Question

A bar of steel has the minimum properties Se = 40 kpsi, Sy = 60 kpsi, and Sut = 80 kpsi. The bar is subjected to steady torsional stress (τm) of 15.3 kpsi, alternating torsional stress (τa) of 10 kpsi, and alternating bending stress (σa) of 12.4 kpsi. Find the factor of safety guarding against a static failure, and either the factor of safety guarding against a fatigue failure or the expected life of the part.

Answer 1

Answer:

The safety factor against the static failure as per Von Mises criteria is 2.05 whereas The factor of safety guarding the fatigue failure as per modified Goodman Criteria is 1.56.

Explanation:

For the bar $\sigma_{max}$ as Von Mises criteria is given as

$\sigma_{max}=\sqrt{\sigma_b^2+3T^2$

Here

$\sigma_b$ is the bending stress given as 12.4 kpsi

T is the torsional stress which is given as 15.3 kpsi

$\sigma_{max}=\sqrt{\sigma_b^2+3T^2}\\\sigma_{max}=\sqrt{12.4^2+3*15.3^2}\\\sigma_{max}=29.26 kpsi$

So the safety factor against static failure is given as

$n_s=\frac{S_y}{\sigma_{max}}\\n_s=\frac{60}{29.26}\\n_s=2.05$

The safety factor against the static failure as per Von Mises criteria is 2.05.

For fatigue failure,as per Modified Goodman criteria

$n_f=\frac{1}{\frac{\sigma_a}{S_e}+\frac{\sigma_m}{S_{ut}}}$

Here

$\sigma_m=\sqrt{3}T\\\sigma_m=\sqrt{3}*15.3\\\sigma_m=26.50 kpsi$

$n_f=\frac{1}{\frac{\sigma_a}{S_e}+\frac{\sigma_m}{S_{ut}}}\\n_f=\frac{1}{\frac{12.4}{40}+\frac{26.50}{80}}\\n_f=\frac{1}{0.6412}\\n_f=1.559 \approx 1.56$

The factor of safety guarding the fatigue failure as per modified Goodman Criteria is 1.56