solution method for a plate with springs at corners 1

[EngForm78]

I have a problem with a suspension system were the center of gravity is not at the geometric center. So, under static load each spring is loaded differently. I would like to find the deflection at each spring.

To analyze the problem I set up a very stiff plate and put a spring element at each corner. I used Pro/E Mechanica for the analysis and I am not aware of a rigid element, so I made a material with very high stiffness and made sure the results showed little to no bending in the plate.

Is there a way to hand check(solve) a problem like this. A three spring problem is straight forward, but with four springs and the constraint they must remain in plane complicates the problem. I thought there might be a solution using a plane equation and iterating on a solution but I am not sure how to start.

Thanks in advance.

 

[rb1957]

actaully i solved a very similar problem. i had to work out the landing gear loads for the A340, in the early days of the design when they were still deciding between 3 or 4 MLG. I had to account for different spring rates at each gear, different ground profiles (sloping one way or the other, or both ways.

it sounds as though your problem is a little simpler in that i'm assuming each of your springs has the same stiffness (doesn't matter too much, the method can be expanded to do this.

i wrote a simple FORTRAN program (yes, it was back in those days) to iterate a balance between the four spring forces, their required deflections, and the requirement for these to be planar. these days you could easily do this on a spreadsheet.

the approach i'd take would be take any three springs and determine the static balance, the initial force in each spring. this will also set up the initial displaced plate. this in turn determines the displacement and force in the fourth spring. this is the out-of-balance force, since it wasn't part of your original balance. this force is equivalent to a force and moment acting at the CG of the original three springs. this'll modify the original static balance loads; hence the iteration until the out-of-balance is zero.

of course, this assumes a rigid plate, but you could expand this solution to include a flexible plate.

of course, the FE solution is trival.

 

[Denial]

The rigid plate lies in the XY plane, and the springs run in the Z direction. The only applied force (which happens to be the plate's self weight) is vertical.

This problem has three degrees of freedom: the vertical movement of some designated, but essentially arbitrary, point on the plate (DZ); the rotation of the plate about the X axis (RX); and the rotation of the plate about the Y axis (RY). From the (unknown) values of these three degrees of freedom you can calculate the vertical displacements of the four corner points using simple geometry, because you know that the plate is rigid. Once you know these vertical displacements you know the forces in each of the springs. If you assume small displacements and rotations, the equations you derive for the spring forces as functions of (DZ,RX,RY) will be linear.

Corresponding to these three degrees of freedom, you have three equations of equilibrium: vertical equilibrium; moment equilibrium about the X axis; and moment equilibrium about the Y axis.

Three unknowns, and three linear equations. Dust off your high school mathematics and solve for DZ, RX and RY. Substitute back into your spring-force equatins and you have your answer.

Ideal for a spreadsheet. No iterations are necessarily required to solve linear equations, although some approaches (such as the one rb1957 suggests above) can be iterative.

 

[rb1957]

don't you have four unknowns and three equations ?

 

[crisb]

I think there is a simple analogy to a structural problem. Calculate the centre of gravity, area, and moments of inertia about 2 axes (for the springs). Then one can calculate deflection of the c of g and rotation about the 2 axes.

 

[Denial]

rb1957. It is the three components of the plate's displacement that are the unknowns, not the spring forces. The method I outlined could be applied to a problem with five springs, or even a thousand springs, and it would still involve only three unknowns.

 

[rb1957]

that's fair ... transfer the load to the spring group CG, so you have a force and two moments. the moments are reacted by a couple, and the force equally distributed.

it seems almost too straight forward !

i think this'll accept different spring constants, it'd move the CG towards the stiffer spring, but it might fall apart if the springs aren't laid out on a square ... probably not, it just the bolt group analogy (reaction is proportional to the distance from the bolt group CG)

good one, denial

 

[GregLocock]

The reason there are only 3 unknowns is that if you know the lengths of three of the springs the length of the fourth one is implied.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[crisb]

There are only 2 variables as the plate will rotate about an axis normal to a line drawn between the load and the spring CG.

 

[Denial]

Crisb. At the risk of appearing pedantic, I believe that your approach still admits the problem is a three-variable one. All you have done is change two of the variables from RX & RY to Rtheta & theta (where theta is the orientation of the axis about which the plate will rotate, and Rtheta is the amount of that rotation).

You do appear to have uncoupled one of the three equations from the other two, thereby making a manual solution easier.

However, I am not sure that your approach actually works, even in the simple case where the four springs are linear and of equal stiffness. Bear with me here. Envisage the plate externally retrained against both rotations. Under the action of the load it will deflect vertically to the point where the total of the four (equal) spring forces equals the vertical load. The resultant of the spring forces will be acting through the "centre of gravity" of the springs. Therefore (as you indirectly assert) the moment disequilibrium for this unrotated displaced position will act about "an axis normal to a line drawn between the load and the spring CG". But the assumption in your approach is that this moment will cause the plate's rotation to be about this same axis.

I do not believe this assumption is correct. If you view the four springs as united to form a single beam-column, then the moment axis and the rotation axis are parallel only if they are also parallel to one of the principal axes of the beam's cross-section.

 

[crisb]

Denial. Thanks for pulling me up there!
Taking a structural analogy the shear centre (also known as flexural centre) is only coincident with the CG for doubly symmmetric sections. I believe that the moment and rotation axes will be parallel only if the section is doubly symmetric, or if the moment axis is parallel to a principal axis.
So I think my statement was correct if the four springs were at the corners of a rectangular plate (and springs have equal stiffness) and not if they were not.
As for the number of variables I take your point but guess it could be anything up to 10 (x and y coordinates of the 4 springs plus loading loint).

 

[crisb]

Denial. Ignore the above comment. Thinking about a long narrow plate loaded 45 degrees from the CG its clear my statement was incorrect.

 

[Denial]

Crisb. I realised yesterday, about an hour after my post, that I owed you at least half of an apology. But I could not get back onto the site to post an addendum.

In fact we are both right. In the more general case where the location of the springs is arbitrary, it is correct that the moment axis and the rotation axis can not be assumed to be parallel. In the specific case that I believe you were discussing, where the springs are located at the corners of a square, you are correct.

In my post I said that the moment axis and the rotation axis are parallel only if they are also parallel to one of the principal axes of the beam-column's cross-section. This statement needs to be further generalised. The requirement is actually that the moment axis be parallel to an axis about which the beam-column has zero product of inertia (ie I[sub]XY[/sub]=0). For the square arrangement, this condition is satisfied for any axis orientation. Not so for a rectangular arrangement.

I went back and re-read EngForm78's original post carefully. He never actually gave us the arrangement of his four springs. The notion of a square arrangement crept in as the discussion built up, perhaps as a result of EngForm78's use of the word "plate".

 

[prex]

The question raised by the OP, whether there's a way to solve the problem by hand calcs, has not been answered. Here is my answer.
The three equations of equilibrium may be written as:
[Σ]k[sub]i[/sub]z[sub]i[/sub]=P
[Σ]k[sub]i[/sub]z[sub]i[/sub]x[sub]i[/sub]=Px[sub]o[/sub]
[Σ]k[sub]i[/sub]z[sub]i[/sub]y[sub]i[/sub]=Py[sub]o[/sub]
where the simbols are easy to figure out (the plate lies in the xy plane before applying the load).
The condition of planarity is written by choosing three base vertices not colinear and writing an equation for each of the remaining ones. The equation consists in equating to zero the determinant of a matrix formed by rows in the form
|x[sub]i[/sub] y[sub]i[/sub] z[sub]i[/sub] 1|
where i is for the three base vertices and the other one.
Of course in this general form, with a plate of any polygonal shape, any number of vertices and different spring constants, only a spreadsheet may be of help for a quick solution.
Let's take now the special case of a rectangular plate of sides 2a and 2b, assuming the sides are parallelel to x and y, the origin is in the center, vertex 1 is in the first quadrant and the vertices are numbered counterclockwise.
The condition of planarity is now simply:
z[sub]1[/sub]-z[sub]2[/sub]+z[sub]3[/sub]-z[sub]4[/sub]=0
Now assuming the four springs have equal rates, and letting
H=P/4k
X=x[sub]o[/sub]/a
Y=y[sub]o[/sub]/b
the equations of equilibrium become
z[sub]1[/sub]+z[sub]2[/sub]+z[sub]3[/sub]+z[sub]4[/sub]=H
z[sub]1[/sub]-z[sub]2[/sub]-z[sub]3[/sub]+z[sub]4[/sub]=HX
z[sub]1[/sub]+z[sub]2[/sub]-z[sub]3[/sub]-z[sub]4[/sub]=HY
and the solution is easily calculated as:
z[sub]1[/sub]=H(1+X+Y)
z[sub]2[/sub]=H(1-X+Y)
z[sub]3[/sub]=H(1-X-Y)
z[sub]4[/sub]=H(1+X-Y)

It's that simple!

prex

http://www.xcalcs.com

Online tools for structural design

 

[rb1957]

personally, i'd thought we had laid a trail of bread crumbs, and i'd left it to the OP to sign off ("thx, got it" or "help, didn't get it") or show some sign of something.

but i do agree with you prex, it is pretty simple.

 

[EngForm78]

Yes, Thanks for all the help. Very helpful. Sorry about the delay in response, I was gone for a few days.

EngForm78