Question 2.6: A solid circular shaft transmits 736 kW power at 200 rpm. If...

The power transmitted is

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

Here, power P = 736 kW = 736 ×10³ W and N = 200 rpm. Substituting these, we get

736 \times 10^3=\frac{2 \pi \times 200 \times T}{60}

or        T=35.14 \times 10^3  Nm =35.14 \times 10^6  N mm

Now, if we consider maximum angle of twist, θ = TL/GJ which is allowed (here, L = 15d, where d is the diameter of the shaft)

\theta=0.0348=\frac{35.14 \times 10^3 \times 15 d}{80 \times 10^3 \times \frac{\pi}{32} \times d^4}

or        d = 156.66 mm           (1)

Similarly, if we consider allowable shear stress,

\tau=\frac{T(d / 2)}{J}=\frac{16 T}{\pi d^3}

or    80=\frac{35.14 \times 10^6 \times 16}{\pi d^3}

or      d = 130.7 mm             (2)

From the above-mentioned two cases, the suitable diameter is obviously, maximum of the values in Eqs. (1) and (2). that is, 156.66 mm.