aluminum DOM

Whatever...what do you need to see to qualify me?

My 20 year old 4-5 mil a year grossing fabrication company?

You guys can argue up a tree...point is section modulus is more important than YIELD OR E!

So your welcome for the real numbers to aid you in your selection...or you could just suck on your finger...stick it up in the air to see which way the wind is blowing and make your "informed" decision just like all the other know it alls....of which 20 of them work for me.

Lol...this is why I rarely add anything to comp drama anymore.


Second...I am not just saying...I am also doing a lot of doing.
 
cquestad said:
Whatever...what do you need to see to qualify me?

My 20 year old 4-5 mil a year grossing fabrication company?

You guys can argue up a tree...point is section modulus is more important than YIELD OR E!

So your welcome for the real numbers to aid you in your selection...or you could just suck on your finger...stick it up in the air to see which way the wind is blowing and make your "informed" decision just like all the other know it alls....of which 20 of them work for me.

Lol...this is why I rarely add anything to comp drama anymore.


Second...I am not just saying...I am also doing a lot of doing.
Click to expand...


No the proper use of the equation is all that is needed to be presented.
I am not hiring you, so I don't need your "resume"
 
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cquestad said:
You guys can argue up a tree...point is section modulus is more important than YIELD OR E!
Click to expand...

Easy Cowboy...hope that wasn't directed at me, thought we were being productive. Although I would say in this case that E and I are equally important in this case since Al is 1/3 that of steel. If it were closer, you could say it was all about the I. Anywho...

I am getting numbers higher than yours but not sure it matters all that much.

So, let's say for a minute we're both on the same page.

If my bars are flexing and the formula says mine are good for 40k lbs...then what? Are the loads far higher than I think? Or are they not flexing and people need to put down their beers?

Here, just for kix, enter L = 80" and 2" x 2" x 1/4" wall steel and see whatcha get.

One thing that is probably contributing to my particular case is that the rear connection is rigid. I am not saying that's a good way to do it, it's just the way I fabbed them back when I knew less than I know now. And it worked, so I never messed with it.
 
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The proper equation was used...and I did give you the example to show that it did not make the allowable load 1/3 as much as steel because of the reduced E. Length and I matter most...as I proved. 80 will drive those loads down alot...will run it again when I am back in the office. Shorten your bars some...no need for them to be so long.
 
cquestad said:
The proper equation was used...and I did give you the example to show that it did not make the allowable load 1/3 as much as steel because of the reduced E. Length and I matter most...as I proved. 80 will drive those loads down alot...will run it again when I am back in the office. Shorten your bars some...no need for them to be so long.
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Look, we agreed on the formula, right?

E and I are both directly proportional to Fcr, right?

Fcr is inversely proportional to the square of the length, right?

I think you're using a formula that's got another factor in it, and I'd like to learn what it is. If I look at your numbers correctly, I think it is taking an additional test into account or applying an additional criteria for Fcr. That's a good thing, just want to know more about it.

I'm not saying you're an idiot or wrong, so I'd prefer if you didn't insinuate that I am, OK?
 
Well, Mr. Questad is saying you have to make them much shorter than you proposed, and if you do that, aluminum is fine.
 
I went back over the calcs one more time, using two 12k-rated pieces at 48" L.

Steel 1.99 OD x 0.144 wall
I = 0.36
Fcr = (29e6 * 0.36 * 3.14^2) / 48^2 = 44676 lb

Aluminum 2 OD x 0.25 wall
I = 0.53
Fcr = (10e6 * 0.53 * 3.14^2) / 48^2 = 22680 lb

Both are way off 12k... and yes I fully understand that this is the most basic application of the formula so I'm not saying it's all encompassing.

This is why I'm saying the program you're using must be taking something else into consideration. I don't do this stuff every day so I'm just asking what I'm missing.
 
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nwpadmax said:
I went back over the calcs one more time, using two 12k-rated pieces at 48" L.

Steel 1.99 OD x 0.144 wall
I = 0.36
Fcr = (29e6 * 0.36 * 3.14^2) / 48^2 = 44676 lb

Aluminum 2 OD x 0.25 wall
I = 0.53
Fcr = (10e6 * 0.53 * 3.14^2) / 48^2 = 22680 lb

Both are way off 12k... and yes I fully understand that this is the most basic application of the formula so I'm not saying it's all encompassing.

This is why I'm saying the program you're using must be taking something else into consideration. I don't do this stuff every day so I'm just asking what I'm missing.
Click to expand...


I see your point, and the best guess I have is that he is using a safety factor..
 
nwpadmax said:
10-4, but why different safety factors based on materials with assumed equal yield strengths?
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That's a good question. Maybe a little higher safety factor for the aluminum. I believe you and I had a similar discussion in another thread. I am hoping he chimes in and gives us a clear answer, my curiosity has been struck.
 
"in this thread". mostly referring to cquestad, btw. (and yes, as an engineer I too appreciate the use of engineering calcs on CompD.)

IMHO, the material necessary to make up the difference in strength between steel & Aluminium appears to make the attempted weight savings negligible, no?
 
Aluminum is approx half the weight of steel...so it would be lighter and it has some really nice dampening characteristics that would be an advantage in this application. If I redo mine...I would make them aluminum....no rust too!
 
I will appologize for the dick swinging...but it gets really old when I contribute something and somebody that doesn't know jack comments in challenging it...so cummins forum like!
 
There's a significant quantitative difference between relatively high theoretical numbers generated by buckling formulae and much lower compressive strength ratings derived from empirical use (i.e. established/accepted design curves).
Safety factor widens the gap even further, the extent of which varies widely according to application & assumptions.

IMO the torsional component of slender column buckling demands a view of material selection both in regard to specific modulus (no weight difference between aluminum or steel for equal torsional stiffness) and cross-section (round tube superior to square tube).

For driveshafts & heim-jointed traction bars, assuming there aren't any packaging contraints - aluminum & steel are equally effective.

However, if the column is primarily modeled as an interconnected group of plates of high L/W ratio, aluminum does offer a 50% weight reduction over steel as far as torsional stiffness - one could make a case for aluminum's preference in a square or rectangular tube form (most certainly for webbed column members).
 
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