## 2 Methodology
### The Stress Response of the Structure by a Simplified Model  
In order to understand the structural behaviour of the system, based on the loads considered, a simplified modelling is proposed based on the study of the mechanical behaviour of the splay arches, which are the final receivers of the load system. Considering the arch as a tri-articulated isostatic structure (Figure 13), we can calculate the support reactions from the three equations of plane statics [39; 40; 41]. For this purpose, we model uniformly distributed loads in each of the three cases studied, choosing the most unfavourable situation. The system will reflect maximum horizontal reactions in the keystone and vertical reactions in the brackets, so that the resultant at the start of the arch originates the stresses that it must absorb [42]. Where Q (KN/m) is the uniformly distributed load considered in each case; L (m) is the span of the arch and f (m) is the deflection or edge of the arch; H (KN) (horizontal reaction at the keystone, which is equal to that of the supports so that the system remains in equilibrium) and V (KN) (equal vertical reaction at the two supports), at the three articulated points of the system (keystone and two supports); R (KN) (resultant of the vertical and horizontal components at the supports); and according to the equations of statics, the horizontal reactions are obtained:  
Figure 13: Tri-articulated isostatic arch. Graphical sketch of tri-articulated isostatic arch. (S= salmer, support stone; K= keystone; L= light of the arch; H= horizontal reaction; V=vertical reaction; R= Resulting reaction; Q= linear distributed load; f= deflection of the arch). Q. \(\frac{L}{2}\frac{L}{4}\) = H.f; H = Q. \(\frac{L^{2}}{8f}\) and the vertical reactions would be V = Q. \(\frac{L}{2}\) We can obtain the Resultant on the supports R = \(\sqrt{H^{2}+V^{2}}\) For each of the stages analysed and studying the worst case for each of them,  
**STAGE 1. ** The largest uniformly distributed load is \(Q_{1}\)= 35 KN/m, so the following is obtained  
H \({}_{1}\) = \(Q_{1}\):\(\frac{L^{2}}{8.7}\) = 83.12 KN; and the vertical reactions would be \(V_{1}\) = \(Q_{1}\):\(\frac{L}{2}\) = 166.25 KN  
We can obtain the resultant in the supports \(R_{1}\) = \(\sqrt{H_{1}^{2}+V_{1}^{2}}\) = 185.871 KN, so that the maximum stress to which the segment is subjected at the support (salmer) during the initial state of the works, that is, in its four hundred years of service life, is \(\sigma 1=\frac{185.871}{400\times 200}\) = 2.32 MPa = 2.32 (N/mm\({}^{2}\)). **STAGE 2. ** The largest uniformly distributed load is \(Q_{2}\)= 13 KN/m, so we obtain  
H \({}_{2}\) = \(Q_{2}\):\(\frac{L^{2}}{8.7}\) = 30.87 KN; and the vertical reactions would be H\({}_{2}\) = \(Q_{2}\):\(\frac{L}{2}\) = 61.75 KN  
We can obtain the resultant at the supports \(R_{1}\) = 69.036 KN. \(\sqrt{H_{1}^{2}+V_{1}^{2}}\) = 69.036 KN, so that the maximum stress to which the segment is subjected at the support (salmer) at the moment when the slabs have been demolished is \(\sigma 1=\frac{69.036}{400\times 200}\) = 0.86 MPa = 0.86 (N/mm\({}^{2}\)). **STAGE 3. ** The highest uniformly distributed load is \(Q_{3}\)= 35 KN/m, which coincides with one of the most unfavourable states in which the initial state of loading (Stage 1) was found, so the same results are obtained, and, therefore, once the restoration work has been completed, the maximum stress to which the voussoir is subjected at the support (salmer) is \(\sigma 3=\frac{185.871}{400\times 200}\) = 2.32 MPa = 2.32 (N/mm\({}^{2}\)). The compressive strength of the limestone used to make the arches has not been tested, but according to the examples consulted [43; 44], we can consider a minimum admissible stress of 10 N/mm\({}^{2}\) (MPa), and for the lime mortar in the joint of the voussoirs of 10 N/mm\({}^{2}\).