Fourth Edition
CHAPTER
11
MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Energy Methods
Lecture Notes:
J. Walt Oler
Texas Tech University
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Energy Methods
Strain Energy
Strain Energy Density
Elastic Strain Energy for Normal Stresses
Strain Energy For Shearing Stresses
Sample Problem 11.2
Strain Energy for a General State of Stress
Impact Loading
Example 11.06
Example 11.07
Design for Impact Loads
Work and Energy Under a Single Load
Deflection Under a Single Load
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
Sample Problem 11.4
Work and Energy Under Several Loads
Castigliano’s Theorem
Deflections by Castigliano’s Theorem
Sample Problem 11.5
11 - 2
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Strain Energy
• A uniform rod is subjected to a slowly increasing load
• The elementary work done by the load P as the rod
elongates by a small dx is
dU  P dx  elementary work
which is equal to the area of width dx under the loaddeformation diagram.
• The total work done by the load for a deformation x1,
x1
U   P dx  total work  strain energy
0
which results in an increase of strain energy in the rod.
• In the case of a linear elastic deformation,
x1
U   kx dx  12 kx12  12 P1x1
0
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11 - 3
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Strain Energy Density
• To eliminate the effects of size, evaluate the strainenergy per unit volume,
x
1
U
P dx
 
V 0A L
1
u    x d x  strain energy density
0
• The total strain energy density resulting from the
deformation is equal to the area under the curve to 1.
• As the material is unloaded, the stress returns to zero
but there is a permanent deformation. Only the strain
energy represented by the triangular area is recovered.
• Remainder of the energy spent in deforming the material
is dissipated as heat.
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11 - 4
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Strain-Energy Density
• The strain energy density resulting from
setting 1  R is the modulus of toughness.
• The energy per unit volume required to cause
the material to rupture is related to its ductility
as well as its ultimate strength.
• If the stress remains within the proportional
limit,
1
E12  12
u   E x d x 

2
2E
0
• The strain energy density resulting from
setting 1  Y is the modulus of resilience.
uY 
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 Y2
2E
 modulus of resilience
11 - 5
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Elastic Strain Energy for Normal Stresses
• In an element with a nonuniform stress distribution,
U dU

dV
V 0 V
u  lim
U   u dV  total strain energy
• For values of u < uY , i.e., below the proportional
limit,
U 
 x2
2E
dV  elastic strain energy
• Under axial loading,  x  P A
dV  A dx
L
P2
U 
dx
2 AE
0
• For a rod of uniform cross-section,
P2L
U
2 AE
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11 - 6
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Elastic Strain Energy for Normal Stresses
• For a beam subjected to a bending load,
U 
 x2
2E
dV  
M 2 y2
2 EI
2
dV
• Setting dV = dA dx,
M 2  2 
U  
dA dx  
y dA dx
2
2

2 EI
2 EI  A

0 A
0
L
x 
My
I
L

0
M 2 y2
L
M2
dx
2 EI
• For an end-loaded cantilever beam,
M   Px
L
P2 x2
P 2 L3
U 
dx 
2 EI
6 EI
0
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11 - 7
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Strain Energy For Shearing Stresses
• For a material subjected to plane shearing
stresses,
 xy
u
 xy d xy
0
• For values of xy within the proportional limit,
2

xy
2
u  12 G xy  12  xy  xy 
2G
• The total strain energy is found from
U   u dV

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
2
 xy
2G
dV
11 - 8
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Strain Energy For Shearing Stresses
• For a shaft subjected to a torsional load,
2
 xy
T 2 2
U 
2G
dV  
2GJ
2
dV
• Setting dV = dA dx,
T 2  2 
U  
dA dx  
 dA dx
2
2

2GJ
2GJ  A

0A
0
L
 xy 
T
J
T 2 2
L
L
T2

dx
2GJ
0
• In the case of a uniform shaft,
T 2L
U
2GJ
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11 - 9
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 11.2
SOLUTION:
• Determine the reactions at A and B
from a free-body diagram of the
complete beam.
• Develop a diagram of the bending
moment distribution.
a) Taking into account only the normal
stresses due to bending, determine the
strain energy of the beam for the
loading shown.
b) Evaluate the strain energy knowing
that the beam is a W10x45, P = 40
kips, L = 12 ft, a = 3 ft, b = 9 ft, and E
= 29x106 psi.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
• Integrate over the volume of the
beam to find the strain energy.
• Apply the particular given
conditions to evaluate the strain
energy.
11 - 10
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 11.2
SOLUTION:
• Determine the reactions at A and B
from a free-body diagram of the
complete beam.
RA 
Pb
L
RB 
Pa
L
• Develop a diagram of the bending
moment distribution.
M1 
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Pb
x
L
M2 
Pa
v
L
11 - 11
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 11.2
• Integrate over the volume of the beam to find
the strain energy.
a
b
0
0
M12
M 22
U 
dx  
dv
2 EI
2 EI
a
Over the portion AD,
0
Pb
M1 
x
L
Over the portion BD,
M2 
Pa
v
L
2
b
2
1  Pb 
1  Pa 

x
dx

x  dx





2 EI  L 
2 EI  L 
0
1 P 2  b 2a3 a 2b3  P 2a 2b 2
a  b 



2 EI L2  3
3  6 EIL2
P 2a 2b2
U
6 EIL
P  45 kips
L  144 in.
a  36 in.
b  108 in.

40 kips 2 36 in 2 108 in 2
U
629  103 ksi 248 in 4 144 in 
E  29  103 ksi
I  248 in 4
U  3.89 in  kips
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 12
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Strain Energy for a General State of Stress
• Previously found strain energy due to uniaxial stress and plane
shearing stress. For a general state of stress,

u  12  x x   y y   z z   xy xy   yz yz   zx zx

• With respect to the principal axes for an elastic, isotropic body,
u


1 2
 a   b2   c2  2  a b   b c   c a 
2E
 uv  ud
uv 
1  2v
 a   b   c 2  due to volume change
6E
ud 
1
 a   b 2   b   c 2   c   a 2  due to distortion
12G


• Basis for the maximum distortion energy failure criteria,
ud  ud Y 
 Y2
6G
for a tensile test specimen
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11 - 13
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Impact Loading
• To determine the maximum stress m
- Assume that the kinetic energy is
transferred entirely to the
structure,
U m  12 mv02
- Assume that the stress-strain
diagram obtained from a static test
is also valid under impact loading.
• Consider a rod which is hit at its
end with a body of mass m moving
with a velocity v0.
• Rod deforms under impact. Stresses
reach a maximum value m and then
disappear.
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• Maximum value of the strain energy,
Um  
2
m
2E
dV
• For the case of a uniform rod,
2U m E
mv02 E
m 

V
V
11 - 14
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 11.06
SOLUTION:
• Due to the change in diameter, the
normal stress distribution is nonuniform.
• Find the static load Pm which produces
the same strain energy as the impact.
• Evaluate the maximum stress
resulting from the static load Pm
Body of mass m with velocity v0 hits
the end of the nonuniform rod BCD.
Knowing that the diameter of the
portion BC is twice the diameter of
portion CD, determine the maximum
value of the normal stress in the rod.
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11 - 15
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 11.06
• Find the static load Pm which produces
the same strain energy as the impact.
Pm2 L 2 Pm2 L 2 5 Pm2 L
Um 


AE
4 AE
16 AE
Pm 
SOLUTION:
• Due to the change in diameter,
the normal stress distribution is
nonuniform.
• Evaluate the maximum stress resulting
from the static load Pm
m 

U m  12 mv02
2
2
m
m
V

dV 
 2E
16 U m AE
5 L
2E
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
Pm
A
16 U m E
5 AL
8 mv02 E

5 AL
11 - 16
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 11.07
SOLUTION:
• The normal stress varies linearly along
the length of the beam and across a
transverse section.
• Find the static load Pm which produces
the same strain energy as the impact.
• Evaluate the maximum stress
A block of weight W is dropped from a
resulting from the static load Pm
height h onto the free end of the
cantilever beam. Determine the
maximum value of the stresses in the
beam.
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11 - 17
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 11.07
• Find the static load Pm which produces
the same strain energy as the impact.
For an end-loaded cantilever beam,
Pm2 L3
Um 
6 EI
Pm 
SOLUTION:
• The normal stress varies linearly
along the length of the beam and
across a transverse section.

2E
dV 
L3
• Evaluate the maximum stress
resulting from the static load Pm
U m  Wh
2
m
6U m EI
2
m
V
2E
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
M m c Pm Lc
m 

I
I

6U m E
 

2
LI c
6WhE
 
L I c2
11 - 18
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Design for Impact Loads
• For the case of a uniform rod,
m 
2U m E
V
• For the case of the nonuniform rod,
m 
16 U m E
5 AL
V  4 AL / 2   AL / 2  5 AL / 2
m 
8U m E
V
• For the case of the cantilever beam
Maximum stress reduced by:
• uniformity of stress
• low modulus of elasticity with
high yield strength
• high volume
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
m 
6U m E
 
L I / c 2   L 14 c 4 / c 2   14 c 2 L   14 V
m 
L I c2
24U m E
V
11 - 19
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Work and Energy Under a Single Load
• Strain energy may also be found from
the work of the single load P1,
x1
U   P dx
0
• For an elastic deformation,
• Previously, we found the strain
energy by integrating the energy
density over the volume.
For a uniform rod,
U   u dV  
L

0

2
2E
dV
P1 A2 Adx 
2E
x1
x1
0
0
U   P dx   kx dx  12 k x12  12 P1x1
• Knowing the relationship between
force and displacement,
P12 L
2 AE
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PL
x1  1
AE
2
P
L
P
L


U  12 P1 1   1
 AE  2 AE
11 - 20
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Work and Energy Under a Single Load
• Strain energy may be found from the work of other types
of single concentrated loads.
• Transverse load
U

y1
1Py
P
dy


2 1 1
• Bending couple
1
U   M d  12 M11
0
0
2 3
 3
1 P  P1L   P1 L
2 1 3EI 
6 EI
1 M  M1L  
2 1 EI 



© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
• Torsional couple
1
U   T d  12 T11
0
M12 L
2 EI

2
1 T  T1L   T1 L
2 1 JG  2 JG
11 - 21
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Deflection Under a Single Load
• If the strain energy of a structure due to a
single concentrated load is known, then the
equality between the work of the load and
energy may be used to find the deflection.
• Strain energy of the structure,
2
2
FBC
LBC FBD
LBD
U

2 AE
2 AE

From the given geometry,
LBC  0.6 l
LBD  0.8l
From statics,
FBC  0.6P FBD  0.8P

P 2l 0.6 3  0.83
P 2l

 0.364
2 AE
AE
• Equating work and strain energy,
P2L 1
U  0.364
 P yB
AE 2
y B  0.728
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Pl
AE
11 - 22
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 11.4
SOLUTION:
• Find the reactions at A and B from a
free-body diagram of the entire truss.
• Apply the method of joints to
determine the axial force in each
member.
Members of the truss shown consist of
sections of aluminum pipe with the
cross-sectional areas indicated. Using
E = 73 GPa, determine the vertical
deflection of the point E caused by the
load P.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
• Evaluate the strain energy of the
truss due to the load P.
• Equate the strain energy to the work
of P and solve for the displacement.
11 - 23
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 11.4
SOLUTION:
• Find the reactions at A and B from a free-body
diagram of the entire truss.
Ax  21 P 8
Ay  P
B  21 P 8
• Apply the method of joints to determine the
axial force in each member.
FDE   17
P
8
FAC   15
P
8
FDE  54 P
FCE   15
P
8
FCD  0
FCE   21
P
8
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FAB  0
11 - 24
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 11.4
• Evaluate the strain energy of the
truss due to the load P.
Fi2 Li
1
Fi2 Li
U 

 Ai
2 Ai E 2 E


1
29700 P 2
2E

• Equate the strain energy to the work by P
and solve for the displacement.
1 Py  U
E
2
2U 2  29700 P 2 
yE 

P
P  2 E 

29.7  103 40  103 
yE 
73  10
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
9
yE  16.27mm 
11 - 25
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Work and Energy Under Several Loads
• Deflections of an elastic beam subjected to two
concentrated loads,
x1  x11  x12  11P1  12 P2
x2  x21  x22   21P1   22 P2
• Compute the strain energy in the beam by
evaluating the work done by slowly applying
P1 followed by P2,

U  12 11P12  212 P1P2   22 P22

• Reversing the application sequence yields

U  12  22 P22  2 21P2 P1  11P12

• Strain energy expressions must be equivalent.
It follows that 1221 (Maxwell’s reciprocal
theorem).
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11 - 26
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Castigliano’s Theorem
• Strain energy for any elastic structure
subjected to two concentrated loads,

U  12 11P12  212 P1P2   22 P22

• Differentiating with respect to the loads,
U
 11P1  12 P2  x1
P1
U
 12 P1   22 P2  x2
P2
• Castigliano’s theorem: For an elastic structure
subjected to n loads, the deflection xj of the
point of application of Pj can be expressed as
xj 
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
U
Pj
and  j 
U
M j
j 
U
T j
11 - 27
Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Deflections by Castigliano’s Theorem
• Application of Castigliano’s theorem is
simplified if the differentiation with respect to
the load Pj is performed before the integration
or summation to obtain the strain energy U.
• In the case of a beam,
L
M2
U 
dx
2 EI
0
L
U
M M
xj 

dx
Pj
EI Pj
0
• For a truss,
n
Fi2 Li
U 
2A E
i 1 i
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
n
U
F L F
xj 
 i i i
Pj i 1 Ai E Pj
11 - 28
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 11.5
SOLUTION:
• For application of Castigliano’s theorem,
introduce a dummy vertical load Q at C.
Find the reactions at A and B due to the
dummy load from a free-body diagram of
the entire truss.
Members of the truss shown
consist of sections of aluminum
pipe with the cross-sectional areas
indicated. Using E = 73 GPa,
determine the vertical deflection of
the joint C caused by the load P.
• Apply the method of joints to determine
the axial force in each member due to Q.
• Combine with the results of Sample
Problem 11.4 to evaluate the derivative
with respect to Q of the strain energy of
the truss due to the loads P and Q.
• Setting Q = 0, evaluate the derivative
which is equivalent to the desired
displacement at C.
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11 - 29
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 11.5
SOLUTION:
• Find the reactions at A and B due to a dummy load Q
at C from a free-body diagram of the entire truss.
Ax   34 Q
Ay  Q
B  34 Q
• Apply the method of joints to determine the axial
force in each member due to Q.
FCE  FDE  0
FAC  0; FCD  Q
FAB  0; FBD   34 Q
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11 - 30
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 11.5
• Combine with the results of Sample Problem 11.4 to evaluate the derivative
with respect to Q of the strain energy of the truss due to the loads P and Q.
 F L  F 1
yC    i i  i  4306 P  4263Q 
 Ai E  Q E
• Setting Q = 0, evaluate the derivative which is equivalent to the desired
displacement at C.
yC 

4306 40  103 N
9
73  10 Pa

yC  2.36 mm 
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
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