Isolated column footing

Isolated column footing

1. Isolated Column Footing

2. Content Types of isolated column footing Bearing capacity of soil Critical section Check for development length Reinforcement requirements Example

3. Isolated column footing Pad footing ( uniform thickness ) Sloped footing

4. Bearing capacity of soil The size of footing depends on bearing capacity of soil. The load on unit area below footing should less than bearing capacity of soil to prevent settlement of footing. Gross bearing capacity : Total pressure acting at base of footing including self weight of footing, weight of column , over burden pressure of soil above footing etc. Is called gross bearing capacity. The net pressure at base of footing after deducting the weight of excavated soil is called safe bearing capacity of soil.

5. Safe B.C. = gross B.C. - overburden pressure q n = q - γ . D Where , γ = density of soil D = depth of footing

6. Critical sections Critical section for B.M. : For max. Bending moment critical section is taken on the face of column. B.M. = w l 2 / 2 w = net upward pressure * b f

7. Critical section for one way shear : For one way shear critical section is taken at distance d from the face of column. τ v = V u / bd V u = p * b f

8. Critical section for two way shear : It is taken at 0.5 d from face of column. τ v = V u / b o d V u = p * hatched area

9. Check for the development length : ( IS : 456 -2000 , P.42 ) L d = ɸ * 0.87 f y / 4 τ bd Reinforcement requirement for footing : 1. Minimum reinforcement ( P.66 ) Fe 250 0.15% of gross area Fe 415 0.12% of gross area 2. Spacing of bar ( T – 15 , P.46 ) 3. Cover : Min. 50 mm cover is required. F y N/mm 2 Clear distance between bars 250 300 mm 415 180 mm 500 150 mm

10. Example Design a rectangular isolated sloped footing for a column of size 250 mm × 750 mm carrying an axial characteristic load of 2000 kN and reinforced with 10 nos. 25 ɸ bars in M 30 grade concrete. The allowable bearing on soil is 220 kN /m 2 .The material for footing are grade M 20 concrete and HYSD of grade fe 415.

11. Size of footing : characteristic load on column = 2000 kN Assume self weight of footing = 200 kN 10% of column load Total load = 2200 kN Area of footing required = Total load / SBC = 2200 / 220 = 10 m 2 As footing is rectangular size of footing may be selected such way that effective cantilever projection on both sides equal. Difference between dimensions of column = 0.75 – 0.25 = 0.5 m

12. If b is width of footing, b ( b + 0.5 ) = 10 b = 2.92 d = 2.95 + 0.5 = 3.45 m Provide size of footing = 2.95 m × 3.45 m A = 10.17 m 2 > 10 m 2 ....... OK ( b) Net upward pressure : p = Factored column load / Area of footing provided = 1.5 × 2000 / 10.17 = 294.8 kN /m 2

13. Bending moment : u.d.l along x- direction = p × 2.95 = 294.8 × 2.95 = 869.66 kN.m u.d.l along y- direction = 294.8 × 2.95 = 1017.06 kN.m M uy = wl 2 / 2 = 869.99 × 1.35 2 / 2 = 792.48 kN.m M ux = wl 2 / 2 = 1017.06 × 1.35 2 / 2 = 926.80 kN.m

14. (d) Depth of footing : M ux = 0.138 f ck b d x 2 926.80 × 10 6 = 0.138 × 20 × 900 × d x 2 d x = 610.8 mm [ b = width of resisting section = 750 + 150 = 900 mm ] M uy = 0.138 f ck b d y 2 792.48 × 10 6 = 0.138 × 20 × 400 × d y 2 d y = 847.2 mm [ b = 250 + 150 = 400 mm ] Try overall depth = 950 mm Assume 12 ɸ bars for footing

15. D y = 950 - 50 – 6 = 894 mm D x = 894 - 12 = 882 mm Average d = 888 mm Assume edge depth = 230 mm (e) Reinforcement : Along x : P t = 0.831 % A st = 0.831 / 100 * 400 *894 = 2972 mm 2 provide 12 ɸ - 27 nos. Check for cracking : Clear distance between bar = (2950 -100-12) /26 = 109.15 mm c/c distance = 109.15 – 12 =97.15 mm < 180 mm.....O.K.

16. Along y: p t = 0.399 % A st = 0.399 /100 * 900 * 882 = 3167 mm 2 Reinforcement parallel to shorter direction : β = long side of footing / short side of footing = 3.45 / 2.95 = 1.169 Width of central band = b = 2950 mm A st in central band = 2/( β + 1 ) * total A st in y direction = 2920 mm 2 A st required per meter i central band = 2920 / 2.95 = 990 mm 2 Provide 12 mm ɸ @ 110 mm c/c in central band. Remaining steel = 3167 – 2920 = 247 mm 2 Width of end band = 3.45 – 2.95 = 0.5 m

17. 720 / 1350 = y ’ / 456 y ’ = 243.2 mm D ’ = y ’ + ( 230 – 50 - ɸ/2 ) = 243.2 + ( 230 – 50 – 6) = 417.2 mm b ’ = width of column + 2d = 250 + 2*894 = 2038 mm V u = S.F. At critical section = p * hatched area = 294.8 * 2.95 * 0.456 = 396.56 kN M u at critical section = wl 2 / 2 = 0.456 2 / 2 * 869.66 = 90.42 kn.m tan β = 720 / 1350 = 0.533

18. Τ v = ( v u - M u/d ’ tan β ) / b ’ d ’ = [ 396.56 –(90.42/0.4172) * 0.533 ] *103 / (2038 * 417.2) = 0.33 N/mm 2 P t = 100 * 3051 / ( 2038* 417.2) = 0.358 % τ c = 0.412 N/mm 2 Τ v < τ c .........O.K. (g) Check for two way shear : average depth d = 888 mm Section id critical at d/2 distance , 720 / 1350 = y ’ / 906 y ’ = 483.2 mm d ’ = 483.2 + (230 – 50 – 12 – 6 ) = 645.2 mm V u = 294.8 * ( 3.45 * 2.95 - (1.638 * 1.138 ) = 2450.80 kN

19. τ v = V u / ( b o * d ’ ) = 2450.80 *10 3 / ( 5552 * 645.2 ) = 0.684 N/mm 2 τ c ’ = ks * τ c ks = 0.5 + β c = 0.5 +(250/750 ) = 0.833 τ c = 0.25 √ f ck = 0.25 √ 20 = 1.11 N/mm 2 τ c ’ = 0.833 * 1.11 = 0.93 N/mm 2 τ v < τ c ’ ...........OK Development length : L d = 47 ɸ = 47*12 = 564 mm L d provided from face = 1350 – 50 = 1300 mm > 564....OK

20. Thank you