Page 194 - Shigley's Mechanical Engineering Design

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Deflection and Stiffness 169
where l is the length of the member. So for the torsion in member CD, F i = F, T = Fa,
and l = b. Thus,
2
∂U CD b Fa b
= (Fa)(a) = (2)
∂F torsion GJ GJ
For the bending in CD,
b 3
∂U CD 1 Fb
= (Fx)(x) dx = (3)
∂F bending EI 0 3EI
Member DG is axially loaded and is bending in two planes. The axial loading is
constant, so Eq. (4–29) can be written as
∂U ∂F l
= F
∂F i ∂F i AE
where l is the length of the member. Thus, for the axial loading of DG, F i = F, l = c,
and

∂U DG Fc
= (4)
∂F axial AE
The bending moments in each plane of DG are constant along the length, with
M DG2 = Fb and M DG1 = Fa. Considering each one separately in the form of
Eq. (4–31) gives
c c
∂U DG 1 1
= (Fb)(b) dx + (Fa)(a) dx (5)
∂F EI EI
bending 0 0
2
2
Fc(a + b )
=
EI
4 2
Adding Eqs. (1) to (5), noting that I = πd /64, J = 2I, A = πd /4, and G =
E/[2(1 + ν)], we find that the deflection of B in the direction of F is
4F
Answer (δ B ) = [16(a + b ) + 48c(a + b ) + 48(1 + ν)a b + 3cd ]
3
2
2
2
2
3
F
3π Ed 4
Now that we have completed the solution, see if you can physically account for each
term in the result using an independent method such as superposition.
4–9 Deflection of Curved Members
Machine frames, springs, clips, fasteners, and the like frequently occur as curved
shapes. The determination of stresses in curved members has already been described in
Sec. 3–18. Castigliano’s theorem is particularly useful for the analysis of deflections in
6
curved parts too. Consider, for example, the curved frame of Fig. 4–12a. We are

6 For more solutions than are included here, see Joseph E. Shigley, “Curved Beams and Rings,” Chap. 38 in
Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine
Design, 3rd ed., McGraw-Hill, New York, 2004.