Question 2.9: For the power transmission system (as shown in Figure 2.10),...

Let P be the power which is being transmitted through the system and ω be the angular speed. Then torque acting on the input shaft is

T_{\text {in }}=\frac{P}{\omega_{\text {in }}}

Now, let us draw the free-body diagrams of the shafts

If we consider the free-body diagram shown in Figures 2.11(a) and (b), we can write

F_{ t } \times \frac{d_{ b }}{2}=T_{\text {in }} \quad \text { and } \quad T_{\text {out }}=F_{ t } \times \frac{d_{ c }}{2}

Therefore,

\frac{T_{ out }}{T_{ in }}=\frac{d_{ c }}{d_{ b }}           (1)

This can be verified considering no power loss in the transmission system and writing

\frac{T_{\text {out }}}{T_{\text {in }}}=\frac{\omega_{\text {in }}}{\omega_{\text {out }}}=\frac{\omega_{ B }}{\omega_{ C }}=\frac{d_{ c }}{d_{ b }}

This equality \left(\omega_{ B } / \omega_{ C }\right)=\left(d_{ c } / d_{ b }\right) holds goods as the tangential pitch line velocity is same for both B and C. If we assume shear stress developed in the shafts to be same, we can write

\frac{T_{\text {out }}}{T_{\text {in }}}=\frac{\frac{\pi}{16} d_2^3}{\frac{\pi}{16} d_1^3}=\frac{d_2^3}{d_1^3}          (2)

From Eqs. (1) and (2),

\frac{d_2^3}{d_1^3}=\frac{d_{ c }}{d_{ b }}

or        \left\lgroup \frac{d_2}{d_1} \right\rgroup =\sqrt[3]{\frac{d_{ c }}{d_{ b }}}=\sqrt[3]{2} \quad\left\lgroup \text { as } \frac{d_{ c }}{d_{ b }}=\frac{2}{1} \right\rgroup

Therefore,

\frac{d_1}{d_2}=\frac{1}{\sqrt[3]{2}}=0.7936

Thus, the ratio of shaft diameters is 0.7936 : 1.