Welded joints: butt, fillet and eccentrically loaded welds

Why it matters

Welded frames, brackets, motor mounts and fabricated machine bases are cheaper and lighter than castings or bolted assemblies for one-off and small-batch machines. But the weld is usually the weakest and most fatigue-sensitive part of the structure. Sizing welds correctly - especially when the load is offset from the weld group - prevents brackets from tearing off in service.

Key ideas

Butt welds join plates lying in the same plane, with the edges prepared (square, V, double-V, U) so the weld penetrates the full thickness. The throat of a full-penetration butt weld is taken as the plate thickness t (reinforcement above the surface is ignored). A butt weld is loaded in tension or compression across its throat.

Fillet welds join surfaces at roughly a right angle (lap, T and corner joints) with a triangular cross-section. For an equal-leg 45° fillet of leg (size) s, the throat t is the shortest distance from the root to the face: t = s·cos 45° = 0.707·s. Fillet welds fail across the throat, so all stresses are calculated on the throat area t·l.

• Parallel (longitudinal) fillet weld: load along the weld axis; the throat is in shear.
• Transverse fillet weld: load perpendicular to the weld axis; the stress state is complex, but Indian textbooks conventionally design it on tensile stress over the throat area: P = 0.707·s·l·σt. (Shigley-style texts treat all fillet stresses as shear on the throat.) Follow the convention and permissible stresses of your syllabus or code.

Practical points: add an allowance (commonly 12.5-15 mm) to calculated weld lengths for poor start and end craters; minimum fillet size depends on plate thickness (code table); the weld toe is a strong stress raiser, so welded joints have low fatigue strength and codes give fatigue stress-reduction factors by joint category; residual stresses and distortion can be reduced by preheating, sequence and stress relieving.

Welds treated as lines. For combined loading, treat each weld as a line of throat t, compute section properties of the line pattern (unit area, unit Z, unit J from tables) and multiply by t.

Circular fillet weld (tube or shaft welded to a plate) of diameter d and throat t:

• under torsion T: τ = 2T/(π·t·d²) (from J = π·t·d³/4),
• under bending M: σb = 4M/(π·t·d²) (from Z = π·t·d²/4),
• combined: τmax = √((σb/2)² + τ²) and σmax = σb/2 + τmax.

Eccentric load in the plane of the weld group (bracket welded to a column face). Shift P to the centroid G of the weld group, adding torque P·e:

• Primary shear τ₁ = P/A, uniform, parallel to P, with A = Σt·l.
• Secondary shear τ₂ = P·e·r/J at a point distance r from G, perpendicular to r, where J is the polar moment of the throat area about G.
• Combine vectorially at the critical point (farthest from G, where components reinforce): τ = √(τ₁² + τ₂² + 2·τ₁·τ₂·cos θ).

Eccentric load perpendicular to the plane of the welds produces direct shear plus bending stress σb = M/Z on the throat; combine as above.

Formulas

t = 0.707·s

• t: throat thickness (m); s: leg size of a 45° fillet (m).

P = σt·t_p·l (butt weld)

• P: load (N); σt: permissible tensile stress of the weld (Pa); t_p: plate thickness (m); l: weld length (m).

P = 0.707·s·l·τ (single parallel fillet) · P = 0.707·s·l·σt (single transverse fillet, Indian convention)

• τ: permissible shear stress in the weld (Pa). For double welds use twice the length.

τ = 2T / (π·t·d²) · σb = 4M / (π·t·d²) (circular fillet weld)

• T: torque (N·m); M: bending moment (N·m); d: weld (tube) diameter (m).

τ₁ = P / A · τ₂ = P·e·r / J · J = t·Ju

• A: total throat area (m²); e: eccentricity from the weld centroid (m); r: distance of the point from the centroid (m); J: polar moment of the throat area (m⁴); Ju: unit polar moment of the weld lines (m³), from tables.

τmax = √((σb/2)² + τ²) · σmax = σb/2 + √((σb/2)² + τ²)

Worked examples

Example 1 (standard). A plate 10 mm thick is welded to another by two parallel fillet welds of leg s = 10 mm. Static load P = 120 kN; permissible shear stress in the weld 94 MPa. Find the length of each weld.

1. Two welds: P = 2 × 0.707·s·l·τ.
2. l = P / (2 × 0.707 × s × τ) = 120,000 / (2 × 0.707 × 10 × 94) = 120,000 / 1329.2 = 90.3 mm.
3. Adding a start/stop allowance of about 12.5 mm: about 103 mm per weld (round up as your code allows).

Example 2 (GATE level). A tube of outer diameter 60 mm is welded to a plate by a circular fillet weld of leg 6 mm all round. It carries a torque of 1.2 kN·m and a bending moment of 0.9 kN·m. Find the maximum shear and normal stresses in the weld.

1. Throat: t = 0.707 × 6 = 4.242 mm; π·t·d² = π × 4.242 × 60² = 47,976 mm³.
2. Torsional shear: τ = 2T/(π·t·d²) = 2 × 1.2 × 10⁶ N·mm / 47,976 mm³ = 50.0 MPa.
3. Bending stress: σb = 4M/(π·t·d²) = 4 × 0.9 × 10⁶ / 47,976 = 75.0 MPa.
4. τmax = √((75.0/2)² + 50.0²) = √(1406 + 2500) = 62.5 MPa.
5. σmax = 37.5 + 62.5 = 100.0 MPa.

Common mistakes

• Using the leg size s instead of the throat 0.707·s as the stress-carrying thickness.
• Counting only one weld when the joint has welds on both sides (double fillet).
• Adding primary and secondary shear as scalars without checking directions.
• Forgetting the start/stop allowance on calculated weld lengths.
• Mixing conventions: designing transverse fillets in tension with one book's permissible stress and another's shear formula.
• Ignoring fatigue: a weld that passes a static check can crack at the toe under cyclic load.

For GATE ME

Expect calculations of throat thickness, the load capacity of parallel and transverse fillet welds, and the maximum shear stress in a circular fillet weld under torsion (often combined with bending) or in an eccentrically loaded weld group (primary plus secondary shear). Practise J of simple weld patterns (two parallel lines, a rectangle, a circle) treated as lines of throat t.

Quick check

1. Throat of a 8 mm fillet weld?
2. Which area carries the stress in a fillet weld?
3. Torque 0.5 kN·m on a circular fillet weld, d = 50 mm, t = 5 mm. Find τ.
4. Why are welded joints weak in fatigue? Answers: 1. 5.66 mm. 2. Throat area t·l. 3. τ = 2 × 0.5 × 10⁶ / (π × 5 × 2500) ≈ 25.5 MPa. 4. The weld toe is a sharp stress raiser with residual tensile stress and defects.