Euler's critical load

In the article buckling, from an elastic point of view, the critical load that a column can bear while staying straight is given by

F=\frac{\pi^2 EI}{(KL)^2}

where

F

= maximum or critical force (longitudinal compression load on column),

E

= modulus of elasticity,

I

= area moment of inertia of the cross section of the rod,

L

= unsupported length of column,

K

= column effective length factor

This result was given by the Swiss mathematician Leonhard Euler in 1757.

The goal of this page is to demonstrate this Euler’s formula in the special case of $K$ = 1.0, that is when both ends of the column are pinned (hinged, free to rotate).

Diagram

Preliminary notations

$\overrightarrow F$ = 2 opposite compressive forces acting along the longitudinal axis of the beam
$\ z = w(x)$ ; the curve w(x) describes the deflection of the beam in the z direction at some position x
$\ M(x)$ ; the bending moment of the beam at the position x
$\ w'= \frac{dw(x)}{dx} = \frac{dz}{dx}$ ; the derivative of w(x)
$\ w'' = \frac{d^2w}{d^2x}$ ; the second derivative of w(x)

We will need the relation between the bending moment $\ M(x)$ and $\ w'' = \frac{d^2w}{d^2x}$ ; the second derivative of the deflection $\ w(x)$

To see this relation, look at the end of this paragraphe: Euler–Bernoulli_beam_theory#Static_beam_equation

Where you can read : $M = -EI\cfrac{\mathrm{d}^2w}{\mathrm{d}x^2}$

(Just before Euler–Bernoulli_beam_theory#Derivation_of_bending_moment_equation)

This will be our Formula F1

Demonstration

The bending moment $M(x)$ results of the 2 strengths $\overrightarrow F$ and to the fact that the axis of the beam moved aside of its original position by the distance $w(x)$ .

This bending moment at the position $x$ is :

Formula F2 : $M(x) = F$ x $w(x)$ ; that's the simple product of the intensity of the strength by the lever arm $w(x)$ .

The formulae F1 and F2 give :

$w''(x) = - \frac{F w(x)}{EI}$

To have easier notations, let's write out : $\Omega^2 = \frac{F}{EI}$

We get a classical homogeneous second-order linear constant coefficient ordinary differential equation :

$w''(x) = -\Omega^2 w(x)$

The general solutions of this equation have the following aspect :

$w(x) = A \cos \Omega x + B \sin \Omega x$

Where A et B are constants to be determined by some conditions which are :

$w(0) = 0$ and $w(L) = 0$

because both ends pinned.

Which implies :

First condition : $f(0) = 0$

so $A \cos 0 + B \sin 0 = 0$

but $\cos (0) = 1$

and $\sin (0) = 0$

so : $1A + 0B = 0$

or : $A=0$

Now the expression of w(x) is :

$w(x) = B \sin \Omega x$

Second Condition : $w(L) = 0$ so:

$B \sin \Omega L=0$

There are only two possibilities : $B = 0$ or $\sin \Omega L = 0$

1) If $B =0$ ; then the solution of the differential equation is $w(x) = 0$ , for all x between 0 and L.

2) If $\sin \Omega L$ = 0 , a think that is possible only for $\Omega L = \pi + 2k\pi$ with k = 0 or -1 or +1 etc...

Pour $k = 0$ we get :

$\Omega L = \pi$ that is :

$\Omega = \frac{\pi}{L}$

So we have $\Omega^2= \frac{\pi^2}{L^2}$

Let's not forget that oabove we wrote out :

$\Omega^2 = \frac{F}{EI}$

so we get $\frac{F}{EI} = \frac{\pi^2}{L^2}$

we deduce the critical load of Euler : $F_\text{e} = \frac{\pi^2EI}{L^2}$

More information : For F > π²EI/l² , the shape of the beam is an arc of a sine curve.

Reference

(fr) This article is the translation of a demonstration created in the article named « Flambage » in the Wikipedia in French.

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Euler's critical load

Diagram

Preliminary notations

Demonstration

See also

Reference