Fillet Weld Strength Calculator
Work out fillet weld capacity by AISC 360, in LRFD or ASD, with the base metal check and the minimum and maximum leg sizes the code requires.
How to use this calculator
- 1Enter the leg size, not the throat. A quarter-inch fillet is measured along the plate faces.
- 2Add up all the weld carrying the load - both sides of a lap, all four sides of a plate.
- 3Match the load to the method. LRFD wants factored loads and ASD wants service loads, and confusing them is out by about half again.
- 4Watch the base metal line. On thin material it is very often the limit, and no amount of extra weld changes it.
How the calculation works
Effective throat = 0.707 x leg
Nominal weld strength = 0.60 x Fexx x throat, per inch of length
LRFD: phi = 0.75 ASD: Omega = 2.00
Base metal shear rupture = 0.60 x Fu x thickness, same factors
E70XX: 1.392 kip/in per sixteenth (LRFD), 0.928 (ASD)- Fexx
- Electrode classification strength in ksi - the 70 in E7018
- Fu
- Tensile strength of the base metal. A36 is 58 ksi, A572 Gr 50 and A992 are 65
- 0.707
- The throat of an equal-leg fillet - the shortest path through the weld, at 45 degrees to both legs
- Directional increase
- 1.0 + 0.50 sin^1.5(theta), so 1.5 for a fillet loaded square across its axis
The per-sixteenth coefficients fall straight out of the equation: 0.75 x 0.60 x 70 x 0.707 / 16 = 1.392, and the ASD version is the same nominal figure divided by 2.00 rather than multiplied by 0.75.
Base metal shear rupture is a genuinely separate limit state, not a refinement. On thin material it governs, and past that point extra weld metal buys nothing.
The minimum fillet size in AISC Table J2.4 is a cooling rate requirement rather than a strength one, which is why it grows with the thickness of the material rather than with the load.
Worked example
Six inches of quarter-inch fillet, E70, LRFD
- 1.A 1/4 in leg is four sixteenths, and E70 at LRFD is 1.392 kip per inch per sixteenth.
- 2.So the weld metal carries 4 x 1.392 = 5.568 kip per inch, and over 6 in that is 33.4 kips.
- 3.The base metal check: 0.75 x 0.60 x 58 ksi x 0.375 in = 9.79 kip per inch - well above the weld.
- 4.So the weld governs, and the joint is good for about 33 kips.
Result: 33.4 kips, weld metal governing
The same weld on thin sheet, where the plate gives first
- 1.Same 5.568 kip per inch of weld metal - nothing about the weld has changed.
- 2.But the base metal is now 1/8 in A1011 sheet at 52 ksi: 0.75 x 0.60 x 52 x 0.125 = 2.925 kip per inch.
- 3.That is barely half the weld, so the plate governs and the joint is good for 17.6 kips rather than 33.4.
- 4.Welding it heavier would add heat, distortion and time, and not one pound of capacity - and a 1/4 in fillet already exceeds the 1/8 in maximum leg AISC permits along the edge of 1/8 in material.
Result: 17.6 kips - the sheet governs, and a bigger weld would not help
Sizing the weld for a 40 kip load
- 1.A 5/16 in fillet is five sixteenths, so 5 x 1.392 = 6.96 kip per inch of weld metal.
- 2.Half-inch A572 gives 0.75 x 0.60 x 65 x 0.5 = 14.6 kip per inch of base metal, so the weld governs.
- 3.Ten inches carries 69.6 kips against a 40 kip factored load - 57% utilised.
- 4.Which means the joint could be done with about 5.8 in of weld, or with a smaller leg over the full ten.
Result: 69.6 kips of capacity against 40 - 57% utilised
Where 1.392 comes from
A fillet weld is roughly a right-angled triangle in section, sitting in the corner between two plates. Its legs run along the plate faces, and the shortest path through it - the effective throat - runs from the root to the face at 45 degrees. For an equal-leg fillet the throat is the leg divided by the square root of two, which is 0.707 times the leg.
AISC takes the nominal strength of weld metal in shear as 0.60 times the electrode classification strength, applied to that throat. For E70XX that is 42 ksi. Multiply by the throat, apply the resistance factor of 0.75, and express it per sixteenth of leg: 0.75 x 42 x 0.707 / 16 gives 1.392 kip per inch. The ASD version divides the nominal figure by 2.00 instead and gives 0.928.
Those two numbers are worth memorising because they collapse the whole calculation into multiplication. A 5/16 fillet is five sixteenths, so five times 1.392 is just under 7 kip per inch. Ten inches of it carries 70 kips. It is the kind of arithmetic that can be done standing at the drawing.
The base metal is the other half of the check
A weld does not exist on its own. The load has to get out of one plate, through the weld, and into the other, and the plate has a limit too: shear rupture along the fusion face at 0.60 times its tensile strength, over its thickness.
On heavy plate that limit is generous and the weld governs. On thin material it is not, and it governs quickly. A quarter-inch fillet on eighth-inch sheet has roughly twice the capacity the sheet can deliver, so the joint fails by tearing the sheet beside a weld that is entirely sound - a failure that looks like bad welding and is nothing of the kind.
The practical consequence is that there is a leg size beyond which more weld buys nothing at all. Past the balance point, extra deposition adds cost, arc time, heat and distortion in exchange for capacity the material cannot use. Knowing where that point sits is most of what separates an economical weld from a wasteful one.
- Weld metal governs — Heavy plate, small fillet. More leg size buys more capacity.
- Base metal governs — Thin material. More leg size buys nothing - use more length or thicker stock.
- The balance point — Where the two are equal. Rarely worth exceeding.
- Long welds — Past about 100 times the leg, AISC reduces the effective length - the ends carry more than the middle.
Why there is a minimum size
AISC Table J2.4 sets a minimum fillet size that grows with the thickness of the thinner part joined: an eighth of an inch up to quarter-inch material, three sixteenths to half inch, a quarter to three quarters, and five sixteenths above that.
It looks like a strength rule and it is not. A small weld on heavy plate would often carry its load perfectly well. The problem is thermal: a small bead deposited into a large mass of cold steel has its heat conducted away extremely fast, and rapid cooling in a hardenable steel produces martensite in the heat affected zone. Combine that with the hydrogen every arc introduces and you get cracking - sometimes immediately, often days later.
The larger bead required by the minimum carries more heat, cools more slowly, and survives. It is the same physics that makes preheat necessary on heavy sections, approached from the other end - and it is why the minimum is set by the thickness of the material rather than by the size of the load.
What this assumes, and where it stops
Assumptions
- Equal-leg fillet welds with an effective throat of 0.707 times the leg, which is the standard assumption for shielded processes without deep penetration credit.
- Weld metal strength taken as 0.60 Fexx on the throat, per AISC 360 Table J2.5.
- Base metal limit taken as shear rupture at 0.60 Fu on the fusion face over the thickness of the thinner part.
- The directional strength increase applies only where the whole weld group is loaded transversely.
Limitations
- This is one limit state on one weld, not a connection design. Real connections need the base metal in tension, block shear, bolt interaction, eccentricity and fatigue considered as well.
- It does not handle eccentrically loaded weld groups, which need the instantaneous centre of rotation method and are a substantially harder problem.
- Fatigue is not addressed at all. A weld that is comfortable under static load can fail under a fraction of it if the load cycles.
- Deep penetration credit for some automated processes can increase the effective throat above 0.707 times the leg, but only where the procedure is qualified for it.
- Long welds and intermittent welds have effective length reductions and spacing rules not applied here.
Common questions
How strong is a quarter-inch fillet weld?
About 5.57 kips per inch of length in E70 by LRFD, or 3.71 by ASD - four sixteenths at 1.392 or 0.928 each. Over six inches that is roughly 33 kips factored. But check the base metal: on material thinner than about a quarter inch, the plate will give before the weld does.
Where does the 1.392 number come from?
From 0.75 x 0.60 x 70 ksi x 0.707, divided by 16. The 0.60 Fexx is the nominal shear strength of the weld metal, 0.707 converts leg size to throat, 0.75 is the LRFD resistance factor, and the 16 expresses it per sixteenth of leg. The ASD equivalent, 0.928, is the same nominal value divided by 2.00.
Can a weld be too big?
Yes, in two ways. Beyond the point where the base metal governs, extra leg size adds no capacity at all - only cost, arc time, heat and distortion. And along the edge of a plate, AISC caps the leg at the thickness less a sixteenth so the original edge stays visible for inspection. A weld that has consumed the edge has no measurable leg.
Why is there a minimum fillet size?
Cooling rate, not strength. A small bead on heavy plate loses its heat into the surrounding steel very fast, and fast cooling in hardenable steel makes brittle martensite that cracks in the presence of weld hydrogen. The minimum grows with material thickness because thicker material is a bigger heat sink. It is the same reasoning behind preheat.
Is a transverse fillet really 50% stronger?
It is genuinely stronger, and AISC permits up to a 50% increase through its directional strength equation. The catch is deformation compatibility: in a weld group with welds loaded in different directions, the transverse ones are stiffer and reach capacity first, so the increase cannot simply be applied across the group. For most connections the longitudinal value is the right one to use.
Sources
- AISC 360 Specification for Structural Steel Buildings, chapter J — American Institute of Steel Construction
- AWS D1.1 Structural Welding Code - Steel — American Welding Society
Formula and content last reviewed on .
Results are estimates for information only, not professional advice.
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