2

LRFD Design Philosophy and Load Combinations

01

Engineering story

A century of steel — from concept to skyline

Engineer reviewing blueprints against a steel-frame construction site at sunrise

Every design check starts with φRn ≥ Ru.

ASCE 7 combines dead, live, roof-live, snow, rain, wind and seismic loads through seven strength-level combinations to bound the worst-case demand.

Chapter 2 covers the LRFD philosophy — factored loads on the demand side, resistance factors on the capacity side — and walks through the seven ASCE 7 strength combinations plus the 0.9D counter-cases for wind/seismic uplift.

Iconic steel structures built on engineering excellence
  1. AISC ASD (1st ed.)
    1923
  2. AISC LRFD 1st edition
    1986
  3. ASCE 7 unified combos
    2002
  4. AISC 360-22 + ASCE 7-22 (current)
    2022
Load pathOccupancy (D+L)Environment (Lr, S, R, W, E)ASCE 7 combosFactored RuφRn check
02

Learning objectives

What you will be able to do after finishing Chapter 1 — and why each objective matters in practice

Master φRn ≥ RuObjective 01

Master φRn ≥ Ru

The LRFD design inequality applies at every limit state (yielding, rupture, buckling, connection failure).

Why it matters
Once you internalise φRn ≥ Ru, every AISC check reduces to plugging in φ, Rn, and Ru.
Where it is used
Every strength check in AISC 360-22.
Connects to
AISC 360-22 B3.1.
Apply the seven ASCE 7 combosObjective 02

Apply the seven ASCE 7 combos

1.4D · 1.2D+1.6L+0.5(Lr/S/R) · 1.2D+1.6(Lr/S/R)+(L or 0.5W) · 1.2D+1.0W+L+0.5(Lr/S/R) · 1.2D+1.0E+L+0.2S · 0.9D+1.0W · 0.9D+1.0E.

Why it matters
One combo controls each demand; miss one and you may under-design for uplift.
Where it is used
Every gravity, wind, and seismic design.
Connects to
ASCE 7-22 §2.3.
Identify the governing comboObjective 03

Identify the governing combo

Compute Ru from each applicable combo and take the maximum. For uplift, use 0.9D + 1.0W or 0.9D + 1.0E.

Why it matters
The governing combo is not always Combo 2 — snow, wind, and seismic each control in different regions.
Where it is used
Foundation uplift, roof design, façade attachments.
Connects to
ASCE 7-22 §2.3.1.
Check uplift (0.9D + 1.0W or 1.0E)Objective 04

Check uplift (0.9D + 1.0W or 1.0E)

In hurricanes and earthquakes, dead load resists uplift. 0.9D bounds the case where D is minimum and W/E is maximum.

Why it matters
Neglecting the 0.9 factor is the classic light-frame roof-lift-off failure.
Where it is used
Roof diaphragms, foundation anchors, light-gauge steel.
Connects to
ASCE 7-22 §2.3.6.
Contrast LRFD with ASDObjective 05

Contrast LRFD with ASD

ASD uses Rn/Ω ≥ Ra with service loads and a single safety factor; LRFD splits into load and resistance factors calibrated by probability.

Why it matters
The two methods produce different reserve capacities on the same section.
Where it is used
AISC still permits ASD; some legacy calcs mix.
Connects to
AISC 360-22 B3.2.
Use unfactored loads for serviceabilityObjective 06

Use unfactored loads for serviceability

Deflection and vibration checks use D + L (no load factors) because they are elastic-response limits, not strength limits.

Why it matters
Applying 1.2D + 1.6L to a deflection check gives absurd deflections.
Where it is used
L/360 live-load deflection, L/240 total deflection.
Connects to
AISC Manual Part 3, IBC serviceability.
03

Engineering motivation

What each part of a steel-frame building actually does — and why it exists

Before you design any single member, you have to see the whole system. A steel-frame building is not a collection of independent shapes bolted together — it is a deliberate load path, engineered so that every kilonewton of gravity, wind, or seismic demand has a continuous route from where it starts to the ground where the earth can resist it.

The photograph below shows a typical steel framing detail. Drag each labelled chip onto the structural element it names — the drop is only accepted when it lands inside the correct element's outlined region. A correct answer locks in with a green outline and a short explanation; a wrong answer flashes the region red, tells you what you actually hit, and returns the chip so you can try again. Press Reveal expected placements to see the reference solution (that attempt is then marked as assisted).

Structural steel framing — identify each element by dragging the labels
Fig. 1.3 · Drop a chip inside the outlined element it names. Green = correct and locked; red flash = wrong element, chip returns.
04

Failure mechanisms

Why we design the way we do — six ways steel structures have failed, and what each disaster taught the profession

Every provision in AISC 360 is a scar. Behind each equation, load factor, and detailing rule is a bridge, a walkway, or a tower whose failure cost lives and rewrote the profession. The six case studies below trace the mechanisms that motivate the code you are about to learn.

Read each one as an engineer, not a spectator: identify the load, the limit state, the missing check, and the specific clause that exists today because that check was missed. When you meet those clauses again in Chapters 5–17, they will read as answers, not rules.

Snow-load overload: Roof snow exceeded design pg and the joists collapsed.
Case 01
Fig. 1.4.1 · Snow-load overload
Failure mechanism

Snow-load overload

Roof snow exceeded design pg and the joists collapsed.

Root cause

Ru from 1.2D+1.6(S) not envelope-checked against local Pg.

Lesson learned
Always use site-specific ground snow and importance factor.
§ASCE 7-22 §7
Wind uplift roof-off: Warehouse roof lifted off; anchors under-designed for 0.9D + 1.0W uplift.
Case 02
Fig. 1.4.2 · Wind uplift roof-off
Failure mechanism

Wind uplift roof-off

Warehouse roof lifted off; anchors under-designed for 0.9D + 1.0W uplift.

Root cause

Uplift combo not run; only gravity Ru checked.

Lesson learned
Run 0.9D+1.0W anywhere W creates uplift.
§ASCE 7-22 §2.3.6
Wrong combo governs: Engineer used 1.2D+1.6L when 1.2D+1.0W+L governed for a tall façade column.
Case 03
Fig. 1.4.3 · Wrong combo governs
Failure mechanism

Wrong combo governs

Engineer used 1.2D+1.6L when 1.2D+1.0W+L governed for a tall façade column.

Root cause

Only ran gravity combos.

Lesson learned
Run every applicable ASCE 7 combo; identify the maximum.
§ASCE 7-22 §2.3.1
How failure propagates
The five-stage failure progression
1Service loads
2Factored via ASCE 7
3Envelope max Ru
4Compare to φRn
5Pass/fail decision

Design codes intervene at the transition from yield to instability. Everything before yield is elastic and reversible; everything after instability is a race to collapse. LRFD keeps the demand well below the first transition.

05

Lecture notes

The full textbook chapter — figures, equations, and engineering narrative

Reflection · Think before you read

Why do we multiply loads by factors greater than 1.0 and resistances by factors less than 1.0 instead of using a single global safety factor? What real-world uncertainty is each φ and γ trying to capture?

Chapter 2 — LRFD Design Philosophy & ASCE 7-22 Load Combinations

Chapter focus. Every structural engineer's job — whether the project is a warehouse, a hospital, or a highway bridge — is to make sure the structure carries every load it will ever see with a defensible margin of safety, without over-spending on steel. Designing means balancing five competing considerations at once: safety (nobody gets hurt when the crowd, wind, or earthquake arrives), serviceability (no cracked drywall, no bouncing floors, no doors that jam), economy (the lightest section that still works), constructability (fabricators and erectors can build it), and durability (it survives 50–100 years of weather and use). Safety governs — collapses like Hyatt Regency (Ch 1) are the reason design codes exist and the reason licensure is required to sign drawings.

Two code-legal methods produce those safe designs from the same AISC steel spec:
  • LRFD (Load and Resistance Factor Design) — factored strength method. Increase the loads by load factors γ (e.g. 1.2 D + 1.6 L), reduce the calculated capacity by a resistance factor φ ≤ 1, and check Ru ≤ φRn. Each factor is calibrated separately to real statistical data (mill reports, occupancy surveys, weather records).
  • ASD (Allowable Stress Design) — service-load method. Keep the loads unfactored (D + L), divide the capacity by a single overall safety factor Ω ≥ 1 (typically 1.5–2.0), and check Ra ≤ Rn/Ω. Historically the default; still permitted for hand checks and retrofits.
This chapter builds those definitions into a full design method: probabilistic basis, the LRFD inequality, ASCE 7-22 load combinations, and the φ table you will reference in Chapters 4–19.

1. Why LRFD, Not ASD?

AISC 360-22 publishes both methods side by side, and both are code-legal. This course teaches LRFD because every modern U.S. building, bridge, and industrial project is designed that way. Here is the direct comparison.

ASD vs. LRFD design inequalities
(1)
IssueASD (Allowable Stress)LRFD (Load & Resistance Factor)
How is safety applied?One number Ω hides all uncertainty in the capacity side.γ scales up each type of load (D vs. L vs. W); φ scales down each type of resistance (yielding vs. rupture vs. bolt).
Does it recognize that dead load is more predictable than live load?No — D and L are added at their service values, treated equally.Yes — γD = 1.2 (small, D is well-known) vs. γL = 1.6 (larger, L is more variable).
Does it recognize a brittle failure needs more safety than a ductile one?No — same Ω regardless of failure mode.Yes — φyielding = 0.90 (ductile) vs. φrupture = 0.75 (brittle).
Is it calibrated to a target reliability?Empirical — inherited from 1920s working-stress practice.Yes — β ≈ 3.0 (members), β ≈ 4.5 (connections). Every φ and γ is back-solved from data.
Economy on live-load-dominated members?Conservative — treats highly variable L the same as reliable D.Sharper — larger γL is offset by removing hidden conservatism, so members are typically 5–15% lighter.
Consistency with ACI 318 and ASCE 7?Poor — ACI abandoned working-stress design in 1971. Load combinations differ from ASCE 7 §2.3.Full — LRFD combinations in ASCE 7 §2.3 match ACI 318 strength design. One load path for the whole building.
Seismic and wind design?Requires special ASD combinations (§2.4) that reintroduce factors on E and W anyway.Native — 1.0W and 1.0E already at strength-level in the LRFD combos.
Which one is the standard in industry today?Used for quick hand checks, retrofits of pre-2005 buildings, and some petrochemical clients.The default for essentially all new commercial, institutional, and infrastructure work in the U.S.

Bridge between methods. On the same section, the two safety factors are related by . Plug in the two most common resistance factors:

Ω from φ — bridge equation
(2)

When the live-to-dead ratio is around 3 (a typical office), LRFD and ASD pick essentially the same W-shape — but LRFD tells you why that shape works.

Bottom line. LRFD is a rational, probabilistically calibrated, code-aligned method. ASD is a legacy method preserved for continuity. If you can only master one — master LRFD.

2. Probabilistic Basis

Reliability model — load effect Q and resistance R normal distributions, showing safe region, failure region, reliability index β, and probability of failure Pf
LRFD calibrates φ and γ so the failure-region overlap area gives a target reliability index β ≈ 3.0 (members) and β ≈ 4.5 (connections).

Plain-language version. Nothing in a real building is a single fixed number. Think of a typical office floor beam:

  • The load Q is uncertain. The dead load depends on how thick the concrete slab actually gets poured, what ceiling and MEP the tenant installs, and whether partitions are moved. The live load depends on the occupants — a light office one year could become a filing archive the next. We don't know these numbers exactly; we know a range. Statisticians represent that range as the blue "Load Q" bell curve.
  • The resistance R is also uncertain. Mill-rolled A992 steel is specified as Fy = 50 ksi, but a coupon test typically measures 55–58 ksi. The web thickness of a W18×35 is nominally 0.300 in but ships anywhere from 0.290 to 0.315 in. Fabrication tolerances shift the true capacity a few percent. That range is the green "Resistance R" bell curve.
  • Failure is the overlap. A beam fails when a specific load draw is bigger than that beam's specific capacity — i.e., a value from the right tail of Q lands past a value in the left tail of R. That is the shaded overlap region in the figure.

What LRFD actually does. The code picks the load factors γ (which push the design load up to the right tail of Q) and the resistance factors φ (which pull the design capacity down to the left tail of R) so the two tails almost never overlap. The remaining probability of failure is expressed as a reliability index β. Larger β = safer.

(3)

for members  ⇒  probability of failure — about 1 in 1,000 over the 50-year design life.

for connections  ⇒  — about 3 in 1,000,000.

How is obtained. Define the safety margin . Under the LRFD reliability model both and are treated as normal random variables, so is also normal with mean and standard deviation . Failure means , i.e. resistance is exhausted by the load effect: where is the standard normal CDF. AISC back-solves the and factors so that the resulting hits the target above for every limit state in the Specification.

Why connections get a higher β (and therefore a smaller φ). A single beam has redundancy — the slab, adjacent beams, and neighboring bays can help redistribute load if it starts to yield. A connection has no redundancy: if the bolts shear or the weld cracks, the beam falls. So AISC calibrates φbolt = 0.75 and φrupture = 0.75, versus φyielding = 0.90 and φbending = 0.90 for members. The Hyatt Regency 1981 and FIU 2018 collapses (Chapter 1) are exactly why this asymmetry exists.

Quick numerical feel — W18×35 floor beam (Fy = 50 ksi, Zx = 66.5 in³). A concrete example of how LRFD absorbs both material and load surprises without over-designing.
Nominal moment capacity
(4)
Actual mill data (mean Fy ≈ 55 ksi, +2% Zx)
(5)
Factored demand — LRFD Combo 2 (MD = 90, ML = 60 kip-ft)
(6)
Factored capacity vs. demand — AISC B3-1 check
(7)

3. The Fundamental LRFD Inequality

ASCE 7-22 cover — Minimum Design Loads and Associated Criteria for Buildings and Other Structures
ASCE 7-22 — the standard that gives us γ (load factors) and every load combination in this chapter

Load and Resistance Factor Design (LRFD) requires that the factored demand Ru not exceed the factored capacity φRn for every relevant limit state:

AISC 360-22 Eq. B3-1 — LRFD design inequality
(8)
  • Ru — required strength from factored load combinations (γDD + γLL + …).
  • Rn — nominal strength computed from AISC provisions using nominal material and section properties.
  • φ — resistance factor (≤ 1.0) that penalizes the calculated capacity for the uncertainty of the limit state.

3A. Loads and Actions

Building cross-section showing dead loads, live loads and environmental loads acting simultaneously
The three families of actions on a building: permanent (dead), variable (live) and environmental (wind, snow, seismic).

An action is anything that causes stress, strain or displacement in a structure. Before a single member can be sized you must know every action, its magnitude, and how likely it is to occur at the same time as the others. ASCE 7-22 is the standard that answers all three questions.

Permanent actions — dead load, D

  • Self-weight of the structural frame, slab and deck; superimposed finishes, ceilings, partitions, cladding and fixed MEP equipment.
  • Computed from geometry and unit weights: steel 490 pcf, normal-weight concrete 150 pcf, lightweight concrete 110–120 pcf.
  • Well known, so its load factor is small (γD = 1.2) — and deliberately reduced to 0.9 in the uplift combinations where dead load is the stabilising action.

Variable actions — live load, L and Lr

  • Occupancy loads from ASCE 7-22 Table 4.3-1: offices 50 psf (plus 15 psf partition allowance), corridors above the first floor 80 psf, classrooms 40 psf, assembly with fixed seats 60 psf, light storage 125 psf.
  • Roof live load Lr covers maintenance access, typically 20 psf reduced by the R1R2 factors of §4.8.2.
  • Live-load reduction (§4.7): L = Lo(0.25 + 15/√(KLLAT)). The larger the tributary area, the less likely it is to be fully loaded everywhere at once. Never reduce below 0.50Lo for members supporting one floor, or 0.40Lo for multiple floors.
  • Highly variable, so γL = 1.6 — a third larger than the dead-load factor.

Environmental actions — S, W, E

  • Snow, S (Ch. 7): pf = 0.7 Ce Ct Is pg, then drift and sliding surcharges at parapets and roof steps — drifts, not the flat load, cause most snow failures.
  • Wind, W (Ch. 26–30): qz = 0.00256 KzKztKdKeV², applied as pressure on the main wind-force resisting system and, at higher local values, on components and cladding. Wind governs uplift on light roofs and drift on tall frames.
  • Seismic, E (Ch. 11–12): V = CsW with Cs = SDS/(R/Ie). Note the inertial nature: seismic demand is proportional to mass, so making a building heavier makes the earthquake worse.
  • Both W and E enter the LRFD combinations at a factor of 1.0 because their mapped values are already strength-level events.

Other actions you must not forget

  • Rain, R: ponding on flat roofs is a progressive instability — deflection admits more water, which increases deflection.
  • Lateral earth and fluid pressure, H and F: basement and retaining walls.
  • Restrained thermal, shrinkage and settlement effects, T: only structurally significant when movement is restrained.
  • Construction and erection loads: often the worst case a member ever sees, before the diaphragm and bracing exist.
Tributary thinking. Loads are given as pressures (psf) but members are designed with line loads (klf) and point loads (kips). w = q × tributary width, and P = q × tributary area. Getting the tributary geometry right is the single most common source of error in the load takeoff.

3B. Design Philosophies and Limit States

Comparison of ultimate limit state collapse and serviceability limit state deflection for a steel beam
Strength limit states protect life; serviceability limit states protect function, finishes and comfort.

A limit state is a condition beyond which the structure no longer fulfils an intended function. Design consists of identifying every applicable limit state and demonstrating that the demand stays on the safe side of each one.

  • Strength (ultimate) limit states — yielding, rupture, buckling, sliding, overturning, connection failure. Checked with factored loads: Ru ≤ φRn. Exceeding one risks lives, so the margin is large.
  • Serviceability limit states — deflection (L/360 live, L/240 total), lateral drift (H/400 to H/500 under wind), floor vibration (AISC Design Guide 11), and permanent cracking or damage to finishes. Checked with unfactored service loads, because these are everyday conditions.

Do not mix the two. Applying factored loads to a deflection check produces an absurdly heavy beam; applying service loads to a strength check is unsafe.

3C. Structural Idealization

A real framed bay converted into an idealized simply supported beam line diagram with a uniform load and bending moment diagram
From a real bay to an analysable model: geometry becomes a line, the floor becomes w, and the connections become supports.

No one analyses a building as it is actually built. Engineers replace it with an idealized model that is simple enough to solve and conservative enough to trust. Four decisions define that model:

  1. Geometry. Members become lines along their centroidal axes; spans are measured centre-to-centre of supports.
  2. Supports and connections. A shear tab is idealized as a pin, a fully welded moment connection as fixed, and a bearing seat as a roller. Real joints are partially restrained — the idealization must err toward the conservative demand.
  3. Loads. Surface pressures become a uniform line load w over the tributary width, and beam reactions become point loads on the girder.
  4. Material behaviour. Linear-elastic analysis for demands, with plastic section capacity for resistance — the standard combination in AISC 360-22.

For the simply supported case in the figure the resulting demands are the two expressions used repeatedly throughout this course:

Simply supported beam under a uniform load
(9)
Sanity rule. Every idealization must be checked against reality. If you model a connection as a pin, the detail you draw must actually be able to rotate; if you model it as fixed, the connection must be designed for the moment your model reports.

4. ASCE 7-22 LRFD Combinations (§2.3.1)

ComboExpressionGoverns when…
11.4 DVery heavy dead load, minimal live/environmental.
21.2 D + 1.6 L + 0.5 (Lr or S or R)Typical office / gravity check.
31.2 D + 1.6 (Lr or S or R) + (1.0 L or 0.5 W)Roof / snow-controlled.
41.2 D + 1.0 W + 1.0 L + 0.5 (Lr or S or R)Wind + gravity.
51.2 D + 1.0 E + 1.0 L + 0.2 SSeismic + gravity.
60.9 D + 1.0 WOverturning / uplift — wind.
70.9 D + 1.0 EOverturning / uplift — seismic.
Companion action: L = 0.5 kip/ft² is used only when unfactored L < 100 psf and the occupancy is not garage/assembly (ASCE 7-22 §2.3.1, Exception).

5. Common Resistance Factors φ

Limit stateφAISC ref
Tension — yielding on Ag0.90§D2(a)
Tension — rupture on Ae0.75§D2(b)
Compression0.90§E1
Flexure0.90§F1
Shear (h/tw ≤ 2.24√E/Fy)1.00§G2.1(a)
Bolt shear / bearing / tearout0.75§J3
Fillet weld0.75§J2.4

6. Safety Factor vs Resistance Factor — Practical Difference

Why safety factors exist at all — the practical story. Nothing you build is ever exactly what the drawings say. Steel comes in slightly heavier or lighter than nominal. Concrete cures a bit weaker on a rainy day. Occupancy loads change when a tenant swaps a call center for a filing archive. Wind gusts arrive from directions the code did not model. Safety factors are the code's way of forcing a margin between what you compute and what could actually happen — without requiring the engineer to become a statistician for every project.

A safety factor is a single scalar > 1 that inflates the demand (or deflates the capacity). It rolls together every source of uncertainty — load variability, material variability, workmanship, analysis simplifications — into one number. Historically Allowable Stress Design (ASD) used one big factor of safety Ω (typically 1.5–2.0) applied to Rn: Ra ≤ Rn/Ω. Every uncertainty was averaged into that single Ω, even though live load is genuinely more variable than dead load and steel is more predictable than concrete.

Resistance factors are what LRFD replaced the single safety factor with. Instead of one number for the whole design, LRFD splits the safety into two families:

Load factor γ (ASCE 7)Resistance factor φ (AISC 360)
What it modifiesThe applied load (D, L, W, E, S, R)The computed capacity (Rn)
Direction> 1.0 — pushes design load up< 1.0 — pulls design capacity down
Tuned toVariability of that load typeVariability & consequence of that failure mode
Examples1.2 D (well-known), 1.6 L (more variable), 1.0 W (mapped design wind)0.90 yielding, 0.90 flexure, 0.75 rupture, 0.75 bolt shear, 0.75 weld
Who publishes itASCE 7 (this book →)AISC 360 (steel), ACI 318 (concrete)

Why splitting matters — a practical example. A member has D = 100 kip, L = 40 kip, Rn = 220 kip. Under the old single-factor ASD approach (Ω = 1.67): capacity allowed = 220/1.67 = 132 kip > 140 kip demand? Fails. Under LRFD: Ru = 1.2·100 + 1.6·40 = 184 kip; φRn = 0.90·220 = 198 kip > 184 kip Passes, with a defensible reliability index β. LRFD "found" capacity because the dead load piece is charged a lighter 1.2× (concrete-weight variability is small) while the live-load piece is charged a heavier 1.6× (occupancy is genuinely uncertain). ASD's one-size-fits-all Ω over-penalized the well-known dead load.

Bottom line for the field. Resistance factors are calibrated, not arbitrary. AISC ran Monte-Carlo simulations against real mill data, real load surveys, and real failure statistics to pick φ so the resulting probability of failure is the same across every limit state. A safety factor Ω (still permitted for ASD in AISC 360 §B3.2 via Ω = 1.5/φ) is derived from φ — the code's math starts with resistance factors, not with a global safety number.

7. Design Workflow

  1. Assemble unfactored loads: D, L, Lr, S, R, W, E (ASCE 7 Chapters 3–13).
  2. Apply live-load reduction where KLL·AT ≥ 400 ft² (§4.7).
  3. Evaluate every combo; take the envelope Ru = max over combos.
  4. Compute nominal strength Rn from AISC 360-22.
  5. Verify φRn ≥ Ru; report utilization ratio Ru/(φRn).

⚠ Common mistakes

  • Adding 1.2D + 1.6L + 1.6Lr — only one of Lr/S/R appears at the 1.6 factor in Combo 2.
  • Forgetting the 0.9D uplift combos for wind and seismic (Combos 6, 7).
  • Using ASD load combinations with LRFD φ-factors — the two systems do not mix.
  • Skipping live-load reduction on interior columns with large tributary areas.

Worked Example 2.1 — Governing LRFD Combination for a Roof Girder

Given

  • Interior roof girder, simply supported, L = 30 ft, tributary width bt = 15 ft.
  • Unfactored uniform loads: D = 25 psf, Lr = 20 psf, S = 30 psf, W = ±15 psf (net uplift/downward), E negligible.
  • No floor live load L (this is a roof).

Required

  • Evaluate every applicable ASCE 7-22 §2.3.1 LRFD load combination and identify the governing factored line load wu.
  • Compute the governing factored midspan moment Mu and end shear Vu.
  • Check whether the wind-uplift combination (0.9 D + 1.0 W) produces net anchor tension.
Roof girder tributary plan and details
Figure 2.1 — Problem statement: interior roof girder tributary plan, span 30 ft × tributary width 15 ft, loads D/L_r/S/W
Simply-supported roof girder — L = 30 ft, tributary width b_t = 15 ft w_u = 1.2D + 1.6S + 0.5W (governs) L = 30 ft V_u = w_u L / 2 V_u M_u = w_u L² / 8 at midspan
Figure 2.1a — Structural layout: simply-supported roof girder carrying its tributary strip

Step 1 — Convert to line loads

Formula
(1)
wD = 25 psf · 15 ft = 375 plf = 0.375 klf
wLr = 20 · 15 = 300 plf = 0.300 klf
wS = 30 · 15 = 450 plf = 0.450 klf
wW = ±15 · 15 = ±225 plf = ±0.225 klf
Step 1 — surface loads × tributary width for D, L_r, S, W
Figure 2.1b — Step 1: convert surface loads (D, L_r, S, W) to line loads using b_t = 15 ft

Step 2 — Evaluate each LRFD combo (ASCE 7-22 §2.3.1)

LRFD combinations (ASCE 7-22 §2.3.1)
(2)
#Combinationwu (klf)
11.4 D1.4·0.375 = 0.525
21.2 D + 1.6 L + 0.5 (Lr or S)  L = 01.2·0.375 + 0.5·0.450 = 0.675
3a1.2 D + 1.6 Lr + 0.5 W (L = 0)1.2·0.375 + 1.6·0.300 + 0.5·0.225 = 1.043
3b1.2 D + 1.6 S + 0.5 W1.2·0.375 + 1.6·0.450 + 0.5·0.225 = 1.283
41.2 D + 1.0 W + 0.5 (Lr or S)1.2·0.375 + 1.0·0.225 + 0.5·0.450 = 0.900
60.9 D + 1.0 W (uplift, use −W)0.9·0.375 − 0.225 = 0.113 (net downward, no uplift)
Step 2 — evaluate each LRFD combination table
Figure 2.1c — Step 2: evaluate every ASCE 7-22 §2.3.1 LRFD combination; Combo 3b governs

Step 3 — Governing gravity demand

Governing: Combo 3b → wu = 1.283 klf
Simply-supported
(3)
Mu = wu L² / 8 = 1.283 · 30² / 8 = 144.4 k-ft
Vu = wu L / 2 = 1.283 · 30 / 2 = 19.2 k
Step 3 — governing gravity demand M_u and V_u
Figure 2.1d — Step 3: governing gravity demand from Combo 3b (M_u = 144.4 k-ft, V_u = 19.2 k)

Step 4 — Wind uplift check (Combo 6: 0.9 D + 1.0 W)

Combo 6 is the wind-uplift check. Use the uplift sign of wind (W acts upward, so it enters with a negative sign against gravity). The net line load is:

Net uplift line load — Combo 6
(4)

The sign convention: uplift occurs if . Here , so gravity from the reduced dead load still exceeds the wind suction — no net uplift.

Step 4 — wind uplift check net line load
Figure 2.1e — Step 4: Combo 6 net line load = +0.1125 klf downward → no net uplift

Step 5 — Anchor uplift reaction

End reaction — Combo 6
(5)
No wind-uplift tension develops in the anchors under Combo 6 for this girder. Anchor design is governed by the seismic or gravity envelope, not by wind uplift.
Step 5 — anchor uplift reaction
Figure 2.1f — Step 5: end reaction R = 1.69 k downward ⇒ 0 kips anchor uplift tension
Reflection: Snow governed here (Combo 3b) because S > Lr and W was modest. In a hurricane-prone region the wind-uplift combo (0.9 D − 1.0 W) frequently governs the anchor design even when the gravity envelope is snow-controlled.

Worked Example 2.2 — Gravity Column Pu Envelope (3-Story Interior Column)

Given

  • Interior gravity column supporting a roof + 2 floors. Tributary bay 20 ft × 25 ft = 500 ft² per level.
  • Unfactored surface loads: floors D = 80 psf, L = 60 psf; roof D = 20 psf, Lr = 20 psf, S = 25 psf.
  • Column self-weight and cladding lumped into D. No seismic or wind on this interior column.

Required

  • Compute the cumulative service axial loads (PD, PL, PLr, PS) at the base of the interior column.
  • Evaluate all applicable ASCE 7-22 §2.3.1 LRFD combinations and determine the governing factored axial demand Pu for design of the column and its base plate.
Architectural floor plan showing interior gravity column tributary bay
Figure 2.2 — Problem statement: architectural floor plan, interior column C1 (W10×49) with 20 ft × 25 ft tributary bay
Interior gravity column — 3-story stack, 20 ft × 25 ft tributary bay D = 80 psf, L = 60 psf per floor, L_r = 20 psf roof Roof Floor 3 Floor 2 A_trib = 20 ft × 25 ft = 500 ft² per level
Figure 2.2a — Schematic: 3-story stack with tributary bay per level

Step 1 — Per-level column reactions (unfactored)

Level reaction
(6)
Roof: PD,r = 20·500 / 1000 = 10.0 k; PLr = 20·500/1000 = 10.0 k; PS = 25·500/1000 = 12.5 k
Each floor: PD,f = 80·500/1000 = 40.0 k; PL,f = 60·500/1000 = 30.0 k

Step 2 — Cumulative service loads at column base

Stack the levels
(7)

Step 3 — Evaluate LRFD combos at the column base

#CombinationPu (k)
11.4 D1.4·90 = 126.0
21.2 D + 1.6 L + 0.5 (Lr or S)1.2·90 + 1.6·60 + 0.5·12.5 = 108 + 96 + 6.25 = 210.3
31.2 D + 1.6 (Lr or S) + 1.0 L1.2·90 + 1.6·12.5 + 1.0·60 = 108 + 20 + 60 = 188.0
Governing combo — AISC 360-22 §B3.1 / ASCE 7-22 §2.3.1
(8)
Design Pu = 210 k governs (Combo 2). This is the axial demand the column and its base plate must resist.

Worked Example 2.3 — Net Wind Uplift on a Light-Frame Roof Anchor

Given

  • Hurricane tie anchoring a wood-plate connection to the wall stud. Anchor spacing s = 4 ft o.c.
  • Tributary strip on each anchor: 25 ft (half span, roof) × 4 ft (spacing) — but check per-anchor basis using strip width b = s = 4 ft along the plate.
  • Roof dead load D = 12 psf; ASCE 7-22 Component & Cladding wind uplift W = 30 psf (upward, negative on the roof).

Required

  • Convert the surface D and W loads to line loads on the roof plate.
  • Apply the wind-uplift LRFD combination (ASCE 7-22 §2.3.1 Combo 6: 0.9 D + 1.0 W) and compute the net uplift line load wu,net.
  • Determine the factored tension force per hurricane tie Tu that must be developed at each anchor.
Hurricane tie connection loads and spacing
Figure 2.3 — Problem statement: hurricane tie connecting rafter to wall stud, anchor spacing 4 ft o.c., tributary strip and net uplift calculation
Light-frame roof anchor wind uplift schematic — problem overview
Figure 2.3a — Problem overview: hurricane tie, anchor spacing s = 4 ft o.c., tributary strip 25 ft × 4 ft = 100 ft², roof D = 12 psf (down), wind uplift W = 30 psf (up).

Step 1 — Convert area pressures to line loads

Take a strip along the wall plate of width equal to the anchor spacing, . The area pressures become line loads on that strip:

Dead-load line load
(9)
Wind-uplift line load
(10)
Convert area pressures to line loads
Figure 2.3-S1 — Area pressures on a 4-ft-wide strip → w_D = 48 plf (down) and w_W = 120 plf (up).

Step 2 — Apply the uplift strength combination (Combo 6: 0.9D + 1.0W)

Taking upward as positive, the net factored line load on the anchor strip is:

Net uplift line load — Combo 6
(11)

ASCE 7-22 strength combination is the appropriate basic combination to examine when dead load stabilizes the structure against wind uplift.

Apply Combo 6 uplift strength combination
Figure 2.3-S2 — Combo 6 net uplift: 1.0·120 − 0.9·48 = 76.8 plf (up).

Step 3 — Force on one interior anchor

Each interior anchor picks up the uplift over its tributary length along the plate, (rafter half-span from ridge to eave):

Per-anchor uplift force
(12)
Per-anchor uplift force
Figure 2.3-S3 — Interior anchor picks up w_{u,net} over L_trib = 25 ft → T_u = 1.92 kips.

Step 4 — Area-based cross-check (equivalent method)

Equivalent area-based check using the anchor tributary area :

Area-based cross-check
(13)
Area-based cross-check
Figure 2.3-S4 — Equivalent area method: p_u = 19.2 psf × 100 ft² = 1,920 lb ✓.

Step 5 — Summary of free-body diagrams

Free-body summary of the three load states — dead load (D), wind uplift (W), and net uplift (Combo 6) — collapsing to the per-anchor tension demand.

Design uplift per interior anchor Tu = 1.92 kips (Combo 6 governs). Select a hurricane tie rated ≥ 1,920 lb factored tension (e.g., Simpson H10A or heavier).
Reflection: Combo 6 (0.9 D + 1.0 W) is the uplift envelope because it deliberately reduces the stabilising dead load by 10%. This is not a safety margin trick — it acknowledges that D can be lower than nominal (e.g., during construction before finishes are installed) when a design-level wind event might strike.
Summary of free-body diagrams
Figure 2.3-S5 — FBD summary: w_D = 48 plf down, w_W = 120 plf up, w_{u,net} = 76.8 plf up → T_u = 1.92 kips per interior anchor.

Worked Example 2.4 — Combined Gravity + Seismic Column Demand

Given

  • Two-story steel moment frame. Bay width Lbay = 20 ft, story height h = 12 ft; hfloor = 12 ft, hroof = 24 ft.
  • Gravity loads on the beams: wD = 1.0 klf (dead), wL = 0.6 klf (floor live), wS = 0.3 klf (roof snow).
  • Equivalent Lateral Force (ASCE 7-22 Ch. 12) design lateral forces: Froof = 25 kips, Ffloor = 40 kips.
  • Redundancy factor ρ = 1.0; design spectral acceleration SDS = 0.30.

Required

  • Assemble the seismic load effect E = Eh + Ev per ASCE 7-22 §12.4.2.
  • Compute the seismic overturning moment MOT and the resulting axial PE in the leeward column from the overturning couple.
  • Apply the ASCE 7-22 §2.3.6 seismic LRFD combination (1.2 D + E + L + 0.2 S) and determine the factored column axial Pu.
  • Compare against the gravity-only LRFD envelope (§2.3.1) and identify the governing combination.
Architectural floor plan and two-story moment frame elevation
Figure 2.4 — Problem statement: architectural floor plan with two-story steel moment frame elevation, 20 ft bay × 12 ft story height, gravity loads D/L/S

Step 1 — Determine the Seismic Load Effect

ASCE/SEI 7-22 §12.4.2 splits the seismic load effect into horizontal and vertical components: E = Eh + Ev, where Eh = ρQE (§12.4.2.1) and Ev = 0.2 SDS D (§12.4.2.2).

Seismic load effect — ASCE 7-22 §12.4.2
(14)
Frame elevation and gravity loading
Figure 2.4a — Frame elevation and gravity loading (Step 1)

Step 2 — Gravity Axial at the Leeward Column

Each column supports one-half of the beam gravity load. Tributary reaction length = Lbay/2 = 10 ft. Dead acts at both levels; live only at the floor; snow only at the roof.

Dead-load axial (both levels)
(15)
Floor live axial
(16)
Roof snow axial
(17)
Frame elevation showing gravity loads
Figure 2.4b — Tributary gravity loads on each column (Step 2)

Step 3 — Seismic Overturning Moment

The ELF procedure (ASCE 7-22 §12.8) delivers lateral forces at each diaphragm. Overturning about the base is MOT = Σ Fi hi.

Overturning moment — ASCE 7-22 §12.8
(18)
Story-shear cross-check
(19)
Overturning moment about base
Figure 2.4c — Overturning moment about the base (Step 3)

Step 4 — Approximate Column Axial from Overturning

For a preliminary hand check, assume the overturning is resisted by an equal compression-tension couple across the two columns.

Overturning couple across the bay
(20)

Leeward column: 54 kips compression; windward column: 54 kips reduced compression (or tension, depending on gravity).

Engineering note: This is an equilibrium approximation for preliminary design. The final axial should come from full frame analysis including beam-column stiffness, joint continuity, base restraint, and second-order (P-Δ) effects.
Axial couple across columns due to overturning
Figure 2.4d — Axial couple across columns (Step 4)

Step 5 — Apply the LRFD Seismic Load Combination

ASCE/SEI 7-22 §2.3.6 governing seismic strength combination: U = 1.2D + E + L + 0.2S. Substituting E = QE + 0.06 D collapses the D coefficient to 1.26.

Seismic LRFD combination — ASCE 7-22 §2.3.6
(21)
Factored axial on the leeward column
(22)
Free body diagram of leeward column base
Figure 2.4e — Free-body of the leeward column base (Step 5)

Step 6 — Compare with the Gravity-Only LRFD Combination

Governing gravity strength combination — ASCE/SEI 7-22 §2.3.1: U = 1.2D + 1.6L + 0.5S.

Gravity-only LRFD — ASCE 7-22 §2.3.1
(23)

Step 7 — Design Comparison

Seismic vs. gravity ratio
(24)
The seismic LRFD load combination governs the design of the leeward column: Pu = 85.8 kips (seismic) vs. Pu,g = 35.1 kips (gravity-only) — a 144% increase over the governing gravity envelope.
ASCE 7-22 code references for the seismic combination
Figure 2.4f — Governing ASCE/SEI 7-22 references (§2.3.1, §2.3.6, §12.4.2)
Reflection: On the windward column, ρPE reverses sign, so the counter combination (0.9 − 0.2 SDS) D + ρE frequently produces net uplift — a tension design case that gravity combos never reveal. Every column of every moment-frame line must be envelope-checked in both directions.

Worked Example 2.5 — ASD vs. LRFD on the Same Floor Beam

Given

  • Simply-supported W16×36 floor beam, L = 24 ft, tributary width 8 ft.
  • Loads: D = 60 psf, L = 80 psf (office). No snow, wind, or seismic on this interior beam.
  • W16×36 nominal moment capacity: Mn = Mp = Fy Zx = 50·64 / 12 = 267 kip-ft (compact, fully braced).
  • AISC 360-22: φb = 0.90 (LRFD), Ωb = 1.67 (ASD).

Required

  • Compute the governing LRFD factored moment Mu and verify the AISC 360-22 Eq. B3-1 inequality Mu ≤ φb Mn.
  • Compute the ASD service moment Ma and verify the §B3.2 inequality Ma ≤ Mnb.
  • Report the demand-to-capacity ratios for both methods and comment on which is more economical for this L/D ratio.
W16x36 simply-supported beam with uniform distributed load
Figure 2.5 — Problem statement: W16×36 simply-supported floor beam, L = 24 ft, tributary width 8 ft, D = 60 psf, L = 80 psf (office)
ASD vs. LRFD — same W-shape beam, same loads ASD Demand R_a = D + L Capacity R_n / Ω Check: R_a ≤ R_n / Ω LRFD Demand R_u = 1.2D + 1.6L Capacity φR_n Check: R_u ≤ φR_n vs.
Figure 2.5a — Side-by-side ASD vs. LRFD inequalities on the same nominal capacity

Step 1 — Line loads

Convert surface load to line load
(25)

Step 2 — LRFD demand and check

LRFD Combo 2 — factored line load
(26)
Factored midspan moment
(27)
LRFD design inequality — AISC 360-22 Eq. B3-1
(28)

Step 3 — ASD demand and check

ASD — service line load, no factoring
(29)
Service midspan moment
(30)
ASD design inequality — AISC 360-22 §B3.2
(31)

Step 4 — Demand-to-capacity ratios (both methods must agree)

Unity checks
(32)
Both methods deliver near-identical utilization (≈ 0.48–0.50) — as intended by the AISC calibration Ω = 1.5/φ. LRFD "wins" here by ~5% because 1.2 D is charged less heavily than the effective 1.5×D that ASD implicitly applies to the well-known dead load piece.
Reflection: When L/D ≈ 3 (typical office), LRFD and ASD are within ~5%. As L/D rises (heavy live loads, warehouses) LRFD becomes noticeably more economical because the 1.6 factor is still less than the 1.67 Ω lumped on everything. As L/D falls (heavy dead — a masonry-clad building), ASD becomes slightly more economical.

Worked Example 2.6 — LRFD Load Distribution from an Inclined Gable Roof to Roof Rafters and Supporting Beams

Given

  • 3-bay × 3-bay steel building; plan 60 ft × 75 ft (three 20-ft transverse bays × three 25-ft longitudinal bays).
  • Gable roof, slope 4:12; ridge runs in the 75-ft direction; eave-to-ridge horizontal run = 30 ft.
  • Rafters spaced s = 5 ft o.c., spanning eave beam → ridge beam. Ridge and eave beams span Lb = 25 ft between columns.
  • Service loads (on horizontal projection): D = 20 psf, Lr = 20 psf, S = 30 psf (balanced), L = 0 psf (unoccupied roof) — ASCE/SEI 7-22.

Required

  • Governing LRFD factored roof pressure qu among ASCE 7-22 combinations 1, 2 and 3.
  • Tributary load, sloped line load, perpendicular/parallel components, reactions, and Mu,max for a typical rafter.
  • Factored line load wu, shear Vu, and moment Mu for one 25-ft segment of the ridge beam and the exterior eave beam.

Step 1 — Roof geometry and framing arrangement

Plan dimensions
(33)
Building geometry and framing plan
Figure 2.6a — Building geometry and framing plan (Step 1)

Step 2 — Service-level area loads on horizontal projection

Roof service loads (unoccupied)
(34)
Loads are specified on the horizontal roof projection, so tributary-area calculations use horizontal dimensions.
Service area loads on horizontal projection
Figure 2.6b — Service area loads on horizontal projection (Step 2)

Step 3 — Inclined rafter length and roof angle

Rise from run and slope 4:12
(35)
Inclined rafter length
(36)
Roof angle and direction cosines
(37)
Roof slope and rafter geometry
Figure 2.6c — Roof slope and rafter geometry (Step 3)

Step 4 — LRFD roof load combinations (ASCE 7-22 §2.3.1 & §2.3.6)

Combination 1 — dead load only
(38)
Combination 2 — roof live load governs
(39)
Combination 3 — snow
(40)
Governing roof pressure
(41)
LRFD roof load combinations table
Figure 2.6d — LRFD roof load combinations (Step 4)

Step 5 — Tributary area and load on one typical rafter

Tributary width and area (horizontal projection)
(42)
Load per foot of horizontal run
(43)
Total factored vertical load on one rafter
(44)
Vertical load per foot of sloped rafter
(45)
Tributary area and load on one rafter
Figure 2.6e — Tributary area and load on one rafter (Step 5)

Step 6 — Resolve the vertical load relative to the inclined rafter

Perpendicular component (causes bending)
(46)
Parallel component (causes axial)
(47)
Distributed load on rafter — perpendicular and parallel components
Figure 2.6f — Distributed load on rafter — perpendicular and parallel components (Step 6)

Step 7 — Rafter reactions and maximum bending moment

Symmetric reactions (simple span)
(48)
Maximum bending moment (using w_⊥ and sloped length)
(49)
Verification — using w_h on horizontal projection
(50)
Free-body diagram of one rafter
Figure 2.6g — Free-body diagram of one rafter (Step 7)

Step 8 — Load on the ridge beam

Concentrated load at each paired-rafter location (both slopes)
(51)
Equivalent uniform load — pair load ÷ spacing
(52)
Check via tributary width (half of each slope)
(53)
Loads on ridge and eave beams
Figure 2.6h — Loads on ridge and eave beams (Step 8)

Step 9 — Ridge-beam forces for one 25-ft bay

Total factored load on one 25-ft segment
(54)
Support reactions and maximum shear
(55)
Maximum bending moment
(56)

Step 10 — Load on the exterior eave beam

Tributary width (one slope only, half of run)
(57)
Uniform line load on the eave beam
(58)
Reactions, shear, and maximum moment — 25-ft simple span
(59)

Step 11 — Summary of calculated loads

Rafter: Wu = 8.40 kips; Rridge = Reave = 4.20 kips; Mu,max = 31.5 kip-ft.
Ridge beam (25-ft bay): wu = 1.68 klf; Vu = 21.0 kips; Mu = 131.25 kip-ft.
Exterior eave beam (25-ft bay): wu = 0.840 klf; Vu = 10.50 kips; Mu = 65.63 kip-ft.
Reflection: Loads on horizontal projection convert to sloped line loads by ws = wh cos θ; the perpendicular component w = ws cos θ = wh cos²θ governs bending. Because Mu = w Lr²/8 = wh Lh²/8 (identically), you may design a symmetric gable rafter on its horizontal projection and still get the correct bending demand. Roof live-load reduction, unbalanced/drift snow, wind uplift (Combo 6), rain-on-snow, and ponding must also be evaluated in real design.
06

Professional practice, safety & ethics

Reliability in practice

Professional practice
  • Load combinations are jurisdictional: confirm the adopted ASCE 7 edition (7-16 vs 7-22) before generating combos — roof live and snow provisions changed.
  • Record the governing combination for each member in the calculation package; reviewers check the envelope, not one load case.
  • Keep ASD and LRFD results separate — mixing them within one member check is a common submittal rejection.
Safety in design & construction
  • φ and Ω exist because loads and resistance are random; do not 'borrow' reserve by rounding demand down.
  • Construction-phase loads (wet concrete, stacked material, pump trucks) frequently exceed service live load — check them explicitly.
  • Flag any member with a demand/capacity ratio above 1.00 immediately; there is no acceptable 'slight' overstress.
Engineering ethics
  • Do not tune load factors or importance factors to make a member work — that transfers risk to the occupant.
  • Disclose to the client when an existing structure is under-capacity, in writing (NSPE II.1.a).
  • Calculations must be reproducible by another engineer; hidden spreadsheet overrides are an ethical failure.
Ironworkers bolting a steel beam connection while tied off at height
Erection safety: OSHA Subpart R fall protection and stable temporary bracing.
Engineers reviewing sealed structural drawings across a conference table
Peer review and the engineer's seal: responsible charge and standard of care.

ABET / licensure link. These points map to ABET Student Outcomes 2 and 4 — engineering design within realistic constraints, and recognition of ethical and professional responsibilities. Expect NCEES FE and PE exam questions on the NSPE Code of Ethics, OSHA construction requirements, and the engineer's standard of care.

07

Cost analysis

Cost of conservatism

Approach
  • Every load factor and φ you apply has a price: quantify what an extra 10% design demand costs in tonnage before adding ‘safety’ informally.
  • Compare governing combinations — designing for a non-governing combo silently inflates the whole frame.
  • Optimization pays off on repeated members (typical bay), not on one-off members.
Worked cost example — Cost of a 10% demand increase across 60 beams
Basis: Typical floor framing bay, 1.2D+1.6L governing
Line itemQtyRateCost
Baseline beam weight
60 beams @ 1,050 lb$0$0
Added steel from +10% demand
3.15 ton$2,700$8,505
Extra shop handling
6 hr$95$570
Design re-check labor
8 hr$145$1,160
Estimated total$10,235

Takeaway. A casual 10% over-design on one typical bay type costs roughly $10k — real money for zero code benefit.

Unit rates are representative US averages for teaching purposes. On a real project, price with current local rates (RSMeans, fabricator quotes, or contractor pricing) and state the estimate date.

08

Animated concepts

Key mechanics visualised — watch the strain profile, stress block, or buckled shape evolve

LRFD load combinations
1.4D28.0 k1.2D + 1.6L72.0 k1.2D + 1.0L + 1.0W69.0 k0.9D + 1.0W33.0 kGoverns: 1.2D + 1.6L (72.0 k)

Bars grow as each ASCE 7 combination is evaluated. The governing combo (orange) drives the demand Ru.

09

Engineering figures

Full-page reference diagrams — the visual vocabulary you will use for the rest of the course

§2.6.1

LRFD in the field

Engineers reviewing factored loads
Fig. 2.1Engineers reviewing factored loads

Design decisions are made against Ru from ASCE 7.

§2.6.2

Lab load-test

Beam under factored load
Fig. 2.2Beam under factored load

Testing Ru at the factored load level validates the LRFD inequality.

§2.6.3

Snow overload

Roof collapse under snow
Fig. 2.3Roof collapse under snow

1.2D + 1.6S envelope missed — collapse.

§2.6.4

Wind uplift

Roof torn off in hurricane
Fig. 2.4Roof torn off in hurricane

0.9D + 1.0W uplift ignored.

10

Worked examples

Full textbook solutions — problem, theory, step-by-step, verification, interpretation

Example 2.1

Governing wu for a roof beam

Snow-controlled Combo 2 governs a cold-climate roof beam.

Problem statement

A roof beam carries D = 0.45 k/ft, Lr = 0.30 k/ft, and S = 0.60 k/ft (no L, W, E). Determine the governing wu per ASCE 7-22.

Snow-loaded roof beam
FIG. 2.1 — Roof beam under D + Lr + S; snow governs the envelope.
D = 0.45 k/ftLr = 0.30 k/ftS = 0.60 k/ft (governs)L = 30 ft (assumed)R_A = wL/2R_B = wL/2Simply-supported roof beam — envelope of ASCE 7 combos
DIMSimply-supported roof beam — dimensioned loads (D, Lr, S) and span; snow controls the envelope.
Given
  • D = 0.45 k/ft
  • Lr = 0.30 k/ft
  • S = 0.60 k/ft
  • L = W = E = 0
Find
  • Governing wu
Assumptions
  • Simply supported roof beam
  • Snow > roof live
Code references
  • ASCE 7-22 §2.3.1
Theory & approach

Run every applicable combo; envelope Ru. For gravity-only, Combos 1–3 usually control.

Step-by-step solution
  1. 1

    Combo 1

    FormulaASCE 7 §2.3.1
    1.4 D
    1.4·0.45 = 0.63 k/ft
  2. 2

    Combo 2

    Formula
    1.2 D + 1.6·max(Lr, S, R) + 0.5·L
    1.2·0.45 + 1.6·0.60 + 0 = 0.54 + 0.96 = 1.50 k/ft
  3. 3

    Combo 3

    Formula
    1.2 D + 1.6·max(Lr,S,R) + 0.5 W
    1.2·0.45 + 1.6·0.60 + 0 = 1.50 k/ft
  4. 4

    Envelope

    wu = max(0.63, 1.50, 1.50) = 1.50 k/ft
Verification

Combo 2 with snow is expected to govern in a cold-climate roof — matches intuition.

Final answer
wu = 1.50 k/ft (Combo 2, snow-controlled).
Common mistakes
  • Using Lr and S together — take the max, not the sum.
  • Skipping Combo 1 when D is small.
References
  • · ASCE 7-22 §2.3.1
11

Guided practice

Compute the governing variables — hints unlock as you need them

A column has service axial forces PD = 200 k (dead), PL = 120 k (live), PW = ±80 k (wind, ± means either sign). Screen the ASCE 7 / AISC LRFD combinations and report (a) the governing compression P_u and (b) the net axial force under wind uplift.

Your turn
Hints
  1. 1.List all seven LRFD combinations (AISC 360 §B2 / ASCE 7 §2.3.1) before you plug in numbers.
12

Independent practice

Solve the chapter's design task — compute each governing variable

Design task

An interior floor beam supports a tributary width of 8 ft over a simple span of 24 ft. Compute the factored uniform load, factored moment, and factored shear.

Given
  • DL = 40 psf (self-weight + finishes)
  • LL = 50 psf
  • Tributary width = 8 ft
  • Span L = 24 ft
Approach
  1. wD = 40·8/1000 = 0.320 klf; wL = 50·8/1000 = 0.400 klf.
  2. LRFD Combo 2: wu = 1.2·wD + 1.6·wL.
  3. For a simply supported UDL: Mu = wu·L²/8, Vu = wu·L/2.
Submit your answer
13

Mini design challenge

Pick the LRFD combination that governs — compression and uplift are separate winners

Brief

Screen ALL seven LRFD combinations for a 3-story column: PD = 320 k, PL = 180 k, PLr = 25 k (roof live), PW = ±90 k, PE = ±60 k (seismic). Identify the governing compression demand AND check for net uplift — this is the outcome for Chapter 2 (loads and combinations).

Requirements
  • Evaluate Combos 1 through 7 (AISC 360-22 §B2, ASCE 7-22 §2.3.1)
  • Report the largest compressive P_u (drives column section)
  • Report the smallest (or most tensile) P_u using 0.9D combos (drives anchor rods / base uplift)
  • Explain which live-load reduction is admissible on the roof term
14

Chapter summary

A mind map of how every concept connects

Graded Chapter Quiz(18 FE-style questions · AISC Manual required)

These questions reference AISC Steel Construction Manual (16th ed.) — sections, equations, and tables are cited explicitly. Use a calculator. Each question offers a clue you may reveal before answering. Submissions are recorded to your account once signed in.

C2-01ASCE 7-22 §2.3.1
1. At a section: D=40, L=60, Lr=15, W=25, E=30 kips. Which LRFD combo controls (max compression)?
C2-02ASCE 7-22 §2.3.1 combo 6
2. For a member with D=20, W=−15 k (uplift possible), check net uplift via which combo?
C2-03ASCE 7-22 Eq. 4.7-1
3. Live load reduction (ASCE 7 §4.7.2) for KLL·AT = 800 ft², Lo = 50 psf:
C2-04ASCE 7-22 §4.7.3
4. Which condition disallows any live-load reduction per ASCE 7?
C2-05AISC 360-22 §D2
5. φ (resistance factor) for tensile yielding on gross section is:
C2-06ASCE 7-22 §2.3.1
6. For an OFFICE floor with D=80 psf, L=50 psf, tributary width 10 ft: factored w (k/ft) using 1.2D+1.6L:
Floor beam · trib 10 ft D+L L = 20 ft
C2-07ASCE 7-22 Eq. 7.3-1
7. Nominal snow load pf per ASCE 7 Eq. 7.3-1 uses:
C2-08ASCE 7-22 §2.3.1
8. An interior column PD=200, PL=140, PLr=15, PW=±80, PE=60 k. Max Pu?
C2-09ASCE 7-22 §4.8
9. For roof live load Lr on a tributary area >600 ft², AT=800, Lo=20 psf, F1 (slope)=1: Lr =?
C2-10ASCE 7-22 §2.3.6
10. The load factor on E in the seismic combo (Combo 6) is:
C2-11AISC 360-22 §D2(b)
11. φRn for tensile rupture on effective net section uses φ =:
C2-12AISC 360-22 Commentary B3
12. In LRFD, the target reliability index β for typical gravity members is approximately:
C2-13ASCE 7-22 §2.3.1
13. For a member D=25, L=0, W=+40 k, find controlling Pu:
C2-14ASCE 7-22 §4.7
14. Office floor plan: 30 ft × 30 ft bays, infill beams at s = 10 ft on center, span L = 30 ft. For the interior beam B2, what is the tributary area AT used for live-load reduction?
Floor plan — interior beam B2 tributary B2 (span L) B1 B3 trib width s
C2-15ASCE 7-22 Eq. 4.7-1
15. Same beam, Lo = 50 psf. Compute the reduced live load L per ASCE 7 Eq. 4.7-1 with KLL·AT = 600 ft².
Floor plan — interior beam B2 tributary B2 (span L) B1 B3 trib width s
C2-16Line load = (surface load) · (tributary width).
16. Convert the nominal surface loads to line loads on B2 using its trib width s = 10 ft. Take D = 85 psf and reduced L = 43 psf.
Floor plan — interior beam B2 tributary B2 (span L) B1 B3 trib width s
C2-17ASCE 7-22 §2.3.1 Combos 1–7
17. Apply LRFD load combos to B2. Which combo governs and what is the factored line load wu?
Floor plan — interior beam B2 tributary B2 (span L) B1 B3 trib width s
C2-18Simple-span statics
18. With wu = 1.71 klf on a simply-supported span L = 30 ft, compute the beam end reaction Ru and midspan moment Mu.
B2 — simply-supported, wu = 1.71 klf wu L = 30 ft

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Q1 — Structural steel braced-frame column base anchors. Service axial loads: D = 45 k (compression), = 15 k (compression), W = ±75 k (overturning). Per ASCE 7-22 basic LRFD combinations, determine the maximum governing design tensile (uplift) force the anchor bolts must resist.

◆◆ MediumASCE 7-22 §2.3.1 Combo 6 (0.9D + 1.0W)