are Lecture26Footings
August8,2003
CVEN444
LectureGoals
FootingClassification
FootingDesign
FootingExamples
Footings
Definition
Footings structural members used to support
columns and walls and to transmit and distribute
their loads to the soil in such a way that the load
bearing capacity of the soil is not exceeded,
excessive settlement, differential settlement,or
rotation are prevented and adequate safety
against overturning or sliding is maintained.
TypesofFootings
Wall footings are used to
support structural walls that
carry loads for other floors
or to support nonstructural
walls.
TypesofFootings
Isolated or single footings
are used to support single
columns. This is one of the
most economical types of
footings and is used when
columns are spaced at
relatively long distances.
TypesofFootings
Combined footings usually
support two columns, or three
columns not in a row.
Combined footings are used
when tow columns are so close
that single footings cannot be
used or when one column is
located at or near a property
line.
TypesofFootings
Cantilever or strap footings
consist of two single footings
connected with a beam or a
strap and support two single
columns. This type replaces a
combined footing and is more
economical.
TypesofFootings
Continuous footings
support a row of three or
more columns. They have
limited width and continue
under all columns.
TypesofFootings
Rafted or mat foundation
consists of one footing usually
placed under the entire building
area. They are used, when soil
bearing capacity is low, column
loads are heavy single footings
cannot be used, piles are not used
and differential settlement must
be reduced.
TypesofFootings
Pile caps are thick slabs
used to tie a group of piles
together to support and
transmit column loads to the
piles.
DistributionofSoilPressure
When the column load P is
applied on the centroid of the
footing, a uniform pressure is
assumed to develop on the soil
surface below the footing area.
However the actual distribution of the soil is not uniform,
but depends on may factors especially the composition of
the soil and degree of flexibility of the footing.
DistributionofSoilPressure
Soil pressure distribution in
cohesive soil.
Soil pressure distribution in
cohesionless soil.
DesignConsiderations
Footings must be designed to carry the column loads
and transmit them to the soil safely while satisfying
code limitations.
1. The area of the footing based on the allowable
bearing soil capacity
2. Two-way shear or punch out shear.
3. One-way bearing
4. Bending moment and steel reinforcement
required
DesignConsiderations
Footings must be designed to carry the column loads
and transmit them to the soil safely while satisfying
code limitations.
1. Bearing capacity of columns at their base
2. Dowel requirements
3. Development length of bars
4. Differential settlement
SizeofFootings
The area of footing can be determined from the actual
external loads such that the allowable soil pressure is
not exceeded.
Area of footing =
Total load ( including self - weight )
allowable soil pressure
Strength design requirements
qu =
Pu
area of footing
TwoWayShear(PunchingShear)
For two-way shear in slabs (& footings) Vc is smallest of
2 + 4 f b d
Vc =
c0
c
where, c =
b0 =
ACI 11-35
long side/short side of column concentrated
load or reaction area < 2
length of critical perimeter around the
column
When > 2 the allowable Vc is reduced.
Designoftwowayshear
1. Assume d.
2. Determine b0.
b0 = 4(c+d)
for square columns
where one side = c
b0 = 2(c1+d) +2(c2+d)
for rectangular
columns of sides c1
and c2.
Designoftwowayshear
3. The shear force Vu acts at a
section that has a length
b0 = 4(c+d) or 2(c1+d) +2(c2+d)
and a depth d; the section is
subjected to a vertical downward
load Pu and vertical upward
pressure qu.
Vu = Pu qu ( c + d ) for square columns
Vu = Pu qu ( c1 + d ) ( c2 + d ) for rectangular columns
2
Designoftwowayshear
4. Allowable Vc = 4
Let Vu= Vc
d=
f c b0 d
Vu
4 f c b0
If d is not close to the assumed d,
revise your assumptions
Designofonewayshear
For footings with bending
action in one direction the
critical section is located a
distance d from face of column
Vc = 2 f c b0 d
Designofonewayshear
The ultimate shearing force at
section m-m can be calculated
L c
Vu = qu b d
2 2
If no shear reinforcement is to be
used, then d can be checked
Designofonewayshear
If no shear reinforcement is
to be used, then d can be
checked, assuming Vu = Vc
d=
Vu
2 f c b
FlexuralStrengthandFooting
reinforcement
The bending moment in each
direction of the footing must be
checked and the appropriate
reinforcement must be provided.
As =
Mu
a
f y d
2
FlexuralStrengthandFooting
reinforcement
Another approach is to
calculated Ru = Mu / bd2 and
determine the steel percentage
required . Determine As then
check if assumed a is close to
calculated a
a=
f y As
0. 8 5 f c b
FlexuralStrengthandFooting
reinforcement
The minimum steel percentage
required in flexural members is
200/fy with minimum area and
maximum spacing of steel bars
in the direction of bending shall
be as required for shrinkage
temperature reinforcement.
FlexuralStrengthandFooting
reinforcement
The reinforcement in one-way footings
and two-way footings must be
distributed across the entire width of
the footing.
Reinforcement in band width
Total reinforcement in short direction
where
=
=
2
+1
long side of footing
short side of footing
BearingCapacityofColumnat
Base
The loads from the column act on the footing at the
base of the column, on an area equal to area of the
column cross-section. Compressive forces are
transferred to the footing directly by bearing on the
concrete. Tensile forces must be resisted by
reinforcement, neglecting any contribution by
concrete.
BearingCapacityofColumnat
Base
Force acting on the concrete at the base of the column
must not exceed the bearing strength of the concrete
N1 = ( 0.85 f c A1 )
where = 0.65 and
A1 =bearing area of column
BearingCapacityofColumnat
Base
The value of the bearing strength may be multiplied by a
factor A2 / A1 2.0 for bearing on footing when the
supporting surface is wider on all sides than the loaded
area.
The modified bearing
strength
N 2 ( 0.85 f c A1 ) A2 / A1
N 2 2 ( 0.85 f c A1 )
DowelsinFootings
A minimum steel ratio = 0.005 of the column section
as compared to = 0.01 as minimum reinforcement for
the column itself. The number of dowel bars needed is
four these may be placed at the four corners of the
column. The dowel bars are usually extended into the
footing, bent at the ends, and tied to the main footing
reinforcement. The dowel diameter shall not exceed
the diameter of the longitudinal bars in the column by
more than 0.15 in.
Developmentlengthofthe
ReinforcingBars
The development length for compression bars was given
ld = 0.02 f y d b /
fc
but not less than
0.003 f y d b 8 in.
Dowel bars must be checked for proper development
length.
DifferentialSettlement
Footing usually support the following loads:
1. Dead loads from the substructure and superstructure
2. Live load resulting from material or occupancy
3. Weight of material used in back filling
4. Wind loads
GeneralRequirementsforFooting
Design
1. A site investigation is required to determine the
chemical and physical properties of the soil.
2. Determine the magnitude and distribution of
loads form the superstructure.
3. Establish the criteria and the tolerance for the
total and differential settlements of the structure.
GeneralRequirementsforFooting
Design
4. Determine the most suitable and economic type
of foundation.
5. Determine the depth of the footings below the
ground level and the method of excavation.
6. Establish the allowable bearing pressure to be
used in design.
GeneralRequirementsforFooting
Design
7. Determine the pressure distribution beneath the
footing based on its width
8. Perform a settlement analysis.
ExampleWall
Design a plain concrete footing to support a 16 in.
thick concrete wall. The load on the wall consist of
16 k/ft dead load (including the self-weight of wall)
and a 10 k/ft live load the base of the footing is 4 ft
below final grade. fc = 3 ksi and the allowable soil
pressure = 5 k/ft2
ExampleWall
Assume a depth of footing. (1.5 ft or 18 in.) The
weight of concrete and the soil are:
1 ft.
3
Wc = d = 150 lb/ft *18 in. *
12 in.
= 225 lb/ft 2
1 ft.
Ws = s ds = 100 lb/ft * 4 ft 18 in. *
12 in.
= 250 lb/ft 2
3
ExampleWall
The effective soil pressure is given as:
qeff = qs Wc Ws
= 5000 lb/ft 225 lb/ft 250 lb/ft
2
2
= 4525 lb/ft 2 4.525 k/ft 2
2
ExampleWall
Calculate the size of the footing for 1-ft of wall:
Actual Loads = DL + LL = 16 k/ft + 10 k/ft
= 26 k/ft
26 k/ft
Width of footing =
= 5.75 ft
2
4.525 k/ft
Use 6 ft
ExampleWall
Calculate net upward pressure:
Actual Loads = 1.2 DL + 1.6 LL
= 1.2 ( 16 k/ft ) + 1.6 ( 10 k/ft )
= 35.2 k/ft
35.2 k/ft
Net upward pressure qn =
= 5.87 k / ft 2
6 ft
ExampleWall
Calculate the depth of the reinforcement use # 8 bars
with a crisscrossing layering.
d = h cover 1.5d b
d = 18 in. 3 in 1.5 ( 1.0 in )
= 13.5 in.
ExampleWall
The depth of the footing can be calculated by using
one-way shear
1 ft
6 ft 16 in 12 in
L c
13.5 in 1 ft
d =
22
2
2
12 in
= 1.21 ft
L c
Vu = qn ( l2 ) d
2 2
= 5.87 k/ft 2 ( 1 ft ) ( 1.21 ft ) = 7.1 k
ExampleWall
The depth of the footing can be calculated by using
one-way shear
1000 lb
7.1 k
Vu
1k
d=
=
2 fc b
12 in
0.75 2 3000 1 ft
1 ft
= 7.2 in.
The footing is 13.5 in. > 7.2 in. so it will work.
ExampleWall
Calculate the bending moment of the footing at the
edge of the wall
1 ft
6 ft 16 in 12 in
L c
= 2.33 ft
=
2
2 2 2
L c
( 2.33 ft ) 1 ft
L c 2 2
2
M u = qn
b = 5.87 k/ft ( 2.33 ft )
()
2
2
2 2
= 15.98 k-ft
ExampleWall
Calculate Ru for the footing to find of the footing.
12 in.
15.98 k-ft *
Mu
1 ft
Ru = 2 =
= 0.0877 ksi
2
bd
( 12 in.) * ( 13.5 in )
ExampleWall
From Ru for the footing the value can be found.
1.7 Ru
Ru = f c ( 1 0.59 ) 1.7 +
=0
fc
2
0.0877 ksi
1.7 ( 1.7 ) 4 1.7
0.9 ( 3 ksi )
=
= 0.03312
2
fy
0.03312 ( 3 ksi )
= 0.03312 =
= 0.00166
fc
60 ksi
2
ExampleWall
Compute the area of steel needed
12 in.
As = bd = 0.00166 1 ft
13.5 in.) = 0.27 in 2
(
1 ft
The minimum amount of steel for shrinkage is
As = 0.0018 bh = 0.0018 ( 12 in.) ( 18 in.) = 0.389 in 2
The minimum amount of steel for flexure is
200
200
As =
bd =
12 in.) ( 13.5 in.) = 0.54 in 2 Use
(
fy
60000
ExampleWall
Use a #7 bar (0.60 in2) Compute the number of bars
need
As 0.54 in 2
n=
=
= 0.9 Use 1 bars/ ft
2
Ab 0.60 in
ExampleWall
Check the bearing stress. The bearing strength N1, at
the base of the wall, 16 in x 12 in., = 0.65
N1 = ( 0.85 f c A1 ) = 0.65 ( 0.85 ( 3 ksi ) ( 16 in ) ( 12 in ) )
= 318.2 k
The bearing strength, N2, at the top of the footing is
N 2 = N1
A2
2 N1
A1
ExampleWall
A2 = ( 6 ft ) ( 1 ft ) = 6 ft 2
1 ft
A1 = 16 in
( 1 ft ) = 1.33 ft 2
12 in.
The bearing strength, N2, at the top of the footing is
A2
6 ft 2
=
= 4.5
2
A1
1.33 ft
>
2
N 2 = 2 N1 = 2 ( 318.2 k ) = 636.4 k
ExampleWall
Pu =35.2 k < N1, bearing stress is adequate. The
minimum area of dowels is required.
0.005 A1 = 0.005* ( 16 in ) ( 12 in ) = 0.92 in 2
Use minimum number of bars is 2, so use 4 # 8 bars
placed at the four corners of the column.
ExampleWall
The development length of the dowels in compression
from ACI Code 12.3.2 for compression.
ld =
0.02d b f y
fc
=
0.02 ( 0.875 in ) ( 60000 psi )
3000 psi
= 19.17 in Use 20 in
The minimum ld , which has to be greater than 8 in., is
ld = 0.0003d b f y = 0.0003 ( 0.875 in.) ( 60000 psi )
= 15.75 in. 8 in.
ExampleWall
Therefore, use 4#7 dowels in the corners of
the column extending 20 in. into the column
and the footing. Note that ld is less than the
given d = 15.75 in., which is sufficient
development length.
ExampleWall
The development length, ld for the #7 bars for the
reinforcement of the footing.
fy
f ydb
ld
=
ld =
d b 20 f c
20 f c
( 60000 psi ) ( 0.875 in )
=
20 3000 psi
= 47.9 in
There is not adequate development length provided
need to design a hook.
L
c 72 in.
16 in.
ld = cover =
3 in.
= 25 in.
2
2
2
2
ExampleSquareFooting
Design a square footing to support a 18 in. square
column tied interior column reinforced with 8 #9
bars. The column carries an unfactored axial dead
load of 245 k and an axial live load of 200 k. The
base of the footing is 4 ft. below final grade and
allowable soil pressure is 5 k/ft2 Use fc = 4 ksi and
fy = 60 ksi
ExampleSquareFooting
Assume a depth of footing. (2 ft or 24 in.) The
weight of concrete and the soil are:
Wc = d = 150 lb/ft * 24 in. *
3
1 ft.
12 in.
= 300 lb/ft 2
1 ft.
= 200 lb/ft 2
Ws = s d s = 100 lb/ft * 4 ft 24 in. *
12 in.
3
ExampleSquareFooting
The effective soil pressure is given as:
qeff = qs Wc Ws
= 5000 lb/ft 2 300 lb/ft 2 200 lb/ft 2
= 4500 lb/ft 2 4.5 k/ft 2
ExampleSquareFooting
Calculate the size of the footing:
Actual Loads = DL + LL = 245 k + 200 k = 445 k
Area of footing =
445 k
4.5 k/ft
= 98.9 ft 2
2
Side of footing = 9.94 ft Use 10 ft
ExampleSquareFooting
Calculate net upward pressure:
Actual Loads = 1.2 DL + 1.6 LL
= 1.2( 245 k ) + 1.6( 200 k ) = 614 k
614 k
Net upward pressure qn =
= 6.14 k / ft 2
100 ft 2
ExampleSquareFooting
Calculate the depth of the reinforcement use # 8 bars
with a crisscrossing layering.
d = h cover 1.5d b
d = 24 in. 3 in 1.5(1.0 in )
= 19.5 in.
ExampleSquareFooting
Calculate perimeter for two-way shear or
punch out shear. The column is 18 in.
square.
bo = 4( c + d )
= 4(18 in. + 19.5 in.) = 150 in.
1 ft
= 3.125 ft
c + d = (18 in. + 19.5 in.)
12 in
ExampleSquareFooting
Calculate the shear Vu
2
Vu = Pu qn ( c + d )
= 614 k 6.14 k/ft ( 3.125 ft )
= 554 k
2
The shape parameter
=
10 ft
10 ft
=1
2
ExampleSquareFooting
Calculate d value from the shear capacity according to
11.12.2.1 chose the largest value of d
4
Vc = 2 + f c b0 d
c
d
Vc = s + 2 f c b0 d
bo
s is 40 for interior, 30 for edge
and 20 for corner column
Vc = 4 f c b0 d
ExampleSquareFooting
The depth of the footing can be calculated by using
two way shear
1000 lb
554 k
Vu
1k
d=
=
4 f c b0 0.75 4 4000 (150 in )
(
= 19.47 in.
)
ExampleSquareFooting
The second equation bo is dependent on d so use the
assumed values and you will find that d is smaller and
= 40
Vu
d=
40d
b + 2 f c b0
o
1000 lb
554 k
1k
=
= 10.81 in.
40(19.5 in )
0.75
+ 2 4000 (150 in )
150 in
(
)
ExampleSquareFooting
The depth of the footing can be calculated
by using one-way shear
1 ft
18 in
L c
10 ft
12 in 19.5 in 1 ft
d =
2
2
2 2
12 in
= 2.625 ft
L c
Vu = qn ( l2 ) d
2 2
= 6.14 k/ft 2 (10 ft ) ( 2.625 ft ) = 161.2 k
ExampleSquareFooting
The depth of the footing can be calculated by using
one-way shear
1000 lb
161.2 k
Vu
1k
d=
=
2 fc b
12 in
0.75 2 4000 10 ft
1 ft
= 14.2 in.
The footing is 19.5 in. > 14.2 in. so it will work.
ExampleSquareFooting
Calculate the bending moment of the footing at the
edge of the column
1 ft
18 in
L c 10 ft
12 in = 4.25 ft
=
2
2
2 2
L c
( 4.25 ft ) (10 ft )
L c 2 2
M u = qn
b = 6.14 k/ft ( 4.25 ft )
2
2
2 2
= 554.5 k - ft
ExampleSquareFooting
Calculate Ru for the footing to find of the footing.
12 in.
554.5 k - ft *
Mu
1 ft
Ru = 2 =
bd
(120 in ) * (19.5 in ) 2
= 0.1458 ksi
ExampleSquareFooting
From Ru for the footing the value can be found.
1.7 Ru
Ru = f c (1 0.59 ) 1.7 +
=0
f c
2
0.1458 ksi
1.7 (1.7 ) 41.7
0.9( 4 ksi )
=
= 0.04152
2
fy
0.04152( 4 ksi )
= 0.04152 =
= 0.00277
fc
60 ksi
2
ExampleSquareFooting
Compute the area of steel needed
12 in.
As = bd = 0.0027710 ft
(19.5 in.) = 6.48 in 2
1 ft
The minimum amount of steel for shrinkage is
As = 0.0018 bh = 0.0018(120 in.) ( 24 in.) = 5.18 in 2
The minimum amount of steel for flexure is
200
200
(120 in.) (19.5 in.) = 7.8 in 2
As =
bd =
fy
60000
Use
ExampleSquareFooting
Use a #7 bar (0.60 in2) Compute the number of bars
need
As
7.8 in 2
n=
=
= 13 Use 13 bars
Ab 0.60 in 2
Determine the spacing between bars
s=
L 2 * cover
( n 1)
=
120 in - 2( 3 in )
12
= 9.5 in
ExampleSquareFooting
Check the bearing stress. The bearing strength N1, at
the base of the column, 18 in x 18 in., = 0.65
(
)
N1 = ( 0.85 f c A1 ) = 0.65 0.85( 4 ksi ) (18 in ) = 716 k
2
The bearing strength, N2, at the top of the footing is
N 2 = N1
A2
2 N1
A1
ExampleSquareFooting
A2 = (10 ft ) = 100 ft 2
2
2
1 ft
= 2.25 ft 2
A1 = 18 in
12 in.
The bearing strength, N2, at the top of the footing is
A2
100 ft 2
=
= 6.67 > 2 N 2 = 2 N1 = 2( 716 k ) = 1432 k
2
A1
2.25 ft
ExampleSquareFooting
Pu =614 k < N1, bearing stress is adequate. The
minimum area of dowels is required.
0.005 A1 = 0.005 * (18 in ) = 1.62 in 2
2
Use minimum number of bars is 4, so use 4 # 8 bars
placed at the four corners of the column.
ExampleSquareFooting
The development length of the dowels in compression
from ACI Code 12.3.2 for compression.
ld =
0.02d b f y
fc
=
0.02(1 in ) ( 60000 psi )
4000 psi
= 18.97 in Use 19 in
The minimum ld , which has to be greater than 8 in., is
ld = 0.0003d b f y = 0.0003(1 in ) ( 60000 psi ) = 18 in 8 in
ExampleSquareFooting
Therefore, use 4#8 dowels in the corners of
the column extending 19 in. into the column
and the footing. Note that ld is less than the
given d = 19.5 in., which is sufficient
development length.
ExampleSquareFooting
The development length, ld for the #7 bars for the
reinforcement of the footing.
fy
f ydb
ld
( 60000 psi ) ( 0.875 in ) = 41.5 in
=
ld =
=
d b 20 f c
20 f c
20 4000 psi
There is adequate development length provided.
ld =
L
2
cover
c
2
=
120 in
2
3 in
18 in
2
= 48 i n
ExampleSquareFooting
FinalDesign
ExampleRestrictedFooting
Design a footing to support a 18 in. square column
tied interior column reinforced with 8 #9 bars.
The column carries an unfactored axial dead load
of 245 k and an axial live load of 200 k. The base
of the footing is 4 ft. below final grade and
allowable soil pressure is 5 k/ft2 Use fc = 3 ksi and
fy = 60 ksi. Limit one side of the footing to 8.5 ft.
ExampleRestrictedFooting
Assume a depth of footing. (2 ft or 24 in.) The
weight of concrete and the soil are:
Wc = d = 150 lb/ft * 24 in. *
3
1 ft.
12 in.
= 300 lb/ft 2
1 ft.
= 200 lb/ft 2
Ws = s d s = 100 lb/ft * 4 ft 24 in. *
12 in.
3
ExampleRestrictedFooting
The effective soil pressure is given as:
qeff = qs Wc Ws
= 5000 lb/ft 2 300 lb/ft 2 200 lb/ft 2
= 4500 lb/ft 2 4.5 k/ft 2
ExampleRestrictedFooting
Calculate the size of the footing:
Actual Loads = DL + LL = 245 k + 200 k = 445 k
445 k
Area of footing =
= 98.9 ft 2
4.5 k/ft 2
98.9 ft 2
Side of footing =
= 11.64 ft Use 12 ft
8.5 ft
ExampleRestrictedFooting
Calculate net upward pressure:
Actual Loads = 1.2 DL + 1.6 LL
= 1.2( 245 k ) + 1.6( 200 k )
= 614 k
614 k
Net upward pressure qn =
( 8.5 ft ) (12 ft )
= 6.02 k / ft
2
ExampleRestrictedFooting
Calculate the depth of the reinforcement use # 8 bars
with a crisscrossing layering.
d = h cover 1.5d b
d = 24 in. 3 in 1.5(1.0 in )
= 19.5 in.
ExampleRestrictedFooting
The depth of the footing can be calculated
by using the one-way shear (long direction)
1 ft
18 in
L c
12 ft
12 in 19.5 in 1 ft
d =
2
2
2 2
12 in
= 3.625 ft
Vu =135.5 k in
short direction
L c
Vu = qn ( l2 ) d
2 2
= 6.02 k/ft 2 ( 8.5 ft ) ( 3.625 ft ) = 185.5 k
ExampleRestrictedFooting
The depth of the footing can be calculated by using
one-way shear design
1000 lb
185.5 k
Vu
1k
d=
=
2 fc b
12 in
0.75 2 4000 8.5 ft
1 ft
= 19.2 in.
The footing is 19.5 in. > 19.2 in. so it will work.
ExampleRestrictedFooting
Calculate perimeter for two-way shear or
punch out shear. The column is 18 in.
square.
bo = 4( c + d )
= 4(18 in. + 19.5 in.) = 150 in.
1 ft
= 3.125 ft
c + d = (18 in. + 19.5 in.)
12 in
ExampleRestrictedFooting
Calculate the shear Vu
Vu = Pu qn ( c + d )
2
= 614 k 6.02 k/ft ( 3.125 ft ) = 555.2 k
2
2
The shape parameter
=
12 ft
8.5 ft
= 1.41
ExampleRestrictedFooting
Calculate d from the shear capacity according to
11.12.2.1 chose the largest value of d.
4
Vc = 2 + f c b0 d
c
d
Vc = s + 2 f c b0 d
bo
s is 40 for interior, 30 for edge
and 20 for corner column
Vc = 4 f c b0 d
ExampleRestrictedFooting
The depth of the footing can be calculated for the
two way shear
1000 lb
555.2 k
Vu
1k
d=
=
4
4
2 + f c b0 0.75 2 +
4000 (150 in )
1.41
= 16.13 in.
ExampleRestrictedFooting
The third equation bo is dependent on d so use the
assumed values and you will find that d is smaller and
= 40
V
d=
u
40d
b + 2 f c b0
o
1000 lb
555.2 k
1k
=
40(19.5 in )
0.75
+ 2 4000 (150 in )
150 in
= 10.84 in.
(
)
ExampleRestrictedFooting
The depth of the footing can be calculated by using
the two way shear
1000 lb
555.2 k
Vu
1 k = 19.5 in.
d=
=
4 f c b0 0.75 4 4000 (150 in )
(
)
ExampleRestrictedFooting
Calculate the bending moment of the footing at the
edge of the column (long direction)
1 ft
18 in
L c 12 ft
12 in = 5.25 ft
=
2
2
2 2
L c
( 5.25 ft ) ( 8.5 ft )
L c 2 2
M u = qn
b = 6.02 k/ft ( 5.25 ft )
2
2
2 2
= 705.2 k - ft
ExampleRestrictedFooting
Calculate Ru for the footing to find of the footing.
12 in.
705.2 k - ft *
Mu
1 ft
Ru = 2 =
bd
12 in
2
8.5 ft
* (19.5 in )
1 ft
= 0.2182 ksi
ExampleRestrictedFooting
Use the Ru for the footing to find .
1.7 Ru
Ru = f c (1 0.59 ) 1.7 +
=0
f c
2
0.2182 ksi
1.7 (1.7 ) 41.7
0.9( 4 ksi )
=
= 0.06294
2
fy
0.06294( 4 ksi )
= 0.06294 =
= 0.004196
fc
60 ksi
2
ExampleRestrictedFooting
Compute the amount of steel needed
12 in.
As = bd = 0.004196 8.5 ft
(19.5 in.) = 8.35 in 2
1 ft
The minimum amount of steel for shrinkage is
As = 0.0018 bh = 0.0018(102 in.) ( 24 in.) = 4.41 in 2
The minimum amount of steel for flexure is
200
200
(102 in.) (19.5 in.) = 6.63 in 2
As =
bd =
fy
60000
ExampleRestrictedFooting
Use As =8.36 in2 with #8 bars (0.79 in2). Compute
the number of bars need
As 9.33 in 2
n=
=
= 11.8 Use 12 bars
Ab 0.79 in 2
Determine the spacing between bars
s=
L 2 * cover
( n 1)
=
102 in - 2( 3 in )
11
= 8.73 in
ExampleRestrictedFooting
Calculate the bending moment of the footing at the
edge of the column for short length
1 ft
18 in
L c 8.5 ft
12 in = 3.5 ft
=
2
2
2 2
L c
( 3.5 ft ) (12 ft )
L c 2 2
M u = qn
b = 6.02 k/ft ( 3.5 ft )
2
2
2 2
= 442.5 k - ft
ExampleRestrictedFooting
Calculate Ru for the footing to find of the footing.
12 in.
442.5 k - ft *
Mu
1 ft = 0.0970 ksi
Ru = 2 =
bd
12 in
2
12 ft
* (19.5 in )
1 ft
ExampleRestrictedFooting
Use Ru for the footing to find .
1.7 Ru
Ru = f c (1 0.59 ) 1.7 +
=0
f c
2
0.0970 ksi
1.7 (1.7 ) 41.7
0.9( 4 ksi )
=
= 0.0274
2
fy
0.0274( 4 ksi )
= 0.0274 =
= 0.00183
fc
60 ksi
2
ExampleRestrictedFooting
Compute the amount of steel needed
12 in.
As = bd = 0.0018312 ft
(19.5 in.) = 5.12 in 2
1 ft
The minimum amount of steel for shrinkage is
As = 0.0018 bh = 0.0018(144 in.) ( 24 in.) = 6.22 in 2
The minimum amount of steel for flexure is
200
200
(144 in.) (19.5 in.) = 9.36 in 2
As =
bd =
fy
60000
ExampleRestrictedFooting
Use As =9.36 in2 with #6 bar (0.44 in2) Compute the
number of bars need
As 9.36 in 2
n=
=
= 21.3 Use 22 bars
Ab 0.44 in 2
Calculate the reinforcement bandwidth
Reinforcement in bandwidth
= 2 = 2 = 0.83
Total reinforcement
+ 1 1.41 + 1
ExampleRestrictedFooting
The number of bars in the 8.5 ft band is 0.83(22)=19 bars .
Total # bars - band bars
outside # bar =
2
22 1.5 19
=
= Use 2 bars
2
So place 19 bars in 8.5 ft section and 2 bars in each in
(12ft -8.5ft)/2 =1.75 ft of the band.
ExampleRestrictedFooting
Determine the spacing between bars for the band of 8.5 ft
s=
L
( n 1)
=
102 in
18
= 5.67 in
Determine the spacing between bars outside the band
s=
L cover
n
=
21 in - 3in
2
= 9 in
ExampleRestrictedFooting
Check the bearing stress. The bearing strength N1, at
the base of the column, 18 in x 18 in., = 0.65
(
)
N1 = ( 0.85 f c A1 ) = 0.65 0.85( 4 ksi ) (18 in ) = 716 k
2
The bearing strength, N2, at the top of the footing is
N 2 = N1
A2
2 N1
A1
ExampleRestrictedFooting
A2 = ( 8.5 ft ) (12 ft ) = 102 ft 2
2
1 ft
2
A1 = 18 in
= 2.25 ft
12 in.
The bearing strength, N2, at the top of the footing is
A2
102 ft 2
=
= 6.74 > 2 N 2 = 2 N1 = 2( 716 k ) = 1432 k
2
A1
2.25 ft
ExampleRestrictedFooting
Pu =614 k < N1, bearing stress is adequate. The
minimum area of dowels is required.
0.005 A1 = 0.005 * (18 in ) = 1.62 in 2
2
Use minimum number of bars is 4, so use 4 # 8 bars
placed at the four corners of the column.
ExampleRestrictedFooting
The development length of the dowels in compression
from ACI Code 12.3.2 for compression.
ld =
0.02d b f y
fc
=
0.02(1 in ) ( 60000 psi )
4000 psi
= 18.97 in Use 19 in
The minimum ld , which has to be greater than 8 in., is
ld = 0.0003d b f y = 0.0003(1 in ) ( 60000 psi ) = 18 in 8 in
ExampleRestrictedFooting
Therefore, use 4#8 dowels in the corners of
the column extending 19 in. into the column
and the footing. Note that ld is less than the
given d = 19.5 in., which is sufficient
development length.
ExampleRestrictedFooting
The development length, ld for the #8 bars
ld
db
=
fy
20 f c
ld =
f ydb
20 f c
( 60000 psi ) (1.0 in ) = 47.4 in
=
20 4000 psi
There is adequate development length provided.
ld =
L
2
cover
c
2
=
144 in
2
3 in
18 in
2
= 60 in
ExampleRestrictedFooting
The development length, ld for the #6 bars
ld
db
=
fy
25 f c
ld =
f ydb
25 f c
( 60000 psi ) ( 0.75 in ) = 28.5 in
=
25 4000 psi
There is adequate development length provided.
ld =
L
2
cover
c
2
=
102 in
2
3 in
18 in
2
= 39 in
ExampleRestrictedFooting
Finaldesign
23 #6
12 #8
ExampleMultiColumnFooting
Design a rectangular footing to support two square
columns. The exterior column (I) has a section 16 x
16 in., which carries DL of 180 k and a LL of 120 k.
The interior column (II) has a section of 20 x 20 in.,
which carries a DL of 250 k
and a LL of 140 k. The base of
the footing is 5 ft. below final
grade and allowable soil
pressure is 5 k/ft2 Use fc = 4 ksi
and fy = 60 ksi The external
column is located 2 ft from the
property line.
ExampleMultiColumnFooting
Determine the location of an equivalent point and
its location select the datum at column I
x F
x=
F
i
i
i
=
16 ft ( 250 k + 140 k ) + 0 ft (180 k + 120 k )
( 250 k + 140 k ) + (180 k + 120 k )
= 9.04 ft. Use 9 ft.
Extend the footing up to the property line, so the length
is l = 9 ft + 2 ft. = 11 ft. So the length of the footing is
2(11 ft.) = 22 ft.
ExampleMultiColumnFooting
Assume a depth of footing. (36 in.) The weight
of concrete and the soil are:
Wc = d = 150 lb/ft * 36 in. *
3
1 ft.
12 in.
= 450 lb/ft 2
1 ft.
5 ft 36 in. *
= 200 lb/ft 2
Ws = s d s = 100 lb/ft *
12 in.
3
ExampleMultiColumnFooting
The effective soil pressure is given as:
qeff = qs Wc Ws
= 5000 lb/ft 2 450 lb/ft 2 200 lb/ft 2
= 4350 lb/ft 2 4.35 k/ft 2
ExampleMultiColumnFooting
Calculate the size of the footing:
Actual Loads = DL + LL = 250 k + 140 k = 390 k
Actual Loads = DL + LL = 180 k + 120 k = 300 k
Total Loads = AL1 + AL2 = 390 k + 300 k = 690 k
Area of footing =
Side of footing =
690 k
4.35 k/ft 2
158.6 ft 2
22 ft
= 158.6 ft
2
= 7.21 ft Use 7.5 ft
ExampleMultiColumnFooting
Calculate net upward pressure:
Actual Loads = 1.4 DL + 1.7 LL
= 1.4(180 k ) + 1.7(120 k )
+ 1.4( 250 k ) + 1.7(140 k )
= 456 k + 588 k
= 1044 k
Net upward pressure qn =
1044 k
( 22 ft ) ( 7.5 ft )
= 6.33 k / ft
2
ExampleMultiColumnFooting
Calculate the depth of the reinforcement use # 8 bars
with a crisscrossing layering.
d = h cover 1.5d b
d = 36 in. 3 in 1.5(1.0 in )
= 31.5 in.
ExampleMultiColumnFooting
Compute the shear and bending moment diagrams.
Shear Forces
V ( x ) = qb ( x ) w
400
= 6.32 k/ft 2 ( 7.5 ft ) x w
358.7 k
300
= 47.454 k/ft ( x ) w
Force (kips)
200
63.3 k
100
0
-100
0
2
4
6
8
10
12
16
18
-150.3 k
-200
-300
14
-329.5 k
-400
location (ft)
20
22
The columns are
considered point loads
but shear values are
taken at each side of the
column.
ExampleMultiColumnFooting
The location of the maximum moment is
1 ft
1 ft
10 in
= 14.5 ft
16 ft 8 in
12 in
12 in
x=
329.5 k
329.5 k + 358.7 k
(14.5 ft ) = 6.9 ft
ExampleMultiColumnFooting
Compute the shear and bending moment diagrams.
x2
M ( x ) = qb wi ( x xi )
2
Bending Moment
400
Bending Moment (k-ft)
200
42.2 k-ft
0
-200 0
2
4
6
8
10
12
14
16
18
20
-400
-600
-800
-1000
-1200
-1278.9 k-ft @ 9.61 ft
-1400
Location (ft)
x2
= 6.32 k/ft ( 7.5 ft )
wi ( x xi )
2
x2
= 47.454 k/ft wi ( x xi )
2
2
249.9 k-ft
22
The columns are
considered point loads
but moments are taken
at each side of the
column. It will not
balance because center
is at 9.04 ft
ExampleMultiColumnFooting
The maximum shear force occurs at the edge of
the 20 in. column. So maximum shear is measured
at distance d from the column.
1 ft
Vmax q ( d ) = 358.7 k 47.454 k/ft 31.5 in
12 in
= 234.1 k
ExampleMultiColumnFooting
The depth of the footing can be calculated by using
one-way shear
1000 lb
234.1 k
Vu
1k
d=
=
= 24.2 in.
12 in
2 fc b
0.85 2 4000 7.5 ft
1 ft
The footing is 31.5 in. > 24.2 in. so it will work.
ExampleMultiColumnFooting
Calculate perimeter for two-way shear or
punch out shear. The column is 20 in.
square.
bo = 4( c + d )
= 4( 20 in. + 31.5 in.) = 206 in.
1 ft
= 4.292 ft
c + d = ( 20 in. + 31.5 in.)
12 in
ExampleMultiColumnFooting
Calculate the shear Vu
Vu = Pu qn ( c + d )
2
= 588 k 6.70 k/ft ( 4.292 ft )
= 464.6 k
2
2
The other column will not be critical,
Pu = 456 k for the 16 in. column
ExampleMultiColumnFooting
The depth of the footing can be calculated by using
two way shear
d=
Vu
4 f c b0
=
1000 lb
464.6 k
1k
(
0.85 4 4000 ( 206 in )
)
= 10.5 in.
ExampleMultiColumnFooting
Calculate Ru for the footing to find of the footing.
12 in.
1278.9 k - ft *
Mu
1 ft = 0.1719 ksi
Ru =
=
( 90 in ) * ( 31.5 in ) 2
bd 2
ExampleMultiColumnFooting
From Ru for the footing the value can be found.
Ru = f c (1 0.59 ) 1.7 +
2
1.7 Ru
f c
0.1719 ksi
1.7 (1.7 ) 41.7
0.9( 4 ksi )
=0
2
=
fy
fc
2
= 0.04917 =
0.04917( 4 ksi )
60 ksi
= 0.04917
= 0.00328
ExampleMultiColumnFooting
Compute the area of steel needed
12 in.
( 31.5 in.) = 9.29 in 2
As = bd = 0.00277 7.5 ft
1 ft
The minimum amount of steel for shrinkage is
As = 0.0018 bh = 0.0018( 90 in.) ( 36 in.) = 5.8 3 in 2
The minimum amount of steel for flexure is
200
200
( 90 in.) ( 31.5 in.) = 9.45 in 2
As =
bd =
fy
60000
Use
ExampleMultiColumnFooting
Use a #9 bar (1.00 in2) Compute the number of bars
needed A 9.45 in 2
n= s =
= 9.45 Use 10 bars
Ab 1.0 in 2
Determine the spacing between bars
s=
L 2 * cover
( n 1)
=
90 in - 2( 3 in )
9
= 9.33 in
ExampleMultiColumnFooting
The minimum amount is steel is going to be due to the
flexural restrictions. So below the columns with
positive moment, the reinforcement will be 10 # 9 bars
running longitudinally. The development length will
have to be calculated.
ExampleMultiColumnFooting
The development length, ld for the #7 bars for the
reinforcement of the footing.
fy
f ydb
ld
( 60000 psi ) (1.128 in ) = 53.5 in
=
ld =
=
d b 20 f c
20 f c
20 4000 psi
The bars have more than 12-in. of concrete below
them, therefore ld = 1.3 ld.
ld = 1.3( 53.5 in ) = 69.6 in Use 70 in.
ExampleMultiColumnFooting
To determine the reinforcement in the short direction.
The bandwidth of the two columns must be determined
for the 16 in. column.
2 ft 16 in 1 ft + 31.5 in 1 ft = 5.3 ft Use 5.5 ft
Band = 16 in +
2 12 in.
12 in.
Compute the moment at the edge
qnet =
456 k
7.5 ft
= 60.8 k/ft
L=
7.5 ft
2
1 ft
= 3.08 ft
8 in
12 in
ExampleMultiColumnFooting
The bending moment will be
M u = qnet
l2
2
( 3.08 ft ) 2
= ( 60.8 k/ft )
2
Compute the Ru
= 289.0 k - ft
12 in.
289 k - ft *
Mu
1 ft
Ru =
=
= 0.053 ksi
12 in
bd 2
5.5 ft
* ( 31.5 in ) 2
1 ft
ExampleMultiColumnFooting
From Ru for the footing the value can be found.
Ru = f c (1 0.59 ) 1.7 +
2
1.7 Ru
f c
0.053 ksi
1.7 (1.7 ) 41.7
0.9( 4 ksi )
=0
2
=
fy
fc
2
= 0.01484 =
0.04917( 4 ksi )
60 ksi
= 0.01484
= 0.001
ExampleMultiColumnFooting
Compute the area of steel needed
12 in.
( 31.5 in.) = 2.08 in 2
As = bd = 0.001 5.5 ft
1 ft
The minimum amount of steel for shrinkage is
As = 0.0018 bh = 0.0018( 66 in.) ( 36 in.) = 4.28 in 2
The minimum amount of steel for flexure is
200
200
( 66 in.) ( 31.5 in.) = 6.93 in 2
As =
bd =
fy
60000
Use
ExampleMultiColumnFooting
Use a #9 bar (1.00 in2) Compute the number of bars
needed A 6.93 in 2
n= s =
= 6.93 Use 7 bars
Ab 1.0 in 2
Determine the spacing between bars
s=
L cover
( n 1)
=
66 in - ( 3 in )
6
= 10.5 in
ExampleMultiColumnFooting
To determine the reinforcement in the short direction.
The 20-in. interior column extends beyond 4 ft from the
center therefore the band is 7.5 ft x 7.5 ft. Compute the
moment at the edge
qnet =
588 k
7.5 ft
= 78.4 k/ft
L=
7.5 ft
2
1 ft
= 2.92 ft
10 in
12 in
ExampleMultiColumnFooting
The bending moment will be
M u = qnet
l2
2
( 2.92 ft ) 2
= ( 78.4 k/ft )
2
Compute the Ru
= 334.3 k - ft
12 in.
334.3 k - ft *
Mu
1 ft
Ru =
=
= 0.045 ksi
12 in
bd 2
7.5 ft
* ( 31.5 in ) 2
1 ft
ExampleMultiColumnFooting
From Ru for the footing the value can be found.
Ru = f c (1 0.59 ) 1.7 +
2
1.7 Ru
f c
0.045 ksi
1.7 (1.7 ) 41.7
0.9( 4 ksi )
=0
2
=
fy
fc
2
= 0.01257 =
0.01257( 4 ksi )
60 ksi
= 0.01257
= 0.00084
ExampleMultiColumnFooting
Compute the area of steel needed
12 in.
( 31.5 in.) = 2.38 in 2
As = bd = 0.00084 7.5 ft
1 ft
The minimum amount of steel for shrinkage is
As = 0.0018 bh = 0.0018( 90 in.) ( 36 in.) = 5.83 in 2
The minimum amount of steel for flexure is
200
200
( 90 in.) ( 31.5 in.) = 9.45 in 2
As =
bd =
fy
60000
Use
ExampleMultiColumnFooting
Check the bearing stress. The bearing strength N1, at
the base of the column, 16 in x 16 in., = 0.7
(
)
N1 = ( 0.85 f c A1 ) = 0.7 0.85( 4 ksi ) (16 in ) = 609 k
2
The bearing strength, N2, at the top of the footing is
N 2 = N1
A2
2 N1
A1
ExampleMultiColumnFooting
2
A2 = ( 5.5 ft ) = 30.25 ft 2
2
1 ft
= 1.78 ft 2
A1 = 16 in
12 in.
The bearing strength, N2, at the top of the footing is
A2
=
A1
30.25 ft 2
1.78 ft 2
= 4.125 > 2 N 2 = 2 N1 = 2( 609 k ) = 1218 k
ExampleMultiColumnFooting
Pu =456 k < N1, bearing stress is adequate. The
minimum area of dowels is required.
0.005 A1 = 0.005 * (16 in ) = 1.28 in 2
2
Use minimum number of bars is 4, so use 4 # 7 bars
placed at the four corners of the column.
Note if the Pu > N1 the area of steel will be
As
( Pu N1 )
=
fy
As long as the area of
steel is greater than the
minimum amount.
ExampleMultiColumnFooting
The development length of the dowels in compression
from ACI Code 12.3.2 for compression.
ld =
0.02d b f y
fc
=
0.02( 0.875 in ) ( 60000 psi )
4000 psi
= 16.6 in Use 17 in
The minimum ld , which has to be greater than 8 in., is
ld = 0.0003d b f y = 0.0003( 0.875 in ) ( 60000 psi ) = 15.75 in 8 in
ExampleMultiColumnFooting
Therefore, use 4#7 dowels in the corners of
the column extending 17 in. into the column
and the footing. Note that ld is less than the
given d = 31.5 in., which is sufficient
development length.
ExampleMultiColumnFooting
Use a #9 bar (1.00 in2) Compute the number of bars
need
As 9.45 in 2
n=
=
= 9.45 Use 10 bars
Ab 1.0 in 2
Determine the spacing between bars
s=
L cover
( n 1)
=
90 in - ( 3 in )
9
= 9.67 in
ExampleMultiColumnFooting
Check the bearing stress. The bearing strength N1, at
the base of the column, 20 in x 20 in., = 0.7
(
)
N1 = ( 0.85 f c A1 ) = 0.7 0.85( 4 ksi ) ( 20 in ) = 952 k
2
The bearing strength, N2, at the top of the footing is
N 2 = N1
A2
2 N1
A1
ExampleMultiColumnFooting
2
A2 = ( 7.5 ft ) = 56.25 ft 2
2
1 ft
= 2.78 ft 2
A1 = 20 in
12 in.
The bearing strength, N2, at the top of the footing is
A2
A1
=
56.25 ft 2
2.78 ft 2
= 4.5 > 2 N 2 = 2 N1 = 2( 952 k ) = 1904 k
ExampleMultiColumnFooting
Pu =588 k < N1, bearing stress is adequate. The
minimum area of dowels is required.
0.005 A1 = 0.005 * ( 20 in ) = 2.0 in 2
2
Use minimum number of bars is 4, so use 4 # 8 bars
placed at the four corners of the column.
ExampleMultiColumnFooting
The development length of the dowels in compression
from ACI Code 12.3.2 for compression.
ld =
0.02d b f y
fc
=
0.02(1 in ) ( 60000 psi )
4000 psi
= 18.7 in Use 19 in
The minimum ld , which has to be greater than 8 in., is
ld = 0.0003d b f y = 0.0003(1 in ) ( 60000 psi ) = 18 in 8 in
ExampleMultiColumnFooting
Therefore, use 4#8 dowels in the corners of
the column extending 19 in. into the column
and the footing. Note that ld is less than the
given d = 31.5 in., which is sufficient
development length.
ExampleSettlement
Determine the footing areas required for equal
settlement (balanced footing design) if the usual live
load is 25% for all footings. The footings are subjected
to dead loads and live loads as indicated by the table.
The net soil pressure is 6 ksf.
Footing Number
1
Dead Load (kips)
Live Load (kips)
2
3
4
5
120
180
140
190
210
150
220
200
170
240
ExampleSettlement
Find the ratio of the live load to dead load, the largest
ratio will control the settlement.
Column 3 has the largest ratio.
ExampleSettlement
Compute the usual loading for the footing, DL + 0.25LL
Column 3 has the largest ratio.
ExampleSettlement
Determine the need area for the footing with the
largest LL/DL ratio.
A=
DL + LL
qnet
=
140 k + 200 k
6 k/ft 2
= 56.67 ft 2
The usual net soil pressure acting on the footing is
qnet =
DL + ( % live load ) ( LL )
A
=
140 k + 0.25( 200 k )
56.67 ft 2
= 3.353 k/ft 2
ExampleSettlement
Use the qnet (3.353 k/ft2) to determine need area for each
of the other footings to have the same settlement.
Usual Loading
157.5 k
Area =
=
= 46.97 ft 2
q net
3.353 k/ft 2
Compute the qnet for each of the footings
q net
Total Load (120 k ) + (150 k )
=
=
= 5.75 k/ft 2
New Area
46.97 ft 2
ExampleSettlement
Use the qnet (3.353 k/ft2) to determine need area for each
of the other footings to have the same settlement.
Footing Number
1
3
4
5
Dead Load (kips)
Live Load (kips)
120
150
180
220
140
200
190
170
210
240
Ratio (LL/DL)
1.25
1.22
1.43
0.89
1.14
Usual Loading (kips)
157.5
235.0
190.0
232.5
270.0
Standard Area (ft 2)
45.00
66.67
56.67
60.00
75.00
New Area(ft2)
2
46.97
70.09
56.67
69.34
80.53
New qnet
5.75
5.71
6.00
5.19
5.59
ExampleCombinedLoading
A 12-in. x 24 in. column of an unsymmetrical shed is
subjected to an axial load PD of 220 k and MD = 180 k-ft
due to dead load and an an axial load PL = 165 k and a
moment ML= 140 k-ft due to
live load. The base of the
footing is 5 ft below final
grade, and the allowable soil
bearing pressure is 5 k/ft2.
Design the footing using
fc = 4 ksi and fy = 60 ksi
ExampleCombinedLoading
Find the combined actual loads, P0 and M0
P0 = PDL + PLL = 220 k + 165 k = 385 k
M 0 = M DL + M LL = 180 k - ft + 140 k - ft = 320 k - ft
Determine the eccentricity of the footing
12 in
320 k - ft
M0
1 ft = 9.97 in Use 10 in.
e=
=
P0
385 k
ExampleCombinedLoading
Assume a depth of footing, 24 in. The weight of
concrete and the soil are:
Wc = d = 150 lb/ft * 24 in. *
3
1 ft.
12 in.
= 300 lb/ft 2
1 ft.
5 ft 24 in. *
= 300 lb/ft 2
Ws = s d s = 100 lb/ft *
12 in.
3
ExampleCombinedLoading
The effective soil pressure is given as:
qeff = qs Wc Ws
= 5000 lb/ft 2 300 lb/ft 2 300 lb/ft 2
= 4400 lb/ft 2 4.4 k/ft 2
ExampleCombinedLoading
Calculate the size of the footing:
Actual Loads = DL + LL = 385 k
Area of footing =
385 k
4.4 k/ft
2
= 87.5 ft 2
Compute the sizes of the footing if width is 9 ft.
Side of footing =
87.5 ft 2
9 ft
= 9.72 ft Use 10 ft
ExampleCombinedLoading
Use the long section and place the column 10 in.
off-center for the 10 ft segment
ExampleCombinedLoading
Calculate net upward pressure:
Actual Loads = 1.2 DL + 1.6 LL
= 1.2( 220 k ) + 1.6(165 k )
= 528.0 k
528.0 k
Net upward pressure qn =
( 9 ft ) (10 ft )
= 5.87 k / ft
2
ExampleCombinedLoading
Calculate the depth of the reinforcement use # 8 bars
with a crisscrossing layering.
d = h cover 1.5d b
d = 24 in. 3 in 1.5(1.0 in )
= 19.5 in.
ExampleCombinedLoading
The depth of the footing can be calculated by using the
one-way shear (long direction)
1 ft
24 in
10 ft
L c
12 in 19.5 in 1 ft + 10 in 1 ft
d+e =
2
2
2 2
12 in
12 in
= 3.208 ft
Vu =139.4 k in
short direction
L c
Vu = qn ( l2 ) d + e
2 2
= 5.87 k/ft 2 ( 9 ft ) ( 3.208 ft ) = 169.4 k
ExampleCombinedLoading
The depth of the footing can be calculated by using
one-way shear design
1000 lb
169.4 k
Vu
1k
d=
=
= 16.53 in.
2 fc b
12 in
0.75 2 4000 9 ft
1 ft
The footing is 19.5 in. > 16.53 in. so it will work.
ExampleCombinedLoading
Calculate perimeter for two-way shear or punch out
shear. The column is 12 in. x 24 in.
bo = 2( c1 + d ) + 2( c2 + d )
= 2(12 in. + 19.5 in.) + 2( 24 in. + 19.5 in.) = 150 in.
1 ft
= 2.625 ft
c1 + d = (12 in. + 19.5 in.)
12 in
1 ft
= 3.625 ft
c2 + d = ( 24 in. + 19.5 in.)
12 in
ExampleCombinedLoading
Calculate the shear Vu
(c + d )2
Vu = Pu qn
= 528.0 k 5.87 k/ft 2 ( 2.625 ft ) ( 3.625 ft )
= 472.2 k
The shape parameter
=
10 ft
9 ft
= 1.11
ExampleCombinedLoading
Calculate d from the shear capacity according to
11.12.2.1 chose the largest value of d.
4
Vc = 2 + f c b0 d
c
d
Vc = s + 2 f c b0 d
bo
s is 40 for interior, 30 for edge
and 20 for corner column
Vc = 4 f c b0 d
ExampleCombinedLoading
The depth of the footing can be calculated for the
two way shear
1000 lb
472.2 k
Vu
1k
d=
=
4
4
2 + f c b0 0.75 2 +
4000 (150 in )
1.11
= 11.84 in.
ExampleCombinedLoading
The third equation bo is dependent on d so use the
assumed values and you will find that d is smaller and
= 40
d=
Vu
40d
+ 2 f c b0
b
o
1000 lb
472.2 k
1k
=
= 9.22 in.
40(19.5 in )
0.75
+ 2 4000 (150 in )
150 in
(
)
ExampleCombinedLoading
The depth of the footing can be calculated by using
the two way shear
1000 lb
472.2 k
Vu
1k
d=
=
4 f c b0 0.75 4 4000 (150 in )
(
= 16.59 in.
)
ExampleCombinedLoading
Calculate the bending moment of the footing at the edge of
the column (long direction)
1 ft
24 in
10 ft
L c
12 in + 10 in 1 ft = 4.83 ft
+e =
2
2
2 2
12 in
L c
+ e
L c
2 2
M u = qn + e
b
2
2 2
( 4.83 ft ) ( 9 ft ) = 616.2 k - ft
= 5.87 k/ft ( 4.83 ft )
2
ExampleCombinedLoading
Calculate Ru for the footing to find of the footing.
12 in.
616.2 k - ft *
Mu
1 ft = 0.1801 ksi
Ru = 2 =
bd
12 in
2
9 ft
* (19.5 in )
1 ft
ExampleCombinedLoading
Use the Ru for the footing to find .
1.7 Ru
Ru = f c (1 0.59 ) 1.7 +
=0
f c
2
0.1801 ksi
1.7 (1.7 ) 41.7
0.9( 4 ksi )
=
= 0.05158
2
fy
0.05158( 4 ksi )
= 0.05158 =
= 0.00344
fc
60 ksi
2
ExampleCombinedLoading
Compute the amount of steel needed
12 in.
As = bd = 0.00344 9 ft
(19.5 in.) = 7.24 in 2
1 ft
The minimum amount of steel for shrinkage is
As = 0.0018 bh = 0.0018(108 in.) ( 24 in.) = 4.67 in 2
The minimum amount of steel for flexure is
200
200
(108 in.) (19.5 in.) = 7.02 in 2
As =
bd =
fy
60000
ExampleCombinedLoading
Use As =8.36 in2 with #8 bars (0.79 in2). Compute
the number of bars need
As
8.1 in 2
n=
=
= 10.25 Use 11 bars
Ab 0.79 in 2
Determine the spacing between bars
s=
L 2 * cover
( n 1)
=
108 in - 2( 3 in )
10
= 10.2 in
ExampleCombinedLoading
Calculate the bending moment of the footing at the
edge of the column for short length
1 ft
12 in
L c 9 ft
12 in = 4 ft
=
2
2
2 2
L c
( 4 ft ) (10 ft )
L c 2 2
M u = qn
b = 5.87 k/ft ( 4 ft )
2
2
2 2
= 469.6 k - ft
ExampleCombinedLoading
Calculate Ru for the footing to find of the footing.
12 in.
469.6 k - ft *
Mu
1 ft
Ru = 2 =
bd
12 in
2
10 ft
* (19.5 in )
1 ft
= 0.1235 ksi
ExampleCombinedLoading
Use Ru for the footing to find .
Ru = f c (1 0.59 ) 2 1.7 +
1.7
1.7 Ru
=0
f c
(1.7 ) 2 41.7 0.1235 ksi
4 ksi
=
= 0.03503
2
fy
0.03503( 4 ksi )
= 0.03503 =
= 0.00234
fc
60 ksi
ExampleCombinedLoading
Compute the amount of steel needed
12 in.
As = bd = 0.0023410 ft
(19.5 in.) = 5.46 in 2
1 ft
The minimum amount of steel for shrinkage is
As = 0.0018 bh = 0.0018(120 in.) ( 24 in.) = 5.18 in 2
The minimum amount of steel for flexure is
200
200
(120 in.) (19.5 in.) = 7.80 in 2
As =
bd =
fy
60000
ExampleCombinedLoading
Use As =9.36 in2 with #6 bar (0.44 in2) Compute the
number of bars need
As 7.80 in 2
n=
=
= 17.7 Use 18 bars
Ab 0.44 in 2
Calculate the reinforcement bandwidth
Reinforcement in bandwidth
= 2 = 2 = 0.947
Total reinforcement
+ 1 1.11 + 1
ExampleCombinedLoading
The number of bars in the 9 ft band is 0.947(18)=17 bars .
Total # bars - band bars
outside # bar =
2
18 17
=
= 0.5 Use 1 bars
2
So place 17 bars in 9 ft section and 1 bars in each in
(10ft - 9ft)/2 =0.5 ft of the band.
ExampleCombinedLoading
Determine the spacing between bars for the band of 9 ft
s=
L
( n 1)
=
108 in
16
= 6.75 in
Determine the spacing between bars outside the band
s=
L cover
n
=
6 in - 3in
1
= 3 in
ExampleCombinedLoading
Check the bearing stress. The bearing strength N1, at
the base of the column, 12 in x 24 in., = 0.65
N1 = ( 0.85 f c A1 ) = 0.65( 0.85( 4 ksi ) (12 in ) ( 24 in ) ) = 636.5 k
The bearing strength, N2, at the top of the footing is
N 2 = N1
A2
2 N1
A1
ExampleCombinedLoading
A2 = 9 ft (10 ft ) = 90 ft 2
1 ft
1 ft
24 in
= 2 ft 2
A1 = 12 in
12 in.
12 in.
The bearing strength, N2, at the top of the footing is
A2
90 ft 2
=
= 6.71 > 2 N 2 = 2 N1 = 2( 636.5 k ) = 1273 k
2
A1
2 ft
ExampleCombinedLoading
Pu =628 k < N1, bearing stress is adequate. The
minimum area of dowels is required.
0.005 A1 = 0.005 * (12 in ) ( 24 in ) = 1.44 in 2
Use minimum number of bars is 4, so use 4 # 8 bars
placed at the four corners of the column.
ExampleCombinedLoading
The development length of the dowels in compression
from ACI Code 12.3.2 for compression.
ld =
0.02d b f y
fc
=
0.02(1 in ) ( 60000 psi )
4000 psi
= 18.97 in Use 19 in
The minimum ld , which has to be greater than 8 in., is
ld = 0.0003d b f y = 0.0003(1 in ) ( 60000 psi ) = 18 in 8 in
ExampleCombinedLoading
Therefore, use 4#8 dowels in the corners of
the column extending 19 in. into the column
and the footing. Note that ld is less than the
given d = 19.5 in., which is sufficient
development length.
ExampleCombinedLoading
The development length, ld for the #8 bars
ld
db
=
fy
20 f c
ld =
f ydb
20 f c
( 60000 psi ) (1.0 in ) = 47.4 in
=
20 4000 psi
There is adequate development length provided.
ld =
L
2
cover
c
2
=
144 in
2
3 in
18 in
2
= 60 in
ExampleCombinedLoading
The development length, ld for the #6 bars
ld
db
=
fy
25 f c
ld =
f ydb
25 f c
( 60000 psi ) ( 0.75 in ) = 28.5 in
=
25 4000 psi
There is adequate development length provided.
ld =
L
2
cover
c
2
=
102 in
2
3 in
18 in
2
= 3 9 in