SME
Steel Design · Problem type
Shear strength
Use this for “A W18×50 of 50 ksi steel has d = 18 in and tw = 0.355 in. What factored shear can it carry?”
Diagram
This is an example. Type your own values to start a new problem.
Aw (in²)
=
d (in)
×
tw (in)
φVₙ (kip)
=
0.6
×
Fᵧ (ksi)
×
Aw (in²)
bf
Flange width
example · in
tf
Flange thickness
example · in
Formulas
Aw
=
d
×
tw
6.39 = 18 × 0.355
φVₙ = 0.6 × Fᵧ × Aw
191.7 = 0.6 × 50 × 6.39
Assumptions
•
AISC 360-22 (LRFD): a rolled I-shape with h ÷ tw ≤ 2.24√(E ÷ Fᵧ), so φᵥ = 1.
•
The web yields in shear at 0.6Fᵧ before it can buckle, as it does for nearly every rolled W-shape.
•
The web area is the full depth times the web thickness: Aw = dtw.
Step-by-step
WE KNOW
d = 18 in, tw = 0.355 in, Fᵧ = 50 ksi, bf = 7.5 in, tf = 0.57 in
FIND
web area (Aw), design shear strength (φVₙ)
Step 1 · Find web area (Aw)
Use Aw = d × tw
The
web
runs
the
whole
depth
d
and
is
tw
thick.
Aw = 18 × 0.355
Aw = 6.39 in²
Step 2 · Find design shear strength (φVₙ)
Use φVₙ = 0.6 × Fᵧ × Aw
Steel yields in shear at 0.6Fᵧ; times the web area, with φᵥ = 1.
φVₙ = 0.6 × 50 × 6.39
φVₙ = 30 × 6.39
φVₙ = 191.7 kip
Check
Put the numbers back in. Both sides should match.
6.39 = 18 × 0.355 ✓
191.7 = 0.6 × 50 × 6.39 ✓
Topic check
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