Question

Consider the fillet-welded lap joint shown in the figure (not to scale). The length of the weld shown is the effective length. The welded surfaces meet at right angle. The weld size is 8 mm, and the permissible stress in the weld is 120 MPa. What is the safe load $P$ (in kN, rounded off to one decimal place) that can be transmitted by this welded joint? \includegraphics[width=0.5\linewidth]{image66.png}

Hint:

For fillet welds, always use throat thickness $t = 0.7s$ to calculate shear area. Multiply by effective weld length and permissible shear stress to get load capacity.

Answer 1

Step 1: Identify weld throat thickness.
For a fillet weld of size $s = 8$ mm, the throat thickness is \[ t = 0.7 \times s = 0.7 \times 8 = 5.6 \,\text{mm}. \]

Step 2: Effective weld length.
From the figure: - One vertical weld of length $50$ mm.
- Two horizontal welds, each of length $75$ mm.
Hence, total effective weld length = $75 + 50 + 75 = 200$ mm.

Step 3: Throat area of weld.
\[ A = \text{throat thickness} \times \text{effective length} = 5.6 \times 200 = 1120 \,\text{mm}^2. \]

Step 4: Load capacity.
Permissible shear stress = $120$ MPa = $120 \,\text{N/mm}^2$.
\[ P = \tau \times A = 120 \times 1120 = 134400 \,\text{N}. \]

Step 5: Convert to kN.
\[ P = \frac{134400}{1000} = 134.4 \,\text{kN}. \] \[ \boxed{134.4 \,\text{kN}} \]