Question

A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of $ 10^5 $ N at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement $ \theta $ of the rod axis from its original position would be: (shear moduli $ G = 10^{10} $ N/m$^2$)

• A. \( \frac{1}{160\pi} \)
• B. \( \frac{1}{4\pi} \)
• C. \( \frac{1}{40\pi} \)
• D. \( \frac{1}{2\pi} \)

Hint:

The angular displacement due to shear force can be found using the formula \( \theta = \frac{F L}{G A} \), where \( A \) is the cross-sectional area and \( G \) is the shear modulus.

Answer 1

The Correct Option is A

To determine the angular displacement \( \theta \) of the rod axis from its original position, we need to use the formula for angular displacement due to shear force in a cylindrical body:

\(\theta = \frac{F \cdot L}{G \cdot A}\)

Where:

• \(F\) is the shear force applied at the top of the rod.
• \(L\) is the length of the rod.
• \(G\) is the shear modulus of the material.
• \(A\) is the cross-sectional area of the rod.

Given values are:

• \(F = 10^5\) N
• \(L = 1\) m
• \(G = 10^{10}\) N/m²
• The radius of the rod is 4 cm, which is \(0.04\) m

First, calculate the cross-sectional area \(A\) of the cylindrical rod:

\(A = \pi r^2 = \pi (0.04)^2 = 0.0016\pi\) m²

Substitute these values into the formula:

\(\theta = \frac{10^5 \cdot 1}{10^{10} \cdot 0.0016\pi}\)

Now, simplify:

\(\theta = \frac{10^5}{10^7 \cdot 0.0016\pi} = \frac{10^5}{1.6 \times 10^4 \pi}\)

\(\theta = \frac{10^5}{1.6 \times 10^4 \pi} = \frac{10^5}{1.6 \times 10^4 \pi} = \frac{10^5}{1.6 \times 10^4 \pi} = \frac{1}{160\pi}\)

Thus, the angular displacement \( \theta \) is \(\frac{1}{160\pi}\). Hence, the correct answer is:

\( \frac{1}{160\pi} \)

Answer 2

The angular displacement \( \theta \) due to a shear force is given by: \[ \theta = \frac{F L}{G A} \] where: 
- \( F = 10^5 \, \text{N} \) is the shear force, - \( L = 1 \, \text{m} \) is the length of the rod, 
- \( G = 10^{10} \, \text{N/m}^2 \) is the shear modulus, 
- \( A = \pi r^2 = \pi (0.04)^2 = 5.027 \times 10^{-3} \, \text{m}^2 \) is the cross-sectional area of the rod. 
Substitute the values into the formula: \[ \theta = \frac{10^5 \times 1}{10^{10} \times 5.027 \times 10^{-3}} = \frac{10^5}{5.027 \times 10^7} = \frac{1}{160\pi} \] 
Thus, the angular displacement is \( \frac{1}{160\pi} \), and the correct answer is (1).