Design of RCC Column footing

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1. FOR 5 TH SEMESTER DIPLOMA IN CIVIL ENGINEERING CONCEPTs AND PROBLEMS DESIGN OF RCC COLUMN FOOTING

2. Learning Outcomes: Concept of column footings. Design criteria. Design steps for column footing. Reinforcement detailing.

3. Column footing: RCC columns are supported by the foundation structures which are located below the ground level are called footings. Purpose of Footings: To support the upper structure. To transfer the Loads and moments safely to subsoil. Footings are designed to resist the bending moment and shear forces developed due to soil reaction. S.J.(Govt.) POLYTECHNIC BENGALURU

4. Types of footing:

5. Isolated footing. The footings which are provided below the column independently are called as isolated footings . This type of footing may be square , rectangular or circular in section Isolated footing consists of thick slab which may be flat or sloped or stepped. S.J.(Govt.) POLYTECHNIC BENGALURU P

6. The structural design of the footing includes the design of Depth of footing Reinforcement requirement Check on serviceability S.J.(Govt.) POLYTECHNIC BENGALURU

7. S.J.(Govt.) POLYTECHNIC BENGALURU

8. S.J.(Govt.) POLYTECHNIC BENGALURU

9. Design Considerations: Minimum reinforcement : (As per IS456:2000, clause 26.5.2.1&2 The mild steel reinforcement in either direction in slabs shall not be less than 0.15 percent of the total cross sectional area . However, this value can be reduced to 0.12 percent when high strength deformed bars or welded wire fabric are used The diameter of reinforcing bars shall not exceed one eight of the total thickness of the slab . S.J.(Govt.) POLYTECHNIC BENGALURU

10. Design Considerations:   /,` S.J.(Govt.) POLYTECHNIC BENGALURU

11.   S.J.(Govt.) POLYTECHNIC BENGALURU

12. ..\IS456:2000.pdf When the depth required for the above development length or the other causes is very large, it is more economical to adopt a stepped or sloped footing so as to reduce the amount of concrete that should go into the footing. SHEAR: One way shear(Wide beam Shear): One way shear is similar to Bending shear in slabs considering the footing as a wide beam. Shear is taken along the vertical plane extending the full width of the base Lowest value of allowable shear in Table 13 of IS 456:2000 Is 0.35N/mm 2 is recommended.

13.   S.J.(Govt.) POLYTECHNIC BENGALURU

14.   S.J.(Govt.) POLYTECHNIC BENGALURU

15. SHERA FA I L U R E

16. 3. Bending Moment for Design: Consider the entire footing as cantilever beam from the face of The column and calculate the BM. Calculate span for the cantilever portion (Hashed portion) = plx = Substitute l=[(B-D)/2] M xx = p ( ) 2 x This is BM for 1m width of the beam  

17. DESIGN STEPS: 1. Assume self weight of footing =0.1p Total load w= P+0.1P 2. Area of footing required, A = For square footing, Size of footing = For Rectangular footing, assume L LxB =A Provide L x B square footing, Total Area = _ _ _ _ _ m 2  

18. 3. Bending Moment Pressure P = _ _ _ 4 . Factored moment/m M u = p ( ) 2 x  

19. 5. Effective depth : d required = increase depth for 1.75 to 2 times more than calculated value for shear considerations. 6. Area of tension reinforcement : M u = 0.87f y A st d ( 1- ) This is a quadratic equation, calculate the value for A st and consider the minimum of values Area of steel per m = A st /span = _ _ _mm 2  

20. Assume diameter bars Area of one bar a st = Spacing of reinforcement , S = 7) Check for one way shear : The critical section is taken at a distance “d” away from the face of the column y-y axis. Shear force per m, V u = p x B x [( )-d]  

21. Nominal Shear stress, = =_ _ _N/mm 2 Percentage steel = pt ? Refer table No. 19 of IS 456:2000 for = _ _ _N/mm 2 should be less than , design is safe against one way shear . <  

22. 8) Check for two way shear : The critical section is taken at a distance “d/2” away from the faces of the column Shear force per m , V u = p x [ A-(0.4+0.5) 2 ] Nominal Shear stress, = b = perimeter = 4( D + d ) Maximum shear stress permitted =0.25 should be greater than , Then design is safe against Punching shear / two way shear.  

23. 9) Development Length Ld = For Fe 415 steel and M20 concrete the values substituted to the above equation and Ld = 37.6 Taken to be , Ld = 40 available Ld =( L-D)/ 2 = _ _ _mm This is alright  

24. PLAN OF FOOTING 10) Reinforcement details:

25. 10) Reinforcement details:

26. PROBLEM 1: Design a square footing to carry a column load of 1100kN from a 400mm square column. The bearing capacity of soil is 100kN/mm 2 . Use M20 concrete and Fe 415 steel.

27. 1. Assume self weight of footing =0.1p= 0.1x 1100 =110kN Total load w= P+0.1P = 1100+110= 1210kN 2. Area of footing required, A = = = 12.1m 2 Size of footing = = = 3.478 Provide 3.5m x 3.5m square footing, Total Area = 12.25m 2  

28. 3. Bending Moment Pressure P = = 4 . Factored moment/m  

29. BM about axis x-x passing through face of the Column as shown in fig. M u = p x B x [ ] 2 X 2 X L = B for square footing D = Size of column = 400mm = 0.4m M u =  

30. 5. Effective depth : d required = = = 253.93mm Adopt 500mm effective depth and overall depth 550mm. (increase depth for 1.75 to 2 times more than calculated value for shear considerations) 6. Area of tension reinforcement : M u = 0.87f y A st d ( 1- ) 622.92 x 10 6 = 0.87 x 415 x A st x 500 ( 1- ) 622.92 x 10 6 = 180525A st - 2.14 A st 2  

31. 622.92 x 10 6 = 180525A st - 2.14 A st 2 2.14 A st 2 - 180525A st + 622.92 x 10 6 = 0 This is a quadratic equation, calculate the value for A st and consider the minimum of values There fore, A st = 3604.62mm 2 Area of steel per m = 3604.5/3.5 = 1029.85mm 2 Provide 12mm diameter bars Area of one bar a st =  

32. Spacing of reinforcement , S = = = 109.81mm Providing 12mm dia bars @ 100mm c/c. 7) Check for one way shear : The critical section is taken at a distance “d” away from the face of the column y-y axis. Shear force per m, V u = p x B x [( )-d] = 148.16x 1 x [( )- 0.50] = 155.57kN Nominal Shear stress, = = 0.31N/mm 2  

33. Percentage steel = = 0.20 Refer table No. 19 of IS 456:2000 for Since , % steel 0.15 0.28 0.20 x 0.25 0.36 For at 0.2 = 0.28 + x(0.2-0.15) = 0.32 N/mm 2  

34. is less than , design is safe against one way shear. 8) Check for two way shear : The critical section is taken at a distance “d/2” away from the faces of the column Shear force per m , V u = p x [ A-(0.4+0.5) 2 ] = 148.16 x [12.25- ( 0.4+0.5) 2 ] = 148.16x 11.44 = 1695kN  

35. b = perimeter = 4( D + d) = 4(400+500) =3600mm Nominal Shear stress, = = = 0.941N/mm 2 Maximum shear stress permitted =0.25 =0.25 = 1.11N/mm 2 > , design is safe against Punching / two way shear.  

36. 9) Development Length Ld = For Fe 415 steel and M20 concrete the values substituted to the above equation and Ld = 37.6 Taken to be , Ld = 40 40 x 12 = 480mm available Ld =( 3500-400)/ 2 = 1550mm This is alright  

37. 10) Details of Reinforcement :

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