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O UT LINE 1) Determine the own weight of building 2) Design of mat foundation 3) Design of pile foundation

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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

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if cracks appear in the structure it is assumed that the foundation did move and that this is the sole cause of cracking. INTRODUCTION

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C HOICE OF THE T YPE OF F OUNDATION 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

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C ROSS SECTION OF THE SOIL

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C ALCULATE THE WEIGHT OF THE BUILDING

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KN KN TableTable of loads

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MAT FOUNDATION

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T HE THICKNESS OF MAT FOUNDATION WAS CALCULATED USING CHECK FOR PUNCHING IN THE NEXT CALCULATION. T HE CRITICAL COLUMNS WERE FOUND TO BE : C14, C 13, C12. Critical columns

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T HICKNESS OF MAT F OUNDATION Assume h = 0.95 m, d = 0.88 m For column 12:- Φ Vcp = 075 (0.33) ( )*2*(880/1000) = > Ok

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T HICKNESS OF MAT F OUNDATION For column 14:- h = 0.95 m, d = 0.88m ΦV cp = 0.75 (0.33) *( ) *2*880/1000 = KN > Ok

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T HICKNESS OF MAT F OUNDATION For column 13:- h=0.95 m, d = 0.88 m ΦV cp = 0.75 (0.33) * ( ) *2* = > Ok So we will select depth of the footing equals to 0.95 m

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C HECK PRESSURE UNDER FOOTING F=KΔ Where; K= 210*100= F = 210 kN. Then Δ will be 0.01 m.

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C OLUMN AND M IDDLE STRIP IN X- DIRECTION Middle strip Column strip

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M OMENT DISTRIBUTION IN X- DIRECTION

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R EINFORCEMENT Top steel: ρ 1 = (1 - ) = A s1 = mm 2 → 1Φ25/150 mm ρ 2 = → A s2 = 2640 mm 2 = 1Φ25/150 mm Bottom steel: ρ 1 = ρ min = A s1 = 1710 mm 2 ρ 2 = ρ min = A s2 =1584 mm 2 → 1Φ20/200 mm

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R EINFORCEMENT OF M AT F OUNDATION IN THE HORIZONTAL DIRECTION

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S ETTLEMENT OF M AT FOUNDATION 35*20 Q all 8.3m 7m

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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/m 3 w% = 6.8% G s = 2.65 γ w = 9.81 KN/m 3 e = 0.54 where γ = LL = 21 (avg. of LL for all the layers in the specific depth) C c = 0.009(LL – 10) = 0.099

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Q all = Q ult = KN (sum of the ultimate axial load on the columns) Q all = KN ∆σ = (∆σ t + 4∆σ c + ∆σ b ) (by using 2:1 method) σ = (γ *8.3) + (γ*3.35) = (18*8.3) + (18*3.5) = KN/m 2

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∆σ t = = = KN/m 2 (where z = 0) ∆σ c = = 56.5 KN/m 2 (where z = 3.5) ∆σ b = = 45.1KN/m 2 (where z = 7) ∆σ = KN/m 2

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SETELMENT EQUAL TO:- S c = H c log S c = *8000 log = 4.6 cm < 5

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P ILE FOUNDATION

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B EARING 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

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B EARING CAPACITY OF PILE Using Ø = 21º c = 22 KN/m²

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B EARING CAPACITY OF PILE FROM ALL - PILE :- Diameter / length 60 cm 80 cm 10 m kN kN 12 m kN kN 15 m kN KN

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C AP 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

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T WO PILE CAP

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T HREE PILE CAP

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D ESIGN FOR FLEXURE M+ve = KN.m M-ve = KN.m ρ+ = < Asmin = *1000*1300 = 4290 mm2 (Use 1Ø25/110mm) ρ- = As =1000*1300* = 4290 mm2 (Use 1Ø25/110mm)

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D ESIGN FOR SHEAR V c = (1/6) b*d = (1/6) (1000) (1300)/1000 = KN V n = = = KN Where; KN is the ultimate shear on the pile taken from SAP V s = V n – V c = = KN = = = 2.5 Assume using Ø = 16 for stirrups A v = 2 * 201 = 402 mm 2 Av/S = 2.5 S = 160 mm (Use 1Ø16/250mm)

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P ILE REINFORCEMENT As 0.005( /4)(800) = 2513 mm2 = 13Ø16 13 Ø 16 1Ø 10/ 120 mm

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T HE END Thank you for listening

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