Seventh Edition in SI Units

CHAPTER MECHANICS OF
MATERIALS
2 Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf Stress and Strain
David F. Mazurek
Sanjeev Sanghi
– Axial Loading
Lecture Notes:
Brock E. Barry
U.S. Military Academy
Sanjeev Sanghi
Indian Institute of Technology Delhi

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MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek • Sanghi

Contents

Stress & Strain: Axial Loading Multiaxial Loading: Generalized

Normal Strain Hooke’s Law
Stress-Strain Test Dilatation: Bulk Modulus
Stress-Strain Diagram: Ductile Materials Shearing Strain
Stress-Strain Diagram: Brittle Materials Concept Application 2.10
Hooke’s Law: Modulus of Elasticity Relation Among E, n, and G
Elastic vs. Plastic Behavior Composite Materials
Fatigue Sample Problem 2.5
Deformations Under Axial Loading Saint-Venant’s Principle
Concept Application 2.1 Stress Concentration: Hole
Sample Problem 2.1 Stress Concentration: Fillet
Static Indeterminate Problems Concept Application 2.12
Concept Application 2.4 Elastoplastic Materials
Problems Involving Temperature Change Plastic Deformations
Poisson’s Ratio Residual Stresses
Concept Applications 2.14, 2.15, 2.16
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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.
• Determination of the stress distribution
within a member also requires consideration
of deformations in the members.
• Chapter 2 in concerned with deformation of
a structural member under axial loading.
Later chapters will deal with torsional and
pure bending loads.

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Normal Strain

Fig. 2.1 Undeformed Fig. 2.3 Twice the load is

and deformed axially required to obtain the same
loaded rod. deformation d when the
cross-sectional area is
doubled.

Fig. 2.4 The deformation is

doubled when the rod length
is doubled while keeping the
load P and cross-sectional
area A.
2P P
P s= =
s= = stress 2A A P
s=
A d A
e=
d L 2d d
e= = normal strain e= =
L 2L L
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Stress-Strain Test

Photo 2.2 Universal test machine used to test Photo 2.3 Elongated tensile test
tensile specimens. specimen having load P and deformed
length L > L0.
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Stress-Strain Diagram: Ductile Materials

Fig. 2.6 Stress-strain diagrams of two typical ductile materials.

Photo 2.4 Ductile material tested

specimens: (a) with cross-section
necking, (b) ruptured.

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Stress-Strain Diagram: Ductile Materials

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Stress-Strain Diagram: Brittle Materials

Fig 2.7 Stress-strain diagram for a typical brittle material.

Photo 2.5 Ruptured brittle materials specimen.

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Stress-Strain Diagram: Brittle Materials

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Hooke’s Law: Modulus of Elasticity

• Below the yield stress

s = Ee
E = Youngs Modulus or
Modulus of Elasticity

• Strength is affected by alloying,

heat treating, and manufacturing
process but stiffness (Modulus of
Elasticity) is not.
Fig 2.11 Stress-strain diagrams for iron and
different grades of steel. Movie: Structure of Metals

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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.

• When the strain does not return

to zero after the stress is
removed, plastic deformation
Fig. 2.13 Stress-strain response of ductile
of the material has taken place.
material load beyond yield and unloaded.

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Fatigue

• Fatigue properties are shown on

s-N diagrams.

• A member may fail due to fatigue

at stress levels significantly below
the ultimate strength if subjected
to many loading cycles.

• When the stress is reduced below

the endurance limit, fatigue
failures do not occur for any
Fig. 2.16 Typical s-n curves.
number of cycles.

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Deformations Under Axial Loading

• From Hooke’s Law:

s P
s = Ee e= =
E AE

• From the definition of strain:

d
e=
L
• Equating and solving for the deformation,
PL
d=
AE

• With variations in loading, cross-section or

material properties,
PL
Fig. 2.17 Undeformed and
deformed axially-loaded rod.
d =å i i
Ai Ei
i

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Concept Application 2.1

SOLUTION:
• Divide the rod into components at
the load application points.

• Apply a free-body analysis on each

E = 200 GPa component to determine the
internal force

• Evaluate the total of the component

Determine the deformation of deflections.
the steel rod shown under the
given loads.

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Concept Application 2.1

SOLUTION: • Apply free-body analysis to each
• Divide the rod into three component to determine internal forces,
components: P1 = 240 ´ 103 N
P2 = -60 ´ 103 N
P3 = 120 ´ 103 N

• Evaluate total deflection,

PL 1 æPL PL PL ö
d =å i i = ç 1 1 + 2 2 + 3 3÷
Ai Ei E è A1 A2 A3 ø
i

=
1 ê
ê
( )
é 240 ´ 103 (0.3)
+
( ) ( )
-60 ´ 103 (0.3) 120 ´ 103 (0.4) ùú
+ ú
29 ´ 106 ê 580 ´ 10 -6 580 ´ 10 -6 190 ´ 10 -6 ú
ë û
= 1.729 ´ 10 -3 m

L1 = L2 = 0.3 m L3 = 0.4 m
A1 = A2 = 580 ´ 10 -6 m 2 A3 = 190 ´ 10 -3 m 2 d = 1.729 mm

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Sample Problem 2.1

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 AB
The rigid bar BDE is supported by two and DC or the displacements of B
links AB and CD. and D.

Link AB is made of aluminum (E = 70 • Work out the geometry to find the

GPa) and has a cross-sectional area of 500 deflection at E given the deflections
mm2. Link CD is made of steel (E = 200 at B and D.
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.

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Sample Problem 2.1

SOLUTION: Displacement of B:
Free body: Bar BDE PL
dB =
AE

=
( -60 ´ 10 N ) ( 0.3m )
3

(500 ´ 10 m )(70 ´ 10 Pa )
-6 2 9

= -514 ´ 10 -6 m

d B = 0.514 mm ­
å MB = 0 Displacement of D:
0 = - ( 30 kN)(0.6 m ) + FCD (0.2 m)
PL
FCD = +90 kN FCD = 90 kN tension dD =
AE
å MD = 0 (90 ´ 10 N) (0.4 m)
3
=
0 = - ( 30 kN ´ 0.4 m ) - FAB ´ 0.2 m (600 ´ 10 m )(200 ´ 10 Pa )
-6 2 9

FAB = -60 kN FAB = 60 kN compression = 300 ´ 10 -6 m

d D = 0.300 mm ¯

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Sample Problem 2.1

Displacement of D:

BB ¢ BH
=
DD ¢ HD
0.514 mm ( 200 mm ) - x
=
0.300 mm x
x = 73.7 mm

EE ¢ HE
=
DD ¢ HD
dE
=
( 400 + 73.7) mm
0.300 mm 73.7 mm
d E = 1.928 mm

d E = 1.928 mm ¯

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Static Indeterminate Problems

• 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.
d = dL + dR = 0
Fig. 2.23

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Concept Application 2.4

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:
• 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 RB.

• Require that the displacements due to the loads

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

• Solve for the reaction at RA due to applied loads

and the reaction found at RB.
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Concept Application 2.4

SOLUTION:
• Solve for the displacement at B due to the applied
loads with the redundant constraint released,
P1 = 0 P2 = P3 = 600 ´ 103 N P4 = 900 ´ 103 N
A1 = A2 = 400 ´ 10 -6 m 2 A3 = A4 = 250 ´ 10 -6 m 2
L1 = L2 = L3 = L4 = 0.150 m
Pi Li 1.125 ´ 109
dL = å =
i Ai Ei E

• Solve for the displacement at B due to the redundant

constraint,
P1 = P2 = - RB
A1 = 400 ´ 10 -6 m 2 A2 = 250 ´ 10 -6 m 2
L1 = L2 = 0.300 m

dR = å
Pi Li
=-
(
1.95 ´ 103 RB )
i A E
i i E
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Concept Application 2.4

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

d = dL + dR = 0

d=
(
3
1.125 ´ 109 1.95 ´ 10 RB
- =0
)
E E
RB = 577 ´ 103 N = 577 kN

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

å Fy = 0 = RA - 300 kN - 600 kN + 577 kN

RA = 323 kN

RA = 323 kN
RB = 577 kN

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Problems Involving Temperature Change
• A temperature change results in a change in length or
thermal strain. There is no stress associated with the
Fig. 2.26 (partial)
thermal strain unless the elongation is restrained by
the supports.
• Treat the additional support as redundant and apply
the principle of superposition.
PL
d T = a ( DT ) L dP =
AE
a = coefficient of thermal expansion

• The thermal deformation and the deformation from

the redundant support must be compatible.
d = dT + d P = 0 PL
a ( DT ) L + =0
AE
P = - AEa ( DT )
Fig. 2.27 Superposition method to find force at point B of restrained P
rod AB undergoing thermal expansion. (a) Initial rod length; (b) s= = - Ea ( DT )
thermally expanded rod length; (c) force P pushes point B back to A
zero deformation.
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We will have a ZOOM online class on March 9th.

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