Answer:
0.7315 kN
Explanation:
We have given diameter = 10 mm , so radius =[tex]r=\frac{10}{2}=5\ mm[/tex]
The specimen registered failure under a load of 32.6 kN
So load = 32.6 kN
We know that [tex]\sigma _{max}=\frac{load}{area}=\frac{32.6}{\pi 5^2}=0.415[/tex] here [tex]\sigma _{max}[/tex] is the maximum stress which the material can withstand
Now for diameter d=1.5 mm
radius [tex]r=\frac{d}{2}=\frac{1.5}{2}=0.75\ mm[/tex]
[tex]\sigma _{max}=\frac{load}{area}[/tex]
[tex]0.415=\frac{load}{\pi \times 0.75^2}[/tex]
load =0.7315 kN
