DESIGN OF CANTILEVER FOOTING Design Parametres:-

Footing designation = CF1Concrete mix M20Steel Fe415Cover to Reinforcement 50mmSafe Bearing Capacity(From soil Report)= 80.36KN/sqmUnit weight of RCC = 25.0KN/cumWidth of Column b = 0.23mDepth of Column d = 0.30mCharacteristic compressive strength of concrete = 20.00N/sqmmYield strength of steel = 415.00N/sqmmCentre to centre distance between column loads Lc = 3.89m

Load particulars:-

604.58KN

605.06KN

Assume width of the footing for Column 1 = B1 = 2.10mTry with a ecentricty of 'e' = 0.60m

3.29m

714.84KN

494.80KN

Sizes of footings required:-

Trial 1:-

Area of footing for Col.2 = 4.45Sqm

Area of footing for Col.6 = 3.08Sqm

The size of footing assumed for Col.6 isLength = 1.80m

Breadth = 1.80m

Area of the footing comes to = 3.24Sqm

The size of footing assumed for Col.2 isLength = 2.15m

Breadth = 2.15m

Area of the footing comes to = 4.62Sqm

New value of ececentricity = 0.63m(which is approximately equal to the assumed value)

Load on Column 2 Q1(Including self weight) =

Load on Column 6 Q2 =

Now LR =

Reactions R1 = Q1(1+e/LR) =

Reactions R2 = Q2-(Q1e/LR) =

Design of Strap beam:-

110.26KN

362.76KNm

Size of the footing required:-

Area of the footing required = #VALUE!

Provide square footing of size 1.05mx1.05m,the area comes to 4.41SqmWidth = 2.10mDepth = 2.10m

The net ultimate bearing pressure acting on the footing due to direct load = #VALUE!

Design moment = 247.50KN-m

Section Modulus for the above section = 1.54KN/sqm

Soil pressure due to moment = M/Z = 160.71KN/sqm

Max.Soil pressure = #VALUE!Hence O.K

Min.Soil pressure = #VALUE!Hence O.K

Hence,the design soil pressure = #VALUE!

Depth of the footing required:-

The critical section for bending is at the face of the column.Hence,the

Maximum factored bending moment = #VALUE!

#VALUE!

The strap beam will carry a constant shear force of Q2-R2 =

The moment carried by the strap beam = (Q2-R2)x(Lc-e) =

(1/6)xbd2 =

pu + mu =

pu - mu =

0.36 fckbxumax(d-0.42xumax) =

For Fe 415 steel,xumax = 0.48d

substituting the above value and finding out the effective depth by solving the aboveequation,

9953250.1 #VALUE!

d = 60.06mm

Assuming 50mm effective cover and 10mm dia bars,the over all depth comes to

115.06mm

However provide over all depth of 250mm and the effective depth is 195mm

Check for single shear:-

The critical section for beam shear is at distance of 'd' from the face of the column

#VALUE!

#VALUE! <2.8 N/sqmm(As per Table 20 of 1S 456)

Hence,the section is safe from shear point of view

Assumed percentage area of the steel reinforcement = 0.21%

The design shear strength of concrete for the above steel percentage from Table 19 of IS 456 is

0.33 N/sqmm 64.35KN

0.33>0.13

Hence,the depth provided is safe from beam shear point of view

Check for two way shear:-

The critical section for two-way shear is along the perphery of the square at adistance d/2 from the face of the column

1840mm

#VALUE!

#VALUE!

Permissible shear stress in concrete for two-way shear for M20 grade concrete

Hence,the factored design shear force VFd =

Nominal shear stress Tv =

Hence Vuc =

Hence perimetre of the preriphery b0 =

Hence,the factored shear force VFd = qu(B2-AB2)=

Nominal shear stress Tv = VFd/b0d =

Tc' =ks . Tc

ks = (0.5+l/b)> 1

1

1.12 N/sqmm

Hence,Tc' = 1.12 N/sqmm

1.12>0.33

Hence,the section provided is safe from two-way shear point of view

Depth of neutral axis:-

For the effective depth,find out the depth of neutral axis

#VALUE!

Solving the equation,xu = 7.210mm

9965640.0816 #VALUE!

0.48d = 93.60mm

under reinforced.

#VALUE!

#VALUE!

Area of steel provided 8mm dia bars @ 125mm c/c spacing

Hence area of steel provided = 401.92

Hence Safe

Provide,the same reinforcement in other direction also

Hence ks =

Tc = 0.25(fck)1/2 =

Mfd = 0.36 fckbxu(d-0.42xu) =

Xu.max =

The actual depth of neutral axis is less than the Xumax.Hence,the section remains

The stress in steel fs = 0.87fy

Mfd = 0.87fyAst(d-0.42xu) =

Ast = mm2

mm2

#REF!

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