Need formula for deflection
Need formula for deflection
(OP)
Hello
Does anyone know the formula for a cantilever beam under a sudden/shock/impact load? Machinery's Handbook, Roark & Young, and a couple of old school books did not have it.I have found the formula for stress in this case.
Thanks
Adam
Does anyone know the formula for a cantilever beam under a sudden/shock/impact load? Machinery's Handbook, Roark & Young, and a couple of old school books did not have it.I have found the formula for stress in this case.
Thanks
Adam





RE: Need formula for deflection
Regards,
jm
"Education is what remains after one has forgotten everything he learned in school." Albert Einstein
RE: Need formula for deflection
Regards,
-Mike
RE: Need formula for deflection
What am I doing wrong?
Adam
RE: Need formula for deflection
I think Blodgetts book on structures has some info on impact loads if that might help.
-Mike
RE: Need formula for deflection
Tried looking for it but my company does not have it. We might run FEA tomorrow but I would still like to know how to do it by hand.
Thanks again
Adam
RE: Need formula for deflection
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.
RE: Need formula for deflection
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
RE: Need formula for deflection
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
RE: Need formula for deflection
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
RE: Need formula for deflection
I have found the formula you are using, how about some sizes of your beams and the load your placing on it.
Its difficult to comment without more information.
regards desertfox
RE: Need formula for deflection
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
RE: Need formula for deflection
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
RE: Need formula for deflection
Yeah we saw these discrepancies. Part of the reason why we were not sure what was going on. Also, I do not remember much on beam theory.
Adam
RE: Need formula for deflection
You cannot use a beam that only has a Sy =21ksi if your expected stress is 20 times this.
What is the application of this beam what exactly are you trying to design.
regards desertfox
RE: Need formula for deflection
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.
RE: Need formula for deflection
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
RE: Need formula for deflection
I confirm those dimensions. I apologize for not specifying how the beam was situated. The force will be acting perpendicular to the depth. I=.000295003 in^4
Adam
RE: Need formula for deflection
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
RE: Need formula for deflection
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
RE: Need formula for deflection
Just thought I should point out that the stress I calculated
for the 'T' section only applies if the impact load hits as you originally described.
If the impact is at some oblique angle to the original then the 'T' section will have a different 'I' value and a different value of stress.
The calculation we have done uses the largest 'I' value you need to calculate stress using the lowest 'I' value.
regards desertfox
RE: Need formula for deflection
Aside from doing the dynamic stress calculations, I would also look very closely at the DISCONTINUITY stresses and ensure that there are no sharp corners or notches that could be a failure point.
My post doesn't doesn't answer your specific question but do consider a cantilevered vaulting pole. It would not even carry the STATIC weight of the vaulter but its design and flexibility is such that when in use it bends almost to horizontal and (hopefully) flicks the athlete over the bar.