OUT LINE

1) Determine the own weight of building

2) Design of mat foundation

3) Design of pile foundation

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.

INTRODUCTION

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

ALQUQA

BUILDING

CROSS SECTION OF THE SOIL

CALCULATE THE WEIGHT OF THE BUILDING

3 D On SAP

9

1008.8 KN

7310.13 KN

Table of loads

MAT FOUNDATION

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.

Critical columns

THICKNESS OF MAT FOUNDATION

Assume h = 0.95 m , d = 0.88 mFor column 12:-Φ Vcp = 075 (0.33)

(400+880+1000+880)*2*(880/1000)= 7283.7 > 6704.9 Ok

1.0

0.4

1.28

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

0.6

0.6

1.48

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 OkSo we will select depth of the footing equals to 0.95 m.

1.4

0.4

2.28

1.28

CHECK PRESSURE UNDER FOOTING

F=KΔWhere;K= 210*100= 21000.F = 210 kN.Then Δ will be 0.01 m.

COLUMN AND MIDDLE STRIP IN X-DIRECTION

Middle strip

Column strip

MOMENT DISTRIBUTION IN X-DIRECTION

REINFORCEMENT

Top steel:ρ1 = (1 - ) = 0.00387As1 = 3405.6 mm2 → 1Φ25/150 mmρ2 = 0.003 → As2 = 2640 mm2 = 1Φ25/150 mm 

Bottom steel:ρ1 = ρ min = 0.0018As1 = 1710 mm2

ρ2 = ρ min = 0.0018As2 =1584 mm2 → 1Φ20/200 mm

REINFORCEMENT OF MAT FOUNDATION IN THE HORIZONTAL DIRECTION

SETTLEMENT OF MAT FOUNDATION

35*20

Qall

8.3m

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

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)

σ = (γ *8.3) + (γ*3.35) = (18*8.3) + (18*3.5) = 212.4 KN/m2

∆σt = = = 73.02 KN/m2

(where z = 0)

∆σc = = 56.5 KN/m2 (where z = 3.5)

∆σb = = 45.1KN/m2 (where z = 7)

∆σ = 57.34 KN/m2

SETELMENT EQUAL TO:-

Sc = Hc log

Sc = *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 PILEFROM 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.mM-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) b*d

= (1/6) (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 stirrupsAv = 2 * 201 = 402 mm2

Av/S = 2.5 S = 160 mm (Use 1Ø16/250mm)

PILE REINFORCEMENT

As 0.005( /4)(800) = 2513 mm2 = 13Ø16

13 Ø 16

1Ø 10/ 120 mm

THE END

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