How to Size a Fillet Weld per AISC 360: Effective Throat, Strength, and a Worked Example

A fillet weld is the workhorse of steel connections, and sizing one is mostly arithmetic — once you know where the geometry and the code limits come from. AISC 360 Chapter J gives you the strength equation; the trap is the leg-size limits and the base-metal check that sit around it. Here is the whole method with a worked example.

Why the throat is what matters

A fillet weld fails across its effective throat — the shortest line through the weld cross-section — not across its visible leg. For an equal-leg fillet, the throat is the leg dimension w rotated 45°:

t_e = 0.707 · w

That single 0.707 is the reason a 1/4" weld is not as strong as 1/4" of base metal — you only get the diagonal throat to work with.

Nominal strength (AISC 360 §J2.4)

The nominal strength of weld metal in shear is:

R_n = F_nw · A_we = (0.60 F_EXX)(t_e · L)

where F_EXX is the electrode classification strength (70 ksi for the ubiquitous E70 family), A_we is the effective throat area, and L is the weld length. Apply the resistance/safety factors:

The limits that actually control your design

Minimum size (Table J2.4)

Set by the thinner of the two parts joined — thicker plate sinks more heat and needs a larger weld to avoid cracking. For example, material up to 1/4" needs a 1/8" minimum fillet; over 1/4" to 1/2" needs 3/16".

Maximum size (§J2.2b)

Along an edge of material less than 1/4" thick, the max weld equals the material thickness. For thicker edges, the max is the thickness minus 1/16", so the weld toe does not overrun the plate corner.

Base metal check

The weld is never the only failure path — the connected metal can shear or rupture too. Always check base-metal strength at the weld interface (§J4) and take the lower of weld and base-metal capacity.

Worked example: a bracket weld in LRFD

A 1/2"-thick bracket plate transfers a factored load of P_u = 40 kips in shear along the weld line, using E70 electrodes. The available weld length is two 6" longitudinal fillets (12" total). Find the required leg size.

Step 1 — Required strength per inch

r_u = P_u / L = 40 / 12 = 3.33 kips/in

Step 2 — Solve for sixteenths

n = r_u / 1.392 = 3.33 / 1.392 = 2.39 → use 3 sixteenths

Round up to a 3/16" fillet, which provides 3 × 1.392 = 4.18 kips/in ≥ 3.33 ✓.

Step 3 — Check the limits

For 1/2" material the minimum fillet is 3/16" — our weld meets it exactly. The maximum along the 1/2" edge is 1/2" − 1/16" = 7/16", so 3/16" is well within range. Finally, confirm base-metal shear rupture of the 1/2" plate exceeds 40 kips. Use a 3/16" E70 fillet, 6" each side.

Common mistakes

• Rounding leg size down. Welds are specified in 1/16" increments — always round the required size up.
• Ignoring the minimum. A tiny computed weld on thick plate still has to meet Table J2.4, or it can crack from rapid cooling.
• Skipping the base-metal check. An over-welded thin plate just shifts the failure into the steel.

Longitudinal vs. transverse welds: the directionality bonus

A fillet weld loaded perpendicular to its axis is stronger than one loaded along its axis — up to 50% stronger. AISC 360 §J2.4 captures this with a directional-strength increase:

F_nw = 0.60 F_EXX (1.0 + 0.50 sin^1.5θ)

Here θ is the angle between the load and the weld axis. At θ = 0° (longitudinal) the bracket is 1.0 and you get the familiar 1.392 kips/in per sixteenth. At θ = 90° (transverse) the bracket is 1.5, giving 2.088 kips/in. Using this increase is optional, and there is a catch: when a connection combines longitudinal and transverse welds, the code makes you either ignore the increase entirely or apply the more conservative of two specific combination rules, because the two weld orientations do not reach peak strength at the same deformation.

Weld size limits at a glance

Two more length rules catch people: an end-loaded longitudinal fillet longer than 100 times its leg size needs a reduction factor β (its far end is barely working), and an intermittent fillet must be at least four times its leg size or 1–1/2" long, whichever is greater. Both exist because stress does not distribute evenly along a long or stubby weld.

Eccentric weld groups

When the load does not pass through the centroid of the weld group, the welds carry combined direct shear and a torsional component. The classic hand method is the elastic vector approach: resolve the load into direct shear (P/L) plus a moment-induced stress (Mc/J, using the polar moment of inertia of the weld group treated as a line). The more economical AISC instantaneous center of rotation method recognizes the directional strength increase and typically gives 1.2–1.5× more capacity — which is exactly why the AISC Manual tabulates coefficient C tables for common bracket configurations.

Sizing a fillet weld is one equation wrapped in three checks. Get the throat, multiply by 1.392, then let the minimum, maximum, and base-metal limits decide whether your number survives.