Chapter 11 Energy Method 

-- Utilize the Energy Method to solve engineering mechanics problems.

-- Set aside the Equations of quilibrium

1. Introduction

The relations between forces and deformation :

Fundamental concept of

We will learn:

Strain – Ch 2

Strain Energy – Ch 11

Stress -- Ch 1

1. Modulus of Toughness

2. Modulus of resilience

3. Castigliano Theorem

11.2 Strain Energy

(11.1)dU Pdx

1

0

xU Pdx

1

0 energy

xStrain U Pdx

(11.2)

If the material response is elastic:

(11.3)

P kx

1 210

12

xU kxdx kx

1 1

12

U P x

11.3 Strain-Energy Density

(11.4)

1

0

xU P dxV A L

1

0 x x

Ud

V

1

0 energy density x xStrain u d

Modulus of Toughness = Toughness

= area under the - curve.

Modulus of Resilience:

(11.5)x xE

121

0 2x x

Eu E d

21

2u

E

2

2Y

Yu E

(11.6)

(11.7)

(11.8)

11.4 Elastic strain Energy for Normal Stresses

(11.9)0

limV

Uu

V

dUudV

1

0 x xu d

221 1 1

2 2 2x

x x xu EE

(11.9)x xE

221 1 1

2 2 2x

x x xu EE

2

2xU dVE

2

22P

U dVEA

(11.12)

(11.11)

(11.13)2

22P

U dVEA

2

0 2

L PU dx

AE

dV Adx

2

2P L

UAE

11.5 Elastic Strain Energy for Shearing

Stresses

(11.18)0

xy

xy xyu d

2

21 12 2 2

xyxy xy xyu G

G

dUu or U udVdV

2

2xyU dVG

(11.19)

(11.20)

(11.21)

Strain Energy in Torsion

(11.22)

2 2 2

22 2xy T

U dV dVG GJ

dV dAdx

22

20 2( )

L TU dA dx

GJ

2

0 2

L TU dx

GJ

2

2T L

UGJ

(11.21)

(11.19)

11.6 Strain Energy for a General State of Stress

(11.25)

12

( )x x y y z z xy xy yz yz zx zxu

2 2 2 2 2 21 12

2 2( ( )] ( )x y z x y y z z x xy yz zxuE G

(11.26)

y zXx

y zXy

y zXz

xy yz zxxy yz zx

E E E

E E E

E E E

G G G

(2.38)

From Eq. (2.38)

2 2 2 2 2 21 12

2 2( ( )] ( )x y z x y y z z x xy yz zxuE G

2 2 212

2( ( )]a b c a b b c c auE

(11.26)

(11.27)

If the principal stresses are used:

Where a, b, c = the principal stresses

v du u u

3a b c

(11.28)

(11.29)

Where uv = the part of energy leading to volume change

= hydrostatic stress

ud = deviatoric energy = the part of energy leading to

shape change.

Defining Mean Stress or Average

Stress

(11.30)' ' ' a a b b c c

0' ' '+ +a b c

And set:

'iwhere = mean stress = deviatoric stress

Combining Eqs. (11-29) and (11.30)

(11.31)

1 2 ' ' '( + + )a b ceE

0' ' '+ +a b c (11.31)

-- They only change the shape, but do not lead to the change of the volume of the material.

The dilatation, V/V, caused by the state of stress can be obtained, via Eq. (2.31)

( )X y z X y zeE E

2

asor e = 0

2 2 21 3 1 23 2 3

2 2( )

[ ( )]vu E E

The uv can be obtained by substituting into Eq. (11.27)

to obtain

By means of Eq. (11.29) we have

(11.32)21 2

6( + + )v a b cu

E

2 2 2 2 2 212 2 2

6[( ) ( ) ( )]d a a b b b b c c c c a au

E

12EG

2 2 2

2

13 6

6

1 2

[ ( ) ( )

( )( ) ]

d v a b c a b b c c a

a b c

u u uE

(2.43)

The distortion or deviatoric energy can be obtained as

After simplification:

Recalling Eq. (2.43)

or1 1

2G E

2 2 2112

[( ) ( ) ( ) ]d a b b c c auG

2 21( )

6d a a b buG

(11.34)

(11.33)

Hence, the previous equation takes a new form

For 2-D cases, 0c Eq. (11.33) reduces to

For uniaxial tension, i.e. 1-D cases, b = 0 and a = y

2

6( ) yd yu

G

(7.26)

2 2 2 22( ) ( ) ( )a b b c c a Y

2 2 2a a b b Y

2 2 2 2( ) ( ) ( ) 2a b b c c a Y (11.36)

(11.35)

Substituting this equation to the previous equation, it leads to

Expanding the same operation to a 3-D case, one can have

Replacing “<“ by “=“, it follows

This is a circular cylinder of radius 2 3/ y

The 2- D Yield Locus:

2 2 2a a b b Y (7.26)

The 3- D Yield Locus:

2 2 2 22( ) ( ) ( )a b b c c a Y (11.35)

11.7 Impact Loading

(11.37)

20

12

. .mU K E mv

2

2m

mU dVE

(11.38)

K. E. of the ball = K. E. = 212 omv

The strain energy in the bar:

Assuming:

1. No heat dissipation

2. The ball sticks with the rod after impact.

2 2

2 2m m

m

VU dV

E E

202 m

m

U E mv EV V

(11.39)

If the stress is uniform within the rod:

Therefore, m can be determined as:

11.8 Design for Impact Loads

8 mm

U EV

2 mm

U EV

4 52 2 2L L L

V A A A

(11.45a)

(11.45b)

Case A: For a Uniform-Diameter Rod:

Case B: For a Multiple-Section Rod:

2 4 2 21 1 14 4 4

( / ) ( / )L I c L c c c L V

2

6( / )m

m

U EL I c

414

I c

(11.45c)

(11.44)

Case C: For a Circular Cantilever Rod:

However,

Hence,

Eq. (11.44) can be reduced to

24 mm

U EV

m

EV

Since

We conclude that:

--- in order to develop lower m in the rod, the rod

should have

1. Lower E

2. Larger V

3. Uniform m

11.9 Work and Energy under a Single Load

(11.2)1

0 energy

xStrain U Pdx

1 1

12

U P x

1 1

12

U P y

3 2 31 1

1

12 3 6

( )PL P L

U PEI EI

(11.3)

(11.46)

Case A: For an Uniaxial Load:

Case B: For a Cantilever Beam:

(11.47)1

1 10

12

U Md M

21 1

1

12 2

( )M L M L

U MEI EI

1

1 10

12

U Td T

2

1 11

12 2

( )T L T L

U TJG JG

Case C: For a Beam in Bending:

(11.48)

Case D: For a Beam in Torsion:

(11.49)

11.10 Deflection under a Single Load by the

Work-Energy Method (11.3)1 1 1

1

1 22

UU P x x

P

1 1 11

1 22

UU M

M (11.47)

where x1 = deflection due to P1

where 1 = deflection due to single moment M1

Case A: For an Uniaxial Load:

Case B: For a Beam in Bending:

11.11 Work and Energy under Several Loads

(11.54)

(11.54)

(11.54)

(11.54)

11 11 1 21 21 1 x P x P

12 12 2 22 22 2 x P x P

1 11 12 11 1 12 2x x x P P

2 21 22 21 1 22 2x x x P P

Deflection due to P1:

Deflection due to P2:

The combining effect of P1 and P2

ij = influence coefficients

Calculating the Work Done by P1 and P2:

(11.58)2

1 11 1 11 1 11 1

1 1 12 2 2

( )P x P P P

22 22 2 22 2 22 2

1 1 12 2 2

( )P x P P P

Case I: Assuming P1 is applied first ---

At Point 1, the work done by P1 is

At Point 2, the work done by P1 is zero.

P2 is applied next ---

At Point 2, the work done by P2 is

(11.59)

(11.60)

At Point 1, the work done by P1 due to additional defection [caused by P2] is

1 12 1 12 2 12 1 2( )P x P P PP

(11.58) + (11.59) + (11.60), the total strain energy is

2 211 1 12 2 1 22 2

12

2( )U P P P P (11.61)

(11.62)

Case II: Assuming P2 is applied first ---

2 222 2 12 2 1 11 1

12

2( )U P P P P

Equating Eqs. (11.61) and (11.62) leads to

12 12

-- Maxwell Reciprocal Theorem

11.12 Castigliano’s Theorem

(11.61)2 211 1 12 2 1 22 2

12

2( )U P P P P

11 1 12 2 11

UP P x

P

12 1 22 2 22

UP P x

P

jj

Ux

P

(11.63)

(11.64)

Or, in general

(11.65)

jj

UM

(11.66)

[ Castigliano’s Theorem ]

General Formulation for Castigliano’s Theorem

For multiple loading, P1, P2, …., Pn

the deflection of the point of application of Pi can be expressed as

j jk kk

x P

12 ik i ki k

U PP

1 12 2jk k ji ik ij

UP P

P

(11.66)

The total strain energy of the structure is

(11.67)

Differentiating U w.r.to Pj

Since ij = ji, the above equation becomes

1 12 2jk k ji i jk k

kk ij

UP P P

P

jj

Ux

P

jj

UT

(11.65)

jj

UM

-- for concentrated loads

-- for moment loads

-- for torsion

(11.70)

11.13 Deflections by Castigliano’s Theorem

2

0 2

L MU dx

EI

0

L

jj j

U M Mx dx

P EI P

2

1 2

ni i

i i

F LU

A E

1

ni i i

jij i j

U F L Fx

P A E P

(11.17)

(11.71)

(11.72)

Total strain energy of a beam subjected to bending

Deflection at point Pj

Total strain energy of a truss member

Deflection at point Pj

But, the differentiation can be applied prior to integration.

[ Castigliano’s Theoem ]

(11.76)jj

Ux

Q

If no load is applied to a point, where we desire to obtain a deflection:

-- Apply a dummy (fictitious) load Q at that point, determine

Then, set Q = 0.

11.14 Statically Indeterminate Structures

Structure indeterminate to the 1st degree:

Procedures:

1. Assume one support as redundant

2. Replace it with an unknown force

3. y = U/RA = 0 solving for RA

0

L

AA A

U M My dx

R EI R

21

2AM R x wx

A

Mx

R

2 3

0

3 4

1 1( )

2

1( ) 0

3 8

L

A A

A

y R x wx dxEI

R L wL

EI

3 38 8

A AR wL R wL

25 18 8

M B BR wL wL