The solid shaft as shown in Figure 2.9 is made of steel. It has a uniform diameter of 40 mm and it runs at 525 rpm. A driving belt feeds 50 hp to pulley A; whereas 30 hp and 20 hp, respectively, are taken off from the other two pulleys, B and C. Find out the maximum shear stress induced in the

Question

The solid shaft as shown in Figure 2.9 is made of steel. It has a uniform diameter of 40 mm and it runs at 525 rpm. A driving belt feeds 50 hp to pulley A; whereas 30 hp and 20 hp, respectively, are taken off from the other two pulleys, B and C. Find out the maximum shear stress induced in the shaft and the total angle of twist. Take G = 100 GPa.

Accepted Answer

We know that 1 horse power (hp) = 736 W. Therefore, power input at A is

50 hp = 50 × 736 = 36.8 kW = 36800 W

Similarly, power outputs at B and C are 22.08 × 10³ W and 14.72 × 10³ W, respectively. As the shaft is of uniform cross-section, we can try to find out maximum shear stress by selecting the portion of the shaft having maximum torque. Now torque at A is

[latex] T_{ A }=\frac{P_{ A } imes 60}{2 \pi N} [/latex]

where [latex] T_{ A } ext { is torque at A, } P_{ A } [/latex] is the power input at A and N is rpm. Substituting the values, we get

[latex] T_{ A }=\frac{36800 imes 60}{2 imes \pi imes 525}=669.6 Nm [/latex]

If we consider the portion ‘AB’, then the torque acting is 669.6 N m. For the portion ‘BC’, obviously the torque acting is less and it will be evident if we approach from right side. Torque at C will be definitely (20/50) or (2/5) times the torque at A. So, the shear stress is maximum in the portion ‘AB’.

[latex] au_{ AB }= au_{\max }=\frac{T_{ A } r}{J}=\frac{669.6 imes 20 imes 10^3}{\frac{\pi}{32} imes 40^4}=53.3 MPa [/latex]

Now, the angles of twist for portions ‘AB’ and ‘BC’ can be summed up to get the total angle of twist.
It is because with respect to A, both C and B points will rotate in the same direction. For portion ‘AB’, angle of twist of B with respect to A is [latex] heta_{ AB } [/latex]:

[latex] heta_{ AB }=\frac{T_{ AB } L_{ AB }}{G J} \quad\left( ext { here } T_{ AB }=T_{ A } ight) [/latex]

For portion ‘BC’, angle of twist of C with respect to B is [latex] heta_{ BC } [/latex]:

[latex] heta_{ BC }=\frac{T_{ BC } L_{ BC }}{G J} \quad\left( ext { here } T_{ BC }=T_{ C } ight) [/latex]

Total angle of twist, that is angle of twist of C with respect to [latex] A = heta_{ AB }+ heta_{ BC } [/latex]. Therefore, total angle of twist is

[latex] \begin{aligned} heta_{ AC } & = heta_{ AB }+ heta_{ BC } \ & =\frac{669.6 imes 10^3 imes 300}{100 imes 10^3 imes \frac{\pi}{32} imes 40^4}+\frac{267.8 imes 10^3 imes 600}{100 imes 10^3 imes \frac{\pi}{32} imes 40^4} \ & =0.146 rad =8^{\circ} 23^{\prime} \end{aligned} [/latex]

Thus, the maximum shear stress is 53.3 MPa and the total angle of twist is 8°23'.

Question 2.8: The solid shaft as shown in Figure 2.9 is made of steel. It ...

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