Strut and tie. Betongbyggnad VBKN05

Transcription

1 Strut and tie Betongbyggnad VBKN05

2 Strut and Tie Design the deep beam (hög balk) in the figure. qd = 200 kn/m 3 m 3 m

3 Strut and Tie

4 Strut and Tie Describe the stresses in a concrete structure with a strut and tie model Are normally used in the ULS Based on plasticity theory

5 B- and D-regions The structure is divided into B- and D-regions

6 B- regions (continuity region) Plane sections remain plane under loading. Plane strain distribution is assumed to be valid (Bernoulli hypothesis)

7 D- region (discontinuity region) Non-linear strain distribution due to geometry and external load. Wall on concentrated supports, deep beam (hög balk)

8 Extension of D-region - St Venant s princip Study a body that is subjected to a system of forces in self equilibrium. On the global level the body is not subjected to a force resultant and will not move. Locally, stresses will occur in the body. The extension d of the stressed region is equal to the maximum distance h between the forces.

9 Example of forces/stresses in a body = =

10 Example of forces/stresses in a body

11 B- and D-regions in simply supported beam Distributed load Concentrated load

12 Example on D-regions

13 Principles for Strut and tie method Describes the stress field in a structure with strut and tie modell (fackverksanalogi) Compressive struts in concrete Reinforcement that acts as ties, subjected to tension Struts and ties are connected in nodes

14 Deep beam without cracks The stresses can be determined with linear elastic analysis under the assumption that the material structure is homogeneous, concrete. Stress field Simplifed stress condition Strut and tie modell Concrete arch

15 Deep beam with cracks When cracks have occurred the stresses will be redistributed since the cracked regions will have a decreased stiffness.

16 Deep beam ultimate state Ultimate state = non-linear material behaviour. Advanced analysing methods are needed. These models show examples of stress fields that can be realistic in the plastic state. Large cracks in tensile zone

17 Deep beam ultimate state Zones that are critical for the capacity of the deep beam

18 Plasticity theory Assumption The stress field is supposed to be in equilibrium with the external loads. The plastic capacity of the structure is large enough so that a redistribution of stresses are possible. Indefinite plastic deformation capacity

19 Plastic analisys Reality Concrete doesn t have indenfinite deformation capacity Ultimate strain in concrete is low, that is there is only a limited capacity for redistribution of stresses Use a strut and tie model that is close to the linear elastic stress field.

20 Design 1. Determine the external loads and the reactions at supports. 2. Divide the structure into B- and D-regions 3. Design the B-regions 4. Determine the contact forces between the B- and D-regions. 5. A stress field in the discontinuity region is assumed by sketching and further analyzed by the load path method. Alternatively it can be determined by linear FE analysis. 6. Choose a suitable strut and tie model based on stress field 7. Determine the forces in the struts and ties 8. Design the reinforcement and control the capacity of the struts and the nodes.

21 Forces between B- and D-regions Prestressed beam Column with two concentrated loads Q

22 Choice for strut and tie model Two different methods 1. Load path method 2. Linear FE-analysis

23 Load path method Streamlined load paths are inserted to simulate the stress field in a simplified way. Two load paths cannot cross Forces on one side of the structure must be in equilibrium with the forces on the other side of the structure

24 Load paths load dividers Load dividers = section where the shear force is zero

25 Load paths in prestressed beam Prestressed beam Q

26 Load paths in prestressed beam U-sväng The curvature of the paths leads to transverse forces

27 Strut and tie model (prestressed beam) Node Ties (solid line) and struts (dotted line)

28 Column with two concentrated loads D-region Load paths Strut and tie model

29 Strut and tie-models

30 Strut and tie-models

31 Strut and tie models based on FEM Result from FE-analysis presented as main stresses.

32 Strut and tie model based on FEM Result of FE-analysis, stress field.

33 FEM - problems Required anchorage capacity can be underestimated if the design is based only on FE-analysis FEM shows tensile stresses (forces) distributed over a large area. In reality the reinforcement will take the tensile forces and the concrete will not contribute to any capacity. The forces must then be anchored in the end of the reinforcement.

34 Collapse of the oil platform Sleipner A 1991 The oil platform collapsed when it was towed to the oil field. Failure took place in one of the walls. Collins et.al. Concrete International, v 19, n 8, p 28-35, Aug 1997

35 Collapse of the oil platform Sleipner A 1991 Failure was initiated near the corner Investigations showed that the designer had used a FE-analysis and that there were flaws in the model that lead to failure. A new platform was designed, basically with hand calculations. Collins et.al. Concrete International, v 19, n 8, p 28-35, Aug 1997

36 Collapse of the oil platform Sleipner A 1991 Initial design New design

37 Design Strut and tie Design 1. The angle between the tie and the strut should be larger than 45, often 60 is appropriate 2. When two ties and a strut meets the angles should be larger than 30, 45 is recommended 3. The stresses from concentrated loads are to distributed in the structure, recommended angles are 30, not larger than

38 Design - The forces in the ties and the struts are determined based on equilibrium - Some of the angles between ties and struts are chosen according to the requirements. Other angles and lengths of struts and ties are given by geometry. - The reinforcement area is given from the forces in the ties. - The stresses in the nodes and the struts are controlled.

39 Nodes Concentrated node: Nodes where concentrated forces meet, normally near the boundary of the D-region Due to the limited area the stresses has to be checked Distributed node: Node where distributed stress fields meet. No control is needed since there is a capacity for stress redistribution within a larger area.

40 Concentrated nodes Compression node (CCC) - node where three compression struts meet Stress should be limited: (EC2) k f Rd,max 1 cd where: fck 1, k = 1,0

41 Concentrated nodes Compression tension node (CCT) two struts and one tie meet Stress should be limited: (EC2) k f Rd, max 2 cd where: fck 1, k = 0,85

42 Concentrated nodes Compression-tension node(ctt) one strut and two (or more) ties meet Stress should be limited: (EC2) k f Rd, max 3 cd where: fck 1, k = 0,75

43 Ties Tie = normal or prestressed reinforcement Distributed ties: Distribute the reinforcement evenly over this height

44 Ties Concentrated ties: Place the reinforcement close to each other

45 Ties Anchorage of reinforcement. Important to control. a l bd a lbd

46 Anchorage of ties Different ways to increase the anchorage capacity

47 Struts The stresses in the struts should fulfill the following requirements (EC 2): Uni-axial conditions: 1. 0 f Rd,max cd For a concrete strut in a cracked compression zone: 0. 6 f Rd,max cd

48 Optimization of models Try to find the model that gives the least amount of reinforcement. Preferred model Less appropriate model Optimization criterion: F i l i mi min Fi = force in part i li = length on part i i εmi = average strain in part i

49 Optimization of reinforcement Place reinforcement parallell or transverse to the boundaries of the structure Use straight bars

50 Serviceability limit state Normally, the same stress field can be used in both serviceability and ultimate limit state. The structure that is controlled in serviceability limit state (tension and crack widths), with the same strut and tie model used for the ultimate limit. Different strut and tie models can be used if the linear elastic stress field gives uneconomical or impractical reinforcement.

51 Strut and tie models, B-regions Shear reinforcement

52 Example Design the deep beam with the help of a strut and tie-model qd = 200 kn/m Concrete C25 Reinf. B500B ᴓ12 Thickness of beam = 300 mm Width of support = 120 mm 3 m 3 m

53 Design the deep beam with the help of a strut and tie-model Concrete C25 Reinf. B500B ᴓ12 Thickness of beam = 300 mm Width of support = 120 mm

54 fcd = 16,7 MPa fyd = 435 MPa Use = 60 Node N1 T = 300/tan30 = 173,2 kn C1 = 300/sin60 = 346,4 kn C1 T 300 kn Reinforcement: A = 173,2 /435*10 3 = 398 mm 2 i.e. 4 bars Controll Node CCT-node: Upper node contact pressure σ Rd,max = k 2 f cd k2 = 0,85 = 1 - fck /250 = 1-25/250 = 0,9 σ Rd,max = 0,85 0,9 16,7 = 12,8 MPa σ c = ,3 0,12 Control strut C1 = 8,3 MPa < 12,8 MPa OK! s = /2 = 28 mm (with respect to cover) u = 2*28 = 56 mm C1 a s T u 120

55 x1 = 56cot60 = 32 mm x2 = = 152 mm a = 152 cos 30 =132 mm Requirement, 0. 6 = 0,6*0,9*16,7 = 9 MPa Rd max f cd σ c1 = 346, ,3 0,132 = 8,7 MPa < 9 MPa OK!

56 Height to next node: h h0 = ( /2)*tan60 = 1195 mm h = h0 + s = = 1223 mm Node N2 C2 = 346,4*cos60 = 173,2 kn C2 = T i.e. C2 = T The stresses are not concentrated in this node, no control is needed.