# A differential equation of Buckling Rod.

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

The problem is to solve the differentail equation 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 diffierential equation: $$v(x) = \frac{c_1\cos(x\sqrt{P})+c_2\sin(x\sqrt{P})}{P}+c_4 x+c_3$$ 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 ones. Could you please help me to find the mistake or explain what's wrong in my equation?

There is nothing wrong with your equation. Let's think about it in a different way:

Your solution is $$v(x) = c_1 + c_2 x + c_3 \cos(\lambda x) + c_4 \sin(\lambda x)$$ where $\lambda = \sqrt{\frac{P}{E I_x}}$.

And your boundary conditions state that \begin{align} v(0) &= c_1 + c_3 = 0\\ v(l) &= c_1 + c_2 l + c_3 \cos(\lambda l) + c_4 \sin(\lambda l) = 0\\ v'(0) &= c_2 + c_4 \lambda = 0\\ v'(l) &= c_2 - c_3 \lambda \sin(\lambda l) + c_4 \lambda \cos(\lambda l) = 0 \end{align}

An easy way to see what's going on is to write the system in its matrix representation:

$$\begin{pmatrix} 1 & 0 & 1 & 0 \\ 1 & l & \cos(\lambda l) & \sin(\lambda l)\\ 0 & 1 & 0 & \lambda\\ 0 & 1 & -\lambda \sin(\lambda l) & \lambda \cos(\lambda l) \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3 \\ c_4\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0\end{pmatrix}$$

Indeed, one posibility for solving the system is $c_1 = c_2 = c_3 = c_4 = 0$, which leads to the trivial solution. There is another posibility though, if the determinant of the matrix equals to zero, then there are nontrivial solutions to the linear equation.

Of course, the determinant of the matrix -which I will call $M$-, is dependent on $\lambda$, i.e. $$\det M(\lambda) = \lambda \big(\lambda l \sin(\lambda l) + 2 \cos(\lambda l) -2\big)$$ hence if $$\tag{*} \lambda \big(\lambda l \sin(\lambda l) + 2 \cos(\lambda l) -2\big) = 0$$ then there are non-trivial solutions to the Boundary Value Problem.

Lets take a look at $\lambda$. The greek letter is dependent on the bending modulus $E I_x$ of the bar, and the force $P$ exerted over it. Evidently, unless you add mass, exchange the chemical composition, or modify the area of the bar, the bending modulus is constant (some books denote it by $B$), and the only way to modify the value of $\lambda$ is to change the force acting on the bar. That said, lets return to the equation $(*)$. The most obvious solution for it is when $\lambda = 0$, ($P = 0$). This means a clamped bar, under no transversal force whatsoever. In such case, the bar is undeformed (i.e. $v(x) = 0$).

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The red arrow denotes the force $P$.

A good question would be if this is the only solution. To answer it, we must look again to equation $(*)$. If $\lambda \neq 0$, then the condition to have nontrivial solutions is that $$\lambda l \sin(\lambda l) + 2 \cos(\lambda l) -2 = 0$$ which can be written as $$\lambda l \sin(\lambda l) - 4 \sin^2\big(\tfrac{\lambda l}{2}\big) = 0.$$ This equation will not have a solution for all values of $\lambda > 0$. In fact, the only way to have a solution is if both sines are zero simultaneously (prove it), meaning that $\lambda = \frac{2\pi n}{l}$, where $n = 1,\,2,\,...\,$. In terms of the force, there are nontrivial solutions to the problem if $$P = \frac{4 \pi^2 n^2 E I_x}{l^2}.$$

What to read of all this?

Suppose you have a clamped bar with no stress whatsoever ($P = 0$) and start applying pressure on one end by pushing the wall. You know that the only solution for $0 \le P < \frac{4 \pi^2 E I_x}{l^2}$ is the trivial one. When you reach the threshold $P = \frac{4 \pi^2 E I_x}{l^2}$, there is another two solutions $$v(x) = \pm A\Big(\cos\big(\tfrac{2\pi}{l} x\big) - 1\Big)$$ where $A$ is a given amplitude.

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What sign will the solution take can only be determined in real life conditions (where inhomogeneity will play a role).

The first buckling will occur at $P_1 = \frac{4 \pi^2 E I_x}{l^2}$. In engineering terms, this will be the ammount of load a column can support before giving up (assuming linear behavior).

If we continue applying force, we will reach a second threshold $P_2 = \frac{8 \pi^2 E I_x}{l^2}$. The second buckling:

$\hskip1.5in$

And so on.

What's behind all this phenomena is the theory of Eigenvalues and Eigenvectors for ODE's, which you really should learn if you want to understand a large portion of mathematics and physics. If you want to learn more about this, you can look at Churchill and Brown Fourier Series and Boundary Value Problems for a good introduction to the subject.

For a thorough exposition of the buckling problem, I recommend you to read A Treatise on the Mathematical Theory of Elasticity by A. E. H. Love.

Finally, I urge you to do all calculations by yourself. I cannot claim that I doublechecked my algebra. The main idea is right though.

Not an answer, but too long for a comment. Some points: In your solution, the $\sqrt {(P)}$'s should be $\sqrt{\frac P{EI_x}}$'s, but this doesn't really change anything. It would be easier to define a new variable $z=v''$, which gets rid of two orders and the $c_3, c_4$ terms. You have an eigensolution restritction coming from the constraint at $l$. In this case you will have some nonzero $z$, but no nonzero $z'(0)$, so only the cosine terms contribute and only those that have wavelengths an integral fraction of $l$.

Intiuitively, to find the buckling point you need to calculate the energy stored in the straight rod by the fact that your force has compressed it, then compare with the energy stored in a rod of the correct length but bent by the cosine waves. When the bend energy is less than the compression energy you get buckling.

• But how can i apply the boundary conditions for $v(x)$ to a variable $z$? I think that the bounday conditions gives: $v'(0)=0 \mapsto (\int z \mathrm{dx})|_{z=0}$ Oct 15, 2012 at 16:52