Q.A metal block used as a shear-test specimen has height $25\ \text{cm}$. A tangential force of $3.0\times10^{4}\ \text{N}$ is applied uniformly across its upper face, of area $0.05\ \text{m}^2$, while its lower face stays fixed. If the modulus of rigidity of the material is $\eta = 6.0\times10^{10}\ \text{Pa}$, find

Given: height $L = 0.25\ \text{m}$, tangential force $F = 3.0\times10^4\ \text{N}$ on the top face of area $A = 0.05\ \text{m}^2$, $\eta = 6.0\times10^{10}\ \text{Pa}$.

1. Shearing stress:

   $\text{Shear stress} = \frac{F}{A} = \frac{3.0\times10^4}{0.05} = 6.0\times10^5\ \text{Pa}$

2. Shearing strain:

   $\text{Shear strain} = \frac{\text{Shear stress}}{\eta} = \frac{6.0\times10^5}{6.0\times10^{10}} = 1.0\times10^{-5}$

3. Lateral displacement: …

Divide the tangential force by the face area to get shear stress; divide by the given modulus of rigidity to get shear strain; multip …

• Using the block's height as the shear-stress area instead of the actual face area the tangential force acts on. …