A question about leverage and stability of shelf support

[confusedlabs]

TL;DR

How do the principals of leverage apply to the stability of a bookshelf when the length of the shelf supports changes?

In this picture, I depicted a shelf that has two sides (black). On each side, there are holes for the shelf supports (red). They can be inserted into the holes and are supposed to hold a bookshelf (blue).

My question is the following: While the depth of the holes stays the same, does the length of the part of the shelf supports that stick out affect the stability of the construction?

On one hand, I thought that when weight is applied to just the point of the shelf support that is furthest from the hole, there will be more leverage, like when you sit on the outermost part of a seesaw.
However, I just had a hunch that this situation might be different from that.

 

[.Scott]

A vertical load applied to a shelf supported by four pins (one on each corner) will be distributed among those pins in proportion to the position of the load on the shelf. So a load placed directly over a pin will be supported only by that pin. A load placed 1/3 of the way towards center from the right and 1/4 of the way back will will be supported as follows:
2/3 of the load will we supported by the pins on the left 1/3 by the two pins on right.
3/4 of the load will supported by the front pins, 1/4 by the back.
For individual pins, you would multiply those values. So, for example, the front left pin would support (2/3)(3/4) = 1/2 the load.

"Stability" suggests that the load may not be properly supported. For example, the pins are probably not along the front or back of the edge. Rather, they are probably set in by 3 or 4 cm. So any load placed beyond those bounds will tend raise the opposite edge of the shelf. If that load is heavy enough, it will tip the shelf.

 

[confusedlabs]

A vertical load applied to a shelf supported by four pins (one on each corner) will be distributed among those pins in proportion to the position of the load on the shelf. So a load placed directly over a pin will be supported only by that pin. A load placed 1/3 of the way towards center from the right and 1/4 of the way back will will be supported as follows:
2/3 of the load will we supported by the pins on the left 1/3 by the two pins on right.
3/4 of the load will supported by the front pins, 1/4 by the back.
For individual pins, you would multiply those values. So, for example, the front left pin would support (2/3)(3/4) = 1/2 the load.

"Stability" suggests that the load may not be properly supported. For example, the pins are probably not along the front or back of the edge. Rather, they are probably set in by 3 or 4 cm. So any load placed beyond those bounds will tend raise the opposite edge of the shelf. If that load is heavy enough, it will tip the shelf.

Hi, Thanks for the interesting info. However, I’m not sure if this answers my question. I want to know if the length of the pins (marked with a double arrow) affects their ability to hold the shelf. I think it doesn’t, but my friend said that if they are longer, leverage can cause them to break or cause damage to the holes in which they are put.

 

[.Scott]

For the most part, the load will be applied as a shear force - which is not dependent on the length of the pin. However, as a practical matter, the edge of the shelf against the top of the pin will not align perfectly with the edge of the hole on the bottom of the pin - and that will tend to turn the pin. So, the pin needs to be long enough and extent far enough into the hole to keep that from happening. It would depend on the strength of the wall material (black in your drawing) and how snugly the pin fits into the hole, but as an approximation, if the pin extends into the hole by 3 times its diameter, you're probably all set. Further than that likely won't help or hurt.
In the other direction, the pin needs to extend out from the wall to the shelf far enough to give the shelf a secure perch. Assuming the pin itself is very strong, the shelf material and the load will determine this. Expect the bottom of the shelf to deform slightly to allow the load to be distributed over a small area in contact with the pin. Without this deformation, the contact area would be zero and the contact pressure would be infinite. Of course, those pins are often made with a flat surface that can be faced up to the bottom of the shelf to reduce this deformation.
As far as the distance that the pin extends inward (your main point, I believe), there are two cases:
1) if the pin extends more than a few diameters inward, either the extra distance will be unused (supporting no or very little of the load), or it will be supporting load (perhaps because the shelf is too narrow or the walls are flexing outward). If it is supporting load, the hole supporting that pin is at a mechanical disadvantage in supporting the load. In either case, the extra length is either unneeded or it may lower the life of the shelf.
2) in the extreme case, the pin extends inward all across the bottom of the shelf to the hole on the opposite side. It would still be best for the shelf to fit snugly between the walls. But, so long as that rod does not bend, this configuration would be better than long pins that do not connect.

 

[berkeman]

I want to know if the length of the pins (marked with a double arrow) affects their ability to hold the shelf.

From looking at typical shelf support pins, it looks like the supporting part (usually flat) extends out about the same amount as the pin that goes into the sidewall. Making the supporting part longer would not have any advantage (as long as the shelf material is strong enough), and would just cost more money.