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Need formula for deflection
- Thread starter aepeszel
- Start date
If everything is linear and you know the stress from the impact load could not back out the strain from this as a first order approximation?
Regards,
jm
"Education is what remains after one has forgotten everything he learned in school." Albert Einstein
Regards,
jm
"Education is what remains after one has forgotten everything he learned in school." Albert Einstein
It depends on the magnitude and shape of the time history curve describing the impact load. A load applied suddenly (no initial velocity) will produce deflections and stresses twice the static value for the same magnitude load. Ramp loading, pulses, and other dynamic or impact loads will of course produce different results. You will need to provide more information.
Regards,
-Mike
Regards,
-Mike
- Thread starter
- #4
We are looking at a member that is cantilevered, when the body is dropped, this member takes the initial impact. I used the approximate formula from Machinery's Handbook for our condition and am now finding an odd development. When I try to use a "T" section with the added rib perp. to bearing face, and parallel to load, I get twice the stress.
What am I doing wrong?
Adam
What am I doing wrong?
Adam
I don't have a Machinery's Handbook with me today so I can't really comment on your approach. However, the equation you are using could be adddressing the same effect that I mentioned in my other post, so the stress should be twice. But I'm only guessing at this point.
I think Blodgetts book on structures has some info on impact loads if that might help.
-Mike
I think Blodgetts book on structures has some info on impact loads if that might help.
-Mike
- Thread starter
- #6
If you know the height, h, from which you are dropping the mass, you simply equate the loss of potential energy of the mass equal to the strain energy of the beam.
The strain energy is given by the integral of;
M^2/2EI*dx over the length of the beam
Assuming the shape of the deflected beam is that of a load on its end, I get
M=k(L-x), where k is some constant m( the equivalent end load) to be determined. The integral becomes
k^2*(L-x)^2/2EI*dx over L which integrates to
-k^2*(L-x)^3/6EI from 0 to L=k^2*L^3/6EI which is the strain energy. Equating this to mg*h where
m= mass
h= dropping height,
you can get the value of k from which you get the maximum bending moment of kL and thence the stress. If the end deflection is not insignificant next to the dropping height you would add this to the height and iterate the solution.
The strain energy is given by the integral of;
M^2/2EI*dx over the length of the beam
Assuming the shape of the deflected beam is that of a load on its end, I get
M=k(L-x), where k is some constant m( the equivalent end load) to be determined. The integral becomes
k^2*(L-x)^2/2EI*dx over L which integrates to
-k^2*(L-x)^3/6EI from 0 to L=k^2*L^3/6EI which is the strain energy. Equating this to mg*h where
m= mass
h= dropping height,
you can get the value of k from which you get the maximum bending moment of kL and thence the stress. If the end deflection is not insignificant next to the dropping height you would add this to the height and iterate the solution.
Hi aepeszel
I have a formula here from a book called "Mechanics of Materials" by E.J.Hearn for deflection of a beam due to an impact load:-
X=Xs*[1 +/- (1+(2*h/Xs))^0.5]
where X=deflection
h= height of falling mass
Xs= static deflection
If you calculate your impact load as a static load and calculate the normal static deflection which you can then use as Xs to calculate X.
However if your beam is subject only to a sudden load without impact then as stated earlier the beam would see double the normal static stress and double the normal deflection.
regards desertfox
I have a formula here from a book called "Mechanics of Materials" by E.J.Hearn for deflection of a beam due to an impact load:-
X=Xs*[1 +/- (1+(2*h/Xs))^0.5]
where X=deflection
h= height of falling mass
Xs= static deflection
If you calculate your impact load as a static load and calculate the normal static deflection which you can then use as Xs to calculate X.
However if your beam is subject only to a sudden load without impact then as stated earlier the beam would see double the normal static stress and double the normal deflection.
regards desertfox
- Thread starter
- #9
Desertfox,
Actually it is a true impact loading. I found some sources list it as sudden or shock, which is why I added those. We already looked at the static load and have set up in Excel so your formula was calculated quickly. Thanks.
I looked for this book here but did not find it. Would E.J. Hearn's book have any additional info on the stress formulas? I used the "approximate" formula from Machinery's Handbook, but it does not seem to work out when we go from a rectangular cross section to one of a "T". Is there another formula that is needed for this shape?
Thanks for the help.
Adam
Actually it is a true impact loading. I found some sources list it as sudden or shock, which is why I added those. We already looked at the static load and have set up in Excel so your formula was calculated quickly. Thanks.
I looked for this book here but did not find it. Would E.J. Hearn's book have any additional info on the stress formulas? I used the "approximate" formula from Machinery's Handbook, but it does not seem to work out when we go from a rectangular cross section to one of a "T". Is there another formula that is needed for this shape?
Thanks for the help.
Adam
Hi aepeszel
Which formula are you using from machinery's all my book says is stresses can be calculated in the normal way.
By this I mean you find the deflection X from my earlier formula and work backwards from the deflection to find the stress.
Also bear in mind that going from rectangular block to a T setion will also change the second moment of area value.
regards desertfox
Which formula are you using from machinery's all my book says is stresses can be calculated in the normal way.
By this I mean you find the deflection X from my earlier formula and work backwards from the deflection to find the stress.
Also bear in mind that going from rectangular block to a T setion will also change the second moment of area value.
regards desertfox
- Thread starter
- #12
Desertfox,
The beam is 1/8" X 1.8125" and 1.8755" long.
The impact is 14 lbs from 30" height. We use
Aluminum A13/413 alloy with a Sy=21ksi, E=10.3X10^6 psi.
When we go to a "T" shape, the Head is the same dims with the leg 1/8" wide and overall height of 1/2".
Using the approximate formula p=a * sqrt(6QhE/LI) we
get 428 ksi. When using the "T", this stress nearly doubles.
I am going to work thru these again by hand to see if I missed something.
Thank you
Adam
The beam is 1/8" X 1.8125" and 1.8755" long.
The impact is 14 lbs from 30" height. We use
Aluminum A13/413 alloy with a Sy=21ksi, E=10.3X10^6 psi.
When we go to a "T" shape, the Head is the same dims with the leg 1/8" wide and overall height of 1/2".
Using the approximate formula p=a * sqrt(6QhE/LI) we
get 428 ksi. When using the "T", this stress nearly doubles.
I am going to work thru these again by hand to see if I missed something.
Thank you
Adam
Hi aepeszel
I need to point out that if your Sy is only 21ksi how can you have a stress of 428ksi?
I take it your rectangular beam is 1.8125" wide by 1/8" deep?
In addition your beam physically might fall outside normal beam theory ie it might be classed as a wide beam, also its length to depth ratio is only 15:1 should be nearer 20 for beam theory.
regards desertfox
I need to point out that if your Sy is only 21ksi how can you have a stress of 428ksi?
I take it your rectangular beam is 1.8125" wide by 1/8" deep?
In addition your beam physically might fall outside normal beam theory ie it might be classed as a wide beam, also its length to depth ratio is only 15:1 should be nearer 20 for beam theory.
regards desertfox
- Thread starter
- #14
- Thread starter
- #16
Desertfox,
It is a foot on a electronics box. It was designed, built and now breaking. We are trying to analyze it and redesign it. Being a die cast part, we want to be able to back up our reasoning behind the redesign.
Adam
p.s. I was not here when they designed the initial product.
It is a foot on a electronics box. It was designed, built and now breaking. We are trying to analyze it and redesign it. Being a die cast part, we want to be able to back up our reasoning behind the redesign.
Adam
p.s. I was not here when they designed the initial product.
Hi aepeszel
I cannot get sensible answers from the figures you have given me even considering it as a beam.
Please provide the I value for the beam and confirm that
currently the depth of the beam is only .125" in depth and 1.8125" wide before going to the T section.
regards desertfox
I cannot get sensible answers from the figures you have given me even considering it as a beam.
Please provide the I value for the beam and confirm that
currently the depth of the beam is only .125" in depth and 1.8125" wide before going to the T section.
regards desertfox
- Thread starter
- #18
Hi aepeszel
Okay I have run the maths and get the same answer as you
for the rectangular beam.For the 'T' section however I obtain a stress of 32psi which is a much lower stress but something I would of expected as the increase in the second moment of area is huge.
I can only think your mathmatical error is in either your
second moment of area for the 'T' shape or the distance from the neutral axis to the extreme fibre.
I am also puzzled as to why there is a requirement to drop a 14lb weight from 30" onto a leg of an electronic box could you expand on this?
I assume the box as three or four legs yet you seem to have only the one failing? is it the same leg that fails on all boxes?
If it where my job to sort this out I would think I would try to find why that leg was failing and the cause before making any design changes.
regards desertfox
Okay I have run the maths and get the same answer as you
for the rectangular beam.For the 'T' section however I obtain a stress of 32psi which is a much lower stress but something I would of expected as the increase in the second moment of area is huge.
I can only think your mathmatical error is in either your
second moment of area for the 'T' shape or the distance from the neutral axis to the extreme fibre.
I am also puzzled as to why there is a requirement to drop a 14lb weight from 30" onto a leg of an electronic box could you expand on this?
I assume the box as three or four legs yet you seem to have only the one failing? is it the same leg that fails on all boxes?
If it where my job to sort this out I would think I would try to find why that leg was failing and the cause before making any design changes.
regards desertfox
- Thread starter
- #20
Desertfox,
Actually the box is 14 lbs. UPS drop test is 30" height. It tends to be whatever leg impacts that breaks off. We looked at the castings, their method of production is not the cause nor the material. It seems to be the design itself.
Thank you much
Adam
Actually the box is 14 lbs. UPS drop test is 30" height. It tends to be whatever leg impacts that breaks off. We looked at the castings, their method of production is not the cause nor the material. It seems to be the design itself.
Thank you much
Adam
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