Question

A bar is subjected to fluctuating tensile load from 20 kN to 100 kN. The material has yield strength of 240 MPa and endurance limit in reversed bending is 160 MPa. Accordong to the Soderberg principle, the area of cross-section in $m{m}^2$ of the bar for a factor of safety of 2 is

• A. 750
• B. 600
• C. 400
• D. 1000

Answer 1

The correct option is D 1000

$P_1=20kN;P_2=100kN$

${\sigma }_{yt}=240MPa;{\sigma }_e=160MPa$

$P_m=\frac{20+100}{2}=60kN$

Let A is area in $m{m}^2$
$P_v=\frac{100-20}{2}=40kN$

${\sigma }_v=\frac{40}{A}kN/m{m}^2$

By Soderberg principle
$\frac{1}{FOS}=\frac{{\sigma }_m}{{\sigma }_{yt}}+\frac{{\sigma }_v}{{\sigma }_e}$

$\frac{1}{2}=\frac{60\times {10}^3}{A{\sigma }_{yt}}+\frac{40\times {10}^3}{A{\sigma }_v}$

$A=[\frac{60}{240}+\frac{40}{160}]2\times {10}^3$

$A=\left[\frac{2}{4}\right]\times 2\times {10}^3=1000m{m}^2$