PRACTICE 2 PROYECTO Y CONSTRUCCIÓN DE PUENTES. 1º Máster Ingeniería de Caminos. E.T.S.I. Caminos, canales y puertos (Ciudad Real) 01/06/2016

Transcription

1 PRACTICE 2 PROYECTO Y CONSTRUCCIÓN DE PUENTES 1º Máster Ingeniería de Caminos E.T.S.I. Caminos, canales y puertos (Ciudad Real) 01/06/2016 AUTHOR:

2 CONTENT 1. INTRODUCTION BRIDGE GEOMETRY AND MATERIAL BRIDGE GEOMETRY MATERIAL MECHANICAL CHARACTERISTICS OF THE SECTION ANALYTICAL RESULTS ANALYTICAL RESULTS MODEL I GEOMETRY OF THE MODEL LOAD SIMPLIFICATIONS RESULTS MODEL I MODEL III LOAD CASES DIVISIONS ANALYSIS CORRECTIONS OF THE MODEL RESULTS MODEL III MODEL II GEOMETRY OF THE GRILLAGE LONGITUDINAL BEAMS TRANSVERSAL BEAMS LOAD CASES RESULTS MODEL II COMPARISON OF RESULTS COMPARISON OF THE MODELS BY LOAD CASES CONCLUSIONS REFERENCES

3 1. INTRODUCTION The aim of this practice is modeling the deck of a concrete bridge by using different models. First of all an analytical solution is going to be obtained, that allows comparing the output results of the three other models that consist of: Model I: Single Beam. The deck is going to be modeled, by using just a single beam that will be located at the centre of gravity of the section. Model II: 2D Grillage. The bridge is modeled by using longitudinal and transversal beams. Model III: 3D Model. The bridge is modeled using shell pieces. Once we have implemented the models in the program SAP 2000 v.15, we have to test them for the different load cases that are exposed in the enunciated. 2. BRIDGE GEOMETRY AND MATERIAL 2.1. BRIDGE GEOMETRY It is pretend to study the strategies for modeling the deck. Therefore, the solution for a portion of deck of 31 m length is going to be studied with the boundaries of a cantilever. The models to be carried out are: The cross section is as the next figure shows. The geometrical features are: concrete= 2500 kg/m3 E= 36,3 Gpa It is asked to obtain and compare the answers in terms of movements from the free extreme of the cantilever model type to three kinds of loads. 3

4 1.- Selfweight. 2.- Vertical loads in Z- direction of 1000kN applied in the points A and B. 3.- Vertical loads of 1000kN applied in the points A with Z- direction and B with Z+ direction. 4.- Horizontal loads in the Y-direction of 1000kN applied in the points A and B. 5.- Horizontal loads in the X+ direction of 1000kN applied in the points A, B, C and D MATERIAL The material that has been used is the same in the three models. It is concrete and its characteristics are: Material Characteristic: Concrete Mass per Unit Volume (kg/m3) Weight per Unit Volume (N/m3) Modulus of Elasticity E (Pa) 3,63E+10 Poisson s Ratio 0,2 Shear Modulus, G (Pa) 1,513 E+10 Table 1. Material properties. The Poisson s ratio has been supposed 0,2 as the normal value in concrete MECHANICAL CHARACTERISTICS OF THE SECTION To calculate the mechanical features of the section, it has been necessary to draw the section in Autocad, which has allowed obtaining the following data: Cross Section Characteristic: Concrete Area (m2) 5,203 Distance to Center of gravitiy from the end of section (m) 6,5 Distance to Center of gravitiy from upper edge (m) 0,5785 Moment of inertia about Y (m4) 1,6538 Moment of inertia about Z (m4) 61,976 Shear Area in Y axis* (m2) 4,196 Shear Area in Z axis* (m2) 1,168 Table 2. Cross section properties. *Notice that to calculate the shear area in both axis, has been followed the recommendation of Hambly (1991), of considering: Shear area in Y axis: Has been obtained, taking into account the whole area of the horizontal parts and half of the inclined parts. Shear area in Z axis: Has been obtained, taking into account the whole area of the vertical webs and half of the inclined parts. 4

5 TORSIONAL STIFFNESS To calculate the torsional stiffness (J), the Manterloa s book Puentes has been followed. So, in a multicellular section the torsional stiffness J is defined: For using this equation it is essential to take into account the shear flow, considering that in each cell there is a flow and that adjacent cells have a common wall. Where: i= area enclosed by the cell (considering the midline) i = thickness of the cell G = shear modulus i = stress flow ds = the length of each section of the cell a This is a system of the kind Ax=b, with three unknowns and three equations that has been solved in MATLAB, giving the following solution: J=4,531 5

6 3. ANALYTICAL RESULTS Before starting with the models, I consider that an analytical assessment has to be done in order to know how the results given by the program should be ANALYTICAL RESULTS For comparing the different results of each model, it s necessary to do it regarding the analytic result of the cantilever using the exact formulation for each load case and then these data are compared with the result of each model in SAP The mechanics characteristics used are as follow: The exact formulation for each case is: A) Self weight. The exact formulation is: Notice that the first term corresponds to the bending moment, while the second is due to the shear stress Where: q: Is a distributed load equal to the self weight, which is calculated from multiplying the specific weight by the cross section area. Substituting each variable with the data shown previously, the following value for the displacement is achieved: In this case, since there is no torsion, the displacement of all points of the cross section, are the same. The deflection can also be determined by: B) Vertical loads in Z- direction in the points A and B, with a value of 1,000 kn each one. The resultant load is kn and as they are symmetric respect to the centre of gravity (where the displacements are calculated), no bending moments are generated. 6

7 Where the first term corresponds to the displacement from the bending moment, and the second term to the shear stress. The deflection can also be determined by: C) Both vertical loads of 1000 kn applied in Z- direction on the point A and Z+ direction on the point B. The resultant load is 0, since the pair of loads has same value but opposite direction. However this produces a torsion moment in the centre of gravity, so the displacement along the cross section is going to change. It is obvious that the displacement in the centre of gravity is going to be zero, because the resultant force acting there is zero: But as the torsion moment is acting, an angle of deflection in the section is going to occur: D) Horizontal loads in Y- direction applied on the points A and B, with a value of kn each one. The total load is 2000 kn in Y- direction, which is going to generate a displacement in Y direction, and also a torsion moment in the cross section around X axis, due to its eccentricity. Where the first term corresponds to the displacement from the bending moment, and the second term to the shear stress. The deflection angle due to the displacement is: 7

8 As there is torsion moment acting, an angle of deflection in the section is going to occur: E) Horizontal loads in X+ direction in the points A, B, C and D, with a value of kn each one. The resultant load to be applied in the centre of gravity is 4000 kn, in X direction. So the deck in going to be subjected to elongation and also a bending moment which tends to push up the bridge. The total displacement is due to bending moment (there is not shear stress): Elongation along X axis: Angle of deflection: All the results obtained in the previous paragraphs are summarized in the next table: LOAD CASE ANALYTICAL RESULTS Ux (m) Uy (m) Uz (m) R1 (rad) R2 (rad) R3 (rad) A 0 0-0, , B 0 0-0, , C , D 0-0, , ,00043 E 0, , , Table 3. Analytical results 8

9 4. MODEL I This model has been carried out in SAP 2000 v.15. In this part of the practice, the whole deck is going to be modeled by a single bar, what means that many simplifications have to be undertaken, giving us the worst solution of the three proposed methods GEOMETRY OF THE MODEL The model has been simplified into just one element bar with the total length of the deck (31 m), since the analysis is static. This simplification has been done applying the material and section features of the deck shown previously. If a dynamical analysis was required, the length should be divided, as we saw in the first practice, where the model was subjected to dynamic loads. As we can see in the above image, the movements in the node1, which is the support, have been restricted in both directions in order to simulate a cantilever LOAD SIMPLIFICATIONS As the deck is going to be simplified to one bar in the center of gravity, some transformations must be undertaken when the loads are applied, since these have eccentricity with respect to the center of gravity. The loads and the regarding moments have to be summed to be applied in the bar as follow: Case A. Self-weight: For the case of self weight, it has been introduced in the material, Concrete which has a specific weight of N/m3, and how the area of the section has been also introduced in the model, the programme is able to calculate the self weight by itself. 9

10 Case B. Vertical loads in the Z direction of 1000 kn applied at points A and B. Now the loads have to be summed, in order to consider them in the centre of gravity bar that we are modelling. So it is going to be a load equal to kn acting in the Z direction. As the two forces are equally separated, the moment in the model is annulated. Case C. Vertical loads in the Z direction in the point A and Z- direction, and in point B and Z+ direction, of 1000 kn applied. In this case the resultant load to be implemented in the model is going to be zero, since the loads are the same value, but opposite direction. However a bending moment is going to be generated by these forces, because they are eccentric: Case D. Horizontal loads in Y- direction in the points A and B, with a value of 1,000 kn each one. In this case the there are two horizontal loads that have to be summed to be applied in the model, in the negative direction of Y axis. But also there is a vertical eccentricity that is going to generate a moment over the model which is equal to the resultant force (2000 ) multiplied by the eccentricity: Case E. Horizontal loads in X+ direction in the points A, B, C and D, with a value of 1,000 kn each one. For this load case, the four loads acting in the X axis has to be summed and implemented in the model. So the resultant is going to be kn, but also bending moments are generated in Y and Z direction. As the section is symmetric respect to Z axis, the bending moment in Z is annulated, but the one in Y axis must be calculated: 10

11 Finally the resultant Moment that is acting over the model in Y axis is: 4.3. RESULTS MODEL I Case A The self weight produces a deflection of the bar, reaching the maximum displacement in the end of the beam. As we will see in the other models, the edge keeps parallel, to the initial position, because there is no torsion moment. Case B The same deformation than in the previous case, but with a different value (load is different, but also vertical). The mean stress along the deck is bending moment (as it has been checked in the analytical assessment that the greater proportion of the deflections comes from the bending). Case C This is a disadvantage of this of this model, because there is a torsion moment affecting the beam, but it is not represented in the figure, since in the centre of gravity it cannot be appreciated (ant symmetric rotation around the centre of gravity). 11

12 Case D In this case the figure has been disposed in YX axis to notice the horizontal deflection, produced by the horizontal loads. Also there is torsion that rotates the section, but again it cannot be appreciated with this model, because the turn is around the centre of gravity. Case E These loads produce the elongation along X axis, and a bending moment which push up the deck, giving a positive deformation in Z direction. In the following table the results for the model I of a single beam are summarized. Notice that they are almost equal than the analytical results. LOAD CASE RESULTS MODEL I Ux (m) Uy (m) Uz (m) R1 (rad) R2 (rad) R3 (rad) A 0 0-0, , B 0 0-0, , C , D 0-0, , , E 0, , , Table 4. Results model I 12

13 5. MODEL III Now, I am going to follow with the 3D Model, which is expected to give us the more accurate solution. I am going to do this model before model II, because an analysis with different divisions of the deck is going to be carried out, in order to find out the more appropriated number of divisions. This number means enough accuracy and not too much computational time. Before starting with the analysis, it is necessary to establish the points that are going to be modeled. These points will be on the middle line of the shells, and in their junctions. So the disposition of nodes is as next figure shows LOAD CASES In this model the load cases are implemented in their corresponding nodes, without any calculation of resultant, given that the loads nodes coincide exactly with the nodes to be simulated DIVISIONS ANALYSIS The analysis has been done just for three load cases, trying to have in them the different loads behaviors (vertical loads, horizontal, torsion ). And only have been analyzed the four central nodes, that is enough to take this decision, because the rest of nodes will behave in the same way. Load case A NODE Displacement Relative Error (%) 1 division 10 divisions 20 division 40 divisions Uz 0,5 0,26 0,16 0 Ry 51,27 0,77 0,24 0 Uz 0,5 0,26 0,16 0 Ry 51,27 0,77 0,24 0 Uz 0,51 0,26 0,16 0 Ry 51,81 0,7 0,24 0 Uz 0,51 0,26 0,16 0 Ry 51,81 0,7 0,24 0 Table 5. Relative error load case A 13

14 Load Case C NODE Displacement Relative Error (%) 1 division 10 divisions 20 division 40 divisions Uz 29,55 3,31 0,18 0 Ry 22,26 3,57 1,42 0 Uz 29,55 3,31 0,18 0 Ry 22,26 3,57 1,42 0 Uz 28,45 2,19 0,3 0 Ry 20,36 2,17 0,61 0 Uz 28,45 2,19 0,3 0 Ry 20,36 2,17 0,61 0 Table 6. Relative error load case C Load Case D NODE A B C D Relative Error (%) Displacement 10 1 division divisions 20 division 40 divisions Uy 23,1 0,35 0,05 0 Uz 8,77 2,44 0,86 0 Rx 11,99 0,65 0,17 0 Rz 6,36 4,19 1,97 0 Uy 23,1 0,35 0,05 0 Uz 8,77 2,44 0,86 0 Rx 11,99 0,65 0,17 0 Rz 6,36 4,19 1,97 0 Uy 23,06 0,05 0,05 0 Uz 8,77 2, Rx 11,99 1,13 0,33 0 Rz 6,36 3,29 1,42 0 Uy 23,06 0,05 0,05 0 Uz 8,77 2, Rx 11,99 1,13 0,33 0 Rz 6,36 3,29 1,42 0 Table 7. Relative error Load Case D As it can be seen in the previous tables, the error is reducing progressively at the time that more divisions are undertaken. I have considered that a good option to be chosen is the model with 20 divisions that allows me to get an accurate solution, while the computational time is quite low. 14

15 5.3. CORRECTIONS OF THE MODEL This model has an inconvenient that has to be solved to increase its accuracy, and it is that the shells that have been used to create the 3D model are overlapping between them. Therefore an increment of the area is taking place, what means more self weight and more moment of inertia. There are different manners to solve this problem: As we know the correct area of the section, and its moment of inertia, one procedure is changing the width of the shells, making different attempts until the reaction in the support at the beginning of the beam is equal to the theoretical self weight, and the moment of inertia is close or equal to the one of the original section. But this is a difficult task that requires many tests until a good solution is obtained, so I have gone for another solution. First of all, the reaction in the support has been checked to obtain the value of the new self weight. Once we have got that, it is compared with the original weight, and the percentage of variation is determined. With this is possible to reduce the weight in each shell, because SAP allows modifying the self weight by means of a multiplier. So the weight must be modified in every shell of the 3D model, by reducing 3,16 % in the command Weight Modifier of SAP And with this it is ensured that the self weight of the deck is correct again. Another thing to take into account is the section stiffness that has been modified, so the way to solve this is similar to what has been done for the self weigh, but this time with the moment of inertia. This has been checked for the new section (with superimposed areas), and compared with the moment of inertia of the original one. From this comparison I obtain a rate that is used to reduce the area of the section. By checking the new section in Autocad to determine the moment of inertia, the next value is obtained: Comparing with the original moment of inertia given by the section: 15

16 Finally the area has to be reduce in order to get that decrement of 2,89% in the inertia, but this is possible thanks to the command bending in the data of each shell, that allows decreasing the bending thickness without altering other characteristics. This way the modified self weight keeps constant with the correct value. Eventually, I have to decide which parts of the section have to be reduced to try to not change too much any other feature of the section. As far as I am concerned, the best manner to do this is reducing the thickness of each shell of the section in the same proportion, as follows: Element Type Thickness (m) Modification (m) (reduction of 2,89%) Slab Sell thick 0,200 0,1942 Web Sell thick 0,210 0,2039 Oblique web Sell thick 0,16 0,1554 Table 8. Modification of the shells 5.4. RESULTS MODEL III Case A Due to the self weight there is a bending moment that produces a deflection in the deck. The bridge goes downwards, reaching the maximum displacement in its end. Case B This is similar to the previous case, but for a lower load that produces smaller displacements. The main stress that is acting in this case and in the previous is the bending moment over the Y Axis. Case C The cross section rotates as effect of the torsion moment generated about X axis. And the rotation is higher in the central cell than in the edges. 16

17 Case D The next figure shows the deformation of the bridge for two horizontal loads applied in the nodes 3 and 5. There will be torsion around the X axis that produces the rotation of the section, and the loads induce a displacement of the deck horizontally and produce bending in Z axis. Case E In this load case an elongation along X axis takes place, as it can be check in the figure below. But also the deck is subjected to bending moment in Y axis that pushes the deck up. 17

18 In the next table, I summarize the results of the model III for the eight nodes of the end section: RESULTS MODEL III JOINT LOAD CASE Ux (m) Uy (m) Uz (m) R1 (rad) R2 (rad) R3 (rad) 1 A 0, , , , , , B 0, , , , , , C 0, , , , , , D -0, , , , , , E 0, , , , , , A 0, , , , , ,132E-06 2 B 0, , , , , , C 0, , , , , , D -0, , , , , , E 0, , , ,374E-06-0, , A 0, , , ,451E-06 0, ,971E-06 3 B 0, , , , , , C 0, , , ,0023 0, , D -0, , , , , , E 0, , , ,481E-06-0, , A -0, ,753E-06-0, , , ,575E-06 4 B -0, , , , , , C -0, , , , , ,211E-06 4 D -0, , , , , , E 0, , , , , , A 0, , , ,451E-06 0, ,971E-06 5 B 0, , , , , , C -0, , , ,0023-0, , D 0, , , , , , E 0, , , ,481E-06-0, , A -0, ,753E-06-0, , , ,575E-06 6 B -0, , , , , , C 0, , , , , ,211E-06 6 D 0, , , , , , E 0, , , , , , A 0, , , , , ,132E-06 7 B 0, , , , , , C -0, , , , , , D 0, , , , , , E 0, , , ,374E-06-0, , A 0, , , , , , B 0, , , , , , C -0, , , , , , D 0, , , , , , E 0, , , , , , Table 9. Results model III 18

19 6. MODEL II This model consists of a plan grillage that tries to simulate the whole deck by means of longitudinal and transversal bars with the associated properties of that deck. It has more complexity than the two other models, since many assumptions are done in order to transfer the properties of parts of the deck to the corresponding longitudinal or transversal bar GEOMETRY OF THE GRILLAGE This second model is going to be divided into 20 parts of 1,55 m along the X axis, because this number of cuts offers accuracy enough solution (as it has been seen in the analysis for model III) and it is more comfortable and rapid to operate (the computational time is lower). On the other hand the transversal bars along Y axis are going to be cut into three elements, one element between two longitudinal bars, and to get the properties for those bars, the average properties of the corresponding cell will be analysed. The grillage is going to be located in a plane on the center of gravity of the deck section, with the longitudinal bars disposed in the points shown in the figure below. This disposition has been taken from the book Hambley (1991) pag. 126, which shows the divisions of a multicellular deck with inclined webs. Now the characteristics of the deck must be transferred to the longitudinal and transversal beams of the grillage model LONGITUDINAL BEAMS Following Hambley (1991) for the case of multicellular decks with inclined webs, he recommends to cut the section by the middle of a cell below figure. Once the section has been chopped in parts, is the moment of transferring the features of those sections into the longitudinal bars. In this way two longitudinal beam sections are going to be established, one with the mechanical properties of the extreme of the deck and another one for the central parts, as the upper scheme shows (notice that the deck is symmetrical). 19

20 The mechanical characteristics have been obtained for each longitudinal beam from the corresponding part of the section as follows: Area: The area of the model beam is the area of the corresponding part of the deck section. Moment of Inertia of Y: This has to be determined with respect to the global Y axis, which is going through the centre of gravity of the whole deck, as Hambly (1991) pag. 125 says. This is due to the fact of having a beam whose centre of gravity does not coincide with the centre of gravity of the whole bridge section, but it is located at the level of the global centre of gravity. Moment of Inertia of Z: This is obtained respect to Z axis of each beam Shear area in Y axis: As Hambly (1991) suggests, the shear area in Y axis has been calculated by considering the full area of the horizontal slabs of the section and one half of the inclined webs. Shear area in Z axis: The procedure is the same than for Y axis, but now, full area of vertical webs and half of the inclined. Torsion stiffness: Hambly (1991) proposes the following expression, and states that can be adapted to multicellular decks with inclined webs, by considering an average value of the cell height. Extreme Beam With the last expression of Hambly, the Torsion constant for the two longitudinal beams is going to be got: 20

21 Where h is the average value of height in the cell, in this case h=0,675/2. To summarize, the characteristics of the extreme beam, calculated with the before methods, are the following: Central Beam LONGITUDINAL EXTREME BEAM Mechanical Properties Value Area (m 2 ) 0,7042 Moment of inertia about Y (m4) 0,1128 Moment of inertia about Z (m4) 0,2571 Shear Area in Y axis* (m2) 0,5750 Shear Area in Z axis* (m2) 0,1291 Torsion Stiffness (m4) 0,0358 Table 10. Longitudinal extreme beam. Where h1=0,675m h2=1,35 m t1=0,2 t2=0,16 t3=0,2 To summarize, the characteristics of the central beam, calculated with the before methods, are the following: LONGITUDINAL CENTRAL BEAM Mechanical Properties Value Area (m 2 ) 1,8979 Moment of inertia about Y (m4) 0,7144 Moment of inertia about Z (m4) 2,5350 Shear Area in Y axis* (m2) 1,5023 Shear Area in Z axis* (m2) 0,3972 Torsion Stiffness (m4) 1,234 Table 11. Longitudinal Central beam 21

22 6.3. TRANSVERSAL BEAMS There are two different transversal bars, since the section is symmetric, but in the extremes of the deck as the tributary area is half of the others, the properties are going to changes as it can be checked below. The mechanical characteristics have been obtained for each transversal beam from the corresponding part of the section as follows: Area: For the transversal beams, the area has not to be considered because the full area of the section is already implemented in the longitudinal bars. Otherwise the area would be double and the grillage model would have different properties. Also the self weight of the section has been introduced in the longitudinal bars (specific weight multiplied by area), so considering zero the area in each transversal bar, the self weigh for them would be zero too. Moment of Inertia of Y: For the transversal beams, Hambly (1991) pag. 127, suggests in multicellular decks with inclined webs, to obtain the moment of inertia with the following formula, but considering the average value of height across the cell. Moment of Inertia of Z: This is obtained respect to Z axis of each beam Shear area in Y axis: As the deck has triangular cells, we cannot use the formula for rectangular cells, but Hambly (1991) pag.127 suggests that the shear area can be derived from a plane frame analysis. This is possible thanks to the relationship between vertical shear across a cell and the effective shear displacement (ws). Where: S= Unitary Load Sa= Shear Area G= Shear modulus of the material ws= Displacement along cell l= length of the cell As Hambly recommends a cross section analysis has been done, which gives us the shearstiffness of each cell (Sa), by substituting in the previous formula. 22

23 The section has been modelled in SAP 2000, restricting the horizontal movements of the nodes, and loading with a unitary load in the extreme, that allow me to calculate the relative displacement between the cells, and in this way obtaining the required shear area from the above formula. CELL Relative displacement between nodes of the cell Extreme Vertical 7e-6 m Horizontal 1e-6 m Central Vertical 2,5e-4 m Horizontal 1,3e-6 m Table 12. Relative displacement in cells With the data shown in the last table, the shear areas for the transversal beam are going to be calculated later. Shear area in Z axis: The procedure is the same than for Y axis, but now, the restrictions are changed and the unitary load is horizontal. Torsion stiffness: For transverse grillage members Hambly (1991) proposes the following expression for the torsion constant, and states that can be adapted to multicellular decks with inclined webs, by considering an average value of height across the cell. Where, h= average of the cell height 23

24 Extreme Transversal beam To determine the properties of this beam, a cut in the middle of the triangle cell has been undertaken: EXTREME TRANSVERSAL BEAM Mechanical Properties Value Area (m 2 )* 0 Moment of inertia about Y (m4) 0,0627 Moment of inertia about Z (m4) 0,110 Shear Area in Y axis* (m2) 0,363 Shear Area in Z axis* (m2) 0,051 Torsion Stiffness (m4) 0,1256 *Notice that the area has been equaled to 0 to avoid counting the area twice. The extreme bars that are at the end and at the beginning of the cantilever deck are going to have the half of the previous characteristics since the beam is cut by the middle. But this changes the centre of gravity of the beam, and the Moment of inertia in Z (Iz) has to be recalculated by means of the formula b /12. Table 13. Extreme transversal beam 24

25 Central Transversal beam EXTREME TRANSVERSAL BEAM (HALF) Mechanical Properties Value Area (m 2 )* 0 Moment of inertia about Y (m4) 0,0314 Moment of inertia about Z (m4) 0,0139 Shear Area in Y axis* (m2) 0,1815 Shear Area in Z axis* (m2) 0,0255 Torsion Stiffness (m4) 0,0628 Table 14. Extreme transversal beam (half). To determine the properties of this beam, a cut in central cell has been undertaken. And following the same procedure than for the extreme transversal beam, the resultsa are: CENTRAL TRANSVERSAL BEAM Mechanical Properties Value Area (m 2 )* 0 Moment of inertia about Y (m4) 0,2825 Moment of inertia about Z (m4) 0,1241 Shear Area in Y axis* (m2) 0,315 Shear Area in Z axis* (m2) 0,0016 Torsion Stiffness (m4) 0,564 Table 15. Central transversal beam 25

26 *Notice that the area has been equaled to 0 to avoid counting the area twice. The central bars that are at the end and at the beginning of the cantilever deck are going to have the half of the previous characteristics since the beam is cut by the middle. But this changes the centre of gravity of the beam, and the Moment of inertia in Z (Iz) has to be recalculated by means of the formula b / LOAD CASES CENTRAL TRANSVERSAL BEAM (HALF) Mechanical Properties Value Area (m 2 )* 0 Moment of inertia about Y (m4) 0,1413 Moment of inertia about Z (m4) 0,0155 Shear Area in Y axis* (m2) 0,1575 Shear Area in Z axis* (m2) 0,00082 Torsion Stiffness (m4) 0,282 Table 16. Central transversal beam (half). Like in the first model, now some modification to implement the loads must be undertaken, since the grillage is placed on the plane of the centre of gravity of the section and not in the points where the loads are applied. Case A. Self-weight: For the case of self weight, it has been introduced in the material, Concrete which has an specific weight of N/m3, and how the area of the beams have been also introduced in the model, the programme is able to calculate the self weight by itself. The distributed load equivalent to the self-weight of each bar is as follows: - Extreme Longitudinal Beam: - Central Longitudinal Beam: By multipliying the previous loads by the length of the deck and by two (there is one more symetric beam of each type) the result is the total self weight of the deck (3956,3 kn) Note that the transversal bars, as they have not area implemented in the model, the self weight is zero. All the self weight is apported by the longitudinal beams. - Transversal Bars: Case B. Vertical loads of 1000 kn applied in Z direction at the points A and B. In this case it is not necessary to do any extra modification when applying the loads, because there is not eccentricity with respect the points A and B of the model. Case C. Vertical loads of 1000kN applied in the points A with Z- direction and B with Z+ direction. Again, no extra input load has to be implemented in the model, since the loads are applied without eccentricity in the points A and B of the model. The program itself calculates the torsion moment generated by this loads. 26

27 Cad D. Horizontal loads in the Y-direction of 1000kN applied in the points A and B. Now the model needs to be implemented with the moment generated by the eccentricity between the real nodes and the model ones. Case E. Horizontal loads in the X+ direction of 1000kN applied in the points A, B, C and D. This case is similar to the same case in model I, but now the total load and the resultant moment over the Y axis is applied half in each intermediate grillage node, as the following images shows. The moments around Z axis are annulated, but even if they were not, should not be necessary to calculate them RESULTS MODEL II Case A The following images show how the movement of the grillage is for a self weight load. As we can see the section goes down by the bending moment of the load, and each transversal section keeps parallel to the original position. 27

28 Case B Since this is a similar load case to the previous one, but with different values, the deformations is almost the same. As we will see later, in this 2D model there is no displacement along X nor Y axis, because the grid is over centre of gravity, however in the 3D model movement in X and Y axis can be appreciated. Case C The two loads with opposite direction create a torsion moment, which pushes the cross section to rotate around X axis in the centre of gravity. Also note that the largest rotation is in the centre of the section. Case D This deflection is originated by two horizontal loads that produce horizontal displacement caused by bending moment, and also the torsion (eccentricity) around X axis that causes the rotation of the section. Note that the central nodes have been displaced farther than the other two. 28

29 Case E The disposition of the loads produces the elongation along X and Y axis as the figure below shows, but also a displacement upwards originated by the moment around Y axis. MODEL II RESULTS NODE LOAD CASE Ux (m) Uy (m) Uz (m) R1 (rad) R2 (rad) R3 (rad) 1 A 0 0-0, , , B 0 0-0, , , C 0 0-0, , , D -0, , , , , , E 0, , , , , , A 0 0-0, , , B 0 0-0, , , C 0 0-0, , , D -0, , , , ,064E-06-0, E 0, , , , , , A 0 0-0, , , B 0 0-0, , , C 0 0 0, , , D 0, , , , ,064E-06-0, E 0, , , , , , A 0 0-0, , , B 0 0-0, , , C 0 0 0, , , D 0, , , , , , E 0, , , , , ,00013 Table 17. Model II results 29

30 7. COMPARISON OF RESULTS To compare, it is necessary to have the results in the same nodes, because in each model, the analysis has been done with a different number of nodes in the cross section. So a good manner to check the results, is contrasting them in the centre of gravity, where the analytical results have been obtained. LOAD CASE ANALYTICAL RESULTS Ux (m) Uy (m) Uz (m) R1 (rad) R2 (rad) R3 (rad) A 0 0-0, , B 0 0-0, , C , D 0-0, , ,00043 E 0, , , MODEL I Table 18. Analytical Results The results coming from the Model I, where the deck is simulated by means of a single beam, are: LOAD CASE MODEL I RESULTS Ux (m) Uy (m) Uz (m) R1 (rad) R2 (rad) R3 (rad) A 0 0-0, , B 0 0-0, , C , D 0-0, , , E 0, , , Table 19. Model I results As it can be checked in the above tables, the results between both analytical and beam solution, are almost exact, what means that SAP 2000 uses this formulation to solve the single bar. But we are dealing with a whole bridge and not only with a beam, so it is evident that other methods are required to get the displacements in more points apart from the central one. 30

31 MODEL II The results for the centre of gravity in the second model can be easily obtained, since the nodes of the grillage are on the plane XY of the centre of gravity. So a normal average has been used to obtain the next values: GRILLAGE RESULTS CENTRE OF GRAVITY LOAD CASE Ux (m) Uy (m) Uz (m) R1 (rad) R2 (rad) R3 (rad) A 0, , , , , , B 0, , , , , , C 0, , , , , , D 0, , , , , , E 0, , , , , , To check the errors Table 20. Grillage results in gravity centre RELATIVE ERRORS OF MODEL II (COMPARING WITH ANALYTICAL IN CG) LOAD CASE Ux (m) Uy (m) Uz (m) R1 (rad) R2 (rad) R3 (rad) A 0,00% 0,00% 0,16% 0,00% 0,07% 0,00% B 0,00% 0,00% 0,15% 0,00% 0,02% 0,00% C 0,00% 0,00% 0,00% 51,02% 0,00% 0,00% D 0,00% 40,91% 0,00% 17,82% 0,00% 31,85% E 19,18% 0,00% 0,79% 0,00% 1,97% 0,00% Table 21. Relative errors of Model II As it can be seen in the comparison, for the load cases A and B, the response of the grillage model is almost the same than the analytical or the one of the model I. There are just a little discrepancies, always lower than 0,16%, what means that this model works well for vertical loads, such as the case A and B. The largest discordance is given by the angle induced by the torsion moment in the load case C, because the way to implement that in the beam model was by means of a torsion moment without any force (they were annulated). So in my opinion this error can come of that simplification, and also because the value of the angle for the model I is an average of the whole section. Checking why the error is so high in the displacement along Y axis for the load Case D, it is found out that the displacement of the two central nodes is much higher than the displacement in the extreme nodes. This can be due to the extreme transversal beam, which has been considered with half of the properties, as it has half area. Also the fact of applying the loads over the central nodes can derivate into a deformation of this transversal beam that affects more in the centre. However if we consider the movement of the four nodes of the grillage section to see the total displacement of the deck, the error will be reduced remarkably. 31

32 Another point to comment is the high error in the displacement along X axis. This is because the elongation of the central nodes (where the loads are applied) is larger than in the edge nodes. If we consider an average of all the nodes of the grillage section, then the error is not significant. MODEL III To obtain the results in the centre of gravity for the 3D model, it has been used a weighted average, giving more weight to the solution of the closer nodes. In this way, the following table has been obtained. LOAD CASE 3D MODEL RESULTS IN CENTRE OF GRAVITY Ux (m) Uy (m) Uz (m) R1 (rad) R2 (rad) R3 (rad) A -0, , , , , ,00000 B -0, , , , , ,00000 C 0, , , , , ,00005 D 0, , , , , ,00044 E 0, , , , , ,00000 Table 22. Results of model III in gravity centre The table below shows the relative error of the 3D model respect to the analytical results. RELATIVE ERROR OF MODEL III (COMPARING WITH ANALYTICAL IN CG) LOAD Ux (m) Uy (m) Uz (m) R1 (rad) R2 (rad) R3 (rad) CASE A 0,00% 0,00% 3,56% 0,00% 4,20% 0,00% B 0,00% 0,00% 3,41% 0,00% 2,44% 0,00% C 0,00% 0,00% 0,00% 1,79% 0,00% 0,00% D 0,00% 3,22% 0,00% 44,73% 0,00% 2,28% E 4,72% 0,00% 3,11% 0,00% 13,04% 0,00% Table 23. Relative error of Model III This model is the more accurate, and the relative errors that are shown in the table can come from the assumptions taken to carry out the analytical calculation, such as the estimation of the shear area (that don t have to be made to implement the model III). Again the bigger errors occur with the angles, specially the one produced by torsion moment. This can be an effect of the beam model or theoretical model that are averages of the whole section. And not only that, but also to make this comparison possible, a weighted average of the nodes of the central cell of the 3D section was done, to get a solution for the centre of gravity. And as it can be checked in the model III solution table, the values of rotation angle are in the central cell, are higher than in the border cells. 32

33 8. COMPARISON OF THE MODELS BY LOAD CASES Now a comparison between the three models in each case of load is going to be carried out. Especially I am going to focus in model II and III, since model I as we have seen already has a results almost equal to the theoretical. Case A Looking at the previous tables, we can see that the larger error committed respect to the analytical solution is in the model III, with more than 3,5%, while in the grillage method (Model II) the maximum error is 0,16%. It is supposed model III to be the more accurate, therefore this difference can be a cause of the simplifications that have been implemented in the analytical formulas and in the model I and II, for instance, the shear area. Case B This situation is very similar to the previous, because the loads are applied vertically (as self weight) so the relative errors are very close to last case. In this case as in the before, there exist elongation along X axis in the 3D model, although this is very small. But this is totally logical, given that the lower fiber of the deck is subjected to compression while the upper fiber is under traction efforts. This effect could not be appreciated with model I and model II, however model III gives a more precise anwer. Case C Comparing the relative errors for the torsion angle in this load case, we see how the 3D model (error=1,79%) is closer to the theoretical solution than the grillage model (error=51%). It is probably that the fact of having so many torsion constants in model II that comes from estimation gives us worse results, although these torsion constants have been determined following the suggestions of Hambly (1991). Case D In general the relative errors in the model II solutions are in this load case higher than in the 3D model, which also presents a high error in the rotation angle around X axis, but this error can be because of the small values of deformation that occurs in this load case. Case E The 3D model has a lower error in the elongation along X axis than model II, this is probably because each of the four loads are applied on its corresponding node, while in the plane grillage, there are two loads in each central node, that is going to produce bigger displacements of that nodes. 33

34 9. CONCLUSIONS As I have checked, the simpler the model is, the more assumptions have to be done. What means that, for instance, in the model of a bar you need to implement many estimations of the real properties of the deck if you want to get a similar answer to the real one. And not only that, but also to consider the different efforts that the applied loads cause, it is necessary to calculate the resultants at just one point, since all the points of the section are simulated together in one (centre of gravity).however in the 3D model, no estimation of resultant loads was done, because the loads are applied directly on their corresponding node. In the grillage model more assumptions of the mechanical properties have to be undertaken, but there are fewer simplifications to make in the application of loads, since there are more nodes in the real position. Looking at the angles of rotation, we see that there are many discrepancies between the model I and the other two models, so it makes me think that the single beam model is not the more suitable to do a torsion analysis, since it gives an angle which is the average of the section, calculated in the centre of gravity. The 2D model (grillage) includes the longitudinal edge cantilever (the small part of the deck, the edge beyond the triangular cell) in the same longitudinal bar that the half of the triangular cell. So it is expected to work a bit worse to torsion than if the edge beam is included in another longitudinal bar. However this model can be used to study the transversal behavior of the whole deck, with less detail than the 3D model. As it can be checked, the more number of nodes we have in our solution, the more precision and more information that model offers. By looking at the output graphs of the model I, is very difficult to see, how the total section is behaving under a determined load. For instance the torsion rotation in some models cannot be noticed, since we are dealing with a single beam in the centre of gravity. This problem is solved with the 2D model and for sure in the 3D model. The 3D Model gives us a more precise solution, showing even displacements that do not appear in the other model. For example, in the load case A and B, in the models I and II there is no displacement along X nor Y axis, since these models are set in the centre of gravity. So when bending moments take place in a deck, the section face is going to rotate around Y axis because of compression in the lower fiber and pull in the upper. And this phenomenon can be realized thanks to the 3D model that gives us information about the upper and lower nodes of the section. The displacements of the deck because of the bending moment are larger than the shear stress ones, as it could be seen in the analytical obtaining of results, where both were calculated by hand. This has to do with the length of the deck, as the formula reveals, given that in the bending the length is powered. 34

35 The more number of division you do in your model, the more accuracy and definition you get in order to achieve the solutions. I estimated that the right number to do it was 20 divisions that allow me to get enough precision and also makes easier and quicker getting the results (computational time). The lowest errors for the 3D and 2D models, occur for the vertical load cases (case A and B). This is because in these cases we don t have to deal with the rotation due to torsion moment, which produces the worst errors. These errors can be originated because the theoretical model gives us a value for the whole section, as average. Or, as the rotation angles are bigger in the central part of the deck, in the centre of gravity it is going to be much larger than in the edges. Also we have to keep in mind that, to get the torsion constants, approximation formulas have been taken, which means a loss of accuracy, and the same for shear areas, although everything has been done following Hambly s proposals. With this practice I have reinforced the concept of the importance of choosing a good method to solve the deck, depending on how much information you want to get from the model and also the accuracy that you need. The 3D model can be used for complex solutions, while the beam method, gives us a quick good approximation of the whole deck behavior. 10. REFERENCES Manterola, J. (2000). Puentes. Madrid. Hambly,E.C. (1991). Bridge Deck Behavior. London: E and FN SPON. 35