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LecturesModule 2

LRFD Design Philosophy and Load Combinations

ASCE 7 load combinations, factored loads, and the φRn ≥ Ru inequality.

AISC Reference Box
  • AISC 360-22 — Specification chapter governing this topic
  • AISC Manual 16th Ed. — Design tables and worked examples

Lecture Notes

This module introduces lrfd design philosophy and load combinations. Lecture content here covers the governing physics, LRFD philosophy, and how the relevant AISC 360-22 chapter organizes the limit states.

Instructors can replace this text in Admin Mode. Each section is structured around: (1) behavior, (2) failure modes, (3) AISC limit-state equations, (4) design workflow, (5) detailing requirements.

A short comparison to ASD is included only where the resistance factor / safety factor relationship clarifies the LRFD design check.

Project case study — Cardinal Square — 4-story braced-frame office

Every chapter's worked example is one step in the design of the same building: Plan: 4 bays N–S × 3 bays E–W, each 30 ft × 30 ft. Stories: 4 @ 13 ft (52 ft roof). Composite floor: 4.5 in NW concrete on 3 VLI20 deck. Roof: 1.5 in B-deck + insulation + membrane. Materials: Wide-flange members A992 (Fy = 50 ksi, Fu = 65 ksi). Plates A572 Gr. 50. HSS bracing A500 Gr. C. Bolts A325-N 7/8 in dia. Welds E70XX. Concrete f'c = 4 ksi. Anchor rods F1554 Gr. 36.

Chapter 2 — Loads & LRFD combinations
Controlling load combination at a typical floor filler beam
Demand carried forward
wu = 1.2 D + 1.6 L = 1.2(0.75) + 1.6(0.50) = 0.90 + 0.80 = 1.70 klf Controls over 1.4 D = 1.05 klf.
This chapter contributes
Computes ASCE 7 combos for every level. Reports wu for floor framing, Pu for gravity columns, and lateral V for the CBF bays.
Feeds into next chapter
Filler-beam wu = 1.70 klf carries into Chapters 6 (flexure) and 7 (shear).
Loads (ASCE 7)LRFD CombosRequired RuTrial SectionLimit StatesφRn ≥ Ru?Serviceability
Seven-step LRFD design workflow culminating in φRn ≥ Ru.

Formula Sheet

NameEquationAISC Ref
Basic LRFD inequalityφ Rn ≥ RuAISC 360-22 §B3.1
Combo 11.4 DASCE 7 §2.3.1
Combo 21.2 D + 1.6 L + 0.5 (Lr or S or R)ASCE 7 §2.3.1
Combo 31.2 D + 1.6 (Lr or S or R) + (L or 0.5 W)ASCE 7 §2.3.1
Combo 41.2 D + 1.0 W + L + 0.5 (Lr or S or R)ASCE 7 §2.3.1
Combo 51.2 D + 1.0 E + L + 0.2 SASCE 7 §2.3.1
Combo 60.9 D + 1.0 WASCE 7 §2.3.1
Combo 70.9 D + 1.0 EASCE 7 §2.3.1

Worked Example

LRFD Design Philosophy and Load Combinations

Given
Replace with project-specific given data (loads, geometry, material).
Load combination
Controlling LRFD load combination from ASCE 7.
Required strength
Compute required strength Ru from the controlling combination.
Limit states
  • Limit state 1
  • Limit state 2
AISC reference
AISC 360-22 — applicable chapter
Solution steps
  1. 1. Required strength
    Compute Ru.
  2. 2. Trial section
    Pick a trial from AISC shape tables Instructor should verify with official AISC Manual.
  3. 3. Check each limit state
    Apply φ Rn ≥ Ru for every governing limit state.
  4. 4. Iterate
    Resize until the most economical section satisfies all checks.
Final design decision
Select the lightest section that satisfies all LRFD limit states.
Common mistakes in this example
  • Skipping a limit state
  • Using the wrong φ factor
  • Forgetting serviceability checks

FE-Style Worked Examples (6)

Each example mirrors the NCEES FE Civil Reference Handbook style: brief givens, a labeled figure, AISC section reference, step-by-step numeric solution, and a single boxed answer.

Given
D = 1.5 k/ft, L = 2.0 k/ft, S = 1.2 k/ft (UDL on beam).
AISC Reference
ASCE 7 §2.3.1
Step-by-step solution
  1. Combo 2
    1.2(1.5)+1.6(2.0)+0.5(1.2) = 1.8+3.2+0.6 = 5.6 k/ft
  2. Combo 3
    1.2(1.5)+1.6(1.2)+0.5(2.0) = 1.8+1.92+1.0 = 4.72 k/ft
Answer wu = 5.6 k/ft (Combo 2 governs).
w (kip/ft)LSimply supported beam, UDL w

Textbook — Aghayere & Vigil (2009)(4 worked examples with figures + numerical answers)

Worked examples scanned directly from the CEGR 436 course textbook. Each card shows the original page (figure + full step-by-step solution) and adds an FE-style numerical multiple-choice prompt with answer key.

Chapter summary

Chapters 2 and 3 of the textbook cover the LRFD design philosophy, ASCE 7 load combinations, gravity loads (dead, live, snow, rain, ice), and lateral loads (wind and seismic). Students learn how to compute factored member demand Ru from service-level D, L, S, W, E.

  • LRFD inequality: φRn ≥ Σγi·Qi (AISC §B3.1).
  • ASCE 7 §2.3.1 — seven basic LRFD combinations (1.4D; 1.2D+1.6L+0.5(Lr/S/R); …; 0.9D+1.0W; 0.9D+1.0E).
  • Live-load reduction: L = L₀(0.25 + 15/√(KLL·AT)), 0.5L₀ ≤ L ≤ L₀ for KLL·AT ≥ 400 ft² (ASCE 7 §4.7).
  • Snow load: pf = 0.7·Ce·Ct·Is·pg ; flat-roof pf used for low-slope roofs (ASCE 7 §7).
  • Wind ASD vs LRFD: ASCE 7-22 uses strength-level wind (do NOT multiply by 1.6 again).
Setup
Interior office column supports 4 floors at 30 ft × 25 ft each. L₀ = 50 psf, KLL = 4 for interior columns. Find reduced live load per floor and total column live load.
AISC Reference
ASCE 7-22 §4.7
Numerical practice
Reduced floor live load (closest value)?
Textbook page 55 — Live-load reduction on a column
Aghayere & Vigil (2009), p. 55 — full worked solution & sketch.

FE Practice Bank (3)

Multiple-choice problems pulled from the instructor's CEGR 492 FE Review packet (EasyFEExam © 2025, Steve Efe, PhD). Pick an answer, then click Reveal solution.

P1A. LRFD Load CombinationsEasyFEExam Structural Design – Problem 1
At a critical section the nominal load effects are D = 38 k, L = 54 k, S = 18 k, W = 26 k, E = 40 k. Which LRFD combination governs?
P2A. Live Load ReductionEasyFEExam Structural Design – Problem 2
Office slab panel A is 30 ft × 20 ft (AT = 600 ft²), L₀ = 50 psf, KLL = 1. Reduced live load L?
P3A. Live Load on BeamEasyFEExam Structural Design – Problem 3
6-in NWC slab + interior beam BB′ (sw 0.20 k/ft), span 40 ft, tributary width 10 ft. Slab DL 75 psf, LL 100 psf. Factored w and Mmax (1.2D + 1.6L)?

Practice Problems

  1. [E] Write all seven LRFD combinations from ASCE 7-22 §2.3.1.
  2. [E] State the φ-factor convention (φ < 1) and contrast with ASD Ω (Ω > 1).
  3. [E] Roof beam D = 0.45 k/ft, Lr = 0.30 k/ft. Compute wu from combos 1, 2, 3.
  4. [E] What load combination is used for live-load deflection checks?
  5. [E] Convert ASD Ω = 1.67 to an equivalent φ for tension yielding and explain why they are not interchangeable.
  6. [M] Column loads PD = 320 k, PL = 180 k, PLr = 25 k, PW = ±90 k. Compute Pu from combos 1–6; report max compression and any uplift.
  7. [M] Short beam D = 1.0 k/ft, L = 0. Show Combo 1 (1.4D) governs over Combo 2.
  8. [M] Column with PD = 100 k down and PW = ±150 k. Use Combo 6 to check NET UPLIFT and size anchor rods.
  9. [M] Girder D = 1.2 k/ft, L = 0.8 k/ft, Lr = 0.20 k/ft, S = 0.40 k/ft. Identify controlling combo.
  10. [M] Compare φRn ≥ Ru with Rn/Ω ≥ Ra for a tension rod Pn = 60 k, φt = 0.90, Ωt = 1.67, PD = 12 k, PL = 18 k.
  11. [H] 12-story building, h = 12 ft, seismic E = 0.18W, W = 1500 k tributary. Compute Combo 5 on a perimeter column with PD = 220 k, PL = 110 k, S = 30 k.
  12. [H] Canopy beam D = 0.5 k/ft, Lr = 0.4 k/ft, W = ±1.1 k/ft uplift. Determine if Combo 6 produces net uplift; compute magnitude.
  13. [H] Re-derive Combo 4 starting from the reliability framework; explain why the wind factor is 1.0.
  14. [H] Footing carries PD = 800 k, PL = 400 k, PE = ±300 k. Compute factored bearing for Combos 5 and 7; which governs uplift?
  15. [H] Beam supports D = 1.0 k/ft + L = 2.0 k/ft over 30 ft. Compute Mu and find the lightest W-shape with Zx ≥ Mu/(φFy), Fy = 50 ksi.
Structured Clues
  • Write all seven combinations first, then drop terms that are zero before computing wu.
  • Combo 6 (0.9D + 1.0W) is the uplift check — do it whenever wind is present.
  • Combo 1 (1.4D) usually governs when L < ~0.25D.
Code References
  • AISC 360-22 §B3.1
  • ASCE 7-22 §2.3.1
  • AISC Manual Part 2 — Combinations

Quiz (3 FE-bank + 2 concept)

1. [P1] At a critical section the nominal load effects are D = 38 k, L = 54 k, S = 18 k, W = 26 k, E = 40 k. Which LRFD combination governs?
2. [P2] Office slab panel A is 30 ft × 20 ft (AT = 600 ft²), L₀ = 50 psf, KLL = 1. Reduced live load L?
3. [P3] 6-in NWC slab + interior beam BB′ (sw 0.20 k/ft), span 40 ft, tributary width 10 ft. Slab DL 75 psf, LL 100 psf. Factored w and Mmax (1.2D + 1.6L)?
4. Which AISC 360-22 chapter primarily governs lrfd design philosophy and load combinations?
5. In LRFD, the basic design inequality is:

Common Student Mistakes

  • Mixing ASD and LRFD load combinations in the same problem.
  • Using nominal strength Rn instead of design strength φRn.
  • Forgetting to check every limit state listed in the AISC chapter.

"Professor Explains" Script

Today we're talking about lrfd design philosophy and load combinations. Think of this topic as one step in the LRFD workflow: identify the demand, identify the limit states from the relevant AISC chapter, then check that φ·Rn is at least equal to Ru. We'll walk through the failure modes, the equations, and a worked example. Pay close attention to where the resistance factor changes — that's where students lose points on exams.