SME
Steel Design
Tension and compression members
Use this for “A W10×49 column of 50 ksi steel has KL = 15 ft about its weak axis. Find its design strength φPₙ.”
Diagram
K = 1: pinned at the bottom, pinned on top.
The whole column bends in one half-wave: KL = L.
This is an example. Type your own values to start a new problem.
KL ÷ r
=
KL (ft)
÷
rᵧ (in)
Fₑ (ksi)
=
π²
×
29,000
ksi
÷
(
)
2
φPₙ (kip)
=
0.9
×
Fcr (ksi)
×
Ag (in²)
shape
W-shape
example
Fᵧ
Yield stress
example · ksi
Formulas
Ag = 14.4
14.4 = 14.4
rᵧ = 2.54
2.54 = 2.54
KL
r
=
KL
rᵧ
70.87
=
180
2.54
Fₑ
=
π
2
×
29,000
÷
(
KL
r
)
2
56.99
=
π
2
×
29,000
÷
(70.87)
2
Fcr
=
0.658
Fᵧ ÷ Fₑ
×
Fᵧ
34.63
=
0.658
50 ÷ 56.99
×
50
φPₙ = 0.9 × Fcr × Ag
448.9 = 0.9 × 34.63 × 14.4
Formulas are worked in in. The step-by-step shows the conversions.
Assumptions
•
AISC 360-22 (LRFD) flexural buckling, E = 29,000 ksi, φ = 0.90; the weak axis (rᵧ) governs; AISC 360-22 writes KL as Lc.
•
No slender elements: the flanges and web are stocky enough not to buckle locally first.
•
Ag and rᵧ are read from the W-shape table; pick “other (typed)” to type another shape’s.
•
KL ÷ r ≤ 200 for a compression member; past 4.71√(E ÷ Fᵧ) it buckles elastically.
Step-by-step
WE KNOW
shape: W10×49, KL = 15 ft, Fᵧ = 50 ksi
FIND
gross area (Ag), radius of gyration (weak axis) (rᵧ), slenderness ratio (KL ÷ r), elastic buckling stress (Fₑ), critical stress (Fcr), design strength (φPₙ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to in).
KL = 15 ft = 180 in (1 ft = 12 in)
Step 2 · Find gross area (Ag)
Use Ag = 14.4
Read it from the W-shape table, for a W10×49.
W10×49 chosen: its row in the W-shape table
Ag = 14.4 in²
Step 3 · Find radius of gyration (weak axis): rᵧ
Use rᵧ = 2.54
Read it from the W-shape table, for a W10×49.
rᵧ = 2.54 in
Step 4 · Find slenderness ratio (KL ÷ r)
Use KL ÷ r = KL ÷ rᵧ
KL in inches over the weak-axis radius of gyration: the column’s slenderness.
KL
r
=
180
2.54
KL
r
=
70.866
Step 5 · Find elastic buckling stress (Fₑ)
Use Fₑ = π² × 29,000 ÷ (KL ÷ r)²
Euler’s buckling stress for this slenderness, with E = 29,000 ksi.
Fₑ
=
π
2
×
29,000
÷
(70.866)
2
Fₑ
=
9.87
×
29,000
5,022
Fₑ
=
286,230
5,022
Fₑ = 56.993 ksi
Step 6 · Find critical stress (Fcr)
Use Fcr = 0.658Fᵧ ÷ Fₑ × Fᵧ
Not slender enough to buckle elastically: part of the section yields first, so AISC’s inelastic curve gives Fcr.
70.866
≤
4.71
×
√(
29,000
50
):
inelastic
buckling
Fcr
=
0.658
50 ÷ 56.993
×
50
4.71
×
√(
29,000
50
)
=
113.4
Fcr
=
0.658
0.8773
×
50
Fcr = 0.6927 × 50
Fcr = 34.634 ksi
Step 7 · Find design strength (φPₙ)
Use φPₙ = 0.9 × Fcr × Ag
The critical stress over the whole area is Pₙ; LRFD takes φ = 0.90 of it.
φPₙ = 0.9 × 34.634 × 14.4
φPₙ = 31.1706 × 14.4
φPₙ = 448.9 kip
Check
Put the numbers back in. Both sides should match.
14.4 = 14.4 ✓
2.54 = 2.54 ✓
70.866
=
180
2.54
✓
56.993
=
π
2
×
29,000
÷
(70.866)
2
✓
34.634
=
0.658
50 ÷ 56.993
×
50
✓
448.9 = 0.9 × 34.634 × 14.4 ✓
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