User:Eas4200c.f08.blue.a/Lecture 19

Example of an Ad Hoc method of proving the relation:

$$\theta =\frac{1}{2G\bar{A}}\oint_{}^{}{}\frac{q}{t}ds$$

Uniform bar with non Circular Cross section subjectred to twist:



Here the displacement of PP' due to $$\alpha$$ is:

$$ \frac{PP'}{OP}=tan\alpha \cong \alpha$$

making a small angle assumption for $$\alpha$$. Now project the displacement orthogonal to OP'.

$$PP''=PP'cos\alpha $$

$$PP''=(OPcos\alpha )tan\alpha $$

$$PP=OPtan\alpha$$

Recall: OP = r and OP'' = $$\rho$$

Therefore:

$$PP''=(rcos\alpha)tan\alpha$$

Where PP" is the displacement of P in the direction of the tangent to the surface of the bar, $$rcos\alpha$$ is $$\rho$$ and $$tan\alpha$$ is approximated to be $$\alpha$$ with a small angle assumption.

Strain can be defined as:

$$\gamma =\frac{PP''}{dx}=\frac{\rho\alpha }{dx}=\rho \theta $$

with

$$d\theta =\frac{\alpha }{dx}$$ as the rate of twist