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

Answer 1

Answer:

104,576 cycles

Explanation:

Step 1: identify given parameters

Ultimate strength of steel ($S_{ut}$)= 120 Kpsi

stress amplitude ($\alpha_{a}$)= 70 kpsi

life of the specimen (N) = ?

$N = (\frac{\alpha_{a}}{a})^\frac{1}{b}$

where a and b are coefficient of fatigue cycle

Step 2: calculate the the endurance limit of specimen

$S_{e} = 0.5*S_{ut}$

$S_{e}$ = 0.5*120 = 60 kpsi

Step 3: calculate coefficient 'a'

$a=\frac {(0.8XS_{ut})^2}{S_{e}}$

$a=\frac {(0.8X120)^2}{60}$

$a= 153.6 kpsiStep 4: calculate the coefficient 'b'[tex]b =-\frac{1}{3}log(\frac{f*S_{ut} }{S_{e}})$

$b =-\frac{1}{3}log(\frac{0.8*120}{60})$

$b =-0.0680Step 5: calculate the life of the specimen[tex]N=(\frac{\alpha_{a}}{a})^\frac{1}{b}$

$N=(\frac{70}{153.6})^\frac{1}{-0.068}$

$N=104,576 cycles$

∴ the life (N) of the steel specimen is 104,576 cycles