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

A fillet weld carries load through its throat area, and the design strength per unit throat area stays the same no matter which leg size is chosen. So instead of comparing spacings symbolically, this route plugs in a working number for the weld strength and tracks the actual load in kN.

Take a representative permissible strength of $f_{wd} = 200$ N/mm$^2$ for the throat area (any consistent value works, since it cancels out at the end; this only makes the arithmetic concrete). The effective throat thickness of a fillet weld of leg size $s$ is $t_t = 0.7s$.

For the 5 mm weld of length 150 mm, the throat area is

\[ A_{t1} = 0.7 \times 5 \times 150 = 525 \ \text{mm}^2 \]

so the load it can carry is

\[ P = 200 \times 525 = 105000 \ \text{N} = 105 \ \text{kN} \]

Now the 6 mm weld must carry this same 105 kN. Its throat area must satisfy

\[ 200 \times A_{t2} = 105000 \implies A_{t2} = 525 \ \text{mm}^2 \]

So the throat area needed does not change at all; what changes is how that area gets built up. With leg size 6 mm, the throat thickness per unit length is $0.7 \times 6 = 4.2$ mm/mm, so the length needed is

\[ l_2 = \frac{A_{t2}}{0.7 \times 6} = \frac{525}{4.2} = 125 \ \text{mm} \]

The value chosen for $f_{wd}$ never enters the final answer since it appears on both sides and cancels; only the ratio of throat areas matters. The weld length required with the larger 6 mm weld is shorter, which makes sense since a thicker weld carries more load per millimetre of length.

The required length is 125 mm.

\[ \boxed{125 \ \text{mm}} \]