Question

"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at a completely reversed stress amplitude of 70 kpsi."

Answer 1

Answer:

life (N) of the specimen is 117000  cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su   .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f  = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = $\frac{(f\times Su)^2}{Se}$     ......................1  

put here value and we get

a = $\frac{(0.82\times 120)^2}{60}$  

a = 161.4  kpsi

so coefficient value (b) will be

b = $-\frac{1}{3}log\frac{(f\times Su)}{Se}$  

b =  $-\frac{1}{3}log\frac{(0.82\times 120)}{60}$  

b = −0.0716

so here number of cycle N will be  

N =  $(\frac{ \sigma a}{a})^{1/b}$

put here value  and we get

N =  $(\frac{ 70}{161.4})^{1/-0.0716}$

N = 117000

so life (N) of the specimen is 117000  cycles