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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