Question

Two plates are connected by fillet welds of size $10$ mm and subjected to tension $P=275$ kN (factored). Plate thickness $=12$ mm. Steel: $f_y=250$ MPa, $f_u=410$ MPa. Workshop welding with partial safety factor $\gamma_{mw=1.25$. As per IS 800:2007 (Limit State), what is the minimum length (in mm, rounded off to the nearest higher multiple of $5$ mm) required of each weld to transmit $P$?} \includegraphics[width=0.5\linewidth]{image9.png}

• A. 100
• B. 105
• C. 110
• D. 115

Hint:

For fillet weld design in IS 800 (LSM): use $f_{wd}=f_u/(\sqrt{3}\,\gamma_{mw})$ and throat $t=0.7s$.
Capacity per mm length $= f_{wd}\,t$ (kN/mm after unit conversion). Split the factored load among the weld lines and round up to detailing multiples (often $5$ mm).

Answer 1

The Correct Option is B

Step 1: Design shear strength of a fillet weld.
For shop welds, the design shear strength per unit throat area is
\[ f_{wd}=\frac{f_u}{\sqrt{3}\,\gamma_{mw}} = \frac{410}{\sqrt{3}\times 1.25} = \frac{410}{2.165} \approx 189.4~\text{MPa}. \] Effective throat thickness for a $10$ mm fillet: $t=0.7s=0.7\times 10=7$ mm.

Step 2: Strength per mm length of one weld.
\[ R_{\text{per mm}}=f_{wd}\times t=189.4\times 7 \approx 1326~\text{N/mm} =1.326~\text{kN/mm}. \]

Step 3: Force shared by two side welds.
Each weld carries $P/2=275/2=137.5$ kN.
Required effective length (one weld):
\[ L=\frac{137.5}{1.326}\approx 103.7~\text{mm}. \]

Step 4: Rounding as per question.
Round up to the nearest higher multiple of $5$ mm $\Rightarrow$ $L=105$ mm.
\[ \boxed{L_{\min} = 105~\text{mm per weld}} \]