Eng . Haytham Besaiso 1
6 Geometric design of shallow foundations and Mat footing
1) Geometric design of isolated footing: Most economical type .
Can be rectangular ,circular or square . It is preferred that the footing matches the
shape of its column .
Used in case of light columns loads and when columns are not closely spaced
1- Find the net allowable bearing capacity
(Soil pressure):
scfcc
all
unetu
netall
u
all
hDhq
FS
FS
FS
FS
2- Find the required area of footing:
LD
netall
req
PPQ
q
QA
Note: We can use gross allowable bearing capacity to find the
Area of footing, but the load Q should be also gross so that it
Includes the weight of foundation and soil above it.
scfsoil
ccfoot
soilfootLDGross
all
Gross
req
hDBLW
hBLW
WWPPQ
q
QA
Eng . Haytham Besaiso 2
2) Geometric design of rectangular combined footing: Used under closely spaced and heavily loaded columns where individual footings , if
they were provided , would be either very close or overlap each other .
Used as an alternative to an eccentrically loaded footing that has a property line
restriction so that the edge columns is linked to an interior column .
rectangular combined footing is more preferred than trapezoidal combined footing
due to its simplicity in both design and construction .
Case I) No limitations:
1- Find the required area:
netall
reqq
QQA 21
2- Find the resultant force location (Xr):
RXLQ
QM
QQR
r
22
1
21
00.0@
3- To ensure uniform soil pressure, the resultant force (R) should be in the center of
rectangular footing:
L
AB
XLL
XLL
r
r
1
1
2
2
Eng . Haytham Besaiso 3
Case II) Limitation that we have known length or width of footing:
Find centerM @ :
1- If centerM @ =0.00, uniform soil pressure
And find the required area:
L
AB
q
QQA
netall
req
21
2- if 00.0@ centerM so that there is a value M
by means of eccentricity so that the soil pressure will not
be uniform.
Calculate the soil pressure:
L
e
LB
Rq
L
e
LB
Rq
61
61
min
max
Find the width of footing B, by equating maxq
With netallq .
Check adequacy of footing width B that is 00.0min q
Eng . Haytham Besaiso 4
3) Geometric design of trapezoidal combined footing:
Trapezoidal combined footing is used rather than rectangular combined footing and
will be more economical in the following two cases :
There are limitations on footing's longitudinal projection beyond the two columns.
There is a large difference between the magnitudes of the columns loads .
1- Find the required area:
netall
reqq
QQA 21
As the area of trapezoidal is given by ))((5.0 21 LBBA
So put AAreq to get an equation which is function of 21 & BB .
2- Determine the resultant force location by taking as example 0@ 1 QM
3- Put the resultant force location at the centroid of trapezoid to achieve uniform soil
pressure.
The centroid equation is:
21
21 2
3 BB
BBLX So we will have another equation of 21 & BB , solve them to
get 21 & BB where ,
B1 : the edge from which the centroid is measured .
B2: the other edge .
Please try to derive the centroid equation .
Eng . Haytham Besaiso 5
4) Geometric design of strap footing (Cantilever): Used when there is a property line which disables the footing to be extended beyond the
face of the edge column. In addition to that the edge column is relatively far from the
interior column so that the rectangular combined footing will be too narrow and long
which increases the cost .
There is a "strap beam" which connects two separated footings . The edge footing is
eccentrically loaded and the interior footing is centrically loaded . The purpose of the
beam is to prevent overturning of the eccentrically loaded footing .
1- Find the resultant force location:
RXQ
QM
QQR
r
Lc
00.0@
2
1
21
2- Assume the width of the exterior (edge) footing B1.
3- Find the distance X1 , X2 :
22
111
BcXX r , X2 = Lc – Xr
4- Find the resultant of each soil pressure:
12
22112 1 00.0@
RRR
RFindXRXXRRM
5- Find the required area for each foot:
netall
netall
q
RA
q
RA
22
11
Eng . Haytham Besaiso 6
5) Geometric and structural design of Mat foundation:
Geometric design (Service loads):
1- Find the center of gravity of mat footing:
i
ii
i
ii
A
AY
A
AX
~
Yg
~
Xg
2- Find the resultant force R:
iQR
3- Find the location of the resultant force:
i
rii
R
i
rii
RQ
YQY
Q
XQX
That means, the moment of the resultant equals the sum of forces' moments
iA : shape's area .
iX : distance between y-axis and the centroid of the shape
iY : distance between x-axis and the centroid of the shape
iQ :The load of column i
riX : distance between column's center and y-axis
riY : distance between column's center and y-axis
Note : Xg , Yg ,XR and YR are all measured from the same axis.
Eng . Haytham Besaiso 7
4- Find the eccentricities: ' YYeXXe RyRx
5- Find the following :
iyX QeM ….. moment of the columns loads about x-axis
ixy QeM ….. moment of the columns loads about y-axis
12
1 3LBI x ….. moment of inertia of the mat's area about its centroidal x-axis
12
1 3BLI y ….. moment of inertia of the mat's area about its centroidal y-axis
6- Find the stresses:
gravity ofcenter thepoint to thefrom Distances : ,YX
YI
MX
I
M
A
X
X
Y
y
mat
i
Now check that:
00.0min
max
q
qqnetall
Otherwise increase the dimensions of the mat foundation
Structural design (Ultimate loads):
The mat foundation is divided into strips in both directions. The width of the strip is directly
proportion to the loads of the column included in this strip.
For the previous mat let we take a strip of width B1 for the columns 13-14-15-16
Locate the points E and F at the middle of strip edges.
Find the stresses at E and F and be careful that we use ultimate loads:
uiXuY
uiyuX
X
uX
Y
uY
mat
ui
QeM
QeM
YI
MX
I
M
A
Find the average stress:
2
FEavg
qqq
Find the summation of loads of the strip StripuiQ .
Check that:
StripuiQ = stripavg Aq
Eng . Haytham Besaiso 8
If ok, draw the SFD and BMD and design the mat.
Otherwise;
We have to make adjustment for the loads as follow:
2 load Average
stripavgStripui AqQ
Find the modified column loads:
stripui
uiuiQ
load Average
mod
Find the modified soil pressure:
stripavgu
avguavguAq
,
,mod,
load Average
Now we must have that:
StripuiQ = stripavg Aq
Draw SFD and BMD.
Eng . Haytham Besaiso 9
Example1
Find the Dimensions of the combined footing for the columns A and B that spaced 6.0m
center to center, column A is 40cm x 40cm carrying dead loads of 50tons and 30tons live
load and column B is 40cm x 40cm carrying 70tons dead load and 50 tons live loads.
./15 2mtqnetall
Solution
1- Find the required area:
221 33.13
15
12080m
q
QQA
netall
req
2- Find the resultant force location (Xr):
mXX
QM
tonsQQR
rr 6.32006120
00.0@
20012080
1
21
3- To ensure uniform soil pressure, the resultant force (R) should be in the center of
rectangular footing:
mB
mL
L
76.16.7
333.13
6.78.32
2.06.32
Eng . Haytham Besaiso 10
Example 2.
Design a rectangular combined footing, given that netallq = 5 ksf , Df = 5 feet, the edge of
column 1 is at the property line, and the spacing between columns is 18 feet center-to-
center (c.c.).
Column1 (18'' *18'') Column 2 (24'' *24'')
DL 80 kips 130 kips
LL 175 kips 200 kips
Solution:
1- Find the required area:
221 117
5
330255ft
q
QQA
netall
req
2- Find the resultant force location (Xr):
ftXX
QM
kipsQQR
rr 15.1058518330
00.0@
585330255
1
21
3- To ensure uniform soil pressure, the resultant force (R) should be in the center of
rectangular footing:
L = (0.7
5+10.15) * 2 = 21.8 ft =22 ft (approximately)
Note : This round off isn't required but it can be accepted
B = A/L = 117/22 = 5.31 ft = 5 ft- 4 in
4- Evaluate the net factored soil pressure :
uQ1 = 1.4 *80 + 1.7 *175 = 410 kips
uQ2 =1.4 *130 + 1.7*200 = 522 kips
net
uq
= ksf96.7
117
5225.409
So, the uniform soil pressure along footing's length q' = net
uq
* B = 7.96 * 5.31=42.3 k/ft
Eng . Haytham Besaiso 11
5- Now, The column loads are treated as concentrated loads acting at the centers of the
columns. The shear and moment diagrams are
Eng . Haytham Besaiso 12
Example3
Find the Dimensions of the trapezoidal combined footing for the columns A and B that
spaced 4.0m center to center, column A is 40cm x 40cm carrying dead loads of 80tons
and 40tons live load and column B is 30cm x 30cm carrying 50tons dead load and 25 tons
live loads. ./85.18 2mtqnetall
Solution
1- Find the required area:
221 34.10
85.18
75120m
q
QQA
netall
req
As the area of trapezoidal is given by ))((5.0 21 LBBA
So put AAreq to get an equation which is function of 21 & BB .
75.434.1035.45.0 2121 BBBB ...............1
2- Determine the resultant force
mXX
QM
rr 55.1195475
0@ 1
Eng . Haytham Besaiso 13
3- Put the resultant force location at the centroid of trapezoid to achieve uniform soil
pressure.
The centroid equation is:
2121
21
21 2305.075.4
2
3
35.42
3BBX
BB
BB
BBLX
For uniform soil pressure:
1.55 + 0.2 = X
75.12305.0
75.1
21
BB
mX
75.52 21 BB .........................2
Solve 1 and 2:
mB 75.31
mB 12
Example 4
Design a strap footing to support two columns, that spaced 4.0m center to center exterior
column is 80cm x 80cm carrying 1500 KN and interior column is 80cm x 80cm carrying
2500KN.
./200 2mKNqnetall
Eng . Haytham Besaiso 14
1- Find the resultant force location:
mXX
QM
KNQQR
rr 5400082500
00.0@
400025001500
1
21
2- Assume the length of any foot, let we assume L1=2m.
3- Find the distance X1 , X2 :
m4.42
2
2
8.05X1 , X2 = 8-5 = 3.0 m
4- Find the resultant of each soil pressure:
KNRRR
KNRRRM
4.23786.16214000
6.1621340004.700.0@
12
112
5- Find the required area for each foot:
mBA
mBmAext
45.3892.11892.11200
4.2378
1.42
108.8108.8
200
6.1621
2
1
2
Eng . Haytham Besaiso 15
Example 5
Design a strap-footing for netallq = 2.5 ksf . The edge of column 1 is placed at the
property line, and the center of the columns are 25 feet center-to-center (c.c.).
Column1 (12'' *12'') Column 2(16'' *16'')
DL 80 kips 120 kips
LL 60 kips 110 kips
Solution:
1- Find the resultant force location:
ftXX
QM
KipsQQR
rr 54.1537025230
00.0@
370230140
1
21
2- Assume the length of any foot, let we assume L1=7ft.
3- Find the distance X1 , X2 :
ft54.122
7
2
154.15X1 , X2 = 25-15.54 = 9.46 ft
4- Find the resultant of each soil pressure:
ftRRR
kipsRRRM
9.2101.159370
1.15946.93702200.0@
12
112
5- Find the required area for each footing:
ftBftA
ftLftA
2.936.8436.845.2
9.210
97
64.6364.63
5.2
1.159
2
2
1
2
1
1- Evaluate the net factored soil pressure :
uQ1 = 1.4 *80 + 1.7 *60 = 214 kips
uQ2 =1.4 *120 + 1.7*110 = 355 kips
Repeat the usual steps to find out R1 = 243 kips and R2 = 326 kips
Edge footing : net
uq
1 = ksf85.3
9*7
243
So, the uniform soil pressure along footing's width q' = net
uq
* L = 3.85 * 9=34.65k/ft
Interior footing : net
uq
2 = ksf85.3
2.9*2.9
326
So, the uniform soil pressure along footing's length q' = net
uq
* B = 3.85 * 9=34.65 k/ft
Eng . Haytham Besaiso 16
Eng . Haytham Besaiso 17
Example 6
For the shown mat foundation:
Interior columns Edge columns Corner columns
Column dimension 60cm x 60cm 60cm x 40cm 40cm x 40cm
Service loads 1800KN 1200KN 600KN
Ultimate loads 2700KN 1800KN 900KN
./150 2mKNqnetall
Check the adequacy of the foundation dimensions.
Calculate the modified soil pressure under the strip ABCD which is 2m width.
Draw SFD and BMD for the strip.
Check the adequacy of the foundation dimensions.
1) Find the center of gravity of mat footing:
Xg = 13.4/2 – 0.2 = 6.5 m , Yg = 17.4/2 – 0.2 = 8.5 m
The distances are taken from (x-y) axes shown in the figure.
Eng . Haytham Besaiso 18
2) Find the resultant force R:
KNQR i 200,1312006180026004
3) Find the location of the resultant force:
mY
mX
R
R
18.8200,13
6002120017120021800121200218004
81.5200,13
1312002136002518002512002
4) Find the eccentricities:
me
m.e
y
x
32.05.818.8
6905.681.5
5) Find the following :
mKNM
mKNM
y
X
.108,9200,1369.0
.224,4200,1332.0
43
43
627.882,5 4.134.1712
1
851.488,3 4.174.1312
1
mI
mI
x
y
6) Find the stresses:
2
min
2
max
/87.327.8627.882,5
224,47.6
85.488,3
108,9
4.174.13
200,13
/35.807.8627.882,5
224,47.6
85.488,3
108,9
4.174.13
200,13
gravity ofcenter thepoint to thefrom Distances : ,
627.882,5
224,4
85.488,3
108,9
4.174.13
200,13
mKNq
mKNq
YX
YXq
OKq
OKqqnetall
00.0min
max
Calculate the modified soil pressure under the strip ABCD which is 2m width.
Locate the points E and F at the middle of strip edges.
Find the stresses at E and F and notice that we use ultimate loads:
Eng . Haytham Besaiso 19
2
2
/6.1167.8627.882,5
336,67.5
85.488,3
662,13
4.174.13
800,19
/8.977.8627.882,5
336,67.5
85.488,3
662,13
4.174.13
800,19
627.882,5
336,6
85.488,3
662,13
4.174.13
800,19
336,6800,1932.0
662,13800,1969.0
800,1927002180069004
mKNq
mKNq
YXq
KNM
KNM
KNQ
F
E
uX
uY
ui
Find the average stress:
2/2.1072
6.1168.97
2mKN
qqq FE
avg
KNQStripui 5400180029002 .
KNAq stripavg 76.37304.1722.107
StripuiQ stripavg Aq
We have to make adjustment for the loads as follow:
KN4.45652
76.37305400 load Average
Find the modified column loads:
0.845by loadcolumn each
845.05400
4565.4mod
Multiply
QQQ uiuiui
Find the modified soil pressure:
2
mod, /2.13176.3730
4565.42.107 mKNq avgu
Draw SFD and BMD.
Eng . Haytham Besaiso 20
Example 7
For the mat foundation shown below. All column dimensions are 50 cm x 50 cm with the
load schedule shown below. The allowable soil pressure is qall =60 kPa.
1) Check the adequacy of the foundation dimensions
2) Draw shear and moment diagrams for the strip AMOJ
Eng . Haytham Besaiso 21
DL, kN LL, kN Column
200 200 A
250 250 B
250 200 C
800 700 D
800 700 E
650 550 F
800 700 G
800 700 H
650 550 I
200 200 J
250 250 K
200 150 L
Solution
1- Find the center of gravity of mat footing:
Xg = 8.25 m , Yg =10.75 m
The distances are taken from (x-y) axes shown in the figure.
2- Find the resultant force R:
KNQR i 000,11
3- Find the location of the resultant force:
mY
mX
R
R
85.10000,11
)450500400(25.21)120015002(25.14)120021500(25.7)350500400(25.0
81.7000,11
)12002450350(25.16)150025002(25.8)150024002(25.0
4- Find the eccentricities:
me
m.e
y
x
1.075.1085.10
44052.881.7
5- Find the following :
mKNM
mKNM
y
X
.840,,4000,1144.0
.100,1000,111.0
Eng . Haytham Besaiso 22
43
43
048,8 5.215.1612
1
665,13 5.165.2112
1
mI
mI
x
y
6- Find the stresses:
2
min
2
max
/62.2675.10048,8
110025.8
665,13
840,,4
5.21*5.16
000,11
/4.3575.10048,8
110025.8
665,13
840,,4
5.21*5.16
000,11
gravity ofcenter thepoint to thefrom Distances : ,
048,8
1100
665,13
5390
5.21*5.16
000,11
mKNq
mKNq
YX
YXq
OKq
OKqqnetall
00.0min
max
So, the mat dimensions are ok
Now, to draw shear and moment diagrams, factored forces are considered
2
2
/84.4875.10048,8
1694.5125.6
665,13
8.455,7
5.21*5.16
16945
/37.5375.10048,8
1694.5125.6
665,13
8.455,7
5.21*5.16
16945
048,8
1694.5
665,13
8.455,7
5.21*5.16
16945
1694.5169451.0
8.455,71694544.0
16945
mKNq
mKNq
YXq
KNM
KNM
KNQ
b
a
uX
uY
ui
Find the average stress:
2/105.512
84.4837.53
2mKN
qqq ba
avg
KNQStripui 5860 .
KNAq stripavg 46705.2125.4105.51
StripuiQ stripavg Aq
We have to make adjustment for the loads as follow:
KN52652
46705860 load Average
Eng . Haytham Besaiso 23
Find the modified column loads:
0.9by loadcolumn each
9.05860
5265mod
Multiply
QQQ uiuiui
Find the modified soil pressure:
2
mod, /62.574670
5265105.51 mKNq avgu
The shear and bending moment diagrams for the selected strip in are shown below.
Eng . Haytham Besaiso 24
Note : This moment diagram could be mirrored about the horizontal axis to get
BMD we are used to (i.e, Moment drawn on tension side)
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