Design of Foundations

Fall Semester 2024-25

K Karthikeyan
Introduction
 Superstructure is placed on foundation structure/
substructure as they are placed below the ground level
 Superstructure transfers loads and moments to the
foundation structure, which in turn, transfers them to the
underlying soil or rock
 Major requirements of design are
➢Transmitting the applied load effects to the soil below,
without exceeding SBC
➢Ensuring settlements within tolerable limits, uniform
➢Safe against Pullout, Overturning and Sliding
 Types: Shallow and Deep foundations
Foundations need special attention due to
 Foundations undergo soil-structure interaction, so behaviour
depends on properties of structural materials and soil
 Accurate estimations of all types of loads, moments and
forces are needed for present as well as future expansion
 Foundation have to be housed within property line which
may cause additional forces and moments due to the
eccentricity of foundation
 Foundations are in direct contact with soil and may get
affected due to harmful chemicals and minerals present in
soil and fluctuations of water table
 Foundation structures, while constructing, may affect
adjoining structures forming cracks to total collapse,
particularly during the driving of piles, etc.
Shallow Foundations
 Used when soil has sufficient strength within a short depth
 They need sufficient plan area to transfer loads to base soil
 These loads are sustained by RC columns or walls of much
less areas of cross-section due to high strength of bricks or
reinforced concrete when compared to that of soil
 Strength of soil, expressed as safe bearing capacity (SBC)
 Types: Plain Concrete Footing, Isolated Footing, Stepped
Footing, Sloped Footing, Combined Footing, Strap Footing
Strip footing and Raft Footing
 Isolated Footing
 Combined Footing
 Strap Footing and Strip Footing
 Raft Footing
Deep Foundations
 Used when soil has insufficient strength within a short depth
 Types: Piles and Caissons
 Piles resist load by either end bearing or skin friction or by
both methods
 Piles are small diameter columns which are driven or cast
into the ground by suitable means
 Normally provided in groups with a pile cap at the top
 Piles used in marshy land
 Length of pile depends on the availability of hard soil/rock
or the actual load test
 They resist uplift also in similar to compression forces
Pile Foundations
Safe Bearing Capacity (SBC) of Soil
 SBC (qc) - permissible soil pressure considering safety factors
in the range of 2 to 6 depending on the type of soil,
approximations and assumptions and uncertainties
 This is applicable under service load condition, partial safety
factors λf for different load combinations are to be taken
from those under limit state of serviceability
 acceptable value of qc is supplied by geotechnical consultant
 Safe bearing stress on soil is also related to corresponding
permissible settlement
 Gross and net bearing capacities are the two terms used in
the design
 Gross bearing capacity = total safe bearing pressure just
below the footing due to load of superstructure, self weight
of footing and weight of earth lying over footing.
 Net bearing capacity is the net pressure in excess of the
existing overburden pressure
 Net bearing capacity = Gross bearing capacity - Pressure
due to overburden soil
 While calculating maximum soil pressure q, consider all
loads of superstructure along with self-weight of foundation
and backfill
 For preliminary calculations, weight of foundation and
backfill may be taken as 10 to 15 per cent of the total axial
load on the footing, subjected to verification afterwards
Design Considerations
Minimum nominal cover
(Cl. 26.4.2.2 of IS 456)
 Minimum nominal cover for footings is more than that of
other elements as they are in direct contact with soil
 A minimum cover of 50 mm for footings is taken
 Actual cover can be more depending on presence of harmful
chemicals or minerals, water table, etc.
Thickness at the edge of footings
(Cl. 34.1.2 and 34.1.3 of IS 456)
 Minimum thickness at edge of reinforced and plain concrete
footings shall be at least 150 mm for footings on soils
 At least 300 mm above the top of piles for footings on piles,
as per the stipulation in cl.34.1.2 of IS 456
 For plain concrete pedestals, the angle α between plane
passing through bottom edge of pedestal and corresponding
junction edge of column with pedestal and horizontal plane
shall be determined from the following expression
tanα ≤ 0.9{(100 qa/fck) + 1}1/2
qa = calculated maximum bearing pressure at base of pedestal
in N/mm2
Bending moments (cl. 34.2 of IS 456)
Critical section of maximum bending moment for an isolated
concrete footing shall be:
 at face of column, pedestal or wall for footing supporting a
concrete column, pedestal or reinforced concrete wall
 halfway between the centre-line and the edge of the wall, for
footing under masonry wall
 For round or octagonal concrete column or pedestal, face of
column or pedestal shall be taken as side of a square
inscribed within perimeter of the round or octagonal
column or pedestal
Shear force
(Cl. 31.6 and 34.2.4 of IS 456)
 Shear force is the one of the governing forces in the
determination of depth of a footing
 Footing slabs face one way shear and two way shear
 Footing slabs shall be checked in one-way or two-way shears
depending on the nature of bending
 If the slab bends primarily in one-way, the footing slab shall
be checked in one-way vertical shear
 when the bending is primarily two-way, the footing slab shall
be checked in two-way shear or punching shear
One-way shear
(Cl. 34.2.4 of IS 456)
One-way shear has to be checked across the full width of the
base slab on a vertical section located from the face of the
column, pedestal or wall at a distance equal to
 effective depth of the footing slab in case of footing slab on
soil
 half the effective depth of the footing slab if the footing slab
is on piles
Two-way or punching shear (cls.31.6 and 34.2.4)
 Two-way or punching shear shall be checked around the
column on a perimeter half the effective depth of the footing
slab away from the face of the column or pedestal
 The permissible shear stress, when shear reinforcement is
not provided, shall not exceed ksτc, where ks = (0.5 + βc),
but not greater than one, βc being the ratio of short side to
long side of the column, and τc = 0.25(fck)1/2 in limit state
method of design
 Normally, the thickness of the base slab is governed by shear.
Hence, the necessary thickness of the slab has to be provided
to avoid shear reinforcement
Bond (cl.34.2.4.3 of IS 456)
 The critical section for checking the development length in a
footing slab shall be the same planes as those of bending
moments in part (c) of this section. Moreover, development
length shall be checked at all other sections where they
change abruptly. The critical sections for checking the
development length are given in cl.34.2.4.3 of IS 456,
which further recommends to check the anchorage
requirements if the reinforcement is curtailed, which shall
be done in accordance with cl.26.2.3 of IS 456.
 Tensile reinforcement (cl.34.3 of IS 456)
 The distribution of the total tensile reinforcement, calculated in
accordance with the moment at critical sections, as specified in
part (c) of this section, shall be done as given below for one-way
and two-way footing slabs separately.
 (i) In one-way reinforced footing slabs like wall footings, the
reinforcement shall be distributed uniformly across the full width
of the footing i.e., perpendicular to the direction of wall.
Nominal distribution reinforcement shall be provided as per cl.
34.5 of IS 456 along the length of the wall to take care of the
secondary moment, differential settlement, shrinkage and
temperature effects.
 (ii) In two-way reinforced square footing slabs, the
reinforcement extending in each direction shall be distributed
uniformly across the full width/length of the footing.
 (iii) In two-way reinforced rectangular footing slabs, the
reinforcement in the long direction shall be distributed
uniformly across the full width of the footing slab. In the
short direction, a central band equal to the width of the
footing shall be marked along the length of the footing,
where the portion of the reinforcement shall be determined
as given in the equation below. This portion of the
reinforcement shall be distributed across the central band:
 Reinforcement in the central band = {2/(β+1)} (Total
reinforcement in the short direction) (11.4)
 whereβ is the ratio of longer dimension to shorter
dimension of the footing slab (Fig.11.28.14).
 Each of the two end bands shall be provided with half of the
remaining reinforcement, distributed uniformly across the
respective end band.
 Transfer of load at the base of column (cl.34.4 of IS 456)
 All forces and moments acting at the base of the column must be
transferred to the pedestal, if any, and then from the base of the
pedestal to the footing, (or directly from the base of the column to the
footing if there is no pedestal) by compression in concrete and steel
and tension in steel. Compression forces are transferred through direct
bearing while tension forces are transferred through developed
reinforcement. The permissible bearing stresses on full area of
concrete shall be taken as given below from cl.34.4 of IS 456:
 brσ = 0.25fck, in working stress method, and (11.5)
 brσ = 0.45fck, in limit state method (11.6)
 It has been mentioned in sec. 10.26.5 of Lesson 26 that the stress of
concrete is taken as 0.45fck while designing the column. Since the area
of footing is much larger, this bearing stress of concrete in column may
be increased considering the dispersion of the concentrated load of
column to footing. Accordingly, the permissible bearing stress of
concrete in footing is given by (cl.34.4 of IS 456):
 brσ = 0.45fck (A1/A2)1/2 (11.7)
 with a condition that
 (A1/A2)1/2 2.0 (11.8) ≤
 where A1 = maximum supporting area of footing for bearing
which is geometrically similar to and concentric with the
loaded area A2, as shown in Fig.11.28.15
 A2 = loaded area at the base of the column.
 The above clause further stipulates that in sloped or stepped
footings, A1 may be taken as the area of the lower base of the
largest frustum of a pyramid or cone contained wholly
within the footing and having for its upper base, the area
actually loaded and having side slope of one vertical to two
horizontal, as shown in Fig.11.28.15.
 If the permissible bearing stress on concrete in column or in
footing is exceeded, reinforcement shall be provided for
developing the excess force (cl.34.4.1 of IS 456), either by
extending the longitudinal bars of columns into the footing
(cl.34.4.2 of IS 456) or by providing dowels as stipulated in
cl.34.4.3 of IS 456 and given below:
 (i) Sufficient development length of the reinforcement shall be
provided to transfer the compression or tension to the
supporting member in accordance with column (cl.34.4.2 of IS
456).
 (ii) Minimum area of extended longitudinal bars or dowels shall
be 0.5 per cent of the cross-sectional area of the supported
column or pedestal (cl.34.4.3 of IS 456).
 (iii) A minimum of four bars shall be provided (cl.34.4.3 of IS
456).
 (iv) The diameter of dowels shall not exceed the diameter of
column bars by more than 3 mm.
 (v) Column bars of diameter larger than 36 mm, in
compression only can be doweled at the footings with bars
of smaller size of the necessary area. The dowel shall extend
into the column, a distance equal to the development length
of the column bar and into the footing, a distance equal to
the development length of the dowel, as stipulated in
cl.34.4.4 of IS 456 and as shown in Fig.11.28.16.
 Nominal reinforcement (cl. 34.5 of IS 456)
 1. Clause 34.5.1 of IS 456 stipulates the minimum
reinforcement and spacing of the bars in footing slabs as per
the requirements of solid slab (cls.26.5.2.1 and 26.3.3b(2)
of IS 456, respectively).
 The nominal reinforcement for concrete sections of
thickness greater than 1 m shall be 360 mm2 per metre
length in each direction on each face, as stipulated in
cl.34.5.2 of IS 456. The clause further specifies that this
provision does not supersede the requirement of minimum
tensile reinforcement based on the depth of section.
 Distribution of Base Pressure
 The foundation, assumed to act as a rigid body, is in equilibrium
under the action of applied forces and moments from the
superstructure and the reactions from the stresses in the soil. The
distribution of base pressure is different for different types of
soil. Typical distributions of pressure, for actual foundations, in
sandy and clayey soils are shown in Figs.11.28.17 and 18,
respectively. However, for the sake of simplicity the footing is
assumed to be a perfectly rigid body, the soil is assumed to
behave elastically and the distributions of stress and stain are
linear in the soil just below the base of the foundation, as shown
in Fig.11.28.19. Accordingly, the foundation shall be designed for
the applied loads, moments and induced reactions keeping in
mind that the safe bearing capacity of the soil is within the
prescribed limit. It is worth mentioning that the soil bearing
capacity is in the serviceable limit state and the foundation
structure shall be designed as per the limit state of collapse,
checking for other limit states as well to ensure an adequate
degree of safety and serviceability.
 In the following, the distributions of base pressure are
explained for (i) concentrically loaded footings, (ii)
eccentrically loaded footings and (iii) unsymmetrical (about
both the axes) footings.
 Concentrically loaded footings
 Figure 11.28.20 shows rectangular footing symmetrically
loaded with service load P1 from the superstructure and P2
from the backfill including the weight of the footing. The
assumed uniformly distributed soil pressure at the base of
magnitude q is obtained from:
 q = (P1 + P2)/A (11.9)
 where A is the area of the base of the footing.
 In the design problem, however, A is to be determined from
the condition that the actual gross intensity of soil pressure
does not exceed qc, the bearing capacity of the soil, a known
given data.Thus, we can write from Eq.11.9:
 A = (P1 + P2)/qc (11.10)
 From the known value of A, the dimensions B and L are
determined such that the maximum bending moment in
each of the two adjacent projections is equal, i.e., the ratio
of the dimensions B and L of the footing shall be in the same
order of the ratio of width b and depth D of the column.
 Eccentrically loaded footings
 In most of the practical situations, a column transfers axial
load P and moment M to the footing, which can be
represented as eccentrically loaded footing when a load P is
subjected to an eccentricity e = M/P. This eccentricity may
also be there, either alone or in combined mode, when
 • the column transfers a vertical load at a distance of e from
the centroidal axis of the footing, and

 the column or the pedestal transfers a lateral load above the

level of foundation, in addition to vertical loads.
 Accordingly, the distribution of pressure may be of any one
of the three types, depending on the magnitude of the
eccentricity of the load, as shown in Figs.11.28.21b to d.
The general expression of qa, the intensity of soil pressure at
a distance of y from the origin is:
 qa = P/A (Pe/I±x)y (11.11)
 We would consider a rectangular footing symmetric to the
column. Substituting the values of A = BL, Ix = BL3/12 and y
= L/2, we get the values of qa at the left edge.
 qa at the left edge = (P/BL) {1 - (6e/L)} (11.12)
 It is evident from Eq.11.12, that the three cases are possible:
 (A) when e < L/6, qa at the left edge is compression (+),
 (B) when e = L/6, qa at the left edge is zero, and
 (C) when e > L/6, qa at the left edge is tension (-).
 The three cases are shown in Figs.11.28.21b to d,
respectively. It is to be noted that similar three cases are also
possible when eccentricity of the load is negative resulting
the values of qa at the right edge as compression, zero or
tension. Evidently, these soil reactions, in compression and
tension, should be permissible and attainable.
 Case (A): when | e | ≤ L/6
 Figures 11.28.21b and c show these two cases, when |e| <
L/6 or |e| = L/6, respectively. It is seen that the entire area
of the footing is in compression having minimum and
maximum values of q at the two edges with a linear and non-
uniform variation. The values of q are obtained from
Eq.11.11.
 In the limiting case i.e., when |e| = L/6, the value of qa is
zero at one edge and the other edge is having qa = 2P/BL
(compression) with a linear variation. Similarly, when e = 0,
the footing is subjected to uniform constant pressure of
P/BL. Thus, when |e| = L/6, the maximum pressure under
one edge of the footing is twice of the uniform pressure
when e = 0.
 In a more general case, as in the case of footing for the
corner column of a building, the load may have biaxial
eccentricities. The general expression of qa at a location of
(x,y) of the footing, when the load is having biaxial
eccentricities of ex and ey is,
 qa = P/A ± P exy/Ix ± P eyx/Iy (11.13)
 Similarly, it can be shown that the rectangular footing of
width B and length L will have no tension when the two
eccentricities are such that
 6ex/L + 6ey/B ≤ 1 (11.14)
 Case (B): when | e | > L/6
 Version
 The eccentricity of the load more than L/6 results in
development of tensile stresses in part of the soil. Stability,
in such case, is ensured by either anchoring or weight of
overburden preventing uplift. However, it is to ensure that
maximum compressive pressure on the other face is within
the limit and sufficient factor of safety is available against
over turning. Accordingly, the maximum pressure in such a
case can be determined considering the soil under
compression part only. Further, assuming the line of action
of the eccentric load coincides with that of resultant soil
pressure (Fig.11.28.22) we have:
 qmax = P/L'B + 12P(0.5 C)(1.5 C)/BL' = 2P/L'B (11.15)
 where L' = 3C (11.16)
 (iii) Unsymmetrical footings
 It may be necessary to provide some cutouts in the
foundation to reduce the uplift pressure or otherwise. The
footing in such cases becomes unsymmetrical about both the
axes. It is possible to determine the soil pressure distribution
using the structural mechanics principle as given below.
 qa(x,y) = P/A {(M±yIx - MxIxy)(x)/(IxIy - )} + {(M2xyIxIy -
MyIxy)(y)/(IxIy - )} (11.17) 2xyI
 where Mx = moment about x axis,
 My = moment about y axis,
 Ix = moment of inertia about x axis,
 Iy = moment of inertia about y axis,
 Ixy = product of inertia
Example 1
Design an isolated footing for a column of size 450 mm x 600
mm supported an axial load of 1500 kN (service load) and a
moment parallel to larger dimension of 125 kNm (service
moment). The net SBC is 180 kN/m2. Use M30 grade
concrete and Fe415 grade steel.
Step 1: Area of footing
Total load = Axial load * 1.15 = 1500*1.15 = 1725 kN
(15% for self-weight of footing + soil on the footing)
Since the column is rectangular, we can go with rectangular
footing with same proportions of column,
Column proportion = 450 : 600 = 1 : 1.33
(if 450 = 1, then 600 = 600/450)
e = P/M = 125 x 106 / 1725 x 103 = 72.46 mm
Assuming e < L/6,
P / A + M / Z ≤ SBC
1725 / BL + 120 / (BL2/6) ≤ 180
180 BL2 – 1725L – 120 ≤ 0
Assuming various combinations for B and L
B = 2 m , L = 4.86 m (proportion = 1 : 2.43)
B = 2.5 m , L = 3.9 m (proportion = 1 : 1.56)
B = 2.75 m , L = 3.55 m (proportion = 1 : 1.29)
The last combination is close to 1.33, so we can go for this
proportion
Adapt size 2.75 m x 3.75 m (rounding up)
Soil pressure = q =P / A + M / Z
= 1725 / (2.75 x 3.75) + (120 x6)/(2.75 x 3.752)
= 185.89 kN/m2 > 180 kN/m2

Revising the size, 3 m x 4 m, q = 158.75 kN/m2

Projection on longer side = (4 - 0.6)/2 = 1.7 m

Projection on shorter side = (3 - 0.4)/2 = 1.3 m
Step 2: Depth of footing
Max. Moment = (158.75 x 1.752 ) / 2 = 243.08 kNm/m
Mu = 243.08 x 1.5 = 364.62 kNm/m
Depth required,
Mu,l = 0.36 fck b xu,l (d – 0.416 xu,l)
Equating Mu = Mu,l
364.62 x 106 = 0.36x30x1000x0.48dx(d-(0.416x0.48d))
d = 297 mm
D = 297 + 50 + (32/2) = 363 mm
Provide D = 700 mm (to counter shear force)
dprovided = 700 – 66 = 634 mm
Step 3: Area of steel in Longer Direction
Mu,l = 0.36 fck b xu,l (d – 0.416 xu,l)
Mu,l = 0.36x30x1000x0.48x534x(634-(0.416x0.48x634))
Mu,l = 1182.59 kNm/m > Mu, singly r/f section

Ast = 1900 mm2/m

Bars Spacing
25# 258 mm
20# 165 mm
Provide 20# bars @ 100 mm c/c in longer direction
Ast, provided = 2093 mm2
Pt = 209300 / 1000 x 634 = 0.495

Step 4: One way Shear check

Vu = 158.75 x 1.116 x 1.5 = 265.75 kN
tv = 265750/ (1000 x 634) = 0.419 N/mm2
tc = 0.5 N/mm2
tc > tv, safe.
Step 5:Two way Shear check
Vu2 = 158.75 x ((3 x 4) – ((0.45+0.634)*(0.6+0.634))) x
1.5 = 2539 kN
tv = 2539000/ (1000 x (2*((0.45+0.634)*(0.6+0.634)))) =
0.494 N/mm2
ks tc = 0.5 N/mm2
tc = 0.25 sqrt(fck) = 0.25* √30 = 1.37 N/mm2
ks = (0.5 + βc), βc = 450/600 = 0.75
ks = 1
ks tc = 1.37 N/mm2
tc > tv, safe.
Step 6: Area of Steel in Shorter Direction
M = 158.75 x 1.32 / 2 = 134.144 kNm/m
Mu = 134.144 x 1.5 = 201.21 kNm/m
Mu,l = 1182.59 kNm/m > Mu, singly r/f section
Ast = 897 mm2

Bar Spacing
20# 350
16# 224

Provide 16# bars @ 200 mm c/c along shorter direction