pile cap -3-1
Post on 03-Apr-2018
214 Views
Preview:
TRANSCRIPT
-
7/28/2019 pile cap -3-1
1/5
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
-
7/28/2019 pile cap -3-1
2/5
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
-
7/28/2019 pile cap -3-1
3/5
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
-
7/28/2019 pile cap -3-1
4/5
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
-
7/28/2019 pile cap -3-1
5/5
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
top related