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 transfer ring the same design load is ______ (in integer).

The correct option is 125.

Step-by-Step Explanation:

The load-carrying capacity (design strength) of a fillet weld is directly proportional to its throat thickness and its effective length. The formula for the design strength of a fillet weld ($P$) is given by:

$P=f_{\mathrm{wd}}$

where:
• $f_{\mathrm{wd}}$ is the design shear strength of the weld metal.
• $t$ is the effective throat thickness of the weld.
• $L$ is the effective length of the weld.

The effective throat thickness ($t$) is related to the size of the weld ($s$) by the formula:

$t=k\times s$

where $k$ is a constant depending on the angle between the fusion faces (typically $0.7$ for a standard angle of 90 degrees).

Substituting $t$ into the load equation gives:

$P=f_{\mathrm{wd}}$

Since the same two plates are being connected to transfer the same design load, the parameters $f_{\mathrm{wd}}$ and $k$ remain constant. Therefore, the product of the weld size and the weld length must remain constant:

$s_1\times L_1=s_2\times L_2$

Given data:
• Initial size of the fillet weld, $s_1=5\text{mm}$
• Initial length of the fillet weld, $L_1=150\text{mm}$
• New size of the fillet weld, $s_2=6\text{mm}$

Let $L_2$ be the new length of the weld needed. Substituting the values into the equation:

$5\times 150=6\times L_2$

$750=6\times L_2$

$L_2=\frac{750}{6}$

$L_2=125\text{mm}$

Thus, the weld length required is 125 mm.