Respuesta :
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 = [tex]\frac{(f\times Su)^2}{Se}[/tex] ......................1
put here value and we get
a = [tex]\frac{(0.82\times 120)^2}{60}[/tex]
a = 161.4 kpsi
so coefficient value (b) will be
b = [tex]-\frac{1}{3}log\frac{(f\times Su)}{Se}[/tex]
b = [tex]-\frac{1}{3}log\frac{(0.82\times 120)}{60}[/tex]
b = −0.0716
so here number of cycle N will be
N = [tex](\frac{ \sigma a}{a})^{1/b}[/tex]
put here value and we get
N = [tex](\frac{ 70}{161.4})^{1/-0.0716}[/tex]
N = 117000
so life (N) of the specimen is 117000 cycles
