030

ksi30in0.1

kips0.3

kips0.3kips7.27.5

kips7.2

6

3

t

2t

,

LE

L

A

P

PPP

PP

t

tt

t

t

rt

rYr

in.1060 3max t

the rod-tube assembly unloads along a line parallel to

0Yr

in.106.4560

in.106.45in.kips125

kips7.5

slopein.kips125in.1036

kips5.4

3maxp

3max

3-

m

P

m

in.104.14 3p

Example 2.14, 2.15, 2.16


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MECHANICS OF MATERIALSrthtion Beer Johnston DeWolf

calculate the residual stresses in the rod and tube.

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

stresses.

ksi2.7ksi8.2230

ksi69ksi6.4536

ksi8.22psi10151052.1

ksi6.45psi10301052.1

in.in.1052.1

in.30

in.106.45

,

,

63

63

33

tttresidual

rrrresidual

tt

rr

.

E

E

L

Example 2.14, 2.15, 2.16
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introduction to axial loading

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