Design Column Strength, Tied Columns

Structural Design · FE Reference Handbook section

Structural Design

35 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 Design Column Strength, Tied 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 design column strength, tied columns describes physically and when it applies.
State every one of the 35 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. Design Column Strength, Tied 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: design column strength, tied columns.
HAER / Library of Congress, public domain

Structural Design — Design Column Strength, Tied 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 35 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

zPn	Quantity produced by "zPn = 0.80z [0.85 fc ' _ Ag - Ast i + Ast fy]" — read its definition and unit from the handbook line directly above the equation.
tfyγ	Quantity produced by "tfyγ = 60 ksi h" — read its definition and unit from the handbook line directly above the equation.
Pρg	Quantity produced by "Pρg = 0.01" — read its definition and unit from the handbook line directly above the equation.
0.6	Quantity produced by "0.6 = 1." — read its definition and unit from the handbook line directly above the equation.
ε tt	Quantity produced by "ε tt = 0.002" — read its definition and unit from the handbook line directly above the equation.
f s	Quantity produced by "f s = tf y" — read its definition and unit from the handbook line directly above the equation.
f'c	Quantity produced by "f'c= 4 ksi h" — read its definition and unit from the handbook line directly above the equation.
γh	Quantity produced by "γh" — read its definition and unit from the handbook line directly above the equation.
e/h	Quantity produced by "e/h = 0.1 f y = 60 ksi" — read its definition and unit from the handbook line directly above the equation.
ρg	Quantity produced by "ρg == 0.05" — read its definition and unit from the handbook line directly above the equation.
Pg	Quantity produced by "Pg == 0.04" — read its definition and unit from the handbook line directly above the equation.
2.5 Pgg	Quantity produced by "2.5 Pgg == 0.02" — 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.

Nominal Column Strength Interaction Diagram for Rectangular Section
INTERACTION
Interaction Diagram
DIAGRAM
0.10
0.20
1.6 0.08
1.4 0.06
0.05
0.04
f 'c Ag
0.03
0.02
f 'c Ag
0.8 /f4
12/21 y
0.4 E 0.002
0.2 E 0.005
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65
Pne Pu e
f 'c Ag h f 'c Ag h
GRAPH A.11
Nilson, Arthur H., David Darwin, and Charles W. Dolan, Design of Concrete Structures, 13th ed., McGraw-Hill, 2004.
E–4–60–0.75

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

Design axial strength of a tied reinforced concrete column — Design Column Strength, Tied Columns

A 16 in × 16 in tied column uses 6-#10 bars (A_st = 7.62 in²) with f′c = 5 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

Ag = 256 in²
A_st = 7.62 in²
f′c = 5 ksi
f_y = 60 ksi
φ = 0.65, α = 0.80 (tied)

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_{n} = 0.80 [0.85 f^{'} c (A_{g} - A_{s} t) + f_{y} A_{s} t]$
Concrete term
$0.85 (5) (256 - 7.62) = 1, 056 k i p s$
Steel term
$60 (7.62) = 457.2 k i p s$
Substituting
$P_{n} = 0.80 [1, 056 + 457.2] = 1, 210 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,210) = 786.7 kips
Steel ratio check
$ρ = 2.98$

Answer: P_n = 1,210 kips, φP_n = 786.7 kips

Why the other options are there

1,545 kips (no 0.80 factor, gross area)
1,210 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Design Column Strength, Tied Columns

Example 3

Design axial strength of a tied reinforced concrete column — Design Column Strength, Tied Columns (2)

A 18 in × 18 in tied column uses 6-#10 bars (A_st = 7.62 in²) with f′c = 5 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

Ag = 324 in²
A_st = 7.62 in²
f′c = 5 ksi
f_y = 60 ksi
φ = 0.65, α = 0.80 (tied)

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_{n} = 0.80 [0.85 f^{'} c (A_{g} - A_{s} t) + f_{y} A_{s} t]$
Concrete term
$0.85 (5) (324 - 7.62) = 1, 345 k i p s$
Steel term
$60 (7.62) = 457.2 k i p s$
Substituting
$P_{n} = 0.80 [1, 345 + 457.2] = 1, 441 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,441) = 936.9 kips
Steel ratio check
$ρ = 2.35$

Answer: P_n = 1,441 kips, φP_n = 936.9 kips

Why the other options are there

1,834 kips (no 0.80 factor, gross area)
1,441 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Design Column Strength, Tied Columns

Example 4

Design axial strength of a tied reinforced concrete column — Design Column Strength, Tied Columns (3)

A 18 in × 18 in tied column uses 8-#10 bars (A_st = 10.16 in²) with f′c = 4 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

Ag = 324 in²
A_st = 10.16 in²
f′c = 4 ksi
f_y = 60 ksi
φ = 0.65, α = 0.80 (tied)

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_{n} = 0.80 [0.85 f^{'} c (A_{g} - A_{s} t) + f_{y} A_{s} t]$
Concrete term
$0.85 (4) (324 - 10.16) = 1, 067 k i p s$
Steel term
$60 (10.16) = 609.6 k i p s$
Substituting
$P_{n} = 0.80 [1, 067 + 609.6] = 1, 341 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,341) = 871.9 kips
Steel ratio check
$ρ = 3.14$

Answer: P_n = 1,341 kips, φP_n = 871.9 kips

Why the other options are there

1,711 kips (no 0.80 factor, gross area)
1,341 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Design Column Strength, Tied Columns

Example 5

Design axial strength of a tied reinforced concrete column — Design Column Strength, Tied Columns (4)

A 18 in × 18 in tied column uses 4-#10 bars (A_st = 5.08 in²) with f′c = 5 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

Ag = 324 in²
A_st = 5.08 in²
f′c = 5 ksi
f_y = 60 ksi
φ = 0.65, α = 0.80 (tied)

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_{n} = 0.80 [0.85 f^{'} c (A_{g} - A_{s} t) + f_{y} A_{s} t]$
Concrete term
$0.85 (5) (324 - 5.08) = 1, 355 k i p s$
Steel term
$60 (5.08) = 304.8 k i p s$
Substituting
$P_{n} = 0.80 [1, 355 + 304.8] = 1, 328 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,328) = 863.3 kips
Steel ratio check
$ρ = 1.57$

Answer: P_n = 1,328 kips, φP_n = 863.3 kips

Why the other options are there

1,682 kips (no 0.80 factor, gross area)
1,328 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Design Column Strength, Tied Columns

Example 6

Design axial strength of a tied reinforced concrete column — Design Column Strength, Tied Columns (5)

A 14 in × 14 in tied column uses 8-#9 bars (A_st = 8.00 in²) with f′c = 5 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

Ag = 196 in²
A_st = 8.00 in²
f′c = 5 ksi
f_y = 60 ksi
φ = 0.65, α = 0.80 (tied)

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_{n} = 0.80 [0.85 f^{'} c (A_{g} - A_{s} t) + f_{y} A_{s} t]$
Concrete term
$0.85 (5) (196 - 8.00) = 799.0 k i p s$
Steel term
$60 (8.00) = 480.0 k i p s$
Substituting
$P_{n} = 0.80 [799.0 + 480.0] = 1, 023 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,023) = 665.1 kips
Steel ratio check
$ρ = 4.08$

Answer: P_n = 1,023 kips, φP_n = 665.1 kips

Why the other options are there

1,313 kips (no 0.80 factor, gross area)
1,023 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Design Column Strength, Tied Columns

Example 7

Design axial strength of a tied reinforced concrete column — Design Column Strength, Tied Columns (6)

A 18 in × 18 in tied column uses 8-#8 bars (A_st = 6.32 in²) with f′c = 6 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

Ag = 324 in²
A_st = 6.32 in²
f′c = 6 ksi
f_y = 60 ksi
φ = 0.65, α = 0.80 (tied)

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_{n} = 0.80 [0.85 f^{'} c (A_{g} - A_{s} t) + f_{y} A_{s} t]$
Concrete term
$0.85 (6) (324 - 6.32) = 1, 620 k i p s$
Steel term
$60 (6.32) = 379.2 k i p s$
Substituting
$P_{n} = 0.80 [1, 620 + 379.2] = 1, 599 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,599) = 1,040 kips
Steel ratio check
$ρ = 1.95$

Answer: P_n = 1,599 kips, φP_n = 1,040 kips

Why the other options are there

2,032 kips (no 0.80 factor, gross area)
1,599 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Design Column Strength, Tied Columns

Example 8

Design axial strength of a tied reinforced concrete column — Design Column Strength, Tied Columns (7)

A 18 in × 18 in tied column uses 6-#10 bars (A_st = 7.62 in²) with f′c = 6 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

Ag = 324 in²
A_st = 7.62 in²
f′c = 6 ksi
f_y = 60 ksi
φ = 0.65, α = 0.80 (tied)

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_{n} = 0.80 [0.85 f^{'} c (A_{g} - A_{s} t) + f_{y} A_{s} t]$
Concrete term
$0.85 (6) (324 - 7.62) = 1, 614 k i p s$
Steel term
$60 (7.62) = 457.2 k i p s$
Substituting
$P_{n} = 0.80 [1, 614 + 457.2] = 1, 657 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,657) = 1,077 kips
Steel ratio check
$ρ = 2.35$

Answer: P_n = 1,657 kips, φP_n = 1,077 kips

Why the other options are there

2,110 kips (no 0.80 factor, gross area)
1,657 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Design Column Strength, Tied Columns

Example 9

Design axial strength of a tied reinforced concrete column — Design Column Strength, Tied Columns (8)

A 20 in × 20 in tied column uses 4-#10 bars (A_st = 5.08 in²) with f′c = 5 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

Ag = 400 in²
A_st = 5.08 in²
f′c = 5 ksi
f_y = 60 ksi
φ = 0.65, α = 0.80 (tied)

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_{n} = 0.80 [0.85 f^{'} c (A_{g} - A_{s} t) + f_{y} A_{s} t]$
Concrete term
$0.85 (5) (400 - 5.08) = 1, 678 k i p s$
Steel term
$60 (5.08) = 304.8 k i p s$
Substituting
$P_{n} = 0.80 [1, 678 + 304.8] = 1, 587 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,587) = 1,031 kips
Steel ratio check
$ρ = 1.27$

Answer: P_n = 1,587 kips, φP_n = 1,031 kips

Why the other options are there

2,005 kips (no 0.80 factor, gross area)
1,587 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Design Column Strength, Tied Columns

Example 10

Design axial strength of a tied reinforced concrete column — Design Column Strength, Tied Columns (9)

A 18 in × 18 in tied column uses 8-#10 bars (A_st = 10.16 in²) with f′c = 6 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

Ag = 324 in²
A_st = 10.16 in²
f′c = 6 ksi
f_y = 60 ksi
φ = 0.65, α = 0.80 (tied)

Find

P_n and φP_n for the tied column

Start with the thinking

The 0.80 factor accounts for the accidental eccentricity permitted in a tied column.
Concrete acts on the net area — the steel area must be deducted.

Step-by-step solution

Formula
$P_{n} = 0.80 [0.85 f^{'} c (A_{g} - A_{s} t) + f_{y} A_{s} t]$
Concrete term
$0.85 (6) (324 - 10.16) = 1, 601 k i p s$
Steel term
$60 (10.16) = 609.6 k i p s$
Substituting
$P_{n} = 0.80 [1, 601 + 609.6] = 1, 768 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,768) = 1,149 kips
Steel ratio check
$ρ = 3.14$

Answer: P_n = 1,768 kips, φP_n = 1,149 kips

Why the other options are there

2,262 kips (no 0.80 factor, gross area)
1,768 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Design Column Strength, Tied 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

Design Column Strength, Tied Columns contains 35 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.