Question

A steel plate of face area $4c{{m}^{2}}$ and thickness of 0.5m is fixed rigidly at the lower surface. A tangential force of 10N is applied on the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel=$8.4\times {{10}^{10}}N{{m}^{-2}}.$

Answer 1

Hint: The face area of the plate is fixed at the bottom. Hence when we apply a tangential stress on the upper surface it will get laterally displaced. The modulus of rigidity for steel is given to us as $8.4\times {{10}^{10}}N{{m}^{-2}}.$ The modulus of rigidity is defined as the ratio of tangential stress by shearing strain. Hence from the definition of modulus of rigidity we can determine the lateral displacement.

Complete step by step answer:
To begin with let us draw the diagram of the above scenario.

In the above figure the lateral displacement is denoted by $\Delta L$, by application of tangential stress on the upper surface. The modulus of rigidity from the above diagram can be defined as,
$\eta =\dfrac{\text{Tangential stress}}{\text{Shear strain}}=\dfrac{F/A}{\theta }$,
where F is the tangential force at the upper surface, A is the area of the bottom fixed surface and $\theta $ is the shear strain as shown in the above diagram.
On the application of tangential stress the shear strain in steel is very small. Therefore we can write $\operatorname{Tan}\theta \cong \theta $ From the above figure we can define the tan of the angle as,
$\operatorname{Tan}\theta =\dfrac{\Delta L}{L}$. Hence the equation for modulus of rigidity now becomes,
$\begin{align}
  & \eta =\dfrac{\text{Tangential stress}}{\text{Shear strain}}=\dfrac{F/A}{\theta } \\
 & \eta =\dfrac{F/A}{\Delta L/L}...(1) \\
\end{align}$
Now let us substitute For each of the given quantities to determine lateral displacement($\Delta L$).
$\begin{align}
  & \eta =\dfrac{F/A}{\Delta L/L} \\
 & \Rightarrow 8.4\times {{10}^{10}}=\dfrac{10/4c{{m}^{2}}}{\Delta L/0.5} \\
 & \Rightarrow 8.4\times {{10}^{10}}=\dfrac{10\times 0.5}{\Delta L\times 4\times {{10}^{-4}}} \\
 & \Rightarrow \Delta L=\dfrac{10\times 0.5}{8.4\times {{10}^{10}}\times 4\times {{10}^{-4}}}=\dfrac{5}{33.6\times {{10}^{6}}} \\
 & \Delta L=0.148\times {{10}^{-6}}=1.48\times {{10}^{-5}}cm \\
\end{align}$
Hence the lateral displacement of the upper surface with respect to the bottom surface is$1.48\times {{10}^{-5}}cm$

Note:
The modulus of rigidity for steel is very high. Hence the lateral displacement is very small when a force of 10N is applied on the upper surface. Steel not only has the highest modulus of rigidity but it also has a very high value of Young’s modulus and bulk modulus.