Pile Cap Design Example as Per Indian Code

Here is a Pile cap design example is solved according to Indian code with step wise procedure.

Problem

Design pilecap foundation with the given data: 

Load Fy = 800 KN, fc = 25 MPa, fy = 415

MPa, Column Dimension = 250 mm x 250 mm. Pedestal ht= 500 mm

Pile Data- Dia of pile= 400 mm.

Vertical capacity = 250 KN,

Horizontal capacity = 100 KN

Uplift capacity = 80 KN

Solution

depth of pilecap= 1.5xpiledia, 

D = 600 mm

Take D=650 mm

c/c pile diatance = 3xpile dia =1200 mm. 

Edge diatance = 350 mm

Assuming 4 pile combination

Pilecap dimension is 1900 mm x1900 mm x 650 mm

Weight of footing = 1.9 x1.9 x 0.65 x 25 KN = 58.66 KN

Weight of pedestal = 0.25 x 0.25 x 0.5 x 25 KN = 0.78 KN

Therefore, total load on the pilecap = (800+58.66 +0.78) KN = 859.44 KN

So Pile reaction = 859.44 /4= 214.86 < 250KN
Hence OK

As there is no lateral load , moment or uplift force, so each pile is safe in lateral And uplift

capacity.

Factored Design

Load factor for self wt is taken =1

Load factor for axial load is taken 1.5

So, Load on pilecap = 1.5 x 800 + 58.66 x 1 + 0.78 x 1 = 1259.44 KN

Load on each pile = 1259.44 /4 = 314.86 KN

Calculation of Moment about Z Axis

For moment wrt X1-X1

Contribution from pile 1 = from pile 2 = 314.86 x 0.475 = 149.56 KN

So Total Mz X1-X1 = 299.12 KNm

For moment wrt X2-X2

Contribution from pile 3 = from pile 4 = 314.86 x 0.475 = 149.56 KNm

So Total Mz X2X2 = 299.12 KNm

So Max value of Mz = 299.12 KNm

Calculation of Moment about X Axis

For moment wrt Z1-Z1

 Contribution from pile 1 = from pile 4 = 314.86 x 0.475 = 149.56 KNm

So Total Mx Z1-Z1 = 299.12 KNm

For moment wrt Z2-Z2

Contribution from pile 2 = from pile 3 = 314.86 x 0.475 = 149.56 KNm

So Total Mx Z2-Z2 = 299.12 KNm

So Max value of MX = 299.12 KNm

Check For Trial Depth against moment

About Z Axis

Bending moment at critical section

Muz = 299.12 KN-m

Assuming 50 mm clear cover, 50 mm pile in pilecap & and 12 mm bar, effective depth

deff = 532 mm

K = 700/(1100+0.87x fy ) = 0.479107

Ru = 0.36 .fck. Kumax .(1-0.42Kumax) = 3.4442 N/mm2

B =1900 mm , deff = 532 mm

Resisting Moment =Ru. B deff2 = 1852 KNm

Hence OK

AboutX Axis

Bending moment at critical section

Mux = 299.12 KN-m

Assuming 50 mm clear cover, 50 mm pile in pilecap & and 12 mm bar, effective depth

deff = 532 mm

K=700/(1100+0.87x fy )= 0.479107

Ru = 0.36 .fck. Kumax .(1-0.42Kumax) = 3.4442 N/mm2

B =1900 mm , D = 532 mm

Resisting Moment =Ru. B deff2 = 1852 KNm

Hence OK

Area of Steel Required

Along X Direction

From IS -456-2000 Annex G, G-1, b:

4.12 IS Pilecap 1

Mu =0.87.fy.Ast.d.(1-Ast.fy/b.d.fck)

So solving equation for Ast,

Astx = 1558 m2m

Minimum area of steel Astmin = 0.0012 x B x D = 1482 mm2 ( as fy>250)

Provided area = 1558 m2m

Along Z Direction

From IS -456-2000 Annex G, G-1, b:

Mu =0.87.fy.Ast.d.(1-Ast.fy/b.d.fck)

So solving equation for Ast,

AstZ = 1558 m2m

Minimum area of steel Astmin = 0.0012 x B x D = 1482 mm2 ( as fy>250)

Provided area = 1558 m2m

Calculation of Shear

Parallel to X Axis

For shear wrt X1X1

Contribution from pile 1 = pile 2 = 317.116 x 0.375 = 113.37 KN

So Total V X1X1 = 226.74 KN

For shear wrt X2X2

Contribution from pile 3 = pile 4 = 317.116 x 0.375 = 113.37 KN

So Total V X2X2 = 226.74 KN

So Maximum V parallel to X direction = 226.74 KN

Parallel to Z Axis

For shear wrt Z1-Z1

Contribution from pile 1 = pile 4 = 317.116 x 0.375 = 113.37 KN

So Total V Z1Z1 = 226.74 KN

For shear wrt Z2-Z2

Contribution from pile 2 = pile 3 = 317.116 x 0.375 = 113.37 KN

So Total V Z2-Z2 = 226.74 KN

So Max V parallel to Z direction = 226.74 KN

Check for One-Way Shear

Along X Direction

Percentage of steel pt = (100 x Ast)/B x de = 0.154

Vumax = 226.74 KN

Developed shear stress V = 226.74 x 1000 / 1900 x 532 = 0.224 N/mm2

Now allowable stress = 0.293 N/mm2

V < τc, Hence Safe

Along Z Direction

Percentage of steel pt = (100 x Ast)/B x de =0.154

Vumax = 226.74 KN

Developed shear stress V = 226.74 x 1000 / 1900 x 532 = 0.224 N/mm2

Now allowable stress = 0.293 N/mm2

V < τc, Hence Safe

Punching Shear

Punching shear is checked on a perimeter 0.5d =266 mm from the column face.

Contribution from pile 1 = from pile 2 = from pile 3 = from pile 4 = 317.12 KN

So total punching shear Vmax= 1268.464 KN

4x(250+532/2+532/2)=3128 mm

τv =Vmax/pm.d  = 0.76 N/mm2

ß=L/B =1900/1900 =1

k=0.5 +ß=1.5 , k<=1

Hence, k=1

Now allowable stress= τc =k.0.25.√fck = 1.25 N/mm2

            τv < τc , Hence safe





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