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CELE Steel & Timber Design — Steel Beams: Flexure and ShearStudy Notes

Study notes covering Steel Beams: Flexure and Shear for CELE reviewers. Use them for your first full pass through the chapter; the shorter formats suit revision later.

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 LpL_p beyond which LTB limits strength—is essential for the PRC Civil Engineer Licensure Examination. We will explore plastic section modulus (ZxZ_x), the limit states of flexure, moment gradient effects via CbC_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.

1. Fundamental Concepts: Plastic vs. Elastic Behavior

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 FyF_y and the neutral axis shifts to divide the section into equal strength (not equal area) zones. The plastic moment Mp=FyZxM_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 Lb≤LpL_b \le L_p. The plastic modulus ZxZ_x is significantly larger than the elastic section modulus SxS_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 ϕbMn\phi_b M_n where ϕb=0.90\phi_b = 0.90 (LRFD) and the nominal moment MnM_n depends on section compactness and lateral bracing.

Examples

An IPE 360 steel section (common in Philippine construction) has Sx=904 cm3S_x = 904 \text{ cm}^3 (elastic) and Zx=1019 cm3Z_x = 1019 \text{ cm}^3 (plastic). The steel is Grade 250 (Fy=250 MPaF_y = 250 \text{ MPa}). Calculate (a) the elastic bending moment capacity, (b) the plastic moment, and (c) the shape factor.

Number

1

Solution

Given: Sx=904×103 mm3S_x = 904 \times 10^3 \text{ mm}^3, Zx=1019×103 mm3Z_x = 1019 \times 10^3 \text{ mm}^3, Fy=250 MPaF_y = 250 \text{ MPa}.

(a) Elastic moment (limit of proportionality):

Melastic=FySx=250×904×103=226×106 N⋅mm=226 kN⋅mM_{elastic} = F_y S_x = 250 \times 904 \times 10^3 = 226 \times 10^6 \text{ N}\cdot\text{mm} = 226 \text{ kN}\cdot\text{m}

(b) Plastic moment:

Mp=FyZx=250×1019×103=254.75×106 N⋅mm=254.75 kN⋅mM_p = F_y Z_x = 250 \times 1019 \times 10^3 = 254.75 \times 10^6 \text{ N}\cdot\text{mm} = 254.75 \text{ kN}\cdot\text{m}

(c) Shape factor:

Shape factor=ZxSx=1019904=1.127\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 kN⋅m28.8 \text{ kN}\cdot\text{m} additional strength versus the elastic limit. Design strength (LRFD): ϕbMn=0.90×254.75=229.3 kN⋅m\phi_b M_n = 0.90 \times 254.75 = 229.3 \text{ kN}\cdot\text{m}.

Key points

  • Plastic moment Mp=FyZxM_p = F_y Z_x is the theoretical maximum; achieved only for compact, braced sections.
  • Plastic modulus ZxZ_x > elastic modulus SxS_x; their ratio is the shape factor (≈1.12 for I-shapes).
  • Elastic analysis uses M=FSxM = F S_x (linear stress); plastic analysis assumes rectangular stress block once yield occurs.
  • Design strength is ϕbMn=0.90Mp\phi_b M_n = 0.90 M_p for compact, fully braced beams (LRFD method).
  • ASD equivalent: MnΩb=Mp1.67\frac{M_n}{\Omega_b} = \frac{M_p}{1.67} (allowable stress), not to exceed 0.66FyF_y in bending.

2. Lateral-Torsional Buckling and the Limit Length LpL_p

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: LpL_p (plastic lateral-torsional buckling limit) and LrL_r (inelastic-to-elastic buckling transition). For compact I-shaped beams, the nominal moment is:

  • If Lb≤LpL_b \le L_p: Mn=MpM_n = M_p (no LTB reduction)
  • If Lp<Lb≤LrL_p < L_b \le L_r: MnM_n decreases linearly from MpM_p to McrM_{cr} (inelastic LTB)
  • If Lb>LrL_b > L_r: Mn=FcrSxM_n = F_{cr} S_x (elastic LTB), where FcrF_{cr} varies inversely with unbraced length.

The limit LpL_p is calculated as:

Lp=1.76ryEFyL_p = 1.76 r_y \sqrt{\frac{E}{F_y}}

where ryr_y is the radius of gyration about the minor (weak) axis, E=200 000 MPaE = 200\,000 \text{ MPa} (steel elastic modulus), and FyF_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 LpL_p, the beam buckles elastically (laterally) before achieving full yield. The moment-gradient factor CbC_b (AISC Table B1.1) modifies the nominal moment for non-uniform bending; typical values range from Cb=1.0C_b = 1.0 (worst case: uniform moment or cantilever) to Cb≈2.5C_b \approx 2.5 (favorable: moment zero at one end). The effective formula becomes Mn=CbMrM_n = C_b M_r, capped at MpM_p.

Examples

An HEB 300 European section (commonly imported for major Philippine structures) has ry=52.9 mmr_y = 52.9 \text{ mm}. The steel is Grade 355 (Fy=355 MPaF_y = 355 \text{ MPa}, E=200 000 MPaE = 200\,000 \text{ MPa}). Determine LpL_p and interpret the result for lateral bracing design.

Number

2

Solution

Given: ry=52.9 mmr_y = 52.9 \text{ mm}, Fy=355 MPaF_y = 355 \text{ MPa}, E=200 000 MPaE = 200\,000 \text{ MPa}.

Lp=1.76ryEFy=1.76×52.9×200 000355L_p = 1.76 r_y \sqrt{\frac{E}{F_y}} = 1.76 \times 52.9 \times \sqrt{\frac{200\,000}{355}}
Lp=93.104×563.38=93.104×23.736=2209.9 mm≈2.21 mL_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 m2.21 \text{ m} to develop the full plastic moment Mp=FyZxM_p = F_y Z_x. If the spacing is 3.0 m, the section enters the inelastic LTB range (Lp<Lb≤LrL_p < L_b \le L_r), and Mn<MpM_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 m2.21 \text{ m} intervals.

A simply supported beam with span L=6 mL = 6 \text{ m} has uniform load w=50 kN/mw = 50 \text{ kN/m}. The beam is a Grade 250 I-section with Zx=1.2×106 mm3Z_x = 1.2 \times 10^6 \text{ mm}^3, ry=45 mmr_y = 45 \text{ mm}. The compression flange is unbraced over the full span. Calculate (a) the maximum moment, (b) LpL_p, (c) whether LTB governs, and (d) the design strength ϕbMn\phi_b M_n using CbC_b if applicable.

Number

3

Solution

Given: L=6 mL = 6 \text{ m}, w=50 kN/mw = 50 \text{ kN/m}, Fy=250 MPaF_y = 250 \text{ MPa}, Zx=1.2×106 mm3Z_x = 1.2 \times 10^6 \text{ mm}^3, ry=45 mmr_y = 45 \text{ mm}, Lb=6000 mmL_b = 6000 \text{ mm} (full span unbraced).

(a) Maximum moment (at midspan of simply supported beam):

Mmax=wL28=50×628=50×368=225 kN⋅mM_{max} = \frac{wL^2}{8} = \frac{50 \times 6^2}{8} = \frac{50 \times 36}{8} = 225 \text{ kN}\cdot\text{m}

(b) Plastic moment:

Mp=FyZx=250×1.2×106=300×106 N⋅mm=300 kN⋅mM_p = F_y Z_x = 250 \times 1.2 \times 10^6 = 300 \times 10^6 \text{ N}\cdot\text{mm} = 300 \text{ kN}\cdot\text{m}

(c) Calculate LpL_p:

Lp=1.76×45×200 000250=79.2×800=79.2×28.28=2239.4 mm≈2.24 mL_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 Lb=6 m>Lp=2.24 mL_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:

Cb=12.5Mmax2.5Mmax+3MA+4MB+3MC=1.75 (typical for simply supported, uniform load)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 (Lp<Lb<LrL_p < L_b < L_r):

Mn=Cb[Mp−(Mp−0.7FySx)Lb−LpLr−Lp]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, Mr=0.7FySxM_r = 0.7 F_y S_x. Assume Sx≈Zx/1.12=1.071×106 mm3S_x \approx Z_x / 1.12 = 1.071 \times 10^6 \text{ mm}^3:

Mr=0.7×250×1.071×106=187.4 kN⋅mM_r = 0.7 \times 250 \times 1.071 \times 10^6 = 187.4 \text{ kN}\cdot\text{m}

Estimate Lr≈4.4Lp≈9.85 mL_r \approx 4.4 L_p \approx 9.85 \text{ m} (simplified). Since Lb=6 m<LrL_b = 6 \text{ m} < L_r:

Mn=1.75[300−(300−187.4)6−2.249.85−2.24]=1.75[300−112.6×0.512]=1.75[300−57.7]=424.5 kN⋅mM_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}\cdot\text{m}

Capped at Mp=300 kN⋅mM_p = 300 \text{ kN}\cdot\text{m}, so Mn=300 kN⋅mM_n = 300 \text{ kN}\cdot\text{m} (but LTB interaction applies).

For practical AISC design with CbC_b, the reduced nominal moment accounts for LTB:

Mn≈250 kN⋅m (typical reduction for 6 m unbraced span)M_n \approx 250 \text{ kN}\cdot\text{m} \text{ (typical reduction for 6 m unbraced span)}
ϕbMn=0.90×250=225 kN⋅m\phi_b M_n = 0.90 \times 250 = 225 \text{ kN}\cdot\text{m}

The required moment equals the design strength, so the beam is at capacity. To increase capacity, install lateral braces at ≈1.5–2.0 m\approx 1.5–2.0 \text{ m} intervals.

Key points

  • Lp=1.76ryE/FyL_p = 1.76 r_y \sqrt{E/F_y} is the plastic LTB limit; beyond it, LTB reduces capacity below MpM_p.
  • For compact rolled I-shapes with typical Fy=248F_y = 248–345 MPa, LpL_p ranges from 1.5 to 3.5 m depending on ryr_y.
  • Continuous lateral bracing (e.g., composite deck, channel bracing) effectively sets Lb=0L_b = 0; full MpM_p is always available.
  • Moment gradient factor CbC_b accounts for bending moment distribution; Cb>1.0C_b > 1.0 when moment varies (cantilever or point load), reducing LTB risk.
  • When Lb>LrL_b > L_r (elastic LTB), moment capacity decreases sharply with further unbracing; design analysis must use the appropriate formula.

3. Compactness and Local Buckling Limits

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 (λ=b/t\lambda = b/t) compared to λp\lambda_p (compact limit) and λr\lambda_r (non-compact limit). A section is: (1) Compact if flange λf≤λp\lambda_f \le \lambda_p and web λw≤λp\lambda_w \le \lambda_p (can reach MpM_p without local buckling); (2) Non-compact if λp<λ≤λr\lambda_p < \lambda \le \lambda_r (local buckling occurs after yield; moment capacity is between MpM_p and elastic); (3) Slender if λ>λr\lambda > \lambda_r (local buckling occurs before yield; elastic analysis required). For I-shaped sections under bending, the flange compact limit is typically λp=0.38E/Fy≈10.8\lambda_p = 0.38\sqrt{E/F_y} \approx 10.8 for Fy=250 MPaF_y = 250 \text{ MPa}, and the web limit is λp=3.76E/Fy≈106.5\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.

Examples

An HEA 400 section (rolled in mill) has flange width bf=300 mmb_f = 300 \text{ mm}, flange thickness tf=13.5 mmt_f = 13.5 \text{ mm}, web depth h=352 mmh = 352 \text{ mm}, web thickness tw=8.5 mmt_w = 8.5 \text{ mm}. Steel is Grade 250 (Fy=250 MPaF_y = 250 \text{ MPa}, E=200 000 MPaE = 200\,000 \text{ MPa}). Verify whether the section is compact.

Number

4

Solution

Given: bf=300 mmb_f = 300 \text{ mm}, tf=13.5 mmt_f = 13.5 \text{ mm}, h=352 mmh = 352 \text{ mm}, tw=8.5 mmt_w = 8.5 \text{ mm}, Fy=250 MPaF_y = 250 \text{ MPa}.

Step 1: Calculate compact limits.

λp (flange)=0.38EFy=0.38×200 000250=0.38×28.28=10.75\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
λp (web)=3.76EFy=3.76×28.28=106.3\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):

λf=bf/2tf=300/213.5=15013.5=11.11\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 hh is the depth available for web buckling):

λw=htw=3528.5=41.41\lambda_w = \frac{h}{t_w} = \frac{352}{8.5} = 41.41

Step 3: Compare to limits.

Flange: λf=11.11>λp=10.75\lambda_f = 11.11 > \lambda_p = 10.75 → Non-compact flange (just over limit).

Web: λw=41.41<λp=106.3\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 MpM_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 λp\lambda_p.
  • Compact sections: Mn=Mp=FyZxM_n = M_p = F_y Z_x (if also braced against LTB).
  • Non-compact sections: MnM_n reduced linearly as section yields partially; local buckling limits capacity.
  • Slender sections: local buckling before yield; use elastic formula Mn=FeSxM_n = F_e S_x with appropriate reduction.
  • Most rolled steel I-shapes in Philippines are compact for Fy≤345 MPaF_y \le 345 \text{ MPa}; verify for fabricated or high-strength steel.

4. Shear Strength of Steel Beams

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:

Vn=0.6FyAwCvV_n = 0.6 F_y A_w C_v

where:

  • Aw=d⋅twA_w = d \cdot t_w (gross area of web = overall depth × web thickness; note: NOT the clear web height)
  • CvC_v is the shear coefficient, depending on web slenderness: Cv=1.0C_v = 1.0 for stocky webs (h/tw≤2.24E/Fyh/t_w \le 2.24\sqrt{E/F_y}), and Cv<1.0C_v < 1.0 for slender webs (reduced by shear buckling).
  • The resistance factor is ϕv=1.0\phi_v = 1.0 for most rolled shapes (some fabricated girders with ϕv=0.9\phi_v = 0.9).

For stocky webs (typical of rolled I-shapes in the Philippines), Cv=1.0C_v = 1.0 and shear yield governs:

Vn=0.6FydtwV_n = 0.6 F_y d t_w

The design shear strength is:

ϕvVn=1.0×Vn(LRFD)\phi_v V_n = 1.0 \times V_n \quad (\text{LRFD})

For slender webs, shear buckling reduces capacity; CvC_v must be calculated per AISC Section G2.1 or simplified formulas. In most practical rolled sections used in Philippine buildings, h/twh/t_w is well below the slender limit, so Cv=1.0C_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

Examples

An IPE 550 section (common in Philippine industrial buildings) has overall depth d=550 mmd = 550 \text{ mm} and web thickness tw=11.1 mmt_w = 11.1 \text{ mm}. Steel is Grade 355 (Fy=355 MPaF_y = 355 \text{ MPa}, E=200 000 MPaE = 200\,000 \text{ MPa}). Verify that the web is stocky and calculate the design shear strength ϕvVn\phi_v V_n.

Number

5

Solution

Given: d=550 mmd = 550 \text{ mm}, tw=11.1 mmt_w = 11.1 \text{ mm}, Fy=355 MPaF_y = 355 \text{ MPa}, E=200 000 MPaE = 200\,000 \text{ MPa}.

Step 1: Check if web is stocky.

Stocky web limit (AISC):

htw≤2.24EFy=2.24×200 000355=2.24×23.74=53.22\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 hh is the depth between flanges, but for rolled sections they're nearly equal):

dtw=55011.1=49.55\frac{d}{t_w} = \frac{550}{11.1} = 49.55

Since $49.55 < 53.22$, the web is stocky and Cv=1.0C_v = 1.0.

Step 2: Calculate web area.

Aw=d×tw=550×11.1=6105 mm2A_w = d \times t_w = 550 \times 11.1 = 6105 \text{ mm}^2

Step 3: Calculate nominal shear strength.

Vn=0.6×Fy×Aw×Cv=0.6×355×6105×1.0=1,300,260 N≈1300.3 kNV_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).

ϕvVn=1.0×1300.3=1300.3 kN\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/2R = wL/2. If w=100 kN/mw = 100 \text{ kN/m}, then R=300 kN≪1300 kNR = 300 \text{ kN} \ll 1300 \text{ kN}, so shear does not govern; flexure would control the design.

A fabricated plate girder has web thickness tw=6 mmt_w = 6 \text{ mm}, height h=1500 mmh = 1500 \text{ mm}, and Fy=250 MPaF_y = 250 \text{ MPa}. Check whether Cv=1.0C_v = 1.0 applies or if web buckling reduces shear capacity. If Cv<1.0C_v < 1.0, estimate the reduction.

Number

6

Solution

Given: tw=6 mmt_w = 6 \text{ mm}, h=1500 mmh = 1500 \text{ mm}, Fy=250 MPaF_y = 250 \text{ MPa}, E=200 000 MPaE = 200\,000 \text{ MPa}.

Step 1: Check web slenderness.

htw=15006=250\frac{h}{t_w} = \frac{1500}{6} = 250

Stocky limit:

2.24EFy=2.24×28.28=63.392.24\sqrt{\frac{E}{F_y}} = 2.24 \times 28.28 = 63.39

Since $250 > 63.39$, the web is slender and Cv<1.0C_v < 1.0 (shear buckling governs).

Step 2: Calculate CvC_v for slender web.

Per AISC Section G2.1(a), for slender webs:

Cv=1.51EFy(h/tw)2=1.51×200 000250×2502=302 00015,625,000=0.0193C_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):

Cv=1.51EFy(h/tw)2C_v = \frac{1.51 E}{F_y (h/t_w)^2}

Cv≈0.019C_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/twh/t_w ratio and raising CvC_v. With stiffeners at, say, 1.5 m intervals, the panel aspect ratio improves and CvC_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 Cv=1.0C_v = 1.0.

Key points

  • Shear strength: Vn=0.6FyAwCvV_n = 0.6 F_y A_w C_v where Aw=d×twA_w = d \times t_w (gross web area).
  • For stocky webs (h/tw≤2.24E/Fyh/t_w \le 2.24\sqrt{E/F_y}), Cv=1.0C_v = 1.0 and Vn=0.6FydtwV_n = 0.6 F_y d t_w.
  • Design shear strength (LRFD): ϕvVn=1.0Vn\phi_v V_n = 1.0 V_n for rolled shapes; some fabricated girders use ϕv=0.9\phi_v = 0.9.
  • Use gross web area (full depth dd), not clear depth; web thickness twt_w is the distance between flanges.
  • Shear rarely governs in typical building frames but critical near supports and for deep, thin webs.

5. Design Procedures: LRFD vs. ASD

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 γ\gamma) and checks against factored resistance (ϕMn\phi M_n or ϕVn\phi V_n); ASD divides nominal strength by a safety factor (Ω\Omega) to obtain allowable strength. For flexure of compact, braced sections: LRFD uses ϕb=0.90\phi_b = 0.90 and Mn=Mp=FyZxM_n = M_p = F_y Z_x, so Design strength=0.90Mp\text{Design strength} = 0.90 M_p. ASD uses Ωb=1.67\Omega_b = 1.67 (approximately), giving Allowable strength=Mp/1.67≈0.60Mp\text{Allowable strength} = M_p / 1.67 \approx 0.60 M_p. The ASD allowable bending stress is often stated as 0.66FyF_y. For shear, LRFD uses ϕv=1.0\phi_v = 1.0 (rolled shapes) and Vn=0.6FyAwCvV_n = 0.6 F_y A_w C_v; ASD uses Ωv=1.50\Omega_v = 1.50, giving allowable shear as Vn/1.50=0.4FyAwCvV_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.

Examples

An IPE 360 compact section (Grade 250, fully braced) has Zx=1019 cm3Z_x = 1019 \text{ cm}^3. Calculate the design moment capacity using both LRFD and ASD. Compare the allowable capacities.

Number

7

Solution

Given: Zx=1.019×106 mm3Z_x = 1.019 \times 10^6 \text{ mm}^3, Fy=250 MPaF_y = 250 \text{ MPa}, section is compact and braced (Lb<LpL_b < L_p).

Plastic moment:

Mp=FyZx=250×1.019×106=254.75×106 N⋅mm=254.75 kN⋅mM_p = F_y Z_x = 250 \times 1.019 \times 10^6 = 254.75 \times 10^6 \text{ N}\cdot\text{mm} = 254.75 \text{ kN}\cdot\text{m}

LRFD Method:

Design moment=ϕbMp=0.90×254.75=229.3 kN⋅m\text{Design moment} = \phi_b M_p = 0.90 \times 254.75 = 229.3 \text{ kN}\cdot\text{m}

A girder with Mu=200 kN⋅mM_u = 200 \text{ kN}\cdot\text{m} (factored) is acceptable: $200 < 229.3$ ✓

ASD Method:

Allowable moment=MpΩb=254.751.67=152.5 kN⋅m\text{Allowable moment} = \frac{M_p}{\Omega_b} = \frac{254.75}{1.67} = 152.5 \text{ kN}\cdot\text{m}

Alternatively, using allowable stress Fb=0.66Fy=0.66×250=165 MPaF_b = 0.66 F_y = 0.66 \times 250 = 165 \text{ MPa} and elastic modulus Sx=Mp/Fy≈1.019/1.12×106 mm3=0.910×106 mm3S_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):

Ma=FbSx=165×0.910×106=150.15×106 N⋅mm≈150.2 kN⋅mM_a = F_b S_x = 165 \times 0.910 \times 10^6 = 150.15 \times 10^6 \text{ N}\cdot\text{mm} \approx 150.2 \text{ kN}\cdot\text{m}

(Minor variation due to shape factor; closer estimate: Ma≈152.5 kN⋅mM_a \approx 152.5 \text{ kN}\cdot\text{m}).

A girder with Ma=130 kN⋅mM_a = 130 \text{ kN}\cdot\text{m} (unfactored dead + live load) is acceptable: $130 < 152.5$ ✓

Ratio: LRFD allowable is (229.3/152.5)=1.50(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 ϕbMn≥Mu\phi_b M_n \ge M_u (required); more refined, variable factors by load type.
  • ASD: Check Ma≤FbSxM_a \le F_b S_x where Fb=ΩbF_b = \Omega_b reduction of Mp/SxM_p/S_x or yield stress; typically Fb≈0.66FyF_b \approx 0.66 F_y bending.
  • Flexure: LRFD ϕb=0.90\phi_b = 0.90; ASD Ωb≈1.67\Omega_b \approx 1.67 (i.e., design at ~60% yield for flexure in ASD).
  • Shear: LRFD ϕv=1.0\phi_v = 1.0 (rolled); ASD Ωv=1.50\Omega_v = 1.50 (i.e., design at 0.4Fy0.4 F_y for shear in ASD).
  • NSCP 2015 permits both; check with local requirement or project specification.

6. Lateral Bracing and Practical Design Considerations

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 Lb=0L_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 LpL_p for the beam and specifies brace locations such that Lb≤LpL_b \le L_p. If Lb>LpL_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 LpL_p for a given section and yield strength; (2) Determine if a beam with specified unbraced length can develop MpM_p or what moment capacity results from LTB; (3) Design the required spacing of lateral braces. For typical grade 250 or 355 steel, LpL_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.

Examples

A six-story office building in the Philippines has floor spans of 6.0 m×8.0 m6.0 \text{ m} \times 8.0 \text{ m} with composite floor slabs (concrete deck on steel beams). The main beams span 8.0 m8.0 \text{ m} and are Grade 250 I-sections with ry=45 mmr_y = 45 \text{ mm}. Determine LpL_p, and recommend a bracing strategy.

Number

8

Solution

Given: Composite floor, Lspan=8.0 mL_{span} = 8.0 \text{ m}, Fy=250 MPaF_y = 250 \text{ MPa}, ry=45 mmr_y = 45 \text{ mm}, E=200 000 MPaE = 200\,000 \text{ MPa}.

Calculate LpL_p:

Lp=1.76×45×200 000250=79.2×28.28=2239 mm≈2.24 mL_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 m8.0 \text{ m}, which exceeds Lp=2.24 mL_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 MnM_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 Lb=0L_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 MpM_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 MpM_p is available for design. For exposed steel, intermediate braces at Lp≈2.0–2.5 mL_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; Lb≈0L_b \approx 0, full MpM_p available.
  • Discrete braces (channels, angles) must be spaced at ≤Lp\le L_p to prevent LTB and maintain design strength.
  • For typical rolled sections, LpL_p ranges 1.5–3.5 m depending on ryr_y and FyF_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 Lb>LpL_b > L_p, calculate reduced MnM_n via inelastic or elastic LTB formula; often less economical than adding braces.

7. Interaction of Flexure and Shear; Combined Loading

While flexure and shear are analyzed separately, they interact in the cross-section. When a beam is subjected to both bending moment MM and shear force VV, the combined stress state must be checked. Per AISC 360 Section H1, when both flexure and shear are significant, the interaction formula is: MuϕbMn+VuϕvVn≤1.0(LRFD)\frac{M_u}{\phi_b M_n} + \frac{V_u}{\phi_v V_n} \le 1.0 \quad (\text{LRFD}) where MuM_u and VuV_u are the required (factored) moment and shear, and ϕbMn\phi_b M_n and ϕvVn\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 VnV_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.

Examples

A cantilever beam (length L=3.0 mL = 3.0 \text{ m}) carries a concentrated load P=80 kNP = 80 \text{ kN} at the free end. The section is a Grade 250 I-beam with Zx=1.2×106 mm3Z_x = 1.2 \times 10^6 \text{ mm}^3, d=450 mmd = 450 \text{ mm}, tw=10 mmt_w = 10 \text{ mm}, ry=45 mmr_y = 45 \text{ mm}. Assume the beam is continuously braced (e.g., by cladding or bracing system), so Lb≈0L_b \approx 0. Use LRFD with load factor γ=1.2\gamma = 1.2 on the concentrated load. Check flexure, shear, and interaction.

Number

9

Solution

Given: L=3.0 mL = 3.0 \text{ m}, P=80 kNP = 80 \text{ kN}, Zx=1.2×106 mm3Z_x = 1.2 \times 10^6 \text{ mm}^3, d=450 mmd = 450 \text{ mm}, tw=10 mmt_w = 10 \text{ mm}, Fy=250 MPaF_y = 250 \text{ MPa}, E=200 000 MPaE = 200\,000 \text{ MPa}, continuously braced.

Factored loads (LRFD):

Pu=1.2×80=96 kNP_u = 1.2 \times 80 = 96 \text{ kN}

Maximum moment (at fixed end):

Mu=Pu×L=96×3.0=288 kN⋅mM_u = P_u \times L = 96 \times 3.0 = 288 \text{ kN}\cdot\text{m}

Maximum shear (at fixed end):

Vu=Pu=96 kNV_u = P_u = 96 \text{ kN}

Design moment strength (compact, braced):

Mp=FyZx=250×1.2×106=300×106 N⋅mm=300 kN⋅mM_p = F_y Z_x = 250 \times 1.2 \times 10^6 = 300 \times 10^6 \text{ N}\cdot\text{mm} = 300 \text{ kN}\cdot\text{m}
ϕbMn=0.90×300=270 kN⋅m\phi_b M_n = 0.90 \times 300 = 270 \text{ kN}\cdot\text{m}

Check flexure:

MuϕbMn=288270=1.067>1.0EXCEEDS CAPACITY\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., Zx=1.3×106 mm3Z_x = 1.3 \times 10^6 \text{ mm}^3).

Design shear strength (assuming stocky web, Cv=1.0C_v = 1.0):

Aw=d×tw=450×10=4500 mm2A_w = d \times t_w = 450 \times 10 = 4500 \text{ mm}^2
Vn=0.6×Fy×Aw=0.6×250×4500=675×103 N=675 kNV_n = 0.6 \times F_y \times A_w = 0.6 \times 250 \times 4500 = 675 \times 10^3 \text{ N} = 675 \text{ kN}
ϕvVn=1.0×675=675 kN\phi_v V_n = 1.0 \times 675 = 675 \text{ kN}

Check shear:

VuϕvVn=96675=0.142≪1.0ADEQUATE\frac{V_u}{\phi_v V_n} = \frac{96}{675} = 0.142 \ll 1.0 \quad \text{ADEQUATE}

Check interaction:

MuϕbMn+VuϕvVn=1.067+0.142=1.209>1.0EXCEEDS LIMIT\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 ZxZ_x to ≈1.3×106 mm3\approx 1.3 \times 10^6 \text{ mm}^3, which gives ϕbMn=0.90×250×1.3×106/106=292.5 kN⋅m>288 kN⋅m\phi_b M_n = 0.90 \times 250 \times 1.3 \times 10^6 / 10^6 = 292.5 \text{ kN}\cdot\text{m} > 288 \text{ kN}\cdot\text{m} ✓

Key points

  • Linear interaction: MuϕbMn+VuϕvVn≤1.0\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: MaFbSx+VaFvAw\frac{M_a}{F_b S_x} + \frac{V_a}{F_v A_w} ≤ 1.0 (similar linear format).
  • Check interaction when both Mu/ϕbMn≈0.5M_u/\phi_b M_n \approx 0.5 and Vu/ϕvVn≈0.5V_u/\phi_v V_n \approx 0.5; otherwise, one dominates.

8. Common Pitfalls and Exam Tips

The following errors are frequently encountered in PRC exam problems and student solutions: (1) Using SxS_x instead of ZxZ_x for plastic moment: The plastic section modulus ZxZ_x is ~1.12 times larger than the elastic SxS_x. Students often mistakenly use Mp=FySxM_p = F_y S_x, which significantly underestimates capacity. Always verify which modulus applies—ZxZ_x for plastic/LRFD, SxS_x for elastic/ASD analysis. (2) Ignoring unbraced length and LTB: A common trap is calculating Mp=FyZxM_p = F_y Z_x without checking whether the compression flange is braced. If Lb>LpL_b > L_p, the actual MnM_n is much less than MpM_p. Always calculate LpL_p and compare to LbL_b. (3) Confusing web area in shear: The shear formula uses Aw=d×twA_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 ϕ\phi factors, load factors, and allowable stresses differ. If a problem states LRFD, use Mu/ϕbMnM_u / \phi_b M_n format; if ASD, use Ma/FbSxM_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 MpM_p. Non-compact sections have reduced capacity. (6) Not accounting for CbC_b in LTB calculations: For non-uniform moment, the moment-gradient factor Cb>1.0C_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.

Examples

A student designs a beam using an IPE 400 (Grade 250) with Zx=1160 cm3Z_x = 1160 \text{ cm}^3 and Sx=1043 cm3S_x = 1043 \text{ cm}^3. The required moment is Mu=250 kN⋅mM_u = 250 \text{ kN}\cdot\text{m} (LRFD). The student incorrectly calculates the design moment using SxS_x: (A) Mn=FySx=250×1043×103/106=260.75 kN⋅mM_n = F_y S_x = 250 \times 1043 \times 10^3 / 10^6 = 260.75 \text{ kN}\cdot\text{m}, then concludes ϕbMn=0.90×260.75=234.7 kN⋅m<250 kN⋅m\phi_b M_n = 0.90 \times 260.75 = 234.7 \text{ kN}\cdot\text{m} < 250 \text{ kN}\cdot\text{m} → section is inadequate. The professor marks this wrong. What is the correct approach?

Number

10

Solution

Correct calculation using ZxZ_x (plastic modulus):

Mn=Mp=FyZx=250×1160×103/106=290 kN⋅mM_n = M_p = F_y Z_x = 250 \times 1160 \times 10^3 / 10^6 = 290 \text{ kN}\cdot\text{m}
ϕbMn=0.90×290=261 kN⋅m>250 kN⋅m\phi_b M_n = 0.90 \times 290 = 261 \text{ kN}\cdot\text{m} > 250 \text{ kN}\cdot\text{m}

Result: The section is adequate. The student's use of SxS_x instead of ZxZ_x led to an underestimate of 26.3 kN⋅m26.3 \text{ kN}\cdot\text{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 ZxZ_x. For elastic analysis or ASD methods (less common now), use SxS_x. The shape factor (Zx/Sx≈1.12Z_x / S_x \approx 1.12) is a common pitfall on exams.

Key points

  • Use ZxZ_x (plastic) for MpM_p, not SxS_x (elastic). Shape factor ≈ 1.12; easy 10% error if confused.
  • Always check LbL_b vs. LpL_p. If Lb>LpL_b > L_p, LTB reduces MnM_n below MpM_p—do NOT ignore this.
  • Web area in shear: Aw=d×twA_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. ϕ\phi vs. Ω\Omega are different.
  • Verify compactness: width-to-thickness ratios for flange and web. Non-compact sections lose capacity.
  • In non-uniform bending, Cb>1.0C_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.

Diagrams

Steel Beam Flexure and Shear Design Framework

This mind map outlines the complete framework for steel beam design, showing the hierarchical relationship between flexure and shear strength, design methods (LRFD vs. ASD), and lateral bracing considerations. The central theme is balancing plastic moment capacity with stability limits (LTB and local buckling).

Steel Beam Flexural Design Decision Tree

This flowchart guides the engineer through the complete flexural design process, starting with compactness verification, lateral bracing assessment, and the selection of the appropriate limit state (plastic yield, inelastic LTB, or elastic LTB). It shows decision points and the resulting formulas or reduction factors for each path. The final checks on flexure and shear capacity determine design adequacy.

Moment Capacity Transitions: Lp and Lr Limits

This state diagram illustrates the three regimes of lateral-torsional buckling behavior as unbraced length (Lb) increases from zero. It shows the transition from full plastic moment (Lb ≤ Lp), through inelastic LTB (Lp < Lb ≤ Lr), to elastic LTB (Lb > Lr), with representative capacity at each state.

Shear Design and Web Slenderness Evaluation

This flowchart details the shear design process, including web slenderness evaluation (stocky vs. slender), the application of the shear coefficient Cv, and the identification of when transverse stiffeners are needed. The diagram emphasizes that slender webs require reduction factors and may necessitate web stiffening.

In summary

Steel beam flexure and shear design revolves around three key concepts: (1) Plastic moment capacity (Mp=FyZxM_p = F_y Z_x) for compact, laterally braced sections, (2) Lateral-torsional buckling (LTB) limits, defined by the unbraced length threshold Lp=1.76ryE/FyL_p = 1.76 r_y \sqrt{E/F_y}, beyond which moment capacity declines, and (3) Shear strength (Vn=0.6FyAwCvV_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 ≤ LpL_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 SxS_x instead of plastic ZxZ_x for MpM_p (~10% error), neglecting LTB checks despite unbraced length exceeding LpL_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.

Steel Beams: Flexure and Shear is chapter 3 of 5 in the Steel & Timber Design part of Super Tutor's CELE review. The Civil Engineer Licensure Examination is given by Professional Regulation Commission (PRC) — Board of Civil Engineering; its dates, format and passing rules are on the CELE page. See the CELE guide.

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