liam.mccaffrey
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| posted on 6/10/05 at 03:55 PM |
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Beam Bending
I have a deflected cantilever and I need to know the amplitude of the oscillations when it is released.
I know that if there is no energy loss then it will be twice the initial deflection. This beam is damped though isn't it? I.E the oscillations
will decay
anyone have any thoughts?
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ned
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| posted on 6/10/05 at 04:00 PM |
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quote: Originally posted by liam.mccaffrey
anyone have any thoughts?
make mine a double, it may come to me
beware, I've got yellow skin
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donut
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| posted on 6/10/05 at 04:07 PM |
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quote:
deflected cantilever, amplitude, oscillations when it is released.
What the Dickens are you on about
-moves hand over head in swoosh motion-
Andy
When I die, I want to go peacefully like my Grandfather did, in his sleep -- not screaming, like the passengers in his car.
http://www.flickr.com/photos/andywest1/
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DaveFJ
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| posted on 6/10/05 at 04:13 PM |
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Ahh yes
was contemplating the oscillations of my bendy beam just last night.......
there was quite an energy loss and my beam was definately damped though..   
[Edited on 6/10/05 by DaveFJ]
Dave
"In Support of Help the Heroes" - Always
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mookaloid
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| posted on 6/10/05 at 04:21 PM |
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You're twanging your ruler on the edge of your desk aren't you?
I had a formula for this from my university days. Bu**ered if I can remember what it was though
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flak monkey
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| posted on 6/10/05 at 04:21 PM |
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A serious post...
Can you model it as a second order system? (Ie mass, spring, damper).
The formula being:
F = M (d^2t/dt^2) + B(dx/dt) + kx
Where M = mass, B = damping coefficient, k = spring constant, x = displacement, F = force...
Use the transfer function to find out the natural frequency and damping factor.
You would need to know a fair bit about the system, but if you could do some experiments to get some numbers, you can work out all the coefficients
(natural frequency, damping factor etc) by working backwards.
Its all to do with system dynamics. I am sure you could find something on the net about second order systems.
Try the trusty google...
David
[Edited on 6/10/05 by flak monkey]
Sera
http://www.motosera.com
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zilspeed
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| posted on 6/10/05 at 04:43 PM |
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I gave up on mechanics when they asked me to believe in CM4 i.e centimetres to the 4th
I can understand CM, can believe in CM2, I can even get CM3.
CM4 is taking it too far though. I gave up after that. I refuse to believe in things I can't picture.
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Cita
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| posted on 6/10/05 at 05:00 PM |
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quote:
The formula being:
F = M (d^2t/dt^2) + B(dx/dt) + kx
Where M = mass, B = damping coefficient, k = spring constant, x = displacement, F = force...
Oh that,just a piece of cake...if I can find the cake
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liam.mccaffrey
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| posted on 6/10/05 at 05:17 PM |
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flack
The problem I have is that the system obviously loses energy though damping. But I can't do any experiments its not practical not sure how to
get the damping coeff. or the spring constant. Is there an straightforward way of calculating K and B from dimensional data, beam is a tube by the
way
I have already calculated the deflection and max stress
I was thinking about using an energy method given that I have calculated the above stuff. What do you think?
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Triton
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| posted on 6/10/05 at 05:46 PM |
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Over my head and back again that one
My Daughter has taken over production of the damn fine Triton race seats and her contact email is emmatrs@live.co.uk.
www.tritonraceseats.com
www.hairyhedgehog.com
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steve_gus
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| posted on 6/10/05 at 06:36 PM |
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thats pretty heavy stuff for a red haired muppet!
atb
steve
http://www.locostbuilder.co.uk
Just knock off the 's'!
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dnmalc
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| posted on 6/10/05 at 07:06 PM |
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If you cant do the simple experiments then I would suggest that you get a CAE package with modal analysis material data and you should be able to set
up a simple mesh that will determin the natural frequency and hence the damping.
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dnmalc
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| posted on 6/10/05 at 07:06 PM |
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If you cant do the simple experiments then I would suggest that you get a CAE package with modal analysis material data and you should be able to set
up a simple mesh that will determin the natural frequency and hence the damping.
[Edited on 6-10-05 by dnmalc]
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britishtrident
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| posted on 6/10/05 at 08:32 PM |
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quote: Originally posted by dnmalc
If you cant do the simple experiments then I would suggest that you get a CAE package with modal analysis material data and you should be able to set
up a simple mesh that will determin the natural frequency and hence the damping.
Last thing you should do if you can't do the basics -- FEM is dangerous in the wrong hands.
[Edited on 6/10/05 by britishtrident]
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britishtrident
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| posted on 6/10/05 at 08:41 PM |
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The initial amplitude must be somewhere between the initial delflection and twice the initial delflection -- as it won't have much damping
somewhere between 1.6 and 1.9 times the initial deflection won't be far out.
[Edited on 6/10/05 by britishtrident]
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liam.mccaffrey
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| posted on 6/10/05 at 10:39 PM |
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i agree BT thats what i thought, I don't think there is an easy way to do this is there?
Many thanks for the help guys!
And all the sarky comments
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NS Dev
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| posted on 7/10/05 at 07:26 AM |
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no easy way out I'm afraid, no.
If you want some really mind boggling answers to stuff like this you need the eng tips forum - Eng Tips
but still doesn't get around the fact that what you are trying to do is actually quite complex to do in any detail.
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Coose
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| posted on 7/10/05 at 03:14 PM |
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I think you'd need to apply Hookes Law, therefore you'll need to know the Young's Modulus of the material which will entail
calculating stress and strain.
Stress = force to produce extension of the bar/cross sectional area of the bar.
Strain = total elongation of bar/length of bar.
Youngs modulus = stress/strain.
Hookes law = (stress x length of bar)/(cross sectional area of bar x Young's Modulus).
I can't be bothered to type any more, so look here instead....
http://www.ntmdt.ru/SPM-Techniques/Basics/2_SFM/text184.html
Spin 'er off Well...
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flak monkey
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| posted on 7/10/05 at 03:25 PM |
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quote: Originally posted by Coose
I think you'd need to apply Hookes Law, therefore you'll need to know the Young's Modulus of the material which will entail
calculating stress and strain.
Stress = force to produce extension of the bar/cross sectional area of the bar.
Strain = total elongation of bar/length of bar.
Youngs modulus = stress/strain.
Hookes law = (stress x length of bar)/(cross sectional area of bar x Young's Modulus).
I can't be bothered to type any more, so look here instead....
http://www.ntmdt.ru/SPM-Techniques/Basics/2_SFM/text184.html
Sorry....but it is much more complex than Hookes law. You need to apply the system equation that I mentioned, which includes hookes law, but adds in
the effect of damping from air resistance etc, along with the effect of the mass of the object dynamically. Then by using the transfer function you
can apply some real coefficients to the equation if you can do some experiments. Otherwise, its just an estimate.
David
Sera
http://www.motosera.com
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clbarclay
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| posted on 7/10/05 at 10:31 PM |
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Beam Bending
From my distant past I seem to remember that:-
The angle of the dangle equals the throb of the knob if the lust renains constant.
Hope that is of some use.
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liam.mccaffrey
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| posted on 8/10/05 at 11:54 AM |
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I found in a old text book a table giving % amplitude decreases per cycle which i used to give an estimate. It turns out though that we didn't
need to do it. What we have to work out now is more complex but is possible considering the info we have. Thanks for all your help guys! (especially
Flakmonkey)
This little community has a hell of a lot of knowledge
Thanks again
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Coose
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| posted on 10/10/05 at 08:00 AM |
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quote: Originally posted by flak monkey
quote: Originally posted by Coose
I think you'd need to apply Hookes Law, therefore you'll need to know the Young's Modulus of the material which will entail
calculating stress and strain.
Stress = force to produce extension of the bar/cross sectional area of the bar.
Strain = total elongation of bar/length of bar.
Youngs modulus = stress/strain.
Hookes law = (stress x length of bar)/(cross sectional area of bar x Young's Modulus).
I can't be bothered to type any more, so look here instead....
http://www.ntmdt.ru/SPM-Techniques/Basics/2_SFM/text184.html
Sorry....but it is much more complex than Hookes law. You need to apply the system equation that I mentioned, which includes hookes law, but adds in
the effect of damping from air resistance etc, along with the effect of the mass of the object dynamically. Then by using the transfer function you
can apply some real coefficients to the equation if you can do some experiments. Otherwise, its just an estimate.
David
You're correct, and I was building up to it by giving the calcs for Hookes Law and then I was going to move on but lost interest..... Hence
giving the link!
I always think it's best to source these things yourself and then you get more of an understanding....
Spin 'er off Well...
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