Page 190 - Shigley's Mechanical Engineering Design

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Deflection and Stiffness 165
terms force and displacement in this statement are broadly interpreted to apply equally
to moments and angular displacements. Mathematically, the theorem of Castigliano is
∂U
δ i = (4–26)
∂F i
where δ i is the displacement of the point of application of the force F i in the direction
of F i . For rotational displacement Eq. (4–26) can be written as
∂U
θ i = (4–27)
∂M i
where θ i is the rotational displacement, in radians, of the beam where the moment
M i exists and in the direction of M i .
As an example, apply Castigliano’s theorem using Eqs. (4–16) and (4–18) to get
the axial and torsional deflections. The results are
2
∂ F l Fl
δ = = (a)
∂F 2AE AE
2
∂ T l Tl
θ = = (b)
∂T 2GJ GJ
Compare Eqs. (a) and (b) with Eqs. (4–3) and (4–5).

EXAMPLE 4–9 The cantilever of Ex. 4–8 is a carbon steel bar 10 in long with a 1-in diameter and is
loaded by a force F = 100 lbf.
(a) Find the maximum deflection using Castigliano’s theorem, including that due to shear.
(b) What error is introduced if shear is neglected?
Solution (a) From Ex. 4–8, the total energy of the beam is
2
2 3
F l 1.11F l
U = + (1)
6EI 2AG
Then, according to Castigliano’s theorem, the deflection of the end is
∂U Fl 3 1.11Fl
y max = = + (2)
∂F 3EI AG
We also find that
πd 4 π(1) 4 4
I = = = 0.0491 in
64 64
πd 2 π(1) 2 2
A = = = 0.7854 in
4 4
Substituting these values, together with F = 100 lbf, l = 10 in, E = 30 Mpsi, and
G = 11.5 Mpsi, in Eq. (3) gives
Answer y max = 0.022 63 + 0.000 12 = 0.022 75 in

Note that the result is positive because it is in the same direction as the force F.
Answer (b) The error in neglecting shear for this problem is (0.02275 − 0.02263)/0.02275 =
0.0053 = 0.53 percent.