Elasticity
8.0 Modulus of Rigidity
8.0 Modulus of Rigidity
Within the elastic limit, it is defined as the ratio of tangential stress to the shearing strain within the elastic limit. It is also called shear modulus of rigidity. It is represented $\eta $(eta).
$$\eta = \frac{{tangential\ or\ shearing\ stress}}{{shearing\ strain}}$$
Let us consider a cube whose lower face is fixed and a tangential force $F$ acts on the upper face whose area is $A$.
$$Tangential\ Stress = \frac{{Force}}{{Area}} = \frac{F}{A}$$
Let the vertical sides of the cube shifts through an angle $\phi $, called as shear strain.
Therefore, $$ShearStrain = \phi $$
So, the modulus of rigidity can be written as, $$\eta = \frac{{\frac{F}{A}}}{\phi } = \frac{F}{{A\phi }}$$
Example 5. A steel plate of face area $4c{m^2}$ and thickness $0.5\ cm$ is fixed rigidly at the lower surface. A tangential force of $10\ N$ is applied on the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel $\left( {{\eta _{steel}}} \right) = 8.4 \times {10^{10}}N{m^{ - 2}}$
Solution: Our aim is to find lateral displacement of upper surface with respect to lower surface i.e. $x$.
$$Shearing\ Force = 10\ N$$
$$Shearing\ Stress = \frac{{Shrearing\ Force}}{{Area}} = \frac{{10}}{{4 \times {{10}^{ - 4}}}} = 2.5 \times {10^4}N{m^{ - 2}}$$
$$Shearing\ Strain = \tan \phi \approx \phi $$
Modulus of rigidity is given by,
$$\begin{equation} \begin{aligned} \eta = \frac{{Shearing\ Stress}}{{Shearing\ Strain}} = \frac{{2.5 \times {{10}^4}}}{\phi } = \frac{{2.5 \times {{10}^4}}}{{\frac{x}{t}}} \\ \eta = \frac{{2.5 \times {{10}^4} \times t}}{x} \\ x = \frac{{2.5 \times {{10}^4} \times t}}{\eta } \\ x = \frac{{2.5 \times {{10}^4} \times 5 \times {{10}^{ - 3}}}}{{8.4 \times {{10}^{10}}}} \\ x = 1.488 \times {10^{ - 9}}\ m\ or\ 1.488\ nm \\\end{aligned} \end{equation} $$
