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 𝑀z 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 𝑇₁ = 300 N and 𝑇₂ = 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).

The correct answer is 23.93 (or 23.94, with 23.93 being the provided key).

1. Calculation of Twisting Moment (Torque)
The shaft carries a thin pulley of radius ^{r = 0.4 m = 400 mm} at end C.
The belt tensions are given as:
Taut side tension, _{T1 = 300 N}
Slack side tension, _{T2 = 100 N}
The twisting moment (torque) _Tmax is caused by the difference in these belt tensions acting at the radius of the pulley:
$$T_{\max }=(T_1-T_2)\times r$$
$$T_{\max }=(300-100)\times 0.4=80\text{N·m}=80\times {10}^3\text{N·mm}$$

2. Calculation of Bending Moment
The belt tensions also act as vertical downward forces on the pulley at C. The total vertical load at C is:
$$T=T_1+T_2=300+100=400\text{N (downward)}$$
The shaft is supported on frictionless bearings at A and B, with span lengths _{AB = BC = L = 0.5 m}.
Let _RA and _RB be the vertical reactions at bearings A and B respectively.
Taking the moment about point A to satisfy equilibrium:
$$\sum M_A=0\Rightarrow R_B\times 0.5-T\times (0.5+0.5)=0$$
$$R_B\times 0.5=400\times 1.0\Rightarrow R_B=800\text{N (upward)}$$
From vertical force equilibrium:
$$R_A+R_B-T=0\Rightarrow R_A=400-800=-400\text{N (downward)}$$
The maximum bending moment in the shaft occurs at bearing B:
$$M_{\max }=|-T\times L_{\mathrm{BC}}|=400\times 0.5=200\text{N·m}=200\times {10}^3\text{N·mm}$$

3. Application of Maximum Shear Stress Theory (Tresca's Criterion)
According to the maximum shear stress theory, the equivalent shear stress must not exceed the allowable shear stress (_{Sys = 80 MPa = 80 N/mm²}):
$${\tau }_{\max }=\frac{16}{\pi d^3}\sqrt{M_{\max }^2+T_{\max }^2}=S_{\mathrm{ys}}$$
Substituting the calculated values:
$$\frac{16}{\pi d^3}\sqrt{{(200\times {10}^3)}^2+{(80\times {10}^3)}^2}=80$$
$$\frac{16}{\pi d^3}\times 215406.59=80$$
$$d^3=\frac{16\times 215406.59}{80\times \pi }\approx 13713.21$$
$$d\approx 23.9357\text{mm}$$
Rounding to two decimal places, the minimum required shaft diameter is 23.93 mm.