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Currently, I'm working on a mechanical mechanism where nodes are connected via beams. This is very comparable to planar truss mechanism analysis, but in my case, the deformations are large (relative to the mechanism). Instead of deforming beams, for which Hooke's law is often used, my beams act as (linear) springs, which can have positive or negative stiffness. I'll include an example with my question.

In the end, I want to know what force is needed (in x,y-direction) to have the purple node (N4, see figure) at every point on the grid. The location of N4 is therefore known and independent of stiffnesses/prestresses.

enter image description here

The green beams act as the springs, the yellow nodes are fixed and the blue node is free. The purple node (N4) is displaced over the grid, and the rest of the free nodes are moving accordingly, which is calculated via the well-known $\mathbf{f} = \mathbf{K}\mathbf{d}$:

enter image description here

Herein, the unknown displacement of the free nodes will be calculated first. In this example, we've two unknown forces and two unknown displacements, which can be solved with the four equations. When all the displacements are known, the same formula will be used to calculate the forces in all nodes.

I'm interested in the force needed (in x,y-direction) of the node which got the predefined translation. Because the deformations are relatively high, this process is done incrementally. The stiffness matrix $\mathbf{K}$ will therefore be changed at every step, corresponding to the new configuration of the mechanism.

Now, I want to include a pre-tension in the mechanism. As trivial as it sounds, I didn't find much information about that in various papers.

From intuition, I would suggest that the total force in the nodes is just a summation of $\mathbf{f_{ext}} = \mathbf{K} \Delta \mathbf{d}$ and the global pretension vector: $\mathbf{f_{int}} = \mathbf{T}^T \mathbf{f_{loc}}$, so that:

$\mathbf{f_{ext}} = \mathbf{K} \Delta \mathbf{d} + \mathbf{T}^T \mathbf{f_{loc}}$ for the first step (the incremental process is out of scope, for now).

However, the global stiffness matrix should be modified by the pretension, since the pretension makes the whole mechanism stiffer. But using the aforementioned reasoning, this is not the case, since the stiffness matrix stays unaffected.

Therefore, I'm wondering if I make any mistake in this reasoning. Any help would be greatly appreciated. In addition to that, any tips about papers/books involving pre-tensioning using the direct stiffness method would be very helpful.

Thank you in advance!

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  • $\begingroup$ Do I understand correctly that the pre-tension in some members is due to this deformation analysis that you describe? I assume that some members are in tension and some are in compression after you finish prescribing these displacements? Are all the members in your mechanism (truss?) joined by essentially pin joints? It would be very helpful if you can include a picture of the initial geometry and one after the displacements have been prescribed/calculated. $\endgroup$ Commented Jun 1, 2023 at 18:21
  • $\begingroup$ Thank you for your suggestion, @BillGreene, I've added two figures now. I have the intention to add the pretension before deformation happens. All members are indeed coupled with (frictionless) pin joints. $\endgroup$
    – Patrick
    Commented Jun 1, 2023 at 19:36

1 Answer 1

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So presumably you intend to pretension the truss members using some other prescribed loading condition?

When you say you want to account for the effects of this pretension, I assume you are thinking of something like a thin cable which can resist transverse loads after it has been pretensioned? Accounting for such an effect requires including the nonlinear strain-displacement relations in the formulation so using just the simple FEM linear stiffness matrix is not sufficient.

To include this type of effect in your analysis, a geometrically non-linear formulation for the truss elements is needed. You are attempting to account for one particular aspect of such an analysis in your current procedure. You are incrementally updating the stiffness matrix based on the deformed geometry. Such a procedure is, at best, a partial approximation to including the nonlinear strain-displacement terms in the truss element formulation.

If you search for geometrically-nonlinear truss finite element you will find many books and papers on the topic. For example, you might find this book to be useful.

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  • $\begingroup$ Thank you for your answer. I'm not really sure if I understand the analogy with the thin cables. It might be useful to rephrase the problem a bit: the mechanism must be seen as a frame where the nodes (blue) are connected with linear springs (green). Therefore, we won't deal with buckling/nonlinear effects due to plastic deformations and suchlike. Pretension happens in the direction of the spring (axial pretension). A program generates a certain design, and the pretension is applied. The question is then: what force is needed (at N4) to hold the mechanism into place? (1/2) $\endgroup$
    – Patrick
    Commented Jun 4, 2023 at 12:10
  • $\begingroup$ When the pretension is applied, node N4 undergoes a prescribed displacement. Again, it must be calculated what force is needed to hold the mechanism in place. I indeed found in various papers that some nonlinear effects might take place when applying a prestress on pin joint frames, but those papers mainly dealt with structures not really comparable with my mechanism. (2/2) $\endgroup$
    – Patrick
    Commented Jun 4, 2023 at 12:16
  • $\begingroup$ You say "we won't deal with buckling/nonlinear effects due to plastic deformations and suchlike". Plastic deformation and other forms of material nonlinearity have nothing to do with geometric nonlinearity. If you say geometric nonlinearity is unimportant for your truss, why are you attempting to account for it in an ad-hoc way by performing the analysis using the deformed geometry? If nonlinearity is unimportant, just include the so-called pre-tensioning displacement along with your other applied displacements and you're done. $\endgroup$ Commented Jun 4, 2023 at 22:55

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