Question 10.31: A slender curved bar AB lying in a horizontal plane with its...
A slender curved bar AB lying in a horizontal plane with its centre line forming a quarter circle of radius R shown in Figure 10.63 is subjected to a vertical force P along the negative z-direction at its free end. The bar has flexural rigidity = EI and torsional rigidity = GJ. Calculate the vertical deflection \left(\delta_v\right)_B and angle of twist \phi_B at end B.
Let us consider the free-body diagram of the bar as shown in Figure 10.64:
In Figure 10.64(a), we show the free-body diagram of the bar. We take an arbitrary section defined by the angle \theta(0 \leq \theta \leq \pi / 2) \text {. Clearly at } B^{\prime} , bending moment, torsional moment and shear force in z-direction act. In Figure 10.64(b), we show the free-body diagram of the portion BB’ of the slender bar in x-y plane. To compute bending moment \left(M_{ b }\right)_{ B ^{\prime}} \text { at } B ^{\prime} , we need to calculate moment of P about OB’ [refer to Figure 10.64(b)]. Similarly, torsional moment \left(M_{ t }\right)_{ B ^{\prime}} \text { at } B ^{\prime} will be the moment of P about BC [refer to Figure 10.64(b)]. Clearly from Figure 10.64(b), BC = R sin θ and B’C = R(1−cos θ). Therefore,
\left(M_{ b }\right)_{ B ^{\prime}}=(P)( BC )=P R \sin \theta
and \left(M_{ t }\right)_{ B ^{\prime}}=(P)\left( B ^{\prime} C \right)=P R(1-\cos \theta)
Therefore, strain energy in the slender bar AB will be due to bending as well as due to torsion:
U_{ AB }=U_{\text {bending }}+U_{\text {torsion }}
=\left\lgroup \frac{1}{2 E I} \right\rgroup \int_0^{\pi / 2}\left(M_{ b }\right)_{ B ^{\prime}}^2(R d \theta)+\left\lgroup \frac{1}{2 G J} \right\rgroup \int_0^{\pi / 2} \left(M_{ t }\right)_{ B ^{\prime}}^2(R d \theta)
The vertical deflection is
\begin{aligned} \frac{\partial U_{ AB }}{\partial P}=\left(\delta_{ v }\right)_{ B }= & \left\lgroup\frac{1}{E I} \right\rgroup\int_0^{\pi / 2}\left(M_{ b }\right)_{ B ^{\prime}}\left\lgroup\frac{\partial M_{ b }}{\partial P} \right\rgroup (R d \theta) \\ & +\left\lgroup\frac{1}{G J} \right\rgroup \int_0^{\pi / 2} \left(M_{ t }\right)_{ B ^{\prime}}\left\lgroup\frac{\partial M_{ t }}{\partial P} \right\rgroup (R d \theta) \end{aligned}
Therefore,
\begin{aligned} \left(\delta_{ v }\right)_{ B } & =\left\lgroup \frac{1}{E I} \right\rgroup\left[\int_0^{\pi / 2} P R^3 \sin ^2 \theta d \theta\right]+\left\lgroup \frac{1}{G J} \right\rgroup\left[\int_0^{\pi / 2} P R^3(1-\cos \theta)^2 d \theta\right] \\ & =\left\lgroup \frac{1}{E I} \right\rgroup\left\lgroup \frac{P R^3}{2} \right\rgroup\left[\int_0^{\pi / 2}(1-\cos 2 \theta) d \theta\right]+\left\lgroup \frac{P R^3}{G J} \right\rgroup\left[\int_0^{\pi / 2}(1-\cos \theta)^2 d \theta\right] \\ & =\frac{P R^3}{2 E I}\left[\frac{\pi}{2}\right]+\left\lgroup \frac{P R^3}{G J} \right\rgroup\left\lgroup \frac{3 \pi-8}{4} \right\rgroup \end{aligned}
as \int_0^{\pi / 2}(1-\cos \theta)^2 d \theta=\left\lgroup\frac{3 \pi-8}{4} \right\rgroup
using our previous result in Example 10.30. Therefore,
\left(\delta_{ v }\right)_{ B }=\frac{\pi P R^3}{4 E I}+\frac{(3 \pi-8) P R^3}{4 G J}(\downarrow)
To find \phi_{ B } , we need to apply a dummy moment, T_{ o } at end B as shown in Figure 10.65:
Clearly,
\begin{aligned} \left(M_{ b }\right)_{B^{\prime}} & =\left[P R \sin \theta+T_{ o } \cos \left\lgroup \frac{\pi}{2}-\theta \right\rgroup\right] \\ & =\left[P R \sin \theta+T_{ o } \sin \theta\right] \end{aligned}
which is the bending moment at B’ and
\left(M_{ t }\right)_{ B ^{\prime}}=P R(1-\cos \theta)-T_{ o } \cos \theta
which is the torsional moment at B’. Therefore,
\left\lgroup \frac{\partial M_{ b }}{\partial T_{ o }} \right\rgroup_{T_{ o }=0}=\sin \theta ; \quad\left\lgroup \frac{\partial M_{ t }}{\partial T_{ o }} \right\rgroup_{T_{ o }=0}=-\cos \theta
and so,
\phi_{ B }=\left\lgroup \frac{\partial U}{\partial T_{ o }} \right\rgroup_{T_{ o }=0}
Thus,
\begin{aligned} \phi_{ B } & =\left\lgroup \frac{1}{E I} \right\rgroup\left[\int_0^{\pi / 2}\left(M_{ b }\right)\left\lgroup \frac{\partial M_{ b }}{\partial T_{ o }} \right\rgroup (R d \theta)\right]_{T_{ o }=0}+\left\lgroup \frac{1}{G J} \right\rgroup\left[\int_0^{\pi / 2}\left(M_{ t }\right)\left\lgroup \frac{\partial M_{ t }}{\partial T_{ o }} \right\rgroup (R d \theta)\right]_{T_{ o }=0} \\ & =\left\lgroup \frac{1}{E I} \right\rgroup\left[\int_0^{\pi / 2} P R^2 \sin ^2 \theta d \theta\right]+\left\lgroup \frac{1}{G J} \right\rgroup\left[\int_0^{\pi / 2} P R^2(1-\cos \theta)(-\cos \theta) d \theta\right] \\ & =\left\lgroup \frac{1}{E I} \right\rgroup\left\lgroup \frac{\pi P R^2}{4} \right\rgroup -\frac{P R^2}{G J}\left\lgroup 1-\frac{\pi}{4} \right\rgroup \\ & =\frac{\pi P R^2}{4 E I}-\frac{(4-\pi) P R^2}{4 G J} \end{aligned}
or \phi_{ B }=\left[\left\lgroup \frac{\pi}{E I} \right\rgroup-\left\lgroup \frac{4-\pi}{G J} \right\rgroup\right] P R^2
Direction of twist depend on relative strengths (EI ) and (GJ ).
