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Eng . Haytham Besaiso2
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:
net all
reqq
QQ A 21
2- Find the resultant force location (Xr):
R X LQ
Q M
QQ R
r
22
1
21
00.0@
3- To ensure uniform soil pressure, the resultant force (R) should be in the center ofrectangular footing:
L
A B
X L L
X L L
r
r
1
1
2
2
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Eng . Haytham Besaiso3
Case II) Limitation that we have known length or width of footing:
Find center M @ :
1- If center M @ =0.00, uniform soil pressure
And find the required area:
L
A B
q
QQ A
net all
req
21
2- if 00.0@ center M 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
L B
Rq
L
e
L B
Rq
61
61
min
max
Find the width of footing B, by equating maxq
With net all q .
Check adequacy of footing width B that is 00.0min q
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Eng . Haytham Besaiso4
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:
net all
reqq
QQ A 21
As the area of trapezoidal is given by ))((5.0 21 L B B A
So put A Areq to get an equation which is function of 21 & B B .
2- Determine the resultant force location by taking as example 0@ 1 Q M
3- Put the resultant force location at the centroid of trapezoid to achieve uniform soil
pressure.
The centroid equation is:
21
21 2
3 B B
B B L
X So we will have another equation of 21 & B B , solve them to
get 21 & B B where ,
B1 : the edge from which the centroid is measured .
B2: the other edge .
Please try to derive the centroid equation .
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Eng . Haytham Besaiso5
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:
R X Q
Q M
QQ R
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
Bc X X r , X2 = Lc – Xr
4- Find the resultant of each soil pressure:
12
22112 100.0@
R R R
R Find X R X X R R M
5- Find the required area for each foot:
net all
net all
q
R A
q
R A
2
2
1
1
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Eng . Haytham Besaiso6
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
A X
~
Yg
~
Xg
2- Find the resultant force R:
iQ R
3- Find the location of the resultant force:
i
rii
R
i
rii
RQ
Y QY
Q
X Q X
That means, the moment of the resultant equals the sum of forces' moments
i A : shape's area . i
X : distance between y-axis and the centroid of the shape
iY :
distance between x-axis and the centroid of the shapeiQ :The load of column i
ri X : 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.
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Eng . Haytham Besaiso7
4- Find the eccentricities:' Y Y e X X e R y R x
5- Find the following :
i y X Qe M ….. moment of the columns loads about x-axis
i x y Qe M ….. moment of the columns loads about y-axis
12
1
3 L B I x ….. moment of inertia of the mat's area about its centroidal x-axis
12
1 3 B L I y ….. moment of inertia of the mat's area about its centroidal y-axis
6- Find the stresses:
gravityof centerthe point tothefromDistances:,Y X
Y I
M X
I
M
A
Qq
X
X
Y
y
mat
i
Now check that:
00.0min
max
q
qqnet all
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:
ui X uY
ui yu X
X
u X
Y
uY
mat
ui
Qe M
Qe M
Y I
M X
I
M
A
Qq
Find the average stress:
2
F E avg
qqq
Find the summation of loads of the strip StripuiQ .
Check that:
StripuiQ = stripavg Aq
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Eng . Haytham Besaiso8
If ok, draw the SFD and BMD and design the mat.
Otherwise;
We have to make adjustment for the loads as follow:
2 loadAverage
stripavg Stripui AqQ
Find the modified column loads:
stripui
uiuiQ
QQ
loadAverage
mod
Find the modified soil pressure:
stripavg u
avg uavg u Aq
qq
,
,mod,
loadAverage
Now we must have that:
StripuiQ = stripavg Aq
Draw SFD and BMD.
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Eng . Haytham Besaiso9
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 2mt q net all
Solution
1- Find the required area:
2
21 33.1315
12080mq
QQ A
net all
req
2- Find the resultant force location (Xr):
m X X
Q M
tonsQQ R
r r 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:
m B
m L
L
76.16.7
333.13
6.78.32
2.06.3
2
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Eng . Haytham Besaiso10
Example 2. Design a rectangular combined footing, given that
net all q = 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
330255 ft
q
QQ A
net all
req
2- Find the resultant force location (Xr):
ft X X
Q M
kipsQQ R
r r 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
= ksf 96.7
1175225.409
So, the uniform soil pressure along footing's length q' =net
uq
* B = 7.96 * 5.31=42.3 k/ft
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Eng . Haytham Besaiso11
5- Now, The column loads are treated as concentrated loads acting at the centers of the
columns. The shear and moment diagrams are
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Eng . Haytham Besaiso12
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 2mt q net all
Solution
1- Find the required area:
221 34.10
85.18
75120m
q
QQ A
net all
req
As the area of trapezoidal is given by ))((5.0 21 L B B A
So put A Areq to get an equation which is function of 21 & B B .
75.434.1035.45.0 2121 B B B B
...............1
2- Determine the resultant force
m X X
Q M
r r 55.1195475
0@ 1
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Eng . Haytham Besaiso13
3- Put the resultant force location at the centroid of trapezoid to achieve uniform soil
pressure.
The centroid equation is:
21
21
21
21 2305.075.4
2
3
35.42
3 B B X
B B
B B
B B L X
For uniform soil pressure:
1.55 + 0.2 = X
75.12305.0
75.1
21
B B
m X
75.52 21 B B .........................2
Solve 1 and 2:
m B 75.31
m B 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 2m KN qnet all
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Eng . Haytham Besaiso14
1- Find the resultant force location:
m X X
Q M
KN QQ R
r r 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:
KN R R R
KN R R R M
4.23786.16214000
6.1621340004.700.0@
12
112
5- Find the required area for each foot:
m B A
m Bm Aext
45.3892.11892.11200
4.2378
1.42
108.8108.8
200
6.1621
2
1
2
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Eng . Haytham Besaiso15
Example 5
Design a strap-footing for net all q = 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:
ft X X
Q M
KipsQQ R
r r 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 : ft 54.12
2
7
2
154.15X1 , X2 = 25-15.54 = 9.46 ft
4- Find the resultant of each soil pressure:
ft R R R
kips R R R M
9.2101.159370
1.15946.93702200.0@
12
112
5- Find the required area for each footing:
ft B ft A
ft L ft A
2.936.8436.845.2
9.210
97
64.6364.63
5.2
1.159
22
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 R 1 = 243 kips and R 2 = 326 kips
Edge footing :net
uq
1 = ksf 85.3
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
= ksf 85.32.9*2.9
326
So, the uniform soil pressure along footing's length q' =net
uq
* B = 3.85 * 9=34.65 k/ft
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Eng . Haytham Besaiso16
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Eng . Haytham Besaiso17
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 2m KN qnet all
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 foundati on 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.
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Eng . Haytham Besaiso18
2) Find the resultant force R:
KN Q R i 200,1312006180026004
3) Find the location of the resultant force:
mY
m X
R
R
18.8200,13
6002120017120021800121200218004
81.5
200,13
1312002136002518002512002
4) Find the eccentricities:
me
m.e
y
x
32.05.818.8
6905.681.5
5) Find the following :
m KN M
m KN M
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
m I
m I
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
gravityof centerthe point tothefromDistances:,
627.882,5
224,4
85.488,3
108,9
4.174.13
200,13
m KN q
m KN q
Y X
Y X q
OK q
OK qqnet all
00.0min
max
Calculate the modif ied soil pressure under the str ip 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:
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Eng . Haytham Besaiso19
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
800,19627.882,5
336,6
85.488,3
662,13
4.174.13
800,19
336,6800,1932.0
662,13800,1969.0
800,1927002180069004
m KN q
m KN q
Y X q
KN M
KN M
KN Q
F
E
u X
uY
ui
Find the average stress:
2/2.1072
6.1168.97
2m KN
qqq F E avg
KN QStripui 5400180029002 .
KN Aq stripavg 76.37304.1722.107
StripuiQ stripavg Aq
We have to make adjustment for the loads as follow:
KN 4.45652
76.37305400 loadAverage
Find the modified column loads:
0.845 byloadcolumneach
845.05400
4565.4mod
Multiply
QQQ uiuiui
Find the modified soil pressure:
2
mod, /2.13176.3730
4565.42.107 m KN q avg u
Draw SFD and BMD.
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Eng . Haytham Besaiso20
Example 7For 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
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Eng . Haytham Besaiso21
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:
KN Q R i 000,11
3- Find the location of the resultant force:
Y
m X
R
R
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 :
m KN M
m KN M
y
X
.840,,4000,1144.0
.100,1000,111.0
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Eng . Haytham Besaiso22
43
43
048,8 5.215.1612
1
665,13 5.165.2112
1
m I
m I
x
y
6- Find the stresses:
2
min
2
max
/62.2675.10
048,8
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
gravityof centerthe point tothefromDistances:,
048,8
1100
665,13
5390
5.21*5.16
000,11
m KN q
m KN q
Y X
Y X q
OK q
OK qqnet all
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
m KN q
m KN q
Y X q
KN M
KN M
KN Q
b
a
u X
uY
ui
Find the average stress:
2/105.512
84.4837.53
2m KN
qqq baavg
KN QStripui 5860 .
KN Aq stripavg 46705.2125.4105.51
StripuiQ stripavg Aq
We have to make adjustment for the loads as follow:
KN 52652
46705860 loadAverage
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Eng . Haytham Besaiso23
Find the modified column loads:
0.9 byloadcolumneach
9.05860
5265mod
Multiply
QQQ uiuiui
Find the modified soil pressure:
2
mod, /62.574670
5265105.51 m KN q avg u
The shear and bending moment diagrams for the selected strip in are shown below.
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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)