2
$\begingroup$

I'd like to mount a shelf on the wall using three shelf supports. Where should I place the three supports for best distribution of load across the shelf?

Edit: To clarify, the load will be distributed and I want to optimize for maximum load, not minimum sag.

$\endgroup$
4
  • 3
    $\begingroup$ Depends on what your load looks like $\endgroup$
    – towe
    Feb 8, 2022 at 11:53
  • 1
    $\begingroup$ to minimize sag: put supports directly under your load... or put your load on top of your supports... $\endgroup$
    – Abel
    Feb 8, 2022 at 11:55
  • 1
    $\begingroup$ What is it that you are trying to optimize? is it deflection (visible sag) or strength (maximum load). Also is your load distributed or is it concentrated? $\endgroup$
    – NMech
    Feb 8, 2022 at 12:33
  • $\begingroup$ divide the shelf in 4 segments, put the brackets at the division points. $\endgroup$
    – Tiger Guy
    Feb 8, 2022 at 14:03

4 Answers 4

1
$\begingroup$

Assuming a uniformly distributed load throughout the entire shelf, you want to position your supports such that your cantilevers are 40-45% of your central spans.

Now to show my work. For starters, we're obviously going to want our supports to be symmetrically placed. That means we know that one of our supports will be at the middle of the shelf. Shelf supports are usually best described as hinged supports (allowing for a bit of rotation), but since we're dealing with a symmetrical structure, we also know that the rotation at the middle of the shelf must also be zero.

This means we can simplify our structure and think of only one half of it, which we can model as a fixed-and-hinged beam with a cantilever. The image below exemplifies how a three-support beam is identical to two fixed-and-hinged beams:

enter image description here

So now we only have one degree of freedom: where we put that hinged support. And we want to optimize the structure's load-bearing capacity. In this case, there are really two ways for the shelf to collapse: the shelf itself breaks apart (almost always due to excessive bending moment) or the supports break off.

To maximize the shelf's bending moment capacity, we need to minimize its largest bending moment. As it happens, a fixed-and-hinged beam with a cantilever will always have a negative bending moment at the hinged support and either a positive or negative bending moment at the fixed support (our central support in the real beam).

However, it is mathematically impossible for the positive bending moment to be greater (in magnitude) than the negative at the hinged support, so really we just want to position our supports such that the bending moment at both supports are equal.

The math for this is a pain in the neck, so I'm going to skim over it here. The first thing we need to determine is the vertical force generated by the hinged support, which we eventually derive is equal to:

$$\begin{align} P &= w\dfrac{a^4 - 4\ell^3a + 3\ell^4}{8(\ell - a)^3} \\ &= w\dfrac{a^2 + 2a\ell + 3 \ell^2}{8(\ell - a)} \end{align}$$

where $w$ is the applied load; $a$, the length of the cantilever; and $\ell$, the total beam span (cantilever plus the rest of the beam).

We can then calculate the bending moments at both supports:

$$\begin{align} M^-_a &= -\dfrac{wa^2}{2} &&\text{ (hinged support)}\\ M^-_b &= -\dfrac{w\ell^2}{2} - P(\ell - a) &&\text{ (fixed support)} \end{align}$$

Now we just need to define the value of $a$ such that those two are equal. This is more annoying algebra, but we end up getting that the optimal value is

$$a = \dfrac{\ell}{1 + \sqrt6} \approx 0.29\ell$$

Since $\ell$ is the entire beam span, we can also conclude that $a = \dfrac{a}{1 - a} = \dfrac{1}{\sqrt6} \approx 41\%$ of the central span.

We can therefore see that indeed, with these dimensions the maximum bending moment is equal at all critical points (forgiving rounding error), ensuring the shelf is resisting bending moment optimally:

enter image description here

However, as I mentioned at the start, we also need to think about how much the supports resist. If our layout is such that the shelf doesn't collapse under bending moment, it may still be overloading one of our supports. So another constraint we have to think about is minimizing the maximum load each support must resist.

As it happens, however, that 41% is still a pretty good choice for this constraint as well. As the image above demonstrates, the load resisted by each of the supports is already quite similar* (~10% difference). I won't get into the math for this case, but $a$ has to be ~44% of the central span for the supports to all have equal reactions.

Obviously, if your shelf has to deal with other cases such as concentrated loads in only a few positions, etc, then the conclusion may be significantly altered, but an $a$ between 40-45% of the central span means your shelf will be well prepared to withstand the greatest uniformly distributed load possible.


* The bending moments should be identical and are slightly different merely due to rounding (the cantilever is 41% of the central span, when it should actually be $1/\sqrt6$). The support reactions are in fact different.

$\endgroup$
1
$\begingroup$

As a quick and dirty solution, I would install the supports equally spaced with the two ends hanging out ( cantilever) 1/3 of the distance between the supports.

This way you can put maximum load on the shelf. Because the max positive moments at the middle of the spans are approximately $0.25 \ \frac{\omega L^2}{8} \ and \ $

The moments at supports are going to be approximately $0.1 \ \frac{\omega L^2}{8} $ Which is optimal use of material and not too much of deflection.

$\endgroup$
1
  • $\begingroup$ Yeah, this is the best approach to reduce the reaction of the middle support, in turn, it maximizes the applied load without overstressing the shelf board. $\endgroup$
    – r13
    Feb 8, 2022 at 21:44
0
$\begingroup$

Consider the shelf as three hinged or loosely connected parts:

   L/3       L/3       L/3
---------+---------+---------
    ^         ^         ^
   L/3       L/3       L/3

Figure 1. Most even.

You would support each section in its middle. This gives balanced load with no turning moment onto any of the other supports.

      L/2            L/2
--------------+--------------
^             ^             ^
L/4          L/2          L/4

Figure 2. An extreme example.

Any other arrangement will result in unbalanced loading.

$\endgroup$
0
$\begingroup$

Well, what do you mean by 'optimize it for maximum load'? It can mean a variety of things here. For example, determining the support locations to optimize it for maximum load can mean you want to do it for the max load which will not result in plastic failure of the shelf, or such that the shelf will not encounter any crack formation or fracture during loading, or such that the deformation of the shelf can be reduced to a minimum, or such that the support locations will result in minimum weight of the shelf by taking maximum load etc.

Anyhow, if you want to optimize it for any of the above cases, it is needed to know the initial dimensions of the shelf, material that the shelf is made up of, and the direction/location/type of loads that this shelf will be subjected to.

I am going to assume that you meant a shelf like this:

enter image description here

The supports are always going to be like this, i.e. one on left end, one in the middle and one on the right end of the shelf. And also, the material used for the shelf is ductile (so plastic deformation is a failure for you, not crack formation). Basically, now you want to find out how much more you can reduce the length of the shelf such that it can take the maximum distributed load (kinda like books of equal mass spread out to cover the complete surface of the shelf). I am also going to assume that you are BOLTING/GLUING the shelf's brackets (i.e. supports) to the walls. [This is important].

Now, the plastic deformation (failure for your case) occurs when the stresses within a certain region of the shelf exceeds the yield strength of the material that the shelf is made up of. So this yield strength is the maximum stress that any region within the shelf can take, and is a known constant value for ductile materials. The stresses within the shelf become higher when you increase the amount of force acting on the shelf or when you increase the distance between the supports. [you can study the relation between stresses with forces and support distances from various books to find out why, prominent being Roark's Formulas for Stress and Strain]. Now, you want to keep the max stress within any region of the shelf to be constant at yield strength (since you don't want it to plastically fail), but by increasing the force acting on the shelf, so what do you do? You reduce the length of the shelf. This way, the supports at the shelf's ends come closer and closer to each other. This means that the max stresses within the shelf will remain the same but the maximum distributed load acting on the shelf's surface can be increased, since the distance has decreased. Ofcourse, there is a practical limit to how much you can decrease the length of the shelf, for example, it cannot be lower than the dimensions of one book. This implies that you can put less number of books side by side on a more lengthy shelf, but more books [like on top of each other] in total (or in other words, more load) on a less lengthy shelf.

You are optimizing the shelf to take more distributed load, but indirectly, you are also optimizing it for the weight of the shelf by reducing it, but this is only when the supports are attached to the shelf as I have mentioned.

In this case, the maximum deformations won't change since the maximum stresses are not changing. However, if you want to optimize it for deformation, then you have to adjust how the load is put on the shelf. It cannot be distributed anymore; it has to concentrated and exactly concentrated at the locations where the supports are attached. The support locations doesn't matter then.

Now, lets come to the other option when you cannot change the length of the shelf but can change the location of the supports on the shelf. You have to now basically perform an iteration to check supports at which locations will not give stresses within the shelf to be bigger than the yield strength of the material. You have to try multiple locations on the shelf for all the three supports, keep on finding the max distributed force that this system can take without suffering plastic failure. This becomes an extremely gruelsome task.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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