Welded joints: Shear strength of a double parallel fillet weld (two sides) of length l and effective throat s, using allowable shear stress τ, is equal to?

Introduction / Context:
Designing fillet welds requires computing load capacity from effective throat area and allowable shear stress. A double parallel fillet weld places fillets on two opposite sides of a joint, doubling the effective area.

Given Data / Assumptions:

• Two parallel fillet welds of length l each.
• Effective throat size = s (for a right-angled fillet, s ≈ 0.707 * leg size).
• Allowable shear stress in weld metal = τ.

Concept / Approach:
Shear strength = allowable shear stress * effective shear area. For parallel fillets on two sides, total effective throat area equals 2 * s * l (neglecting end returns). Therefore capacity equals P_allow = 2 * s * l * τ.

Step-by-Step Solution:

Verification / Alternative check:
Compare with code formulas where effective throat = 0.707 * leg; substituting s = 0.707 a gives the same scaling factor of 2 for two weld lines.

Why Other Options Are Wrong:
0.5 s l τ and s l τ underestimate by ignoring the second fillet or using an incorrect factor.4 s l τ overestimates; there are only two parallel fillets, not four.Option “2.s.l.τ” is the same as the correct expression but duplicated; we select the standard formatting in option (c).

Common Pitfalls:
Confusing leg size with throat size; forgetting that two-sided fillets double the area; neglecting reduction factors for intermittent welds or eccentric loading (not applicable here).

Final Answer:

2 s l τ