Which metal would flex more?

[magicniner]

Please accept my apologies, I assumed that you had really basic access to Internet reference and the required brain power with which you could work out from the prolific use of Aluminium alloys in Aircraft that for a similar weight a suitable Aluminium Alloy can provide greater stiffness for a given weight than any Iron based Alloy.
From there it should be a basic mathematical job to compare size and strength, and hey, if you're lazy then someone else might do the work for you:D

[magicniner]

And block me, unless you are going to make an effort, I will point out everywhere that you (and anyone else) is making no effort to resolve their own issues.

[PaisleyPCdoctor]

Please accept my apologies, I assumed that you had really basic access to Internet reference and the required brain power with which you could work out from the prolific use of Aluminium alloys in Aircraft that for a similar weight a suitable Aluminium Alloy can provide greater stiffness for a given weight than any Iron based Alloy.
From there it should be a basic mathematical job to compare size and strength, and hey, if you're lazy then someone else might do the work for you:D

Yes, I'm an uneducated, illiterate, retarded, deaf, blind, quadraplegic, emotionally unstable, little girly with serious mental health problems including hallucinations and psychosis. I have no Internet access and I'm banned from google for asking too many silly questions and overloading their servers (not easy as I have no Internet and my electricity was cut off when I was an infant and is still off).
I frequently soil my pants and I'm a total burden on society due to my asylum status and unwillingness to work but still letting the nhs replace my heart and lungs and my liver twice.

Does that qualify me as a forum member you could manage to ignore and possibly even better, not dignify with responses?

[magicniner]

Yes, I'm an uneducated, illiterate, retarded, deaf, blind, quadraplegic, emotionally unstable, little girly with serious mental health problems including hallucinations and psychosis. I have no Internet access and I'm banned from google for asking too many silly questions and overloading their servers (not easy as I have no Internet and my electricity was cut off when I was an infant and is still off).
I frequently soil my pants and I'm a total burden on society due to my asylum status and unwillingness to work but still letting the nhs replace my heart and lungs and my liver twice.

Does that qualify me as a forum member you could manage to ignore and possibly even better, not dignify with responses?

No, you can type, don't be lazy, learn how to block responses you don't want to see, this is a further demonstration of your lazy requirement for others to solve your problems! ;-)
Asylum status is a Red Light, you're actually obviously an Illegal!
If you're too lazy then you can ask other productive users on the forum to tell you how to block responses from someone you don't like!
To be honest you are becoming mildly endearing, make an effort and I may come to love you! :D

[hanermo2]

The better A. is that it depends.

By mass the alu is 3x less rigid - and importantly it likes to ring and vibrate.
But when you get the same mass in alu, it is thicker, and thickness scales pwr 3 or pwr 4 by section size.

Steel is == 3x better than alu, in rigidity for the same mass and == 3x cheaper by mass.
Cast iron is cheaper than steel and nearly as rigid.
But cast iron is 3x better than steel in vibration dampening, 10x better than alu.

There are effectively zero commercial machines using alu in any modern machine tool bodies, if they cut something via contact (vs laser/plasma etc.).

All modern machine tool bodies are thin-skinned stressed-skin structures. All.
This means the frames are as big as possible, and very thin, and used ribs/preload/mass/anchors.
Typical thickness is around 10 mm on cast iron machine bodies, in the centres, give or take.

An extreme example are modern optical tables.
They are very very light, for what they do.
They are very very rigid, for their work volume or free span length.

Typically, an optical table might be 40 cm thick for a 1 m table, and == 20x more rigid than the best japanese mazak/integrex mill-turn of similar size, or 200-400k$ cost for the mill turn.
My numbers from tlar.

The optical table is two thin skins, and tubes inside.
When I researched them they seemed to be mostly vertical-only tubes, sometimes with various goops/ballast/members/unicorn snot/magic.
The guaranteed rigidity of optical tables is totally unbelievable.

My point.
No-one uses alu.

Here is a data point - I use Thorlabs stuff for some things.
They are a top-shelf supplier of excellent rep.

https://www.thorlabs.com/newgrouppag...tgroup_id=9222

NOTE: !!!
1.7 um / 150 kg load deflection (constraints).
= 1500 N.
882 N / um rigidity.

The best machine tools might do 30-100 N/um, or 8-24x worse.
Iirc the industrial criteria for VMCs == 30N/um... on textbooks.

[Zeeflyboy]

The better A. is that it depends.

A better answer indeed, but an answer to a different question.

[Desertboy]

I was going bed but now I have to work out how to do simulation tests in fusion grrrr.

;)

Couldn't we apply antivibration (Like headphones) to our routers (This is something I've been thinking about for a while) seems like it would be simple enough to do.

I was thinking either hack a pair of headphones (Might not be able to get the frequencies you need) or a raspberry pi.

[PaisleyPCdoctor]

I hadn't even considered vibration as a factor. Thanks.

For what I'm doing, I'll go with the alu, but when I get to the stage of building my own, I'll seriously consider a steel construction.

Thanks again!

[routercnc]

If you want to work it out by hand it is not too tricky for a simple shape like that.

1. The geometry of the part (how the material is arranged, into what shape). This is the cross section property and is called the second moment of area:
Ixx = (bd^3)/12
where:
b is the width
d is the depth
For aluminium part this is (100x25^3)/12 = 130208 mm^4
For the steel part this is (100x10^10)/12 = 8333 mm^4

2. The end condition of the beam (how it is restrained at the ends)
Ec = 192 for fully welded
Ec = 48 for simply supported (e.g. pin joint)

3. The material properties for the material being analysed (Young' modulus)
For aluminium Ym = 69000 N/mm2
For steel Ym = 200000 N/mm2

4. The force being applied in the middle of the beam
For the example given this is 200 N

5. The calculation:

Deflection = (force (N) x length^3) / (Ec x Ym x Ixx)

For welded supported ends this is:
Deflection (Aluminium) = (200 x 500^3) / (192 x 69000 x 130208) = 0.01449 mm
Deflection (Steel) = (200 x 500 ^3) / (192 x 200000 x 8333) = 0.07813 mm

These are pretty close to the Fusion FEA results from Zeeflyboy (for fully supported end conditions):
25mm Alu deflection = 0.01471 mm
10mm Steel deflection = 0.06861 mm

If the ends are simply supported pin joints it makes a big difference:
Deflection (Aluminium) = (200 x 500^3) / (48 x 69000 x 130208) =0.0579 mm
Deflection (Steel) = (200 x 500 ^3) / (48 x 200000 x 8333) = 0.312 mm

So there you have it !

[Desertboy]

If you want to work it out by hand it is not too tricky for a simple shape like that.

1. The geometry of the part (how the material is arranged, into what shape). This is the cross section property and is called the second moment of area:
Ixx = (bd^3)/12
where:
b is the width
d is the depth
For aluminium part this is (100x25^3)/12 = 130208 mm^4
For the steel part this is (100x10^10)/12 = 8333 mm^4

2. The end condition of the beam (how it is restrained at the ends)
Ec = 192 for fully welded
Ec = 48 for simply supported (e.g. pin joint)

3. The material properties for the material being analysed (Young' modulus)
For aluminium Ym = 69000 N/mm2
For steel Ym = 200000 N/mm2

4. The force being applied in the middle of the beam
For the example given this is 200 N

5. The calculation:

Deflection = (force (N) x length^3) / (Ec x Ym x Ixx)

For welded supported ends this is:
Deflection (Aluminium) = (200 x 500^3) / (192 x 69000 x 130208) = 0.01449 mm
Deflection (Steel) = (200 x 500 ^3) / (192 x 200000 x 8333) = 0.07813 mm

These are pretty close to the Fusion FEA results from Zeeflyboy (for fully supported end conditions):
25mm Alu deflection = 0.01471 mm
10mm Steel deflection = 0.06861 mm

If the ends are simply supported pin joints it makes a big difference:
Deflection (Aluminium) = (200 x 500^3) / (48 x 69000 x 130208) =0.0579 mm
Deflection (Steel) = (200 x 500 ^3) / (48 x 200000 x 8333) = 0.312 mm

So there you have it !

Nice one! I always want work it out by hand if I can! I normally do my trig on paper then check it in cad so I already know what numbers I should be getting back.