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: d = 50 mm, L = 2 m, axial compression P.
Geometry and section properties used in calculations.
Given Data
Input properties for the column buckling example
| INPUT PROPERTIES SUMMARY |
| Parameter | Value |
| Diameter of column [d] | 50 | mm |
| Length of column [L] | 2 | m |
| End condition | Pinned–Pinned |
| Design factor [nd] | 5 | — |
| Yield strength [Sy] | 345 | MPa |
| Elastic modulus [E] | 205 | GPa |
| Density [ρ] | 7.87 | g/cm³ |
Step-by-Step Solution (Formulas)
-
Cross-sectional area (solid round):
A = (π d2)/4 = (π · 502)/4 = 1963.495 mm²
.
-
Second moment of area (solid round):
I = (π d4)/64 = (π · 504)/64 = 306,796.158 mm⁴
.
-
Radius of gyration:
r = √(I/A) = √(306,796.158 / 1,963.495) = 12.50 mm
.
-
Slenderness ratio:
S = L / r = 2000 / 12.5 = 160.0
.
-
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.
-
Effective length factor for pinned–pinned: K = 1.0.
-
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.
-
Allowable load with design factor nd=5:
Fa = Pcr / nd = 155.18 / 5 = 31.04 kN
.
-
(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 |
| Quantity | Formula Result | Calculator Result* |
| Critical load Pcr | 155.18 kN | 155.18 kN |
| Allowable load Fa | 31.04 kN | 31.04 kN |
| Classification | Long 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 = 160 → long 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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