Chapter Eight

Energy Method 8.1 Introduction

8.2 Strain energy expressions

8.3 Principal of stationary potential energy; several degrees of

freedom ------ Castigliano’s first theorem ---- Examples

8.4 Principal of stationary complementary energy ------

Castigliano;s second theorem ----Examples

8.5 Statically indeterminate problems ----Examples

Review and Summary

8.1 Introduction

Two methods in solid mechanics ---- vectorial methods

and energy methods or variational methods

(1) vectorial methods ---- emphasized in elementary courses

and are formulated in terms of vector quantities such as

forces and displacement

(2) Energy methods---- formulated in terms of scalar quantities

such as work and energy. Advantages o f energy methods --

---- ability to avoid some extraneous detail and to yield

approximate solution for complicated problems.

Degree of freedom (d.o.f.) ------ the number of independent

quantities needed to define uniquely the configuration of a

system ----generalized coordinates.

m

d

8.2 Strain energy expression

Work is done by a force as it moves through a distance,

and by a moment as it turns through an angle.

dufWdufdWB

A cos)cos(

(the work done by a moment : dmcos )

If the effect of force is to distort an elastic body (such as a linear

spring), work done by f id stored as strain energy U (expressed

in terms of displacement). Complementary strain energy U*

(expressed in terms of force). For a linear elastic material U and

U* are numerically equal.

(1) The strain energy U=U()

2

2

0

kkudufduU

B

A

(2) The complementary strain energy U*=U

*(F)

k

Fdf

k

fudfU

FB

A 2

2

0

*

f=ku

U=f/k

A unit column of linear elastic material can be viewed as a

linear spring, so the complementary strain energy U* is

The complementary strain energy U

* (expressed in terms

of stress or force) for a bar of length L under general complex

loading is (by volume integral)

Expression for the strain energy U (in terms of displacements)

of a slender straight bar is (replacing My by ),/( 22 dxwdEIy )

)/(),/( 22 dxdGJTdxwdEIM zz

Slender circular rings

Rdvd

vd

R

EIU

2

2

0 2

2

2

1

2

( UR , strain energy for straight bar)

Energy of transverse shear

dxGA

V

bdzdxG

U

bhAzh

Ah

VE

Lz

zx

L h

h

zzx

0

2

0

2/

2/

*

22

2

22.1

2

1

),4

(6

Moment-curvature relations for the

ring

)(1

2

2

2v

d

vd

REI

M

8.3 Principal of stationary potential energy; several degrees

of freedom ---- Castigliano’s first theorem ---- Examples

Admissible or kinematically admissible configurations

Potential energy of a structure ---- =U+

U ---- the strain energy

---- the potential of the loads

For linear spring, d.o.f.=

FkU

FkU

2/

,2/

2

2

The principal of stationary potential energy: Among all

admissible configuration (that satisfied static equilibrium

conditions) makes the potential energy stationary with

respect to small admissible variations of displacement. If

the stationary condition is a minimum, the equilibrium

state is stable.

0)(

)(00

dFkd

k

FkFFk

d

d

Several degrees of freedom ---- Castigliano’s first theorem

(1) the potential energy for n degree of freedom Di

n

ii

n

P

DDDUU

U

1

21 ),,(

(2) then principal of stationary potential energy gives

n

n

dDD

dDD

dDD

d

2

2

1

1

0

For any and all of these dDi, d must vanish, this is

possible only if

),,2,1(0 niDi

(3) Castigilano’s first theorem: If strain energy U is

expressed in terms of independent displacement d.o.f.,

then the load Pi that corresponds to d.o.f. Di is given by

the partial derivative of U with respect to Di

From above 0

iD (for ni ,,2,1 )

We have

i

i

PD

U

(for ni ,,2,1 )

For example : 21 , DD then MU

FU

,

Examples (Determine the displacement d.o.f. that define the static

equilibrium configuration)

(1) Two-Bar Linkage

The only d.o.f. is , that =0 as reference state,

we have (neglecting strain energy U)

W

F

d

d

LFL

W

2tan0

)sin2()cos1(2

2

( is the value for static equilibrium.)

(2) Rigid Bar

The d.o.f. is B (C=2B), the rotation of the rigid bar is

=B/b

bM

L

AE

L

AEU B

CBB

22 )2(22

CB

B

BCB

MAEb

L

d

d

bM

L

AE

50

52

2

the forces in the wires

b

M

L

AEP

b

M

L

AEP C

BCC

BB5

2)2(,

5

(3) Spring in Series

Two d.o.f. are needed to define the configuration:

A, B. Initially , A=B=0.

2

2

2

1 )(2

1

2

1ABA kkU

Using Castigliano’s First theorem, we have

211

2

21

,k

F

k

FF

k

FF

FkFU

FkkFU

BBAB

BAA

BABB

B

AABAA

A

8.4 Principal of stationary complementary energy ---------

castigliano’s second theorem ---- Examples

Complementary energy of a structure ---- U*+

The principal of stationary complementary energy: among

all statically admissible stress fields, the actual stress field

(that yields kinematically admissible displacements) makes

the complementary energy stationary with respect small

statically variations of stress.

(1)

kFk

F

dF

d,00

(2) If there are several forces, we have

0*

iP for i=1,2,……,n

In previous example of two springs in series, we have

0

0

21

1

BBBA

B

ABA

A

k

F

k

FF

F

k

FF

F

For linear spring

Fk

F

F

kFU

2

2/ 22

The complementary energy of a structure loaded by

concentrated forces and/or moments is

n

ii DPU1

),......,2,1(0 niDP

U

Pi

ii

Castigliaon’s second theorem: the partial derivative of

complementary energy U* with respect to a load yields the

displacement component of the loaded point in the direction

of that load.

),2,1(*

niDP

Ui

i

Pi can be a force F or moment M.

i

i

i

i M

U

F

U

**

,

Unit load method (A convient format of Castigliano’s

second theorem) Using complementary strain energy expression U

* for curved

bar and i

iP

UD

, we find

L

i

zzz

i

z

z

z

i

y

y

y

i dxP

V

GA

Vk

P

M

EI

M

P

M

EI

MD

0

introducing my mz Vz

dx

GJ

Tt

EI

mM

EI

mM

z

zz

y

yy

u

y

yP

Mm

is a moment produced by a unit load Pi (a unit

force or a unit moment)

Examples

(1) Cantilever beam

(a) First determine the vertical deflection of B.

The bending moment is (neglect shearing)

2)(

2qxFxxM

using unit load methods, ymxFM

EI

qL

EI

FLdxx

qxFx

EIB

83))(

2(

1 432

(b) Find the deflection at B due to q along (F=0), two method:

(1) simply set F=0 in above expression for B,

(2) temporarily apply a load (such as unit load) in the

desired dorection, after using unit load method, then

set this load to zero: ((case (b))

EI

qLdxxx

qx

EIB

8)()(1

2

1 42

(c) the horizontal deflection at C

Lh

HEI

hqLhdx

qx

EIdss

EI 0

32

0 62

1))(0(

1

Set this load=0

(2) Split Ring

(a) Bending and twisting moments (in ring) MR and T are:

cos)cos1(

sinsin

CFRT

CFRM R

here C is the unit couple for calculation of rotation

(b) Calculate deflection, we set C=0

GJ

FR

EI

FR

RdRGJ

FR

RdREI

FR

dxF

T

GJ

T

F

M

EI

M RRz

33

2

0

2

0

2

0

3

)cos1()cos1(

)sin(sin

(c) Calculate the rotation, C=1 mR=-sin and T=-cos, in

the above equation, we use 0CRM , 0C

T , C

M R

,

C

T

, we

obtain GJ

FR

EI

FRy

22

Determine the Z-direction displacement of the

loaded end and its rotation component about the y

axis

(3) Truss Analysis

(a) Calculate the vertical deflection at C by unit force method:

first calculate the internal force N in each bar due to Q;

then the force n due to unit force. Nn is nonzero only bars

DB and DC, so

DCDBi i

C nEA

NLn

EA

NLn

EA

NL

011

EA

QL

EA

LQ

(b) Relative motion of points C. G. : we apply collinear force

(unit load) as shown. Then

EA

QLn

EA

NL

FDFCCD

CG 414.3,,

(c) Rotation of bar FG: we apply couple forces 1/L (unit

moment), then

CFACBD

FG nEA

NL

,,

8.5 Statically indeterminate problems ---- Examples

(1) propped cantilever beam

(a) Find the redundant and regard it as a known load on the

structure.

(b) The bending moment is

2/2qxRxM

L

R dxR

M

EI

Mor

R

U

0

*

00

i.e.,

8/302

1

0

2

qLRdxxqx

RxEI

L

(c) we can choose ML as redundant, then by

L

LL

dsM

M

EI

Mor

M

U

0

*

00 we

Obtain the same result as in (b).

Determine the bending moment in

the statically indeterminate beam

(3) Elastically support cantilever beam

We use two methods to solve this problem:

(a) Consider the U* of the beam only, then

dxEI

MU

L

0

2

2 where

2

2qxRxM

L

mdxEI

Mor

k

U

R

U

0)(

when is known, then R=k

(b) Consider the U* of whole system (beam and spring).

dxEI

M

k

RU

L

0

22

22 where 22qxRxM

R is the reaction at the base of the spring, by Castigliano’s

second theorem or unit load method:

L

mdxEI

M

k

Ror

R

U

000

we can get kRR

k

-x

xR

M

(3) Semicircular arch

This is the statically indeterminate to the third degree. There are

several methods to solve the problem:

(a) Find U* of the arch as U

*= U

*(V, h, M0), then by using

0,0,00

M

U

H

U

V

U

we obtain V, H, M0.

(b) By using symmetry, only have 2 unknows, H , 2

M as

shown in (c), then

2

, MHUU , by

0

H

U , 0

2

M

U , we solve the problem.

(c) As shown in (d), ),( 0MHUU , by

0,00

M

U

H

U

Find the support reaction of the arch

(4) Internally indeterminate truss

we elect to use forces in bars AD and CF as redundants. These

forces FAD and FCF are exposed if bars AD and CF are cut. Then

the truss is rendered as statically determinate, i.e., all forces in

bars can be written in terms of Q1, Q2, FAD and FCF. We can

write U* as

11

1

2

2i ii

i

EA

LNU

FAD and FCF can be obtain by

0

ADF

U , 0

CFF

U ,

Since in the imagined cut in each bar (before and after loads FAD,

FCF), the relative approach or separation of the cut ends is zero.

Find the forces in

all bars