Q11-36E - University Physics with Modern Physics

Question

A square steel plate is 10.0 cm on a side and 0.500 cm thick. (a) Find the shear strain that results if a force of magnitude $\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}$ N is applied to each of the four sides, parallel to the side. (b) Find the displacement x in centimeters.

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right)Shear stain is 2.4\times {10}^{-2}. \\ \left(b\right)Displacement is 0.24 cm \end{gathered}$$

Step 1: The given data

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

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

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

Step 2: Formula used

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

$\mathit{S}\mathbf{=}\frac{\mathbf{S}\mathbf{h}\mathbf{e}\mathbf{a}\mathbf{r}\mathbf{ }\mathbf{s}\mathbf{t}\mathbf{r}\mathbf{e}\mathbf{s}\mathbf{s}}{\mathbf{S}\mathbf{h}\mathbf{e}\mathbf{a}\mathbf{r}\mathbf{ }\mathbf{s}\mathbf{t}\mathbf{r}\mathbf{a}\mathbf{i}\mathbf{n}}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{I}\mathbf{I}}\mathbf{/}\mathbf{A}}{\mathbf{x}\mathbf{/}\mathbf{h}}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{I}\mathbf{I}}\mathbf{h}}{\mathbf{A}\mathbf{x}}$ ……………(1)

Where F_II 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.

Step 2: Find Shear Strain (a)

$Shear strain=\frac{Shear stress}{Shear modulus}=\frac{F_{II}/A}{S}=\frac{F_{II}}{AS}.....................\left(2\right)$

Where F_II 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

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

$\begin{array}{rcl}Shear Strain& =& \frac{9\times {10}^{5 }N}{\left[\left(0.1 m\right)\left(0.005 m\right)\right]\left[7.5\times {10}^5 Pa\right]}\\ & =& 2.4\times {10}^{-2}\end{array}$

Hence, the Shear stain is $2.4\times {10}^{-2}$.

Step 3: Find the displacement (b)

$=\frac{x}{h}$Shear Strain  ………………..(3)

Where x is deformation and h is the transverse dimension

Therefore, using equation (3), we get

$\begin{array}{rcl}x& =& \left(Shearstrain\right)\left(h\right)\\ & =& \left(0.024\right)\left(10 cm\right)\\ & =& 0.24 cm\end{array}$

Hence, displacement is 0.24 cm.