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Design Of Foundation for a Commercial and Residential Building Under the Supervision of: Dr. Mohammad Ghazal Prepared by: Moayad Qadarah Raja’e Omar Lu’ai.

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Presentation on theme: "Design Of Foundation for a Commercial and Residential Building Under the Supervision of: Dr. Mohammad Ghazal Prepared by: Moayad Qadarah Raja’e Omar Lu’ai."— Presentation transcript:

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2 Design Of Foundation for a Commercial and Residential Building Under the Supervision of: Dr. Mohammad Ghazal Prepared by: Moayad Qadarah Raja’e Omar Lu’ai Abu Sharshuh May 2011 An-Najah National University Engineering Collage Civil Engineering Department

3 Project Description  Name: Eisheh Taha Oudeh Building.  Type: commercial and Residential Building.  Location: Nablus City, Rafedia Main Street, opposite to Ben Qutaiba School.  Number Of Floors: 12 floor, 2 under ground level and 10 over.  Plane Area: 550 m 2 for the floor.

4 SCOPE OF PROJECT  Evaluation of foundations.  Selection of the proper foundations.  Design of foundations.

5 Literature Review  Foundations are the part of an engineered system to receive & transmit loads from superstructure to the underlying soil or rock.  There are two types of foundations : shallow & deep foundations.  Many factors should be taken into consideration in choosing foundation types such as soil properties, economic factors, engineering practice,....etc

6  Isolated Footing.  Combined Footing.  Mat or Raft Foundations.  Strap or Cantilever Footings.  Pile Foundations. Types of footing

7 Isolated Footings Isolated Footings  Are used to support single columns.  This is one of the most economical types of footings and is used when columns are spaced at relatively long distances.  Its function is to spread the column load to the soil, so that the stress intensity is reduced.

8 Are used in the following cases: 1) When there are two columns so close to each other & in turn the two isolated footing areas would overlap. 2) When the combined stresses are more than the allowable bearing capacity of the soil. 3) When columns are placed at the property line. Combined Foundations

9  Are used to spread the load from a structure over a large area, normally the entire are of the structure.  They often needed on soft or loose soils with low bearing capacity as they can spread the loads over a larger area.  They have the advantage of reducing differential settlements. Mat or Raft Foundations

10  Cantilever footing construction uses a strap beam to connect an eccentrically loaded column foundation to the foundation of an interior column.  Are used when the allowable soil bearing capacity is high, and the distances between the columns are large. Strap or Cantilever Footings

11 Pile Foundations  They are long & slender members that are used to carry & transfer the load of the structure to deeper soil or rocks of high bearing capacity, when the upper soil layer are too weak to support the loads from the structure.  Piles costs more than shallow foundations; so the geotechnical engineer should know in depth the properties & conditions of the soil to decide whether piles are needed or not.

12  Bearing Capacity : is the ability of a soil to support the loads applied to the ground.  Ultimate bearing capacity is the theoretical maximum pressure which can be supported without failure;  Allowable bearing capacity is the ultimate bearing capacity qu divided by a factor of safety (F.S).  There are three modes of failure that limit bearing capacity: general shear failure, local shear failure, and punching shear failure. Bearing Capacity

13  When building structures on top of soils, one needs to have some knowledge of how settlement occurs & how fast settlement will occur in a given situation.  So, There are three types of settlement: 1. Initial settlement 2. Primary settlement 3. Secondary settlement Settlement

14 Settlement (Cont.)  Total Settlement = SI + SC + SS  The allowable bearing capacity and the type of foundations provided later are evaluated based on the settlements limits. This means that the settlement of the proposed foundation would be within the acceptable limits if the allowable bearing capacity provided is used.

15 Geotechnical Investigation  The studied area is approximately flat with slight difference in the three existing elevations. The general soil formation within Highly fragmented weathered limestone and marlstone of soft to medium strength with cavities filled with marl soil  The geotechnical engineer decided to drill Three boreholes trying to cover the whole construction area.

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17  The depths of the drilled boreholes were as follows: Borehole No.Depth (m)  laboratory test results: γ= 20 KN/m³ qall. = 3.5 kg/cm 2

18  Taking the lowest compressive strength value of rock core specimens with test results and applying the percentage of 5%, the strength will be: b.h1 – Qall = 5% * 75 = 3.75 kg\cm2 b.h2 – Qall = 5% * 70 = 3.5 kg\cm2 b.h3 – Qall = 5% * 78 = 3.9 kg\cm2  But considering the fact that rocks is some areas may be encountered in fragmented conditions, as well as the presence of there fracture rocks and marls, it is recommended to consider the bearing capacity value of the rock formations countered in the site of not more than 3.5 kg\cm2 within the described rock layers after the removal of all loose fill materials over the rock.

19 Structuraldesign Structural design Column loads are calculated using (sap program), the structure subjected to the following loads: 1) Dead Load (own weight). 2) Super imposed dead load =350 kg/m2. 3) Live load =500 kg/m2. Using ACI code, the ultimate loads are calculated considering load combination : Pu =1.2Dead + 1.6Live. Material characteristics used in this project are: f’c =240kg/cm2 (B 300) Where: f’c is the compressive strength of concrete fy = 4200 kg/cm2 Where: fy is the yield strength of steel

20 Manual Design steps: 1) 1) Area of footing = Total service loads on column / net soil pressure 2) 2) Determine footing dimensions B & H. 3) 3) Assume depth for footing. 4) 4) Check soil pressure. 5) 5) Check wide beam shear : V c > V ult 6) 6) Check punching shear : V cp > P ult, punching 7) 7) Determine reinforcement steel in the two directions. 8) 8) Check development length. IsolatedFootingDesign Isolated Footing Design

21 Area (m 2 ) L(m)B(m)Footingcolumn F1c F2c F3c F4c F5c F6c F15C F16c F19c F20c F28c F31c F34c F35c F36C36 DIMENSIONSOFSINGLE FOOTING DIMENSIONS OF SINGLE FOOTING

22 THICKNESSES OF FOOTINGS Depth of footing will be controlled by wide beam shear (one way action) and punching shear (two way action). Wide Beam Shear: Shear cracks are form at distance “d” from the face of column, and extend to the compression zone, the compression zone will be fails due to combination of compression and shear stress. Punching Shear: Formation of inclined cracks around the perimeter of the concentrated load may cause failure of footing. Max, formation of these cracks occurred at distance “d\2” from each face of he column.

23 THICKNESSES OF SINGLE FOOTINGS h(m)d(m)Footingcolumn F1c F2c F3c F4c F5c F6c6 10.9F15C F16c F19c F20c F28c F31c F34c F35c F36C36

24 Steel reinforcement : Isolated footing represented as cantilever, so the max moment occurs at the face of the column: Ultimate moment at the face of the column (Mult) =(qult*ln2)/2 Mn =Mu\ Φ, where, Φ =0.9 Mn =Rnbd2 ρ = 1\m(1-( 1-2mRn\ fy ).5) WHERE ρ : Steel ratio m= fy\0.85 f ′ c As = ρ bd

25 single footing reinforcement: Reinforcement in long direction/cm Reinforcement in short direction/cm Footing 1 ø 16/271 ø 14/21 F1 1 ø 16/281 ø 14/20 F2 1 ø 16/281 ø 14/20 F3 1 ø 16/271 ø 14/21 F4 1 ø 14/271 ø 14/21 F5 1 ø 16/271 ø 14/21 F6 1 ø 18/121 ø 18/10 F15 1 ø 18/12 F16 1 ø 18/13 F19 1 ø 16/11 F20 1 ø 18/101 ø 16/11 F28 1 ø 18/101 ø 16/11 F31 1 ø 16/121 ø 14/10 F34 1 ø 16/121 ø 14/10 F35 1 ø 14/12 F36

26 DESIGN OF COMBINED FOOTING footingcolumn Fc1C7,C8 Fc4C23,C24 Fc5C26,C27 Fc6C29,C30 Fc7C32,C33 SUMMARY OF DIMENSIONS h(m)d(m)L(m)B(m)Footing Fc Fc Fc Fc Fc Fc Fc7

27 Steel Reinforcement (Flexural) By using sap program to get the maximum negative and positive Moment Mn =Mu\Φ, where, Φ=0.9 Mn =Rnbd2 As = ρbd Steel reinforcement for Fc2 The figure below show bending moment in x-direction using SAP2000:

28 Mu =93.175ton.m Mn =93.175\.9 =103.52ton.m Mn =Rnbd2 Rn =21.1kg\cm2 Ρ =.0054 As =37.8cm2 Use 1 Φ16\12cm The figure below show bending moment in y-direction using SAP2000:

29 Steel reinforcement for Fc2 Mu = ton.m Mn = / ton.m Mn =Rnbd2 Rn=20.1kg\cm2 Ρ =.005 As =37.5cm2 Use 1 Φ 16\12cm

30 Reinforcement in y direction Reinforcement in x direction Footing # 1Ф16/12cm Fc 1 1Ф18/10cm1Ф22/10cmFc2 1Ф18/10cm1Ф22/10cmFc 3 1Ф32/10cm1Ф22/10cmFc 4 1Ф32/10cm1Ф22/10cmFc 5 Steel reinforcement for Combined footing

31 Mat foundation Design In this project the mat foundation was designed using Sap2000 with the following data: ( fc= 240 kg \cm2, fy= 4200 kg\cm2 ) Calculating the Thickness for mat The thickness of mat foundation was calculated using check for punching in the next calculation. (the most critical for determining the thickness for mat in the punching shear ) To calculate the thickness, it was used the next equation: Pu =.75*1.06(fc).5 *bo*d

32 Where : Pu : the load at the column bo: parameter of the bunching area d: thickness of the mat foundation for mat foundation 1which include (col 21,22,25) Col 21 has the critical load,Pu = ton *1000 =.75*1.06*(240).5*(2(d+50)+(2(d+70))*d d = 80 cm

33 Design of mat foundation using sap 2000 Deflection shape When we do the analysis using sap 2000 it was found that the maximum settlement was equal to m

34 Reinforcement reinforcement in x direction we take the maximum moment at the face of columns and the maximum between the columns. Then,the area of reinforcement will calculate by the Equation: As =p*b*d 1)at the face of column Mu =123 ton.m p= As = *100*80 As = 43 cm2 use 10Ф25 mm/m 2)between the columns Mu = 40 ton.m P = ˂ So use pmin = As=26.4 cm2 use 10Ф18 mm/m

35 reinforcement in y direction 1)at the face of column Mu = 75 ton.m ρ= ˂ use ρ min for all the zone. As = ρ * b * d = * 100 * 80 = 26.4 cm2 Use 10 Ф 18 mm / m secondary reinforcement (Negative moment) the max. moment equals to 20 ton.m Ρ = ˂ use ρ min for all the zone. As = ρ * b * d = * 100 * 80 = 26.4 cm2 Use 10 Ф 18 mm / m

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