Steel Beams: Flexure and Shear - Revision Notes
Steel beams are the workhorses of structural frames — they carry gravity loads primarily through bending and shear. For the PRC Civil Engineer Licensure Examination, you must master three interconnected limit states: (1) plastic moment capacity Mp for compact, adequately braced beams; (2) lateral-torsional buckling (LTB) when the compression flange is insufficiently braced; and (3) shear yielding of the web. These topics appear consistently in board exams under Steel & Timber Design (NSCP 2015 / AISC 360-10 provisions adopted in the Philippines). This chapter arms you with every formula, concept, and board-exam trap you need to score high.
1. Plastic Moment Capacity — Compact, Fully Braced Beams
Formulas
Mn = Mp = Fy × Zx
Example
Zx = 1.2×10⁶ mm³, Fy = 248 MPa → Mp = 248 × 1.2×10⁶ = 297.6×10⁶ N·mm = 297.6 kN·m; φbMn = 0.90 × 297.6 = 267.8 kN·m
Variables
Fy = specified minimum yield stress (MPa); Zx = plastic section modulus about strong axis (mm³)
Application
Compact section, Lb ≤ Lp. This is the maximum possible nominal moment strength.
φbMn = 0.90 Mp
Example
If Mp = 297.6 kN·m, then φbMn = 267.8 kN·m. The factored moment demand Mu must not exceed 267.8 kN·m.
Variables
φb = 0.90 (LRFD flexure resistance factor); Mp = plastic moment (N·mm or kN·m)
Application
LRFD design check: Mu ≤ φbMn
Shape factor f = Zx / Sx
Example
If Zx = 1 200 000 mm³ and Sx = 1 070 000 mm³, then f = 1.2/1.07 = 1.12 — typical of I-sections.
Variables
Zx = plastic section modulus (mm³); Sx = elastic section modulus (mm³)
Application
Quantifies the reserve capacity beyond first yield. For W-shapes, f ≈ 1.10–1.18.
Exam tips
- Memorize: Mp = FyZx and φb = 0.90. These two lines solve most straightforward board problems.
- If the problem gives Sx only, compute Zx = f × Sx using f ≈ 1.12 for W-shapes (unless exact value is given).
- Double-check whether the problem asks for Mn (nominal) or φbMn (design strength) — boards ask both.
- For Fy = 248 MPa (A36 equivalent) and Fy = 345 MPa (A572 Gr.50) — two most common values in Philippine board exams.
Key points
- The nominal flexural strength of a compact, fully braced steel beam equals the plastic moment: Mn = Mp = Fy × Zx.
- The plastic section modulus Zx is always larger than the elastic section modulus Sx; their ratio is the shape factor f = Zx/Sx ≈ 1.12 for standard W-shapes.
- The LRFD resistance factor for flexure is φb = 0.90; the ASD safety factor is Ωb = 1.67.
- Design flexural strength (LRFD): φbMn = 0.90 FyZx.
- Allowable flexural strength (ASD): Mn/Ωb = FyZx/1.67.
- Full Mp is only achievable when the section is compact AND the compression flange is braced at intervals Lb ≤ Lp.
- A section is compact if both its flange and web width-to-thickness ratios are at or below the compact limit λp.
- For I-shapes: flange compactness requires bf/(2tf) ≤ λp = 0.38√(E/Fy); web compactness requires h/tw ≤ λp = 3.76√(E/Fy).
Definitions
Plastic Section Modulus (Zx)
The first moment of area of the cross-section about the plastic neutral axis (PNA), computed by summing the product of each area segment and its distance from the PNA. For doubly-symmetric I-shapes, Zx = bf tf (d − tf) + tw(d − 2tf)²/4.
Importance
Used directly in Mp = FyZx. Confusing Zx with Sx (elastic modulus) is the single most common board-exam error.
Compact Section
A cross-section whose flange and web width-to-thickness ratios are both below the compact limit λp, ensuring the section can develop its full plastic moment without local buckling.
Importance
Only compact sections can achieve Mp. Non-compact or slender sections have reduced Mn due to local buckling.
Shape Factor (f)
Ratio f = Zx/Sx, representing how much additional moment capacity exists beyond first yield (My = FySx) up to full plastification (Mp = FyZx).
Importance
For I-shapes f ≈ 1.12, meaning Mp is about 12% larger than My — this reserve is what LRFD harnesses.
Common mistakes
- Using Sx (elastic modulus) instead of Zx (plastic modulus) when computing Mp — this under-estimates capacity by the shape factor (~12% for W-sections).
- Applying φb = 0.90 to Mp without first checking compactness and Lb ≤ Lp.
- Forgetting to convert N·mm to kN·m (divide by 10⁶) — unit errors cause wrong answers in board exams.
- Treating all steel beams as compact without verifying flange and web λ ratios.
2. Lateral-Torsional Buckling (LTB) and Unbraced Length Limits
Formulas
Lp = 1.76 ry √(E / Fy)
Example
ry = 40 mm, Fy = 248 MPa, E = 200,000 MPa → Lp = 1.76 × 40 × √(200,000/248) = 70.4 × 28.40 = 2,000 mm = 2.0 m
Variables
ry = radius of gyration about weak axis (mm); E = modulus of elasticity = 200,000 MPa; Fy = yield stress (MPa)
Application
Maximum unbraced length for full plastic moment. If Lb ≤ Lp, LTB does not govern.
Mn = Cb [Mp − (Mp − 0.7FySx)(Lb − Lp)/(Lr − Lp)] ≤ Mp
Example
With Cb = 1.0 (uniform moment), this gives the minimum strength for the inelastic LTB range.
Variables
Cb = moment gradient factor; Mp = FyZx; Sx = elastic section modulus; Lb = unbraced length; Lp, Lr = limit lengths
Application
Inelastic LTB zone (Lp < Lb ≤ Lr). Linear interpolation between Mp and 0.7FySx.
Mn = Fcr Sx ≤ Mp (Elastic LTB, Lb > Lr)
Example
When Lb is very large (say 8 m for a typical W-section), Mn may be only 50–60% of Mp.
Variables
Fcr = elastic LTB critical stress; Sx = elastic section modulus
Application
Elastic LTB zone — Mn is governed by elastic buckling, significantly reduced. Compute Fcr from AISC Table or full formula.
Cb = 12.5Mmax / (2.5Mmax + 3MA + 4MB + 3MC)
Example
For a simply supported beam with midspan point load: Cb ≈ 1.32 — increases allowable unbraced length effectively.
Variables
Mmax = maximum moment in unbraced segment; MA = moment at quarter point; MB = midspan moment; MC = three-quarter point moment
Application
Moment gradient factor. Cb > 1.0 when moment varies along Lb (e.g., midspan load). Cb = 1.0 for uniform moment (conservative).
Exam tips
- Board-exam LTB problems almost always fall in Zone 1 (Lb ≤ Lp) or require computing Lp — master Lp = 1.76ry√(E/Fy) cold.
- When Lb is given but Lp is not, compute Lp first and compare — this sets up the rest of the solution.
- Remember: 0.7FySx is the lower bound of the inelastic LTB zone — it equals 70% of My, not 70% of Mp.
- For quick estimates: at Fy = 248 MPa and typical ry = 35–50 mm, Lp falls in the 1.7–2.5 m range — beams longer than this need checking.
- Cb is always ≥ 1.0; if your computed Cb < 1.0, recheck — it cannot be less than 1.0 by definition.
Key points
- LTB occurs when an insufficiently braced compression flange moves laterally and twists, preventing full plastic moment development.
- Three LTB zones are defined by comparing the unbraced length Lb to limit lengths Lp and Lr.
- Zone 1 — No LTB (Lb ≤ Lp): Mn = Mp. Full plastic moment is available.
- Zone 2 — Inelastic LTB (Lp < Lb ≤ Lr): Mn reduces linearly from Mp to 0.7FySx.
- Zone 3 — Elastic LTB (Lb > Lr): Mn = FcrSx, governed by elastic buckling; Mn can fall well below Mp.
- The moment gradient factor Cb amplifies Mn for non-uniform moment diagrams; Cb = 1.0 for uniform moment (most conservative). Cb always caps at Mp.
- For doubly-symmetric I-shapes: Lp = 1.76 ry√(E/Fy) and Lr depends on section warping properties.
- Increasing brace frequency (reducing Lb) is the most direct way to restore Mp capacity.
Definitions
Unbraced Length (Lb)
The distance between points that are braced against lateral displacement of the compression flange or against twist of the cross-section.
Importance
Determines which LTB zone governs. Underestimating Lb (missing a brace point) leads to unconservative designs.
Lp — Plastic Limit Unbraced Length
The maximum Lb below which the full plastic moment Mp can be developed without reduction. Lp = 1.76ry√(E/Fy).
Importance
Key threshold for board problems — if Lb ≤ Lp, the beam achieves full Mp and LTB need not be checked further.
Lr — Elastic Limit Unbraced Length
The unbraced length above which LTB becomes fully elastic. Between Lp and Lr, buckling is inelastic.
Importance
Defines the boundary between inelastic LTB (linear interpolation) and elastic LTB (Fcr formula).
Moment Gradient Factor (Cb)
A factor ≥ 1.0 that accounts for the beneficial effect of non-uniform moment along the unbraced segment. Uniform moment gives Cb = 1.0 (most critical). Higher Cb means less LTB risk.
Importance
Using Cb = 1.0 when a higher value is justified is conservative but wastes steel. Board problems sometimes test whether you can compute Cb correctly.
Common mistakes
- Forgetting to check whether Lb ≤ Lp before jumping to LTB calculations — always locate the LTB zone first.
- Using Cb = 1.0 in problems that explicitly give moment values at quarter points — compute Cb; it can significantly raise Mn.
- Confusing ry (weak-axis radius of gyration) with rx (strong-axis) when computing Lp.
- Not capping Mn at Mp even when Cb > 1.0 raises the calculated value above Mp.
- Mixing up the inelastic and elastic LTB formulas — the linear interpolation formula applies only between Lp and Lr.
3. Local Buckling — Compact vs. Non-Compact vs. Slender Sections
Formulas
λpf = 0.38 √(E / Fy) [compact flange limit]
Example
Fy = 248 MPa → λpf = 0.38√(200,000/248) = 0.38 × 28.40 = 10.79. A W-flange with bf/(2tf) = 8.5 is compact.
Variables
E = 200,000 MPa; Fy = yield stress (MPa); λf = bf/(2tf) = actual flange slenderness
Application
If bf/(2tf) ≤ 0.38√(E/Fy), the flange is compact — no local buckling before full plastification.
λpw = 3.76 √(E / Fy) [compact web limit]
Example
Fy = 248 MPa → λpw = 3.76 × 28.40 = 106.8. Most standard W-shapes satisfy this easily.
Variables
E = 200,000 MPa; Fy = yield stress (MPa); λw = h/tw = actual web slenderness
Application
If h/tw ≤ 3.76√(E/Fy), the web is compact — no web local buckling before Mp.
Exam tips
- Most PRC board problems involving standard W-sections implicitly assume compact sections — but if the problem states 'non-compact' or gives unusually thin flanges, reduce Mn accordingly.
- Quick check: if Fy = 248 MPa and the section is a standard W-shape from AISC tables, it is almost certainly compact.
- For Fy = 345 MPa, recompute λp = 0.38√(200,000/345) = 9.15 for flanges — some W-shapes fall non-compact.
Key points
- Local buckling of flange or web plate elements can reduce Mn below Mp before LTB even occurs.
- Three classification tiers: Compact (λ ≤ λp) → full Mp; Non-compact (λp < λ ≤ λr) → Mp reduced; Slender (λ > λr) → Mn governed by elastic local buckling.
- Flange slenderness: λf = bf / (2tf) compared to λp = 0.38√(E/Fy) and λr = 1.0√(E/Fy).
- Web slenderness: λw = h / tw compared to λp = 3.76√(E/Fy) and λr = 5.70√(E/Fy).
- Most standard W-sections (W-shapes) from AISC tables are compact for Fy = 248 MPa — verify for Fy = 345 MPa.
- Non-compact flange: Mn interpolates linearly between Mp (at λp) and 0.7FySx (at λr) — same form as inelastic LTB.
- Slender sections use critical stress formulas similar to plate buckling.
Definitions
Width-to-Thickness Ratio (λ)
A dimensionless plate slenderness ratio — for flanges: λf = bf/(2tf); for webs: λw = h/tw — compared to code limits λp and λr to classify a section.
Importance
Governs whether local buckling reduces Mn. Must be checked before applying Mp formulas.
Non-Compact Section
A section where at least one plate element has λp < λ ≤ λr. The section can yield locally but cannot develop full strain hardening needed for Mp.
Importance
Non-compact sections require strength reduction — ignoring this is unconservative and a common board-exam trap.
Common mistakes
- Assuming all W-sections are compact — for higher grades (Fy = 345 MPa or above), λp values are lower and some sections become non-compact.
- Using the full depth d instead of the clear web height h in computing web slenderness λw = h/tw.
- Forgetting that both flange AND web must be compact for the section to qualify as compact overall.
4. Shear Strength of Steel Beams
Formulas
Vn = 0.6 Fy Aw Cv
Example
d = 450 mm, tw = 10 mm, Fy = 248 MPa, Cv = 1.0 → Aw = 450×10 = 4,500 mm²; Vn = 0.6×248×4,500×1.0 = 669,600 N = 669.6 kN
Variables
Fy = yield stress (MPa); Aw = dtw = web area (mm²); d = overall section depth (mm); tw = web thickness (mm); Cv = shear buckling coefficient (1.0 for stocky webs)
Application
Primary shear strength formula for I-shaped beams. Governs shear design check.
φvVn = 1.0 × 0.6 Fy Aw (for stocky webs, Cv = 1.0, φv = 1.0)
Example
From above: φvVn = 1.0 × 669.6 = 669.6 kN. If Vu ≤ 669.6 kN, the web is adequate in shear.
Variables
φv = 1.0 for rolled I-shapes with h/tw ≤ 2.24√(E/Fy); otherwise φv = 0.90
Application
LRFD design shear strength. Most common board-exam scenario uses φv = 1.0.
Aw = d × tw
Example
W530×74 has d = 529 mm, tw = 9.65 mm → Aw = 529 × 9.65 = 5,105 mm²
Variables
d = total nominal depth of section (mm); tw = web thickness (mm)
Application
Web area for shear calculations. Note: this is total depth × tw, NOT clear web height × tw.
Stocky web limit: h/tw ≤ 2.24 √(E/Fy)
Example
Fy = 248 MPa → limit = 2.24√(200,000/248) = 2.24 × 28.40 = 63.6. A web with h/tw = 50 is stocky → Cv = 1.0, φv = 1.0.
Variables
h = clear distance between flanges; tw = web thickness; E = 200,000 MPa; Fy = yield stress
Application
If this condition is met → Cv = 1.0, φv = 1.0. Most rolled W-shapes satisfy this for Fy ≤ 345 MPa.
Exam tips
- Shear problems are usually straightforward in board exams: φvVn = 1.0 × 0.6 × Fy × d × tw. Three numbers multiplied — do not overthink it.
- The factor 0.6 comes from von Mises yield criterion: shear yield stress τy = 0.577Fy ≈ 0.6Fy.
- When a problem says 'compact web' or 'stocky web', immediately write φv = 1.0 and Cv = 1.0.
- For Fy = 248 MPa: φvVn = 1.0 × 0.6 × 248 × Aw = 148.8 Aw (N) where Aw in mm². Memorize this shortcut.
Key points
- Steel beam webs resist shear primarily through shear yielding — the web acts like a shear panel.
- Nominal shear strength: Vn = 0.6 Fy Aw Cv.
- Web area: Aw = d × tw (overall depth d × web thickness tw).
- For stocky webs where h/tw ≤ 2.24√(E/Fy): Cv = 1.0 and φv = 1.0 (LRFD).
- For webs with h/tw > 2.24√(E/Fy): Cv < 1.0 (shear buckling governs) and φv = 0.90.
- Most standard rolled W-shapes satisfy h/tw ≤ 2.24√(E/Fy) for Fy = 248 MPa — use Cv = 1.0 and φv = 1.0.
- LRFD check: Vu ≤ φvVn; ASD check: Va ≤ Vn/Ωv where Ωv = 1.50 (or 1.67 for non-stocky webs).
- Do NOT use net web height for Aw in the AISC shear formula — use full depth d.
Definitions
Shear Coefficient Cv
A dimensionless factor (0 < Cv ≤ 1.0) accounting for web shear buckling. Cv = 1.0 when the web is stocky enough to develop full shear yielding without buckling.
Importance
Using Cv = 1.0 when the web is actually slender is unconservative. Always verify the h/tw limit.
Web Area (Aw)
Product of the overall section depth d and the web thickness tw: Aw = d × tw. This differs from the clear web area used in some other formulas.
Importance
Using net web height instead of full depth d underestimates Vn by about 10–15% for typical W-sections.
Shear Resistance Factor (φv)
LRFD factor for shear. φv = 1.0 for rolled I-shapes with stocky webs (most common); φv = 0.90 for all other cases.
Importance
φv = 1.0 is a special, more liberal provision unique to shear — unlike the universal 0.90 in earlier AISC editions.
Common mistakes
- Using φv = 0.90 for all shear calculations — for rolled W-shapes with stocky webs, φv = 1.0 (AISC 360-10 Chapter G).
- Computing Aw using clear web height instead of total depth d — the AISC shear formula explicitly uses d × tw.
- Forgetting to check the stocky web limit h/tw ≤ 2.24√(E/Fy) before assuming Cv = 1.0.
- Applying shear reduction (Cv < 1.0) to standard rolled W-shapes — they are almost always stocky enough for Cv = 1.0.
5. Design Summary and Step-by-Step Procedure
Formulas
LRFD flexure check: Mu ≤ φbMn = 0.90 FyZx (when Lb ≤ Lp, compact section)
Example
If Mu = 250 kN·m and φbMn = 267.8 kN·m → 250 ≤ 267.8 ✓ Section is adequate.
Variables
Mu = factored moment demand (kN·m); φbMn = design flexural strength (kN·m)
Application
Final design adequacy check for flexure under LRFD.
LRFD shear check: Vu ≤ φvVn = 1.0 × 0.6FyAw (stocky web)
Example
If Vu = 300 kN and φvVn = 669.6 kN → 300 ≤ 669.6 ✓ Web is adequate in shear.
Variables
Vu = factored shear demand (kN); φvVn = design shear strength (kN)
Application
Final design adequacy check for shear under LRFD.
Exam tips
- In multi-part board problems, part (a) often asks for flexural strength, part (b) for shear strength — solve systematically, do not mix formulas.
- When a W-section is given by designation (e.g., W310×97), look up its properties from AISC tables — boards provide these in the problem appendix.
- If Lb is not mentioned in a board problem and the section is compact, assume full bracing (Lb ≤ Lp) → use Mn = Mp.
- Always state units clearly: N·mm vs kN·m, mm vs m — unit errors are the #1 arithmetic pitfall.
Key points
- Steel beam design follows a logical sequence: classify section → determine governing limit state → compute Mn → apply φb → check Mu ≤ φbMn.
- Shear is checked separately: compute φvVn → check Vu ≤ φvVn.
- Deflection checks (serviceability) use unfactored loads and elastic section modulus Ix — separate from strength checks.
- In Philippine practice (NSCP 2015 Section 502), AISC 360 provisions are adopted by reference.
- Beam selection from tables uses φbMpx values directly — verify Lb ≤ Lp and then pick the lightest section satisfying φbMpx ≥ Mu.
- Always specify: Fy, E, section properties (Zx, Sx, ry, d, tw, bf, tf), Lb, and loading pattern before starting calculations.
Definitions
Limit State
A condition beyond which a structural member no longer fulfills its intended function. For steel beams: yielding/plastic moment, LTB, flange local buckling, web local buckling, and shear yielding are the primary limit states.
Importance
AISC 360 and NSCP 2015 design is limit-state based — every check corresponds to a specific limit state.
Common mistakes
- Skipping the compactness classification and LTB check — assuming all beams achieve Mp without verification.
- Checking flexure but forgetting the shear check, or vice versa — both must be satisfied.
- Using factored loads for deflection calculations — deflection uses service (unfactored) loads.
- Selecting a section from tables without verifying Lb ≤ Lp — table φbMp values assume full bracing.
Connections
- Plastic moment (Mp = FyZx) connects to structural analysis — the plastic hinge moment used in plastic analysis of indeterminate frames equals Mp, forming the basis for plastic design and pushover analysis.
- Lateral-torsional buckling (LTB) is conceptually analogous to Euler column buckling — both are elastic instability phenomena; a slender column buckles under compression just as an unbraced beam buckles under bending.
- The compact section classification (λ ≤ λp) links directly to section classification in concrete (ductile vs. brittle failure) and timber (allowable stress vs. LRFD approaches) — the concept of ensuring yielding before buckling is universal.
- Shear yield stress τy = 0.577Fy (von Mises) → approximated as 0.6Fy in AISC, connecting beam shear design to Material Strength & Mechanics of Materials (Mohr's circle, principal stresses).
- The moment gradient factor Cb ties beam design to structural analysis — moment diagrams from different load patterns (uniform load, point load, cantilever) directly affect Cb and thus LTB capacity.
- Deflection serviceability checks (separate from strength) use Ix and elastic analysis — connecting beam design to Structural Theory (moment-area method, conjugate beam, virtual work).
- NSCP 2015 Section 502 adopts AISC 360 by reference — understanding the Philippine code framework (RA 544, the Civil Engineering Law) contextualizes why these design provisions apply in practice.
- Connection design (bolts, welds at beam ends) must transmit the shear force Vu computed here — linking beam shear capacity to connection design in the next chapter.
Quick review questions
A compact W-section with Zx = 2.0×10⁶ mm³ and Fy = 345 MPa is fully braced. What is the LRFD design flexural strength φbMn?
Answer
φbMn = 0.90 × 345 × 2.0×10⁶ = 621×10⁶ N·mm = 621 kN·m
Since the section is compact and fully braced (Lb ≤ Lp implied), Mn = Mp = FyZx = 345 × 2.0×10⁶ = 690×10⁶ N·mm = 690 kN·m. Applying φb = 0.90: φbMn = 0.90 × 690 = 621 kN·m.
Compute Lp for a section with ry = 50 mm, Fy = 345 MPa, E = 200,000 MPa.
Answer
Lp = 1.76 × 50 × √(200,000/345) = 88 × 24.07 = 2,118 mm ≈ 2.12 m
Lp = 1.76 ry√(E/Fy) = 1.76 × 50 × √(200,000/345). √(200,000/345) = √579.7 = 24.07. So Lp = 88 × 24.07 = 2,118 mm = 2.12 m. This means if braces are spaced at ≤ 2.12 m, the beam achieves full Mp.
A steel beam has d = 600 mm, tw = 12 mm, Fy = 248 MPa, compact/stocky web. Find the LRFD design shear strength φvVn.
Answer
φvVn = 1.0 × 0.6 × 248 × (600 × 12) = 1.0 × 0.6 × 248 × 7,200 = 1,071,360 N = 1,071.4 kN
Aw = d × tw = 600 × 12 = 7,200 mm². Since the web is stocky, Cv = 1.0 and φv = 1.0. Vn = 0.6 × 248 × 7,200 × 1.0 = 1,071,360 N = 1,071.4 kN. Design shear strength = 1.0 × 1,071.4 = 1,071.4 kN.
A W-section has Zx = 1.5×10⁶ mm³ and Sx = 1.32×10⁶ mm³. What is the shape factor, and what is My if Fy = 248 MPa?
Answer
Shape factor f = Zx/Sx = 1.5/1.32 = 1.136. My = FySx = 248 × 1.32×10⁶ = 327.4×10⁶ N·mm = 327.4 kN·m.
Shape factor f = Zx/Sx = 1.5×10⁶/1.32×10⁶ = 1.136 (about 13.6% reserve above first yield). My = FySx = 248 × 1.32×10⁶ = 327.4 kN·m. Mp = FyZx = 248 × 1.5×10⁶ = 372 kN·m — 13.6% larger than My, confirming f.
If a beam's unbraced length Lb = 3.5 m and Lp = 2.0 m, does the beam achieve full Mp? What limit state governs?
Answer
No. Since Lb = 3.5 m > Lp = 2.0 m, LTB governs and Mn < Mp. The beam is in either the inelastic or elastic LTB zone depending on comparison with Lr.
The LTB zones are: Lb ≤ Lp → full Mp; Lp < Lb ≤ Lr → inelastic LTB (linear reduction); Lb > Lr → elastic LTB. Since Lb = 3.5 m > Lp = 2.0 m, we next compare Lb to Lr. If 2.0 < 3.5 ≤ Lr, use inelastic LTB formula: Mn = Cb[Mp − (Mp − 0.7FySx)(Lb − Lp)/(Lr − Lp)] ≤ Mp.
Why is φv = 1.0 used for rolled I-shapes in shear, while φb = 0.90 is used for flexure?
Answer
AISC 360 Chapter G assigns φv = 1.0 for rolled I-shapes with stocky webs (Cv = 1.0) because shear yielding is a very ductile, predictable failure mode with lower variability than flexural buckling, justifying the higher resistance factor.
The resistance factor φ reflects uncertainty and failure mode ductility. Shear yielding in stocky webs is highly ductile and predictable, so AISC 360 allows φv = 1.0. Flexural strength involves more complex interactions (LTB, local buckling), justifying the more conservative φb = 0.90. This is a frequently asked conceptual point in board exams.
For Fy = 248 MPa and E = 200,000 MPa, compute the compact web limit λpw = 3.76√(E/Fy).
Answer
λpw = 3.76 × √(200,000/248) = 3.76 × 28.40 = 106.8
λpw = 3.76√(E/Fy) = 3.76 × √(200,000/248) = 3.76 × 28.40 = 106.8. This means a web must have h/tw ≤ 106.8 to be compact. Most standard W-shape webs satisfy this limit by a wide margin for Fy = 248 MPa.
A beam segment has Mmax = 400 kN·m, MA = 200 kN·m (quarter point), MB = 350 kN·m (midspan), MC = 250 kN·m (3/4 point). Compute Cb.
Answer
Cb = 12.5 × 400 / (2.5 × 400 + 3 × 200 + 4 × 350 + 3 × 250) = 5,000 / (1,000 + 600 + 1,400 + 750) = 5,000 / 3,750 = 1.333
Using Cb = 12.5Mmax / (2.5Mmax + 3MA + 4MB + 3MC): numerator = 12.5 × 400 = 5,000; denominator = 2.5(400) + 3(200) + 4(350) + 3(250) = 1,000 + 600 + 1,400 + 750 = 3,750. Cb = 5,000/3,750 = 1.333. Since Cb > 1.0, Mn for LTB is amplified (more favorable than uniform moment).
Diagrams
Steel Beam Flexure and Shear — Chapter Concept Map
A comprehensive mind map showing the hierarchical relationships between all major concepts in steel beam flexure and shear design, from plastic moment capacity through LTB zones, local buckling classification, and shear strength. Use this as a quick visual reference before exam to confirm you have covered all branches.
Steel Beam Flexural Strength — Step-by-Step Design Flowchart
A complete decision flowchart for determining the LRFD design flexural strength of a steel beam. Follow this flowchart systematically in board exams: start with compactness classification, then locate the LTB zone, compute the appropriate Mn, apply φb, and verify the demand-to-capacity check.
LTB Zone Identification and Mn Computation
A focused flowchart showing how to identify which of the three LTB zones applies based on comparing Lb with Lp and Lr, and which formula to use in each zone. This is the critical decision tree for LTB — the most commonly tested topic in steel beam board problems.
Shear Strength Design Procedure for Rolled I-Shapes
Step-by-step shear design flowchart for rolled I-shaped beams. Note the key distinction: stocky webs (most rolled W-shapes) use Cv = 1.0 and φv = 1.0, giving a simpler and slightly more liberal design strength than the general case.
Typical Distribution of Steel Beam Board Exam Questions by Limit State
Approximate distribution of steel beam flexure and shear questions in PRC Civil Engineer Licensure Examinations based on historical board exam patterns. Focus 65% of your preparation on plastic moment and LTB zone identification — these dominate the exam. Shear problems are usually simpler but appear frequently.
Exam strategy
For PRC board exam problems on steel beam flexure and shear, follow this disciplined sequence: (1) IDENTIFY given data — Fy, E, section properties (Zx, Sx, ry, d, tw), Lb, and load type. (2) CLASSIFY the section — check flange and web λ against λp; most board problems use compact W-sections so this is usually fast. (3) LOCATE the LTB zone — compute Lp = 1.76ry√(E/Fy) and compare to Lb; if Lb ≤ Lp, write Mn = Mp = FyZx immediately. (4) COMPUTE flexural strength — apply φb = 0.90 for LRFD or divide by Ωb = 1.67 for ASD. (5) COMPUTE shear strength — Aw = d×tw, Vn = 0.6FyAwCv, φv = 1.0 for stocky webs. (6) CHECK — Mu ≤ φbMn and Vu ≤ φvVn. Time allocation: these problems should take 3–5 minutes each. The most points-per-minute strategy is to memorize Mp = FyZx, φb = 0.90, Lp = 1.76ry√(E/Fy), and φvVn = 0.6FydtwCv (Cv=1, φv=1) — these four results cover roughly 70% of steel beam board questions. Avoid spending time on elastic LTB (Lb > Lr) unless the problem explicitly asks for it, as it requires lengthy Fcr calculations rarely tested at the licensure level.