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# calculating braking point of test-tube

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# calculating braking point of test-tube

05-27-23, 07:10 AM
#1
CrowSeph
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calculating breaking point

I have repaired a test-tube made from carbon, Now i want to know how much force (N) is required to break the repaired point.
Is correct to proceed in this way?
Since is a tube and is pretty short i was thinking to hang up from the roof with a rope passing inside it. Then apply a bunch of weight until it breaks.
In this way the force required should be calculated
F= G*M
Were
F stand by force required
G stand for gravity acceleration
M stand for the final weight to break the tube.

It is a been a while since last time i read a physics book, would be nice if someone will help me out.
05-27-23, 07:38 AM
#2
unterhausen

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Generally weight is in kilograms force, not mass. So you don't need the acceleration of gravity unless you are doing this test on the moon.
05-27-23, 08:21 AM
#3
Andrew R Stewart
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The classic way to test stuff was to make a few and try various manors of destruction.

Is the test tube a vessel a tube with one end closed to be able to hold a substance? I think of blood and a centrifuge spinning the test tube. Reason why I ask is that what this test tube is to do will have a significant impact of what type of break you are testing for. Which stress are you concerned about? Torsion, tension, shear, bending? What forces will be acting on the test tube when in use?

An interesting question but lacks much context as yet. Andy
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05-27-23, 09:07 AM
#4
guy153
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Originally Posted by CrowSeph
I have repaired a test-tube made from carbon, Now i want to know how much force (N) is required to break the repaired point.
Is correct to proceed in this way?
Since is a tube and is pretty short i was thinking to hang up from the roof with a rope passing inside it. Then apply a bunch of weight until it breaks.
In this way the force required should be calculated
F= G*M
Were
F stand by force required
G stand for gravity acceleration
M stand for the final weight to break the tube.

It is a been a while since last time i read a physics book, would be nice if someone will help me out.
That's the right formula for the force in Newtons if G is in m/s-2 and M is in kilograms. But what you have here is a bending force, and that depends also on how long the tube is and where it's supported. And really for strength you need to know the force per unit area (or the "stress"). It all gets quite complicated. You'd have to find equations on wiki and all sorts. But you don't need any of this. Just break the tube. If it breaks where you've repaired it then your repair isn't good enough. If it breaks somewhere else on the original tube then you're good to go. Any repair or joint should be at least as strong as the original tube.
05-27-23, 10:08 AM
#5
unterhausen

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I thought about testing things on my fatigue machine, but the fixturing was always the problem that stopped me from doing that. In the end it will probably break where you hang the weight. I think even with carbon, bike parts fail in fatigue. If they fail in rupture under reasonable forces, then they are actively dangerous. A fundamental aspect of fatigue is that failure occurs at loads that are smaller than yield.
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05-27-23, 11:24 AM
#6
CrowSeph
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Originally Posted by Andrew R Stewart
The classic way to test stuff was to make a few and try various manors of destruction.

Is the test tube a vessel a tube with one end closed to be able to hold a substance? I think of blood and a centrifuge spinning the test tube. Reason why I ask is that what this test tube is to do will have a significant impact of what type of break you are testing for. Which stress are you concerned about? Torsion, tension, shear, bending? What forces will be acting on the test tube when in use?

An interesting question but lacks much context as yet. Andy
Since its a steering tube i think the correct force may be shear and bending (probably bending is the optimal one).
05-27-23, 01:19 PM
#7
unterhausen

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Steering tubes take a lot of bending forces.
05-27-23, 01:47 PM
#8
CrowSeph
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Originally Posted by unterhausen
Steering tubes take a lot of bending forces.
i know. But since we are experimenting with repair methods we want to know the differences.
05-28-23, 06:23 PM
#9
Andrew R Stewart
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My non engineering seat of pants degree says that there's more bending forces acting on a steerer (glad to actually know what kind of test tube we are talking about) than shear ones.

Torsional forces would seem to be small too, what with a headset bearing allowing significant rotational freedom. Hub brakes however will produce some, unsure how significant WRT to the steerer as it's the blades that are usually talked about.

There is some compressive forces due to the stem clamp. A well fitted pressure plug will help counter them.

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05-29-23, 09:01 AM
#10
guy153
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Originally Posted by Andrew R Stewart

Hopefully not to the emergency room
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