Weld design (double fillet, lap/tee joint): Using the conventional leg size s and weld length l, the tensile strength (capacity) of a double-fillet welded joint is approximated as:

Introduction / Context:
Design of fillet welds often uses simplified strength expressions in terms of leg size s and weld length l for quick sizing. For symmetric double-fillet joints (e.g., lap or tee with fillets on both sides), the effective area is roughly doubled compared to a single fillet, leading to a doubled tensile capacity under direct load sharing.

Given Data / Assumptions:

• Double fillet of leg size s on both sides, length l.
• Allowable tensile stress for weld metal = σt.
• Conventional simplified design (leg-size basis), ignoring detailed throat coefficient in this question's formulation.

Concept / Approach:
In basic textbook treatments aligned with these options, the tensile capacity is taken proportional to weld effective area. For a single fillet, capacity is approximately s * l * σt; for a double fillet, area doubles, so capacity ≈ 2 * s * l * σt. (More rigorous practice uses the throat thickness t_throat = 0.707 s, giving 2 * 0.707 s * l * σ_allow, but such detail is not reflected in the given options.)

Step-by-Step Solution:

Verification / Alternative check:
Throat-based check: using t_throat = 0.707 s gives capacity ≈ 2 * 0.707 s * l * σ_allow = 1.414 s l σ_allow, close to but slightly less than 2 s l σt when σ_allow definitions differ. The problem's answer set aligns with the simplified leg-size expression.

Why Other Options Are Wrong:

• 0.5 s l σt and s l σt correspond to smaller capacities (single fillet or half value).
• 0.707 s l σt is a single-fillet throat approximation, not double fillet.
• 2.0 s l σt duplicates option C; both indicate the same correct value.

Common Pitfalls:
Confusing leg size with throat thickness; mixing allowable weld metal stress with base metal allowable; forgetting that two fillets share the load.

Final Answer:
2 s * l * σt