CELE Steel & Timber Design — Steel Beams: Flexure and ShearDetailed Explanation
Detailed explanation of Steel Beams: Flexure and Shear for the CELE 2026. Full depth, full reasoning — exactly what you need when Professional Regulation Commission (PRC) — Board of Civil Engineering tests this chapter with applied or scenario-based questions in the CELE Steel & Timber Design subtest.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Steel & Timber Design section sits under a "Core" weighting, and Steel Beams: Flexure and Shear is the 3rd chapter in the 5-chapter CELE Steel & Timber Design rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Steel & Timber Design.
Steel Beams: Flexure and Shear - Detailed Explanation
Steel beams are the workhorses of structural framing — they carry gravity loads primarily through bending and shear. In the PRC Civil Engineer Licensure Examination, flexure and shear of steel beams consistently appear as computational problems worth 3–5 points each. This chapter covers the three governing limit states you must master: (1) plastic moment capacity M_p for compact, fully braced beams, (2) lateral-torsional buckling (LTB) when the compression flange is unbraced over a significant length L_b, and (3) shear yielding of the web. All formulas follow AISC 360-16 (Load and Resistance Factor Design, LRFD), which is adopted by reference in the NSCP 2015 (Section 502). Mastery of these concepts — especially knowing when to use Z_x versus S_x and how L_b compares with L_p — separates passers from repeaters on board exam day.
Concepts
Compact Sections and the Plastic Moment M_p
A steel beam cross-section is classified as COMPACT, NON-COMPACT, or SLENDER based on the width-to-thickness (λ) ratios of its compression flange and web elements, compared with limiting ratios λ_p (compact limit) and λ_r (non-compact limit). For a section to be COMPACT, both of the following must be satisfied: Flange: b_f / (2t_f) ≤ λ_pf = 0.38√(E/F_y) Web: h / t_w ≤ λ_pw = 3.76√(E/F_y) When a section is compact AND the compression flange is adequately braced (L_b ≤ L_p), the entire cross-section can yield in flexure without any local buckling occurring first. In this ideal condition, the stress distribution transitions from a triangular (elastic) pattern to a fully rectangular (plastic) pattern, and the beam develops its maximum bending resistance — the PLASTIC MOMENT: M_p = F_y × Z_x where: F_y = specified minimum yield stress of steel (MPa) Z_x = plastic section modulus about the x-axis (mm³) The plastic section modulus Z_x is found by locating the plastic neutral axis (PNA) — which divides the cross-section into two equal areas — and summing the first moments of those two half-areas about the PNA: Z_x = A/2 × ȳ_top + A/2 × ȳ_bot For standard W-shapes (wide-flange sections), Z_x values are tabulated in the AISC Steel Construction Manual. Note that Z_x > S_x always; their ratio is the SHAPE FACTOR f: f = Z_x / S_x ≈ 1.12 for most W-shapes (ranges 1.10 to 1.18) The elastic section modulus S_x = I_x / c gives only the YIELD MOMENT M_y = F_y × S_x, which is about 10–15% less than M_p. Using M_y instead of M_p is conservative and wrong for a compact, braced beam on the board exam. In LRFD, the design flexural strength is: φ_b M_n = 0.90 M_p = 0.90 F_y Z_x (φ_b = 0.90) In ASD, the allowable flexural strength is: M_n / Ω_b = M_p / 1.67 (Ω_b = 1.67)
Examples
This is the most straightforward board exam problem type. The key action is using Z_x (plastic modulus), not S_x. The 0.90 factor is φ_b for flexure in LRFD. Note the unit chain: MPa × mm³ = N·mm; divide by 10⁶ to get kN·m.
Scenario
A compact W-shape beam with Z_x = 1.20 × 10⁶ mm³ is fully braced (L_b ≤ L_p). Steel is A36, F_y = 248 MPa, E = 200,000 MPa. Determine the LRFD design flexural strength φ_b M_n.
Solution
Step 1: Confirm compact section (given — assume compact). Step 2: Since L_b ≤ L_p, M_n = M_p = F_y Z_x M_p = 248 MPa × 1.20 × 10⁶ mm³ M_p = 2.976 × 10⁸ N·mm = 297.6 kN·m Step 3: Apply resistance factor: φ_b M_n = 0.90 × 297.6 = 267.8 kN·m Answer: φ_b M_n = 267.8 kN·m
For a solid rectangle, Z_x = bd²/4 (half-area × distance between centroids of half-areas). Shape factor = 1.50 is exact for rectangles. For W-shapes, the flanges concentrate area far from the NA, making f closer to 1.12 — less reserve beyond yield moment.
Scenario
A rectangular steel bar 100 mm wide × 200 mm deep, F_y = 250 MPa. Compute M_y, M_p, and the shape factor.
Solution
Section properties: I_x = (100)(200)³/12 = 66.67 × 10⁶ mm⁴ S_x = I_x / c = 66.67 × 10⁶ / 100 = 666,700 mm³ Z_x = b × d² / 4 = 100 × 200² / 4 = 1,000,000 mm³ Elastic (yield) moment: M_y = F_y S_x = 250 × 666,700 = 166.67 × 10⁶ N·mm = 166.7 kN·m Plastic moment: M_p = F_y Z_x = 250 × 1,000,000 = 250 × 10⁶ N·mm = 250.0 kN·m Shape factor: f = Z_x / S_x = 1,000,000 / 666,700 = 1.50 Answer: M_y = 166.7 kN·m, M_p = 250.0 kN·m, f = 1.50
Applications
- Design of floor beams in commercial buildings (e.g., SM malls, BPO offices) where continuous floor bracing from the slab makes L_b effectively zero.
- Checking adequacy of existing steel beams during renovation or change-of-occupancy assessment under NSCP 2015.
- Sizing of runway beams in industrial plants where crane loads govern and compact sections are preferred for ductility.
- Verification of purlins on industrial roofs using light-gauge W-sections with closely spaced roof panels providing lateral bracing.
Misconceptions
- WRONG: Using M_n = F_y S_x for a compact, braced beam — this underestimates by ~12% and loses points on board exams.
- WRONG: Applying φ_b = 1.0 to flexure — φ_b = 0.90 for bending; φ_v = 1.0 is for shear (rolled I-shapes).
- WRONG: Assuming any W-shape is automatically compact without checking — verify λ_f and λ_w, especially for high-yield steels (F_y = 485 MPa) where λ_p is smaller.
- WRONG: Using the full cross-sectional area to compute Z_x — only half the area is used per half (the plastic neutral axis divides total area equally).
Related Concepts
- Lateral-Torsional Buckling (LTB) — what happens when M_p cannot be reached due to unbraced length
- Local Buckling — flange and web compactness limits λ_p and λ_r
- Elastic Section Modulus S_x — governs yield moment and non-compact/slender sections
- Plastic Neutral Axis (PNA) — divides cross-section into equal areas, basis for Z_x calculation
Common Exam Questions
Example
A W350×90 beam (Z_x = 1,760 × 10³ mm³) of A572 Gr. 50 (F_y = 345 MPa) is compact and fully braced. Find φ_b M_n. → M_p = 345 × 1,760,000 = 607.2 × 10⁶ N·mm = 607.2 kN·m; φ_b M_n = 0.90 × 607.2 = 546.5 kN·m.
Approach
Given F_y, Z_x (or dimensions to compute Z_x), compute M_p = F_y Z_x, then multiply by 0.90. Watch unit conversions: N·mm → kN·m requires dividing by 10⁶.
Question Type
Direct computation of φ_b M_n
Example
If the problem states 'the beam just begins to yield at the extreme fiber,' use M_y = F_y S_x, not M_p.
Approach
Read the problem carefully: 'plastic moment' → Z_x; 'yield moment' or 'moment at first yield' → S_x. Examiners frequently provide both Z_x and S_x as distractors.
Question Type
Identifying which modulus to use
Example
For a W-shape with S_x = 1,200 × 10³ mm³ and Z_x = 1,350 × 10³ mm³: f = 1,350/1,200 = 1.125.
Approach
f = Z_x / S_x. Board exams may ask you to compute Z_x from scratch for a T-section or built-up section, then find f.
Question Type
Shape factor computation
Key Points To Remember
- M_p = F_y × Z_x — always use the PLASTIC modulus Z_x, not the elastic S_x, for compact beams.
- Shape factor f = Z_x/S_x ≈ 1.12 for W-shapes; for a rectangle f = 1.50.
- Compact section requires BOTH flange AND web λ ≤ λ_p; check both.
- LRFD design strength: φ_b M_n = 0.90 F_y Z_x; ASD allowable: M_p/1.67.
- M_p is achievable only when L_b ≤ L_p (fully braced or within plastic zone).
- Most standard W-shapes in Philippine construction (A36 or A572 Gr. 50 steel) are compact by default — verify this for non-standard sections.
Lateral-Torsional Buckling (LTB) and Unbraced Length
When a beam's compression flange is not continuously supported and the unbraced length L_b is large, the beam may fail by LATERAL-TORSIONAL BUCKLING (LTB) — the compression flange buckles sideways while the tension flange remains in place, causing the beam to twist and lose capacity before reaching M_p. LTB is the most critical limit state for long-span beams without adequate lateral bracing. AISC 360 (adopted by NSCP 2015) defines three LTB zones based on L_b: ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ ZONE 1 — No LTB (Plastic Zone): L_b ≤ L_p M_n = M_p = F_y Z_x L_p = 1.76 r_y √(E/F_y) ZONE 2 — Inelastic LTB: L_p < L_b ≤ L_r M_n = C_b [M_p - (M_p - 0.7 F_y S_x)(L_b - L_p)/(L_r - L_p)] ≤ M_p ZONE 3 — Elastic LTB: L_b > L_r M_n = F_cr S_x ≤ M_p F_cr = C_b π² E / (L_b/r_ts)² × √[1 + 0.078(J c)/(S_x h_o)(L_b/r_ts)²] ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ Key parameters: r_y = radius of gyration about the weak (y) axis (mm) r_ts = effective radius of gyration for LTB (≈ r_y for doubly symmetric I-shapes; tabulated in AISC Manual) L_p = limiting unbraced length for full plastic moment (mm) L_r = limiting unbraced length for inelastic LTB (mm) J = St. Venant torsional constant (mm⁴) h_o = distance between flange centroids (mm) c = 1.0 for doubly symmetric I-shapes S_x = elastic section modulus (mm³) THE MOMENT GRADIENT FACTOR C_b: C_b accounts for the fact that non-uniform moment diagrams are less severe than uniform moment (pure bending). For uniform moment (worst case), C_b = 1.0. For any other loading, C_b > 1.0, increasing M_n (but the result is always capped at M_p). C_b = 12.5 M_max / (2.5 M_max + 3 M_A + 4 M_B + 3 M_C) where M_max = maximum moment in the unbraced segment; M_A, M_B, M_C = moments at quarter-point, midpoint, and three-quarter-point of the unbraced segment, respectively (all absolute values). For a simply supported beam with uniform load: C_b ≈ 1.14 For a simply supported beam with midpoint load: C_b ≈ 1.32 For cantilever with tip load (worst case): C_b = 1.0 (conservative) CRITICAL BOARD EXAM FORMULA — L_p COMPUTATION: L_p = 1.76 r_y √(E/F_y) For A36 (F_y = 248 MPa, E = 200,000 MPa): √(E/F_y) = √(200,000/248) = √806.45 = 28.40 L_p = 1.76 × r_y × 28.40 = 49.98 r_y ≈ 50 r_y For A572 Gr.50 (F_y = 345 MPa): √(E/F_y) = √(200,000/345) = √579.71 = 24.08 L_p = 1.76 × r_y × 24.08 = 42.38 r_y ≈ 42 r_y
Examples
This is a classic board exam problem. Note the arithmetic: √(200,000/248) = 28.40. Multiply by 1.76 × 40 = 70.4, giving L_p = 2,000 mm exactly. The examiner may provide L_r and ask you to classify the beam — always compare L_b with L_p first, then L_r.
Scenario
A beam has r_y = 40 mm, F_y = 248 MPa, E = 200,000 MPa. Find L_p. If the actual unbraced length is L_b = 2.5 m, is the beam in Zone 1?
Solution
L_p = 1.76 × r_y × √(E/F_y) = 1.76 × 40 × √(200,000/248) = 70.4 × √806.45 = 70.4 × 28.40 = 2,000 mm = 2.00 m Since L_b = 2.5 m > L_p = 2.0 m, the beam is NOT in Zone 1. The beam is in Zone 2 (inelastic LTB) if L_b ≤ L_r, or Zone 3 if L_b > L_r. LTB REDUCES M_n below M_p. Answer: L_p = 2.00 m; beam is in the inelastic LTB zone.
Zone 2 is a LINEAR INTERPOLATION between M_p (at L_b = L_p) and 0.7 F_y S_x (at L_b = L_r). The ratio (L_b - L_p)/(L_r - L_p) = (3.5-1.8)/(5.2-1.8) = 1.7/3.4 = 0.50 means the beam is at the midpoint of the inelastic range — M_n is the average of M_p and 0.7 F_y S_x. With C_b = 1.0, no bonus is applied.
Scenario
A compact W-shape beam: Z_x = 1,500 × 10³ mm³, S_x = 1,340 × 10³ mm³, F_y = 248 MPa, E = 200,000 MPa, L_p = 1.8 m, L_r = 5.2 m, C_b = 1.0. The unbraced length L_b = 3.5 m. Find M_n.
Solution
Check zone: L_p = 1.8 m < L_b = 3.5 m ≤ L_r = 5.2 m → Zone 2 (Inelastic LTB) Compute reference moments: M_p = F_y Z_x = 248 × 1,500,000 = 372 × 10⁶ N·mm = 372 kN·m 0.7 F_y S_x = 0.7 × 248 × 1,340,000 = 232.6 × 10⁶ N·mm = 232.6 kN·m Inelastic LTB formula: M_n = C_b [M_p - (M_p - 0.7 F_y S_x)(L_b - L_p)/(L_r - L_p)] = 1.0 [372 - (372 - 232.6)(3.5 - 1.8)/(5.2 - 1.8)] = 1.0 [372 - (139.4)(1.7/3.4)] = 1.0 [372 - 139.4 × 0.50] = 1.0 [372 - 69.7] = 302.3 kN·m Design strength: φ_b M_n = 0.90 × 302.3 = 272.1 kN·m Answer: M_n = 302.3 kN·m; φ_b M_n = 272.1 kN·m
Applications
- Specifying brace spacing for roof beams in warehouses — the distance between purlins or cross-frames must not exceed L_p to maintain full M_p.
- Design of transfer beams in high-rise buildings where long unbraced spans may trigger elastic LTB, requiring composite action or lateral bracing.
- Checking LTB for crane girder top flanges in industrial buildings — if the crane rail does not provide adequate lateral restraint, L_b may be large.
- Bridge girder design (although bridge design uses AASHTO LRFD, the LTB concept is identical and appears in CE licensure exams).
Misconceptions
- WRONG: Ignoring LTB and always using M_n = M_p — for an unbraced beam with L_b > L_p, this severely overestimates capacity.
- WRONG: Thinking that C_b > 1.0 can increase M_n beyond M_p — the cap is always M_p regardless of C_b.
- WRONG: Using r_x (strong-axis) instead of r_y (weak-axis) in the L_p formula — LTB involves weak-axis bending and torsion.
- WRONG: Computing 0.7 F_y Z_x for the lower reference moment — it is 0.7 F_y S_x (elastic modulus), representing first yield with residual stresses.
- WRONG: Assuming lateral bracing to the TENSION flange prevents LTB — bracing the COMPRESSION flange is what matters.
Related Concepts
- Plastic Moment M_p — the upper-bound capacity that LTB prevents from being reached
- Compact Section Classification — compactness is prerequisite for Zone 1 capacity
- Moment Gradient Factor C_b — bonus factor for non-uniform bending diagrams
- Elastic Buckling — Zone 3 LTB is analogous to Euler column buckling but involves lateral-torsional mode
- Lateral Bracing Requirements — minimum brace stiffness and strength requirements (AISC Appendix 6)
Common Exam Questions
Example
r_y = 35 mm, F_y = 345 MPa: L_p = 1.76(35)√(200000/345) = 61.6 × 24.08 = 1,483 mm = 1.48 m.
Approach
Direct substitution: L_p = 1.76 r_y √(E/F_y). Use E = 200,000 MPa for steel. Express answer in meters by dividing by 1,000.
Question Type
Compute L_p given r_y and F_y
Example
L_b = 4.0 m, L_p = 2.5 m, L_r = 7.0 m → Zone 2 (inelastic LTB). Apply interpolation formula.
Approach
Compare: if L_b ≤ L_p → Zone 1; if L_p < L_b ≤ L_r → Zone 2; if L_b > L_r → Zone 3. Then apply the correct M_n formula for that zone.
Question Type
Zone classification given L_b, L_p, L_r
Example
Common trap: forgetting to cap C_b × M_n,computed at M_p. If C_b = 1.3 makes the computed value exceed M_p, the answer is simply M_p.
Approach
Compute M_p and 0.7 F_y S_x first. Then apply the linear interpolation formula. Do not forget C_b (often given as 1.0 if not specified). Cap result at M_p.
Question Type
Inelastic LTB — compute M_n in Zone 2
Key Points To Remember
- Three LTB zones: Zone 1 (L_b ≤ L_p) → M_n = M_p; Zone 2 (L_p < L_b ≤ L_r) → linear interpolation; Zone 3 (L_b > L_r) → elastic buckling.
- L_p = 1.76 r_y √(E/F_y) — memorize this; it is the most frequently tested LTB formula on board exams.
- C_b ≥ 1.0 always; never reduces M_n; result capped at M_p.
- For uniform moment (constant M along unbraced length), C_b = 1.0 — conservative and simple.
- LTB is a COMPRESSION FLANGE problem — providing lateral bracing to the compression flange (not the tension flange) prevents LTB.
- r_y is the weak-axis radius of gyration — a small r_y means early onset of LTB.
- In Zone 2, the 0.7 F_y S_x term represents the moment at which residual stresses cause first yielding to begin — the onset of inelastic behavior.
Shear Strength of Steel Beams
Shear in steel beams is resisted primarily by the WEB (the vertical plate element connecting the two flanges). The nominal shear strength of a rolled I-shape is: V_n = 0.6 F_y A_w C_v where: F_y = yield strength of steel (MPa) A_w = gross web area = d × t_w (mm²) d = overall depth of section (mm) t_w = web thickness (mm) C_v = web shear coefficient (dimensionless) IMPORTANT: A_w uses the FULL depth d, not the clear height of the web (h) between flanges. This is a deliberate simplification in the AISC shear formula that has trapped many examinees. THE SHEAR COEFFICIENT C_v: For most standard rolled I-shapes in A36 or A572 steel (where the web slenderness h/t_w is small), C_v = 1.0 — meaning the web yields in shear before shear buckling occurs. The condition for C_v = 1.0 (shear yielding governs): h/t_w ≤ 2.24 √(E/F_y) For A36: 2.24√(200,000/248) = 2.24 × 28.40 = 63.6 For Gr.50: 2.24√(200,000/345) = 2.24 × 24.08 = 53.9 Almost all standard W-shapes satisfy this limit, so C_v = 1.0 in typical board exam problems. When C_v = 1.0: V_n = 0.6 F_y A_w (full shear yielding capacity) THE RESISTANCE FACTOR FOR SHEAR φ_v: φ_v = 1.00 for hot-rolled I-shapes satisfying h/t_w ≤ 2.24√(E/F_y) φ_v = 0.90 for all other cases (plate girders, slender webs, C-shapes) Design shear strength: φ_v V_n = 1.0 × 0.6 F_y A_w = 0.60 F_y A_w (most rolled I-shapes) ASD allowable shear: V_n / Ω_v = V_n / 1.67 (when φ_v = 1.0, Ω_v = 1.50 → note: AISC uses Ω_v = 1.67/1.0 × 0.9 = 1.50 for this case) V_n / Ω_v = V_n / 1.67 (when φ_v = 0.90) SHEAR FLOW AND CONNECTIONS: For composite beams or built-up sections, shear flow q = VQ/I must be checked at the interface between elements. This is not the primary LRFD shear limit state but appears in bolt/weld design problems. PHYSICAL PICTURE: The von Mises yield criterion gives the shear yield stress τ_y = F_y/√3 ≈ 0.577 F_y. AISC rounds this to 0.6 F_y (slightly conservative). The full nominal shear capacity assumes the web yields uniformly at 0.6 F_y across the area A_w = d × t_w.
Examples
Straightforward application. Note φ_v = 1.0 (not 0.90). The formula is simply 0.6 × F_y × d × t_w for standard rolled sections — combine 0.6 × 1.0 × 1.0 = 0.60, so φ_v V_n = 0.60 × 248 × 4,500 = 669.6 kN. Quick mental computation: 0.6 × 248 = 148.8; 148.8 × 4,500 = 669,600 N.
Scenario
A W450×90 beam (d = 450 mm, t_w = 10 mm) of A36 steel (F_y = 248 MPa) has a stocky web (C_v = 1.0). Find the LRFD design shear strength φ_v V_n.
Solution
Step 1: Compute web area: A_w = d × t_w = 450 × 10 = 4,500 mm² Step 2: Compute nominal shear strength: V_n = 0.6 F_y A_w C_v = 0.6 × 248 × 4,500 × 1.0 = 669,600 N = 669.6 kN Step 3: Apply resistance factor (φ_v = 1.0 for compact web): φ_v V_n = 1.0 × 669.6 = 669.6 kN Answer: φ_v V_n = 669.6 kN
This problem distinguishes standard rolled sections from plate girders. The slender web (h/t_w = 75 > 63.6) means both C_v < 1.0 and φ_v = 0.90 apply. Board exams may specify C_v directly; if not, you must use the full AISC shear coefficient computation involving web slenderness zones.
Scenario
A plate girder web: d = 600 mm, t_w = 8 mm, F_y = 248 MPa, E = 200,000 MPa. Check if C_v = 1.0 applies, then find φ_v V_n.
Solution
Step 1: Check web slenderness: h/t_w ≈ 600/8 = 75.0 (using d ≈ h for approximation) Limit: 2.24√(E/F_y) = 2.24√(200,000/248) = 2.24 × 28.40 = 63.6 75.0 > 63.6 → C_v ≠ 1.0 necessarily; φ_v = 0.90 Step 2: For this problem, assume the examiner provides C_v = 0.85 (from web shear buckling table). A_w = 600 × 8 = 4,800 mm² V_n = 0.6 × 248 × 4,800 × 0.85 = 0.6 × 248 × 4,080 = 607,104 N = 607.1 kN φ_v V_n = 0.90 × 607.1 = 546.4 kN Answer: φ_v V_n = 546.4 kN (φ_v = 0.90 since h/t_w > 2.24√(E/F_y))
Applications
- Checking shear capacity of short-span transfer beams in multi-story buildings (shear-critical members).
- Design of coped beams at connections where web depth is reduced, increasing shear stress on the remaining web.
- Verifying shear capacity of crane runway girders at support reactions where high point loads are applied.
- Web stiffener design for plate girders where slender webs require transverse stiffeners to prevent shear buckling.
Misconceptions
- WRONG: Using φ_v = 0.90 for standard rolled I-shapes — φ_v = 1.00 for compact web sections, giving 10% more capacity.
- WRONG: Computing A_w = (d - 2t_f) × t_w (clear height only) — AISC explicitly uses full depth d in A_w = d × t_w.
- WRONG: Using 0.577 F_y instead of 0.6 F_y — AISC uses the rounded value 0.6 F_y (slightly conservative vs. von Mises).
- WRONG: Neglecting to check C_v for non-standard or built-up sections — C_v = 1.0 is not automatic for plate girders.
Related Concepts
- Web Crippling — localized bearing failure at concentrated loads, separate from distributed shear
- Shear Buckling — elastic shear buckling of slender webs (C_v < 1.0 regime)
- Transverse Stiffeners — used to increase shear capacity of plate girder webs via tension field action
- Block Shear — failure mode at bolted connections combining shear yielding and tension rupture
Common Exam Questions
Example
d = 500 mm, t_w = 9 mm, F_y = 248 MPa: φ_v V_n = 0.6 × 248 × 500 × 9 = 669,600 N = 669.6 kN.
Approach
Given d, t_w, F_y. Compute A_w = d × t_w, then V_n = 0.6 F_y A_w (assuming C_v = 1.0), then multiply by φ_v = 1.0. Total: φ_v V_n = 0.6 F_y d t_w.
Question Type
Direct computation of φ_v V_n
Example
A beam has φ_b M_n = 350 kN·m and φ_v V_n = 580 kN. If M_u = 280 kN·m (< 350, OK) and V_u = 150 kN (< 580, OK), the beam is adequate for both.
Approach
Compute both φ_b M_n and φ_v V_n. Compare with the factored loads. The governing limit state is the one that is more critical relative to demand.
Question Type
Checking if shear or flexure governs
Example
W400×85 with d = 399 mm, t_f = 18 mm, t_w = 10 mm: A_w = 399 × 10 = 3,990 mm² (not (399 - 2×18) × 10 = 3,630 mm²).
Approach
Always use A_w = d × t_w (overall depth × web thickness). Do not subtract flange thicknesses. This is the AISC definition.
Question Type
Identifying the correct A_w
Key Points To Remember
- V_n = 0.6 F_y A_w C_v — memorize; this appears on virtually every board exam session.
- A_w = d × t_w — use FULL depth d, NOT the clear web height h.
- C_v = 1.0 for most standard rolled W-shapes (compact web condition satisfied).
- φ_v = 1.00 (not 0.90) for rolled I-shapes with compact webs — this is a common trap.
- ASD: Ω_v = 1.50 when φ_v = 1.0; Ω_v = 1.67 when φ_v = 0.90.
- Shear rarely governs for normal-span beams (flexure usually governs); shear may govern for short, heavily loaded beams or transfer beams.
Compact vs. Non-Compact vs. Slender Section Classification
Before computing M_n, you MUST classify the section's cross-section elements (flanges and web) using the width-to-thickness ratio λ compared with the plastic limit λ_p and the non-compact limit λ_r. FOR THE COMPRESSION FLANGE (half-flange width to flange thickness): λ_f = b_f / (2 t_f) λ_pf = 0.38 √(E/F_y) [compact flange limit] λ_rf = 1.00 √(E/F_y) [non-compact flange limit] FOR THE WEB (clear height to web thickness): λ_w = h / t_w λ_pw = 3.76 √(E/F_y) [compact web limit] λ_rw = 5.70 √(E/F_y) [non-compact web limit] For A36 (F_y = 248 MPa, E = 200,000 MPa): √(E/F_y) = 28.40 λ_pf = 0.38 × 28.40 = 10.79 λ_rf = 28.40 λ_pw = 3.76 × 28.40 = 106.8 λ_rw = 5.70 × 28.40 = 161.9 For A572 Gr.50 (F_y = 345 MPa): √(E/F_y) = 24.08 λ_pf = 0.38 × 24.08 = 9.15 λ_rf = 24.08 λ_pw = 3.76 × 24.08 = 90.5 λ_rw = 5.70 × 24.08 = 137.3 CLASSIFICATION RULES: COMPACT: both λ_f ≤ λ_pf AND λ_w ≤ λ_pw → M_n = M_p (if braced) NON-COMPACT: λ_pf < λ_f ≤ λ_rf OR λ_pw < λ_w ≤ λ_rw → M_n = linear interpolation between M_p and 0.7 F_y S_x (similar to LTB Zone 2 but for local buckling) SLENDER: λ_f > λ_rf OR λ_w > λ_rw → M_n = F_cr S_x (elastic local buckling; significant capacity reduction) Note: For most standard rolled W-shapes with F_y ≤ 345 MPa, the flanges and web are compact. For higher-strength steels (F_y = 485 MPa or higher) or built-up sections with wide thin flanges, non-compact or slender elements may occur. PRACTICAL IMPLICATION FOR BOARD EXAMS: Problem statements that say 'compact section' relieve you of the compactness check. If the problem says 'a W350×90 beam of A572 Gr.50,' you should verify compactness using tabulated properties or by checking λ_f and λ_w against the limits.
Examples
Both elements pass the compact limit. Note that even at Gr.50 steel, typical W-shapes remain compact. The web limit λ_pw = 90.5 is very lenient — only very slender plate girder webs fail this check.
Scenario
A W350×90 section: b_f = 250 mm, t_f = 16 mm, h = 290 mm (clear web height), t_w = 9.5 mm, F_y = 345 MPa. Classify the section.
Solution
√(E/F_y) = √(200,000/345) = 24.08 Flange check: λ_f = b_f/(2t_f) = 250/(2×16) = 7.81 λ_pf = 0.38 × 24.08 = 9.15 7.81 < 9.15 → COMPACT FLANGE ✓ Web check: λ_w = h/t_w = 290/9.5 = 30.5 λ_pw = 3.76 × 24.08 = 90.5 30.5 < 90.5 → COMPACT WEB ✓ Conclusion: Section is COMPACT for F_y = 345 MPa. M_n = M_p (if L_b ≤ L_p).
Applications
- Selection of W-shapes for beam design — compact sections are preferred to utilize M_p and provide ductility for seismic design.
- Assessment of hybrid and built-up girders where individual plates may not meet compactness requirements.
- Seismic design provisions (AISC 341 Seismic Provisions) require 'highly ductile' sections with even tighter λ limits than the standard compact limits.
Misconceptions
- WRONG: Using full flange width b_f without dividing by 2 for λ_f — only the unsupported projection (half-flange) is the outstanding element.
- WRONG: Assuming all W-shapes are automatically compact for any F_y — at very high yield strengths, λ_p decreases and some sections become non-compact.
- WRONG: Using h = d (total depth) instead of h (clear web height between flanges) for the web check — h = d - 2(t_f + k) where k is the fillet size (use tabulated h/t_w from AISC Manual).
Related Concepts
- Local Buckling — flange local buckling (FLB) and web local buckling (WLB) limit states
- Plastic Moment M_p — only achievable for compact sections with adequate bracing
- Seismic Compact (Highly Ductile) Limits — tighter than standard compact, required for SMF and IMF in seismic design
- Plate Girder Design — slender web classification leads to tension field action (Vierendeel) shear design
Common Exam Questions
Example
b_f = 200 mm, t_f = 12 mm, t_w = 8 mm, h = 300 mm, F_y = 248 MPa: λ_f = 200/(2×12) = 8.33 < 10.79 (compact); λ_w = 300/8 = 37.5 < 106.8 (compact). → Compact section.
Approach
Compute λ_f = b_f/(2t_f) and λ_w = h/t_w. Compare with 0.38√(E/F_y) and 3.76√(E/F_y) respectively. State the classification.
Question Type
Compactness check given section dimensions
Key Points To Remember
- Compact = both flange AND web λ ≤ λ_p (two conditions, both must be met).
- λ_pf = 0.38√(E/F_y) for flanges; λ_pw = 3.76√(E/F_y) for webs — memorize the coefficients 0.38 and 3.76.
- A single slender element (flange OR web) downgrades the entire section.
- Most standard W-shapes with F_y ≤ 345 MPa are compact — but verify for high-strength steels.
- Non-compact sections use interpolation between M_p and 0.7 F_y S_x for flange local buckling (FLB) or web local buckling (WLB).
- λ_f = b_f/(2t_f) — the factor 2 in the denominator is often missed (half-flange width is the unsupported projection).
Practice Problems
Fully braced compact beam → use M_n = M_p = F_y Z_x. The shape factor of 1.137 means the beam can carry 13.7% more moment beyond first yield before fully plasticizing. φ_b = 0.90 for all flexural limit states in LRFD. Note the unit conversion: N·mm ÷ 10⁶ = kN·m.
Problem
PROBLEM 1 — Plastic Moment (Board Exam Type) A W530×82 compact steel beam (Z_x = 2,080 × 10³ mm³, S_x = 1,830 × 10³ mm³) is made of A36 steel (F_y = 248 MPa). The compression flange is continuously braced. Determine: (a) the plastic moment M_p, (b) the LRFD design flexural strength φ_b M_n, and (c) the shape factor.
Solution
(a) Plastic Moment: M_p = F_y × Z_x = 248 × 2,080,000 = 516,240,000 N·mm = 516.2 kN·m (b) LRFD Design Flexural Strength: φ_b M_n = 0.90 × M_p = 0.90 × 516.2 = 464.6 kN·m (c) Shape Factor: f = Z_x / S_x = 2,080,000 / 1,830,000 = 1.137 Final Answers: (a) M_p = 516.2 kN·m (b) φ_b M_n = 464.6 kN·m (c) f = 1.137
The computation of L_p is a must-know formula: 1.76 × r_y × √(E/F_y). For A36 steel, √(200,000/248) = 28.40 — memorize this value. The comparison L_b vs L_p is a YES/NO decision: if L_b ≤ L_p, full M_p is achieved; if L_b > L_p, LTB reduces M_n.
Problem
PROBLEM 2 — Limit Length L_p (Board Exam Type) For a W-shape beam with weak-axis radius of gyration r_y = 45.5 mm, F_y = 248 MPa, and E = 200,000 MPa: (a) Compute L_p, the maximum unbraced length for full plastic moment. (b) If the actual brace spacing is L_b = 1.8 m, does the beam develop M_p?
Solution
(a) Compute L_p: L_p = 1.76 r_y √(E/F_y) = 1.76 × 45.5 × √(200,000/248) = 80.08 × √(806.45) = 80.08 × 28.40 = 2,274 mm = 2.27 m (b) Compare L_b vs L_p: L_b = 1.8 m < L_p = 2.27 m → L_b is within Zone 1 (Plastic Zone) → The beam CAN develop M_p = F_y Z_x fully. Final Answers: (a) L_p = 2,274 mm ≈ 2.27 m (b) YES — since L_b < L_p, the beam develops full M_p with no LTB reduction.
Key points: (1) A_w = d × t_w = 600 × 12 (full depth, no subtraction of flanges). (2) 0.6 × 248 = 148.8; 148.8 × 7,200 = 1,071,360 N = 1,071.4 kN. (3) φ_v = 1.0 for compact web — applying 0.90 here would be incorrect and would underestimate capacity by 10%. This is a classic board exam trap.
Problem
PROBLEM 3 — Shear Strength (Board Exam Type) A simply supported W600×110 beam (d = 600 mm, t_w = 12 mm) of A36 steel (F_y = 248 MPa) carries a factored uniform load. The web is compact (C_v = 1.0). Find: (a) the gross web area A_w, (b) the nominal shear strength V_n, and (c) the LRFD design shear strength φ_v V_n.
Solution
(a) Gross Web Area: A_w = d × t_w = 600 × 12 = 7,200 mm² (b) Nominal Shear Strength: V_n = 0.6 F_y A_w C_v = 0.6 × 248 × 7,200 × 1.0 = 1,071,360 N = 1,071.4 kN (c) Design Shear Strength: φ_v = 1.0 (compact web, rolled I-shape) φ_v V_n = 1.0 × 1,071.4 = 1,071.4 kN Final Answers: (a) A_w = 7,200 mm² (b) V_n = 1,071.4 kN (c) φ_v V_n = 1,071.4 kN
Zone 2 procedure: (1) Compute M_p and 0.7F_yS_x. (2) Compute the interpolation ratio. (3) Apply the interpolation formula. (4) Multiply by C_b = 1.14 (this gives a 14% bonus for the non-uniform moment gradient). (5) Cap at M_p. (6) Apply φ_b = 0.90. The C_b factor raised M_n from 362.1 to 412.8 kN·m — significant but not enough to reach M_p = 446.4 kN·m.
Problem
PROBLEM 4 — Inelastic LTB (Board Exam Type) A compact W-shape beam has: Z_x = 1,800 × 10³ mm³, S_x = 1,600 × 10³ mm³, F_y = 248 MPa, L_p = 2.0 m, L_r = 6.0 m. The unbraced length is L_b = 4.0 m and C_b = 1.14 (simply supported with UDL). Find: (a) M_p, (b) 0.7 F_y S_x, (c) M_n for this unbraced segment, (d) φ_b M_n.
Solution
(a) Plastic Moment: M_p = F_y Z_x = 248 × 1,800,000 = 446,400,000 N·mm = 446.4 kN·m (b) Reference Moment at L_r: 0.7 F_y S_x = 0.7 × 248 × 1,600,000 = 277,760,000 N·mm = 277.8 kN·m (c) Zone Check: L_p = 2.0 m < L_b = 4.0 m ≤ L_r = 6.0 m → ZONE 2 (Inelastic LTB) M_n = C_b [M_p - (M_p - 0.7 F_y S_x)(L_b - L_p)/(L_r - L_p)] ≤ M_p (L_b - L_p)/(L_r - L_p) = (4.0 - 2.0)/(6.0 - 2.0) = 2.0/4.0 = 0.50 M_n = 1.14 × [446.4 - (446.4 - 277.8)(0.50)] = 1.14 × [446.4 - (168.6)(0.50)] = 1.14 × [446.4 - 84.3] = 1.14 × 362.1 = 412.8 kN·m Cap check: 412.8 kN·m < M_p = 446.4 kN·m ✓ (no cap needed) (d) Design Flexural Strength: φ_b M_n = 0.90 × 412.8 = 371.5 kN·m Final Answers: (a) M_p = 446.4 kN·m (b) 0.7 F_y S_x = 277.8 kN·m (c) M_n = 412.8 kN·m (d) φ_b M_n = 371.5 kN·m
This combined check is the most realistic board exam scenario. Flexure governs here (as it typically does for normal spans). The minimum required Z_x can be back-computed: Z_x,req = M_u/(φ_b F_y) = 360 × 10⁶/(0.90 × 248) = 1,613,600 mm³ = 1,614 × 10³ mm³. The given W410×85 with Z_x = 1,510 × 10³ mm³ is insufficient — select the next heavier section.
Problem
PROBLEM 5 — Combined Check (Board Exam Type) A simply supported beam spans 6 m and carries a factored uniform load w_u = 80 kN/m. The beam is W410×85 (d = 417 mm, t_w = 10.9 mm, Z_x = 1,510 × 10³ mm³, F_y = 248 MPa). The section is compact and fully braced. Check adequacy for BOTH flexure and shear.
Solution
STEP 1: Factored Internal Forces Maximum moment (midspan): M_u = w_u L²/8 = 80 × 6²/8 = 80 × 4.5 = 360 kN·m Maximum shear (at support): V_u = w_u L/2 = 80 × 6/2 = 240 kN STEP 2: Flexure Check M_p = F_y Z_x = 248 × 1,510,000 = 374,480,000 N·mm = 374.5 kN·m φ_b M_n = 0.90 × 374.5 = 337.0 kN·m Demand M_u = 360 kN·m Check: M_u = 360 kN·m > φ_b M_n = 337.0 kN·m → FLEXURE FAILS ✗ STEP 3: Shear Check A_w = d × t_w = 417 × 10.9 = 4,545 mm² V_n = 0.6 × 248 × 4,545 × 1.0 = 675,504 N = 675.5 kN φ_v V_n = 1.0 × 675.5 = 675.5 kN Check: V_u = 240 kN < φ_v V_n = 675.5 kN → SHEAR ADEQUATE ✓ Conclusion: W410×85 is INADEQUATE in FLEXURE but adequate in shear. A heavier section with larger Z_x must be selected (e.g., W410×100 or W460×82 with Z_x ≥ 360/0.9/248 × 10⁶ = 1,613 × 10³ mm³).
Exam Preparation Tips
- MEMORIZE THE FOUR KEY FORMULAS in order: (1) M_p = F_y Z_x, (2) L_p = 1.76 r_y √(E/F_y), (3) Zone 2 LTB interpolation for M_n, (4) V_n = 0.6 F_y A_w C_v. These appear on nearly every board exam session covering Steel Design.
- PRECOMPUTE √(E/F_y) FOR COMMON STEELS: A36 (F_y = 248): √(200,000/248) = 28.40; A572 Gr.50 (F_y = 345): √(200,000/345) = 24.08. Storing these in your memory saves critical exam time.
- KNOW YOUR φ FACTORS by limit state: Flexure φ_b = 0.90; Shear (compact rolled I) φ_v = 1.00; Shear (others) φ_v = 0.90. Using 0.90 for shear of rolled I-shapes is a -5% to -10% error.
- ALWAYS CLASSIFY THE ZONE FIRST in LTB problems: compute or read L_p and L_r, compare with L_b, then apply only the correct zone formula. Mixing up Zone 1 and Zone 2 equations is the #1 source of LTB errors.
- USE FULL DEPTH d IN SHEAR: A_w = d × t_w, not (d - 2t_f) × t_w. This is explicitly stated in AISC 360 Section G2.1 and has been tested explicitly in board exams.
- WATCH UNITS CAREFULLY: Forces in kN and dimensions in mm → result in N·mm → divide by 10⁶ for kN·m. A common error is computing in N·mm and reporting as kN·m without converting, giving an answer 10⁶ times too large.
- FOR C_b: Unless the problem specifically provides C_b or gives moment diagram data to compute it, use C_b = 1.0 (conservative). Never apply C_b < 1.0 — it is always ≥ 1.0.
- REMEMBER THE CAP: C_b × M_n (computed from Zone 2 formula) must never exceed M_p. Always check this cap — examiners deliberately set up problems where C_b pushes the value above M_p.
- IN MULTIPLE-CHOICE EXAMS, eliminate answers that use S_x (elastic modulus) when computing M_p for compact beams — Z_x is always larger than S_x, so M_p > M_y. If two answers differ by ~12%, the larger one using Z_x is likely correct.
- REVIEW NSCP 2015 SECTION 502 which adopts AISC 360 by reference. Understanding this adoption means all AISC formulas have legal force in Philippine practice. The PRC board exam tests AISC 360 provisions as the governing standard for structural steel design.
- BACK-COMPUTE REQUIRED PROPERTIES: Know how to find Z_x,req = M_u/(φ_b F_y) for beam selection problems, and A_w,req = V_u/(φ_v × 0.6 F_y) for shear adequacy problems. Selection-type questions are common board exam formats.
In summary
Steel beam design for flexure and shear can be distilled to four governing equations and a simple decision tree. For the PRC Civil Engineer Licensure Examination, you must be able to (1) compute M_p = F_y Z_x for compact, fully braced beams — always using Z_x, never S_x; (2) compute L_p = 1.76 r_y √(E/F_y) and classify a beam's unbraced length into Zone 1, 2, or 3 to determine the appropriate M_n; (3) apply the LTB interpolation formula in Zone 2, multiply by C_b ≥ 1.0, and cap the result at M_p; and (4) compute V_n = 0.6 F_y A_w C_v using A_w = d × t_w (full depth) with φ_v = 1.00 for standard rolled I-shapes. The resistance factors φ_b = 0.90 (flexure) and φ_v = 1.00 (shear, compact web) are non-negotiable — confusing them costs points. Most real-world W-shapes in A36 or A572 Gr.50 are compact by default, making Zone 1 computation straightforward. The critical challenge is Zone 2 LTB, which requires careful arithmetic with M_p, 0.7 F_y S_x, and the length ratios. Practice the five worked problems in this chapter until you can solve each type in under 5 minutes — that is the pace required for board exam success. Under Philippine law (RA 544, as amended), registered civil engineers bear professional responsibility for the structural adequacy of their designs, making rigorous mastery of these provisions not only exam-essential but a lifelong professional obligation.
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