shallow foundations ( combined, strap, raft foundation)
TRANSCRIPT
SHALLOW FOUNDATIONS 2DESIGN OF COMBINED, STRAP & RAFT FOOTING
INTRODUCTION
Contents :
Combined Footing
Strap Footing
Raft Foundation
1
2
3
1. Combined Footing
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.
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.
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 :
x
y
Eccentricity = ZERO
STEPS OF DESIGN
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
C1 C2
L
RC
MaxS.F.D
B.M.D
Zero shear
c1 c2
c1 c2
H.B.1 H.B.2
c1 c2
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
Solution
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
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/㎡
Design of Footing in Longitudinal Direction
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㎜
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
c1 c2
1.23 1.43
0.83
0.83
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
Design of Footing in Short Direction
c1 c2
1.73 m 1.96 m
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
Reinforcement
Asmin =1.5 X d = 1.5 x 530 = 795 mm2 Asmin = 4 1 16
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
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
7ϕ18
4ϕ1611ϕ22 10ϕ25
4ϕ16
c1 c2
7ϕ18
4ϕ16
4ϕ16
11ϕ2
2
4ϕ16
4ϕ16
10ϕ2
5
4ϕ16
2. Strap Footing
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
C2C1
0.5 0.9
4.9 m
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
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
C2C10.5 0.9
0.50.90.50.5 2.11.6
M = 602720
695
432
262
B.M.D
S.F.D
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
Reinforcement:Astop = =1646 = 7/m’
Asbot = = 295
Asmin = 0.15% b*d = 0.15%*400*1230 =738
= 4
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
Design of Footings
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
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
Footing 2:qu==455kN/
L2==0.75m
Mu=455*=128kN.m
d=5 =358mm
t=45cm d=38cm
Critical Sec
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
3ϕ18 + 4ϕ18 = 7ϕ18
4ϕ16
5ϕ10
2ϕ12
2ϕ12
0.4
m
2.3
m1.
5 m
2.1 m2.5 m 2.7 m
1.9 m
5ϕ12 5ϕ12
5ϕ14
5ϕ16
3. Raft Foundation