FOOTING

Footing are structural elements that transmit to the soil column loads, wall loads or lateral loads. If these loads are to be properly transmitted, footing must be designed to prevent excessive settlement or rotation and to minimize differential settlement.

Introduction

To limit settlement within tolerably small amount, it is necessary.

i. To transmit the load of the structure to a soil stratum of sufficient strength and

ii. To spread the load over a sufficiently large area of that stratum to minimize bearing pressure.

If satisfactory soil directly underlies the structure, it is merely necessary to spread load by footings. Such substructure are known as “spread foundation.”

Introduction If adequate soil is not found immediately below the structure, if becomes necessary to use deep foundations such as piles to transmit the load to deeper, firmer layers.

Type of Spread Footing

i. Wall Footing

A wall footing is simply a strip of reinforced or plain concrete, wider than the wall that distributes its pressure.

ii. Isolated spread footing under single column

These may be usually square, sometimes rectangular or occasionally circular in plan.

Type of Spread Footing iii. Combined Footing Supporting two or more column loads. These may be continuous with a rectangular or trapezoidal plan or isolated column footing joined by a beam which is referred to as strap footing.

iii. Mat or Raft Foundation Which is one large continuous footing supporting all columns of the structure. This is used when the soil is poor.

Type of Spread Footing

Load, bearing Pressures & Footing Size

When the soil is coarse grained

When the soil is clayey or fine grained

areq q

LDA

35.2q

q ulta

Load, bearing Pressures & Footing Size D and L=loads (service) at the level of base of base of footing

Eccentric load but e does not exceed the kern distance.

IMC

AP

qminmax

PM

e

Load, bearing Pressures & Footing Size D and L=loads (service) at the level of base of base of footing If eccentricity falls outside the kern distance

mb3P2

q

bm3P

2qo

max

max

amax qq

Design of “Plain Concrete” Wall Footing

Problem Design a wall footing of plain concrete to carry 20 k/ft dead load including the wall weight and 8 k/ft live load. Use fc’=3000, an allowable soil pressure of 6 k/ft. The bottom of the footing is 4 ft below the final grade.

Solution W =20 + 8 = 28 k/ft

Assume 24” thick footing

Footing weight 2ft/lb 290)145(

1224

Soil Fill on footing 2ft/lb 200)100(

1224

qe = Effective soil pressure = 6000 – 490 = 5510 lb/ft2

Width of footing required "' 3-5 say ft 08.5510.528

Wu = (20) (1.4) + (8) (1.7) = 41.6 k/ft Net soil pressure under factored loads ksf 923.7

125.56.41

2

u 215.2

)923.7(M

ft/ft.k 89.17

Moment of Inertia 43

in 1064812

)22)(12(

2” bottom concrete is neglected

= modulus of rupture and ø = 0.65 ct f5f

)12000(2

22)10648)(30005)(65.0(

2h

IfM t

n

= 14.36 k.ft/ft < Mu

2h

IfM t

u

Putting

Solution

2h

12bh

)30005)(65.0()12000)(89.17(

3

h = 24.56”

with poor concrete h = 26.56”

Say total depth of footing = 27”

Now revise with new thickness

Footing weight 2ft/lb 25.326)145(

1227

Soil back filling 2ft/lb 175)100(

1221

2e ft/lb 75.5498)17525.326(6000q

Width of footing required ft/lb 09.549875.528

Say 5’ – 3”

Solution

43

in 1562512

)25)(12(I

)12000(2

25)15625)(30005)(65.0(

2h

IfM t

n

= 18.54 k.ft/ft So O.K.

Shear Check dbf2V wcc

251230002 = 32863.35 lb

øVc = (0.85) (32863.35)

= 27933.85 lb

d

2ab

)q(V uu

1225

2125.5

)1000)(923.7(V So u

= 330.125 lb So O.K.

Design of wall footing of reinforced concreteProblem Design a Wall footing to support a 12” wide wall with a dead load D = 20 k/ft and a live load L=15 k/ft. The bottom of footing is 4’ below the final grade and the soil weighs 100 lb/ft3. The allowable soil pressure qa is 4 k/ft2, fy = 40 ksi , fc’ = 4 ksi

SolutionAssume 12” thick footing

Footing weight 2ft/lb 150)150(

1212

Soil fill2ft/lb 300)100(

1236

Total = 450 lb/ft2

qe = qa -450 = 4000 – 450 = 3550 lb/ft2

ft 10 say ft 86.9550.335

footingofWidh

k29.20

125.8

2110

35.5

d2

ab35.5V

ftk/ 35.5)1)(10(

)15)(7.1()20)(4.1(reactionloadFactored

u

2

k97.10V

k9.121000

15.812 40002bd f2V

c

cc

footingthick02Assume32.19dSo

5.3let37.15d1000

1d12400085.0229.20Let

Solution

Since Vc < Vu, so depth should be increased.

k29.211000

15.16124000285.0

bdf2V

k72.16

25.16

29

35.5V

ft/k35.5110

157.1204.1q

01say59.951667.335

requiredwidth

ft/lb67.3516

100x1228

1501220

4000q

cc

u

2u

2

e

Solution

Therefore002.0or005.0000,40

200f

200

00572.0000,40

1.22177.11211

77.111

fmR2

11m1

77.11400085.0

000,40f85.0

fm

psi1.2215.16129.0

1200017.54

bdM

bdM

RquiredRe

ftk17.5425.4

35.5M

miny

min

y

n

c

y

2

2u

2n

n

2

u

Solution

2

s

2s

in33.1Ac/c9@9#Use

in133.15.161200572.0bdA

Solution

Development Length Length required for # 9 bars = 25” (Table A.11)

Length available = (4.5) (12) – 3 = 51” So O.K

Longitudinal & Temp. Reinforcement

At least As = (0.002)(b)(h)

As =(0.002)(12)(20) = 0.48 in2

Use #5 @ c/c (As = 0.49 in2)"

21

7

Design of Column Footing

Shear Two Shear conditions

i. One way or beam shear

ii. Two way or punching shear

i. Beam Shear At a distance d from face of column Similar as in beams and one-way slabs

bdf5.3bdM

dV2500f9.1V c

u

ucc

b = width of footing at a distance d from face of column

Mu = Moment of Vu about section xxVu = Total factored shear force on section xx

= qu times footing area outside that section

In footing design, the simpler and more conservative eq. is usedbdf2V cc

ii. Punching ShearACI code stated that the critical section for two-way shear is

located at a distance from the face of column2d

Part that tends to punch out

bo=Perimeter

=4(a+d)

ii. Punching Shear The ACI code give equation for nominal

punching shear on this perimeter.)i( dbf4V occ

For columns of very elongated x-section

)ii( dbf4

2V occ

c

Where column of side Shortside Long

c

Test show that Vc decreases as the ratio increases. dbo

According ACI code states that Vc in punching shear must not be taken greater than

)iii( dbf2b

dV oc

o

sc

Where

αs = 40 for interior columns

αs = 30 for edge loading

αs = 20 for corner loading of a footing

Punching shear strength should be taken smallest of all three equations.

ii. Punching Shear

ba

c

Bearing-Load transfer for column to footing

At the base of the column, the permitted design bearing strength is

2A)f85.0(AA

A)f85.0(P 1c1

21cn

A1 is column cross-sectional area

A2 is the area of the portion of the supporting footing as defined in Fig.

At least four dowels having area not less that 0.005 times the x-sectional area of the column and the diameter of dowel should not exceed the diameter of the column by more than 0.15”.

Development Length Dowel should have sufficient development length

For compression bars, that is, byc

byd df0003.0

f

df02.0l

The lapped length must be at least that required for a lap splice in compression. i.e. the length of lap must not be less than the usual development length in compression and must not be less than 0.0005 fydb

Where bars of different sizes are la spliced, the splice length should be the larger of the development length of larger bar or the splice length of the smaller bar.

Splice length for compression

For bar with 0.0005 fydbpsi 000,60fy

For bar with fy > 60,000 psi (0.0009fy-24)db

Not less than 12”

For fc’ <3000 psi lap is increased by one third.

ProblemDesign a square column footing to carry a column dead load of

197 k and a live load of 160 k from a 16” squired tied column containing # 11 bars as the principal column steal. The allowable soil pressure is 4.5 k/ft2. Use fc’=3000 psi fy=40000 psi , 4’ below final grade.

Solution Assume 2’ thick column footing

qe = 4500 – (2)(150) - (2)(100) = 4000 lb/ft2

Required area 2ft 25.894

160197

Let 9’ – 6” = 90.5 ft2

Pu = (1.4) (197) + (1.7) (160) = 547.8 k

2u ftk/ 07.6

25.908.547

q

Let depth required for punching shear d=20”

16+20=36”=3’

Vu = (qu) (Area)

=(6.07){(9.5)2-32} = 493.19 kβc=1

k3.53698.63085.0V

k98.6301000

120364 30004dbf4V

c

occ

Solution

The other equations are checked and give larger value so this value is valid and for this øVc

øVc > Vu Therefore no shear reinforcement is required.

93.18

)36()(4) 3000)(4)(85.0()1000)(19.493(

bf4V

doc

u

So d=20” is O.K. w.r.t two-way shear.

For one-way shearSolution

k55.1395.942.207.6V

24.21220

2167.8

1216

5.9

u

According to ACI code

k3.21276.24985.0V

k76.2491000

120125.930002

dbf2V

c

wcc

Since øVc>Vu therefore no shear reinforcement is required

51.13

)125.9)(3000)(2)(85.0(100055.139

bf2V

dwc

u

Therefore depth d=20” is O.K. w.r.t. one-way shear.Bending strength

083.4ft.k66.480

2167.8

216

5.9)5.9(2083.4

)083.4)(07.6(Mu

Solution

2

s

c

y

y

n

22u

n

in76.8

2011400384.0bdAquiredRe

00384.0

000,4081.14869.152

1169.15

1

69.15

300085.0000,40

f85.0

fm

fmR2

11m1

psi81.148

)20)(114)(85.0()12000)(66.480(

bdM

RquiredRe

Minimum As = (0.002)(114)(22) = 5.016 in2

Use As = 8.76 in2

Try 15 # 7 having As = 9 in2

Solution

C = 0.85fc’ba = (0.85)(3)(9.5)(12)(a) = 290.7 a

T = Asfy = (9)(40) = 360 k

C = T

"2384.17.290

360a

ft.k66.480MM

ft.k29.523

121

22384.1

204099.0

2a

dfA9.0M

un

ysn

So use 15 # 7 bars each way

Bearing strength

øPn = ø(0.85fc’) Ag

= (0.7)(0.85 x 3000)(16)2

= 456960 lb = 456.96 k

Pu = 547.8 k

Solution

øPn < Pu thus column load can not be transferred by concrete bearing alone and dowels bars are required.

Excess Pu = 547.8 – 456.96 = 90.84 k

2

y

us in 25.3

)40)(7.0(84.90

fP Excess

A

Required

2

cy

us

in47.3

)385.040)(7.0(84.90

)f85.0f(P Excess

A

More correctly Required

Use 4 # 9 bars As = 4 in2 Min As=(0.005)(16)2=1.28 in2

The # 9 dowels must be developed below and above the junction of column and footing.

Development length in compression

74.163000

128.1000,4002.0

f

df02.0

c

bydb

Bearing strengthSolution

If should not be less than ldb = 0.0003 fydb

=(0.0003)(40000)(1.128)=13.56” and can not be less than 8”

24” thick footing is adequate for straight dowels.

24-3.0(cover)-2x0.875(footing bars)-1.128(dowels)= 18.122” > 16.47”

alternate solutions (i) thicker footing(ii) Larger number of smaller bars(iii) pedestal

Development length of main bars

for # 7 bars from table ld = 18”

Available length )er(cov32

16)12)(5.9( "

If bends are provided in dowels then the available length is

Solution

=46” > ld=18”

Solution

Design of Rectangular Footing Use in restricted locations Identical procedure for design, except that one-way shear

action and bending moment must be considered in both principal directions.

In short direction, larger % age of reinforcement in band width

side shortside Long

1

2direction short in orcementinfre Total

width band in orcementinfRe

Rest of reinforcement should meet at least temperature reinforcement requirement.

Problem Design a rectangular footing for an 18” square column with

a dead load of 185 k and a live load of 150 k. Make the length of the long side equal to twice the width of the short side. fc’=4000 psi , fy=40,000 psi, qa = 4000 lb/ft2. Assume the base of the footing is 5ft below grade.

Solution Assume 24” thick footing d = 19.5”

2

e

ft/lb3400

1001236

1502

244000q

D + L = 185 + 150 = 335 k

Area Required 2ft53.984.3

335

b=4 L=24.6

b=5 L=19.7

b=6 L=16.4

b=7 L=14.08

So Use 7’ x 14’ = 98 ft2

Solution

2u ft/k 25.5

98)150)(7.1()185)(4.1(

q

One-Way Shear

b = 7’ = 84”

562.412

5.1952.6

52.625.12

5.141

Vu = (7) (4.625) (5.25) = 169.97 k

.K.Oso28.18

844000285.01000x97.169

bf2V

dc

u

Vu = (1.125)(14)(5.25)

= 82.69 k

SolutionOne-Way Shear

b = 14’ = 168”

512.112

5.1957.2

25.5

5.17

.K.Oso85.4

1684000285.0100069.82

bf2V

dc

u

Two-Way Shear

bo = (4)(18 + 19.5) = 150”

k23.463

5.375.191825.512

5.37714V

2

u

Steel in Long-Direction

ft.k 77.717225.6

)25.5)(7)(25.6(Mu

psi 63.299

(19.5)840.91200077.717

bdM

RquiredRe 22u

n

Solution

.K.Oso63.14

1504000485.0100023.463

bf2V

dc

u

)in13A(9#13Use

in696.32284002.0.minA

in87.12

5.1984007854.0bdA

7854.00

000,4063.299765.112

11765.111

fmR2

11m1

765.11400085.0

000,40f85.0

fm

2s

2s

2

s

y

n

c

y

Solution

Dev. Length

Required for # 9 bars ld = 25”

Available = 6.25 x 12 = 75” – 3” = 72” O.K

Longitudinal Steel in Short Direction

765.11mNowpsi58

2.1912149.0

12000922.277bdM

RquiredRe

ft.k 922.277

275.2

)25.5)(14)(75.2(M

22u

n

u

Solution

005.0000,40

200f

200

in79.4

5.19121400146.0bd00146.0A

00146.0

000,4058765.112

11765.111

fmR2

11m1

ymin

2

s

y

n

Longitudinal Steel in Short DirectionSolution

)in96.18A(bars8#24Use

in48.18121214005.0.minA2

s

2s

32

122

1

2 orcementinfre Totalwidth band in orcementinfRe

bars81624stRe

widthbandinbars1632

24Use

Dev. Length For # 8 bar ld = 20” Available (2.75)(12)-3=30” O.K

k12.771

10001

18400085.07.0

A)f85.0(P

2

gcn

Pu = (1.4)(185) + (1.7)(150) = 514 k

Since øPn > Pu therefore only four bars will be used as dowels

As(min) = (0.005)(18)2 = 1.62 in2

Solution

Bearing Strength

Use 4 # 6 (As = 1.77 in2) Required Dev. Length = 9” Available is more than 9” O.K

Solution

Design of Combined Footing Design a rectangular combined footing for the two columns

shown in fig. qa = 6000 lb/ft2 .The bottom of footing is 6’ below grade. fc’ = 3000 psi fy = 60,000 psi

Solution

Assume 3’ – 6” thickness h = 42” d=37.5”

qe = 6000 – (3.5)(150) - (2.5)(100) = 5225 lb/ft2

2req ft 54.143

225.5200250130170

A

Location of Resultant

59.7750

25.1

12)450(25.1

)300(x

Length of footing = (7.95)(2) = 15.9’ Try 16’

The required width is use 9’ width 97.816

54.143

The net upward soil pressure

2u k/ft 98.7

169)200130)(7.1()250170)(4.1(

q

= 7.98 x 9= 71.82 k/ft

Location of ResultantVa = 0

Vb = -(306)(1.5)+(1.5)(71.82) = -351.27 k

Vc = - 351.27 + (71.82)(10.25) = 384.885 k

Vd = 384.885 – (345)(2)+(71.82)(2) = -161.475 k

Ve = -161.475 + (2.25)(71.82) = 0.12

Vbc = -(x)(306)(1.5) + (71.82)(x)

Vdc = -(x)(71.82) + (x-2.25)(345)

x = 2.84

Ma = 0

ft/k 45.26325.1

)82.71)(5.1(25.1

)105)(306(Mb

ft/k 48.1122239.6

)39.6)(82.71(25.1

39.6)105)(306(MF

ft/k 38.41)1)(2)(345(225.4

)25.4)(82.71(Mc

ft/k 59.2292

)25.284.2()345(

284.2

)84.2)(82.71(M2

G

ft/k 79.181225.2

)25.2)(82.71(MD

Solution

Shear At a distance d to the left of the left face of the interior column.

k 45.160)82.71(12

5.37885.384Vu

At a distance d from right of point B

k 83.12682.7112

5.3727.351Vu

Now design shear strength

dbf285.0V cc

k1.3771

lb 98.377106

5.371083000285.0

Therefore d = 37.5” is O.K

Punching Shear For exterior column

ft25.11

125.5)0625.3(2

125.37

21225.37

5.12bo

Vu = column load – soil pressure within the perimeter

= 459-3.0625 x 2.125 x 7.98 = 333.75 k Design shear strength

k76.942

lb45.942767

5.371225.114000485.0

db)f4)(85.0(V occ

Solution

Longitudinal Steel

53.23

300085.0000,60

f85.0

fm

psi544.98

)5.37)(129)(9.0(1200048.1122

bdM

RquiredRe

c

y

22u

n

Solution

0033.0000,60

200f

200

001675.0

000,60544.9853.232

1153.23

1

fmR2

11m1

ymin

y

n

So min holds

As = bd = 0.0033 x 9 x 12 x 37.5 = 13.365 in2

Use 17 # 8 bars

54

3000000,6079.004.03.1

f

fA04.03.1l

c

ybd

Longitudinal SteelSolution

Available length = 6.39 x 12 – 3 = 73.68 in So O.K

For bottom bars ld = 34.62 in

Available 2.84 ft 34.08 – 3 31.08” For hooked bars

9.213000

11200f

d1200quiredRe

c

bd

Transverse Beams The ACI code recommends band width from the face

of the column 2d

Interior Column Width of beam 5.615.3724

2d

224

Upward load /ft

in.lb5635000

25.3

5.367.76666M

ft/lb 67.7666691000x690

u

Solution

Transverse bars are placed on top of the longitudinal bars, with actual d = 37.5 – 1 = 36.5”

Minimum requirement of steel criteria holds

2

ys in 48.75.365.61

f200

A

Use 13 # 7 bars with As = 7.8 in2

Exterior Column Width of beam= "75.36

2

5.3718

Minimum steel criteria holds

2

ys in 47.45.3675.35

f200

A

Use 8 # 7 (As = 4.8 in2)

Solution

Development Length

for # 7 bars in tension ld = 26” (Table)

Available 61.5” So O.K

Bearing Strength Interior Columns

øPn = ø(0.85fc’) A1 = (0.7)(0.85fc’)(24 x 24)

= 1028160lb 1028.16 k

Pu = 690 k < øPn So nominal dowels

As = 0.005 x 24” x 24” = 2.88 in2

Use 4 # 8 dowels (As = 3.14 in2) dev.length 22” So O.K

Solution

Pu = 459 k < øPn So use nominal dowels

As = 0.005 x 18 x 24 = 2.16 in2

Use 4 # 7 (As = 2.41 in2) dev.length = 19” O.K

Edge column

øPu = (0.7)(0.85)(3000)(18 x 24) = 771120 lb = 771.12 k

SolutionBearing Strength