# How to size a rectangular steel tube to hold certain weight?

I would like to build a shelf (1m width and 0.5m depth) using only four rectangular steel tubes on for legs. Each legs are at the end of the rectangular tube. I would like to put 500kg on the tubes, which are equally distributed. How can I choose the width, height and wall thickness to ensure that the steel tubes are within a certain deflection?

There are design aids which tabulate in detail all the section properties of hollow structural sections, HSS, and have online calculators and you find many if you google HSS sections.

If we disregard the roundness on the corners, which is not a big factor:

We assume a maximum deflection of $$\dfrac{\ell}{180} = \dfrac{1000}{180} = 5.5\text{ mm}$$

\begin{align} M &=\frac{\omega l^2}{8} \\ &= 500/1000 * 1000^2/8 \\ &=62500kg-mm2 \end{align}

Adopting ASTM A36 steel with the following properties:

\begin{align} E &= 200\text{ GPa} = 20400\text{ kg-mm}^2\\ F_a &= \dfrac{1}{2} F_y = \dfrac{1}{2} 210\text{ MPa} = \dfrac{1}{2} 21.4 = 10.7\text{ kg-mm}^2 \end{align}

Lets pick up an HSS section with dimensions $$A_{base}, 2A_{height}, A/10_{thickness}$$

\begin{align} \sigma &= \frac{mc}{I} \\ I &= \frac{A\cdot(2A)^3 - 0.8A\cdot (1.6A)^3)}{12} = \frac{0.72A^4}{12} \\ \sigma &= 10.7 = \frac{62500A \cdot 12}{0.72A^4} = \frac{ 62500 \cdot 12}{0.72A^3} \\ A^3&=9786.7 \\ A &= 21\text{ mm, pick } 25\text{ mm} \end{align}

Now we need to plug this $$A$$ and check its deflection:

\begin{align} y &= \dfrac{5\omega\ell^4}{384EI} \\ &=\frac{5 \cdot 0.5 \cdot 1000^4}{384 \cdot 20400 \cdot 23437} \\ &= 13.6\text{ mm} > 5.5\text{ mm} \end{align}

That deflection is not okay.

So we pick one size larger and plug the numbers again until the deflection is less than 5.5 mm.

Note: We Used kg-force as opposed to N and millimeter versus meter and also we did not use a factor of safety, even though taking $$F_a=1/2F_y$$ is some measure of that. Also we picked thickness as 1/10 base, which is not necessarily true, just as a roadmap. Check my numbers please.

• Aren't HSS sections simply hollow rectangular sections? Why is the "hollow height" in your inertia calculation equal to $1.6A$ instead of $(2 - 2\cdot0.1)A = 1.8A$? – Wasabi Oct 9 at 1:13
• @Wasabi, my apologies. you are correct in both cases. I will edit my answer. As I Said in my answer I just tried to set the steps, but I really appreciate you spending time and combing through the admittedly not a transparent answer. – kamran Oct 9 at 3:50