About Weld Group Calculator
The weld group calculator analyzes an eccentrically loaded fillet weld group by the elastic (vector) method. The weld is treated as a line of unit throat, so each straight segment contributes a line area equal to its length. The tool returns the total weld length, the length-weighted centroid, the line moments of inertia Ix and Iy about the centroidal axes, and the polar moment J = Ix + Iy.
An applied vertical load P at a horizontal eccentricity e produces both a direct shear flow f_v = P / Lw and a torsion T = P·e about the centroid. The torsional shear flow varies linearly with distance from the centroid, so the calculator locates the critical point (the farthest corner), combines the direct and torsional components vectorially, and reports the resultant shear flow together with the required fillet leg size leg = f_r / (0.707·Fw).
How It Works
- Pick a shape: two parallel vertical welds, a rectangular weld box, or build a custom group from straight segments defined by their endpoints.
- Enter the applied vertical load P, the horizontal eccentricity e of the load from the weld centroid, and the allowable weld stress Fw.
- The calculator sums the segment lengths for Lw, finds the length-weighted centroid, and computes Ix, Iy and J about the centroidal axes using the parallel-axis theorem on each segment as a line.
- It then evaluates the direct shear plus the torsional shear flow at every segment endpoint, takes the largest resultant as the critical stress, and converts it to a required fillet leg size.
Worked Example
Two vertical fillet welds, each h = 200 mm long, are spaced d = 100 mm apart (one at x = -50 mm, one at x = +50 mm). A vertical load P = 50 kN = 50,000 N acts at an eccentricity e = 150 mm from the centroid, with an allowable weld stress Fw = 200 MPa. The total weld length is Lw = 2h = 400 mm and the centroid is at the origin. Treating each weld as a line: Ix = 2·(h³/12) = 1,333,333 mm³, Iy = 2·h·(d/2)² = 1,000,000 mm³, so J = Ix + Iy = 2,333,333 mm³. The direct shear flow is f_v = P/Lw = 50,000/400 = 125 N/mm and the torsion is T = P·e = 7,500,000 N·mm. At the critical corner (50, 100): f_tx = T·y/J = 7,500,000·100/2,333,333 = 321.43 N/mm and f_ty = T·x/J = 7,500,000·50/2,333,333 = 160.71 N/mm. The resultant is f_r = sqrt((125 + 160.71)² + 321.43²) = sqrt(285.71² + 321.43²) = 430.06 N/mm, giving a required leg size of 430.06 / (0.707·200) = 3.04 mm.
Formulas
- Total weld length and centroid
Lw = sum(L_i), xc = sum(L_i*xm_i)/Lw, yc = sum(L_i*ym_i)/Lw- Line moments of inertia (parallel-axis)
Ix = sum( Ix_local_i + L_i*(ym_i - yc)^2 ), Iy = sum( Iy_local_i + L_i*(xm_i - xc)^2 ), J = Ix + Iy- Direct and torsional shear flow
f_v = P / Lw, T = P * e, f_tx = T*y / J, f_ty = T*x / J- Resultant stress and required leg
f_r = sqrt( (f_v + f_ty)^2 + f_tx^2 ), leg = f_r / (0.707 * Fw)
Standards & References
- AISC 360 (Specification for Structural Steel Buildings)
- Elastic (vector) method for eccentrically loaded weld groups
- Eurocode 3 EN 1993-1-8
Frequently Asked Questions
What is the elastic (vector) method for weld groups?
It treats the fillet weld as a line of unit throat and superimposes the direct shear flow (P divided by the total weld length) with the torsional shear flow (T·r/J) as vectors at the most highly stressed point. It is conservative compared with the instantaneous-centre method.
Why is the weld treated as a line of unit throat?
Computing properties per unit throat keeps the analysis independent of the leg size. The resulting shear flow has units of force per unit length, which is then divided by 0.707·Fw to find the leg size that provides enough throat area to resist it.
Where is the critical point in a weld group?
The torsional shear flow grows linearly with distance from the centroid, so the critical point is normally the corner farthest from the centroid where the direct and torsional components add most directly. The calculator checks every segment endpoint and reports the largest resultant.
How is the required fillet leg size obtained?
The throat of a fillet weld is 0.707 times the leg size, so the required leg equals the resultant shear flow divided by (0.707 times the allowable weld stress Fw). Larger eccentricity or load increases the resultant and therefore the required leg.