Castigliano’s TheoremAlberto Castigliano, 1879

Castigliano’s second theorem• Applies only to structures that have constant temperature, unyielding supports, and linear elastic material response.

• The displacement to a point is equal to the first partial derivative of the strain energy in the structure with respect to a force acting at the point and in direction of the displacement.

• The slope at a point in a structure with respect to a couple moment acting at the point and in the direction of rotation.

• Method of least work

Castigliano’s Theorem for Trusses

• Where• = external joint displacement of the truss

• P = external force applied to the truss joint in the direction of

• N = internal force in a member caused by both the force P and the loads on the truss

• L = length of member• A = cross-sectional area of a member

• E = modulus of elasticity of a member

∆=∑ 𝑁 (𝜕𝑁𝜕 𝑃 ) 𝐿𝐴𝐸

Castigliano’s Theorem for Beams and Frames

• Where• = external displacement of the point caused by the real loads acting on the beam or frame

• P = external force applied to the beam or frame in the direction of

• M = internal moment in the beam or frame, expressed as a function of x and caused by both the force P and real loads on the beam.

• E = modulus of elasticity of a member

• I = moment of inertia of cross-sectional area computed about the neutral axis.If the slope at a point is to be determined, we must find the partial

derivative of the internal moment with respect to an external couple moment M’ acting at the point.

Use Castigliano’s Theorem to determine the horizontal and vertical displacement of joint B

Example 1

For horizontal displacement

P

member N N(P=0) LAB 0.625P -

66.6670.625 - 66.667 5 - 208.334

BC 0.625P + 66.667

0.625 66.667 5 208.334

AC -0.375P + 40 -0.375 40 6 -90=-90 kN2m

∆=−90

300(200)∆=−1.5𝑚𝑚

For vertical displacement

P

member N N(P=80) LAB - 0.8333P -0.8333 - 66.664 5 277.756BC 0.8333P 0.8333 66.664 5 277.756AC 0.5P 0.5 40 6 120

=675.512 kN2m

∆=∑ 𝑁 (𝜕𝑁𝜕 𝑃 ) 𝐿𝐴𝐸

∆=675.512300(200) ∆=11.26𝑚𝑚

Determine the max. deflection at point D

15 kN/m

180 kN

5 m 5 m 4 m

E = 70 GpaI = 2340 x 10 6 mm 4

segment

origin

M

AB A 0-5 -0.4x

78x

BC A 5-10 -0.4x

-102x + 90

DC D 0-4-x

∆=1𝐸𝐼∫0

5

78 𝑥 (−0.4 𝑥 ) 𝑑𝑥+ 1𝐸𝐼 ∫5

10

(−102 𝑥+900 ) (−0.4 𝑥 ) 𝑑𝑥+ 1𝐸𝐼∫0

4

−7.5 𝑥2 (− 𝑥 ) 𝑑𝑥

∆=∫0

𝐿

𝑀 (𝜕𝑀𝜕 𝑃 ) 𝑑𝑥𝐸𝐼

= -14.77 mm