Question

A tensile test on a 10 mm diameter specimen registered failure under a load of 32.6 kN. What force would be sufficient to break a 1.5 mm diameter wire made from this material?

Answer 1

Answer:

0.7315 kN

Explanation:

We have given diameter = 10 mm , so radius =$r=\frac{10}{2}=5\ mm$

The specimen registered failure under a load of 32.6 kN

So load = 32.6 kN

We know that $\sigma _{max}=\frac{load}{area}=\frac{32.6}{\pi 5^2}=0.415$ here $\sigma _{max}$ is the maximum stress which the material can withstand

Now for diameter d=1.5 mm

radius $r=\frac{d}{2}=\frac{1.5}{2}=0.75\ mm$

$\sigma _{max}=\frac{load}{area}$

$0.415=\frac{load}{\pi \times 0.75^2}$

load =0.7315 kN