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Reinforcement for Plates

Reinforcement for Plates . Course CT4150 Lecture12 5 Jan. 2010. Normal forces in concrete plates. Only rebars in the x and y directions Equilibrium of a plate part. Equations Design Choose q such that n s x + n sy is minimal (lower bound theorem). Solution. Example 1.

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Reinforcement for Plates

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  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 q such that nsx+nsy is minimal (lower bound theorem)

  4. Solution

  5. Example 1 • Situation nxx=1200, nyy=-200, nxy=-400 kN/m fy = 500 N/mm², f’c= 30 N/mm² Thickness = 100 mm • Reinforcement nsx= 1200+400 = 1600 kN/m nsy= -200+400 = 200 • Concrete nc= -2x400 = -800 kN/m

  6. Asx=1600/500 = 3.20 mm = 3200 mm²/m 212–280 = 2p/4x12²x1000/280 = 3231 OK Asy=200/500 = 0.40 mm = 400 mm²/m 26–500 = 2p/4x6²x1000/500 = 452 OK sc= 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, fcd = 16.67 N/mm², fyd = 435 N/mm²

  8. Forces nxx, nyy, nxy (linear elastic analysis) • Principal stresses

  9. Reinforcement requirements (software)

  10. Reinforcement (engineer)

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

  12. Example 3 • Moments in a point mxx = 13, myy = -8, mxy = 5 kNm/m • Moment capacities mpx = 17, mpy = 0, m’px =0, m’py = 10 • Is the capacity sufficient? • Yes

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

  14. Solution 2 (when mpx would be < 0) Solution 3 (when mpy would be < 0)

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

  16. Solution 6 (when mpx and mpy would be < 0) Solution 7 (when m’px and m’py would be < 0)

  17. Example 4 • Moments in a point (as in example 1) mxx = 13, myy = -8, mxy = 5 kNm/m • Moment capacities mpx = 13+5²/8 = 16.13 m’px = 0 mpy = 0 m’py = 8+5²/13 = 9.92 • Amount of reinforcement is proportional to 16.13+0+0+9.92 = 26 • Amount of reinforcement in example 3 17+0+0+10 = 27 (larger, so not optimal)

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

  19. Example 5 continued • Decomposition of the load

  20. Example 5 Torsion moment

  21. Example 5 All moments • Moments in the bridge middle • Moments at the bridge support

  22. Example 5 FEM moments

  23. Example 5 Reinforcement Middle Support Designed

  24. Example 5 Upper bound check Result > 1 OK

  25. Computed requirements

  26. 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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