Interactive Lecture
Euler Critical Load, Effective Length, and Slenderness Ratio
by Dr Arnaldo Delli Carri
Elastic instability, commonly known as buckling, is a structural phenomenon where a slender member under compression suddenly deflects laterally when the applied load reaches a critical value. It sets the load limit for slender compression members, and that limit is reached long before the material itself would yield or crush.
Elastic instability occurs when a compression member loses its ability to stay straight and deflects sideways — even though the material is still within its elastic range. The deflections are large and grow rapidly for small load increases.
The foundational theory for understanding column buckling was developed by Leonhard Euler in 1744. The derivation starts with the fundamental differential equation for the elastic curve of a beam-column.
Starting Point - Differential Equation:
For a column under axial load \(P\) with lateral deflection \(y\):
\(EI\frac{d^2y}{dx^2} + Py = 0\)
Rearranging: \(\frac{d^2y}{dx^2} + \frac{P}{EI}y = 0\)
Let \(k^2 = \frac{P}{EI}\), then: \(\frac{d^2y}{dx^2} + k^2y = 0\)
General Solution:
The general solution to this differential equation is:
\(y = C_1\sin(kx) + C_2\cos(kx)\)
Where \(C_1\) and \(C_2\) are constants determined by boundary conditions
For a column with pinned ends (simply supported), the boundary conditions are \(y = 0\) at \(x = 0\) and \(x = L\):
Applying Boundary Conditions:
At \(x = 0\): \(y = 0 \Rightarrow C_2 = 0\)
At \(x = L\): \(y = 0 \Rightarrow C_1\sin(kL) = 0\)
For non-trivial solutions (\(C_1 \neq 0\)): \(\sin(kL) = 0\)
This gives: \(kL = m\pi\) where \(m = 1, 2, 3, ...\)
Critical Loads and Mode Shapes:
Substituting \(k^2 = \frac{P}{EI}\) and \(k = \frac{m\pi}{L}\):
\(P_{cr,m} = \frac{m^2\pi^2 EI}{L^2}\) for \(m = 1, 2, 3, ...\)
The deflected shape: \(y = C_1\sin\left(\frac{m\pi x}{L}\right)\)
First mode (\(m = 1\)): \(P_{cr} = \frac{\pi^2 EI}{L^2}\) (Euler's formula)
Higher modes occur at \(4P_{cr}\), \(9P_{cr}\), \(16P_{cr}\), etc.
The first buckling mode (\(m = 1\)) is the lowest critical load and is a single half-sine wave. Higher modes (\(m > 1\)) need far larger loads, so in practice a column reaches the first mode first and buckles there — the higher modes only appear if the column is braced to suppress the first.
The effective length \(L_e\) depends on the support conditions at the column ends:
\(L_e = K \cdot L\)
Where \(K\) is the effective length factor:
The tendency of a column to buckle is characterised by its slenderness ratio:
\(\lambda = \frac{L_e}{r}\)
Where \(r = \sqrt{\frac{I}{A}}\) is the radius of gyration
Higher slenderness ratios indicate greater susceptibility to buckling
Adjust the parameters to see how they affect the critical buckling load and column behaviour.
Occurs in thin-walled sections where individual plates (flanges or webs) buckle before the overall member fails. Common in steel I-beams and hollow sections.
Affects beams where the compression flange becomes unstable and deflects laterally whilst twisting. Critical in unbraced steel beams.
The overall buckling of the entire structural member, as described by Euler's theory.
Some structures can carry additional load after initial buckling. A buckled aircraft skin panel keeps carrying load by shedding it into the surrounding stiffeners, so thin plates and shells are routinely designed to work past their first buckling point.
In slender structures — tall buildings, bridges, aerospace frames — buckling limits the design long before the material yields. It's often the governing failure mode, not material strength.
Real structures have initial imperfections (geometrical and material) that reduce the actual buckling load below the theoretical Euler load.
The Rankine-Gordon formula provides a practical approach for design by combining crushing failure and Euler buckling:
\(\frac{1}{P_{design}} = \frac{1}{P_{crushing}} + \frac{1}{P_{Euler}}\)
Where:
This can be written as:
\(P_{design} = \frac{\sigma_c A}{1 + \frac{\sigma_c L_e^2}{\pi^2 E r^2}} = \frac{\sigma_c A}{1 + a\lambda^2}\)
Where \(a = \frac{\sigma_c}{\pi^2 E}\) is the Rankine constant and \(\lambda = \frac{L_e}{r}\) is the slenderness ratio
Modern design codes (Eurocode 3, BS 5950, AISC) provide buckling curves that account for:
Leonhard Euler first derived the critical load formula for pin-ended columns, laying the mathematical foundation for stability analysis.
Stephen Timoshenko extended buckling theory to include shear deformation effects and developed solutions for different end conditions.
Contemporary research focuses on:
You'll need buckling analysis wherever slender members carry compression:
| Concept | Equation | Description |
|---|---|---|
| Euler Buckling Load | \(P_{cr} = \frac{\pi^2 EI}{(KL)^2}\) | Critical load for slender columns |
| Effective Length Factor | \(K\) | Depends on end conditions (0.5 to 2.0) |
| Slenderness Ratio | \(\lambda = \frac{KL}{r}\) | Ratio of effective length to radius of gyration |
| Critical Stress | \(\sigma_{cr} = \frac{P_{cr}}{A} = \frac{\pi^2 E}{(KL/r)^2}\) | Buckling stress for slender columns |
| Radius of Gyration | \(r = \sqrt{\frac{I}{A}}\) | Geometric property of cross-section |
A steel column with pinned ends has the following properties:
Calculate the critical buckling load.
Show Solution
Given: \(L = 3 \text{ m}, E = 200 \times 10^9 \text{ Pa}, I = 8.5 \times 10^{-6} \text{ m}^4\)
For pinned ends: \(K = 1.0\), so \(L_e = 3 \text{ m}\)
Using Euler's formula:
\(P_{cr} = \frac{\pi^2 EI}{L_e^2} = \frac{\pi^2 \times 200 \times 10^9 \times 8.5 \times 10^{-6}}{3^2}\)
\(P_{cr} = \frac{\pi^2 \times 1700}{9} = \frac{16778.3}{9} = 1864 \text{ kN}\)
Answer: \(P_{cr} \approx 1864 \text{ kN}\)
Compare the critical buckling loads for the same column from Problem 1 with different end conditions:
Show Solution
Using the same column properties from Problem 1:
a) Fixed-fixed ends (\(K = 0.5\)):
\(L_e = 0.5 \times 3 = 1.5 \text{ m}\)
\(P_{cr} = \frac{\pi^2 \times 200 \times 10^9 \times 8.5 \times 10^{-6}}{1.5^2} = 7457 \text{ kN}\)
b) Fixed-free (\(K = 2.0\)):
\(L_e = 2.0 \times 3 = 6 \text{ m}\)
\(P_{cr} = \frac{\pi^2 \times 200 \times 10^9 \times 8.5 \times 10^{-6}}{6^2} = 466 \text{ kN}\)
The fixed-fixed condition provides 4 times the load capacity of pinned-pinned, whilst cantilever reduces it to 1/4.
Answer: fixed-fixed \(P_{cr} \approx 7457 \text{ kN}\); fixed-free \(P_{cr} \approx 466 \text{ kN}\)
A hollow rectangular section column has:
Calculate the slenderness ratio about both axes.
Show Solution
For a hollow rectangular section:
External: 100 × 50 mm, Internal: 90 × 40 mm
Area: \(A = 100 \times 50 - 90 \times 40 = 1400 \text{ mm}^2\)
Second moments of area:
\(I_{xx} = \frac{100 \times 50^3 - 90 \times 40^3}{12} = \frac{12,500,000 - 5,760,000}{12} = 561,667 \text{ mm}^4\)
\(I_{yy} = \frac{50 \times 100^3 - 40 \times 90^3}{12} = \frac{50,000,000 - 29,160,000}{12} = 1,736,667 \text{ mm}^4\)
Radii of gyration:
\(r_x = \sqrt{\frac{I_{xx}}{A}} = \sqrt{\frac{561,667}{1400}} = 20.03 \text{ mm}\)
\(r_y = \sqrt{\frac{I_{yy}}{A}} = \sqrt{\frac{1,736,667}{1400}} = 35.23 \text{ mm}\)
Slenderness ratios:
\(\lambda_x = \frac{2500}{20.03} = 125\) (critical - weak axis)
\(\lambda_y = \frac{2500}{35.23} = 71\)
The column will buckle about the x-x axis (weak axis) as it has the higher slenderness ratio.
Answer: \(\lambda_x \approx 125\) (governs, x-x weak axis), \(\lambda_y \approx 71\)
A steel column with the following properties needs to be designed using the Rankine-Gordon formula:
Calculate the design load using the Rankine-Gordon formula.
Show Solution
Given: \(L = 4 \text{ m}, A = 2500 \text{ mm}^2 = 0.0025 \text{ m}^2, r = 25 \text{ mm} = 0.025 \text{ m}\)
\(\sigma_c = 250 \text{ MPa} = 250 \times 10^6 \text{ Pa}, E = 200 \text{ GPa} = 200 \times 10^9 \text{ Pa}\)
Step 1: Calculate slenderness ratio
\(\lambda = \frac{L_e}{r} = \frac{4}{0.025} = 160\)
Step 2: Calculate Rankine constant
\(a = \frac{\sigma_c}{\pi^2 E} = \frac{250 \times 10^6}{\pi^2 \times 200 \times 10^9} = \frac{1}{800\pi^2} = 1.2665 \times 10^{-4}\)
Step 3: Apply Rankine-Gordon formula
\(P_{design} = \frac{\sigma_c A}{1 + a\lambda^2} = \frac{250 \times 10^6 \times 0.0025}{1 + a \times 160^2}\)
\(a\lambda^2 = \frac{160^2}{800\pi^2} = \frac{32}{\pi^2} = 3.2423\)
\(P_{design} = \frac{625,000}{1 + 32/\pi^2} = \frac{625,000}{4.2423} = 1.473 \times 10^5 \text{ N} = 147.3 \text{ kN}\)
Comparison with pure Euler:
\(P_{Euler} = \frac{\pi^2 EI}{L^2} = \frac{\pi^2 \times 200 \times 10^9 \times (0.025^2 \times 0.0025)}{4^2} = 192.8 \text{ kN}\)
The Rankine-Gordon formula gives a more conservative (lower) design load that accounts for material failure as well as buckling.
Answer: \(P_{design} \approx 147.3 \text{ kN}\) (Rankine-Gordon), versus \(P_{Euler} \approx 192.8 \text{ kN}\)