Question

A horizontal aluminum rod $4.8cm$ in diameter projects $5.3cm$ from a wall. An $1200kg$ object is suspended from the end of the rod. The shear modulus of aluminum rod is $3.01010N{m}^{-2}$ neglecting the rod’s mass, find
(a) The shear stress on the rod and
(b) The vertical deflection of the end of the rod.

Answer 1

Hint: In order to solve this question we need to understand the definition of shear modulus and shear stress. Shear modulus or modulus of rigidity is defined as ratio of shear stress and shear strain and its measure of elastic shear stiffness of material, denoted by G. Shear stress is defined as the amount of deformation force applied per unit area.

Complete step by step answer:
According to the given question, the mass of the object hung with the rod is actually creating a deformation force in the rod and it is given as $1200kg$. So the deformation force on the rod is $F=mg$ where $m$ is mass and $g$ is acceleration due to gravity.
$$\begin{gathered} F=(1200\times 9.8) \\ \Rightarrow F=11760kgm{s}^{-2} \end{gathered}$$
Also diameter of rod is $d=4.8cm=0.048m$
So the radius of rod is $R={\displaystyle \frac{d}{2}}$ putting values we get
$R={\displaystyle \frac{0.048}{2}}m=0.024m$
So the area of rod is given by $A=\pi R^2$
Putting values we get
$$\begin{gathered} A=\pi \times (0.024{)}^2{m}^2 \\ \Rightarrow A=1.80\times {10}^{-3}{m}^2 \end{gathered}$$
So the shear stress is given by $S={\displaystyle \frac{F}{A}}$
Putting values we get
$$\begin{gathered} S={\displaystyle \frac{11760N}{1.8\times {10}^{-3}{m}^2}} \\ \therefore S=6.5\times {10}^{6}N{m}^{-2} \end{gathered}$$

Hence, the shear stress on the rod is $6.5\times {10}^{6}N{m}^{-2}$.

(b) Let the vertical deflection of the rod be $\mathrm{\Delta }x$.
Given the length of rod $L=5.3cm=0.053m$.
Also given the shear modulus of the aluminum rod is $G=3.01010N{m}^{-2}$.
So by using formula of shear modulus which is
$G={\displaystyle \frac{({\displaystyle \frac{F}{A}})}{({\displaystyle \frac{\mathrm{\Delta }x}{L}})}}$
$\Rightarrow \mathrm{\Delta }x={\displaystyle \frac{FL}{AG}}$
Putting values we get
$$\begin{gathered} \mathrm{\Delta }x={\displaystyle \frac{(11760\times 0.053)}{(1.8\times {10}^{-3}\times 3.01)}} \\ \therefore \mathrm{\Delta }x=0.115m \end{gathered}$$

Hence, the vertical deflection of the end of rod is $0.115m$.

Note: It should be remembered that deflection in the rod is only vertical in nature because the aluminum rod is under a force which is vertical in nature so the deformation produced is vertical. Also shear stress and strain are the reason why rail lines in summer got extended and in winter got shrink. Here if the rod’s mass is also considered then deformation force increases which in turn increases shear stress.