Shallow Foundations ( Combined, Strap, Raft foundation)
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Transcript of Shallow Foundations ( Combined, Strap, Raft foundation)
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SHALLOW FOUNDATIONS 2DESIGN OF COMBINED, STRAP & RAFT FOOTING
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INTRODUCTION
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Contents :
Combined Footing
Strap Footing
Raft Foundation
1
2
3
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1. Combined Footing
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Combined Footing
A combined Footing is a long footing supporting two or more columns in (typically two) one row.
A combined Footing is a rectangular or Trapezoidal shaped footing.
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Using of Combined Footing
Construction Practice may dictate using only one footing for two or more columns due to:
a) Closeness of Columns
b) Due to property line constraint, which may limit the size of footing at boundary.
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The Design of Combined Footing requires that the centroid of the area be as close as possible to the resultant of the 2 column loads for uniform pressure and settlement.
Which Means :
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x
y
Eccentricity = ZERO
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STEPS OF DESIGN
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Dim
ensi
onin
g fo
r P.C
Find Area“L” & “B”
Des
ign
of R
.C F
ootin
g (lo
ngitu
dina
l)
Find “Max Moment”Find “d”
Check Shear
Check Punching
Des
ign
of R
.C F
ootin
g (S
hort)
Find Moment @ Hidden beam 1
Find Moment @ hidden beam 2
Ensure That C1 > 2.3
RFT
RFT in Long Direction
RFT in Short Direction
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C1 C2
L
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RC
MaxS.F.D
B.M.D
Zero shear
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c1 c2
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c1 c2
H.B.1 H.B.2
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c1 c2
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EXAMPLE
C1 (30*70) C2 (30*90)
4 m
Design a combined footing to support a working load of p1=160 ton & p2=220 ton. Bearing capacity: 12.5 ton/m2Thickness of plain concrete 15 cm
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Solution
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1- Dimensions of footing :R = p1+p2 = 160 + 220 = 380 ton R . X = p2 . S 380 x X = 220 x 4 X= 2.32 m
Lpc / 2 = X + b1 / 2 + 0.5 + tpc
Lpc / 2 = 2.32 + 0.7 / 2 + 0.5 + 0.3 = 3.47m Lpc = 6.94 m LRc = LPc - 2 x tPC = 6.94 - 2 X 0.3 = 6.34m
APC = 380 / 12.5 = 30.4 LPc X BPc = 30.4 6.34 x BPc = 30.4BPc = 4.8mBRc = 4.8 - 2 x 0.3 = 4.2m
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2Design of R.c footing
P1u = 160 x 1.5 = 240 ton P2u = 220 x 1.5 = 330 ton Ru = 380 x 1.5 = 570 ton Wu = 570 / 6.34 = 89.9 ton qu = 570 / 6.34 x 4.2 = 21.4 t/㎡
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Design of Footing in Longitudinal Direction
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At point of zero shear :
P1u = wu . X
240 = 89.9 . X X = 2.67m Mmax = 89.9 x 7.1289 / 2 - 240 x 1.82 = 116.3 D =5 × √116.35×10^7 / 25 × 4200 = 526.3 ㎜ D = 530㎜
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2- check shear
Qsu = 0.16√25/1.5 = 0.653 N/m㎡
Qsumax = 155.56 - 89.99 × 0.53/2 = 131.7365
Qs = 131.73×10^4 / 530×4200 = 0.59 ≤ qsu
Safe
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c1 c2
1.23 1.43
0.83
0.83
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For c1 (30x70) Qpcu1 = 0.316 ( 0.5 + 300 / 700) √25/1.5 = 1.197
Qpu1 = 240 - 21.4 x (1.23x0.83)=218.15 ton Qpu1 = 218.15 x 10^4 / 530x2 (1230+830) =0.99 ≤ qpcu1 Safe
For c2 (30*90)
Qpcu2 = 0.316 x (0.5 + 300/900) √25/1.5 = 1.07 Qpu2 = 220 -21.4 (1.43x0.83) = 194.6 Qpu2 = 194.6 x 10^4 / 530 (1430+830)x2 = 0.81 ≤ 1.07 safe
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Design of Footing in Short Direction
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c1 c2
1.73 m 1.96 m
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For hidden Beam1 Qu1 = 240 / 4.2 x 1.73 = 33.03t/㎡M1 = 33.03 * 1.95^2 /2 = 62.9 t.m
For hidden Beam 2 Qu2 = 330 / 4.2 x 1.96 = 40.1t/㎡ M2 = 40.1 x 1.95^2 / 2 = 76.2 t.m d = 530 = c1 √76.2x10^7 / 25 x 1000
C1 = 3.1≥ 2.3
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Reinforcement
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Asmin =1.5 X d = 1.5 x 530 = 795 mm2 Asmin = 4 1 16
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RFT in long direction
Astop = 116.35 x 107 / 360 x 0.826 x 530 = 72382mm2/4.2m =1757.7 mm2 7 ϕ18
Asbottom = 48.61 x 107 / 360x 0.826 x 530 = 3084mm2 /4.2 = 734.3 mm2 4ϕ16
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RFT in short direction
As1 = 62.79 X 107 / 360 X 0.826 X 530 = 3984 mm2 11ϕ22
As2 = 76.2 X 107 / 360 X 0.826 X 530 = 4835 mm2 10ϕ25
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7ϕ18
4ϕ1611ϕ22 10ϕ25
4ϕ16
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c1 c2
7ϕ18
4ϕ16
4ϕ16
11ϕ2
2
4ϕ16
4ϕ16
10ϕ2
5
4ϕ16
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2. Strap Footing
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Design a strap footing to support an exterior column (30*50cm) and an interior column (30*90cm). The un factored Loads C1 =685 KN, C2 =1270 kN . Assume the allowable Bearing capacity is 150 kN/m2, fcu=25 N/mm2 and Fy = 360 N/mm2 , P.C. thickness = 40 cm
EXAMPLE
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C2C1
0.5 0.9
4.9 m
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Calculation of reaction:
Assume e=0.1 to 0.2 (L)
e = 0.5 to 1m (take it 1m)
R1= = 860.6 KN
R2=685 + 1270 - 860.6 = 1094.4 KN
C2C10.5 0.9
e
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Area of plain concrete footing:A1 = = 5.73 Dimension ((1+0.25)*2) = 2.5 (2.5*2.3)
A2 = = 7.29
Dimension = (2.7*2.7) = 7.29
R.C dimensions:F1 = (2.1*1.5) m
F2 = (1.9*1.9) m
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C2C10.5 0.9
0.50.90.50.5 2.11.6
M = 602720
695
432
262
B.M.D
S.F.D
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Design of strap beam:qu1= = 614 kN/m’
qu2= = 864 kN/m’
Point of zero shear:614X-1027=0 , X=1.67m , Mmax=602kN.m
d = 5 * =1227m
t=130cm d=123cm
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Reinforcement:Astop = =1646 = 7/m’
Asbot = = 295
Asmin = 0.15% b*d = 0.15%*400*1230 =738
= 4
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CHECK SHEAR:
Qs = (720*0.55)/1.17 = 341.53kN
qs = = 0.694N/
qcu =0.24 = 0.98N/
qs use min stirrups 5
0.550.62
720
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Design of Footings
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Critical Sec
Footing 1: qu = = 409.5 kN/
L1 = (1.5-0.4)/2 = 0.55 m
Mu = 409.5*(/2) = 61.9kN.m
d = 5 = 249mm
t = 35cm d = 28cm
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Check shear:Qs = 409.5*(0.55-(0.28/2)) = 168kN
qs = = 0.6 MPa
qcu = 0.16 = 0.65 MPa
qs safe
As = =743.44 = 5
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Footing 2:qu==455kN/
L2==0.75m
Mu=455*=128kN.m
d=5 =358mm
t=45cm d=38cm
Critical Sec
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Check shear:Qs = 455*(0.75-) = 255KN
qs ==0.67N/
qcu =0.65N/ qs unsafe
0.65=d=392mm
t=50cm d=43cm
As ==1001=5
Asmin = 1.5d = 1.5*430 = 645
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3ϕ18 + 4ϕ18 = 7ϕ18
4ϕ16
5ϕ10
2ϕ12
2ϕ12
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0.4
m
2.3
m1.
5 m
2.1 m2.5 m 2.7 m
1.9 m
5ϕ12 5ϕ12
5ϕ14
5ϕ16
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3. Raft Foundation
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