02GF74
|
| posted on 23/9/14 at 08:48 PM |
|
|
mechanics question
Is there an equation that gives the diameter and wall thickness of a tube for a given cross section area.
For example you start with s 20 mm dia solid metal rod, if it was a tube with say a wall thickness of 8 mm, it would resist bending better thsn the
rod and if the wall thickness is further reduced it would be stronger still.... and if this is continued, there will be a point when the walls are so
thin and diameter so large its strength has gone.
[Edited on 25/9/14 by 02GF74]
|
|
|
|
|
bi22le
|
| posted on 23/9/14 at 09:08 PM |
|
|
A lovely excel spread sheet would give you the answer to your main question.
Regarding the strength of tube vs dia vs wall thickness. You would need to add more detail including youngs mod ( I think is applicable here) and
other material property.
Track days ARE the best thing since sliced bread, until I get a supercharger that is!
Please read my ring story:
http://www.locostbuilders.co.uk/forum/13/viewthread.php?tid=139152&page=1
Me doing a sub 56sec lap around Brands Indy. I need a geo set up! http://www.youtube.com/watch?v=EHksfvIGB3I
|
|
|
britishtrident
|
| posted on 23/9/14 at 09:38 PM |
|
|
(1). An 8mm dia tube is less stiff and fails at a lower load bending than an 8mm rod.
(2) A larger dia tube of the same length and mass as the 8mm rod will be be stiffer in bending and fail at a higher load in bending than 8mm rod.
(3) When wall thickness becomes very thin in comparison to diameter the failure mode in compression changes to a buckling mode.
(4) Long slender load bearing members fail at lower load when loaded in compression than in tension due to Euler Buckling.
To illustrate this consider a drinking straw or length of spaghetti both are relatively strong in tension but buckle when subject to compressive
loads.
[Edited on 23/9/14 by britishtrident]
[I] “ What use our work, Bennet, if we cannot care for those we love? .”
― From BBC TV/Amazon's Ripper Street.
[/I]
|
|
|
HowardB
|
| posted on 24/9/14 at 06:34 AM |
|
|
Moments of area will give the ratio... have a look here...linky as already said
failure mode is subject to the application of the load.
Do you have a specific application in mind..?
Howard
Fisher Fury was 2000 Zetec - now a 1600 (it Lives again and goes zoom)
|
|
|
mcerd1
|
| posted on 24/9/14 at 09:17 AM |
|
|
quote: Originally posted by 02GF74
Is there an equation that gives the diameter and wall thickness of a tube for a given surface area.
do you mean cross section area rather than
surface area ?
as above if you have a 20mm solid rod and any 20mm tube - then rod will be stiffer
but if you compare a tube with the same x-sectional area as your solid rod it will be stiffer in bending (i.e. bigger diameter but the same weight per
unit length)
think of it like this - the further you move the material away from the centre of the x-section the stiffer it gets
however once you get to the point where the wall thickness is to thin it will buckle quite easily and if its really thin it could be impractical to
handle / weld / bolt to - and could get damaged easily depending on the application
as BT said the likelihood of it buckling is defined by the sections slenderness (i.e. in tubes this is the ratio of the outside diameter to it
thickness)
as an extreme example imagine a tube made of tinfoil that had the same x-section as your solid rod - in theory it would be a very stiff section except
that it would be so slender that it would buckle as soon as any load was applied...
this is also the main reason that larger diameter tubes tend to come with thicker walls so as to maintain a reasonable slenderness ratio
Area's of the rod / tube are easy enough:
A-Rod = Pi * r²
A-Tube = Pi * (R² - r²), were R = outside radius and r = inside radius
as above the second moments of area are a reasonable way to compare the overall stiffness:
I-rod = ( Pi * D^4 ) / 64
I-tube = ( Pi * (D^4 - d^4) ) / 64
http://www.roymechx.co.uk/Useful_Tables/Maths/M_of_Inertia_2.html
as for the slenderness / buckling its often quite dependant on the material its made from as well as the loads applied and the length of the
section...
[Edited on 24/9/2014 by mcerd1]
-
|
|
|
02GF74
|
| posted on 25/9/14 at 06:07 PM |
|
|
Arggghhh. Yes ofcourse i meant to write cross sectional area.
An applicatiom could be a bicycle frame. Typically a steel frame would have 0.7 mm walled tube that is 5 mm in diameter.
How are the dimensions determined? Why not use 0.3 mm with larger dia or 2.0 mm with smaller dia tubing??
|
|
|
theprisioner
|
| posted on 25/9/14 at 06:26 PM |
|
|
Steel framed Bicycles are usually made out of a special Reynolds tube that can be thinner in the middle thn at either end. The ends can only be silver
soldered or brazed because the Reynolds has a grain for strength. I imagine this information may change your calculations.
http://sylvabuild.blogspot.com/
http://austin7special.blogspot.co.uk/
|
|
|
britishtrident
|
| posted on 25/9/14 at 07:38 PM |
|
|
If you make the walls too thin they will buckle, the walls of a alloy Coke can will easily buckle under a crushing load they will also buckle easily
under torsion.
[I] “ What use our work, Bennet, if we cannot care for those we love? .”
― From BBC TV/Amazon's Ripper Street.
[/I]
|
|
|