**Introduction**

It is common to find complex structures in both engineering and architecture. We understand complex structures that are formed by various type of structural elements: walls, bars (pillars and beams), slabs, membranes, cables, etc.

To tackle with these structures, regardless of which is done by hand or with the help of computer programs, the structure usually is broken down to simple substructures, i.e. with a single structural type, and analyze each of these substructures. Obviously we have to start by those that do not depend directly on the other, and go by applying the reactions of each substructure as actions on the next.

It would be, for example, the case of a metal truss resting on two pillars of walls. First of all we analyze the metal truss with any method of articulated bars and then consider the reactions obtained as loads applied to the wall.

This type of analysis greatly facilitates the process of calculating, but in some cases the effects of the deformations produced in each stage are wasted, resulting in a structure designed on safety side, but obviously less economic than it could be.

To illustrate this problem, we will perform the calculation of the following structure:

It’s a ten-meter light cable, with a sag of 50 cm hanging from the top of two pillars formed by steel tubes. A load of 5 kN every 1m are applied on the cable, so there is 9 forces applied with a total of 45 kN.

**Manual calculation**

First we will analyze this structure manually. A cable of 10 m long and 50 cm of sag has an initial form of catenary. If we apply 9 point loads, the cable shall take a form of funicular polygon much like a parabola. In fact if we consider a uniformly distributed load and don’t take the weight of the cable into consideration, the final shape would be really a parabole. We must add that a parabole and a catenary of 10 m and 50 cm of sag are virtually identical at first glance.

So we are going to find the cable tension assuming that we apply a distributed load Q = 45kN / 10 m = 4.5 kN/m

The vertical tension of the cable at the supports will be Rv = 45/2 = 22,5 kN

The horizontal tension is equivalent to Rh = Q · L2 / (8 · f) = 4,5 · 100 / (8 · 0,5) = 112,5 kN

Thus the overall tension of the cable will be Rt = √ (Rv2+Rh2) = 114,73 kN

For the calculation of the pillar ends (still do not take into account the weights) we consider a pillar that is able to withstand a vertical load of 27.5 kN and a bending moment at base

M = Rh · H = 112,5 · 5 = 562,5 kNm

A possible sizing for these efforts, applying usual safety factors, would be:

- Steel cable 1×37 diameter 20 mm
- Steel tube S355 Ø500.12 (diameter500 mm, thickness 12 mm)

**Accurate calculation of cable**

We will, now, analyze the cable with a special software for this type of structures (WinTess), and in principle we look forward to benefiting from the predictable deformation of the cable under the influence of the loads. If the cable is bent, the sag will be higher and therefore the load that the cable will be subjected to will be lower.

At first glance it shows that the deformation of a cable Ø20 mm, with its own weight and the loads described above, is not negligible. Specifically 110.7 mm. This increase in the sag causes a tension on the cable less than calculated previously.

Indeed, we had 115,81 kN and now have only 104,72 kN. The difference is not even very large.

**Integrated calculation cable-pillars**

Let’s take a step further in the integration of the calculation. So we will consider the balance due to the deformation of the pillar at bottom subjected to a horizontal force on the top. It is a second-order calculation, since the balance is based on an unknown end position.

To do so will have to develop a process of iterative calculation that recalculate each time the tension of the cable (light cable will be shorter due to the deformation of the pillar), and then verify the moment and the deformation that undergoes the pillar. We now see that the deformation of the cable is visibly higher (257,6 mm at the center point). This means that the tension of the cable will be lower and also the moment the side pillars are subjected to. The program indicates that the cable tension is now 84,79 kN. We see that, against the 115,81 kN we had at the beginning, the integrated calculation makes the difference.

The same thing happens with the pillars. Instead of a moment at the base of 562,5 kNm calculated manually, we see now we have 409,4 kNm through the integrated calculation. The deformation of the head of the pillar is 29.6 mm

Saving is 36%, both cable and for the moment at the pillar.

**P-Delta effect**

We could still give one more turn in the integrated calculation. It would be to see what is the importance of the deformation of the head of the pillar on the vertical reaction of cable. This deformation, along with the vertical reaction of the cable, will cause an increase in the moment at the base. So, it will again increase the deformation of the head of the pillar and, again, will begin an iterative process of deformation of the pillar and loss of tension on the cable.

However, in this case, given that the pillar has a considerable section and a deformation is very small, effects are practically nil:

Displacement of the head of the pillar: 29,6 mm to 29,7 mm Moment at the base: 409,4 kNm to 409,36 kNm

**Readjustment of the dimensions**

If really we have obtained smaller values in the tension of the cable and the moments of the pillar, we may consider re-sizing these elements. Following the same criteria as we have previously used, we get:

- Steel cable 1×37 diameter 18 mm
- Steel tube S355 Ø470.10 (diameter 470 mm, thickness 10 mm)

A new calculation gives us a sag of 314 mm on the cable, a force of 79 kN on the cable and a maximum moment of 379.8 kNm on the pillars.

That is to say the efforts have again lowered and we could still continue to adjust the dimensions of the cable and the pillars.

**Conclusion**

Especially in those structures in which the deformations are significant, it is very interesting to perform the calculation in an integrated manner, i.e. with all the elements at the same time, since the results are taking advantage of the deformation of the structure to get smaller dimensions.

However, in complex structures, the need for a tool that will analyze all the structural types at the same time, will force us to have a powerful software that will surely be expensive and difficult to master.

Ramon Sastre,

June 2012