Unified Design Provisions

Structural Design · FE Reference Handbook section

Structural Design

15 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 Unified Design Provisions 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 unified design provisions describes physically and when it applies.
State every one of the 15 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. Unified Design Provisions 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: unified design provisions.
HAER / Library of Congress, public domain

Structural Design — Unified Design Provisions: 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 15 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

c ≤ 0.375 dt	Quantity produced by "c ≤ 0.375 dt" — read its definition and unit from the handbook line directly above the equation.
c ≥ 0.6 dt	Quantity produced by "c ≥ 0.6 dt" — read its definition and unit from the handbook line directly above the equation.
s	Quantity produced by "s= Vs = − Vc" — read its definition and unit from the handbook line directly above the equation.
spacing s	Quantity produced by "spacing s= Av f y d" — read its definition and unit from the handbook line directly above the equation.
2 s	Quantity produced by "2 s = 24"" — 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.

Internal Stress, Strain, and Forces due to Posi�ve Moment Loading
Compressive Strain
εt fs T
Strain Condi�ons
0.003 0.003 0.003
c c
Tension- Transition Compression-
controlled section controlled
section: section:
Smaller of:
Av f y Vu
50b w φ
Required Av f y
0.75 b w fc '
Smaller of:
Smaller of: d
Maximum d 2
permitted
spacing
Smaller of:

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

Flexural capacity of a singly reinforced beam

A beam has b = 12 in., d = 21.5 in., As = 3.16 in² (4-#8), f′c = 4,000 psi and fy = 60,000 psi. Compute φMn.

Given

b = 12 in., d = 21.5 in.
As = 3.16 in²
f′c = 4 ksi, fy = 60 ksi

Find

φMn (kip-ft)

Start with the thinking

Find the compression block depth from force equilibrium first.
Check that the section is tension controlled before using φ = 0.90.

Figure for Flexural capacity of a singly reinforced beam

Step-by-step solution

Equilibrium
$a = A s f y / (0.85 f^{'} c b)$
Substitute
$a = 3.16 (60) / (0.85 \times 4 \times 12) = 189.6/40.8 = 4.65 in$
Nominal moment
$M n = A s f y (d - a /2)$
Substitute
$M n = 3.16 (60) (21.5 - 2.32) = 189.6 (19.18) = 3, 637 k i p - in$
Convert
$M n = 3, 637/12 = 303 k i p - f t$
Strength — φMn = 0.90(303) = 273 kip-ft

Answer: φMn ≈ 273 kip-ft

Why the other options are there

303 kip-ft (φ omitted)
326 kip-ft (a/2 not subtracted)

Reference: FE Reference Handbook — Structural Design — Reinforced concrete flexure

Example 2

One-way shear check

The same 12 × 24 in. beam (d = 21.5 in., f′c = 4,000 psi) carries Vu = 42 kips at the critical section. Is stirrup reinforcement required, and what φVc applies?

Given

b_w = 12 in., d = 21.5 in.
f′c = 4,000 psi
Vu = 42 kips
φ = 0.75

Find

φVc and whether stirrups are required

Start with the thinking

Concrete shear strength uses 2√f′c b_w d in psi units.
Compare Vu with φVc and with φVc/2.

Step-by-step solution

Concrete shear — Vc = 2√f′c b_w d
Root term — √4,000 = 63.2 psi
Substitute
$V c = 2 (63.2) (12) (21.5) = 32, 620 l b = 32.6 k i p s$
Design strength — φVc = 0.75(32.6) = 24.5 kips
Compare — Vu = 42 kips > φVc = 24.5 kips, so stirrups must carry Vs = (42 − 24.5)/0.75 = 23.3 kips

Answer: φVc = 24.5 kips; stirrups required for Vs ≈ 23.3 kips

Why the other options are there

32.6 kips (φ omitted)
No stirrups (Vu compared with Vc rather than φVc)

Reference: FE Reference Handbook — Structural Design — Shear in concrete beams

Example 3

Design axial strength of a tied reinforced concrete column — Unified Design Provisions

A 14 in × 14 in tied column uses 8-#8 bars (A_st = 6.32 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 = 6.32 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
$0.85 (5) (196 - 6.32) = 806.1 k i p s$
Steel term
$60 (6.32) = 379.2 k i p s$
Substituting
$P_{n} = 0.80 [806.1 + 379.2] = 948.3 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(948.3) = 616.4 kips
Steel ratio check
$ρ = 3.22$

Answer: P_n = 948.3 kips, φP_n = 616.4 kips

Why the other options are there

1,212 kips (no 0.80 factor, gross area)
948.3 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 4

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (2)

A 14 in × 14 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 = 196 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
$0.85 (6) (196 - 7.62) = 960.7 k i p s$
Steel term
$60 (7.62) = 457.2 k i p s$
Substituting
$P_{n} = 0.80 [960.7 + 457.2] = 1, 134 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,134) = 737.3 kips
Steel ratio check
$ρ = 3.89$

Answer: P_n = 1,134 kips, φP_n = 737.3 kips

Why the other options are there

1,457 kips (no 0.80 factor, gross area)
1,134 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 5

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (3)

A 20 in × 20 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 = 400 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
$0.85 (6) (400 - 6.32) = 2, 008 k i p s$
Steel term
$60 (6.32) = 379.2 k i p s$
Substituting
$P_{n} = 0.80 [2, 008 + 379.2] = 1, 910 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,910) = 1,241 kips
Steel ratio check
$ρ = 1.58$

Answer: P_n = 1,910 kips, φP_n = 1,241 kips

Why the other options are there

2,419 kips (no 0.80 factor, gross area)
1,910 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 6

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (4)

A 18 in × 18 in tied column uses 4-#8 bars (A_st = 3.16 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 = 3.16 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
$0.85 (5) (324 - 3.16) = 1, 364 k i p s$
Steel term
$60 (3.16) = 189.6 k i p s$
Substituting
$P_{n} = 0.80 [1, 364 + 189.6] = 1, 243 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,243) = 807.6 kips
Steel ratio check
$ρ = 0.98$

Answer: P_n = 1,243 kips, φP_n = 807.6 kips

Why the other options are there

1,567 kips (no 0.80 factor, gross area)
1,243 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 7

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (5)

A 12 in × 12 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 = 144 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
$0.85 (4) (144 - 10.16) = 455.1 k i p s$
Steel term
$60 (10.16) = 609.6 k i p s$
Substituting
$P_{n} = 0.80 [455.1 + 609.6] = 851.7 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(851.7) = 553.6 kips
Steel ratio check
$ρ = 7.06$

Answer: P_n = 851.7 kips, φP_n = 553.6 kips

Why the other options are there

1,099 kips (no 0.80 factor, gross area)
851.7 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 8

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (6)

A 18 in × 18 in tied column uses 6-#9 bars (A_st = 6.00 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 = 6.00 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
$0.85 (4) (324 - 6.00) = 1, 081 k i p s$
Steel term
$60 (6.00) = 360.0 k i p s$
Substituting
$P_{n} = 0.80 [1, 081 + 360.0] = 1, 153 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,153) = 749.4 kips
Steel ratio check
$ρ = 1.85$

Answer: P_n = 1,153 kips, φP_n = 749.4 kips

Why the other options are there

1,462 kips (no 0.80 factor, gross area)
1,153 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 9

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (7)

A 18 in × 18 in tied column uses 4-#9 bars (A_st = 4.00 in²) with f′c = 5 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

Ag = 324 in²
A_st = 4.00 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
$0.85 (5) (324 - 4.00) = 1, 360 k i p s$
Steel term
$60 (4.00) = 240.0 k i p s$
Substituting
$P_{n} = 0.80 [1, 360 + 240.0] = 1, 280 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,280) = 832.0 kips
Steel ratio check
$ρ = 1.23$

Answer: P_n = 1,280 kips, φP_n = 832.0 kips

Why the other options are there

1,617 kips (no 0.80 factor, gross area)
1,280 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

Example 10

Design axial strength of a tied reinforced concrete column — Unified Design Provisions (8)

A 16 in × 16 in tied column uses 4-#8 bars (A_st = 3.16 in²) with f′c = 6 ksi and f_y = 60 ksi. Compute the nominal and design column strength under concentric axial load.

Given

Ag = 256 in²
A_st = 3.16 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
$0.85 (6) (256 - 3.16) = 1, 289 k i p s$
Steel term
$60 (3.16) = 189.6 k i p s$
Substituting
$P_{n} = 0.80 [1, 289 + 189.6] = 1, 183 k i p s$
Formula — φP_n = 0.65 P_n
Substituting — φP_n = 0.65(1,183) = 769.1 kips
Steel ratio check
$ρ = 1.23$

Answer: P_n = 1,183 kips, φP_n = 769.1 kips

Why the other options are there

1,495 kips (no 0.80 factor, gross area)
1,183 kips (φ never applied)

Reference: FE Reference Handbook — Structural Design → Unified Design Provisions

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

Unified Design Provisions contains 15 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.