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## Lecture # 5

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**Design of Concrete Structure II**University of Palestine بسم الله الرحمن الرحيم Lecture # 5 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine B B L L P P M Page 1 Footing Introduction Footings are structural elements used to support columns and walls and transmit their loads to the underlying soil without exceeding its safe bearing capacity below the structure. Loads Column Beam Footing Soil Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 2 Footing Introduction The design of footings calls for the combined efforts of geotechnical and structural engineers. The geotechnical engineer, on one hand, conducts the site investigation and on the light of his findings, recommends the most suitable type of foundation and the allowable bearing capacity of the soil at the suggested foundation level. The structural engineer, on the other hand, determines the concrete dimensions and reinforcement details of the approved foundation Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 3 Types of Footing Wall Footings Wall footing are used to support structural walls that carry loads for other floors or to support nonstructural walls. W kN/m W kN/m Wall Secondary reinft Footing Main reinft. Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 4 Types of Footing Isolated Footings Isolated or single 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. P kN B C2 C1 L P Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 5 Types of Footing Combined Footings Combined footings are used when two columns are so close that single footings cannot be used or when one column is located at or near a property line. P1 P2 P2 kN P1 kN L B C2 C2 C1 C1 L2 L1 L2 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine P1 P2 P3 P4 Page 6 Types of Footing Continuous Footings Continuous footings support a row of three or more columns P4 kN P3 kN P2 kN L P1 kN B Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 7 Types of Footing Strap (Cantilever ) footings Strap footings consists of two separate footings, one under each column, connected together by a beam called “strap beam”. The purpose of the strap beam is to prevent overturning of the eccentrically loaded footing. P2 kN P1 P2 Strap Beam property line P1 kN L1 L2 C2 B1 B2 C2 C1 C1 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 8 Types of Footing Mat (Raft) Footings Mat Footings consists of one footing usually placed under the entire building area. They are used, when soil bearing capacity is low, column loads are heavy and differential settlement for single footings are very large. B L Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 9 Types of Footing Pile caps Pile caps are thick slabs used to tie a group of piles together to support and transmit column loads to the piles. P B L Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 10 Footing Distribution of Soil Pressure The distribution of soil pressure under a footing is a function of the type of soil, the relative rigidity of the soil and the footing, and the depth of foundation at level of contact between footing and soil P P P L L L Footing on sand Footing on clay Equivalent uniform distribution Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 11 Footing Concentrically loaded Footings If the resultant of the loads acting at the base of the footing coincides with the centroid of the footing area, the footing is concentrically loaded and a uniform distribution of soil pressure is assumed in design, as shown in Figure P Centroidal axis L P/A L B Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 12 Footing Eccentrically Loaded Footings Footings are often designed for both axial load and moment. Moment may be caused by lateral forces due to wind or earthquake, and by lateral soil pressures. Footing is eccentrically Loaded if the supported column is not concentric with the footing area or if the column transmits at its juncture with the footing not only a vertical load but also a bending moment. P P e M Centroidal axis Centroidal axis y y L L P/A P/A My/I My/I Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 13 Footing Eccentrically Loaded Footings The pressure distribution on the base of footing that support combined concentric load (P) and moment (M) is given by P e Centroidal axis Where e The eccentricity of the load relative to centroidal axis of area. I Moment of inertia about centroidal axis y the distance from the centroidal axis to point where the stresses are being calculated y L P/A My/I Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 14 Footing Eccentrically Loaded Footings Large eccentricities cause tensile stresses on part of the base area of the footing. Since soil cannot resist tensile stresses For rectangular footing, this occurs when the eccentricity exceeds ek= L/6 This is referred to as Kern distance Loads applied within the kern, then pressure distribution will be compression over the entire area of the footing Various pressure distribution for rectangular footing are shown in the following page P e Centroidal axis y L P/A My/I L Kern B/6 B L/6 L/6 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page 15 Footing Eccentrically Loaded Footings P e P ek q=P/A=qavg q<P/A q>P/A a) Concentric load, e =0 b) e < ek P Resultant of load on footing P e ek ek q=0 q>2qavg q=2qavg c) e > ek c) e = ek Resultant of soil pressure Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX1-1 Footing Example # 1 Design an isolated footing to support an interior column 0.5m×0.3m PD = 1000 kN and PL = 600 kN MD = 120 kN.m and ML = 70 kN.m Use fc’= 25 MPa , fy = 420 MPa, and qall net = 200 kpa P M 0.5 0.3 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX1-2 Footing Solution 1- Select a trial footing depth Assume that the footing is 0.6 m thick 2- Establish the required base area of the footing Check stress Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page Ex1-3 Footing Solution 3- Evaluate the net factored soil pressure Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine L 177.6 kPa 254.4 kPa Page Ex1-4 Footing Solution P M 254.4 kPa 177.6 kPa 254.4 kPa B 254.4 kPa Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page Ex1-5 Footing Solution 4- Check footing thickness for punching shear 2.5m C1+d C2+d 4.0 m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page Ex1-6 Footing Solution 5- Check footing thickness for beam shear In short direction (sec 1-1) In long direction (sec 2-2) 1 2 2 d d 2.5m 1 4.0 m d 231 kPa 177.6 kPa 255.4 kPa d 216 kPa Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page Ex1-7 Footing Solution 6- Compute the area of flexural reinforcement in each direction In long direction (Sec 1-1) 1 2.5m 1 4.0 m 177.6 254.4 221 kPa 1.75 237.6 x 2.5 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page Ex1-8 Footing Solution 6- Compute the area of flexural reinforcement in each direction In short direction (Sec 2-2) 2.5m 2 2 216x4 4.0 m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page Ex1-9 Footing Solution 7- Check for anchorage of the reinforcement Bottom longitudinal reinforcement (Φ18mm) α=1.0 for bottom bars, β=1.0 for uncoated bars αβ =1.0 <1.7 OK γ=0.8 for Φ18mm, λ=1.0 for normal weight concrete C the smallest of 50+18=68mm [2500-2(50)-18]/(18)(2)=66.17mm i.e., C is taken as 66.17mm Available length =1750-50=1700> 450 mm Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page Ex1-10 Footing Solution 7- Check for anchorage of the reinforcement Bottom longitudinal reinforcement (Φ16mm) α=1.0 for bottom bars, β=1.0 for uncoated bars αβ =1.0 <1.7 OK γ=0.8 for Φ16mm, λ=1.0 for normal weight concrete C the smallest of 50+16=66mm [2500-2(50)-16]/(16)(2)=74.5mm i.e., C is taken as 66mm Available length =1250-50=1200> 400 mm Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page ex1-11 Footing Solution 8- Prepare neat design drawings showing footing dimensions and provided reinforcement 0.60 m 0.75m 2.50 m 0.75m 2.50 m 3Φ14 3Φ14 17Φ16 19Φ18 4.00 m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-1 Footing Example # 2 Design a combined footing, to support two columns A and B spaced at distance 6.0 m center-to-center as shown in Figure Column A is 0.4mx0.4m and carries a dead load of 500kNs and a live load of 300kNs Column B is also 0.4mx0.4m in cross section but carries a dead load of 700 kNs and a live load of 500 kNs. Use fc’= 25 MPa , fy = 420 MPa, and qall net = 150 kpa A B 0.4 0.4 0.4 0.4 Property limit PD=500kN PD=750kN PL=300kN PL=450kN ?? m 6 m ?? m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-2 Footing Solution 1- Establish the required base area of the footing To locate the resultant of the column forces ∑M@start =0.0 → 800(0.2)+1200 (6) =2000 (x)→ x= 3.8 m Ps=2000kN B A Ps=1200kN Ps=800kN X=3.8 m 0.2 m 6 m Ps=2000kN B A Ps=1200kN Ps=800kN X=3.8 m X=3.8 m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-3 Footing Solution Length of footing L=2 (3.80) = 7.60m Width of footing B =13.33/7.6 =1.754 m , taken as 1.80 m. 2- Evaluate the net factored soil pressure 3- Select a trial footing depth Assume that the footing is 0.8 m thick Effective depth d = 800 – 75 – 10 = 715 mm Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-4 Footing Solution 4- Check footing thickness for punching shear Column A The factored shear force Vu =1080−197.4(1.115)(0.758) =914 kN b =2(400+715/2)+400+715=2630 mm ΦVc = 2350 kN > Vu= 914 OK B A 1115 1115 1.8 m 1115 758 7.6 m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-5 Footing Solution 4- Check footing thickness for punching shear [contd.] Column B The factored shear force Vu =1620−197.4(1.115)2 =1375 kN b =4(400+715)=4460 mm ΦVc = 3986 kN > Vu= 1375 OK B A 1115 1115 1.8 m 1115 758 7.6 m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-6 Footing Solution 5- Draw S.F.D and B.M.D for footing 1620 kN 1080kN 197.37 x1.8= 355.26 kN/m 1122.6 2.84 m 798 0.915 71 497.4 1009 1425.7 7.10 348 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-7 Footing Solution 6- Check footing thickness for beam shear Effective depth d= 800–75–10=715 mm (lower layer) Maximum factored shear force Vu is located at distance d from the face of column B, Vu,critical =798 kN 7- Compute the areas of flexural reinforcement a) Top longitudinal reinforcement Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-8 Footing Solution 7- Compute the areas of flexural reinforcement [contd.] b) Bottom longitudinal reinforcement c) Short Direction : Effective depth d= 800–75–16-8=700 mm (Upper layer) 7.6 m 0.4 0.4 1.8 m 0.4 0.4 400+700 400+350 1100 750 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine 1080 1.8 Page EX2-9 Footing Solution Under Column A 7.6 m 0.7 0.4 0.4 1.8 m 0.4 0.4 400+700 400+350 1100 750 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine 1620 1.8 Page EX2-10 Footing Solution Under Column B 7.6 m 0.7 0.4 0.4 1.8 m 0.4 0.4 400+700 400+350 1100 750 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-11 Footing Solution Shrinkage Reinforcement in the short direction 9- Check for anchorage of the reinforcement a) Top reinforcement (Φ20mm) α=1.3 for top bars, β=1.0 for uncoated bars γ=0.8 for Φ20mm, λ=1.0 for normal weight concrete C the smallest of 75+10=85 mm [1800-2(75)-20]/(17)(2)=48 mm i.e., C is taken as 48mm Available length =2840+200-75=2965> 655 mm Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-12 Footing Solution b) Bottom reinforcement (Φ16mm) α = β= 1.0 γ=0.8 for Φ16mm, λ=1.0 for normal weight concrete C the smallest of 75+8=83 mm [1800-2(75)-16]/(12)(2)=68.1 mm i.e., C is taken as 68.1mm Available length =1400-75=1325 > 387 mm 10- Prepare neat design drawings showing footing dimensions and provided reinforcement Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-13 Footing Solution 0.4m 5.6 m 0.4m 1.2 m Φ14@100 Φ14@100 18Φ20 0.80 m 13Φ16 6Φ16 Φ14@100 Φ14@100 8Φ16 0.75m 1.10 m 18Φ20 T 1.80 m 8Φ16 B 6Φ16 B Φ14@100 Φ14@100 13Φ16 B 7.60 m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-1 Footing Example # 3 Design a combined footing, to support two columns A and B spaced at distance 4.0 m center-to-center as shown in Figure Column A is 0.4mx0.4mand carries a dead load of 800kNsand a live load of 400kNs Column B is also 0.3mx0.3min cross section but carries a dead load of 500 kNs and a live load of 250 kNs. Use fc’= 28 MPa , fy = 420 MPa, and qall net = 180 kpa A B 0.3 0.4 0.4 0.3 Property limit PD=800kN PD=500kN PL=400kN PL=250kN ?? m 4 m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-2 Footing Solution 1- Establish the required base area of the footing To locate the resultant of the column forces ∑M@start =0.0 → 750(4) =1950 (x)→ x =1.538 m Ps=1950kN A B Ps=750kN Ps=1200kN 0.2 m 4 m Ps=1950kN A B Ps=750kN Ps=1200kN X=1.54 m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-3 Footing Solution Try a rectangular combined footing Length of footing L=2(1.54+ 0.20)=3.48m i.e. this type of footing is not possible if uniform soil pressure is to be maintained below the footing. Try a trapezoidal combined footing Assume no projection of footing beyond column B. L=0.2+4+0.15=4.35m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-3 Footing Solution 2- Evaluate the net factored soil pressure 3- Select a trial footing depth Assume that the footing is 0.75 m thick Effective depth d = 750 – 75 – 10 = 665 mm (lower layer) A=10.55 m2 A B 4.0 m 1.0 m 0.3 c.g 0.4 0.3 0.4 Xbar=1.74 m 4.35 m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-4 Footing Solution 4- Check footing thickness for punching shear Column A The factored shear force, Vu =1600−240(1.065)(0.733)=1412.8 kN bo =2(400+665/2)+400+665 =2530 mm ΦVc = 2225 kN > Vu= 1413 OK 0.633 B 0.965 A 0.3 0.4 1.065 0.3 0.4 0.733 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-4 Footing Solution 4- Check footing thickness for punching shear [contd.] Column B The factored shear force, Vu =1000−240(0.965)(0.633)=854 kN bo =2(300+665/2)+300+665 =2230 mm ΦVc = 1962 kN > Vu= 854 OK 0.633 B A 0.3 0.4 0.965 1.065 0.3 0.4 0.733 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-6 Footing Solution 5- Draw S.F.D and B.M.D for footing q1= 240 x 1.0= 240 kN/m & q2= 240 x 4= 960 kN/m The intensity of soil pressure at distance x from the left side of footing is qux=960-165.5x The shear force is given as Vux=960x-82.76x2+C1 ,Where C1 is the constant of integration accounting for shear due to column loads The bending moment is given by Mux=480x2-27.59x3+C2 ,Where C2is the constant of integration accounting for bending due to column loads. The section of maximum bending moment corresponds to the section of zero shear force, or 960x-82.76x2-1600=0.0 Solving this equation gives x1= 2.02 m, and x2 =9.71m (rejected). Substituting x1=2.05m in moment equation gives Mu,max=1187kN.m Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-6 Footing Solution 6- Check footing thickness for beam shear Section A-A Effective depth d=750–75–10= 665mm Critical section for beam shear is located at distance 665+200= 865mm to the right side of center of column A. X=0.2+0.865=1.065 m Vu=960x-82.76x2-1600=671 kN The width of footing at @ x=1.065 is equal to 3266 mm 1600 kN 1000 kN qu (x) 240 960 x 962 701 2.02 0.815 188 0.865 38 671 1187 1412 2.8 18.9 Instructor: Eng. Mazen Alshorafa**Design of Concrete Structure II**University of Palestine Page EX2-6 Footing Solution 6- Check footing thickness for beam shear Section B-B Effective depth d=750–75–10= 665mm Critical section for beam shear is located at distance 665+150= 815mm to the left side of center of column B. X = 4.2 - 0.815 = 3.385 m Vu=960x-82.76x2-1600=701 kN The width of footing at @ x=3.385 is equal to 1666 mm 1600 kN 1000 kN qu (x) 240 960 x 962 701 2.02 0.815 188 0.865 38 671 1187 1412 2.8 18.9 Instructor: Eng. Mazen Alshorafa