Question

A tensile test is carried out on a bar of mild steel of diameter 2 cm. The bar yields under a load of 80 kN. It reaches a maximum load of 150 kN and finally breaks at a load of 70 kN. If the diameter of fracture neck is 1 cm. Calculate the ratio of ultimate tensile stress and the breaking point stress.

• A. 1.80
• B. 2.02
• C. 0.53
• D. 1.55

Answer 1

The correct option is C 0.53

$d=2cm$ , $P_{max}=150kN$ ,

$P_{break}=70kN$

$d_{fracture}=1cm$

${\sigma }_{ut}=\frac{P_{max}}{A_o}=\frac{150}{\frac{\pi }{4}(2{)}^2}$

${\sigma }_{break}=\frac{P_{fracture}}{A_{fracture}}=\frac{70}{\frac{\pi }{4}(1{)}^2}$

$\therefore \frac{{\sigma }_{ut}}{{\sigma }_{break}}=\frac{\frac{150}{\frac{\pi }{4}(2{)}^2}}{\frac{70}{\frac{\pi }{4}(1{)}^2}}=0.535$