buckling

[logistic_guy]

A \(\displaystyle 50\)-\(\displaystyle \text{in.}\)-\(\displaystyle \text{long}\) rod is made from a \(\displaystyle 1\)-\(\displaystyle \text{in.}\)-\(\displaystyle \text{diameter}\) steel rod. Determine the critical buckling load if the ends are fixed supported. \(\displaystyle E = 29(10^3) \ \text{ksi}\), \(\displaystyle \sigma_Y = 36 \ \text{ksi}\).

 

[khansaheb]

A \(\displaystyle 50\)-\(\displaystyle \text{in.}\)-\(\displaystyle \text{long}\) rod is made from a \(\displaystyle 1\)-\(\displaystyle \text{in.}\)-\(\displaystyle \text{diameter}\) steel rod. Determine the critical buckling load if the ends are fixed supported. \(\displaystyle E = 29(10^3) \ \text{ksi}\), \(\displaystyle \sigma_Y = 36 \ \text{ksi}\).

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[logistic_guy]

To warm up the audience, we start with Euler’s buckling formula:

\(\displaystyle P_{cr} = \frac{\pi^2EI}{L_e} = \frac{\pi^2EI}{KL}\)

And the most beautiful thing about this formula is the presence of \(\displaystyle I\) which is the second moment of area. You should know from now and on that any problem involves moment of inertia is fun to solve.

 

[logistic_guy]

We start by finding the second moment of area, \(\displaystyle I\).

\(\displaystyle I = \int y^2 \ dA = \int\int r^2\sin^2\theta \ r \ dr \ d\theta\)

Or

\(\displaystyle I = \int_{0}^{2\pi}\int_{0}^{r} r^3\sin^2\theta \ dr \ d\theta = \frac{r^4}{4}\int_{0}^{2\pi} \sin^2\theta \ d\theta\)


\(\displaystyle = \frac{r^4}{8}\int_{0}^{2\pi} (1 - \cos 2\theta) \ d\theta = \frac{r^4}{8}\left(\theta - \frac{\sin 2\theta}{2}\right)\bigg|_{0}^{2\pi}\)


\(\displaystyle = \frac{r^4}{8}\left(2\pi - \frac{0}{2}\right)\)

Or

\(\displaystyle I = \frac{\pi r^4}{4}\)

 

[logistic_guy]

For a fixed rod, we take \(\displaystyle K = 0.5\).

Determine the critical buckling load

Plug in numbers.

\(\displaystyle P_{cr} = \frac{\pi^2EI}{K^2L^2} = \frac{\pi^2E\pi r^4}{4K^2L^2} =\frac{\pi^2\left(29 \times 10^3\right)\pi\left(\frac{1}{2}\right)^4}{4(0.5^2)50^2} = \textcolor{blue}{22.47955 \ \text{kip}}\)