First slide

Elastic behaviour of solids

Question

A metal cube of side 10 cm is subjected to a shearing stress of $10^{6} N / m^{2}$ . Calculate the modulus of rigidity $(in \times 10^{8} {Nm}^{- 2})$ if the top of the cube is displaced by 0.05 cm with respect to its bottom.

Moderate

Solution

Shearing stress , $F / A = 10^{6} N / m^{2}$
Length of side, $L = 10 cm = 0 .1 m$
Shearing displacement , $Δx = 0 .05 cm = 0 .0005 m$
The modulus of rigidity is
$\begin{matrix} η = \frac{F}{Aθ} = \frac{FL}{AΔx} (∵ θ ≅ \tan θ = \frac{Δx}{L}) \\ = \frac{10^{6} \times 0 .1}{0 .0005} = 2 \times 10^{8} N / m^{2} \end{matrix}$

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