Load Combinations using Allowable Stress Design (ASD)
Structural Design · FE Reference Handbook section
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.
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
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
Combination 2
Combination 3
Compare — governing value is 180.3 psf from 1.2D + 1.6L + 0.5Lr
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)
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
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
Combination 1
Combination 2
Combination 3
Governing — 127.0 psf from D + L
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)
A floor beam carries dead load 1.30 kip/ft and live load 1.30 kip/ft. Determine the governing factored load.
Given
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
Combination 1
Combination 2
Governing
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)
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
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
Combination 2
Combination 3
Governing
Formula — φR_n ≥ R_u → R_n ≥ R_u/φ
Substituting
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)
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
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
Combination 2
Combination 3
Compare — governing value is 253.2 psf from 1.2D + 1.6L + 0.5Lr
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)
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
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
Combination 1
Combination 2
Combination 3
Governing — 165.0 psf from D + L
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)
A floor beam carries dead load 0.60 kip/ft and live load 2.10 kip/ft. Determine the governing factored load.
Given
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
Combination 1
Combination 2
Governing
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)
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
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
Combination 2
Combination 3
Governing
Formula — φR_n ≥ R_u → R_n ≥ R_u/φ
Substituting
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)
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
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
Combination 2
Combination 3
Compare — governing value is 223.9 psf from 1.2D + 1.6L + 0.5Lr
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)
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
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
Combination 1
Combination 2
Combination 3
Governing — 134.8 psf from D + 0.75L + 0.75S
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)