# How to calculate what size steel RHS to use?

I am welding up a rack to support a distributed load of up to 1000kg, what is the process to work out what steel section size I should use? Many Thanks.

• Hi Swag_B. Welcome to Engineering. your question requires more details like span of the shelves, are the shelves welded, what type of material are you using etc. – NMech May 4 at 6:17

Because the weight is quite high, and the use is for supports, I will assume that you will use the same cross-section for the shelve columns (if there are any). In you use the same cross-section, you probably will not have a problem with buckling

I am assuming RHS refers to Rectangular Hollow Cross-section.

Regarding the process that you need to follow IMHO for the shelves is:

1. Define the allowable deflection $$\delta_{all}$$ (usually its about $$\frac{1}{200}L_{\text{shelves span}}$$ and the allowable stress (usually its safety factor times the yield stress $$N\cdot \sigma_y$$)

2. Determine the support type. Are the shelves simply supported or are they welded on the frame. In any case I'll be using the formulas for simply supported beams.

• max deflection: $$\delta_{max} = \frac{5 wl^4}{384EI}$$
• max operating stress: $$\sigma = \frac{wl^2}{8I}\cdot \frac{y_{max}}{2}$$

where:

• $$E$$ is the young's Modulus
• $$I$$ is the second moment of area which you can find in tables or calculate
• $$y_{max}$$ is the height of the RHS.
1. you can solve for I (which is depended on the geometric properties of the cross-section), for both constraints (Deflection and stress) and you will obtain two parameters:

$$I_{Deflection} = \frac{5 wl^4}{384E\delta_{max}}, \qquad \frac{2 I}{y_{max}}= \frac{wl^2}{8\sigma_{Allowed}}$$

UPDATE:

The equations for the fixed ends (see welded shelves on the frame) are:

• max deflection: $$\delta_{max} = \frac{wl^4}{384EI}$$
• max operating stress: $$\sigma = \frac{wl^2}{12 I}\cdot \frac{y_{max}}{2}$$

As you can see they are less stringent (that is why I opted for the simple beam), because that would built in a safety factor.

• Hi NMech. That's great, I appreciate you taking the time to help out. I'll work through the process you have described. Just one extra question, is there a way to factor in the welded joints? Thanks again! – Swag_B May 5 at 20:34
• @Swag_B I update the post and at the end I used the updated formulas for the fixed end conditions. As you can see they are less stringent than the simple supported case. You can solve for the parameters in a similar manner as in the original. – NMech May 5 at 20:45