Which is stronger and why?
Thanks,
Steve
93' 325iS
2JZ-GTE Powered BMW is complete.
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Given reasonable wall thickness, hollow is stronger. It's an engineering thing...
Which means, I'm too lazy to break out an old text book and look.
No matter where you go, there you are...
Yes, it's a physics thing....Originally Posted by RRSperry
If two bars are the same outside diameter, and same material, the tube will always be stronger than the solid bar.
///Max
How.. i dont see it.
i'm not an engineer, but i was always under the impression that a solid bar has 1 outside edge, whereas a hollow bar has 2 "outside edges" thus creating more surface area for added strength.
someone please correct me if i'm wrong.
-Brent
Ya made me go look it up didn't ya...
Maximum shear stress (tm)=(Torque [T])/Polar moment of inertia (Ip)
tm=T/Ip
substituting r=d/2 and Ip=Pi*d^4/32 you get
tm=(16T)/(pi*d^3) for a solid bar
For a hollow shaft, Ip =(pi*r^4)/2
Assume an inner raduis of .6r for the tube.
for the solid bar, the Ip = .5*pi*r^4
and for the tube Ip= (pi*r^4)/2 - (Pi*(.6r)^4)/2
= .4352*Pi*r^4
therefor the ratio of tm is .5/.4352 = 1.15 or, more plainly, the tube is 1.15 times stronger than a solid shaft, given the same outer diameter, and material.
Not only are they stronger, they are lighter too.
No matter where you go, there you are...
ok what you just showed me looks like ancient hebrew to me. is there a simpler explanation haha. basically hollow is better right?
<img src=http://img29.photobucket.com/albums/v87/denverlicious/000_0272.jpg>
M3= Multiply $$'s x 3
M3 SEDAN#5
God I love having an engineering degree...
No matter where you go, there you are...
Now you know for a fact no one knows what the heck you are talking about.
Originally Posted by RRSperry
“The bitterness of poor quality remains long after the sweetness of low price is forgotten” – Benjamin Franklin
You should pay more attention to the instructor in mechanics of materials.
No matter where you go, there you are...
i know what hes talking aboutOriginally Posted by Jean-Claude
and i'm not a rocket scientist... yet
Thanks RRSperry, nothing like a real life example to motivate me to get to work on my engineering homework.
Originally Posted by RRSperry
Let me start by saying that I think the original poster's question is a little ambiguous, and that may be why this discussion took the course it did...
Stronger?
For a given diameter, a solid bar is stronger in torsion than a hollow bar. A greater diameter hollow bar can equal or exceed the torsional strength of a solid bar. It just depends on the wall thickness (i.e. how much material is missing out of the middle).
Tau Max isn't the bar's ability to resist torsion, it is the max shear stress that will result from an applied Torque. If you apply 5Nm of torque to a plastic straw and to a .25" diameter steel bolt, the straw will see a higher Tau Max. Doesn't mean it is better able to resist torsion.Originally Posted by RRSperry
Before you say it, I know.... 2 different materials. But that's my point; Tau Max is just a measurement of stress, not an indicator of the ability to resist stress. It doesn't take into account things like materials.
With constant OD, a tubular cross section has a lower polar moment of inertia than a solid one. We agree on that part. Keeping the applied torque a constant, dividing by the lower polar moment is how you end up with the higher Tau Max. Do it the other way; assume Tau Max is a constant related to the material's plastic deformation or ultimate strength point. Now the higher polar moment multiplied by the constant Tau Max gives a higher Max Torque before failure.
To actually see how the solid bar handles that stress compared to the tube, you need to picture the cross section. At the center point in the diameter of the bar, Tau=0. At the outer radius, Tau is at its Max. Tau increases linearly from 0 to max. Since the tubular cross section is missing a large amount of material that would otherwise do the job of resisting the stress near the neutral axis, the smaller annular area of the tubular cross section has to carry the same total load as the circular area of the solid cross section.
Image found via Google image search! Awesome!
Admittedly, the part of the solid bar that is missing with a hollow cross section is the part that resists a Tau closer to zero, so it does not have an overwhelming effect on the bar's ability to resist applied torque.
So....
Even though the tubular bar isn't actually stronger, the advantage of the hollow cross section is a significant weight reduction without significant loss of resistance to torsion. In some (most) cases, this trade-off is worthwhile....they are lighter too.
When comparing a solid sway bar with a tubular bar of identical material and arm geometry, you need to subtract the inside diameter (i.e. wall thickness times 2) to the fourth power from the outside diameter to the 4th power, and then take the fourth root of the whole thing.
In "Excel-speak", think (SQRT(SQRT((OD^4)-(ID^4))).
In practical-speak, I haven't seen a sway bar on a BMW break in years. If the original question about "strength" was due to a concern about breakage, don't worry. If you want to compare a tubular bar to a solid bar, then you just need to keep in mind what I said at the beginning: For a given diameter, a solid bar is stronger in torsion than a hollow bar. A greater diameter hollow bar can equal or exceed the torsional strength of a solid bar. It just depends on the wall thickness (i.e. how much material is missing out of the middle).
For a look at how a tubular bar's measurements will compare to a smaller-but-heavier solid bar, check out the page on UUC's E46 M3 sway bars: http://www.uucmotorwerks.com/html_pr...scription2.htm
If the manufacturer's whose tubular bars you are looking at doesn't have this kind of comparison page, email me and I will help you write an Excel spreadsheet to calculate it for yourself (TK Solver and Mathcad requests also OK).
BTW, while the bar geometry (width of straight portion, length of arms, position of adjustment holes/blade length, angle of arm to straight portion) has a huge effect on overall stiffness (i.e. effective spring rate), the material does not. The modulus of elasticity of different steels doesn't vary significantly.
--Matt
Sig removed by user.
That's what i'm sayin'.
Originally Posted by Matt M.
“The bitterness of poor quality remains long after the sweetness of low price is forgotten” – Benjamin Franklin
Right out of the Mechanics of Materials text.
Seeing as how shear stress, Tau, is related to shear strain (y) by Hook's law, in a linearly elistic material, we get Tau=G*y = G*r*Theta
Where G is the shear modulas of elasticity, and Theta is the angle of twist.
Torsional rigidity = G*Ip
G*Ip/L ((length) is the torsional stiffness, torsional flexability is the inverse.
Anyway, for bars made of the same material, and with the same outer diameter, assuming that the wall thickness isn't too thin. The internal stresses will be higher, in the hollow bar, but still below the allowable.
A hollow bar will resist torsion more efficiently that a solid bar! And has the added benifit of being lighter.
Last edited by RRSperry; 03-25-2005 at 08:24 AM.
No matter where you go, there you are...
I don't have any disagreement with your equations. But be careful about what you're actually solving for.Originally Posted by RRSperry
Here's where you started...
Nothing wrong with that. That equation defines Tau Max as a function of the applied torque and the polar moment.
Maximum shear stress (tm)=(Torque [T])/Polar moment of inertia (Ip)
tm=T/Ip
But Tau Max is just a stress in a single plane. Lower is better, with respect to the material's stress-strain curve. And you'd need to use Tresca or Von Mises to get a more accurate idea of the total distortion energy a sway bar is actually subjected to.
Then you wrote:
We still agree on the polar moments...
Assume an inner raduis of .6r for the tube.
for the solid bar, the Ip = .5*pi*r^4
and for the tube Ip= (pi*r^4)/2 - (Pi*(.6r)^4)/2
= .4352*Pi*r^4
...but look at what you are actually calculating - you solved for Tau Max, not "strength". The tube isn't 1.15 times stronger, it has 1.15 times higher max shear stress at the outer radius.therefor the ratio of tm is .5/.4352 = 1.15 or, more plainly, the tube is 1.15 times stronger than a solid shaft, given the same outer diameter, and material.
This is also true, but it doesn't take the applied torque into account. All this equation does is prove once again that the max shear stress at the OD is higher for a tubular member.Seeing as how shear stress, Tau, is related to shear strain (y) by Hook's law, in a linearly elistic material, we get Tau=G*y = G*r*Theta
Where G is the shear modulas of elasticity, and Theta is the angle of twist.
For a constant torque, i.e. the load applied by a car under cornering, R*theta will be greater with a tubular bar than a solid one. The tubular bar needs to twist farther to do the same job.... because it is weaker in torsion than a solid bar of the same OD!
Once again, no disagreement from me.Torsional rigidity = G*Ip
But we already established that at a given OD, a tube has a lower polar moment than a solid bar, right?
So for the same material (constant G), the lower polar moment of the tubular member produces a lower torsional ridigity. And you need to increase the diameter and/or wall thickness of the tube to get the polar moment and therefore the torsional ridigity to meet or exceed the behavior of the solid bar.
Assume an inner raduis of .6r for the tube.
for the solid bar, the Ip = .5*pi*r^4
and for the tube Ip= (pi*r^4)/2 - (Pi*(.6r)^4)/2
= .4352*Pi*r^4
True, regardless of the tubular bar's wall thickness...Anyway, for bars made of the same material, and with the same outer diameter, assuming that the wall thickness isn't too thin. The internal stresses will be higher, in the hollow bar,
Whoa! Not necessarily true! Guys who design stuff for a living feel their stomachs turn when you say stuff like this.but still below the allowable.
If what you wrote were the case, then every engineer in the world could just design their driveshafts, halfshafts, crankshafts, propshafts, torsion bars, sway bars and other torsionally stressed members as though they were solid and then at the last minute substitute a hollow bar. I'm sure that the guys at Boeing will be thrilled when you call them with this news!
There are definitely cases where the max stress is so far below the deformation point for a solid torsion bar that the tubular bar will also not deform, and this is called bad design. It means that the engineer grossly oversized the bar. Any weight savings you would get from switching to a tubular bar of the same OD is a band-aid. The right thing to do is go back and resize the correct bar for the application.
Richard, for two members of the same OD, one solid and one tubular, loaded in pure torsion, this just isn't true. Your equations aren't wrong, but the conclusions you're reaching with them are.A hollow bar will resist torsion more efficiently that a solid bar!
For what its worth, a tubular member is more resistant to buckling when loaded in bending. And a sway bar does see some bending loads, but without doing a VM calculation I am confident that they are insignificant compared to the Torsional stresses.
I am *not* saying there are no advantages to tubular torsion bars. But greater torsional strength than a comparable solid bar isn't one of them.
The Moment Of Inertia for a cross section indicates the member's ability to resist bending. But Polar Moment of Inertia, which we've been working with here, indicates the member's ability to resist torsion. We already agree on the fact that for a given OD, the tube has a lower polar moment. If we could just agree that the polar moment is an indicator of the member's ability to resist torsion, we'd be all set!
Hibbeler? If so, I'll point you to some example problems so you see what I mean.Right out of the Mechanics of Materials text.
--Matt
PS - If the discussion above still isn't cutting it for you, here's another piece of marketing literature that illustrates solid vs tubular sway bars: http://www.whiteline.com.au/docs/bul...%20Swaybar.pdf
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Damn, you guys are GOOD! Nice detail.
///Max
There is more to it than just torsional force that the sway sees.
Consider Loads:
Torsion
bending moment
shear
axial
You have to use the principle of superposition to break down all the forces and combine all the loads in one diagram to *see* the actual loading on the bar.
Material properties also need to be considered.
97 M3/4
ya'll is so smart. I wes never mech fer yer book lernin and sech.
lolOriginally Posted by tmoeller
i hated mechanics of materials...took it twice because i never went to class the first time...
NOW for a more practical thought:
Just buy the sway bar...who cares if it's hollow or solid. You should buy the hollow just because it is has a plain and obvious weight advantage.
The people that designed the sway bar have supposedly done all the engineering calculations. If they haven't and it breaks...claim warranty
///M3 Cc#65
95 M3 Alpine White Acquired 6/17/03
Cosmos Racing 3.5" CAI / AA Stage II Euro HFM Kit / UUC Stage II Flywheel / M5 Clutch / AA Gen III Exhaust / Bilstein + Eibach Prokit / OEM X-Brace / AA Strut Brace / ZKW + OEM M5 Projectors / Philips 85123 upgrade
97 Integra LS-Turbo Sold 6/14/03
10 psi of boost
Weight smate. I buy the shiniest oneOriginally Posted by 95MMM
Originally Posted by 95beema
Are you a raccoon?
Sure are some fart smellers on this board. Seriously though, great write up. Glad we have peeps like you guys on this board. Finally, a board where people know their sh*t!
D
Assuming the same material, it all comes down to geometry, specifically sectional properties. And assuming the same diameter, a solid cylinder always has higher sectional properties (area, axial and torsional inertias) than a hollow cylinder. Simple.
Originally Posted by Duy325is
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