The last lesson checked a weld somebody else had drawn. This one draws it. And weld design is done per unit LENGTH, because a continuous weld's length is set by the joint, not by the designer — you get the length the bracket gives you, and the leg size is the only dial you hold.

$w = \dfrac{f}{0.707\,\tau_a}$ — read aloud w equals f over nought-point-seven-nought-seven tau-a. $w$ is the required fillet leg size in millimetres. $f$ is the load per unit length of weld, in newtons per millimetre — the total load divided by the length carrying it. $\tau_a$ is the allowable shear stress on the throat in megapascals, subscript a for allowable. Notice what has vanished: the weld length. Once the load is expressed per millimetre, the leg follows from the allowable alone, and that is exactly why weld capacity tables are published per millimetre of leg.

The check direction is the same relation wearing different clothes: $\tau = \dfrac{F}{0.707\,w L}$, where $F$ is the total load on the weld in newtons, $w$ the leg in millimetres and $L$ the effective weld length in millimetres. The denominator is just the throat area written out in one step.

Then the part the arithmetic cannot do. Your answer will land between stocked sizes, and it rounds UP, never to the nearest — a weld rounded down is a weld that does not carry the load. Round up to a size a welder can gauge and deposit in one pass, in the position the joint will actually be welded in. Two code tables then bracket the answer: a MINIMUM fillet on thick material, because a small weld on a heavy plate is chilled by the mass around it and cools too fast, and a MAXIMUM along a plate edge, so the edge stays visible. Neither is arithmetic. Both are mandatory.

Last nugget, and it decides real drawings: weld metal costs as the SQUARE of the leg. Going from 6 mm to 8 mm is not a third more weld, it is nearly twice as much, with the heat and the distortion to match. When a weld is short of capacity, run it LONGER before you run it bigger.