Question

A steel rotating-beam test specimen has an ultimate strength Sut of 1600 MPa. Estimate the life (N) of the specimen if it is tested at a completely reversed stress amplitude σa, of 900 MPa.

Answer 1

Answer:

the life (N) of the specimen is 46400 cycles

Explanation:

given data

ultimate strength Su = 1600 MPa

stress amplitude σa = 900 MPa

to find out

life (N) of the specimen

solution

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

Se = 0.5× Su   .............1

Se = 0.5 × 1600

Se = 800 Mpa

and we know

Se for steel is 700 Mpa for Su ≥ 1400 Mpa

so we take endurance limit Se is = 700 Mpa

and strength of friction f  = 0.77 for 232 ksi

because for Se 0.5 Su at $10^{6}$ cycle = (1600 × 0.145 ksi ) = 232

so here coefficient value (a) will be

a = $\frac{(f*Su)^2}{Se}$    

a = $\frac{(0.77*1600)^2}{700}$  

a = 2168.3 Mpa

so

coefficient value (b) will be

a = -$\frac{1}{3}$log$\frac{(f*Su)}{Se}$

b =  -$\frac{1}{3}$log$\frac{(0.77*1600)}{700}$

b = -0.0818

so no of cycle N is

N =  $(\frac{ \sigma a}{a})^{1/b}$

put here value

N =  $(\frac{ 900}{2168.3})^{1/-0.0818}$

N = 46400

the life (N) of the specimen is 46400 cycles