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Check Beam Shear

Check Beam Shear. 245. 166. 245. Bending moment design. 2.5m. Short direction. Central band ratio =. Central band of short direction = 0.83 As = 0.83 (10.1)=8.6cm 2. Footing Design Part II Combined footing. Example 1. Design a combined footing As shown. Dimension calculation

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Check Beam Shear

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  1. Check Beam Shear

  2. 245 166 245 Bending moment design 2.5m Short direction

  3. Central band ratio = Central band of short direction = 0.83 As = 0.83 (10.1)=8.6cm2

  4. Footing Design Part II Combined footing

  5. Example 1 Design a combined footing As shown

  6. Dimension calculation The base dimension to get uniform distributed load 2000kN 800 kN 1200 kN A x1=0.2m x2=6.2m x 800(0.2)+1200(6.2)=2000(x) x = 3.8m 800 kN 1200 kN Try thickness =80cm 2x =7.6 m

  7. Area required

  8. Check for punching Shear d = 730 mm 1.13m 0.765 A

  9. B

  10. Draw S.F.D & B.M.D Stress under footing = 190 *1.8 = 342 kN/m

  11. Check for beam shear b = 1800mm, d = 730mm

  12. Bending moment Long direction

  13. Bending moment Short direction Under Column A

  14. Under Column B Shrinkage Reinforcement in short direction

  15. Footing Design Part III Combined footing, strip footing, & Mat foundation

  16. Example 2 Design a combined footing As shown

  17. 1950kN 1200 kN 750 kN A x2=4.2m x Dimension calculation The base dimension to get uniform distributed load 750(4.2)+1200(0.2)=1950 (x) x = 1.75m x1=0.2m

  18. Area required

  19. Check for punching Shear h= 750mm d = 732 mm A B2=1m B1=4m

  20. B

  21. Draw S.F.D & B.M.D

  22. Empirical S.F.D & B.M.D m Convert trapezoidal load to rectangle Mmax Clear distance between column B in moment design = ave. width = 2.5m

  23. b Y=1.5m 1m x 4m Check for beam shear d = 665mm

  24. Bending moment Long direction Top Bottom

  25. Bending moment Short direction Under Column A

  26. Under Column B Shrinkage Reinforcement in short direction

  27. Reinforcement details

  28. Example 3 (Strip footing) Design a combined footing As shown

  29. Dimension calculation The base dimension to get uniform distributed load 3040kN 800 kN 1280 kN 960 kN A Assume L1=0.6 x1=5.2m L2 x2=10.7m x 800(0.6)+1280(5.1)+960(10.6)= 3040 (x) x = 5.65m, 2(x)=11.3m L2=11.3 -(10.6)=0.7

  30. Check for punching Shear h = 700 mm d=630mm Example B You can check other columns

  31. Draw S.F.D & B.M.D Stress under footing = 195 *1.8 = 351 kN/m

  32. Check for beam shear b = 1800mm, d = 630mm

  33. Bending moment Long direction Design Short direction as example 1 (lecture 11)

  34. Reinforcement details

  35. Mat Foundation

  36. Check for punching Shear

  37. General Example, Ref. 2 Modified load

  38. General reinforcement details

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