Pile design Numerical

 

Numerical

Design a precast concrete RCC pile of size 350 mm X 350 mm carrying a service load of 700 KN. The total length of the pile is 6.5 m. use

1)   Grade of concrete = M25

2)   Grade of steel Fe415

3)   Assume effective length = 0.8 L

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1)   Given data

a.     P = 700 Kn

b.    D= 350 mm (D = b) (0.35 M)

c.     Fck =  25 N/MM^2

d.    Fy = 415 N/MM^2

e.     L = 6.5 m

Size of the column is < 400 mm it will be designed as an eccentrically loaded column subjected to minimum moment.

Minimum eccentricity: (e min) (According to IS 456)

          e min =(L/ 500) + (D/30)

                   = (6500 / 500) + (350 / 30) =24.667mm > 20 mm

Minimum moment (M min) = Pu x e min.

          Factored load (Pu) = (1.5 X 700) =1050 Kn

Therefore (M min) = 1050 X 24.667

 M min = 25900.35 Kn-mm

          = 25.90 KN-M

Now effective length (L eff) = 0.8 x 6500

          L eff = 5200 mm

 (Leff / width) = (5200 /350)

                   =14.857   > 12 -----------column is slender column

Additional moment (Ma) = ((Pu. b)/200) x ((Leff / b) ^2) X k     

Reduction factor K= ((Puz – Pu) / (Puz – Pub)) <= 1   --------- (clause 39 –IS 456-2000)

Ma = ((1050 X 0.35)/ 2000) x (14.85) ^2

 Ma = 40.52 Kn-m

 Total Moment = 40.52 + 25.90

          = 66.45 Kn-m

 Assuming diameter of bar = 25 mm and lateral ties 8 mm

Effective cover = 50 + (25/2) + 8

          =       70.5 mm

(d’ /D) = (70.5 / 350) = 0.2

Fck.b.D = (25 x 350x 350) = 3062500 n/M

Therefor 1 Length

fck.b.D = (25 x 350x 350) / 1000 = 3062.500 Kn

Pu / fck.b.D = 1050 / (25 x 350x 350) = 0.34

Mu / fck.b.D = 66.45 / (25 x 350x 350) = 0.06

For (d’ / D) =0.2 --------- (From chart 9 Appendix C, Page A-18 or SP16 chart 50)

P / fck = 0.03 and

p = 0.03 X 25 = 0.75 % <1.25 %

Minimum % of reinforcement = 1.25 % (since L < 30 x 0.35) = 10.5M)

Asc = (1.25 / 100) x 350 x 350 = 1531.25 mm^2

Provide 4 bars 25 mm Φ in diameter (Provided Asc =1963.4 mm^2)

Lateral Reinforcement:-

In the body of piles:-

Volume of lateral reinforcement = 0.2 % of BD

          = (0.2 / 100) X 350 X 350 X 1000

          = 245000 mm^3

Using 8 mm Φ in diameter ties the total length of ties = 4 x (350-(50 x 2))           =1000 mm

Volume of one ties = area x length

                   = (π/4) x 8² x 1000

                   = 50265.5 mm ^ 3

Number of ties per meter length = (245000 / 50265.5) = 4.874

Spacing of ties = (1000 / 4.874) = 205.17 mm

Maximum permissible spacing = (350 / 2) = 175 mm

Therefore provide 8 mm Φ in diameter ties at 175 mm c/c in the body of the pile.

Bottom lateral reinforcement

Length = 3 x 350 =1050 mm

Volume of lateral ties at 0.6 % = 245000 x (0.6 / 0.2) = 735000 mm^3

Spacing required = 204 x (0.6 / 0.2) = 68 mm Say 70 mm

Provide 8 mm Φ ties foe a length of 1050 m at 70 mm c/c

 Lateral spiral reinforcement at the pile head

Length = 3 x 350 = 1050 mm

The volume of spiral = circumferential of spiral   x area of 8 mm Φ

          = π (350 – (50 x 2)) x ((π/4) x 8 ^2) = 39478.18 mm^3

Number of spiral = 735000 / 39478.18

          = 18.61

Spacing = 1000 / 18.61

          =53.72 mm (say 55 mm)

Provide 8 mm Φ spiral at 55 mm pitch for top 1050 mm length.

Provide 6 additional bar of 12 mm Φ vertically within the spiral of support.

Spacer forks and links

Provide two pairs of 12 mm, Φ spacer fork with 6 mm links at 105m c/c along the whole length.

Hole for the handling of the pile.

Provide two holes at each end at a distance of 0.207L = (0.207 6.5) 1.35m

For the purpose of lifting, transporting and stacking at the site.

Provide one hole from either side at a distance of 0.293L for hosting = (0.293 x 6.5) =1.9045 m.