Euler's critical load

The following assumptions are made while deriving Euler's formula: • The material of the column is homogeneous and isotropic. • The compressive load on the column is axial only. • The column is free from initial stress. • The weight of the column is neglected. • The column is initially straight (no eccentricity of the axial load). • Pin joints are friction-less (no moment constraint) and fixed ends are rigid (no rotation deflection). • The cross-section of the column is uniform throughout its length. • The direct stress is very small as compared to the bending stress (the material is compressed only within the elastic range of strains). • The length of the column is very large as compared to the cross-sectional dimensions of the column. • The column fails only by buckling. This is true if the compressive stress in the column does not exceed the yield strength \sigma_y (see figure 1): \sigma = \frac{P_{cr}}{A} = \frac{\pi^2E}{(L_e/r)^2} where: • {L_e}/{r} is the slenderness ratio, • L_e = KL is the effective length, • r = \sqrt{I / A} is the radius of gyration, • I is the second moment of area (area moment of inertia), • A is the area cross section. For slender columns, the critical buckling stress is usually lower than the yield stress. In contrast, a stocky column can have a critical buckling stress higher than the yield, i.e. it yields prior to buckling. == Mathematical derivation ==

Pin ended column The following model applies to columns simply supported at each end (K = 1). Firstly, we will put attention to the fact there are no reactions in the hinged ends, so we also have no shear force in any cross-section of the column. The reason for no reactions can be obtained from symmetry (so the reactions should be in the same direction) and from moment equilibrium (so the reactions should be in opposite directions). Using the free body diagram in the right side of figure 3, and making a summation of moments about point : \Sigma M = 0 \Rightarrow M(x) + Pw = 0 where is the lateral deflection. According to Euler–Bernoulli beam theory, the deflection of a beam is related with its bending moment by: M = -EI\frac{d^2 w}{dx^2}. so: EI\frac{d^2w}{dx^2} + Pw = 0 Let \lambda^2 = \frac{P}{EI}, so: \frac{d^2w}{dx^2} + \lambda^2 w = 0 We get a classical homogeneous second-order ordinary differential equation. The general solutions of this equation is: w(x) = A \cos(\lambda x) + B \sin(\lambda x), where A and B are constants to be determined by boundary conditions, which are: • Left end pinned: w(0) = 0 \rightarrow A = 0 • Right end pinned: w(\ell) = 0 \rightarrow B \sin(\lambda \ell) = 0 If B = 0, no bending moment exists and we get the trivial solution of w(x) = 0. However, from the other solution \sin(\lambda \ell) = 0 we get \lambda_n \ell = n\pi, for n = 0, 1, 2, \ldots Together with \lambda^2 = \frac{P}{EI} as defined before, the various critical loads are: P_{n} = \frac{n^2 \pi^2 EI}{\ell^2} \; , \quad \text{ for } n = 0, 1, 2, \ldots and depending upon the value of n , different buckling modes are produced as shown in figure 4. The load and mode for n=0 is the nonbuckled mode. Theoretically, any buckling mode is possible, but in the case of a slowly applied load only the first modal shape is likely to be produced. The critical load of Euler for a pin ended column is therefore: P_{cr} = \frac{\pi^2 EI}{\ell^2} and the obtained shape of the buckled column in the first mode is: w(x) = B \sin \left({\pi \over \ell} x\right) . General approach The differential equation of the axis of a beam is: \frac{d^4 w}{dx^4} + \frac{P}{EI}\frac{d^2 w}{dx^2} = \frac{q}{EI} For a column with axial load only, the lateral load q(x) vanishes and substituting \lambda^2 = \frac{P}{EI}, we get: \frac{d^4 w}{dx^4} + \lambda^2\frac{d^2 w}{dx^2} = 0 This is a homogeneous fourth-order differential equation and its general solution is w(x) = A\sin(\lambda x) + B\cos(\lambda x) + Cx + D The four constants A, B, C, D are determined by the boundary conditions (end constraints) on w(x) , at each end. There are three cases: • Pinned end: • : w = 0 and M = 0 \rightarrow {d^2w \over dx^2} = 0 • Fixed end: • : w = 0 and {dw \over dx} = 0 • Free end: • : M = 0 \rightarrow {d^2w \over dx^2} = 0 and V = 0 \rightarrow {d^3w \over dx^3} + \lambda^2{dw \over dx} = 0 For each combination of these boundary conditions, an eigenvalue problem is obtained. Solving those, we get the values of Euler's critical load for each one of the cases presented in Figure 2. == See also ==