Question

A bracket plate is welded to the flange of a column section as shown in figure. The size of weld is 8 mm. The distance of centroid of weld group is 60 mm from left edge as shown. The polar moment of inertia of weld is $6250\times {10}^{4}m{m}^4$. The maximum force (in kN), rounded off to two decimal place) that can be applied is____.

• A. 184.37

Answer 1

The correct option is A 184.37
Let the maximum force can be applied be P kN.

$t_t=kS=0.7\times 8=5.6mm$

Direct Shear Stress

$q_1=\frac{P\times {10}^3}{\left[\right(200\times 2)+250]\times 5.6}=0.275PN/m{m}^2$

Shear stress due to twisting

$q_2=\frac{P\times {10}^3\times (70+200-60)\times r}{6250\times {10}^4}$

where,
$r=\sqrt{{125}^2+(200-60{)}^2}=187.68mm$

$q_2=0.631PN/m{m}^2$

$\cos \theta =\frac{140}{187.68}=0.746$

Combined stress
$=\sqrt{q_1^2+q_2^2+2q_1{q}_2\cos \theta }$

$=\sqrt{(0.275P{)}^2+(0.631P{)}^2+2\times 0.275\times 0.631P^2\times 0.746}$

$=0.856PN/m{m}^2$

$\therefore 0.856P\le \frac{f_u}{\sqrt{3}{\gamma }_{mw}}$
$[{\gamma }_{mw}=1.5\text{for site weld}]$

$0.856P\le \frac{410}{\sqrt{3}\times 1.5}$

$\Rightarrow P\le 184.37kN$