DESIGN OF TWO PILE CAP
DATA:-
Pile Diameter: 400 mm
Spacing of piles 2 hp = 2 x 400 : 800 mm
Column Dimension B x D : 300 x 450 mm
Factored Load: 1072.8 KN
Factored Moment Mxu:51.29 KN.m
Safe Load on Single Pile:500KN
Concrete Mix: M20
Steel Grade: Fe 415
DESIGN: -
1. Pile Cap Dimension :
Breadth of Pile Cap = C/c of Pile + hp /2+ 150 + hp /2 + 150
= 800 + 400/2+ 150 +400/2+ 150 =1500 mm
Width of pile cap = hp + 150 + 150 = 700 mm
Depth of Pile cap = 2 hp + 100 = 2 x 400 + 100 = 900 mm.
2. Check for Pile Load capacity :-
Total factored axial compressive load
= Pu +/- Mxy +/- Mxx
n Σy2 Σx2
Self weight of Pile Cap = (1.5 x0.7 x 0.9 x 25 ) x1.5 = 35.45 KN
Factored load from column Pu = 1072.80 KN
------------
Total Factored Load Pu = 1108.25 KN
No. of Piles along one side of axis = 2
y coordinate of Pile cap = 0.4 m
Mx = Moment about x axis = 51.29 KN.m
Compressive load in A1 & A2 about x – x axis
= (1108.25 / 2) + ((51.29 x 0.4) /(2 x 0.42))
= 554.13 + 64.11
=618.24 KN
Design working load = 618.24 /1.5 = 412.16 KN < Safe Load on Pile i.e 500KN. O.K.
3. Bending Moment :-
Factored Moment in section Y-Y
Mu = 618.24 x 0.5 x (0.8-0.3)= 154.56 KN.m
4. Check for effective depth :
Mu = 0.138 fck b d2 = 154.56 x 106
d required = √ (154.56 x 106 ) / 2.76 x700 =282.84 mm
D provided = 900 mm
d available = 900 – 60 -12- 6 = 822 mm > d required i.e. 282.84 mm
5. Check for Punching Shear (Two way shear) : -
Punching shear at a distance d/2 (i.e.822/2= 411mm) from face of column = 1072.80 KN
The critical section of punching comes the centre of the pile.
Hence the net load is to be taken. However, the depth is checked for factored axial load from
column = 1072.80 KN
b= 700 x 822 mm
d= 822 mm
Perimeter of critical section = 2 (700 + 822) = 3044 mm
Punching shear stress = (1072.80 x 103) / ( 3044 x 822) = 0. 43 N/mm2
Allowable shear stress for M20
= 0.25 √fck = 0.25 √20 = 1.12 N/mm2
Hence safe.
6. Main Reinforcement : -
Mu = 154.56 x 106 KN.m
K = Mu / bd2 = (154.56 x 10^6) / (700 x 822) = 0.33
Pt from Table 2 of Design Aid=0.11
Minimum Ast = (0.12/ 100) x 700 x 822 = 690.48 mm^2
Provide 7 Nos. 12 Φ RTS at the bottom on both ways.
(Ast = 791 mm2 > 690.48 mm2)
Reinforcement at top :-
Minimum Ast = (0.12/ 100) x 700 x 822 = 690.48 mm^2
Provide 7 Nos. 12 mm Dia RTS at top.
(Ast = 791 mm2 > 690.48 mm2)
7. Check for one way shear :-
Maximum Shear force at face of column = 618.24 KN
Shear stress = (618.24 x 10^3) / ( 700 x 822) = 1.07 N/mm2
For Pt = 0.20%
ζc from Table 61 of Design Aid to IS 456 -1978 = 0.33 N /mm2
Shear to be carried by stirrups shear
Vus =(1.07– 0.33) x700 x822 x 10^-3= 425.80 KN.
Vus /d = 425.80 / 82.2 = 5.18 KN/cm
Provide 8 Φ RTS 4 legged stirrups @ 120 mm c/c.
(Vus /d =5.58 KN/m > 5.18 KN/cm ).
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