Question

The diameter of a solid shaft made of mild steel rotating at 200 RPM is 40 mm. The shaft is designed to transmit 40 kW. What will be the factor of safety if the ultimate shear stress at yield is 417 N/mm²?
1. 4.5
2. 3.5
3. 2.7
4. 3.8

Answer 1

Correct Answer - Option 3 : 2.7

Concept:

\(factor\;of\;safety (N) = \frac{{maximum\;stress\;permissible}}{{maximum \;stress\;obtained}}\)      (1)

Here,

maximum stress permissible = ultimate shear stress at yield = 417 N/mm²

maximum stress obtained is torsional shear stress calculated as follows

Calculation:

Given:

P = 40 kW

D = 40 mm

\(\omega = \frac{{200 \times 2 \times \pi }}{{60}}\frac{{rad}}{s}\)

\(\begin{array}{l} P = T\omega \\ \end{array}\)    (2)

By substituting the given values in (1), we get 

T = 1909.859 Nm

\(\begin{array}{l} \mathop \tau \nolimits_{\max } = \frac{{16T}}{{\pi \mathop D\nolimits^3 }}\\ \mathop \tau \nolimits_{\max } = \frac{{16 \times 1909.859}}{{\pi \mathop {0.04}\nolimits^3 }}\\ \mathop \tau \nolimits_{\max } = 151.98MPa\\ \end{array}\) 

From (1),

\(N = \frac{{417}}{{151.98}}\)

N = 2.743

Factor of safety = 2.743