SME
Steel Design · Problem type
Design and allowable strength
Use this for “A member yields at a nominal strength of 100 kips. Find its LRFD design strength (φ = 0.9) and ASD allowable strength (Ω = 1.67).”
Chart
Resistance factor φ = 0.9
Safety factor Ω = 1.67
This is an example. Type your own values to start a new problem.
φRₙ (kip)
=
φ
×
Rₙ (kip)
Rallow (kip)
=
Rₙ (kip)
÷
Ω
Formulas
φRₙ
=
φ
×
Rₙ
90 = 0.9 × 100
Rallow
=
Rₙ
Ω
59.8802
=
100
1.67
Assumptions
•
AISC 360-22: φ and Ω come from the limit state, such as yielding 0.90 and 1.67, rupture 0.75 and 2.00.
•
LRFD checks φRₙ ≥ Rᵤ from factored loads; ASD checks Rₙ ÷ Ω ≥ Rₐ from service loads.
•
Rₙ is the nominal strength the specification’s formula gives for that limit state.
Step-by-step
WE KNOW
Rₙ = 100 kip, φ = 0.9, Ω = 1.67
FIND
design strength (φRₙ), allowable strength (Rallow)
Step 1 · Find design strength (φRₙ)
Use φRₙ = φ × Rₙ
LRFD
trims
the
nominal
strength
by
φ
below
1,
covering
weak
material
and
loose
fabrication.
φRₙ = 0.9 × 100
φRₙ = 90 kip
Step 2 · Find allowable strength (Rallow)
Use Rallow = Rₙ ÷ Ω
ASD
divides
the
nominal
strength
by
the
safety
factor
Ω
,
above
1,
to
keep
service
loads
well
below
it.
Rallow
=
100
1.67
Rallow = 59.8802 kip
Check
Put the numbers back in. Both sides should match.
90 = 0.9 × 100 ✓
59.8802
=
100
1.67
✓
Topic check
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