MECHANICS OF MATERIALS
Third Edition
Ferdinand P. BeerE. Russell Johnston, Jr.John T. DeWolf
Lecture Notes:J. Walt OlerTexas Tech University
CHAPTER
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Energy Methods
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Energy MethodsStrain EnergyStrain Energy DensityElastic Strain Energy for Normal StressesStrain Energy For Shearing StressesSample Problem 11.2Strain Energy for a General State of StressImpact LoadingExample 11.06Example 11.07Design for Impact LoadsWork and Energy Under a Single LoadDeflection Under a Single Load
Sample Problem 11.4Work and Energy Under Several LoadsCastigliano’s TheoremDeflections by Castigliano’s TheoremSample Problem 11.5
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• 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.
energystrainworktotaldxPUx
=== ∫1
0
11212
121
0
1
xPkxdxkxUx
=== ∫
• In the case of a linear elastic deformation,
Strain Energy
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Strain Energy Density• To eliminate the effects of size, evaluate the strain-
energy per unit volume,
densityenergy straindu
Ldx
AP
VU
x
x
==
=
∫
∫
1
1
0
0
εεσ
• The total strain energy density resulting from the deformation is equal to the area under the curve to ε1.
• 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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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,
EEdEu x 22
21
21
01
1 σεεεε
=== ∫
• The strain energy density resulting from setting σ1 = σY is the modulus of resilience.
resilience of modulusE
u YY ==
2
2σ
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Elastic Strain Energy for Normal Stresses• In an element with a nonuniform stress distribution,
energystrain totallim0
===∆∆
= ∫→∆dVuU
dVdU
VUu
V
• For values of u < uY , i.e., below the proportional limit,
energy strainelasticdVE
U x 2
2
∫ ==σ
• Under axial loading, dxAdVAPx ==σ
∫=L
dxAEPU
0
2
2
AELPU
2
2=
• For a rod of uniform cross-section,
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Elastic Strain Energy for Normal Stresses
IyM
x =σ
• For a beam subjected to a bending load,
∫∫ == dVEI
yMdVE
U x2
222
22σ
• Setting dV = dA dx,
dxEI
M
dxdAyEIMdxdA
EIyMU
L
L
A
L
A
∫
∫ ∫∫ ∫
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛==
0
2
0
22
2
02
22
2
22
• For an end-loaded cantilever beam,
EILPdx
EIxPU
PxML
62
32
0
22==
−=
∫
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Strain Energy For Shearing Stresses• For a material subjected to plane shearing
stresses,
∫=xy
xyxy duγ
γτ0
• For values of τxy within the proportional limit,
GGu xy
xyxyxy 2
2
212
21 τ
γτγ ===
• The total strain energy is found from
∫
∫
=
=
dVG
dVuU
xy2
2τ
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Strain Energy For Shearing Stresses
JT
xyρτ =
∫∫ == dVGJ
TdVG
U xy2
222
22ρτ
• For a shaft subjected to a torsional load,
• Setting dV = dA dx,
∫
∫ ∫∫ ∫
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛==
L
L
A
L
A
dxGJT
dxdAGJTdxdA
GJTU
0
2
0
22
2
02
22
2
22ρρ
• In the case of a uniform shaft,
GJLTU
2
2=
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Sample Problem 11.2SOLUTION:
• Determine the reactions at A and Bfrom 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.
• Integrate over the volume of the beam to find the strain energy.
• Apply the particular given conditions to evaluate the strain energy.
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Sample Problem 11.2SOLUTION:
• Determine the reactions at A and Bfrom a free-body diagram of the complete beam.
LPaR
LPbR BA ==
• Develop a diagram of the bending moment distribution.
vL
PaMxL
PbM == 21
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Sample Problem 11.2• Integrate over the volume of the beam to find
the strain energy.
( )baEIL
baPbaabLP
EI
dxxL
PaEI
dxxL
PbEI
dvEI
MdxEI
MU
ba
ba
+=⎟⎟⎠
⎞⎜⎜⎝
⎛+=
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=
+=
∫∫
∫∫
2
2223232
2
2
0
2
0
2
0
22
0
21
63321
21
21
22
EILbaPU
6
222=v
LPaM
xL
PbM
=
=
2
1
BD,portion Over the
AD,portion Over the
43 in 248ksi1029
in. 108in. 36a
in. 144kips45
=×=
==
==
IE
b
LP ( ) ( ) ( )( )( )( )in 144in 248ksi 10296
in 108in 36kips4043
222
×=U
kipsin 89.3 ⋅=U
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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 +++++= 21
• With respect to the principal axes for an elastic, isotropic body,
( )[ ]
( )
( ) ( ) ( )[ ] distortion todue 12
1
change volume todue 6
21
221
222
2
222
=−+−+−=
=++−
=
+=
++−++=
accbbad
cbav
dv
accbbacba
Gu
Evu
uuE
u
σσσσσσ
σσσ
σσσσσσνσσσ
• Basis for the maximum distortion energy failure criteria,
( ) specimen test tensileafor 6
2
Guu Y
Yddσ
=<
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Impact Loading• To determine the maximum stress σm
- Assume that the kinetic energy is transferred entirely to the structure,
202
1 mvUm =
- 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. ∫= dV
EU m
m 2
2σ
• Maximum value of the strain energy,
• Rod deforms under impact. Stresses reach a maximum value σm and then disappear.
• For the case of a uniform rod,
VEmv
VEUm
m202
==σ
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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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Example 11.06• Find the static load Pm which produces
the same strain energy as the impact.
( ) ( )
LAEUP
AELP
AELP
AELPU
mm
mmmm
516
165
422 222
=
=+=
• Evaluate the maximum stress resulting from the static load Pm
ALEmv
ALEU
AP
m
mm
20
58
516
=
=
=σ
SOLUTION:
• Due to the change in diameter, the normal stress distribution is nonuniform.
EVdV
E
mvU
mm
m
22
22
202
1
σσ≠=
=
∫
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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 as 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
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Example 11.07• Find the static load Pm which produces
the same strain energy as the impact.
For an end-loaded cantilever beam,
3
32
6
6
LEIUP
EILPU
mm
mm
=
=
SOLUTION:
• The normal stress varies linearly along the length of the beam as across a transverse section.
EVdV
E
WhU
mm
m
22
22 σσ≠=
=
∫
• Evaluate the maximum stress resulting from the static load Pm
( ) ( )2266
cILWhE
cILEU
ILcP
IcM
m
mmm
==
==σ
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Design for Impact Loads• For the case of a uniform rod,
VEUm
m2
=σ
( )( ) ( ) ( )
VEU
VLcccLcIL
cILEU
mm
mm
24
//
6
412
4124
412
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,
( ) ( )
VEU
ALLALAVAL
EU
mm
mm
8
2/52/2/45
16
=
=+=
=
σ
σ
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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
LPdxAEAP
dVE
dVuU
L
22
22
1
0
21
2
==
==
∫
∫ ∫σ
• Strain energy may also be found from the work of the single load P1,
∫=1
0
xdxPU
• For an elastic deformation,
11212
121
00
11
xPxkdxkxdxPUxx
==== ∫∫
• Knowing the relationship between force and displacement,
AELP
AELPPU
AELPx
2
211
121
11
=⎟⎠⎞
⎜⎝⎛=
=
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Work and Energy Under a Single Load
• Strain energy may be found from the work of other types of single concentrated loads.
EILP
EILPP
yPdyPUy
63
321
31
121
1121
0
1
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
== ∫
• Transverse load
EILM
EILMM
MdMU
2
211
121
1121
0
1
=⎟⎠⎞
⎜⎝⎛=
== ∫ θθθ
• Bending couple
JGLT
JGLTT
TdTU
2
211
121
1121
0
1
=⎟⎠⎞
⎜⎝⎛=
== ∫ φφφ
• Torsional couple
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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
lPAE
lP
AELF
AELFU BDBDBCBC
2332
22
364.02
8.06.0
22
=+
=
+=
• Equating work and strain energy,
AEPly
yPAE
LPU
B
B
728.0
364.0 21
2
=
==
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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.
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Sample Problem 11.4SOLUTION:
• 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
815
817
+=
−=
08
15
=
+=
CD
AC
F
PF
PF
PF
CE
DE
821
45
−=
= 0=ABF
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Sample Problem 11.4
• Evaluate the strain energy of the truss due to the load P.
( )2
22
2970021
21
2
PE
ALF
EEALFU
i
ii
i
ii
=
== ∑∑
• Equate the strain energy to the work by Pand solve for the displacement.
( )( )9
33
2
21
10731040107.29
22970022
×
××=
⎟⎟⎠
⎞⎜⎜⎝
⎛==
=
E
E
E
y
EP
PPUy
UPy
↓= mm27.16Ey
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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
( )21111221
22222
1 2 PPPPU ααα ++=
• Strain energy expressions must be equivalent. It follows that α12=α21 (Maxwell’s reciprocal theorem).
( )22222112
21112
1 2 PPPPU ααα ++=
• Compute the strain energy in the beam by evaluating the work done by slowly applying P1 followed by P2,
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Castigliano’s Theorem
( )22222112
21112
1 2 PPPPU ααα ++=
• Strain energy for any elastic structure subjected to two concentrated loads,
• Differentiating with respect to the loads,
22221122
12121111
xPPPU
xPPPU
=+=∂∂
=+=∂∂
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
jj
jj
j TU
MU
PUx
∂∂
=∂∂
=∂∂
= φθ
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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
jjj
Ldx
PM
EIM
PUxdx
EIMU
00
2
2
• For a truss,
j
in
i i
ii
jj
n
i i
iiPF
EALF
PUx
EALFU
∂∂
=∂∂
== ∑∑== 11
2
2
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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.
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Sample Problem 11.5SOLUTION:
• Find the reactions at A and B due to a dummy load Qat C from a free-body diagram of the entire truss.
QBQAQA yx 43
43 ==−=
• Apply the method of joints to determine the axial force in each member due to Q.
QFF
QFF
FF
BDAB
CDAC
DECE
43;0
;0
0
−==
−==
==
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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.
( )QPEQ
FEALFy i
i
iiC 426343061
+=∂∂
⎟⎟⎠
⎞⎜⎜⎝
⎛= ∑
• Setting Q = 0, evaluate the derivative which is equivalent to the desireddisplacement at C.
( )Pa1073
104043069
3
×
×=
NyC ↓= mm 36.2Cy