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Start with the differential equation giving the deflected shape of an elastic member subjected to bending.
Set equal to zero.
Divide everything by $EI$ .
Set the variable, $\alpha ^{2}$
then, plug that in to get:
Since this is a second order, linear, ordinary differential equation with constant coefficients, it solves to:
Take the boundary condition that $x=0$ and $y=0$ to solve for $B$
Now, take the boundary conditions $x=L$ and $y=0$ .
Since $A$ cannot equal zero:
Take the sine inverse of both sides, and $\alpha L$ can be 0, $\pi $ , $2\pi $ , etc. So...
Solve for $\alpha ^{2}$
Set the two $\alpha ^{2}$ 's equal and solve for $P$ .
Assume that $n=1$
Now we can solve for ${F}_{cr}$ using this equation.
where:
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