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Design of Pile Cap of Column at C (Intermediate Column

)
Assume, Diameter of pile cap = 500 mm
Assume, Capacity of pile cap = 450 mm
Axial load = 3278.538 KN MX = 54.47 KN-m MY = 40.71 KN-m
3278.538
∴ No. of piles = = 7.29 ≈ 8
450

Y
750 1250 625 625 1250 750

750
5 6 7 8
625
X X
625
1 2 3 4
750

Y
Reaction on The Pile:

P Mxx My y
R= n + ∑x
2 + ∑y
2

P1 = P2 = P3 =
Pile P Mxx My y R = R1 + R2
x y x2 y2
No. n ∑x
2
∑y
2 + R3

- - 409.81
1 3.516 0.391 -6.536 -8.142 395.139
1.875 0.625 7
- - 409.81
2 0.391 0.391 -2.179 -8.142 399.496
0.625 0.625 7
- 409.81
3 0.625 0.391 0.391 2.179 -8.142 403.854
0.625 7
- 409.81
4 1.875 3.516 0.391 6.536 -8.142 408.212
0.625 7
- 409.81
5 0.625 3.516 0.391 -6.536 8.142 411.423
1.875 7
- 409.81
6 0.625 0.391 0.391 -2.179 8.142 415.780
0.625 7
409.81
7 0.625 0.625 0.391 0.391 2.179 8.142 420.138
7
8 1.875 0.625 3.516 0.391 409.81 6.536 8.142 424.496

8 = (403. 5. B.689 & fck = 25 N/mm2.541 mm 16 ∴ Dreqd.8 = (411.496+403.139+399.125 Bending Moment about X-X: a. about Y-Y due to pile no.625 3.094 0.898 2. B. about Y-Y due to pile no.6 = (395.875 = 2076.423+415.4 = (395.496) X 0.188 KN-m b.625 + (408.496) X 1.780) X 0. 3.7.2. 2 = 426 mm AST Calculation: Direction Mu MU/bd2 pt AST Spacing X 2076. 2076.625 = 2021.138fckbd2 [From Note of Cl. 1.851 KN-m Bending Moment about Y-Y: a.625 = 1004.496+415. d = 378.1(c)] Or.212+424.838 X 1000 τV Nominal shear stress = bd = 5250 X 426 = 0.6.625 = 1044.875 + (399.554 N/mm2 < .323 X 106 = 0.M.870 Check for One Way Shear: Maximum shear force = 1621.689 15409. 7 2 2 ∑x = ∑y = 15.138+424.179 0.323 KN-m Check for Required Depth: For M20 & Fe415 material combination.M.780+420. 16 ∴ dprovided = 500 – 50 – 16 .898 KN-m MU max = 2076.854+420.854+408.541 + 50 + 16 + 2 = 452.138) X 0.485 Y 1044. about X-X due to pile no.212) X 0.54 mm Let us provide 500 mm depth.838 KN VU 1621.139+411. G-1.323 KN-m b. = 0.5.M.1 & Cl.3. B. about X-X due to pile no. B.323 2.7. = 378. 1.725 N/mm2 τC τV For pt = 0. From Table 19.618 7239.4. 38.M.2.138 X 20 X 5250 X d2 Or.423) X 1. Mu max = 0.

Spacing = 60 mm Let us provide 10 ∅ 1-legged stirrup @ 60 mm c/c Design of Pile Cap of Column at C (Corner Column) Assume. τC VUS = VU .838 X 103 – 0.95 ≈ 4 450 Y 750 625 625 750 750 3 4 625 X X 625 1 2 750 Y Reaction on The Pile: P Mxx My y R= n + ∑x 2 + ∑ y2 Pile x y x2 y2 P1 = P2 = P3 = R = R1 + R2 . bd = 1621.98 KN/cm From Table 62 of SP 16.436 ∴ No. for ∅ = 10 mm.817 KN V US 382.817 d = 42.554 X 5250 X 426 = 382.436 KN MX = 122. of piles = = 3.49 KN-m 1778. Capacity of pile cap = 450 mm Axial load = 1778.6 = 8. ∴ Shear design required. Diameter of pile cap = 500 mm Assume.

613+493.000 493.605 5 09 5 - 0.761 1. 617.996 0.391 0.62 444. 1.613 X 0. about X-X due to pile no. - 444.391 0.6 1 0.006 1.761 KN-m MU max = 617.62 0.006 KN-m Bending Moment about Y-Y: B.M.62 0.391 -48.12 mm 16 ∴ Dreqd. 16 ∴ dprovided = 400 – 50 – 16 .6 3 0.030 Y 555. G-1.12 + 50 + 16 + 2 = 359. 1.62 0.M. 38. d = 285.000 395.516 KN-m b. B.62 0.3 = 2 X 395.613 09 5 5 - 0.2 or 3.1 & Cl.996 0.902 0.006 KN-m Check for Required Depth: For M20 & Fe415 material combination.605 X 0. 2.000 493.6 2 0.391 -48.605) X 0.4 = 2 X 493.563 1.12 mm Let us provide 400 mm depth.138 X 20 X 5250 X d2 Or.625 = 555. n ∑x 2 ∑y 2 + R3 .M.1(c)] Or.4 = (395.62 444. = 285.595 .583 5226.000 395.625 = 617.605 5 5 09 2 2 ∑x = ∑y = 1.322 5511.391 0.996 0. Mu max = 0. P Mxx My y No. B.106 0.391 0.625 = 494.391 48.613 5 09 5 0. about X-X due to pile no.62 444.391 48.996 0. 2 = 326 mm AST Calculation: Direction Mu MU/bd2 pt AST Spacing X 617.6 4 0. about Y-Y due to pile no.62 0.563 Bending Moment about X-X: a.138fckbd2 [From Note of Cl.006 X 106 = 0.

583 & fck = 25 N/mm2. for ∅ = 10 mm.1 KN/cm From Table 62 of SP 16.218 KN VU 889. = 0.218 X 103 – 0. Spacing = 50 mm Let us provide 10 ∅ 1-legged stirrup @ 50 mm c/c .516 X 2750 X 326 = 426.992 N/mm2 τC τV For pt = 0. τC VUS = VU . bd = 889.6 = 13. From Table 19.516 N/mm2 < ∴ Shear design required.Check for One Way Shear: Maximum shear force = 889.983 KN V US 426.983 d = 32.218 X 1000 τV Nominal shear stress = bd = 2750 X 326 = 0.