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MECHANICS OF MATERIALS
Sixth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
CHAPTER
2Stress and Strain
David F. Mazurek
Lecture Notes:
J. Walt Oler
Texas Tech University
© 2012 The McGraw-Hill Companies, Inc. All rights reserved.
– Axial Loading
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Stress & Strain: Axial Loading
• Suitability of a structure or machine may depend on the deformations in the 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.
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 2
• 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 under axial loading. Later chapters will deal with torsional and pure bending loads.
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Normal Strain
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 3
strain normal
stress
==
==
L
A
P
δε
σ
Fig. 2.1
L
A
P
A
P
δε
σ
=
==22
Fig. 2.3
LL
A
P
δδε
σ
==
=
22
Fig. 2.4
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Stress-Strain Test
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 4
Fig 2.7 This machine is used to test tensile test specimens, such as those shown in this chapter.
Fig 2.8 Test specimen with tensile load.
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Stress-Strain Diagram: Ductile Materials
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 5
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Stress-Strain Diagram: Brittle Materials
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 6
Fig 2.1 Stress-strain diagram for a typical brittle material.
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Hooke’s Law: Modulus of Elasticity
• Below the yield stress
Elasticity of Modulus or Modulus Youngs=
=E
Eεσ
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 7
• Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of Elasticity) is not.
Fig 2.16 Stress-strain diagrams for iron and different grades of steel.
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Elastic vs. Plastic Behavior
• 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.
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• When the strain does not return to zero after the stress is removed, the material is said to behave plastically.
occurs is called the elastic limit.
Fig. 2.18
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Fatigue
• Fatigue properties are shown on S-N diagrams.
• A member may fail due to fatigueat stress levels significantly below the ultimate strength if subjected
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 9
• When the stress is reduced below the endurance limit, fatigue failures do not occur for any number of cycles.
the ultimate strength if subjected to many loading cycles.
Fig. 2.21
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Factor of Safety
stress ultimate
safety ofFactor
u ==
=
σσ
FS
FS
Structural members or machines must be designed such that the working stresses are less than the ultimate strength of the material.
Factor of safety considerations:
• uncertainty in material properties • uncertainty of loadings• uncertainty of analyses• number of loading cycles• types of failure• maintenance requirements and
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 10
stress allowableall==
σFS • maintenance requirements and
deterioration effects• importance of member to integrity of
whole structure• risk to life and property• influence on machine function
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Deformations Under Axial Loading
AE
P
EE === σεεσ
• From Hooke’s Law:
• From the definition of strain:
L
δε =
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 11
• Equating and solving for the deformation,
AE
PL=δ
• With variations in loading, cross-section or material properties,
∑=i ii
iiEA
LPδFig. 2.22
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Example 2.01
psi1029 6
==
×= −E
SOLUTION:
• Divide the rod into components at the load application points.
• Apply a free-body analysis on each component to determine the internal force
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 12
Determine the total deformation of the steel rod shown under the given loads.
in. 618.0 in. 07.1 == dD internal force
• Evaluate the total of the component deflections.
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
SOLUTION:
• Divide the rod into three components:
• Apply free-body analysis to each component to determine internal forces,
lb1030
lb1015
lb1060
33
32
31
×=
×−=
×=
P
P
P
• Evaluate total deflection,
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 13
221
21
in 9.0
in. 12
==
==
AA
LL
23
3
in 3.0
in. 16
=
=
A
L
( ) ( ) ( )
in.109.75
3.0
161030
9.0
121015
9.0
121060
1029
1
1
3
333
6
3
33
2
22
1
11
−×=
×+×−+××
=
++=∑=
A
LP
A
LP
A
LP
EEA
LP
i ii
iiδ
in. 109.75 3−×=δ
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Sample Problem 2.1
The rigid bar BDE is supported by two links AB and CD.
SOLUTION:
• Apply a free-body analysis to the bar BDE to find the forces exerted by links AB and DC.
• Evaluate the deformation of links ABand DC or the displacements of Band D.
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 14
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 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.
and D.
• Work out the geometry to find the deflection at E given the deflections at B and D.
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Sample Problem 2.1
Free body: Bar BDE
SOLUTION: Displacement of B:
( )( )( )( )
m10514
Pa1070m10500
m3.0N1060
6
926-
3
−×−=
×××−=
=AE
PLBδ
↑= mm 514.0Bδ
© 2012 The McGraw-Hill Companies, Inc. All rights reserved. 3- 15
( )
( )ncompressioF
F
tensionF
F
M
AB
AB
CD
CD
B
kN60
m2.0m4.0kN300
0M
kN90
m2.0m6.0kN300
0
D
−=
×−×−=
=
+=
×+×−=
=
∑
∑↑= mm 514.0Bδ
Displacement of D:
( )( )( )( )
m10300
Pa10200m10600
m4.0N1090
6
926-
3
−×=
×××=
=AE
PLDδ
↓= mm 300.0Dδ
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MECHANICS OF MATERIALS
Sixth
Edition
Beer • Johnston • DeWolf • Mazurek
Displacement of D:
( )
mm 7.73
mm 200mm 0.300mm 514.0
=
−=
=′′
x
x
x
HD
BH
DD
BB
Sample Problem 2.1
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↓= mm 928.1Eδ
( )
mm 928.1
mm 7.73
mm7.73400
mm 300.0
=
+=
=′′
E
E
HD
HE
DD
EE
δ
δ