AISC 360 — Compression Members (Chapter E)

Column buckling per AISC 360-16: flexural, torsional, and flexural-torsional buckling with solved examples

Column design is arguably the most critical check in structural steel design. Unlike tension members that can develop the full yield strength of the cross section, compression members are governed by stability — the tendency to buckle before reaching material strength. Chapter E of AISC 360-16 provides a unified framework for computing the available compressive strength considering flexural buckling, torsional buckling, flexural-torsional buckling, and the effects of local buckling in slender elements.

1. Elastic (Euler) Buckling Stress

The foundation of all column design is the Euler critical stress — the theoretical maximum stress that a perfectly straight, perfectly elastic column can sustain before buckling laterally:

F_e = \frac{\pi^2 E}{\left(\dfrac{KL}{r}\right)^2}

where $E = 29{,}000$ ksi (200,000 MPa) is the modulus of elasticity, $K$ is the effective length factor, $L$ is the unbraced length, and $r$ is the radius of gyration about the buckling axis. The product $KL/r$ is the slenderness ratio — the single most important parameter in column design.

Real columns never reach $F_e$ because of residual stresses from the rolling/welding process, initial out-of-straightness (L/1000 per ASTM A6), and material inelasticity as portions of the cross section begin to yield. The AISC column curve (Section E3) accounts for all of these effects empirically.

2. Effective Length Factor K

The effective length factor $K$ transforms the actual column length into an equivalent pin-ended column length. The idealized values for common end conditions are:

End Conditions	Theoretical K	Recommended K	Buckled Shape
Fixed–Fixed	0.50	0.65	S-curve, no sway
Fixed–Pinned	0.70	0.80	Inflection near fixed end
Pinned–Pinned	1.00	1.00	Half sine wave
Fixed–Free (cantilever)	2.00	2.10	Quarter sine wave, sway
Fixed–Fixed (sway permitted)	1.00	1.20	Full sine, lateral sway
Fixed–Pinned (sway permitted)	2.00	2.00	Sway with inflection

Recommended vs theoretical: The "recommended" values from AISC Commentary Table C-A-7.1 are higher than theoretical because perfect fixity is never achieved in practice. Use theoretical K only if you can demonstrate true fixity.

[DIAGRAM: Six column buckled shapes showing the end conditions and corresponding K values. Side-by-side comparison of no-sway (K ≤ 1.0) and sway-permitted (K ≥ 1.0) cases.]

2.1. Alignment Charts (Nomographs)

For columns in frames, K depends on the relative stiffness of the beams and columns meeting at each joint. The AISC Commentary provides two alignment charts (Figures C-A-7.1 and C-A-7.2):

Sidesway Inhibited (braced frame): K ranges from 0.5 to 1.0. The chart uses G factors at top and bottom joints: $G = \frac{\sum (EI/L)_{\text{columns}}}{\sum (EI/L)_{\text{beams}}}$
Sidesway Uninhibited (unbraced/moment frame): K ranges from 1.0 to infinity. Same G formula, but buckled shape involves lateral sway.

Special cases: $G = 10$ for a pinned support (theoretically infinity, but 10 is used); $G = 1.0$ for a fixed support (theoretically 0, but 1.0 accounts for partial fixity).

2.2. Direct Analysis Method (Chapter C) — K = 1.0 Always

The Direct Analysis Method (DAM) is the primary stability method in AISC 360-16. Its key advantage: you always use K = 1.0, regardless of the frame type. The method accounts for stability effects through:

1. Notional loads: Apply $N_i = 0.002 \, Y_i$ horizontally at each level, where $Y_i$ is the total gravity load at that level. These simulate initial out-of-plumbness (H/500).

2. Reduced stiffness:

EI^* = 0.8 \, \tau_b \, EI \quad\quad EA^* = 0.8 \, EA

where the stiffness reduction factor $\tau_b$ accounts for inelasticity:

\tau_b = \begin{cases} 1.0 & \text{if } \alpha P_r / P_y \leq 0.5 \\ 4 \cdot \dfrac{\alpha P_r}{P_y} \left(1 - \dfrac{\alpha P_r}{P_y}\right) & \text{if } \alpha P_r / P_y > 0.5 \end{cases}

with $\alpha = 1.0$ for LRFD and $\alpha = 1.6$ for ASD, and $P_y = F_y \cdot A_g$ .

3. Second-order analysis: The analysis must capture P-Δ (frame sway) and P-δ (member curvature) effects. Any software that performs geometric nonlinear analysis satisfies this requirement. CalcSteel's FEM solver does this automatically.

When to use DAM vs Effective Length: The DAM is required unless you can show the structure satisfies the conditions for the Effective Length Method (Appendix 7) — specifically, that the second-order amplification (B2) is ≤ 1.5 for all stories. In practice, use DAM by default. It is simpler (K = 1.0), more general, and the method that CalcSteel and most modern software implement.

3. Flexural Buckling — Section E3

Flexural buckling is the most common buckling mode for doubly symmetric shapes (W, HSS, pipe). The critical stress $F_{cr}$ is determined by comparing the slenderness ratio with the transition point at $4.71\sqrt{E/F_y}$ :

3.1. Inelastic Buckling (short to intermediate columns)

When $KL/r \leq 4.71\sqrt{E/F_y}$ (equivalently, $F_e \geq 0.44 F_y$ ):

F_{cr} = \left(0.658^{F_y/F_e}\right) F_y

This exponential curve was calibrated to match the SSRC Column Curve 2P, which accounts for the effects of residual stresses (typically 10–15 ksi in hot-rolled shapes) and initial geometric imperfections. At low slenderness, $F_{cr}$ approaches $F_y$ but never reaches it — a stocky W column with KL/r = 20 achieves about 95% of $F_y$ .

3.2. Elastic Buckling (slender columns)

When $KL/r > 4.71\sqrt{E/F_y}$ (equivalently, $F_e < 0.44 F_y$ ):

F_{cr} = 0.877 \, F_e

The 0.877 factor (approximately 1/1.14) accounts for initial out-of-straightness. In this regime, residual stresses have minimal effect because the entire cross section is elastic at buckling.

3.3. Available Compressive Strength

The nominal compressive strength is:

P_n = F_{cr} \cdot A_g

LRFD: $\phi_c P_n = 0.90 \, F_{cr} \cdot A_g$

ASD: $P_n / \Omega_c = F_{cr} \cdot A_g / 1.67$

For A992 steel ( $F_y = 50$ ksi), the transition slenderness is:

\frac{KL}{r} = 4.71 \sqrt{\frac{29{,}000}{50}} = 113.4

Columns with $KL/r \leq 113.4$ use the inelastic curve; those above use the elastic curve. Most building columns fall in the inelastic range.

[DIAGRAM: AISC column curve plotting Fcr/Fy vs KL/r, showing the inelastic branch (0.658 curve) and elastic branch (0.877Fe), with the transition at KL/r = 4.71√(E/Fy). Overlay the Euler curve for comparison.]

4. Torsional and Flexural-Torsional Buckling — Section E4

Doubly symmetric shapes (W, HSS) can only buckle by flexure or pure torsion. But singly symmetric shapes (WT, channels loaded through the web) and unsymmetric shapes (single angles) can experience flexural-torsional buckling — a coupled mode where the member simultaneously bends and twists. Section E4 requires checking this mode whenever the cross section is not doubly symmetric.

4.1. Doubly Symmetric — Torsional Buckling

For doubly symmetric I-shapes (only critical for very short columns or cruciform sections):

F_e = \left(\frac{\pi^2 E C_w}{(K_z L)^2} + GJ\right) \frac{1}{I_x + I_y}

4.2. Singly Symmetric — Flexural-Torsional Buckling

For singly symmetric shapes (axis of symmetry = y-axis):

F_e = \frac{F_{ey} + F_{ez}}{2H} \left[1 - \sqrt{1 - \frac{4 F_{ey} F_{ez} H}{(F_{ey} + F_{ez})^2}}\right]

where $F_{ey}$ is the flexural buckling stress about the y-axis, $F_{ez}$ is the torsional buckling stress, and $H = 1 - (x_o^2 + y_o^2)/r_o^2$ with $x_o, y_o$ being the coordinates of the shear center relative to the centroid.

Once $F_e$ is determined from E4, the same $F_{cr}$ equations from E3 apply — you simply substitute this $F_e$ instead of the flexural $F_e$ .

5. Single Angle Compression — Section E5

Single angles are unique because their principal axes do not align with the geometric axes. AISC 360 provides a simplified approach where you compute an equivalent slenderness ratio that accounts for the eccentricity of loading and the end restraint conditions. For equal-leg angles with the ratio $L/r_x$ :

\frac{KL}{r} = 72 + 0.75 \frac{L}{r_x} \quad \text{for } \frac{L}{r_x} \leq 80

\frac{KL}{r} = 32 + 1.25 \frac{L}{r_x} \quad \text{for } \frac{L}{r_x} > 80

This simplified method avoids the need for a full flexural-torsional buckling analysis and is valid when the angle is loaded through one leg with bolt connections at each end.

6. Built-up Members — Section E6

Built-up compression members (double angles, double channels back-to-back, laced columns) require a modified slenderness ratio that accounts for the shear deformation of the connectors:

\left(\frac{KL}{r}\right)_m = \sqrt{\left(\frac{KL}{r}\right)_o^2 + \left(\frac{a}{r_i}\right)^2}

where $(KL/r)_o$ is the slenderness ratio of the built-up member as a whole, $a$ is the distance between intermediate connectors, and $r_i$ is the minimum radius of gyration of an individual component. The individual components must also satisfy $a/r_i \leq 3/4 \cdot (KL/r)_o$ .

7. Slender Elements in Compression — Section E7

Before computing $F_{cr}$ , you must check whether the cross section has any slender compression elements. Table B4.1a defines width-to-thickness limits for compression members:

Element	Description	Width	$\lambda_r$ (Nonslender Limit)	A992 Value
Flanges of I-shapes	Unstiffened	$b/t = b_f/(2t_f)$	$0.56\sqrt{E/F_y}$	13.5
Webs of I-shapes	Stiffened	$h/t_w$	$1.49\sqrt{E/F_y}$	35.9
HSS walls	Stiffened	$b/t$ or $h/t$	$1.40\sqrt{E/F_y}$	33.7
Angles	Unstiffened	$b/t$	$0.45\sqrt{E/F_y}$	10.8
Round HSS / Pipe	Stiffened	$D/t$	$0.11 \, E/F_y$	63.8

If $\lambda \leq \lambda_r$ , the element is nonslender and you use the full $A_g$ with $F_{cr}$ from E3. If $\lambda > \lambda_r$ , the element is slender and you must reduce the effective area using the $Q$ factor approach (Section E7):

F_{cr} = Q \left(0.658^{Q F_y / F_e}\right) F_y \quad \text{when } \frac{KL}{r} \leq 4.71\sqrt{\frac{E}{QF_y}}

F_{cr} = 0.877 \, F_e \quad \text{when } \frac{KL}{r} > 4.71\sqrt{\frac{E}{QF_y}}

where $Q = Q_s \cdot Q_a$ . $Q_s$ applies to unstiffened elements (flanges, angle legs) and $Q_a$ applies to stiffened elements (webs, HSS walls). For a cross section with no slender elements, $Q = 1.0$ and the formulas reduce to the standard E3 equations.

Practical note: Nearly all standard W shapes with

F_y \leq 50

ksi have nonslender elements for compression. Slender elements are most commonly encountered in HSS (especially thinner walls), welded built-up sections, and high-strength steels (A913 Gr 65/70).

Solved Example 1 — W10×49 Column

Given: W10×49 (W250×73) column, A992 steel ( $F_y = 50$ ksi / 345 MPa), unbraced length $L = 20$ ft (6.10 m), pinned-pinned ( $K = 1.0$ both axes).

Properties from AISC Manual: $A_g = 14.4$ in², $r_x = 4.35$ in, $r_y = 2.54$ in, $d = 10.0$ in, $b_f = 10.0$ in, $t_f = 0.560$ in, $t_w = 0.340$ in.

Step 1 — Slenderness Ratios

\frac{KL}{r_x} = \frac{1.0 \times 20 \times 12}{4.35} = 55.2

\frac{KL}{r_y} = \frac{1.0 \times 20 \times 12}{2.54} = 94.5 \quad \leftarrow \text{governs}

The weak-axis slenderness (94.5) governs, as expected for a W shape with approximately equal flange width and depth.

Step 2 — Check for Slender Elements

Flange: $b_f/(2t_f) = 10.0/(2 \times 0.560) = 8.93 \leq 13.5$ → nonslender ✓

Web: $h/t_w \approx (10.0 - 2 \times 0.560)/0.340 = 26.1 \leq 35.9$ → nonslender ✓

No slender elements → $Q = 1.0$

Step 3 — Elastic Buckling Stress

F_e = \frac{\pi^2 \times 29{,}000}{94.5^2} = \frac{286{,}216}{8{,}930} = 32.1 \text{ ksi}

Step 4 — Determine Which Fcr Equation

Transition: $4.71\sqrt{E/F_y} = 4.71\sqrt{29{,}000/50} = 113.4$

Since $KL/r = 94.5 \leq 113.4$ → use inelastic equation

Check: $F_e = 32.1 \geq 0.44 \times 50 = 22.0$ ksi ✓ (confirms inelastic)

Step 5 — Critical Stress

F_{cr} = \left(0.658^{F_y/F_e}\right) F_y = \left(0.658^{50/32.1}\right) \times 50 = 0.658^{1.558} \times 50

0.658^{1.558} = e^{1.558 \ln(0.658)} = e^{1.558 \times (-0.4193)} = e^{-0.6533} = 0.5203

F_{cr} = 0.5203 \times 50 = 26.0 \text{ ksi}

Step 6 — Available Compressive Strength

LRFD:

\phi_c P_n = 0.90 \times 26.0 \times 14.4 = 337 \text{ kips (1,500 kN)}

ASD:

\frac{P_n}{\Omega_c} = \frac{26.0 \times 14.4}{1.67} = 224 \text{ kips (997 kN)}

Manual check: AISC Steel Construction Manual Table 4-1 lists φPn = 338 kips for W10×49 at KL = 20 ft. Our result of 337 kips matches within rounding — confirming the calculation.

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Solved Example 2 — HSS 8×8×1/2

Given: HSS 8×8×1/2 (HSS 200×200×12.5), A500 Gr C ( $F_y = 50$ ksi / 345 MPa), $L = 15$ ft (4.57 m), pinned-pinned.

Properties: $A_g = 13.5$ in², $r = 3.04$ in, $t_{\text{design}} = 0.465$ in (93% of nominal wall thickness per AISC convention for ERW HSS), $b/t = (8.00 - 3 \times 0.465)/0.465 = 14.2$ .

Step 1 — Slenderness Check

\frac{KL}{r} = \frac{1.0 \times 15 \times 12}{3.04} = 59.2

Both axes equal for square HSS.

Step 2 — Check for Slender Walls

Wall slenderness: $b/t = 14.2$

Limit (Table B4.1a): $\lambda_r = 1.40\sqrt{E/F_y} = 1.40\sqrt{29{,}000/50} = 33.7$

Since $14.2 \leq 33.7$ → nonslender ✓ ( $Q = 1.0$ )

Step 3 — Fcr and Available Strength (LRFD)

F_e = \frac{\pi^2 \times 29{,}000}{59.2^2} = 81.6 \text{ ksi}

Since $59.2 \leq 113.4$ → inelastic:

F_{cr} = 0.658^{50/81.6} \times 50 = 0.658^{0.613} \times 50 = 0.770 \times 50 = 38.5 \text{ ksi}

\phi_c P_n = 0.90 \times 38.5 \times 13.5 = 468 \text{ kips (2,082 kN)}

The HSS 8×8×1/2 achieves 468 kips at 15 ft — about 39% stronger than the W10×49 at 20 ft (337 kips), despite having a similar cross-sectional area (13.5 vs 14.4 in²). This reflects the inherent efficiency of closed sections in compression: HSS shapes have nearly equal radii of gyration in both axes, eliminating the weak-axis penalty that governs W shapes.

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Available Strength Table — W10×49 (A992)

The following table shows the available compressive strength $\phi_c P_n$ (LRFD) for a W10×49 at various effective lengths, similar to AISC Manual Table 4-1:

KL (ft)	KL/r_y	F_e (ksi)	F_cr (ksi)	φ_cP_n (kips)
6	28.3	356	46.6	604
10	47.2	128	42.0	544
14	66.1	65.5	35.1	455
20	94.5	32.1	26.0	337
24	113.4	22.2	19.5	253
30	141.7	14.2	12.5	162

At KL = 24 ft, the column is right at the transition point (KL/r = 113.4). Above this, the elastic curve governs and capacity drops rapidly. This illustrates why slender columns are inefficient — at KL = 30 ft the capacity is only 27% of the short-column strength.

8. International Comparison — Column Curves

The treatment of column buckling is one of the areas where major steel codes differ most significantly:

Aspect	AISC 360	Eurocode 3	NBR 8800
Number of curves	1 (SSRC 2P)	5 (a0, a, b, c, d)	1 (same as AISC)
Curve selection	Automatic (single curve)	Based on cross-section type, axis, and manufacturing (Table 6.2)	Automatic (single curve)
Inelastic formula	$0.658^{F_y/F_e} \cdot F_y$	$\chi \cdot F_y$ with imperfection factor $\alpha$	$0.658^{F_y/F_e} \cdot F_y$
Elastic formula	$0.877 F_e$	$\chi \cdot F_y$ (same χ formula)	$0.877 F_e$
Stability method	DAM (K=1.0) or ELM	GNA, GMNIA, or equivalent imperfections	DAM or ELM
φ (LRFD) / γ	φ_c = 0.90	γ_M1 = 1.00	γ_a1 = 1.10 (≈ φ = 0.91)

The Eurocode 3 approach with 5 curves is more precise — it assigns different curves to different cross-section types. For example, hot-rolled H-shapes buckling about the strong axis get curve "a" (optimistic), while the same shape buckling about the weak axis gets curve "b" (more conservative). Welded sections get curves "c" or "d". The AISC single curve sits approximately at the EC3 "b" curve, which means:

AISC is slightly conservative for strong-axis buckling of hot-rolled wide-flange shapes (EC3 gives more capacity with curve "a")
AISC is slightly unconservative for welded sections and weak-axis buckling of heavy shapes (EC3 uses curves "c" or "d")
NBR 8800 is identical to AISC — same single curve, same formulas, same philosophy

CalcSteel implements all three approaches: the AISC single curve, the EC3 five-curve system, and the NBR 8800 curve (which produces the same results as AISC). You can switch standards and instantly see how the capacity changes.

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