Lateral-Torsional Buckling

Structural Design · FE Reference Handbook section

Structural Design

8 formulas

10 exam-style examples

~60 min

All Structural Design lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Lateral-Torsional Buckling within Structural Design. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what lateral-torsional buckling describes physically and when it applies.
State every one of the 8 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: f'c and Fy in ksi with areas in in² give kips.

Lecture

Why this section exists. Lateral-Torsional Buckling is the part of Structural Design that lets you connect a reinforced concrete or steel member being checked to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a factored demand compared against φ times a nominal capacity. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. f'c and Fy in ksi with areas in in² give kips. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Bolted connection at the bottom chord of a steel truss bridge. — Photo 1. Where this shows up in practice: lateral-torsional buckling.
HAER / Library of Congress, public domain

Structural Design — Lateral-Torsional Buckling: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a reinforced concrete or steel member being checked. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 8 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

When Lp < Lb ≤ Lr	Quantity produced by "When Lp < Lb ≤ Lr" — read its definition and unit from the handbook line directly above the equation.
Cb	Quantity produced by "Cb = 2.5M" — read its definition and unit from the handbook line directly above the equation.
Mmax	Quantity produced by "Mmax = absolute value of maximum moment in the unbraced segment" — read its definition and unit from the handbook line directly above the equation.
MA	Quantity produced by "MA = absolute value of maximum moment at quarter point of the unbraced segment" — read its definition and unit from the handbook line directly above the equation.
MB	Quantity produced by "MB = absolute value of maximum moment at centerline of the unbraced segment" — read its definition and unit from the handbook line directly above the equation.
MC	Quantity produced by "MC = absolute value of maximum moment at three-quarter of the unbraced segment" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Based on bracing where Lb is the length between points that are either braced against lateral displacement of the compression
flange or braced against twist of the cross section with respect to the length limits Lp and Lr:
Lb − Lp
Lr − Lp
where
12.5Mmax
max + 3MA + 4MB + 3MC

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Euler buckling load of a pinned column

A pinned-pinned steel column is 4.5 m long with I = 22.2 × 10⁶ mm⁴ and E = 200 GPa. What is its Euler critical load?

Given

L = 4.5 m
K = 1.0 (pinned-pinned)
I = 22.2 × 10⁶ mm⁴
E = 200 GPa

Find

P_cr

Start with the thinking

Effective length depends on end conditions — pinned gives K = 1.
Weak-axis I governs; the exam gives the value to use.

Figure for Euler buckling load of a pinned column

Step-by-step solution

Euler load
$P_{c} r = π^{2} E I / (K L)^{2}$
Effective length
$K L = 1.0 (4, 500) = 4, 500 mm$
Numerator
$π^{2} (200, 000) (22.2 \times 1 0^{6}) = 4.383 \times 1 0^{13}$
Denominator
$(4, 500)^{2} = 2.025 \times 1 0^{7}$
Result
$P_{c} r = 2.164 \times 1 0^{6} N = 2, 160 k N$

Answer: P_cr ≈ 2,160 kN

Why the other options are there

8,660 kN (K = 0.5 used)
540 kN (KL squared incorrectly as 4KL)

Reference: FE Reference Handbook — Mechanics of Materials — Columns

Example 2

Tension member capacity: yielding versus rupture

An A36 plate 6 in. × ½ in. has two ⅞ in. bolt holes in the critical section. With Fy = 36 ksi, Fu = 58 ksi and U = 1.0, what governs the design tensile strength?

Given

Ag = 6(0.5) = 3.00 in²
Two holes, d_h = 0.875 + 0.125 = 1.00 in.
Fy = 36 ksi, Fu = 58 ksi

Find

φPn (governing limit state)

Start with the thinking

Two limit states: gross-section yielding (φ = 0.90) and net-section rupture (φ = 0.75).
Hole diameter includes the 1/16 in. damage allowance.

Figure for Tension member capacity: yielding versus rupture

Step-by-step solution

Gross yielding — φPn = 0.90 Fy Ag = 0.90(36)(3.00) = 97.2 kips
Net area
$A n = 3.00 - 2 (1.00) (0.5) = 2.00 i n^{2}$
Effective net
$A e = U A n = 1.0 (2.00) = 2.00 i n^{2}$
Rupture — φPn = 0.75 Fu Ae = 0.75(58)(2.00) = 87.0 kips
Governing — the smaller value, 87.0 kips (net-section rupture)

Answer: φPn = 87.0 kips, governed by rupture

Why the other options are there

97.2 kips (yielding taken as governing)
104 kips (holes ignored)

Reference: FE Reference Handbook — Structural Design — Steel tension members

Example 3

Plastic moment capacity of a compact W-shape — Lateral-Torsional Buckling

A compact, fully braced W-shape has Zx = 150.0 in³ and Fy = 50 ksi, and must carry Mu = 393.8 kip·ft. Check φMp.

Given

Zx = 150.0 in³
Fy = 50 ksi
Mu = 393.8 kip·ft
φ = 0.90

Find

φMp and the check

Start with the thinking

Compact and fully braced means the plastic moment governs.
Zx·Fy is in kip·in — divide by 12.

Figure for Plastic moment capacity of a compact W-shape — Lateral-Torsional Buckling

Step-by-step solution

Plastic moment — Mp = Zx·Fy
Substituting — Mp = 150.0(50) = 7,500 kip·in
Convert — Mp = 625.0 kip·ft
Design — φMp = 0.90(625.0) = 562.5 kip·ft
Check — Mu = 393.8 ≤ φMp = 562.5 → adequate

Answer: φMp ≈ 562.5 kip·ft → section is adequate

Why the other options are there

7,500 kip·ft (kip·in reported as kip·ft)
375.0 kip·ft (ASD factor applied to LRFD)

Reference: FE Reference Handbook — Structural Design → Lateral-Torsional Buckling

Example 4

Available moment φM_n of a compact steel beam — Lateral-Torsional Buckling

A compact, fully braced A992 beam has Z_x = 110 in³. Compute the plastic moment, the available moment φM_n in foot-kips, and the uniform factored load it can carry over a 36 ft simple span.

Given

Z_x = 110 in³
F_y = 50 ksi
φ_b = 0.90
L = 36 ft

Find

M_p, φM_n and the allowable w_u

Start with the thinking

For a compact, fully braced section the nominal moment equals the plastic moment.
Dividing by 12 converts kip-inches to the foot-kips used in the available moment tables.

Step-by-step solution

Formula
$M_{p} = F_{y} Z_{x}$
Substituting — M_p = 50(110) = 5,500 kip·in = 458.3 kip·ft
Formula — φM_n = 0.90 M_p
Substituting — φM_n = 0.90(458.3) = 412.5 kip·ft
Formula — M_u = w_u L²/8 ≤ φM_n → w_u = 8φM_n/L²
Substituting
$w_{u} = 8 (412.5) / (36)^{2} = 2.546 k i p / f t$

Answer: φM_n = 412.5 kip·ft, permitting w_u = 2.55 kip/ft over 36 ft

Why the other options are there

5,500 kip·ft (inches never converted)
458.3 kip·ft (φ omitted)

Reference: FE Reference Handbook — Structural Design → Lateral-Torsional Buckling

Example 5

Plastic moment capacity of a compact W-shape — Lateral-Torsional Buckling (2)

A compact, fully braced W-shape has Zx = 105.0 in³ and Fy = 50 ksi, and must carry Mu = 216.6 kip·ft. Check φMp.

Given

Zx = 105.0 in³
Fy = 50 ksi
Mu = 216.6 kip·ft
φ = 0.90

Find

φMp and the check

Start with the thinking

Compact and fully braced means the plastic moment governs.
Zx·Fy is in kip·in — divide by 12.

Figure for Plastic moment capacity of a compact W-shape — Lateral-Torsional Buckling (2)

Step-by-step solution

Plastic moment — Mp = Zx·Fy
Substituting — Mp = 105.0(50) = 5,250 kip·in
Convert — Mp = 437.5 kip·ft
Design — φMp = 0.90(437.5) = 393.8 kip·ft
Check — Mu = 216.6 ≤ φMp = 393.8 → adequate

Answer: φMp ≈ 393.8 kip·ft → section is adequate

Why the other options are there

5,250 kip·ft (kip·in reported as kip·ft)
262.5 kip·ft (ASD factor applied to LRFD)

Reference: FE Reference Handbook — Structural Design → Lateral-Torsional Buckling

Example 6

Available moment φM_n of a compact steel beam — Lateral-Torsional Buckling (2)

A compact, fully braced A992 beam has Z_x = 77 in³. Compute the plastic moment, the available moment φM_n in foot-kips, and the uniform factored load it can carry over a 24 ft simple span.

Given

Z_x = 77 in³
F_y = 50 ksi
φ_b = 0.90
L = 24 ft

Find

M_p, φM_n and the allowable w_u

Start with the thinking

For a compact, fully braced section the nominal moment equals the plastic moment.
Dividing by 12 converts kip-inches to the foot-kips used in the available moment tables.

Step-by-step solution

Formula
$M_{p} = F_{y} Z_{x}$
Substituting — M_p = 50(77) = 3,850 kip·in = 320.8 kip·ft
Formula — φM_n = 0.90 M_p
Substituting — φM_n = 0.90(320.8) = 288.8 kip·ft
Formula — M_u = w_u L²/8 ≤ φM_n → w_u = 8φM_n/L²
Substituting
$w_{u} = 8 (288.8) / (24)^{2} = 4.010 k i p / f t$

Answer: φM_n = 288.8 kip·ft, permitting w_u = 4.01 kip/ft over 24 ft

Why the other options are there

3,850 kip·ft (inches never converted)
320.8 kip·ft (φ omitted)

Reference: FE Reference Handbook — Structural Design → Lateral-Torsional Buckling

Example 7

Plastic moment capacity of a compact W-shape — Lateral-Torsional Buckling (3)

A compact, fully braced W-shape has Zx = 155.0 in³ and Fy = 50 ksi, and must carry Mu = 465.0 kip·ft. Check φMp.

Given

Zx = 155.0 in³
Fy = 50 ksi
Mu = 465.0 kip·ft
φ = 0.90

Find

φMp and the check

Start with the thinking

Compact and fully braced means the plastic moment governs.
Zx·Fy is in kip·in — divide by 12.

Figure for Plastic moment capacity of a compact W-shape — Lateral-Torsional Buckling (3)

Step-by-step solution

Plastic moment — Mp = Zx·Fy
Substituting — Mp = 155.0(50) = 7,750 kip·in
Convert — Mp = 645.8 kip·ft
Design — φMp = 0.90(645.8) = 581.3 kip·ft
Check — Mu = 465.0 ≤ φMp = 581.3 → adequate

Answer: φMp ≈ 581.3 kip·ft → section is adequate

Why the other options are there

7,750 kip·ft (kip·in reported as kip·ft)
387.5 kip·ft (ASD factor applied to LRFD)

Reference: FE Reference Handbook — Structural Design → Lateral-Torsional Buckling

Example 8

Available moment φM_n of a compact steel beam — Lateral-Torsional Buckling (3)

A compact, fully braced A992 beam has Z_x = 146 in³. Compute the plastic moment, the available moment φM_n in foot-kips, and the uniform factored load it can carry over a 22 ft simple span.

Given

Z_x = 146 in³
F_y = 50 ksi
φ_b = 0.90
L = 22 ft

Find

M_p, φM_n and the allowable w_u

Start with the thinking

For a compact, fully braced section the nominal moment equals the plastic moment.
Dividing by 12 converts kip-inches to the foot-kips used in the available moment tables.

Step-by-step solution

Formula
$M_{p} = F_{y} Z_{x}$
Substituting — M_p = 50(146) = 7,300 kip·in = 608.3 kip·ft
Formula — φM_n = 0.90 M_p
Substituting — φM_n = 0.90(608.3) = 547.5 kip·ft
Formula — M_u = w_u L²/8 ≤ φM_n → w_u = 8φM_n/L²
Substituting
$w_{u} = 8 (547.5) / (22)^{2} = 9.050 k i p / f t$

Answer: φM_n = 547.5 kip·ft, permitting w_u = 9.05 kip/ft over 22 ft

Why the other options are there

7,300 kip·ft (inches never converted)
608.3 kip·ft (φ omitted)

Reference: FE Reference Handbook — Structural Design → Lateral-Torsional Buckling

Example 9

Plastic moment capacity of a compact W-shape — Lateral-Torsional Buckling (4)

A compact, fully braced W-shape has Zx = 185.0 in³ and Fy = 50 ksi, and must carry Mu = 381.6 kip·ft. Check φMp.

Given

Zx = 185.0 in³
Fy = 50 ksi
Mu = 381.6 kip·ft
φ = 0.90

Find

φMp and the check

Start with the thinking

Compact and fully braced means the plastic moment governs.
Zx·Fy is in kip·in — divide by 12.

Figure for Plastic moment capacity of a compact W-shape — Lateral-Torsional Buckling (4)

Step-by-step solution

Plastic moment — Mp = Zx·Fy
Substituting — Mp = 185.0(50) = 9,250 kip·in
Convert — Mp = 770.8 kip·ft
Design — φMp = 0.90(770.8) = 693.8 kip·ft
Check — Mu = 381.6 ≤ φMp = 693.8 → adequate

Answer: φMp ≈ 693.8 kip·ft → section is adequate

Why the other options are there

9,250 kip·ft (kip·in reported as kip·ft)
462.5 kip·ft (ASD factor applied to LRFD)

Reference: FE Reference Handbook — Structural Design → Lateral-Torsional Buckling

Example 10

Available moment φM_n of a compact steel beam — Lateral-Torsional Buckling (4)

A compact, fully braced A992 beam has Z_x = 234 in³. Compute the plastic moment, the available moment φM_n in foot-kips, and the uniform factored load it can carry over a 40 ft simple span.

Given

Z_x = 234 in³
F_y = 50 ksi
φ_b = 0.90
L = 40 ft

Find

M_p, φM_n and the allowable w_u

Start with the thinking

For a compact, fully braced section the nominal moment equals the plastic moment.
Dividing by 12 converts kip-inches to the foot-kips used in the available moment tables.

Step-by-step solution

Formula
$M_{p} = F_{y} Z_{x}$
Substituting — M_p = 50(234) = 11,700 kip·in = 975.0 kip·ft
Formula — φM_n = 0.90 M_p
Substituting — φM_n = 0.90(975.0) = 877.5 kip·ft
Formula — M_u = w_u L²/8 ≤ φM_n → w_u = 8φM_n/L²
Substituting
$w_{u} = 8 (877.5) / (40)^{2} = 4.388 k i p / f t$

Answer: φM_n = 877.5 kip·ft, permitting w_u = 4.39 kip/ft over 40 ft

Why the other options are there

11,700 kip·ft (inches never converted)
975.0 kip·ft (φ omitted)

Reference: FE Reference Handbook — Structural Design → Lateral-Torsional Buckling

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a reinforced concrete or steel member being checked, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Lateral-Torsional Buckling contains 8 relations; you must be able to find this page in under 15 seconds.
Exam style: a factored demand compared against φ times a nominal capacity.
Unit rule: f'c and Fy in ksi with areas in in² give kips.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

f'c and Fy in ksi with areas in in² give kips
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.