Question

Under fatigue load, the life of a component is 0.3 million cycles under the load ${\sigma }_1$ and its life is 0.9 million cycles under the load ${\sigma }_2$. The component completes 0.1 million revolutions under load ${\sigma }_1$ and then the load is reduced to ${\sigma }_2$. What will be the expected life left in millions of revolutions?

• A. 0.8
• B. 0.75
• C. 0.6
• D. 0.45

Answer 1

The correct option is C 0.6

$N_1$ = Life under ${\sigma }_1$ = 0.3 million

$N_2$ = Life under ${\sigma }_2$ = 0.9 million

$n_1$ = 0.1 million

Now, $\frac{n_1}{N_1}+\frac{n_2}{N_2}=1$

$\frac{0.1}{0.3}+\frac{n_2}{0.9}=1$

$\Rightarrow n_2=0.6$ million