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）

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