I'd like to mount a shelf on the wall. I have two shelf supports to do so, like this:

Relative to the shelf, where do I place the (blue) supports to achieve the best distribution of load?

  • 5
    $\begingroup$ There are quite a few details that would be useful to add to this problem statement. How do the supports attach to the wall? Are they physically attached to the shelf or is it resting freely on top of them? Do they extend out to the full depth of the shelf? What is the intended use of the shelf (i.e., light things, heavy things, a single item, your cat)? Of what material are the shelf and supports made? Are the supports rated to handle a certain load? What are the dimensions of the shelf (or at least its own weight)? $\endgroup$ – Air Dec 27 '16 at 20:26

As with all good things, it depends.

If you can assume that your supports are totally stiff and that the loading on the shelf will be approximately uniform, then you basically have the following structure:

enter image description here

A rectangular cross-section (such as a plank) will behave equally under positive or negative bending moment, so your objective should be to balance both. To do so, you want your main span to be $2\sqrt2 \approx 2.83$ times the cantilevers. This is found by calculating the cantilever required to offset half of the bending moment due to a uniform load along a simply supported beam:

$$\begin{align} M_{mid} &= \dfrac{qL_{mid}^2}{8} - M_{cant} \\ M_{cant} &= \dfrac{qL_{cant}^2}{2} = \dfrac{M_{mid}}{2} = \dfrac{qL_{mid}^2}{16} \\ \therefore 8L_{cant}^2 &= L_{mid}^2 \\ 2\sqrt2 L_{cant} &= L_{mid} \end{align}$$

If your shelf may be empty, except for a single concentrated load anywhere, then you have to find the points for the supports where the maximum bending moment due to a concentrated load at the edges will be equal to that of a load at midspan.

enter image description here

$$\begin{align} M_{mid} &= \dfrac{PL_{mid}}{4} \\ M_{cant} &= PL_{cant} = M_{mid} = \dfrac{PL_{mid}}{4} \\ \therefore 4L_{cant} &= L_{mid} \\ \end{align}$$

Taking an average of the two cases above, I'd adopt a main span some 3.5 times larger than the cantilevers as a first guess.

If your supports are loose, meaning they won't hold the shelf if it is raised, so there's the risk of it toppling over with too much load on one cantilever, then you need to calculate the maximum load the weight of your shelf can resist. There's too many variables to figure this out trivially, though (linear weight of the shelf, maximum adopted weight, etc). And, honestly, if your supports are loose, you really should just put the supports at the ends, eliminating this risk entirely (but therefore reducing the maximum load the shelf will resist).

Diagrams obtained with Ftool, a free 2D frame analysis program.


There are several criteria you need to consider:

  1. To distribute the load most evenly between the two supports, put them as far apart as possible. That would be at the ends in this case.

  2. The worst case heavy object placed at one end with nothing else on the shelf has to not flip the shelf over. This puts a limit on how far from the ends the supports can be.

  3. The strength of the shelf needs to be higher to support the same load as the supports are moved farther apart. In this case, consider the worst case load placed exactly at center. Even if the shelf is strong enough to not be damaged by such a load, the resulting bend may be objectionable to you.

Only you can decide how these various competing criteria trade off in your case.

  • $\begingroup$ A correction: contrary to #1, so long as the supports are placed symmetrically, the load will always be evenly distributed between them. And #2 only applies if the supports are loose (can't handle tension). $\endgroup$ – Wasabi Dec 27 '16 at 19:00
  • 2
    $\begingroup$ @Was: For your #1 to be true, the load on the shelf has to be evenly distributed on the shelf. This is unrealistic for a unspecified "shelf". I agree about #2. However, many shelf supports are just something for the shelf to rest on, so tipping is a issue. Even if not, then the support that doesn't have negative load gets more than the full load, so has to be designed more robustly. I kept it simple and didn't go into all that, which seemed to be more in line with the level of the question. $\endgroup$ – Olin Lathrop Dec 27 '16 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.