Unformatted text preview: Lecture 26 – Footings
August 8, 2003
CVEN444 Lecture Goals
Footing Classification
Footing Design
Footing Examples Footings
Definition
Footings are 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. Types of Footings
Wall footings are used to
support structural walls that
carry loads for other floors
or to support nonstructural
walls. Types of Footings
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. Types of Footings
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. Types of Footings
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. Types of Footings
Continuous footings
support a row of three or
more columns. They have
limited width and continue
under all columns. Types of Footings
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. Types of Footings
Pile caps are thick slabs
used to tie a group of piles
together to support and
transmit column loads to the
piles. Distribution of Soil Pressure
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. Distribution of Soil Pressure Soil pressure distribution in
cohesive soil. Soil pressure distribution in
cohesionless soil. Design Considerations
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. Twoway shear or punch out shear.
3. Oneway bearing
4. Bending moment and steel reinforcement
required Design Considerations
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 Size of Footings
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 TwoWay Shear (Punching Shear)
For twoway shear in slabs (& footings) Vc is smallest of 2 + 4 f b d
Vc = c0
βc where, β c =
b0 = ACI 1135 long side/short side of column concentrated
load or reaction area < 2
length of critical perimeter around the
column When β > 2 the allowable Vc is reduced. Design of twoway shear
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. Design of twoway shear
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 Design of twoway shear
4. Allowable φVc = 4φ
Let Vu=φ Vc d= f c b0 d Vu
4φ f c b0 If d is not close to the assumed d,
revise your assumptions Design of oneway shear
For footings with bending
action in one direction the
critical section is located a
distance d from face of column φVc = 2φ f c b0 d Design of oneway shear
The ultimate shearing force at
section mm can be calculated L c Vu = qu b − − d 2 2 If no shear reinforcement is to be
used, then d can be checked Design of oneway shear
If no shear reinforcement is
to be used, then d can be
checked, assuming Vu = φ Vc d= Vu
2φ f c b Flexural Strength and Footing reinforcement The bending moment in each
direction of the footing must be
checked and the appropriate
reinforcement must be provided. As = Mu a
φf y d − 2 Flexural Strength and Footing reinforcement Another approach is to
calculated Ru = Mu / bd2 and
determine the steel percentage
required ρ . Determine As then
check if assumed a is close to
calculated a a= f y As 0. 8 5 f c b Flexural Strength and Footing 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. Flexural Strength and Footing reinforcement The reinforcement in oneway footings
and twoway footings must be
distributed across the entire width of
the footing.
Reinforcement in band width
Total reinforcement in short direction where β= = 2 β +1 long side of footing
short side of footing Bearing Capacity of Column at 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 crosssection. 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. Bearing Capacity of Column at Base
Force acting on the concrete at the base of the column
must not exceed the bearing strength of the concrete N1 = φ ( 0.85 f c A1 )
where φ = 0.65 and
A1 =bearing area of column Bearing Capacity of Column at 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 ≤ φ ( 0.85 f c A1 ) A2 / A1
N 2 ≤ 2φ ( 0.85 f c A1 ) Dowels in Footings
A minimum steel ratio ρ = 0.005 of the column section
as compared to ρ = 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. Development length of the Reinforcing Bars
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. Differential Settlement
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 General Requirements for Footing 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. General Requirements for Footing 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. General Requirements for Footing Design
7. Determine the pressure distribution beneath the
footing based on its width
8. Perform a settlement analysis. Example – Wall 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 selfweight 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 Example Wall
Assume a depth of footing. (1.5 ft or 18 in.) The
weight of concrete and the soil are:
1 ft.
3
Wc = γ d = 150 lb/ft *18 in. *
12 in.
= 225 lb/ft 2
1 ft. Ws = γ s ds = 100 lb/ft * 4 ft − 18 in. * 12 in. = 250 lb/ft 2
3 Example Wall
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 Example Wall
Calculate the size of the footing for 1ft 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 Example – Wall 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 Example – Wall 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. Example – Wall The depth of the footing can be calculated by using
oneway 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 Example – Wall
The depth of the footing can be calculated by using
oneway shear 1000 lb 7.1 k Vu 1k d=
= φ 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. Example – Wall 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 kft Example – Wall Calculate Ru for the footing to find ρ of the footing. 12 in. 15.98 kft * Mu
1 ft Ru = 2 =
= 0.0877 ksi
2
bd
( 12 in.) * ( 13.5 in ) Example – Wall From Ru for the footing the ρ value can be found.
1.7 Ru
Ru = ω f c ( 1 − 0.59ω ) ⇒ ω − 1.7ω +
=0
φ fc
2 0.0877 ksi 1.7 − ( 1.7 ) − 4 1.7 0.9 ( 3 ksi ) ω=
= 0.03312
2
ρ fy
0.03312 ( 3 ksi )
= 0.03312 ⇒ ρ =
= 0.00166
fc
60 ksi
2 Example – Wall Compute the area of steel needed 12 in. As = ρ 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 Example – Wall 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 Example – Wall Check the bearing stress. The bearing strength N1, at
the base of the wall, 16 in x 12 in., φ = 0.65
N1 = φ ( 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 Example – Wall 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 Example – Wall 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. Example – Wall 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. Example – Wall 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. Example – Wall 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 Example – Square Footing 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 Example – Square Footing
Assume a depth of footing. (2 ft or 24 in.) The
weight of concrete and the soil are: Wc = γd = 150 lb/ft * 24 in. *
3 1 ft.
12 in. = 300 lb/ft 2 1 ft. = 200 lb/ft 2
Ws = γ s d s = 100 lb/ft * 4 ft − 24 in. * 12 in. 3 Example – Square Footing
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 Example – Square Footing
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 Example – Square Footing
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 Example – Square Footing
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. Example – Square Footing
Calculate perimeter for twoway 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 Example –Square Footing
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 β= 10 ft
10 ft =1 2 Example – Square Footing
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 βc α d Vc = s + 2 f c b0 d bo αs is 40 for interior, 30 for edge
and 20 for corner column Vc = 4 f c b0 d Example – Square Footing
The depth of the footing can be calculated by using
two way shear 1000 lb 554 k Vu 1k d=
=
φ 4 f c b0 0.75 4 4000 (150 in ) ( = 19.47 in. ) Example – Square Footing
The second equation bo is dependent on d so use the
assumed values and you will find that d is smaller and
α = 40
Vu
d= 40d φ 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 ( ) Example – Square Footing
The depth of the footing can be calculated
by using oneway 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 Example – Square Footing
The depth of the footing can be calculated by using
oneway shear 1000 lb 161.2 k Vu 1k d=
= φ 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. Example –Square Footing
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 Example –Square Footing
Calculate Ru for the footing to find ρ of the footing. 12 in. 554.5 k  ft * Mu 1 ft Ru = 2 =
bd
(120 in ) * (19.5 in ) 2
= 0.1458 ksi Example – Square Footing From Ru for the footing the ρ value can be found.
1.7 Ru
Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω +
=0
φf c
2 0.1458 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) ω=
= 0.04152
2
ρ fy
0.04152( 4 ksi )
= 0.04152 ⇒ ρ =
= 0.00277
fc
60 ksi
2 Example – Square Footing
Compute the area of steel needed 12 in. As = ρ 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 Example – Square Footing
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 Example – Square Footing
Check the bearing stress. The bearing strength N1, at
the base of the column, 18 in x 18 in., φ = 0.65 ( ) N1 = φ ( 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 Example – Square Footing
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 Example – Square Footing
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. Example – Square Footing
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 Example – Square Footing
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. Example – Square Footing
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 Example – Square Footing Final Design Example – Restricted Footing
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. Example – Restricted Footing
Assume a depth of footing. (2 ft or 24 in.) The
weight of concrete and the soil are: Wc = γd = 150 lb/ft * 24 in. *
3 1 ft.
12 in. = 300 lb/ft 2 1 ft. = 200 lb/ft 2
Ws = γ s d s = 100 lb/ft * 4 ft − 24 in. * 12 in. 3 Example – Restricted Footing
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 Example – Restricted Footing
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 Example – Restricted Footing
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 Example – Restricted Footing
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. Example – Restricted Footing
The depth of the footing can be calculated
by using the oneway 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 Example – Restricted Footing
The depth of the footing can be calculated by using
oneway shear design 1000 lb 185.5 k Vu 1k d=
= φ 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. Example – Restricted Footing
Calculate perimeter for twoway 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 Example – Restricted Footing
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 β= 12 ft
8.5 ft = 1.41 Example – Restricted Footing
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 βc α d Vc = s + 2 f c b0 d bo αs is 40 for interior, 30 for edge
and 20 for corner column Vc = 4 f c b0 d Example – Restricted Footing
The depth of the footing can be calculated for the
two way shear 1000 lb 555.2 k Vu 1k d=
= 4
4
φ 2 + f c b0 0.75 2 + 4000 (150 in ) β 1.41 = 16.13 in. Example – Restricted Footing
The third equation bo is dependent on d so use the
assumed values and you will find that d is smaller and
α = 40
V
d= u 40d φ 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. ( ) Example – Restricted Footing
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=
=
φ 4 f c b0 0.75 4 4000 (150 in ) ( ) Example – Restricted Footing
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 Example – Restricted Footing
Calculate Ru for the footing to find ρ 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 Example – Restricted Footing
Use the Ru for the footing to find ρ . 1.7 Ru
Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω +
=0
φf c
2 0.2182 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) ω=
= 0.06294
2
ρ fy
0.06294( 4 ksi )
= 0.06294 ⇒ ρ =
= 0.004196
fc
60 ksi
2 Example – Restricted Footing
Compute the amount of steel needed 12 in. As = ρ 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 Example – Restricted Footing
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 Example – Restricted Footing
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 Example – Restricted Footing
Calculate Ru for the footing to find ρ 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 Example – Restricted Footing
Use Ru for the footing to find ρ . 1.7 Ru
Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω +
=0
φf c
2 0.0970 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) ω=
= 0.0274
2
ρ fy
0.0274( 4 ksi )
= 0.0274 ⇒ ρ =
= 0.00183
fc
60 ksi
2 Example – Restricted Footing
Compute the amount of steel needed 12 in. As = ρ 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 Example – Restricted Footing
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 β + 1 1.41 + 1 Example – Restricted Footing
The number of bars in the 8.5 ft band is 0.83(22)=19 bars .
Total # bars  band bars
outside # bar =
2
22 − 19
=
= 1.5 ⇒ 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. Example – Restricted Footing
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 Example – Restricted Footing
Check the bearing stress. The bearing strength N1, at
the base of the column, 18 in x 18 in., φ = 0.65 ( ) N1 = φ ( 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 Example – Restricted Footing
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 Example – Restricted Footing
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. Example – Restricted Footing
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 Example – Restricted Footing
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. Example – Restricted Footing
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 Example – Restricted Footing
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 Example – Restricted Footing Final design 23 #6 12 #8 Example – MultiColumn Footing
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. Example – MultiColumn Footing
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. Example – MultiColumn Footing
Assume a depth of footing. (36 in.) The weight
of concrete and the soil are: Wc = γd = 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 = γ s d s = 100 lb/ft * 12 in. 3 Example – MultiColumn Footing
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 Example – MultiColumn Footing
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 Example – MultiColumn Footing
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 Example – MultiColumn Footing
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. Example – MultiColumn Footing 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. Example – MultiColumn Footing
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 Example – MultiColumn Footing
Compute the shear and bending moment diagrams.
x2
M ( x ) = qb − wi ( x − xi )
2 Bending Moment
400 Bending Moment (kft) 200 42.2 kft 0
200 0 2 4 6 8 10 12 14 16 18 20 400
600
800
1000
1200 1278.9 kft @ 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 kft 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 Example – MultiColumn Footing
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 Example – MultiColumn Footing
The depth of the footing can be calculated by using
oneway shear 1000 lb 234.1 k Vu 1k d=
=
= 24.2 in. 12 in φ 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. Example – MultiColumn Footing
Calculate perimeter for twoway 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 Example – MultiColumn Footing
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 Example – MultiColumn Footing
The depth of the footing can be calculated by using
two way shear d= Vu φ 4 f c b0 = 1000 lb 464.6 k 1k ( 0.85 4 4000 ( 206 in ) ) = 10.5 in. Example – MultiColumn Footing
Calculate Ru for the footing to find ρ of the footing. 12 in. 1278.9 k  ft * Mu 1 ft = 0.1719 ksi
Ru =
=
( 90 in ) * ( 31.5 in ) 2
bd 2 Example – MultiColumn Footing
From Ru for the footing the ρ value can be found.
Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω +
2 1.7 Ru φf c 0.1719 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) =0 2 ω= ρ fy
fc 2
= 0.04917 ⇒ ρ = 0.04917( 4 ksi )
60 ksi = 0.04917
= 0.00328 Example – MultiColumn Footing
Compute the area of steel needed 12 in. ( 31.5 in.) = 9.29 in 2
As = ρ 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 Example – MultiColumn Footing
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 Example – MultiColumn Footing
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. Example – MultiColumn Footing
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 12in. of concrete below
them, therefore ld = 1.3 ld.
ld = 1.3( 53.5 in ) = 69.6 in ⇒ Use 70 in. Example – MultiColumn Footing
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 Example – MultiColumn Footing
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 Example – MultiColumn Footing
From Ru for the footing the ρ value can be found.
Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω +
2 1.7 Ru φf c 0.053 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) =0 2 ω= ρ fy
fc 2
= 0.01484 ⇒ ρ = 0.04917( 4 ksi )
60 ksi = 0.01484
= 0.001 Example – MultiColumn Footing
Compute the area of steel needed 12 in. ( 31.5 in.) = 2.08 in 2
As = ρ 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 Example – MultiColumn Footing
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 Example – MultiColumn Footing
To determine the reinforcement in the short direction.
The 20in. 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 Example – MultiColumn Footing
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 Example – MultiColumn Footing
From Ru for the footing the ρ value can be found.
Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω +
2 1.7 Ru φf c 0.045 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) =0 2 ω= ρ fy
fc 2
= 0.01257 ⇒ ρ = 0.01257( 4 ksi )
60 ksi = 0.01257
= 0.00084 Example – MultiColumn Footing
Compute the area of steel needed 12 in. ( 31.5 in.) = 2.38 in 2
As = ρ 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 Example – MultiColumn Footing
Check the bearing stress. The bearing strength N1, at
the base of the column, 16 in x 16 in., φ = 0.7 ( ) N1 = φ ( 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 Example – MultiColumn Footing
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 Example – MultiColumn Footing
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. Example – MultiColumn Footing
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 Example – MultiColumn Footing
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. Example – MultiColumn Footing
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 Example – MultiColumn Footing
Check the bearing stress. The bearing strength N1, at
the base of the column, 20 in x 20 in., φ = 0.7 ( ) N1 = φ ( 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 Example – MultiColumn Footing
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 Example – MultiColumn Footing
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. Example – MultiColumn Footing
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 Example – MultiColumn Footing
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. Example – Settlement 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 Example – Settlement Find the ratio of the live load to dead load, the largest
ratio will control the settlement. Column 3 has the largest ratio. Example – Settlement Compute the usual loading for the footing, DL + 0.25LL Column 3 has the largest ratio. Example – Settlement 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 Example – Settlement 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 Example – Settlement 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 Example – Combined Loading
A 12in. x 24 in. column of an unsymmetrical shed is
subjected to an axial load PD of 220 k and MD = 180 kft
due to dead load and an an axial load PL = 165 k and a
moment ML= 140 kft 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 Example – Combined Loading
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 Example – Combined Loading
Assume a depth of footing, 24 in. The weight of
concrete and the soil are: Wc = γd = 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 = γ s d s = 100 lb/ft * 12 in. 3 Example – Combined Loading
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 Example – Combined Loading
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 Example – Combined Loading
Use the long section and place the column 10 in.
offcenter for the 10 ft segment Example – Combined Loading
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 Example – Combined Loading
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. Example – Combined Loading
The depth of the footing can be calculated by using the
oneway 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 Example – Combined Loading
The depth of the footing can be calculated by using
oneway shear design 1000 lb 169.4 k Vu 1k d=
=
= 16.53 in. φ 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. Example – Combined Loading
Calculate perimeter for twoway 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 Example – Combined Loading
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 β= 10 ft
9 ft = 1.11 Example – Combined Loading
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 βc α d Vc = s + 2 f c b0 d bo αs is 40 for interior, 30 for edge
and 20 for corner column Vc = 4 f c b0 d Example – Combined Loading
The depth of the footing can be calculated for the
two way shear 1000 lb 472.2 k Vu 1k d=
= 4
4
φ 2 + f c b0 0.75 2 + 4000 (150 in ) β 1.11 = 11.84 in. Example – Combined Loading
The third equation bo is dependent on d so use the
assumed values and you will find that d is smaller and
α = 40
d= Vu 40d φ
+ 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 ( ) Example – Combined Loading
The depth of the footing can be calculated by using
the two way shear 1000 lb 472.2 k Vu 1k d=
=
φ 4 f c b0 0.75 4 4000 (150 in ) ( = 16.59 in. ) Example – Combined Loading
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 Example – Combined Loading
Calculate Ru for the footing to find ρ 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 Example – Combined Loading
Use the Ru for the footing to find ρ .
1.7 Ru
Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω +
=0
φf c
2 0.1801 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) ω=
= 0.05158
2
ρ fy
0.05158( 4 ksi )
= 0.05158 ⇒ ρ =
= 0.00344
fc
60 ksi
2 Example – Combined Loading
Compute the amount of steel needed 12 in. As = ρ 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 Example – Combined Loading
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 Example – Combined Loading
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 Example – Combined Loading
Calculate Ru for the footing to find ρ 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 Example – Combined Loading
Use Ru for the footing to find ρ .
Ru = ωf c (1 − 0.59ω ) ⇒ ω 2 − 1.7ω +
1.7 − 1.7 Ru
=0
φf c (1.7 ) 2 − 41.7 0.1235 ksi 4 ksi ω=
= 0.03503
2
ρ fy
0.03503( 4 ksi )
= 0.03503 ⇒ ρ =
= 0.00234
fc
60 ksi Example – Combined Loading
Compute the amount of steel needed 12 in. As = ρ 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 Example – Combined Loading
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 β + 1 1.11 + 1 Example – Combined Loading
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. Example – Combined Loading
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 Example – Combined Loading
Check the bearing stress. The bearing strength N1, at
the base of the column, 12 in x 24 in., φ = 0.65
N1 = φ ( 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 Example – Combined Loading
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 Example – Combined Loading
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. Example – Combined Loading
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 Example – Combined Loading
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. Example – Combined Loading
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 Example – Combined Loading
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 ...
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This note was uploaded on 10/03/2011 for the course CVEN 444 taught by Professor Staff during the Summer '08 term at Texas A&M.
 Summer '08
 Staff

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