PE Exam (Civil) Steel Design: Flexure, Shear, Axial, Beam-column Interaction (AISC 360 LRFD/ASD)

Last updated: May 2, 2026

Steel Design: Flexure, Shear, Axial, Beam-column Interaction (AISC 360 LRFD/ASD) questions are one of the highest-leverage areas to study for the PE Exam (Civil). This guide breaks down the rule, the elements you need to recognize, the named traps that catch most students, and a memory aid that scales to test day. Read it once, then practice the same sub-topic adaptively in the app.

The rule

Under AISC 360, every steel member is checked against limit states using either LRFD ($\phi R_n \ge R_u$) or ASD ($R_n / \Omega \ge R_a$). For flexure (Chapter F), the nominal moment $M_n$ is the smallest of yielding, lateral-torsional buckling (LTB), and local buckling; for shear (Chapter G), $V_n = 0.6 F_y A_w C_v$; for compression (Chapter E), $P_n = F_{cr} A_g$ with $F_{cr}$ depending on slenderness; for tension (Chapter D), the lesser of yielding on $A_g$ and rupture on $A_e$ governs. Beam-columns combine these via the Chapter H interaction equations $\frac{P_r}{P_c} + \frac{8}{9}\left(\frac{M_{rx}}{M_{cx}}+\frac{M_{ry}}{M_{cy}}\right) \le 1.0$ when $\frac{P_r}{P_c} \ge 0.2$, else $\frac{P_r}{2P_c} + \left(\frac{M_{rx}}{M_{cx}}+\frac{M_{ry}}{M_{cy}}\right) \le 1.0$. The $\phi$ and $\Omega$ factors differ by limit state; you must pick the right pair.

Elements breakdown

Flexure (AISC 360 Chapter F)

Beam moment capacity is the smallest $M_n$ across all applicable limit states for the shape and bracing condition.

  • Compute $M_p = F_y Z_x$ for compact W-shapes
  • Determine $L_p$ and $L_r$ unbraced lengths
  • If $L_b \le L_p$: $M_n = M_p$
  • If $L_p < L_b \le L_r$: linear interpolation with $C_b$
  • If $L_b > L_r$: elastic LTB controls
  • Apply $\phi_b = 0.90$ (LRFD) or $\Omega_b = 1.67$ (ASD)
  • Check flange and web local buckling for non-compact

Common examples:

  • W21x44 with $L_b = 8 \text{ ft}$
  • Continuous beam with $C_b > 1.0$

Shear (AISC 360 Chapter G)

Web shear capacity for I-shapes; almost always governed by yielding for rolled W-shapes.

  • $V_n = 0.6 F_y A_w C_v$
  • $A_w = d \cdot t_w$ for I-shapes
  • $C_v = 1.0$ when $h/t_w \le 2.24\sqrt{E/F_y}$ (most rolled W)
  • $\phi_v = 1.00$ (LRFD), $\Omega_v = 1.50$ (ASD) for rolled I
  • Check at face of support, not centerline

Compression (AISC 360 Chapter E)

Axial compressive capacity governed by flexural buckling about the weaker axis (or torsional/flexural-torsional for singly symmetric).

  • Compute $\frac{KL}{r}$ about each axis, take larger
  • $F_e = \frac{\pi^2 E}{(KL/r)^2}$ (Euler)
  • If $\frac{KL}{r} \le 4.71\sqrt{E/F_y}$: inelastic, $F_{cr} = [0.658^{F_y/F_e}] F_y$
  • Else elastic: $F_{cr} = 0.877 F_e$
  • $P_n = F_{cr} A_g$
  • $\phi_c = 0.90$, $\Omega_c = 1.67$

Tension (AISC 360 Chapter D)

Two limit states: yielding on the gross section and rupture on the net effective area.

  • Yielding: $P_n = F_y A_g$, $\phi = 0.90$, $\Omega = 1.67$
  • Rupture: $P_n = F_u A_e$, $\phi = 0.75$, $\Omega = 2.00$
  • $A_e = U A_n$ where $U$ is shear lag factor
  • Smaller controls

Beam-Column Interaction (AISC 360 Chapter H)

Combined axial and flexure interaction equations H1-1a and H1-1b.

  • Compute $P_r/P_c$ ratio
  • If $\ge 0.2$: use H1-1a (P-dominant form)
  • If $< 0.2$: use H1-1b (M-dominant form)
  • Include second-order effects ($B_1$, $B_2$ multipliers)
  • Sum must be $\le 1.0$

Common patterns and traps

The $S_x$-for-$Z_x$ Substitution

A distractor that swaps elastic modulus $S_x$ for plastic modulus $Z_x$ when computing $M_p$. The numeric error is about 10-15% low because the shape factor $Z_x/S_x \approx 1.10$ to $1.15$ for typical W-shapes. This trap appears in nearly every flexure problem on the depth module.

A choice value about 88-90% of the correct factored moment, e.g., $\phi M_n = 340 \text{ kip-ft}$ when the right answer is $379 \text{ kip-ft}$.

The Wrong-$\phi$ Swap

A distractor that uses $\phi = 0.90$ for shear (correct value is $\phi_v = 1.00$ for rolled I-shapes) or uses $\phi = 0.75$ for tension yielding (correct is $0.90$; only rupture is $0.75$). This trap punishes candidates who memorize one $\phi$ and apply it everywhere.

A choice 10% lower than the correct shear capacity, e.g., $173 \text{ kips}$ instead of $192 \text{ kips}$.

The H1-1a / H1-1b Threshold Slip

For beam-columns, candidates use the wrong interaction equation by misjudging the $P_r/P_c \ge 0.2$ threshold or by using H1-1a regardless. Using H1-1a when H1-1b applies dramatically overestimates demand because of the $\frac{8}{9}$ multiplier on the moment terms.

An interaction ratio reported around 1.05 (apparent fail) when the correct equation gives 0.78 (passes).

The KL/r Wrong-Axis Pick

For compression, candidates compute $KL/r$ about the strong axis and forget that the weak axis ($r_y$) typically governs unless lateral bracing differs. The result is a much higher (incorrect) capacity.

A compressive capacity 30-50% higher than the correct value because $r_x$ was used in place of $r_y$.

The Unit-Inconsistency Multiplier

$F_y$ in ksi multiplied by $Z_x$ in $\text{in}^3$ gives kip-in, not kip-ft. Candidates forget to divide by 12 and report a moment 12$\times$ too large. Or they convert too early and lose a factor.

A choice value approximately 12$\times$ the correct kip-ft answer, e.g., $5{,}050 \text{ kip-ft}$ instead of $421 \text{ kip-ft}$.

How it works

Pick the limit state that controls before grabbing the calculator. Suppose you have a W18x50 ($Z_x = 101 \text{ in}^3$, $F_y = 50 \text{ ksi}$) braced at $L_b = 6 \text{ ft} < L_p \approx 5.83 \text{ ft}$ — actually slightly above $L_p$, so technically inelastic LTB applies, but for $L_b \le L_p$ you get $M_p = F_y Z_x = 50(101) = 5{,}050 \text{ kip-in} = 421 \text{ kip-ft}$. Then $\phi_b M_n = 0.90(421) = 379 \text{ kip-ft}$ for LRFD. If a candidate slips and uses $S_x$ instead of $Z_x$, they compute $M_y = F_y S_x$ and underestimate by about $\frac{Z_x}{S_x} \approx 1.12$. The shape factor matters. For shear on the same beam, $A_w = d \cdot t_w = 18.0 \times 0.355 = 6.39 \text{ in}^2$, so $V_n = 0.6(50)(6.39)(1.0) = 192 \text{ kips}$ and $\phi_v V_n = 192 \text{ kips}$ since $\phi_v = 1.00$. The trap nearly everyone falls for: applying $\phi = 0.90$ to shear out of habit.

Worked examples

Worked Example 1

On the Reyes Bridge Replacement Project, a simply supported floor beam is part of the elevated walkway framing. The beam is a W21x44 ($Z_x = 95.4 \text{ in}^3$, $S_x = 81.6 \text{ in}^3$, $r_y = 1.26 \text{ in}$, $L_p = 4.45 \text{ ft}$, $L_r = 13.0 \text{ ft}$) with $F_y = 50 \text{ ksi}$ and length $L = 24 \text{ ft}$. The compression flange is fully braced by the deck, so $L_b$ is effectively zero for LTB. The beam carries a uniformly distributed factored load $w_u$. Use AISC 360 LRFD.

Most nearly, what is the maximum factored uniform load $w_u$ the beam can support based on flexural strength?

  • A $w_u = 4.4 \text{ kip/ft}$
  • B $w_u = 5.0 \text{ kip/ft}$ ✓ Correct
  • C $w_u = 5.6 \text{ kip/ft}$
  • D $w_u = 6.2 \text{ kip/ft}$

Why B is correct: Because the deck fully braces the compression flange, $L_b \le L_p$, so $M_n = M_p = F_y Z_x = 50 \text{ ksi} \times 95.4 \text{ in}^3 = 4{,}770 \text{ kip-in} = 397.5 \text{ kip-ft}$. With $\phi_b = 0.90$, $\phi_b M_n = 0.90 \times 397.5 = 358 \text{ kip-ft}$. For a simply supported beam with uniform load, $M_u = \frac{w_u L^2}{8}$, so $w_u = \frac{8 M_u}{L^2} = \frac{8 \times 358}{24^2} = \frac{2{,}864}{576} = 4.97 \text{ kip/ft}$, rounding to $5.0 \text{ kip/ft}$.

Why each wrong choice fails:

  • A: This uses $S_x = 81.6 \text{ in}^3$ in place of $Z_x$, computing $M_y = 50(81.6)/12 = 340 \text{ kip-ft}$ and $\phi M = 306 \text{ kip-ft}$, then $w_u = 8(306)/576 = 4.25 \text{ kip/ft}$. The plastic modulus governs for a compact compact W-shape. (The $S_x$-for-$Z_x$ Substitution)
  • C: This forgets the $\phi_b = 0.90$ resistance factor and uses the nominal moment $M_n = 397.5 \text{ kip-ft}$ directly, giving $w = 8(397.5)/576 = 5.52 \text{ kip/ft}$. LRFD demands the factored capacity. (The Wrong-$\phi$ Swap)
  • D: This uses $\phi = 1.00$ (mistaken shear value) instead of $0.90$ AND rounds aggressively, yielding roughly $6.2 \text{ kip/ft}$. Misapplying the shear $\phi$ to flexure inflates capacity. (The Wrong-$\phi$ Swap)
Worked Example 2

A W12x65 column ($A_g = 19.1 \text{ in}^2$, $r_x = 5.28 \text{ in}$, $r_y = 3.02 \text{ in}$) anchors the central bay of the Liu Civic Center. It is pinned top and bottom about both axes ($K = 1.0$), with unbraced length $L = 14 \text{ ft}$ for both axes. Steel is A992 ($F_y = 50 \text{ ksi}$, $E = 29{,}000 \text{ ksi}$). Use AISC 360 LRFD with $\phi_c = 0.90$.

Most nearly, what is the design axial compressive strength $\phi_c P_n$?

  • A $\phi_c P_n = 540 \text{ kips}$
  • B $\phi_c P_n = 660 \text{ kips}$ ✓ Correct
  • C $\phi_c P_n = 730 \text{ kips}$
  • D $\phi_c P_n = 860 \text{ kips}$

Why B is correct: Weak axis governs: $\frac{KL}{r_y} = \frac{1.0 \times 14 \times 12}{3.02} = 55.6$. Limit: $4.71\sqrt{E/F_y} = 4.71\sqrt{29000/50} = 113$, so we are in the inelastic range. $F_e = \frac{\pi^2 (29000)}{55.6^2} = \frac{286{,}220}{3{,}091} = 92.6 \text{ ksi}$. Then $F_{cr} = [0.658^{50/92.6}] \times 50 = [0.658^{0.540}] \times 50 = 0.798 \times 50 = 39.9 \text{ ksi}$. $P_n = F_{cr} A_g = 39.9 \times 19.1 = 762 \text{ kips}$, and $\phi_c P_n = 0.90 \times 762 = 686 \text{ kips}$, closest to $660 \text{ kips}$.

Why each wrong choice fails:

  • A: This uses $r_y$ but applies the elastic Euler formula directly: $F_{cr} = 0.877 F_e = 0.877(92.6) = 81.2 \text{ ksi}$, capped at $F_y = 50$, then $0.877(50) = 43.9$ — but a candidate who botches the inelastic transition often lands near $540 \text{ kips}$ by using $0.658^1$ or similar misstep. (The Wrong-$\phi$ Swap)
  • C: This forgets the $\phi_c = 0.90$ factor and reports nominal $P_n = 762 \text{ kips}$, rounded to $730 \text{ kips}$. LRFD demands the factored capacity. (The Wrong-$\phi$ Swap)
  • D: This uses the strong axis $r_x = 5.28 \text{ in}$, giving $KL/r = 31.8$, $F_e = 283 \text{ ksi}$, $F_{cr} = 0.658^{0.177}(50) = 0.926(50) = 46.3 \text{ ksi}$, $P_n = 884 \text{ kips}$, and $\phi P_n \approx 796 \text{ kips}$, rounding into the $860 \text{ kips}$ neighborhood with rounding errors. Weak axis governs when bracing is equal. (The KL/r Wrong-Axis Pick)
Worked Example 3

A W14x82 beam-column ($A_g = 24.0 \text{ in}^2$, $\phi_c P_n = 870 \text{ kips}$, $\phi_b M_{nx} = 480 \text{ kip-ft}$) carries factored axial load $P_u = 230 \text{ kips}$ and factored strong-axis moment $M_{ux} = 280 \text{ kip-ft}$ at the Patel Distribution Center mezzanine. There is no weak-axis moment, and second-order effects are already included in the demands. Use AISC 360 Chapter H interaction.

Most nearly, what is the controlling interaction ratio, and does the member pass?

  • A Ratio $= 0.72$; passes
  • B Ratio $= 0.78$; passes ✓ Correct
  • C Ratio $= 0.85$; passes
  • D Ratio $= 1.05$; fails

Why B is correct: Compute $\frac{P_r}{P_c} = \frac{230}{870} = 0.264$. Since $0.264 \ge 0.2$, use Equation H1-1a: $\frac{P_r}{P_c} + \frac{8}{9}\left(\frac{M_{rx}}{M_{cx}}\right) = 0.264 + \frac{8}{9}\left(\frac{280}{480}\right) = 0.264 + 0.889 \times 0.583 = 0.264 + 0.519 = 0.783$. Since $0.78 \le 1.0$, the member passes.

Why each wrong choice fails:

  • A: This omits the $\frac{8}{9}$ multiplier and computes $0.264 + 0.583 = 0.847$, then mis-rounds to $0.72$. The H1-1a equation requires the $\frac{8}{9}$ factor on the moment terms. (The H1-1a / H1-1b Threshold Slip)
  • C: This uses the simpler sum $\frac{P_r}{P_c} + \frac{M_{rx}}{M_{cx}} = 0.264 + 0.583 = 0.847 \approx 0.85$, treating the interaction as straight addition. AISC requires the $\frac{8}{9}$ weighting when $P_r/P_c \ge 0.2$. (The H1-1a / H1-1b Threshold Slip)
  • D: This applies H1-1b incorrectly with a doubled axial term: $\frac{P_r}{P_c} + \frac{M_{rx}}{M_{cx}} + \frac{P_r}{P_c} = 1.11$, rounding to $1.05$. H1-1b uses $\frac{P_r}{2P_c}$, not $2 \frac{P_r}{P_c}$, and only when $P_r/P_c < 0.2$. (The H1-1a / H1-1b Threshold Slip)

Memory aid

FSCAB: Flexure ($\phi = 0.90$), Shear ($\phi = 1.00$ for rolled I), Compression ($\phi = 0.90$), Axial-tension yielding ($\phi = 0.90$) / rupture ($\phi = 0.75$), Beam-column (H1-1a if $P_r/P_c \ge 0.2$).

Key distinction

Plastic section modulus $Z_x$ governs flexure capacity for compact shapes, while elastic section modulus $S_x$ only enters for non-compact or slender flanges. Always reach for $Z_x$ first on a W-shape unless local buckling forces you to use $S_x$.

Summary

Identify the limit state, pull the matching $\phi$ or $\Omega$, plug Handbook formulas with consistent units, and for beam-columns choose H1-1a or H1-1b based on the $P_r/P_c$ threshold of 0.2.

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Frequently asked questions

What is steel design: flexure, shear, axial, beam-column interaction (aisc 360 lrfd/asd) on the PE Exam (Civil)?

Under AISC 360, every steel member is checked against limit states using either LRFD ($\phi R_n \ge R_u$) or ASD ($R_n / \Omega \ge R_a$). For flexure (Chapter F), the nominal moment $M_n$ is the smallest of yielding, lateral-torsional buckling (LTB), and local buckling; for shear (Chapter G), $V_n = 0.6 F_y A_w C_v$; for compression (Chapter E), $P_n = F_{cr} A_g$ with $F_{cr}$ depending on slenderness; for tension (Chapter D), the lesser of yielding on $A_g$ and rupture on $A_e$ governs. Beam-columns combine these via the Chapter H interaction equations $\frac{P_r}{P_c} + \frac{8}{9}\left(\frac{M_{rx}}{M_{cx}}+\frac{M_{ry}}{M_{cy}}\right) \le 1.0$ when $\frac{P_r}{P_c} \ge 0.2$, else $\frac{P_r}{2P_c} + \left(\frac{M_{rx}}{M_{cx}}+\frac{M_{ry}}{M_{cy}}\right) \le 1.0$. The $\phi$ and $\Omega$ factors differ by limit state; you must pick the right pair.

How do I practice steel design: flexure, shear, axial, beam-column interaction (aisc 360 lrfd/asd) questions?

The fastest way to improve on steel design: flexure, shear, axial, beam-column interaction (aisc 360 lrfd/asd) is targeted, adaptive practice — working questions that focus on your specific weak spots within this sub-topic, getting immediate feedback, and revisiting items you missed on a spaced-repetition schedule. Neureto's adaptive engine does this automatically across the PE Exam (Civil); start a free 7-day trial to see your sub-topic mastery climb in real time.

What's the most important distinction to remember for steel design: flexure, shear, axial, beam-column interaction (aisc 360 lrfd/asd)?

Plastic section modulus $Z_x$ governs flexure capacity for compact shapes, while elastic section modulus $S_x$ only enters for non-compact or slender flanges. Always reach for $Z_x$ first on a W-shape unless local buckling forces you to use $S_x$.

Is there a memory aid for steel design: flexure, shear, axial, beam-column interaction (aisc 360 lrfd/asd) questions?

FSCAB: Flexure ($\phi = 0.90$), Shear ($\phi = 1.00$ for rolled I), Compression ($\phi = 0.90$), Axial-tension yielding ($\phi = 0.90$) / rupture ($\phi = 0.75$), Beam-column (H1-1a if $P_r/P_c \ge 0.2$).

What's a common trap on steel design: flexure, shear, axial, beam-column interaction (aisc 360 lrfd/asd) questions?

Using $S_x$ instead of $Z_x$ for plastic moment $M_p$

What's a common trap on steel design: flexure, shear, axial, beam-column interaction (aisc 360 lrfd/asd) questions?

Confusing $\phi$ values across limit states (shear is 1.00, not 0.90)

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