Fillet Weld Throat Shear Check — Worked Example

1. Problem

A steel bracket is fillet-welded to a vertical plate. The weld is a rectangular pattern: two vertical welds of length 100 mm, connected by a horizontal weld of 80 mm at the top. The bracket carries a vertical load of 25 kN applied at a horizontal eccentricity of 120 mm from the weld centroid. Check the fillet weld throat shear stress for a 6 mm leg size.

2. Given

3. Required

• Weld group properties (area, centroid, polar moment)
• Direct shear stress
• Torsional shear stress
• Combined throat shear stress
• Margin of safety

4. Assumptions

• The weld is treated as a line (unit throat) for property calculation
• Linear elastic stress distribution
• The load is static — no fatigue assessment
• The weld is uniform and of specified leg size
• The bracket is rigid relative to the weld

5. Governing Equations

Weld group area (unit throat):
  A_w  =  Σ L_i

Polar moment (unit throat):
  J_w  =  I_wx + I_wy  (parallel axis)

Direct shear:
  τ_direct  =  P / A_w

Torsional shear:
  τ_torsion  =  P × e × r / J_w

Combined:
  τ_max  =  √[(τ_direct_x + τ_torsion_x)² + (τ_direct_y + τ_torsion_y)²]

Throat stress (multiply by 1/a for actual throat):
  σ_throat  =  τ_max / a

Margin of safety:
  MS  =  (f_w / σ_throat) − 1

6. Calculation

Step 1: Weld group properties (unit throat, origin at bottom-left).

A_w  =  2 × 100 + 80  =  280 mm   (total weld length)

Centroid:
  x̄  =  (2 × 100 × 0 + 80 × 40) / 280  =  3200 / 280  =  11.4 mm
  ȳ  =  (2 × 100 × 50 + 80 × 100) / 280  =  (10,000 + 8000) / 280  =  64.3 mm

I_wx  =  2 × [100³/12 + 100 × (50 − 64.3)²] + 80 × (100 − 64.3)²
      =  2 × [83,333 + 100 × 204.5] + 80 × 1277
      =  2 × 103,783 + 102,160  =  309,726 mm³

I_wy  =  2 × [100 × (0 − 11.4)²] + [80³/12 + 80 × (40 − 11.4)²]
      =  2 × [100 × 130] + [42,667 + 80 × 818]
      =  26,000 + 108,107  =  134,107 mm³

J_w  =  I_wx + I_wy  =  309,726 + 134,107  =  443,833 mm³

6. Calculation (continued)

Step 2: Direct and torsional shear at the critical point (bottom-right weld, maximum distance from centroid).

Critical point: bottom-right, coordinates (0, 0) relative to weld group
  r_x  =  0 − 11.4  =  −11.4 mm
  r_y  =  0 − 64.3  =  −64.3 mm
  r  =  √(11.4² + 64.3²)  =  √(130 + 4134)  =  √4264  =  65.3 mm

Direct shear (uniform):
  τ_dx  =  0   (no horizontal load)
  τ_dy  =  P / A_w  =  25,000 / 280  =  89.3 N/mm  (per unit throat)

Torsional shear (P × e = 25,000 × 120 = 3,000,000 N·mm):
  τ_tx  =  P × e × r_y / J_w  =  3,000,000 × 64.3 / 443,833  =  434.4 N/mm
  τ_ty  =  P × e × r_x / J_w  =  3,000,000 × 11.4 / 443,833  =  77.1 N/mm

Combined (unit throat):
  τ_x  =  0 + 434.4  =  434.4 N/mm
  τ_y  =  89.3 + 77.1  =  166.4 N/mm
  τ_max  =  √(434.4² + 166.4²)  =  √(188,699 + 27,689)  =  √216,388  =  465.2 N/mm

6. Calculation (continued)

Step 3: Convert to actual throat stress.

σ_throat  =  τ_max / a  =  465.2 / 4.24  =  109.7 MPa

MS  =  (215 / 109.7) − 1  =  1.96 − 1  =  +0.96

7. Result

Maximum throat shear stress = 110 MPa. MS = +0.96 (RF = 1.96). The weld meets the static strength requirement. The critical point is at the bottom-right of the weld group where torsional and direct shear combine.

8. Check

• The torsional component (434 N/mm) dominates over the direct shear (89 N/mm) — expected for a large eccentricity
• Dimensional check: N·mm × mm / mm³ = N/mm (force per unit throat) → divide by mm = MPa — consistent
• The critical point at the bottom-right corner is the furthest from the centroid — physically correct
• Sanity: 25 kN on a 280 mm weld with 4.24 mm throat gives an average shear of ~21 MPa, and the peak of 110 MPa (5× average) is plausible with torsion

9. Interpretation

The weld has a positive margin (+0.96) but the torsional shear dominates. If the eccentricity were reduced (e.g. by moving the load closer to the weld), the torsional component would decrease significantly. The result is most sensitive to the eccentricity e and the weld group polar moment J_w. Increasing the vertical weld length or adding a bottom horizontal weld would increase J_w and reduce the stress.

10. Limitations

• No fatigue assessment — weld toe fatigue requires a different approach (nominal stress, hot-spot, or notch stress)
• The weld is assumed uniform — in practice, start/stop craters and weld profile variation affect the stress
• No weld toe stress concentration — this is a throat stress check, not a local toe stress
• The bracket is assumed rigid — flexibility would redistribute the load
• Ultimate strength only — no plastic redistribution considered

11. Related Resources

• Weld Throat and Weld-Group Properties (Reference Library)
• How to Assess a Welded Joint
• How to Assess Weld Fatigue