91Ó°ÊÓ

Given that a square steel plate is 10.0 cm on a side and 0.500 cm thick.

The force of magnitude $\mathrm{F}=9×{10}^5\mathrm{N}$

Area $\mathrm{A}=\left(0.1\mathrm{m}\right)\left(0.005\mathrm{m}\right)=5.00×{10}^{-4}{\mathrm{m}}^2$

Shear Modulus is defined as the ratio of the shear stress and shear strain.

\(S = \frac{{{\rm{Shear stress}}}}{{{\rm{Shear strain}}}} = \frac{{{{{F_\parallel }} \mathord{\left/{\vphantom {{{F_\parallel }} A}} \right.} A}}}{{{x \mathord{\left/ {\vphantom {x h}} \right.} h}}} = \frac{{{F_\parallel }h}}{{Ax}}\)

..............(1)

Where is the force applied tangent to the surface of the object, h is the transverse dimension, A is the area over which force is exerted, and x is deformation

Where F is the force applied tangent to the surface of the object, A is the area over which force is exerted, S is the shear modulus

\(S = \frac{{{\rm{Shear stress}}}}{{{\rm{Shear strain}}}} = \frac{{{{{F_\parallel }} \mathord{\left/{\vphantom {{{F_\parallel }} A}} \right.} A}}}{{{x \mathord{\left/ {\vphantom {x h}} \right.} h}}} = \frac{{{F_\parallel }h}}{{Ax}}\)

For steel, $\mathrm{S}=7.5×{10}^{}5\mathrm{Pa}$, and using equation (2), we get,

$\begin{array}{r}\text{Shear Strain}=\frac{9×{10}^5\mathrm{N}}{\left[\right(0.1\mathrm{m}\left)\right(0.005\mathrm{m}\left)\right]\left[7.5×{10}^5\mathrm{Pa}\right]}\\ =2.4×{10}^{−2}\end{array}$

Hence, the Shear stain is $2.4×{10}^{-2}$

Shear Strain =$\frac{x}{h}$ …â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(3)

Where x is deformation and h is the transverse dimension

Therefore, using equation (3), we get

$\begin{array}{r}x=\left(\text{Shear strain}\right)\left(\mathrm{h}\right)\\ =\left(0.024\right)\left(10\mathrm{cm}\right)\\ =0.24\mathrm{cm}\end{array}$

Hence, displacement is 0.24 cm.