Question

A horizontal aluminum rod of diameter 4.8 cm projects 5.3 cm from a wall. A 1200 kg object is suspended from the end of the rod. The shear modulus of aluminum is $3.0 \times {10^{10}}N/{m^2}$. Neglecting the mass of the rod, find the vertical deflection of the end of the rod.
A. Thus, $\Delta x = 2.1 \times {10^{ - 5}}m$
B. Thus, $\Delta x = 1.1 \times {10^{ - 5}}m$
C. Thus, $\Delta x = 11.1 \times {10^{ - 5}}m$
D. Thus, $\Delta x = 12.1 \times {10^{ - 5}}m$

Answer 1

Hint: We are given to calculate the vertical deflection of the end of rod, which is the change in displacement of the plane. This can be calculated using the shear strain formula. Shear modulus is the ratio of shear stress over shear strain. Shear stress is the ratio of force over area of the rod. Shear strain is the ratio of displacement of the plane over the distance of that surface from the plane opposite to it. Use this info to solve the question.

Formula used:
Shear modulus is equal to $G = \dfrac{{Stres{s_{shear}}}}{{Strai{n_{shear}}}}$, where $Stres{s_{shear}} = \dfrac{F}{A},F = m \times g,A = \pi {r^2}$ and $Strai{n_{shear}} = \dfrac{{\Delta x}}{L}$, where r is the radius of the rod, m is the mass of the object, L is the distance of rod from the wall, g is acceleration due to gravity and $\Delta x$ is the vertical deflection of the end of the rod.

Complete step by step answer:
We are given that a horizontal aluminum rod of diameter 4.8 cm projects 5.3 cm from a wall and a 1200 kg object is suspended from the end of the rod and the shear modulus of aluminum is $3.0 \times {10^{10}}N/{m^2}$.
We have to calculate the vertical deflection of the rod when the object is suspended.
Shear modulus G is equal to $G = \dfrac{{Stres{s_{shear}}}}{{Strai{n_{shear}}}}$
The base of the rod is in circle shape. So we are using the area of circle formula.
$
  Stres{s_{shear}} = \dfrac{F}{A} \\
  F = m \times g \\
  A = \pi {r^2} \\
  Strai{n_{shear}} = \dfrac{{\Delta x}}{L} \\
   \implies G = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\left( {\dfrac{{\Delta x}}{L}} \right)}} = \dfrac{{\left( {\dfrac{{mg}}{{\pi {r^2}}}} \right)}}{{\left( {\dfrac{{\Delta x}}{L}} \right)}} \\
   \implies G = \dfrac{{mgL}}{{\pi {r^2}\Delta x}} \\
   \implies \Delta x = \dfrac{{mgL}}{{\pi {r^2}G}} \\
 $
On substituting the values of mass, acceleration due to gravity, shear modulus, Length of the rod and the radius in the above equation, we get
$
  \Delta x = \dfrac{{1200 \times 9.8 \times 5.3 \times {{10}^{ - 2}}}}{{\pi \times {{\left( {2.4 \times {{10}^{ - 2}}} \right)}^2} \times 3 \times {{10}^{10}}}} \\
   \implies \Delta x = \dfrac{{62328 \times {{10}^{ - 2}}}}{{54286.7 \times {{10}^3}}} \\
  \therefore \Delta x = 1.1 \times {10^{ - 5}}m \\
 $
Therefore, the vertical deflection is $\Delta x = 1.1 \times {10^{ - 5}}m$.

So, the correct answer is “Option B”.

Note:
Do not confuse shear strain with normal strain. A strain is simply a ratio of change in length to the original length where the stress applied is perpendicular to the cross section in normal strain but in shear strain, the stress is applied parallel to the cross section. And strain is a ratio of length to length, so it will have no units.