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!

 

[GregLocock]

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

Assuming B is at or above the step, OK, if the ratio of T2 to T1 is large.

"And for stress:

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

No. That is not true for beams of varying cross section.

IF B is ///exactly/// at the transition of T2 to T1 then the maximum stress in T1 at the wall is Stress = (3*(Deflection at B)*Modulus of Elasticity*T1)/(2*B^2)

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

Well we're already wildly unsafe as the situation has not been correctly analysed. Did you do elastic beam theory at uni? If so revisit your notes. If not then it is interesting, it is not all that difficult, but it isn't trivial.




Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[rb1957]

you say you've got a plate ... how wide ? how are the sides supported ? it looks like you've done a section thru the applied load (at the middle of the plate ?).

if your plate is wide, then you can model the plate as a cantilever (if you've got a fixed reaction, two rows of fasteners), and you need to decide on an effective width (probably something less than a 45deg angle projected from the load).

From roark, deflection of a cantilever with a point load is P*L^3/(3*EI) for a beam with a constant I. For a beam with changing I, read up beam analysis in a strength of materials text ... it's too long to explain here, but it ain't hard math (just a lot of it).

 

[KevinH673]

Seems I is mass moment of area. Would I use (1/12)bh^3? With b being the width (I didnt give this, assuming the plate is ~200 mm wide, and the height of the cut is "B", would the I value be I=(1/12)(200)(B)^3?

As far as plate being wide, yes Im going to assume >200mm

 

[rb1957]

i probably wouldn't use the entire width of the plate, 'cause the sides would react some of the load. maybe use the distance between the load and cantilevered edge, ie L.

but notice my two cavets ... is the edge fixed ? and how are you accounting for the different thickness ? the formula is for constant thickness, if you're worried about bending stress in the plate, use the thinner t (near the cantilever edge).

btw, your FE is probably giving you way high stresses and displacements.

 

[KevinH673]

I'm still hung up on the I calculation.

It says, I=(1/12)b*h^3

Is the "b", the width (or say 200mm, or less if we dont take the whole thing), and the h is the thickness (say .5mm?, what I designated T1).

Then, in the deflection calculation, the L would be the height of the cutout I drew in the cross section, labeled "B"?

 

[rb1957]

if this "plate" is 0.5mm thick, it ain't a plate ... formulaes provided are good for deflection < 1/2 the thickness. if you apply a plate formula you get (other than the wrong answer) a very large displacement and corresponding stress. the thin membrane deflects and reacts the applied load with in-plane tension. you Might assume that a strip the width of the plate (200mm i think)is bending in an arc and the applied load is reacted by a component of tension, something like T = P/2/sin(theta) and you determine theta from a right triangle with sides R, R-d, and W/2 (=100mm).

but this is for school, isn't it ?

 

[KevinH673]

It's not for school, for a project at work.

Well, the plate would be something like 10 mm thick, it would just have that cut/groove put in to make it bend under less force than a 10mm thick plate would.

 

[MikeHalloran]

You need to get an engineer to help you, right there in person.


Mike Halloran
Pembroke Pines, FL, USA

 

[Twoballcane]

"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?"

I agree with Mike, if you don't understand the difference between mass moment of inertia (lbm*in^2) and inertia for stiffness (in^4) and why width is important. A friendly suggestion would be to get a Mechanical Engineer to help you out. Are you a designer or engineer?

Tobalcane
"If you avoid failure, you also avoid success."

 

[Twoballcane]

I would also like to add that the stress you are calculating is just one piece of the analysis. You should continue to calculate the principal stresses and ultimately the Von Missies stress so you can use Sy and Su correctly. So be carful.

Tobalcane
"If you avoid failure, you also avoid success."

 

[KevinH673]

I am an engineer, I just havent dealt with this since school.

If I make a plate .5mm thick, 3 mm high, and 10 mm wide in CAD, run it through FEA, I get a deflection of 5.21 µm and a stress of 27.24 MPa.

Using the equations I've mentioned, I get 5.57 µm and a stress of 32.03 MPa.

So, these seem accurate. Its the step that throws me off.

The bending should only occur (in a marginal scale) at the "thin" part. The part above that, at 10 mm thick, should not flex.

 

[Twoballcane]

Sorry I ment Von Mises

Tobalcane
"If you avoid failure, you also avoid success."

 

[rb1957]

kevin,

i'm sure we don't mean to sound like "pissy little sh!ts" but you're asking a whole bunch of really basic questions and you're using some very complicated pieces of software (FEA) that will mislead you quicker than you can imagine.

And the job you're trying to do is really quite tricky ... if you're trying to control displacement with an undercut, why not use a thinner plate to start with ?

you're making little test models that you can check with hand calcs whihc is good, but it'll only get you so far; you're using complicated tools 'cause you're trying to solve a complicated problem, and it's hard to hand calc a complicated problem.

i imagine that you've tried your full size plate in the FE and gotten a ridiculous answer. this is alomt certainly because the FEM doesn't apply membrane forces to react the applied load.

 

[KevinH673]

rb, yes it is a bit silly. Its frustrating applying some of these basic equations when I'm not quite sure if they apply in the geometry I'm using. The reason I ask them is for some validation that I may use them in the manner here (due to the "step" in thickness). Some of the responses have made me feel a quite foolish for asking, perhaps I should have better thought out my questions.

Yes, the FEA maybe making things worse but I'm hoping its not so far off I cant use it for a reference. Perhaps it is.

I wouldve liked to use one solid, thinner plate, but I need the majority of the plate to be thicker (~10mm) so I can tap it, to hold some alignment optics. I would like to use something like a bolt at the top of this plate, so that I can get very little movement at the bottom (where the optic would be) to "tune" it with relative precision.

 

[dhengr]

Kevin:
Yes, you probably should feel a bit foolish for that design, and most certainly you should understand your problem better, and think through your question better before asking. Don’t be too embarrassed though, there are plenty of people here trying to do problems that they don’t have the foggiest idea about what they are doing, applying formulas they’ve picked out of the air and don’t understand, and if all else fails trying to apply FEA, assuming that’ll solve all their problems. The CAD drawing you show is not a particularly good design for what you are trying to do, but you also still leave so much design info unsaid that it’s kinda tough to comment. How much length do you have for the canti.? What’s the weight of your optics, I’ll bet its base is actually fairly stiff to support its own weight. Where are you aiming it and how do you plan to tune it, for what alignment? Can you add a 5 or 10 pound wt. under it to help tune the bm. and how much deflection do you want the bm. to have? How is this contraption fixed to the wall, there will be deflection there too. Does vibration of the wall effect your equip.? I’m sure there are another dozen questions, but they are further into the process. A .5mm strip will not do what you want, machined into the edge of a 10mm pl. You would want spring steel for that, and to design it as a canti. leaf spring or flat spring.

The equations you want are for a canti. bm. with self wt. on it, plus a point load at the optics, and a length L. Any Mechanics of Materials book should show you those, but you must understand the subject well enough to apply them. And, you shouldn’t be puzzling about the moment of inertial of the canti. pl. ( I = b d^3/12 ) for a stress and deflection calc., and mass moment of inertia because this isn’t a dynamics problem. That puzzlement shows a fair lack of understanding of the basics of the problem, and prompted the “get an engineer to help you” reply. Our basic beam formulas and theory are based on small deflection theory, and generally go to hell when you exceed yield or have large deflection to thickness ratios, thus the membrane or thin shell theory comment.

Not knowing any of the answers to the questions above, I’m shooting from the hip, but try this on for size. My dimensions are intended to suggest general proportions, I haven’t run any of the numbers which I would do long hand, not with FEA, and you want to work well within the elastic range so you don’t cause a permanent deformation. You want a welded tee section with the base drilled for bolts to the wall. Say 100mm high x 200mm long & 5mm thick. Your canti. bm. is 3 - 4mm thick (that’s d above) x 200mm wide ( b above) & 300 or 400mm long ( L above). Slot holes in this pl. to match your optics base, with slots running from the tip toward the wall, so you can move your load and change the deflection. Make part of you added weight a base pl. tapped to match the optics base, but sliding with the optics, under the canti. pl. Provide a means of adding weights out at the tip of the canti. so you can further change its curvature and delta, in the vert. plane perpendicular to the wall. If you wish you could add more load to one corner tip of the canti. pl., that would cause it to twist and change the alignment of the optics in a vert. plane parallel to the wall.

Work well within the elastic limit of the canti. pl., an approx. stress calc.; you want a fair approx. of delta, so you can set up your equip., but then you are going to fine tune it anyway, then an exact delta calc. isn’t that important. Not even worth turning on the computer. You gotta study a little and really start to understand your own problem. The rest is your’s to do.

 

[prex]

KevinH673, these should be the equations you are looking for, calling W the width where the thickness is T[sub]1[/sub]:
f[sub]B[/sub]=FB[sup]2[/sup](B/3+(A-B)/2)/(EI) deflection at 'B' due to load F and end moment F(A-B)
E=modulus of elasticity
I=WT[sub]1[/sub][sup]3[/sup]/12 moment of inertia of section
[&alpha;]=FB(B/2+A-B)/(EI) slope at 'B' due to load F and end moment F(A-B)
f[sub]A[/sub]=f[sub]B[/sub]+[&alpha;](A-B) total deflection at 'A'

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

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: Air bearing pads

 

[rb1957]

prex, those equations account for two thicknesses ?

 

[prex]

No, they account for only T[sub]1[/sub] thickness (the only one that appears in them). With a ratio of 0.5 (T[sub]1[/sub]) to 10 (T[sub]2[/sub]) as proposed by OP, the deflection of the thicker portion is clearly negligible (and for the same reason the full width of the thinner portion should be taken into account, no need for a 45° angle spread from load).

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]

denghr, thanks for suggestions but this needs to be air tight. I can not go in and add weights, as the box this will go in needs to be sealed. I'm trying to use hand calculations, not FEA, to design it. I am here for guidance on using the correct equations.

prex, thank you for the help. I have tried your equations under 1/2 lbf load (2.18 N). For a plate 59 mm high (point of F), 5 mm high "cut", .75" thickness, and 20 mm wide, my deflection seems to be 7.75 micrometers at the top, and 6.65 micrometers at the point "B". These seem a bit low? Also, wouldnt the deflection at the top part (point A) be much higher than point B?

Could I not take the load at A, and solve for the load at B using moments, F@B=F@B*(A/B), or is this not valid due to change in thickness?

Then, take the load@B, and apply it to a beam of thickness T1, width "W", Length B

Where deflection=(F@B*B^3)/(3*E*I)
where I = W*T1^3/12

I'm hesitant to analyze the whole thing in one piece using deflection equations is that the top, thicker part will not bend (or will bend a negligable amount), all bending will happen at the bottom, thinner part.