# Deriving euler's equation

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#### Derivation of euler's equation

Start with the differential equation giving the deflected shape of an elastic member subjected to bending.

$M=-(EI\frac{d y}{d }})=Py$

Set equal to zero.

$EI\frac{d y}{d }}+Py=0$

Divide everything by $EI$ .

$\frac{d y}{d }}+\frac{P}{EI}y=0$

Set the variable, $\alpha ^{2}$

$\alpha ^{2}=\frac{P}{EI}$

then, plug that in to get:

$\frac{d y}{d }}+\alpha ^{2}y=0$

Since this is a second order, linear, ordinary differential equation with constant coefficients, it solves to:

$y=A\sin (\alpha x)=B\cos (\alpha x)$

Take the boundary condition that $x=0$ and $y=0$ to solve for $B$

$y(0)=0=A(0)=B(1)$
$B=0$

Now, take the boundary conditions $x=L$ and $y=0$ .

$y(L)=0=A\sin (\alpha L)$

Since $A$ cannot equal zero:

$\sin (\alpha L)=0$

Take the sine inverse of both sides, and $\alpha L$ can be 0, $\pi$ , $2\pi$ , etc. So...

$\alpha L=n\pi$

Solve for $\alpha ^{2}$

$\alpha ^{2}=\frac{n^{2}\pi ^{2}}{L^{2}}$

Set the two $\alpha ^{2}$ 's equal and solve for $P$ .

${P}_{cr}=\frac{n^{2}\pi ^{2}EI}{L}$

Assume that $n=1$

Now we can solve for ${F}_{cr}$ using this equation.

${F}_{cr}=\frac{{P}_{cr}}{{A}_{g}}=\frac{\pi ^{2}E}{L^{2}}\frac{I}{A}=\frac{\pi ^{2}Er^{2}}{(kL)^{2}}$

where:

$r=\sqrt{\frac{I}{A}}$

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