Live Load Reduction

Structural Design · FE Reference Handbook section

Structural Design

8 formulas

10 exam-style examples

~60 min

All Structural Design lectures

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
For members supporting two or more floors

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

$Lo = 100 psf$
$AT = 543 ft^{2}$
$KLL = 4$
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
$factor used = 0.572$
Reduced load
$L = 100(0.572) = 57.2 psf$
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

$Lo = 50 psf$
$AT = 696 ft^{2}$
$KLL = 4$
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
$factor used = 0.534$
Reduced load
$L = 50(0.534) = 26.7 psf$
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

$Lo = 40 psf$
$AT = 460 ft^{2}$
$KLL = 2$
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
$\sqrt920.0 = 30.33$
Factor
$0.25 + 15/30.33 = 0.745$
Cap at 1.0
$factor used = 0.745$
Reduced load
$L = 40(0.745) = 29.8 psf$
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

$Lo = 50 psf$
$AT = 860 ft^{2}$
$KLL = 4$
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
$factor used = 0.506$
Reduced load
$L = 50(0.506) = 25.3 psf$
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

$Lo = 80 psf$
$AT = 202 ft^{2}$
$KLL = 4$
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
$\sqrt808.0 = 28.43$
Factor
$0.25 + 15/28.43 = 0.778$
Cap at 1.0
$factor used = 0.778$
Reduced load
$L = 80(0.778) = 62.2 psf$
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

$Lo = 40 psf$
$AT = 552 ft^{2}$
$KLL = 2$
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
$factor used = 0.701$
Reduced load
$L = 40(0.701) = 28.1 psf$
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

$Lo = 80 psf$
$AT = 556 ft^{2}$
$KLL = 4$
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
$factor used = 0.568$
Reduced load
$L = 80(0.568) = 45.4 psf$
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

$Lo = 80 psf$
$AT = 594 ft^{2}$
$KLL = 4$
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
$factor used = 0.558$
Reduced load
$L = 80(0.558) = 44.6 psf$
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

$Lo = 80 psf$
$AT = 842 ft^{2}$
$KLL = 4$
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
$factor used = 0.508$
Reduced load
$L = 80(0.508) = 40.7 psf$
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

$Lo = 100 psf$
$AT = 789 ft^{2}$
$KLL = 2$
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
$factor used = 0.628$
Reduced load
$L = 100(0.628) = 62.8 psf$
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