GATE Civil Engineering (CE) 2018 Shift 2 Solved Paper

Factored normal stress, ${\mathit{f}}_{\mathit{a}}=120\mathit{M}\mathit{P}\mathit{a}$
Factored shear stress, $\mathit{q}=50\mathit{M}\mathit{P}\mathit{a}$
According to IS-800 : 2007, clause 10.5.10.1.1
The equivalent stress,
${\mathit{f}}_{\mathit{e}}=\sqrt{{\mathit{f}}_{\mathit{a}}^2+3{\mathit{q}}^2}\le \frac{{\mathit{f}}_{\mathit{u}}}{\sqrt{3}{\mathit{\gamma }}_{\mathit{m}\mathit{w}}}$
${\mathit{f}}_{\mathit{e}}=\sqrt{{120}^2+3\times {50}^2}=147.99$
Note: The above check for combination of stresses need not be done for fillet welds where the sum of normal and shear stresses does not exceed [clause 10.5.10.1.2(b)] Hence, sum of normal and shear stresses So, the above check need not be done. So, weld is designed for a resultant shear stress of $\left({\mathit{f}}_{\mathit{r}}\right)$
So, the design stress is $130\mathit{N}/{\mathit{m}\mathit{m}}^2;$ however equivalent stress $147.99\mathit{M}\mathit{P}\mathit{a}$.