Page 191 - Shigley's Mechanical Engineering Design

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166 Mechanical Engineering Design
The relative contribution of transverse shear to beam deflection decreases as the
length-to-height ratio of the beam increases, and is generally considered negligible for
l/d > 10. Note that the deflection equations for the beams in Table A–9 do not include
the effects of transverse shear.
Castigliano’s theorem can be used to find the deflection at a point even though no
force or moment acts there. The procedure is:
1 Set up the equation for the total strain energy U by including the energy due to a
fictitious force or moment Q acting at the point whose deflection is to be found.
2 Find an expression for the desired deflection δ, in the direction of Q, by taking the
derivative of the total strain energy with respect to Q.
3 Since Q is a fictitious force, solve the expression obtained in step 2 by setting Q
equal to zero. Thus, the displacement at the point of application of the fictitious
force Q is

∂U
δ = (4–28)
∂Q Q=0
In cases where integration is necessary to obtain the strain energy, it is more effi-
cient to obtain the deflection directly without explicitly finding the strain energy, by
moving the partial derivative inside the integral. For the example of the bending case,
∂M
2M
∂U ∂ M 2 ∂ M 2 ∂F i

δ i = = dx = dx = dx
∂F i ∂F i 2EI ∂F i 2EI 2EI
1 ∂M

= M dx
EI ∂F i
This allows the derivative to be taken before integration, simplifying the mathematics.
This method is especially helpful if the force is a fictitious force Q, since it can be set
to zero as soon as the derivative is taken. The expressions for the common cases in
Eqs. (4–17), (4–19), and (4–23) are rewritten as

∂U 1 ∂F
δ i = = F dx tension and compression (4–29)
∂F i AE ∂F i
∂U 1 ∂T
θ i = = T dx torsion (4–30)
∂M i GJ ∂M i
∂U 1 ∂M
δ i = = M dx bending (4–31)
∂F i EI ∂F i

EXAMPLE 4–10 Using Castigliano’s method, determine the deflections of points A and B due to the
force F applied at the end of the step shaft shown in Fig. 4–10. The second area
moments for sections AB and BC are I 1 and 2I 1 , respectively.

Solution To avoid the need to determine the ground reaction forces, define the origin of x at the
left end of the beam as shown. For 0 ≤ x ≤ l, the bending moment is
M =−Fx (1)