Question

A steel rod of radius 20 mm and length of 2 m is acted upon by a force of 400 kN along the length. The values of stress and strain are respectively (Y\(_{\text{steel}}\) = 2 \( \times \) 10\(^{11}\) Nm\(^{-2} \))

• A. \( 1.96 \times 10^8 \) Nm\(^{-2} \), \( 0.16\% \)
• B. \( 3.18 \times 10^8 \) Nm\(^{-2} \), \( 0.16\% \)
• C. \( 3.18 \times 10^8 \) Nm\(^{-2} \), \( 0.32\% \)
• D. \( 4 \times 10^8 \) Nm\(^{-2} \), \( 0.2\% \)

Hint:

Calculate the cross-sectional area of the rod using its radius. Then, calculate the stress by dividing the applied force by the cross-sectional area. Use Young's modulus to find the strain, which is the ratio of stress to Young's modulus. Finally, convert the strain to percentage if required. Ensure consistent units throughout the calculations.

Answer 1

The Correct Option is B

The radius of the steel rod is \( r = 20 \) mm \( = 20 \times 10^{-3} \) m.
The length of the steel rod is \( L = 2 \) m.
The applied force is \( F = 400 \) kN \( = 400 \times 10^3 \) N.
Young's modulus of steel is \( Y = 2 \times 10^{11} \) Nm\(^{-2} \).
First, calculate the cross-sectional area \( A \) of the rod: $$ A = \pi r^2 = \pi (20 \times 10^{-3})^2 = \pi (400 \times 10^{-6}) = 4\pi \times 10^{-4} \text{ m}^2 $$ Using \( \pi \approx 3.
14 \), \( A = 4 \times 3.
14 \times 10^{-4} = 12.
56 \times 10^{-4} \text{ m}^2 \).
Next, calculate the stress \( \sigma \): $$ \sigma = \frac{F}{A} = \frac{400 \times 10^3}{12.
56 \times 10^{-4}} = \frac{4 \times 10^5}{1.
256 \times 10^{-3}} = \frac{4}{1.
256} \times 10^8 \approx 3.
18 \times 10^8 \text{ Nm}^{-2} $$ Now, calculate the strain \( \epsilon \) using Young's modulus \( Y = \frac{\sigma}{\epsilon} \): $$ \epsilon = \frac{\sigma}{Y} = \frac{3.
18 \times 10^8}{2 \times 10^{11}} = 1.
59 \times 10^{-3} $$ Convert the strain to percentage: $$ \text{Strain percentage} = \epsilon \times 100\
= 1.
59 \times 10^{-3} \times 100\
= 0.
159\
\approx 0.
16\
$$ The values of stress and strain are approximately \( 3.
18 \times 10^8 \) Nm\(^{-2} \) and \( 0.
16\
\) respectively.