Question

Two thin-walled tubular members made of the same material have the same length, same wall thickness and same total weight and are both subjected to the same torque of magnitude T. If the individual cross-section are circular and square, respectively, as in the figures, then the ratios of the shear stresses reckoned for the circular member in relation to the square member will be

• A. 0.785
• B. 0.905
• C. 0.616
• D. 0.513

Answer 1

The correct option is A 0.785

For tubular cross-section

${\tau }_{max}=\frac{T}{2A_{m}t}$

$A_m=$ Cross-sectional area under mean line.

$\therefore \frac{{\tau }_{ma{x}_c}}{{\tau }_{ma{x}_s}}=\frac{A_{m_s}}{A_{m_c}}$


Since weight is same for both tubes

$\therefore A_C=A_S$

$\frac{\pi }{4}[d_0^2-d_i^2]=b_0^2-b_i^2$

$\frac{\pi }{4}\left[\right(d+2t{)}^2-d^2]=(b+2t{)}^2-b^2$

$\frac{\pi }{4}.4t(t+d)=4t(t+b)$

$\therefore \frac{t+b}{t+d}=\frac{\pi }{4}$

$\therefore \frac{A_{m_s}}{A_{m_c}}=\frac{(t+b{)}^2}{\frac{\pi }{4}(d+t{)}^2}=\frac{\pi }{4}=0.7853$