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MECHANICS OF
MATERIALS
Fourth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Lecture Notes:
J. Walt Oler
Texas Tech University
CHAPTER
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
11 Energy Methods
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
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
Sample Problem 11.4
Work and Energy Under Several Loads
Castigliano’s Theorem
Deflections by Castigliano’s Theorem
Sample Problem 11.5
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
• 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
which is equal to the area of width dx under the load-
deformation diagram.
workelementarydxPdU ==
• The total work done by the load for a deformation x1,
which results in an increase of strain energy in the rod.
energystrainworktotaldxPU
x
=== ∫
1
0
112
12
12
1
0
1
xPkxdxkxU
x
=== ∫
• In the case of a linear elastic deformation,
Strain Energy
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Strain Energy Density
• To eliminate the effects of size, evaluate the strain-
energy per unit volume,
densityenergystraindu
L
dx
A
P
V
U
xx
x
==
=
∫
∫
1
1
0
0
ε
εσ
• 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.
• The total strain energy density resulting from the
deformation is equal to the area under the curve to ε1.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Strain-Energy Density
• The strain energy density resulting from
setting ε1 = εR 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,
E
E
dEu xx
22
2
1
2
1
0
1
σε
εε
ε
=== ∫
• The strain energy density resulting from
setting σ1 = σY is the modulus of resilience.
resilienceofmodulus
E
u Y
Y ==
2
2
σ
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Elastic Strain Energy for Normal Stresses
• In an element with a nonuniform stress distribution,
energystraintotallim
0
===
∆
∆
= ∫→∆
dVuU
dV
dU
V
U
u
V
• For values of u < uY , i.e., below the proportional
limit,
energystrainelasticdV
E
U x
2
2
∫ ==
σ
• Under axial loading, dxAdVAPx ==σ
∫=
L
dx
AE
P
U
0
2
2
AE
LP
U
2
2
=
• For a rod of uniform cross-section,
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Elastic Strain Energy for Normal Stresses
I
yM
x =σ
• For a beam subjected to a bending load,
∫∫ == dV
EI
yM
dV
E
U x
2
222
22
σ
• Setting dV = dA dx,
dx
EI
M
dxdAy
EI
M
dxdA
EI
yM
U
L
L
A
L
A
∫
∫ ∫∫ ∫
=








==
0
2
0
2
2
2
0
2
22
2
22
• For an end-loaded cantilever beam,
EI
LP
dx
EI
xP
U
PxM
L
62
32
0
22
==
−=
∫
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Strain Energy For Shearing Stresses
• For a material subjected to plane shearing
stresses,
∫=
xy
xyxy du
γ
γτ
0
• For values of τxy within the proportional limit,
G
Gu
xy
xyxyxy
2
2
2
12
2
1
τ
γτγ ===
• The total strain energy is found from
∫
∫
=
=
dV
G
dVuU
xy
2
2
τ
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Strain Energy For Shearing Stresses
J
T
xy
ρ
τ =
∫∫ == dV
GJ
T
dV
G
U
xy
2
222
22
ρτ
• For a shaft subjected to a torsional load,
• Setting dV = dA dx,
∫
∫ ∫∫∫
=








==
L
L
A
L
A
dx
GJ
T
dxdA
GJ
T
dxdA
GJ
T
U
0
2
0
2
2
2
0
2
22
2
22
ρ
ρ
• In the case of a uniform shaft,
GJ
LT
U
2
2
=
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Sample Problem 11.2
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.
SOLUTION:
• Determine the reactions at A and B
from a free-body diagram of the
complete beam.
• Integrate over the volume of the
beam to find the strain energy.
• Apply the particular given
conditions to evaluate the strain
energy.
• Develop a diagram of the bending
moment distribution.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Sample Problem 11.2
SOLUTION:
• Determine the reactions at A and B
from a free-body diagram of the
complete beam.
L
Pa
R
L
Pb
R BA ==
• Develop a diagram of the bending
moment distribution.
v
L
Pa
Mx
L
Pb
M == 21
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Sample Problem 11.2
v
L
Pa
M
x
L
Pb
M
=
=
2
1
BD,portionOver the
AD,portionOver the
43
in248ksi1029
in.108in.36a
in.144kips45
=×=
==
==
IE
b
LP
• Integrate over the volume of the beam to find
the strain energy.
( )ba
EIL
baPbaab
L
P
EI
dxx
L
Pa
EI
dxx
L
Pb
EI
dv
EI
M
dx
EI
M
U
ba
ba
+=








+=






+





=
+=
∫∫
∫∫
2
2223232
2
2
0
2
0
2
0
2
2
0
2
1
6332
1
2
1
2
1
22
EIL
baP
U
6
222
=
( ) ( ) ( )
( )( )( )in144in248ksi10296
in108in36kips40
43
222
×
=U
kipsin89.3 ⋅=U
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
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,
( )zxzxyzyzxyxyzzyyxxu γτγτγτεσεσεσ +++++= 2
1
• With respect to the principal axes for an elastic, isotropic body,
( )[ ]
( )
( ) ( ) ( )[ ] distortiontodue
12
1
changevolumetodue
6
21
2
2
1
222
2
222
=−+−+−=
=++
−
=
+=
++−++=
accbbad
cbav
dv
accbbacba
G
u
E
v
u
uu
E
u
σσσσσσ
σσσ
σσσσσσνσσσ
• Basis for the maximum distortion energy failure criteria,
( ) specimentesttensileafor
6
2
G
uu Y
Ydd
σ
=<
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
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 σm and then
disappear.
• To determine the maximum stress σm
- Assume that the kinetic energy is
transferred entirely to the
structure,
2
02
1 mvUm =
- Assume that the stress-strain
diagram obtained from a static test
is also valid under impact loading.
∫= dV
E
U m
m
2
2
σ
• Maximum value of the strain energy,
• For the case of a uniform rod,
V
Emv
V
EUm
m
2
02
==σ
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Example 11.06
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.
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
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Example 11.06
SOLUTION:
• Due to the change in diameter,
the normal stress distribution is
nonuniform.
E
V
dV
E
mvU
mm
m
22
22
2
02
1
σσ
≠=
=
∫
• Find the static load Pm which produces
the same strain energy as the impact.
( ) ( )
L
AEU
P
AE
LP
AE
LP
AE
LP
U
m
m
mmm
m
5
16
16
5
4
22 222
=
=+=
• Evaluate the maximum stress resulting
from the static load Pm
AL
Emv
AL
EU
A
P
m
m
m
2
0
5
8
5
16
=
=
=σ
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Example 11.07
A block of weight W is dropped from a
height h onto the free end of the
cantilever beam. Determine the
maximum value of the stresses in the
beam.
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
resulting from the static load Pm
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Example 11.07
SOLUTION:
• The normal stress varies linearly
along the length of the beam and
across a transverse section.
E
V
dV
E
WhU
mm
m
22
22
σσ
≠=
=
∫
• Find the static load Pm which produces
the same strain energy as the impact.
For an end-loaded cantilever beam,
3
32
6
6
L
EIU
P
EI
LP
U
m
m
m
m
=
=
• Evaluate the maximum stress
resulting from the static load Pm
( ) ( )22
66
cIL
WhE
cIL
EU
I
LcP
I
cM
m
mm
m
==
==σ
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Design for Impact Loads
• For the case of a uniform rod,
V
EUm
m
2
=σ
( )
( ) ( ) ( )
V
EU
VLcccLcIL
cIL
EU
m
m
m
m
24
//
6
4
12
4
124
4
12
2
=
===
=
σ
ππ
σ
• For the case of the cantilever beam
Maximum stress reduced by:
• uniformity of stress
• low modulus of elasticity with
high yield strength
• high volume
• For the case of the nonuniform rod,
( ) ( )
V
EU
ALLALAV
AL
EU
m
m
m
m
8
2/52/2/4
5
16
=
=+=
=
σ
σ
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Work and Energy Under a Single Load
• Previously, we found the strain
energy by integrating the energy
density over the volume.
For a uniform rod,
( )
AE
LP
dxA
E
AP
dV
E
dVuU
L
22
2
2
1
0
2
1
2
==
==
∫
∫ ∫
σ
• Strain energy may also be found from
the work of the single load P1,
∫=
1
0
x
dxPU
• For an elastic deformation,
112
12
12
1
00
11
xPxkdxkxdxPU
xx
==== ∫∫
• Knowing the relationship between
force and displacement,
AE
LP
AE
LP
PU
AE
LP
x
2
2
11
12
1
1
1
=





=
=
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
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.
EI
LP
EI
LP
P
yPdyPU
y
63
32
1
3
1
12
1
112
1
0
1
=








=
== ∫
• Transverse load
EI
LM
EI
LM
M
MdMU
2
2
11
12
1
112
1
0
1
=





=
== ∫ θθ
θ
• Bending couple
JG
LT
JG
LT
T
TdTU
2
2
11
12
1
112
1
0
1
=





=
== ∫ φφ
φ
• Torsional couple
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
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.
lLlL BDBC 8.06.0 ==
From statics,
PFPF BDBC 8.06.0 −=+=
From the given geometry,
• Strain energy of the structure,
( ) ( )[ ]
AE
lP
AE
lP
AE
LF
AE
LF
U BDBDBCBC
2332
22
364.0
2
8.06.0
22
=
+
=
+=
• Equating work and strain energy,
AE
Pl
y
yP
AE
LP
U
B
B
728.0
364.0 2
1
2
=
==
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Sample Problem 11.4
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.
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.
• 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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Sample Problem 11.4
SOLUTION:
• Find the reactions at A and B from a free-
body diagram of the entire truss.
821821 PBPAPA yx ==−=
• Apply the method of joints to determine
the axial force in each member.
PF
PF
CE
DE
8
15
8
17
+=
−=
0
8
15
=
+=
CD
AC
F
PF
PF
PF
CE
DE
8
21
4
5
−=
= 0=ABF
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Sample Problem 11.4
• Evaluate the strain energy of the
truss due to the load P.
( )2
22
29700
2
1
2
1
2
P
E
A
LF
EEA
LF
U
i
ii
i
ii
=
== ∑∑
• Equate the strain energy to the work by P
and solve for the displacement.
( )( )
9
33
2
2
1
1073
1040107.29
2
2970022
×
××
=








==
=
E
E
E
y
E
P
PP
U
y
UPy
↓= mm27.16Ey
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Work and Energy Under Several Loads
• Deflections of an elastic beam subjected to two
concentrated loads,
22212122212
21211112111
PPxxx
PPxxx
αα
αα
+=+=
+=+=
• Reversing the application sequence yields
( )2
1111221
2
2222
1 2 PPPPU ααα ++=
• Strain energy expressions must be equivalent.
It follows that α12=α21 (Maxwell’s reciprocal
theorem).
( )2
2222112
2
1112
1 2 PPPPU ααα ++=
• Compute the strain energy in the beam by
evaluating the work done by slowly applying
P1 followed by P2,
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Castigliano’s Theorem
( )2
2222112
2
1112
1 2 PPPPU ααα ++=
• Strain energy for any elastic structure
subjected to two concentrated loads,
• Differentiating with respect to the loads,
2222112
2
1212111
1
xPP
P
U
xPP
P
U
=+=
∂
∂
=+=
∂
∂
αα
αα
• 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
and
j
j
j
j
j
j
T
U
M
U
P
U
x
∂
∂
=
∂
∂
=
∂
∂
= φθ
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
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
jj
j
L
dx
P
M
EI
M
P
U
xdx
EI
M
U
00
2
2
• For a truss,
j
i
n
i i
ii
j
j
n
i i
ii
P
F
EA
LF
P
U
x
EA
LF
U
∂
∂
=
∂
∂
== ∑∑
== 11
2
2
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
Beer • Johnston • DeWolf
Sample Problem 11.5
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.
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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
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.
QBQAQA yx 4
3
4
3 ==−=
• Apply the method of joints to determine the axial
force in each member due to Q.
QFF
QFF
FF
BDAB
CDAC
DECE
4
3;0
;0
0
−==
−==
==
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFourth
Edition
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.
( )QP
EQ
F
EA
LF
y i
i
ii
C 42634306
1
+=
∂
∂






= ∑
• Setting Q = 0, evaluate the derivative which is equivalent to the desired
displacement at C.
( )
Pa1073
10404306
9
3
×
×
=
N
yC ↓= mm36.2Cy

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11 energy methods- Mechanics of Materials - 4th - Beer

  • 1. MECHANICS OF MATERIALS Fourth Edition Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University CHAPTER © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 11 Energy Methods
  • 2. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition 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 Sample Problem 11.4 Work and Energy Under Several Loads Castigliano’s Theorem Deflections by Castigliano’s Theorem Sample Problem 11.5
  • 3. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf • 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 which is equal to the area of width dx under the load- deformation diagram. workelementarydxPdU == • The total work done by the load for a deformation x1, which results in an increase of strain energy in the rod. energystrainworktotaldxPU x === ∫ 1 0 112 12 12 1 0 1 xPkxdxkxU x === ∫ • In the case of a linear elastic deformation, Strain Energy
  • 4. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Strain Energy Density • To eliminate the effects of size, evaluate the strain- energy per unit volume, densityenergystraindu L dx A P V U xx x == = ∫ ∫ 1 1 0 0 ε εσ • 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. • The total strain energy density resulting from the deformation is equal to the area under the curve to ε1.
  • 5. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Strain-Energy Density • The strain energy density resulting from setting ε1 = εR 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, E E dEu xx 22 2 1 2 1 0 1 σε εε ε === ∫ • The strain energy density resulting from setting σ1 = σY is the modulus of resilience. resilienceofmodulus E u Y Y == 2 2 σ
  • 6. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Elastic Strain Energy for Normal Stresses • In an element with a nonuniform stress distribution, energystraintotallim 0 === ∆ ∆ = ∫→∆ dVuU dV dU V U u V • For values of u < uY , i.e., below the proportional limit, energystrainelasticdV E U x 2 2 ∫ == σ • Under axial loading, dxAdVAPx ==σ ∫= L dx AE P U 0 2 2 AE LP U 2 2 = • For a rod of uniform cross-section,
  • 7. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Elastic Strain Energy for Normal Stresses I yM x =σ • For a beam subjected to a bending load, ∫∫ == dV EI yM dV E U x 2 222 22 σ • Setting dV = dA dx, dx EI M dxdAy EI M dxdA EI yM U L L A L A ∫ ∫ ∫∫ ∫ =         == 0 2 0 2 2 2 0 2 22 2 22 • For an end-loaded cantilever beam, EI LP dx EI xP U PxM L 62 32 0 22 == −= ∫
  • 8. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Strain Energy For Shearing Stresses • For a material subjected to plane shearing stresses, ∫= xy xyxy du γ γτ 0 • For values of τxy within the proportional limit, G Gu xy xyxyxy 2 2 2 12 2 1 τ γτγ === • The total strain energy is found from ∫ ∫ = = dV G dVuU xy 2 2 τ
  • 9. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Strain Energy For Shearing Stresses J T xy ρ τ = ∫∫ == dV GJ T dV G U xy 2 222 22 ρτ • For a shaft subjected to a torsional load, • Setting dV = dA dx, ∫ ∫ ∫∫∫ =         == L L A L A dx GJ T dxdA GJ T dxdA GJ T U 0 2 0 2 2 2 0 2 22 2 22 ρ ρ • In the case of a uniform shaft, GJ LT U 2 2 =
  • 10. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Sample Problem 11.2 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. SOLUTION: • Determine the reactions at A and B from a free-body diagram of the complete beam. • Integrate over the volume of the beam to find the strain energy. • Apply the particular given conditions to evaluate the strain energy. • Develop a diagram of the bending moment distribution.
  • 11. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Sample Problem 11.2 SOLUTION: • Determine the reactions at A and B from a free-body diagram of the complete beam. L Pa R L Pb R BA == • Develop a diagram of the bending moment distribution. v L Pa Mx L Pb M == 21
  • 12. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Sample Problem 11.2 v L Pa M x L Pb M = = 2 1 BD,portionOver the AD,portionOver the 43 in248ksi1029 in.108in.36a in.144kips45 =×= == == IE b LP • Integrate over the volume of the beam to find the strain energy. ( )ba EIL baPbaab L P EI dxx L Pa EI dxx L Pb EI dv EI M dx EI M U ba ba +=         +=       +      = += ∫∫ ∫∫ 2 2223232 2 2 0 2 0 2 0 2 2 0 2 1 6332 1 2 1 2 1 22 EIL baP U 6 222 = ( ) ( ) ( ) ( )( )( )in144in248ksi10296 in108in36kips40 43 222 × =U kipsin89.3 ⋅=U
  • 13. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition 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, ( )zxzxyzyzxyxyzzyyxxu γτγτγτεσεσεσ +++++= 2 1 • With respect to the principal axes for an elastic, isotropic body, ( )[ ] ( ) ( ) ( ) ( )[ ] distortiontodue 12 1 changevolumetodue 6 21 2 2 1 222 2 222 =−+−+−= =++ − = += ++−++= accbbad cbav dv accbbacba G u E v u uu E u σσσσσσ σσσ σσσσσσνσσσ • Basis for the maximum distortion energy failure criteria, ( ) specimentesttensileafor 6 2 G uu Y Ydd σ =<
  • 14. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 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 σm and then disappear. • To determine the maximum stress σm - Assume that the kinetic energy is transferred entirely to the structure, 2 02 1 mvUm = - Assume that the stress-strain diagram obtained from a static test is also valid under impact loading. ∫= dV E U m m 2 2 σ • Maximum value of the strain energy, • For the case of a uniform rod, V Emv V EUm m 2 02 ==σ
  • 15. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Example 11.06 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. 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
  • 16. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Example 11.06 SOLUTION: • Due to the change in diameter, the normal stress distribution is nonuniform. E V dV E mvU mm m 22 22 2 02 1 σσ ≠= = ∫ • Find the static load Pm which produces the same strain energy as the impact. ( ) ( ) L AEU P AE LP AE LP AE LP U m m mmm m 5 16 16 5 4 22 222 = =+= • Evaluate the maximum stress resulting from the static load Pm AL Emv AL EU A P m m m 2 0 5 8 5 16 = = =σ
  • 17. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Example 11.07 A block of weight W is dropped from a height h onto the free end of the cantilever beam. Determine the maximum value of the stresses in the beam. 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 resulting from the static load Pm
  • 18. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Example 11.07 SOLUTION: • The normal stress varies linearly along the length of the beam and across a transverse section. E V dV E WhU mm m 22 22 σσ ≠= = ∫ • Find the static load Pm which produces the same strain energy as the impact. For an end-loaded cantilever beam, 3 32 6 6 L EIU P EI LP U m m m m = = • Evaluate the maximum stress resulting from the static load Pm ( ) ( )22 66 cIL WhE cIL EU I LcP I cM m mm m == ==σ
  • 19. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Design for Impact Loads • For the case of a uniform rod, V EUm m 2 =σ ( ) ( ) ( ) ( ) V EU VLcccLcIL cIL EU m m m m 24 // 6 4 12 4 124 4 12 2 = === = σ ππ σ • For the case of the cantilever beam Maximum stress reduced by: • uniformity of stress • low modulus of elasticity with high yield strength • high volume • For the case of the nonuniform rod, ( ) ( ) V EU ALLALAV AL EU m m m m 8 2/52/2/4 5 16 = =+= = σ σ
  • 20. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Work and Energy Under a Single Load • Previously, we found the strain energy by integrating the energy density over the volume. For a uniform rod, ( ) AE LP dxA E AP dV E dVuU L 22 2 2 1 0 2 1 2 == == ∫ ∫ ∫ σ • Strain energy may also be found from the work of the single load P1, ∫= 1 0 x dxPU • For an elastic deformation, 112 12 12 1 00 11 xPxkdxkxdxPU xx ==== ∫∫ • Knowing the relationship between force and displacement, AE LP AE LP PU AE LP x 2 2 11 12 1 1 1 =      = =
  • 21. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition 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. EI LP EI LP P yPdyPU y 63 32 1 3 1 12 1 112 1 0 1 =         = == ∫ • Transverse load EI LM EI LM M MdMU 2 2 11 12 1 112 1 0 1 =      = == ∫ θθ θ • Bending couple JG LT JG LT T TdTU 2 2 11 12 1 112 1 0 1 =      = == ∫ φφ φ • Torsional couple
  • 22. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition 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. lLlL BDBC 8.06.0 == From statics, PFPF BDBC 8.06.0 −=+= From the given geometry, • Strain energy of the structure, ( ) ( )[ ] AE lP AE lP AE LF AE LF U BDBDBCBC 2332 22 364.0 2 8.06.0 22 = + = += • Equating work and strain energy, AE Pl y yP AE LP U B B 728.0 364.0 2 1 2 = ==
  • 23. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Sample Problem 11.4 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. 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. • 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.
  • 24. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Sample Problem 11.4 SOLUTION: • Find the reactions at A and B from a free- body diagram of the entire truss. 821821 PBPAPA yx ==−= • Apply the method of joints to determine the axial force in each member. PF PF CE DE 8 15 8 17 += −= 0 8 15 = += CD AC F PF PF PF CE DE 8 21 4 5 −= = 0=ABF
  • 25. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Sample Problem 11.4 • Evaluate the strain energy of the truss due to the load P. ( )2 22 29700 2 1 2 1 2 P E A LF EEA LF U i ii i ii = == ∑∑ • Equate the strain energy to the work by P and solve for the displacement. ( )( ) 9 33 2 2 1 1073 1040107.29 2 2970022 × ×× =         == = E E E y E P PP U y UPy ↓= mm27.16Ey
  • 26. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Work and Energy Under Several Loads • Deflections of an elastic beam subjected to two concentrated loads, 22212122212 21211112111 PPxxx PPxxx αα αα +=+= +=+= • Reversing the application sequence yields ( )2 1111221 2 2222 1 2 PPPPU ααα ++= • Strain energy expressions must be equivalent. It follows that α12=α21 (Maxwell’s reciprocal theorem). ( )2 2222112 2 1112 1 2 PPPPU ααα ++= • Compute the strain energy in the beam by evaluating the work done by slowly applying P1 followed by P2,
  • 27. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Castigliano’s Theorem ( )2 2222112 2 1112 1 2 PPPPU ααα ++= • Strain energy for any elastic structure subjected to two concentrated loads, • Differentiating with respect to the loads, 2222112 2 1212111 1 xPP P U xPP P U =+= ∂ ∂ =+= ∂ ∂ αα αα • 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 and j j j j j j T U M U P U x ∂ ∂ = ∂ ∂ = ∂ ∂ = φθ
  • 28. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition 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 jj j L dx P M EI M P U xdx EI M U 00 2 2 • For a truss, j i n i i ii j j n i i ii P F EA LF P U x EA LF U ∂ ∂ = ∂ ∂ == ∑∑ == 11 2 2
  • 29. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf Sample Problem 11.5 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. 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.
  • 30. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition 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. QBQAQA yx 4 3 4 3 ==−= • Apply the method of joints to determine the axial force in each member due to Q. QFF QFF FF BDAB CDAC DECE 4 3;0 ;0 0 −== −== ==
  • 31. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition 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. ( )QP EQ F EA LF y i i ii C 42634306 1 += ∂ ∂       = ∑ • Setting Q = 0, evaluate the derivative which is equivalent to the desired displacement at C. ( ) Pa1073 10404306 9 3 × × = N yC ↓= mm36.2Cy