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 δh, δv and angular rotation θ of end B.

Question

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 [latex] \delta_h, \delta_v [/latex] and angular rotation θ of end B.

Accepted Answer

Let us consider the free-body diagrams of the bar assuming a fictitious load Q (vertical) and a moment [latex] M_{ o } [/latex] at end B and that of an arbitrary section defined by the angle [latex] \phi [/latex] as shown in Figures 10.60(a) and (b), respectively:

From Figure 10.60(b), taking moment about C, we get the bending moment [latex] M_x [/latex] of the arbitrary section as follows:

[latex] M_x=P R(1-\cos \phi)+Q R \sin \phi+M_{ o } ; \quad 0 \leq \phi \leq \pi / 2 [/latex]

Now,

[latex] heta_{ B }=\left\lgroup\frac{\partial U}{\partial M_{ o }} ight group_{M_{ o }=0}=\left\lgroup\frac{1}{E I} ight group\left[\int_0^{\pi / 2} M_x\left\lgroup\frac{\partial M_x}{\partial M_{ o }} ight group R d \phi ight]_{M_{ o }=0} [/latex]

as differential length of bar, [latex] d s=R d \phi . So [/latex]

[latex] \begin{aligned} heta_{ B } & =\left\lgroup \frac{1}{E I} ight group\left[\int_0^{\pi / 2} P R^2(1-\cos \phi) d \phi ight]=\left\lgroup \frac{P R^2}{E I} ight group [\phi-\sin \phi]_0^{\pi / 2} \ & =\left\lgroup \frac{P R^2}{E I} ight group\left[\frac{\pi}{2}-1 ight] \end{aligned} [/latex]

or [latex] heta_{ B }=\frac{(\pi-2) P R^2}{2 E I} [/latex]⦪

Again vertical component of displacement of end B of the curved bar is

[latex] \left(\delta_{ v } ight)_{ B }=\left\lgroup \frac{\partial U}{\partial Q} ight group_{Q=0} [/latex]

Considering [latex] M_{ o }=0 [/latex], we get

[latex] \left(\delta_{ v } ight)_{ B }=\left\lgroup \frac{1}{E I} ight group\left[\int_0^{\pi / 2} M_x \frac{\partial M_x}{\partial Q}(R d \phi) ight]_{Q=0} [/latex]

[latex] \begin{aligned} & =\left\lgroup \frac{1}{E I} ight group\left[\int_0^{\pi / 2} P R^3(1-\cos \phi) \sin \phi d \phi ight] \ & =\left\lgroup \frac{P R^3}{E I} ight group\left[\int_0^{\pi / 2}\left\lgroup \sin \phi-\frac{\sin 2 \phi}{2} ight group d \phi ight] \ & =\left\lgroup \frac{P R^3}{E I} ight group\left[-\cos \phi+\frac{\cos 2 \phi}{4} ight]_0^{\pi / 2} \ & =\left\lgroup \frac{P R^3}{E I} ight group\left[1+\left\lgroup \frac{-1}{4}-\frac{1}{4} ight group ight]=\frac{P R^3}{E I}\left\lgroup 1-\frac{1}{2} ight group \end{aligned} [/latex]

Therefore,

[latex] \left(\delta_{ v } ight)_{ B }=\frac{P R^3}{2 E I}(\downarrow) [/latex]

Finally, horizontal component of displacement of end B of the curved bar is

[latex] \left(\delta_{ h } ight)_{ B }=\frac{\partial U}{\partial P}, ext { considering } Q=M_{ o }=0 [/latex]

Therefore,

[latex] \begin{aligned} \left(\delta_{ h } ight)_{ B } & =\left\lgroup \frac{1}{E I} ight group\int_0^{\pi / 2} M_x\left\lgroup \frac{\partial M_x}{\partial P} ight group (R d \phi) \ & =\left\lgroup \frac{1}{E I} ight group \int_0^{\pi / 2} P R^3(1-\cos \phi)^2 d \phi \ & =\left\lgroup \frac{P R^3}{E I} ight group \int_0^{\pi / 2}\left(1-2 \cos \phi+\cos ^2 \phi ight) d \phi \ & =\left\lgroup \frac{P R^3}{E I} ight group\int_0^{\pi / 2}\left\lgroup \frac{3}{2}-2 \cos \phi+\frac{\cos 2 \phi}{2} ight group d \phi \ & =\left\lgroup \frac{P R^3}{E I} ight group\left[\frac{3}{2} \phi-2 \sin \phi+\frac{\sin 2 \phi}{4} ight]_0^{\pi / 4} \ & =\left\lgroup \frac{P R^3}{E I} ight group\left\lgroup \frac{3 \pi}{4}-2 ight group =\frac{(3 \pi-8) P R^3}{4 E I} \end{aligned} [/latex]

Thus,

[latex] \left(\delta_{ h } ight)_{ B }=\frac{(3 \pi-8) P R^3}{4 E I}( ightarrow) [/latex]

Question 10.29: A thin curved bar AB has a centreline in the form of a quart...

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