CELE Steel & Timber Design — Steel Compression MembersDetailed Explanation
A detailed, step-by-step explanation of Steel Compression Members for CELE aspirants. This page goes deeper than the summary and study notes, walking through the reasoning behind each concept so you understand why Professional Regulation Commission (PRC) — Board of Civil Engineering tests it the way it does in the CELE Steel & Timber Design subtest.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Steel & Timber Design section sits under a "Core" weighting, and Steel Compression Members is the 2nd chapter in the 5-chapter CELE Steel & Timber Design rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Steel & Timber Design.
Steel Compression Members - Detailed Explanation
Steel compression members — commonly called columns — are structural elements that carry axial compressive loads. Unlike tension members that fail by yielding or fracture, compression members typically fail by **buckling** at stresses well below the yield stress, especially when the member is long or slender. This chapter covers the design philosophy under NSCP 2015 (which adopts AISC 360-10), focusing on the critical buckling stress Fcr, the inelastic-to-elastic transition, effective-length concepts, and the design strength equation φcPn = 0.90 FcrAg. Mastery of these concepts is essential for the PRC Civil Engineer Licensure Examination (CELE), as steel column design problems consistently appear in the Structural Engineering and Construction (SEC) subject. The governing Philippine code is NSCP 2015 Volume I, Section 502, based on AISC 360-10 Chapter E.
Concepts
Euler Elastic Buckling Stress (Fe)
The elastic (Euler) buckling stress is the theoretical compressive stress at which a perfectly straight, pinned-end column of uniform cross-section becomes unstable. Derived by Leonhard Euler in the 18th century, it serves as the backbone of all modern column design equations. The formula is: Fe = π²E / (KL/r)² Where: • E = modulus of elasticity = 200,000 MPa for structural steel • K = effective-length factor (depends on end conditions) • L = unbraced length of the member (mm) • r = radius of gyration of the gross cross-section (mm) • KL/r = effective slenderness ratio (dimensionless) The radius of gyration r = √(I/Ag), where I is the moment of inertia and Ag is the gross cross-sectional area. **Critical rule: Always use the LARGEST KL/r ratio.** For a doubly symmetric section (e.g., W-shapes), check both the strong axis (KL/rx) and the weak axis (KL/ry). The weak axis almost always governs because ry < rx. However, if the weak axis is braced at intermediate points but the strong axis is not, the strong axis may control — always verify both. As KL/r increases (longer or more slender member), Fe decreases rapidly (inverse-square relationship). A column with KL/r = 200 has only (70/200)² = 1/8.16 the elastic buckling capacity of a column with KL/r = 70 — buckling sensitivity to slenderness is dramatic.
Examples
In this case the effective lengths are chosen so both axes have nearly equal slenderness. In typical building frames with continuous bracing at floor levels, KyLy ≪ KxLx, and the weak axis governs overwhelmingly. Always check both — never assume.
Scenario
A W250×73 steel column (Ag = 9,280 mm², rx = 112 mm, ry = 65 mm) has an effective length KxLx = 6.0 m for the strong axis and KyLy = 3.5 m for the weak axis. E = 200,000 MPa. Compute Fe for each axis.
Solution
Strong axis: KxLx/rx = 6000/112 = 53.57 Weak axis: KyLy/ry = 3500/65 = 53.85 Governing KL/r = 53.85 (weak axis, barely controls) Fe = π²(200,000) / (53.85)² = 1,973,921 / 2899.8 = 680.4 MPa
For circular sections (pipe, round HSS), the radius of gyration is the same about all centroidal axes. Only one slenderness ratio needs to be checked. This is a common board-exam simplification.
Scenario
A round pipe column: Ag = 3,600 mm², r = 55 mm (same in all directions), K = 1.0, L = 4.0 m. Find Fe.
Solution
KL/r = (1.0)(4000) / 55 = 72.73 Fe = π²(200,000) / (72.73)² = 1,973,921 / 5289.6 = 373.2 MPa
Applications
- Computing the governing slenderness ratio before selecting the Fcr equation
- Checking adequacy of an existing column under increased loading
- Designing bracing systems to reduce effective length and increase capacity
- Comparing columns of different cross-sections at the same slenderness
- First step in all NSCP 2015/AISC 360 column design calculations
Misconceptions
- Using the smaller KL/r (stronger axis) instead of the larger — this OVERESTIMATES capacity
- Confusing Fe (elastic buckling stress) with Fcr (critical stress used for design) — Fcr accounts for inelastic effects and imperfections
- Forgetting that Fe is purely theoretical (perfect column); real columns use Fe to compute Fcr
- Using L in meters without converting to mm — KL/r becomes 1000× smaller, Fe becomes 10⁶× larger
Related Concepts
- Effective length factor K
- Radius of gyration r = √(I/Ag)
- Critical stress Fcr (inelastic and elastic)
- Bracing and lateral support of columns
- Torsional and flexural-torsional buckling
Common Exam Questions
Example
Find Fe for a column with K=1.0, L=5 m, r=48 mm. Answer: KL/r=104.2, Fe=π²(200000)/10,862=181.7 MPa
Approach
Given K, L, and r, compute KL/r then Fe = π²E/(KL/r)². Watch units — L in mm, r in mm, E in MPa → Fe in MPa.
Question Type
Direct computation
Example
KxLx/rx=80, KyLy/ry=95 → Use KL/r=95 → Fe=π²(200000)/9025=219.0 MPa
Approach
Compute KL/r for each axis separately; the LARGER value gives the LOWER Fe and governs.
Question Type
Identify governing axis
Key Points To Remember
- Fe = π²E/(KL/r)² — the Euler elastic buckling stress, in MPa
- Use E = 200,000 MPa for all structural steel grades
- Always compute KL/r for BOTH axes; the LARGEST value governs design
- r = √(I/Ag); for W-shapes, ry is almost always smaller than rx
- Fe decreases with the SQUARE of (KL/r) — doubling slenderness reduces Fe by 4×
- Fe has no direct dependence on Fy — it is purely a geometric and material stiffness property
- Recommended slenderness limit: KL/r ≤ 200 (NSCP 2015 Section 502.3)
- K values: pin-pin = 1.0, fixed-free = 2.0, fixed-pin = 0.7, fixed-fixed = 0.5
Critical Stress Fcr — Inelastic vs Elastic Buckling
Real steel columns are never perfectly straight, perfectly loaded, or perfectly homogeneous. Residual stresses from rolling and welding further reduce effective stiffness. NSCP 2015 (AISC 360-10 Section E3) accounts for these imperfections through the critical stress Fcr, which lies on a smooth curve between two regimes: **Transition Point:** KL/r_transition = 4.71√(E/Fy) Equivalently: Fy/Fe = 2.25 at the transition For A36 steel (Fy = 248 MPa, E = 200,000 MPa): KL/r_transition = 4.71√(200,000/248) = 4.71 × 28.39 = 133.7 For Grade 50 steel (Fy = 345 MPa): KL/r_transition = 4.71√(200,000/345) = 4.71 × 24.04 = 113.3 **Case 1 — Inelastic Buckling (stocky columns):** Applies when KL/r ≤ 4.71√(E/Fy) [or Fy/Fe ≤ 2.25] Fcr = [0.658^(Fy/Fe)] × Fy The exponent (Fy/Fe) controls how far Fcr falls below Fy. As KL/r decreases toward zero, Fy/Fe → 0 and 0.658^0 → 1.0, so Fcr → Fy (yielding controls). As KL/r increases toward the transition, 0.658^2.25 ≈ 0.394, giving Fcr ≈ 0.394Fy at the boundary. **Case 2 — Elastic Buckling (slender columns):** Applies when KL/r > 4.71√(E/Fy) [or Fy/Fe > 2.25] Fcr = 0.877 × Fe The factor 0.877 (≈ 1/1.14) reduces the ideal Euler stress by about 12% to account for the initial out-of-straightness (geometric imperfection) assumed in AISC as L/1500 of the member length. Note that Fy does not appear in this equation — for very slender columns, material strength is irrelevant; only geometry and stiffness matter. **Continuity at the transition:** Both equations give the same Fcr at KL/r = 4.71√(E/Fy), ensuring a smooth design curve with no sudden jump.
Examples
This is the classic board-exam pattern for an inelastic column. The key calculator step is computing 0.658^(Fy/Fe) using the y^x key: enter 0.658, press y^x, enter 0.616, press =. Result: 0.772. Multiply by Fy.
Scenario
Column: KL/r = 70, Fy = 248 MPa, E = 200,000 MPa. Determine Fcr.
Solution
Step 1: Transition check 4.71√(200,000/248) = 133.7 70 < 133.7 → INELASTIC Step 2: Compute Fe Fe = π²(200,000) / 70² = 1,973,921 / 4,900 = 402.8 MPa Step 3: Compute exponent Fy/Fe = 248/402.8 = 0.616 Step 4: Compute Fcr Fcr = [0.658^0.616](248) 0.658^0.616 = e^(0.616 × ln 0.658) = e^(0.616 × (-0.4193)) = e^(-0.2583) = 0.7724 Fcr = 0.7724 × 248 = 191.6 MPa
For elastic buckling, the computation is straightforward: just multiply Fe by 0.877. Notice Fcr = 76.9 MPa is far below Fy = 248 MPa — the column buckles at only 31% of yield. This illustrates the danger of slender columns.
Scenario
Column: KL/r = 150, Fy = 248 MPa, E = 200,000 MPa. Determine Fcr.
Solution
Step 1: Transition check 4.71√(200,000/248) = 133.7 150 > 133.7 → ELASTIC Step 2: Compute Fe Fe = π²(200,000) / 150² = 1,973,921 / 22,500 = 87.73 MPa Step 3: Compute Fcr Fcr = 0.877 × 87.73 = 76.9 MPa
High-strength steel (Fy = 345 MPa) is more susceptible to buckling at moderate slenderness than A36. At KL/r = 100, Fcr/Fy = 165.9/345 = 48% — the column works at less than half its yield capacity. This is why column selection requires both strength and stiffness considerations.
Scenario
Column: KL/r = 100, Fy = 345 MPa (Grade 50 steel), Ag = 6,000 mm². Find Fcr.
Solution
Step 1: Transition check 4.71√(200,000/345) = 4.71 × 24.04 = 113.3 100 < 113.3 → INELASTIC Step 2: Fe Fe = π²(200,000)/100² = 197.4 MPa Step 3: Fy/Fe = 345/197.4 = 1.748 Step 4: Fcr = [0.658^1.748](345) 0.658^1.748 = e^(1.748 × (−0.4193)) = e^(−0.7329) = 0.4808 Fcr = 0.4808 × 345 = 165.9 MPa
Applications
- Selecting between inelastic and elastic buckling equations for design
- Understanding why increasing Fy alone does not improve very slender columns
- Determining whether adding bracing (reducing KL/r) shifts a column from elastic to inelastic range
- Computing allowable compressive stress for ASD (Fa = Fcr/1.67)
- Foundation of LRFD column design tables in AISC Steel Construction Manual
Misconceptions
- Using 0.685 instead of 0.658 — a common typo that significantly changes the answer
- Using the elastic formula when KL/r < transition (unconservative error) or inelastic formula when KL/r > transition (unconservative error)
- Believing Fy matters for elastic buckling — it does NOT appear in Fcr = 0.877Fe
- Confusing Fy/Fe with Fe/Fy — the exponent must be Fy/Fe (larger than 1 in inelastic range near transition, less than 1 for stocky columns)
- Forgetting to check the transition before applying any formula — always check first
Related Concepts
- Euler elastic buckling stress Fe
- Design strength φcPn
- Residual stresses and initial imperfections
- Local buckling (Q factor for slender elements)
- ASD allowable stress Fa = Fcr/Ωc = Fcr/1.67
Common Exam Questions
Example
KL/r=120, Fy=248, Ag=7500 mm² → transition=133.7 → inelastic → Fe=136.8 MPa → Fy/Fe=1.813 → Fcr=0.658^1.813×248=0.464×248=115.0 MPa → φcPn=0.90×115.0×7500=776kN
Approach
Step 1: Compute transition slenderness 4.71√(E/Fy). Step 2: Compare KL/r. Step 3: Apply correct Fcr equation. Step 4: φcPn = 0.90 FcrAg.
Question Type
Full design strength computation
Example
KL/r=180, Fy=345 MPa → 4.71√(200000/345)=113.3 → 180>113.3 → ELASTIC buckling
Approach
Compute 4.71√(E/Fy) and compare to given KL/r. State clearly: INELASTIC (stocky) or ELASTIC (slender).
Question Type
Identify buckling mode
Example
Given φcPn=800kN, Ag=6000mm², find Fcr needed: Fcr=800000/(0.90×6000)=148.1 MPa
Approach
Set φcPn equal to required capacity, solve for Fcr, then for Fe or Fy/Fe, then for KL/r, then L.
Question Type
Reverse problem — find L given capacity
Key Points To Remember
- Transition slenderness = 4.71√(E/Fy) — memorize or be able to compute quickly
- Equivalently, the transition occurs at Fy/Fe = 2.25
- Inelastic (stocky): Fcr = [0.658^(Fy/Fe)]Fy — exponent is Fy/Fe
- Elastic (slender): Fcr = 0.877 Fe — only Fe appears, Fy is irrelevant
- The factor is 0.658 (not 0.685) and 0.877 — memorize exact values
- For A36 (Fy=248 MPa): transition at KL/r ≈ 133.7
- For Grade 50 (Fy=345 MPa): transition at KL/r ≈ 113.3
- Higher Fy → lower transition slenderness (higher-strength steel enters elastic range sooner)
- Both equations meet continuously at the transition — no discontinuity
Design Strength: φcPn
Once Fcr is determined, the LRFD design (factored-load) strength of the compression member is computed as: φcPn = φc × Fcr × Ag Where: • φc = 0.90 (resistance factor for compression, NSCP 2015 Section 502.1) • Pn = Fcr × Ag = nominal compressive strength (N) • Fcr = critical buckling stress (MPa) • Ag = gross cross-sectional area (mm²) The resistance factor φc = 0.90 (not to be confused with φ = 0.65 or 0.75 used in reinforced concrete columns under ACI 318/NSCP for RC). This distinction is a frequent board-exam trap. **LRFD Design Criterion:** Pu ≤ φcPn where Pu = factored axial load = 1.2PD + 1.6PL (or other NSCP load combinations) **ASD Equivalent (for reference):** Pa ≤ Pn/Ωc where Ωc = 1.67 Allowable stress: Fa = Fcr/1.67 **Practical Steps for CELE Problems:** 1. Identify Ag, r (least), K, L, E, Fy 2. Compute KL/r (largest) 3. Check: KL/r ≤ 200? (serviceability limit) 4. Compute transition: 4.71√(E/Fy) 5. Compare KL/r to transition → select Fcr equation 6. Compute Fe, then Fcr 7. Compute φcPn = 0.90 × Fcr × Ag 8. Check: Pu ≤ φcPn? **Compact, Noncompact, and Slender Elements:** For doubly symmetric, compact W-shapes (the most common in board exams), the above procedure is complete. For sections with slender elements (thin flanges or webs), a reduction factor Q < 1.0 modifies the equations. This is covered in NSCP 2015 Section 502.7 (local buckling). Most board problems use compact sections.
Examples
The column has 13% reserve capacity. In practice, engineers also check local buckling (width-to-thickness ratios) and combined loading (if moments exist). For the board exam, this 7-step procedure covers all standard column problems.
Scenario
Complete design check: A W-section column with Ag = 8,000 mm², least r = 50 mm, K = 1.0, L = 3.5 m, Fy = 248 MPa, E = 200,000 MPa. Factored load Pu = 1,200 kN. Is the column adequate?
Solution
Step 1: KL/r = (1.0)(3500)/50 = 70 Step 2: Transition = 4.71√(200,000/248) = 133.7 → 70 < 133.7 → INELASTIC Step 3: Fe = π²(200,000)/70² = 402.8 MPa Step 4: Fy/Fe = 248/402.8 = 0.616 Step 5: Fcr = [0.658^0.616](248) = 0.7724(248) = 191.6 MPa Step 6: φcPn = 0.90 × 191.6 × 8,000 = 1,379,520 N = 1,380 kN Step 7: Check: Pu = 1,200 kN ≤ φcPn = 1,380 kN ✓ ADEQUATE Utilization ratio = 1,200/1,380 = 87% (well-utilized)
This reverse-type problem tests both design-strength calculation and LRFD load combination knowledge. The key is recognizing the load combination pattern before solving for the unknown service load.
Scenario
Find the maximum service dead load PD a column can carry if PL = 0.5PD, given: Ag = 9,500 mm², KL/r = 76.9, Fy = 248 MPa.
Solution
Step 1: Fe = π²(200,000)/76.9² = 1,973,921/5913.6 = 333.8 MPa Transition = 133.7; 76.9 < 133.7 → INELASTIC Step 2: Fy/Fe = 248/333.8 = 0.743 Fcr = [0.658^0.743](248) = 0.7481 × 248 = 185.5 MPa Step 3: φcPn = 0.90 × 185.5 × 9,500 = 1,586,025 N = 1,586 kN Step 4: Load combination: Pu = 1.2PD + 1.6PL = 1.2PD + 1.6(0.5PD) = 1.2PD + 0.8PD = 2.0PD Step 5: Set Pu = φcPn: 2.0PD = 1,586 kN → PD = 793 kN
Applications
- Final design check for all steel column problems
- Verifying adequacy of existing columns under new loading
- Selecting column size from AISC load tables
- Computing required Ag given Pu, KL/r, and steel grade
- Comparing LRFD and ASD results for the same member
Misconceptions
- Using φ = 0.65 (RC column value) instead of φc = 0.90 for steel — a critical error that underestimates capacity by ~28%
- Using net area An instead of gross area Ag — holes are NOT deducted for compression members
- Forgetting to convert kN to N or mm to m when computing φcPn
- Treating φcPn as the service load capacity — it is the FACTORED (ultimate) load capacity
- Omitting the 0.90 factor and using φcPn = FcrAg directly
Related Concepts
- LRFD load combinations (1.2D + 1.6L, etc.)
- Nominal strength Pn vs design strength φcPn
- ASD allowable strength Pn/Ωc
- Net vs gross area (only gross used for compression)
- Local buckling and Q-factor reduction for slender elements
Common Exam Questions
Example
Ag=10000mm², r=70mm, KL=6m, Fy=248MPa → KL/r=85.7 → inelastic → Fe=269.3MPa → Fy/Fe=0.921 → Fcr=[0.658^0.921](248)=0.741×248=183.8MPa → φcPn=0.90×183.8×10000=1654kN
Approach
Follow the 7-step procedure: KL/r → transition → inelastic/elastic → Fe → Fcr → φcPn = 0.90FcrAg
Question Type
Find φcPn given all properties
Example
If PD=400kN, PL=250kN → Pu=1.2(400)+1.6(250)=880kN; compare to φcPn
Approach
Compute φcPn, compute Pu from load combination, compare.
Question Type
Adequacy check with factored loads
Example
Pu=900kN, Fcr=150MPa → Ag=900,000/(0.90×150)=6,667mm² minimum
Approach
Set Pu = φcPn = 0.90FcrAg; assume or iterate KL/r to find Fcr, then solve for Ag.
Question Type
Find required Ag
Key Points To Remember
- φc = 0.90 for steel columns (NOT 0.65 or 0.75 as in RC columns)
- Ωc = 1.67 for ASD — both φc and Ωc are unique to steel compression members
- φcPn = 0.90 × Fcr × Ag — three quantities, all must be correct
- Ag is the GROSS area — do not subtract holes for bolts in compression members
- Pn = FcrAg is the NOMINAL strength; φcPn is the DESIGN strength
- Slenderness limit KL/r ≤ 200 is recommended but not mandatory (NSCP Commentary)
- For slender-element sections, use Q-modified equations (beyond standard board scope)
- Pu ≤ φcPn is the LRFD design check
- NSCP 2015 adopts AISC 360-10; cite both in practice
Effective Length Factor K and End Conditions
The effective length KL represents the equivalent pinned-end length that produces the same buckling load as the actual column with its true end conditions. The factor K accounts for the rotational and translational restraint at column ends. **Theoretical K values (NSCP 2015 Table C-A-7.1 / AISC Commentary):** | End Conditions | Theoretical K | Recommended K (Design) | |---|---|---| | Both ends pinned (pin-pin) | 1.00 | 1.00 | | Both ends fixed (fix-fix) | 0.50 | 0.65 | | Fixed-pinned (one end) | 0.70 | 0.80 | | Fixed-free (flag pole) | 2.00 | 2.10 | | Fixed-fixed, one end slides | 1.00 | 1.20 | | Pinned-fixed, one end slides | 2.00 | 2.00 | The recommended design values are slightly larger than theoretical to account for real-world imperfect fixity. **Braced vs Unbraced Frames:** • Braced frames (with diagonal bracing, shear walls): K ≤ 1.0. Sway is prevented, so columns benefit from double curvature effect. • Unbraced frames (moment frames only): K > 1.0. Sway amplifies buckling tendency. For most board exam problems, K is given directly (K = 1.0 is most common for pin-ended members in trusses or braced frames). Always use K as stated in the problem. **Different K values for different axes:** A column may have different bracing in two planes. For example, a column in a braced frame with floor bracing every 3 m on one axis but not on the other may have KyLy ≠ KxLx. Always compute both slenderness ratios.
Examples
The cantilever column with K=2.0 effectively behaves like a pin-pin column twice as long. This is why unbraced cantilever columns are the most slender and should be avoided or heavily braced in practice.
Scenario
A 6-m column is fixed at the base and free at the top (cantilever column). r = 60 mm. Find KL/r.
Solution
End condition: Fixed-free → K = 2.0 (use recommended value) KL = 2.0 × 6,000 = 12,000 mm KL/r = 12,000/60 = 200 This column is at the very limit of the recommended slenderness of 200.
This example shows that bracing the weak axis at mid-height (halving Ly) can shift the governing axis from weak to strong. This is an important design concept: intermediate bracing must reduce KL/r sufficiently on the weak axis.
Scenario
A column in a braced frame: Lx = 6 m (Kx = 1.0), Ly = 3 m (Ky = 1.0). W-section: rx = 100 mm, ry = 55 mm. Find governing KL/r.
Solution
Strong axis: KxLx/rx = (1.0)(6000)/100 = 60.0 Weak axis: KyLy/ry = (1.0)(3000)/55 = 54.5 Governing KL/r = 60.0 (strong axis governs — bracing at mid-height reduced weak-axis slenderness significantly)
Applications
- Determining effective column lengths in building frames
- Designing lateral bracing systems for columns
- Analysis of cantilever poles, flagpoles, and unbraced piers
- Using AISC alignment charts (nomographs) for K in frames
- Board exam problems specifying K directly for simple column checks
Misconceptions
- Using K=1.0 for all columns regardless of end conditions
- Confusing theoretical K with recommended design K — always use the recommended (larger) value
- Assuming K<1.0 is always achievable in real structures — true fixity is rare; use judgment
- Forgetting that K>1.0 for unbraced frames, which dramatically increases effective length
Related Concepts
- Euler buckling load PE = π²EI/(KL)²
- Alignment charts (Jackson-Moreland nomographs) for frame columns
- Braced vs unbraced frames
- Lateral bracing requirements
- Effective length of compression members in trusses
Common Exam Questions
Example
Fixed base, pinned top, L=5m, r=65mm → K=0.80 (recommended) → KL=4000mm → KL/r=4000/65=61.5
Approach
Identify end fixity → select K from table → KL = K×L → KL/r = KL divided by r
Question Type
Given end conditions, find K and KL/r
Example
K=1.0, r=50mm → L_max = 200×50/1.0 = 10,000mm = 10m
Approach
Set KL/r = 200 (limit) → KL = 200r → L = 200r/K
Question Type
Find maximum unbraced length
Key Points To Remember
- K = 1.0 for pin-pin (most common in board exams)
- K = 2.0 for cantilever / flagpole column (fixed-free) — most vulnerable
- K = 0.5 for fixed-fixed (theoretical) — most stable end condition
- Braced frames: K ≤ 1.0; Unbraced frames: K > 1.0
- Recommended design K values are always ≥ theoretical K (conservative)
- KL is the effective length; use it directly in KL/r
- Different K may apply to strong and weak axes — check both KxLx/rx and KyLy/ry
- NSCP 2015 Section 502.3 and Commentary provide K alignment charts for frames
Local Buckling and Slender Elements
Global (flexural) buckling involves the entire column bowing sideways. **Local buckling** involves the individual plate elements (flanges, web) of the cross-section buckling locally — forming wrinkles or waves — before the global buckling load is reached. NSCP 2015 Section 502.4 and Table 502.4.1 define **width-to-thickness ratios (λ)** for compression elements: **Limiting ratios for uniform compression:** | Element | λ (actual) | λr (limit for non-slender) | |---|---|---| | Flange of W-shape | bf/(2tf) | 0.56√(E/Fy) | | Web of W-shape | h/tw | 1.49√(E/Fy) | | Outstanding leg of angle | b/t | 0.45√(E/Fy) | | HSS wall | b/t | 1.40√(E/Fy) | | Round HSS | D/t | 0.15E/Fy | For A36 steel (Fy=248 MPa, E=200,000 MPa): • Flange limit: 0.56√(200,000/248) = 15.9 • Web limit: 1.49√(200,000/248) = 42.3 **If λ ≤ λr (non-slender):** No local buckling reduction needed. The standard Fcr equations apply directly. **If λ > λr (slender element):** Local buckling reduces capacity. A reduction factor Q < 1.0 is applied. Fcr equations are modified to use QFy in place of Fy (NSCP 2015 Section 502.7). Q = QaQs where Qa accounts for web slenderness (effective area) and Qs accounts for flange slenderness. For most standard board exam problems involving W-sections with given properties, sections are assumed compact/non-slender. Problems involving HSS or built-up sections may explicitly test local buckling limits. For torsional and flexural-torsional buckling of singly symmetric shapes (T-sections, channels, angles), additional equations from NSCP 2015 Section 502.4 apply. These are tested occasionally at the advanced level.
Examples
W200×100 is a stocky, heavy section. Its width-to-thickness ratios are well within limits. Lighter W-sections (e.g., W200×27) may have thinner flanges and could approach the limit — always check for unfamiliar sections.
Scenario
Check if a W200×100 (bf = 210 mm, tf = 23.7 mm, h = 152 mm, tw = 14.5 mm) is non-slender for Fy = 248 MPa.
Solution
Flange: λ = bf/(2tf) = 210/(2×23.7) = 4.43 Limit: 0.56√(200,000/248) = 15.9 4.43 < 15.9 ✓ Non-slender flange Web: λ = h/tw = 152/14.5 = 10.5 Limit: 1.49√(200,000/248) = 42.3 10.5 < 42.3 ✓ Non-slender web Conclusion: W200×100 is non-slender (compact for compression). No Q reduction needed.
Higher-strength steels have lower slenderness limits. A flange that is non-slender for A36 (Fy=248) may become slender for Grade 50 (Fy=345). Always recheck limits when changing steel grade.
Scenario
Determine the flange slenderness limit for Fy = 345 MPa and check if λ = 14.0 requires Q reduction.
Solution
Flange limit = 0.56√(200,000/345) = 0.56 × 24.04 = 13.5 Actual λ = 14.0 > 13.5 → SLENDER flange → Q < 1.0 applies This section has a slender flange for Grade 50 steel and requires Q-factor reduction per NSCP 2015 Section 502.7.
Applications
- Verifying that W-sections selected from tables are non-slender for given Fy
- Design of built-up columns with thin plates
- Checking HSS sections for local buckling under compression
- Applying Q-factor reduction for cold-formed steel sections
- Advanced board problems involving non-standard sections
Misconceptions
- Assuming all W-shapes from AISC tables are always non-slender — this is true for most but not all, especially for higher Fy
- Confusing compact limits for beams (flexure) with non-slender limits for columns (compression) — different table, different values
- Ignoring local buckling entirely in column design problems — Q < 1.0 can reduce capacity significantly for thin-walled sections
Related Concepts
- Compact vs non-compact vs slender classification for beams
- Q-factor and effective area for slender-element columns
- Width-to-thickness ratios for HSS, pipes, angles, tees
- Torsional and flexural-torsional buckling
- Local buckling in beam compression flanges
Common Exam Questions
Example
bf/(2tf)=12.0 vs limit 0.56√(200000/248)=15.9 → 12.0<15.9 → Non-slender ✓
Approach
Compute λ = b/t for flanges and web; compare to λr = 0.56√(E/Fy) for flanges and 1.49√(E/Fy) for web.
Question Type
Check if section is slender
Example
A board question may ask: 'According to NSCP 2015, what is the limiting λr for outstanding flanges of W-shapes in compression?' Answer: 0.56√(E/Fy)
Approach
NSCP 2015 Section 502.4, Table 502.4.1 (based on AISC 360-10 Table B4.1a)
Question Type
Identify which standard states local buckling limits
Key Points To Remember
- Local buckling involves individual plate elements, not the whole member
- Width-to-thickness ratio λ must be checked against limiting λr
- For flanges of W-shapes: λ = bf/(2tf); limit = 0.56√(E/Fy)
- For webs of W-shapes: λ = h/tw; limit = 1.49√(E/Fy)
- Non-slender (λ ≤ λr): use standard Fcr equations without modification
- Slender (λ > λr): use Q-factor reduction (Q = QaQs < 1.0)
- For A36: flange limit ≈ 15.9, web limit ≈ 42.3 (memorize these)
- Standard W-shapes in AISC database are mostly non-slender for Fy ≤ 345 MPa
- Singly symmetric shapes may also fail by torsional or flexural-torsional buckling
Practice Problems
A pipe column (circular hollow section) has the same r in all directions, so only one KL/r needs to be computed. This is a classic 'inelastic column' problem. Note: 0.658^0.559 is evaluated using the y^x function on a scientific calculator. The answer 883 kN is the maximum factored load the column can resist.
Problem
Problem 1 (Basic — Inelastic Column): A pipe column has the following properties: Ag = 5,000 mm², r = 60 mm, K = 1.0, L = 4.0 m, Fy = 248 MPa, E = 200,000 MPa. Find the LRFD design strength φcPn.
Solution
Step 1: Compute KL/r KL/r = (1.0)(4000)/60 = 66.67 Step 2: Compute transition slenderness 4.71√(E/Fy) = 4.71√(200,000/248) = 4.71 × 28.39 = 133.7 Step 3: Compare → 66.67 < 133.7 → INELASTIC BUCKLING Step 4: Compute Fe Fe = π²(200,000)/(66.67)² = 1,973,921/4444.9 = 443.8 MPa Step 5: Compute Fy/Fe Fy/Fe = 248/443.8 = 0.559 Step 6: Compute Fcr Fcr = [0.658^0.559](248) 0.658^0.559 = e^(0.559 × ln 0.658) = e^(0.559 × (−0.4193)) = e^(−0.2344) = 0.7912 Fcr = 0.7912 × 248 = 196.2 MPa Step 7: Compute φcPn φcPn = 0.90 × 196.2 × 5,000 = 882,900 N ≈ 883 kN
This problem reinforces the key concept: very slender columns waste high-strength steel. The elastic Fcr = 0.877Fe depends only on E and KL/r, not on Fy. Switching from A36 to Grade 50 steel would not improve this column's capacity at all — the governing equation does not include Fy.
Problem
Problem 2 (Elastic Column): A slender column has KL/r = 180, Fy = 248 MPa, Ag = 4,200 mm². Find Fcr and φcPn. Confirm it is in the elastic range.
Solution
Step 1: Transition check 4.71√(200,000/248) = 133.7 180 > 133.7 → ELASTIC BUCKLING ✓ Step 2: Compute Fe Fe = π²(200,000)/180² = 1,973,921/32,400 = 60.92 MPa Step 3: Compute Fcr Fcr = 0.877 × 60.92 = 53.43 MPa Step 4: Compute φcPn φcPn = 0.90 × 53.43 × 4,200 = 201,974 N ≈ 202 kN Comparison: Fcr/Fy = 53.43/248 = 21.5% — column fails at only 22% of yield. Very inefficient use of material.
This is a complete board-exam type problem. All steps are applied systematically. Always present your work in numbered steps during the CELE to maximize partial credit. The answer 1,552 kN is the LRFD design strength of this column.
Problem
Problem 3 (High-strength steel): A W-column with Ag = 9,500 mm², r (least) = 65 mm, K = 1.0, L = 5.0 m, Fy = 248 MPa. Find φcPn. [This problem also appears in NSCP reference materials as Example 4 in this chapter.]
Solution
Step 1: KL/r = (1.0)(5000)/65 = 76.92 Step 2: Transition = 4.71√(200,000/248) = 133.7 → 76.92 < 133.7 → INELASTIC Step 3: Fe = π²(200,000)/76.92² = 1,973,921/5916.7 = 333.6 MPa Step 4: Fy/Fe = 248/333.6 = 0.744 Step 5: Fcr = [0.658^0.744](248) 0.658^0.744 = e^(0.744 × (−0.4193)) = e^(−0.3120) = 0.7319 Fcr = 0.7319 × 248 = 181.5 MPa Step 6: φcPn = 0.90 × 181.5 × 9,500 = 1,551,825 N ≈ 1,552 kN
Higher yield strength steel has a LOWER transition slenderness. This means columns of a given KL/r are more likely to be in the elastic (slender) range if made of higher-strength steel. For KL/r = 120: A36 → inelastic (120 < 133.7), Grade 50 → elastic (120 > 113.3). Choosing higher Fy does NOT help elastic columns!
Problem
Problem 4 (Transition slenderness comparison): Compute the transition slenderness KL/r for (a) Fy = 248 MPa, (b) Fy = 345 MPa, and (c) Fy = 415 MPa. All use E = 200,000 MPa.
Solution
Formula: KL/r_transition = 4.71√(E/Fy) (a) Fy = 248 MPa: 4.71√(200,000/248) = 4.71 × 28.39 = 133.7 (b) Fy = 345 MPa: 4.71√(200,000/345) = 4.71 × 24.04 = 113.3 (c) Fy = 415 MPa: 4.71√(200,000/415) = 4.71 × 21.94 = 103.4 Summary: Fy = 248 MPa → transition at KL/r = 133.7 Fy = 345 MPa → transition at KL/r = 113.3 Fy = 415 MPa → transition at KL/r = 103.4
λ = 18.0 exceeds the limit 15.9, so local buckling will reduce capacity below what the standard Fcr equations predict. In real design, Qs is computed from the actual b/t ratio and applied as a reduction. Board problems typically state Q directly if needed; this problem tests whether you can identify when Q < 1.0 is required.
Problem
Problem 5 (Local buckling check): A built-up box column has outstanding flange element b/t = 18.0. Fy = 248 MPa. Determine if local buckling reduction (Q factor) is needed.
Solution
Limiting ratio for outstanding legs/flanges (uniform compression, NSCP 2015 Table 502.4.1): λr = 0.56√(E/Fy) = 0.56√(200,000/248) = 0.56 × 28.39 = 15.9 Actual λ = 18.0 > λr = 15.9 → SLENDER ELEMENT Conclusion: Q < 1.0 is required. The column capacity must be reduced per NSCP 2015 Section 502.7. The standard Fcr equations with Fy replaced by QFy apply.
This problem integrates LRFD load combinations with column design. Note the load combination 1.2D + 1.6L is the governing pattern when live load dominates. The column utilization of 72.4% suggests a lighter section could be chosen — but without additional load cases (wind, seismic), this cannot be confirmed.
Problem
Problem 6 (Design adequacy with load combinations): A steel column carries PD = 500 kN (dead load) and PL = 350 kN (live load). The column has Ag = 10,500 mm², KL/r = 85, Fy = 248 MPa. Is the column adequate under NSCP 2015 LRFD?
Solution
Step 1: Factored load (NSCP 2015 / ASCE 7 combination) Pu = 1.2PD + 1.6PL = 1.2(500) + 1.6(350) = 600 + 560 = 1,160 kN Step 2: Transition check 4.71√(200,000/248) = 133.7 → 85 < 133.7 → INELASTIC Step 3: Fe = π²(200,000)/85² = 1,973,921/7,225 = 273.2 MPa Step 4: Fy/Fe = 248/273.2 = 0.908 Step 5: Fcr = [0.658^0.908](248) 0.658^0.908 = e^(0.908 × (−0.4193)) = e^(−0.3807) = 0.6833 Fcr = 0.6833 × 248 = 169.5 MPa Step 6: φcPn = 0.90 × 169.5 × 10,500 = 1,601,775 N = 1,602 kN Step 7: Check: Pu = 1,160 kN ≤ φcPn = 1,602 kN ✓ ADEQUATE Utilization = 1,160/1,602 = 72.4% — column has significant reserve
Exam Preparation Tips
- MEMORIZE the four key formulas exactly: Fe = π²E/(KL/r)², transition = 4.71√(E/Fy), inelastic Fcr = [0.658^(Fy/Fe)]Fy, elastic Fcr = 0.877Fe, and φcPn = 0.90FcrAg. Write them on your formula sheet before the exam starts.
- ALWAYS CHECK THE TRANSITION FIRST. Every column problem begins with comparing KL/r to 4.71√(E/Fy). Circle or underline which range applies before proceeding. Wrong range selection = wrong answer, no partial credit.
- CALCULATOR DRILL: Practice computing 0.658^(Fy/Fe) using the y^x key on your Casio fx-991 or fx-570 ES Plus. Example: 0.658 [y^x] 0.616 [=] → 0.7724. A 30-second calculator fluency drill daily for two weeks eliminates this source of error.
- MEMORIZE COMMON TRANSITION VALUES: Fy=248→133.7, Fy=345→113.3, Fy=415→103.4. These appear so frequently that knowing them without computation saves 2–3 minutes per problem.
- UNIT CONSISTENCY: Keep all dimensions in mm and forces in N. KL in mm, r in mm → KL/r is dimensionless. Fe in MPa = N/mm². φcPn in N — divide by 1,000 for kN. One unit error = wrong answer.
- φc = 0.90 FOR STEEL COLUMNS, NOT 0.65 or 0.75. The RC column resistance factors (ACI 318/NSCP for RC) are completely different. Steel uses φc = 0.90. This is the single most common source of confusion between steel and concrete design.
- GROSS AREA Ag for compression: Unlike tension members where you deduct bolt holes for net area, compression members use the FULL GROSS AREA. The loads spread around holes in compression.
- WEAK AXIS GOVERNS unless stated otherwise or unless weak-axis bracing is provided. For W-shapes, ry << rx. Always compute both KL/r values and use the LARGER one.
- SLENDERNESS LIMIT KL/r ≤ 200: If a problem gives a column with KL/r > 200, it is technically outside the recommended range. Note this in your solution but still apply the formulas if asked for capacity.
- BOARD EXAM PATTERN: The most common column problem type gives Ag, r, K, L, Fy and asks for φcPn. The second most common gives Pu and asks if the column is adequate. Practice these two patterns until they take under 4 minutes each.
- ASD vs LRFD: The CELE primarily tests LRFD. But if an ASD problem appears: Fa = Fcr/1.67 and Pa ≤ FaAg. The Fcr computation is identical — only the safety check changes.
- STUDY THE THREE WORKED EXAMPLES from the NSCP reference: KL/r=70 (inelastic, A36), KL/r=150 (elastic, A36), KL/r=100 (inelastic, Grade 50). Reproduce each without looking at the solution. These exact parameter sets have appeared on past board exams.
- AVOID THE 0.685 TYPO: The base is 0.658, not 0.685. Compute both to see the difference: 0.658^1.0 = 0.658, but 0.685^1.0 = 0.685 — a 4% overestimate of capacity. In exams, this leads to wrong answer choices.
- FOR REVIEW SESSIONS: Use the transition equivalence Fy/Fe = 2.25 as a cross-check. After computing Fe, verify: if KL/r < transition, then Fy/Fe should be < 2.25. If KL/r > transition, then Fy/Fe > 2.25. A quick sanity check.
- NSCP 2015 CITATION: In any written exam or structural design report, cite 'NSCP 2015 Volume I, Section 502 (based on AISC 360-10 Chapter E)' for steel column design provisions. RA 544 (Civil Engineering Law) governs professional practice; NSCP governs technical standards.
In summary
Steel compression member design under NSCP 2015 (AISC 360-10 Chapter E) centers on five interconnected ideas: (1) the elastic buckling stress Fe = π²E/(KL/r)², governed by the largest slenderness ratio; (2) the transition slenderness 4.71√(E/Fy) that separates inelastic from elastic behavior; (3) the critical stress Fcr using either the 0.658-base exponential (inelastic) or 0.877 multiplier (elastic); (4) the design strength φcPn = 0.90 FcrAg; and (5) local buckling checks via width-to-thickness ratios. For the PRC Civil Engineer Licensure Examination, the board-exam pattern is highly predictable: given Ag, r, K, L, Fy — find φcPn, or given loads — check adequacy. Mastery requires: (a) memorizing the exact formulas (especially 0.658 not 0.685, and φc = 0.90 not 0.65); (b) calculator fluency with the y^x function for the exponential Fcr; (c) systematic step-by-step problem solving with unit consistency in mm and N; and (d) knowing the transition values for A36 (133.7) and Grade 50 (113.3) by heart. Beyond the examination, steel column design is fundamental to Philippine structural engineering practice. Building columns, industrial tower legs, transmission poles, and bridge compression chords all rely on these equations. Competence in this topic reflects the technical capability expected of a licensed Civil Engineer under RA 544, responsible for public safety in the design of structures that serve Filipino communities. A thorough understanding of NSCP 2015 provisions — grounded in the mechanics of buckling and the probabilistic basis of LRFD — distinguishes a knowledgeable practitioner from one who merely memorizes formulas.
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