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1
Energy Method
mi@seu.edu.cn
• Work and Strain Energy(功与应变能)
• Strain Energy Density(应变能密度)
• Strain Energy due to Normal Stresses(正应力所致应变能)
• Strain Energy due to Shearing Stresses(切应力所致应
变能)
• Strain Energy due to Bending and Transverse Shear
(弯矩和横力所致应变能对比)
• Strain Energy due to a General State of Stress(一般应力状态所致应变能)
• Work and Energy under a Single Load(单载下的功能
互等原理)2
Contents
• Strain Energy cannot be Superposed(应变能的不可叠加性)
• Work and Energy under Several Loads(多载下的功能原理)
• Castigliano’s Second Theorem(卡氏第二定理)
• Statically Indeterminate Truss(超静定桁架)
• Statically Indeterminate Shafts(超静定扭转轴)
• Statically Indeterminate Beams(超静定梁)
• Method of Dummy Load(虚力法)
• Method of Unit Dummy Load (单位虚力法)
Contents
3
• A uniform rod is subjected to a slowly increasing load
• The total work done by the load for a deformation x1,
which results in an increase of strain energy in the rod.
energystrainworktotaldxPW
x
1
0
Work done by External Loads and Strain Energy
• 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.
workelementarydxPdW
1 1
21 11 1 12 2
0 0
x x
W P dx kx dx kx Px U
• In the case of a linear elastic deformation,
4
Strain Energy Density
• To eliminate the effects of size, evaluate the strain-
energy per unit volume,
densityenergy straindu
L
dx
A
P
V
U
xx
x
1
1
0
0
• 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 1.
5
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.
• The strain energy density resulting from
setting 1 Y is the modulus of resilience.
resilience of modulusE
u YY
2
2
• If the stress remains within the proportional
limit,
E
EdEu xx
22
21
21
0
1
6
Strain Energy due to Normal Stresses
• In an element with a nonuniform stress distribution,
energystrain totallim0
dVuU
dV
dU
V
Uu
V
• For values of u < uY , i.e., below the proportional
limit,
energy strainelasticdVE
U x 2
2
• Under axial loading, dxAdVAPx
L
dxAE
PU
0
2
2
AE
LPU
2
2
• For a rod of uniform cross-section,
7
Strain Energy due to Normal Stresses
I
yMx
• For a beam subjected to a bending load,
dVEI
yMdV
EU x
2
222
22
• Setting dV = dA dx,
dxEI
M
dxdAyEI
MdxdA
EI
yMU
L
L
A
L
A
0
2
0
22
2
02
22
2
22
• For an end-loaded cantilever beam,
EI
LPdx
EI
xPU
PxM
L
62
32
0
22
8
Strain Energy due to Shearing Stresses
• For a material subjected to plane shearing
stresses,
xy
xyxy du
0
• For values of xy within the proportional limit,
GGu
xyxyxyxy
2
2
212
21
• The total strain energy is found from
dVG
dVuU
xy
2
2
9
Strain Energy due to Shearing Stresses
2 2 2
22 2
x
p
TU dV dV
G GI
• For a shaft subjected to a torsional load,
• Setting dV = dA dx,
2 2 22
2 2
0 0
2
0
2 2
2
L L
p pA A
L
p
T TU dAdx dA dx
GI GI
Tdx
GI
• In the case of a uniform shaft,
2
2 p
T LU
GI
10
Strain Energy due to Shearing Stresses
• For a beam subjected to a bending load,
• Setting dV = dA dx,
• In terms of the form factor
22 *
S1d
2 2
xy z
z
F SU dV V
G G I b
2*
S
0
22 *
S
20
1d d
2
d d2
Lz
Az
Lz
Az
F SU A x
G I b
F SA x
GI b
2
SS
0d
2
L FU f x
GA
*
S zxy
z
F S
I b
• Define the form factor for shear
• Dimensionless
• Unique for each specific cross-
sectional area
2*
2
zS
Az
SAf dA
I b
11
Form Factors for Shear
• The form factor for rectangular cross-section
2*
S 2
22
22
2 23
d
1d 1.20
2 412
z
Az
h
h
SAf A
I b
bh b hb y y
bbh
hz
b
ydy
• The form factor for circular cross-section
2 2
2 2
23/2
22
22 '
4
S 2 1/22 ' 24 4 2
24 43 d d 1.08
64 24
dd y
dd y
d ydf z y
dd y
z
y
d
y
dy• For thin-walled circular tubes: fS = 1.95.
• For thin-walled square tubes: fS = 2.35.12
• For an end-loaded cantilever beam
Strain Energy due to Bending & Transverse Shear
• The total strain energy due to both
bending and transverse shear2 2 2 2 3 2
00
2 3 2 3 2
2 2
6 3d
2 5 2 6 5
18 31 1
6 5 6 10
LL
b s
P x P P L P LU U U dx x
EI GA EI GA
P L EI P L Eh
EI GAL EI GL
• For steel, take E/G ≈ 2.6
• The strain energy due to transverse shear is of importance only
in the case of very short deep beams, i.e., for large h/L ratios.13
Strain Energy due to a General State of Stress
• Previously found strain energy due to uniaxial stress and plane
shearing stress. For a general state of stress,
zxzxyzyzxyxyzzyyxxu 21
• With respect to the principal axes for an elastic, isotropic body,
distortion todue 12
1
change volume todue 6
21
22
1
222
2
222
accbbad
cbav
dv
accbbacba
Gu
E
vu
uu
Eu
• Basis for the maximum distortion energy failure criteria,
specimen test tensileafor 6
2
Guu Y
Ydd
14
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,
2
2 21 1
0
2
2 2
L
U u dV dVE
P A P LAdx
E EA
• Strain energy may also be found from
the work of the single load P1,
1
0
x
dxPW
• For an elastic deformation,
11212
121
00
11
xPxkdxkxdxPW
xx
• Knowing the equivalence between
strain energy and work,
15
11
1
2
PLWU W x
P AE
• Strain energy may be found from the work of other types of
single concentrated loads.
• Transverse load
1
11 12
0
3 2 3
1 1112
3 6
y
W P dy P y
PL P LU P
EI EI
• Bending Moment
1
11 12
0
2
1 1112
2
W M d M
M L M LU M
EI EI
1
11 12
0
2
1 1112
2p p
W T d T
T L T LU T
GI GI
• Twisting Moment
16
Work and Energy under a Single Load
• The strain energy produced by the two loads
acting simultaneously is not equal to the sum
of the strain energies produced by the loads
acting separately.
• Strain energy is a quadratic function of the
loads, not a linear function.
• A solid circular bar is fixed at one end and free at
the other. Three different loading conditions are to
be considered. For each case of loading, obtain a
formula for the strain energy stored in the bar.
Strain Energy cannot be Superposed
2
2 2
22 2 2
2
2
2 4
22
2 2 2 42
aa
p
b bb
p p
a ba a bac
p p ppp
b
T LU
GI
T L T LU
GI GI
T T LT L T L T LU
GI GI GI
T T L
GI GI
17
• Strain energy of the structure,
AE
lP
AE
lP
AE
LF
AE
LFU BDBDBCBC
2332
22
364.02
8.06.0
22
• Equating work and strain energy,
AE
Ply
yPAE
LPU
B
B
728.0
364.021
2
• 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,
18
Sample Problem
Sample Problem
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.
19
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
815
817
0
815
CD
AC
F
PF
PF
PF
CE
DE
821
45
0ABF
20
• Evaluate the strain energy of the
truss due to the load P.
2
22
297002
1
2
1
2
PE
A
LF
EEA
LFU
i
ii
i
ii
• Equate the strain energy to the work by P
and solve for the displacement.
9
33
2
21
1073
1040107.29
2
2970022
E
E
E
y
E
P
PP
Uy
UPy
mm27.16Ey
21
• Determine the angle of twist
at end D of the shaft by
equating the strain energy to
the work done by the load.
Sample Problem
2 2
4
2
4
4
2 2
2 2
1 11
2 2
1 11
2 2
n DC CB
p p
p
nn
p
T L T LU U U
GI G n I
T L
n GI
U TL
T n GI
• For a given allowable stress,
increasing the diameter of portion
BC of the shaft results in a decrease
of the overall energy-absorbing
capacity of the shaft.
2 4
1 2 1 3 1 14
4
1 2 1 3 1 14
17 41 1, , ,
2 32 81 2
17 41 1, , ,
32 81 2
n
p
n
p
T L nU U U U U U U
GI n
TL n
GI n
• Solution
22
Sample Problem
• Determine the angle of twist
at end A of the shaft by
equating the strain energy to
the work done by the load.
44
2
0
2
40
2
3 3
3 3
32 32
2
232
16 1 1
3
32 1 1
2 3
x A B A
xpx A B A
L
px
L
A B A
B A A B
A
B A A B
xd d d d
L
d xI d d d
L
T dxU
GI
T dx
xG d d d
L
T L
G d d d d
U TL
T G d d d d
• Solution
23
Sample Problem
Taking into account only the normal
stresses due to bending, determine the
vertical displacement at cross-section D of
the beam for the loading shown.
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.
• Develop a diagram of the bending
moment distribution.
24
• Find the vertical displacement at D
by equating the work done by the
transverse force to the strain energy.
Solution:
• Determine the reactions at A and B
from a free-body diagram of the
complete beam.
L
PaR
L
PbR BA
• Develop a diagram of the bending
moment distribution.
vL
PaMx
L
PbM 21
25
• Integrate over the volume of the beam to find
the strain energy.
2 2
1 2
0 0
2 2
0 0
2 2 3 2 3 2 2 2
2 2
2 2 2
2 2
1 1
2 2
1
2 3 3 6
6
a b
a b
M MU dx dv
EI EI
Pb Pax dx x dx
EI L EI L
P b a a b P a ba b
EI L EIL
P a bU
EIL
1
2
Over the portion ,
Over the portion ,
AD
PbM x
LBD
PaM x
L
26
• Find the vertical displacement at D by equating the work done by the transverse
force to the strain energy.
2 2
2 3D
U Pa by
P EIL
Sample Problem
• Determine the vertical
displacement at A. Only
consider the strain energy
due to bending. Assume
constant flexural rigidity
EI.
22
0 0
2 32
0
2 3
3
1 cos
1 cos
2 2
1 cos2
3
4
3
2 2
L
A
M Pr
Pr rdM dsU
EI EI
P rd
EI
P r
EI
U Pry
P EI
• Solution
27
Work and Energy under Several Loads
• Reversing the application sequence yields
21111221
22222
1 2 PPPPU
• Strain energy expressions must be equivalent.
It follows that 1221 (Maxwell’s reciprocal
theorem).
• Deflections of an elastic beam subjected to two
concentrated loads,
22212122212
21211112111
PPxxx
PPxxx
22222112
21112
1 2 PPPPU
• Compute the strain energy in the beam by
evaluating the work done by slowly applying
P1 followed by P2,
28
Castigliano’s Second Theorem
22222112
21112
1 2 PPPPU
• Strain energy for any elastic structure
subjected to two concentrated loads,
• Differentiating with respect to the loads,
22221122
12121111
xPPP
U
xPPP
U
• Castigliano’s theorem: For an elastic structure
subjected to n loads, the deflection yj of the
point of application of Pj can be expressed as
and j j j
j j j
U U Uy
P M T
29
Carlo Alberto Castigliano (9
November 1847 – 25 October 1884)
Italian mathematician and physicist.
• Castigliano’s theorem is simplified if the differentiation w.r.t. the
load is performed before the integration or summation.
• For bending
L
jj
j
L
dxP
M
EI
M
P
Uydx
EI
MU
00
2
2
• For tension / compression
j
in
i i
ii
j
j
n
i i
ii
P
F
EA
LF
P
Uy
EA
LFU
11
2
2
• For torsion
2
0 0,
2
L L
j
p j p j
T T TUU dx dx
GI T GI T
30
Castigliano’s Second Theorem
Sample Problem
• Solution
22
0 0
2 2 2
0
2 2 2 3
2
2 2
12
2
3
2
2
x
L Lx
p p
L
p
p
B
p
T T tx
T tx dxT dxU
GI GI
T Ttx t x dxGI
T L TtL t L
GI
U TL tL
T GI
xB
A
2
0 0
2L Lx x
B
p p p
T txT TU TL tLdx dx
T GI T GI GI
• Alternatively
31
• Determine the angle of twist
at end A of the shaft.
Sample Problem
• The cantilever beam AB supports a uniformly distributed load w and a
concentrated load P as shown. Knowing that L = 2 m, w = 4 kN/m, P = 6 kN,
and EI = 5 MNm2, determine the deflection at A.
• Solution
2
3 42
0 0
3 3 3 4
6
1
2
1 1 1
2 3 8
1 6 10 2 4 10 24.8
5 10 3 8
L L
A
MM Px wx x
P
U M M PL wLy dx Px wx xdx
P EI P EI EI
mm
32
• A load P is supported at B by three rods of the same
material and the same cross-sectional area A. Determine the
axial force in each rod.
• Solution
1
0.6 0.6 0.6
0.8 0.8
0.8
BH
HBH H
BCBC H
HBD H
BD
H
F
RF R
FF P R
RF R P
F
R
33
Statically Indeterminate Truss
1
0.5 11
0.6 0.6 0.6 0.6
0.8 0.8 0.8 0.8
0.593
0.593 0.244
0 26
0
.3
ni i i
H
iH i H
BC BC BCBH BH BH BD BD BD
H H H
H
H
H
BH
H BC
BD
F L FUy
R A E R
F L FF L F F L F
AE R AE R AE R
R L
P R LAE
R P L
F P
R P F P
F P
34
Statically Indeterminate Shafts
• Determine (a) the reactive torques at the ends, (b) the angle of rotation at the cross section where the load T0 is applied.
• Solution:
0 0 0
0 0 0
0
0
i i i A A A B B Bj
ij pi j pA j pB j
A B A A BA A A A AA
pA A pB A pA pB
B A B B AB B B B BB
pA B pB B pA pB
T L T T L T T L TU
T GI T GI T GI T
T T L T T T T LT L T T L
GI T GI T GI GI
T T L T T T T LT L T T L
GI T GI T GI GI
35
Statically Indeterminate Shafts
• For the special case of dA = dB:
0 0
0
, B AA B
A B A A B BC
p p p
L LT T T T
L L
L L T T L T L
LGI GI GI
0 0
0
0 0
, B pA A pB
A B
B pA A pB B pA A pB
A BA A A B B B A A B BC
pA pB pA pBB pA A pB
L I L IT T T T
L I L I L I L I
L L TT L T T L T T L T L
GI T GI T GI GIG L I L I
36
• Determine the reactions at the supports for the
prismatic beam and loading shown.
• Solution
2
0
3 42
0
2
1
2
1 1 1
2 3 8
5
3 81
0
8
8
A
A
L
A
A A
LA
A
B
A
B
MM R x wx x
R
U M My dx
R EI R
R L wLR x wx xdx
EI EI
R wL
R wL
M wL
37
Statically Indeterminate Beams
• For the uniform beam and loading shown,
determine the reactions at the supports.
2 11
11
0
3 42
0
1
2
1
1 1 1
2 3 8
A
A
L
A
LA
A
MM R x wx x
R
MM dx
EI R
R L wLR x wx xdx
EI EI
Solution:
1. Basic determinate system:
3. Portion AB of the beam:
2. From static equilibrium:
9 33 ; 2
4 4B A C AR wL R R wL R
38
Statically Indeterminate Beams
x
2x wx
2 22
2 222
0 0
3 4
3 12 2
4 2
1 1 3 12 2
4 2
1 5
6 64
A
A
LL
A
A
A
MM R wL x wx x
R
MM dx R x wLx wx x dx
EI R EI
R L wL
EI
4. Portion CB of the beam:
5. Reaction at A:
3 34 41 1 5
3 8 6 64
9 333
13 4 323 132
24
0
16
A AA
B A
A
C A
R L R LwL wLy
EI EI
R wL R wL
R wL
R wL R wL
39
2
0
2
0
1
2
1 1
2
A
A
L
A
A A
L
A
MM Q x wx x
Q
U M My dx
Q EI A
Q x wx x dxEI
Method of Dummy Load
• The cantilever beam AB supports a uniformly
distributed load w. Determine the deflection and
slope at A.
Solution:
1. Apply a dummy force QA at A:
2. Set the dummy force QA as zero:
1A Ay Q
EI
42 2
0 0
1 1 1
2 2 8
L L wLx wx x dx wx x dx
EI EI
Note: since the dummy load points downward, + indicates downward deflection at A.
40
3. Apply a dummy moment MA at A:
2
2
0 0
11
2
1 11
2
A
A
L L
A A
A A
MM M wx
M
U M Mdx M wx dx
M EI M EI
4. Set the dummy moment MA as zero:
1A AM
EI
32 2
0 0
1 1 11 1
2 2 6
L L wLwx dx wx dx
EI EI
Note: since the dummy moment acts counter clockwise, + indicates counter
clockwise rotation of cross-section A.
2 2
0 0 0 0
1 1 1 1; 1
1 1
2 2
L L L L
A Ay dx dx dx dxEI EI EI E
MwI
Mx M Mx wx
: bending moment in beam developed by real loads.
: fictitious moment in beam developed by a unit dummy load (force/moment)
applied at a point of interest.
M
M
41
Sample Problem
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 previous
example 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.
42
• Apply the method of joints to determine the axial
force in each member due to Q.
3 54 4
0
0;
0; ;
CE DE
AC CD
AB BD AD
F F
F F Q
F F Q F Q
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 43
43
43
• Combine with the results of previous example to evaluate the derivative
with respect to Q of the strain energy of the truss due to the loads P and Q.
14306 4263
i i i i ii i iC
i i
P Q L P QF L Fy P Q
EA Q EA Q E
• Setting Q = 0, evaluate the derivative which is equivalent to the desired
displacement at C.
3
9
0
4306 40 10
73 10 Pa
i i iC
i Q
NPL Qy
EA Q
mm 36.2Cy
44
i iC i
i
F Ly F
A E
axial forces developed in individual members under a unit
load applied at joint C.0
:ii
Q
QF
Q
Method of Unit Dummy Load
• For an elastic structure, the deflection of a particular point can be found by
applying a unit dummy load at the point of interest
• Bending
2
0
0
2
L
L
MU dx
EI
MMy dx
EI
• Tension/compression
2
1 2
ni i
i i
i ii
i
F LU
EA
F Ly F
EA
• Torsion
2
0
0
2
L
p
L
p
TU dx
GI
TTdx
GI
45
• Determine the angle of
twist at cross-section B
of the shaft.
2
0 0
1
2
L L
B
p p p
txTT tLdx dx
GI GI GI
Sample Problem
46
• Solution
• Work and Strain Energy(功与应变能)
• Strain Energy Density(应变能密度)
• Strain Energy due to Normal Stresses(正应力所致应变能)
• Strain Energy due to Shearing Stresses(切应力所致应
变能)
• Strain Energy due to Bending and Transverse Shear
(弯矩和横力所致应变能对比)
• Strain Energy due to a General State of Stress(一般应力状态所致应变能)
• Work and Energy under a Single Load(单载下的功能
互等原理)47
Contents
• Strain Energy cannot be Superposed(应变能的不可叠加性)
• Work and Energy under Several Loads(多载下的功能原理)
• Castigliano’s Second Theorem(卡氏第二定理)
• Statically Indeterminate Truss(超静定桁架)
• Statically Indeterminate Shafts(超静定扭转轴)
• Statically Indeterminate Beams(超静定梁)
• Method of Dummy Load(虚力法)
• Method of Unit Dummy Load (单位虚力法)
Contents
48
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