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Footing

Example # 1

Design a combined footing, to support two columns A and B spacedat distance 6.0 m center-to-center as shown in Figure Column A is

0.4mx0.4m and carries a dead load of 500kNs and a live load of

300kNs Column B is also 0.4mx0.4m in cross section but carries a

dead load of 750 kNs and a live load of 450 kNs.

Use fc= 25 MPa , fy= 420 MPa, and qallnet = 150 kpa

A B

0.4

0.4

0.4

0.4Property

limit

6 m ?? m

?? m

PD=500kN

PL=300kN

PD=750kN

PL=450kN

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Footing

Solution

1- Establish the required base area of the footing

To locate the resultant of the column forces

M@start =0.0 800(0.2)+1200 (6,2) =2000 (x) x= 3.8 m

2

all(net)

Ag m33.31

150

450750300500

q

PA

B

P

Ps=800kN Ps=1200kN

Ps=2000kN

6 m0.2 m

X=3.8 m

AB

Ps=800kNPs=1200kN

Ps=2000kN

X=3.8 m

A B

X=3.8 m

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Footing

Solution

Length of footing L=2 (3.80) = 7.60m

Width of footing B =13.33/7.6 =1.754 m , taken as 1.80 m.

2- Evaluate the net factored soil pressure

3- Select a trial footing depth

Assume that the footing is 0.8 m thickEffective depth d = 800 75 10 = 715 mm

kPa197.41.87.6

16201080

A

PPq

kN16201.6(450)7501.2P

kN10801.6(300)5001.2P

uB,uA,

u

uB,

uA,

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Footing

Solution

4- Check footing thickness for punching shear

Column A

The factored shear force Vu =1080197.4(1.115)(0.758) =914 kN

b =2(400+715/2)+400+715=2630 mm

Vc = 2350 kN > Vu= 914 OK

kN59681000

7152630

2630

71530225

12

10.75db

b

d2'f

12

1

controlkN23501000

7152630

253

1

0.75db'f3

1

ofsmallesttheisV

o

o

sc

oc

C

7.6 m

1.8m

758 1115

1115

1115

A B

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Footing

Solution

4- Check footing thickness for punching shear [contd.]

Column B

The factored shear force Vu =1620197.4(1.115)2 =1375 kN

b =4(400+715)=4460 mm

Vc = 3986 kN > Vu= 1375 OK

kN83831000

7154460

4460

71540225

12

10.75db

b

d2'f

12

1

controlkN39861000

7154460

253

1

0.75db'f3

1

ofsmallesttheisV

o

o

sc

oc

C

7.6 m

1.8m

758 1115

1115

1115

A B

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Footing

Solution

5- Draw S.F.D and B.M.D for footing1080kN 1620 kN

197.37 x1.8=355.26 kN/m

71

1009

1122.6

497.4

2.84 m

7.10

1425.7

348

0.915

798

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Footing

Solution

6- Compute the areas of flexural reinforcement

a) Top longitudinal reinforcement

20mm18Use,mm550871518000.00428A

0.004281800*715*250.850.9

1425.7*102-1-1

420

250.85

2

reqs,

2

6

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Footing

Solution

b

) Bottom longitudinal reinforcement

1613Use,mm259280018000.0018AA

0.001011800*715*250.850.9

348*102-1-1

420

250.85

2

mins,reqs,

min2

6

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Footing

Under Column A 7.6 m

1.8m

400+715400+375.51115757.5

0.4

0.4

0.4

0.4

16Use6.mm1080757.58000.0018A

757.5*715*250.850.9

147*102-1-1

420

25*0.85

mm715dmm,757.5b

kN.m1472

0.41.8

2

0.76

0.76)*(1.8

1080M

2

mins,

min2

6

2

uA,

10801.8

0.7

c) Short Direction :

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Footing

Solution

Under Column B

168Use.1584mm11158000.0018A

1115*715*250.850.9

220.5*102-1-1420

25*0.85

715mmdmm,1115b

kN.m220.52

0.41.8

2

1.1

1.1)*(1.8

1620M

2

mins,

min2

6

2

uB,

7.6 m

1.8m

400+715400+375.51115757.5

0.4

0.4

0.4

0.4

10801.8

0.7

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Footing

Solution

Shrinkage Reinforcement in the short direction14@100mmUse.1440mm10008000.0018A 2mins,

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Footing

Solution

14@1

00

1316 B

0.75m 1.10 m

1.80m

0.80m

816

B

616

B

14@1

00

1820 T

7.60 m

0.4m 0.4m5.6 m 1.2 m

616 816

14@100

1316

1820

14@100 14@100

14@100