COLUMN BUCKLING EXAMPLE OF A SOLID ROUND BAR

Goal: Determine the maximum buckling load for a round column with diameter d = 50 mm and length L = 2 m. Ends are pinned–pinned. Use design factor nd = 5 and material ASTM A588 steel: Sy = 345 MPa, E = 205 GPa, density ρ = 7.87 g/cm³.

Pinned–pinned round column with axial load P, d=50 mm, L=2 m
Pinned–pinned round column: d = 50 mm, L = 2 m, axial compression P.
Column geometry: length L, diameter d; A and I referenced at the cross-section
Geometry and section properties used in calculations.

Given Data

Input properties for the column buckling example
INPUT PROPERTIES SUMMARY
ParameterValue
Diameter of column [d]50mm
Length of column [L]2m
End conditionPinned–Pinned
Design factor [nd]5
Yield strength [Sy]345MPa
Elastic modulus [E]205GPa
Density [ρ]7.87g/cm³
Tip: You can verify these numbers with the Column Buckling Calculator.

Step-by-Step Solution (Formulas)

  1. Cross-sectional area (solid round): A = (π d2)/4 = (π · 502)/4 = 1963.495 mm² .
  2. Second moment of area (solid round): I = (π d4)/64 = (π · 504)/64 = 306,796.158 mm⁴ .
  3. Radius of gyration: r = √(I/A) = √(306,796.158 / 1,963.495) = 12.50 mm .
  4. Slenderness ratio: S = L / r = 2000 / 12.5 = 160.0 .
  5. Johnson/Euler boundary: Scr = √(2 π2 E / Sy) = √(2 π2 · 205,000 / 345) = 108.30 . Since S = 160 > Scr, this is a long column → use Euler.
  6. Effective length factor for pinned–pinned: K = 1.0.
  7. Euler critical load:
    Pcr = (π2 E I) / (K L)2 = 9.8696 · 205,000 · 306,796.158 / (1 · 2000)2 = 155,182.78 N = 155.18 kN.
  8. Allowable load with design factor nd=5: Fa = Pcr / nd = 155.18 / 5 = 31.04 kN .
  9. (Optional) Mass check: Volume = A · L = 1,963.495 mm² × 2000 mm = 3.92699×106 mm³ = 3926.99 cm³,  M = ρ·V = 7.87 × 3926.99 ≈ 30.905 kg.

Calculator Cross-Check

Comparison of formula results with calculator results
RESULTS COMPARISON
QuantityFormula ResultCalculator Result*
Critical load Pcr155.18 kN155.18 kN
Allowable load Fa31.04 kN31.04 kN
ClassificationLong column (Euler)Long column (Euler)

* Using the same inputs in the Column Buckling Calculator. Minor differences (< 0.01) may appear due to rounding.

Is This Example Realistic?

Yes. A slender, pin-ended round member in axial compression is a very common real-world case. You’ll see close analogs in practice:

  • Truss/space-frame members: bars or rods connected to gusset plates with pins/bolts → behavior close to pinned–pinned (K ≈ 1).
  • Construction props/shoring: temporary steel props typically checked as pin-ended columns.
  • Machine struts/links: members with spherical rod ends (Heim/rose joints) approximate pinned ends.
  • Architectural/bridge bracing: tension/compression bars often designed as pin-ended compression members.

Sanity Check of the Numbers

  • Slenderness S = L/r = 2000/12.5 = 160long column regime, so Euler buckling governs (as shown).
  • Computed critical load ≈ 155.18 kN and allowable ≈ 31.04 kN (with nd=5) are plausible.
  • Self-weight of the bar is small (≈ 0.31 kN) compared to the capacity → negligible influence here.

Design Caveats & Good Practice

  • End restraint isn’t perfectly “pinned” in reality. Effective length factor K depends on joint/frame stiffness. Use the appropriate K from your design code or a stability analysis.
  • Initial crookedness & load eccentricity always exist. Codes (e.g., AISC/EC3) handle this via column curves or secant/Perry–Robertson style formulas. Our calculator also supports eccentricity checks.
  • Use your governing code’s resistance format. For final design, prefer code provisions (AISC, EC3, etc.) over a generic “design factor”.
  • Section choice matters. Real columns are often tubes/rolled shapes; check local buckling limits and slender-element rules where applicable.

Bottom line: this worked example is a sound baseline and matches the calculator. For a project deliverable, apply your code’s prescribed K, imperfection model, and resistance factors/curves.

FAQ

Is pinned–pinned a realistic end condition?
Yes—truss members, construction props, and rod-end struts behave close to K≈1, which is well approximated by pinned–pinned.
Why Euler instead of Johnson here?
The slenderness S=160 exceeds the Johnson/Euler boundary Scr≈108, so Euler buckling governs.

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