Basic combinations

Structural Design · FE Reference Handbook section

Structural Design

2 formulas

10 exam-style examples

~49 min

All Structural Design lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Nominal loads used in the following combinations

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

Governing LRFD (strength) load combination — Basic combinations

A floor system carries D = 48 psf, L = 80 psf and Lr = 22 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

$D = 48 psf$
$L = 80 psf$
$Lr = 22 psf$

Find

Governing factored load wu

Start with the thinking

Evaluate every applicable combination — the largest result governs.
Companion live load enters at 0.5L or 1.0L depending on the combination.

Step-by-step solution

Combination 1
$1.4D = 1.4(48) = 67.2 psf$
Combination 2
$1.2D + 1.6L + 0.5Lr = 1.2(48) + 1.6(80) + 0.5(22) = 196.6 psf$
Combination 3
$1.2D + 1.6Lr + 1.0L = 1.2(48) + 1.6(22) + 1.0(80) = 172.8 psf$
Compare — governing value is 196.6 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 196.6 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

150.0 psf (service loads added without factors)
67.2 psf (only 1.4D evaluated)

Reference: FE Reference Handbook — Structural Design → Basic combinations

Example 2

Governing LRFD (strength) load combination — Basic combinations (2)

A floor system carries D = 66 psf, L = 97 psf and Lr = 25 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

$D = 66 psf$
$L = 97 psf$
$Lr = 25 psf$

Find

Governing factored load wu

Start with the thinking

Evaluate every applicable combination — the largest result governs.
Companion live load enters at 0.5L or 1.0L depending on the combination.

Step-by-step solution

Combination 1
$1.4D = 1.4(66) = 92.4 psf$
Combination 2
$1.2D + 1.6L + 0.5Lr = 1.2(66) + 1.6(97) + 0.5(25) = 246.9 psf$
Combination 3
$1.2D + 1.6Lr + 1.0L = 1.2(66) + 1.6(25) + 1.0(97) = 216.2 psf$
Compare — governing value is 246.9 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 246.9 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

188.0 psf (service loads added without factors)
92.4 psf (only 1.4D evaluated)

Reference: FE Reference Handbook — Structural Design → Basic combinations

Example 3

Governing LRFD (strength) load combination — Basic combinations (3)

A floor system carries D = 78 psf, L = 64 psf and Lr = 21 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

$D = 78 psf$
$L = 64 psf$
$Lr = 21 psf$

Find

Governing factored load wu

Start with the thinking

Evaluate every applicable combination — the largest result governs.
Companion live load enters at 0.5L or 1.0L depending on the combination.

Step-by-step solution

Combination 1
$1.4D = 1.4(78) = 109.2 psf$
Combination 2
$1.2D + 1.6L + 0.5Lr = 1.2(78) + 1.6(64) + 0.5(21) = 206.5 psf$
Combination 3
$1.2D + 1.6Lr + 1.0L = 1.2(78) + 1.6(21) + 1.0(64) = 191.2 psf$
Compare — governing value is 206.5 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 206.5 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

163.0 psf (service loads added without factors)
109.2 psf (only 1.4D evaluated)

Reference: FE Reference Handbook — Structural Design → Basic combinations

Example 4

Governing LRFD (strength) load combination — Basic combinations (4)

A floor system carries D = 47 psf, L = 100 psf and Lr = 24 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

$D = 47 psf$
$L = 100 psf$
$Lr = 24 psf$

Find

Governing factored load wu

Start with the thinking

Evaluate every applicable combination — the largest result governs.
Companion live load enters at 0.5L or 1.0L depending on the combination.

Step-by-step solution

Combination 1
$1.4D = 1.4(47) = 65.8 psf$
Combination 2
$1.2D + 1.6L + 0.5Lr = 1.2(47) + 1.6(100) + 0.5(24) = 228.4 psf$
Combination 3
$1.2D + 1.6Lr + 1.0L = 1.2(47) + 1.6(24) + 1.0(100) = 194.8 psf$
Compare — governing value is 228.4 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 228.4 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

171.0 psf (service loads added without factors)
65.8 psf (only 1.4D evaluated)

Reference: FE Reference Handbook — Structural Design → Basic combinations

Example 5

Governing LRFD (strength) load combination — Basic combinations (5)

A floor system carries D = 77 psf, L = 64 psf and Lr = 22 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

$D = 77 psf$
$L = 64 psf$
$Lr = 22 psf$

Find

Governing factored load wu

Start with the thinking

Evaluate every applicable combination — the largest result governs.
Companion live load enters at 0.5L or 1.0L depending on the combination.

Step-by-step solution

Combination 1
$1.4D = 1.4(77) = 107.8 psf$
Combination 2
$1.2D + 1.6L + 0.5Lr = 1.2(77) + 1.6(64) + 0.5(22) = 205.8 psf$
Combination 3
$1.2D + 1.6Lr + 1.0L = 1.2(77) + 1.6(22) + 1.0(64) = 191.6 psf$
Compare — governing value is 205.8 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 205.8 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

163.0 psf (service loads added without factors)
107.8 psf (only 1.4D evaluated)

Reference: FE Reference Handbook — Structural Design → Basic combinations

Example 6

Governing LRFD (strength) load combination — Basic combinations (6)

A floor system carries D = 76 psf, L = 99 psf and Lr = 20 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

$D = 76 psf$
$L = 99 psf$
$Lr = 20 psf$

Find

Governing factored load wu

Start with the thinking

Evaluate every applicable combination — the largest result governs.
Companion live load enters at 0.5L or 1.0L depending on the combination.

Step-by-step solution

Combination 1
$1.4D = 1.4(76) = 106.4 psf$
Combination 2
$1.2D + 1.6L + 0.5Lr = 1.2(76) + 1.6(99) + 0.5(20) = 259.6 psf$
Combination 3
$1.2D + 1.6Lr + 1.0L = 1.2(76) + 1.6(20) + 1.0(99) = 222.2 psf$
Compare — governing value is 259.6 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 259.6 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

195.0 psf (service loads added without factors)
106.4 psf (only 1.4D evaluated)

Reference: FE Reference Handbook — Structural Design → Basic combinations

Example 7

Governing LRFD (strength) load combination — Basic combinations (7)

A floor system carries D = 54 psf, L = 73 psf and Lr = 22 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

$D = 54 psf$
$L = 73 psf$
$Lr = 22 psf$

Find

Governing factored load wu

Start with the thinking

Evaluate every applicable combination — the largest result governs.
Companion live load enters at 0.5L or 1.0L depending on the combination.

Step-by-step solution

Combination 1
$1.4D = 1.4(54) = 75.6 psf$
Combination 2
$1.2D + 1.6L + 0.5Lr = 1.2(54) + 1.6(73) + 0.5(22) = 192.6 psf$
Combination 3
$1.2D + 1.6Lr + 1.0L = 1.2(54) + 1.6(22) + 1.0(73) = 173.0 psf$
Compare — governing value is 192.6 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 192.6 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

149.0 psf (service loads added without factors)
75.6 psf (only 1.4D evaluated)

Reference: FE Reference Handbook — Structural Design → Basic combinations

Example 8

Governing LRFD (strength) load combination — Basic combinations (8)

A floor system carries D = 40 psf, L = 61 psf and Lr = 15 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

$D = 40 psf$
$L = 61 psf$
$Lr = 15 psf$

Find

Governing factored load wu

Start with the thinking

Evaluate every applicable combination — the largest result governs.
Companion live load enters at 0.5L or 1.0L depending on the combination.

Step-by-step solution

Combination 1
$1.4D = 1.4(40) = 56.0 psf$
Combination 2
$1.2D + 1.6L + 0.5Lr = 1.2(40) + 1.6(61) + 0.5(15) = 153.1 psf$
Combination 3
$1.2D + 1.6Lr + 1.0L = 1.2(40) + 1.6(15) + 1.0(61) = 133.0 psf$
Compare — governing value is 153.1 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 153.1 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

116.0 psf (service loads added without factors)
56.0 psf (only 1.4D evaluated)

Reference: FE Reference Handbook — Structural Design → Basic combinations

Example 9

Governing LRFD (strength) load combination — Basic combinations (9)

A floor system carries D = 48 psf, L = 57 psf and Lr = 15 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

$D = 48 psf$
$L = 57 psf$
$Lr = 15 psf$

Find

Governing factored load wu

Start with the thinking

Evaluate every applicable combination — the largest result governs.
Companion live load enters at 0.5L or 1.0L depending on the combination.

Step-by-step solution

Combination 1
$1.4D = 1.4(48) = 67.2 psf$
Combination 2
$1.2D + 1.6L + 0.5Lr = 1.2(48) + 1.6(57) + 0.5(15) = 156.3 psf$
Combination 3
$1.2D + 1.6Lr + 1.0L = 1.2(48) + 1.6(15) + 1.0(57) = 138.6 psf$
Compare — governing value is 156.3 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 156.3 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

120.0 psf (service loads added without factors)
67.2 psf (only 1.4D evaluated)

Reference: FE Reference Handbook — Structural Design → Basic combinations

Example 10

Governing LRFD (strength) load combination — Basic combinations (10)

A floor system carries D = 42 psf, L = 54 psf and Lr = 18 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

$D = 42 psf$
$L = 54 psf$
$Lr = 18 psf$

Find

Governing factored load wu

Start with the thinking

Evaluate every applicable combination — the largest result governs.
Companion live load enters at 0.5L or 1.0L depending on the combination.

Step-by-step solution

Combination 1
$1.4D = 1.4(42) = 58.8 psf$
Combination 2
$1.2D + 1.6L + 0.5Lr = 1.2(42) + 1.6(54) + 0.5(18) = 145.8 psf$
Combination 3
$1.2D + 1.6Lr + 1.0L = 1.2(42) + 1.6(18) + 1.0(54) = 133.2 psf$
Compare — governing value is 145.8 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 145.8 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

114.0 psf (service loads added without factors)
58.8 psf (only 1.4D evaluated)

Reference: FE Reference Handbook — Structural Design → Basic combinations