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

# Derivation of τ

F = total weight of coil below turn = volume x density x acceleration due to gravity

$F = \left( {\frac{{\pi {w^2}}}{4} \times \pi D \times N} \right) \times \rho \times g$

T = Torque = $$\left( {\frac{{\pi {w^2}}}{4} \times \frac{{\pi {D^2}}}{2} \times N} \right)$$$$\times \rho \times g$$

The equation for the shear stress caused by torsion is,$$\tau =$$ $$\frac{{Tr}}{J} = \frac{{Tr}}{{2I}}$$

where r is the radial distance from the centre of the wire ($$0 \le r \le \frac{w}{2}$$) and J is the polar moment of inertia (m4) which, in the case of a circular cross section, is equal to twice I, the second moment of area ( = $$\frac{{\pi {w^4}}}{{64}}$$)

Therefore the shear stress varies from zero at the centre of the wire to a maximum value at its surface:

${\tau _{_{\max }}} = \frac{{16T}}{{\pi {w^3}}} = 2\pi \rho g{D^2}\frac{N}{w}$

An important note is that the average shear stress, τ, is proportional to the peak shear stress:

$\tau \propto \frac{{N{D^2}}}{w}$

Furthermore, by noting that D and w are control variables in our experiment we can simplify our expression to:

$\tau = BN$

where B is a constant for this experiment