Question

Buckling of a tapered column.

In example 4 of section 5.2 we saw that when a constant vertical compressive force or load P was applied to a thin column of uniform cross section, the deflection y(x) was a solution of the boundary-value problem.

$\mathbf{El}\frac{{\mathbf{dy}}^{\mathbf{2}}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{Py}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{L}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}$

The assumption here is that the column is hinged at both ends. The column will buckle or deflect only when the compressive force is a critical load P_n.

1. In this problem let us assume that the column is of length L, is hinged at both ends, has circular cross sections, and is tapered as shown in Figure 6.3.1(a). If the column, a truncated cone, has a linear taper y = cxas shown in cross section in Figure 6.3.1(b), the moment of inertia of a cross section with respect to an axis perpendicular to the-plane isxy, $\mathbf{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}{\mathbf{\pi r}}^{\mathbf{4}}$where r = yand y =cx. Hence we can write$\mathbf{l}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{l}}_{\mathbf{0}}{\left(\frac{x}{b}\right)}^{\mathbf{4}}$ where${\mathbf{l}}_{\mathbf{0}}\mathbf{=}\mathbf{l}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{\pi }\mathbf{(}\mathbf{cb}{\mathbf{)}}^{\mathbf{4}}$.Substitutinginto the differential equation in (24), we see that the deflection in this case is determined from the.

${\mathbf{x}}^{\mathbf{4}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{\lambda y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}$

where $\mathbf{\lambda }\mathbf{=}\frac{{\mathbf{Pb}}^{\mathbf{4}}}{{\mathbf{El}}_{\mathbf{0}}}$. Use the results of problem 33 to find the critical loads P_n for the tapered column. Use an appropriate identity to express the buckling modes y_n(x) as a single function.

b) Use a CAS to plot the graph of the first buckling mode y(x) corresponding to the Euler load when $\mathbf{b}\mathbf{=}\mathbf{11}\mathbf{,}\mathbf{a}\mathbf{=}\mathbf{1}$.

Short answer

The answer is:

${\mathrm{P}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2{\mathrm{\pi }}^2{\mathrm{a}}^2\mathrm{El}}{{\mathrm{L}}^2{\mathrm{b}}^2}{\mathrm{y}}_{\mathrm{n}}\left(\mathrm{x}\right)={\mathrm{C}}_3\mathrm{xsin}\frac{\mathrm{n\pi ab}}{\mathrm{L}}\left(\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{a}}\right)$

Step-by-step solution

Step 1: Part a):

In problem 33, we found that

$\mathrm{y}\left(\mathrm{x}\right)={\mathrm{C}}_1\mathrm{xsin}\left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{x}}\right)+{\mathrm{C}}_2\mathrm{xcos}\left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{x}}\right)...\left(1\right)$

Is the solution of the equation

$x^{4}y\text{'}\text{'}+\lambda y=0...\left(2\right)$

Substituting the boundary values

$y\left(a\right)=0,y\left(b\right)=0...\left(3\right)$

Into (1), we obtain

$$\begin{gathered} C_{1}a\sin \left(\frac{\sqrt{\lambda }}{a}\right)+C_{2}a\cos \left(\frac{\sqrt{\lambda }}{a}\right)=0...\left(4\right) \\ {C}_{1}b\sin \left(\frac{\sqrt{\lambda }}{b}\right)+C_{2}b\cos \left(\frac{\sqrt{\lambda }}{b}\right)=0...\left(5\right) \end{gathered}$$

We would like to find non-trivial constants C₁, c₂ such that the previous two equations are true. (4) and (5) form a homogeneous linear system. Such system has a non-trivial solution if the determinant of its matrix of coefficients is equal to zero, that is if

$$\begin{gathered} \left|\sin \left(\frac{\sqrt{\lambda }}{a}\right)a\cos \left(\frac{\sqrt{\lambda }}{a}\right) \\ b\sin \left(\frac{\sqrt{\lambda }}{b}\right)b\cos \left(\frac{\sqrt{\lambda }}{b}\right)\right|=0 \end{gathered}$$

…(6)

Equation (6) is equivalent to

localid="1664865592899" $\begin{array}{rcl}0& =& \mathrm{absin}\left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{a}}\right)\cos \left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{b}}\right)-\mathrm{abcos}\left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{a}}\right)\sin \left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{b}}\right)\\ & =& \mathrm{absin}\left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{a}}-\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{b}}\right)=\mathrm{absin}\left(\frac{\mathrm{b}-\mathrm{a}}{\mathrm{a}}\sqrt{\mathrm{\lambda }}\right)\end{array}$

Assuming$a\ne b\ne 0,$the previous condition will be true if

$\frac{b-a}{ab}\sqrt{\lambda }=n\pi ,n$is integer.

Which is equivalent to

$\sqrt{\lambda }=\frac{n\pi ab}{L}◃▹\lambda =\frac{n^2{\pi }^2{a}^2{b}^2}{L^2}=\frac{P_n{b}^4}{EI},n=1,2,...\left(7\right)$

Therefore, the critical loads are given by

$P_n=\frac{n^2{\pi }^2{a}^{2}EI}{L^2{b}^2}...\left(8\right)$

From (4), we get

$C_2=-C_1\frac{\sin \left(\frac{\sqrt{\lambda }}{a}\right)}{\cos \left(\frac{\sqrt{\lambda }}{a}\right)}$

Substituting it into (1), rewriting it, and using the fact that $\cos \left(\frac{\sqrt{\lambda }}{a}\right)$ is actually a constant, we obtain:

localid="1664865490185" $\begin{array}{rcl}\mathrm{y}\left(\mathrm{x}\right)& =& {\mathrm{C}}_1\left[\mathrm{xsin}\left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{x}}\right)-\mathrm{x}\frac{\sin \left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{a}}\right)}{\cos \left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{b}}\right)}\cos \left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{x}}\right)\right]\\ & =& {\mathrm{C}}_3\mathrm{x}\left[\sin \left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{x}}\right)\cos \left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{a}}\right)-\cos \left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{x}}\right)\sin \left(\frac{\sqrt{\mathrm{\lambda }}}{\mathrm{a}}\right)\right]\\ & =& {\mathrm{C}}_3\mathrm{xsin}\sqrt{\mathrm{\lambda }}\left(\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{a}}\right)\end{array}$

Using (7), we have

localid="1664865521540" role="math" ${\mathrm{y}}_{\mathrm{n}}\left(\mathrm{x}\right)={\mathrm{C}}_3\mathrm{xsin}\frac{\mathrm{n\pi ab}}{\mathrm{L}}\left(\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{a}}\right)$

Step 2: Part (b):

Here is the plot of y₁(x) for b = 11, a = 1, C₃=-1