AISC 360 — Flexural Members (Chapter F)

Beam design per AISC 360-16: plastic moment, LTB, FLB, Cb factor, and section classification

Chapter F of AISC 360-16 is the most detailed chapter in the specification, containing 12 sections (F2 through F13) covering every conceivable cross-section shape. The central challenge in beam design is lateral-torsional buckling (LTB) — the tendency of a beam's compression flange to buckle sideways and twist when the unbraced length exceeds certain limits. This guide covers the complete flexural design framework with two fully solved examples.

1. Plastic Moment — The Upper Bound

The maximum moment a cross section can develop is the plastic moment:

M_p = F_y \cdot Z_x

where $Z_x$ is the plastic section modulus — the first moment of area of each half of the cross section about the plastic neutral axis. For a compact, fully braced beam, the nominal flexural strength is simply $M_n = M_p$ . No beam can exceed $M_p$ regardless of how much bracing is provided; any LTB, FLB, or WLB check that computes $M_n > M_p$ is capped at $M_p$ .

LRFD: $\phi_b M_n = 0.90 \, M_p = 0.90 \, F_y \, Z_x$

ASD: $M_n / \Omega_b = M_p / 1.67 = F_y \, Z_x / 1.67$

2. Section Classification for Flexure (Table B4.1b)

Unlike compression (where elements are either "nonslender" or "slender"), flexural members have a three-tier classification that determines which limit states apply:

Class	Criterion	Behavior
Compact	$\lambda \leq \lambda_p$	Full plastic capacity ( $M_p$ ). No local buckling before full plastification.
Noncompact	$\lambda_p < \lambda \leq \lambda_r$	Partial yielding. Capacity between $M_p$ and $0.7 F_y S_x$ (linear interpolation).
Slender	$\lambda > \lambda_r$	Elastic local buckling governs. Capacity below $0.7 F_y S_x$ .

Width-to-Thickness Limits

Element	$\lambda$	$\lambda_p$ (compact)	$\lambda_r$ (noncompact)	A992 values
Flange (I-shape)	$b_f/(2t_f)$	$0.38\sqrt{E/F_y}$	$1.0\sqrt{E/F_y}$	9.15 / 24.1
Web (I-shape, flexure)	$h/t_w$	$3.76\sqrt{E/F_y}$	$5.70\sqrt{E/F_y}$	90.6 / 137.3
HSS flange	$b/t$	$1.12\sqrt{E/F_y}$	$1.40\sqrt{E/F_y}$	27.0 / 33.7
Round HSS / Pipe	$D/t$	$0.07 E/F_y$	$0.31 E/F_y$	40.6 / 179.8

Key fact: Nearly all standard W shapes in A992 (

F_y = 50

ksi) are compact for both flange and web. The exceptions are a handful of shapes like W21×48, W14×99 in high-strength steel, and some light W shapes. For A992, you can generally assume compact and verify with a quick b/t check.

3. Lateral-Torsional Buckling (LTB) — Section F2.2

LTB is the dominant limit state for beams. It occurs when the compression flange, acting like a column on an elastic foundation (the web), buckles laterally between brace points. The unbraced length $L_b$ (distance between points of lateral support of the compression flange) determines the behavior zone:

3.1. Three Zones of Behavior

Zone 1: Full Plasticity ( $L_b \leq L_p$ )

M_n = M_p = F_y \cdot Z_x

The limiting unbraced length for full plastic capacity:

L_p = 1.76 \, r_y \sqrt{\frac{E}{F_y}}

For A992 steel: $L_p = 1.76 \, r_y \sqrt{29{,}000/50} = 42.4 \, r_y$ (in inches).

Zone 2: Inelastic LTB ( $L_p < L_b \leq L_r$ )

M_n = C_b \left[M_p - (M_p - 0.7 F_y S_x)\left(\frac{L_b - L_p}{L_r - L_p}\right)\right] \leq M_p

This is a linear interpolation between $M_p$ (at $L_b = L_p$ ) and $0.7 F_y S_x$ (at $L_b = L_r$ ), multiplied by the moment gradient factor $C_b$ . The limiting unbraced length for inelastic LTB:

L_r = 1.95 \, r_{ts} \frac{E}{0.7 F_y} \sqrt{\frac{Jc}{S_x h_o} + \sqrt{\left(\frac{Jc}{S_x h_o}\right)^2 + 6.76 \left(\frac{0.7 F_y}{E}\right)^2}}

where $r_{ts}$ is the effective radius of gyration for LTB (tabulated in AISC Manual), $J$ is the torsional constant, $c = 1.0$ for doubly symmetric I-shapes, $S_x$ is the elastic section modulus, and $h_o$ is the distance between flange centroids.

Zone 3: Elastic LTB ( $L_b > L_r$ )

M_n = F_{cr} \cdot S_x \leq M_p

where the elastic critical stress is:

F_{cr} = \frac{C_b \pi^2 E}{\left(\dfrac{L_b}{r_{ts}}\right)^2} \sqrt{1 + 0.078 \frac{Jc}{S_x h_o} \left(\frac{L_b}{r_{ts}}\right)^2}

4. Moment Gradient Factor Cb

The $C_b$ factor accounts for the fact that LTB is more critical under uniform moment than under moment gradient. A beam with end moments causing reverse curvature is much more stable than one with uniform moment:

C_b = \frac{12.5 \, M_{\max}}{2.5 \, M_{\max} + 3 M_A + 4 M_B + 3 M_C} \cdot R_m

where $M_{\max}$ = absolute maximum moment in the unbraced segment, $M_A, M_B, M_C$ = absolute moments at the quarter point, midpoint, and three-quarter point, respectively, and $R_m = 1.0$ for doubly symmetric shapes.

Common Cb Values

Loading / Moment Diagram	C_b
Uniform moment (worst case)	1.00
Simply supported, uniform load	1.14
Simply supported, concentrated load at midspan	1.32
Cantilever, concentrated load at free end	1.00 (use 1.0 for cantilevers)
Equal and opposite end moments (reverse curvature)	2.27
One end moment only (M at one end, 0 at other)	1.67

Critical note:

C_b

can significantly increase the available moment. A beam that fails with

C_b = 1.0

may be adequate with the actual

C_b

. Always compute the actual

C_b

— using 1.0 is conservative but wasteful. However,

M_n

is always capped at

M_p

regardless of how large

C_b

is.

5. Flange Local Buckling (FLB) — Section F3

FLB is checked only for noncompact or slender flanges. For compact flanges (all standard W shapes in A992), FLB does not govern.

Noncompact Flanges

M_n = M_p - (M_p - 0.7 F_y S_x) \left(\frac{\lambda - \lambda_{pf}}{\lambda_{rf} - \lambda_{pf}}\right)

Slender Flanges

M_n = \frac{0.9 E k_c S_x}{\lambda^2}

where $k_c = 4/\sqrt{h/t_w}$ and $0.35 \leq k_c \leq 0.76$ .

6. Other Cross-Section Types (F4–F13)

Chapter F provides specific provisions for every shape type. The applicable section is determined by the cross-section geometry:

Section	Cross Section	Key Difference from F2
F2	Doubly symmetric compact I-shapes	Primary section — LTB and yielding only
F3	Doubly symmetric I-shapes with noncompact/slender flanges	Adds FLB check
F4	Other I-shapes with compact/noncompact webs	Singly symmetric, uses $R_{pc}$ and $R_{pt}$
F5	Doubly symmetric I-shapes with slender webs	Plate girders, $R_{pg}$ factor
F6	I-shapes bent about minor axis	No LTB — only yielding and FLB
F7	Square and rectangular HSS	Yielding + FLB + WLB (no LTB for closed sections)
F8	Round HSS and Pipes	Yielding + local buckling based on D/t
F9	Tees and double angles	LTB with stem in compression, $M_n$ reduced
F10	Single angles	Yielding + LTB about geometric axis, with $M_y$ = 1.5 My for unequal legs
F11	Rectangular bars and rounds	LTB and yielding, $M_p = F_y Z \leq 1.6 F_y S$
F12	Unsymmetric shapes	General yielding and LTB provisions
F13	Proportioning limits	$h/t_w \leq 260$ , $I_{yc}/I_y \geq 0.1$

Solved Example 1 — W14×22 Simply Supported Beam

Given: W14×22 (W360×32.9), A992 steel ( $F_y = 50$ ksi), simply supported span $L = 20$ ft (6.10 m), uniform load, lateral bracing at supports and midspan ( $L_b = 10$ ft = 120 in).

Properties: $Z_x = 33.2$ in³, $S_x = 29.0$ in³, $r_y = 1.04$ in, $r_{ts} = 1.22$ in, $J = 0.208$ in&sup4;, $h_o = 13.7$ in, $b_f = 5.00$ in, $t_f = 0.335$ in, $c = 1.0$ .

Step 1 — Section Classification

Flange: $b_f/(2t_f) = 5.00/(2 \times 0.335) = 7.46 \leq \lambda_{pf} = 9.15$ → compact ✓

Web: $h/t_w = 53.3 \leq \lambda_{pw} = 90.6$ → compact ✓

Section is compact → use Section F2 (yielding + LTB only, no FLB).

Step 2 — Plastic Moment

M_p = F_y \cdot Z_x = 50 \times 33.2 = 1{,}660 \text{ in·kips} = 138.3 \text{ ft·kips}

Step 3 — Limiting Unbraced Lengths

L_p = 1.76 \times 1.04 \times \sqrt{\frac{29{,}000}{50}} = 1.76 \times 1.04 \times 24.08 = 44.1 \text{ in} = 3.67 \text{ ft}

L_r = 1.95 \times 1.22 \times \frac{29{,}000}{35} \sqrt{\frac{0.208 \times 1.0}{29.0 \times 13.7} + \sqrt{\left(\frac{0.208}{29.0 \times 13.7}\right)^2 + 6.76\left(\frac{35}{29{,}000}\right)^2}}

Computing the terms: $Jc/(S_x h_o) = 0.208/(29.0 \times 13.7) = 5.23 \times 10^{-4}$

$(0.7F_y/E)^2 = (35/29{,}000)^2 = 1.457 \times 10^{-6}$

Inner sqrt: $\sqrt{(5.23 \times 10^{-4})^2 + 6.76 \times 1.457 \times 10^{-6}} = \sqrt{2.74 \times 10^{-7} + 9.85 \times 10^{-6}} = 3.18 \times 10^{-3}$

Outer sqrt: $\sqrt{5.23 \times 10^{-4} + 3.18 \times 10^{-3}} = \sqrt{3.70 \times 10^{-3}} = 0.0608$

L_r = 1.95 \times 1.22 \times 828.6 \times 0.0608 = 119.8 \text{ in} = 9.98 \text{ ft}

Step 4 — Determine LTB Zone

$L_b = 10$ ft = 120 in. Since $L_r = 9.98$ ft and $L_b = 10.0 > L_r$ , the beam is just barely in Zone 3 (elastic LTB).

Step 5 — Cb Factor

For a simply supported beam with uniform load, braced at midspan, the critical unbraced segment is one half of the span. The moment diagram in each half is a parabola. Using quarter-point moments in the segment:

C_b = \frac{12.5 \times M_{\max}}{2.5 M_{\max} + 3 M_A + 4 M_B + 3 M_C}

For the half-span segment with maximum moment at the midspan brace point: $C_b = 1.30$ (from AISC Manual Table 3-1 or calculated from quarter-point moments).

Step 6 — Nominal Flexural Strength (LRFD)

Since $L_b > L_r$ , use elastic LTB (F2-4):

F_{cr} = \frac{1.30 \times \pi^2 \times 29{,}000}{(120/1.22)^2} \sqrt{1 + 0.078 \times \frac{0.208}{29.0 \times 13.7} \times (120/1.22)^2}

$(L_b/r_{ts})^2 = (120/1.22)^2 = 9{,}672$

$0.078 \times 5.23 \times 10^{-4} \times 9{,}672 = 0.394$

F_{cr} = \frac{1.30 \times 286{,}216}{9{,}672} \sqrt{1 + 0.394} = 38.47 \times 1.181 = 45.4 \text{ ksi}

M_n = F_{cr} \cdot S_x = 45.4 \times 29.0 = 1{,}317 \text{ in·kips} = 109.8 \text{ ft·kips}

Check: $M_n = 109.8 \leq M_p = 138.3$ → OK, cap not reached

\phi_b M_n = 0.90 \times 109.8 = 98.8 \text{ ft·kips}

Impact of Lb: With

L_b = 10

ft,

\phi_b M_n = 98.8

ft·kips (71% of

\phi_b M_p

). If the beam were fully braced (

L_b \leq L_p = 3.67

ft),

\phi_b M_n = 124.5

ft·kips — a 26% increase. This illustrates the dramatic impact of inadequate lateral bracing. Adding one more brace point (L_b = 6.67 ft) would put the beam in the inelastic zone and recover most of the capacity.

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Solved Example 2 — W21×44, Concentrated Load

Given: W21×44 (W530×65.6), A992, span 24 ft (7.32 m), concentrated load at midspan, lateral bracing at supports and load point ( $L_b = 12$ ft = 144 in). Find $\phi_b M_n$ .

Properties: $Z_x = 95.4$ in³, $S_x = 81.6$ in³, $r_y = 1.26$ in, $r_{ts} = 1.48$ in, $J = 0.843$ in&sup4;, $h_o = 20.7$ in.

Step 1 — Lp and Lr

L_p = 1.76 \times 1.26 \times 24.08 = 53.4 \text{ in} = 4.45 \text{ ft}

$L_r$ (from AISC Manual Table 3-2 or computed): approximately 12.5 ft (150 in).

Since $L_b = 12.0$ ft and $L_p = 4.45 < L_b = 12.0 < L_r = 12.5$ , the beam is in Zone 2 (inelastic LTB).

Step 2 — Cb for Concentrated Load at Midspan

With a point load at midspan and bracing at midspan, each half-span has a linear moment diagram going from $M_{\max}$ at midspan to 0 at the support. For this triangle diagram:

C_b = \frac{12.5 \times 1.0}{2.5(1.0) + 3(0.75) + 4(0.50) + 3(0.25)} = \frac{12.5}{7.0} = 1.79

(Using normalized moments: Mmax = 1.0, MA = 0.75, MB = 0.50, MC = 0.25.) However, the AISC Manual lists $C_b = 1.67$ for this case per the full formula with Rm. We'll use $C_b = 1.67$ .

Step 3 — Inelastic LTB

M_p = 50 \times 95.4 = 4{,}770 \text{ in·kips} = 397.5 \text{ ft·kips}

0.7 F_y S_x = 35 \times 81.6 = 2{,}856 \text{ in·kips} = 238.0 \text{ ft·kips}

M_n = 1.67 \left[397.5 - (397.5 - 238.0)\left(\frac{12.0 - 4.45}{12.5 - 4.45}\right)\right]

= 1.67 \left[397.5 - 159.5 \times 0.938\right] = 1.67 \times [397.5 - 149.6]

= 1.67 \times 247.9 = 414.0 \text{ ft·kips}

Check cap: $M_n = 414.0 > M_p = 397.5$ → use $M_n = M_p = 397.5$ ft·kips

\phi_b M_n = 0.90 \times 397.5 = 357.8 \text{ ft·kips}

Cb impact: Without the Cb factor (

C_b = 1.0

), the inelastic LTB capacity would be

M_n = 247.9

ft·kips (62% of

M_p

). With

C_b = 1.67

, the Mn exceeds Mp and is capped at full plastic capacity. The Cb factor alone recovered 150 ft·kips of capacity — this is why computing the actual Cb is essential for economical beam design.