The cantilevered hook shown in Fig. 4–13a is formed from a round steel wire with a diameter of 2 mm. The hook dimensions are l = 40 \ and \ R = 50 \ mm.
A force P of 1 N is applied at point C.
Use Castigliano’s theorem to estimate the deflection at point D at the tip.
Since {l}/{d} \ and \ {R}/{d} are significantly greater than 10, only the contributions due to bending will be considered.
To obtain the vertical deflection at D, a fictitious force Q will be applied there. Free-body diagrams are shown in Figs. 4–13b, c, and d, with breaks in sections AB, BC, and CD, respectively. The normal and shear forces, N and V respectively, are shown but are considered negligible in the deflection analysis.
For section AB, with the variable of integration x defined as shown in Fig. 4–13b,summing moments about the break gives an equation for the moment in section AB,
M_{AB}= P\left(R + x\right) +Q\left(2R+x\right) (1)
{∂M_{AB}}/{∂Q}=2R+x (2)
Since the derivative with respect to Q has been taken, we can set Q equal to zero. From Eq. (4–31) inserting Eqs. (1) and (2),
\delta _i=\frac{∂U}{∂F_i} \int{\frac{1}{EI}\left(M\frac{∂M}{∂F_i}\right) dx } \ \ bending (4–31)
\begin{aligned}\left(\delta_{D}\right)_{A B} &=\int_{0}^{l} \frac{1}{E I}\left(M_{A B} \frac{\partial M_{A B}}{\partial Q}\right) d x=\frac{1}{E I} \int_{0}^{l} P(R+x)(2 R+x) d x \\\\&=\frac{P}{E I} \int_{0}^{l}\left(2 R^{2}+3 R x+x^{2}\right) d x=\frac{P}{E I}\left(2 R^{2} l+\frac{3}{2} l^{2} R+\frac{1}{3} l^{3}\right)\end{aligned} (3)
For section BC, with the variable of integration θ defined as shown in Fig. 4–13c, summing moments about the break gives the moment equation for section BC.
M_{BC}=Q\left(R+R\sin \theta \right) +PR\sin \theta (4)
{∂M_{BC}}/{∂Q}=R\left(1+\sin \theta \right) (5)
From Eq. (4–41), inserting Eqs. (4) and (5) and setting Q = 0, we get
\left(δ_D\right)_{BC}=\int_{0}^{{\pi }/{2}}{\frac{1}{EI}\left(M_{BC}\frac{∂M_{BC}}{∂Q} \right)Rd\theta }=\frac{R}{EI} \int_{0}^{{\pi }/{2}}{\left(PR\sin \theta \right)\left[R\left(1+\sin \theta \right) \right]dx}
=\frac{PR^3}{EI} \left(1+\frac{\pi }{4} \right) (6)
Noting that the break in section CD contains nothing but Q, and after setting Q = 0, we can conclude that there is no actual strain energy contribution in this section. Combining terms from Eqs. (3) and (6) to get the total vertical deflection at D,
\begin{aligned}\delta_{D} &=\left(\delta_{D}\right)_{A B}+\left(\delta_{D}\right)_{B C}=\frac{P}{E I}\left(2 R^{2} l+\frac{3}{2} l^{2} R+\frac{1}{3} l^{3}\right)+\frac{P R^{3}}{E I}\left(1+\frac{\pi}{4}\right) \\\\&=\frac{P}{E I}\left(1.785 R^{3}+2 R^{2} l+1.5 R l^{2}+0.333 l^{3}\right)\end{aligned} (7)Substituting values, and noting I = {\pi d^4}/{64} \ , \ and \ E = 207 \ GPa for steel, we get
\begin{aligned}\delta_{D}=& \frac{1}{207\left(10^{9}\right)\left[\pi\left(0.002^{4}\right) / 64\right]}\left[1.785\left(0.05^{3}\right)+2\left(0.05^{2}\right) 0.04\right.\\\\&\left.+1.5(0.05) 0.04^{2}+0.333\left(0.04^{3}\right)\right] \\\\=& 3.47\left(10^{-3}\right) \mathrm{m}=3.47 \mathrm{~mm}\end{aligned}
