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Design Parameters :

Grade of Concrete =

Grade of Reinforcement =

Characteristic Strength of Concrete fck = mpa

Yield Strength of Reinforcement fy = mpa

Density of concrete = kN/m

Dry Density of soil dry = kN/m3

Submerged Density of soil sub = kN/m3

Specific weight of water w = kN/m3

Co-efficient of internal friction =

Dia of pile m

No of piles in X direction Nos

No of piles inY direction Nosprojection of pile cap from pile m

Length of pile cap (0.75*3.0+0.75+2*0.15) L = m

Breadth of Pile ca (0.75*3.0) SQRT3/2)+0.75+2*0.15) B = m

B' = m

Thickness of pile cap H = m

VNCRev. Prep. By Date Chkd. By Date

Project : BINA COAL SHED

Client :

PILE CAP FOUNDATION CALCULATIONS 1 RK 28.02.13 HM 28.02.13

Subject : WARE HOUSE

2.0

M30

Fe415

30

415

25

19

9

10

3

1.05

1.50

0.40

0.75

2

10.15

3.3

ear cover o oo ng = mm

side cover to pilecap = mm

Effective thickness of footing d = m

Bottom of pile below soil h = m

Breadth of the column parallel to breadth of footing a = m

Length of the column parallel to length of footing b = m

Safe vertical load of pile = KN s = m

Safe Lateral load of pile = KN

Safe uplift load of pile = KN

Assume edestal hei ht = m

2.25

100.00

50

1.395

500.00

0.00

1.00

1.06

1.06

1500.00

1

23

D

C

A B

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Design forces for pile cap :

Axial force = KN

Fx = KN

Fz = KN

Mx = FZ * hei ht of edestal

161.8 X(0+1.395) = KN-m

Mz= FX * height of pedestal

0 X 0+1.395 = KN-m

Final moments :

Mx = (225.72+-1517) = KN-m

Mz = (0-0) = KN-m

No of ies re considerin verticle load (1470/1500)+2 = Nos

Check pile Capacity : I xx= Izz = M^4

1470

0

161.8

225.72

0

1742.72

0

3

0.0155

S No Pile dia X1 Y1 X^2 Y^2 iX^2 i ^21 0.75 0 0.98 0 0.95 0 0.01473

2 0.75 -1.13 -0.98 1.27 0.95 0.01962 0.01473

0.01962 0.01473 0.03923 0.0442

3 0.75 1.13 -0.98 1.27 0.95

Pile loads : unfactored load case

each Pile load P/N +Mx 1/x 2 +M x1 /y^2

1) P1 = (1470/3)+(1742.72*0.0155*0.975/0.039234375)+(0*0.0155*0/0.0442040625)

= KN

2) P2 = (1470/3)+(1742.72*0.0155*-0.975/0.039234375)+(0*0.0155*-1.125/0.0442040625)

= KN

3) P3 = (1470/3)+(1742.72*0.0155*-0.975/0.039234375)+(0*0.0155*1.125/0.0442040625)

= KN

safe

Critical section along Y-Y direction :

Design of Beam DC :

Maximun column load 1/3 L from AB line

1161.27

-181.27

-181.27

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Width of the Beam = Dia of the pile = m

KN

= M 2L/3 = M

Rd = KN L (3 /2) m Rc = KN

(1.732/2)*2.25Max Reaction load of piles (unfactored) = KN

Max bending along Y-Y direction

unfactored moment M xx = (980*0.65) = KN-M

Factored moment (M u xx = (1.5*637) = KN-M

Mu / Bd^2 = (955.5/(1.05*1.395^2)) =

Percentage of tension reinforcement (pt) from sp-16 =

Ast = (0.28*0.75*1.39510^4)

= Sq mm

Assume dia of bar = mm

Area of Bar = sq mm

0.75

1470

L/3 0.65 1.3

980 1.95

2930

25.00

491

490

1470

637

956

0.47

0.28

Spacing along X-X directionn (491/2930)*0.75 = mm

Provided spacing = mm

Oneway shear check :

if shear line within pile , shear force will be reduceing enhanced by percentage

Shear line distance from column face (1.395/2) = M

Shear line with in pile

Shear line with in pile dia (0.698-0.395) = M

shear contribution in pile area (0.75-0.303) = M

Max shear force along X-X direction (1161.27*0.45/0.75) = KN

Or reaction Rc or Rd MAX(696.76,980,490) = KN

Shear stress ( v = Vu /Bd

= (1.5*98010^3/1.05*1.395*10^6) = N/s mm

Max shear stress c (max) = N/sq mm

Percentage of steel provided (100*1.05*1000/150*491) =

= 15

shear stress c = N/sq mm

provide shear rein in pile cap

126

150

0.698

0.23

0.356

0.303

0.45

696.76

980

1.004

3.5

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Shear capacity of concrete (0.356*1.05*1.39510^3) = KN

Shear to be resisted by stirrups V us = (1.5*980-521.45) = KN

Assume dia of vertical stirrups = mm

Area of stirrup bar = Sq mm

Spacing of stirrups (0.87*415*78.55*4*1.395/948.55) = MM

Provide &10-Dia-4le ed stirru s 100mm

Critical section along X-X direction :

Design of Beam AB :

Max column reaction at centre of AB = KN

Max bending will be at pedestal face along X-X direction

KN(center)

KN M KN

Ra Rb

unfactored moment M xx = (980*2.25/4) = KN-M

*

521

949

10

490 2.25 490

78.6

167

551

980

980

. . -

Mu / Bd^2 = (826.875/(1.05*1.395^2)) =

Percentage of tension reinforcement (pt) from sp-16 =

Ast = (0.23*1.05*1.39510^4)

= Sq mmAssume dia of bar = mm

Area of Bar = sq mm

Spacing along y-y directionn (314/2406)*1.05 = MM

Provided spacing = mm

Oneway shear check : ( at pile face)

Max shear force along X-X direction MAX(490&-181.27&-181.27) = KN

Shear stress ( v = Vu /Bd

= (1.5*49010^3/1.05*1.395*10^6) = N/s mm

Max shear stress c (max) = N/sq mm

Percentage of steel provided (100*1000/100*314) =

= 15

shear stress c = N/sq mm

Provide Shear rein inpile cap

0.41

0.23

240620

490

0.502

3.5

0.23

0.356

314

137

100

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Shear capacity of concrete (0.356*1.05*1.39510^3) = KN

Shear to be resisted by stirrups V us = (1.5*490-521.45) = KN

Assume dia of vertical stirrups = mm

Area of stirrup bar = Sq mm

Spacing of stirrups (0.87*415*78.55*4*1.395/213.55) = MM

Provide &10-Dia-4le ed stirru s 700mm

Two way shear check :

punching shear will be checked at " d/2 " distance from column = (1.395/2) m

Punching shear load (factored) (KN) = KN

Perimeter of punching line 2*((1.06+1.395)+(1.06+1.395)) = m

Effective depth at punching line = m

Punching shear stress (T v) = V u / Bo D = N/sq mm

(1470/(1.395*9.82))

741

1470

9.82

521

214

10

78.6

1.395

0.11

Permissible shear stress for Grade TC = Ks Tc

Ks = 0.5+

= short side /long side =

= (1.06/1.06)

Ks = (0.5+1) = or 1

Ks = should not more than "1" =

Tc = 0.25 Fck (0.25*sqrt30) = N/sq mmSafe

Check for Develo ed len th :

Dia of the Max bar = MM

T bd value (plain bar ) = N/sq mm

Development length required L d = (0.87 *Fy *)/(4*T bd)

= (0.87*415*25)/(4*1.5*1.6) = MM

Pile Cap depth (1.395-0.05) = M

OK

25

1.5

940

1345

1

1.5

1

1.37