Question

Suppose that a uniform thin elastic column is hinged at the end$\mathit{x}\mathbf{=}\mathbf{0}$and embedded at the end$\mathit{x}\mathbf{=}\mathit{L}\mathbf{.}$

(a) Use the fourth-order differential equation given in Problem 25 to find the eigenvalues${\mathit{\lambda }}_{\mathbf{n}}\mathbf{,}$the critical loads${\mathit{P}}_{\mathbf{n}}\mathbf{,}$the Euler load${\mathit{P}}_{\mathbf{1}},$and the deflections

${\mathit{y}}_{\mathbf{n}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{.}$

(b) Use a graphing utility to graph the first buckling mode.

Short answer

Therefore, the solution is

$$\begin{gathered} \left(a\right)20.1907EI/L^2. \\ \left(b\right)y_n\left(x\right)=c_2\left(\sin \frac{{\beta }_{n}x}{L}-\frac{{\beta }_n\cos {\beta }_n}{L}x\right) \end{gathered}$$

Step-by-step solution

Use the boundary value problem

We have the boundary-value problem is

$\frac{d^{4}y}{d{x}^4}+\frac{P}{EI}\frac{d^{2}y}{d{x}^2}=0,y\left(0\right)=y^{\text{'}\text{'}}\left(0\right)=y\left(L\right)=y^{\text{'}}\left(L\right)=0$

As in problem 25,

$y\left(x\right)=c_1\cos \sqrt{\frac{P}{EI}}x+c_2\sin \sqrt{\frac{P}{EI}}x+c_{3}x+c_4$

Using the initial conditions $y\left(0\right)=y^{\text{'}\text{'}}\left(0\right),$we get $c_1=c_4=0.$Simply

$y\left(x\right)=c_2\sin \sqrt{\frac{P}{EI}}x+c_{3}x.$

Since$y\left(L\right)=y^{\text{'}}\left(L\right)=0,$ hence

$c_2\sin \sqrt{\frac{P}{EI}}L+c_{3}L=0$

And

$c_2\frac{P}{EI}\cos \sqrt{\frac{P}{EI}}L+c_3=0$

Using crammer method

Using crammer method to solve the previous equations, we shall makes

$$\begin{gathered} \left|\sin \sqrt{\frac{P}{EI}}LL \\ \frac{P}{EI}\cos \sqrt{\frac{P}{EI}}L1\right|=0 \end{gathered}$$

to get a nontrivial solution. So that $\tan \frac{P}{EI}L=\frac{P}{EI}L.$Let${\beta }_n,n=1,2,3\cdots$be the positive roots of nonlinear equation $\tan x-x=0.$Therefore,

$P_n=\frac{{\beta }_n^{2}EI}{L^2},n=1,2,3\cdots$

The eigenvaluesis given by

${\lambda }_n=\frac{P_n}{EI}=\frac{{\beta }_n^2}{L^2},n=1,2,3\cdots$

With the aid of a CAS we find that the first positive root of $\tan x-x=0$is (approximately) ${\beta }_1$= 4.4934,and so the Euler load is (approximately)$P_1=20.1907EI/L^2.$

Mapping the graph

(b) Finally, if we use$c_3=-c_2\frac{P}{EI}\cos \frac{P}{EI}L,$then the deflection curves are

$y_n\left(x\right)=c_2\left(\sin \frac{{\beta }_{n}x}{L}-\frac{{\beta }_n\cos {\beta }_n}{L}x\right)$

Choose $L={\beta }_1$and $c_2=1,$we get

$y_1\left(x\right)=\sin x+0.2172x$