FOR 5TH SEMESTER DIPLOMA IN CIVIL ENGINEERING S.J.(Govt.) Polytechnic, Bangalore FOR 5TH SEMESTER DIPLOMA IN CIVIL ENGINEERING CONCEPTs AND PROBLEMS DESIGN OF RCC COLUMN FOOTING

S.J.(Govt.) Polytechnic, Bangalore Learning Outcomes: Concept of column footings. Design criteria. Design steps for column footing. Reinforcement detailing.

S.J.(Govt.) POLYTECHNIC BENGALURU 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

Types of footing:

S.J.(Govt.) POLYTECHNIC BENGALURU 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

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

S.J.(Govt.) POLYTECHNIC BENGALURU

S.J.(Govt.) Polytechnic, Bangalore S.J.(Govt.) POLYTECHNIC BENGALURU

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

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

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

S.J.(Govt.) Polytechnic, Bangalore ..\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/mm2 is recommended. Having sloped footing is much economical and also safer in overturning too, Hence many design engineers adopt stepped/slopped footing rather square or rectangular

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

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

SHERA FA I L U R E

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 𝑙 2 = 𝑝𝑙2 2 Substitute l=[(B-D)/2] Mxx= p ( 𝐵−𝐷 2 )2 x 1 2 This is BM for 1m width of the beam

DESIGN STEPS: 1. Assume self weight of footing =0.1p Total load w= P+0.1P 2. Area of footing required, A = 𝑤 𝑆𝐵𝐶 For square footing, Size of footing = 𝐴 For Rectangular footing, assume L LxB=A Provide L x B square footing, Total Area = _ _ _ _ _ m2

3. Bending Moment Pressure P = 𝑃𝑢 𝐴 _ _ _kN/m2 4 3. Bending Moment Pressure P = 𝑃𝑢 𝐴 _ _ _kN/m2 4. Factored moment/m Mu= p ( 𝐵−𝐷 2 )2 x 1 2

5. Effective depth : d required = Mu 0 5. Effective depth : d required = Mu 0.138 x fckx b increase depth for 1.75 to 2 times more than calculated value for shear considerations. 6. Area of tension reinforcement : Mu = 0.87fy Astd(1- Ast fy bdfck ) This is a quadratic equation, calculate the value for Ast and consider the minimum of values Area of steel per m = Ast /span = _ _ _mm2 Hello all shall

Assume diameter bars Area of one bar ast= π x 𝑑𝑖𝑎2 4 = _ _ _ mm2 Spacing of reinforcement , S = 1000 ast Ast 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, Vu = p x B x [( L−D 2 )-d]

Nominal Shear stress, 𝜏𝑣 = 𝑉𝑢 𝑏𝑑 =_ _ _N/mm2 Percentage steel = 100Ast 𝐵𝑑 = pt? Refer table No. 19 of IS 456:2000 for 𝜏𝑐 𝜏𝑐 = _ _ _N/mm2 𝜏v should be less than 𝜏𝑐, design is safe against one way shear. 𝜏v < 𝜏𝑐

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, Vu = p x [A-(0.4+0.5)2] Nominal Shear stress, 𝜏𝑣= Vu b0d b0 = perimeter = 4( D + d) Maximum shear stress permitted 𝜏𝑐=0.25 𝑓𝑐𝑘 𝜏𝑐 should be greater than 𝜏v , Then design is safe against Punching shear / two way shear.

9) Development Length Ld = fsx dia of bar 4𝜏𝑏𝑑 = 0.87x415x dia of bar 4𝑥2.4 = 37.6∅ 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

10) Reinforcement details: PLAN OF FOOTING

10) Reinforcement details:

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/mm2. Use M20 concrete and Fe 415 steel.

1. Assume self weight of footing =0. 1p= 0 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 = 𝑤 𝑆𝐵𝐶 = 1210 100 = 12.1m2 Size of footing = 𝐴 = 12.1 = 3.478 Provide 3.5m x 3.5m square footing, Total Area = 12.25m2

=148.16kN/m2 3. Bending Moment Pressure P = 𝑃𝑢 𝐴 = 1.5 x 1210 12.25 4. Factored moment/m

BM about axis x-x passing through face of the Column as shown in fig BM about axis x-x passing through face of the Column as shown in fig. Mu= p x B x [ L−D 2 ]2 X 1 2 =148.16x3.5 x[ 3.5−0.4 2 ]2 X 1 2 =622.92kN−m L = B for square footing D = Size of column = 400mm = 0.4m Mu =622.92kN−m

5. Effective depth : d required = Mu 0. 138 x fckx b = 622. 92x106 0 5. Effective depth : d required = Mu 0.138 x fckx b = 622.92x106 0.138 x 20x 3500 = 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 : Mu = 0.87fy Astd(1- Ast fy bdfck ) 622.92 x 106 = 0.87 x 415 x Astx 500(1- Ast x 415 3500x500x20 ) 622.92 x 106 = 180525Ast- 2.14 Ast2 Hello all shall

622.92 x 106 = 180525Ast- 2.14 Ast2 2.14 Ast2 - 180525Ast+ 622.92 x 106 = 0 This is a quadratic equation, calculate the value for Ast and consider the minimum of values There fore, Ast = 3604.62mm2 Area of steel per m = 3604.5/3.5 = 1029.85mm2 Provide 12mm diameter bars Area of one bar ast= π x 122 4 = 113.09mm2

Spacing of reinforcement , S = 1000 ast Ast = 1000x113. 09 1029 Spacing of reinforcement , S = 1000 ast Ast = 1000x113.09 1029.85 = 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, Vu = p x B x [( L−D 2 )-d] = 148.16x 1 x [( 3.5−0.40 2 )- 0.50] = 155.57kN Nominal Shear stress, 𝜏𝑣 = 𝑉𝑢 𝑏𝑑 = 155.57x103 1000x500 = 0.31N/mm2

Percentage steel = 100Ast 𝐵𝑑 = 100x3604. 62 3500x 500 = 0 Percentage steel = 100Ast 𝐵𝑑 = 100x3604.62 3500x 500 = 0.20 Refer table No. 19 of IS 456:2000 for 𝜏𝑐 Since , % steel 𝜏𝑐 0.15 0.28 0.20 x 0.25 0.36 For 𝜏𝑐 at 0.2 = 0.28 + (0.36−0.28) (0.25−0.15) x(0.2-0.15) 𝜏𝑐 = 0.32N/mm2 Formulae used throughout

𝜏v is less than 𝜏𝑐, design is safe against one way shear 𝜏v is less than 𝜏𝑐, 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, Vu = p x [A-(0.4+0.5)2] = 148.16 x [12.25-(0.4+0.5)2] = 148.16x 11.44 =1695kN

b0 = perimeter = 4( D + d) = 4(400+500) =3600mm Nominal Shear stress, 𝜏𝑣= Vu b0d = 1695 x103 3600x 500 𝜏𝑣 = 0.941N/mm2 Maximum shear stress permitted 𝜏𝑐=0.25 𝑓𝑐𝑘 =0.25 20 = 1.11N/mm2 since , 𝜏𝑐 >𝜏v, design is safe against Punching / two way shear. Formulae used throughout

9) Development Length Ld = fsx dia of bar 4𝜏𝑏𝑑 = 0.87x415x dia of bar 4𝑥2.4 = 37.6∅ 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

10) Details of Reinforcement :

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