Question

Determine the length of a 10mm diameter copper wire so that it can extend through one complete revolution without exceeding through a sheering stress of 42mN/m^2. Take G - 90GN/m^2

Answer 1

$$\frac{T}{J}=\frac{G\theta}{l}\\T=\frac{\pi}{16}\tau d^3=\frac{\pi}{16}(42) 10^3=8246.7\ Nmm\\J=\frac{\pi}{32} d^4=\frac{\pi}{32} 10^4=981.75\ mm^4\\l=G\theta\frac{J}{T}=(90\cdot10^3)(2\pi)\frac{981.75}{8246.7}=67300\ mm=67.3\ m$$