OUT LINE
1) Determine the own
weight of building
2) Design of mat foundation
3) Design of pile foundation
INTRODUCTION
A foundation is the lower part of a
structure which transmits loads to the
underlying soil without causing a
shear failure of soil or excessive
settlement.
INTRODUCTION
if cracks appear in the structure it is
assumed that the foundation did move
and that this is the sole cause of cracking.
CHOICE OF THE TYPE OF FOUNDATION

The choice of the appropriate type of foundation
is governed by some important factors such as

1. The nature of the structure

2. The loads exerted by the structure

3. The subsoil characteristics

4. The allotted cost of foundations
CROSS SECTION OF THE SOIL
CALCULATE THE WEIGHT OF THE
BUILDING
Table of loads
9
7310.13 KN
1008.8 KN
MAT FOUNDATION
Critical columns
THE THICKNESS OF MAT
FOUNDATION WAS
CALCULATED USING CHECK
FOR PUNCHING IN THE
NEXT CALCULATION. THE
CRITICAL COLUMNS WERE
FOUND TO BE: C14, C 13,
C12.
THICKNESS OF MAT FOUNDATION
Assume h = 0.95 m , d = 0.88 m
For column 12:Φ Vcp = 075 (0.33) (400+880+1000+880)*2*(880/1000)
1.28
= 7283.7 > 6704.9 Ok
0.4
1.0
2.28
THICKNESS OF MAT FOUNDATION
For column 14:h = 0.95 m , d = 0.88m
ΦVcp = 0.75 (0.33) *(1480+1480) *2*880/1000
= 6822.74 KN > 5531.7
Ok
1.48
0.6
0.6
1.48
THICKNESS OF MAT FOUNDATION
For column 13:h=0.95 m
, d = 0.88 m
ΦVcp = 0.75 (0.33) * (2280+1280) *2*
= 8205.7 > 8015.9
Ok
So we will select depth of the
footing equals to 0.95 m.
2.28
1.4
0.4
1.28
CHECK PRESSURE UNDER FOOTING
F=KΔ
Where;
K= 210*100= 21000.
F = 210 kN.
Then Δ will be 0.01 m.
COLUMN AND MIDDLE STRIP IN XDIRECTION
Column
strip
Middle strip
MOMENT DISTRIBUTION IN X-DIRECTION
REINFORCEMENT
Top steel:
ρ1 = (1 - ) = 0.00387
As1 = 3405.6 mm2 → 1Φ25/150
mm
ρ2 = 0.003 → As2 = 2640 mm2 =
1Φ25/150 mm
Bottom steel:
ρ1 = ρ min = 0.0018
As1 = 1710 mm2
ρ2 = ρ min = 0.0018
As2 =1584 mm2 → 1Φ20/200 mm
REINFORCEMENT OF MAT FOUNDATION
IN THE HORIZONTAL DIRECTION
SETTLEMENT OF MAT FOUNDATION
Qall
8.3m
35*20
7m
Also the soil has been assumed to act as a Normally
Consolidated Clay (NCC) and some of its characteristics
were taken from the soil report to compute the settlement :
γ = 18 KN/m3
w% = 6.8%
Gs = 2.65
γw = 9.81 KN/m3
e = 0.54
where
γ=
LL = 21 (avg. of LL for all the layers in the specific depth)
Cc = 0.009(LL – 10) = 0.099
σ = (γ *8.3) + (γ*3.35)
= (18*8.3) + (18*3.5) = 212.4 KN/m2
Qall =
Qult = 71556.9 KN
(sum of the ultimate axial load on the columns)
Qall = 51112.07 KN
∆σ =
(∆σt + 4∆σc + ∆σb)
(by using 2:1 method)
∆σt =
=
= 73.02 KN/m2
(where z = 0)
∆σc =
∆σb =
= 56.5 KN/m2
= 45.1KN/m2
∆σ = 57.34 KN/m2
(where z = 3.5)
(where z = 7)
SETELMENT EQUAL TO:-
Sc =
Sc =
Hc log
*8000 log
= 4.6 cm < 5
PILE FOUNDATION
BEARING CAPACITY OF PILE
Qu = Qs + Qp
Where;
Qs = ultimate capacity of single pile due to point
bearing
Qp = ultimate capacity of single pile due to skin
friction
BEARING CAPACITY OF PILE

Using
Ø = 21º
c = 22 KN/m²
BEARING CAPACITY OF PILE
FROM ALL-PILE:Diameter / length
60 cm
80 cm
10 m
506.30 kN
715.17 kN
12 m
635.42 kN
892.86 kN
15 m
818.81 kN
1158.34 KN
CAP DESIGN



Principles:
Distance between piles = 3D
Where D is the diameter of pile
Minimum distance between edge of cap & edge of
pile is more that 15 cm
TWO PILE CAP
THREE PILE CAP
DESIGN FOR FLEXURE
M+ve = 1852.5 KN.m
M-ve = 285.52 KN.m
ρ+ = 0.0029 < 0.0033
Asmin = 0.0033*1000*1300
= 4290 mm2
(Use 1Ø25/110mm)
ρ- = 0.003
As =1000*1300*0.0033
= 4290 mm2
(Use 1Ø25/110mm)
DESIGN FOR SHEAR
Vc = (1/6)
= (1/6)
b*d
(1000) (1300)/1000 = 1146.5 KN
Vn =
=
= 2535.3 KN
Where; 1901.5 KN is the ultimate shear on the pile taken
from SAP
Vs = Vn – Vc = 2535.3-1146.5= 1388.8 KN
=
=
= 2.5
Assume using Ø = 16 for stirrups
Av = 2 * 201 = 402 mm2
Av/S = 2.5
S = 160 mm
(Use 1Ø16/250mm)
PILE REINFORCEMENT
13 Ø 16
As 0.005( /4)(800) = 2513 mm2
= 13Ø16
1Ø 10/ 120 mm
THE END
Thank you for
listening