A fillet welded joint is subjected to transverse loading F as shown in the figure. Both legs of the fillets are of 10 mm size and the weld length is 30 mm. If the allowable shear stress of the weld is 94 MPa, considering the minimum throat area of the weld, the maximum allowable transverse load in kN is
GATE ME - 2012,
Answer: C
The key's answer is 19.93 kN.
For a fillet weld with equal legs of size s, the minimum throat is \(t = s\cos 45^{\circ} = 0.707s\). The key's value comes from taking the stated weld length of 30 mm as the total length of weld resisting the load:
\[ \begin{aligned} A &= 0.707 \times 10 \times 30 = 212.1\ \text{mm}^{2} \\ F &= 94 \times 212.1 = 19\,937\ \text{N} \approx 19.93\ \text{kN} \end{aligned} \]
The figure shows a fillet along the edge of each plate. If each of the two fillets were itself 30 mm long, the throat area and the load would double, to about 39.9 kN, which is not among the options, so the question intends 30 mm as the whole length of weld.
Using the full leg size of 10 mm instead of the throat gives 28.2 kN, also not offered; the other options do not follow from the throat-area method.
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