Question

Buckling of a Thin Vertical Column In Example 4 ofSection 5.2 we saw that when a constant vertical compressiveforce, or load,was applied to a thin column of uniform crosssection and hinged at both ends, the deflectionis a solutionof the BVP:

$El\frac{d^{2}y}{d{x}^2}+Py=0,y\left(0\right)=0,y\left(L\right)=0$

Short answer

Let the Bessel equation be $x^{2}y\text{'}\text{'}+xy\text{'}+\left(x^2-n^2\right)y=0$. This equation has two linearly independent solutions for a fixed value of n. A Bessel equation of the first kind, indicated by J_n(X), is one of these solutions that may be derived using Frobinous approach.

$$\begin{gathered} y_1=x^a{J}_p\left(b{x}^c\right) \\ {y}_2=x^a{J}_{-p}\left(b{x}^c\right) \end{gathered}$$

At X=0, this solution is regular. The second solution, which is singular at X=0 , is represented by Y_n(X) and is called a Bessel function of the second kind.$y_3=x^a\left[\frac{cosp\pi J_p\left(b{x}^c\right)-J_{-p}\left(b{x}^c\right)}{sinp\pi }\right]$

Step-by-step solution

Define Bessel’s equation.

Let the Bessel equation be $x^{2}y\text{'}\text{'}+xy\text{'}+\left(x^2-n^2\right)y=0$. This equation has two linearly independent solutions for a fixed value of n. A Bessel equation of the first kind, indicated by J_n(X), is one of these solutions that may be derived using Frobinous approach.

$$\begin{gathered} y_1=x^a{J}_p\left(b{x}^c\right) \\ {y}_2=x^a{J}_{-p}\left(b{x}^c\right) \end{gathered}$$

At X=0, this solution is regular. The second solution, which is singular at X=0 , is represented by Y_n(X) and is called a Bessel function of the second kind.$y_3=x^a\left[\frac{cosp\pi J_p\left(b{x}^c\right)-J_{-p}\left(b{x}^c\right)}{sinp\pi }\right]$

Determine the general solution of the DE.

Find the solution of the BVP.

(b)

Substitute the value Y((L)=0 in the equation (5).

$$\begin{gathered} c_1\sqrt{x}{J}_1\left(2\sqrt{PL\times L/M}\right)=0 \\ {J}_1\left(2L\sqrt{P/M}\right)=0 \end{gathered}$$

From table 6.4.1, the first positive zero of J₁ is 3.8317 . Therefore,

$$\begin{gathered} 2L\sqrt{P_1/M}=3.8317 \\ {P}_1=\frac{3.{8317}^{2}M}{4L^2} \\ {P}_1=\frac{3.6705M}{L^2} \end{gathered}$$

Step 4:Find the critical length of the solid steel rod.

(c)

Substitute the value $P_1=\frac{3.6705M}{L^2}$ For c₁=L=1 , the equation (6)

becomes $y_1\left(x\right)=\sqrt{x}{J}_1(3.8317\sqrt{x})$. Using GNU Octave, the code to plot the graph is,

Let the graph be,