Engineering story

A century of steel — from concept to skyline

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

Almost every real column is a beam-column.

AISC 360-22 Chapter H closes the loop: axial + biaxial bending in a single interaction equation, with second-order (B1, B2) amplification.

Chapter 8 covers AISC 360-22 Chapter H: the H1-1a / H1-1b interaction equations, second-order amplification B1 (member) and B2 (story sway), the Cm reduction factor, and how to combine them into a code-compliant beam-column check.

Iconic steel structures built on engineering excellence

AISC ASD interaction (fa/Fa + fb/Fb)
1963
LRFD H1 introduced
1986
AISC 360-05 direct analysis method (DAM)
2005
AISC 360-22 Chapter H (current)
2022

Load pathRoof/floor gravityWind momentColumn axial + momentBase plateFoundation

02

Learning objectives

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

Objective 01

Apply H1-1a (Pr/Pc ≥ 0.2)

Pr/Pc + (8/9)(Mrx/Mcx + Mry/Mcy) ≤ 1.0 governs when axial is the primary demand.

Why it matters: Most gravity columns fall here; a single line captures failure.
Where it is used: Interior gravity columns in moment frames.
Connects to: AISC H1.1a.

Objective 02

Apply H1-1b (Pr/Pc < 0.2)

Pr/(2Pc) + (Mrx/Mcx + Mry/Mcy) ≤ 1.0 governs when bending dominates.

Why it matters: Perimeter/moment-frame columns with light axial load.
Where it is used: Perimeter moment-frame columns.
Connects to: AISC H1.1b.

Objective 03

Compute B1 amplifier

B1 = Cm / (1 − α·Pr/Pe1) ≥ 1.0, applied to non-translation moments Mnt.

Why it matters: Captures P-δ effect on member curvature.
Where it is used: Every beam-column when second-order effects are non-negligible.
Connects to: AISC App. 8; C2.1(b).

Objective 04

Compute B2 sway amplifier

B2 = 1 / (1 − α·Pstory/Pe,story) applied to lateral-translation moments Mlt.

Why it matters: Captures P-Δ story sway in moment/braced frames.
Where it is used: Story-level analysis of moment frames.
Connects to: AISC App. 8; C2.1(b).

Objective 05

Select Cm for the load pattern

Cm = 0.6 − 0.4(M1/M2) for members without transverse loads between supports; Cm = 1.0 otherwise (conservative).

Why it matters: Reverse curvature reduces B1; single curvature increases it.
Where it is used: Every B1 calculation.
Connects to: AISC App. 8.2.1.

Objective 06

Read Manual Table 6-2 (φPn, φMnx)

Table 6-2 collects φPn and φMn for common W-shapes — the fastest H1 check.

Why it matters: Turns a full beam-column check into two lookups + one line.
Where it is used: Every gravity/wind column check in practice.
Connects to: AISC Manual Part 6.

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.

Failure mechanism

In-plane interaction collapse

The column reaches the H1-1a envelope — plastic hinge forms as P and M both climb.

Root cause

Pr/Pc + (8/9)(Mr/Mc) exceeds 1.0.

Lesson learned

Either upsize the shape or reduce Mr via bracing / stiffer frame.

§AISC H1.1a

Failure mechanism

Out-of-plane LTB of beam-column

Column buckles laterally out-of-plane before H1-1a is reached because Mcy governs.

Root cause

Lb too long about the weak axis; Mn drops.

Lesson learned

Add weak-axis bracing to raise Mcy and unlock H1.

§AISC H1.3 / F2

Failure mechanism

P-Δ sway instability

Story drift amplifies moments until B2 → ∞.

Root cause

α·Pstory / Pe,story approaches 1.0.

Lesson learned

Increase story lateral stiffness (bracing / larger columns).

§AISC App. 8 / DAM C2

How failure propagates

The five-stage failure progression

1Gravity axial

2Add wind moment

3B1/B2 amplification

4Reach H1 = 1.0

5Plastic hinge / sway collapse

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

A column in a moment frame carries axial gravity load plus wind moment. Why isn't it safe to check axial and bending separately and add the demand/capacity ratios? What does the AISC interaction equation capture that a simple sum misses?

Chapter 8 — Beam-Columns (AISC 360-22 Chapter H)

Chapter focus. Almost every real column also carries moment — from gravity eccentricity, wind, or seismic — making it a beam-column. AISC Chapter H combines axial and flexural demand in a single H1 interaction equation (with two branches at Pr/φPn = 0.2). The second-order Pr and Mr you plug in come from either B1/B2 (Ch 9) or the Direct Analysis Method (Ch 17).

1. Behavior

A beam-column is any member subjected to combined axial force and bending moment. Nearly every real column is a beam-column because of gravity eccentricity, wind, or seismic. Design uses an interaction equation that combines the axial and flexural demand ratios.

2. AISC H1 Interaction Equations

Decision box — which H1 equation applies?

Compute the axial demand ratio P_r/(φP_n).
If ≥ 0.2 → use H1-1a (upper, 8/9-slope line). The member is axial-dominated; bending is penalized by 8/9.
If < 0.2 → use H1-1b (lower, gentler line). The member is moment-dominated; axial is halved before adding.
Bi-axial bending? Both M_rx/φM_nx and M_ry/φM_ny appear in the same equation — do not check axes separately.
If P_r/(φP_n) is right at 0.2, run both equations and take the larger utilization (they meet at 0.2 by construction, but numerical rounding can flip which governs).

Two-part linear/8-9 slope envelope

H1-1a: Pr/(φPn) ≥ 0.2

(1)

P_{r} / (ϕ P_{n}) + (8/9) \cdot [M_{r x} / (ϕ M_{n x}) + M_{r y} / (ϕ M_{n y})] \leq 1.0

H1-1b: Pr/(φPn) < 0.2

(2)

P_{r} / (2 \cdot ϕ P_{n}) + [M_{r x} / (ϕ M_{n x}) + M_{r y} / (ϕ M_{n y})] \leq 1.0

P_r = required 2nd-order axial force; M_r = required 2nd-order moment. φP_n from Ch. E, φM_n from Ch. F. Both about strong and weak axes if biaxial bending applies.

3. Second-Order Effects — Direct Analysis Method (Ch. C) or B1/B2 (§ App. 8)

Column moments increase because axial load acts through the deflected shape (P-δ, P-Δ). Before applying H1 you must extract the 2nd-order P_r and M_r. Two allowable paths, both fully derived elsewhere in the course:

B₁/B₂ amplification of a 1st-order analysis → see Chapter 9 for the full B₁, B₂, C_m, and P_e,story equations, plus the α·P_r/P_e1 threshold ladder that tells you when B₁/B₂ is even required.
Direct Analysis Method (Ch. C) → reduced stiffness (0.8·τ_b·EI, 0.8·EA), notional loads N_i = 0.002·α·Y_i, K = 1.0. Full ingredients + worked notional-load example in Chapter 17.

For this chapter, treat P_r and M_rx, M_ry as inputs — whichever second-order path you used, plug the final amplified values straight into H1-1a / H1-1b.

4. Preliminary Design

Use AISC Manual Table 6-2 (Available Combined Force Design) or the classic equivalent-axial-load trick:

(3)

P_{u, e q} \approx P_{u} + m \cdot M_{ux} + m \cdot U \cdot M_{u y}

where m and U are chart coefficients from the Manual. Enter Table 4-1a with P_u,eq to pick a trial W-shape, then verify with H1.

5. Design Procedure

Extract P_r, M_rx, M_ry from 2nd-order analysis or amplified 1st-order results.
Compute φP_n (Ch. E) and φM_n (Ch. F) for trial section.
Compute P_r/φP_n; pick H1-1a or H1-1b.
Verify ≤ 1.0; if 0.85–1.00 accept; if < 0.7 lighten.

Additional Design Aids & Stratified Equations

H1 utilization envelope shaded — engineering intuition.

Utilization ratio

(4)

U = P_{r} / (ϕ P_{n}) + (8/9) \cdot M_{r} / (ϕ M_{n}) (H 1 - 1 a)

Reserve for combined loads

(5)

Δ = 1.0 - U

a dd i t i o na l m o m e n t M_{r}, e x t r a = (9/8) \cdot Δ \cdot ϕ M_{n}

⚠ Common mistakes

Using 1st-order moments without B1/B2 amplification (unless doing DAM).
Applying H1-1a when P_r/φP_n is actually < 0.2 (H1-1b is less conservative).
Forgetting L_b ≠ KL — flexural braces are often different from axial-buckling braces.
Ignoring biaxial bending when small M_ry exists at corner columns.

Worked Example 8.1 — Beam-Column, Cardinal Square 1st-story Perimeter Column

Given: First-story perimeter column of the Cardinal Square 4-story building. From ASCE 7 governing combination 1.2D + 1.0W + 1.0L + 0.5L_r: P_u = 182 k, M_ux = 103 k·ft, M_uy = 12 k·ft (2nd-order values, DAM). Story 14 ft; K = 1.0 both axes (DAM); L_b = 14 ft (compression flange braced only at floors). Try W14×82, A992.

Figure 8.1a — Beam-column free-body: axial P_u with biaxial moments M_ux and M_uy at the joint

Step 1 — Axial capacity φ_cP_n (from Ch. 5 solution)

φ_c P_n = 772 k (weak-axis governs at KL/r = 67.7)

Step 2 — Flexural capacity φ_bM_nx

Eq. F2-2 (inelastic LTB)

(1)

M_{n} = C_{b} [M_{p} - (M_{p} - 0.7 F_{y} S_{x}) (L_{b} - L_{p}) / (L_{r} - L_{p})] \leq M_{p}

M_{p} = F_{y} Z_{x}

W14×82: Z_x = 139 in³, S_x = 123 in³, r_y = 2.48 in, L_p = 8.76 ft, L_r = 33.2 ft (Manual Table 3-2).
L_b = 14 ft → L_p < L_b < L_r → inelastic LTB (F2-2).
Reverse-curvature bending in column: C_b ≈ 1.67 (assume M₁/M₂ = −0.5).
M_p = F_y·Z_x = 50·139/12 = 579 k·ft; 0.7 F_y S_x = 0.7·50·123/12 = 359 k·ft.
M_n = C_b[M_p − (M_p − 0.7F_yS_x)(L_b−L_p)/(L_r−L_p)]
= 1.67[579 − (579 − 359)(14 − 8.76)/(33.2 − 8.76)]
= 1.67[579 − 220·(5.24/24.44)] = 1.67[579 − 47.2] = 1.67·531.8 = 888 k·ft > M_p → cap at M_p
φ_bM_nx = 0.9·579 = 521 k·ft

Step 3 — Weak-axis flexural capacity φ_bM_ny

Eq. F6-1

(2)

M_{n y} = min (F_{y} Z_{y}, 1.6 F_{y} S_{y})

W14×82: Z_y = 44.8 in³. LTB does not apply about weak axis → M_ny = min(F_yZ_y, 1.6 F_yS_y).
S_y = 29.3 in³ → 1.6·50·29.3/12 = 195 k·ft; F_yZ_y/12 = 50·44.8/12 = 187 k·ft → governs.
φ_bM_ny = 0.9·187 = 168 k·ft

Step 4 — Choose interaction equation (H1)

§H1.1 selector

(3)

I f P_{r} / ϕ P_{n} \geq 0.2

u seH 1 - 1 a, e l se u seH 1 - 1 b

P_r/φP_n = 182/772 = 0.236 ≥ 0.2 → use Eq. H1-1a

Step 5 — Apply H1-1a

Eq. H1-1a

(4)

P_{r} / (ϕ P_{n}) + (8/9) \cdot [M_{r x} / (ϕ M_{n x}) + M_{r y} / (ϕ M_{n y})] \leq 1.0

= 0.236 + (8/9)·[103/521 + 12/168]
= 0.236 + 0.889·[0.198 + 0.0714]
= 0.236 + 0.889·0.269
= 0.236 + 0.239
= 0.475 ≤ 1.00 ✓

DESIGN: W14×82 (A992). H1-1a demand ratio = 0.48 — section is efficient with reserve for future load path changes. Adopt for the first two stories, resize upper stories with Table 6-2.

Sanity check via AISC Manual Table 6-2

For W14×82 with L_c = 14 ft the tabulated coefficients give p·P_r + b_x·M_rx + b_y·M_ry ≈ 0.47 — matches the hand calculation.

Engineering tips

Why interaction equations exist. The 8/9 coefficient linearises the plastic P–M envelope for W-shapes.
H1-1a vs H1-1b branch. Always compute r FIRST, then pick the branch.
Second-order amplifiers B1 and B2. For sidesway-inhibited frames, only B1 matters.
Cm reduction. AISC uses M1/M2 positive for reverse curvature.

Common mistakes

Why interaction equations exist. Adding utilisations Pr/Pc + Mr/Mc — that only matches the low-axial branch.
H1-1a vs H1-1b branch. Using H1-1a for a light-axial column — over-conservative but common on the FE.
Second-order amplifiers B1 and B2. Setting Mr = Mnt + Mlt without amplification.
Cm reduction. Sign error on M1/M2 that flips Cm.

Real-world use

Why interaction equations exist. Every column carrying moment.
H1-1a vs H1-1b branch. All Chapter H checks.
Second-order amplifiers B1 and B2. Any second-order analysis using AISC App. 8.
Cm reduction. Every B1 amplification.

06

Professional practice, safety & ethics

Combined-force practice

Professional practice

State clearly which interaction equation (H1-1a / H1-1b) governs and the second-order method used.
Report Pr/Pc and Mr/Mc separately in the calculation so a reviewer can trace the governing term.
Coordinate with the analysis model: interaction results are only as good as the stiffness assumptions.

Safety in design & construction

Members near the interaction limit have little reserve for any load path change — treat 0.95+ ratios as a design flag.
Beam-columns in moment frames carry seismic drift demands; check the deformation-compatibility case too.
Never neglect the moment from an eccentric connection just because the member is 'axially loaded'.

Engineering ethics

Do not ignore small moments to keep a member 'axial only'; that is fabricating a favorable assumption.
Disclose modeling simplifications to the reviewer.
If the analysis software output is not understood, do not seal it.

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 — Design review: documenting assumptions before the drawings are sealed.

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

Beam-column trade-offs

Approach

Interaction-governed members are sensitive to moment: reducing end eccentricity is cheaper than upsizing.
Simple (pinned) connections are far cheaper than moment connections — push moments to the dedicated lateral system.
Round to available sections; a theoretically optimal size that is not stocked adds lead time cost.

Worked cost example — Moment connection vs simple connection + brace

Basis: Pu = 320 kip, Mu = 180 k-ft

Line item	Qty	Rate	Cost
Field-welded moment connection	2 ea	$2,400	$4,800
Simple shear tab connection	2 ea	$260	$520
Brace + gusset added to bay	1 ls	$3,200	$3,200
Estimated total			$8,520

Takeaway. Two moment connections (≈$4,800) cost more than a brace plus simple connections (≈$3,720) — architecture decides whether that brace fits.

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

P–M interaction (H1-1)

Combined axial + flexural demand traces a path in P–M space. Cross the envelope and the beam-column fails.

09

Engineering figures

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

§8.6.1

Beam-column in service

Fig. 8.1Multi-story frame column carrying axial + moment
A perimeter column in a moment frame — the classic beam-column.

§8.6.2

Interaction failure

Fig. 8.2Buckled beam-column
H1-1a envelope reached — hinge forms combining axial and flexural yield.

§8.6.3

Second-order sway

Fig. 8.3Sway frame under lateral load
B2 amplifies Mlt as story drift grows.

§8.6.4

Out-of-plane LTB

Fig. 8.4Beam-column with weak-axis buckle
Out-of-plane LTB governs when Lb (weak axis) is too long.

10

Worked examples

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

Example 8.1

W14x82 beam-column — H1-1a interaction check

A W14x82 (A992) interior column, KL = 14 ft, carries Pu = 400 k and Mux = 220 k-ft (Muy = 0). Manual Table 6-2 gives φPn = 900 k and φMnx = 480 k-ft. Verify AISC H1.

Problem statement

Interior column: W14x82 (A992), KL = 14 ft, Lb = 14 ft. Factored demands Pu = 400 kips, Mux = 220 kip-ft, Muy = 0. From Manual Table 6-2, φPn = 900 k and φMnx = 480 k-ft. Check Chapter H.

Steel moment-frame column — FIG. 8.1 — Perimeter W-shape column carrying axial + wind moment.

DIMH1 P–M envelope with the H1-1a / H1-1b branches (Pr/Pc = 0.20 kink).

Given

Pu = 400 k, Mux = 220 k-ft, Muy = 0
φPn = 900 k, φMnx = 480 k-ft
A992: Fy = 50 ksi

Find

H1 utilisation and pass/fail

Assumptions

Braced-frame column, Mlt = 0
B1 already included in Mux (or B1 = 1.0 for this problem)

Code references

AISC 360-22 H1-1a
AISC Manual Table 6-2

Theory & approach

H1 picks the branch by r = Pr/φPn: r ≥ 0.20 → H1-1a; r < 0.20 → H1-1b. The 8/9 factor linearises the plastic P–M envelope for W-shapes.

Step-by-step solution

1
Pick branch
FormulaAISC H1
r = P_r / (φP_n)
r = 400 / 900 = 0.444
0.444 ≥ 0.20 → use H1-1a
2
H1-1a formula
FormulaH1-1a
P_r/(φP_n) + (8/9)·[M_rx/(φM_nx) + M_ry/(φM_ny)] ≤ 1.0
3
Substitute
H1 = 0.444 + (8/9)·(220/480 + 0)
H1 = 0.444 + (8/9)·(0.4583)
H1 = 0.444 + 0.4074
H1 = 0.851 ≤ 1.0 ✓
4
Margin
Margin = 1.00 − 0.851 = 0.15 (~15%). The section is adequate.

Verification

H1 < 1.0 with reasonable margin, and the branch selection agrees with Manual Table 6-2 hand-calc for W14x82.

Final answer

H1-1a = 0.85 ≤ 1.0 — OK. W14x82 works.

Design interpretation

Adding any Muy or increasing Pu by ~15% pushes utilisation past 1.0. In-plane weak-axis bracing is essential.

Common mistakes

Using H1-1a when r < 0.20 (over-conservative).
Adding utilisations linearly without the 8/9 coefficient.
Forgetting to amplify Mnt by B1 before feeding into H1.

Engineering insight

The 8/9 factor is why H1-1a curves smoothly rather than being a straight sum — it matches the true P–M envelope for compact W-shapes.

References

· AISC 360-22 Chapter H
· AISC 360-22 App. 8
· AISC Manual 16th ed., Table 6-2

11

Guided practice

Compute the governing variables — hints unlock as you need them

A column has Pu = 150 k, φPn = 800 k, Mux = 300 k-ft, φMnx = 500 k-ft (Muy = 0). Which H1 branch governs, and what is the utilisation?

Your turn

P_{r/phi*P_n} =H1_util =

Hints

1.Compute r = Pr/φPn.

12

Independent practice

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

Design task

A W14×22 (Ix = 199 in⁴) simply-supported floor beam spans L = 22 ft under a service live load of wL = 0.35 klf. Compute the mid-span live-load deflection and compare to the AISC/IBC serviceability limit L/360.

Given

E = 29,000 ksi
Ix = 199 in⁴
L = 22 ft = 264 in
wL = 0.35 klf = 0.0292 k/in
Limit: Δ ≤ L/360

Approach

For a simple beam under UDL: Δ = 5·w·L⁴ / (384·E·I).
Keep units consistent — convert w to k/in and L to in.
Compare Δcalc to Δlim = L/360.

Submit your answer

Delta_LL =inDelta_lim =in

13

Mini design challenge

Select the option that satisfies every code and serviceability requirement in the brief

Brief

Select a W14 (A992) beam-column for Pu = 400 k, Mux = 220 k-ft, KL = 14 ft, Lb = 14 ft. Aim for H1 ≤ 1.0 with reasonable margin.

Requirements

Pu = 400 k, Mux = 220 k-ft
KL = Lb = 14 ft
H1 (Chapter H) ≤ 1.0 with margin

Section

Wt (lb/ft)

Δ (in)

Ru/Rn

Cost

Pick

14

Chapter summary

A mind map of how every concept connects

Graded Chapter Quiz(13 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.

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16

FE exam preparation

NCEES-style practice with timer, equation sheet, and mastery tracking

Exam mode

30:00 Calculator

Question 1 / 9

Pu = 400 k, φPn = 900 k. The ratio Pu/φPn is closest to and which H1 branch governs?

◆ EasyAISC H1.1

A. 0.30, H1-1b
B. 0.44, H1-1a
C. 0.55, H1-1a
D. 0.15, H1-1b

Combined Axial Load and Bending

Apply H1-1a (Pr/Pc ≥ 0.2)

Apply H1-1b (Pr/Pc < 0.2)

Compute B1 amplifier

Compute B2 sway amplifier

Select Cm for the load pattern

Read Manual Table 6-2 (φPn, φMnx)

In-plane interaction collapse

Out-of-plane LTB of beam-column

P-Δ sway instability

Chapter 8 — Beam-Columns (AISC 360-22 Chapter H)

1. Behavior

2. AISC H1 Interaction Equations

3. Second-Order Effects — Direct Analysis Method (Ch. C) or B1/B2 (§ App. 8)

4. Preliminary Design

5. Design Procedure

Additional Design Aids & Stratified Equations

⚠ Common mistakes

Worked Example 8.1 — Beam-Column, Cardinal Square 1st-story Perimeter Column

Step 1 — Axial capacity φcPn (from Ch. 5 solution)

Step 2 — Flexural capacity φbMnx

Step 3 — Weak-axis flexural capacity φbMny

Step 4 — Choose interaction equation (H1)

Step 5 — Apply H1-1a

Sanity check via AISC Manual Table 6-2

Beam-column in service

Interaction failure

Second-order sway

Out-of-plane LTB

W14x82 beam-column — H1-1a interaction check

Pick branch

H1-1a formula

Substitute

Margin

Graded Chapter Quiz(13 FE-style questions · AISC Manual required)

Upload your worked solution (PDF)

Step 1 — Axial capacity φ_cP_n (from Ch. 5 solution)

Step 2 — Flexural capacity φ_bM_nx

Step 3 — Weak-axis flexural capacity φ_bM_ny