Fillet weld design – maximum direct load on a single fillet weld of length L A flat plate is joined by a single fillet weld (size = leg length). If the permissible shear stress in the weld metal is S, what is the maximum direct load the fillet joint of length L can safely carry (neglecting eccentricity)?

Introduction / Context:
Design of fillet welds in steel structures often requires calculating the direct load a weld can carry. A fillet weld develops shear on its effective throat, not on the full leg size. Recognizing the relationship between leg size and throat is critical for accurate and safe design.

Given Data / Assumptions:

• Single straight fillet weld of length L.
• Fillet weld leg length (commonly called the “size”).
• Permissible (allowable) shear stress in weld metal = S.
• Load is in-plane and concentric; no bending or eccentricity.

Concept / Approach:
The effective throat thickness t_throat of a 45° equal-leg fillet weld equals 0.707 × (fillet size). The shear capacity equals shear stress × resisting area. Resisting area for a straight weld line is (throat) × (length L). Hence, capacity = S × (0.707 × size) × L.

Step-by-Step Solution:
Identify effective throat: t_throat = 0.707 × (fillet size).Resisting area = t_throat × L.Capacity = S × t_throat × L = S × 0.707 × (fillet size) × L.Therefore, the correct expression is 0.707 × S × (fillet size) × L.

Verification / Alternative check:
Many steel codes and handbooks tabulate fillet weld throat as 0.7 times the leg size. Multiplying by allowable shear stress and weld length directly yields the simple design formula.

Why Other Options Are Wrong:

• (b) and (d) use arbitrary multipliers unrelated to the throat relation.
• (c) ignores the 0.707 factor and overestimates capacity.
• (e) is incorrect since a valid option exists.

Common Pitfalls:
Confusing leg size with throat thickness; forgetting to reduce for returns or end crater. For intermittent or staggered welds, use the effective total length.

Final Answer:
0.707 × S × (fillet size) × L