CIVL471 DESIGN OF RC
STRUCTURES
LECTURE NOTE #14
CHAPTER V
FOOTINGS and FOUNDATIONS
(CONT.)
5.6.2 DESIGN PRINCIPLES OF COMBINED FOOTINGS (TS500
REQUIREMENTS)
In combined footings there may be beam parts and plate parts. If there is a
beam part, total height of the beam should not be less than 1/10 of the clear span.
Plate part of a continuous footing may not be thinner than 20 cm. Design shear
force should be calculated at the face of the column. If there is not a beam part in
a continuous footing, plate thickness should not be less than 30 cm.
Requirements of TS500 for the reinforcement of the beams and the plates can
also apply to the beam and plate parts of the footings. There must be
reinforcement in the compression part of the `continuous footing equal to at least
one third of the tension reinforcement.
5.6.3 TWO-COLUMN COMBINED FOOTINGS
N
1
N2
R
n
m
q
ef
b
L / 2
L / 2
L
L = 2(n + m)
b = R / (q
ef
L)
C
C
b
2
b
1
L
c
1
c
2
C
b
1
b
2
b
1
+ b
2
=
Lq
R2
ef
c
1
=
)b+b(3
)b2+b(L
21
21
c
2
=
)b+b(3
)b+b2(L
21
21
L
L
1
L
2
b
1
=
)L+L(L
L-m)+2(n
q
R
211
2
ef
b
2
=
ef2
qL
R
−
1
2
1
b
L
L
L
1
b
1
+ L
2
b
2
=
ef
q
R
Figure 5.25
These footings generally combine two very close columns or two columns
one of which is very close to the next building (or property line). They are
usually designed such that the resultant of the column loads passes through the
centroid of the bearing area. Thus it is assumed that the soil stresses under the
footing are uniformly distributed. The shape of the bearing area may be
rectangular, trapezoidal or T shaped depending on the relative values of the
column loads. In Fig. 5.25 some equations are given to simplify the
determination of the bearing area dimensions for these shapes. These equations
are based on the assumption that the resultant (R) of the column loads passes
through the centroid of the bearing area.
Example 5.4
One end of the footing is limited by the property line which is 30 cm away from
the column center as shown in Fig.5.26. Design this combined footing.
N
1
N2
h
30cm
420 cm
L
40cm
40cm
40cm
40cm
Figure 5.26
b
Dead loads: N
d1
= 235 kN
N
d2
= 365 kN
Live loads: N
l1
= 180 kN
N
l2
= 280 kN
q
a
= 245 kN/m
2
Materials: C18 S420
Solution:
Let h =100 cm >380/10=38cm q
ef
= 245−1*25 = 245 − 25 = 220 kN/m
2
Resultant of the column loads and the location of the resultant force:
N
1
= 235 + 180 =415 kN N
2
= 365 + 280 =645 kN
R = 415 + 645 = 1060 kN
415 kN
645 kN
R
n
m
4.20 m
L
Moment equilibrium:
R.n = 1060n = 645*4.2 = 2709 kN-m
n = 2709 / 1060 = 2.55 m. m + n = 0.3 + 2.55 = 2.85 m
L = 2(n +m) = 2*2.85 = 5.70 m.
Required minimum bearing area and minimum b:
min (bL) =
220
1060
=
q
R
ef
= 4.82 m
2
min b =
85.0=
7.5
82.4
=
L
82.4
m
Selected b = 95 cm.
Two column combined footings may be analyzed like beams. The soil pressure
corresponds to distributed load and column loads correspond to support
reactions. First shear force diagram is drawn and then moments are calculated
from the shear areas. In the following these calculations are given.
Design in long direction:
N
1u
= 1.4*235 + 1.6*180 = 617 kN N
2u
= 1.4*365 + 1.6*280 = 959 kN
R
u
= 617 + 959 = 1576 kN
Design soil pressure per linear meter:
q
u
=
m/kN5.276=
7.5
1576
=
L
R
u
Shear forces and bending moments:
12.44 − 1.93*534.05/2 = − 502.92 kN-m
− 502.92 + 627.25* (4.2 − 1.93) /2 = 209.01 kN-m
0.3m
4.20m
1.20m
617 kN
959 kN
−
−
+
−
+
−
Shears (kN)
Moments (kN-m)
82.95
-534.05
627.25
-331.75
x
o
-502.92
209.01
12.44
Shear forces:
276.5*.0.30 = 82.95 kN
82.95 − 617 = − 534.05 kN
− 534.05 + 276.5*4.2 = 627.25 kN
627.25 − 959 = − 331.75 kN
− 331.75 + 276.5*1.2 = 0
x
o
=
93.1=
5.276
05.534
m
Moments:
82.95*0.3/2 = 12.44 kN-m
276.5 kN/m
Punching shear:
Punching perimeter can not develop in this narrow footing. Therefore punching
shear failure is not possible.
One-way shear check:
Design shear force will be calculated at the face of the second column since
maximum shear force is 627.35 kN in this column. If d = 100 − 7 = 93 cm
V
d
= 627.25 − 0.20*276.5 = 572.2 kN
V
cr
= ( 0.65*1*950*930 ) / 1000 = 574.28 kN > V
d
Shear reinforcement is not necessary but stirrups will be provided for assembling
purposes.
Bending design:
- M = 12.44 kN-m is too small to be considered.
- M = 502.92 kN-m = 5029200 kg-cm R =
12.6=
93*95
5029200
2
kg/cm
2
ρ < ρ
min
= 0.0022 A
s
= 0.0022*95*93 = 19.44 cm
2
Selected: 6Ø22 (22.81 cm
2
) (At the top)
- M = 209.10 kN-m < 502.92 kN-m Select : 6Ø22 (At the bottom)
Four of these six bottom bars (more than one third of the tension reinforcement)
will continue across the compression zone and be cut at the other end of the
footing. The height of the beam is more than 60 cm; therefore longitudinal web
reinforcement should be provided. According to TS500 minimum area of these
bars is 0.001b
w
d.
0.001*95*93 = 8.84 cm
2
Selected: 4Ø18 (10.18 cm
2
)
For assembling the reinforcement stirrups with 30 cm spacing will be provided.
Two stirrups per set will be used since the footing is rather wide. Details are
shown in Fig. 5.27. Development lengths should be checked especially for the
bars provided for the positive moments.
If column dimensions are large and support moments are high adjustments
may be made on support moments. Shear diagrams between the faces of columns
may be drawn more accurately taking the column loads as uniformly distributed
loads as shown in Fig.5.28. Then maximum moment is calculated at the section
where shear force is zero. In the figure shear forces and the moments at the
bottom of the right column are shown just as an example.
6Ø22
4Ø22
2Ø18
2Ø18
100cm
95cm
10cm
40cm
380cm
40cm
100cm
6Ø22
4Ø18
2Ø22
4Ø22
A
A
A-A
2Ø10/30
2Ø10/30
stirrups
Figure 5.27
40 cm
N
572.2
−331.75
2.07m
0.27m
M
max
2397.5 kN/m
276.5
+
−
Distribured column load:
40.0
959
=
40.0
N
= 2397.5 kN/m
Uniformly distributed load across the
column width:
2397.5 − 276.5 = 2121 kN/m
Distance of the section where shear force
is zero:
m27.0=2121/2.572
M
max
= −502.92 +
2
2.572*)27.0+07.2(
= 166.55 kN-m
+
Figure 5.28
Footing designed in this example was a narrow one. Therefore in the
short direction no calculation was necessary. There may be cases where the
width of the footing is much larger than the column width. In such cases in
column areas reinforcement parallel to the short sides should also be provided.
In Fig. 5.29 these areas are shown. They are transverse strips and designed as
cantilevers. The widths of these cantilevers are equal to the widths of columns
plus d/2 at two sides. The pressure acting below the cantilever is found dividing
the column load by b. Shear force and bending moment are computed at section
I-I. Calculated reinforcement is placed over the main reinforcement.
u
q
=
b
N
b
1
1
2
d
2
d
2
d
2
d
Figure 5.29
1
1
5.7 CONTINUOUS STRIP FOOTINGS
Continuous strip footings are the combined footings that support some or
all columns of a frame. As mentioned earlier if soil is not suitable for making
single column footings, continuous footings may be the solution for the
foundation problem. Besides, instead of making single column footings and
tying them with tie-beams, making strip footings in one direction and tying them
in the other direction may be more suitable in earthquake zones. Soil pressure
distribution under a strip footing depends on the rigidity of the footing and on the
compressibility of the soil. For compressible soils it is possible to assume that
the soil pressure and the settlement of the footing at any point are proportional to
each other. By this assumption and by using the theory of the beams on elastic
foundations, a very reasonable solution can be obtained. However this method is
not very practical and the results will not be realistic if true soil properties are not
defined very well. For this reason in practice generally some approximate and
easier methods are used.
If number of columns are small and spans are not too large footings
behave rigidly. Deformations of the rigid footings are linear. Therefore soil stress
distribution may also be assumed as linearly varying. As a special case uniformly
distributed pressures may be assumed if the resultant of the column loads is
passing across the centroid of the footing. Rigid continuous strip footings may be
designed exactly in the same way as the two-column combined footings.
On the other hand if spans are large, footing may not behave as a rigid
one. It behaves as a flexible footing. Under the flexible footings soil pressures
vary as shown in Fig.5.30b. This distribution may be approximated by triangular
or uniform distributions as shown in Figs.5.30c and 5.30d.
N
1
N
2
b) Real distribution
c) Approximation as linear distribution
d) Approximation as uniform distribution
a) Deformation of flexible footing
Figure 5.30
The following approximate design method assumes uniform soil
distributions under the columns as shown in Fig. 5.30d.
At first, it is assumed that each column has an effective length on the footing.
This length is equal to the distance between the middle points of the spans. If
there are cantilevering parts at the ends, they are included in the effective lengths
of the external columns as shown in the Fig. 5.31. For each column uniformly
distributed linear soil load is determined dividing the column load by the
effective length. Thus q
1
, q
2
, q
3
etc. are calculated. The width of the footing is
computed dividing maximum of them by q
ef
. For design q
u1
, q
u2
, q
u3
etc. are
calculated like q
1
, q
2
, q
3
etc. are calculated but this time by using design column
loads (factored loads). At last, average soil loads are found as shown in Fig.5.31
and shear forces and bending moments are calculated by using these average
loads.
N
1
N
2
N
3
a
L
1
L
2
c
q
1
q
2
q
3
(q
u1
)
(q
u2
)
(q
u3
)
q
a
= q
u1
q
12
=
2
qq
2u+1u
q
23
=
2
qq
3u+2u
q
c
= q
u3
Figure 5.31
5.8 GRID FOUNDATIONS
Grid foundations are essentially strip footings arranged in two directions.
They are more effective than one-way strip footings for the prevention of
differential settlement. The theory of beams on elastic foundations may also be
applied to grid foundations. However in practice generally approximate, simple
but sufficiently accurate methods are used. As an example, one of these widely
used methods is given below. In Fig.5.34 a grid foundation example supporting
9 columns is shown. Design method will be described on this example.
c
y
L
y2
L
x1
L
y1
L
x2
a
x
c
x
a
y
b
b
Figure 5.34
C1
C2
C3
C4
C5
C6
C7
C8
C9
It is assumed in this method that while one part of the column load is spread in
one direction the remaining part is spread in the other direction. After
determining these portions of the column loads, each strip footing is analyzed
and designed independently. To simplify the computations it is advised to select
the footing widths equal in both directions. For each column effective lengths are
defined in both directions as the lengths between the midspans. For example
effective lengths for the column C1 shown in Fig.5.34 are:
L
ex
= a
x
+
2
L
1x
and L
ey
= c
y
+
2
L
2y
Effective lengths for the column C2:
L
ex
=
2
L+L
2x1x
and L
ey
= c
y
+
2
L
2y
Effective lengths for the column C5:
L
ex
=
2
L+L
2x1x
and L
ey
=
2
L+L
2y1y
Effective lengths for all other columns are found similarly. The bearing area
under the effective lengths of a column is defined as “effective area” of the
column. If “b” is the widths of the strips the effective area “A” is:
A = L
ex
b + L
ey
b − b
2
(5.9)
If a uniform pressure distribution is assumed under the effective area,
A =
ef
q
N
(5.10)
By equating the right sides of two equations:
b)L+L(=
q
N
eyex
ef
− b
2
(5 .11)
By solving Eq.(5.11) the value of “b” for this particular column can be
calculated. It is most likely that for each column a different “b” will be
calculated. Maximum of them is selected for the widths of all the strips. The
portion of the column loads considered in x and y direction strips are found as
follows:
N
x
=
N
L+L
L
=N
bL+bL
bL
=N
A+A
A
eyex
ex
eyex
ex
yx
x
(5.12)
N
y
=
N
L+L
L
=N
bL+bL
bL
=N
A+A
A
eyex
ey
eyex
ey
yx
y
(5.13)
Using factored loads in Eqs.(5.12) and (5.13) column loads can be computed for
each footing. Then they can be designed by the methods given in the previous
section. That is either rigid beam or flexible beam approach is employed
according to the rigidity of the footing.
5.9 MAT FOUNDATIONS
Mat foundations are thick solid plates as
mentioned earlier. The design of them also depends on
the rigidities of the plates. Under the rigid foundations
soil pressure distributions may be assumed as uniform or
linearly varying. They may be designed like beamless
slabs. Sometimes mat foundations may have beam parts
as shown in Fig. 5.35. These foundations may be
designed like slabs supported by the beams. If soil
pressure is not uniform because of eccentricity, average
soil pressure may be used in design.
L
x
L
y
Figure 5.35
These foundations may be designed like slabs supported by the beams. If soil
pressure is not uniform because of eccentricity, average soil pressure may be
used in design.
If foundation is assumed flexible, analysis may be done as follows:
Effective areas are defined for columns which are the areas within the middle
lines of the slab parts as shown in Fig.5.35 (shaded area). The soil pressure for
each effective area is calculated. Thus four different pressure values are
computed for every corner quarter of a slab, but slabs are designed by using
average of these four pressure values. The design of the beam parts is similar to
those of the grid foundations.
Bearing area of mat foundation may be computed by using average soil
pressure:
q
av
=
yx
LL
NΣ
≤ q
ef
(5.14)
where ΣN is the sum of the column loads. From this equation,
L
x
L
y
≥ ( ΣN /q
ef
) (5.15)
During the calculation of q
ef
usual reductions should be made in allowable soil
pressure. However if there is basement floor over the foundation, live load of the
basement floor should also be substracted from the allowable soil pressure.