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

0.7315 kN

Step-by-step explanation:

We have given diameter = 10 mm , so radius =
`r=(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)=(load)/(area)=(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=(d)/(2)=(1.5)/(2)=0.75\ mm`

`\sigma _(max)=(load)/(area)`

`0.415=(load)/(\pi * 0.75^2)`

load =0.7315 kN