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Fillet Weld Capacity

AISC LRFD fillet weld design — strength per unit length and total capacity, with optional load-orientation increase factor and base metal shear check.

Inputs

mm
mm
deg from weld axis (0=long, 90=transverse)
MPa (A36 = 400, A992 = 450)
mm (thinner connected part)
kN
Method & assumptions
  • Throat thickness te = 0.707·w (for equal-leg fillet).
  • φRw per unit length = φ·Fnw·te; φ = 0.75; Fnw = 0.6·FEXX·(1.0 + 0.50·sin¹·⁵θ) (J2-5).
  • Base metal shear rupture per length = φ·0.6·Fu·t; φ = 0.75. Controls if base metal is thinner than ~1.41·w (rule of thumb).
  • Use lower of (weld) and (base metal) capacities.
  • For full strength of base metal (Type J2): wmin based on AISC J2.2b plate thickness table (not handled here).

Results

Throat te
Weld Strength Fnw
φRw per length
Base Metal φRbm per length
Controlling per length
Total Capacity
Demand / Capacity
StatusOK
Required Length Lreq

About this calculator

This tool determines the LRFD design strength of a fillet weld, comparing the weld metal's own shear strength against the base-metal shear rupture strength of the thinner connected part, and reports the controlling capacity, the demand/capacity ratio, and the weld length required to carry a given factored load. It is intended for structural engineers and fabricators sizing fillet welds on steel connections.

Method & formulas

Effective throat: te = 0.707·w, for an equal-leg fillet weld of leg size w.

Weld metal nominal strength (AISC LRFD): FEXX is the electrode classification strength; θ is the angle between the load line and the weld's axis (0° = parallel, 90° = transverse — transverse welds carry up to 50% more per unit length).

te = 0.707·w
Fnw = 0.6·FEXX·(1 + 0.50·sin1.5θ)
φRw = φ·Fnw·te  (φ = 0.75)

Base metal shear rupture per unit length: φRbm = φ·0.6·Fu·t (φ = 0.75), with Fu = base metal tensile strength, t = thickness of the thinner connected part.

Governing strength. Controlling strength = min(φRw, φRbm). Total capacity = controlling × L; demand/capacity = Pu / (controlling × L); required length Lreq = Pu / controlling.

Worked example

Inputs: w = 8 mm fillet, E70XX electrode (FEXX = 485 MPa), θ = 90° (transverse), Fu = 450 MPa (A992 base metal), t = 12 mm, Pu = 250 kN.

te = 0.707 × 8 ≈ 5.66 mm.
Fnw = 0.6 × 485 × (1 + 0.50×1) = 436.5 MPa (sin 90° = 1).
φRw = 0.75 × 436.5 × 5.66 ≈ 1852 N/mm.
φRbm = 0.75 × 0.6 × 450 × 12 = 2430 N/mm → weld controls.
Required length: Lreq = 250,000 / 1852 ≈ 135 mm (e.g. two 70 mm-long fillet welds, one each side of the connected element).

Assumptions & limitations

FAQ

Why is a transverse (90°) fillet weld stronger per unit length than a longitudinal (0°) weld of the same size?
Test data shows fillet welds loaded transverse to their axis fail at a higher unit stress than those loaded longitudinally; the (1 + 0.50 sin1.5θ) factor captures that directional strength increase, up to 50% higher at θ = 90°.
When does the base metal control instead of the weld?
When the connected material is thick relative to the weld leg size — once the connected part's own shear rupture strength drops below the weld's strength, undersized fillets rarely control and the plate or shape becomes the weak link.
How does "required length" differ from "total capacity"?
Total capacity multiplies the controlling unit strength by whatever length you entered; required length instead solves backward for the length needed so controlling strength × length just equals your applied factored load — it sizes the weld for you.