Question 2.11: Two shafts of same diameter are used to transmit same power....

Let the diameter of both the shafts be ‘d’ and the power transmitted by each of them be P. As we know power is

P=\frac{2 \pi N T}{60}

Therefore,

\text { Torque, } T=\frac{60 P}{2 \pi N}=\frac{K}{N}

where K = 60P/2π. Now, the stress developed in the shaft is shear stress which varies linearly following the equation τ = Tρ /J with ρ as radius. At ρ = 0, that is, at the centre, shear stress is zero and it is maximum at the circumference

\tau_{\max }=\frac{16 T}{\pi d^3}

Replacing T with K /N, we get

\tau_{\max }=\frac{16}{\pi d^3} \times \frac{K}{N}

If diameter is the same for two shafts, then 16K/πd³ is also same. Therefore,

\tau_{\max }=\frac{K^{\prime}}{N}          (1)

where

K^{\prime}=\frac{16 K}{\pi d^3}=\frac{16}{\pi d^3} \times \frac{60 P}{2 \pi}=\text { constant }

which is a constant for given diameter and transmitted power.
From Eq. (1), we get maximum shear stress occurring at the circumference varies inversely with rpm, so for the above-mentioned two shafts shear stresses will be different and will be generally more for the shaft with lower rpm.