GATE 2022 ME (ME2) – Question 53
A shaft AC rotating at a constant speed carries a thin pulley of radius $r=0.4$ m at the end C which drives a belt. A motor is coupled at the end A of the shaft such that it applies a torque $M_z$ about the shaft axis without causing any bending moment. The shaft is mounted on narrow frictionless bearings at A and B where AB = BC = L = 0.5 m. The taut and slack side tensions of the belt are $T_1=300$ N and $T_2=100$ N, respectively. The allowable shear stress for the shaft material is 80 MPa. The self-weights of the pulley and the shaft are negligible. Use the value of π available in the on-screen virtual calculator. Neglecting shock and fatigue loading and assuming maximum shear stress theory, the minimum required shaft diameter is ______ mm (round off to 2 decimal places).
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Correct answer: 23.82 to 24.06
Explanation
The shaft carries a torque from the motor and is bent by the belt tensions at the overhung pulley.
**Step 1: loads on the pulley.** Both belt tensions act vertically downward, so the load on the shaft is
$$P=T_1+T_2=300+100=400\text{ N}.$$
**Step 2: bending moment.** The pulley is at C, an overhang of $L=0.5$ m beyond the bearing B, so the maximum bending moment is at B:
$$M=P\,L=400\times0.5=200\text{ N m}.$$
**Step 3: torque.**
$$T=(T_1-T_2)\,r=(300-100)\times0.4=80\text{ N m}.$$
**Step 4: maximum shear stress theory.** For a solid shaft of diameter $d$,
$$\tau_{max}=\frac{16}{\pi d^3}\sqrt{M^2+T^2}\leq80\times10^6\text{ Pa}.$$
$$\sqrt{200^2+80^2}=215.4\text{ N m}$$
$$d^3=\frac{16(215.4)}{\pi(80\times10^6)}=1.3714\times10^{-5}\text{ m}^3\;\Rightarrow\;d=0.02394\text{ m}.$$
The minimum diameter is **23.94 mm**.