Why is a fillet weld measured by its leg but checked at its throat?

The leg is what a welder can gauge directly at the joint, so drawings specify it — but shear failure cuts across the weld’s narrowest section, the diagonal throat, which for a 45-degree fillet is 0.707 times the leg. All strength math happens on the throat, and the conversion back to a drawing callout is the final division by 0.707·Fw. Confusing the two overstates capacity by 41%.

Does doubling the eccentricity double the stress in the weld?

It doubles the torsional part only. The twist T = P·e scales linearly with eccentricity, and with it the T·r/J components at the far corner — but the direct shear P/Lw is unchanged, and the two combine as vectors. So the resultant grows steeply yet somewhat less than double. In the bracket example, going from zero eccentricity to 3.5 in doubled the required leg; a further doubling of e would raise it again, but by less than a factor of two.

What does a shear flow in pounds per inch mean physically?

It is the force each inch of weld line must transmit, computed on a weld of unit throat so the answer is independent of size. A resultant of 2,486 lb/in means every inch of weld at the critical corner carries that much shear; dividing by the throat capacity per inch of leg, 0.707·Fw, converts the demand into the leg size that supplies enough throat area.

Is the elastic vector method the most accurate way to size a weld group?

It is the conservative one. The elastic method assumes stress grows linearly from the centroid and caps the group at first yield of the worst corner, while the instantaneous-center method accounts for load-deformation behavior and typically demonstrates more capacity from the same weld layout. Design practice per AISC 360 accepts both; the elastic method’s appeal is that it is transparent, hand-checkable, and always errs toward a bigger weld.

How Strong Is a Fillet Weld? Throat Size, Weld Lines, and Eccentric Loads

Leg, throat, and the 0.707 that connects them

A fillet weld is the triangle of metal laid into the corner where two plates meet, and it is measured by its leg — the length of each side of that triangle along the plates. But the weld does not fail across its leg. Load shears a fillet across its narrowest section, the throat, which runs from the root of the corner diagonally out to the face. For the standard 45-degree fillet, that diagonal is the leg times sin 45°, so throat = 0.707 × leg. A 1/4-in fillet is, structurally, a 0.177-in-thick strip of weld metal; the quoted size always flatters the working section by a factor of √2.

That one geometric constant runs through every fillet calculation in this guide. When a code or a calculator asks whether the weld metal’s allowable stress Fw is exceeded, the demand is compared against the throat area, not the leg — and when a required strength has been computed, converting it back into a size on the drawing means dividing by 0.707·Fw. Get comfortable with the throat and the rest of fillet-weld design becomes bookkeeping about where the load goes; forget it and every weld you size is 41% weaker than you think.

Treating a weld as a line

The elegant trick behind practical weld design is to postpone choosing the size. Instead of analyzing a weld of some particular throat, treat each weld as a line of unit throat: a straight segment whose "area" is simply its length. Every property a cross-section has, a weld line has too — a total length Lw standing in for area, a length-weighted centroid, line moments of inertia Ix and Iy computed segment by segment with the parallel-axis theorem, and a polar moment J = Ix + Iy for twisting. The units come out one power of length short of the familiar section properties, which is the tell that a line, not an area, is doing the work.

Analyzing per unit throat means the stress result arrives as a shear flow — force per unit length of weld, pounds per inch rather than psi — and that number is independent of any leg size you might pick. Sizing then happens in a single closing step: required leg = f_r / (0.707·Fw), where f_r is the worst shear flow found anywhere on the group. The separation is what makes the method so usable — geometry and loading are analyzed once, and the leg size falls out at the end, rather than being guessed up front and iterated.

When the load is off-center: direct shear plus torsion

If a load P passed exactly through the centroid of the weld group, every inch of weld would share it equally: a uniform direct shear flow f_v = P / Lw. Brackets almost never cooperate. A load hanging at a horizontal eccentricity e from the centroid also applies a torsion T = P·e, and the weld group must resist both at once. The elastic (vector) method handles this by superposition: the direct component is spread uniformly, the torsional component varies over the group, and at any point the two are added as vectors.

The torsional shear flow behaves exactly like torsion on a shaft: it grows linearly with distance from the centroid and acts perpendicular to the radius, with components f_tx = T·y/J and f_ty = T·x/J at a point (x, y) measured from the centroid. Near the centroid the twist contributes almost nothing; at the extremities it can dwarf the direct shear. Eccentricity is therefore expensive twice over — it adds a whole new load component, and it concentrates that component at the weld’s farthest fibers instead of sharing it evenly.

The critical corner

Since the torsional component increases with radius, the governing point of a weld group is normally the corner farthest from the centroid — and specifically the far corner where the torsional vector lines up most directly with the direct shear, so the two add rather than partially cancel. On the near side of the group, the twist actually opposes the direct shear and relieves it; the far side pays for that relief with interest. A weld group fails from its worst-loaded fiber, not its average, so the design check lives entirely at that corner: f_r = sqrt((f_v + f_ty)² + f_tx²).

In practice you do not need to intuit which corner governs — the honest procedure evaluates the resultant at every segment endpoint and keeps the largest, which is exactly what software does. But the geometry intuition still earns its keep when shaping the joint: anything that moves weld length farther from the centroid raises J faster than it raises the worst radius, which is why spreading two welds farther apart, or boxing a bracket’s perimeter, tames an eccentric load far more effectively than simply laying a fatter bead in the same place.

A two-weld bracket, worked to a leg size

The running example: a plate bracket hung on two vertical fillet welds, each 8 in long, spaced 6 in apart, carrying P = 20,000 lb at an eccentricity of 3.5 in from the weld centroid, with an allowable weld stress taken as Fw = 20 ksi for the illustration. Line properties first: Lw = 2 × 8 = 16 in, centroid midway between the welds. Ix = 2·(8³/12) = 85.3 in³, Iy = 2·8·(3)² = 144 in³, so J = 229.3 in³. The direct shear flow is f_v = 20,000/16 = 1,250 lb/in, and the torsion is T = 20,000 × 3.5 = 70,000 lb·in.

Now the critical corner, at (3, 4) from the centroid — top of the right-hand weld. Torsional components: f_tx = 70,000×4/229.3 = 1,221 lb/in and f_ty = 70,000×3/229.3 = 916 lb/in. The resultant: f_r = sqrt((1,250 + 916)² + 1,221²) = sqrt(2,166² + 1,221²) = 2,486 lb/in. Required leg: 2,486/(0.707 × 20,000) = 0.176 in — a 3/16-in fillet covers it. The instructive comparison: had the same 20,000 lb passed through the centroid, the demand would have been the bare 1,250 lb/in, needing only a 0.088-in leg. Three and a half inches of eccentricity exactly doubled the weld. The weld group calculator runs this whole analysis for parallel welds, boxes, or any custom segment layout — centroid, Ix, Iy, J, the critical-corner resultant, and the required leg — with the layout and centroid plotted so you can see which corner is doing the suffering.

Sizing in one pass

Recap the bracket: two 8-in welds 6 in apart give Lw = 16 in and J = 229.3 in³; 20,000 lb at 3.5 in of eccentricity produces 1,250 lb/in of direct shear plus a 70,000 lb·in twist; the far corner sees the vector sum, 2,486 lb/in; and dividing by 0.707 × 20,000 psi turns that into a 0.176-in leg — a 3/16-in fillet, twice what the centered load would have needed. That is the entire elastic method in five moves: treat welds as lines, find the group’s centroid and J, split the load into direct and torsional shear flows, add them as vectors at the farthest corner, and convert the worst resultant to a leg through the 0.707 throat factor. It is deliberately conservative — the instantaneous-center method extracts more capacity from the same welds when you need it — but as a first sizing it is fast, transparent, and safely on the strong side.