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?

Question

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 [latex]F = 100 lbf[/latex].
(a) Find the maximum deflection using Castigliano’s theorem, including that due to shear.
(b) What error is introduced if shear is neglected?

Accepted Answer

(a) From Ex. 4–8, the total energy of the beam is

[latex]U=\frac{F^2l^3}{6EI} +\frac{1.11F^2l}{2AG} [/latex] (1)

Then, according to Castigliano’s theorem, the deflection of the end is

[latex]y_{max}=\frac{∂U}{∂F} =\frac{Fl^3}{3EI} +\frac{1.11Fl}{AG} [/latex] (2)

We also find that

[latex]I=\frac{\pi d^4}{64} =\frac{\pi \left(1\right)^4 }{64} =0.0491 \ in^4[/latex]

[latex]A=\frac{\pi d^2}{4} =\frac{\pi \left(1\right)^2 }{4} =0.7854 \ in^2[/latex]

Substituting these values, together with [latex]F = 100 \ lbf[/latex] , [latex] l = 10 \ in \ , \ E = 30 \ Mpsi[/latex], and [latex]G = 11.5 [/latex] Mpsi, in Eq. (3) gives

[latex]y_{max} = 0.022 63 + 0.000 12 = 0.022 75[/latex] in

Note that the result is positive because it is in the same direction as the force F.

(b) The error in neglecting shear for this problem is [latex]\left(0.02275-0.02263\right) /{0.02275}=0.0053=0.53[/latex] percent.

Question 1.4.9: The cantilever of Ex. 4–8 is a carbon steel bar 10 in long w...

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