Answer:
250 kpsi
Explanation:
Given:
Width of the specimen, w = 0.45 in
Thickness of the specimen, t = 0.20 in
length of the specimen between supports, L = 2.5 in
Failure load, F = 1200 lb
Now,
The transverse rupture strength [tex]\sigma_t=\frac{1.5FL}{wt^2}[/tex]
on substituting the respective values, we get
[tex]\sigma_t=\frac{1.5\times1200\times2.5}{0.45\times0.2^2}[/tex]
or
[tex]\sigma_t=250,000\ psi\ =\ 250 kpsi[/tex]
