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Reinforced Concrete Suspended Floors and beam design
<ul><li>Ribbed Floors are used to…. </li></ul><ul><li>reduce the overall depth  of a traditional cast insitu reinforced co...
The basic concept is to replace the wide spaced deep  beams with narrow spaced shallow beams or ribs which will carry only...
<ul><li>Ribbed floors are usually cast against metal, glass fibre or polypropylene They are made in preformed moulds which...
From the following diagram make sure that you  note down the joist data  then work out the total load per joist
Answer: Total load (W) per joist = 5m X 0.4m X 2.25 kN/m 2  = 45 kN <ul><li>On the following page you should  </li></ul><u...
Answer The formula for the Bending moment of a beam is  ….   Transpose the BM formula to find, d.   Answer from standard s...
<ul><li>Joist and Beam Sizing for calculating overall dimensions alone is insufficient. </li></ul><ul><li>checks should al...
(Use b and d from before) The formula for calculating deflection due to a uniformly distributed load  is: Answer Answer TN...
Answer
Important take note   The formula is
The formula is   Answer   in cm TN3   I= 9504   from tables of values
Permissible deflection is 1/360 of 4 m = 11.1 cm. Therefore actual deflection of  0.835 cm is acceptable. Ref. BS 5950: St...
Structural Steelwork—Column Design <ul><li>Steel columns or stanchions have a tendency to….  </li></ul><ul><li>buckle or b...
(b) and (d) are incorporated into a geometric property of section, known as the radius of gyration (r).   It can be calcul...
 
Answer Answer Answer TN3
 
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Construction Insitu Rc Suspended Floors Using Bm Bending Moment Formula Maths

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Practical maths
Construction Insitu RC Suspended Floors using BM Bending moment formula maths

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Construction Insitu Rc Suspended Floors Using Bm Bending Moment Formula Maths

  1. 1. Reinforced Concrete Suspended Floors and beam design
  2. 2. <ul><li>Ribbed Floors are used to…. </li></ul><ul><li>reduce the overall depth of a traditional cast insitu reinforced concrete beam and slab suspended floor </li></ul>
  3. 3. The basic concept is to replace the wide spaced deep beams with narrow spaced shallow beams or ribs which will carry only a small amount of slab loading . <ul><li>These floors can be designed as one or two way spanning floors. </li></ul><ul><li>One way spanning ribbed floors are sometimes called trough floors </li></ul><ul><li>the two way spanning ribbed floors are called coffered or waffle floors. </li></ul>
  4. 4. <ul><li>Ribbed floors are usually cast against metal, glass fibre or polypropylene They are made in preformed moulds which are temporarily supported on plywood decking joists and props </li></ul>
  5. 5. From the following diagram make sure that you note down the joist data then work out the total load per joist
  6. 6. Answer: Total load (W) per joist = 5m X 0.4m X 2.25 kN/m 2 = 45 kN <ul><li>On the following page you should </li></ul><ul><li>Make a note of the bending moment (BM) formula </li></ul><ul><li>Make a note of the data </li></ul><ul><li>Find the depth d , of the joist </li></ul>
  7. 7. Answer The formula for the Bending moment of a beam is …. Transpose the BM formula to find, d. Answer from standard sizes TN4 Work slowly through the following
  8. 8. <ul><li>Joist and Beam Sizing for calculating overall dimensions alone is insufficient. </li></ul><ul><li>checks should also be made to satisfy: resistance to deflection </li></ul><ul><li>adequate safe bearing and </li></ul><ul><li>resistance to shear . </li></ul><ul><li>Deflection — should be minimal to prevent damage to…. </li></ul><ul><li>plastered ceilings. </li></ul><ul><li>An allowance of up to 0003 X span is normally acceptable; for the last example this will be:- 0003 X 5000mm = 15mm </li></ul>
  9. 9. (Use b and d from before) The formula for calculating deflection due to a uniformly distributed load is: Answer Answer TN4 Use the nearest commercial size to find I then find ,d. Answer
  10. 10. Answer
  11. 11. Important take note The formula is
  12. 12. The formula is Answer in cm TN3 I= 9504 from tables of values
  13. 13. Permissible deflection is 1/360 of 4 m = 11.1 cm. Therefore actual deflection of 0.835 cm is acceptable. Ref. BS 5950: Structural use of steelwork in building.
  14. 14. Structural Steelwork—Column Design <ul><li>Steel columns or stanchions have a tendency to…. </li></ul><ul><li>buckle or bend under extreme loading. This can be attributed to: </li></ul>(a) length (b) cross sectional area (c) method of end fixing (d) the shape of section
  15. 15. (b) and (d) are incorporated into a geometric property of section, known as the radius of gyration (r). It can be calculated:- r = SQRT (1/A) where: I = (Inertia) -2nd moment of area A = cross sectional area Note: r,I and A are all listed in steel design tables, e.g.. BS4:1980.
  16. 17. Answer Answer Answer TN3

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