Shallow Foundations ( Combined, Strap, Raft foundation)

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