##### Introduction

In both engineering and architecture, it is common to find complex structures that are formed by various types of structural elements: walls, bars (pillars and beams), slabs, membranes, cables, etc.

To address these structures, regardless of done by hand or with the help of computer programs, we usually break down the structure into simple substructures, i.e. a single structural type, and calculate each of these substructures. Obviously we have to start by those that do not depend directly on the other, and apply the reactions of each substructure as actions of the next.

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

This type of analysis greatly facilitates the process of calculation, but in some cases we end up in a structure designed on the safe side due to the effects of the deformations produced in each stage, but obviously less economic than it could be.

To illustrate this, we are going to perform the calculation of the following structure:

It is a light cable of ten meters with a sag of 50 cm that hangs from the top of two masts made of steel tubes. We place a load of 5 kN at points 1 m apart. So, there are a total of 9 point loads 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, it adapts to a form of polygon much like a parabola. In fact if we consider it as uniformly distributed load and don’t take the weight of the cable into account, the final shape would be really a parabola. We must add that a parabola and a catenary of 10 m with sag of 50 cm are virtually identical at first glance.

Thus we are going to find the cable tension assuming that distributed apply load is applied Q = 45kN/10m = 4,5 kN/m

The vertical load of the cable at the supports will be Rv = 45/2 = 27,5 kN

The horizontal load is equivalent to Rh = Q · L^{2} / (8 · f) = 4,5 · 100 / (8 · 0,5) = 112,5 kN

Thus the overall tension on the cable will be Rt = √ (Rv^{2}+Rh^{2}) = 115,81 kN

For the calculation of the mast ends (still we do not take the weights into account) we consider that a mast should be able to withstand a vertical load of 27.5 kN and a bending moment on the base M = Rh · H = 112.5 · 5 = 562,5 kNm.

Applying usual safety factors, a possible sizing for these efforts would be:

- Steel cable 1×37 diameter 20 mm
- Steel tube S355 Ø500.12 (diameter 500 mm, thicness 12 mm)

##### Accurate calculation of cable

We will, now, calculate the cable using 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 on the cable will be lower.

At a glance you can see that the deformation of a cable of Ø20 mm with its own weight and loads as 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, before we had 115.81 kN and now only have 104.72 kN. The difference is not yet very large.

##### Integrated calculation of cable-masts

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

To do this we need to develop an iterative process of calculation to recalculate the tension of the wire each time (projected length of the cable will be shorter due to the deformation of the pillar) and the moment created on the mast and then check that the deformation is the same.

We now see that the deformation of the cable is visibly higher (257.6 mm at the center). This means that the tension on the cable and the moment created by the cable on the mast are lower.

The program tells us that the cable tension is 84.79 kN. We see that, compared to the 115.81 kN that we had at the beginning, the integrated calculation makes a difference.

The same thing happens with the masts. Instead of manually calculated the moment at the base of 562,5 kNm, we now see 409,4 kNm through the integrated calculation. The deformation at the top of the mast is 29.6 mm.

The saving is 36 %, both for the cable and the moment on the mast.

##### P-Delta effect

We could still give one more turn in the integrated calculation. It would be good to see the importance of the deformation at the top of the mast by vertical reaction of the cable. This deformation along with vertical reaction of the cable will cause an increase in the moment at the bottom. This, in turn, will again increase the deformation of the top of the mast. Again, an iterative process of calculation will begin to find deformation of the pillar and loss of tension on the cable.

However, in this case, given that the mast has a considerable cross section, effects of deformation are very small, practically nil:

Displacement of the top of the mast: 29,6 mm to 29,7 mm

Moment at the bottom: 409,4 kNm to 409,36 kNm

##### Readjustment of dimensions

If really we have obtained a few smaller values in the tension of the cable and the moments of the mast, we may consider re-sizing these elements. Following the same criteria that 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 load on the wire of 79 kN and a maximum moment on the masts of 379.8 kNm.

That is to say that the loads have lowered again and we could still continue to adjust the dimensions of the cable and the masts.

##### Conclusions

Especially in those structures in which deformations are significant, it is interesting to perform calculation in an integrated way, i.e. with all the elements at the same time, since the results take advantage of the deformation of the structure to give values clearly smaller .

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