Question

A solid circular bar of steel (G = 78 GPa) transmits a torque $T=360N-m$. The allowable stresses in tension, compression, and shear are 90 MPa, 70 MPa and 40 MPa respectively. Also the allowable tensile strain is $220\times {10}^{-6}$. The minimum required diameter(in mm) of the bar is__________. (Rounded to one decimal place).

• A. 37.7

Answer 1

The correct option is A 37.7
$T=360N-m,G=78\times {10}^{3}MPa,({\sigma }_t{)}_{allow}=90MPa$
$({\sigma }_c{)}_{allow}=70MPa,{\tau }_{allow}=40MPa,{\epsilon }_{max}=220\times {10}^{-6}$

Based on Allowable Stress:

When bar is under pure torsion, the maximum tensile,compressive and shear stresses are numerically equal .
$\therefore$ Design should be on lowest allowable stress (i.e. shear stress)

$\therefore {\tau }_{max}=\frac{16T}{\pi d^3}$

$d^3=\frac{16T}{\pi {\tau }_{allow}}=\frac{16\times 360\times {10}^3}{\pi \times 40}$

$d=35.8mm$

Based on Allowable Tensile Strain:

$\frac{{\varphi }_{xy}}{2}={\epsilon }_{max}\Rightarrow {\varphi }_{max}=2{\epsilon }_{max}$

and ${\tau }_{max}=G{\varphi }_{max}=G\left(2{\epsilon }_{max}\right)$

$\therefore {\tau }_{max}=\frac{16T}{\pi d^3}$

$d^3=\frac{16T}{\pi {\tau }_{max}}=\frac{16T}{\pi .2G{\epsilon }_{max}}=\frac{16\times 360\times {10}^3}{2\pi \times 78\times {10}^3\times 220\times {10}^{-6}}$

$\therefore d=37.7mm$