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Load Combinations using Allowable Stress Design (ASD)

Structural Design · FE Reference Handbook section

Structural Design
0 formulas
10 exam-style examples
~45 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.

This section is conceptual; there are no equations to memorise.

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 — Load Combinations using Allowable Stress Design (ASD)

A floor system carries D = 85 psf, L = 43 psf and Lr = 19 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

  • D=85psfD = 85 psf
  • L=43psfL = 43 psf
  • Lr=19psfLr = 19 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

  1. Combination 1

    1.4D=1.4(85)=119.0psf1.4D = 1.4(85) = 119.0 psf
  2. Combination 2

    1.2D+1.6L+0.5Lr=1.2(85)+1.6(43)+0.5(19)=180.3psf1.2D + 1.6L + 0.5Lr = 1.2(85) + 1.6(43) + 0.5(19) = 180.3 psf
  3. Combination 3

    1.2D+1.6Lr+1.0L=1.2(85)+1.6(19)+1.0(43)=175.4psf1.2D + 1.6Lr + 1.0L = 1.2(85) + 1.6(19) + 1.0(43) = 175.4 psf
  4. Compare — governing value is 180.3 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 180.3 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

  • 147.0 psf (service loads added without factors)
  • 119.0 psf (only 1.4D evaluated)

Reference: FE Reference Handbook — Structural Design → Load Combinations using Allowable Stress Design (ASD)

Example 2
Governing ASD load combination — Load Combinations using Allowable Stress Design (ASD)

A roof beam supports D = 32 psf, L = 95 psf and snow S = 27 psf. Evaluate the ASD combinations and find the governing service load.

Given

  • D=32psfD = 32 psf
  • L=95psfL = 95 psf
  • S=27psfS = 27 psf

Find

Governing ASD service load wa

Start with the thinking

  • ASD combinations use unfactored loads with companion factors of 0.75.
  • Do not mix ASD demands with LRFD (φ) capacities.

Step-by-step solution

  1. Combination 1

    D=32.0psfD = 32.0 psf
  2. Combination 2

    D+L=32+95=127.0psfD + L = 32 + 95 = 127.0 psf
  3. Combination 3

    D+0.75L+0.75S=32+0.75(95)+0.75(27)=123.5psfD + 0.75L + 0.75S = 32 + 0.75(95) + 0.75(27) = 123.5 psf
  4. Governing — 127.0 psf from D + L

Answer:

wa ≈ 127.0 psf (D + L governs)

Why the other options are there

  • 190.4 psf (LRFD factors applied in an ASD check)
  • 154.0 psf (0.75 companion factors omitted)

Reference: FE Reference Handbook — Structural Design → Load Combinations using Allowable Stress Design (ASD)

Example 3
Governing LRFD load combination — Load Combinations using Allowable Stress Design (ASD)

A floor beam carries dead load 1.30 kip/ft and live load 1.30 kip/ft. Determine the governing factored load.

Given

  • D=1.30kip/ftD = 1.30 kip/ft
  • L=1.30kip/ftL = 1.30 kip/ft

Find

w_u

Start with the thinking

  • Check 1.4D and 1.2D + 1.6L; the larger governs.
  • Service loads are never used for strength design.

Step-by-step solution

  1. Combination 1

    1.4D=1.4(1.30)=1.820kip/ft1.4D = 1.4(1.30) = 1.820 kip/ft
  2. Combination 2

    1.2D+1.6L=1.2(1.30)+1.6(1.30)=3.640kip/ft1.2D + 1.6L = 1.2(1.30) + 1.6(1.30) = 3.640 kip/ft
  3. Governing

    wu=max⁡=3.640kip/ftw_u = \max = 3.640 kip/ft
Answer:

w_u ≈ 3.64 kip/ft

Why the other options are there

  • 2.60 kip/ft (service load used)
  • 1.82 kip/ft (only the dead-load case checked)

Reference: FE Reference Handbook — Structural Design → Load Combinations using Allowable Stress Design (ASD)

Example 4
ASCE 7-16 load combinations and the required nominal strength — Load Combinations using Allowable Stress Design (ASD)

A floor beam carries D = 2.1, L = 2.2, L_r = 0.8 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=2.1kip/ftD = 2.1 kip/ft
  • L=2.2kip/ftL = 2.2 kip/ft
  • Lr=0.8kip/ftL_r = 0.8 kip/ft
  • W=1.0kip/ftW = 1.0 kip/ft
  • ϕ=0.90\phi = 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

  1. Combination 1

    1.4D=1.4(2.1)=2.940kip/ft1.4D = 1.4(2.1) = 2.940 kip/ft
  2. Combination 2

    1.2D+1.6L+0.5Lr=6.440kip/ft1.2D + 1.6L + 0.5L_r = 6.440 kip/ft
  3. Combination 3

    1.2D+1.0W+1.0L+0.5Lr=6.120kip/ft1.2D + 1.0W + 1.0L + 0.5L_r = 6.120 kip/ft
  4. Governing

    wu=6.440kip/ftw_u = 6.440 kip/ft
  5. Formula — φR_n ≥ R_u → R_n ≥ R_u/φ

  6. Substituting

    Rn=6.440/0.90=7.156kip/ftR_n = 6.440/0.90 = 7.156 kip/ft
Answer:
wu=6.44kip/ftgoverns;requirednominalcapacity=7.16kip/ftw_u = 6.44 kip/ft governs; required nominal capacity = 7.16 kip/ft

Why the other options are there

  • 5.10 kip/ft (service loads, unfactored)
  • 5.80 kip/ft (multiplied by φ instead of dividing)

Reference: FE Reference Handbook — Structural Design → Load Combinations using Allowable Stress Design (ASD)

Example 5
Governing LRFD (strength) load combination — Load Combinations using Allowable Stress Design (ASD) (2)

A floor system carries D = 76 psf, L = 95 psf and Lr = 20 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

  • D=76psfD = 76 psf
  • L=95psfL = 95 psf
  • Lr=20psfLr = 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

  1. Combination 1

    1.4D=1.4(76)=106.4psf1.4D = 1.4(76) = 106.4 psf
  2. Combination 2

    1.2D+1.6L+0.5Lr=1.2(76)+1.6(95)+0.5(20)=253.2psf1.2D + 1.6L + 0.5Lr = 1.2(76) + 1.6(95) + 0.5(20) = 253.2 psf
  3. Combination 3

    1.2D+1.6Lr+1.0L=1.2(76)+1.6(20)+1.0(95)=218.2psf1.2D + 1.6Lr + 1.0L = 1.2(76) + 1.6(20) + 1.0(95) = 218.2 psf
  4. Compare — governing value is 253.2 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 253.2 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

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

Reference: FE Reference Handbook — Structural Design → Load Combinations using Allowable Stress Design (ASD)

Example 6
Governing ASD load combination — Load Combinations using Allowable Stress Design (ASD) (2)

A roof beam supports D = 81 psf, L = 84 psf and snow S = 18 psf. Evaluate the ASD combinations and find the governing service load.

Given

  • D=81psfD = 81 psf
  • L=84psfL = 84 psf
  • S=18psfS = 18 psf

Find

Governing ASD service load wa

Start with the thinking

  • ASD combinations use unfactored loads with companion factors of 0.75.
  • Do not mix ASD demands with LRFD (φ) capacities.

Step-by-step solution

  1. Combination 1

    D=81.0psfD = 81.0 psf
  2. Combination 2

    D+L=81+84=165.0psfD + L = 81 + 84 = 165.0 psf
  3. Combination 3

    D+0.75L+0.75S=81+0.75(84)+0.75(18)=157.5psfD + 0.75L + 0.75S = 81 + 0.75(84) + 0.75(18) = 157.5 psf
  4. Governing — 165.0 psf from D + L

Answer:

wa ≈ 165.0 psf (D + L governs)

Why the other options are there

  • 231.6 psf (LRFD factors applied in an ASD check)
  • 183.0 psf (0.75 companion factors omitted)

Reference: FE Reference Handbook — Structural Design → Load Combinations using Allowable Stress Design (ASD)

Example 7
Governing LRFD load combination — Load Combinations using Allowable Stress Design (ASD) (2)

A floor beam carries dead load 0.60 kip/ft and live load 2.10 kip/ft. Determine the governing factored load.

Given

  • D=0.60kip/ftD = 0.60 kip/ft
  • L=2.10kip/ftL = 2.10 kip/ft

Find

w_u

Start with the thinking

  • Check 1.4D and 1.2D + 1.6L; the larger governs.
  • Service loads are never used for strength design.

Step-by-step solution

  1. Combination 1

    1.4D=1.4(0.60)=0.840kip/ft1.4D = 1.4(0.60) = 0.840 kip/ft
  2. Combination 2

    1.2D+1.6L=1.2(0.60)+1.6(2.10)=4.080kip/ft1.2D + 1.6L = 1.2(0.60) + 1.6(2.10) = 4.080 kip/ft
  3. Governing

    wu=max⁡=4.080kip/ftw_u = \max = 4.080 kip/ft
Answer:

w_u ≈ 4.08 kip/ft

Why the other options are there

  • 2.70 kip/ft (service load used)
  • 0.84 kip/ft (only the dead-load case checked)

Reference: FE Reference Handbook — Structural Design → Load Combinations using Allowable Stress Design (ASD)

Example 8
ASCE 7-16 load combinations and the required nominal strength — Load Combinations using Allowable Stress Design (ASD) (2)

A floor beam carries D = 2.2, L = 1.6, L_r = 0.5 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=2.2kip/ftD = 2.2 kip/ft
  • L=1.6kip/ftL = 1.6 kip/ft
  • Lr=0.5kip/ftL_r = 0.5 kip/ft
  • W=0.3kip/ftW = 0.3 kip/ft
  • ϕ=0.90\phi = 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

  1. Combination 1

    1.4D=1.4(2.2)=3.080kip/ft1.4D = 1.4(2.2) = 3.080 kip/ft
  2. Combination 2

    1.2D+1.6L+0.5Lr=5.450kip/ft1.2D + 1.6L + 0.5L_r = 5.450 kip/ft
  3. Combination 3

    1.2D+1.0W+1.0L+0.5Lr=4.790kip/ft1.2D + 1.0W + 1.0L + 0.5L_r = 4.790 kip/ft
  4. Governing

    wu=5.450kip/ftw_u = 5.450 kip/ft
  5. Formula — φR_n ≥ R_u → R_n ≥ R_u/φ

  6. Substituting

    Rn=5.450/0.90=6.056kip/ftR_n = 5.450/0.90 = 6.056 kip/ft
Answer:
wu=5.45kip/ftgoverns;requirednominalcapacity=6.06kip/ftw_u = 5.45 kip/ft governs; required nominal capacity = 6.06 kip/ft

Why the other options are there

  • 4.30 kip/ft (service loads, unfactored)
  • 4.91 kip/ft (multiplied by φ instead of dividing)

Reference: FE Reference Handbook — Structural Design → Load Combinations using Allowable Stress Design (ASD)

Example 9
Governing LRFD (strength) load combination — Load Combinations using Allowable Stress Design (ASD) (3)

A floor system carries D = 79 psf, L = 76 psf and Lr = 15 psf. Evaluate the basic strength combinations and identify the governing factored load.

Given

  • D=79psfD = 79 psf
  • L=76psfL = 76 psf
  • Lr=15psfLr = 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

  1. Combination 1

    1.4D=1.4(79)=110.6psf1.4D = 1.4(79) = 110.6 psf
  2. Combination 2

    1.2D+1.6L+0.5Lr=1.2(79)+1.6(76)+0.5(15)=223.9psf1.2D + 1.6L + 0.5Lr = 1.2(79) + 1.6(76) + 0.5(15) = 223.9 psf
  3. Combination 3

    1.2D+1.6Lr+1.0L=1.2(79)+1.6(15)+1.0(76)=194.8psf1.2D + 1.6Lr + 1.0L = 1.2(79) + 1.6(15) + 1.0(76) = 194.8 psf
  4. Compare — governing value is 223.9 psf from 1.2D + 1.6L + 0.5Lr

Answer:

wu ≈ 223.9 psf (1.2D + 1.6L + 0.5Lr governs)

Why the other options are there

  • 170.0 psf (service loads added without factors)
  • 110.6 psf (only 1.4D evaluated)

Reference: FE Reference Handbook — Structural Design → Load Combinations using Allowable Stress Design (ASD)

Example 10
Governing ASD load combination — Load Combinations using Allowable Stress Design (ASD) (3)

A roof beam supports D = 89 psf, L = 45 psf and snow S = 16 psf. Evaluate the ASD combinations and find the governing service load.

Given

  • D=89psfD = 89 psf
  • L=45psfL = 45 psf
  • S=16psfS = 16 psf

Find

Governing ASD service load wa

Start with the thinking

  • ASD combinations use unfactored loads with companion factors of 0.75.
  • Do not mix ASD demands with LRFD (φ) capacities.

Step-by-step solution

  1. Combination 1

    D=89.0psfD = 89.0 psf
  2. Combination 2

    D+L=89+45=134.0psfD + L = 89 + 45 = 134.0 psf
  3. Combination 3

    D+0.75L+0.75S=89+0.75(45)+0.75(16)=134.8psfD + 0.75L + 0.75S = 89 + 0.75(45) + 0.75(16) = 134.8 psf
  4. Governing — 134.8 psf from D + 0.75L + 0.75S

Answer:

wa ≈ 134.8 psf (D + 0.75L + 0.75S governs)

Why the other options are there

  • 178.8 psf (LRFD factors applied in an ASD check)
  • 150.0 psf (0.75 companion factors omitted)

Reference: FE Reference Handbook — Structural Design → Load Combinations using Allowable Stress Design (ASD)

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