Introduction

In some cases, especially in the perimeter of tensile constructions we find solutions based on bipods (a mast and a stay cable) or tripods (one or two masts with two or one stay cables respectively) and even more complex sets of masts and stay cables.

Stay cables usually hinder, since they are usually very thin, not easily seen and can cause accidents. They also occupy a space that is not usable, so very often you tend to eliminate them.

To do this, it is necessary to have an element embedded in the base, able to work as a cantilever and support the force that is applied at its top, as seen in the attached image.

The typical triangular shape responds to the distribution of forces in a cantilever. However, sometimes for aesthetic or economical reasons it is preferred to have a vertical element of constant section, as a column or a tube, although it means we use more material than necessary.

In a first approximation it could seem that this is a more economical solution than using a bipod or tripod, but this can be misleading, since one of the great differences between these solutions is precisely the foundation. This foundation will have to be able to transmit the following forces to the ground:

Fh = horizontal force
Fv = vertical force (pulling or pushing force)
Mh = moment due to the horizontal force
Mv = moment due to the vertical force (if the element is not completely vertical)

 
Forces

The force applied at the top of the element is transmitted directly to the base and therefore to the foundation. So a horizontal and a vertical force (pushing or pulling) will exist. These forces have already been studied in previous sections  and we cannot add very much.

There is, however, a particularity that we must take into account. It is the weight of the element. It can be a non-negligible value and therefore, it will be necessary to add the weight of the element to the pushing force or subtract the weight of the element from the pulling force, before applying the forces to the foundation.

As for the balance of these forces, we must remember the balance mechanisms:

 Vertical force.

  • self weight: foundation, pavement (if any), of the element itself.
  • vertical friction of the lateral faces of the footing against the surrounding soil (provided there are no problems of high humidity or clay soils retraction).
  • Shera force of concrete pavement slab (if any).

Horizontal force.

  • at rest earth pressure resistance, (we  will only use passive earth pressure resistance in order to accomplish with the safety factor and after evaluating the effects of the necessary displacement to generate this passive earth pressure resistance).
  • horizontal friction forces on both sides of the foundation.
  • friction force on the base of the footing (as long as there is no danger of overturning, due to a significant moment)
 
Moment

Especially the horizontal force, but also the vertical force if the element is inclined, applied at the top of the element will cause a moment on the foundation that can become the most decisive action to be balanced. This moment will have to be combined with the moment that is generated when there is a horizontal force at the top of the footing (as it has been studied in previous sections).

As we have done in previous cases, for horizontal forces and vertical forces, we will analyse what are the foundation resources to balance this moment.

  • Foundation weight, embedded element weight and vertical force (when it is a pushing force).
  • Vertical friction of the opposite side to the one that has the turning point.
  • Vertical friction of the lateral sides.

If we take into account that the second and third resources may not exist (retraction of the soil if it is clayey or high humidity in the contact between the foundation and the soil), we only have the first resource as a reliable factor.

In this sense, balancing moment of the weight of the foundation depends not only on the weight itself but also on the distance from the centre of gravity of the foundation to the overturning centre point. We have already mentioned that it will be somewhere close to the point of application of the resultant of the soil pressure at rest. Therefore, it seems clear that not only will we have to provide a sufficiently heavy foundation, but it will be good to use extended foundations (in the direction of the horizontal force) in order to improve the balancing moment.

It is difficult to recommend an ideal geometry, since it depends on many things, but it seems reasonable to suggest proportions similar to 3x1x1 (axbxh) since it has an important section to deal with the moment and a length that will allow to contribute with a balancing element, with a sufficient self weight.

However, we must note a different fact regarding the behaviour of the foundation. Especially in the case of vertical forces, both pushing and pulling, the behaviour of the foundation has been commented through reinforced concrete mechanisms as struts and rebars. In the case that concerns us now, a new action appears and can be very important, it is the bending moment applied to the footing and has to be countered by the cross section of the footing and its reinforcement.

So, in this case more than in the others, it will be necessary to complete the design of the foundation with the analysis of reinforced concrete. It will not be enough to give dimensions a, b, h to the foundation but it will be necessary to decide which reinforcement has to be incorporated in order to behave like an element under a significant bending moment.

 
Examples

The aim is to design a foundation able to withstand an inclined force of 11.3º with the following values

Ft = 5,01 t (Fx = 5 t ; Fy = +1 t )

This force is applied to an inclined metal mast 2.4 m high and an inclination such that the top point is  0.5 m behind  according to the vertical line that passes through the axis of the anchor of the mast with the foundation. The weight of the mast is 0.2 t.

Soil data in which the foundation is located are:

  • soil bearing capacity at 1m depth = 2,5 kg/cm² (25 t/m²)
  • soil density: γ = 1,9 t/m³
  • soil internal angle of friction: φ = 35º
  • roughness of the side faces of the footing = normal
  • density of concrete = 2,3
  • soil-concrete angle of friction:  f(f) = tg (2φ /3) = 0,43
  • soil pressure at rest coefficient:  K0 = 0,4

 

Case 1

We assume that the footing is located in an area where there is no type of pavement above the foundation. Therefore, the rain directly affects the soil and can nullify the friction resistance between the footing and the adjacent soil. However, the depth of the shoe protects the bottom of the footing and in this case we will take into account the friction.

We can pre-dimension the footing with proportions 3:1:1. To be able to choose initial values you need some preliminary scores or rely on previous experiences. We assume a footing with dimensions 4×1,3×1,3.

Balancing overturning:

  • volume = 4 · 1,3 · 1,3 = 6,76 m³
  • footing weight = 6,76 · 2,3 = 15, 55 t
  • total weight = 15,55 + 0,2 – 1 = 14,75 t
  • overturning moment = 5 · (2,4 + 2/3 · 1,3) + 1 · (½ · 4 + 0,5) = 18,85 mt
  • balancing moment = 14,75 · 2 = 29, 5 mt
  • safety factor for overturning = Me / Mb = 29,5 / 18,85 = 1,56

It seems, then, that in this aspect dimensions are appropriate. Let’s check now  what happens with forces.

Vertical force:

  • total weight = 14,75 t
  • safety factor = 14,75 / 1 = 14,75 >> more than enough

Horizontal force:

  • Fx = 5 t
  • soil pressure at rest reaction:  REo = (b · h) · ½ · K0 · g . h² =
  • (1,3 · 1,3)· ½ · 0,4 · 1,9 · 1,3² = 1,085 t
  • friction at the bottom of the footing = Pt · f(f) = 14,75 · 0,43 = 6,34 t
  • safety factor = (1,085 + 6.34) / 5 = 1, 485 >> enough