10-Analysis of Beam Foundation-En-Part III

Beam Foundations after Kany and El Gendy
by GEO Tools
(Analysis and Design)
Part III: Numerical Examples

P 1=1000 [kN] P 2=3000 [kN] P 3=1000 [kN]
Ground surface
Df =2.0 [m]
qall = 210 [kN/m2]

γa = 20 [kN/m3]
0.25 0.25
4.75 [m] 4.75 [m]
B = 3.0 [m]
C1 C2 C3
1000 [kN] 3000 [kN] 1000 [kN]
A =10.0 [m]
Mahmoud El Gendy
Mohamed El Gendy
Copyright ©
GEOTEC Software Inc.
PO Box 14001 Richmond Road PO, Calgary AB, Canada T3E 7Y7
Tele.:+1(587) 332-3323
geotec@geotecsoftware.com
www.geotecsoftware.com
2023
Analysis of Beam Foundations
Content
Page
10 Analysis and Design of Beam Foundations after Kany and El Gendy ........................ 4
10.1 Introduction ............................................................................................................ 4
10.2 Numerical Examples............................................................................................... 5
10.2.1 Calculation methods ........................................................................................... 5
10.2.2 Material and section for concrete design ............................................................. 6
10.2.3 Example 1: Design of a combined footing for two unequal columns ................... 7
10.2.4 Example 2: Design of a combined footing for two equal columns ..................... 16
10.2.5 Example 3: Design of a combined footing for three equal columns ................... 23
10.2.6 Example 4: Design of a combined footing for three unequal columns ............... 31
10.2.7 Example 5: Design of a combined footing for three equal columns ................... 40
-10.2-
GEO Tools
Preface
Various problems in Geotechnical Engineering can be investigated by the program GEO Tools. The
original version of GEO Tools in ELPLA package was developed by M. Kany, M. El Gendy, and A.
El Gendy to determine the contact pressure, settlements, and moments and shear forces of beam
foundations. After the death of Kany, (M. & A.) El Gendy further developed the program to meet the
needs of the practice.
This book describes the essential methods used in GEO Tools to analyze beam foundations with
verification examples. GEO Tools is a simple user interface program and needs little information to
define a problem.
There are three soil models with five methods available in GEO Tools for analyzing beam
foundations. Many test examples are presented to verify and illustrate the soil models and methods
for analyzing beam foundations available in GEO Tools.
-10.3-
10 Analysis and Design of Beam Foundations after Kany and El Gendy
10.1 Introduction
Different calculation methods are known in the literature for the calculation of shallow foundations.
The early one is that assumes a uniform contact pressure distribution under shallow foundations. This
assumption is too far from the reality, Winkler (1867) and Zimmermann (1930) developed the
Modulus of subgrade method. In this method, the subsoil is simulated by isolated springs. The
settlement of the spring is only dependent on the loading at the same point on the subsoil surface at
the spring location. This also applies to possible refinements with springs of different stiffness.
However, Boussinesq (1885) had already recognized that when the subsoil is loaded at one point, the
subsoil also settles outside the load point. Therefore, it does not behave like a spring. Because of this
finding, Ohde (1942) developed a calculation method for the first time, with which shallow
foundations can be analyzed, considering the soil structure interaction. This method, which is called
Modulus of compressibility method, was later further developed by different authors (Graßhoff
(1966-1978), Kany (1974), Graßhoff/Kany (1992)). The program GEO Tools is based on the Modulus
of compressibility method after Kany (1974) and the Modulus of subgrade reaction method after
Kany/ El Gendy (1995). However, some refinements are included, some of which are new and have
not yet been dealt with in detail in the literature. It is therefore necessary to explain the calculation
method in more detail than usual in order to be able to check the results and compare them with other
results.
-10.4-
GEO Tools
10.2 Numerical Examples
10.2.1 Calculation methods
It is possible by GEO Tools to use the same data for analyzing beam foundations by five different
conventional and refined calculation methods. The interaction between the beam and the subsoil can
be analyzed by:
1 Linear contact pressure method

2 Modulus of subgrade reaction method after Kany/ El Gendy (1995)
3 Modulus of compressibility method after Kany (1974)
4 Rigid beam foundation
5 Flexible beam foundation
It is also possible to consider irregular soil layers and the thickness of the base beam that varies in
each element. Furthermore, the influence of temperature changes and additional settlement on the
beam foundation can be taken into account. With the help of GEO Tools, an analysis of different
examples was carried out to verify and test the methods and the program for analyzing the problems
of beam on elastic foundation.
In the analysis, the beam foundation is divided into equal elements according to Figure 10.1. Using
the available five calculation methods, the settlement and the contact pressure can be determined in
each element.
Concentrated load P 1 Uniform load pF 1 P2 GS
Tf
d1 i d2 d3
Edge moment
MRl (+) Beam thickness and loads MRr (+)
B 1 2 3 i n-1 n
A= n× a
Beam foundation with element division
Figure 10.1 Loads, beam thickness und beam foundation with element division
-10.5-
10.2.2 Material and section for concrete design
The concrete design of the beam foundations sections is carried out according to EC 2, DIN 1045,
ACI and ECP. The material and section for concrete design are supposed to have the following
parameters:
10.2.2.1 Material properties
Concrete grade according to ECP C 250
Steel grade according to ECP S 36/52
Concrete cube strength fcu = 250 [kg/ cm2] = 25 [MN/ m2]
Concrete cylinder strength fc = 0.8 fcu [-] = 20 [MN/ m2]
Compressive stress of concrete fc = 95 [kg/ cm2] = 9.5 [MN/ m2]
Tensile stress of steel fs = 2000 [kg/ cm2] = 200 [MN/ m2]
Reinforcement yield strength fy = 3600 [kg/ cm2] = 360 [MN/ m2]
Young's modulus of concrete Eb = 3×107 [kN/ m2] = 30000 [MN/ m2]
Poisson's ratio of concrete νb = 0.15 [-]
Unit weight of concrete γb = 25 [kN/ m3]
In some examples, unit weight of concrete is chosen γb = 0.0 to neglect the own weight of the beam
foundation.
10.2.2.2 Section properties
Width of the section to be designed b = 1.0 [m]
Section thickness t [m]
Concrete cover + 1/2 bar diameter c=5 [cm]
Effective depth of the section d = t - c = 0.45 [m]
Steel bar diameter Φ = 16 to 22 [mm]
-10.6-
GEO Tools
10.2.3 Example 1: Design of a combined footing for two unequal columns
10.2.3.1 Description of the problem

It is required to find the contact pressure distribution, settlements, moment, and shear force diagrams
for a rectangular combined footing for an exterior column C1 of a load of 1050 [kN] and an interior
column C2 of a load of 1800 [kN] as shown in Figure 10.2. C1 is 0.50×0.50 [m2] and reinforced by
6Φ22, while column C2 is 0.60×0.50 [m2] and reinforced by 10Φ22. The allowable soil pressure qall
= 185 [kN/m2] at a depth of Df = 1.90 [m] and average unit weight of the soil and concrete is γa = 20
[kN/m3]. Modulus of elasticity of the footing concrete is Eb = 2×107, while the Modulus of subgrade
reaction of the soil is ks = 25000 [kN/m3].
Ground surface
Df = 1.9 [m]
qall = 185 [kN/m2]

γa = 20 [kN/m3]
0.25 L= 4.75 [m]
C1 C2
Property line 1050 [kN] 1800 [kN]
Figure 10.2 Combined rectangular footing
10.2.3.1.1 Determining point of application of the resultant force

Resultant force R equals:
R = P1 + P2 = 1050 + 1800 = 2850 [kN]
Take the moment about C1 to get the distance S:
P2A = RS
1800×4.75 = 2850×S
S = 3.0 [m]
10.2.3.1.2 Determining footing sides A and B
A =2 Xp = 2 (0.25 + S) = 2 (0.25 + 3.0) = 6.5 [m]
Allowable net soil pressure qnet is given by:

-10.7-
qnet = qall – γaDf = 185 – 20×1.9 = 147 [kN/m2]
Area of footing Af is obtained from:
R 2850
Af = = = 19.39 [m 2 ]
qnet 147
Take Af = 6.5×3.0 = 19.50 [m2] rectangular combined footing, as shown in Figure 10.3.
Xp=3.25 [m]
P 1=1050 [kN]
R = 2850 [kN]
S=3.0 [m]
P 2=1800 [kN]
Ground surface
Df =1.9 [m]
0.25 L=4.75 [m] 1.5 [m]
C1 C2
B =3.0 [m]
Property line 1050 [kN] 1800 [kN]

Figure 10.3 The combined rectangular footing with property line and column loads
-10.8-
GEO Tools
10.2.3.1.3 Analysis of the combined footing

The footing can be regarded as a beam on elastic foundation. In the analysis, the beam is divided into
four equal elements, each 1.625 [m] long (Figure 10.4). The self-weight of the footing is neglected.
According to Kany/ El Gendy (1995), the analysis of combined footing as a beam on elastic
foundation is carried out in the following steps:
10.2.3.1.4 Calculation of ui, vi and wi:

For a constant beam moment of inertia Ii =I
1 𝐼𝑖
𝑢𝑖 = (1 + )
2 𝐼𝑖−1
1 𝐼 1
𝑢𝑖 = (1 + ) = × 2 = 1
2 𝐼 2
1 𝐼𝑖 𝐼𝑖
𝑣𝑖 = ( + 14 + )
4 𝐼𝑖−1 𝐼𝑖+1
1 𝐼 𝐼 1
𝑣𝑖 = ( + 14 + ) = × 16 = 4
4 𝐼 𝐼 4
1 𝐼𝑖
𝑤𝑖 = (1 + )
2 𝐼𝑖+1
1 𝐼 1
𝑤𝑖 = (1 + ) = × 2 = 1
2 𝐼 2
Moment of inertia Ii =I:
𝐵𝑑𝑖 3 3 × 0.553
𝐼𝑖 = 𝐼 = = = 0.0416 [m4 ]
12 12
𝑎 4𝐵 1.6254 ×3
𝛼= = 20000000×0.0416 = 2.514 × 10−5 [m3/kN]
𝐸𝑝 𝐼
10.2.3.1.5 Determining external moments M i(l )

The external moments M i(l ) at points 1, 2, 3 and 4 are:
(𝑙)
𝑀1 = 0 [kN.m]
(𝑙)
𝑀2 = 1050 × (1.5 × 1.625 − 0.25) = 2296.875 [kN.m]
(𝑙)
𝑀3 = 1050 × (2.5 × 1.625 − 0.25) = 4003.125 [kN.m]
(𝑙) 1.625
𝑀4 = 1050 × (3.5 × 1.625 − 0.25) + 1800(1.5 − ) = 6946.875 [kN.m]
2
-10.9-
GS P 1=1050 [kN]
P 2=1800 [kN]
1 2 3 4
k1 k2 k3 k4
Beam foundation with loads and springs
a= 1.625 [m]
B =3 [m]
1 2 3 4
A =4×1.625= 6.5 [m]

Plan of beam foundation with elements
1 2 3 4
si
Settlement
a= 1.625 [m]
1 2 3 4
q2
Contact pressure
Figure 10.4 Combined footing as a beam on elastic foundation
-10.10-
GEO Tools
10.2.3.1.6 Determining the right-hand side Ri

The right-hand side Ri of the contact pressure equation is:
(𝑙) (𝑙) (𝑙) 𝑎2

𝑅𝑖 = (𝑢𝑖 𝑀𝑖−1 + 𝑣𝑖 𝑀𝑖 + 𝑤𝑖 𝑀𝑖+1 )
6𝐸𝐼𝑖
(𝑙) (𝑙) (𝑙) 1.6252

𝑅𝑖 = (1 × 𝑀𝑖−1 + 4 × 𝑀𝑖 + 1 × 𝑀𝑖+1 )
6 × 20000000 × 0.0416
(𝑙) (𝑙) (𝑙)
𝑅𝑖 = 5.29 × 10−7 (𝑀𝑖−1 + 4𝑀𝑖 + 𝑀𝑖+1 )
Apply the above equation at points 2 and 3:
At point 2:
(𝑙) (𝑙) (𝑙)
𝑅2 = 5.29 × 10−7 (𝑀1 + 4𝑀2 + 𝑀3 )
𝑅2 = 5.29 × 10−7 (0 + 4 × 2296.875 + 4003.125) = 0.00698
At point 3:
(𝑙) (𝑙) (𝑙)
𝑅3 = 5.29 × 10−7 (𝑀2 + 4𝑀3 + 𝑀4 )
𝑅3 = 5.29 × 10−7 (2296.875 + 4 × 4003.125 + 6946.875 ) = 0.0134
10.2.3.1.7 Determining contact pressures

The contact pressure equation is:
1 2 𝛼𝑖 1 𝛼𝑖
( ) 𝑞𝑖+1 − ( − 𝑤𝑖 ) 𝑞𝑖 + ( + (𝑣𝑖 + 2𝑤𝑖 )) 𝑞𝑖−1
𝑘𝑖+1 𝑘𝑖 6 𝑘𝑖−1 6
𝑖−2
𝛼𝑖
+ (∑[(𝑖 − 𝑗 − 1)𝑢𝑖 + (𝑖 − 𝑗)𝑣𝑖 + (𝑖 − 𝑗 + 1)𝑤𝑖 ] 𝑞𝑗 ) = 𝑅𝑖
6
𝑗=1
1 2 2.514 × 10−5 1 2.514 × 10−5

( ) 𝑞𝑖+1 − ( − ) 𝑞𝑖 + ( + (4 + 2)) 𝑞𝑖−1
25000 25000 6 25000 6
𝑖−2
2.514 × 10−5
+ (∑[(𝑖 − 𝑗 − 1) + 4(𝑖 − 𝑗) + (𝑖 − 𝑗 + 1)] 𝑞𝑗 ) = 𝑅𝑖
6
𝑗=1
-10.11-
or
𝑖−2
𝑞𝑖+1 − 1.895 𝑞𝑖 + 1.629 𝑞𝑖−1 + 0.629 (∑(𝑖 − 𝑗) 𝑞𝑗 ) = 25000𝑅𝑖

𝑗=1
Apply the above equation at points 2 and 3:
At point 2:
𝑞3 − 1.895 𝑞2 + 1.629 𝑞1 = 25000 × 0.00698
𝑞3 − 1.895 𝑞2 + 1.629 𝑞1 = 174.5

At point 3:
𝑞4 − 1.895 𝑞3 + 1.629 𝑞2 + 0.629 × 2 𝑞1 = 25000 × 0.0134
𝑞4 − 1.895 𝑞3 + 1.629 𝑞2 + 1.258 𝑞1 = 334.014
There are four unknown q1, q2, q3, and q4, so a farther two equations are required. This can be obtained
by considering the overall equilibrium of vertical forces and moments.
Overall equilibrium of vertical forces
∑𝑉 = 0
𝑎𝐵( 𝑞1 + 𝑞2 + 𝑞3 + 𝑞4 ) = 𝑃1 + 𝑃2
1.625 × 3( 𝑞1 + 𝑞2 + 𝑞3 + 𝑞4 ) = 1050 + 1800
𝑞1 + 𝑞2 + 𝑞3 + 𝑞4 = 584.615
Overall equilibrium of vertical forces
∑𝑀 = 0
𝑎𝐵( 3.5𝑎𝑞1 + 2.5𝑎𝑞2 + 1.5𝑎𝑞3 + 0.5𝑎𝑞4 ) = 𝑃1 (𝐴 − 0.25) + 𝑃2 1.5
1.625 × 3( 𝑞1 (3.5 × 1.625) + 𝑞2 (2.5 × 1.625) + 𝑞3 (1.5 × 1.625) + 𝑞4 (0.5 × 1.625))

= 1050(6.5 − 0.25) + 1800 × 1.5
3.5𝑞1 + 2.5𝑞2 + 1.5𝑞3 + 0.5𝑞4 = 1169.231
-10.12-
GEO Tools
Contact pressure equations in matrix form:
1 1 1 1 𝑞1 584.615
3.5 2.5 1.5 0.5 𝑞2 1169.231
[ ][ 𝑞 ] = [ ]
1.629 −1.895 1 0 3 174.5
1.258 1.629 −1.895 1 𝑞4 334.014
Solving the above system of linear equations to obtain the contact pressures q1, q2, q3, and q4.
𝑞1 169.188
𝑞2 120.775
[𝑞 ]=[ ] [kN/m2]
3 127.808
𝑞4 166.844
10.2.3.1.8 Determining settlements si

The settlement si can be given by:
𝑞 𝑞
𝑠𝑖 = 𝑘𝑖 = 25000
𝑖
[m]
𝑖
s1 = 0.68 [cm]
s2 = 0.48 [cm]
s3 = 0.51 [cm]
s4 = 0.67 [cm]
The contact pressure distribution, settlement, moment, and shear force diagrams for the raft are shown
in Figure 10.5 to Figure 10.8. Once the internal forces are obtained at various sections, the design of
the raft can be completed in the normal manner.
10.2.3.1.9 Computer calculation

The input data and results of GEO Tools are presented on the next pages. By comparison, one can see
an agreement with the hand calculation.
-10.13-
Distance x [m]
0 1.625 3.25 4.875 6.5
0
20
40
Contact pressure q [kN/m2]
60
80
100
120
140
160
180
Figure 10.5 Contact pressures
Distance x [m]
0 1.625 3.25 4.875 6.5
0
0.1
0.2
Settlement s [cm]
0.3
0.4
0.5
0.6
0.7
0.8
Figure 10.6 Settlements
-10.14-
GEO Tools
Distance x [m]
-1000
-800
-600
-400
Moment M [kN.m]
-200
0 1.625 3.25 4.875 6.5
0
200
400
600
Figure 10.7 Moments
Distance x [m]
-1200
-800
-400
Shear force Qk [kN]
0 1.625 3.25 4.875 6.5

0
400
800
1200
Figure 10.8 Shear forces
-10.15-
10.2.4 Example 2: Design of a combined footing for two equal columns

It is required to find the contact pressure distribution, settlements, moment and shear force diagrams
for a rectangular combined footing for for two equal edge columns, each carrying a load of P=1125
[kN] as shown in Figure 10.9. Column sides are 0.50×0.50 [m2] while column reinforcement is 4Ф19.
The allowable soil pressure is qall=210 [kN/m2] at a depth of Df=1.5 [m] and average unit weight of
the soil and concrete is γa=20 [kN/m3]. Modulus of elasticity of the footing concrete is Eb=3×107,
while the Modulus of subgrade reaction of the soil is ks=35000 [kN/m3].
P R P =1125 [kN]
0.5×0.5 [m]
Ground surface
Df =1.5 [m] qall =210 [kN/m2]

γa =20 [kN/m3]
B =2.5 [m]
C.G
A =5.0 [m]
Figure 10.9 Combined rectangular footing
1. Determining footing sides A and B

Resultant of loads R at the ground surface level is given by:
R = 2 P = 2×1125 = 2250 [kN]
qnet = qall – γa Df= 210 – 20×1.5 = 180 [kN/m2]
R 2250
=Af = = 12 .5 [m 2 ]
qnet 180
2
Take Af = 2.5×5.0 = 12.5 [m ] rectangular combined footing
-10.16-
GEO Tools
10.2.4.1.1 Analysis of the combined footing


1 𝐼𝑖
𝑢𝑖 = (1 + )
2 𝐼𝑖−1
1 𝐼 1
𝑢𝑖 = (1 + ) = × 2 = 1
2 𝐼 2
1 𝐼𝑖 𝐼𝑖
𝑣𝑖 = ( + 14 + )
4 𝐼𝑖−1 𝐼𝑖+1
1 𝐼 𝐼 1
𝑣𝑖 = ( + 14 + ) = × 16 = 4
4 𝐼 𝐼 4
1 𝐼𝑖
𝑤𝑖 = (1 + )
2 𝐼𝑖+1
1 𝐼 1
𝑤𝑖 = (1 + ) = × 2 = 1
2 𝐼 2
𝐵𝑑𝑖 3 2.5 × 0.63

𝐼𝑖 = 𝐼 = = = 0.045 [m4 ]
12 12
𝑎 4𝐵 1.254 ×2.5
𝛼= = 30000000×0.045 = 4.52 × 10−6 [m3/kN]
𝐸𝑏 𝐼

The external moments M i(l ) at points 2 and 3 are:
(𝑙)
𝑀2 = 1125 × (1.875 − 0.25) = 1828.125 [kN.m]
(𝑙)
𝑀3 = 1125 × (3.125 − 0.25) = 3234.375 [kN.m]
-10.17-
P =1125 [kN] P =1125 [kN]

GS
1 2 3 4
k1 k2 k3 k4
a= 1.25 [m]
B =2.5 [m]
1 2 3 4
A =4×1.25=5 [m]
1 2 3 4
si
Settlement
a= 1.25 [m]
1 2 3 4
q2
Contact pressure
-10.18-
GEO Tools
10.2.4.1.4 Determining the right hand side Ri

The right hand side Ri of the contact pressure equation is:
(𝑙) (𝑙) (𝑙) 𝑎2

6𝐸𝐼𝑖
(𝑙) (𝑙) (𝑙) 1.252

𝑅𝑖 = (1 × 𝑀𝑖−1 + 4 × 𝑀𝑖 + 1 × 𝑀𝑖+1 )
6 × 30000000 × 0.045
(𝑙) (𝑙) (𝑙)
𝑅𝑖 = 1.93 × 10−7 (𝑀𝑖−1 + 4𝑀𝑖 + 𝑀𝑖+1 )
Apply the above equation at point 2:
(𝑙) (𝑙) (𝑙)

𝑅2 = 1.93 × 10−7 (𝑀1 + 4𝑀2 + 𝑀3 )
𝑅2 = 1.93 × 10−7 (0 + 4 × 1828.125 + 3234.375) = 2.036 × 10−3

1 2 𝛼𝑖 1 𝛼𝑖
( ) 𝑞𝑖+1 − ( − 𝑤𝑖 ) 𝑞𝑖 + ( + (𝑣𝑖 + 2𝑤𝑖 )) 𝑞𝑖−1
𝑘𝑖+1 𝑘𝑖 6 𝑘𝑖−1 6
𝑖−2
𝛼𝑖
6
𝑗=1
1 2 4.52 × 10−6 1 4.52 × 10−6

( ) 𝑞𝑖+1 − ( − ) 𝑞𝑖 + ( + (4 + 2)) 𝑞𝑖−1
35000 35000 6 35000 6
𝑖−2
4.52 × 10−6
+ (∑[(𝑖 − 𝑗 − 1) + 4 × (𝑖 − 𝑗) + (𝑖 − 𝑗 + 1)] 𝑞𝑗 ) = 𝑅𝑖
6
𝑗=1
or
𝑖−2
𝑞𝑖+1 − 1.974 𝑞𝑖 + 1.158 𝑞𝑖−1 + 0.026 (∑(𝑖 − 𝑗) 𝑞𝑗 ) = 35000𝑅𝑖

𝑗=1
Apply the above equation at points 2:
𝑞3 − 1.974 𝑞2 + 1.158 𝑞1 = 35000 × 2.036 × 10−3
-10.19-
−0.974 𝑞2 + 1.158 𝑞1 = 71.244
− 𝑞2 + 1.189 𝑞1 = 73.146
There are four unknown q1, q2, q3, and q4, so a farther equation is required. This can be obtained by
considering the overall equilibrium of vertical forces.
∑𝑉 = 0
𝑎𝐵( 𝑞1 + 𝑞2 + 𝑞3 + 𝑞4 ) = 𝑃1 + 𝑃2
or
𝑞1 + 𝑞2 = 360
1.189 −1 𝑞1 73.146
[ ][ 𝑞 ] = [ ]
1 1 2 360
𝑞1 197.874
[𝑞 ]=[ ] [kN/m2]
2 162.126

𝑞𝑖 𝑞𝑖
𝑠𝑖 = = [m]
𝑘𝑖 35000
s1 = 0.57 [cm]
s2 = 0.46 [cm]
The contact pressure distribution, settlement, moment and shear force diagrams for the raft are shown
in Figure 10.11 to Figure 10.14. Once the internal forces are obtained at various sections, the design
of the raft can be completed in the normal manner.

-10.20-
GEO Tools
Distance x [m]
0 1.25 2.5 3.75 5
0
50
100
150
200
250
Distance x [m]
0 1.25 2.5 3.75 5
0
0.1
Settlement s [cm]
0.2
0.3
0.4
0.5
0.6
-10.21-
Distance x [m]
-1200
-1000
-800
Moment M [kN.m]
-600
-400
-200
0 1.25 2.5 3.75 5

0
200
Figure 10.13 Moments
Distance x [m]
-1500
-1000
-500
Shear force Qk [kN]
0 1.25 2.5 3.75 5

0
500
1000
1500
-10.22-
GEO Tools
10.2.5 Example 3: Design of a combined footing for three equal columns

An example is carried out to design a combined footing for three equal columns according to EC 2,
DIN 1045, ACI and ECP.
A rectangular combined footing of 0.5 [m] thickness with dimensions of 7.8 [m]×2.6 [m] is chosen
(Figure 10.15). The footing is supported to three equal columns, each of dimension 0.4 [m]×0.4 [m],
reinforced by 8Φ16 and carries a load of 1276 [kN]. The footing rests on Winkler springs that have
Modulus of Subgrade Reaction of ks=40000 [kN/ m3]. A thin plain concrete of thickness 0.15 [m] is
chosen under the footing and is not considered in any calculations.
10.2.5.2 Footing material and section
The footing material and section are supposed to have the following parameters:
Unit weight of concrete γb = 0.0 [kN/ m3]
Unit weight of concrete is chosen γb = 0.0 to neglect the own weight of the footing.
Section thickness t = 0.50 [m]
Steel bar diameter Φ = 18 [mm]
-10.23-
P = 1276 [kN] P = 1276 [kN] P = 1276 [kN]
Ground surface
1.3 [m] 2.6 [m] 2.6 [m] 1.3 [m]

B =2.6 [m]
A =7.8 [m]
Figure 10.15 Combined rectangular footing for three columns
10.2.5.5 Analysis of the combined footing


1 𝐼𝑖
𝑢𝑖 = (1 + )
2 𝐼𝑖−1
1 𝐼 1
𝑢𝑖 = (1 + ) = × 2 = 1
2 𝐼 2
1 𝐼𝑖 𝐼𝑖
𝑣𝑖 = ( + 14 + )
4 𝐼𝑖−1 𝐼𝑖+1
1 𝐼 𝐼 1
𝑣𝑖 = ( + 14 + ) = × 16 = 4
4 𝐼 𝐼 4
1 𝐼𝑖
𝑤𝑖 = (1 + )
2 𝐼𝑖+1
-10.24-
GEO Tools
1 𝐼 1
𝑤𝑖 = (1 + ) = × 2 = 1
2 𝐼 2
𝐵𝑑𝑖 3 2.6 × 0.53

𝐼𝑖 = 𝐼 = = = 0.027 [m4 ]
12 12
𝑎 4𝐵 1.954 ×2.6
𝛼= = 30000000×0.027 = 4.64 × 10−5 [m3/kN]
𝐸𝑝 𝐼

The external moments M i(l ) at points 2 and 3 are:
(𝑙)
𝑀2 = 1276(2.925 − 1.3) = 2073.5 [kN.m]
(𝑙) 1.95
𝑀3 = 1276(4.875 − 1.3) + 1276 = 5805.8 [kN.m]
2
-10.25-
P =1276 [kN] P =1276 [kN] P =1276 [kN]

GS
1 2 3 4
k1 k2 k3 k4
a= 1.95 [m]
B =2.6 [m]
1 2 3 4
A =4×1.95=7.8 [m]
1 2 3 4
si
Settlement
a= 1.95 [m]
1 2 3 4
q2
Contact pressure
-10.26-
GEO Tools

(𝑙) (𝑙) (𝑙) 𝑎2
6𝐸𝐼𝑖
(𝑙) (𝑙) (𝑙) 1.952

𝑅𝑖 = (1 × 𝑀𝑖−1 +4× 𝑀𝑖 +1× 𝑀𝑖+1 )
6 × 30000000 × 0.027
(𝑙) (𝑙) (𝑙)
𝑅𝑖 = 7.82 × 10−7 (𝑀𝑖−1 + 4𝑀𝑖 + 𝑀𝑖+1 )
Apply the above equation at point 2:
(𝑙) (𝑙) (𝑙)

𝑅2 = 7.82 × 10−7 (𝑀1 + 4𝑀2 + 𝑀3 )
𝑅2 = 7.82 × 10−7 (0 + 4 × 2073.5 + 5805.8) = 0.011

1 2 𝛼𝑖 1 𝛼𝑖
( ) 𝑞𝑖+1 − ( − 𝑤𝑖 ) 𝑞𝑖 + ( + (𝑣𝑖 + 2𝑤𝑖 )) 𝑞𝑖−1
𝑘𝑖+1 𝑘𝑖 6 𝑘𝑖−1 6
𝑖−2
𝛼𝑖
6
𝑗=1
1 2 4.64 × 10−5 1 4.64 × 10−5

( )𝑞 − ( − ) 𝑞𝑖 + ( + (4 + 2)) 𝑞𝑖−1
40000 𝑖+1 40000 6 40000 6
𝑖−2
4.64 × 10−5
+ (∑[(𝑖 − 𝑗 − 1) + 4 × (𝑖 − 𝑗) + (𝑖 − 𝑗 + 1)] 𝑞𝑗 ) = 𝑅𝑖
6
𝑗=1
or
𝑖−2
𝑞𝑖+1 − 1.691 𝑞𝑖 + 2.856 𝑞𝑖−1 + 1.856 (∑(𝑖 − 𝑗) 𝑞𝑗 ) = 40000𝑅𝑖

𝑗=1
Apply the above equation at points 2:
𝑞3 − 1.691 𝑞2 + 2.856 𝑞1 = 40000 × 0.011
−0.691 𝑞2 + 2.856 𝑞1 = 440
-10.27-
− 𝑞2 + 4.133 𝑞1 = 636.758
∑𝑉 = 0
𝑎𝐵( 𝑞1 + 𝑞2 + 𝑞3 + 𝑞4 ) = 𝑃1 + 𝑃2 + 𝑃3
or
𝑞1 + 𝑞2 = 377.515
4.133 −1 𝑞1
[ ] [ 𝑞 ] = [636.758]
1 1 2 377.515
𝑞1 197.600
[𝑞 ]=[ ] [kN/m2]
2 179.917

𝑞 𝑞
𝑠𝑖 = 𝑘𝑖 = 40000
𝑖
[m]
𝑖
s1 = 0.49 [cm]
s2 = 0.45 [cm]

For the different codes, the footing is designed to resist the bending moment and punching shear.
Then, the required reinforcement is obtained. Finally, a comparison among the results of the four
codes is presented.
-10.28-
GEO Tools
Distance x [m]
0 1.95 3.9 5.85 7.8
0
50
100
150
200
250
Distance x [m]
0 1.95 3.9 5.85 7.8
0
0.1
Settlement s [cm]
0.2
0.3
0.4
0.5
0.6
-10.29-
Distance x [m]
0 1.95 3.9 5.85 7.8
0
100
200
Moment M [kN.m]
300
400
500
600
Distance x [m]
-800
-600
-400
-200
Shear force Qk [kN]
0 1.95 3.9 5.85 7.8

0
200
400
600
800
-10.30-
GEO Tools
10.2.6 Example 4: Design of a combined footing for three unequal columns

Design a combined rectangular footing for three unequal columns. Edge columns C1 and C3 have
dimensions of 0.3×0.5 [m2], reinforcement 6Φ19 and a load of 1000 [kN], while the central column
C2 has dimensions of 0.5×0.75 [m2], reinforcement 10Φ25 and a load of 3000 [kN] as shown in Figure
10.1Figure 10.1. The allowable soil pressure qall = 210 [kN/m2] at a depth of Df = 2.0 [m] and the
average unit weight of the soil and concrete is γa = 20 [kN/m3]. The footing rests on Winkler springs
that have Modulus of Subgrade Reaction of ks=30000 [kN/ m3]. A thin plain concrete of thickness
0.15 [m] is chosen under the footing and is not considered in any calculations. The loading and the
footing are symmetrical.
10.2.6.2 Footing material and section
The footing material and section are supposed to have the following parameters:
Unit weight of concrete γb = 0.0 [kN/ m3]
Unit weight of concrete is chosen γb = 0.0 to neglect the own weight of the footing.
Section thickness t = 0.8 [m]
Steel bar diameter Φ = 18 [mm]
-10.31-
P 1=1000 [kN] P 2=3000 [kN] P 3=1000 [kN]
Ground surface
Df =2.0 [m]
qall = 210 [kN/m2]

γa = 20 [kN/m3]
0.25 0.25
4.75 [m] 4.75 [m]
B = 3.0 [m]
C1 C2 C3
1000 [kN] 3000 [kN] 1000 [kN]
A =10.0 [m]
Figure 10.1 Combined rectangular footing for three unequal columns

1. Determining footing sides A and B
R = 2 P1 + P2 = 2 × 1000 + 3000 = 5000 [kN]
qnet = qall – γaDf= 210 – 20 × 2.0 = 170 [kN/m2]
R 5000
Af = = = 29 .41 [m 2 ]
qnet 170
take Af = 10.0 × 3.0 = 30.0 [m2] rectangular combined footing
-10.32-
GEO Tools
10.2.6.5 Hand calculation

The footing can be regarded as a beam on elastic foundation subjected to three unequal concentrated
forces. The beam is divided into eight equal elements, each 1.25 [m] long (Figure 10.21). Because of
the symmetry of the system, the analysis can be carried out by considering only half of the beam.
Hence, the total number of equations is reduced to four.
According to Kany/ El Gendy (1995), the analysis of beam on elastic foundation is carried out in the
following steps:

1 𝐼𝑖
𝑢𝑖 = (1 + )
2 𝐼𝑖−1
1 𝐼 1
𝑢𝑖 = (1 + ) = × 2 = 1
2 𝐼 2
1 𝐼𝑖 𝐼𝑖
𝑣𝑖 = ( + 14 + )
4 𝐼𝑖−1 𝐼𝑖+1
1 𝐼 𝐼 1
𝑣𝑖 = ( + 14 + ) = × 16 = 4
4 𝐼 𝐼 4
1 𝐼𝑖
𝑤𝑖 = (1 + )
2 𝐼𝑖+1
1 𝐼 1
𝑤𝑖 = (1 + ) = × 2 = 1
2 𝐼 2
𝐵𝑑𝑖 3 3 × 0.83
𝐼𝑖 = 𝐼 = = = 0.128 [𝑚4 ]
12 12
𝑎 4𝐵 1.254 ×3
𝛼= = 30000000×0.128 = 1.91 × 10−6 [m3/kN]
𝐸𝑝 𝐼
-10.33-
P 3=1000 [kN]
GS P 1=1000 [kN] P 2=3000 [kN]
1 2 3 4
k1 k2 k3 k4
a= 1.25 [m]
B =3 [m] 2 6
1 3 4 5 7 8
A =8×1.25=10 [m]
si
Settlement
a= 1.25 [m]
q4
Contact pressure
-10.34-
GEO Tools
(𝑙)
𝑀1 = 𝑧𝑒𝑟𝑜
(𝑙)
𝑀2 = 1000(1.5 × 1.25 − 0.25) = 1625 [kN.m]
(𝑙)
𝑀3 = 1000(2.5 × 1.25 − 0.25) = 2875 [kN.m]
(𝑙)
𝑀4 = 1000(3.5 × 1.25 − 0.25) = 4125 [kN.m]
(𝑙)
𝑀5 = 1000(4.5 × 1.25 − 0.25) + 3000 × 0.5 × 1.25 = 7250 [kN.m]

(𝑙) (𝑙) (𝑙) 𝑎2

6𝐸𝐼𝑖
(𝑙) (𝑙) (𝑙) 1.252

𝑅𝑖 = (1 × 𝑀𝑖−1 + 4 × 𝑀𝑖 + 1 × 𝑀𝑖+1 )
6 × 30000000 × 0.128
(𝑙) (𝑙) (𝑙)
𝑅𝑖 = 6.8 × 10−8 (𝑀𝑖−1 + 4𝑀𝑖 + 𝑀𝑖+1 )
Apply the above equation at points 2, 3 and 4:
At point 2:
(𝑙) (𝑙) (𝑙)
𝑅2 = 6.8 × 10−8 (𝑀1 + 4𝑀2 + 𝑀3 )
𝑅2 = 6.782 × 10−8 (0 + 4 × 1625 + 2875) = 6.358 × 10−4
At point 3:
(𝑙) (𝑙) (𝑙)
𝑅3 = 6.782 × 10−8 (𝑀2 + 4𝑀3 + 𝑀4 )
𝑅3 = 6.782 × 10−8 (1625 + 4 × 2875 + 4125) = 1.17 × 10−3
At point 4:
(𝑙) (𝑙) (𝑙)
𝑅4 = 6.782 × 10−8 (𝑀3 + 4𝑀4 + 𝑀5 )
𝑅4 = 6.782 × 10−8 (2875 + 4 × 4125 + 7250) = 1.806 × 10−3
-10.35-

𝑖−2
1 2 𝛼 1
( ) 𝑞𝑖+1 − ( − ) 𝑞𝑖 + ( + 𝛼) 𝑞𝑖−1 + 𝛼 (∑(𝑖 − 𝑗) 𝑞𝑗 ) = 𝑅𝑖
𝑘 𝑘 6 𝑘
𝑗=1
1 2 1.91 × 10−6 1
( ) 𝑞𝑖+1 − ( − ) 𝑞𝑖 + ( + 1.91 × 10−6 ) 𝑞𝑖−1
30000 30000 6 30000
𝑖−2
+ 1.91 × 10−6 (∑(𝑖 − 𝑗) 𝑞𝑗 ) = 𝑅𝑖

𝑗=1
or
𝑖−2
𝑞𝑖+1 − 1.991 𝑞𝑖 + 1.0573 𝑞𝑖−1 + 0.0573 (∑(𝑖 − 𝑗) 𝑞𝑗 ) = 30000𝑅𝑖

𝑗=1
Apply the above equation at points 2, 3 and 4:
𝑞3 − 1.991 𝑞2 + 1.0573 𝑞1 = 19.074
𝑞4 − 1.991 𝑞3 + 1.0573 𝑞2 + 0.115 𝑞1 = 35.1
−0.991 𝑞4 + 1.0573 𝑞3 + 0.115 𝑞2 + 0.172 𝑞1 = 54.18
∑𝑉 = 0
𝑎𝐵( 𝑞1 + 𝑞2 + 𝑞3 + 𝑞4 + 𝑞5 + 𝑞6 + 𝑞7 + 𝑞8 ) = 𝑃1 + 𝑃2 + 𝑃3
or
𝑞1 + 𝑞2 + 𝑞3 + 𝑞4 = 666.667
1.0573 −1.991 1 0 𝑞1 19.074

0.115 1.0573 −1.991 1 𝑞2 35.1
[ ][ 𝑞 ] = [ ]
0.172 0.115 1.0573 −0.991 3 54.18
1 1 1 1 𝑞4 666.667
-10.36-
GEO Tools
𝑞1 171.021
𝑞2 163.505
[𝑞 ]=[ ] [kN/m2]
3 163.719
𝑞4 168.539

𝑞 𝑞
𝑠𝑖 = 𝑘𝑖 = 30000
𝑖
[m]
𝑖
s1 = 0.57 [cm]
s2 = 0.55 [cm]
s3 = 0.55 [cm]
s4 = 0.56 [cm]

-10.37-
Distance x [m]
0 1.25 2.5 3.75 5 6.25 7.5 8.75 10
0
20
40
60
80
100
120
140
160
180
Distance x [m]
0 1.25 2.5 3.75 5 6.25 7.5 8.75 10
0
0.1
Settlement s [cm]
0.2
0.3
0.4
0.5
0.6
-10.38-
GEO Tools
Distance x [m]
-1000
-500
0 1.25 2.5 3.75 5 6.25 7.5 8.75 10

0
Moment M [kN.m]
500
1000
1500
2000
Distance x [m]
-2000
-1500
-1000
-500
Shear force Qk [kN]
0 1.25 2.5 3.75 5 6.25 7.5 8.75 10

0
500
1000
1500
2000
-10.39-
10.2.7 Example 5: Design of a combined footing for three equal columns
Design a combined footing for three equal columns, each of 0.50×0.50 [m], reinforced by 4Ф19,
carrying a load of P = 1500 [kN] as shown in Figure 10.26. The distance center to center of columns
is 3.0 [m]. The allowable soil pressure is qall = 210 [kN/m2] at a depth of Df= 2.0 [m] and average
unit weight of the soil and concrete is γa = 20 [kN/m3].
P = 1500 [kN] P = 1500 [kN] P = 1500 [kN]
0.5×0.5 [m]
Ground surface
Df = 2.0 [m]
qall = 210 [kN/m2]

γa = 20 [kN/m3]
1.5 [m] 3.0 [m] 3.0 [m] 1.5 [m]

B = 3.0 [m]
L = 9.0 [m]
Figure 10.26 Combined rectangular footing for three columns
10.2.7.1.1 Determining footing sides L and B

R = 3 P = 3×1500 = 4500 [kN]
qnet = qall – γaDf= 210 – 20×2.0 = 170 [kN/m2]
R 4500
Af = = = 26 .47 [m 2 ]
qnet 170
take Af = 3.0×9.0 = 27.0 [m2] rectangular combined footing

10.2.7.2 Computing the contact pressure
The contact pressure per meter square under the base of the footing will be uniform. It is given by:
-10.40-
GEO Tools
R 4500
qo = = = 166 .67 [kN/m 2 ] = 0.167 [MN/m 2 ]
Af 27
Now, the rectangular footing is treated as a beam footing. The contact pressure per meter is given by:
q = qo B = 0.167×3.0 = 0.5 [MN/m]
Figure 10.27 shows the load diagram and moment diagram for the beam footing.
10.2.7.3 Determining the maximum moment Mmax
From Figure 10.27, the maximum bending moment in the longitudinal direction occurs at the middle
of the footing, which is:
c2 1.25 2
M max =P = 0.5 = 0.391 [MN.m]
2 2
P = 1.5 [MN] P = 1.5 [MN] P = 1.5 [MN]

t = 0.5[m]
d = 0.45
Asbx Asbx Asbx

Asty Astx Asty Astx
q = 0.5 [MN/m]
c= 1.25 m
Load diagram
Mmax = 0.391 [MN.m]
Chosen 6Ф19/m
Bending moment diagram
Bottom reinforcement Asxb
Figure 10.27 Contact pressure, bending moment and critical sections
10.2.7.4 Determining the depth required to resist the moment dm
From Table 2 for fc = 9.5 [MN/m2] and fs = 200 [MN/m2], the coefficient k1 to obtain the section depth
at balanced condition is k1 = 0.766, while the coefficient k2 [MN/m2] to obtain the tensile
reinforcement for singly reinforced section is k2 = 172 [MN/m2].
The maximum depth dm as a singly reinforced section is given by:
-10.41-
M max
d m = k1
B
0.391
d m = 0.766 = 0.28 [m]
3.0
Take d = 0.45 [m] >dm = 0.28 [m], then the section is designed as singly reinforced section.
The corresponding k1 for d = 0.45 [m] is given by:
3.0
k1 = 0.45 = 1.25
0.391
From Table 2 at k1 = 1.27, fc = 5.0 [MN/m2] and k2 = 182

10.2.7.5 Check for punching shear
The critical punching shear section lies on a perimeter at a distance d/2 = 0.225 [m] from the face of
the column as shown in Figure 10.28.
P = 1.5 [MN] P = 1.5 [MN] P = 1.5 [MN]
0.5×0.5
t = 0.6 [m]
d = 0.55
1:2 1:2 1:2
qo= 0.167 [MN/m2] qo qo

a ) Section I-I
0.95 [m]
B = 3.0 [m]
0.95 [m]
I I
b) Plan L = 6.0 [m]

Figure 10.28 Critical section for punching shear
10.2.7.5.1 Geometry (Figure 10.28)

Effective depth of the section d = 0.45 [m]
Column side a = b = 0.5 [m]
Area of critical punching shear section Ap = (a + d)2 = 0.9025 [m2]
Perimeter of critical punching shear section bo = 4 (a + d) = 3.8 [m]
-10.42-
GEO Tools
10.2.7.5.2 Loads and stresses

Column load P = 1. 5 [MN]
Soil pressure under the column qo = 0.167 [MN/m2]
Main value of shear strength for concrete C 250 qcp = 0.9 [MN/m2]
10.2.7.5.3 Check for section capacity

The punching shear force Qp is:
Qp = P – qo Ap
Qp = 1.5 – 0.167×0.9025 = 1.35 [MN]
The punching shear stress qp is given by:
Qp
qp =
bo d
1.35
qp = = 0.79 [MN/m 2 ]
3.8  0.45
The allowable concrete punching strength qpall [MN/m2] is given by:
 a
q pall =  0.5 +  qcp  qcp
 b
 0.5 
q pall =  0.5 +  0.9  0.9
 0.5 
q pall = 0.9 [MN/m 2 ]
qpall = 0.9 [MN/m2] >qp = 0.79 [MN/m2], the section is safe for punching shear.
10.2.7.6 Computing the area of steel reinforcement
Minimum area of steel reinforcement Asmin = 0.15% Ac = 0.0015×50×100 = 7.5 [cm2/ m]
Take As min = 5Φ16 = 10.1 [cm2/ m].
The required area of steel reinforcement As is:
M max
Asxb =
k2 d
0.391
Asxb = = 0.004774 [m 2 /3.0 m]
182  0.45
Asxt = 47 .74 [cm 2 /3.0 m] = 15 .91 [cm 2 /m]
Chosen steel 6Φ19/m = 17.0 [cm2/m]
-10.43-
10.2.7.7 Computing the steel reinforcement in transverse direction

The transverse bending moment may be approximately determined at each column by assuming that
the column load is distributed outward at 45 [°] from the face of the column over an appropriate area
as shown in Figure 10.29. Both sides of this area should not be greater than the breadth of the footing.
P = 1.5 [MN] P P
t=0.5
0.5×0.5
t = 0.5 [m]
1:1 1:1 1:1
P w= 0.667 [MN/m2] Pw Pw
a ) Elevation 2×0.5+0.5=1.5 m
a
B = 3.0 [m]
LT = 1.5 [m]
a a
I I
b) Plan L = 6.0 [m]

Figure 10.29 Sections for determining transverse moments
Side length of the loaded area LT is given by:
LT = a + 2 t = 0.5 + 2×0.5 = 1.5 [m]
The transverse bending moment MT under the column is given by:
P (LT − a ) 1.5 (1.5 − 0.5)

2 2
MT = = = 0.125 [MN.m]
8 LT 8 1.5
The required area of steel reinforcement in transverse direction under the column AsT is:
MT
AsT =
k2 d
0.125
AsT = = 0.001526 [m2 /1.5m]
182  0.45
AsT = 15.26 [cm2/1.5m] = 10.17 [cm2/m]
-10.44-
GEO Tools
Chosen steel 6Ф16/m = 12.1 [cm2/m]. The details of reinforcement in plan and section a-a through
the footing are shown Figure 10.30.
0.5×0.5
5Ф16/m
0.5 [m]
0.15
6Ф19/m 6Ф16/m 6Ф16/m 6Ф16/m

6Ф19/m 6Ф19/m
1.5 [m] 3 [m] 3 [m] 1.5 [m]

a ) Section I-I
0.15
5Ф16/m
5Ф16/m 5Ф16/m
3.0 [m]
6Ф16/m
I 4Ф19 I
6Ф19/m
6Ф19/m 6Ф19/m
0.15
0.15 [m] 9 [m] 0.15 [m]

b) Plan
Figure 10.30 Details of reinforcement in plan and section a-a through the footing
-10.45-

10-Analysis of Beam Foundation-En-Part III

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Beam Foundations after Kany and El Gendy

Part III: Numerical Examples

qall = 210 [kN/m2]

1000 [kN] 3000 [kN] 1000 [kN]

10 Analysis and Design of Beam Foundations after Kany and El Gendy

10.2 Numerical Examples

10.2.1 Calculation methods

1 Linear contact pressure method

Concentrated load P 1 Uniform load pF 1 P2 GS

10.2.2 Material and section for concrete design

10.2.3 Example 1: Design of a combined footing for two unequal columns

10.2.3.1 Description of the problem

qall = 185 [kN/m2]

0.25 L= 4.75 [m]

Property line 1050 [kN] 1800 [kN]

Figure 10.2 Combined rectangular footing

10.2.3.1.1 Determining point of application of the resultant force

R = P1 + P2 = 1050 + 1800 = 2850 [kN]

Take the moment about C1 to get the distance S:

10.2.3.1.2 Determining footing sides A and B

A =2 Xp = 2 (0.25 + S) = 2 (0.25 + 3.0) = 6.5 [m]

Allowable net soil pressure qnet is given by:

qnet = qall – γaDf = 185 – 20×1.9 = 147 [kN/m2]

Area of footing Af is obtained from:

0.25 L=4.75 [m] 1.5 [m]

Property line 1050 [kN] 1800 [kN]

10.2.3.1.3 Analysis of the combined footing

10.2.3.1.4 Calculation of ui, vi and wi:

10.2.3.1.5 Determining external moments M i(l )

Beam foundation with loads and springs

A =4×1.625= 6.5 [m]

Figure 10.4 Combined footing as a beam on elastic foundation

10.2.3.1.6 Determining the right-hand side Ri

(𝑙) (𝑙) (𝑙) 𝑎2

(𝑙) (𝑙) (𝑙) 1.6252

Apply the above equation at points 2 and 3:

𝑅2 = 5.29 × 10−7 (0 + 4 × 2296.875 + 4003.125) = 0.00698

𝑅3 = 5.29 × 10−7 (2296.875 + 4 × 4003.125 + 6946.875 ) = 0.0134

10.2.3.1.7 Determining contact pressures

1 2 2.514 × 10−5 1 2.514 × 10−5

𝑞𝑖+1 − 1.895 𝑞𝑖 + 1.629 𝑞𝑖−1 + 0.629 (∑(𝑖 − 𝑗) 𝑞𝑗 ) = 25000𝑅𝑖

Apply the above equation at points 2 and 3:

𝑞3 − 1.895 𝑞2 + 1.629 𝑞1 = 174.5

𝑞4 − 1.895 𝑞3 + 1.629 𝑞2 + 1.258 𝑞1 = 334.014

Overall equilibrium of vertical forces

𝑎𝐵( 3.5𝑎𝑞1 + 2.5𝑎𝑞2 + 1.5𝑎𝑞3 + 0.5𝑎𝑞4 ) = 𝑃1 (𝐴 − 0.25) + 𝑃2 1.5

1.625 × 3( 𝑞1 (3.5 × 1.625) + 𝑞2 (2.5 × 1.625) + 𝑞3 (1.5 × 1.625) + 𝑞4 (0.5 × 1.625))

3.5𝑞1 + 2.5𝑞2 + 1.5𝑞3 + 0.5𝑞4 = 1169.231

Contact pressure equations in matrix form:

10.2.3.1.8 Determining settlements si

10.2.3.1.9 Computer calculation

Figure 10.5 Contact pressures

Figure 10.6 Settlements

Figure 10.7 Moments

0 1.625 3.25 4.875 6.5

Figure 10.8 Shear forces

10.2.4 Example 2: Design of a combined footing for two equal columns

10.2.4.1 Description of the problem

Df =1.5 [m] qall =210 [kN/m2]

1. Determining footing sides A and B