CELE Steel & Timber Design — Steel Compression MembersRevision Notes
Condensed revision notes for Steel Compression Members, built for the final weeks before the CELE 2026. These are the distilled key points you need when there is no time left for full study notes — just the concepts, formulas, and traps Professional Regulation Commission (PRC) — Board of Civil Engineering tests.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Steel & Timber Design subtest is marked as "Core" in the official pattern, and Steel Compression Members appears in position 2nd of 5 in the CELE Steel & Timber Design review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Steel Compression Members - Revision Notes
Steel compression members (columns, struts, chord members) fail by buckling long before the steel yields — this is the fundamental difference from tension members. The PRC Board Exam consistently tests the NSCP 2015 / AISC 360-10 column design procedure: computing the elastic buckling stress Fe, determining which branch of the Fcr curve applies, and calculating the LRFD design strength φcPn. Mastery of three equations and the transition slenderness check is the core of this chapter. All Philippine structural steel practice references NSCP 2015 Vol. I, Section 502 (formerly AISC 360), which adopts the same column curve. Slenderness limits, the φc = 0.90 resistance factor, and correct axis selection are the most frequent sources of board-exam errors.
Sections
Formulas
Example
KL/r = 70, E = 200,000 MPa → Fe = π²(200,000)/70² = 1,973,921/4,900 = 402.8 MPa
Formula
Fe = π²E / (KL/r)²
Variables
Fe = elastic (Euler) buckling stress (MPa); E = modulus of elasticity = 200,000 MPa for steel; K = effective-length factor; L = unbraced length (mm); r = radius of gyration about the buckling axis (mm)
Application
Compute Fe first using the controlling (largest) KL/r. This is the Euler critical stress that separates inelastic from elastic buckling.
Example
Fy = 248 MPa: 4.71√(200,000/248) = 4.71 × 28.40 = 133.7. Fy = 345 MPa: 4.71√(200,000/345) = 4.71 × 24.07 = 113.4.
Formula
Transition slenderness: (KL/r)_transition = 4.71√(E/Fy)
Variables
E = 200,000 MPa; Fy = specified yield stress (MPa). Equivalently, the transition occurs when Fy/Fe = 2.25.
Application
Compare the actual KL/r to this value to decide which Fcr formula to use. If KL/r ≤ 4.71√(E/Fy) → inelastic; if KL/r > 4.71√(E/Fy) → elastic.
Example
Fy/Fe = 248/402.8 = 0.616 → Fcr = 0.658^0.616 × 248 = 0.773 × 248 = 191.7 MPa
Formula
Fcr (inelastic) = [0.658^(Fy/Fe)] × Fy
Variables
Valid when KL/r ≤ 4.71√(E/Fy) or equivalently Fy/Fe ≤ 2.25. The exponent Fy/Fe is a dimensionless ratio.
Application
Used for stocky-to-intermediate columns (the most common range for practical building columns). The exponential form accounts for residual stresses and initial imperfections.
Example
KL/r = 150: Fe = π²(200,000)/150² = 87.73 MPa → Fcr = 0.877 × 87.73 = 76.94 MPa
Formula
Fcr (elastic) = 0.877 × Fe
Variables
Valid when KL/r > 4.71√(E/Fy). The factor 0.877 accounts for initial out-of-straightness (L/1500 assumed imperfection).
Application
Used for slender columns with large KL/r. Note that Fcr is simply 87.7% of the ideal Euler stress.
Example
Fcr = 191.7 MPa, Ag = 8,000 mm² → φcPn = 0.90 × 191.7 × 8,000 = 1,380,240 N ≈ 1,380 kN
Formula
φcPn = 0.90 × Fcr × Ag
Variables
φc = 0.90 (LRFD resistance factor for compression); Pn = nominal compressive strength (N); Fcr = critical stress from inelastic or elastic formula (MPa); Ag = gross cross-sectional area (mm²)
Application
Final LRFD design strength. The required factored load Pu must satisfy Pu ≤ φcPn. For ASD: Pn/Ωc where Ωc = 1.67.
Exam Tips
- Memorize the three key numbers for A36 steel (Fy = 248 MPa): transition KL/r = 133.7, E/Fy = 806.5. For A572 Gr 50 (Fy = 345 MPa): transition KL/r = 113.4.
- The five-step board-exam procedure: (1) Compute KL/r. (2) Compute 4.71√(E/Fy). (3) Compare → select Fcr formula. (4) Compute Fcr. (5) Compute φcPn = 0.90FcrAg.
- When the problem gives different K values for strong and weak axes, compute BOTH KL/r values and use the larger one.
- If only one KL is given with no axis specified, assume it controls weak-axis buckling (conservative).
- The 0.658 base: think of it as 'six-five-eight goes up from zero to one as Fy/Fe decreases from 2.25 to 0.' At the transition, 0.658^2.25 × Fy ≈ 0.39Fy... confirming Fcr is far below Fy for the elastic branch (check: 0.877Fe at transition = 0.877 × Fy/2.25 ≈ 0.39Fy ✓ — both branches give the same Fcr at the transition).
Key Points
- A compression member fails by buckling (lateral deflection) at a load far below the squash load Fy × Ag for slender members.
- Slenderness ratio KL/r governs the buckling load — the higher KL/r, the lower the critical stress Fcr.
- Always use the LARGEST KL/r (i.e., the smallest radius of gyration r, typically the weak axis rmin) unless weak-axis buckling is restrained by bracing.
- K is the effective-length factor: K = 1.0 for pin-pin, K = 0.5 for fixed-fixed, K = 0.7 for fixed-pin, K = 2.0 for fixed-free (flagpole). Use Table C-A-7.1 of AISC 360 or NSCP Table 502.2.
- Doubly symmetric compact sections (standard W-shapes, pipes, HSS) fail by flexural buckling. Singly symmetric or unsymmetric sections may also fail by torsional or flexural-torsional buckling.
- Recommended maximum KL/r ≤ 200 (NSCP 2015 Section 502.2 / AISC 360 Section E2 commentary).
- Gross area Ag is used for compression (no hole deductions), unlike tension members.
Definitions
Term
Slenderness Ratio (KL/r)
Definition
The ratio of effective column length KL to the radius of gyration r about the axis of buckling. It is the primary parameter controlling column strength.
Importance
Determines which buckling regime (inelastic or elastic) governs and directly computes Fe. Exam problems always require this as the first step.
Term
Effective Length Factor (K)
Definition
A dimensionless multiplier that converts the actual unbraced length L into the equivalent pin-pin length for buckling. K depends on the end boundary conditions (fixity).
Importance
Using K = 1.0 when fixity is present is conservative but may be too conservative; using K = 0.5 without proper justification is unconservative. Board exam problems usually specify K or the end conditions explicitly.
Term
Radius of Gyration (r)
Definition
r = √(I/A), where I is the moment of inertia and A is the cross-sectional area. It measures how efficiently a cross-section resists buckling.
Importance
Use the MINIMUM r (weak axis) unless otherwise noted. For W-shapes, ry < rx; for circular sections, rx = ry.
Term
Elastic Buckling Stress (Fe)
Definition
The theoretical Euler critical stress π²E/(KL/r)² at which a perfectly straight, pin-ended column would buckle elastically.
Importance
Fe is computed first and serves as the reference stress in both Fcr equations. It also defines the transition: inelastic if Fy/Fe ≤ 2.25, elastic if Fy/Fe > 2.25.
Term
Critical Stress (Fcr)
Definition
The actual compressive stress at which a column buckles, accounting for residual stresses (inelastic range) and initial imperfections (elastic range). Fcr < Fy always.
Importance
Fcr times Ag gives the nominal strength Pn. Every column design problem ends with computing Fcr.
Term
Gross Area (Ag)
Definition
The total cross-sectional area of the compression member without deductions for bolt holes or other openings.
Importance
Unlike tension members where net area Ae governs at connections, compression design uses Ag throughout the member length.
Section Title
1. Fundamental Concepts of Column Buckling
Common Mistakes
- Using the strong-axis radius rx instead of the weak-axis ry when computing KL/r (always use the LARGEST KL/r, which comes from the smallest r).
- Forgetting to check the transition: students sometimes apply the inelastic formula for slender columns (KL/r > 133.7 for A36 steel), getting an unconservatively high Fcr.
- Writing 0.685 instead of 0.658 as the base of the exponential — a common typographical error that leads to wrong Fcr.
- Using φc = 0.65 or 0.75 (the RC column factors from concrete design) instead of the steel φc = 0.90.
- Substituting KL in meters instead of millimeters in the slenderness ratio (r is in mm, so L must also be in mm).
- Forgetting that Pn = FcrAg and that φcPn = 0.90FcrAg — not φcPn = 0.90FyAg.
Formulas
Example
A W-column: KLx = 6 m, rx = 120 mm → KL/r|x = 50; KLy = 3 m, ry = 40 mm → KL/r|y = 75. Governing KL/r = 75 (weak axis governs despite shorter unbraced length).
Formula
(KL/r)_controlling = max[(KLx/rx), (KLy/ry)]
Variables
KLx, KLy = effective lengths about strong and weak axes respectively; rx, ry = radii of gyration about strong (x) and weak (y) axes
Application
When a column is braced at intermediate points about one axis but not the other, each axis has a different effective length. The larger KL/r governs.
Example
Fcr = 191.7 MPa, Ag = 8,000 mm² → Pn = 191.7 × 8,000 = 1,533,600 N; Allowable P = 1,533,600/1.67 = 918,323 N ≈ 918 kN
Formula
Pn/Ωc where Ωc = 1.67 (ASD)
Variables
Ωc = safety factor for compression (ASD); allowable strength = Pn/1.67 = FcrAg/1.67
Application
Some board exam questions still use ASD (older editions). The same Fcr equations apply; only the strength reduction changes from φcPn to Pn/Ωc.
Exam Tips
- Write out all five steps clearly in board exam solutions — partial credit is awarded even if arithmetic errors occur.
- When KL/r is exactly at the transition, either formula gives the same Fcr (they are continuous). Use the inelastic formula (it is the default boundary).
- For pipe and HSS circular sections, rx = ry, so only one KL/r needs to be checked.
- Quick mental check: if KL/r ≈ 100 and Fy = 248 MPa, the column is inelastic (100 < 133.7). If KL/r ≈ 150, it is elastic (150 > 133.7).
Key Points
- Step 1 — Determine the effective slenderness: Identify K for each axis. Compute (KL/r)x and (KL/r)y. Use the LARGER value.
- Step 2 — Verify slenderness limit: KL/r ≤ 200 (advisory limit per NSCP 2015 Commentary).
- Step 3 — Compute the transition slenderness: 4.71√(E/Fy). Compare with the actual KL/r.
- Step 4 — Compute Fe: Fe = π²E/(KL/r)².
- Step 5 — Compute Fcr: If KL/r ≤ 4.71√(E/Fy), use Fcr = [0.658^(Fy/Fe)]Fy. If KL/r > 4.71√(E/Fy), use Fcr = 0.877Fe.
- Step 6 — Compute design strength: φcPn = 0.90 × Fcr × Ag.
- Step 7 — Check adequacy: Pu ≤ φcPn.
- For selection problems (find required section), assume a trial Fcr (typically 50–70% of Fy for intermediate columns), compute required Ag, then verify the chosen section.
Definitions
Term
Inelastic Buckling Range
Definition
The range where KL/r ≤ 4.71√(E/Fy). Residual stresses and partial yielding reduce the column strength below the Euler load but above the elastic-range prediction. The Fcr curve bows inward from the yield stress.
Importance
Most practical building columns (3–6 m story heights, W-shapes) fall in this range. The exponential Fcr formula must be applied.
Term
Elastic Buckling Range
Definition
The range where KL/r > 4.71√(E/Fy). The column buckles at stresses well below yield, and material behavior is essentially linear-elastic. Only the 0.877 imperfection factor reduces the Euler stress.
Importance
Slender bracing members, long struts, and undersized columns may fall here. Fcr = 0.877Fe is simple to compute.
Term
Squash Load
Definition
The maximum compressive load a short column can carry before yielding: Py = FyAg. The column curve starts at Py for KL/r = 0.
Importance
Upper bound reference. Actual column strength Fcr is always less than Fy due to buckling and residual stress effects.
Section Title
2. Step-by-Step Column Design Procedure (LRFD)
Common Mistakes
- Computing only one KL/r (usually the strong axis) and missing the weak-axis governing case.
- In step-by-step solutions, forgetting to square KL/r when computing Fe — writing π²E/(KL/r) instead of π²E/(KL/r)².
- Treating the 0.658 base computation as multiplication instead of exponentiation: Fcr ≠ 0.658 × (Fy/Fe) × Fy.
Formulas
Example
Fy = 248 MPa: limit = 0.56√(200,000/248) = 0.56 × 28.40 = 15.90. If bf/2t = 14 < 15.90 → nonslender flange, Q = 1.0.
Formula
λ_flange = b/t ≤ 0.56√(E/Fy) for nonslender unstiffened element
Variables
b = half flange width for W-shapes (b = bf/2); t = flange thickness tf; limit = 0.56√(E/Fy)
Application
Check if the flanges of a compression member qualify as nonslender. If b/t > limit, Q < 1.0 and a Q-factor reduction is needed.
Example
Fy = 248 MPa: limit = 1.49√(200,000/248) = 1.49 × 28.40 = 42.3. Standard W-shapes rarely exceed this.
Formula
λ_web = h/tw ≤ 1.49√(E/Fy) for nonslender stiffened web
Variables
h = clear distance between flanges; tw = web thickness; limit = 1.49√(E/Fy)
Application
Check web slenderness for compression members. Most rolled W-shapes satisfy this limit.
Exam Tips
- For standard rolled W, HP, pipe, and HSS shapes listed in AISC tables, Q = 1.0 unless explicitly stated otherwise.
- If a board exam problem provides b/t ratios and asks you to check local buckling, compare against 0.56√(E/Fy) for flanges and 1.49√(E/Fy) for webs.
- The Q-factor topic is a secondary check — ensure you first master the global buckling check (KL/r → Fcr → φcPn).
Key Points
- The Fcr equations above assume the cross-section elements (flanges, webs) are compact or noncompact — they do not buckle locally before the member buckles globally.
- For sections with slender elements (λ > λr), local buckling reduces strength via the Q factor: Fcr_effective = Q × [0.658^(QFy/Fe)]Fy for the inelastic range.
- Q = QsQa where Qs applies to unstiffened elements (flanges) and Qa applies to stiffened elements (webs), computed from effective widths.
- Standard W-shapes from AISC tables are almost always noncompact or compact in compression; Q = 1.0 is typical for rolled sections.
- Board exams occasionally test the concept (checking if a section is slender) but rarely require a full Q-factor computation — know the concept and the slenderness limits.
- Width-to-thickness limit for unstiffened flanges: λr = 0.56√(E/Fy). For stiffened webs: λr = 1.49√(E/Fy).
- WT-shapes, double angles, and built-up sections are more likely to have slender elements.
Definitions
Term
Compact Section (for compression)
Definition
A section whose elements have b/t ratios below the nonslender limit λr, so local buckling does not reduce global buckling strength (Q = 1.0).
Importance
Most board exam problems use compact W-sections where Q = 1.0 and the standard Fcr formulas apply directly.
Term
Slender Element Section
Definition
A section where at least one element (flange or web) has b/t > λr. The Q-factor (< 1.0) must be applied, reducing the effective yield stress used in the column curve.
Importance
Identifies when the basic column formulas are insufficient. Signals need for Q-factor modification.
Section Title
3. Local Buckling and the Q-Factor (Slender Element Sections)
Common Mistakes
- Assuming all W-sections automatically have Q = 1.0 without checking — always verify if the problem specifies a non-standard (built-up or thin-walled) section.
- Confusing the local buckling check for compression (using λr limits) with the compactness check for flexure (which uses λp and λr for bending).
Formulas
Example
For most practical W-column checks, flexural Fe < torsional Fe, so flexural buckling governs. The section properties Cw and J are tabulated in AISC Steel Construction Manual.
Formula
Fe (torsional, doubly symmetric) = [π²ECw/(KzL)² + GJ] × 1/(Ix+Iy)
Variables
Cw = warping constant (mm⁶); G = shear modulus = 77,200 MPa; J = St. Venant torsional constant (mm⁴); Kz = effective length factor for torsional buckling; Ix, Iy = moments of inertia (mm⁴)
Application
Applies to doubly symmetric W-shapes when checking torsional buckling (rarely governs over flexural buckling for compact sections, but testable in theory).
Exam Tips
- Board exam problems with standard W-shapes: use flexural buckling only (Fe = π²E/(KL/r)²).
- If the problem involves a channel (C or MC), WT-section, or single angle, note 'flexural-torsional buckling governs — use AISC 360 E4/E5' even if you compute only flexural buckling in the solution steps (acknowledging it demonstrates exam awareness).
- Section properties Cw and J are never computed by hand in board exams — they are provided or tabulated.
Key Points
- Doubly symmetric shapes (W, HP, HSS, pipe) buckle only by flexural (Euler) buckling — no torsional component.
- Singly symmetric shapes (channels C, WT-tees, single angles) may buckle by FLEXURAL-TORSIONAL buckling, which can control over flexural buckling.
- Unsymmetric shapes (single unequal leg angles, Z-sections) must always be checked for torsional-flexural effects.
- NSCP 2015 / AISC 360 Section E4 covers torsional and flexural-torsional buckling with modified Fe expressions involving Cw (warping constant), J (torsional constant), and Ix+Iy.
- For board exam purposes: know which section types require the torsional check and that torsional Fe replaces Euler Fe in the same Fcr formulas.
- Single angles used as bracing members in trusses are particularly prone to flexural-torsional buckling — AISC 360 E5 provides simplified equations.
Definitions
Term
Flexural Buckling
Definition
The classical Euler mode — the column bends (deflects laterally) about its weak axis under axial compression. Governs for doubly symmetric sections.
Importance
This is the primary mode covered in board exams. All standard column design examples use flexural buckling.
Term
Torsional Buckling
Definition
Buckling by twisting about the member's longitudinal axis without lateral bending. Can occur in doubly symmetric shapes with open thin sections.
Importance
Rarely governs for standard W-shapes but is conceptually important. Know when to check it.
Term
Flexural-Torsional Buckling
Definition
Combined bending and twisting buckling mode. Governs for singly symmetric and unsymmetric sections under axial compression.
Importance
Singly symmetric shapes (channels, WT-tees) commonly appear in board exam questions — flag them for this check.
Section Title
4. Torsional and Flexural-Torsional Buckling
Common Mistakes
- Applying only the flexural buckling check (KL/r → Fe) to a channel or WT-section without noting that flexural-torsional buckling may govern with a lower Fe.
- Overlooking the torsional check for very short columns with open thin cross-sections.
Formulas
Example
Board Problem B (Elastic): same section, KL/r = 150 > 133.7 → elastic. Fe = π²(200,000)/150² = 87.73 MPa. Fcr = 0.877 × 87.73 = 76.94 MPa. φcPn = 0.90 × 76.94 × 8,000 = 554 kN. Board Problem C (Gr 50): KL/r = 100, Ag = 6,000 mm², Fy = 345 MPa. Transition = 113.4 > 100 → inelastic. Fe = 197.4 MPa. Fy/Fe = 1.748. Fcr = 0.658^1.748 × 345 = 0.481 × 345 = 166.0 MPa. φcPn = 0.90 × 166.0 × 6,000 = 896 kN.
Formula
Complete LRFD Column Design: φcPn = 0.90 × Fcr × Ag
Variables
Fcr selected based on KL/r vs. 4.71√(E/Fy); Ag in mm²; result in N, divide by 1,000 for kN
Application
Board Problem A (Inelastic): Ag = 8,000 mm², r = 50 mm, KL = 3,500 mm, Fy = 248 MPa. KL/r = 70 < 133.7 → inelastic. Fe = π²(200,000)/70² = 402.8 MPa. Fy/Fe = 0.616. Fcr = 0.658^0.616 × 248 = 191.7 MPa. φcPn = 0.90 × 191.7 × 8,000 = 1,380 kN.
Example
Exercise 4: W-column Ag=9,500mm², r=65mm, KL=5,000mm, Fy=248MPa. KL/r=5000/65=76.9 < 133.7 → inelastic. Fe=π²(200,000)/76.9²=335.0 MPa. Fy/Fe=0.740. 0.658^0.740=e^(0.740×(−0.4193))=e^(−0.3103)=0.7331. Fcr=0.733×248=181.8MPa. φcPn=0.90×181.8×9,500=1,554,390 N≈1,554 kN.
Formula
Exercise 1 solution: Pipe column K=1, L=4m, r=60mm, Ag=5,000mm², Fy=248MPa
Variables
KL/r = 4,000/60 = 66.7; Transition = 133.7 → inelastic. Fe = π²(200,000)/66.7² = 443.6 MPa. Fy/Fe = 248/443.6 = 0.559. Fcr = 0.658^0.559 × 248.
Application
0.658^0.559: use logarithm → 0.559 × ln(0.658) = 0.559 × (−0.4193) = −0.2344 → e^(−0.2344) = 0.7910. Fcr = 0.791 × 248 = 196.2 MPa. φcPn = 0.90 × 196.2 × 5,000 = 882,900 N ≈ 883 kN.
Exam Tips
- Scientific calculator proficiency is essential: practice computing 0.658^x using the [^] or [y^x] key. Board exam allows use of non-programmable scientific calculators.
- Alternative computation: 0.658^(Fy/Fe) = e^[(Fy/Fe) × ln(0.658)] = e^[−0.4193 × (Fy/Fe)]. Memorize ln(0.658) ≈ −0.419.
- Always box your final answer and state the unit (kN or N) — board examiners deduct for missing units.
- Time management: a column design problem should take 4–6 minutes. If you spend more, move on and return.
Key Points
- Three representative board problems covering: (A) A36 inelastic column, (B) A36 elastic column, (C) high-strength steel A572 Gr 50.
- All solutions follow the five-step procedure.
- SI units throughout: lengths in mm, areas in mm², stresses in MPa, forces in N then converted to kN.
- Standard steel material properties: E = 200,000 MPa; for A36, Fy = 248 MPa; for A572 Gr 50, Fy = 345 MPa; for A572 Gr 60, Fy = 415 MPa.
Definitions
Term
A36 Steel
Definition
The most common structural steel grade in Philippine construction: Fy = 248 MPa (36 ksi), Fu = 400 MPa, E = 200,000 MPa. Transition slenderness = 133.7.
Importance
Default steel grade for most board exam column problems unless otherwise specified.
Term
A572 Grade 50 / Grade 60
Definition
High-strength low-alloy steel: Gr 50 has Fy = 345 MPa, Gr 60 has Fy = 415 MPa. Higher Fy shifts the transition to lower KL/r (113.4 and 103.7 respectively), meaning more columns fall in the inelastic range.
Importance
Increasingly used in Philippine high-rise and long-span structures. Board exams test whether examinees can adapt the transition check for different Fy values.
Section Title
5. Worked Board-Exam Problems
Common Mistakes
- Computing 0.658^(Fy/Fe) by multiplying instead of using the exponent: 0.658^1.748 ≠ 0.658 × 1.748. Use the formula x^n = e^(n × ln x).
- In Exercise 4 (or similar), forgetting to convert KL from meters to mm before dividing by r in mm.
- Rounding Fe too early — keep at least 4 significant figures in Fe before computing Fy/Fe to avoid cascading rounding errors.
Connections
- Steel Tension Members: Both use Ag for area; but tension uses net area An for fracture checks while compression uses Ag throughout. The radius of gyration r appears only in compression (buckling), not in tension.
- Beam-Column Design (Combined Axial + Bending): The column design strength φcPn from this chapter becomes the denominator in the AISC 360 H1-1 interaction equations — master Fcr before attempting beam-column problems.
- Euler Buckling Theory (Engineering Mechanics): The Euler critical load Pe = π²EI/(KL)² is the basis of Fe. Fe = Pe/Ag = π²E/(KL/r)². Buckling concepts from mechanics of materials apply directly.
- Effective Length Factors (K): K values require knowledge of structural stability theory and boundary conditions. Frame stability, sidesway, and alignment charts (Chapter C of AISC 360) connect to column effective length.
- Connection Design: Columns transfer load to beams and foundations at connections. The factored load Pu that the connection must carry equals the column's axial demand — connected to bolted/welded connection design.
- Foundation Design: The ultimate column load (Pu from LRFD) transfers to footings. Column design completes the load path from beam → column → footing.
- Local Buckling / Plate Buckling: Width-to-thickness limits (λ < λr) for compression members parallel the compactness criteria for flexural members — both stem from plate buckling theory.
- Timber Compression Members (Column Design): The Fc' adjusted capacity for timber columns uses a similar column stability factor Cp that parallels the steel Fcr/Fy ratio — conceptual similarity aids cross-subject review.
Exam Strategy
For Steel Compression Member problems in the PRC Civil Engineering Board Exam (typically 3–5 questions per exam from the Steel Design set): FIRST, identify the section type (W, pipe, channel) and whether standard flexural buckling applies. SECOND, execute the five-step procedure: (1) KL/r, (2) transition check 4.71√(E/Fy), (3) select Fcr formula, (4) compute Fe then Fcr, (5) compute φcPn = 0.90FcrAg. Focus memorization on three items: the two Fcr equations, the transition formula, and φc = 0.90. Common board exam traps include: wrong axis for KL/r, units error (m vs mm), using 0.685 instead of 0.658, and applying φc = 0.65. Allocate 5–7 minutes per column problem. If the problem involves a channel or WT, note the torsional-flexural concern but proceed with flexural buckling unless section properties are given. Practice computing 0.658^x on your calculator (use y^x key) before the exam day. Know transition slenderness for Fy = 248 (133.7), 345 (113.4), and 415 (103.7) MPa by heart — these appear repeatedly on Philippine board exams.
Quick Review Questions
A W-column has Ag = 10,000 mm², ry = 45 mm, rx = 95 mm, K = 1.0, L = 4 m, Fy = 248 MPa, E = 200,000 MPa. Which axis governs and what is the controlling KL/r?
Always compute KL/r for both axes and use the LARGER value. The weak axis (smaller r) almost always governs for W-shapes under uniform end conditions. KL/r = 88.9 < transition (133.7) → inelastic buckling applies.
For Fy = 248 MPa, what is the transition slenderness ratio, and what does it mean physically?
The transition corresponds to Fy/Fe = 2.25. At KL/r = 133.7: Fe = π²(200,000)/133.7² = 110.3 MPa, and Fy/Fe = 248/110.3 = 2.25. Both Fcr formulas give the same value at the transition: 0.658^2.25 × 248 = 0.877 × 110.3 ≈ 96.8 MPa (approximately equal, confirming continuity of the curve).
A column has KL/r = 160 and Fy = 248 MPa. Which Fcr formula applies and what is Fcr?
Since KL/r = 160 exceeds the transition of 133.7, use the elastic formula Fcr = 0.877Fe. The 0.877 factor reduces the ideal Euler stress by 12.3% to account for initial out-of-straightness (assumed L/1,500 imperfection per AISC column research).
What is φcPn for a column with Fcr = 180 MPa and Ag = 7,500 mm²?
The resistance factor for compression is φc = 0.90 (LRFD, AISC 360 / NSCP 2015 Section 502.3). Multiply by Fcr and Ag in consistent units (MPa × mm² = N). Convert to kN by dividing by 1,000.
A column with KL/r = 100 and Fy = 345 MPa (A572 Gr 50): is buckling inelastic or elastic?
Higher Fy shifts the transition to a lower KL/r compared to A36 (133.7 for Fy = 248 MPa vs. 113.4 for Fy = 345 MPa). A KL/r = 100 would be elastic for a hypothetical very high Fy but is inelastic for both A36 and A572 Gr 50 under standard conditions.
What is the recommended maximum slenderness ratio for compression members per NSCP 2015?
NSCP 2015 Section 502.2 (AISC 360 E2 commentary) recommends KL/r ≤ 200 for main compression members. This is a serviceability / practicality limit — above KL/r = 200, the column becomes so slender that it is impractical, not necessarily that the equations break down. Bracing members may have higher slenderness in some codes.
Why is Ag used for compression design instead of net area An?
This contrasts directly with tension member design (NSCP 502.1 / AISC 360 D) where An and Ae govern. For compression, the full Ag participates in resisting the compressive force along the full member length. Only at the connection end where shear lag or local effects occur may reductions be considered in special cases.
A pipe column (r = 60 mm, Ag = 5,000 mm², Fy = 248 MPa) has K = 1.0 and L = 4 m. Find φcPn.
For a pipe (circular hollow section), rx = ry, so there is only one KL/r to check. Steps: compute KL/r → compare with transition → compute Fe → compute Fy/Fe → compute 0.658^(Fy/Fe) using logarithms → Fcr → φcPn.
Ready to practise for the CELE 2026?
Super Tutor's AI review plan adapts to your weak areas and builds a weekly practice schedule around your target CELE exam date.