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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.
MECHANICS OF MATERIALS
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
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 3
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Normal Strain
stress==AP
δ
σAP
AP
δ
σ ==22
AP
δδ
σ =
2
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 4
strain normal==Lδε
Lδε =
LLδδε ==
22
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Stress-Strain Test
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 5
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Stress-Strain Diagram: Ductile Materials
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 6
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Stress-Strain Diagram: Brittle Materials
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 7
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 8
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 9
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 10
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 11
i ii
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 12
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 13
221 in 9.0== AA 2
3 in 3.0=A
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 14
deflection a) of B, b) of D, and c) of E.
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
−=
×−×−=
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 15
↓= mm 300.0Dδ
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 16
↓9 8.Eδ
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 17
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 18
• Solve for the reaction at A due to applied loads and the reaction found at B.
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 19
( )∑
×−==
i
B
ii
iiR E
REALPδ
31095.1
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 20
B
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 21
( )TEAP
∆−== ασ