Question

A solid rod of 12 mm diameter was tested for tensile strength with the gauge length of 50 mm. Final length = 80 mm; Final diameter = 4 mm; Yield load = 1130 N. What is the nearest yield stress and % reduction in area respectively?
1. 10 MPa and 10%
2. 90 MPa and 90%
3. 10 MPa and 90%
4. 90 MPa and 10%

Answer 1

Correct Answer - Option 3 : 10 MPa and 90%

Concept:

Percentage reduction in area is, \({\rm{\Delta }}A = \frac{{\frac{\pi }{4}d_1^2 - \frac{\pi }{4}d_2^2}}{{\frac{\pi }{4}d_1^2}} \)

And yield stress is, \(\sigma=\frac{P}{A_1}\)

Calculation:

Given:

L₁ = 50 mm          d₁ = 12 mm

L₂ = 80 mm          d₂ = 4 mm

ΔL = 30 mm         Δd = 8 mm

P = 1130 N

\({\rm{\Delta }}A = \frac{{\frac{\pi }{4}d_1^2 - \frac{\pi }{4}d_2^2}}{{\frac{\pi }{4}d_1^2}} = \frac{{144 - 16}}{{144}} = 0.88 \approx 90\% \)

Nearest yield stress \( = \frac{{1130}}{{\frac{\pi }{4} \times {{\left( {12} \right)}^2}}} = 9.99\;MPa \approx 10MPa\)