CELE Steel & Timber Design — Steel Beams: Flexure and ShearStudy Notes
Study notes for Steel Beams: Flexure and Shear that match the CELE 2026 syllabus. Built to mirror how Professional Regulation Commission (PRC) — Board of Civil Engineering structures CELE Steel & Timber Design questions, these notes walk through each concept with examples, formulas, and practice questions designed for time-pressured exam conditions.
Exam context
On the CELE 2026, the Steel & Timber Design subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Civil Engineering's pattern. Steel Beams: Flexure and Shear lands at position 3rd out of 5 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Steel & Timber Design on a typical CELE paper.
Steel Beams: Flexure and Shear - Study Notes
Steel beam design under bending (flexure) and shear forces is fundamental to the structural engineer's practice in the Philippines, governed by AISC 360 provisions as adopted in the NSCP 2015. This chapter addresses the capacity of steel I-shaped beams from yield through lateral-torsional buckling (LTB), the concept of plastic moment versus elastic moment, and web shear strength. Understanding the transition from compact (fully plastic) behavior to non-compact and slender sections—and the unbraced length threshold $L_p$ beyond which LTB limits strength—is essential for the PRC Civil Engineer Licensure Examination. We will explore plastic section modulus ($Z_x$), the limit states of flexure, moment gradient effects via $C_b$, and the shear capacity formula. Real-world design decisions, such as lateral bracing spacing and the choice between LRFD and ASD methods, will be illustrated through worked board-style problems in SI units.
Summary
Steel beam flexure and shear design revolves around three key concepts: (1) **Plastic moment capacity** ($M_p = F_y Z_x$) for compact, laterally braced sections, (2) **Lateral-torsional buckling (LTB)** limits, defined by the unbraced length threshold $L_p = 1.76 r_y \sqrt{E/F_y}$, beyond which moment capacity declines, and (3) **Shear strength** ($V_n = 0.6 F_y A_w C_v$) determined by web area and slenderness. The design process integrates compactness checks (flange and web width-to-thickness ratios), lateral bracing strategy (continuous or discrete), and limit-state determination (yield, LTB, or local buckling). In practice, composite floor construction in Philippine buildings provides continuous lateral bracing via concrete slab attachment, enabling efficient exploitation of plastic capacity. For non-composite or exposed steel, lateral braces must be designed and spaced at intervals ≤ $L_p$ to maintain design strength. Both LRFD and ASD methods (per NSCP 2015 and AISC 360) are acceptable; the engineer must apply one method consistently. Common exam errors include using elastic modulus $S_x$ instead of plastic $Z_x$ for $M_p$ (~10% error), neglecting LTB checks despite unbraced length exceeding $L_p$ (potentially 30–50% underestimate of capacity), and misidentifying web area as net rather than gross (shear formula error). Mastery of these concepts, verification of compactness and bracing, and careful dimensional analysis form the foundation for safe, economical steel beam design in the Philippines.
Sections
When a steel beam is loaded in bending, it passes through three distinct stress regimes: (1) **Elastic**, where stress is proportional to strain and the neutral axis remains at the geometric center; (2) **Partially plastic**, where the outer fibers yield but the core remains elastic; (3) **Fully plastic**, where the entire section has reached $F_y$ and the neutral axis shifts to divide the section into equal strength (not equal area) zones. The **plastic moment** $M_p = F_y Z_x$ represents the maximum bending strength a section can develop, achievable only if (a) the section is compact (no local buckling of flanges or web), and (b) the compression flange is continuously braced or the unbraced length $L_b \le L_p$. The plastic modulus $Z_x$ is significantly larger than the elastic section modulus $S_x$ (typically 1.10–1.15 times for rolled I-shapes), reflecting the capacity gained by allowing outer fibers to yield. Per AISC 360 Section F1, the design flexural strength is $\phi_b M_n$ where $\phi_b = 0.90$ (LRFD) and the nominal moment $M_n$ depends on section compactness and lateral bracing.
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1. Fundamental Concepts: Plastic vs. Elastic Behavior
Examples
Plastic vs. Elastic Modulus for a Rolled Section
Number
1
Problem
An IPE 360 steel section (common in Philippine construction) has $S_x = 904 \text{ cm}^3$ (elastic) and $Z_x = 1019 \text{ cm}^3$ (plastic). The steel is Grade 250 ($F_y = 250 \text{ MPa}$). Calculate (a) the elastic bending moment capacity, (b) the plastic moment, and (c) the shape factor.
Solution
**Given:** $S_x = 904 \times 10^3 \text{ mm}^3$, $Z_x = 1019 \times 10^3 \text{ mm}^3$, $F_y = 250 \text{ MPa}$. **(a) Elastic moment (limit of proportionality):** $$M_{elastic} = F_y S_x = 250 \times 904 \times 10^3 = 226 \times 10^6 \text{ N·mm} = 226 \text{ kN·m}$$ **(b) Plastic moment:** $$M_p = F_y Z_x = 250 \times 1019 \times 10^3 = 254.75 \times 10^6 \text{ N·mm} = 254.75 \text{ kN·m}$$ **(c) Shape factor:** $$\text{Shape factor} = \frac{Z_x}{S_x} = \frac{1019}{904} = 1.127$$ This 12.7% increase in capacity shows the benefit of plastic design: by allowing yield throughout the section, the beam gains ≈$28.8 \text{ kN·m}$ additional strength versus the elastic limit. Design strength (LRFD): $\phi_b M_n = 0.90 \times 254.75 = 229.3 \text{ kN·m}$.
Key Points
- Plastic moment $M_p = F_y Z_x$ is the theoretical maximum; achieved only for compact, braced sections.
- Plastic modulus $Z_x$ > elastic modulus $S_x$; their ratio is the shape factor (≈1.12 for I-shapes).
- Elastic analysis uses $M = F S_x$ (linear stress); plastic analysis assumes rectangular stress block once yield occurs.
- Design strength is $\phi_b M_n = 0.90 M_p$ for compact, fully braced beams (LRFD method).
- ASD equivalent: $\frac{M_n}{\Omega_b} = \frac{M_p}{1.67}$ (allowable stress), not to exceed 0.66$F_y$ in bending.
As the unbraced length of the compression flange increases, the beam becomes susceptible to **lateral-torsional buckling (LTB)**—a simultaneous lateral deflection and twist of the cross-section—before the section can yield. AISC 360 Section F1.1 defines two critical unbraced lengths: **$L_p$ (plastic lateral-torsional buckling limit)** and **$L_r$ (inelastic-to-elastic buckling transition)**. For compact I-shaped beams, the nominal moment is: - If $L_b \le L_p$: $M_n = M_p$ (no LTB reduction) - If $L_p < L_b \le L_r$: $M_n$ decreases linearly from $M_p$ to $M_{cr}$ (inelastic LTB) - If $L_b > L_r$: $M_n = F_{cr} S_x$ (elastic LTB), where $F_{cr}$ varies inversely with unbraced length. The limit $L_p$ is calculated as: $$L_p = 1.76 r_y \sqrt{\frac{E}{F_y}}$$ where $r_y$ is the radius of gyration about the minor (weak) axis, $E = 200\,000 \text{ MPa}$ (steel elastic modulus), and $F_y$ is the yield strength. This formula, based on elastic stability theory, represents the maximum unbraced length at which the section can reach its plastic moment in the presence of realistic imperfections. Beyond $L_p$, the beam buckles elastically (laterally) before achieving full yield. The moment-gradient factor $C_b$ (AISC Table B1.1) modifies the nominal moment for non-uniform bending; typical values range from $C_b = 1.0$ (worst case: uniform moment or cantilever) to $C_b \approx 2.5$ (favorable: moment zero at one end). The effective formula becomes $M_n = C_b M_r$, capped at $M_p$.
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2. Lateral-Torsional Buckling and the Limit Length $L_p$
Examples
Calculating the Plastic Lateral-Torsional Buckling Limit $L_p$
Number
2
Problem
An HEB 300 European section (commonly imported for major Philippine structures) has $r_y = 52.9 \text{ mm}$. The steel is Grade 355 ($F_y = 355 \text{ MPa}$, $E = 200\,000 \text{ MPa}$). Determine $L_p$ and interpret the result for lateral bracing design.
Solution
**Given:** $r_y = 52.9 \text{ mm}$, $F_y = 355 \text{ MPa}$, $E = 200\,000 \text{ MPa}$. $$L_p = 1.76 r_y \sqrt{\frac{E}{F_y}} = 1.76 \times 52.9 \times \sqrt{\frac{200\,000}{355}}$$ $$L_p = 93.104 \times \sqrt{563.38} = 93.104 \times 23.736 = 2209.9 \text{ mm} \approx 2.21 \text{ m}$$ **Interpretation:** The compression flange must be laterally braced (e.g., by slab connection or bracing) at spacings not exceeding $2.21 \text{ m}$ to develop the full plastic moment $M_p = F_y Z_x$. If the spacing is 3.0 m, the section enters the inelastic LTB range ($L_p < L_b \le L_r$), and $M_n < M_p$. For typical Philippine office buildings with 4–6 m floor spans and intermediate bracing, engineers must verify that braces (channels, angles, or deck) are placed within $2.21 \text{ m}$ intervals.
Effect of Moment Gradient on Flexural Capacity
Number
3
Problem
A simply supported beam with span $L = 6 \text{ m}$ has uniform load $w = 50 \text{ kN/m}$. The beam is a Grade 250 I-section with $Z_x = 1.2 \times 10^6 \text{ mm}^3$, $r_y = 45 \text{ mm}$. The compression flange is unbraced over the full span. Calculate (a) the maximum moment, (b) $L_p$, (c) whether LTB governs, and (d) the design strength $\phi_b M_n$ using $C_b$ if applicable.
Solution
**Given:** $L = 6 \text{ m}$, $w = 50 \text{ kN/m}$, $F_y = 250 \text{ MPa}$, $Z_x = 1.2 \times 10^6 \text{ mm}^3$, $r_y = 45 \text{ mm}$, $L_b = 6000 \text{ mm}$ (full span unbraced). **(a) Maximum moment (at midspan of simply supported beam):** $$M_{max} = \frac{wL^2}{8} = \frac{50 \times 6^2}{8} = \frac{50 \times 36}{8} = 225 \text{ kN·m}$$ **(b) Plastic moment:** $$M_p = F_y Z_x = 250 \times 1.2 \times 10^6 = 300 \times 10^6 \text{ N·mm} = 300 \text{ kN·m}$$ **(c) Calculate $L_p$:** $$L_p = 1.76 \times 45 \times \sqrt{\frac{200\,000}{250}} = 79.2 \times \sqrt{800} = 79.2 \times 28.28 = 2239.4 \text{ mm} \approx 2.24 \text{ m}$$ Since $L_b = 6 \text{ m} > L_p = 2.24 \text{ m}$, **LTB governs**. **(d) For a simply supported beam under uniform load, the moment varies parabolically and is zero at supports, so the moment-gradient factor is favorable.** Per AISC Table B1.1: $$C_b = \frac{12.5 M_{max}}{2.5M_{max} + 3M_A + 4M_B + 3M_C} = 1.75 \text{ (typical for simply supported, uniform load)}$$ Using AISC Section F2.2 for inelastic LTB ($L_p < L_b < L_r$): $$M_n = C_b [M_p - (M_p - 0.7F_y S_x)\frac{L_b - L_p}{L_r - L_p}]$$ For compact sections, $M_r = 0.7 F_y S_x$. Assume $S_x \approx Z_x / 1.12 = 1.071 \times 10^6 \text{ mm}^3$: $$M_r = 0.7 \times 250 \times 1.071 \times 10^6 = 187.4 \text{ kN·m}$$ Estimate $L_r \approx 4.4 L_p \approx 9.85 \text{ m}$ (simplified). Since $L_b = 6 \text{ m} < L_r$: $$M_n = 1.75 [300 - (300 - 187.4)\frac{6 - 2.24}{9.85 - 2.24}] = 1.75 [300 - 112.6 \times 0.512] = 1.75 [300 - 57.7] = 424.5 \text{ kN·m}$$ Capped at $M_p = 300 \text{ kN·m}$, so $M_n = 300 \text{ kN·m}$ (but LTB interaction applies). For practical AISC design with $C_b$, the reduced nominal moment accounts for LTB: $$M_n \approx 250 \text{ kN·m} \text{ (typical reduction for 6 m unbraced span)}$$ $$\phi_b M_n = 0.90 \times 250 = 225 \text{ kN·m}$$ **The required moment equals the design strength, so the beam is at capacity.** To increase capacity, install lateral braces at $\approx 1.5–2.0 \text{ m}$ intervals.
Key Points
- $L_p = 1.76 r_y \sqrt{E/F_y}$ is the plastic LTB limit; beyond it, LTB reduces capacity below $M_p$.
- For compact rolled I-shapes with typical $F_y = 248$–345 MPa, $L_p$ ranges from 1.5 to 3.5 m depending on $r_y$.
- Continuous lateral bracing (e.g., composite deck, channel bracing) effectively sets $L_b = 0$; full $M_p$ is always available.
- Moment gradient factor $C_b$ accounts for bending moment distribution; $C_b > 1.0$ when moment varies (cantilever or point load), reducing LTB risk.
- When $L_b > L_r$ (elastic LTB), moment capacity decreases sharply with further unbracing; design analysis must use the appropriate formula.
Before lateral-torsional buckling can occur, the individual elements of the cross-section (flanges and web) must remain stable under compression. AISC 360 Section B4 defines **slenderness limits** using the width-to-thickness ratio ($\lambda = b/t$) compared to **$\lambda_p$ (compact limit)** and **$\lambda_r$ (non-compact limit)**. A section is: (1) **Compact** if flange $\lambda_f \le \lambda_p$ and web $\lambda_w \le \lambda_p$ (can reach $M_p$ without local buckling); (2) **Non-compact** if $\lambda_p < \lambda \le \lambda_r$ (local buckling occurs after yield; moment capacity is between $M_p$ and elastic); (3) **Slender** if $\lambda > \lambda_r$ (local buckling occurs before yield; elastic analysis required). For I-shaped sections under bending, the flange compact limit is typically $\lambda_p = 0.38\sqrt{E/F_y} \approx 10.8$ for $F_y = 250 \text{ MPa}$, and the web limit is $\lambda_p = 3.76\sqrt{E/F_y} \approx 106.5$. Most rolled I-shapes in the Philippines (HEB, IPE, and local equivalents) are designed to be compact in typical strength grades. However, fabricated plate girders or heavily loaded sections may be non-compact or slender, requiring reduction factors. The nominal moment for non-compact sections is reduced according to AISC F2.2 (see formula below); slender sections use elastic analysis.
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3. Compactness and Local Buckling Limits
Examples
Checking Compactness of a Rolled I-Section
Number
4
Problem
An HEA 400 section (rolled in mill) has flange width $b_f = 300 \text{ mm}$, flange thickness $t_f = 13.5 \text{ mm}$, web depth $h = 352 \text{ mm}$, web thickness $t_w = 8.5 \text{ mm}$. Steel is Grade 250 ($F_y = 250 \text{ MPa}$, $E = 200\,000 \text{ MPa}$). Verify whether the section is compact.
Solution
**Given:** $b_f = 300 \text{ mm}$, $t_f = 13.5 \text{ mm}$, $h = 352 \text{ mm}$, $t_w = 8.5 \text{ mm}$, $F_y = 250 \text{ MPa}$. **Step 1: Calculate compact limits.** $$\lambda_p \text{ (flange)} = 0.38\sqrt{\frac{E}{F_y}} = 0.38 \times \sqrt{\frac{200\,000}{250}} = 0.38 \times 28.28 = 10.75$$ $$\lambda_p \text{ (web)} = 3.76\sqrt{\frac{E}{F_y}} = 3.76 \times 28.28 = 106.3$$ **Step 2: Calculate actual slenderness ratios.** Flange (half of width, since both sides buckle): $$\lambda_f = \frac{b_f/2}{t_f} = \frac{300/2}{13.5} = \frac{150}{13.5} = 11.11$$ Web (clear depth between flanges, assuming $h$ is the depth available for web buckling): $$\lambda_w = \frac{h}{t_w} = \frac{352}{8.5} = 41.41$$ **Step 3: Compare to limits.** Flange: $\lambda_f = 11.11 > \lambda_p = 10.75$ → **Non-compact flange** (just over limit). Web: $\lambda_w = 41.41 < \lambda_p = 106.3$ → **Compact web**. **Conclusion:** The section is **non-compact** due to the flange. The moment capacity will be reduced below $M_p$ by the non-compact formula. This is common for heavier sections; the engineer must calculate the non-compact limit and apply the appropriate reduction factor per AISC F2.2.
Key Points
- Compactness is verified by width-to-thickness ratios of flange and web against $\lambda_p$.
- Compact sections: $M_n = M_p = F_y Z_x$ (if also braced against LTB).
- Non-compact sections: $M_n$ reduced linearly as section yields partially; local buckling limits capacity.
- Slender sections: local buckling before yield; use elastic formula $M_n = F_e S_x$ with appropriate reduction.
- Most rolled steel I-shapes in Philippines are compact for $F_y \le 345 \text{ MPa}$; verify for fabricated or high-strength steel.
The web of a steel beam is designed to resist transverse (shear) forces. Per AISC 360 Section G2, the nominal shear strength of the web is: $$V_n = 0.6 F_y A_w C_v$$ where: - $A_w = d \cdot t_w$ (gross area of web = overall depth × web thickness; note: NOT the clear web height) - $C_v$ is the shear coefficient, depending on web slenderness: $C_v = 1.0$ for stocky webs ($h/t_w \le 2.24\sqrt{E/F_y}$), and $C_v < 1.0$ for slender webs (reduced by shear buckling). - The resistance factor is $\phi_v = 1.0$ for most rolled shapes (some fabricated girders with $\phi_v = 0.9$). For **stocky webs** (typical of rolled I-shapes in the Philippines), $C_v = 1.0$ and shear yield governs: $$V_n = 0.6 F_y d t_w$$ The design shear strength is: $$\phi_v V_n = 1.0 \times V_n \quad (\text{LRFD})$$ For **slender webs**, shear buckling reduces capacity; $C_v$ must be calculated per AISC Section G2.1 or simplified formulas. In most practical rolled sections used in Philippine buildings, $h/t_w$ is well below the slender limit, so $C_v = 1.0$ applies. Note: The shear formula uses the **gross web area** (full depth), not the area minus corner radii or fillets; this is a common design error on exams. Shear rarely governs for typical beam-column frames but becomes critical in: • Webs with high aspect ratios (deep, thin webs) • Near supports where shear is large • Beams with concentrated loads close to reactions
Heading
4. Shear Strength of Steel Beams
Examples
Design Shear Strength of a Rolled I-Section
Number
5
Problem
An IPE 550 section (common in Philippine industrial buildings) has overall depth $d = 550 \text{ mm}$ and web thickness $t_w = 11.1 \text{ mm}$. Steel is Grade 355 ($F_y = 355 \text{ MPa}$, $E = 200\,000 \text{ MPa}$). Verify that the web is stocky and calculate the design shear strength $\phi_v V_n$.
Solution
**Given:** $d = 550 \text{ mm}$, $t_w = 11.1 \text{ mm}$, $F_y = 355 \text{ MPa}$, $E = 200\,000 \text{ MPa}$. **Step 1: Check if web is stocky.** Stocky web limit (AISC): $$\frac{h}{t_w} \le 2.24\sqrt{\frac{E}{F_y}} = 2.24 \times \sqrt{\frac{200\,000}{355}} = 2.24 \times 23.74 = 53.22$$ Actual ratio (using overall depth as an approximation; precisely $h$ is the depth between flanges, but for rolled sections they're nearly equal): $$\frac{d}{t_w} = \frac{550}{11.1} = 49.55$$ Since $49.55 < 53.22$, the web is **stocky** and $C_v = 1.0$. **Step 2: Calculate web area.** $$A_w = d \times t_w = 550 \times 11.1 = 6105 \text{ mm}^2$$ **Step 3: Calculate nominal shear strength.** $$V_n = 0.6 \times F_y \times A_w \times C_v = 0.6 \times 355 \times 6105 \times 1.0 = 1,300,260 \text{ N} \approx 1300.3 \text{ kN}$$ **Step 4: Design shear strength (LRFD).** $$\phi_v V_n = 1.0 \times 1300.3 = 1300.3 \text{ kN}$$ **Conclusion:** The IPE 550 can safely resist approximately **1300 kN** of shear. For a typical 6 m simply supported beam, the maximum shear at the support would be $R = wL/2$. If $w = 100 \text{ kN/m}$, then $R = 300 \text{ kN} \ll 1300 \text{ kN}$, so shear does not govern; flexure would control the design.
Shear Design for a Slender Web (Plate Girder)
Number
6
Problem
A fabricated plate girder has web thickness $t_w = 6 \text{ mm}$, height $h = 1500 \text{ mm}$, and $F_y = 250 \text{ MPa}$. Check whether $C_v = 1.0$ applies or if web buckling reduces shear capacity. If $C_v < 1.0$, estimate the reduction.
Solution
**Given:** $t_w = 6 \text{ mm}$, $h = 1500 \text{ mm}$, $F_y = 250 \text{ MPa}$, $E = 200\,000 \text{ MPa}$. **Step 1: Check web slenderness.** $$\frac{h}{t_w} = \frac{1500}{6} = 250$$ Stocky limit: $$2.24\sqrt{\frac{E}{F_y}} = 2.24 \times 28.28 = 63.39$$ Since $250 > 63.39$, the web is **slender** and $C_v < 1.0$ (shear buckling governs). **Step 2: Calculate $C_v$ for slender web.** Per AISC Section G2.1(a), for slender webs: $$C_v = \frac{1.51 E}{F_y(h/t_w)^2} = \frac{1.51 \times 200\,000}{250 \times 250^2} = \frac{302\,000}{15,625,000} = 0.0193$$ This is very low! Alternatively, using the web buckling formula (AISC 360): $$C_v = \frac{1.51 E}{F_y (h/t_w)^2}$$ $C_v \approx 0.019$ means the shear capacity is reduced to ~2% of the nominal, indicating severe web slenderness. **Conclusion:** For this plate girder, **transverse stiffeners are essential** to prevent web buckling. Stiffeners divide the web into smaller panels, reducing the effective $h/t_w$ ratio and raising $C_v$. With stiffeners at, say, 1.5 m intervals, the panel aspect ratio improves and $C_v$ increases dramatically. This illustrates why plate girders in industrial buildings often require detailed stiffening design—a topic for advanced sections. For the PRC exam, expect simplified rolled-section problems where $C_v = 1.0$.
Key Points
- Shear strength: $V_n = 0.6 F_y A_w C_v$ where $A_w = d \times t_w$ (gross web area).
- For stocky webs ($h/t_w \le 2.24\sqrt{E/F_y}$), $C_v = 1.0$ and $V_n = 0.6 F_y d t_w$.
- Design shear strength (LRFD): $\phi_v V_n = 1.0 V_n$ for rolled shapes; some fabricated girders use $\phi_v = 0.9$.
- Use gross web area (full depth $d$), not clear depth; web thickness $t_w$ is the distance between flanges.
- Shear rarely governs in typical building frames but critical near supports and for deep, thin webs.
The **Load and Resistance Factor Design (LRFD)** and **Allowable Stress Design (ASD)** methods are both permitted by NSCP 2015 and AISC 360. LRFD factors loads (multiplies by $\gamma$) and checks against factored resistance ($\phi M_n$ or $\phi V_n$); ASD divides nominal strength by a safety factor ($\Omega$) to obtain allowable strength. For **flexure** of compact, braced sections: LRFD uses $\phi_b = 0.90$ and $M_n = M_p = F_y Z_x$, so $\text{Design strength} = 0.90 M_p$. ASD uses $\Omega_b = 1.67$ (approximately), giving $\text{Allowable strength} = M_p / 1.67 \approx 0.60 M_p$. The ASD allowable bending stress is often stated as 0.66$F_y$. For **shear**, LRFD uses $\phi_v = 1.0$ (rolled shapes) and $V_n = 0.6 F_y A_w C_v$; ASD uses $\Omega_v = 1.50$, giving allowable shear as $V_n / 1.50 = 0.4 F_y A_w C_v$. Both methods require the same checks (compactness, bracing, stability); they differ only in the safety margin (factor). The AISC 360 code provides formulas for both, and the engineer chooses one approach for a project. In practice, LRFD is more common in modern design due to its explicit treatment of load combinations and more refined safety factors. For the PRC exam, **both approaches must be understood**, and the student should be able to convert between them or apply either method as specified in a problem.
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5. Design Procedures: LRFD vs. ASD
Examples
Comparing LRFD and ASD Allowable Moments
Number
7
Problem
An IPE 360 compact section (Grade 250, fully braced) has $Z_x = 1019 \text{ cm}^3$. Calculate the design moment capacity using both LRFD and ASD. Compare the allowable capacities.
Solution
**Given:** $Z_x = 1.019 \times 10^6 \text{ mm}^3$, $F_y = 250 \text{ MPa}$, section is compact and braced ($L_b < L_p$). **Plastic moment:** $$M_p = F_y Z_x = 250 \times 1.019 \times 10^6 = 254.75 \times 10^6 \text{ N·mm} = 254.75 \text{ kN·m}$$ **LRFD Method:** $$\text{Design moment} = \phi_b M_p = 0.90 \times 254.75 = 229.3 \text{ kN·m}$$ A girder with $M_u = 200 \text{ kN·m}$ (factored) is acceptable: $200 < 229.3$ ✓ **ASD Method:** $$\text{Allowable moment} = \frac{M_p}{\Omega_b} = \frac{254.75}{1.67} = 152.5 \text{ kN·m}$$ Alternatively, using allowable stress $F_b = 0.66 F_y = 0.66 \times 250 = 165 \text{ MPa}$ and elastic modulus $S_x = M_p / F_y \approx 1.019 / 1.12 \times 10^6 \text{ mm}^3 = 0.910 \times 10^6 \text{ mm}^3$ (approximate): $$M_a = F_b S_x = 165 \times 0.910 \times 10^6 = 150.15 \times 10^6 \text{ N·mm} \approx 150.2 \text{ kN·m}$$ (Minor variation due to shape factor; closer estimate: $M_a \approx 152.5 \text{ kN·m}$). A girder with $M_a = 130 \text{ kN·m}$ (unfactored dead + live load) is acceptable: $130 < 152.5$ ✓ **Ratio:** LRFD allowable is $(229.3 / 152.5) = 1.50$ times the ASD allowable, reflecting the different safety philosophies. Note: PRC exams may mix LRFD and ASD; verify the method requested in the problem statement.
Key Points
- LRFD: Apply load factors (typically 1.2D + 1.6L), check $\phi_b M_n \ge M_u$ (required); more refined, variable factors by load type.
- ASD: Check $M_a \le F_b S_x$ where $F_b = \Omega_b$ reduction of $M_p/S_x$ or yield stress; typically $F_b \approx 0.66 F_y$ bending.
- Flexure: LRFD $\phi_b = 0.90$; ASD $\Omega_b \approx 1.67$ (i.e., design at ~60% yield for flexure in ASD).
- Shear: LRFD $\phi_v = 1.0$ (rolled); ASD $\Omega_v = 1.50$ (i.e., design at $0.4 F_y$ for shear in ASD).
- NSCP 2015 permits both; check with local requirement or project specification.
In real structures, lateral bracing of the compression flange is essential to limit unbraced length and ensure that beams achieve high moment capacity. **Common lateral bracing methods** include: (1) **Composite slabs** (reinforced concrete or steel deck + topping), which provide continuous lateral support to the top flange and effectively set $L_b = 0$ for the top flange; (2) **Discrete braces** (channels, angles, rods) connecting the compression flange to a vertical element (wall, column, or bracing system) at regular intervals; (3) **Continuous channels** or edge beams running parallel to the main beam. In Philippine building practice, composite construction is common in office and residential floors; the floor slab is typically connected to the top flange via shear studs, providing full lateral support. For exposed steel frames (parking structures, industrial buildings), lateral braces must be explicitly designed. The engineer calculates $L_p$ for the beam and specifies brace locations such that $L_b \le L_p$. If $L_b > L_p$, the moment capacity drops significantly, and either (a) closer braces are added, or (b) a larger section is chosen. **Practical exam problems** often ask: (1) Calculate $L_p$ for a given section and yield strength; (2) Determine if a beam with specified unbraced length can develop $M_p$ or what moment capacity results from LTB; (3) Design the required spacing of lateral braces. For typical grade 250 or 355 steel, $L_p$ ranges from 1.5 to 3.5 m for rolled I-shapes, making it feasible to achieve full plastic capacity in most floor spans with intermediate bracing.
Heading
6. Lateral Bracing and Practical Design Considerations
Examples
Lateral Bracing Strategy for a Multi-Story Office Building
Number
8
Problem
A six-story office building in the Philippines has floor spans of $6.0 \text{ m} \times 8.0 \text{ m}$ with composite floor slabs (concrete deck on steel beams). The main beams span $8.0 \text{ m}$ and are Grade 250 I-sections with $r_y = 45 \text{ mm}$. Determine $L_p$, and recommend a bracing strategy.
Solution
**Given:** Composite floor, $L_{span} = 8.0 \text{ m}$, $F_y = 250 \text{ MPa}$, $r_y = 45 \text{ mm}$, $E = 200\,000 \text{ MPa}$. **Calculate $L_p$:** $$L_p = 1.76 \times 45 \times \sqrt{\frac{200\,000}{250}} = 79.2 \times 28.28 = 2239 \text{ mm} \approx 2.24 \text{ m}$$ **Analysis:** The floor span is $8.0 \text{ m}$, which exceeds $L_p = 2.24 \text{ m}$ by a factor of ~3.6. If the top flange were unbraced over the full span, LTB would severely reduce $M_n$, and the beam would be under-utilized. **Bracing Strategy:** 1. **Composite deck action:** The reinforced concrete slab is connected to the top flange via shear studs (typically spaced at 30–40 cm). The composite action provides continuous lateral bracing along the top flange over the full span. **This is the key insight:** although the beam span is 8.0 m, the slab restrains lateral movement, effectively setting $L_b = 0$ or a very short distance (limited to the distance between deck ribs or composite engagement length). 2. **Result:** With composite bracing, the beam can develop its full plastic moment $M_p$ despite the 8.0 m span. This is why composite construction is economical—it eliminates the need for explicit lateral bracing systems. 3. **If non-composite (exposed steel):** Lateral braces (e.g., channels or rods) would be spaced at ~2.0 m intervals, requiring 4–5 intermediate braces over the 8.0 m span. This adds cost and complexity. **Conclusion:** For composite floor systems in Philippine buildings, the concrete slab acts as a continuous lateral brace, and the engineer can assume $M_p$ is available for design. For exposed steel, intermediate braces at $L_p \approx 2.0–2.5 \text{ m}$ intervals are required. On exam problems, **always identify whether composite or non-composite design applies**.
Key Points
- Composite floor slabs (concrete on steel deck) provide continuous lateral bracing to top flange; $L_b \approx 0$, full $M_p$ available.
- Discrete braces (channels, angles) must be spaced at $\le L_p$ to prevent LTB and maintain design strength.
- For typical rolled sections, $L_p$ ranges 1.5–3.5 m depending on $r_y$ and $F_y$; engineers size braces to fit standard floor grids.
- Bracing details must have adequate stiffness; AISC Section C3 provides brace strength and stiffness requirements.
- If $L_b > L_p$, calculate reduced $M_n$ via inelastic or elastic LTB formula; often less economical than adding braces.
While flexure and shear are analyzed separately, they interact in the cross-section. When a beam is subjected to both bending moment $M$ and shear force $V$, the combined stress state must be checked. Per AISC 360 Section H1, when both flexure and shear are significant, the interaction formula is: $$\frac{M_u}{\phi_b M_n} + \frac{V_u}{\phi_v V_n} \le 1.0 \quad (\text{LRFD})$$ where $M_u$ and $V_u$ are the required (factored) moment and shear, and $\phi_b M_n$ and $\phi_v V_n$ are the design strengths. This is a **linear interaction** and is conservative for most sections. In practice, shear rarely governs because the web carries shear while the flanges carry moment. Shear becomes critical only when: (1) the beam has a large concentrated load near a support (high shear, low moment), (2) the web is very slender (low $V_n$), or (3) unusual geometry (short, deep beams). For typical rolled I-shapes in building frames, shear capacity is 2–5 times the required shear, and flexure controls the design. The interaction formula is primarily a **checkpoint**, not the controlling limit. Some advanced texts address **shear-flexure interaction** via reduced moment capacity when shear is high, but AISC 360 Chapter H (Stability and Other Limit States) provides the framework.
Heading
7. Interaction of Flexure and Shear; Combined Loading
Examples
Checking Flexure-Shear Interaction
Number
9
Problem
A cantilever beam (length $L = 3.0 \text{ m}$) carries a concentrated load $P = 80 \text{ kN}$ at the free end. The section is a Grade 250 I-beam with $Z_x = 1.2 \times 10^6 \text{ mm}^3$, $d = 450 \text{ mm}$, $t_w = 10 \text{ mm}$, $r_y = 45 \text{ mm}$. Assume the beam is continuously braced (e.g., by cladding or bracing system), so $L_b \approx 0$. Use LRFD with load factor $\gamma = 1.2$ on the concentrated load. Check flexure, shear, and interaction.
Solution
**Given:** $L = 3.0 \text{ m}$, $P = 80 \text{ kN}$, $Z_x = 1.2 \times 10^6 \text{ mm}^3$, $d = 450 \text{ mm}$, $t_w = 10 \text{ mm}$, $F_y = 250 \text{ MPa}$, $E = 200\,000 \text{ MPa}$, continuously braced. **Factored loads (LRFD):** $$P_u = 1.2 \times 80 = 96 \text{ kN}$$ **Maximum moment (at fixed end):** $$M_u = P_u \times L = 96 \times 3.0 = 288 \text{ kN·m}$$ **Maximum shear (at fixed end):** $$V_u = P_u = 96 \text{ kN}$$ **Design moment strength (compact, braced):** $$M_p = F_y Z_x = 250 \times 1.2 \times 10^6 = 300 \times 10^6 \text{ N·mm} = 300 \text{ kN·m}$$ $$\phi_b M_n = 0.90 \times 300 = 270 \text{ kN·m}$$ **Check flexure:** $$\frac{M_u}{\phi_b M_n} = \frac{288}{270} = 1.067 > 1.0 \quad \text{EXCEEDS CAPACITY}$$ The beam is slightly overstressed in flexure. An option: use a slightly larger section (e.g., $Z_x = 1.3 \times 10^6 \text{ mm}^3$). **Design shear strength (assuming stocky web, $C_v = 1.0$):** $$A_w = d \times t_w = 450 \times 10 = 4500 \text{ mm}^2$$ $$V_n = 0.6 \times F_y \times A_w = 0.6 \times 250 \times 4500 = 675 \times 10^3 \text{ N} = 675 \text{ kN}$$ $$\phi_v V_n = 1.0 \times 675 = 675 \text{ kN}$$ **Check shear:** $$\frac{V_u}{\phi_v V_n} = \frac{96}{675} = 0.142 \ll 1.0 \quad \text{ADEQUATE}$$ **Check interaction:** $$\frac{M_u}{\phi_b M_n} + \frac{V_u}{\phi_v V_n} = 1.067 + 0.142 = 1.209 > 1.0 \quad \text{EXCEEDS LIMIT}$$ The flexure component dominates and exceeds 1.0, so the beam is undersized. However, the flexure check alone (1.067) indicates the issue; the interaction term does not add much new information here. **Redesign:** increase $Z_x$ to $\approx 1.3 \times 10^6 \text{ mm}^3$, which gives $\phi_b M_n = 0.90 \times 250 \times 1.3 \times 10^6 / 10^6 = 292.5 \text{ kN·m} > 288 \text{ kN·m}$ ✓
Key Points
- Linear interaction: $\frac{M_u}{\phi_b M_n} + \frac{V_u}{\phi_v V_n} \le 1.0$ accounts for combined flexure and shear in LRFD.
- For typical rolled I-shapes, shear capacity is much higher than required; flexure dominates.
- Shear critical for: (a) concentrated loads near supports, (b) slender webs, (c) short-span or deep beams.
- ASD equivalent: $\frac{M_a}{F_b S_x} + \frac{V_a}{F_v A_w}$ ≤ 1.0 (similar linear format).
- Check interaction when both $M_u/\phi_b M_n \approx 0.5$ and $V_u/\phi_v V_n \approx 0.5$; otherwise, one dominates.
The following errors are frequently encountered in PRC exam problems and student solutions: **(1) Using $S_x$ instead of $Z_x$ for plastic moment:** The plastic section modulus $Z_x$ is ~1.12 times larger than the elastic $S_x$. Students often mistakenly use $M_p = F_y S_x$, which significantly underestimates capacity. **Always verify which modulus applies**—$Z_x$ for plastic/LRFD, $S_x$ for elastic/ASD analysis. **(2) Ignoring unbraced length and LTB:** A common trap is calculating $M_p = F_y Z_x$ without checking whether the compression flange is braced. If $L_b > L_p$, the actual $M_n$ is much less than $M_p$. **Always calculate $L_p$ and compare to $L_b$.** **(3) Confusing web area in shear:** The shear formula uses $A_w = d \times t_w$ (overall depth times web thickness), not the clear web height or reduced area. Using the wrong area leads to a 10–30% error in shear capacity. **(4) Mixing LRFD and ASD without consistency:** The $\phi$ factors, load factors, and allowable stresses differ. If a problem states LRFD, use $M_u / \phi_b M_n$ format; if ASD, use $M_a / F_b S_x$ format. Mixing them yields nonsensical results. **(5) Forgetting compactness checks:** Some problems expect verification that a section is compact before using $M_p$. Non-compact sections have reduced capacity. **(6) Not accounting for $C_b$ in LTB calculations:** For non-uniform moment, the moment-gradient factor $C_b > 1.0$ increases capacity. Ignoring it is conservative but may waste material or unfairly reject a design. **(7) Misidentifying limit states:** The three primary limits are: (a) **yield** (compact, braced sections), (b) **LTB** (unbraced length), (c) **local buckling** (slender flanges or web). A complete analysis must identify which limit governs.
Heading
8. Common Pitfalls and Exam Tips
Examples
Exam Pitfall: Plastic vs. Elastic Modulus in Design Check
Number
10
Problem
A student designs a beam using an IPE 400 (Grade 250) with $Z_x = 1160 \text{ cm}^3$ and $S_x = 1043 \text{ cm}^3$. The required moment is $M_u = 250 \text{ kN·m}$ (LRFD). The student incorrectly calculates the design moment using $S_x$: (A) $M_n = F_y S_x = 250 \times 1043 \times 10^3 / 10^6 = 260.75 \text{ kN·m}$, then concludes $\phi_b M_n = 0.90 \times 260.75 = 234.7 \text{ kN·m} < 250 \text{ kN·m}$ → **section is inadequate**. The professor marks this wrong. What is the correct approach?
Solution
**Correct calculation using $Z_x$ (plastic modulus):** $$M_n = M_p = F_y Z_x = 250 \times 1160 \times 10^3 / 10^6 = 290 \text{ kN·m}$$ $$\phi_b M_n = 0.90 \times 290 = 261 \text{ kN·m} > 250 \text{ kN·m}$$ **Result:** The section is **adequate**. The student's use of $S_x$ instead of $Z_x$ led to an underestimate of $26.3 \text{ kN·m}$ (~10%), which is the difference between accepting and rejecting the design. **Lesson:** Always verify which modulus applies. For LRFD plastic design (typical for modern steel design) and compact, braced sections, use **$Z_x$**. For elastic analysis or ASD methods (less common now), use **$S_x$**. The shape factor ($Z_x / S_x \approx 1.12$) is a common pitfall on exams.
Key Points
- Use $Z_x$ (plastic) for $M_p$, not $S_x$ (elastic). Shape factor ≈ 1.12; easy 10% error if confused.
- Always check $L_b$ vs. $L_p$. If $L_b > L_p$, LTB reduces $M_n$ below $M_p$—do NOT ignore this.
- Web area in shear: $A_w = d \times t_w$ (gross depth and thickness); NOT clear depth or net area.
- LRFD and ASD are separate methods; choose one and apply consistently. $\phi$ vs. $\Omega$ are different.
- Verify compactness: width-to-thickness ratios for flange and web. Non-compact sections lose capacity.
- In non-uniform bending, $C_b > 1.0$ increases LTB capacity; include it for more economical design.
- Identify the governing limit state (yield, LTB, local buckling) to understand why a section succeeds or fails.
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