10.7 Column Buckling (Euler Formula) and Yield/Failure Criteria

10.7 Column Buckling (Euler Formula) and Yield/Failure Criteria

Core Engineering Principle: Structural safety requires checking both structural stability (buckling under compressive loads) and material strength limits (yielding or fracture under combined stress states). The choice of appropriate failure criteria depends directly on material ductility versus brittleness.

Column Buckling and Euler's Critical Load Formula

Buckling is a sudden lateral displacement collapse of a slender structural column subjected to axial compression, occurring at a stress level often far below the material's compressive yield strength $S_y$.

Euler's Critical Buckling Load ($P_{cr}$)

For a long, ideal, elastic column, the critical buckling load $P_{cr}$ is defined by Euler's Formula:

$$P_{cr} = \frac{\pi^2 E I}{(K L)^2}$$

where:

• $E$ is the Young's Modulus of the material
• $I$ is the minimum area moment of inertia of the cross-section ($I_{min} = \min(I_x, I_y)$)
• $L$ is the unbraced physical length of the column
• $K$ is the Effective Length Factor depending on support end conditions

Effective Length Factor ($K$) and End Conditions

Critical Buckling Stress ($\sigma_{cr}$) and Slenderness Ratio ($KL/r$)

Dividing Euler's critical load $P_{cr}$ by cross-sectional area $A$ yields the critical buckling stress $\sigma_{cr}$:

$$\sigma_{cr} = \frac{P_{cr}}{A} = \frac{\pi^2 E I}{A (K L)^2} = \frac{\pi^2 E}{\left( \frac{K L}{r} \right)^2}$$

where $r = \sqrt{\frac{I}{A}}$ is the Radius of Gyration of the cross-section about the governing buckling axis, and $\frac{K L}{r}$ is the dimensionless Slenderness Ratio.

Critical Slenderness Ratio and Column Classification

• Long (Slender) Columns: $\frac{KL}{r} \ge \left( \frac{KL}{r} \right)_{crit} = \sqrt{\frac{2 \pi^2 E}{S_y}}$. Governed by elastic Euler buckling ($\sigma_{cr} < S_y$).
• Intermediate / Short Columns: Governed by inelastic buckling (Johnson parabolic criterion) or pure compressive material yield ($P_{yield} = A S_y$).

Failure Theories for Ductile Materials

Ductile materials (such as structural steel, aluminum, and copper) fail primarily by excessive plastic yielding driven by shear stresses.

1. Maximum Shear Stress Theory (Tresca Criterion)

The Maximum Shear Stress Theory asserts that yielding occurs when the maximum shear stress $\tau_{max}$ in a multi-axial stress state equals the maximum shear stress at yielding in a simple uniaxial tension test ($S_y / 2$).

$$\tau_{max} = \frac{\sigma_1 - \sigma_3}{2} \ge \frac{S_y}{2} \implies \sigma_1 - \sigma_3 \ge S_y$$

The factor of safety $N_{Tresca}$ according to Tresca is:

$$N_{Tresca} = \frac{S_y}{\sigma_1 - \sigma_3}$$

For pure torsion (shear stress $\tau$ only, $\sigma_1 = \tau, \sigma_3 = -\tau$), Tresca predicts a shear yield strength of:

$$S_{sy} = 0.50 S_y$$

2. Distortion Energy Theory (von Mises Criterion)

The Distortion Energy Theory asserts that yielding occurs when the distortion energy per unit volume in a multi-axial stress state reaches the distortion energy at yield in uniaxial tension.

The von Mises Equivalent Stress $\sigma_{vm}$ in 3D is:

$$\sigma_{vm} = \frac{1}{\sqrt{2}} \sqrt{(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2}$$

For 2D Plane Stress ($\sigma_z = 0, \sigma_3 = 0$):

$$\sigma_{vm} = \sqrt{\sigma_1^2 - \sigma_1 \sigma_2 + \sigma_2^2} = \sqrt{\sigma_x^2 - \sigma_x \sigma_y + \sigma_y^2 + 3 \tau_{xy}^2}$$

Yielding occurs when $\sigma_{vm} \ge S_y$, and the factor of safety $N_{vonMises}$ is:

$$N_{vonMises} = \frac{S_y}{\sigma_{vm}}$$

For pure torsion, von Mises predicts a shear yield strength of:

$$S_{sy} = \frac{S_y}{\sqrt{3}} \approx 0.577 S_y$$

Comparison Note: Von Mises is up to $15.5\%$ less conservative than Tresca and provides the closest agreement with experimental yield tests for ductile metals.

Failure Theories for Brittle Materials

Brittle materials (such as cast iron, concrete, ceramics, and glass) exhibit little to no plastic deformation and fail suddenly by fracture across planes of maximum normal tensile stress.

Maximum Normal Stress Theory (Rankine Criterion)

The Maximum Normal Stress Theory states that failure occurs whenever the maximum principal normal stress reaches the ultimate tensile strength $S_{ut}$ or ultimate compressive strength $S_{uc}$:

$$\sigma_1 \ge S_{ut} \quad \text{or} \quad |\sigma_3| \ge S_{uc}$$

The factor of safety $N$ is:

$$N = \frac{S_{ut}}{\sigma_1}$$

Mohr-Coulomb Failure Criterion

For brittle materials where ultimate compressive strength $S_{uc}$ is significantly larger than ultimate tensile strength $S_{ut}$ ($S_{uc} > S_{ut}$), the Mohr-Coulomb Criterion defines the failure envelope:

$$\frac{\sigma_1}{S_{ut}} - \frac{\sigma_3}{S_{uc}} \ge 1$$

Worked Engineering Problems

Problem 1: Column Buckling Capacity and Slenderness Ratio

Scenario: A solid circular structural steel column ($E = 200\text{ GPa}$, yield strength $S_y = 250\text{ MPa}$) has length $L = 4.0\text{ m}$ and diameter $d = 80\text{ mm} = 0.08\text{ m}$. The column is pinned at both ends ($K = 1.0$). Calculate (a) area $A$, (b) moment of inertia $I$, (c) radius of gyration $r$, (d) slenderness ratio $KL/r$, (e) Euler critical buckling load $P_{cr}$, and (f) verify whether elastic buckling or material yielding governs.

Solution:

1. Calculate cross-sectional properties:

$$A = \frac{\pi d^2}{4} = \frac{\pi (0.08)^2}{4} = 5.0265 \times 10^{-3}\text{ m}^2$$ $$I = \frac{\pi d^4}{64} = \frac{\pi (0.08)^4}{64} = \frac{\pi (4.096 \times 10^{-5})}{64} = 2.0106 \times 10^{-6}\text{ m}^4$$ $$r = \sqrt{\frac{I}{A}} = \sqrt{\frac{2.0106 \times 10^{-6}}{5.0265 \times 10^{-3}}} = \sqrt{4.0 \times 10^{-4}} = 0.020\text{ m} = 20\text{ mm}$$

1. Calculate slenderness ratio:

$$\frac{K L}{r} = \frac{(1.0)(4.0\text{ m})}{0.020\text{ m}} = 200$$

1. Calculate Euler critical buckling load $P_{cr}$:

$$P_{cr} = \frac{\pi^2 E I}{(K L)^2} = \frac{\pi^2 (200 \times 10^9\text{ Pa})(2.0106 \times 10^{-6}\text{ m}^4)}{(1.0 \times 4.0\text{ m})^2}$$ $$P_{cr} = \frac{\pi^2 \times 402,120}{16.0} = \frac{3,968,858}{16.0} = 248,054\text{ N} = 248.05\text{ kN}$$

1. Calculate critical buckling stress $\sigma_{cr}$ and check governing mode:

$$\sigma_{cr} = \frac{P_{cr}}{A} = \frac{248,054\text{ N}}{5.0265 \times 10^{-3}\text{ m}^2} = 49.35 \times 10^6\text{ Pa} = 49.35\text{ MPa}$$

Since $\sigma_{cr} = 49.35\text{ MPa} < S_y = 250\text{ MPa}$, elastic Euler buckling governs column failure at $P_{cr} = 248.05\text{ kN}$.

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Problem 2: Ductile Yield Analysis using Von Mises and Tresca Criteria

Scenario: A machine component made of structural steel with yield strength $S_y = 300\text{ MPa}$ is subjected to plane stress conditions with $\sigma_x = 180\text{ MPa}$, $\sigma_y = 60\text{ MPa}$, and $\tau_{xy} = 40\text{ MPa}$. Calculate the factor of safety $N$ against yielding using (a) Maximum Shear Stress Theory (Tresca) and (b) Distortion Energy Theory (von Mises).

Solution:

1. Calculate principal normal stresses $\sigma_1$ and $\sigma_2$:

$$\sigma_{avg} = \frac{180 + 60}{2} = 120\text{ MPa}$$ $$R = \sqrt{\left( \frac{180 - 60}{2} \right)^2 + (40)^2} = \sqrt{(60)^2 + (40)^2} = \sqrt{3600 + 1600} = \sqrt{5200} = 72.11\text{ MPa}$$ $$\sigma_1 = 120 + 72.11 = 192.11\text{ MPa}, \quad \sigma_2 = 120 - 72.11 = 47.89\text{ MPa}, \quad \sigma_3 = 0\text{ MPa}$$

1. Factor of Safety according to Tresca (Max Shear Stress):

$$\tau_{max} = \frac{\sigma_1 - \sigma_3}{2} = \frac{192.11 - 0}{2} = 96.06\text{ MPa}$$ $$N_{Tresca} = \frac{S_y}{\sigma_1 - \sigma_3} = \frac{300\text{ MPa}}{192.11\text{ MPa}} = 1.56$$

1. Factor of Safety according to Von Mises (Distortion Energy):

$$\sigma_{vm} = \sqrt{\sigma_x^2 - \sigma_x \sigma_y + \sigma_y^2 + 3 \tau_{xy}^2}$$ $$\sigma_{vm} = \sqrt{(180)^2 - (180)(60) + (60)^2 + 3(40)^2}$$ $$\sigma_{vm} = \sqrt{32,400 - 10,800 + 3,600 + 4,800} = \sqrt{30,000} = 173.21\text{ MPa}$$ $$N_{vonMises} = \frac{S_y}{\sigma_{vm}} = \frac{300\text{ MPa}}{173.21\text{ MPa}} = 1.73$$