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Practical maths

Construction Insitu RC Suspended Floors using BM Bending moment formula maths

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- 1. Reinforced Concrete Suspended Floors and beam design
- 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. 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. <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. From the following diagram make sure that you note down the joist data then work out the total load per joist
- 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. 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. <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. (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. Answer
- 11. Important take note The formula is
- 12. The formula is Answer in cm TN3 I= 9504 from tables of values
- 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. 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. (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.
- 17. Answer Answer Answer TN3

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