Dissemination of IT for the Promotion of Materials Science (DoITPoMS)

# Derivation of γ

This small element of the wire subtends an angle dβ at the centre of the coil.

The shear strain produced by torsion is given by

$\gamma = \frac{{{\rm{distance\; through\; which\; we\; twist}}}}{{{\rm{distance\; over\; which\; twist\; occurs}}}}$

Which can be simplified, algebraically, to

$\gamma = \frac{r}{2}{\theta _L}$

where θL is the angular twist per unit length induced by the torque and r is the radial distance from the centre ($$0 \le r \le \frac{w}{2}$$). The torque acting on this element causes it to twist through an angle dθ.

Therefore, the twist per unit length is:

${\theta _{\rm{L}}} = \frac{{{\rm{d}}\theta }}{{(D/2){\rm{d}}\beta }}$ The vertical deflection of the coil due to the twisting of this element is:

${\rm{d}}s = \frac{D}{2}{\rm{d}}\theta = {\left( {\frac{D}{2}} \right)^2}{\theta _{\rm{L}}}{\rm{d}}\beta$

The vertical deflection associated with the twist in one turn of the coil is:

$s = \int_0^{2\pi } {{{\left( {\frac{D}{2}} \right)}^2}{\theta _{\rm{L}}}{\rm{d}}\beta = \frac{{\pi {D^2}}}{2}{\theta _{\rm{L}}}}$

The peak local shear strain is thus given by:

${\gamma _{{\rm{max}}}} = \frac{{sw}}{{\pi {D^2}}}$

It may be noted that the average local shear strain, γ, is proportional to the peak strain,

$\gamma \propto \frac{{sw}}{{ {D^2}}}$

As we did for the stress, we can note that D and w are control variables, and so simplify our expression to:

$\gamma = Cs$

where C is a constant for this experiment.