Steel Column Design Examples

Three steel checks from the AISC Steel Construction Manual, worked by hand to the AISC 360-16 specification: a column in compression, a wide flange beam's bending strength as its bracing changes, and a shape whose flanges are too thin to reach full plastic strength.

Every number on this page comes from the code behind the free column designer, so you can follow each step and see it land on the value in the manual. All three shapes are ASTM A992 steel, F_y = 50 ksi and E = 29,000 ksi.

The AISC values and the column designer’s, side by side
Example	AISC	Here
1. W14×132 at 30 ft, φP_n	893 kips	893 kips
1. W14×132 at 30 ft, P_n/Ω	594 kips	594 kips
2. W18×50, φM_p	379 kip-ft	379 kip-ft
2. W18×50, L_p	5.83 ft	5.83 ft
2. W18×50, L_r	16.9 ft	16.96 ft
3. W14×90, φM_n	574 kip-ft	574 kip-ft

Example 1: a W14×132 column, 30 ft tall

This is AISC Design Example E.1A. A W14×132 column is pinned at the top and bottom and braced only at its ends, so it can buckle over the full 30 ft about either axis (K = 1.0). It carries a dead load of 140 kips and a live load of 420 kips.

Loads

LRFD (load and resistance factor design) and ASD (allowable strength design) each combine the loads their own way:

LRFD: P_u = 1.2D + 1.6L = 1.2(140) + 1.6(420) = 840 kipsASCE 7-16 2.3.1

ASD: P_a = D + L = 140 + 420 = 560 kipsASCE 7-16 2.4.1

Are the flanges and web thick enough?

If a flange or the web is thin for its width, it wrinkles before the column buckles as a whole, and section E7 cuts the area that counts. Compare each width-to-thickness ratio with its limit from Table B4.1a:

Flange: b_f/2t_f = 7.15 < 0.56√(E/F_y) = 13.5Table B4.1a

Web: h/t_w = 17.7 < 1.49√(E/F_y) = 35.9Table B4.1a

Both pass, so the full area A = 38.8 in² works.

Which axis buckles?

A column buckles about the axis with the larger slenderness, L_c/r. With the same 30 ft (360 in) length both ways, that's the weak axis, the one with the smaller radius of gyration:

Strong axis: L_cx/r_x = 360 / 6.28 = 57.3

Weak axis: L_cy/r_y = 360 / 3.76 = 95.7 (governs)

Under 200, so the column isn't too slenderE2 user note

Buckling stress

F_e = π²E / (L_c/r)² = π²(29,000) / 95.7² = 31.2 ksiE3-4

F_y/F_e = 50 / 31.2 = 1.60 ≤ 2.25, so inelastic bucklingE3-2

F_cr = 0.658^F_y/F_e F_y = 0.658^1.60(50) = 25.6 ksiE3-2

A W shape can also twist instead of bending sideways. Section E4 gives the twisting buckling stress; for a stocky shape like this one it's far higher, so it doesn't govern:

F_e (twisting) = [π²EC_w/L_cz² + GJ] / (I_x + I_y) = 93.4 ksi > 31.2 ksiE4-2

Strength

P_n = F_crA_g = 25.58(38.8) = 992 kipsE3-1

LRFD: φ_cP_n = 0.90(992) = 893 kips > 840 kips, OKE1

ASD: P_n/Ω_c = 992 / 1.67 = 594 kips > 560 kips, OKE1

These match the manual's 893 kips and 594 kips, which are also the values in its column load table (Table 4-1a) for a W14×132 at 30 ft. The column uses 94% of its strength under LRFD.

Open this column in the column designer

Example 2: how bracing changes a W18×50's bending strength

A wide flange bent about its strong axis can fail two ways: it yields across the whole section, or its compression flange buckles sideways and twists the beam (lateral-torsional buckling). Which one happens depends on L_b, the length between the points that stop the compression flange moving sideways. The column designer runs this check for every W shape column with a moment about its strong axis.

Fully braced: the plastic moment

M_p = F_yZ_x = 50(101) = 5,050 kip-in = 420.8 kip-ftF2-1

φ_bM_p = 0.90(420.8) = 379 kip-ftF1

The flanges (b_f/2t_f = 6.57) are under the compact limit of 9.15, so they don't buckle first.

The two bracing limits

Up to L_p the beam reaches its full plastic moment. Past L_r it buckles elastically. Between them the strength drops along a straight line.

L_p = 1.76r_y√(E/F_y) = 1.76(1.65)(24.1) = 69.9 in = 5.83 ftF2-5

L_r = 1.95r_ts(E/0.7F_y)√[Jc/S_xh_o + √((Jc/S_xh_o)² + 6.76(0.7F_y/E)²)] = 203.5 in = 16.96 ftF2-6

with r_ts = 1.98 in, J = 1.24 in⁴, c = 1, h_o = 17.43 inF2-7

Braced every 11.67 ft

Design Example F.1-2A takes this beam on a 35 ft span, braced at its ends and third points, so L_b = 11.67 ft (140 in), between L_p and L_r:

M_n = C_b[M_p − (M_p − 0.7F_yS_x)(L_b − L_p)/(L_r − L_p)]F2-2

= 1.0[5,050 − (5,050 − 3,112)(140 − 69.9)/(203.5 − 69.9)] = 4,033 kip-in

φ_bM_n = 0.90(4,033) / 12 = 302 kip-ftF1

Bracing at 11.67 ft instead of every 5.83 ft costs the beam 20% of its strength. Each extra foot of unbraced length takes φBF = 0.90(5,050 − 3,112) / 133.5 = 13.1 kip-ft off, the slope of the straight line.

C_b raises the strength when the moment isn't constant along the unbraced length. The design example works out C_b for its middle segment; the column designer takes C_b = 1.0, which is the lowest it can be and always safe.

Example 3: a W14×90 with noncompact flanges

Most wide flange shapes have flanges thick enough to reach the plastic moment. A few of the wide W14 column shapes don't: their flanges are so wide for their thickness that they buckle locally a little before the whole section yields. The manual marks these shapes in its beam tables, and the W14×90 is one of them.

λ = b_f/2t_f = 14.5 / 2(0.71) = 10.2Table B4.1b

λ_p = 0.38√(E/F_y) = 9.15Table B4.1b

λ_r = 1.0√(E/F_y) = 24.08Table B4.1b

9.15 < 10.2 < 24.08, so the flanges are noncompact

The strength drops along a straight line between M_p at λ_p and 0.7F_yS_x at λ_r:

M_p = F_yZ_x = 50(157) = 7,850 kip-inF2-1

M_n = M_p − (M_p − 0.7F_yS_x)(λ − λ_p)/(λ_r − λ_p)F3-1

= 7,850 − (7,850 − 5,005)(10.2 − 9.15)/(24.08 − 9.15) = 7,650 kip-in

φ_bM_n = 0.90(7,650) / 12 = 574 kip-ftF1

That's the 574 kip-ft in Table 3-2, 2.5% under the plastic moment. A program that skips the flange check would report 589 kip-ft and overstate the strength. For weak axis bending, section F6 makes the same flange check.

Putting them together

A real column usually carries load and bending at once. The column designer works out the compression strength (example 1) and the bending strength about each axis (examples 2 and 3), increases the moments for the extra bending the axial load causes as the column bows (the B₁ factor in Appendix 8), and adds them up with the interaction equations in section H1:

P_r/P_c ≥ 0.2: P_r/P_c + 8/9 (M_rx/M_cx + M_ry/M_cy) ≤ 1.0H1-1a

P_r/P_c < 0.2: P_r/2P_c + (M_rx/M_cx + M_ry/M_cy) ≤ 1.0H1-1b

It also checks shear, and runs every ASCE 7-16 load combination. Try a column of your own in the free column designer; the worked calculations for the combination that governs are under its results.

Column designer: steel and wood columns under load and bending
How to design a steel beam: bending, bracing, shear and deflection
Steel beam sizes: W-shape dimensions and properties
Load cases and load combinations