A force 'F' of same magnitude is applied tangentially on upper and lower face of a cube, in opposite…
A force 'F' of same magnitude is applied tangentially on upper and lower face of a
cube, in opposite directions. Side of the cube is 'L'. The upper face of the cube shifts
parallel to itself by a distance ' $x_{1}$ '. If another cube of same material but side '2L' is
subjected to the above condition, then the displacement of the top layer is
$\frac{x_{1}}{6}$
$\frac{x_{1}}{2}$
$\frac{x_{1}}{8}$
$\frac{x_{1}}{4}$
Solution
When the force $F$ is applied tangentially on the upper and lower faces of a cube, the displacement $x$ of the upper face depends on the side length $L$ of the cube.
For the first cube with side $L$, the displacement is $x_1$.
For the second cube with side $2 L$, shear strain remains constant because the material and force are the same.
Using the relation for shear strain:
$\frac{x_1}{L}=\frac{x_2}{2 L}$
From this:
$x_2=\frac{x_1}{2}$
Thus, the displacement of the top layer of the second cube is $\frac{x_1}{2}$.