CELE Strength of Materials — Columns and BucklingRevision Notes

Formulas

Exam Tips

• When a problem states 'which axis does the column buckle about?', always answer: the axis with the LESSER moment of inertia (lesser radius of gyration r).
• If bracing is provided about one axis, check the slenderness ratio for both axes separately and use the governing (larger) value.
• Board questions often give I_x and I_y — always use the SMALLER one unless told the weak axis is braced.

Key Points

• A column is a structural member loaded primarily in axial compression along its longitudinal axis.
• Short columns fail by material yielding or crushing: P_crush = σy × A.
• Long (slender) columns fail by elastic buckling at P_cr << P_crush — stiffness governs, not strength.
• Intermediate columns fail by inelastic buckling — partial yielding reduces stiffness before elastic buckling occurs.
• Buckling always occurs about the axis of LEAST moment of inertia (weakest axis) unless that axis is braced.
• Buckling load is independent of material strength for elastic (Euler) columns — only E and I matter.
• The transition from long to intermediate columns is defined by the critical slenderness ratio Cc.
• Real columns always have initial imperfections, load eccentricities, and residual stresses — design codes account for these through capacity reduction factors and empirical formulas.

Definitions

Section Title

1. Column Behavior and Failure Modes

Common Mistakes

• Confusing the failure mode: short columns crush (σ controls), long columns buckle (E and I control).
• Forgetting that buckling is always about the WEAK axis — always use the MINIMUM I and minimum r.
• Assuming that stronger steel (higher σy) automatically gives a higher buckling load — for long elastic columns, P_cr depends on E (200 GPa for all structural steel grades), not σy.

Formulas

Exam Tips

• Memorize: For pin-ended steel column, P_cr = π²(200,000)I / L² — precompute π²×200,000 = 1,973,921 ≈ 1.974×10⁶ N·mm⁻² to speed up calculations.
• Quick ratio check: If two columns differ only in length (L₁ and L₂, same section), then P_cr1/P_cr2 = (L₂/L₁)². E.g., column B is twice as long → P_crB = P_crA/4.
• When comparing end conditions: P_cr(fixed-fixed) = 4 × P_cr(pin-pin); P_cr(fixed-free) = P_cr(pin-pin)/4.

Key Points

• Derived by Leonhard Euler in 1744 for a perfectly straight, pin-ended, elastic column with centrically applied load.
• The critical load: P_cr = π²EI / L²
• The critical stress: σ_cr = π²E / (L/r)²
• The radius of gyration r = √(I/A) converts the formula from load to stress form.
• Euler's formula is ONLY valid when σ_cr is below the proportional limit (elastic range).
• P_cr increases with: higher E, larger I (use bigger sections or move material away from centroid), shorter effective length.
• P_cr is INDEPENDENT of yield strength σy — switching from A36 to A572 Gr.50 steel does NOT change the Euler buckling load.
• For a given cross-section and material, P_cr varies as 1/L² — doubling the length reduces P_cr by 4×.

Definitions

Section Title

2. Euler's Buckling Formula

Common Mistakes

• Using L in meters instead of millimeters — L² enormously magnifies unit errors. Always convert to mm when using MPa (N/mm²) and mm⁴.
• Using the LARGER moment of inertia — buckling is about the weak axis (minimum I, minimum r).
• Applying Euler's formula to an intermediate or short column without first checking KL/r vs. Cc — this can over-predict the capacity by 50–100%.
• Forgetting to square the effective length: P_cr ∝ 1/(KL)², not 1/(KL).

Formulas

Exam Tips

• Memorize the four K values as a mental table: PP=1.0, FP=0.7, FF=0.5, Ff=2.0 (where F=fixed, P=pinned, f=free). Many board problems test these directly.
• Capacity ratio between end conditions (all same L, E, I): P(FF) : P(FP) : P(PP) : P(Ff) = 4 : 2.04 : 1 : 0.25.
• When a problem says 'one end is fixed against rotation and translation, the other end is free to rotate but not translate' — that is a PINNED end, not a fixed end.

Key Points

• The effective length L_e = KL is the length of the equivalent pin-ended column that would buckle at the same load.
• K is the effective length factor — it depends on the degree of rotational and translational fixity at each end.
• Theoretical K values: Pin-Pin = 1.0; Fixed-Fixed = 0.5; Fixed-Pin = 0.7; Fixed-Free = 2.0.
• Design (recommended) K values per AISC/NSCP are slightly more conservative: Pin-Pin = 1.0; Fixed-Fixed = 0.65; Fixed-Pin = 0.80; Fixed-Free = 2.10.
• A fixed-fixed column carries 4× the load of an equivalent pin-ended column (since K = 0.5, and P_cr ∝ 1/K²).
• A fixed-free (cantilever) column carries only 1/4 the load of a pin-ended column of the same length.
• In sway (unbraced) frames, K > 1.0 is possible for columns — effective length exceeds actual length.
• In braced frames, K ≤ 1.0 — sidesway is prevented and the column is more stable.
• NSCP 2015 Section 502.3 provides alignment charts (nomographs) for determining K in frames with partial fixity.

Definitions

Section Title

3. Effective Length and End Conditions

Common Mistakes

• Using theoretical K instead of design (recommended) K when a problem asks for safe design — use recommended values unless problem specifies theoretical.
• Confusing fixed-free with fixed-pinned: Fixed-free (cantilever) has K = 2.0; fixed-pinned has K = 0.7. These are often confused on exams.
• Forgetting that P_cr scales as 1/K² — when K doubles (e.g., from 1.0 to 2.0), P_cr reduces to 1/4, not 1/2.

Formulas

Exam Tips

• Memorize Cc ≈ 126 for A36 steel (σy = 250 MPa) — this value appears frequently in Philippine board exam problems.
• For LRFD problems: λ_c < 1.5 means inelastic; λ_c ≥ 1.5 means elastic. λ_c = 1.5 corresponds exactly to KL/r = Cc.
• Quick check: If a problem says 'slender column' or 'long column', assume Euler governs and skip the Cc check — but verify by computing σ_cr < σy.

Key Points

• The slenderness ratio KL/r is the primary parameter for classifying and designing columns.
• Three column classes: Short (low KL/r) → fails by yielding; Intermediate → fails by inelastic buckling; Long (high KL/r) → fails by elastic (Euler) buckling.
• The transition slenderness Cc separates intermediate from long columns: Cc = √(2π²E/σy).
• For A36 steel (σy = 250 MPa, E = 200,000 MPa): Cc = √(2π²×200,000/250) = √15,791 ≈ 125.7.
• For A572 Gr.50 (σy = 345 MPa): Cc = √(2π²×200,000/345) = √11,468 ≈ 107.1.
• Euler's formula is valid only when KL/r > Cc (long columns) — σ_cr computed is below proportional limit.
• For KL/r < Cc, use NSCP/AISC empirical formulas or Rankine formula.
• Higher-strength steel has a lower Cc — more columns fall in the intermediate range where strength matters.
• NSCP 2015 LRFD uses λ_c = (KL/πr)√(σy/E) = KL/r × √(σy/(π²E)) — λ_c < 1.5 is intermediate; λ_c ≥ 1.5 is elastic (Euler).

Definitions

Section Title

4. Slenderness Ratio and Column Classification

Common Mistakes

• Applying Euler's formula when KL/r < Cc — this is the most consequential error, potentially overestimating column capacity by a large margin.
• Using the same Cc value for all steel grades — Cc depends on σy, so A36 and A572 Gr.50 have different Cc values.
• Computing only one slenderness ratio when the column is unsupported in both planes — compute both KxLx/rx and KyLy/ry, and use the LARGER value.

Formulas

Exam Tips

• For Rankine problems, set up the formula as: P_cr = (Numerator) / (1 + Denominator) where Numerator = σyA and Denominator = a×(L_e/r)². This systematic layout prevents arithmetic errors.
• When both Euler and Rankine are computed for the same column, Rankine gives the lower (safer) value for intermediate columns — use Rankine for intermediate and Euler only for confirmed long columns.
• NSCP LRFD design capacity: φ_cP_n = 0.90 × F_cr × A_g. Remember φ_c = 0.90 for compression.

Key Points

• Most practical structural columns (KL/r = 50 to 120) are intermediate — pure Euler overestimates their capacity.
• Intermediate columns undergo inelastic buckling: partial yielding reduces effective EI before buckling occurs.
• NSCP 2015 / AISC 360-16 Chapter E provides the standard design equations for steel columns.
• NSCP LRFD compressive strength: φ_c P_n where φ_c = 0.90 and P_n = F_cr × A_g.
• For λ_c ≤ 1.5 (inelastic): F_cr = (0.658^(λc²)) × σy — an exponential decay form.
• For λ_c > 1.5 (elastic/Euler): F_cr = [0.877 / λ_c²] × σy — Euler-based with a 0.877 imperfection factor.
• The Rankine-Gordon formula is a classic empirical approach still common on Philippine board exams: P_cr = σyA / [1 + a(L_e/r)²].
• Rankine constant a = σy / (π²E) for the theoretical value; a = 1/7500 is commonly used for structural steel on Philippine boards.
• Rankine formula reduces to P_crush = σyA for short columns (L_e/r → 0) and approaches Euler for long columns (L_e/r → ∞).
• For intermediate columns, Rankine < Euler < Crush — Euler alone is dangerously unconservative.

Definitions

Section Title

5. Intermediate-Column Formulas

Common Mistakes

• Using Euler for an intermediate column (KL/r < Cc or λ_c < 1.5) — Euler overestimates by potentially 85% as shown in Example 4.
• Misidentifying which NSCP formula applies — always compute λ_c first: if λ_c ≤ 1.5 use exponential form; if λ_c > 1.5 use 0.877/λ_c² form.
• In Rankine formula: forgetting to square the slenderness ratio — it is a(L_e/r)², not a(L_e/r).

Formulas

Exam Tips

• For simple board exam eccentricity problems, use σ = P/A + Mc/I with M = Pe. Check if σ_max ≤ σ_allowable.
• The term ec/r² is called the eccentricity ratio. Higher ec/r² → greater bending contribution relative to axial stress.
• When a problem mentions 'load applied at eccentricity e from centroid', immediately write M = Pe and proceed with combined stress analysis.

Key Points

• A perfectly concentric axial load is an idealization — real columns always have some eccentricity e due to load positioning, initial curvature, or connection details.
• An eccentric load P at eccentricity e creates both axial compression (P/A) and bending moment (M = Pe) simultaneously.
• For stocky columns with small eccentricity: use the simple combined-stress formula σ = P/A + Mc/I (superposition method).
• For slender columns, the bending moment is amplified because the lateral deflection δ creates additional moment P×δ (P-δ effect or second-order effect).
• The secant formula gives the exact maximum stress for an eccentrically loaded elastic column.
• The secant formula shows that even tiny eccentricities dramatically amplify stress as P approaches P_cr — the column effectively has no reserve capacity beyond P_cr.
• In NSCP/AISC beam-column design, combined axial + bending is handled by interaction equations (Section 502.8 / AISC 360-16 Chapter H).
• The moment magnification factor B1 = 1/[1 - P/P_cr] accounts for P-δ amplification in LRFD analysis.

Definitions

Section Title

6. Eccentrically Loaded Columns

Common Mistakes

• Using the first-order formula σ = P/A + Mc/I for slender columns under large eccentric load — this ignores P-δ amplification and is unconservative.
• Computing the secant formula in degrees instead of radians — the argument of sec() must be in RADIANS.
• Ignoring the direction of eccentricity — eccentricity adds to compression on one face and reduces it on the other; σ_max occurs on the compression side.