I need help.
I need to do some metal works, but I am in trouble with a specific
characteristic.
I need to select a round aluminium tube. Due to other reasons I need a
tube from 20 to 25 mm diameter (no idea about wall thickness).
Now, suppose I have a 1m length aluminium tube fixed in one of the
extremities.
Now suppose I will apply a lateral force of 5Kg, every second, for one
year, to the other extremity of the tube.
How do I select a tube able to withstand such load without breaking it
due to metal fatigue (or other problem)?
How do I know the amount of tube bending?
I have spent a couple days browsing Internet, but I can't find this
type of aluminium tube specifications.
Any help will be highly appreciated (links, tables, etc).
Regards,
H. Martins

Once a second for a year is:
1 * 60 * 60 * 24 * 365 = 31000000.
36 millions. Realy worth considering!
Basicaly, you need a diagram that shows stress vs. number of load
changes. These diagrams distinguish between swelling and changing load.
I don't know how you call them, but in German(y), the are called Woehler
(or Wöhler)-diagrams. Might help you to google.
It is calculated this way:
f = F * l^3 / (3 * E * I)
f: deflection (in mm)
F: Force (in Newton) 10N roughly are 1kg
l: length (of the tube; in mm)
E: elasticity modulus (in N / mm^2) it is (about) 0.7 * 10^5 N / mm^2
I: Umm, don't know how this translates. "Inertial moment"?
You get I with: (for a tube)
I = pi / 64 * (D^4 - d^4)
pi: 3.14 :-))
D: outer diameter (in mm)
d: inner diameter (in mm)
If (to explain the math) you take a tube with OD = 25mm, ID = 20mm and
length = 1000mm you get:
I = 3.14 / 64 * (25^4 - 20^4)
= 0.049 * (390625 - 160000)
= 11300 [mm^4]
f = 50 * 1000^3 / (3 * 0.7E5 * 11300)
= 0.21mm
Hope I didn't screw up the numbers, the calculater I had at hand was
quite stupid. Don F. will jump in and correct me. I know. :-)
OK, deflection is very little, so will be the stress. If you want to
know the stress, that's a different formula. Come back if you want.
Now the deflection is so little, that you will not have problems with
stress, but only with fatique.
I can't find that number right now, only for steel. And it is at about
2000000 ... 20000000 changes of load and reducing the allowable stress
by about half. If you get more loads, it no longer matters. Now if I
only could find:
a) that limit-number of load changes.
b) allowable stress with fatique.
But by guestimating, I'd like to say no problem with fatique. If your
construction is critical, you need to investigate more on this and also
need a certified engineer who is allowed do do the calculations in your
state and will sign it.
HTH, for now,
Nick

Ah! That helps a lot. I don't have the time today to do the rest of the
math, but I'll come back tomorrow.
Just a short addition to my other posting(s):
Alumin(i)um continues to fatique with the number of cycles.
Nick

Once you have confirmed that the fatigue life is OK you have really got to
think about fretting fatigue at the points where the tube is clamped and/or
the load is applied. Aluminium is is a pig, especially for the home
constructor. In my biker days I used to use Ally plate for fairing brackets
usually clamped to the fibreglass with a sheet of rubber between, and they
typically lasted 6 months.

Good question, Jerry. Based on the sort of thing 7075 is typically used
for I assumed so, but have never personally had to shop for 7075 tubing.
It appears to be available, but I'm sure it'll be quite a bit more
expensive than 6061.

Hi Ned
I rely on you and Don Foreman to supply the answers to questions I'm too
dumb to be able to answer without studdying.
I once worked with a smart guy who told me that 7075 was available in
sheet *only*. I'm too lazy to reserch it so I'm depending on you guys who
studdy to answer this 7075 tube question.
Jerry

According to my ALCOA Aluminum Handbook (a little dated at 1957
copyright), 7075 T6 is indeed drawn into tubes.
Physical properties for 7075 T6:
UTS- 83,000 PSI with extruded products having values 10% higher.
YPS- 73,000 PSI with extruded products having values 10% higher.
Elongation in 2 inches - 11%.
Shear strength- 48,000 PSI.
Endurance Limit (Fatigue Strength) - 23,000 PSI based on 500,000,000
(five hundred million!) cycles of completely reversed stress, using
R.R. Moore type of (test) machine and specimen.
"These results are typical properties, average for various forms,
sizes, and methods of manufacture and may not exactly describe any
particular product." (quoted from handbook).
I must admit that I believed the endurance limit for aluminum to be
much lower than this; but then it does not state the preparation of
the test specimen such as surface finish, direction of lay of finish,
shot peening, etc.
Wolfgang
Jerry Martes wrote:

I am carefully reading all your help, by the way, I deeply appreciate.
I can't understand a couple things:
Nick M=FCller says:
------------------------
f =3D 50 * 1000^3 / (3 * 0.7E5 * 11300)
=3D 0.21mm
-----------------------
1) There is 50 coming from? Shouldn't it be 10 (force in N)?
2) 0.7E5 (70000) ? Shouldn't it be 0.007?
E: elasticity modulus (in N / mm^2) it is (about) 0.7 * 10^5 N / mm^2:
0=2E7*Force^5/Length^2
0=2E7*10^5/1000^2
0=2E7*100000/1000000
0=2E07
Of course, the result according to my calculations is 4.2Km J
I need help again.
I will come back with problems concerning the points where the forces
are applied.
Martins

That is the "force" (it ain't force, no nickpicking) of 5kg = 50N
No, 0.7 * 10^5. Thats 70000. You can understand that number like a
spring rate. Quite a stiff spring out of solid.
0.7E5 is the engineering notation for 0.7 * 10^5. Maybe that confused
you.
Huh!? I didn't write that.
f = F * l^3 / (3 * E * I)
f = 50 * 1000^3 / (3 * 7000 * 11300)
= 21 [mm]
Sorry, I misstyped something. Don't know where.
Check that by yourself, I cant find my beloved pocket calculator and
have to use the crap on the PC.
Checking the units:
N * mm^3 / (N / mm^2 * mm^4)
= N * mm^3 / (N * mm^2)
= mm (looks good :-))
4.2 Km What is that? Did you mean 4.2 km? And J? Where do the Joule come
from?
While I'm at it, I'll have a cup of coffe and calculate the stress:
s = M / W
s: stress in [N/mm^2]
M: moment in [Nmm]
W: don't know how you call that (Resistance moment?) in [mm^3]
M = F * l
F: Force (prev. posting)
l: length (prev. posting)
W = pi / 32 * (D^4 - d^4) / D
pi, D, d: prev. posting
Now with the real numbers (as I used in the prev. posting):
W = 3.14 / 32 * (25^4 - 20^4) / 25
= 0.098 * 9225
= 905 [mm^3]
M = 1000 * 50
= 50000 [Nmm]
s = 50000 / 905
= 55 [N/mm^2]
Now I know that you don't want that. I'm sorry for my wrong guestimate
in the previous posting, but 0.2mm deflection is quite different to the
21mm.
Now to the pdf about technical data Ned posted for the 6061-T6.
Its elastic modulus (E) is 0.97E5. Quite different ('bout 30% off for
the better) to the 0.7E5 that I assumed. Recalc with that number. You'll
get a different deflection.
The allowable stress for 6061-T6 is only 30N/mm^2 with just 1000 load
cycles. **CRACK**
With 10,000,000 (10Mio) about 15 N/mm^2 etc. see page 5, Fig 5. Thanks
Ned for finding that.
An other comment from Newshound is also really important. The clamping.
Your bar will break there. Don't use a tube at that place, but a solid.
Make radii smooth and make a security factor of 2.
Remember that I still used that stupid calculator!
Nick

Sorry, but (I hope I am not wrong),
f = 50 * 1000^3 / (3 * 7000 * 11300)
= 210 !!!
The 'J' is just noise. I wrote a smilling face and something somewhere
changed it to a 'J'.
I knew that, My point was that I according to my calculations,
Elasticity calculations result should be 0.07 not 0.7E5.
Elasticity=0.7*Force^5/Length^2
=0.7*10^5/1000^2
=0.7*100000/1000000
=0.07
Regards
H. Martins
(from correct e-mail now - the other was a browser mistake)

There is a mistake I previous post.
The Elasticity, according to my calculations, should be:
0.7*Force^5/length^2
0.7*50^5/1000^2
=218.75
... but you say it is 70000 !!
H. Martins

Damned! My 7000 are a typo and really are 70000 as I initialy wrote.
Now slowly:
f = 50 * 1000^3 / (3 * 70000 * 11300)
f = 50 * 1,000,000,000 / 2373000000
= 21[mm]
Any zeros left out again?
The elasticity modulus (0.7E5) is a constant for the material. For
6061-T6 it is 0.97E5 according to the PDF. 0.7E5 is for "some"
aluminium.
Where does that formula come from?
What is Elasticity by your definition? Spring rate?
There is a difference between spring rate (if you mean that by
elasticity) and elasticity modulus. And in this case, it really matters.
Maybe I confused you that you can look at the elasticity module as being
a spring rate. It is, but only if you compress a solid bar and do not
bend it.
Nick

I ran the numbers you used and came up with the same result.
1 m cantilever, 25 mm OD x 25 mm ID, 5 kg load applied at the end
Deflection = 21 mm
Max stress = 54 MPa
I don't trust the 97 GPa figure for elastic modulus - I used 69 GPa.
By the way, your "W" is "section modulus" in English, usually denoted
"S".
Ned Simmons

I'm flattered you'd group me with Don, especially knowing that you're
one of the folks here who limits his posts to those subjects on which
you can speak with authority.
Not that I've noticed anyone posting riffs based on unfounded
extrapolation and wild speculation.
Ned Simmons

Well, from your initial post:
You had wrote before:
I assume the above 'E' must be replaced by '(about) 0.7 * 10^5 N /
mm^2' which you refer as elasticity modulus.
The 70000 figure seems do come from this formula, but a member of the
formula is missing: 0.7*10^5 is indeed 70000, but we still have to
divide by the square of the length which is 1 million. The result is
0=2E07. I am assuming " 0.7 * 10^5 N" is some sort of constant for this
type of material.
Thanks
H=2E- Martins

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