Engineering story

A century of steel — from concept to skyline

Engineer reviewing blueprints against a steel-frame construction site at sunrise

A column is only as strong as its slenderness lets it be.

Yield strength alone never designs a real column — buckling does.

Chapter 5 covers concentrically loaded columns under AISC 360-22 Chapter E. Flexural buckling (§E3), effective length KL = Lc, elastic vs inelastic branches at Lc/r = 4.71·√(E/Fy), local buckling of slender elements (§E7), and the AISC Manual Table 4-1a design-strength look-ups.

Iconic steel structures built on engineering excellence

Euler's buckling equation
St. Petersburg1757
Engesser tangent-modulus theory
1889
Shanley — inelastic column theory
1947
AISC 360-22 Chapter E (current)
2022

Load pathBeam reactionColumn topFlexural buckling checkBase plateFoundation

02

Learning objectives

What you will be able to do after finishing Chapter 1 — and why each objective matters in practice

Objective 01

Compute effective length Lc = K·L

Choose K from Commentary Table C-A-7.1 (0.65 fixed–fixed, 1.0 pinned–pinned, 2.0 cantilever) or the sidesway alignment chart.

Why it matters: K sets the entire buckling equation. Underestimating K underdesigns the column.
Where it is used: Every steel column in every steel-framed building.
Connects to: AISC §C2, App. 7; Direct Analysis Method reappears in Chapter 17.

Objective 02

Classify elastic vs inelastic buckling

Compare Lc/r to 4.71·√(E/Fy) ≈ 113 for A992. Below → §E3 inelastic; above → §E3 elastic (Fcr = 0.877·Fe).

Why it matters: The Fcr equation is a piecewise curve. Using the wrong branch changes φPn by tens of percent.
Where it is used: Every §E3 flexural-buckling check.
Connects to: AISC §E3; Fe = π²E/(Lc/r)² is Euler.

Objective 03

Apply Fcr → φcPn = 0.90·Fcr·Ag

For non-slender sections, φc = 0.90. Slender-element sections require §E7 with Ae < Ag.

Why it matters: This is the number that governs the design of every gravity column.
Where it is used: Gravity columns, brace-frame columns, truss verticals in compression.
Connects to: AISC §E3, §E7; Manual Table 4-1a tabulates the result.

Objective 04

Read AISC Manual Table 4-1a

Look up φcPn vs KL for A992 W-shapes; interpolate between tabulated KL values.

Why it matters: Fluent Manual reading turns a 20-minute buckling calc into a 30-second design decision.
Where it is used: Every design office, every FE exam column question.
Connects to: AISC Manual Part 4 (Design of Compression Members).

Objective 05

Screen slenderness KL/r ≤ 200

AISC §E2 recommends KL/r ≤ 200 for compression members. Above this, φPn is very small and details govern.

Why it matters: Long slender columns lose most of their capacity to Euler; screen early.
Where it is used: Braces, secondary columns, truss verticals.
Connects to: AISC §E2 (User Note); contrast with tension L/r ≤ 300 (Ch. 4).

Objective 06

Handle slender-element (E7) columns

When λ = b/t > λr from Table B4.1a, use §E7 with an effective area Ae based on Fcr and the plate stress ratio.

Why it matters: Slender-element checks catch local buckling of thin HSS walls and slender flanges — missing this over-predicts strength.
Where it is used: HSS with thin walls, built-up columns, unstiffened plate compression elements.
Connects to: AISC §E7 + Table B4.1a.

03

Engineering motivation

What each part of a steel-frame building actually does — and why it exists

Before you design any single member, you have to see the whole system. A steel-frame building is not a collection of independent shapes bolted together — it is a deliberate load path, engineered so that every kilonewton of gravity, wind, or seismic demand has a continuous route from where it starts to the ground where the earth can resist it.

The photograph below shows a typical steel framing detail. Drag each labelled chip onto the structural element it names — the drop is only accepted when it lands inside the correct element's outlined region. A correct answer locks in with a green outline and a short explanation; a wrong answer flashes the region red, tells you what you actually hit, and returns the chip so you can try again. Press Reveal expected placements to see the reference solution (that attempt is then marked as assisted).

Structural steel framing — identify each element by dragging the labels — Fig. 1.3 · Drop a chip inside the outlined element it names. Green = correct and locked; red flash = wrong element, chip returns.

04

Failure mechanisms

Why we design the way we do — six ways steel structures have failed, and what each disaster taught the profession

Every provision in AISC 360 is a scar. Behind each equation, load factor, and detailing rule is a bridge, a walkway, or a tower whose failure cost lives and rewrote the profession. The six case studies below trace the mechanisms that motivate the code you are about to learn.

Read each one as an engineer, not a spectator: identify the load, the limit state, the missing check, and the specific clause that exists today because that check was missed. When you meet those clauses again in Chapters 5–17, they will read as answers, not rules.

Failure mechanism

Global flexural buckling

A slender column bows laterally under axial load once the applied stress reaches Fcr, well below Fy.

Root cause

KL/r too high — Fe = π²E/(KL/r)² drops below Fy and the elastic branch governs.

Lesson learned

Screen KL/r first; if greater than 4.71·√(E/Fy), the elastic Fcr = 0.877 Fe applies.

§AISC 360-22 §E3

Failure mechanism

Local plate / flange buckling

A slender flange or HSS wall crinkles locally before global buckling can develop, cutting φcPn.

Root cause

λ = b/t > λr from Table B4.1a; the plate is a slender element.

Lesson learned

Use §E7 with reduced effective area Ae — do not rely on §E3 alone for slender-element shapes.

§AISC 360-22 §E7 + Table B4.1a

Failure mechanism

Hartford Civic Center roof (1978)

A space-truss roof collapsed under snow when slender top-chord compression members buckled before the truss reached its design load.

Root cause

As-built KL/r of top-chord compression struts was far above the assumed value used in design.

Historical case

Hartford, CT, January 1978.

Lesson learned

Buckling is unforgiving of construction tolerances — the as-built K matters, not the assumed K.

§AISC 360-22 §E3, §C2

Failure mechanism

Quebec Bridge (1907)

Compression chord members of a cantilever bridge buckled during construction — 75 workers killed. Root cause: an unaccounted increase in dead load pushing slender chords past their buckling capacity.

Root cause

Latticed built-up compression chords with high effective slenderness were used at loads exceeding the buckling capacity assumed by the designer.

Historical case

Quebec City, 29 August 1907.

Lesson learned

Always verify actual dead load and effective length; slender built-up compression members demand extreme care.

§AISC 360-22 §E6 (built-up members)

How failure propagates

The five-stage failure progression

1Applied load

2Elastic shortening

3Bifurcation

4Lateral bow

5Collapse

Design codes intervene at the transition from yield to instability. Everything before yield is elastic and reversible; everything after instability is a race to collapse. LRFD keeps the demand well below the first transition.

05

Lecture notes

The full textbook chapter — figures, equations, and engineering narrative

Reflection · Think before you read

Two identical W-shapes carry the same axial load, but one is braced at mid-height about its weak axis and the other is not. Predict — before opening AISC §E3 — which one governs, and by roughly how much.

Chapter 5 — Compression Members & Column Buckling (AISC 360-22 Chapter E)

Chapter focus. Columns fail by buckling long before the steel yields — a slender pin-ended column at 30 ft can carry a small fraction of its A·Fy. This chapter turns Euler's classical result into AISC Chapter E: effective length KL, slenderness ratio KL/r, the inelastic Fcr curve, and the resulting φPn you look up in Manual Table 4-1a.

1. Behavior

Unlike tension, compression capacity is governed by stability — a slender member can fail at a stress well below F_y due to elastic buckling. The 1757 Euler solution is the theoretical benchmark:

P_e = π² · E · I / (K L)² ⇒ F_e = π² E / (K L / r)²

AISC modifies this classic result to account for residual stresses, initial crookedness, and inelastic behavior at intermediate slenderness.

Representative Structures

Where compression members carry the load in real structures. Each highlighted element is stressed primarily in axial compression — and its capacity is governed by stability (buckling), not by yielding.

Willis Tower Chicago — Willis Tower — Steel Building Columns
The bundled-tube frame collects every floor's gravity load into perimeter and interior columns. At the base, each column carries thousands of kips of pure compression. Slabs brace the weak axis at every floor, so KL is set by story height — a classic AISC E3 flexural-buckling design.

New River Gorge Bridge steel arch — New River Gorge Bridge — Steel Arch Ribs
The twin steel arch ribs carry the entire deck load into the canyon walls as axial compression. In-plane buckling is prevented by cross-bracing between ribs; out-of-plane buckling drives the design — a real-world (KL/r)_y problem.

Sydney Harbour Bridge piers — Bridge Piers — Concrete Compression Members
Massive granite-faced concrete piers receive the horizontal thrust and vertical reaction of the steel arch. Concrete piers of this size are compression-dominated and typically governed by short-column crushing plus P–δ effects, not slender buckling.

Transmission tower legs — Transmission Tower Legs
Each of the four battered corner legs is a built-up angle column carrying the conductor and self-weight in compression. Diagonal lacing shortens the unbraced length between panel points — the reason a "slender" lattice tower is stable at all.

Petronas Towers Kuala Lumpur — Petronas Towers — High-Rise Perimeter Columns
Ring of high-strength concrete-encased columns at each floor. The composite section increases both A_g and stiffness (EI), pushing the effective slenderness KL/r far below 4.71√(E/F_y) — well inside the inelastic range where F_cr ≈ F_y.

Multi-story parking garage columns — Parking Garage — Repetitive Interior Columns
Regular grid of columns accumulates gravity load one floor at a time. Because slabs brace both axes at every level, the story-height unbraced length usually places these members in the inelastic buckling zone — Manual Table 4-1a is used directly.

2. Effective Length K L

Theoretical (recommended) K values for common end conditions

In real frames, K is computed from stiffness ratios (nomograph) or by rigorous stability analysis. For braced (non-sway) frames K ≤ 1.0; for unbraced (sway) frames K > 1.0.

3. Slenderness Ratio KL/r — the Single Number that Drives Everything

Once you have K L, divide by the radius of gyration r for each axis. The governing slenderness is the larger of (KL/r)_x and (KL/r)_y — that is the axis about which the column buckles first.

Slenderness

(1)

(K L / r)_{g o v} = ma x [(K L / r)_{x}, (K L / r)_{y}]

Important subtlety: the unbraced lengths about the two axes are often different (a slab may brace weak-axis at every floor while strong-axis only at the top and bottom of the column). Compute (KL)_x and (KL)_y with their own unbraced lengths before dividing by r_x or r_y.

Recommended limit (§E2): KL/r ≤ 200. Above 200 the column is too flexible for practical construction.

4. AISC Flexural-Buckling Equations (§E3)

Slenderness limit

(2)

K L / r \leq 4.71 (E / F_{y}) \Leftrightarrow F_{e} \geq 0.44 F_{y} (in e l a s t i cr an g e)

E3-2 Inelastic

(3)

F_{cr} = [0.65 8^{F y / F e}] \cdot F_{y}

E3-3 Elastic (long)

(4)

F_{cr} = 0.877 \cdot F_{e}

Nominal strength

(5)

P_{n} = F_{cr} \cdot A_{g}

ϕ_{c} = 0.90

The 4.71√(E/F_y) boundary is the only decision you have to make: short/stocky → E3-2 (inelastic); long/slender → E3-3 (elastic Euler-like). Both formulas return F_cr, and P_n = F_cr·A_g. For A992 (F_y = 50 ksi) the boundary is KL/r ≈ 113.

5. Torsional & Flexural-Torsional Buckling (§E4)

Doubly-symmetric shapes rarely control on torsion. Singly-symmetric (e.g., tees, angles, channels) and cruciform sections require checking F_e from E4-2 through E4-5.

6. Slender-Element Sections (§E7)

If any element's b/t exceeds the λ_r limit in Table B4.1a, the section is slender. Nominal strength becomes P_n = F_cr · A_e, where A_e uses effective widths. Most rolled W-shapes at F_y = 50 ksi are non-slender.

7. Two Solution Methods — Companion Tables vs. AISC Specification Formulas

Every compression-member problem in this course can be solved by either of two equivalent routes. Learn both — the Specification route explains why the number is what it is, and the Manual (companion-table) route is what you use in practice and on the FE / PE exam.

Method A — AISC Specification Formulas (Chapter E, "the long way")

Compute effective length KL using the K from Table C-A-7.1 or the alignment charts (Fig. C-A-7.1 / C-A-7.2).
Compute slenderness (KL/r)_x, (KL/r)_y; keep the larger.
Compare KL/r to 4.71√(E/F_y) to decide inelastic (Eq. E3-2) or elastic (Eq. E3-3).
Compute F_e = π²E/(KL/r)² → F_cr → φ_cP_n = 0.90·F_cr·A_g.

Use when: non-standard K, non-A992 steel, slender elements requiring E7, or when you must show every step in a report or on an exam. See Worked Examples 5.10 and 5.11.

AISC Manual Table 4-1a — Available Strength in Axial Compression, W-shapes, Fy = 50 ksi — AISC Manual **Table 4-1a** — Available Strength in Axial Compression (W-shapes, F_y = 50 ksi). Enter with the effective length KL (ft); read φ_cP_n (LRFD) or P_n/Ω_c (ASD).

AISC Commentary Fig. C-A-7.1 / C-A-7.2 — nomograph for K

Mini-Examples — Reading Between the Two Methods

The following short examples show Method A (Chapter E formulas) and Method B (Manual Table 4-1a) side-by-side for realistic W and HSS columns. Each is a self-contained AISC 360-22 check.

Example 5-1 — Critical buckling load by Euler

Setup: W10×33 column, KL = 14 ft, A992 (F_y = 50 ksi), r_y = 1.94 in. Compute F_e and classify. AISC §E3

X X X \frac{K L}{r _{y}} = \frac{1.0 \cdot 14 \cdot 12}{1.94}

X X X = 86.6

X X X < 4.71 \frac{29000}{50} = 113 \Rightarrow inelastic

F_{e} = \frac{π ^{2} E}{( K L / r ) ^{2}}

X X X = \frac{π ^{2} ( 29000 )}{( 86.6 ) ^{2}}

X X X = 38.1 ksi

Answer. KL/r_y = 86.6 < 113 → inelastic; F_e = 38.1 ksi.

FE trap. Memorize 4.71√(E/F_y) = 113 for F_y = 50 ksi — the most-missed transition slenderness in Chapter E.

Example 5-4 — Effective length factor by alignment chart

Setup: interior column of an unbraced frame with G_A = 1.5, G_B = 0.8. Use the sidesway-permitted nomograph. AISC App. 7, Fig. C-A-7.2

Reading the sidesway-permitted (unbraced) nomograph:

G_{A} = 1.5, G_{B} = 0.8

X X X \Rightarrow K \approx 1.4

Answer. K ≈ 1.4. For an unbraced frame K ≥ 1.0 always.

FE trap. Braced frames → Fig. C-A-7.1 (K ≤ 1); moment frames → Fig. C-A-7.2 (K ≥ 1).

Example 5-7 — Pick a W column for a given load (Method B)

Setup: P_u = 600 k, KL = 13 ft, A992. Find the lightest W12 that works. AISC Manual Table 4-1a

Enter Table 4-1a at KL = 13 ft and scan W12 shapes by increasing weight:

X X X W12 \times 53 : ϕ_{c} P_{n} \approx 542 k

X X X < P_{u} = 600 k \Rightarrow NG

X X X W12 \times 58 : ϕ_{c} P_{n} \approx 605 k

X X X \geq P_{u} = 600 k \Rightarrow OK

Answer. Use W12×58 (lightest W12 with φ_cP_n ≥ 600 k at KL = 13 ft).

FE trap. "Lightest" = smallest lb/ft, not smallest depth.

Example 5-9 — HSS column check (Method A)

Setup: HSS 6×6×3/8, A500 Gr. B (F_y = 46 ksi), KL = 12 ft, A_g = 7.58 in², r_y = 2.28 in. Check φ_cP_n. AISC §E3

X X X \frac{K L}{r} = \frac{12 \cdot 12}{2.28}

X X X = 63.2

X X X < 4.71 \frac{29000}{46} = 118 \Rightarrow inelastic

F_{e} = \frac{π ^{2} ( 29000 )}{( 63.2 ) ^{2}}

X X X = 71.6 ksi

F_{cr} = (0.65 8^{F_{y} / F_{e}}) F_{y}

X X X = 0.65 8^{46/71.6} (46)

X X X = 35.8 ksi

X X X ϕ_{c} P_{n} = 0.90 F_{cr} A_{g}

X X X = 0.90 (35.8) (7.58)

X X X = 244 k

Answer. φ_cP_n ≈ 244 k.

FE trap. HSS with slender walls (b/t > λ_r) trigger §E7 — check b/t before applying §E3.

FE Ex 1 — Effective length for a pinned–pinned column

Given: W10×49, L = 14 ft, K = 1.0, r_y = 2.54 in. AISC Commentary Table C-A-7.1

X X X \frac{L _{c}}{r _{y}} = \frac{K L}{r _{y}}

X X X = \frac{1.0 ( 14 \cdot 12 )}{2.54}

X X X = 66.1

Answer. L_c/r_y = 66.1.

FE trap. Convert L to inches before dividing by r (r is tabulated in inches).

Method B — AISC Manual Companion Tables (Part 4, "the fast way")

Determine the governing effective length. Manual Table 4-1a is tabulated on KL using r_y (weak-axis). If the strong axis governs, enter with the equivalent length KL_x/(r_x/r_y) instead.
Read φ_cP_n directly from Table 4-1a (W-shapes, F_y = 50 ksi) or Table 4-1b (HSS, pipes, etc.). Interpolate linearly on KL.
Verify φ_cP_n ≥ P_u. The table already embeds F_cr, φ, weak-axis check, and non-slender-element classification for standard rolled shapes.

Use when: rolled W or HSS in A992 (or Manual-listed grade), K ≈ 1.0 or you have an equivalent KL. See Worked Example 5.1 (Table 4-1a cross-check).

AISC Manual Table 4-1a (excerpt) — φ_c P_n vs. KL for common W-shapes, F_y = 50 ksi

Which method for which problem?

Braced/unbraced frame with alignment-chart K: Method A to get K, then either method for φ_cP_n. When K_x ≠ K_y, convert to an equivalent KL before entering Table 4-1a.
Trial section for a known P_u: always start with Method B — it lets you scan many shapes in seconds. Then verify with Method A only if K, F_y, or slenderness is non-standard.
Exam problems: use whichever your reference allows. The FE Reference Handbook gives the Chapter E equations (Method A); the PE exam expects Manual tables (Method B).

The two methods agree to within round-off (~1 %) for standard cases — every worked example in this chapter cross-checks one against the other.

8. Design Procedure (put it all together)

Compute P_u from ASCE 7 combinations.
Estimate required φP_n; enter AISC Manual Table 4-1a / 4-1b (KL vs. φP_n) to pick a trial W-shape (Method B).
Compute (KL/r)_x, (KL/r)_y; take the larger. Verify KL/r ≤ 200.
Compare to 4.71√(E/F_y); compute F_e, then F_cr from E3-2 or E3-3 (Method A cross-check).
Verify φ_c P_n ≥ P_u; check flange & web slenderness (§E7) if the trial shape is not tabulated as non-slender.

⚠ Common mistakes

Using L instead of KL — always multiply by K.
Using r_x when weak-axis (r_y) governs.
Confusing braced vs unbraced length — bracing about y-axis may differ from x-axis.
Applying φ = 0.75 instead of 0.90 for compression.

Worked Example 5.1 — Interior Column, W14 (A992)

Method: AISC Specification (Chapter E, Eq. E3-2) — hand solution | Cross-check: AISC Manual Table 4-1a (Companion Table)

Given: Interior 1st-story column of the Cardinal Square 4-story braced-frame office. Factored axial P_u = 640 k. Story height 14 ft; braced in both directions at top & bottom (K = 1.0). Material A992 (F_y = 50 ksi, E = 29,000 ksi). Try W14×82: A = 24.0 in², r_x = 6.05 in, r_y = 2.48 in, b_f/2t_f = 5.92, h/t_w = 22.4.

Column model and unbraced length

Step 1 — Effective lengths

(1)

K L_{x} = K L_{y} = 1.0 \cdot 14 \cdot 12 = 168 in

Step 2 — Governing slenderness

(KL/r)_x = 168/6.05 = 27.8
(KL/r)_y = 168/2.48 = 67.7 ← governs
KL/r < 200 ✓

Step 3 — Classify range (E3)

(2)

4.71 (E / F_{y}) = 4.71 (29000/50) = 113.4

Since 67.7 < 113.4 → inelastic buckling, use Eq. E3-2

Step 4 — Compute F_e and F_cr

(3)

F_{e} = π^{2} E / (K L / r)^{2} = π^{2} \cdot 29000/67. 7^{2}

F_e = 286,220 / 4583 = 62.45 ksi

(4)

F_{cr} = [0.65 8^{F y / F e}] \cdot F_{y}

F_y/F_e = 50/62.45 = 0.8007
0.658^0.8007 = e^{0.8007·ln 0.658} = e^−0.3352 = 0.7151
F_cr = 0.7151 · 50 = 35.76 ksi

Step 5 — Element slenderness (Table B4.1a)

Flange: 0.56 √(E/F_y) = 0.56·24.08 = 13.5; b_f/2t_f = 5.92 < 13.5 ✓ (non-slender)
Web: 1.49 √(E/F_y) = 35.9; h/t_w = 22.4 < 35.9 ✓ (non-slender)

Step 6 — Design strength

(5)

ϕ_{c} P_{n} = 0.90 \cdot F_{cr} \cdot A_{g}

φ_c P_n = 0.90 · 35.76 · 24.0 = 772.4 k

Check: φ_cP_n = 772 k ≥ P_u = 640 k ✓ Demand/Capacity = 0.83. Use W14×82, A992.

Verify with AISC Manual Table 4-1a

For KL_y = 14 ft the manual tabulates φ_cP_n ≈ 770 k for W14×82 — matches within round-off. Table 4-1 columns already embed F_cr, φ, and the weak-axis check; use the hand solution above only to confirm the tabulated value or when K deviates from 1.0.

Worked Example 5.10 — Intermediate-Floor Column, Braced Frame (Alignment Chart)

Method: AISC Specification — alignment chart (Fig. C-A-7.1) for K + Eq. E3-2 for F_cr | Cross-check: Manual Table 4-1a at the equivalent KL

Given: Interior column on the 2nd story of a 4-story braced office frame. All stories 12 ft; typical bay 25 ft. All columns W12×72 (I_x = 597 in⁴, A = 21.1 in², r_x = 5.31 in, r_y = 3.04 in). Girders framing to the joint on both sides: W18×50 (I_x = 800 in⁴). Column loaded to P_u = 500 k. Beams provide weak-axis bracing at every floor (K_yL_y = 1.0·12 ft). A992 steel (F_y=50 ksi, E=29,000 ksi).

Figure 5.10a — Frame elevation. The highlighted red column is the design member; joints A (top) and B (bottom) each receive one column above/below and one beam on each side.

Step 1 — Alignment-chart G at each joint

Definition

(6)

G = Σ (I_{c} / L_{c}) \div Σ (I_{g} / L_{g})

Formula first, then substitute per the drawing above.
Columns at joint A: one above (W12×72, 12 ft) + one below (W12×72, 12 ft).
Σ(I_c/L_c) = 2 · 597/(12·12) = 2 · 4.146 = 8.29 in³.
Beams at joint A: two W18×50 girders spanning 25 ft each side.
Σ(I_g/L_g) = 2 · 800/(25·12) = 2 · 2.667 = 5.33 in³.
G_A = 8.29 / 5.33 = 1.55.
By symmetry the framing at joint B is identical → G_B = 1.55.

Step 2 — K from the braced-frame (sidesway-inhibited) chart

Figure 5.10b — AISC Commentary Fig. C-A-7.1 (sidesway-inhibited). Mark G_A = 1.55 on the left axis and G_B = 1.55 on the right axis; draw a straight line between them. Where that line crosses the middle K axis, read K ≈ 0.83. The yellow dot shows the reading.

How to use the chart: (1) locate G_A on the left G-scale, (2) locate G_B on the right G-scale, (3) join the two points with a straight edge, (4) read K where the line intersects the central K-scale. Values increase upward (0 at the bottom, ∞ at the top).

Numeric cross-check (French approx.)

(7)

K = [3 G_{A} G_{B} + 1.4 (G_{A} + G_{B}) + 0.64] \div [3 G_{A} G_{B} + 2 (G_{A} + G_{B}) + 1.28]

Numerator: 3·(1.55)(1.55) + 1.4·(3.10) + 0.64 = 7.24 + 4.35 + 0.64 = 12.23.
Denominator: 3·(1.55)(1.55) + 2·(3.10) + 1.28 = 7.24 + 6.20 + 1.28 = 14.72.
K = 12.23/14.72 = 0.83 — matches the chart reading above.
Because of weak-axis floor bracing, use K_x = 0.83 for in-plane buckling and K_y = 1.0 for weak-axis.

Step 3 — Effective lengths

(8)

K_{x} L = 0.83 \cdot 12 \cdot 12 = 119.5 in

K_{y} L = 1.0 \cdot 12 \cdot 12 = 144 in

(KL/r)_x = 119.5 / 5.31 = 22.5.
(KL/r)_y = 144 / 3.04 = 47.4 ← governs (weak-axis still controls).

Step 4 — Buckling range & F_cr (AISC §E3)

Range check

(9)

4.71 (E / F_{y}) = 113.4 > 47.4

in e l a s t i c

u se E q . E 3 - 2

(10)

F_{e} = π^{2} E / (K L / r)^{2}

F_{cr} = [0.65 8^{F_{y} / F_{e}}] \cdot F_{y}

F_e = π²·29000 / 47.4² = 286,220 / 2246 = 127.4 ksi.
F_y/F_e = 50/127.4 = 0.3925.
0.658^0.3925 = e^{0.3925·ln 0.658} = e^−0.1643 = 0.8485.
F_cr = 0.8485 · 50 = 42.42 ksi.

Step 5 — Design strength

(11)

ϕ_{c} P_{n} = 0.90 \cdot F_{cr} \cdot A_{g}

φ_cP_n = 0.90 · 42.42 · 21.1 = 805 k.

Check: φ_cP_n = 805 k ≥ P_u = 500 k ✓ (D/C = 0.62). Use W12×72. Cross-check: AISC Manual Table 4-1a at KL_y = 12 ft gives φ_cP_n ≈ 800 k for W12×72 — matches.

Worked Example 5.11 — Lowest-Floor Column, Unbraced Moment Frame (Pinned Base)

Method: AISC Specification — alignment chart (Fig. C-A-7.2, sidesway-uninhibited) for K + Eq. E3-2 | Cross-check: Manual Table 4-1a with equivalent KL_x/(r_x/r_y)

Given: Interior column on the ground story of an unbraced (moment-resisting) 4-story frame. Story height 14 ft; typical bay 30 ft. Column W14×90 (I_x = 999 in⁴, A = 26.5 in², r_x = 6.14 in, r_y = 3.70 in). Column above at joint A is also W14×90, L=14 ft. Girders at joint A: W21×62 (I_x = 1330 in⁴), 30 ft each side. Base at joint B is pinned. Weak axis braced at each floor by the slab (K_yL_y = 1.0·14 ft). P_u = 750 k. A992 steel.

Figure 5.11a — Unbraced frame. Rigid joints shown as filled squares; the pinned column base is drawn as an open circle on a ground line. Sway freedom → use the sidesway-uninhibited chart.

Step 1 — Alignment-chart G at each joint

Definition

(12)

G = Σ (I_{c} / L_{c}) \div Σ (I_{g} / L_{g})

Joint A (top, rigid): one column above + one column below, both W14×90 at 14 ft.
Σ(I_c/L_c) = 2 · 999/(14·12) = 2 · 5.946 = 11.89 in³.
Beams: two W21×62 girders, 30 ft each side.
Σ(I_g/L_g) = 2 · 1330/(30·12) = 2 · 3.694 = 7.39 in³.
G_A = 11.89/7.39 = 1.61.
Joint B (pinned base): AISC Commentary C-A-7.2 — a column truly pinned to the foundation is theoretically G = ∞; use G_B = 10 for design (a truly fixed base uses G = 1).

Step 2 — K from the unbraced-frame (sidesway-uninhibited) chart

Figure 5.11b — AISC Commentary Fig. C-A-7.2 (sidesway-uninhibited). Mark G_A = 1.61 on the left axis and G_B = 10 on the right axis (pinned base → use 10, not ∞); connect with a straight edge. The line crosses the K-scale at K ≈ 2.0 — the yellow dot shows the reading. Notice that on the unbraced chart the K-scale runs from 1 (bottom) to ∞ (top), so K ≥ 1 always.

How to use the chart: same three-step procedure — plot G_A, plot G_B, connect, read K on the middle scale. Only the K-axis differs from the braced chart (K = 1…∞ here vs. K = 0.5…1.0 for braced).

Numeric cross-check (Dumonteil approx.)

(13)

K = [1.6 G_{A} G_{B} + 4 (G_{A} + G_{B}) + 7.5] \div [G_{A} + G_{B} + 7.5]

Numerator: 1.6·(1.61)(10) + 4·(11.61) + 7.5 = 25.76 + 46.44 + 7.5 = 79.70.
Denominator: 11.61 + 7.5 = 19.11.
K_x = √(79.70/19.11) = √4.17 = 2.04 — matches the chart reading (~2.0) above.
Weak axis (slab-braced at every floor): K_y = 1.0.

Step 3 — Effective lengths

(14)

K_{x} L = 2.04 \cdot 14 \cdot 12 = 342.7 in

K_{y} L = 1.0 \cdot 14 \cdot 12 = 168 in

(KL/r)_x = 342.7 / 6.14 = 55.8 ← governs (large K on strong axis beats small K on weak axis).
(KL/r)_y = 168 / 3.70 = 45.4.
KL/r < 200 ✓.

Step 4 — Buckling range & F_cr

(15)

4.71 (E / F_{y}) = 113.4 > 55.8

in e l a s t i c

E q . E 3 - 2

(16)

F_{e} = π^{2} E / (K L / r)^{2}

F_{cr} = [0.65 8^{F_{y} / F_{e}}] \cdot F_{y}

F_e = π²·29000 / 55.8² = 286,220 / 3114 = 91.9 ksi.
F_y/F_e = 50/91.9 = 0.544.
0.658^0.544 = e^{0.544·(−0.4187)} = e^−0.2278 = 0.7963.
F_cr = 0.7963 · 50 = 39.82 ksi.

Step 5 — Design strength

(17)

ϕ_{c} P_{n} = 0.90 \cdot F_{cr} \cdot A_{g}

φ_cP_n = 0.90 · 39.82 · 26.5 = 950 k.

Check: φ_cP_n = 950 k ≥ P_u = 750 k ✓ (D/C = 0.79). Use W14×90. Cross-check: AISC Manual Table 4-1a with the equivalent KL_x/(r_x/r_y) = 342.7/(6.14/3.70) = 206.5 in / r_y-basis → interpolate on the W14×90 column at KL ≈ 17.2 ft to confirm ≈ 950 k.

⚠ Common student mistakes on the alignment chart

Using the braced chart when the frame can sway (or vice versa). Look at the drawing first: diagonal braces or shear walls → braced; only moment joints → unbraced.
Forgetting to add the column above to Σ(I_c/L_c) at an interior joint.
Using G = 0 for a pinned base or G = ∞ for a fixed base — the recommended design values are G = 10 (pinned) and G = 1 (fixed).
Applying the strong-axis K to the weak-axis slenderness. Track K_x, K_y separately.

Engineering tips

Euler buckling — the elastic ceiling. Fe drops as (KL/r)² — halving unbraced length quadruples Fe.
Inelastic branch (short/intermediate). The 0.658 exponent is empirical, fitted to test data; it is not derived from mechanics.
Effective length & the alignment chart. K = 1.0 is a safe default; K ≪ 1.0 requires real fixity that must be detailed and inspected.
Slender-element (§E7) columns. Thin-wall HSS and slender flanges are the usual culprits.

Common mistakes

Euler buckling — the elastic ceiling. Applying Fe as if it were the design stress. Fe is a benchmark; Fcr comes from §E3.
Inelastic branch (short/intermediate). Using the elastic 0.877·Fe branch when KL/r < 113.
Effective length & the alignment chart. Confusing braced and sway-permitted charts.
Slender-element (§E7) columns. Applying §E3 to a slender-element shape and reporting an inflated φPn.

Real-world use

Euler buckling — the elastic ceiling. Any slender column, especially braces and truss verticals.
Inelastic branch (short/intermediate). Most gravity columns in low- and mid-rise buildings.
Effective length & the alignment chart. Moment frames, braced frames, mixed systems.
Slender-element (§E7) columns. HSS8×8×3/16, cold-formed built-ups, thin-flange W-shapes.

06

Professional practice, safety & ethics

Compression member practice

Professional practice

The effective length KL is a judgment call — state the assumed end conditions and bracing for every column on the drawings.
Verify the brace actually exists in the architectural layout; a 'braced' column with no wall or beam framing to it is unbraced.
Coordinate column splices: splice location changes the unbraced length for the segment above.

Safety in design & construction

Buckling is sudden and unforgiving; unlike yielding, it gives no warning and no ductility.
During erection, columns are far more slender than the final condition — require temporary guys until the diaphragm is complete.
Never allow field modification (coping, cutting) of a compression member without recheck.

Engineering ethics

Do not assume K = 0.5 or 0.7 to make a column work unless the restraint is real and detailed.
Report to the client any column found overloaded during a renovation survey.
Hyatt Regency (1981) began as an unreviewed connection change — always re-analyze the whole load path after a change request.

Ironworkers bolting a steel beam connection while tied off at height — Erection safety: OSHA Subpart R fall protection and stable temporary bracing.

Engineers reviewing sealed structural drawings across a conference table — Design review: documenting assumptions before the drawings are sealed.

ABET / licensure link. These points map to ABET Student Outcomes 2 and 4 — engineering design within realistic constraints, and recognition of ethical and professional responsibilities. Expect NCEES FE and PE exam questions on the NSPE Code of Ethics, OSHA construction requirements, and the engineer's standard of care.

07

Cost analysis

Column cost and slenderness

Approach

Column cost is nearly linear in weight, so reducing KL by adding bracing is often cheaper than upsizing the column.
Keep column sizes constant over 2–3 stories — splices cost more than the steel saved by resizing every level.
Include fireproofing: it is priced per surface area, so a heavy compact column can be cheaper to protect than a light spread one.

Worked cost example — Add a brace vs upsize the column

Basis: Pu = 500 kip, story height 26 ft unbraced

Line item	Qty	Rate	Cost
Option A — W14×109 unbraced, 26 ft	1.42 ton	$2,700	$3,834
Option B — W14×74 braced at mid-height	0.96 ton	$2,700	$2,592
Option B added brace + connections	1 ls	$1,400	$1,400
Estimated total			$7,826

Takeaway. Bracing beats upsizing here by roughly $100 — essentially a wash, so choose on constructability and floor-plan impact.

Unit rates are representative US averages for teaching purposes. On a real project, price with current local rates (RSMeans, fabricator quotes, or contractor pricing) and state the estimate date.

08

Animated concepts

Key mechanics visualised — watch the strain profile, stress block, or buckled shape evolve

Euler column buckling

Amplitude grows with KL/r. Below KL/r ≈ 113 (A992) inelastic E3-2 governs; above, elastic Euler E3-3 governs.

09

Engineering figures

Full-page reference diagrams — the visual vocabulary you will use for the rest of the course

§5.6.1

Column erection

Fig. 5.1W-shape column being set on baseplate
Gravity columns transfer beam reactions vertically into the base. Concentric alignment is essential for §E3 to apply.

§5.6.2

Base condition

Fig. 5.2Base plate + anchor rods
The fixity of the base sets K. A grouted, fully-detailed base can approach K = 0.65; a poorly detailed base is closer to K = 1.0.

§5.6.3

Global buckling

Fig. 5.3Slender column — Euler mode
First-mode buckling of a pinned-pinned column: single half-sine wave, peak lateral displacement at mid-height.

§5.6.4

Local buckling

Fig. 5.4Local flange / wall crinkle
Slender plate elements buckle locally before global flexural buckling can develop — §E7 territory.

10

Worked examples

Full textbook solutions — problem, theory, step-by-step, verification, interpretation

Example 5.1

Interior gravity column — φcPn per AISC §E3

An interior gravity column of a mid-rise steel-framed building is a W10×49, A992 steel, pinned–pinned, unbraced length 14 ft. Compute φcPn using AISC §E3.

Problem statement

An interior gravity column is a W10×49 (A992), pinned at both ends, with unbraced length L = 14 ft in both axes. Compute the design axial compressive strength φcPn using AISC 360-22 §E3 and verify against Manual Table 4-1a.

Steel wide-flange column being erected on site — FIG. 5.1 — Wide-flange gravity column typical of the W10×49 in this example.

Floor plan

3-D isometric

Element detail

DIMFloor plan locates the interior gravity column, isometric shows its story, detail carries the axial load — φcPn per §E3.

Given

W10×49: Ag = 14.4 in², ry = 2.54 in, rx = 4.35 in
A992 steel: Fy = 50 ksi, E = 29,000 ksi
K = 1.0 (pinned–pinned)
L = 14 ft = 168 in (both axes)

Find

φcPn per §E3
Governing axis (rx vs ry)

Assumptions

Concentric axial load — no bending
Non-slender section (verify against Table B4.1a)
No slender-element reduction (§E7 not triggered)

Code references

AISC 360-22 §E1 — φc = 0.90
AISC 360-22 §E3 — flexural buckling
AISC Commentary Table C-A-7.1 — K values
AISC Manual Table 4-1a — φcPn table

Theory & approach

§E3 uses Fe = π²E/(Lc/r)² and a piecewise Fcr equation. Below Lc/r = 4.71·√(E/Fy) (inelastic) the curve fits residual-stress-affected test data; above, elastic Euler controls with a 0.877 knock-down.

Step-by-step solution

1
Effective slenderness
FormulaAISC §E3
L_c / r = K · L / r_min
L_c / r_y = 1.0 · 168 / 2.54 = 66.1 (weak axis governs)
L_c / r_x = 1.0 · 168 / 4.35 = 38.6 (strong axis, non-critical)
2
Classify branch
Formula§E3
Limit = 4.71·√(E / F_y)
Limit = 4.71·√(29,000 / 50) = 4.71·√580 = 113.4
66.1 < 113.4 → INELASTIC branch
3
Euler stress Fe
Formula§E3-4
F_e = π² E / (L_c/r)²
F_e = π² · 29,000 / (66.1)² = 286,220 / 4,369.2 = 65.5 ksi
4
Critical stress Fcr
Formula§E3-2
F_cr = 0.658^(F_y / F_e) · F_y
F_y / F_e = 50 / 65.5 = 0.764
0.658^(0.764) = 0.726
F_cr = 0.726 · 50 = 36.3 ksi
5
Design strength φcPn
Formula§E1
φ_c P_n = 0.90 · F_cr · A_g
φ_c P_n = 0.90 · 36.3 · 14.4 = 470 kips
6
Manual cross-check
AISC Manual Table 4-1a for W10×49 at KL = 14 ft (y-axis) gives φcPn ≈ 470 k — matches within rounding.

Verification

All three checks (slenderness, regime, and computed φcPn) agree with the AISC Manual table. Non-slender per Table B4.1a — §E7 does not apply.

Final answer

φcPn ≈ 470 kips — inelastic §E3, weak-axis controls.

Design interpretation

A W10×49 gravity column at 14 ft is a textbook example of the inelastic branch — Lc/r sits well below the 113 transition. If the unbraced length grew past ~24 ft the same column would cross into the elastic branch and lose more than half its capacity.

Common mistakes

Using rx (strong axis) instead of ry when there is no weak-axis bracing.
Applying the elastic 0.877·Fe branch when Lc/r is below 113 for A992.
Forgetting to check Table B4.1a for slender elements before invoking §E3.

Engineering insight

Weak-axis bracing at mid-height reduces Lc/ry from 66 to ~33, driving Fe up by a factor of 4 and pushing Fcr close to Fy. A single kicker brace can lift φcPn by 25–35% at essentially no material cost.

References

· AISC 360-22 Chapter E
· AISC Manual 16th ed., Part 4 & Table 4-1a

11

Guided practice

Compute the governing variables — hints unlock as you need them

A W8×24 (A992) pinned–pinned column has KL = 12 ft. Compute Lc/ry (ry = 1.61 in), classify (elastic vs inelastic vs the 113.4 limit), and estimate φcPn using §E3 with Ag = 7.08 in².

Your turn

L_{c/r_y} =phi_{c*P_n} =kip

Hints

1.Lc/ry = KL(in)/ry.

12

Independent practice

Solve the chapter's design task — compute each governing variable

Design task

A W10×49 (A992) column has KL = 14 ft about the weak axis. Compute the slenderness KL/ry, the critical stress Fcr, and the design axial strength φc Pn. Verify whether the column is in the elastic or inelastic buckling range.

Given

Fy = 50 ksi, E = 29,000 ksi
W10×49: Ag = 14.4 in², ry = 2.54 in
KL = 14 ft = 168 in
Transition: 4.71·√(E/Fy) = 113.4

Approach

Compute KL/ry and compare to 4.71·√(E/Fy) — inelastic if less.
Elastic buckling stress: Fe = π²·E / (KL/r)².
Inelastic: Fcr = 0.658^(Fy/Fe)·Fy. Design: φc Pn = 0.90·Fcr·Ag.

Submit your answer

KL/r_y =F_cr =ksiphi_{c*P_n} =kip

13

Mini design challenge

Select the option that satisfies every code and serviceability requirement in the brief

Brief

Select the lightest W12 A992 gravity column for Pu = 600 k, KL = 13 ft (weak axis controls). Use AISC Manual Table 4-1a and confirm §E3 non-slender.

Requirements

Factored demand Pu = 600 kips
A992 (Fy = 50 ksi, E = 29,000 ksi)
KL = 13 ft weak axis
Non-slender per Table B4.1a
Lightest W12 satisfying φcPn ≥ Pu

Section

Wt (lb/ft)

Δ (in)

Ru/Rn

Cost

Pick

14

Chapter summary

A mind map of how every concept connects

Graded Chapter Quiz(19 FE-style questions · AISC Manual required)

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16

FE exam preparation

NCEES-style practice with timer, equation sheet, and mastery tracking

Exam mode

30:00 Calculator

Question 1 / 11

A W14×53 steel column in a braced frame has pinned connections at both ends and an unsupported length of 18 ft. Determine the effective length factor K and effective length KL.

◆ EasyAISC 360-22 Commentary Table C-A-7.1 / NCEES FE Handbook

A. K = 0.50, KL = 9 ft
B. K = 0.65, KL = 11.7 ft
C. K = 1.00, KL = 18 ft
D. K = 2.00, KL = 36 ft

Compression Members and Column Buckling

Compute effective length Lc = K·L

Classify elastic vs inelastic buckling

Apply Fcr → φcPn = 0.90·Fcr·Ag

Read AISC Manual Table 4-1a

Screen slenderness KL/r ≤ 200

Handle slender-element (E7) columns

Global flexural buckling

Local plate / flange buckling

Hartford Civic Center roof (1978)

Quebec Bridge (1907)

Chapter 5 — Compression Members & Column Buckling (AISC 360-22 Chapter E)

1. Behavior

Representative Structures

2. Effective Length K L

3. Slenderness Ratio KL/r — the Single Number that Drives Everything

4. AISC Flexural-Buckling Equations (§E3)

5. Torsional & Flexural-Torsional Buckling (§E4)

6. Slender-Element Sections (§E7)

7. Two Solution Methods — Companion Tables vs. AISC Specification Formulas

Method A — AISC Specification Formulas (Chapter E, "the long way")

Method B — AISC Manual Companion Tables (Part 4, "the fast way")

Which method for which problem?

8. Design Procedure (put it all together)

⚠ Common mistakes

Worked Example 5.1 — Interior Column, W14 (A992)

Step 1 — Effective lengths

Step 2 — Governing slenderness

Step 3 — Classify range (E3)

Step 4 — Compute Fe and Fcr

Step 5 — Element slenderness (Table B4.1a)

Step 6 — Design strength

Verify with AISC Manual Table 4-1a

Worked Example 5.10 — Intermediate-Floor Column, Braced Frame (Alignment Chart)

Step 1 — Alignment-chart G at each joint

Step 2 — K from the braced-frame (sidesway-inhibited) chart

Step 3 — Effective lengths

Step 4 — Buckling range & Fcr (AISC §E3)

Step 5 — Design strength

Worked Example 5.11 — Lowest-Floor Column, Unbraced Moment Frame (Pinned Base)

Step 1 — Alignment-chart G at each joint

Step 2 — K from the unbraced-frame (sidesway-uninhibited) chart

Step 3 — Effective lengths

Step 4 — Buckling range & Fcr

Step 5 — Design strength

⚠ Common student mistakes on the alignment chart

Column erection

Base condition

Global buckling

Local buckling

Interior gravity column — φcPn per AISC §E3

Effective slenderness

Classify branch

Euler stress Fe

Critical stress Fcr

Design strength φcPn

Manual cross-check

Graded Chapter Quiz(19 FE-style questions · AISC Manual required)

Upload your worked solution (PDF)

Step 4 — Compute F_e and F_cr

Step 4 — Buckling range & F_cr (AISC §E3)

Step 4 — Buckling range & F_cr