DESIGN OF FOUNDATIONS
Foundations are required primarily to carry the dead and imposed loads
due to the structure’s floors, beams, walls, columns etc. and transmit and
distribute the loads safely to the ground.
The purpose of distributing the loads is to avoid the safe bearing capacity of
the soil being exceeded otherwise the excessive settlement of the structure
may occur.
FOUNDATION TYPES
There are many types of foundations, namely strip, pad and raft. The
foundations may bear directly on the ground or be supported on piles.
The choice of foundation type will largely depend upon (1) ground
condition i.e. strength and type of soil. (2) type of structure i.e. layout and
level of loading.
Diagram of pad and continuous footings
Pad footings are usually square or rectangular slabs and used to support a
single column. Continuous strip footings are used to support load bearing
walls or under a line of closely spaced columns.
DESIGN OF PAD FOOTINGS
The general procedure to be adopted for the design of pad footings is as
follows:
1. Calculate the plan area using the serviceability loads.
I.e Serviceability axial load (N) = 1.0 Gk + 1.0 Qk, Where Gk is characteristic
dead load and Qk is the characteristic live load.
2. Determine the reinforcement areas required for bending using ultimate
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loads.
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3. Check for bunching shear failures.
Example
A 400mm square column carries a dead load (Gk) of 1050 KN and an
Imposed load Qk) of 300KN. The safe bearing capacity of the soil is
170KNm-2. Design a square footing to resist the loading assuming the
following material strengths: fcu = 35Nm-2 fy = 500Nmm-2.
Solution
i. Plan area of the base.
Loading
Dead load
Assume a footing weight of 130KN.
Total dead load Gk = 1150 + 130 = 1080 KN
ii. Serviceability load
Design load (N) = 1.0Gk = + 1.0 Qk = 1.0X 1180 + 1.0 X 300 = 1480KN
N 1480
iii. Plan area of the base = Bearing capacity of the soil = 170 = 8.70 m2
Hence provide a 3m square base ( Plan area = 9m2)
iv. Design reinforced concrete to BS 8110
Weight of the footing
Assume the overall depth of the footing (h) = 600mm
2
Weight of the footing = Area X h x Density of the concrete.
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= 9 x 0.6 x 24 = 129.6 KN < assumed (130KN)
v. Bending reinforcement. M
Ultimate load (W) = 1.4 Gk + 1.6 Qk
= 1.4 X 1050 + 1.6 X 300 = 1950 KN
W 1950
Pressure = Plan area of the base = 9 = 217 KN/m
2
1.3
The design moment occurs at the face of the column (Mu) = 217 X 1.3 X 2
= 183 KNm per m length of
the footing
vi. Effective Depth
Footing is to be cast against blinding layer. Cover to reinforcement = 50mm,
assume 20mm diameter reinforcement.
Effective depth = 600 – 50 -20 = 530mm
M 183 X 10
6
vii. K = b d2 f = 2 = 0.0186 < 0.156. Therefore, design as singly
cu 1000 X 530 X 35
reinforced.
viii. [ √
The lever arm; Z = d 0.5+ 0.25− K
0.9 ]
√
= d [ 0.5 + 0.25− 0.0186 ]
0.9
= 0.97 < 0.95 d, therefore adopt 0.95 d
Therefore Z = 0.95d = 0.96 X 530mm = 503.5mm
M 183 X 10
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ix. Area of steel As = 0.87 f z = = 908.19 mm2/ m
y 0.87 X 460 X 503.5
Minimum steel area is 0.13% bh = 0.13/100 x 1000 x 600 = 780 mm 2 < As ,
Ok
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Hence provide 16mm diameter bars at 200mm centres. (As prov = 1010
mm2) across the full width of the footing parallel to the xx axis and the yy
axis.
x. Critical shear stresses
Punching shear.
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