SME
Manufacturing Processes · Problem type
Strength of a fillet weld
Use this for “A 5/16 in fillet weld 6 in long has an allowable shear of 21 ksi on the throat. What load can it carry?”
Diagram
The throat, the weld’s thinnest part, is 0.707w = 0.707 × 0.313 = 0.221 in.
The section is enlarged; the plan is to scale along the welds.
This is an example. Type your own values to start a new problem.
t (in)
=
0.707
×
w (in)
F (kip)
=
τₐ (ksi)
×
t (in)
×
L (in)
Formulas
t
=
0.707
×
w
0.2209 = 0.707 × 0.3125
F
=
τₐ
×
t
×
L
27.8381 = 21 × 0.2209 × 6
Assumptions
•
The legs are equal and the face sits at 45°, so the throat is w ÷ √2 ≈ 0.707w.
•
The load counts as shear on the throat whatever its direction: the usual design rule.
•
τₐ already holds the safety factor (for instance 0.3 times the filler metal’s strength).
Step-by-step
WE KNOW
w = 0.3125 in, L = 6 in, τₐ = 21 ksi
FIND
throat (t), allowable load (F)
Step 1 · Find throat (t)
Use t = 0.707 × w
The throat runs from the root to the 45° face: the leg times sin 45°.
t
=
0.707
×
0.3125
t
=
0.2209375
in
Step 2 · Find allowable load (F)
Use F = τₐ × t × L
The
throat
area
t
×
L
in
in²
times
the
stress
in
ksi
gives
kips.
F
=
21
×
0.2209375
×
6
F
=
4.64
×
6
F
=
27.8381
kip
Check
Put the numbers back in. Both sides should match.
0.2209375 = 0.707 × 0.3125 ✓
27.8381 = 21 × 0.2209375 × 6 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.