Question

Consider a solid round elastic bar with constant shear modulus, G, and cross-sectional area, A. The bar is built-in at both ends and subject to a spatially varying distributed torsional load t(x) = p sin( 2π L x) , where p is a constant with units of torque per unit length. Determine the location and magnitude of the maximum internal torque in the bar.

Answer 1

Answer:

$\t(x)_{max} =\dfrac{p\times L}{2\times \pi}$

Explanation:

Given that

Shear modulus= G

Sectional area = A

Torsional load,

$t(x) = p sin( \frac{2\pi}{ L} x)$

For the maximum value of internal torque

$\dfrac{dt(x)}{dx}=0$

Therefore

$\dfrac{dt(x)}{dx} = p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}\\ p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}=0\\cos( \frac{2\pi}{ L} x)=0\\ \dfrac{2\pi}{ L} x=\dfrac{\pi}{2}\\\\x=\dfrac{L}{4}$

Thus the maximum internal torque will be at x= 0.25 L

$t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}} \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}$