# 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 *θ _{L}* 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 d*θ*.

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.