You are on page 1of 5

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

139+399. about X-X due to pile no.M. 3. about Y-Y due to pile no. 38.138 X 20 X 5250 X d2 Or.139+411.M.423+415.870 Check for One Way Shear: Maximum shear force = 1621.496) X 1. 1.541 + 50 + 16 + 2 = 452.496+415. 5.625 + (408. B.875 + (399.780+420. 7 2 2 ∑x = ∑y = 15. 1.094 0.2. 2 = 426 mm AST Calculation: Direction Mu MU/bd2 pt AST Spacing X 2076.618 7239.838 X 1000 τV Nominal shear stress = bd = 5250 X 426 = 0.725 N/mm2 τC τV For pt = 0. 2076. about X-X due to pile no.54 mm Let us provide 500 mm depth.212+424.423) X 1.3. B.138) X 0.541 mm 16 ∴ Dreqd.496) X 0.323 2.625 3. B.323 X 106 = 0.125 Bending Moment about X-X: a.7.4 = (395.854+420.898 KN-m MU max = 2076. From Table 19. about Y-Y due to pile no.554 N/mm2 < .851 KN-m Bending Moment about Y-Y: a.M.212) X 0.625 = 1004.2.179 0.323 KN-m Check for Required Depth: For M20 & Fe415 material combination.7.6 = (395.875 = 2076.689 15409.625 = 1044.8 = (403.M.496+403. Mu max = 0.689 & fck = 25 N/mm2.138+424.1(c)] Or. 16 ∴ dprovided = 500 – 50 – 16 .138fckbd2 [From Note of Cl. B. d = 378.485 Y 1044.898 2.838 KN VU 1621. = 378.323 KN-m b.780) X 0.8 = (411.5.854+408. = 0.1 & Cl. G-1.4.6.188 KN-m b.625 = 2021.

∴ Shear design required.6 = 8. bd = 1621. 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.838 X 103 – 0. Diameter of pile cap = 500 mm Assume.817 d = 42. for ∅ = 10 mm. of piles = = 3. τC VUS = VU .98 KN/cm From Table 62 of SP 16.554 X 5250 X 426 = 382.49 KN-m 1778.817 KN V US 382.436 KN MX = 122.95 ≈ 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 .436 ∴ No. Capacity of pile cap = 450 mm Axial load = 1778.

563 1. = 285.6 3 0.391 0.391 -48.391 48.761 KN-m MU max = 617. 2.1 & Cl.391 -48.62 0.M. G-1. P Mxx My y No.583 5226.62 444.322 5511. about X-X due to pile no.138fckbd2 [From Note of Cl. about X-X due to pile no.12 mm Let us provide 400 mm depth.391 0. d = 285.563 Bending Moment about X-X: a.006 X 106 = 0.605 X 0.613 09 5 5 - 0.391 48.006 1.M.62 0.391 0.613 5 09 5 0.2 or 3.138 X 20 X 5250 X d2 Or.625 = 617. 16 ∴ dprovided = 400 – 50 – 16 .M. 1. B. about Y-Y due to pile no. 617. 38.761 1.996 0.996 0. B.3 = 2 X 395.000 395. 1.6 1 0. 2 = 326 mm AST Calculation: Direction Mu MU/bd2 pt AST Spacing X 617.000 493.030 Y 555.605) X 0.625 = 494.605 5 5 09 2 2 ∑x = ∑y = 1.000 395. - 444.000 493.996 0.62 0. Mu max = 0.62 444.595 .6 4 0.4 = (395.516 KN-m b.006 KN-m Bending Moment about Y-Y: B.605 5 09 5 - 0.1(c)] Or.6 2 0.902 0.62 0. n ∑x 2 ∑y 2 + R3 .62 0.996 0.613 X 0.625 = 555.106 0.006 KN-m Check for Required Depth: For M20 & Fe415 material combination.4 = 2 X 493.62 444.613+493.12 mm 16 ∴ Dreqd.12 + 50 + 16 + 2 = 359.391 0.

516 N/mm2 < ∴ Shear design required.983 KN V US 426.583 & fck = 25 N/mm2. for ∅ = 10 mm.Check for One Way Shear: Maximum shear force = 889.218 X 1000 τV Nominal shear stress = bd = 2750 X 326 = 0.1 KN/cm From Table 62 of SP 16. = 0.992 N/mm2 τC τV For pt = 0.218 KN VU 889.516 X 2750 X 326 = 426. bd = 889.218 X 103 – 0. Spacing = 50 mm Let us provide 10 ∅ 1-legged stirrup @ 50 mm c/c .6 = 13. τC VUS = VU .983 d = 32. From Table 19.