CELE Steel & Timber Design — Steel Compression MembersStudy Notes
Study notes for Steel Compression Members 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 Compression Members lands at position 2nd 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 Compression Members - Study Notes
Steel compression members (columns) are fundamental load-bearing elements in structural frames. Unlike tension members that fail when material yields, steel columns fail by buckling—a stability phenomenon where the member becomes unstable and deflects laterally before the material reaches yield stress. This chapter covers the critical stress calculations, design strength formulas, and practical applications governed by NSCP 2015 (based on AISC 360-16). Understanding buckling is essential for safe and economical column design in Philippine construction practice.
Summary
Steel compression members fail by buckling, a stability phenomenon governed by slenderness ratio KL/r and the critical stress Fcr. NSCP 2015 (based on AISC 360) defines Fcr using a single smooth column curve: inelastic buckling (KL/r below transition) follows Fcr = [0.658^(Fy/Fe)]·Fy, while elastic buckling (KL/r above transition) follows Fcr = 0.877·Fe. The transition occurs at (KL/r)trans = 4.71√(E/Fy). Design strength is φcPn = 0.90·Fcr·Ag (LRFD). The controlling slenderness is the largest KL/r, typically the weak axis of an I-section. Designers must verify section compactness (Q-factor for slender elements), confirm K and L values from the structural system, and ensure KL/r ≤ 200. Singly symmetric sections require additional checks for torsional or flexural-torsional buckling. Mastering the column design procedure—estimate, select, calculate KL/r, determine inelastic/elastic range, compute Fcr, check local buckling, verify φcPn—is essential for exam success and safe design practice in the Philippines.
Sections
Steel columns differ fundamentally from tension members because they fail by elastic or inelastic buckling rather than material rupture. When a column is loaded axially, any small lateral deflection creates a bending moment (P-Δ effect) that increases deflection, eventually leading to collapse. The slenderness ratio KL/r (where K is effective length factor, L is unsupported length, and r is radius of gyration) controls whether buckling occurs at low stress (elastic) or after some inelastic deformation (inelastic). The design philosophy in NSCP 2015 uses a single smooth column curve that transitions from inelastic to elastic behavior. This differs from older methods that used separate Euler and parabolic formulas. The critical stress Fcr depends entirely on slenderness—a stocky column may safely reach 0.5Fy, while a very slender column might fail at only 0.05Fy.
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1. Introduction to Steel Compression Members and Buckling Behavior
Examples
Conceptual example: Comparing stocky vs. slender columns
Problem
Two identical W-shape steel sections are loaded axially. Section A has KL/r = 60 (stocky); Section B has KL/r = 180 (slender). Both have Fy = 248 MPa. Which fails at higher stress and why?
Solution
Section A (KL/r = 60) is in the inelastic range and reaches higher critical stress (~180 MPa); Section B (KL/r = 180) is in the elastic range and fails at much lower stress (~60 MPa). Slenderness dramatically reduces capacity. The transition occurs at KL/r = 4.71√(E/Fy) ≈ 133.7 for Fy = 248 MPa.
Key Points
- Buckling is a stability failure, not a material failure; occurs well below yield stress in slender members
- Slenderness ratio KL/r is the controlling parameter; largest KL/r (weakest axis) governs design
- NSCP 2015 uses a single smooth curve equation rather than separate inelastic/elastic formulas
- Design strength φcPn = 0.90·Fcr·Ag, where Ag is gross section area
- Recommended slenderness limit is KL/r ≤ 200 for main compression members
- Local buckling of slender elements is addressed separately via effective width or Q-factor reduction
The elastic (Euler) buckling stress is the theoretical stress at which a perfectly straight, initially unstressed column becomes unstable under axial load. Derived from differential equations of beam deflection, the Euler formula is: Fe = π²E/(KL/r)². This formula applies only to very slender members where stress remains below the proportional limit. In practice, NSCP 2015 applies a reduction factor of 0.877 to Euler stress in the elastic range to account for initial out-of-straightness and residual stresses. The elastic range begins at the transition slenderness (KL/r)trans = 4.71√(E/Fy), which corresponds to Fy/Fe = 2.25. For standard structural steel with E = 200,000 MPa and Fy = 248 MPa (A36 grade equivalent in Philippines), this transition occurs at KL/r ≈ 133.7. For higher-grade steel (Fy = 345 MPa), the transition is lower (~113.4), meaning thinner sections enter the elastic range sooner.
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2. Elastic Buckling Stress (Euler Buckling)
Examples
Calculate Euler stress and critical stress for a slender column
Problem
A pipe column has KL/r = 160, E = 200,000 MPa, Fy = 248 MPa. Calculate Fe and Fcr (elastic range).
Solution
Step 1: Verify elastic range. (KL/r)trans = 4.71√(200,000/248) = 133.7. Since 160 > 133.7, column is elastic. ✓ Step 2: Calculate Fe. Fe = π²(200,000)/(160)² = 39,478,418/25,600 = 77.26 MPa. Step 3: Calculate Fcr. Fcr = 0.877(77.26) = 67.78 MPa. Note: The 0.877 factor reflects practical reductions from ideal Euler theory.
Transition slenderness for different steel grades
Problem
Calculate (KL/r)trans for Fy = 248, 345, and 415 MPa (common Philippine grades).
Solution
Using (KL/r)trans = 4.71√(E/Fy) with E = 200,000 MPa: • Fy = 248 MPa: (KL/r)trans = 4.71√(200,000/248) = 133.7 • Fy = 345 MPa: (KL/r)trans = 4.71√(200,000/345) = 113.4 • Fy = 415 MPa: (KL/r)trans = 4.71√(200,000/415) = 104.2 Higher-grade steel transitions to elastic range at lower slenderness; thinner sections must be used carefully.
Key Points
- Euler formula: Fe = π²E/(KL/r)²; valid for elastic buckling only
- Transition slenderness: (KL/r)trans = 4.71√(E/Fy); equivalently, Fy/Fe = 2.25 at transition
- Applied Fcr in elastic range: Fcr = 0.877·Fe (NSCP 2015); accounts for imperfections and residual stress
- Elastic buckling dominates for KL/r > (KL/r)trans
- Fe decreases with square of KL/r; even small increases in slenderness greatly reduce capacity
- For E = 200,000 MPa: Fe = 81,680/(KL/r)² (in MPa when KL/r is dimensionless)
Inelastic buckling occurs in stocky members (KL/r below transition) where some fibers have yielded or reached the plastic region before overall buckling. NSCP 2015 uses the Johnson-modified formula in the inelastic range: Fcr = [0.658^(Fy/Fe)]·Fy. This single smooth equation, derived from regression of test data, eliminates the need to specify separate parabolic or Johnson formulas. The exponent Fy/Fe represents the ratio of yield stress to elastic buckling stress; when Fy/Fe is large (short, stocky members), 0.658^(Fy/Fe) approaches a minimum, and Fcr approaches zero for very short members (though Fcr ≤ Fy always). The column curve smoothly transitions from Fcr ≈ Fy (when KL/r = 0) through inelastic reduction, crossing at the transition point (Fy/Fe = 2.25), and then following 0.877·Fe in the elastic range. This single curve eliminates discontinuities present in older codes. For design, the governing value is always the minimum Fcr across all possible member lengths and orientations.
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3. Inelastic Buckling Stress and the Column Curve
Examples
Inelastic column design (stocky member)
Problem
A W360×110 section has Ag = 14,000 mm², least r = 90 mm, K = 1.0, L = 4.0 m, Fy = 248 MPa. Calculate Fcr and design strength φcPn.
Solution
Step 1: Calculate KL/r. KL/r = (1.0)(4,000)/(90) = 44.4. Step 2: Check range. (KL/r)trans = 133.7 > 44.4, so inelastic. ✓ Step 3: Calculate Fe. Fe = π²(200,000)/(44.4)² = 1,007.4 MPa. Step 4: Calculate Fy/Fe. Fy/Fe = 248/1,007.4 = 0.246. Step 5: Calculate Fcr. Fcr = [0.658^0.246](248) = [0.923](248) = 228.9 MPa. Step 6: Calculate design strength. φcPn = 0.90(228.9)(14,000) = 2,883.6 kN. Note: Critical stress remains high because KL/r is low (stocky member).
Comparing inelastic and elastic regions at transition
Problem
At transition (KL/r)trans ≈ 133.7 for Fy = 248 MPa, verify Fcr using both formulas.
Solution
At transition, Fy/Fe = 2.25 by definition. Inelastic formula: Fcr = [0.658^2.25](248) = [0.441](248) = 109.4 MPa. Elastic formula: Fe = π²(200,000)/(133.7)² = 111.1 MPa, so Fcr = 0.877(111.1) = 97.4 MPa. Note: Slight discrepancy (109 vs. 97 MPa) because transition is defined at Fy/Fe = 2.25, not at exact crossover. In practice, use (KL/r)trans and pick appropriate formula; the smooth curve ensures no real discontinuity.
Key Points
- Inelastic buckling formula: Fcr = [0.658^(Fy/Fe)]·Fy; applies when KL/r ≤ 4.71√(E/Fy)
- Exponent 0.658 is dimensionless; represents reduction from test data correlation
- When Fy/Fe is small (slender), exponent is small and Fcr ≈ 0.877·Fe (transitions to elastic)
- When Fy/Fe is large (stocky, KL/r → 0), exponent → large and Fcr → Fy
- Single smooth curve replaces older split inelastic/elastic formulas; no discontinuity at transition
- Fy/Fe = 2.25 marks the transition point: 0.658^2.25 ≈ 0.441, so Fcr ≈ 0.441·Fy at transition
The nominal compressive strength is Pn = Fcr·Ag, where Fcr is the critical stress from either the inelastic or elastic formula. Design (LRFD) strength is φcPn = 0.90·Fcr·Ag, with resistance factor φc = 0.90 (per NSCP 2015, based on AISC 360). For ASD (allowable stress design, less common in modern Philippine practice but may appear on exams), use Ωc = 1.67, giving allowable strength Pa = Pn/Ωc. The designer must evaluate both axes of the section and use the axis with the largest KL/r (smallest radius of gyration r), because that axis governs. For doubly symmetric I-sections, this is typically the weak axis (y-axis); for pipes, both axes are equal. The practical design procedure is: (1) calculate KL/r for both axes; (2) use the largest; (3) determine inelastic or elastic range; (4) calculate Fcr; (5) compute φcPn. A design check is: required strength ≤ φcPn (LRFD) or ≤ Pa (ASD). The NSCP 2015 recommends limiting KL/r ≤ 200 for main compression members to reduce risk of accidental overload and minimize second-order effects.
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4. Design Strength and Capacity Determination
Examples
Complete design capacity calculation for a built-up column
Problem
A built-up column is formed by connecting two C-channels back-to-back. Data: Ag = 8,500 mm², rx = 65 mm (x-axis, strong), ry = 45 mm (y-axis, weak), K = 0.9 (braced frame), L = 5.0 m, Fy = 248 MPa. Determine φcPn.
Solution
Step 1: Calculate slenderness for both axes. KLx/rx = (0.9)(5,000)/(65) = 69.2 KLy/ry = (0.9)(5,000)/(45) = 100.0 (governs; use this) Step 2: Check range. (KL/r)trans = 133.7 > 100, so inelastic. ✓ Step 3: Calculate Fe and Fcr. Fe = π²(200,000)/(100)² = 197.4 MPa Fy/Fe = 248/197.4 = 1.256 Fcr = [0.658^1.256](248) = [0.612](248) = 151.8 MPa Step 4: Calculate design strength. φcPn = 0.90(151.8)(8,500) = 1,160.5 kN Note: Weak axis (y-axis) controls due to smaller ry and larger KL/ry; design capacity is limited by this.
Verification of column slenderness limits
Problem
A column design yields φcPn = 800 kN with KL/r = 220. Is this acceptable per NSCP 2015?
Solution
NSCP 2015 recommends KL/r ≤ 200 for main compression members. Here, KL/r = 220 > 200, which exceeds the guideline. Although the calculation is technically correct, the excess slenderness raises concerns about: • Accidental overload sensitivity (small load increases cause large deflections) • Second-order P-Δ effects not fully captured in the column formula • Practical considerations (vibration, bolted connections) Recommendation: Increase section size or reduce unsupported length (add bracing) to achieve KL/r ≤ 200. If economically infeasible, document justification and verify P-Δ analysis in overall frame design.
Key Points
- Design strength: φcPn = 0.90·Fcr·Ag (LRFD, NSCP 2015); φc = 0.90 for compression
- Allowable strength: Pa = Pn/Ωc = Fcr·Ag/1.67 (ASD, if used)
- Always use largest KL/r (weakest axis) to determine Fcr; typically the minor axis for I-shapes
- For doubly symmetric sections, both axes equal in geometry; check minor axis y-axis (Iyy < Ixx)
- For unsymmetric sections (channels, angles), check both principal axes and torsional buckling
- Recommended slenderness limit: KL/r ≤ 200; exceeding this may indicate over-slenderness
- Design capacity is independent of applied load; determined entirely by geometry and Fy
The effective length factor K accounts for end conditions and boundary constraints. In the column formula KL/r, the product KL is the effective length Le, which may differ from geometric length L depending on how the member is supported. NSCP 2015 recognizes four classical pin-pin (K ≈ 1.0), fixed-pin (K ≈ 0.8), fixed-fixed (K ≈ 0.65), and fixed-free (K = 2.0) conditions. For braced frames (where lateral movement is prevented at floors by floor diaphragms and bracing), K is smaller (0.65 to 1.0) because both ends are partially or fully prevented from sidesway. For unbraced frames (moment-resistant frames where lateral stability depends on frame stiffness), K can exceed 1.0 because both ends may deflect in the same direction. Sideways buckling in unbraced frames is a critical design consideration. The effective length is used only for the flexural buckling mode; torsional buckling of singly symmetric or non-symmetric sections is handled separately with different formulas and effective length factors. For doubly symmetric sections in typical frames, flexural buckling about the weak axis usually governs.
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5. Effective Length and Buckling Modes
Examples
Effective length in braced vs. unbraced frames
Problem
An interior column in a multi-story building is 4.0 m tall. Estimate K for: (a) braced frame with concrete floor diaphragm, and (b) unbraced (moment-resistant) steel frame.
Solution
(a) Braced frame: Floors provide lateral bracing at each level. Column ends are partially restrained against rotation. Typical K = 0.8 to 0.9. Conservative estimate: K = 0.85. Le = 0.85(4,000) = 3,400 mm (b) Unbraced frame: Column must resist lateral load through frame bending. Frame sway causes both ends to deflect. Typical K = 1.2 to 1.5 depending on relative stiffness of columns and girders. Conservative estimate: K = 1.3. Le = 1.3(4,000) = 5,200 mm Note: Unbraced frame column has much larger effective length, resulting in larger KL/r and reduced capacity. This is why unbraced frames are more costly—they require larger columns or lateral bracing systems.
Impact of K on slenderness and capacity
Problem
Same W360×110 as Example 1 (r = 90 mm, Ag = 14,000 mm²), L = 3.5 m, Fy = 248 MPa. Calculate Fcr and φcPn for K = 0.8 (braced) and K = 1.2 (unbraced).
Solution
K = 0.8 (braced): KL/r = (0.8)(3,500)/(90) = 31.1 (stocky) Fe = π²(200,000)/(31.1)² = 2,034 MPa Fy/Fe = 248/2,034 = 0.122 Fcr = [0.658^0.122](248) = [0.965](248) = 239.4 MPa φcPn = 0.90(239.4)(14,000) = 3,015 kN K = 1.2 (unbraced): KL/r = (1.2)(3,500)/(90) = 46.7 (still stocky) Fe = π²(200,000)/(46.7)² = 905.4 MPa Fy/Fe = 248/905.4 = 0.274 Fcr = [0.658^0.274](248) = [0.903](248) = 224.0 MPa φcPn = 0.90(224.0)(14,000) = 2,822 kN Conclusion: Increasing K from 0.8 to 1.2 reduces capacity by ~193 kN (~6%). The unbraced frame requires either a larger column or different configuration.
Key Points
- Effective length Le = K·L; K varies from 0.65 (fixed-fixed) to 2.0 (fixed-free)
- Braced frames: K ≤ 1.0, often 0.8-0.9 for interior columns; sidesway prevented by floor system
- Unbraced frames: K > 1.0, typically 1.2-1.5 for moment-resistant frames; buckling involves frame sideways
- K ≈ 1.0 for pin-pin and simple span conditions; commonly used when in doubt
- K ≈ 0.8 for fixed-pin; K ≈ 0.65 for fixed-fixed; K = 2.0 for cantilever (fixed-free)
- Torsional and flexural-torsional buckling may govern for singly symmetric or non-symmetric shapes
- Alignment charts (Jackson and Moreland) available in references for non-standard end conditions
Even if a column avoids flexural buckling, local buckling of thin-walled elements (flanges, webs) can cause failure at lower stresses. Local buckling occurs when individual plate elements buckle under compressive stress before the overall member becomes unstable. NSCP 2015 addresses this through width-thickness limits for compact, non-compact, and slender classifications. Compact sections (stockier elements) can reach full Fy without local buckling; non-compact sections reach Fy but may lose capacity after yielding; slender sections fail by local buckling at stress below Fy. The critical stress for slender elements is reduced by a Q-factor: Fcr,reduced = Q·Fcr,calculated. The parameter Q is the product of reduction factors for flange local buckling (Qf) and web local buckling (Qw), each typically determined from element width-thickness ratios. For a rolled I-section with standard proportions, Q = 1.0 (compact); for a heavily built-up section or welded plate girder with thin flanges, Q may be 0.8 or lower. Designers must check width-thickness ratios during member selection and adjust Fcr if slender elements are present. This is especially critical in Philippine practice where high-strength materials and economic pressures may lead to thinner sections.
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6. Local Buckling and Slender Elements
Examples
Local buckling check for a built-up column
Problem
A welded column is built from two PL305×6 mm flanges connected by a PL305×4 mm web. Total section: 305×305 mm, Ag ≈ 4,200 mm², Fy = 248 MPa. Check if local buckling governs. (Compact limit for flange: b/t ≤ 0.381√(E/Fy) ≈ 10.8 for Fy = 248 MPa.)
Solution
Flange check: b/t = 305/(2×6) = 25.4 > 10.8 → slender flange Flange local buckling reduction: For b/t = 25.4, use AISC formula. Qf = 1.34 - 0.76(b/t)/√(E/Fy) = 1.34 - 0.76(25.4)/35.6 = 1.34 - 0.544 = 0.796 Web check: h/t = 305/(4) = 76.25 (very thin relative to height) For slender webs: Qw = 0.415(E)/(Fy)(t/h)² = 0.415(200,000)/(248)(4/305)² ≈ 0.334 Overall Q = Qf·Qw = 0.796(0.334) ≈ 0.266 Conclusion: Severe local buckling. If calculated Fcr (flexural) = 150 MPa, applied Fcr = 0.266(150) ≈ 40 MPa. Local buckling drastically reduces capacity. This section is unsuitable; use rolled I-section or increase thickness.
Avoiding local buckling through section selection
Problem
Specify a column section for an axial load of 1,200 kN, L = 3.5 m, K = 0.9, Fy = 248 MPa, avoiding local buckling complications.
Solution
Required Ag ≥ 1,200,000/(0.9×0.90×248) = 5,993 mm² (assuming mid-range Fcr ≈ 0.9Fy for conservative estimate) Select a rolled section: W310×143 (Ag = 18,200 mm², rx = 147 mm, ry = 89 mm). Verify: A rolled I-section is compact by design (AISC shapes are compact in both axes for typical Fy). KL/r = (0.9)(3,500)/(89) = 35.4 (inelastic, stocky) Fe = π²(200,000)/(35.4)² = 1,573 MPa Fy/Fe = 248/1,573 = 0.158 Fcr = [0.658^0.158](248) = [0.973](248) = 241.3 MPa φcPn = 0.90(241.3)(18,200) = 3,956 kN >> 1,200 kN ✓ Q = 1.0 (no local buckling for compact section) ✓ Conclusion: Rolled W-section avoids local buckling concerns and provides substantial margin.
Key Points
- Local buckling of plate elements can reduce strength below flexural buckling capacity
- Width-thickness (b/t or h/t) ratios determine element classification: compact, non-compact, or slender
- Compact elements: b/t ≤ λp; reach Fy; no reduction in design strength for local buckling
- Non-compact elements: λp < b/t ≤ λr; reduced inelastic capacity; Fcr reduced based on b/t
- Slender elements: b/t > λr; Fcr = Q·Fcr,calculated; Q is reduction factor from local buckling analysis
- Q = Qf·Qw (product of flange and web reduction factors); typical formulas in AISC 360 Appendix B
- Design procedure: Check section classification first; if slender, apply Q reduction to calculated Fcr
For singly symmetric or non-symmetric sections (channels, angles, T-shapes), flexural buckling about the weak axis may not govern. Instead, torsional buckling (twisting without bending) or flexural-torsional buckling (combined twisting and bending) can be the critical mode. Torsional buckling stress is: Ft = (1/r₀²)[GJ/A + (π²EIw)/(KL)²], where G is shear modulus, J is torsional constant, Iw is warping constant, r₀ is polar radius of gyration, and A is area. For most hot-rolled shapes, the radius of gyration r₀ and warping constant Iw are tabulated in steel manual appendices. The critical buckling mode is the minimum of flexural, torsional, and flexural-torsional stresses. In practice, for typical building columns: (1) I-sections with two axes of symmetry buckle in flexure (weak axis usually governs); (2) rectangular tubes, pipes buckle in flexure or torsion (both equal for circles); (3) channels and angles require torsional analysis because weak-axis flexural buckling may be less critical than torsional twisting. The NSCP 2015 and AISC 360 provide simplified expressions for singly symmetric sections, but full torsional analysis is often deferred to advanced references or software. For exam purposes, recognize that singly symmetric shapes require additional checks; problem statements typically specify whether to consider only flexural or to check torsional buckling.
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7. Torsional and Flexural-Torsional Buckling
Examples
Identifying buckling mode for a channel section
Problem
A C380×50 channel (singly symmetric about y-axis) is used as a column. Data: A = 6,400 mm², Ixx = 68×10⁶ mm⁴, Iyy = 4×10⁶ mm⁴, J = 0.12×10⁶ mm⁴, Iw = 2,200×10⁹ mm⁶, x₀ (to shear center) = 12 mm. Length L = 3.5 m, K = 1.0, Fy = 248 MPa, E = 200,000 MPa, G = 77,000 MPa. Determine the likely buckling mode.
Solution
Calculate slenderness for each axis: rx = √(Ixx/A) = √(68×10⁶/6,400) = 103.3 mm ry = √(Iyy/A) = √(4×10⁶/6,400) = 25.0 mm KL/rx = (1.0)(3,500)/(103.3) = 33.9 KL/ry = (1.0)(3,500)/(25.0) = 140 (largest; but check torsional) Flexural buckling about y-axis: Fe,flex = π²(200,000)/(140)² = 101 MPa Torsional buckling: r₀² = rx² + ry² + x₀² = 103.3² + 25.0² + 12² = 11,400 + 625 + 144 = 12,169 mm² Ft ≈ (1/12,169)[77,000(0.12×10⁶)/6,400 + π²(200,000)(2,200×10⁹)/(3,500)²] = (1/12,169)[1,440 + 35,470] ≈ 3.03 MPa (very low!) Conclusion: Torsional buckling (Ft ≈ 3 MPa) is far more critical than flexural (Fe,flex ≈ 101 MPa). The channel cannot be used as a simple compression member; it must be braced against torsion (e.g., wrapped in a bracing frame or restrained by adjacent elements) or a symmetric section must be used.
Recognizing when symmetric vs. singly symmetric matters in exam context
Problem
An exam question provides: 'Design a column for 500 kN axial load, L = 4.0 m, K = 0.8. Use Fy = 248 MPa. A.Specify a W-section. B. If using a C-channel, what additional consideration is necessary?'
Solution
A. W-section (doubly symmetric): Estimate required area: Ag ≈ 500,000/(0.9×200) = 2,778 mm² (crude estimate with Fcr ≈ 200 MPa) Select W310×67 (Ag = 8,500 mm², ry = 64 mm) KL/ry = (0.8)(4,000)/(64) = 50 Calculate Fcr (inelastic, stocky) Fcr ≈ 230 MPa → φcPn ≈ 1,770 kN >> 500 kN ✓ Design is straightforward. B. C-channel (singly symmetric): Before selecting a channel section, must verify that torsional buckling stress is not critical. Calculate Ft using available section properties. If Ft is low (< Fcr,flexural), the channel is inadequate unless externally braced against torsion. Recommendation: Use symmetric section (W, pipe, or built-up box) for simplicity, unless problem explicitly permits and provides torsional properties.
Key Points
- Singly symmetric and non-symmetric sections may buckle by torsion or flexural-torsion, not just flexure
- Torsional buckling stress: Ft = (1/r₀²)[GJ/A + (π²EIw)/(KL)²]; depends on section properties not in standard beam-column tables
- For singly symmetric about y-axis (e.g., channel): check flexural buckling (x and y) and torsional/flexural-torsional buckling
- Critical stress is the minimum of all possible buckling modes; governs design
- Polar radius r₀ = √(x₀² + y₀²) where (x₀, y₀) is shear center location
- Warping constant Iw available in steel manual; represents resistance to differential bending of flanges during torsion
- For exam: Recognize when torsional check is needed (singly symmetric shapes); detailed calculations may be simplified or eliminated
Steel column design in practice follows a systematic workflow. First, estimate required section size based on preliminary capacity (assuming mid-range critical stress, e.g., 0.5Fy or 100 MPa for a rough check). Second, select candidate sections from a steel shape manual (rolled shapes are preferred for ease of fabrication and connection). Third, calculate slenderness ratio(s) using the appropriate K and L from structural arrangement. Fourth, determine whether the member is in the inelastic or elastic range by comparing KL/r to the transition value. Fifth, compute Fcr using the corresponding formula. Sixth, check for local buckling by verifying section compactness or applying Q-factor if needed. Seventh, calculate design strength φcPn = 0.90·Fcr·Ag and verify against required strength. Eighth, if capacity is insufficient, increase section size and repeat; if capacity is excessive, consider a smaller section to optimize cost. This iteration is tedious by hand but essential for proper design. In Philippines practice, checking minimum and maximum slenderness limits (KL/r ≤ 200) and ensuring proper connection design are equally critical as stress calculation. Always verify that selected sections meet NSCP 2015 material grade requirements and are available from local suppliers.
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8. Practical Design Examples and Problem-Solving Strategy
Examples
Complete column design example: Multi-step approach
Problem
Design a vertical compression member for a braced office building. Given: Factored load Pu = 2,000 kN; unsupported height L = 3.5 m; bracing at mid-height (K = 0.85); Fy = 248 MPa; steel grade A36-equivalent. Select a suitable W-shape.
Solution
Step 1—Preliminary estimate: Assume Fcr ≈ 150 MPa (typical mid-range for KL/r ≈ 50). Required Ag ≥ 2,000,000 / (0.90 × 150) = 14,815 mm² Round up to ~15,000 mm². Step 2—Select candidate section: From steel manual (or NSCP Annex), choose W360×179 (Ag = 22,800 mm²; rx = 162 mm; ry = 112 mm). Slightly conservative but easier to source. Step 3—Calculate slenderness: With bracing at mid-height: - For lower half-story: K = 0.85, L = 1.75 m → KL/ry = (0.85)(1,750)/(112) = 13.3 - For upper half-story: K = 0.85, L = 1.75 m → KL/ry = 13.3 (Weak axis governs; if unbraced top, would use full L = 3.5 m) KL/ry = (0.85)(3,500)/(112) = 26.3 (full height, if no mid-bracing) Assume worst case: KL/r = 26.3 (very stocky). Step 4—Determine Fcr: (KL/r)trans = 4.71√(200,000/248) = 133.7 >> 26.3 → inelastic. Fe = π²(200,000)/(26.3)² = 2,853 MPa Fy/Fe = 248/2,853 = 0.087 Fcr = [0.658^0.087](248) = [0.994](248) = 246.6 MPa Step 5—Check local buckling: W360×179 is a rolled section, compact by design. Q = 1.0 ✓ Step 6—Calculate design strength: φcPn = 0.90(246.6)(22,800) = 5,061 kN Step 7—Verify: Required: 2,000 kN Provided: 5,061 kN Margin: (5,061 - 2,000) / 5,061 = 60.5% excess capacity The section is oversize. Step 8—Optimize (optional): Try W310×158 (Ag = 20,100 mm²; ry = 86 mm): KL/r = (0.85)(3,500)/(86) = 34.6 Fe = π²(200,000)/(34.6)² = 1,649 MPa Fy/Fe = 0.150 Fcr = [0.658^0.150](248) = [0.976](248) = 242.1 MPa φcPn = 0.90(242.1)(20,100) = 4,381 kN > 2,000 kN ✓ (43% excess) Try W310×143 (Ag = 18,200 mm²; ry = 89 mm): KL/r = (0.85)(3,500)/(89) = 33.4 Fe = π²(200,000)/(33.4)² = 1,772 MPa Fy/Fe = 0.140 Fcr = [0.658^0.140](248) = [0.977](248) = 242.3 MPa φcPn = 0.90(242.3)(18,200) = 3,963 kN > 2,000 kN ✓ (49% excess) Recommendation: Select W310×143 (good balance of strength and economy). Verify availability in Philippines market and connection design for flanges and web.
Exam-style short-answer problem
Problem
A pin-ended column (K = 1.0) is 6.0 m tall. It must carry 1,200 kN factored axial load. Use Fy = 345 MPa (higher-grade steel). Specify minimum section dimensions (approx. Ag and r_min required).
Solution
Preliminary: Assume Fcr ≈ 120 MPa (higher grade, so transition is lower) Required Ag ≥ 1,200,000 / (0.90 × 120) = 11,111 mm² For a tall, pin-ended column with L = 6.0 m, expect KL/r ≈ 80-100 (inelastic transition ≈ 113.4 for Fy = 345). If KL/r = 90: Fe = π²(200,000)/(90)² = 244 MPa Fy/Fe = 345/244 = 1.414 Fcr = [0.658^1.414](345) = [0.573](345) = 197.6 MPa φcPn = 0.90(197.6)(11,111) = 1,973 kN > 1,200 kN ✓ Minimum r_min ≥ KL/KL/r = (1.0)(6,000)/90 = 66.7 mm Candidate: W310×117 (Ag = 14,900 mm², ry = 88 mm) KL/r = 6,000/88 = 68.2 < 90 (even better, lower slenderness) Capacity would be higher than preliminary estimate. Answer: Approximate requirements are Ag ≥ 11,000 mm² and r_min ≥ 65 mm; a W310×117 or similar mid-range section should suffice. Detailed check required with exact section properties.
Key Points
- Design workflow: Estimate → Select → Calculate KL/r → Determine Fcr → Check local buckling → Verify φcPn ≥ required
- Preliminary estimate: Ag ≥ required load / (0.90 × 150), assuming Fcr ≈ 150 MPa for mid-range member
- Always calculate slenderness ratio for both axes; use the larger KL/r (governs design)
- Transition test is critical: if KL/r is near 4.71√(E/Fy), small changes dramatically alter Fcr
- Verify compactness of selected section; if slender, apply Q-reduction
- Check practical limits: KL/r ≤ 200 for main members; minimize KL/r for load-bearing columns
- In Philippines, confirm steel grade (A36 equiv., A50 equiv., A70 equiv. per ASTM or local standards) and availability
Exam questions on steel columns frequently trip up even well-prepared reviewees due to subtle definitional and computational errors. Common pitfalls include: (1) using the wrong (smaller) radius of gyration—always verify which axis has the smallest r, and thus largest KL/r; (2) incorrectly applying the transition test—remembering that the transition slenderness is 4.71√(E/Fy), not a fixed number; (3) confusing the exponent in Fcr = [0.658^(Fy/Fe)]Fy—ensure the exponent is Fy/Fe, not Fe/Fy or something else; (4) forgetting the 0.877 factor in the elastic range; (5) misinterpreting K—over-conservative (too large) or under-conservative (too small) values based on frame type; (6) neglecting local buckling for thin-walled shapes—always check b/t ratios; (7) ignoring torsional buckling for singly symmetric sections; (8) not verifying that KL/r ≤ 200 (NSCP 2015 recommendation); (9) calculating flexural stress when torsional stress governs; (10) mixing LRFD and ASD in the same problem—be consistent. For exam success: (a) always show a clear slenderness calculation; (b) explicitly state whether inelastic or elastic; (c) show Fcr formula and numerical result; (d) compute φcPn = 0.90 × Fcr × Ag step-by-step; (e) verify answer against known limits (Fcr should be between 0 and Fy). Multiple-choice exams often use distractor answers based on these common errors, so understanding the reasoning (not just memorizing formulas) is critical.
Heading
9. Common Pitfalls and Exam Tips for Board Reviewers
Examples
Common error: Wrong axis slenderness
Problem
A W310×79 section is selected for a column. Manual data: rx = 134 mm, ry = 71 mm, L = 4.0 m, K = 1.0. A careless reviewer calculates KL/r = 4,000/134 = 29.9 and proceeds with inelastic Fcr ≈ 0.95Fy ≈ 235 MPa. What is the error?
Solution
Error: Used strong-axis radius rx = 134 mm instead of weak-axis ry = 71 mm. Correct calculation: KL/r = 4,000/71 = 56.3 (not 29.9) Fy/Fe = 248 / [π²(200,000)/(56.3)²] = 248 / 625 = 0.396 Fcr = [0.658^0.396](248) = [0.833](248) = 206.6 MPa (not 235 MPa) Error in Fcr: 235 - 206.6 = 28.4 MPa (~14% overestimate of capacity) Exam consequence: Design margin overstated; actual safety factor is lower than believed. Lesson: Always identify the critical (weak) axis, which has the smallest radius of gyration r_min; use that in KL/r calculations.
Common error: Forgetting the transition test
Problem
A column has KL/r = 140, Fy = 248 MPa. A reviewer directly uses inelastic formula without checking transition. Transition is (KL/r)trans = 133.7. What is the impact?
Solution
Error: KL/r = 140 > 133.7, so the column is actually elastic, not inelastic. Incorrect (inelastic) calculation: Fe = π²(200,000)/(140)² = 101.0 MPa Fy/Fe = 248/101.0 = 2.455 Fcr = [0.658^2.455](248) = [0.376](248) = 93.2 MPa (incorrect) Correct (elastic) calculation: Fcr = 0.877 × 101.0 = 88.6 MPa Error magnitude: 93.2 - 88.6 = 4.6 MPa (~5% overestimate), but the form of the answer is wrong. Exam consequence: Mixing formulas suggests incomplete understanding; may trigger review/partial credit. Lesson: Always calculate (KL/r)trans = 4.71√(E/Fy) and compare KL/r to the threshold before picking the Fcr formula.
Board exam problem with intentional distractor
Problem
Multiple-choice: A W360×122 column (Ag = 15,500 mm², ry = 100 mm) with KL = 3.5 m, Fy = 248 MPa, has design strength φcPn approximately: A) 2,100 kN B) 2,300 kN C) 2,700 kN D) 3,100 kN
Solution
KL/r = 3,500/100 = 35 (stocky, inelastic) Fe = π²(200,000)/(35)² = 1,613 MPa Fy/Fe = 248/1,613 = 0.154 Fcr = [0.658^0.154](248) = [0.977](248) = 242.2 MPa φcPn = 0.90(242.2)(15,500) = 3,379 kN Nearest answer: D) 3,100 kN (reasonable for rounding) Distractor analysis: A) 2,100 kN → Fcr ≈ 150 MPa (typical for KL/r ≈ 75, wrong slenderness) B) 2,300 kN → Fcr ≈ 165 MPa (typical for KL/r ≈ 65) C) 2,700 kN → Fcr ≈ 195 MPa (typical for KL/r ≈ 50) D) 3,100 kN → Fcr ≈ 223 MPa (matches KL/r ≈ 35) ✓ Exam tip: If answer doesn't match any option, recalculate; likely an arithmetic error or wrong axis used. Distractor answers often correspond to off-by-one mistakes or wrong formula application.
Key Points
- Most common error: using strong-axis radius (larger r) instead of weak-axis (smaller r), yielding wrong KL/r
- Transition slenderness 4.71√(E/Fy) depends on Fy; high-strength steel transitions at smaller KL/r than mild steel
- Exponent in inelastic formula is Fy/Fe (yield divided by elastic), not reciprocal
- Elastic range always includes 0.877 factor on Euler stress (accounts for imperfections)
- K = 0.85 is not 'safe default'—verify from structural system (braced vs. unbraced frame)
- Section compactness affects Q-factor; rolled shapes are usually compact, built-up shapes may not be
- Torsional buckling critical for channels, angles, T-shapes; not relevant for pipes and closed sections
- KL/r > 200 is red flag; indicates over-slender member even if calculation shows adequate capacity
- ASD uses Ωc = 1.67 (not 1.5); LRFD uses φc = 0.90; never mix in same problem
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