Question 12.10: For a torsional pendulum, length of the shaft = 500 mm, shaf...
For a torsional pendulum, length of the shaft = 500 mm, shaft diameter = 3 mm whereas for the shaft material, G = 84 GPa. What is the maximum weight of the pendulous disc in order that pendulum may have an amplitude, \phi=12^{\circ} during torsional oscillation without exceeding the allowable shear of 56 MPa.
Let the length of the pendulum shaft be L. When it is subjected to torque T, then its angle of twist is
\phi=\frac{T L}{G J}
where J is the polar moment of inertia of the shaft. Now,
T=\frac{G J \phi}{L}
Therefore, shear stress due to T is
\tau_{x y}=G \frac{R \phi}{L}
\text { as } \tau_{x y}=G \gamma_{x y} \text { and } \gamma_{x y}=R \phi / L . Putting the values, we get
\tau_{x y}=84\left(10^3\right) \frac{(1.5)}{500}\left\lgroup\frac{\pi}{180} \times 12 \right\rgroup MPa
= 52.78 MPa
and normal stress due to axial load (W ) is
\sigma_{x x}=\frac{W}{A}=\frac{4 W}{\pi(3)^2}=0.141 W MPa
Maximum shear stress on the shaft material is
\tau_{\max }=\sqrt{\left\lgroup \frac{\sigma_{x x}}{2} \right\rgroup^2+\tau_{x y}^2}
or \tau_{\max }=\sqrt{5\left(10^{-3}\right) w^2+52.78^2}=56
⇒ w = 264.68N
Thus, the safe weight of the pendulous mass is 264.68 N.