5.3 Castigliano’s theorem on deflection for linear load-deflection relations

• For this case complementary strain energy is equal to strain energy and we get

• For general case

and j j j jj j j

U U Uq xP M T

θ φ∂ ∂ ∂= = = =

∂ ∂ ∂

ii i i i i

ii i i i i

ii i i i i

U N N kV V M M T Tx dx dx dx dxF EA F GA F EI F GJ FU N N kV V M M T Tdx dx dx dxM EA M GA M EI M GJ MU N N kV V M M T Tdx dx dx dxM EA M GA M EI M GJ M

∂ ∂ ∂ ∂ ∂= = + + +∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂

θ = = + + +∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂

θ = = + + +∂ ∂ ∂ ∂

∫ ∫ ∫ ∫

∫ ∫ ∫ ∫

∫ ∫ ∫ ∫

Procedure

• Write an expression for each of the internal actions (axial force, bending moment, shear force, and torque) in each member of the structure in terms of external loads.

• Take derivatives of strain energy to get deflections and/or rotations

Example

• What displacement will we get if we differentiate with respect to w?

2

3 42 3

0 0

( 0.5 )

1 1( 0.5 )3 8

L L

A

MM Px wx xP

M M PL wLy dx Px wx dxEI P EI EI

∂= − + = −

∂⎛ ⎞∂

= = + = +⎜ ⎟∂ ⎝ ⎠∫ ∫

From 2005 qualifying exam

Caution: Note that ABCD is in the horizontal plane perpendicular to the direction of F!

Solution

• Member CD acts like cantilever beam with end load

• Member BC has in addition torque FL

• Member AB has the end force plus a clockwise moment FL plus torque FL

• Altogether

2 2 3

0 2 6

L

CDM dx F LM Fx U

EI EI= = =∫

2 2 3 2 3 2 3

2 6 2 6BCT L F L F L F LUGJ EI GJ EI

= + = +

2 3 2 3

3 2ABF L F LM Fx FL U

EI GJ= − = +

2 3 2 3 3 32 4 23 3DF L F L U FL FLU yEI GJ F EI GJ

∂= + = = +

∂How can we solve this problem as easily without castigliano?

Dummy load method

• When we want to calculate displacements or rotation at a point where no forces are applied use the following procedure

1. Apply a “dummy” load at the point2. Calculate the energy in term of actual forces

plus dummy load3. Take derivative with respect to magnitude of

dummy load4. Set magnitude of dummy load to zero

Example

• What is the rotation at point D?• Add clockwise moment m at point D• Examine contribution of member CD

• Member BC will have the same contribution in bending and in addition a torque m

( )2 2

00 02 2

L LCD

CDm

Fx m dx U Fxdx FLM Fx m UEI m EI EI=

+ ∂= + = = =

∂∫ ∫

( ) ( ) ( )2 2

02

BC torsionBC torsion

m

UT m L TL FLUGJ m GJ GJ

=

∂+= = =

∂

Example continued

• Member AB

• Altogether

• Please check!

( ) ( ) ( )

( ) ( )

2

0

2

00

2

2

L

AB AB bending

LAB bending

m

Fx m FL dxM Fx FL m U

EI

U Fx FL dx FLm EI EI

=

+ −= − − =

∂ −= = −

∂

∫

∫

2

DFLGJ

θ =

Maxwell Reciprocal Theorem(?James Clerk Maxwell 1831-1879)

• If we have two generalized loads Q1 and Q2, Strain energy will be a quadratic form in both

• Generalized coordinate q2 when only Q1 applies

• Generalized coordinate q1 when only Q2 applies

• Instead of measuring rotation due to force may be more convenient to measure displacement due to moment

2 21 1 2 2U aQ bQ Q cQ= + +

2

12 0Q

U bQQ

=

∂=

∂

1

21 0Q

U bQQ

=

∂=

∂

Simple example

• What is the rotation of a cantilever beam under end force?– Requires dealing with bending moment that is function

of location

• Instead calculate displacement at tip due to end moment (set M=P)

2

2

22

( ) (0) '(0) 0

( / 2) '( )2

d wEI P L x w wdx

dw P PLLx x w Ldx EI EI

= − = =

= − =

2

2

22

(0) '(0) 0

( )

d wEI M w wdx

M MLw x w L2 2EI EI

= = =

= =Beware of units!

Reading assignmentSection 5.5: What is more difficult about applying Castigliano’s

theorem to statically indeterminate structures compared to statically determinate ones?

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