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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 E I 2 y P y

Set equal to zero.

E I 2 y P y 0

Divide everything by E I .

2 y P E I y 0

Set the variable, α 2

α 2 P E I

then, plug that in to get:

2 y α 2 y 0

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

y A α x B α 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 α L

Since A cannot equal zero:

α L 0

Take the sine inverse of both sides, and α L can be 0, , 2 , etc. So...

α L n

Solve for α 2

α 2 n 2 2 L 2

Set the two α 2 's equal and solve for P .

P cr n 2 2 E I L

Assume that n 1

Now we can solve for F cr using this equation.

F cr P cr A g 2 E L 2 I A 2 E r 2 k L 2

where:

r I A

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Source:  OpenStax, Steel design (civi 306). OpenStax CNX. Jan 22, 2004 Download for free at http://cnx.org/content/col10153/1.3
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