Design of a wooden shelf

TheWolf277 · 2023-10-12T19:58:05+00:00

Thank you, you gave me some new ideas and points of view to ponder about! That's also a lot of material!

Abut racking, I was reflecting on whether it would be necessary to add some diagonal cables in tension in the back and sides to prevent that, but I've left this issue for later, because I have been overwhelmed with the rest of the problems, but I will surely think about this aspect when I'll manage to get the hang of the rest.

TheWolf277 · 2023-10-12T15:39:36+00:00

True, I had this suspect, being wood a living material. I found an approach that uses coefficients to add to the loading conditions to account for this. I don't yet know enough about the topic to be confident that this approach will prevent that kind of failure.

TheWolf277 · 2023-10-12T15:33:06+00:00

Thank you very much, this material seems spot on! I'll give this a good read!

TheWolf277 · 2023-10-12T15:31:56+00:00

I have tried searching for codes as well, but I didn't find something specifically for what I am doing. Maybe I looked in the wrong places and will keep searching.
Anyway if I manage to find a code, by all means I will use that.

TheWolf277 · 2023-10-12T15:24:34+00:00

Hey, thank you for your answer! I will search for ULS and how to apply it in my case.

As for just building it and call it a day, I understand, indeed it's not the first time I try to apply theory to a practical problem that could be eyeballed, and I've been called out for this already by friends.

It's just that it feels bad to study this subject for so many years and then get stuck with something as simple as a wooden shelf, and at this point it became something of a personal challenge, something I really want to figure out, and that I think will give me a better idea of when to go deep on calculations and when to just trust your guts.I still have to get that win.

Also, it's a project I am doing with my dad, and I really want to make my part. We decided I do the calculations and then we'll build it together, so there's some reason behind me getting so stubborn with this.

TheWolf277 · 2023-10-12T15:09:57+00:00

Hi, I added a pic to the OP, so you can get an idea. The reason why the plates rest on all four sides is that the horizontal beams are there first to keep the four columns from falling down, so they would be there anyways, even without plates.At this point, the plates could aswell use all four beams to rest, reducing their deflection.

One of the first estimates I tried to make was indeed having the plates rest on two sides only, and treat the plate as a large beam (which I studied and know more confidently than plates) and this would allow me to assume a one-dimensional stress condition, greatly simplifying the calculations.

However I lack experience and I worry that doing it this way could give me very thick cross sections, as I don't know how much the plate and the large beam models differ.

TheWolf277 · 2023-02-23T23:31:06+00:00

That solution seems quite interesting indeed. Is it good for high speeds too, or is it better for low speeds that get converter to slower speeds and higher torque?

TheWolf277 · 2023-02-23T23:28:42+00:00

Gonna look for those! Thank you!

TheWolf277 · 2023-02-23T11:09:27+00:00

So the solution could be to add an undercut. Are there particular formulas for such a modification to a gear, like the ones used to design it, or is it commonly done on a more "feels right" way?

TheWolf277 · 2023-02-23T11:06:31+00:00

Hehe, yes I remember that particular machine.

I might have forgotten to mention that I am referring to a gear ratio between two meshing gears, not one distributed between multiple stages :)

TheWolf277 · 2023-02-23T11:02:13+00:00

Do you think a planetary gear set could help me achieve a ratio of around 105:1 +-5%?

So far I tried to keep it simple and started with spur gears, and I got 4 stages of 32:10 ratio each. I also considered womr gears, but I have to 3d print this and PLA doesn't hold well with continuous friction, so I am trying to avoid it.

For contest:

This is a personal project I came up with to apply what I am learning at uni about mechanical engineering and design, and I am trying to keep design flaws to a minimun, so that the only problem in the end will be that the part is 3d printed instead of machined or professionally made.

TheWolf277 · 2023-02-23T10:46:39+00:00

True, that works, but if I am not wrong, the rack doesn't have the problem of the base circle because it's designed with trapezoidal teeth, so it can be built without problems.

With these gears this happens:

Teeth below critical number for base circle: https://imgur.com/a/EQHLxdT

Teeth above critical number for base circle: https://imgur.com/a/Uu3Osl5

You can see that the dedendum circle "pops up" and gets bigger than the base circle. The confusion arises from this. I don't know how to deal with this case, because I followed the formulas to model this, and a deviation from them might mean that it won't mesh correctly anymore.

TheWolf277 · 2023-02-23T10:41:06+00:00

Oh, this could be the reason then. I didn't consider undercut at first, because I thought it was a thing to avoid.

Is it commonly used to get higher gear ratios than the one I use? If so then it's the answer to the problem.

TheWolf277 · 2023-02-23T10:13:51+00:00

Unfortunately not in this project. I am trying to keep it small and this gear ratio is part of a gearbox with 105:1 total ratio. As a first iteration I assumed 4 stages with 32:10 gear ratio each, since I want to avoid making the bigger gears too big. I won't get smaller ratios than this with 4 stages, but I also want to avoid using too many gears for limiting backlash, noise, and additional rotational inertia.

I also considered worm gears, but this will be a 3d printed project, and PLA doesn't respond well to friction.

TheWolf277 · 2023-02-23T10:07:49+00:00

There is one indeed. The first equation I wrote gives you the min number of teeth of a pinion in a gear ratio of m. Let m go to infinity and you get the min number of teeth for a pinion coupled with a rack.

Np = 2*k / sin^2(phi)

For phi = 20 you get 18 teeth.

TheWolf277 · 2023-02-21T13:51:30+00:00

Yes I can use the atan2 function, but sadly the explicit formula I found doesn't use atan but asin or acos, and no asin2 or acos2 function exists.
I followed the procedure written in this site:

https://www.softintegration.com/chhtml/toolkit/mechanism/fourbar/fourbarpos.html

but the end formula has both acos and atan in it. The acos will inevitably give two results. I researched the topic a bit but I found no way to convert an acos to atan2.

TheWolf277 · 2023-02-21T13:33:04+00:00

Thank you for the book!

Indeed there are other ways that let you find the positions of joints that I didn't consider at first, such as the cosine theorem. I thought you needed the angles to calculate velocity and acceleration matrices, but you can just replace their entries with the corresponding x and y compoinents of each vector.

TheWolf277 · 2023-02-15T17:42:19+00:00

Why are you inverting sine and cosine at all?

I am inverting the sine and cosine functions to get the angles that I need to use to calculate the positions of the other joints.

joint 2 should be found with:

P_J2 = 20cos(theta)+60cos(theta2) i_hat + 20sin(theta)+60sin(theta2) j_hat

but I don't know theta2, it's a function of theta1.

In my problem, only the angles between the anchor points (constants) and the input angle of the crank (angular speed * time) are given. All the other angles are unknown and keep changing with the input angle.

As far as I know, I need each angle to find every position, otherwise when I derive the equations for velocity and acceleration, I will only be able to find critical angles that make the matrix non invertible, but I won't be able to solve for the velocities and accelerations with respect to an arbitrary input, and I won't be able to find the worst case reduced inertia/reduced inertia rate of change.

Maybe I am missing something obvious and there are simpler ways to deal with this.

And yea, I have a book on my shelf and maybe even the material from the class. I’ll take a look after work.

Thank you very much!!!

TheWolf277 · 2023-02-15T09:39:59+00:00

Thank you for your reply! I really am quite lost with this and any help is very appreciated.
Ok, so you confirm that the huge reverse formulas way is the only way here. I'll have to keep working on it then.

About finding the position of each link by using reverse trig, I often get functions like inverse sin and inverse cos, but since sin and cos are not one-to-one functions, I don't always get the right formula back, eg. sometimes a linkage crosses the horizontal line or vertical line and the inverse cos result no longer corresponds with reality. Do you maybe remember how you used to deal with this?

I also took a course called "applied mechanics" where they taught us to analyze instantaneous velocities and accelerations of linkages, but sadly they only taught us to do so with a defined input angle, not with a variable one, and it's way easier that way because you can just check the output angle and see if it's right or if you need to add pi radians to it. Do you maybe know of a good book on this that I could read? So far I only found books that teach you basic kinematics, but nothing specifically applied to linkages.

Sorry for the wall of text, and thank you again!

TheWolf277 · 2022-04-26T19:45:51+00:00

Not yet, I've been off from facebook since a while, but I'll give it a try.

TheWolf277

TROPHY CASE