Question

The critical loads of thin columns depend on the end conditions of the column. The value of the Euler load ${\mathit{P}}_{\mathbf{1}}$in Example 4 was derived under the assumption that the column was hinged at both ends. Suppose that a thin vertical homogeneous column is embedded at its base $(x=0)$and free at its top$(x=L)$and that a constant axial load P is applied to its free end. This load either causes a small deflection as shown in Figure 5.2.9 or does not cause such a detection $\mathit{\delta }$. In either case the differential equation for the detection $\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is

$\mathit{E}\mathit{I}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\mathit{p}\mathit{y}\mathbf{=}\mathit{p}\mathit{\delta }$

FIGURE 5.2.9 Deflection of vertical column in Problem 24

(a) What is the predicted deflection when$\mathit{\delta }\mathbf{=}\mathbf{0}$?

(b) When$\mathit{\delta }\mathbf{\pm }\mathbf{0}$, show that the Euler load for this column is

one-fourth of the Euler load for the hinged column in

Example 4.

Short answer

Therefore, the final answer is:

1. No deflection
2. $P_0=\frac{1}{4}\left(\frac{{\pi }^{2}EI}{L^2}\right)$
   

Step-by-step solution

to find predicted deflection when δ=0. ?

$c_2=0$(a)According the given

$\frac{d^{2}y}{d{x}^2}+\frac{P}{EI}y=\frac{P}{EI}\delta$

The characteristic equation is

$m^2+\frac{P}{EI}=0$

The roots$\pm \sqrt{\frac{P}{EI}i}$

$y_c\left(x\right)=c_{1}cos\sqrt{\frac{P}{EI}x}+c_{2}sin\sqrt{\frac{P}{EI}x}$

$y_p=\delta$

The general solution is

$y\left(x\right)=c_{1}cos\sqrt{\frac{P}{EI}x}+c_{2}sin\sqrt{\frac{P}{EI}x}+\delta$

Column is embedded at$x=0$,then the initial condition are$y\left(0\right)=y\text{'}\left(0\right)=0$. So that $c_1=-\delta$and

$y\left(x\right)=\delta \left(1-cos\sqrt{\frac{P}{EI}x}\right)$

If $\delta =0$, then $y\left(x\right)=0$and so there is no deflection.

When δ=0 , show that the Euler load for this column isone-fourth of the Euler load for the hinged column

(b)If$\delta \ne 0$

$y\left(x\right)=\delta \left(1-cos\sqrt{\frac{P}{EI}x}\right)$

The boundary condition$y\left(L\right)=\delta$

$$\begin{gathered} \delta =\delta \left(1-cos\sqrt{\frac{P}{EI}}L\right) \\ cos\sqrt{\frac{P}{EI}}L=1 \end{gathered}$$

This give the

$\sqrt{\frac{P}{EI}}L=\frac{(}{n}+\frac{1}{2})\pi$

Where$n=0,1,2,...$ the smallest value$P_n$ is$P_0$

$\sqrt{\frac{P_0}{EI}}L=\frac{\pi }{2}$and$P_0=\frac{1}{4}\left(\frac{{\pi }^{2}EI}{L^2}\right)$

Which is one fourth of the Euler load for the hinged column in example 4

1. No deflection
2. $P_0=\frac{1}{4}\left(\frac{{\pi }^{2}EI}{L^2}\right)$