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Reinforcement for Plates Course CT4150 Lecture12 5 Jan. 2010.

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Presentation on theme: "Reinforcement for Plates Course CT4150 Lecture12 5 Jan. 2010."— Presentation transcript:

1 Reinforcement for Plates Course CT4150 Lecture12 5 Jan. 2010

2 Normal forces in concrete plates Only rebars in the x and y directions Equilibrium of a plate part

3 Equations Design Choose  such that n sx +n sy is minimal (lower bound theorem)

4 Solution

5 Example 1 Situation n xx =1200, n yy =-200, n xy =-400 kN/m f y = 500 N/mm², f’ c = 30 N/mm² Thickness = 100 mm Reinforcement n sx = = 1600 kN/m n sy = = 200 Concrete n c = -2x400 = -800 kN/m

6 A sx =1600/500 = 3.20 mm = 3200 mm²/m 2  12–280 = 2  /4x12²x1000/280 = 3231 OK A sy =200/500 = 0.40 mm = 400 mm²/m 2  6–500 = 2  /4x6²x1000/500 = 452 OK  c = 800/100 = 8 N/mm² < f’ c OK (safety factors omitted)

7 Example 2 Deep beam 4.7x7.5 m, 2 supports, opening 1.5x1.5 m, point load 3000 kN, f cd = N/mm², f yd = 435 N/mm²

8 Forces n xx, n yy, n xy (linear elastic analysis) Principal stresses

9 Reinforcement requirements (software)

10

11 Reinforcement (engineer)

12 Moments in concrete plates Only rebars in the x and y directions Equilibrium of a plate part Result (Yield contour)

13

14 Example 3 Moments in a point m xx = 13, m yy = -8, m xy = 5 kNm/m Moment capacities m px = 17, m py = 0, m’ px =0, m’ py = 10 Is the capacity sufficient? Yes

15 Design of moment reinforcement Carry the moments with the least amount of reinforcement. So, minimize m px + m py + m’ px + m’ py 5 constraints m px, m py, m’ px, m’ py ≥ 0 Solution 1 (Wood-Armer moments) Crack direction 45º to the reinforcing bars

16 Solution 2 (when m px would be < 0) Solution 3 (when m py would be < 0)

17 Solution 4 (when m’ px would be < 0) Solution 5 (when m’ py would be < 0)

18 Solution 6 (when m px and m py would be < 0) Solution 7 (when m’ px and m’ py would be < 0)

19 Example 4 Moments in a point (as in example 1) m xx = 13, m yy = -8, m xy = 5 kNm/m Moment capacities m px = 13+5²/8 = 16.13m’ px = 0 m py = 0m’ py = 8+5²/13 = 9.92 Amount of reinforcement is proportional to = 26 Amount of reinforcement in example = 27 (larger, so not optimal)

20 Example 5 Plate bridge, simply supported 4 x 8 m, point load 80 kN, thick 0.25 m

21 Example 5 continued Decomposition of the load

22 Example 5 Torsion moment

23 Example 5 All moments Moments in the bridge middle Moments at the bridge support

24 Example 5 FEM moments

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27 Example 5 Reinforcement Middle Support Designed

28 Example 5 Upper bound check Result > 1OK

29 Computed requirements

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33 Conclusions The design procedure used is 1 Compute the force flow linear elastically 2 Choose the dimensions plastically The reason for the linear elastic analysis in the first step is that it shows us how an as yet imperfect design can be improved. A plastic (or nonlinear) analysis in step 1 would shows us how the structure would collapse; but that is not what we want to know in design. This procedure is applied to design many types of structure for the ULS.


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