Live Load Reduction

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 Live Load Reduction 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 live load reduction 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. Live Load Reduction 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: live load reduction.
HAER / Library of Congress, public domain

Structural Design — Live Load Reduction: 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

L	Quantity produced by "L = L o e 0.25 + o" — read its definition and unit from the handbook line directly above the equation.
Lo	Quantity produced by "Lo = unreduced design live load per ft2 (m2) of area supported by the member" — read its definition and unit from the handbook line directly above the equation.
KLL	Quantity produced by "KLL = live load element factor" — read its definition and unit from the handbook line directly above the equation.
AT	Quantity produced by "AT = tributary area (ft2 or m2)" — 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.

The effect on a building member of nominal occupancy live loads may often be reduced based on the loaded floor area support-
ed by the member. A typical model used for computing reduced live load (as found in ASCE 7 and many building codes) is:
For members supporting one floor
K LL AT
L $ 0.5 L o
For members supporting two or more floors
KLL AT
L $ 0.4 Lo
where

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 live load reduction for a supporting member — Live Load Reduction

A member with KLL = 4 supports a tributary area of 543 ft². The unreduced live load is Lo = 100 psf. Determine the reduced design live load.

Given

Lo = 100 psf
AT = 543 ft²
KLL = 4
L = Lo(0.25 + 15/√(KLL·AT))

Find

Reduced live load L (psf)

Start with the thinking

Reduction is permitted only when KLL·AT ≥ 400 ft².
The reduction may not drop below 40% of Lo for this member type.

Step-by-step solution

Influence area — KLL·AT = 4(543) = 2,172 ft² ≥ 400 ft² → reduction permitted
Reduction equation — L = Lo(0.25 + 15/√(KLL·AT))
Radical — √2,172 = 46.60
Factor
$0.25 + 15/46.60 = 0.572$
Cap at 1.0
$f a c t or u se d = 0.572$
Reduced load
$L = 100 (0.572) = 57.2 p s f$
Floor check — 0.572 ≥ 0.40 → governs as computed

Answer: L ≈ 57.2 psf

Why the other options are there

89.4 psf (KLL omitted from the influence area)
100 psf (reduction not taken although KLL·AT ≥ 400 ft²)

Reference: FE Reference Handbook — Structural Design → Live Load Reduction

Example 2

ASCE 7 live load reduction for a supporting member — Live Load Reduction (2)

A member with KLL = 4 supports a tributary area of 696 ft². The unreduced live load is Lo = 50 psf. Determine the reduced design live load.

Given

Lo = 50 psf
AT = 696 ft²
KLL = 4
L = Lo(0.25 + 15/√(KLL·AT))

Find

Reduced live load L (psf)

Start with the thinking

Reduction is permitted only when KLL·AT ≥ 400 ft².
The reduction may not drop below 40% of Lo for this member type.

Step-by-step solution

Influence area — KLL·AT = 4(696) = 2,784 ft² ≥ 400 ft² → reduction permitted
Reduction equation — L = Lo(0.25 + 15/√(KLL·AT))
Radical — √2,784 = 52.76
Factor
$0.25 + 15/52.76 = 0.534$
Cap at 1.0
$f a c t or u se d = 0.534$
Reduced load
$L = 50 (0.534) = 26.7 p s f$
Floor check — 0.534 ≥ 0.40 → governs as computed

Answer: L ≈ 26.7 psf

Why the other options are there

40.9 psf (KLL omitted from the influence area)
50 psf (reduction not taken although KLL·AT ≥ 400 ft²)

Reference: FE Reference Handbook — Structural Design → Live Load Reduction

Example 3

ASCE 7 live load reduction for a supporting member — Live Load Reduction (3)

A member with KLL = 2 supports a tributary area of 460 ft². The unreduced live load is Lo = 40 psf. Determine the reduced design live load.

Given

Lo = 40 psf
AT = 460 ft²
KLL = 2
L = Lo(0.25 + 15/√(KLL·AT))

Find

Reduced live load L (psf)

Start with the thinking

Reduction is permitted only when KLL·AT ≥ 400 ft².
The reduction may not drop below 50% of Lo for this member type.

Step-by-step solution

Influence area — KLL·AT = 2(460) = 920.0 ft² ≥ 400 ft² → reduction permitted
Reduction equation — L = Lo(0.25 + 15/√(KLL·AT))
Radical
$9 20.0 = 30.33$
Factor
$0.25 + 15/30.33 = 0.745$
Cap at 1.0
$f a c t or u se d = 0.745$
Reduced load
$L = 40 (0.745) = 29.8 p s f$
Floor check — 0.745 ≥ 0.50 → governs as computed

Answer: L ≈ 29.8 psf

Why the other options are there

38.0 psf (KLL omitted from the influence area)
40 psf (reduction not taken although KLL·AT ≥ 400 ft²)

Reference: FE Reference Handbook — Structural Design → Live Load Reduction

Example 4

ASCE 7 live load reduction for a supporting member — Live Load Reduction (4)

A member with KLL = 4 supports a tributary area of 860 ft². The unreduced live load is Lo = 50 psf. Determine the reduced design live load.

Given

Lo = 50 psf
AT = 860 ft²
KLL = 4
L = Lo(0.25 + 15/√(KLL·AT))

Find

Reduced live load L (psf)

Start with the thinking

Reduction is permitted only when KLL·AT ≥ 400 ft².
The reduction may not drop below 40% of Lo for this member type.

Step-by-step solution

Influence area — KLL·AT = 4(860) = 3,440 ft² ≥ 400 ft² → reduction permitted
Reduction equation — L = Lo(0.25 + 15/√(KLL·AT))
Radical — √3,440 = 58.65
Factor
$0.25 + 15/58.65 = 0.506$
Cap at 1.0
$f a c t or u se d = 0.506$
Reduced load
$L = 50 (0.506) = 25.3 p s f$
Floor check — 0.506 ≥ 0.40 → governs as computed

Answer: L ≈ 25.3 psf

Why the other options are there

38.1 psf (KLL omitted from the influence area)
50 psf (reduction not taken although KLL·AT ≥ 400 ft²)

Reference: FE Reference Handbook — Structural Design → Live Load Reduction

Example 5

ASCE 7 live load reduction for a supporting member — Live Load Reduction (5)

A member with KLL = 4 supports a tributary area of 202 ft². The unreduced live load is Lo = 80 psf. Determine the reduced design live load.

Given

Lo = 80 psf
AT = 202 ft²
KLL = 4
L = Lo(0.25 + 15/√(KLL·AT))

Find

Reduced live load L (psf)

Start with the thinking

Reduction is permitted only when KLL·AT ≥ 400 ft².
The reduction may not drop below 40% of Lo for this member type.

Step-by-step solution

Influence area — KLL·AT = 4(202) = 808.0 ft² ≥ 400 ft² → reduction permitted
Reduction equation — L = Lo(0.25 + 15/√(KLL·AT))
Radical
$8 08.0 = 28.43$
Factor
$0.25 + 15/28.43 = 0.778$
Cap at 1.0
$f a c t or u se d = 0.778$
Reduced load
$L = 80 (0.778) = 62.2 p s f$
Floor check — 0.778 ≥ 0.40 → governs as computed

Answer: L ≈ 62.2 psf

Why the other options are there

104.4 psf (KLL omitted from the influence area)
80 psf (reduction not taken although KLL·AT ≥ 400 ft²)

Reference: FE Reference Handbook — Structural Design → Live Load Reduction

Example 6

ASCE 7 live load reduction for a supporting member — Live Load Reduction (6)

A member with KLL = 2 supports a tributary area of 552 ft². The unreduced live load is Lo = 40 psf. Determine the reduced design live load.

Given

Lo = 40 psf
AT = 552 ft²
KLL = 2
L = Lo(0.25 + 15/√(KLL·AT))

Find

Reduced live load L (psf)

Start with the thinking

Reduction is permitted only when KLL·AT ≥ 400 ft².
The reduction may not drop below 50% of Lo for this member type.

Step-by-step solution

Influence area — KLL·AT = 2(552) = 1,104 ft² ≥ 400 ft² → reduction permitted
Reduction equation — L = Lo(0.25 + 15/√(KLL·AT))
Radical — √1,104 = 33.23
Factor
$0.25 + 15/33.23 = 0.701$
Cap at 1.0
$f a c t or u se d = 0.701$
Reduced load
$L = 40 (0.701) = 28.1 p s f$
Floor check — 0.701 ≥ 0.50 → governs as computed

Answer: L ≈ 28.1 psf

Why the other options are there

35.5 psf (KLL omitted from the influence area)
40 psf (reduction not taken although KLL·AT ≥ 400 ft²)

Reference: FE Reference Handbook — Structural Design → Live Load Reduction

Example 7

ASCE 7 live load reduction for a supporting member — Live Load Reduction (7)

A member with KLL = 4 supports a tributary area of 556 ft². The unreduced live load is Lo = 80 psf. Determine the reduced design live load.

Given

Lo = 80 psf
AT = 556 ft²
KLL = 4
L = Lo(0.25 + 15/√(KLL·AT))

Find

Reduced live load L (psf)

Start with the thinking

Reduction is permitted only when KLL·AT ≥ 400 ft².
The reduction may not drop below 40% of Lo for this member type.

Step-by-step solution

Influence area — KLL·AT = 4(556) = 2,224 ft² ≥ 400 ft² → reduction permitted
Reduction equation — L = Lo(0.25 + 15/√(KLL·AT))
Radical — √2,224 = 47.16
Factor
$0.25 + 15/47.16 = 0.568$
Cap at 1.0
$f a c t or u se d = 0.568$
Reduced load
$L = 80 (0.568) = 45.4 p s f$
Floor check — 0.568 ≥ 0.40 → governs as computed

Answer: L ≈ 45.4 psf

Why the other options are there

70.9 psf (KLL omitted from the influence area)
80 psf (reduction not taken although KLL·AT ≥ 400 ft²)

Reference: FE Reference Handbook — Structural Design → Live Load Reduction

Example 8

ASCE 7 live load reduction for a supporting member — Live Load Reduction (8)

A member with KLL = 4 supports a tributary area of 594 ft². The unreduced live load is Lo = 80 psf. Determine the reduced design live load.

Given

Lo = 80 psf
AT = 594 ft²
KLL = 4
L = Lo(0.25 + 15/√(KLL·AT))

Find

Reduced live load L (psf)

Start with the thinking

Reduction is permitted only when KLL·AT ≥ 400 ft².
The reduction may not drop below 40% of Lo for this member type.

Step-by-step solution

Influence area — KLL·AT = 4(594) = 2,376 ft² ≥ 400 ft² → reduction permitted
Reduction equation — L = Lo(0.25 + 15/√(KLL·AT))
Radical — √2,376 = 48.74
Factor
$0.25 + 15/48.74 = 0.558$
Cap at 1.0
$f a c t or u se d = 0.558$
Reduced load
$L = 80 (0.558) = 44.6 p s f$
Floor check — 0.558 ≥ 0.40 → governs as computed

Answer: L ≈ 44.6 psf

Why the other options are there

69.2 psf (KLL omitted from the influence area)
80 psf (reduction not taken although KLL·AT ≥ 400 ft²)

Reference: FE Reference Handbook — Structural Design → Live Load Reduction

Example 9

ASCE 7 live load reduction for a supporting member — Live Load Reduction (9)

A member with KLL = 4 supports a tributary area of 842 ft². The unreduced live load is Lo = 80 psf. Determine the reduced design live load.

Given

Lo = 80 psf
AT = 842 ft²
KLL = 4
L = Lo(0.25 + 15/√(KLL·AT))

Find

Reduced live load L (psf)

Start with the thinking

Reduction is permitted only when KLL·AT ≥ 400 ft².
The reduction may not drop below 40% of Lo for this member type.

Step-by-step solution

Influence area — KLL·AT = 4(842) = 3,368 ft² ≥ 400 ft² → reduction permitted
Reduction equation — L = Lo(0.25 + 15/√(KLL·AT))
Radical — √3,368 = 58.03
Factor
$0.25 + 15/58.03 = 0.508$
Cap at 1.0
$f a c t or u se d = 0.508$
Reduced load
$L = 80 (0.508) = 40.7 p s f$
Floor check — 0.508 ≥ 0.40 → governs as computed

Answer: L ≈ 40.7 psf

Why the other options are there

61.4 psf (KLL omitted from the influence area)
80 psf (reduction not taken although KLL·AT ≥ 400 ft²)

Reference: FE Reference Handbook — Structural Design → Live Load Reduction

Example 10

ASCE 7 live load reduction for a supporting member — Live Load Reduction (10)

A member with KLL = 2 supports a tributary area of 789 ft². The unreduced live load is Lo = 100 psf. Determine the reduced design live load.

Given

Lo = 100 psf
AT = 789 ft²
KLL = 2
L = Lo(0.25 + 15/√(KLL·AT))

Find

Reduced live load L (psf)

Start with the thinking

Reduction is permitted only when KLL·AT ≥ 400 ft².
The reduction may not drop below 50% of Lo for this member type.

Step-by-step solution

Influence area — KLL·AT = 2(789) = 1,578 ft² ≥ 400 ft² → reduction permitted
Reduction equation — L = Lo(0.25 + 15/√(KLL·AT))
Radical — √1,578 = 39.72
Factor
$0.25 + 15/39.72 = 0.628$
Cap at 1.0
$f a c t or u se d = 0.628$
Reduced load
$L = 100 (0.628) = 62.8 p s f$
Floor check — 0.628 ≥ 0.50 → governs as computed

Answer: L ≈ 62.8 psf

Why the other options are there

78.4 psf (KLL omitted from the influence area)
100 psf (reduction not taken although KLL·AT ≥ 400 ft²)

Reference: FE Reference Handbook — Structural Design → Live Load Reduction

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

Live Load Reduction 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.