Question

Two steel plates are to be connected together by a 5 mm fillet weld of length 150 mm to transfer a design load. If the size of the fillet weld used to connect the same two plates is changed to 6 mm, the weld length (in mm) needed for transferring the same design load is (in integer).

Hint:

Fillet weld strength is proportional to size times length; equate size1 x length1 = size2 x length2.

Answer 1

Correct Answer: 125

Step 1: Recall the strength formula of a fillet weld.
The design strength of a fillet weld connection is
\[ P = f_{wd} \times t_t \times l_w \]
where \(f_{wd}\) is the design strength of the weld per unit throat area (a material and code constant, unchanged here since the plates, grade and electrode stay the same), \(t_t = 0.7s\) is the effective throat thickness for a fillet weld of leg size \(s\), and \(l_w\) is the effective length of the weld.

Step 2: Express the design load carried by the 5 mm weld.
For \(s_1 = 5\) mm and \(l_1 = 150\) mm,
\[ P = f_{wd} \times (0.7 \times 5) \times 150 = f_{wd} \times 525 \]

Step 3: Express the load carried by the 6 mm weld and equate.
For \(s_2 = 6\) mm and unknown length \(l_2\),
\[ P = f_{wd} \times (0.7 \times 6) \times l_2 = f_{wd} \times 4.2 \times l_2 \]
Since both welds carry the same design load \(P\),
\[ f_{wd} \times 525 = f_{wd} \times 4.2 \times l_2 \]
\[ l_2 = \frac{525}{4.2} = 125 \ \text{mm} \]

Step 4: Sanity check using the direct proportion.
Since \(P \propto s \times l\) here, holding \(P\) constant gives \(s_1 l_1 = s_2 l_2\).
\[ 5 \times 150 = 6 \times l_2 \implies l_2 = \frac{750}{6} = 125 \ \text{mm} \]
Both routes agree.

Final Answer:
The required weld length with the 6 mm fillet weld is 125 mm.
\[ \boxed{l_2 = 125 \ \text{mm}} \]