fourth edition mechanics of materials - hacettepeyunus.hacettepe.edu.tr/~boray/2_1_axial_loading...

MECHANICS OFFourth Edition

MECHANICS OF MATERIALS

CHAPTER

MATERIALSFerdinand P. BeerE. Russell Johnston, Jr.John T DeWolf

Stress and Strain John T. DeWolf

Lecture Notes:– Axial Loading

J. Walt OlerTexas Tech University

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.


FourthEdition

Beer • Johnston • DeWolf

Contents

Stress & Strain: Axial LoadingNormal Strain

Generalized Hooke’s LawDilatation: Bulk ModulusNormal Strain

Stress-Strain TestStress-Strain Diagram: Ductile Materials

Dilatation: Bulk ModulusShearing StrainExample 2.10

Stress-Strain Diagram: Brittle Materials Hooke’s Law: Modulus of ElasticityElastic vs. Plastic Behavior

Relation Among E, ν, and GSample Problem 2.5Composite MaterialsElastic vs. Plastic Behavior

FatigueDeformations Under Axial LoadingE l 2 01

Composite MaterialsSaint-Venant’s PrincipleStress Concentration: HoleS C i FillExample 2.01

Sample Problem 2.1Static Indeterminacy

Stress Concentration: FilletExample 2.12Elastoplastic Materialsy

Example 2.04Thermal StressesPoisson’s Ratio

pPlastic DeformationsResidual StressesE l 2 14 2 15 2 16

© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 2

Poisson s Ratio Example 2.14, 2.15, 2.16


FourthEdition


Stress & Strain: Axial Loading

• Suitability of a structure or machine may depend on the deformations in y y pthe 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 id ti f d f ti i th bconsideration of deformations in the member.

• Chapter 2 is concerned with deformation of a structural member under paxial loading. Later chapters will deal with torsional and pure bending loads.



FourthEdition


Normal Strain

stress==AP

δ

σAP

AP

δ

σ ==22

AP

δδ

σ =

2


strain normal==Lδε

Lδε =

LLδδε ==

22


FourthEdition


Stress-Strain Test



FourthEdition


Stress-Strain Diagram: Ductile Materials



FourthEdition


Stress-Strain Diagram: Brittle Materials



FourthEdition


Hooke’s Law: Modulus of Elasticity

• Below the yield stress

M d lY=

EEεσ

Elasticity of Modulus orModulusYoungs=E

• Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of Elasticity) is not.



FourthEdition


Elastic vs. Plastic Behavior

• If the strain disappears when the• If the strain disappears when the stress is removed, the material is said to behave elastically.

• The largest stress for which this occurs is called the elastic limit

• When the strain does not return

occurs is called the elastic limit.

to zero after the stress is removed, the material is said to behave plastically.behave plastically.



FourthEdition


Fatigue

• Fatigue properties are shown on g p pS-N diagrams.

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

• When the stress is reduced below the endurance limit, fatigue failures do not occur for any ynumber of cycles.



FourthEdition


Deformations Under Axial Loading

Pσ• From Hooke’s Law:

AEP

EE ===

σεεσ

• From the definition of strain:• From the definition of strain:

Lδε =

• Equating and solving for the deformation,PL

=δAE

δ

• With variations in loading, cross-section or i l imaterial properties,

∑=i ii

iiEALPδ


i ii


FourthEdition


Example 2.01

SOLUTION:SOLUTION:• Divide the rod into components at

the load application points.

psi1029 6×= −E• Apply a free-body analysis on each

component to determine the i t l f

Determine the deformation of

in.618.0 in. 07.1 == dD internal force

• Evaluate the total of the component d fl tiDetermine the deformation of

the steel rod shown under the given loads.

deflections.



FourthEdition


SOLUTION: • Apply free-body analysis to each

• Divide the rod into three components:

component to determine internal forces,

lb1060 31 ×=P

lb1030

lb1015

33

32

×=

×−=

P

P

• Evaluate total deflection,

( ) ( ) ( )

1

333

3

33

2

22

1

11

⎤⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛++=∑=

ALP

ALP

ALP

EEALP

i ii

iiδ

( ) ( ) ( )

i109

3.0161030

9.0121015

9.0121060

10291

3

333

6

−

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×+

×−+

×

×=

2

21 in. 12== LL 3 in. 16=L

in.109.75 3−×=

in.109.75 3−×=δ


221 in 9.0== AA 2

3 in 3.0=A


FourthEdition


Sample Problem 2.1

SOLUTION:

• Apply a free-body analysis to the bar BDE to find the forces exerted by links AB and DC.

The rigid bar BDE is supported by two li k AB d CD

• Evaluate the deformation of links ABand DC or the displacements of Band Dlinks AB and CD.

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

and D.

• Work out the geometry to find the deflection at E given the deflections GPa) and has a cross sectional area of 500

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

gat B and D.

mm2).

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


deflection a) of B, b) of D, and c) of E.


FourthEdition


Sample Problem 2.1Displacement of B:

=PL

BδFree body: Bar BDE

SOLUTION:

( )( )( )( )Pa1070m10500

m3.0N1060926-

3

××

×−=

AEBδFree body: Bar BDE

( )( )m10514

Pa1070m10500

6−×−=

××

↑= mm5140Bδ ↑= mm514.0BδDisplacement of D:

=PL

Dδ( ) F

M

CD

B

m2.0m6.0kN300

0

×+×−=

=∑

( )( )( )( )Pa10200m10600

m4.0N1090926-

3

××

×=

=AEDδ

( ) F

tensionFCD

m20m40kN300

0M

kN90

D

××

=

+=

∑( )( )

m10300

Pa10200m10600

6−×=

××( )ncompressioF

F

AB

AB

kN60

m2.0m4.0kN300

−=

×−×−=


↓= mm 300.0Dδ


FourthEdition


Sample Problem 2.1

Displacement of D:

( )mm200mm 514.0 −

=′′

xHDBH

DDBB

( )

mm 7.73mm 0.300

=

=

xx

=′′

HDHE

DDEE

( )

mm9281mm 7.73

mm7.73400mm 300.0

=

+=

E

E

δ

δ

↓= mm928.1Eδ

mm928.1=Eδ


↓9 8.Eδ


FourthEdition


Static Indeterminacy• Structures for which internal forces and reactions

cannot be determined from statics alone are said to be statically indeterminate.

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

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

• Deformations due to actual loads and redundant reactions are determined separately and then added

0=+= RL δδδ

p yor superposed.



FourthEdition


Example 2.04Determine the reactions at A and B for the steel bar and loading shown, assuming a close fit at both supports before the loads are appliedboth supports before the loads are applied.

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.

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

p pp

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

redundant reaction at B.

• Solve for the reaction at A due to applied loads

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


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


FourthEdition


Example 2.04SOLUTION:• Solve for the displacement at B due to the applied

loads with the redundant constraint releasedloads with the redundant constraint released,

AAAA

PPPP

2643

2621

34

3321

m10250m10400

N10900N106000

×==×==

×=×===

−−

LP

LLLL

ii9

4321

4321

10125.1

m 150.0

×∑

====

δEEAi ii

iiL =∑=δ

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

−==

−−

B

AA

RPP

2626

21

1025010400

( )==

×=×=

RLP

LL

AA

3

21

262

261

10951

m 300.0

m10250m10400


( )∑

×−==

i

B

ii

iiR E

REALPδ

31095.1


FourthEdition


Example 2.04

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

( ) 01095.110125.1

0

39=

×−

×=

=+=

B

RL

Rδ

δδδ

( )

kN 577N10577

0

3 =×=

==

BR

EEδ

• Find the reaction at A due to the loads and the reaction at B

kN323

kN577kN600kN 3000

=

∑ +−−==

A

Ay

R

RF

kN577

kN323

=

=

B

A

R

R


B


FourthEdition


Thermal 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.

h ddi i l d d d l

( ) =∆= δαδ PLLT

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

( )coef.expansion thermal=

=∆=

α

δαδAE

LT PT

Th th l d f ti d th d f ti f

0=+= PT δδδ

• The thermal deformation and the deformation from the redundant support must be compatible.

( ) PLLT +∆ 00=+= PT δδδ ( )

( )

( )PTAEP

AELT

∆−=

=+∆

α

α 0


( )TEAP

∆−== ασ

fourth edition mechanics of materials - hacettepeyunus.hacettepe.edu.tr/~boray/2_1_axial_loading...

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