Question

A hot-rolled steel has a yield strength of Syr-Sye-170 kpsi and a true strain at fracture of 0.55. Estimate the factor of safety for the following stress state, by using a) the maximum shear stress theory b) the distortion energy theory. (30 points) S1. Ox-30 kpsi, o,' =-15 kpsi, τ,--45 kpsi.

Answer 1

a) Factor of safety ≤ 1.689

b) Factor of safety ≤ 1.944

Step-by-step explanation:

Given data:

Yield strength,
`S_(yr)=S_(ye)=170\ \textup{kpsi}`

True strain factor = 0.55

`\sigma_x=30\ \textup{kpsi}`

`\sigma_y=-15\ \textup{kpsi}`

`\tau_(xy)=-45\ \textup{kpsi}`

Now, principle stress is given as:

`\sigma_1,\sigma_2 = (\sigma_x+\sigma_y)/(2)\pm\sqrt{((\sigma_x-\sigma_y)/(2))^2+\tau_(xy)^2`

on substituting the values we get

`\sigma_1,\sigma_2 = (30-15)/(2)\pm\sqrt{((30-(-15))/(2))^2+(-45)^2`

`\sigma_1,\sigma_2 =7.5\pm50.312`

or

`\sigma_1 =7.5+50.312=57.812\ \textup{kpsi}`

and

`\sigma_2 =7.5-50.312=-42.81\ \textup{kpsi}`

a) By maximum shear stress theory

we have

`\sigma_1-\sigma_2\leqslant \frac{\textup{Yield\ strength}}{\textup{Factor\ of\ safety}}`

on substituting the values, we get

`57.812-(-42.81)\leqslant\frac{170}{\textup{Factor\ of\ safety}}`

or

Factor of safety ≤ 1.689

b) By the distortion energy theory

`(1)/(2)[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2]\leqslant\frac{S_(yr)}{\textup{Factor\ of\ safety}}`

since no force is acting in the z- direction, thus
`\sigma_3 = 0`

on substituting the values, we get

`(1)/(2)[(30-(-15))^2+(-15-0)^2+(0-30)^2]\leqslant\frac{170}{\textup{Factor\ of\ safety}}`

or

Factor of safety ≤ 1.944