GATE Civil Engineering (CE) 2018 Shift 1 Solved Paper

Direct bending tensile stress,
${\mathit{f}}_{\mathit{a}}=150\mathit{M}\mathit{P}\mathit{a}$
Direct 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 ${\mathit{f}}_{\mathit{e}}=\sqrt{{\mathit{f}}_{\mathit{a}}^2+3{\mathit{q}}^2}\le \frac{{\mathit{f}}_{\mathit{u}}}{\sqrt{3}{\mathit{\gamma }}_{\mathit{m}\mathit{w}}}$
Note : The above check for combination of stress need not be done for fillet welds where the sum of normal and shear stresses does not exceed ${\mathit{f}}_{\mathit{w}\mathit{d}}$ [Clause 10.5.10.1.2. (b)]
Thus, ${\mathit{f}}_{\mathit{a}}+\mathit{q}=150+50=200\mathit{M}\mathit{P}\mathit{a}$
${\mathit{f}}_{\mathit{a}}+\mathit{q}>{\mathit{f}}_{\mathit{w}\mathit{d}}\left(=184.75\mathit{M}\mathit{P}\mathit{a}\right)$
$\Rightarrow {\mathit{r}}_{\mathit{e}}=173.21\mathit{M}\mathit{P}\mathit{a}$
Thus, the weld is designed for an equivalent stress of $173.21\mathit{M}\mathit{P}\mathit{a}$.