Columns

Structural Design · FE Reference Handbook section

Structural Design

6 formulas

10 exam-style examples

~57 min

All Structural Design lectures

Learning objectives

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

This chapter section covers Columns 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 columns describes physically and when it applies.
State every one of the 6 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. Columns 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: columns.
HAER / Library of Congress, public domain

Structural Design — Columns: 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 6 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

Pn	Quantity produced by "Pn = Fcr Ag" — read its definition and unit from the handbook line directly above the equation.
Lc	Quantity produced by "Lc = KL = effective length of member (in.)" — read its definition and unit from the handbook line directly above the equation.
Fe	Quantity produced by "Fe = elastic buckling stress = π2E/(KL/r)2" — 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.

slender elements and is determined as follows:
where the critical stress Fcr is determined as follows:
70.658 Fe A Fy
Lc E
Lc E
where
VALUES OF Cb FOR SIMPLY SUPPORTED BEAMS
LATERAL BRACING
LOAD Cb
ALONG SPAN
P NONE
LOAD AT MIDPOINT 1.32
AT LOAD POINT 1.67 1.67
P P NONE
LOADS AT THIRD POINTS 1.14
AT LOAD POINTS
LOADS SYMMETRICALLY PLACED 1.67 1.00 1.67
P P P NONE
LOADS AT QUARTER POINTS 1.14
AT LOAD POINTS
LOADS AT QUARTER POINTS 1.67 1.11 1.11 1.67
NONE 1.14
AT MIDPOINT 1.30 1.30
AT THIRD POINTS 1.45 1.01 1.45

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

Axial compressive strength of a steel column by AISC flexural buckling — Columns

A 50 ksi steel column has A_g = 8.5 in.², r_min = 3.40 in. and an effective length KL = 19.5 ft. Compute the slenderness ratio, elastic buckling stress, critical stress and the design compressive strength φ_cP_n.

Given

F_y = 50 ksi, E = 29 000 ksi
A_g = 8.5 in.²
r_min = 3.40 in.
KL = 234.0 in.

Find

KL/r, F_e, F_cr and φ_cP_n

Start with the thinking

The 4.71√(E/F_y) limit separates inelastic (0.658 exponential) from elastic (0.877F_e) buckling.
Always use the smallest radius of gyration — the column buckles about its weak axis.

Step-by-step solution

Formula
$λ = \frac{K L}{r}$
Substituting
$λ = 234.0/3.40 = 68.8$
Formula
$F_{e} = \frac{π ^{2} E}{( K L / r ) ^{2}}$
Substituting
$F_{e} = π^{2} (29000) / (68.8)^{2} = 60.43 k s i$
Limit
$4.71 (E / F_{y}) = 4.71 (29000/50) = 113.4 \to in e l a s t i c b u c k l in g co n t r o l s$
Formula
$F_{cr} = (0.65 8^{F_{y} / F_{e}}) F_{y} = (0.65 8^{0.827}) (50) = 35.36 k s i$
Nominal strength
$P_{n} = F_{cr} A_{g} = 35.36 (8.5) = 300.6 k i p s$
Design strength
$ϕ_{c} P_{n} = 0.90 (300.6) = 270.5 k i p s$

Answer: KL/r = 68.8, F_cr = 35.36 ksi, φ_cP_n = 270.5 kips

Why the other options are there

425.0 kips (yield squash load, buckling ignored)
450.4 kips (elastic equation used in the inelastic range)

Reference: FE Reference Handbook — Structural Design → Columns

Example 3

Axial compressive strength of a steel column by AISC flexural buckling — Columns (2)

A 46 ksi steel column has A_g = 25.0 in.², r_min = 2.70 in. and an effective length KL = 21.0 ft. Compute the slenderness ratio, elastic buckling stress, critical stress and the design compressive strength φ_cP_n.

Given

F_y = 46 ksi, E = 29 000 ksi
A_g = 25.0 in.²
r_min = 2.70 in.
KL = 252.0 in.

Find

KL/r, F_e, F_cr and φ_cP_n

Start with the thinking

The 4.71√(E/F_y) limit separates inelastic (0.658 exponential) from elastic (0.877F_e) buckling.
Always use the smallest radius of gyration — the column buckles about its weak axis.

Step-by-step solution

Formula
$λ = \frac{K L}{r}$
Substituting
$λ = 252.0/2.70 = 93.3$
Formula
$F_{e} = \frac{π ^{2} E}{( K L / r ) ^{2}}$
Substituting
$F_{e} = π^{2} (29000) / (93.3)^{2} = 32.86 k s i$
Limit
$4.71 (E / F_{y}) = 4.71 (29000/46) = 118.3 \to in e l a s t i c b u c k l in g co n t r o l s$
Formula
$F_{cr} = (0.65 8^{F_{y} / F_{e}}) F_{y} = (0.65 8^{1.400}) (46) = 25.60 k s i$
Nominal strength
$P_{n} = F_{cr} A_{g} = 25.60 (25.0) = 640.0 k i p s$
Design strength
$ϕ_{c} P_{n} = 0.90 (640.0) = 576.0 k i p s$

Answer: KL/r = 93.3, F_cr = 25.60 ksi, φ_cP_n = 576.0 kips

Why the other options are there

1,150 kips (yield squash load, buckling ignored)
720.4 kips (elastic equation used in the inelastic range)

Reference: FE Reference Handbook — Structural Design → Columns

Example 4

Axial compressive strength of a steel column by AISC flexural buckling — Columns (3)

A 50 ksi steel column has A_g = 25.0 in.², r_min = 1.60 in. and an effective length KL = 12.0 ft. Compute the slenderness ratio, elastic buckling stress, critical stress and the design compressive strength φ_cP_n.

Given

F_y = 50 ksi, E = 29 000 ksi
A_g = 25.0 in.²
r_min = 1.60 in.
KL = 144.0 in.

Find

KL/r, F_e, F_cr and φ_cP_n

Start with the thinking

The 4.71√(E/F_y) limit separates inelastic (0.658 exponential) from elastic (0.877F_e) buckling.
Always use the smallest radius of gyration — the column buckles about its weak axis.

Step-by-step solution

Formula
$λ = \frac{K L}{r}$
Substituting
$λ = 144.0/1.60 = 90.0$
Formula
$F_{e} = \frac{π ^{2} E}{( K L / r ) ^{2}}$
Substituting
$F_{e} = π^{2} (29000) / (90.0)^{2} = 35.34 k s i$
Limit
$4.71 (E / F_{y}) = 4.71 (29000/50) = 113.4 \to in e l a s t i c b u c k l in g co n t r o l s$
Formula
$F_{cr} = (0.65 8^{F_{y} / F_{e}}) F_{y} = (0.65 8^{1.415}) (50) = 27.65 k s i$
Nominal strength
$P_{n} = F_{cr} A_{g} = 27.65 (25.0) = 691.4 k i p s$
Design strength
$ϕ_{c} P_{n} = 0.90 (691.4) = 622.2 k i p s$

Answer: KL/r = 90.0, F_cr = 27.65 ksi, φ_cP_n = 622.2 kips

Why the other options are there

1,250 kips (yield squash load, buckling ignored)
774.7 kips (elastic equation used in the inelastic range)

Reference: FE Reference Handbook — Structural Design → Columns

Example 5

Axial compressive strength of a steel column by AISC flexural buckling — Columns (4)

A 50 ksi steel column has A_g = 23.0 in.², r_min = 2.10 in. and an effective length KL = 15.5 ft. Compute the slenderness ratio, elastic buckling stress, critical stress and the design compressive strength φ_cP_n.

Given

F_y = 50 ksi, E = 29 000 ksi
A_g = 23.0 in.²
r_min = 2.10 in.
KL = 186.0 in.

Find

KL/r, F_e, F_cr and φ_cP_n

Start with the thinking

The 4.71√(E/F_y) limit separates inelastic (0.658 exponential) from elastic (0.877F_e) buckling.
Always use the smallest radius of gyration — the column buckles about its weak axis.

Step-by-step solution

Formula
$λ = \frac{K L}{r}$
Substituting
$λ = 186.0/2.10 = 88.6$
Formula
$F_{e} = \frac{π ^{2} E}{( K L / r ) ^{2}}$
Substituting
$F_{e} = π^{2} (29000) / (88.6)^{2} = 36.48 k s i$
Limit
$4.71 (E / F_{y}) = 4.71 (29000/50) = 113.4 \to in e l a s t i c b u c k l in g co n t r o l s$
Formula
$F_{cr} = (0.65 8^{F_{y} / F_{e}}) F_{y} = (0.65 8^{1.370}) (50) = 28.17 k s i$
Nominal strength
$P_{n} = F_{cr} A_{g} = 28.17 (23.0) = 648.0 k i p s$
Design strength
$ϕ_{c} P_{n} = 0.90 (648.0) = 583.2 k i p s$

Answer: KL/r = 88.6, F_cr = 28.17 ksi, φ_cP_n = 583.2 kips

Why the other options are there

1,150 kips (yield squash load, buckling ignored)
735.9 kips (elastic equation used in the inelastic range)

Reference: FE Reference Handbook — Structural Design → Columns

Example 6

Axial compressive strength of a steel column by AISC flexural buckling — Columns (5)

A 36 ksi steel column has A_g = 29.0 in.², r_min = 2.20 in. and an effective length KL = 25.5 ft. Compute the slenderness ratio, elastic buckling stress, critical stress and the design compressive strength φ_cP_n.

Given

F_y = 36 ksi, E = 29 000 ksi
A_g = 29.0 in.²
r_min = 2.20 in.
KL = 306.0 in.

Find

KL/r, F_e, F_cr and φ_cP_n

Start with the thinking

The 4.71√(E/F_y) limit separates inelastic (0.658 exponential) from elastic (0.877F_e) buckling.
Always use the smallest radius of gyration — the column buckles about its weak axis.

Step-by-step solution

Formula
$λ = \frac{K L}{r}$
Substituting
$λ = 306.0/2.20 = 139.1$
Formula
$F_{e} = \frac{π ^{2} E}{( K L / r ) ^{2}}$
Substituting
$F_{e} = π^{2} (29000) / (139.1)^{2} = 14.79 k s i$
Limit
$4.71 (E / F_{y}) = 4.71 (29000/36) = 133.7 \to e l a s t i c b u c k l in g co n t r o l s$
Formula
$F_{cr} = 0.877 F_{e} = 0.877 (14.79) = 12.97 k s i$
Nominal strength
$P_{n} = F_{cr} A_{g} = 12.97 (29.0) = 376.3 k i p s$
Design strength
$ϕ_{c} P_{n} = 0.90 (376.3) = 338.6 k i p s$

Answer: KL/r = 139.1, F_cr = 12.97 ksi, φ_cP_n = 338.6 kips

Why the other options are there

1,044 kips (yield squash load, buckling ignored)
376.3 kips (elastic equation used in the inelastic range)

Reference: FE Reference Handbook — Structural Design → Columns

Example 7

Axial compressive strength of a steel column by AISC flexural buckling — Columns (6)

A 50 ksi steel column has A_g = 12.0 in.², r_min = 1.90 in. and an effective length KL = 24.0 ft. Compute the slenderness ratio, elastic buckling stress, critical stress and the design compressive strength φ_cP_n.

Given

F_y = 50 ksi, E = 29 000 ksi
A_g = 12.0 in.²
r_min = 1.90 in.
KL = 288.0 in.

Find

KL/r, F_e, F_cr and φ_cP_n

Start with the thinking

The 4.71√(E/F_y) limit separates inelastic (0.658 exponential) from elastic (0.877F_e) buckling.
Always use the smallest radius of gyration — the column buckles about its weak axis.

Step-by-step solution

Formula
$λ = \frac{K L}{r}$
Substituting
$λ = 288.0/1.90 = 151.6$
Formula
$F_{e} = \frac{π ^{2} E}{( K L / r ) ^{2}}$
Substituting
$F_{e} = π^{2} (29000) / (151.6)^{2} = 12.46 k s i$
Limit
$4.71 (E / F_{y}) = 4.71 (29000/50) = 113.4 \to e l a s t i c b u c k l in g co n t r o l s$
Formula
$F_{cr} = 0.877 F_{e} = 0.877 (12.46) = 10.92 k s i$
Nominal strength
$P_{n} = F_{cr} A_{g} = 10.92 (12.0) = 131.1 k i p s$
Design strength
$ϕ_{c} P_{n} = 0.90 (131.1) = 118.0 k i p s$

Answer: KL/r = 151.6, F_cr = 10.92 ksi, φ_cP_n = 118.0 kips

Why the other options are there

600.0 kips (yield squash load, buckling ignored)
131.1 kips (elastic equation used in the inelastic range)

Reference: FE Reference Handbook — Structural Design → Columns

Example 8

Axial compressive strength of a steel column by AISC flexural buckling — Columns (7)

A 50 ksi steel column has A_g = 24.0 in.², r_min = 3.50 in. and an effective length KL = 23.5 ft. Compute the slenderness ratio, elastic buckling stress, critical stress and the design compressive strength φ_cP_n.

Given

F_y = 50 ksi, E = 29 000 ksi
A_g = 24.0 in.²
r_min = 3.50 in.
KL = 282.0 in.

Find

KL/r, F_e, F_cr and φ_cP_n

Start with the thinking

The 4.71√(E/F_y) limit separates inelastic (0.658 exponential) from elastic (0.877F_e) buckling.
Always use the smallest radius of gyration — the column buckles about its weak axis.

Step-by-step solution

Formula
$λ = \frac{K L}{r}$
Substituting
$λ = 282.0/3.50 = 80.6$
Formula
$F_{e} = \frac{π ^{2} E}{( K L / r ) ^{2}}$
Substituting
$F_{e} = π^{2} (29000) / (80.6)^{2} = 44.09 k s i$
Limit
$4.71 (E / F_{y}) = 4.71 (29000/50) = 113.4 \to in e l a s t i c b u c k l in g co n t r o l s$
Formula
$F_{cr} = (0.65 8^{F_{y} / F_{e}}) F_{y} = (0.65 8^{1.134}) (50) = 31.10 k s i$
Nominal strength
$P_{n} = F_{cr} A_{g} = 31.10 (24.0) = 746.5 k i p s$
Design strength
$ϕ_{c} P_{n} = 0.90 (746.5) = 671.9 k i p s$

Answer: KL/r = 80.6, F_cr = 31.10 ksi, φ_cP_n = 671.9 kips

Why the other options are there

1,200 kips (yield squash load, buckling ignored)
928.0 kips (elastic equation used in the inelastic range)

Reference: FE Reference Handbook — Structural Design → Columns

Example 9

Axial compressive strength of a steel column by AISC flexural buckling — Columns (8)

A 50 ksi steel column has A_g = 14.5 in.², r_min = 1.80 in. and an effective length KL = 17.5 ft. Compute the slenderness ratio, elastic buckling stress, critical stress and the design compressive strength φ_cP_n.

Given

F_y = 50 ksi, E = 29 000 ksi
A_g = 14.5 in.²
r_min = 1.80 in.
KL = 210.0 in.

Find

KL/r, F_e, F_cr and φ_cP_n

Start with the thinking

The 4.71√(E/F_y) limit separates inelastic (0.658 exponential) from elastic (0.877F_e) buckling.
Always use the smallest radius of gyration — the column buckles about its weak axis.

Step-by-step solution

Formula
$λ = \frac{K L}{r}$
Substituting
$λ = 210.0/1.80 = 116.7$
Formula
$F_{e} = \frac{π ^{2} E}{( K L / r ) ^{2}}$
Substituting
$F_{e} = π^{2} (29000) / (116.7)^{2} = 21.03 k s i$
Limit
$4.71 (E / F_{y}) = 4.71 (29000/50) = 113.4 \to e l a s t i c b u c k l in g co n t r o l s$
Formula
$F_{cr} = 0.877 F_{e} = 0.877 (21.03) = 18.44 k s i$
Nominal strength
$P_{n} = F_{cr} A_{g} = 18.44 (14.5) = 267.4 k i p s$
Design strength
$ϕ_{c} P_{n} = 0.90 (267.4) = 240.7 k i p s$

Answer: KL/r = 116.7, F_cr = 18.44 ksi, φ_cP_n = 240.7 kips

Why the other options are there

725.0 kips (yield squash load, buckling ignored)
267.4 kips (elastic equation used in the inelastic range)

Reference: FE Reference Handbook — Structural Design → Columns

Example 10

Axial compressive strength of a steel column by AISC flexural buckling — Columns (9)

A 36 ksi steel column has A_g = 21.0 in.², r_min = 1.90 in. and an effective length KL = 19.0 ft. Compute the slenderness ratio, elastic buckling stress, critical stress and the design compressive strength φ_cP_n.

Given

F_y = 36 ksi, E = 29 000 ksi
A_g = 21.0 in.²
r_min = 1.90 in.
KL = 228.0 in.

Find

KL/r, F_e, F_cr and φ_cP_n

Start with the thinking

The 4.71√(E/F_y) limit separates inelastic (0.658 exponential) from elastic (0.877F_e) buckling.
Always use the smallest radius of gyration — the column buckles about its weak axis.

Step-by-step solution

Formula
$λ = \frac{K L}{r}$
Substituting
$λ = 228.0/1.90 = 120.0$
Formula
$F_{e} = \frac{π ^{2} E}{( K L / r ) ^{2}}$
Substituting
$F_{e} = π^{2} (29000) / (120.0)^{2} = 19.88 k s i$
Limit
$4.71 (E / F_{y}) = 4.71 (29000/36) = 133.7 \to in e l a s t i c b u c k l in g co n t r o l s$
Formula
$F_{cr} = (0.65 8^{F_{y} / F_{e}}) F_{y} = (0.65 8^{1.811}) (36) = 16.87 k s i$
Nominal strength
$P_{n} = F_{cr} A_{g} = 16.87 (21.0) = 354.2 k i p s$
Design strength
$ϕ_{c} P_{n} = 0.90 (354.2) = 318.8 k i p s$

Answer: KL/r = 120.0, F_cr = 16.87 ksi, φ_cP_n = 318.8 kips

Why the other options are there

756.0 kips (yield squash load, buckling ignored)
366.1 kips (elastic equation used in the inelastic range)

Reference: FE Reference Handbook — Structural Design → Columns

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

Columns contains 6 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.

Columns

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

The design compressive strength φcPn is determined with φc = 0.90 and Ω = 1.67 for flexural buckling of members withoutNeeds technical review

Pn=FcrAg

(a) When r # 4.71 Fy , Fcr =Needs technical review

(b)Whenr>4.71Fy,Fcr=0.877Fe

Lc=KL=effectivelengthofmember(in.)

Fe=elasticbucklingstress=π2​E/(KL/r)2

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section