Design and analysis of deep beams, plates and other discontinuity regions BJÖRN ENGSTRÖM. q l (SLS) d ef. T SLS A ef. Tie = reinforced concrete Aef

Transcription

1 q l (SLS) d ef T SLS A ef Tie = reinforced concrete Aef Design and analysis of deep beams, plates and other discontinuity regions BJÖRN ENGSTRÖM Department of Civil and Environmental Engineering Division of Structural Engineering Concrete Structures CHALMERS UNIVERSITY OF TECHNOLOGY Göteborg, Sweden 2011 Report 2011:6

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3 REPORT 2011:6 Design and analysis of deep beams, plates and other discontinuity region BJÖRN ENGSTRÖM Department of Civil and Environmental Engineering Division of Structural Engineering Concrete Structures CHALMERS UNIVERSITY OF TECHNOLOGY Göteborg, Sweden 2011 Report 2011:6

4 Design and analysis of deep beams, plates and other discontinuity regions BJÖRN ENGSTRÖM BJÖRN ENGSTRÖM, 2011 Report 2011:6 Department of Civil and Environmental Engineering Division of Structural Engineering Concrete Structures Chalmers University of Technology SE Göteborg Sweden Telephone: + 46 (0) Cover: Strut-and-tie model for deep beam Department of Civil and Environmental Engineering Göteborg, Sweden 2011 Report 2011:6 II

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7 CONTENT 1. INTRODUCTION Discontinuity regions Introductory examples Extension of discontinuity regions Statically determinate and statically indeterminate problems Typical behaviour of discontinuity regions and modelling Multi-axial state of stresses THE STRUT AND TIE METHOD Procedure The load path method Development of the strut and tie model Basic principles Application rules Design of components Forces in struts and ties Design of nodes Design of ties Anchorage of ties Stress limitations in struts DESIGN ON THE BASIS OF PLASTIC ANALYSIS Assumptions Uncertainties Design with regard to the need for ductility Design with regard to the serviceability limit state Alternative strut and tie models APPLICATIONS Transfer of shear and bending moment truss models Shear resisted by inclined struts Bending moment and need for load paths in U-turn Boundary between B- and D-regions Modelling of B-regions Combination of vertical and horizontal loads on supports Problems in 3 D Combination of 2-D models in perpendicular planes General 3-D models DETAILING Bar spacing in ties Primary and secondary reinforcement Deep beams Walls Discontinuity regions 54 REFERENCES 55 I

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9 1. INTRODUCTION 1.1 Discontinuity regions Introductory examples It is generally assumed that normal stresses in a section x of a fully prestressed beam under service load can be determined by means of Navier s formula, for instance by using the tendon force approach. Pi ( x) Pi ( x) e( x) M ( x) c ( z) z A ( x) I ( x) net net This expression for stress calculations has been derived under the assumption that plane sections remain plane in bending, and that the materials have a linear elastic response. It is obvious, however, that stresses near the anchorage of tendons near the end of a posttensioned beam can not be linearly distributed across the section. Near the anchorage the compressive stresses, due to equilibrium conditions, need to be concentrated to the region just under the anchor plate, see Figure 1. Hence, the stress field is essentially influenced by the dimension of the anchor plate, which is not at all considered by Navier s formula. D B P Pe M c( z) A I z Figure 1 Stress field in the end of a post-tensioned beam In the end of the beam these stresses from the anchorage are spread out across the section within a dispersion zone. At a certain distance from the anchorage the conditions are more stable and from this section on Navier s formula gives a reasonable estimate of the stress distribution across the section. Within the zone where the compressive stresses disperse across the section, transverse tensile stresses appear which, in a concrete structure, might result in cracking. To keep equilibrium and prevent uncontrolled development of cracks, transverse reinforcement needs to be designed and provided. The zone where the concentrated force from the anchorage disperses across the section is referred to as a discontinuity region. 1

10 Another example is a column head that supports two beams near the corners, Figure 2. The column head is subjected to the load from each of the beams. If these loads are assumed to be the same, equal to Q, and the beams are symmetrically arranged on the column head, the average compressive stress in the column can be calculated as c 2Q b h Q Q D B b 2Q b Figure 2 Column head subjected to loads from supported beams However, near the top of the column the compressive stresses will be concentrated under the bearings only. At a certain distance from the top the stresses have evened out across the section and the actual stress is found to be the same as the average stress. In the dispersion zone where the concentrated loads are evened out, transverse tensile stresses appear with risk of cracking and corresponding need for reinforcement. In a discontinuity region the assumption concerning plane sections remaining plane under loading is generally not valid. If a region is a discontinuity region or not depends not only on the geometry and location of the region, but also on the load case. For instance a column head is a discontinuity zone when it is subjected to concentrated loads. If the same column head is subjected to a uniformly distributed load q, the top region of the column will not be a discontinuity region, since in this case plane sections remain plane under loading. For all sections of the column the stresses can be calculated as q c h Hence, in the design of reinforced concrete members it is important and appropriate to distinguish discontinuity regions and continuity regions Schlaich et al. (1987). In a discontinuity region the effect of a local discontinuity is evened out. In the adjacent continuity region, the effect of the same local discontinuity is not noticeable. 2

11 Discontinuity regions can result from geometric discontinuities, Figure 3, or static discontinuities, Figure 4 Schlaich et al. (1987). Discontinuity regions are also known as D-regions or disturbed regions. In continuity regions plane sections remain plane under loading. Continuity regions are also known as B-regions, where the Bernoulli hypothesis of a plane strain distribution is assumed to be valid. h 1 h 2 h h 1 h 2 h h h 2 h 2 h 1 h 1 h h Figure 3 Discontinuity regions due to geometric discontinuities, adopted from Scleich et al. (1987) 3

12 h h h h h 2h h h h h h Figure 4 Discontinuity regions due to static discontinuities, adopted from Schlaich et al. (1987) In deep beams the assumption of plane sections remaining plane under loading is often not valid even in the maximum moment section, Figure 5. The proportions of the structural member are such that the beam theory cannot be applied in the design. In this case the whole structural member is a discontinuity region. 4

13 Figure 5 Distribution of normal strain in a beam and in a deep beam and resulting forces to corresponding linear elastic stresses Extension of discontinuity regions The extension of discontinuity regions can be estimated by means of Saint-Vénant s principle Schlaich et al. (1987). Study a body that is subjected to a system of forces in self equilibrium, Figure 6. It means that on the global level the body is not subjected to a force resultant and will not accelerate. d h h 0 d h 0 d h Figure 6 Illustration of Saint-Vénant s principle 5

14 However, according to Saint-Venant s principle, stresses will appear in the body as a local effect of the applied forces and the extension d of the stressed region is equal to the maximum distance h between the forces in the applied system of forces. A structural member with a concentrated compressive force F in the centre of the end section is used to exemplify how the extension of the discontinuity region can be estimated, Figure 7. Further away in the member, in the continuity region, the strain distribution is linear, which results in a uniform stress F A The load case with F acting to the right is now divided in two, which, when they are superimposed, results in the original load case. h F F q h F q h + F 0 0 B-zone D-zone d h B-zone F D-zone d h B-zone Figure 7 Extension of a discontinuity regions determined by Saint-Vénant s principle 6

15 In load case 1 the element is subjected to a uniformly distributed compressive stress = F/A. In this load case the whole element will be a continuity region with the same stress in all sections. In load case 2 the beam element is subjected to a force system in self equilibrium with one centric concentrated force F acting to the right and one uniformly distributed stress acting to the left. According to Saint-Vénant s principle stresses will appear within a region with an extension d equal to the maximum distance between the forces in the system of self equilibrium. In this case this distance corresponds to the height h. Outside of this region no stresses appear. When the load cases 1 and 2 are added, the searched stress field under the concentrated force F is found. Uniformly distributed stresses appear behind the end zone with extension h. This region deep inside the member is the continuity zone. In the region with extension h near the concentrated force, this uniform stress field is disturbed by local effects. This end region is the discontinuity region. According to Eurocode 2 CEN (2004) discontinuity regions extend up to a distance h from the discontinuity, where h is the section depth of the member. However, it is not always that simple to determine the extension of the discontinuity region. The main principle that should be kept in mind is that the discontinuity region is the region where the effect of local discontinuities is evened out. For instance near the end support of a beam with T-section and a wide flange, the effect of the concentrated support reaction will first be evened out into the web and further on across the width of the flange. In this case the extension of the discontinuity region will be equal to the width of the flange, see Figure 8. B-zone b D-zone d b Figure 8 Extension of discontinuity region of a flanged beam, adopted from Schlaich (1987) It is normally not important to define the extension of the discontinuity regions very accurately. The load path method, presented in Section 2.2 will naturally guide the designer to a reasonable estimation of the stress field, and by that the discontinuity region can be defined approximately. The design and arrangement of reinforcement in the discontinuity region will normally not depend on the exact extension of the zone. 7

16 1.2 Statically determinate and statically indeterminate problems A simply supported beam under load is generally characterised as a statically determinate problem. This means that the sectional forces can be determined by means of equilibrium conditions only. This is not the case with a continuous beam on three or more supports. To solve the sectional forces in such case, equilibrium conditions must be combined with compatibility conditions and constitutive relations. Constitutive relations are often depending on assumptions, for instance that there is a linear relation between moment and curvature in a section. Deep beams can also be classified as statically determinate and statically indeterminate by the same rules as for ordinary beams. When the sectional forces are known in beams or deep beams, the internal stresses are still unknown. The sectional moment in a statically determinate beam can be resisted in many different ways depending on the material response in the section. For a given section, for instance, the internal lever arm will be different if the material has elastic or plastic response. Furthermore, in a reinforced concrete beam the moment can be resisted by many alternative combinations of steel area and sectional depth. Hence, to solve the stress distribution in a section subjected to a bending moment is a statically indeterminate problem. As usual, this can be solved by a combination of equilibrium and compatibility conditions and constitutive relations, as exemplified in Figure 9. For beam sections the assumption of a linear strain distribution (plane sections remaining plane) is a very convenient compatibility condition, which can be used also in hand calculations. - 3 cc=3,5. 10 fcc V d ( x ) M d ( x ) x c ( c ) F s = s As x s s f st s Figure 9 Model to determine the statically indeterminate stress distribution in a beam section. Equilibrium, compatibility and constitutive relations are combined. In deep beams and other discontinuity regions the stress field is statically indeterminate, independent of if the structural member on the global level is statically determinate or not. The problem with discontinuity regions is that there exists no simple compatibility condition which can be used to solve the stress field. 8

17 1.3 Typical behaviour of discontinuity regions and modelling The behaviour of a discontinuity region of reinforced concrete under increasing load can be characterised by the following four main stages. Uncracked state As long as the concrete is uncracked the influence of the reinforcement is small and the behaviour under load is almost linear. In this state it is appropriate to analyse the structural member by linear analysis assuming a homogenous material. A linear analysis will result in a unique stress field (stress field configuration) that is independent of the load. The stress field can be used to identify regions with high tensile stress that are prone to cracking and where reinforcement might be needed. In linear analysis, equilibrium, compatibility and constitutive relations are combined to solve statically indeterminate problems, for instance stress fields in discontinuity regions. The constitutive relation is a linear stress-strain relationship for the assumed homogenous material. A linear analysis can be carried out by the finite element method a linear FE analysis. The stress field can be presented in different ways, stress trajectories, principal stresses, normal and shear stresses, see Figure 10. Figure 10 Stress field in deep beam presented by stress trajectories and a simplified interpretation of it arch action and need for reinforcement in the bottom Typical for a linear analysis is that it results in one unique solution. It means that if the stress field is determined for one value of the load, the configuration of the stress field will remain when the load increases (for the same load case). Only the magnitude of the stresses will increase. Stresses and deformations will increase linearly with the load (for the same load case). The reason for this linear behaviour is that the stiffness of the structure is determined by the geometry and the elasticity of the material. The stiffness will not change when the load increases. 9

18 A linear analysis requires little information about the structural element. It is often sufficient with the gross geometry and the load. Any linear elastic material can be assumed. Therefore, linear analysis can be carried out early in the design process. Cracked state When cracking occurs, there will be a drastic change of the stiffness conditions in the structural member, Figure 11. The stiffness will vary between different regions depending on if they are cracked or not. Furthermore, the stiffness of the cracked regions is essentially influenced by the amount and arrangement of the reinforcing steel there. As a result the configuration of the stress field will deviate from the stress field before cracking and it will change successively when cracking develops. Even if structural members have the same geometry and the same load, the stress fields can be different depending on how the reinforcing steel has been designed and arranged. The actual stress field depends on the stiffness distribution. In each load step stiffer regions attract forces from softer regions. Hence, when the load increases and cracking develops there will be a continuous change of the stiffness distribution and, as a result, of the stress field configuration. This is known as stress redistribution due to cracking. In each load step stresses redistribute, since forces are controlled by the stiffness distribution that varies continuously. Figure 11 Deep bream of reinforced concrete in the cracked state When the configuration of the stress field varies under increasing load, stresses and deformations will not increase in proportion to the load. Instead there will be a nonlinear behaviour, in spite of the fact that the materials, concrete and steel, still have linear elastic material responses. It is not possible to predict the stress field or the behaviour in the cracked state by linear analysis. Instead non-linear analysis is needed. This can be carried out as a non-linear FE analysis. For a reinforced concrete member a non-linear analysis considering cracking of concrete by fracture mechanics, reinforcing steel and the interaction between steel and concrete is an advanced analysis. It requires full information about the structural member concerning material properties and reinforcement arrangement and can only be carried out as verification at the end of the design process when the data is available. 10

19 Most reinforced concrete elements will be cracked under service load, but this normal behaviour is seldom predicted accurately in practice. It is commonly assumed that a linear analysis can be used to estimate the behaviour in the service state. However, it should be kept in mind that this is an approximation of the real behaviour. It should also be noted that a reinforced concrete element will have a non-linear behaviour even if it is designed on the basis of linear analysis (using a stress field found by linear analysis). Ultimate state The ultimate state is characterised by non-linear response of the materials. When one of the materials, concrete or steel, starts to have a significant non-linear behaviour, the structural element reaches the ultimate state. A typical event is when the reinforcement starts to yield in a region of the element. The ultimate state lasts, during load increase, until the collapse of the structural member (this is the ultimate limit state). When plastic deformations develop in a region, this is equivalent with a drastic decrease of, or even loss of, stiffness. Since, in each load step, stiffer regions attract forces from softening regions, plastic behaviour in a highly stressed region results in stress redistribution in the structural member and a change of the stress field configuration. When stress redistribution is caused by plastic deformations, this is known as plastic redistribution. Large plastic deformations can result in considerable stress redistribution. Therefore, the stress field configuration obtained late in the ultimate state, can deviate considerably from that in the uncracked state. For instance, when the main tie in the deep beam in Figure 12 starts to yield, the tie loses stiffness and the deflection starts to increase considerably. Consequently, the cracks propagate upwards and crack widths increases. As a result the compressive arch will be forced upwards. Even though the tensile force in the main tie is constant after yielding, the load on the deep beam can increase since the internal lever arm increases. It means that the configuration of the stress field changes due to plastic redistribution in the ultimate state. This can continue as long as there is room enough for the compressive arch to rise and no critical details limit the load-bearing capacity. The anchorage of the main tie and the resistance of the highly compressed regions at the supports are critical issues in the design. Loss of anchorage in node Crushing of compression node Figure 12 Example of plastic redistribution of stress field of a deep beam in the ultimate state, a) before, b) after plastic redistribution and critical issues 11

20 Plastic redistribution results in non-linear behaviour, and as for the cracked state nonlinear analysis is needed to predict the structural response. In this case information about the non-linear behaviour of the materials is needed. Ultimate limit state In the ultimate limit state the structural member is on the limit of collapse. A further small increase of the load can not be resisted, because there are no reserves left that can be mobilised by plastic redistribution. Instead, a collapse mechanism develops. In this stage the plastic resistance is reached in some critical regions that determine the resistance of the whole structural member. The final equilibrium condition can be studied by means of theory of plasticity, which assumes an ideally plastic behaviour of the materials. Such an analysis is known as plastic analysis. The strut and tie method is a method for plastic analysis, lower bound approach, see Section 3.1. Strut and tie models are used to simulate stress fields of cracked reinforced concrete in the ultimate limit state, after considerable plastic redistribution, Figure 13. The strut and tie method requires little information: gross geometry, loads, and strength of materials (plastic resistance). It can be used early in the design process to design and arrange reinforcing steel. When plastic analysis is used for design and the problem is statically indeterminate, there exist many alternative solutions. It means that the load in the ultimate limit state can be carried by many alternative stress fields. One of the alternative models results in a certain reinforcement arrangement, which in turn results in a certain stiffness distribution and corresponding development of the stress field. It means that each design out of possible alternatives results in its own way to carry the load. Figure 13 Idealisation of the stress field in the ultimate limit state by a strut and tie model As mentioned above the behaviour in the ultimate state of a designed structural member can be analysed by non-linear analysis. When failure is reached in the model, this is equivalent with the conditions in the ultimate limit state. Hence, the non-linear analysis can also be used to predict the behaviour and the resistance of a member in the ultimate limit state. 12

21 1.4 Multi-axial state of stresses In Eurocode 2 CEN (2004) the compressive strength of concrete is defined as the strength of concrete cylinders tested according to a standard procedure. The test cylinders should have a diameter of 150 mm and a height of 300 mm. The loading equipment, the loading rate, fabrication of concrete, conditions for curing and storing are all specified in the standard procedure. The characteristic compressive strength specified in the code corresponds to the 5%- fractile in a series of many test samples. The mean compressive strength in the same series of tests is higher. It is assumed in Eurocode 2 that the mean value exceeds the characteristic value with 8 MPa for all specified concrete grades. It is generally assumed that the compressive strength determined by the standard test corresponds to the uniaxial compressive strength of concrete. It means the strength when stresses are applied in one direction only. However, the compressive strength in one (main) direction is influenced by stresses applied in other directions in bi-axial or tri-axial state of stresses. If a test cylinder, when loaded in compression in its main direction, is also subjected to transverse compression, the compressive strength increases in relation to that under uniaxial compression, Figure 14. The strength increases with the transverse compression. If, on the other hand, the specimen is subjected to transverse tension, the compressive strength decreases. In structural members of reinforced concrete the compressive strength of critical regions might also be influenced of cracks, parallel with or in a skew angle relative to the direction of the compressive stress, or transverse reinforcement that is acting in a perpendicular of skew direction. 3 2 (= 2 ) [MPa] = 2 = -28 MPa [ o ] 3 /oo Figure 14 Influence of transverse compressive stress on the compressive strength in the main direction, adopted from Richart et al. (1978). The strut and tie method is a method for plastic analysis. It is based on the theory of plasticity in which an ideally plastic response is assumed in the materials. However, the behaviour of concrete under compression is not ideally plastic, see Figure 15. The compressive strength determined in tests of cylinder specimens corresponds to the peak value. It is obvious that in a plastic design approach the resistance would be overestimated if this peak value is assumed to be constant during the entire deformation. 13

22 cc [ MPa ] ,001 0,002 0,003 0,004 cc Figure 15 Stress-strain relationship for concrete under uniaxial compression. Various strength classes, adopted from Nilson (1987) It is reasonable to assume that an equivalent plastic strength, constant during deformation, must be smaller than the peak value. An effectiveness factor can be used to relate the equivalent plastic strength to the maximum stress (peak value in standard test). f c, pl f cc where f c,pl = equivalent plastic strength to be used in plastic analysis = effectiveness factor f cc = uniaxial compressive strength from cylinder test The effectiveness factor cannot be determined easily, but varies for different types of structural problems, Nielsen (1984). Eurocode 2 gives expressions for estimation of the effectiveness factor for some fundamental cases. In the strut and tie method, the resistance of the member depends among others on the compressive strength of concrete in struts and nodes. Of the reasons mentioned above the design compressive strength of these components must be determined with due regard to the actual state of stresses. This is further discussed in Sections and 2.4.7, where recommendations from Eurocode 2 are presented. 14

23 2. THE STRUT AND TIE METHOD The following presentation of the strut and tie method is based on the concept which was developed by Schlaich and co-workers. An extensive presentation of the method was given by Schlaich et al. (1987). At that time the method was considered for the new CEB-FIP Model Code, which was under preparation. A more detailed presentation of the method was given in the German Concrete Design Handbook Schlaich and Schäfer (1989), and a concise presentation is found in Schlaich and Schäfer (1991). In the CEB- FIP Model Code 1990 CEB-FIP (1993) the method using stress fields with struts, nodes and ties was possible to use for structural analysis and design of deep beams and discontinuity regions. A text book on design according to Model Code 1990 was published as fib bulletins (International Federation for Structural Concrete), where two sections were devoted to design of nodes and design of deep beams and discontinuity regions Schäfer (1999a) and (1999b). Later on the strut and tie method has been introduced in Eurocode 2 CEN (2004). 2.1 Procedure The strut and tie method is based on theory of plasticity. The purpose of the strut and tie model is to simulate the stress field in cracked reinforced concrete in the ultimate limit state. Then the structure is just about to reach collapse and there has been more or less plastic redistribution. In critical regions the materials have reached plastic behaviour. Nevertheless, it is normally recommended to choose a strut and tie model that reminds, in a simplified way, of the linear elastic stress field. The reasons are limited ability to plastic redistribution in reinforced concrete; the design should also fulfil needs concerning appropriate performance in the service state. These items are further discussed in Section 3.2. However, it should be noticed that the linear elastic stress field should not be considered as the true solution, which should be copied by the strut and tie model. This is not at all the case. It is generally recommended, however, that the stress field in the ultimate limit state after cracking and yielding is not too extreme and too different from the stress-field before cracking and yielding. To get an appropriate strut and tie model that reminds, in a simplified way, of the linear elastic stress fields there are two possible approaches. The model can be developed on the basis of: the load path method, Section 2.2, together with application rules for the geometry of the strut and tie model, Section stress trajectories or principal stresses from a linear FE analysis When a structural element is designed using the strut and tie method, it is appropriate to carry out the calculations according to the following steps. 1) Structural analysis is carried out to solve support reactions and sectional forces under the design load, Figure 16. For statically indeterminate problems an appropriate choice of unknown has to be done. 15

24 2) Discontinuity (D-) and continuity (B-) regions are identified, Figure 16. It can generally be assumed that the extension of the discontinuity region is equal to the width (or height) of the structural element to which concentrated forces are spread, compare with Figures 7 and 8. 3) The stress distribution is determined in the boundaries between discontinuity and continuity regions, Figure 16. Q Q d h D-zone 2Q/h B-zone 2Q/h h Figure 16 Analysis of a column with a concentrated load, structural analysis gives the reaction force, discontinuity region is identified, stresses are calculated in the boundary between the D- and B-regions 4) A reasonable stress field in the discontinuity region is assumed by sketching and further analysed by the load path method, see Section 2.2 (Figure 19). Alternatively a reasonable stress field is determined by linear FE analysis, Figure 17. 5) A strut and tie model is chosen (Figure 21) on the basis of the load paths considering the application rules, see Section 2.3. Alternatively, the strut and tie model is chosen on the basis of stress trajectories or principal stresses found in a linear FE analysis. 6) The forces in the components of the strut and tie model are solved by means of equilibrium conditions, see Section ) Ties are designed, Figure 22, and stresses are checked in nodes and struts, see Sections 2.4. When needed dimensions are modified to avoid over-stressing of components. 16

25 Figure 17 Strut and tie model developed on the basis of stress trajectories from a linear analysis, idea from Schlaich et al. (1987) 2.2 The load path method In the load path method, streamlined load paths are inserted to simulate the stress field in a simplified way. The load path should in every section along its way represent the resultant of the corresponding stress field. It is often appropriate to divide a stress field in several parts and introduce a load path in each of them. Otherwise, the model could be over-simplistic and not guide the designer to an appropriate design and detailing, see Figure 18. For a given load case there is a unique relation between the load applied on the structure and the support reactions. It means that each support reaction is caused by a certain part of the load. The load is divided by load-dividers in sections where the shear force is zero, so that each part of the load is carried by the nearest support, Figure 18. R A R B R A R B1 R B2 Figure 18 Stress field and load paths in a deep beam with cantilevering end, a) over-simplistic model with two load paths, b) appropriate model with three load paths In the load path method a smoothly curved force path is inserted between the support reaction and that part of the load that causes this reaction. When necessary, in order to simulate the stress fields better, support reactions are split in two or more parts. It is appropriate to develop the load paths according to the following steps. 17

26 1) Identify the load dividers and the part of the load that belongs to each support reaction. 2) Sketch the stress field. At the boundaries of the discontinuity region the stresses should be parallel to the load or support reactions. When the load or the support reaction is distributed, the stress field should be spread over the same width at the boundary. When the load or the support reaction is concentrated, the stress field should start being concentrated to the same width, but disperse when it goes further into the discontinuity region. 3) Insert a smoothly curved load path that follows the resultant of the corresponding part of the stress field. 4) Check that the pattern of load paths is able to characterise the shape of the stress field. If not, the pattern is over-simplistic and should be further developed by splitting the stress fields into more parts, see Figure 18. Typically bottle-shaped stress fields need to be represented by two load paths. 5) Identify, principally, the transverse forces that are required to change the direction of the load path along its way through the discontinuity region and their location. With regard to equilibrium conditions a load path cannot change its direction without the influence from transverse forces. Where a load path has a sharp bend concentrated transverse forces are needed. When a load path has a soft bend, this is achieved by transverse forces that are distributed within a region where the load paths changes its direction successively, step by step, see Figure 19. If no external forces are added, for instance by prestressing, the transverse forces must be established between load paths, so that the load paths influence each other to change their direction. Sharp bend near concentrated force Q Q Load and load path meet in the same direction Soft bend away from concentrated force Q Q Figure 19 Load paths in a column head, the discontinuity region in Figure 16 18

27 In the load path method the load paths should fulfil the following rules: The load path should in each section along its way represent the resultant of the stress field which it simulated. Load paths cannot cross each other. At the boundary of the discontinuity region the load path should start in the same direction as the load or support reaction. Load paths should have a sharp bend near a concentrated force. When a load path needs to change direction away from a concentrated force the bend should be soft. In certain cases there will be no complete balance between loads and support reactions when load paths are inserted according to the principles listed above. In such cases a load path in U-turn must be added. A load path in U-turn enters and leaves the discontinuity region at the same boundary on the reaction side, see Figure 20. D B D T T Q Q T T Q Q Figure 20 Example of load path in U-turn. Post-tensioned beam with eccentric anchorage 19

28 2.3 Development of the strut and tie model Basic principles When the stress field is characterised by a number of load paths and the transverse forces needed to change their directions, a strut and tie model can be found by a further idealisation. Along the load paths node points are inserted in the centre of the regions where the load paths curve, Figure 21. In between the node points straight struts or ties are inserted. Struts are used to carry compression and ties to carry tension. According to the convention struts are represented by dotted lines in models and ties by full lines. Q Q Q Q Figure 21 Strut and tie model of stress field in column head, compare with Figure 16 The elements of the strut and tie model represent both the load paths where the load is carried, and the transverse forces that are necessary to change the direction of the load paths. The struts and ties that represent the transverse forces should be placed in the centre of the corresponding stress fields. For instance when a load path has a soft bend, this is achieved by transverse forces that are distributed over a certain region and the corresponding strut or tie should be placed in the centre of this region. Ties, struts and nodes can be classified as concentrated or distributed. Concentrated nodes appear where concentrated forces (load, reaction force) act at the boundaries of the discontinuity region. Concentrated nodes are typically limited in their extension by dimensions that are chosen in the design, for instance the width of supports, anchor plates, loading plates etc. Of this reason the stresses in concentrated nodes must be checked to avoid overstressing. Distributed nodes appear where distributed stress fields 20

29 meet. A distributed stress field can be the result of a distributed load or reaction at a boundary or distributed transverse forces needed to softly change the direction of a load path away from concentrated forces. It is not necessary to check stresses in distributed nodes, because there is no limitation of their dimensions. A distributed node cannot be overstressed. Instead, by means of plastic redistribution, the structure will incorporate more and more volume of material into the node, if it becomes highly stressed. A tie which is needed in a concentrated node to abruptly change the direction of load path is a concentrated tie. This should be designed by reinforcement concentrated together near the edge of the element, see Figure 22. Typical examples are ties in deep beams, to enable tied arch action, or the tensile chord in beams. When a wide transverse tensile stress field is needed to softly change the direction of a load path, this is achieved by a distributed tie. A distributed tie should be designed by reinforcement that is distributed on many small diameter bars, which are placed within the corresponding region. Figure 22 Reinforcement for concentrated tie in column head, compare with Figure 21 When a transverse strut between concentrated nodes is needed to change the direction of load paths, this can be a concentrated strut that follows the edge of the element. A compressive zone of a beam is a typically example. Otherwise, concentrated struts will not pass through the interior of a discontinuity region. It is natural that concentrated forces will spread out transversally and incorporate more and more of the material available. This spread of compressive stresses from a concentrated node can be fanshaped or bottle-shaped. It should be checked early in the development of the strut and tie model, before calculations are carried out, that the model is able to fulfil equilibrium conditions. By a visual inspection of the nodes the conditions for obtaining horizontal and vertical equilibrium can be checked and major mistakes can be revealed. 21

30 To achieve a model with perfect equilibrium, the nodes should be positioned carefully with regard to the locations of reaction forces. When a stress field has been divided in parts, each part is represented by one load path. The load path should meet its reaction in the centre of the corresponding stress field, see Figure 23. This means that support reactions in supports with several load paths must be divided in parts proportional to the respective loads and the node should be in the centre of each part. l B1 R R B1 B l B R A R B1 R B2 R B1 l B1 l B2 R B2 R lb2 R R B R B1 B2 B l B R B2 l B Figure 23 Support regions should be divided in proportion to the load paths before node points are inserted in the centre of the respective stress field In the development of a strut and tie model one or more angles between struts and ties need to be chosen. It is important to be aware of the number of possible choices that can be made. When one or some angles are chosen, other angles in the model will be determined and can be solved. They cannot be chosen too. If possible the angles between struts and ties should be close to preferred values, see Section Application rules With regard to the need for ductility and an appropriate behaviour in the service state, see Section 3.3, it is recommended to follow the following application rules. Stresses under concentrated forces (loads, reactions) should be spread out as soon as possible when they enter the discontinuity region. A deviation angle of about 30 is a reasonable choice and it should not exceed 45, Figure 24. The angle between a strut and a tie should not be too small, Figure 25. o In case of a strut meeting a single tie the preferred angle is about 60 and it should not be less than 45. o In case of a strut between two perpendicular ties the preferred angle is about 45 and it should not be less than 30. Concentrated forces should not be carried by concentrated struts across wide elements, with exception for concentrated compressive zones in continuity regions (B-regions). 22

31 Q/2 Q/2 Q Figure 24 Deviation of concentrated forces, according to Schäfer (1999b) 2 > 30 2 > 30 > 45 1 > 30 1 > 30 Figure 25 Recommended minimum angles between struts and ties, adopted from Schäfer (1999b) In strut and tie models with many components it might be difficult and even impossible to fulfil the rules concerning minimum angles for all the components. In such cases it is more important to fulfil the rules for heavily loaded struts, while the rules may be violated for less important struts. 2.4 Design of components Forces in struts and ties Before forces in struts and ties are calculated it is recommended to carry out the following basic checks of the model. Angles that depend on each other cannot be chosen independently, but must be solved with regard to geometrical conditions. The geometry of the strut and tie model must be possible with regard to chosen angles between strut and ties and given dimensions of the discontinuity region. It means that the model must be located fully within the boundaries of the element, with the chosen angles. Mistakes can be made if the model is not drawn in correct proportions. 23

32 Visual inspection of all nodes to check that vertical and horizontal equilibrium can be achieved. In case of impossible nodes, the strut and tie model needs to be rearranged. Struts along edges of the element need a certain width to avoid overstressing. This influences the location of the corresponding node points. For each node the forces in the adjacent struts and ties can be solved by vertical and horizontal equilibrium conditions. The results from one node influence the forces in adjacent nodes. In a correctly developed strut and tie model there should be equilibrium in all nodes without contradictions. However, in statically indeterminate systems, some parameters must be chosen. Forces in struts and ties can also be solved by equilibrium conditions for free bodies separated by cuts through the structural elements. A deep beam with vertical load can serve as an example, Figure 26. A vertical cut through the strut and tie model and the condition of horizontal equilibrium reveal that the force in the horizontal tie (tension) and the force in the horizontal strut (compression) must be of the same magnitude (but opposite signs). Moment equilibrium shows that the same force couple with its internal lever arm must resist the bending moment (load effect) in the section for maximum moment. Such free body conditions can be used to check the forces determined on the basis of node equilibrium. l/2 l/2 q q M E l l q 2 4 C C T M R C z T z T z Figure 26 Equilibrium conditions of deep beam Design of nodes Nodes are classified as concentrated nodes and distributed nodes. Concentrated nodes appear where concentrated forces act, normally at the edges of the structural element. Distributed nodes appear where distributed stress fields meet. Distributed nodes are never critical in the design and are not treated further in this section. The most common types of concentrated nodes are compression nodes (without anchored reinforcement) under bearings, loading plates, etc., see Figure 27 24

33 compression - tension nodes (with anchored reinforcement in one direction) under bearings, loading plates, etc, Figure 28 compression tension nodes with reinforcement in two directions, Figure 29 compression tension nodes with bent reinforcement in two directions, Figure 30 compression nodes with reduced support width The design of nodes includes limitation of the compressive stress applied at the edges of the nodes. The stresses in nodes are influenced by dimensions of support bearings, anchor plates, loading plates etc., but also by the geometry of the strut and tie model. In Eurocode 2 the maximum stress that can be applied at the edges of nodes is specified for the following cases. Compression nodes where no ties are anchored in the node Rd, max k1f cd where k 1 = 1,0 (national parameter, recommended value) fck MPa 1 (national parameter, recommended value) 250 a 2 a 3 a 2 a 3 C 2 c0 C 3 C 2 c0 C 3 c2 2 3 c3 c2 2 3 c3 C 1 C 2 u C 0 C 0 u R c1 c1 R R 1 R 2 R 1 R 2 a 1 R 1 R 2 a 1 Figure 27 Compression node, alternative models, adopted from Schäfer (1999b) Figure 27 shows two alternative idealisations of the node geometry of a typical compression node. With regard to equilibrium conditions and check of critical stresses the two models yield the same result. The two stresses c1 and c0 are the principal stresses of the node. It is sufficient to check the principal stresses, since the other incoming stresses can never be higher. 25

34 Normally, as a first step, the dimension a 1 of the loading plate is chosen such that the stress c1 fulfils the stress limit with regard to the force R. Further, the node points where the idealised struts meet are positioned such that the node height a 0 is sufficient with regard to the horizontal force C 0 within the node. For a certain height of the node region the strut forces C 2 and C 3 act perpendicular to the faces of the triangular node region. This corresponds to plane hydrostatic pressure for which all incoming stresses and the principal stresses are equal. If a 0 is chosen greater than corresponding to plane hydrostatic pressure, the stress c0 will not be critical for the design and only c1 needs to be checked. Compression tension nodes with anchored ties in one direction Rd, max k2f cd where k 2 = 0,85 (national parameter, recommended value) A typical compression tension node with ties anchored in one direction is shown in Figure 28. The tie is considered to be a reinforced concrete tensile member with a formal height u, i.e. the tie steel with surrounding concrete. The node region is defined as the overlap between this tensile member and the inclined compressive strut. a 2 a 2 C 2 C 2 c2 c2 R C 2 T a s a 1 R c1 T u u = 2a s s 0 2s 0 a 1 R c1 u T u = 2s 0 Figure 28 Compression tension node with anchored reinforcement, a) node where the bars are anchored within the node, b) node where the bars are extended beyond the node, adopted from Schäfer (1999b) Both the stress c1 and c2 must be checked with regard to the stress limit. The width a 2 of the face of the node where the inclined compressive force C 2 acts, can be calculated as u a2 a1 sin a1 sin ucos tan When a 2 is calculated, it is recommended to use the following values u for the height of the tension member: 26

35 u 0 for nodes with one layer of reinforcement not extending beyond the node region Schäfer(1999b) u 2s 0 for nodes with one layer of reinforcement extending at least the distance 2s 0 beyond the node region Eurocode 2 u 2s 0 n 1s for nodes with n layers of reinforcement extending at least the distance 2s 0 beyond the node region Eurocode 2 where s 0 = distance from the bottom edge to the centre of the bottom reinforcement layer s = spacing between reinforcement layers (centre to centre) The design of the node includes normally the following steps Choice of support length a 1 such that the stress limit is fulfilled with regard to the force R Arrangement of reinforcement in one or several layers Check of concrete stress in section a 2 with regard to the force C 2 Arrangement of anchorage of the tie bars, see Section Provision of transverse reinforcement in the anchorage and node regions Proper detailing of compression tension nodes is very important. The anchorage of the tie bars results in splitting effects and micro cracking. Transverse reinforcement, in form of stirrups or U-bends around the anchored bars, confines the anchorage zone and balances splitting effects. Furthermore, it is favourable if some longitudinal compression is built up beyond the node region to confine it by compression. Therefore, it is recommended that the tie bars should extend at least a minimum distance 2s 0 beyond the node region, even when it is not required with regard to anchorage (see Section 2.4.4). If the tie bars are not well distributed across the width of the section transverse tensile stresses might develop in the anchorage region. Also for this reason transverse reinforcement is recommended. For the design of compression tension nodes it is favourable to keep the angles between struts and ties greater than 55, arrange the ties in several layers, anchor the reinforcement to a considerable part behind the node, confine the node region by stirrups, bearing details and/or friction. Eurocode 2 CEN (2004) permits higher stress limits in case of improved detailing. Compression tension nodes with anchored ties in more than one direction Rd, max k3f cd where k 3 = 0,75 (national parameter, recommended value) In compression tension nodes with anchored ties in more than one direction, Figure 29, the change of the tensile force from T 1 to T 2 requires inclined compression for 27

36 equilibrium. The width of the inclined compressive strut depends on the design anchorage length l bd which is used to anchor the difference between T 1 and T 2. T 3 C a T 2 T 3 C cc T 1 T 1 T 2 u l bd u = 2s 0 Figure 29 Compression tension node with reinforcement in two directions, adopted form Schäfer (1999b) The design of the node regions consists of the following steps. Arrangement of anchorage for the main reinforcement Check of stress limit in section a for the compressive force C Arrangement of the reinforcement for T 3, normally loops, hooks or stirrups of small bars Arrangement of transverse reinforcement (third direction), normally legs of stirrups of loops. A similar type of compression tension node is provided with bent bars for tensile forces in two directions, Figure 30. T 1 a C cc T 1 1 C 2 T 2 T 2 Figure 30 Compression tension node with bent reinforcement, adopted from Schäfer (1999b) 28