2 sqr ftg wsm

ISOLATED FOOTING OF UNIFORM THICKNESS FOR A RC COLUMN INPUT DATA: LOAD,W = 600kN Column size: size: 500mm*500mm. SBC of soil = 120kN/m^2. Let us take M 20 , Fe415 steel.

Design constants: for given set of materials ,we have kc = 0.289: jc = 0.904 and R C = 0.914 SIZE OF FOOTING : W = 600kN . W’ = 10% of 600kN = 60kN Base area reqd. A = 660/120 = 5.5 m^2 footing of 2.4 m * 2.4 m. √  = 2.345m. provide a square footing

Size B of square =

Net upward upward pressure ,Po ,Po = 600/(2.4*2.4) = 104.17 104.17 kN/m^2

DESIGN OF SECTION : The maximum bending moment acts at the face of column, and its magnitude is given by: M = Po*(B/8)*(B-b)^2 *10^6 N-mm = 104.17*(2.4/8)*(2.4-5)^2 *10^6 = 1.128*10^8 N-mm d=

  =    = 227mm

CHECK FOR SHEAR : The depth found above should be checked for shear. For one way shear, the critical section yy is located at a distance d (=227mm) from the face face of the column , where shear force V is given by V = Po.B {0.5(B-b)  – d} = 104.17 *2.4 {0.5(2.4-0.5) – o.227} = 180.755kN Τc = V/B.d = 180755/(2400*227) = 0.322N/mm^2 Using 12mm Ф bars at at 60mm clear cover, total depth will be about about

227 +60+6 +60+6 = 293 293 mm =

300mm. Hence from IS code .k = 1 .also, for a balanced section having p = 0.44% for M 20 concrete, Fe 415 and hence, τc = 0.28 N/mm^2 from table 3.1 Hence the permissible shear stress = k. τc = 1*0.28 = 0.28 N/mm^2.

Hence the section is unsafe. Required d = 180755/(2400*0.28) = 269 mm

and D = 269+60+6 = 335 mm For two way shear ( punching shear) , the section lies at d/2 from the column face all round. The width b o = (b + d )= 500 + 227 = 727 mm. Also, shear force around the section is F = Po[B^2 – (b + d)^2] = 104.17[(2.4)^2 – (0.727)^2] = 544.962 kN Τv = F/(4.bo.d) = F /4[b + d]d = 544.962/(4*727*227) = 0.826 N/mm^2

Permissible shear stress = ks .τc Ks = (0.5 + β) = (0.5 +1) with max. value of 1 . hence, ks = 1 Also, τc = 0.16

  

ck

= 0.16

√  = 0.715 N/mm^2

Hence, permissible shear stress = 0.715 N/ mm^2 < punching shear stress Hence the section is unsafe ( i.e. depth is insufficient ).The required effective depth from punching shear point of view is found by equating τv to permissible shear stress of 0.715

N/mm^2. Τv = F/ 4(b+d)d ≤ ks.τc

(] = 0.715  d^2 + 517.6d – 193636 = 0 d = 251.7mm < 269mm required for one way shear. Hence, provide a total depth of  350mm. using 12 mm Ф bars and a clear cover of 60 mm, available d = 350-6-60 = 284 mm in

one direction and 284 – 12 = 272 mm in other direction.

DESIGN OF STEEL REINFORCEMENT The area of reinforcement in each direction is given by Ast = M/(σst .j .d) = 1.128*10^8/ (230*0.904*272) = 1995 mm^2 Using 12 mm Ф bars, no. of bars = 1995/113.1 = 18 bars Provide 18 bars of 12 mm Ф uniformly distributed in each direction.

CHECK FOR DEVELOPMENT LENGTH : Ld

= 45Ф = 45 * 12 = 540mm

Providing 60mm side cover, length of bars available 0.5[B-b] – 60 = 0.5[2400 – 500] – 60 = 890 mm > Ld Hence , safe.

TRANSFER OF LOAD AT COLUMN BASE A2 = 500*500 = 250000mm^2 A1 = [500 + 2(2*350)]^2 = (1900)^2 = 3610000 mm^2 for side spread rate of 2: 1

   =    =3.8 Adopt max. value of    

=2

  

Hence permissible bearing stress = 0.25f ck

= 0.25 *20*2 = 10 N/mm^2

Actual bearing pressure = 60000/(500*500) = 2.4 N/mm^2 Hence , safe .

Thus no separate dowel bars are required for transfer of load. However, it is advisable to continue all the bars of the column, into the foundation.