Question

A cube of side $40mm$ has its upper face displaced by $0.1mm$ by a tangential force of $8kN$. The shearing modulus of cube is

• A. $2\times {10}^{9}N{m}^{-2}$
• B. $4\times {10}^{9}N{m}^{-2}$
• C. $8\times {10}^{9}N{m}^{-2}$
• D. $16\times {10}^{9}N{m}^{-2}$

Answer 1

The correct option is A $2\times {10}^{9}N{m}^{-2}$

Shearing stress $=$ Shearing force/Area being sheared.

Force applied $=8kN$,

Area being sheared $=40\times 40=1600m{m}^2$

$\Rightarrow$ Shear stress $=\frac{8000}{1600}=5MPa$ ........(1)

$\Rightarrow$ Shear strain $=\frac{\partial y}{\partial x}=\frac{0.1}{40}=2.5\times {10}^{-3}$ .........(2)

Note that shear strain is merely an angle so it has no dimesions.

Therefore, Shearing Modulus $=$ Shearing stress/Shear strain $=$ $\frac{5}{2.5\times {10}^{-3}}=2\times {10}^{9}N/m^2$