Determine the thickness of a 120 mm wide plate for safe continuous operation if the plate is to be subjected to a tensile load that has a maximum value of 250 KN and a minimum value of 100 KN. The properties of plate materials are as follows:

Endurance limit stress=225 MPa, Yield point stress=300 MPa, FOS based on yeild pooint may be taken as 1.5.

Solution:

$t=?$

$b=120mm$

$W_{max}=250KN$

$W_{min}=100KN$

$\sigma _{-1}=225MPa$

$\sigma _y=300MPa$

$FOS=1.5$

$\sigma _{max}=\frac{W_{max}}{A}=\frac{250\times 10^3}{b\times t}=\frac{250\times 10^3}{120\times t}=\frac{2083.33}{t}$

$\sigma _{min}=\frac{W_{min}}{A}=\frac{100\times 10^3}{120\times t}=\frac{833.33}{t}$

$\sigma_m=\frac{\frac{2083.33}{t}+\frac{833.33}{t}}{2}=\frac{1(1458.33)}{t}$

$\sigma _a=\frac{\frac{2083.33}{t}-\frac{833.33}{t}}{2}=\frac{625}{t}$

Soderberg equation

$\frac{1}{n}=\frac{\sigma _m}{\sigma _y}+\frac{\sigma _a}{\sigma _{-1}}$

$\frac{1}{1.5}=\frac{\frac{1458.33}{t}}{300}+\frac{\frac{625}{t}}{225}$

$t=11.46mm$