reduction of stiffness matrix when spring fixity is involved 3

[rk_19]

hi, I have a vertical cantilever beam which is having spring fixities in all 6 dofs (at the base)... there is a load in vertical and in lateral direction at the free end.... after developing the stiffness matrix, how do i do prepare the reduced K matrix - basically, how do i solve for the displacements/rotations in this case..thanks for any help on this

 

[XR250]

Is this a homework problem? Otherwise, seems like maybe it can be done by hand using superposition.

 

[rk_19]

hi, this is not a homework.. .. i have developed a code which can solve a typical frame structure assuming pinned or fixed foundation.. basically i eliminated the corresponding degrees of freedom from the stiffness matrix depending whether it is fixed or pinned...now i want to solve it considering spring foudnation... so i need some help on how the matrix reduction happens... any notes/reference materials?? could you please elaborate on the superposition method you have hinted here.. thanks

 

[gravityandinertia]

Your software would calculate the reactions at the support from the external loads applied, they all must ultimately be resisted at the base. From there you would use the spring stiffnesses at the base along with those loads to calculate displacement there. From there, you would reduce the matrix per the usual method you use when you have a fixed displacement at a certain location.

 

[XR250]

https://en.wikipedia.org/wiki/Superposition_principle

 

[IDS]

You need to adjust the stiffness matrix to account for the reduced stiffness of the beam at the spring restraints. You can do that without any coding by inserting dummy spring members fixed at the base and with the required stiffness, or add a routine to your code to generate the same matrix for the specified spring stiffnesses.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rk_19]

@gravityandinertia (Structural)

if the structure was a cantilever fixed at base, my reduced K matrix would be of size 6x6, now in this case, where I have springs in 6dof. how do I get to reduced matrix.. if I consider the base fixed and calculate the reactions, based on the stiffness of springs, I can get the displacements at the base (due the reactions I calculated assuming encastre fixity).. so how do I move from here... thanks for your guidance

 

[rk_19]

@IDS (Civil/Environmental), are telling to treat the spring supports as beam with relevant stiffness ? thanks

 

[Denial]

A crude approach, not necessarily applicable in all cases, but applicable for the structure you describe.[ ] Assemble your global stiffness matrix ignoring all external restraints.[ ] In the case you describe this matrix will be 12x12 since you have 6 degrees of freedom at the bottom of the column and 6 at the top.[ ] Now for each of the degrees of freedom that are supported by springs you simply add the value of the spring stiffness to the appropriate term on the leading diagonal of your global stiffness matrix.

 

[rk_19]

Denial (Structural), thanks for the suggestion.. the spring at bottom has got 6 dofs (ie; axial stiffness in 2 horizontal+1 vertical and rotational stiffness in all these 3 cords, total = 6)....how do I add these to the stiffness coeff of the node representing the base?

 

[Denial]

I understood your original post to be saying that the base of your cantilever was supported by six individual springs, three translational springs each aligned with the global Cartesian axes and three rotational springs also each aligned with the global axes.[ ] Each of these springs has its own stiffness value, and because of their orthogonal alignments their actions are completely uncoupled.[ ] You now seem to be suggesting you have only a single spring ("the spring at bottom has got 6 dofs"):[ ] if this is the case then I do not adequately understand your problem, so ignore what I said above and ignore what I am about to say below.

If my original understanding still applies, then the global stiffness matrix you create by ignoring all external restraints is 12x12.[ ] Each row and column represents one of the structure's 12 overall degrees of freedom.[ ] Each one of the 12 terms along the matrix's leading diagonal represents the stiffness "experienced by" one of the degrees of freedom if all the other degrees of freedom were rigidly clamped.[ ] Hence the effect of a spring support at that degree of freedom can be modelled merely by adding the numerical value of the spring's stiffness to that term on the leading diagonal.[ ] What you are doing, in effect, is treating each spring as a very simple member (so simple it has a 1x1 member stiffness matrix), then merging that simple member stiffness matrix into the global stiffness matrix.

This is really basic stuff.[ ] So basic it is actually quite hard to describe.

If you have any follow-up queries I will not be able to field them, as I am going away for a fortnight to one of those increasingly rare places that do not have ready internet access.

 

[IDS]

@IDS (Civil/Environmental), are telling to treat the spring supports as beam with relevant stiffness ?

Yes, that is what I suggested, but reading the recent posts from Denial, I agree with his point that if you want to model separate springs for each freedom, with no interaction, then it will be simpler to just add the relevant stiffness to each element of the leading diagonal for the base node.

One other suggestion is that while you are experimenting with these things it is much easier to work in 2D, and when you have worked out a procedure it is fairly straightforward to extend it to 3D.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rk_19]

Denial (Structural), thanks for your suggestion.. may b I was not well with explaning the problem.. ur understanding is correct - I have 6 independent springs representing an independent dof... thanks for your help, I will try this out...
thanks IDS (Civil/Environmental) too

 

[JasonMcKee71]

You should find the element of the diagonal stiffness matrix that represent the degrees of freedom that are supported by the springs and add to them the spring constant. Then you should inverse the matrix and multiply it with the forces vector in order to get the displacements of all degrees of freedom, including the spring-supported ones.

Jason McKee
proud R&D Manager of
Cross Section Analysis & Design
Software for the structural design of cross sections
Moment Curvature Analysis
Interaction Diagrams
Reifnorcement Design etc.

 

[rk_19]

JasonMcKee71 (Structural), thanks ...

 

[cal91]

Denial, I have a question.

I am writing a nonlinear program with spring supports using the method you described. It all works except the spring support displacement adds to itself with each iteration.

For example if I have a simple column supporting 1 kip and the base of the column has a support stiffness of 1 kip/in, then I'll get the following displacements at the base of the column...

1st iteration: 1 in
2nd iteration: 2 in
3rd iteration: 3 in, etc.

I'm trying to figure out where the error would be and thought you might have an idea based on your responses. Any idea?

 

[cal91]

Important to mention that the column is only resisting 1 kip with each iteration, so thats not the problem. (resistance doesn't add with each iteration)

 

[cal91]

Nevermind, I think typing the question helped me figure it out. I was adding the support stiffness to the global stiffness matrix, but wasn't adding the support stiffness * support displacement to the global resistance vector.

Thanks anyways!

 

[BAretired]

I'm a bit rusty on matrix methods, but perhaps this would help:

thread727-416155

BA

 

[Denial]

Cal91.[ ] I'm please that, by my absence, I was able to help.

(Many a time I have had the same experience as you seem to have had here.[ ] The mere act of attempting to describe a problem in sufficient detail for another person to understand it shines a new light on it for me.)