Analogous to Young’s Modulus, the modulus of rigidity is the proportionality constant in the elastic range for shear.

\begin{equation}
\tau=G\gamma
\end{equation}

\begin{equation}
\gamma_{xy}=\frac{\tau_{xy}}{G} \text{, }\gamma_{yz}=\frac{\tau_{yz}}{G} \text{, }\gamma_{zx}=\frac{\tau_{zx}}{G}
\end{equation}

Take a unit cube under uniaxial tension, and look at a portion cut by a 45 degree diagonal.

mod-of-rigidity-4-dv-a

Figure 1

Taking a cut at a 45 degree angle results in max shear:

\begin{equation}
P_{tan}=P\sin\frac{\pi}{4}=0.707P
\end{equation}

\begin{equation}
A_{tan}=\frac{A}{\cos\frac{\pi}{4}}=\frac{A}{0.707}
\end{equation}

\begin{equation}
\tau=\frac{P\sin(\pi/4)}{\dfrac{A}{\cos(\pi/4)}}=\frac{P\sin(\pi/4)\cos(\pi/4)}{A}=\frac{0.5P}{A}
\end{equation}

\begin{equation}
\tau_{max}=\frac{P}{2A}
\end{equation}

stresses_on_rotated_face_plot

Figure 2

Using the Figure 1 above, the first equation relating the horizontal and vertical strains:

\begin{equation}
\tan(\beta)=\frac{1-\nu\epsilon_x}{1+\epsilon_x}
\end{equation}

The second equation relating the change in angle:

\begin{equation}
\beta=\frac{\pi}{4}-\frac{\gamma_m}{2}
\end{equation}

We will use this trig identity to further the derivation:
\begin{equation}
\tan(\alpha \pm \beta)=\frac{\tan(\alpha)\pm \tan(\beta)}{1 \mp \tan(\alpha)\tan(\beta)}
\end{equation}

Applying the trig identity and simplifying:
\begin{equation}
\tan(\beta)=\tan\left(\frac{\pi}{4}-\frac{\gamma_m}{2}\right)=\frac{\tan(\pi/4)-\tan(\gamma_m/4)}{1+\tan(\pi/4)tan(\gamma_m/2)} \\
=\frac{1-\tan(\gamma_m/2)}{1+\tan(\gamma_m/2)}=\frac{1-\gamma_m/2}{1+\gamma_m/2}
\end{equation}

\begin{equation}
\frac{1-\gamma/2}{1+\gamma/2} = \frac{1-\nu\epsilon_x}{1+\epsilon_x}
\end{equation}

\begin{equation}
(1+\epsilon_x)\left(1-\frac{\gamma_m}{2}\right)=(1-\nu\epsilon_x)\left(1+\frac{\gamma_m}{2}\right)
\end{equation}

\begin{equation}
1-\frac{\gamma_m}{2}+\epsilon_x-\epsilon_x\frac{\gamma_m}{2}=1-\frac{\gamma_m}{2}-\nu\epsilon_x-\nu\epsilon_x\frac{\gamma_m}{2}
\end{equation}

\begin{equation}
\gamma_m=\frac{-\epsilon_x-\nu\epsilon_x}{-1-\frac{\epsilon_x}{2}+\nu\frac{\epsilon_x}{2}}=\frac{\epsilon_x(1+\nu)}{\epsilon_x\left(\frac{1-\nu}{2}\right)+1}
\end{equation}

\begin{equation}
\gamma_m = \epsilon_x(1+\nu)
\end{equation}

Substitute the formulas for normal and shear strain:

\begin{equation}
\frac{\tau_m}{G}=\frac{\sigma_x}{E}(1+\nu)
\end{equation}

\begin{equation}
G=\frac{E}{(1+\nu)\sigma_x}\frac{\tau_m}{\sigma_x}=\frac{e}{(1+\nu)}\frac{P}{2A}\frac{A}{P}
\end{equation}

\begin{equation}
G=\frac{E}{2(1+\nu)}
\end{equation}

References:

  1. Beer, Ferdinand P. and Johnston, Russell E. Mechanics of Materials. s.l. : McGraw-Hill, 1981.