A differential equation of Buckling Rod

Question

I tried to solve a differential equation, but unfortunately got stuck at some point.

The problem is to solve the diff. eq. of hard clamped on both ends rod. And the force compresses the rod at both ends. the solution(v(x)) is the value of bending I need.

I assuming, that the differential equation of buckling rod is $$ EI_{x}v''''+Pv''=0$$ where $$P$$ is a force. and $$EI_x$$ is inflexibility.

Then I find the solution for the diff. eq: $$v(x) = \frac{(\frac{(c_2 \sin(\sqrt(P) x))}{\sqrt(P)}+\frac{(c_1 \cos(\sqrt(P) x))}{\sqrt(P)})}{\sqrt(P)}+c_4 x+c_3$$ the boundary conditions: $$v(0)=v(l)=0=v'(0)=v'(l)$$ gives the trivial solution for $$c_{1},c_{2},c_{3},c_{4}$$ but I need non-trivial solution.

Could you please help me to find the mistake or explain what's wrong in my equation?

Answer 1

First, the solution to your equation is not exactly what you got, but,

$$v(x) = C_1 \cos ax + C_2 \sin ax + C_3x + C_4$$

where $a^2 = \frac{P}{EI_x}$. And then you need to look more carefully at your boundary conditions...

$$v(0) = C_1 + C_4 = 0,\ C_4 = -C_1$$ $$v'(0) = C_2 a + C_3 = 0,\ C_3 = -aC_2$$ $$v(l) = C_1 \cos al + C_2 \sin al + C_3 l + C_4 = C_1(\cos al - 1) +C_2(\sin al -al) = 0$$ $$v'(l) = -C_1 a \sin al + C_2 a \cos al +C_3 = -C_1 a \sin al + C_2 a (\cos al - 1) = 0$$

The last two equations have the trivial solution, but may have a non-trivial solution if the system is degenerate. In this case it equates to the determinant of the coefficient matrix being 0, or:

$$(\cos al -1)^2 + \sin al (\sin al - al)=0$$

Working on this, you can eventually reach that there is a non-trivial solution if

$$\cos al = \frac{4 \pm a^2l^2}{4+a^2l^2}$$.

The simplest of the two solutions comes when $cos al = 1$, then $al = 2\pi n$. For $n=1$, you get a non trivial solution for

$$P = \frac{4 \pi^2 EI}{L^2}$$

which is the critical load for a doubly clamped buckling rod.

The other solution, $\cos al = \frac{4 - a^2l^2}{4+a^2l^2}$ has also inifnitely many solutions, one at $al=0$, the next one around $al \approx 4$, see the graph below where both sides of the equation have been plotted. Since $4 > 2 \pi$, the effective critical load is the other one.