Question

A simply supported steel joist with a 4 m effective span carries a uniformly distribution load of 40 kN over its span inclusive of self weight. The beam is supported laterally throughout the beam is checked for crippling.
Given
$t_w=5.4mm,h_2=17.15mm$
and bearing length = 100 mm

The bearing stress is _______.

Answer 1

The bearing stress =
${\sigma }_p=\frac{Reaction}{(b+h_2\sqrt{3})t}$

Reaction =
$\frac{WL}{2}=\frac{40\times {10}^3}{2}=20\times {10}^{3}N$

${\sigma }_p=\frac{20\times {10}^3}{(100+17.15\times \sqrt{3})\times 5.4}=28.55N/m{m}^2$