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20 Reinforced Concrete Design Design of Footings 2 Flipbook PDF
Combined Footings 2footings close to each other P1close to property line and P 2>P1 property line P1 P2 Centroid of l
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Reinforced Concrete Design
Design of Footings 2 Combined Footings Pile Foundation
Mongkol JIRAVACHARADET
SURANAREE UNIVERSITY OF TECHNOLOGY
INSTITUTE OF ENGINEERING SCHOOL OF CIVIL ENGINEERING
Combined Footings
Centroid of load resultant and footing must coinside property line
P1
P2
property line
P1
P2
2 footings close
P1 close to property
If 1/2 < P2/P1 < 1
to each other
line and P2 > P1
use trapizoidal footing
property line
If P2/P1 < 1/2, use strap combined footing
Centroid of Combined Footings P1
R
P2 n
m (1) Compute centroid C qe
n = P1s / ( P1 + P2 ) = P1s / R (2) Footing area
s
L=2(m+n) b = R / ( qe L )
C
b
qe = allowable soil pressure
n L/2
L/2
m
b2 3(n + m ) − L = b1 2L − 3(n + m )
c2
c1 C
b1
b2
n
m
L
(b1 + b2 ) = c1 =
L(b1 + 2b2 ) 3(b1 + b2 )
c2 =
L(2b1 + b2 ) 3(b1 + b2 )
b1 =
2(n + m ) − L2 L1(L1 + L2 )
b2 =
R Lb − 1 1 qeL2 L2
C
b1
b2
n L1
L2
m
2R qeL
L1 b1 + L2 b2 =
R qe
Reinforcement in Combined Footings
P2
A
b
pu Section A-A
A
Transverse reinforcement
L P1
P2
h pu Vu
Mu
Transverse Reinforcement B
P2 Transverse steel Perimeter of bottom of failure surface
b
45o
45o
b Section B-B B
c d/2
c+d
EXAMPLE 12.5: Design of a Combined Footing. Design a rectangular combined footing to support the two columns. Property lines require that the footing not extend beyond the face of column A more than 20 cm. The allowable soil pressure is 20 t/m2. f’c = 210 kg/cm2, fy = 4000 kg/cm2, and the unit weight of the soil is γs = 2.0 t/m3. A
B
1) Locate R of service load: 40cm x 40cm
(75 + 120) x = 120(5)
45cm x 45cm
D = 50 ton L = 25 ton
x = 3.1 m D = 80 ton L = 40 ton
R
x
2) Length of footing: Set C.G. under R C.G. to left = 3.1 + 0.2 + 0.2 = 3.5 m
20 cm
C.G. 5.0 m
1.0 m
Total length = 2 x 3.5 = 7.0 m
3) Width of footing: assume footing depth = 60 cm Net soil pressure pn = 20 - [0.6(2.4) + 0.4(2.0)] = 17.76 t/m2 Required area = R / pn = (50+25+80+40)/17.76 = 11.0 m2 Width of footing = 11.0 / 7.0 = 1.57 m
USE 1.60 m
4) Shear and moment diagram in longitudinal direction: Column A: Pu = 1.4(50) + 1.7(25) = 112.5 ton Column B: Pu = 1.4(80) + 1.7(40) = 180 ton Factored soil pressure pu = (112.5+180)/(7.0x1.6) = 26.12 t/m2 Factored load per length wu = (112.5+180)/7.0 = 41.79 t/m
112.5 t
180 t
wu = 41.79 t/m 0.4
5.0
1.6 113.2 t
16.7 t
Vu (ton)
2.29 m -95.8 t
-66.8 t 47.0 t-m
3.34 t-m
-106.4 t-m
Mu (t-m) err = 6.4 t-m
5) Flexural design: max Mu = 106.4 t-m
Mu 106.4 × 105 Rn = = = 27.3 kg/cm2 2 2 φ bd 0.9 × 160 × 52
ρ=
0.85fc′ 2Rn 1 − 1 − = 0.0074 < ρmax fy 0.85f ′
OK
As = 0.0074(160)(52) = 61.6 cm2 USE 10DB28 (As = 61.58 cm2) for +M = 47.0 t-m and 3.3 t-m USE As,min 10DB20 (As = 31.42 cm2) 6) Check punching shear: Column A: b0 = 4 (40 + 52) = 368 cm Vu = 112.5 - (0.92)2(26.12) = 90.4 ton
φ Vc = 1.06 φ fc′ b0 d = 250 ton > Vu
OK
Column B: b0 = 4 (45 + 52) = 388 cm Vu = 180 - (0.97)2(26.12) = 155.4 ton
φ Vc = 1.06 φ fc′ b0 d = 263 ton > Vu
OK
7) Check beam-shear: max Vu = 113.2 ton
φ Vc = 0.53φ
f c′ b d
= 0.53 × 0.85 × 210 × 160 × 52 /1, 000 = 54.3 ton < Vu
Stirrup required
8) Stirrup Design:
φ Vs = Vu − φ Vc = 113.2 − 54.3 Vs = 69.3 ton < [1.1 210 × 160 × 52/1,000 = 132.6 ton] max s = d / 2 = 26 cm USE s = 20 cm Av =
Vs s 69.3 × 1, 000 × 20 = = 6.7 cm 2 4, 000 × 52 fy d
min Av =
3.5 bw s 3.5 × 160 × 20 = = 2.8 cm 2 4, 000 fy
USE DB16 stirrup with 4 legs ( Av = 8.04 cm2 )
9) Transverse reinforcement: PA = 112.5 ton DB16 stirrup with 4 legs
0.4 m
0.60 m
wu=112.5/1.6 = 70.3 t/m 1.60 m
1.60 m
Column A: be = 40 + 52 = 92 cm, wu = 112.5/(1.6x0.92) = 76.4 t/m Mu = 0.5(76.4)(0.6)2 = 13.8 t-m
ρ = 0.0016 < [ρ min=0.0035]
USE ρ min
As = 0.0035(92)(52) = 16.7 cm2 USE 6DB20 (As = 18.85 cm2) Column B : be = 45 + 52 = 97 cm, wu = 180/(1.6x0.97) = 116.0 t/m Mu = 0.5(116)(0.6)2 = 20.9 t-m
ρ = 0.0023 < [ρ min=0.0035]
USE ρ min
As = 0.0035(97)(52) = 17.7 cm2 USE 6DB20 (As = 18.85 cm2)
(10) Temperature steel: As = 0.0018(100)(60) = 10.8 cm2 USE DB20 @ 0.15 (As = 12.56 cm2/m)
A
[email protected]
B
10DB28
0.60 m
6DB20 1.0 m
6DB20 1.0 m
10DB20 [email protected] 7.0 m
25 /.. ! " # 1.8 " f’c = 210 ./(.2, fy = 4000 ./(.2 +,- γs = 2.0 /#..
Example 12.6: Strap Footing Design
20 cm
DL = 50 ton Pe{ LL = 25 ton
R 5.0 m
DL = 80 ton Pi { LL = 40 ton
40x40cm
45x45 cm
3.1 m -1.80 m pn
pn
Re
Ri
1.8 m
2.5 m
1) Locate R of service load:
(75 + 120) x = 120(5) x = 3.1 m
2) Size of footing: Assume footing depth = 40 cm Net soil pressure pn = 25 - [0.4(2.4) + 1.4(2.0)] = 21.24 t/m2 Required area = R / pn = (50+25+80+40)/21.24 = 9.18 m2 Assume interior footing 2.5m x 2.5m = 6.25 m2 Assume exterior footing length = 1.8 m Compute the required width b of exterior footing so that the area centroid is located at 3.1 m 0.2 m b
1.8 m
2.5 m
3.1 m C.G.
2.5 m
5.0 m Exterior footing
Interior footing
1.8b (0.9) + 6.25(5 + 0.2) = (3.1 + 0.2)(1.8b + 6.25) b = 2.75 m
3) Shear and moment diagram in longitudinal direction: Exterior col.: Pu = 1.4(50) + 1.7(25) = 112.5 ton Interior col.: Pu = 1.4(80) + 1.7(40) = 180 ton Factored soil pressure:
pu =
112.5 + 180 = 26.12 t/m2 1.8 × 2.75 + 2.5 × 2.5
4) Design strap beam: Since interior footing is designed without strap beam, strap beam will be designed only for exterior footing loading. 5.0 m
20 cm
wu = 2.75 pu = 2.75(26.12) = 71.83 t/m 1.8 m
Pile Foundation Effective pile reaction( Re ): Re = Ra - Wf 15 cm
Ra = allowable bearing of piles Wf = total weight of footing 1.5D
Number of piles( n ):
3D
n=
3D D
1.5D 1.5D
3D
3D
1.5D
DL + LL Re
Factored pile reaction:
Ru =
1.4DL + 1.7LL n
Typical Arrangement of Piles 1.5D
1.5D
1.5D
1.5D 3D
3D
3D 1.5D
3 2D
3D 1.5D
1.5D
1.5D
1.5D 3D
1.5D 1.5D
2 PILES
1.5D
3 PILES
1.5D
1.5D
4 PILES
3 2D
1.5D
5 PILES 1.5D
1.5D
1.5D 3D
3D
3D
3D 3D
1.5D
1.5D
3D 1.5D
1.5D
3D
3D
6 PILES
3D
3D
3D
1.5D
1.5D 1.5D
3 2D 1.5D 3 2D
7 PILES
1.5D 1.5D
3 2D
3 2D
8 PILES
1.5D
1.5D
3D
3D
3D
1.5D
1.5D 1.5D
3D
3 3D 3D 1.5D 1.5D
3D
3D
1.5D
1.5D
3D
9 PILES 1.5D
3D
3D
3D
10 PILES 3D
1.5D 1.5D 1.5D
3D
3 3D 3D 1.5D 3D
3D
11 PILES
1.5D 1.5D
3D
3D
3D
12 PILES
1.5D
( ) Section
Size(m)
Load capacity (ton)
0.18 x 0.18
15
0.22 x 0.22
22
0.26 x 0.26
30
0.30 x 0.30
43
0.35 x 0.35
57
0.40 x 0.40
80
0.16 x 0.16
15
0.18 x 0.18
21
0.22 x 0.22
30
0.26 x 0.26
43
0.30 x 0.30
50
0.35 x 0.35
80
0.40 x 0.40
100
Pa = 0.25(0.85 f c′Ag )
Example 12.7 ##! ##."/0" +"12 #,- 100
+ ,- 50 "34 " 3 " 5 3/ 4040 (. f’c= 240 ./(.2 fy = 4,000 ./(.2 +,- γs = 2.0 /#.. ! 6 1.50 . + # Solution: Assume pile ∅ = 40 cm → Ag =
π 4
× 402 = 1,256 cm2
Pa = 0.25(0.85fc′Ag ) = 0.25 × 0.85 × 240 × 1,256 /1,000 = 64 ton USE 40 cm bored pile with safe load 50 ton Assume surcharge & footing wt. 15%
0.60
Number of piles = 1.15(100+50)/50 = 3.45
1.20 2.40
USE 4 piles 0.60 0.60
1.20 2.40
0.60
35 ! 90 (. 4+3 7 d = 82 (. ,- ! + = (0.9×2.4 + 0.6×2.0)(2.4)2 = 19.4 < 4+89 22.5 ,- # 5 4+ "3 = 1.4(100)+1.7(50) = 225 ,- # 5 4+ "3"/0+
= (225 + 1.4(19.4))/4 = 63.0 /
Check punching shear: Pu = 225 ton 40 cm
d/2=41 cm
Vu = Pu = 225 bo = 4(122) = 488 (.
φ Vc = 0.85(1.06) 240(488)(82) /1,000 122 cm
= 559 ton > Vu
OK
OK
Check beam shear of 2 piles: d=82 cm
40 cm
Vu = 2(63.0) = 126
φ Vc = 0.85(0.53) 240(240)(82) /1,000 = 137 ton > Vu
OK
2(63) ton
Bending moment:
Mu = 2(63.0)(0.4) = 50.4 t-m 40 cm
40 cm
50.4(105 ) Rn = = 3.47 kg/cm2 2 0.9 × 240 × 82
ρ = 0.0009 < [ρ min = 0.0035] 2(63) ton
As = 0.0035(240)(82) = 68.88 cm2 USE 22DB20# (As = 69.08 cm2)
USE ρ min
DB20 # 22DB20# 0.90
0.05 0.10
# # "/0" + ∅ 0.40 . # .. 4 : + 50 - 4