Answer to Question #191463 in Mechanical Engineering for Brian

Question

A solid shaft is to transmit 800kw at 300rpm. if the shaft is not to twist more than 1 degree on a length of 20 diameters and the shear stress is not to exceed 50mn/m2 with a modulus of rigidity of 90gn/m2, design the shaft by determining a suitable diameter

Answer 1

Given:

$\text{Power}(P)=800kW$

RPM(N)=300rpm

Maximum angle of twist $(\theta)$ =$1^0=\frac{\pi}{180}=0.01745\: \text{radians}$

Length (L)=20$\times \text{diameter}(d)$

maximum shear stress($\tau_s$ )=50N/mm²

Modulus of rigidity(G)=90 G N/mm²

To find : Suitable diameter of shaft

Formula : Torsional equation

$\frac{\text{Torque}(T)}{\text{Polar M I}(I_p)}=\frac{\tau_s}{\text{Radius}(R)}=\frac{G\times \theta}{L}$

Power(P) =$\frac{2\pi \times N\times T}{60}$

Calculation:

Power(P) =$\frac{2\pi \times N\times T}{60}$

800x10³=$\frac{2\pi \times 300\times T}{60}$

T=25464.79 N-m

For solid shaft considering strength of shaft,

$T=\frac{\pi}{16}\times \tau_s\times d^3$

After substituting respective values , we get

Diameter(d)=137.40 mm=138 mm (approximately)

Considering angle of twist of shaft,

$T=\frac{\pi}{16}\times \frac{G\theta}{L}\times d^3$

$T=\frac{\pi}{16}\times \frac{90\times1000\times 0.01745}{20d}\times d^3$

$25464.79\times 10^3=\frac{\pi}{32}\times \frac{90\times1000\times 0.01745}{20d}\times d^4$

d=148.93 mm

Considering larger diameter as suitable diameter of shaft(d)=149 mm