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Deriving Euler's equation

Module by: Michael Terk. E-mail the author

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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=-EIy′′=Py M E I 2 y P y (1)

Set equal to zero.

EIy′′+Py=0 E I 2 y P y 0 (2)

Divide everything by EI E I .

y′′+PEIy=0 2 y P E I y 0 (3)

Set the variable, α2 α 2

α2=PEI α 2 P E I (4)

then, plug that in to get:

y′′+α2y=0 2 y α 2 y 0 (5)

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

y=Asinαx=Bcosαx y A α x B α x (6)

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

y0=0=A0=B1 y 0 0 A 0 B 1 (7)
B=0 B 0 (8)

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

yL=0=AsinαL y L 0 A α L (9)

Since A A cannot equal zero:

sinαL=0 α L 0 (10)

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

αL=nπ α L n (11)

Solve for α2 α 2

α2=n2π2L2 α 2 n 2 2 L 2 (12)

Set the two α2 α 2 's equal and solve for P P.

P cr =n2π2EIL P cr n 2 2 E I L (13)

Assume that n=1 n 1

Now we can solve for F cr F cr using this equation.

F cr = P cr A g =π2EL2IA=π2Er2kL2 F cr P cr A g 2 E L 2 I A 2 E r 2 k L 2 (14)

where:

r=IA r I A (15)

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