Welded joints: butt, fillet and eccentrically loaded welds

Why it matters

Chassis frames, axle housings, exhaust brackets, seat frames and suspension arms are welded. A weld is usually the weakest link of a fabricated part, and its stress depends on the throat area, which students often get wrong. Eccentric loads on welded brackets are a standard design calculation and a regular GATE numerical.

Key ideas

Butt weld: the plates are joined edge to edge, and the weld fills the gap. Its effective throat equals the plate thickness t (for a full-penetration weld), so the weld behaves like the parent plate: σ = P/(t·l). Reinforcement above the plate surface is ignored for strength.

Fillet weld: a roughly triangular weld in the corner of a lap, T or corner joint. Its size is the leg s. Failure happens across the throat, the shortest section through the weld, which for an equal-leg 45° fillet is t = s·cos 45° = 0.707·s.

• Parallel (longitudinal) fillet: weld along the direction of the load; the throat is in shear, τ = P/(0.707·s·l).
• Transverse fillet: weld across the load direction. The throat carries combined normal and shear stress; tests show it is somewhat stronger than a parallel fillet. In the usual design method (as in most Indian textbooks) the same allowable shear stress is applied to the throat area for both, which is conservative for transverse welds. Some texts use the allowable tensile stress for transverse welds; follow the convention of your syllabus.
• Circular fillet around a shaft or tube: under torsion τ = 2T/(π·t·d²), under bending σ = 4M/(π·t·d²), with t = 0.707·s.

Effective length: allow for the craters at each end by adding about 12.5–15 mm (or about twice the leg size) to the calculated length.

Eccentric load in the plane of the weld: treat the weld as lines of throat t. The load P at eccentricity e from the weld-group centroid gives:

1. primary shear τ₁ = P/A (A = t × total weld length), parallel to P;
2. secondary (torsional) shear τ₂ = P·e·r/J at a point distance r from the centroid, perpendicular to r, where J = t·J_u and J_u is the polar moment of the weld lines per unit throat. The critical point is the weld end farthest from the centroid on the load side; combine vectorially and compare with the permissible shear stress.

Eccentric load out of the weld plane (bracket welded to a column face, load causing bending): direct shear τ = P/A and bending stress σ_b = M/Z (Z from the throat area), combined by τ_max = √((σ_b/2)² + τ²).

Formulas

Throat: t = 0.707·s Butt weld: σ = P / (t_p·l) Parallel fillet: τ = P / (0.707·s·l) Double parallel fillet: P = 2 × 0.707·s·l·τ Circular fillet, torsion: τ = 2T / (π·t·d²) Circular fillet, bending: σ_b = 4M / (π·t·d²) Primary shear: τ₁ = P / A Secondary shear: τ₂ = P·e·r / J, J = t·J_u Two parallel lines b long, d apart: J_u = b·(3d² + b²) / 6 Resultant: τ = √(τ₁² + τ₂² + 2·τ₁·τ₂·cos θ) Combined: τ_max = √((σ_b/2)² + τ²)

Symbols: s = leg size (mm); t = throat (mm); t_p = plate thickness (mm); l = weld length (mm); P = load (N); T = torque (N·mm); M = bending moment (N·mm); d = shaft diameter or weld-line separation (mm); b = weld-line length (mm); A = throat area (mm²); e = eccentricity (mm); r = distance of the point from the weld centroid (mm); J = polar moment of the throat area (mm⁴); J_u = polar moment per unit throat (mm³); θ = angle between τ₁ and τ₂. Stresses in MPa.

Worked examples

Example 1 (standard). A plate is joined to another by two parallel fillet welds of 10 mm leg and carries a static load of 80 kN along the welds. Permissible shear stress = 56 MPa. Find the length of each weld.

1. Throat: t = 0.707 × 10 = 7.07 mm.
2. Strength of two welds: P = 2 × t × l × τ, so l = 80 000 / (2 × 7.07 × 56) = 101.0 mm.
3. Add about 15 mm for start and stop craters: l ≈ 116 mm, say 120 mm for each weld.

Example 2 (GATE level). A bracket is welded to a column by two horizontal fillet welds, each 100 mm long, 150 mm apart vertically. A vertical load of 20 kN acts 200 mm from the centroid of the weld group. Permissible shear stress = 80 MPa. Find the leg size.

1. Work per unit throat t. Primary shear: τ₁ = 20 000 / (2 × 100 × t) = 100/t MPa.
2. J_u = b(3d² + b²)/6 = 100 × (3 × 150² + 100²)/6 = 1.292 × 10⁶ mm³.
3. Critical points: the ends of both welds on the load side, r = √(50² + 75²) = 90.14 mm.
4. Secondary shear: τ₂ = P·e·r/(t·J_u) = 20 000 × 200 × 90.14 / (1.292 × 10⁶ · t) = 279.1/t MPa.
5. The vertical component of τ₂ there points the same way as τ₁, with cos θ = 50/90.14 = 0.555.
6. Resultant: τ = (1/t) × √(100² + 279.1² + 2 × 100 × 279.1 × 0.555) = 344.8/t MPa.
7. Set τ = 80 MPa: t = 4.31 mm, so s = 4.31/0.707 = 6.10 mm. Use an 8 mm fillet (next standard size above 6.1 mm; check your standard's list).

Common mistakes

• Using the leg size s instead of the throat 0.707·s.
• Using π·d·t for a circular weld under torsion and forgetting the r: the correct shear is 2T/(π·t·d²).
• Writing σ = M/J or τ = M/J without the distance r; that is dimensionally wrong.
• Adding primary and secondary shear as scalars instead of using the angle θ.
• Forgetting the end allowance on weld length.
• Using the polar moment of the bracket plate instead of the weld throat lines.

For GATE ME

Expect: strength of parallel and transverse fillet welds, throat size, weld length for a given load, circular fillet welds under torque, and eccentrically loaded weld groups (primary plus secondary shear). Practise J_u for common weld shapes and the vector addition at the critical point.

Quick check

1. What is the throat of a 12 mm equal-leg fillet weld?
2. A single parallel fillet weld, s = 6 mm, l = 100 mm, carries 21.2 kN. What is the throat shear stress?
3. Which weld is stronger per unit length in tests, transverse or parallel?
4. Why is extra length added to a calculated weld length?
5. For a circular fillet weld under torque, how does τ vary with shaft diameter d?

Answers: 1. 8.48 mm. 2. 21 200/(0.707 × 6 × 100) ≈ 50 MPa. 3. Transverse. 4. To allow for the craters at the start and end of the weld. 5. τ ∝ 1/d².