Funicular arch

This example shows how to find the geometry of an arch which is funicular to an uniformilly applied load case by setting constraints to form and force diagram.

The input geometry is the following circular arch:

We are going to assume that:

• The load in each node is equal to 10 kN.

• The two extremities are pin supports (restrain on x, y).

With this example we want to check if the input geometry is suitable for a uniformilly distributed load, and if not which is the shape corresponding to such load case.

1. Form Diagram with loads and support

We input the FormDiagram with the function Crerate Form Diagram from the lines provided in the Rhino template, applied loads and reactions are translated as lines in the diagram. The result look like this:

We assign the forces with the button Assign Forces. This predefined shape allows the selection of only one independent edge. We select one of the edges representing the applied loads and assign the load -10. Here edge #2 was selected.

We press OK and validate the sign of the applied load, we must also restraint the two extremity vertices assigning it as supports/anchors with the button Identify Anchors. The result will look like the following image:

With this information updated we can go to the drawing of the Force Diagram.

2. Drawing of the Force Diagram

We press the button Create Force Diagram which generated the ForceDiagram highlighted in the following image. We observe that the force 10 kN is only applied to the edge selected as independent and the other applied loads are different to 10 kN. This happens because the original input geometry was not form found to correspond to a uniformilly distributed load. Instead we see forces varying from 10 to 24.6 kN.

If we want to find out the correct geometry of the structure for the uniformilly applied load case we must assign constraints to the problem.

3. Constraint the equilibrium to uniformilly distributed load

To achieve the geometry corresponding to the uniformilly distributed load from our hypothesis we first assign the default constraint by clicking in the button Assign default constraints. We see highlighted the constraints in white to the form and force diagrams:

• Edges in the form and force diagrams with orientation fixed are highlighted in white.

• Vertices in the form and force diagram with line constraint are highlighted in white.

• Anchored vertices are shown in red.

Since white is the default color for the constraints remember to change your backgroud color. Grey is prefered. The visualisation of the arch after the default constrains should look as follows:

On top of the default constraints applied, we assign target forces to the load edges with the same magnitude of the applied load. This reflect as a constraint on the target length of the dual edges in the force diagram. To assign these additional constraints click on the button Assign edge constraints and select the option ForceMagnitude. Select all applied loads and give the target magnitude (10 kN). The sign +/- is not important here since it will always take the same sign of the applied load. The target forces will show in white in the form and force diagrams.

Now, we applied constraints to the form and force diagrams which force them to look for a new equilibrium. On the next section we will perform the bi-directional update.

3. Bi-directional update

Before performing the bi-directional update (Form<->Force) turn on the bi-directional update in the Settings. If you continue without turning it on you will receive an error message.

Now that the bi-directional module is activated you click on Update Both Diagrams. Both diagrams will update and the result should be as displayed below:

The final solution is shalower than the original input and now the geometry is a parabola instead of a circle. The differences among the two solutions are displayed below.

The constraints can be turned on and off, in the latter the original force in the edge is displayed. Aditionally in the Inspect diagrams > ConstraintsTable a table is called showing all constraints and the current force in the edges, as well as the constraints in the vertices applied.

The constraints can be erased from the form on the Menu function IGS> Constraints> Remove all constraints.

4. Additional modifications

If the constraints are not erased, or are reassigned more modificaitions could be done on top of the current design. We will explore two simple ones. In the first modification we move one of the supports as in the image below:

We then press in the button Update both diagrams and both diagrams are matched according to the constraints and the new support position. The resultant structure is still a funicular for the uniformelly distributed load case but with supports in different elevations which make the vertical reaction forces unbalanced: 38.3kN and 21.7 kN as shown in the next figure:

The second additional modification imposes an additional target force to one of the reaction forces. Here we set the horizontal reaction force to have magnitude of 30 kN. As a result, the funicular change its height and in the force polygon the horizontal reaction force has its length decreased. The following images show this modifications.

Many more modifications are possible and playing around with the previous unidirectional is still possible.