For TY Civil Engineering Students

1. 1
Module No. 4: Design of Shallow Foundation
Dr. Mohd. Zameeruddin
[PhD (VJTI), ME (Structure), MIE, AIV]
Associate Professor
Civil Engineering Department
MGM’s College of Engineering, Nanded
Contact:
md_zameeruddin@mgmcen.ac.in
zameerstd1@hotmail.com
M: +919822913231
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2. Classification of Foundation
Shallow
Foundation
Deep
Foundation
According to Terrazaghi
D ≤ B D > B
1. Wall footing
2. Strip footing
3. Spread footing
4. Combined footing
5. Mat or raft footing
6. Pier foundation
7. Pile foundation
8. Well foundation

3. Spread Footing or Simple Footing
 Function is to transfer load of isolated
columns
 To spread pressure to larger area
Pressure distribution depends on rigidity of footing, type of soil and
condition of soil
From pressure distribution – Bending Moments, Shear forces and
thickness of structural member
Design of foundation is done

4. Safe Bearing Pressure
Allowable
bearing capacity
1. Safe bearing capacity based on
ultimate capacity
2. Allowable bearing pressure on
tolerable settlement
Computation of safe bearing pressure based on ultimate capacity
For Strip Footing
For Square Footing

5. For Circular Footing
For Rectangular Footing
Where;
D = Depth of footing, B = Width of footing (Strip or Square) or
diameter of circular footing, L = Length of the footing
Nc, Nq, and Nγ are bearing capacity factor for general shear failure
N’c, N’q, and N’γ are bearing capacity factor for local shear failure
Rw1 and Rw2 are water reduction factors
F is factor of safety (ranging between 2 to 3)

6. Computation of safe bearing pressure based on limiting
settlement of individual footing to 25 mm
Where;
qp is allowable net increase in soil pressure over existing soil pressure
for the settlement of 25 mm
N is the standard penetration number with applicable corrections,
B is the width of footing,
Rw2 is water reduction factor
Rd is depth factor equals to (1+0.2(D/B) ≤ 1.20
If the settlement is ‘S’’’ in place of S=25mm, the corresponding net
pressure q’
p is given by,

7. Settlement in Footing S = Si + Sc + Ss
Where;
Immediate Settlement (Si) is
Primary Consolidation Settlement (Si) is
Secondary Consolidation Settlement (Si) is

8. Numericals
N1: Design a strip footing to carry a load of 750 kN/m at a depth of
1.6m in a C-ϕ soil having a unit weight of 18 kN/m3 and shear
strength parameter as 20 kN/m2 and ϕ equals to 250. Determine the
width of the footing using a factor of safety of 3 against shear failure.
Use Terzaghi equations (assume general shear failure).
Solution:
Given data:
Depth of footing = 1.6 m
Unit weight of soil (γ) = 18 kN/m3
Shear strength (C)= 20 kN/m2
ϕ =250, Nc = 25.1, Nq = 12.7, and Nγ = 9.7
The safe bearing capacity of strip footing is given as;

9. Equation (A)
Now, applied load intensity,
Equation (B)
Now, Equating equation A and B, we get
Numericals
B = 2.04 m say 2.10 m

10. Now, consider that this footing supports a concrete wall of thickness 300 mm.
Then the interest lies to know about the bending moment, shear force attracted
by the footing and the sectional details such as required thickness and rebar to
resists these forces. Material properties are M20 grade of concrete and Fe 415
grade of steel
Evaluation of bending moment attracted by the footing
B
x
Critical
Section
Net upward
Pressure (q’)
The forces applied on footing are
dead load and live load.
The net upward pressure on
footing is
= (750/2.1 x 1)
= 357.14 kN/m

11. Bending Moment (MXX)
B
x
Critical
Section
Net upward
Pressure (q’)
X
X
The critical section under bending lies at the junction of the wall and
the footing.
Let B be the width of footing and b be the width of wall
b
Then the projection x from the
critical section is,
The moment of forces on critical
section will be
= upward pressure x lever arm

12. Bending Moment (MXX)
The thickness of footing required to resist the applied moment is
given by the relationship
In Working Stress Method
In Limit State Method
Since WSM is not in regular practice lets being with LSM approach
For M 20 Concrete and Fe415 steel, Ru = 2.76
d = 275.11 mm say 280 mm
D = 280 + 40 (Cover) = 320 mm

13. Shear Force (V)
Bx
X
X
d
Net upward
Pressure (q’)
The critical section for shear will lie at a distance d
from the edge of wall having one way shear action
This shear force should not exceed
the shear stress capacity of footing
material. In case of concrete this
relationship is given as

14. Shear Force (V)
Hence adopt higher value d = 0.90
Rebar requirement
Hence Provide 12 mm φ @
150 mm c/c

15. N3: A square footing located at a depth of 1.3m below the ground has
to carry a safe load of 800 kN. Find the size of the footing if the
desired factor of safety is 3. The soil has the following properties:
Voids ratio = 0.55; Degree of saturation = 50%; Specific Gravity
2.67; C = 8kN/m2, ϕ = 300
Solution:
The bulk unit weight of soil is given by;
For ϕ = 300, Nc = 37.2, Nq = 22.5, and Nγ = 19.7
Equation (A)

16. Now, applied load intensity,
Equation (B)
Now, Equating equation A and B, we get
B = 1.42 m say 1.45 m

17. Now, consider that this footing supports a concrete column of size 300 x 450
mm. Then the interest lies to know about the bending moment, shear force
attracted by the footing and the sectional details such as required thickness and
rebar to resists these forces. Material properties are M20 grade of concrete and
Fe 415 grade of steel
Evaluation of bending moment attracted by the footing
B
x
Critical
Section
Net upward
Pressure (q’)
The forces applied on footing are
dead load and live load.
The net upward pressure on
footing is
= (800/1.45 x 1.45)
= 380.50 kN/m

18. Bending Moment (MXX)
B
x
Critical
Section
X
X
The critical section under bending lies at
the junction of the Column and the
footing. Let B be the width of footing and
a be the width of column, b transverse
depth of the column. Then the projection x
from the critical section is,
b
The moment of forces on critical
section will be
= upward pressure x lever arm
a
b
Net upward
Pressure (q’)

20. Shear Force (V)
d
One Way Shear (Beam Shear)
d/2
d/2
Two Way Shear (Punching Shear)
d

21. One Way Shear (Beam Shear)
d = 0.573 m say 0.575 m
D = 575+ 40 (Cover) = 615 mm
d = 0.498 m say 0.500 m
D = 500+ 40 (Cover) = 540 mm

22. Two Way Shear (Punching Shear)
d = 1.047 m say 1.05 m
D = 1050+ 40 (Cover) = 1090 mm
Hence adopt maximum depth required = 1090 mm

23. Rebar requirement

24. N4: Determine the depth of at which a circular footing of 2m diameter
Be founded to provide a factor of safety of 3,if it has to carry a load of
1800kN.The foundation soil has ‘c’=12kN/m2 ,ϕ=300 and unit weight
=17.8 kN/m3.Use terazaghi’s analysis:
Solution:
Given data:
for ϕ = 300 , NC = 37.2, Nq = 22.5, Nγ = 19.7
C = 12 kN/m2, γ = 17.8 kN/m3, FOS = 3, Diameter of footing = D
For Circular Footing
Equation (A)

25. Now, applied load intensity,
Equation (B)
Now, Equating equation A and B, we get
D = 2.13 m, Say 2.15 m

26. Combined Footing
Combined rectangular footing
Combined Trapezoidal footing
Strap Footing
StrapBeam
A combined footing is a large footing supporting
two or more column in a row. The combined
footing is mainly provided in following
circumstances;
1. When the foundation of the columns is
overlapped, that is, the distance between the
column is very less.
2. When the safe bearing capacity of soil is too
low.
3. When the exterior column is near about the
property line

27. Design of Combined Footing by Rigid Method (Conventional Method)
Assumptions:
1. The footing or mat is infinitely rigid, and therefore the defection
of the footing or mat does not influence the pressure distribution.
2. The soil pressure is distributed in a straight line or a plane surface
such that the centroid of the soil pressure coincides with the line
of action of the resultant force of all loads acting on the
foundation.

28. Combined rectangular footing
l
V
P1 P2a b
L
P1 = P2C1 C2
Let,
P1 be the load acting on
column C1
P2 be the load acting on
column C2
Let P1 and P2 be equal
in magnitude, hence
one may go with
rectangular shape of
footing.
Let,
B be the width of footing, and L be the length of the footing
Hence total load on footing = P1+ P2 + 10-20% (P1+P2) (as self
weight of the footing)
Then area of footing is;

29. Combined rectangular footing
Now, to obtain the length equate centroid due to area with centroid
due to loading as;
Geometric Centroid
Load Centriod
Required Length
Width of the footing (B)
Select suitable dimensions for the footing
Footing will act as a beam subjected to an upward pressure
distributed over the entire length and two concentrated load from the
columns downwards.
Net upward pressure

30. Combined rectangular footing
l
q1= P1/b1
a b
L
Shear Force Diagram
P1 = P2
C1 C2
The failure modes of footing will be determined by shear force and
bending moment
q2= P2/b2
1 2 3 4 5 6
1 2 3 54 6
Shear Force Calculation
q0
Select maximum value for design

31. Bending Moment Diagram
l
q1= P1/b1
a b
L
P1 = P2
C1 C2
1 2 3 4 5 6
q0
q2= P2/b2 Bending Moment
Calculations
Select maximum value for design

32. Thickness of the footing
For Maximum Bending Moment
For Maximum Shear Force
One way shear
Two way shear
Rebars

33. Combined Trapezoidal footing Geometric Centroid
Load Centriod
Required Length
lP1 P2a b
L
Shear Force Diagram
Bending Moment Diagram
b1 b2
L
P1 ≠ P2

34. Strap footing
Geometric Centroid
l
P1 P2a b
L1
Shear Force Diagram
Bending Moment Diagram
P1 ≠ P2
Strap Beam
L2
Load Centriod

35. Numericals:
N5: Proportion a rectangular footing given the following data:
Column loads: P1 = 1455 kN P2 = 1500 kN
Size of columns: 0.5 m x 0.5 m
c/c Spacing of columns = 6.2 m
Column one is located at a distance of 0.4m from edge of footing
Safe bearing capacity (qs) = 384 kN/m2
P1 P2
l = 6.2 m0.4 m
B
L
Required area of footing

36. Geometric Centroid
Load Centriod
Required Length

37. N6: Proportion a Trapezodial footing given the following data:
Column loads: P1 = 2016 kN P2 = 1560 kN
Size of columns: 0.46 m x 0.46 m
c/c Spacing of columns = 5.48 m
Column P1 is located along the property line
Safe bearing capacity (qs) = 190 kN/m2
B1 B2
l
0.23
Assume between 0.3-0.5 m
Required area of footing

38. Length of footing = 0.23 + 5.48 + 0.5 = 6.21 m
Geometric Centroid
Load Centriod
Substituting respective values we get B1 = 4.87 m and B2 = 1.80 m

39. PROPORTIONING OF CANTILEVER FOOTING
Strap or cantilever footings are designed on the basis of the
following assumptions:
1 . The strap is infinitely stiff. It serves to transfer the column loads
to the soil with equal and uniform soil pressure under both the
footings.
2. The strap is a pure flexural member and does not take soil
reaction. To avoid bearing on the bottom of the strap a few
centimeters of the underlying soil may be loosened prior to the
placement of concrete.

40. PROPORTIONING OF CANTILEVER FOOTING
Design Steps
Step 1: select trial values of “e”
Step 2: Compute R1 and R2 reactions
Step 3: Estimate tentative areas of footing
Step 4: Compute “e” in reference to computed areas, if matches stop

41. N5: Proportion a Starp footing given the following data:
Column loads: P1 = 1455 kN P2 = 1500 kN
Size of columns: 0.5 m x 0.5 m
c/c Spacing of columns = 6.2 m
Column with load P1 is located edge of footing
Safe bearing capacity (qs) = 384 kN/m2
Assuming trial “e” = 0.5 m
Hence LR will be = 6.2 -0.50 = 5.7 m
Compute R1 and R2

42. Compute tentative areas of footing
Provide footing of size 2.03 x 2.03 m
Provide footing of size 1.89 x 1.89 m
Re-compute eccentricity

43. Provide footing of size 2.07 x 2.07 m
Provide footing of size 1.84 x 1.84 m
Hence, provide footing of size 2.1 x 2.1 m for column 1 and
footing of size 1.8 x 1.8 m for column 2

44. MAT or RAFT FOUNDATION
A raft or mat is a combined footing that covers the entire area
beneath a structure and supports all the walls and columns.
The mat or raft footing is preferred;
 When the allowable soil pressure is low, or the building loads
are heavy, the use of spread footing would cover more that one-
half the area.
 Where the soil mass contains compressible lenses or the soil is
sufficiently erratic deposits and eliminates the differential
settlements.

45. MAT or RAFT FOUNDATION

46. MAT or RAFT FOUNDATION
Specifications:
1. The maximum differential settlement in foundations should not
exceeds 40 mm in foundation on clayey soils and 25 mm in
foundations on sandy soils.
2. The maximum settlement should generally be limited to the
following values;
Raft foundation on clay: 65 to 100 mm
Raft foundation on soil: 40 to 65 mm
There are two approaches for design
1. Conventional or Rigid Method
2. Elastic or Soil Line Method

47. MAT or RAFT FOUNDATION
(Rigid Method or Conventional Method IS 2950-1965)
In the conventional rigid method the mat is assumed to be infinitely
rigid and the bearing pressure against the bottom of the mat follows a
planar distribution where the centroid of the bearing pressure
coincides with the line of action of the resultant force of all loads
acting on the mat. The procedure of design is as follows:
1. The column loads of all the columns coming from the
superstructure are calculated as per standard practice. The loads
include live and dead loads.
2. Determine the line of action of the resultant of all the loads.
However, the weight of the mat is not included in the structural
design of the mat because every point of the mat is supported by
the soil under it, causing no flexural stresses.

48. MAT or RAFT FOUNDATION
(Rigid Method or Conventional Method)
3. Calculate the soil pressure at desired locations by the use of Eq.
Where;
q1 and q2 is the allowable soil pressure under the raft in kg/m2 (using
factor of safety of three)
Rw1 and Rw2 are the reduction factors on account of sub-soil water
The smaller of the two values shall be used for design
In case of saturated silts, the equivalent penetration resistance Ne
for the values of N greater than 15 should be taken for design

49. MAT or RAFT FOUNDATION
(Rigid Method or Conventional Method)
4. The pressure distribution under the raft is determined as;
Where;
Q is the total vertical shear on the raft
x, y are coordinates of any point on the raft with respect of x and
y axes passing through the centriod of the area of raft.
A is the total area of raft
are the moment of inertia and eccentricities about the
principal axes through the centroid of the section

50. 4. The mat is treated as a whole in each of two perpendicular
directions. Thus the total shear force acting on any section
cutting across the entire mat is equal to the arithmetic sum of
all forces and reactions (bearing pressure) to the left (or right)
of the section. The total bending moment acting on such a
section is equal to the sum of all the moments to the left (or
right) of the section
MAT or RAFT FOUNDATION
(Rigid Method or Conventional Method)

51. MAT or RAFT FOUNDATION
(Elastic Method or Soil Line Method)
1. Simplified Elastic Foundation
The soil is replaced by an infinite number of isolated springs
2. Truly Elastic Foundation
The soil is assumed to be continuous elastic medium obeying
Hooke’s Law
Limitations:
Foundation is comparatively flexible
Loads tends to concentrate over small areas

52. MAT or RAFT FOUNDATION
(Elastic Method or Soil Line Method)
The Coefficient of Subgrade Reaction (ks)
The coefficient of subgrade reaction is defined as the ratio between the
pressure against the footing or mat and the settlement at a given point
expressed as
Where;
ks = coefficient of subgrade reaction expressed as force/length3 (FL-3),
q = pressure on the footing or mat at a given point expressed as
force/length2 (FL-2),
S = settlement of the same point of the footing or mat in the
corresponding unit of length

53. In other words, the coefficient of subgrade reaction required to
produce a unit settlement.
In clayey soils, settlement under the load takes place over a long
period of time and the coefficient should be determined on the basis
of the final settlement. On purely granular soils, settlement takes
place shortly after load application.
It is based on two simplifying assumptions:
1. The value of ks is independent of the magnitude of pressure
2. The value of ks has the same value for every point on the surface
of the footing.
MAT or RAFT FOUNDATION
(Elastic Method or Soil Line Method)

54. MAT or RAFT FOUNDATION
(Elastic Method or Soil Line Method)
Measurement of ks
A value for ks1 for a particular subgrade can be obtained by
carrying out plate load tests. The standard size of plate used for this
purpose is 0.30 x 0.30 m size.
From experiments it has been found that for sands ks1 = k1.
For clays ks1 varies with the length of the beam. Terzaghi (1955)
gives the formula for clays

55. Soil Characteristics ks (kg/cm2)
Relative density Value of N Dry or moist
state
Submerged
state
Loose < 10 1.5 0.90
Medium 10 - < 30 4.7 2.9
Dense 30 and over 18.0 10.8
MAT or RAFT FOUNDATION
(Elastic Method or Soil Line Method)
Cohessionless
Soil
Soil Characteristics ks
(kg/cm2)Relative density Value of N
Stiff 1-< 2 2.7
Very stiff 2 - < 4 5.4
Hard 4 and over 10.8
Cohesive Soil

56. MAT or RAFT FOUNDATION
(Elastic Method or Soil Line Method)
Effect of Size
For cohessionless soil
For cohesive soil
K is modulus of subgrade reaction for footing of width B cm
ks is modulus of subgrade reaction for a square plate 30 cm x 30 cm
k‘ is modulus of subgrade reactions for footing width ‘x’ cm

57. MAT or RAFT FOUNDATION
(Elastic Method or Soil Line Method)
Effect of Shape
Where:
K1 is modulus of subgrade reaction for a rectangular footing having
a length L and width B
K2 is modulud of subgrade reaction for square footing of side B

58. MAT or RAFT FOUNDATION
(Elastic Method or Soil Line Method)
Soil Line method
The relationship between deflection y. At any point on an elastic
beam and the corresponding bending moment M may be expressed
by the equation
The equations for shear V and reaction q at the same point may be
expressed
where x is the coordinate along the length of the beam.

59. MAT or RAFT FOUNDATION
(Elastic Method or Soil Line Method)
From the basic assumption of an elastic foundation
q = -yBks
Where,
B = width of footing,
ks = coefficient of subgrade reaction
This is a closed form solution, but is not general in their
applications

60. Elastic Plate method
MAT or RAFT FOUNDATION
(Elastic Method or Soil Line Method)
This method is based on the assumption that the subgrade can be
substituted by a bed of uniformly distributed coil springs with a
spring ks which is called the coefficient of subgrade reaction.
The finite difference method uses the fourth order
differential equation.
Where;

61. MAT or RAFT FOUNDATION
(Elastic Method or Soil Line Method)
q = subgrade reaction per unit area,
ks = coefficient of subgrade reaction,
w = deflection
E = modulus of elasticity of the material of the footing
t = thickness of mat
= poisson’s ratio
This equation may be solved by dividing the mat into suitable square
grid elements, and writing difference equations for each of the grid
points. The equation can be solved rapidly with an electronic
computer. In today's modern age software's based on FEM methods
such as SAFE is in common use.
After knowing deflections bending moments and shear force
can be computed using relevant difference equation