3 Engineering Example 5

3.1 Buckling of a strut

The equation governing the buckling load $P$ of a strut with one end fixed and the other end simply supported is given by $tan\mu L=\mu L$ where $\mu =\sqrt{\frac{P}{EI}}$ , $L$ is the length of the strut and $EI$ is the flexural rigidity of the strut. For safe design it is important that the load applied to the strut is less than the lowest buckling load. This equation has no exact solution and we must therefore use the method described in this Workbook to find the lowest buckling load $P$ .

Figure 26

We let $\mu L=x$ and so we need to solve the equation $tanx=x$ . Before starting to apply the Newton-Raphson iteration we must first obtain an approximate solution by plotting graphs of $y=tanx$ and $y=x$ using the same axes.

From the graph it can be seen that the solution is near to but below $x=3\pi ∕2\left(\sim 4.7\right)$ . We therefore start the Newton-Raphson iteration with a value $x_0=4.5$ .

The equation is rewritten as $tanx-x=0$ . Let $f\left(x\right)=tanx-x$ then $f^{\prime }\left(x\right)={sec}^{2}x-1={tan}^{2}x$

The Newton-Raphson iteration is $x_{n+1}=x_n-\frac{tan{x}_n-x_n}{{tan}^2{x}_n},x_0=4.5$

so $x_1=4.5-\frac{tan\left(4.5\right)-4.5}{{tan}^{2}4.5}=4.5-\frac{0.137332}{21.504847}=4.493614$   to 7 sig.fig.

Rounding to 4 sig.fig. and iterating:

$x_2=4.494-\frac{tan\left(4.494\right)-4.494}{{tan}^{2}4.494}=4.494-\frac{0.004132}{20.229717}=4.493410$   to 7 sig.fig.

So we conclude that the value of $x$ is 4.493 to 4 sig.fig. As $x=\mu L=\left(\sqrt{P∕EI}\right)L$ we find, after re-arrangement, that the smallest buckling load is given by $P=20.19\frac{EI}{L^2}$ .

Exercises

1. By sketching the function $f\left(x\right)=x-1-sinx$ show that there is a simple root near $x=2$ . Use two iterations of the Newton-Raphson method to obtain a better estimate of the root.
2. Obtain an estimation accurate to 2 d.p. of the point of intersection of the curves $y=x-1$ and $y=cosx$ .

Answer