Solid round bar versus round tube

Reflecting on this some more, I think one would find it stupidly difficult to calculate the shear load that it takes to remove the pins from a u-joint. Unless you can exactly calculate how much rigidity there is in the yoke ear, you won't be able to factor in how much tip over affects the shear and then there is the issue of angle. Supported on two pins fully with a load straight down on one of the unsupported pins will result in different results that the same two pins and pushing straight down on the body.

The caps trapped in a forged pinion yoke will suffer from tip over much less than those in a cast iron H bar in a Double Cardan. The amount of tip over or the ability to bring the pin into straight shear is not something that will be easy to define due to the vastly different abilities of all the yoke designs and materials.

Yes, it would be difficult to calculate but the shear is there since the pin has some sort of fixity (it is not pinned) but restrained from moving and thus would see traverse shear in addition to the transitional shear due to bending. Again, I am not arguing, I have the utmost respect. I think the point is that there is strong enough and there is overkill.
 
Wouldn't we have to factor in that just being an impulse force wouldn't necessarily be higher loader depending on the particular impulse?

I can go tap on the hood of your rig with a table spoon I robbed out of the cutlery drawer and not dent it. If I grab that big honking serving spoon from the cafeteria and give it a good whack, I'm likely to be paying for some body work.

There will be a g factor associated with the dynamic loading that would be factored into the load. So if your spoon is moving at x mph and weighs y, a g factor would be calculated and added to the load of just the spoon itself. This higher load would result in higher stresses and yes could result in the hood material going into yield and not springing back or past ultimate and result in a hole (ignoring plastic effects).
 
Yes, it would be difficult to calculate but the shear is there since the pin has some sort of fixity (it is not pinned) but restrained from moving and thus would see traverse shear in addition to the transitional shear due to bending. Again, I am not arguing, I have the utmost respect. I think the point is that there is strong enough and there is overkill.

One of the reasons I try and highlight that difference is due to some poorly done u-joint testing that was then published as a valid comparison. The fixture that they set the pins on was flawed.

They built a fixture plate with two roughly 3/4" wide by 2" long pieces of steel rectangular bar welded to it at the same distance the base of the cap would stop. They set a joint in it and pushed on it to see how much it took to break the pins off. The bars each had a 1/2 circle close to the pin diameter that the pins rested in.

The giant glaring problem with the whole test that no one caught but me is as they used the fixture, it deformed the material in the 1/2 circle downward close to the inner edges and the more joints they did, the worse that condition got. When they finally got to testing the CTM which fared relatively poorly against some others, the two pins were resting on the ends of tapered cones at the outer edges of the half circles. That brought the leverage up much higher since it moved it out to the ends of the pins instead of being far more supported like the earlier joints. It was supposedly an "impartial" blind test but it only took about 2 minutes to figure out it really wasn't.
 
There will be a g factor associated with the dynamic loading that would be factored into the load. So if your spoon is moving at x mph and weighs y, a g factor would be calculated and added to the load of just the spoon itself. This higher load would result in higher stresses and yes could result in the hood material going into yield and not springing back or past ultimate and result in a hole (ignoring plastic effects).

I understand that but just being an impulse load doesn't mean that the ultimate load into the structure is higher, correct?
 
Yes, it would be difficult to calculate but the shear is there since the pin has some sort of fixity (it is not pinned) but restrained from moving and thus would see traverse shear in addition to the transitional shear due to bending. Again, I am not arguing, I have the utmost respect. I think the point is that there is strong enough and there is overkill.

That is also one of the reasons why joints fail in high strength shafts and stock shafts fail at the yokes with far fewer broken joints. The yoke strength affects the tip over and brings in a much higher level of fixity in the high strength versions. Thusly, with both pins looking the same.
1702666302210.webp
 
Wouldn't we have to factor in that just being an impulse force wouldn't necessarily be higher loader depending on the particular impulse?

I can go tap on the hood of your rig with a table spoon I robbed out of the cutlery drawer and not dent it. If I grab that big honking serving spoon from the cafeteria and give it a good whack, I'm likely to be paying for some body work.

A tap with a table spoon might not dent the hood, but it may damage the clearcoat in a way that is unseen to the naked eye. At any rate, it seems to illustrate that there would be a level of impulse force that would be negligible, but then as you go up the spectrum you cross a threshold and it becomes critical to figure into the design.
 
A tap with a table spoon might not dent the hood, but it may damage the clearcoat in a way that is unseen to the naked eye. At any rate, it seems to illustrate that there would be a level of impulse force that would be negligible, but then as you go up the spectrum you cross a threshold and it becomes critical to figure into the design.

I'm aware, I'm just trying to clarify the point that just because it is an impulse load, it can be small enough to ignore.
 
I understand that but just being an impulse load doesn't mean that the ultimate load into the structure is higher, correct?

The actual load into the structure would be higher due to the g factor and that higher load would result in higher stresses in the structure; however, those higher stresses may or not be beyond the limit (plastic, ability to bend back) or ultimate (failure) capabilities of the material. Not sure if that answers your question, let me know if that what you were asking/referring.
 
Plus, I’ll probably be building some CA links sometime in the future.

I just did my own CA last year and I never ran a single number (other than length). I just looked at what others have done and what seemed to have worked for the long run and then also looked at availability. Availability and price is a driving factor! Doesn't make a lot of sense for us to design a one off diameter control arm out of an exotic material is er can't afford to have them made or can't readily be made when WOD (or similar) has a dia and material that has been proven to work on similar rigs at similar lengths.
 
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I'm aware, I'm just trying to clarify the point that just because it is an impulse load, it can be small enough to ignore.

Yes. Perfectly stated, much better than my "I stayed at Holiday inn express" response
 
One of my engineering mentors taught me to never trust someone else's program (before Excel, we wrote FORTRAN programs to solve repetitive calculations) without verifying that it works correctly. With those lessons ringing in my head, the first thing I did with Mr. Blaine's first post was try to replicate the results of the calculations. I went to Rogue's site and played with the calculator, and then derived the equations used in the calculator. My calculations don't match for the round tube calculator, but they do for the square tube calculator, and it's driving me crazy. Can one of the other mechanical engineer's verify whether Rogue has an error or not?

Here are the basic equations I used:

1702669970605.png


I've also attached an Excel spreadsheet I created using these formulas to replicate the calculator. If you compare it to the Rogue calculator, you'll get the same results for the square tube calculator, but not for the round tube calculator. The only difference in the two is the calculation of the moment of inertia, so the difference should be there. Either I have an error in my calculations, or they do. I can't find an error in mine, and I can't see their equations.

If there is an error, this tool may still work fine for comparing materials, but it really depends on where the error is.

Edited to add a link to the Rogue calculator: https://www.roguefab.com/tube-calculator/
 

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I wish textbook authors and college profs were better at teaching this stuff in real applications instead of the meaningless generic crap examples they would give. I'd probably have found mechanics a lot more interesting than I did.

I ended up gravitating toward fluids and energy because all the teaching and problems were in the context of real systems like turbine engines and powerplants.
 
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I wish textbook authors and college profs were better at teaching this stuff in real applications instead of the meaningless generic crap examples they would give. I'd probably have found mechanics a lot more interesting than I did.

I ended up gravitating toward fluids and energy because all the teaching and problems were in the context of real systems like turbine engines and powerplants.

After I got my BSME, I started in a direct-to-PhD program at Purdue in fluids, working as a research assistant for a leading jet engine expert. After almost two semesters, I realized that I really had no interest in jet engines and wanted to go racing, so I quit and went racing. I suspect that had I specialized in (land) vehicle dynamics, I wouldn't have quit. However, I'm glad I quit. It made me realize that practicing engineering in the real world, surrounded by smart people with practical experience (not degreed engineers), made me a much better engineer then any advanced degree would have...
 
One of my engineering mentors taught me to never trust someone else's program (before Excel, we wrote FORTRAN programs to solve repetitive calculations) without verifying that it works correctly. With those lessons ringing in my head, the first thing I did with Mr. Blaine's first post was try to replicate the results of the calculations. I went to Rogue's site and played with the calculator, and then derived the equations used in the calculator. My calculations don't match for the round tube calculator, but they do for the square tube calculator, and it's driving me crazy. Can one of the other mechanical engineer's verify whether Rogue has an error or not?

Here are the basic equations I used:

View attachment 482638

I've also attached an Excel spreadsheet I created using these formulas to replicate the calculator. If you compare it to the Rogue calculator, you'll get the same results for the square tube calculator, but not for the round tube calculator. The only difference in the two is the calculation of the moment of inertia, so the difference should be there. Either I have an error in my calculations, or they do. I can't find an error in mine, and I can't see their equations.

If there is an error, this tool may still work fine for comparing materials, but it really depends on where the error is.

Ah good old Fortran, I had one semester of that. I'm too far removed from the calculation side to help you out without spending days looking up formulas.

I wish textbook authors and college profs were better at teaching this stuff in real applications instead of the meaningless generic crap examples they would give. I'd probably have found mechanics a lot more interesting than I did.

I ended up gravitating toward fluids and energy because all the teaching and problems were in the context of real systems like turbine engines and powerplants.

One of the worst professors I ever had also wrote the textbook for our physics class. Horrible experience.

But another wrote this book and he was awesome so it's hit or miss.

1702671297848.webp
 
After I got my BSME, I started in a direct-to-PhD program at Purdue in fluids, working as a research assistant for a leading jet engine expert. After almost two semesters, I realized that I really had no interest in jet engines and wanted to go racing, so I quit and went racing. I suspect that had I specialized in (land) vehicle dynamics, I wouldn't have quit. However, I'm glad I quit. It made me realize that practicing engineering in the real world, surrounded by smart people with practical experience (not degreed engineers), made me a much better engineer then any advanced degree would have...

fortunately I didn't have to start another program to figure that out...to fill out my senior year I had to take an "elective" that had to be an upper division engineering class, and having already taken the aerospace classes that interested me and not having the prerequisities for anything in chemical or electrical, I chose a grad level Advanced Thermodynamics course because it was taught by my favorite prof. I got through it but after a semester talking about entropy I knew I didn't want to be anywhere near a career that had me doing that stuff day after day and an advanced degree would not benefit me at all.
 
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I am not at my desk, but check your Moment of Inertia calculation. I think the R is ^4 not squared. The result has to be inches^4

Yes, that's it - thanks! That's what I get for relying on old-man memory, and not looking it up. I was thinking area, not moment of inertia. Here's the correction and now the spreadsheet works just like Rogue's calculator:

1702673694162.webp


Thanks, @gasiorv, you saved me from a sleepless night. I figured I'd erred somewhere, but couldn't see it...
 

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Oh man...This post gave me flashbacks to college...I headed straight into manufacturing engineering since I don't have a "head" for math. I'm always impressed when someone can whip out the calculations as @sab did, even if he did make a mistake. I would have spent all day looking that stuff up in my old textbooks and trying to remember how to solve the problems.
 
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