91Ó°ÊÓ

1. Diameter of aluminum rod 4.8 cm which is projects 5.3 cm from a wall
2. Object of mass 1200 kg suspended from the rod end
3. Shear modulus of aluminum is $3.0×{10}^{10}N/m^2$

The shear modulus is defined as the ratio of shear stress to the shear strain. We can find the shear stress using the formula for shear stress in terms of force and area. The deflection of the rod is found by using the formula for the shear modulus in terms of shear stress and shear strain.

Consider the formula for the shear stress:

$\text{Shearstress}=\frac{F}{A}$ …… (i)

Here,F is the force, A is the area of cross-section

Consider the expression for the shear modulus:

$G=\frac{F/A}{X/L}$ …… (ii)

Here, G is shear modulus, F is force, A is area of cross-section, x is change in length, and L is original length.

From equation (i), find the shear stress as follows:

$\text{Shearstress}=\frac{F}{A}$

Here,F = Mg and $A=\mathrm{Ï€}{r}^2$.

Determine the shear stress as follows:
$$\begin{gathered} Shearstress=\frac{Mg}{\mathrm{π}{r}^2} \\ =\frac{\left(1200kg\right)\left(9.8\right)}{3.14×{\left(0.024m\right)}^2} \\ =6.5×{10}^6\frac{N}{m^2} \end{gathered}$$

The shear stress on the rod is $6.5×{10}^6\frac{N}{m^2}$

From equation (ii), determine the shear modulus as,

$G=\frac{F/A}{X/L}$

Rewrite the equation for change in displacement and substituting the values and solve as:

$$\begin{gathered} x=\frac{\left(F/A\right)L}{G} \\ =\frac{\left(6.5×{10}^6\right)\left(0.053m\right)}{3.0×{10}^{10}} \\ =1.1×{10}^{-5}m \end{gathered}$$

The vertical deflection of the end of the rod is $1.1×{10}^{-5}m$.