Limit States and Available Strengths

Structural Design · FE Reference Handbook section

Structural Design

19 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 Limit States and Available Strengths 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 limit states and available strengths describes physically and when it applies.
State every one of the 19 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. Limit States and Available Strengths 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: limit states and available strengths.
HAER / Library of Congress, public domain

Structural Design — Limit States and Available Strengths: 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 19 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

Yielding: φy	Quantity produced by "Yielding: φy = 0.90" — read its definition and unit from the handbook line directly above the equation.
Pn	Quantity produced by "Pn = Fy Ag" — read its definition and unit from the handbook line directly above the equation.
Ω	Quantity produced by "Ω = 1.67" — read its definition and unit from the handbook line directly above the equation.
Rupture: φf	Quantity produced by "Rupture: φf = 0.75" — read its definition and unit from the handbook line directly above the equation.
Block shear: φ	Quantity produced by "Block shear: φ = 0.75" — read its definition and unit from the handbook line directly above the equation.
Ubs	Quantity produced by "Ubs = 1.0 (flat bars and angles)" — read its definition and unit from the handbook line directly above the equation.
Agv	Quantity produced by "Agv = gross area for shear" — read its definition and unit from the handbook line directly above the equation.
Anv	Quantity produced by "Anv = net area for shear" — read its definition and unit from the handbook line directly above the equation.
Ant	Quantity produced by "Ant = net area for tension" — read its definition and unit from the handbook line directly above the equation.
Rn	Quantity produced by "Rn = * u" — read its definition and unit from the handbook line directly above the equation.
φb	Quantity produced by "φb = 0.90" — read its definition and unit from the handbook line directly above the equation.
φv	Quantity produced by "φv = 1.00" — 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.

F [0.6Anv + UbsAnt]
[0.6Fy Agv + Ubs Fu Ant]
Smaller
tf Y
Table 1-1: W Shapes Dimensions
and Properties
d X X
Area Depth Web Flange Axis X-X Axis Y-Y
Shape A d tw bf tf I S r Z I r
2 4 3 3 4
In. In. In. In. In. In. In. In. In. In. In.
W24X68 20.1 23.7 0.415 8.97 0.585 1830 154 9.55 177 70.4 1.87
W24X62 18.2 23.7 0.430 7.04 0.590 1550 131 9.23 153 34.5 1.38
W24X55 16.3 23.6 0.395 7.01 0.505 1350 114 9.11 134 29.1 1.34
W21X73 21.5 21.2 0.455 8.30 0.740 1600 151 8.64 172 70.6 1.81
W21X68 20.0 21.1 0.430 8.27 0.685 1480 140 8.60 160 64.7 1.80
W21X62 18.3 21.0 0.400 8.24 0.615 1330 127 8.54 144 57.5 1.77
W21X55 16.2 20.8 0.375 8.22 0.522 1140 110 8.40 126 48.4 1.73
W21X57 16.7 21.1 0.405 6.56 0.650 1170 111 8.36 129 30.6 1.35
W21X50 14.7 20.8 0.380 6.53 0.535 984 94.5 8.18 110 24.9 1.30
W21X48 14.1 20.6 0.350 8.14 0.430 959 93.0 8.24 107 38.7 1.66
W21X44 13.0 20.7 0.350 6.50 0.450 843 81.6 8.06 95.4 20.7 1.26
W18X71 20.8 18.5 0.495 7.64 0.810 1170 127 7.50 146 60.3 1.70
W18X65 19.1 18.4 0.450 7.59 0.750 1070 117 7.49 133 54.8 1.69

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

ASCE 7-16 load combinations and the required nominal strength — Limit States and Available Strengths

A floor beam carries D = 0.7, L = 1.4, L_r = 0.7 and W = 0.9 kip/ft. Evaluate the governing LRFD load combination and, with a resistance factor φ = 0.90, determine the nominal strength the member must provide.

Given

D = 0.7 kip/ft
L = 1.4 kip/ft
L_r = 0.7 kip/ft
W = 0.9 kip/ft
φ = 0.90

Find

Governing factored load and required nominal strength

Start with the thinking

Every ASCE 7-16 combination must be evaluated; the largest result governs the limit state.
LRFD requires φRn ≥ Ru, so the nominal demand is the factored load divided by φ.

Step-by-step solution

Combination 1
$1.4 D = 1.4 (0.7) = 0.980 k i p / f t$
Combination 2
$1.2 D + 1.6 L + 0.5 L_{r} = 3.430 k i p / f t$
Combination 3
$1.2 D + 1.0 W + 1.0 L + 0.5 L_{r} = 3.490 k i p / f t$
Governing
$w_{u} = 3.490 k i p / f t$
Formula — φR_n ≥ R_u → R_n ≥ R_u/φ
Substituting
$R_{n} = 3.490/0.90 = 3.878 k i p / f t$

Answer: w_u = 3.49 kip/ft governs; required nominal capacity = 3.88 kip/ft

Why the other options are there

2.80 kip/ft (service loads, unfactored)
3.14 kip/ft (multiplied by φ instead of dividing)

Reference: FE Reference Handbook — Structural Design → Limit States and Available Strengths

Example 2

ASCE 7-16 load combinations and the required nominal strength — Limit States and Available Strengths (2)

A floor beam carries D = 2.2, L = 1.3, L_r = 0.7 and W = 0.4 kip/ft. Evaluate the governing LRFD load combination and, with a resistance factor φ = 0.90, determine the nominal strength the member must provide.

Given

D = 2.2 kip/ft
L = 1.3 kip/ft
L_r = 0.7 kip/ft
W = 0.4 kip/ft
φ = 0.90

Find

Governing factored load and required nominal strength

Start with the thinking

Every ASCE 7-16 combination must be evaluated; the largest result governs the limit state.
LRFD requires φRn ≥ Ru, so the nominal demand is the factored load divided by φ.

Step-by-step solution

Combination 1
$1.4 D = 1.4 (2.2) = 3.080 k i p / f t$
Combination 2
$1.2 D + 1.6 L + 0.5 L_{r} = 5.070 k i p / f t$
Combination 3
$1.2 D + 1.0 W + 1.0 L + 0.5 L_{r} = 4.690 k i p / f t$
Governing
$w_{u} = 5.070 k i p / f t$
Formula — φR_n ≥ R_u → R_n ≥ R_u/φ
Substituting
$R_{n} = 5.070/0.90 = 5.633 k i p / f t$

Answer: w_u = 5.07 kip/ft governs; required nominal capacity = 5.63 kip/ft

Why the other options are there

4.20 kip/ft (service loads, unfactored)
4.56 kip/ft (multiplied by φ instead of dividing)

Reference: FE Reference Handbook — Structural Design → Limit States and Available Strengths

Example 3

ASCE 7-16 load combinations and the required nominal strength — Limit States and Available Strengths (3)

A floor beam carries D = 0.7, L = 2.0, L_r = 0.6 and W = 0.8 kip/ft. Evaluate the governing LRFD load combination and, with a resistance factor φ = 0.90, determine the nominal strength the member must provide.

Given

D = 0.7 kip/ft
L = 2.0 kip/ft
L_r = 0.6 kip/ft
W = 0.8 kip/ft
φ = 0.90

Find

Governing factored load and required nominal strength

Start with the thinking

Every ASCE 7-16 combination must be evaluated; the largest result governs the limit state.
LRFD requires φRn ≥ Ru, so the nominal demand is the factored load divided by φ.

Step-by-step solution

Combination 1
$1.4 D = 1.4 (0.7) = 0.980 k i p / f t$
Combination 2
$1.2 D + 1.6 L + 0.5 L_{r} = 4.340 k i p / f t$
Combination 3
$1.2 D + 1.0 W + 1.0 L + 0.5 L_{r} = 3.940 k i p / f t$
Governing
$w_{u} = 4.340 k i p / f t$
Formula — φR_n ≥ R_u → R_n ≥ R_u/φ
Substituting
$R_{n} = 4.340/0.90 = 4.822 k i p / f t$

Answer: w_u = 4.34 kip/ft governs; required nominal capacity = 4.82 kip/ft

Why the other options are there

3.30 kip/ft (service loads, unfactored)
3.91 kip/ft (multiplied by φ instead of dividing)

Reference: FE Reference Handbook — Structural Design → Limit States and Available Strengths

Example 4

ASCE 7-16 load combinations and the required nominal strength — Limit States and Available Strengths (4)

A floor beam carries D = 0.8, L = 3.0, L_r = 0.3 and W = 0.3 kip/ft. Evaluate the governing LRFD load combination and, with a resistance factor φ = 0.90, determine the nominal strength the member must provide.

Given

D = 0.8 kip/ft
L = 3.0 kip/ft
L_r = 0.3 kip/ft
W = 0.3 kip/ft
φ = 0.90

Find

Governing factored load and required nominal strength

Start with the thinking

Every ASCE 7-16 combination must be evaluated; the largest result governs the limit state.
LRFD requires φRn ≥ Ru, so the nominal demand is the factored load divided by φ.

Step-by-step solution

Combination 1
$1.4 D = 1.4 (0.8) = 1.120 k i p / f t$
Combination 2
$1.2 D + 1.6 L + 0.5 L_{r} = 5.910 k i p / f t$
Combination 3
$1.2 D + 1.0 W + 1.0 L + 0.5 L_{r} = 4.410 k i p / f t$
Governing
$w_{u} = 5.910 k i p / f t$
Formula — φR_n ≥ R_u → R_n ≥ R_u/φ
Substituting
$R_{n} = 5.910/0.90 = 6.567 k i p / f t$

Answer: w_u = 5.91 kip/ft governs; required nominal capacity = 6.57 kip/ft

Why the other options are there

4.10 kip/ft (service loads, unfactored)
5.32 kip/ft (multiplied by φ instead of dividing)

Reference: FE Reference Handbook — Structural Design → Limit States and Available Strengths

Example 5

ASCE 7-16 load combinations and the required nominal strength — Limit States and Available Strengths (5)

A floor beam carries D = 1.7, L = 2.7, L_r = 0.4 and W = 1.2 kip/ft. Evaluate the governing LRFD load combination and, with a resistance factor φ = 0.90, determine the nominal strength the member must provide.

Given

D = 1.7 kip/ft
L = 2.7 kip/ft
L_r = 0.4 kip/ft
W = 1.2 kip/ft
φ = 0.90

Find

Governing factored load and required nominal strength

Start with the thinking

Every ASCE 7-16 combination must be evaluated; the largest result governs the limit state.
LRFD requires φRn ≥ Ru, so the nominal demand is the factored load divided by φ.

Step-by-step solution

Combination 1
$1.4 D = 1.4 (1.7) = 2.380 k i p / f t$
Combination 2
$1.2 D + 1.6 L + 0.5 L_{r} = 6.560 k i p / f t$
Combination 3
$1.2 D + 1.0 W + 1.0 L + 0.5 L_{r} = 6.140 k i p / f t$
Governing
$w_{u} = 6.560 k i p / f t$
Formula — φR_n ≥ R_u → R_n ≥ R_u/φ
Substituting
$R_{n} = 6.560/0.90 = 7.289 k i p / f t$

Answer: w_u = 6.56 kip/ft governs; required nominal capacity = 7.29 kip/ft

Why the other options are there

4.80 kip/ft (service loads, unfactored)
5.90 kip/ft (multiplied by φ instead of dividing)

Reference: FE Reference Handbook — Structural Design → Limit States and Available Strengths

Example 6

ASCE 7-16 load combinations and the required nominal strength — Limit States and Available Strengths (6)

A floor beam carries D = 1.0, L = 2.3, L_r = 0.8 and W = 1.5 kip/ft. Evaluate the governing LRFD load combination and, with a resistance factor φ = 0.90, determine the nominal strength the member must provide.

Given

D = 1.0 kip/ft
L = 2.3 kip/ft
L_r = 0.8 kip/ft
W = 1.5 kip/ft
φ = 0.90

Find

Governing factored load and required nominal strength

Start with the thinking

Every ASCE 7-16 combination must be evaluated; the largest result governs the limit state.
LRFD requires φRn ≥ Ru, so the nominal demand is the factored load divided by φ.

Step-by-step solution

Combination 1
$1.4 D = 1.4 (1.0) = 1.400 k i p / f t$
Combination 2
$1.2 D + 1.6 L + 0.5 L_{r} = 5.280 k i p / f t$
Combination 3
$1.2 D + 1.0 W + 1.0 L + 0.5 L_{r} = 5.400 k i p / f t$
Governing
$w_{u} = 5.400 k i p / f t$
Formula — φR_n ≥ R_u → R_n ≥ R_u/φ
Substituting
$R_{n} = 5.400/0.90 = 6.000 k i p / f t$

Answer: w_u = 5.40 kip/ft governs; required nominal capacity = 6.00 kip/ft

Why the other options are there

4.10 kip/ft (service loads, unfactored)
4.86 kip/ft (multiplied by φ instead of dividing)

Reference: FE Reference Handbook — Structural Design → Limit States and Available Strengths

Example 7

ASCE 7-16 load combinations and the required nominal strength — Limit States and Available Strengths (7)

A floor beam carries D = 1.0, L = 2.4, L_r = 0.5 and W = 1.5 kip/ft. Evaluate the governing LRFD load combination and, with a resistance factor φ = 0.90, determine the nominal strength the member must provide.

Given

D = 1.0 kip/ft
L = 2.4 kip/ft
L_r = 0.5 kip/ft
W = 1.5 kip/ft
φ = 0.90

Find

Governing factored load and required nominal strength

Start with the thinking

Every ASCE 7-16 combination must be evaluated; the largest result governs the limit state.
LRFD requires φRn ≥ Ru, so the nominal demand is the factored load divided by φ.

Step-by-step solution

Combination 1
$1.4 D = 1.4 (1.0) = 1.400 k i p / f t$
Combination 2
$1.2 D + 1.6 L + 0.5 L_{r} = 5.290 k i p / f t$
Combination 3
$1.2 D + 1.0 W + 1.0 L + 0.5 L_{r} = 5.350 k i p / f t$
Governing
$w_{u} = 5.350 k i p / f t$
Formula — φR_n ≥ R_u → R_n ≥ R_u/φ
Substituting
$R_{n} = 5.350/0.90 = 5.944 k i p / f t$

Answer: w_u = 5.35 kip/ft governs; required nominal capacity = 5.94 kip/ft

Why the other options are there

3.90 kip/ft (service loads, unfactored)
4.81 kip/ft (multiplied by φ instead of dividing)

Reference: FE Reference Handbook — Structural Design → Limit States and Available Strengths

Example 8

ASCE 7-16 load combinations and the required nominal strength — Limit States and Available Strengths (8)

A floor beam carries D = 1.3, L = 2.1, L_r = 0.8 and W = 1.3 kip/ft. Evaluate the governing LRFD load combination and, with a resistance factor φ = 0.90, determine the nominal strength the member must provide.

Given

D = 1.3 kip/ft
L = 2.1 kip/ft
L_r = 0.8 kip/ft
W = 1.3 kip/ft
φ = 0.90

Find

Governing factored load and required nominal strength

Start with the thinking

Every ASCE 7-16 combination must be evaluated; the largest result governs the limit state.
LRFD requires φRn ≥ Ru, so the nominal demand is the factored load divided by φ.

Step-by-step solution

Combination 1
$1.4 D = 1.4 (1.3) = 1.820 k i p / f t$
Combination 2
$1.2 D + 1.6 L + 0.5 L_{r} = 5.320 k i p / f t$
Combination 3
$1.2 D + 1.0 W + 1.0 L + 0.5 L_{r} = 5.360 k i p / f t$
Governing
$w_{u} = 5.360 k i p / f t$
Formula — φR_n ≥ R_u → R_n ≥ R_u/φ
Substituting
$R_{n} = 5.360/0.90 = 5.956 k i p / f t$

Answer: w_u = 5.36 kip/ft governs; required nominal capacity = 5.96 kip/ft

Why the other options are there

4.20 kip/ft (service loads, unfactored)
4.82 kip/ft (multiplied by φ instead of dividing)

Reference: FE Reference Handbook — Structural Design → Limit States and Available Strengths

Example 9

ASCE 7-16 load combinations and the required nominal strength — Limit States and Available Strengths (9)

A floor beam carries D = 1.7, L = 2.0, L_r = 0.2 and W = 1.0 kip/ft. Evaluate the governing LRFD load combination and, with a resistance factor φ = 0.90, determine the nominal strength the member must provide.

Given

D = 1.7 kip/ft
L = 2.0 kip/ft
L_r = 0.2 kip/ft
W = 1.0 kip/ft
φ = 0.90

Find

Governing factored load and required nominal strength

Start with the thinking

Every ASCE 7-16 combination must be evaluated; the largest result governs the limit state.
LRFD requires φRn ≥ Ru, so the nominal demand is the factored load divided by φ.

Step-by-step solution

Combination 1
$1.4 D = 1.4 (1.7) = 2.380 k i p / f t$
Combination 2
$1.2 D + 1.6 L + 0.5 L_{r} = 5.340 k i p / f t$
Combination 3
$1.2 D + 1.0 W + 1.0 L + 0.5 L_{r} = 5.140 k i p / f t$
Governing
$w_{u} = 5.340 k i p / f t$
Formula — φR_n ≥ R_u → R_n ≥ R_u/φ
Substituting
$R_{n} = 5.340/0.90 = 5.933 k i p / f t$

Answer: w_u = 5.34 kip/ft governs; required nominal capacity = 5.93 kip/ft

Why the other options are there

3.90 kip/ft (service loads, unfactored)
4.81 kip/ft (multiplied by φ instead of dividing)

Reference: FE Reference Handbook — Structural Design → Limit States and Available Strengths

Example 10

ASCE 7-16 load combinations and the required nominal strength — Limit States and Available Strengths (10)

A floor beam carries D = 1.0, L = 2.1, L_r = 0.7 and W = 1.0 kip/ft. Evaluate the governing LRFD load combination and, with a resistance factor φ = 0.90, determine the nominal strength the member must provide.

Given

D = 1.0 kip/ft
L = 2.1 kip/ft
L_r = 0.7 kip/ft
W = 1.0 kip/ft
φ = 0.90

Find

Governing factored load and required nominal strength

Start with the thinking

Every ASCE 7-16 combination must be evaluated; the largest result governs the limit state.
LRFD requires φRn ≥ Ru, so the nominal demand is the factored load divided by φ.

Step-by-step solution

Combination 1
$1.4 D = 1.4 (1.0) = 1.400 k i p / f t$
Combination 2
$1.2 D + 1.6 L + 0.5 L_{r} = 4.910 k i p / f t$
Combination 3
$1.2 D + 1.0 W + 1.0 L + 0.5 L_{r} = 4.650 k i p / f t$
Governing
$w_{u} = 4.910 k i p / f t$
Formula — φR_n ≥ R_u → R_n ≥ R_u/φ
Substituting
$R_{n} = 4.910/0.90 = 5.456 k i p / f t$

Answer: w_u = 4.91 kip/ft governs; required nominal capacity = 5.46 kip/ft

Why the other options are there

3.80 kip/ft (service loads, unfactored)
4.42 kip/ft (multiplied by φ instead of dividing)

Reference: FE Reference Handbook — Structural Design → Limit States and Available Strengths

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

Limit States and Available Strengths contains 19 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.