Question 10.29: A thin curved bar AB has a centreline in the form of a quart...
A thin curved bar AB has a centreline in the form of a quarter circle of radius R as shown in Figure 10.59. Find \delta_h, \delta_v and angular rotation θ of end B.
Let us consider the free-body diagrams of the bar assuming a fictitious load Q (vertical) and a moment M_{ o } at end B and that of an arbitrary section defined by the angle \phi as shown in Figures 10.60(a) and (b), respectively:
From Figure 10.60(b), taking moment about C, we get the bending moment M_x of the arbitrary section as follows:
M_x=P R(1-\cos \phi)+Q R \sin \phi+M_{ o } ; \quad 0 \leq \phi \leq \pi / 2
Now,
\theta_{ B }=\left\lgroup\frac{\partial U}{\partial M_{ o }} \right\rgroup_{M_{ o }=0}=\left\lgroup\frac{1}{E I} \right\rgroup\left[\int_0^{\pi / 2} M_x\left\lgroup\frac{\partial M_x}{\partial M_{ o }} \right\rgroup R d \phi\right]_{M_{ o }=0}
as differential length of bar, d s=R d \phi . So
\begin{aligned} \theta_{ B } & =\left\lgroup \frac{1}{E I}\right\rgroup\left[\int_0^{\pi / 2} P R^2(1-\cos \phi) d \phi\right]=\left\lgroup \frac{P R^2}{E I}\right\rgroup [\phi-\sin \phi]_0^{\pi / 2} \\ & =\left\lgroup \frac{P R^2}{E I}\right\rgroup\left[\frac{\pi}{2}-1\right] \end{aligned}
or \theta_{ B }=\frac{(\pi-2) P R^2}{2 E I} ⦪
Again vertical component of displacement of end B of the curved bar is
\left(\delta_{ v }\right)_{ B }=\left\lgroup \frac{\partial U}{\partial Q} \right\rgroup_{Q=0}
Considering M_{ o }=0 , we get
\left(\delta_{ v }\right)_{ B }=\left\lgroup \frac{1}{E I} \right\rgroup\left[\int_0^{\pi / 2} M_x \frac{\partial M_x}{\partial Q}(R d \phi)\right]_{Q=0}
\begin{aligned} & =\left\lgroup \frac{1}{E I} \right\rgroup\left[\int_0^{\pi / 2} P R^3(1-\cos \phi) \sin \phi d \phi\right] \\ & =\left\lgroup \frac{P R^3}{E I} \right\rgroup\left[\int_0^{\pi / 2}\left\lgroup \sin \phi-\frac{\sin 2 \phi}{2} \right\rgroup d \phi\right] \\ & =\left\lgroup \frac{P R^3}{E I} \right\rgroup\left[-\cos \phi+\frac{\cos 2 \phi}{4}\right]_0^{\pi / 2} \\ & =\left\lgroup \frac{P R^3}{E I} \right\rgroup\left[1+\left\lgroup \frac{-1}{4}-\frac{1}{4} \right\rgroup\right]=\frac{P R^3}{E I}\left\lgroup 1-\frac{1}{2} \right\rgroup \end{aligned}
Therefore,
\left(\delta_{ v }\right)_{ B }=\frac{P R^3}{2 E I}(\downarrow)
Finally, horizontal component of displacement of end B of the curved bar is
\left(\delta_{ h }\right)_{ B }=\frac{\partial U}{\partial P}, \text { considering } Q=M_{ o }=0
Therefore,
\begin{aligned} \left(\delta_{ h }\right)_{ B } & =\left\lgroup \frac{1}{E I} \right\rgroup\int_0^{\pi / 2} M_x\left\lgroup \frac{\partial M_x}{\partial P} \right\rgroup (R d \phi) \\ & =\left\lgroup \frac{1}{E I} \right\rgroup \int_0^{\pi / 2} P R^3(1-\cos \phi)^2 d \phi \\ & =\left\lgroup \frac{P R^3}{E I} \right\rgroup \int_0^{\pi / 2}\left(1-2 \cos \phi+\cos ^2 \phi\right) d \phi \\ & =\left\lgroup \frac{P R^3}{E I} \right\rgroup\int_0^{\pi / 2}\left\lgroup \frac{3}{2}-2 \cos \phi+\frac{\cos 2 \phi}{2} \right\rgroup d \phi \\ & =\left\lgroup \frac{P R^3}{E I} \right\rgroup\left[\frac{3}{2} \phi-2 \sin \phi+\frac{\sin 2 \phi}{4}\right]_0^{\pi / 4} \\ & =\left\lgroup \frac{P R^3}{E I} \right\rgroup\left\lgroup \frac{3 \pi}{4}-2 \right\rgroup =\frac{(3 \pi-8) P R^3}{4 E I} \end{aligned}
Thus,
\left(\delta_{ h }\right)_{ B }=\frac{(3 \pi-8) P R^3}{4 E I}(\rightarrow)
