A bar of circular cross-section is subjected to alternating tensile force varying from a minimum of 200 KN to a maximum of 500 KN. It is to be manufactured of a material with an ultimate tensile strength of 900 MPa and an endurance limit of 700 MPa. Determine the diameter of bar using FOS of 3.5 related to ultimate tensile strength and 4 related to endurance limit. Use Goodman straight line as basis of design.

Solution:

$W_{min}=200KN$

$W_{max}=500KN$

$\sigma _u=900MPa$

$\sigma _{-1}=700Mpa$

$\text{FOS related to }\sigma _u=3.5$

$\text{FOS related to }\sigma _{-1}=4$

Find: d=? by using Goodman criteria

Soln. $\sigma _{max}=\frac{W_{max}}{A}=\frac{500\times 10^3}{\frac{\pi}{4}d^2}=\frac{636.61\times 10^3}{d^2}$

$\sigma _{min}=\frac{W_{min}}{A}=\frac{200\times 10^3}{\frac{\pi}{4}d^2}=\frac{254.62\times 10^3}{d^2}$

$\sigma _m=\frac{[\frac{636.61\times 10^3}{d^2}+\frac{254.62\times 10^3}{d^2}]}{2}=\frac{445.61\times 10^3}{d^2}$

$\sigma _a=\frac{[\frac{636.61\times 10^3}{d^2}-\frac{254.62\times 10^3}{d^2}]}{2}=\frac{190.9\times 10^3}{d^2}$

$\therefore \frac{\frac{190.9\times 10^3}{d^2}}{(\frac{700}{4})}=1-\frac{\frac{445.61\times 10^3}{d^2}}{(\frac{900}{3.5})}$

$d=53.1mm$