Stress/Deflection of Plate 1

[KevinH673]

I'm working on some stress analysis for work. I'm designing a plate, part of which is meant to deflect under a load. Due to this sheet metal plate being somewhat non-straight forward geometry, I'm unsure if I'm attacking it correctly.

I'm putting symmetrical cuts into the plate to allow for easier deflection. Since its not one solid thickness, I didnt know if I could use a cantilever equation. I'm trying to make sure I have an adequate safety factor, but I seem to get numbers I dont trust when I do hand calculations. These numbers also dont match the FEA numbers I get (not that I trust those, either....)

I've included pictures, one of the CAD model showing what the actual plate looks like. The other, a cross section view at the cutout (hopefully this is the correct way to analyze).

Any help or suggestions would be greatly appreciated. I'd like to do well on this project!

http://img210.imageshack.us/i/beam1.png/

http://img121.imageshack.us/i/beam2.png/

I've been using the following equations:

Deflection at B = (Deflection at A)(Height of A/Height of B)

And for stress:

Stress = (3*(Deflection at B)*Modulus of Elasticity*T2)/(2*B^2)

And calculating the safety factor by taking the Yield Stength divided by the Stress.


I have seen this equation, though, for cantilever beams, to calculate deflection:

http://www.advancepipeliner.com/Reso...n_Formulae.pdf

However, for the equation of deflection at a specific point:

deflection = (Force*(B^2)*(3A-B))/(6*E*I)

I'm not sure how to calculate the "I" value. Why would there be inertia, and how do I solve it? When I look up equations, they seem to deal with rotation, and also require I plug in a width value for the plate. I didnt think width would be a factor for this?

Sorry for the long post, and any help is appreciated!

 

[rb1957]

kevin,

i don't understand why you're using an undercut to control deflection ... why not a constant thickness ? why have 10mm thick plates when you could probably get the desired result with 2mm thick ? what material ??

you've said the plate is 200mm long. is this the side of a box ? how big are the other sides. you've said a point load. is that an approximation of the applied loads ? no distributed load ? how big ??

i think you're focusing on the wrong part of the problem ... the deflection at A = deflection at B+slope at B*(A/B).

 

[KevinH673]

rb,

We need the thickness to be about 10 mm for threading in the optic holder. The reason the undercut is to such a small thickness is because we arent using a large amount of force to cause deflection, just a bolt (double threaded, so internal thread moves micrometers per turn). The part where the cut is (and respectively the optic) we only desire to move hundreds of micrometers. We dont want hardly any movement in the "forward" direction, merely to change the angle. We also need to make sure that we stay in the elastic range.

 

[KevinH673]

I'm unsure on calculating the slope. I know the deflection at A by how many turns of the bolt I do. I am not sure on the translation to B, though. I dont expect any bending, or not much, between points A and B due to the thickness of it, I would think that would act almost as a straight line. The movement/bending should happen between the base, and point B. How would you go about accounting for both? I thought a translation of the Force from A to B would allow you to calculate the deflection.

How I would expect it to move

http://img188.imageshack.us/i/draw1t.png/

 

[rb1957]

sounds like quite a challenge. how big is the box ? i ask 'cause we're focusing on deflection between the load and the cantilevered edge and there's a trade-off between deflection to the base (as a cantilever) and deflection with the sides (bending like a beam).

have you thought about using a hinge ?

 

[KevinH673]

I have considered a hinge when we were first looking at this concept here. The problem with it would be too much movement in the "forward and backward" directions. The optic needs to stay, as best as possible, in the same position, while only changing angle. In practice, it of course wont work perfectly, but hopefully we can minimize the movement more with bending. Thanks for the suggestion though.

 

[MikeHalloran]

So you're designing a flexure, basically etched/ sawcut/ EDMed from the soid.
In which case, I'd just, at least at first, assume that the part doesn't deflect where it's 10mm thick.
In which case the flexures become cantilever beams with a force or deflection applied at some distance beyond the end of the cantilever beam.
I think I'd treat that as a cantilever beam with a moment on the end.
To avoid cross- coupling, you might need to iteratively move the location of the flexures a little until the axis of the central hole (and the axis of the attached optics?) doesn't translate, but only rotates, as the force/deflection is applied to the projected tip of the cantilever.
It might be illuminative to calculate the curvature of the cantilever, in which case the geometry of the solution becomes analgous to a tube bending problem, just with an extra large radius and extra small angle.
A couple hours iterating the equations in Excel should give you a good feel for it.




Mike Halloran
Pembroke Pines, FL, USA

 

[rb1957]

i think your grove will act just like a hinge ... it'll set up a slope at B ... what if you put the hinge in the mid-depth of the plate, you want rotation without displacement.

 

[prex]

KevinH673, what you plan to do 'The movement/bending should happen between the base, and point B. How would you go about accounting for both?' is exactly what is in my equations and is also suggested by MikeHalloran:
FB[sup]3[/sup]/3EI is the deflection at 'B' due to F translated at 'B'
F(A-B)B[sup]2[/sup]/2EI is the deflection at 'B' due to the moment of F acting at 'A'
FB[sup]2[/sup]/2EI is the rotation at 'B' due to F translated at 'B'
F(A-B)B/EI is the rotation at 'B' due to the moment of F acting at 'A'
Multiplying the rotation by A-B you get the increase in deflection (as a straight line as you say) from 'B' to 'A'.
All this should sum up to:
FB(B[sup]2[/sup]/3+A[sup]2[/sup]-AB)/EI
Hope this is the same as in my preceding post

prex

http://www.xcalcs.com

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http://www.megamag.it

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: Air bearing pads

 

[rb1957]

i get ... deflection @ A =F/EI*([A-B]^3/6 + A^2*B/2 - A^3/6 + ([A-B]^2/2 - A^2/2)*[A-B] ... maybe it's the same

 

[rb1957]

yeah, it's the same as prex's.

but i really don't think you will get the results you're looking for, rotation without translation

 

[KevinH673]

Hi guys. This project was put on the backburner for a little while, and I've been back to it. I took your advice and went to the textbooks. I'm really not trying to use this site as a crutch, but I do find it as a good source to clear things up.

It may prove the design doesn't work. It is not my design, but I'd atleast like to validate it if it doesn't.

prex, back to your equations. Why do you sum up all of these? I thought if you choose one way to analyze it (such as using the moment), why do you also calculate using force such as a cantilever beam equation?

If you guys have some reading you could direct me to, I'd appreciate it. Trying to learn (or re-learn I should say) some of this.

 

[hydtools]

Kevin, as I understand what you are doing you are creating a displacement with a threaded connection. This creates a known displacement. I think you need to start with the known displacement, at A the free end, and determine resulting displacment at some other location, at B. You are describing a displacement input problem rather than a force input problem.
Perhaps this example may serve you:

http://www.engineersedge.com/beam_bending/beam_bending10.htm

Knowing the end displacement, calculate the displacement at the unknown point load and the unknown point load located between the free end and the fixed support. Then calculate the stress in the flexing length to evaluate plastic or elastic result. I assume the threaded adjustment does not introduce a moment to complicate the deflection of the unloaded end of the beam/plate.

Ted

 

[prex]

KevinH673, it's simply substitution of a system of forces with an equivalent one. The force at A is equivalent to a force at B plus the moment F(A-B).
I'm not sure of what you are trying to do exactly (perhaps because I had forgotten everything of this thread): you say you want to (in)validate a design, but against what? You need something to compare with to validate.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[KevinH673]

Well, I'm trying to make sure I have all my "ducks in a row" so that I can predict how this thing will behave. I can control the deflection at A, as there will be a bolt pressing against it controlling it. My goal is to predict how point "B" will deflect relative to "A". I also have FEA to analyze, which I only am using to validate the hand calculation.

 

[KevinH673]

"The force at A is equivalent to a force at B plus the moment F(A-B)."

prex, wouldn't that mean the force at A is larger than the force at B? Shouldn't it be the opposite?

Also, this claims slope is the integral of M/EI with respect to x, so slope = Mx/EI, or F(A-B)*B/EI as has been posted above, and that d^2y/dx^2 = M/EI; however, another site I had read claims d^2y/dx^2 to be Mx/EI. Which is correct? I have the link to one site here:

http://www.roymech.co.uk/Useful_Tables/Beams/Beam_theory.html

and the image that shows the equation:

http://www.roymech.co.uk/images11/beam_21.gif

However, I will attach the picture of the equation that says d^2y/dx^2 = Mx/EI

Which is correct? I assume the former, since it has been mentioned before. I'm just confused by the discrepency.

 

[rb1957]

i believe your posted equation says M is function of x, sort of like f(x).

d2y/dx2 = M/EI ... M is a function of x ... integrate for slope (dy/dx), and again for displacement

 

[KevinH673]

rb, ok this makes sense then. I thought the x was an actual value. Thanks.

 

[prex]

To be clearer:
"The force at A is equivalent to the same force translated at B plus the moment F(A-B)."

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[gadkinsj]

if you have a point load. why don't you make your "cantilevered" section much much narrower, and keep some thickness in your "undercut" area so you can truly apply a straight beam calculation? With the accuracy of measuring devices you should be able to pick up any changes in geometry.

I think the current approach is untenable as far as getting consistent results, because with that thin a piece of material you have much more potential problems with non-homogeneous effects, alot of stresses from machining.

 

[trainguy]

Kevin,

This problem screams for an FEA that incorporates geometric non-linearity. Do you have access to such expertise? If not, you should sub-contract it. This would help validate the complex hand-calcs.

tg