Lecture # 5 Design of Concrete Structure IIUniversity of PalestineInstructor:Eng. Mazen Alshorafa Page 1Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaIntroductionFootingsarestructuralelementsusedtosupportcolumnsandwalls andtransmittheirloadstotheunderlyingsoilwithoutexceedingits safe bearing capacity below the structure.ColumnBeamLoadsFootingSoilMPLBPLB Page 2Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaIntroductionThedesignoffootingscallsforthecombinedeffortsofgeotechnical and structural engineers. Thegeotechnicalengineer,ononehand,conductsthesite investigationandonthelightofhisfindings,recommendsthemost suitable type of foundationand the allowable bearing capacity ofthe soil at the suggested foundation level. Thestructuralengineer,ontheotherhand,determinestheconcrete dimensions and reinforcement details of the approved foundation Page 3Design of Concrete Structure IIUniversity of PalestineTypes of FootingInstructor:Eng. Mazen AlshorafaWall FootingsWall footingare used to support structural wallsthat carryloads for other floors or to support nonstructural walls.WallFootingMain reinft.Secondary reinftW kN/mW kN/m Page 4Design of Concrete Structure IIUniversity of PalestineTypes of FootingInstructor:Eng. Mazen AlshorafaIsolated FootingsIsolated or singlefootingsare used to support single columns.This isoneofthemosteconomicaltypesoffootingsandisusedwhencolumns are spaced at relatively long distances.PLBC 2C 1P kN Page 5Design of Concrete Structure IIUniversity of PalestineTypes of FootingInstructor:Eng. Mazen AlshorafaCombined FootingsCombinedfootingsareusedwhentwocolumnsaresoclosethat singlefootingscannotbeusedorwhenonecolumnislocatedator near a property line.P 2P 1P 1kNP 2kNC 2C 1C 2C 1LBL 2L 1 L 2 Page 6Design of Concrete Structure IIUniversity of PalestineTypes of FootingInstructor:Eng. Mazen AlshorafaContinuous FootingsContinuous footings support a row of three or more columnsP 1P 1kNLBP 2kNP 3kNP 4kNP 2P 3P 4 Page 7Design of Concrete Structure IIUniversity of PalestineTypes of FootingInstructor:Eng. Mazen AlshorafaStrap (Cantilever ) footingsStrapfootingsconsistsoftwoseparatefootings,oneundereachcolumn, connected together by a beam called strap beam.Thepurposeofthestrapbeamistopreventoverturningofthe eccentrically loaded footing. P 2P 1P 1kNC 2C 1C 2C 1B 1P 2kNB 2L 1L 2Strap Beamproperty line Page 8Design of Concrete Structure IIUniversity of PalestineTypes of FootingInstructor:Eng. Mazen AlshorafaMat (Raft) FootingsMatFootingsconsistsofonefootingusuallyplacedundertheentire buildingarea.Theyareused,whensoilbearingcapacityislow,column loads are heavy and differential settlement for single footings are very large.LB Page 9Design of Concrete Structure IIUniversity of PalestineTypes of FootingInstructor:Eng. Mazen AlshorafaPile capsPilecapsarethickslabsusedtotieagroupofpilestogetherto support and transmit column loads to the piles.PLB Page 10Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaDistribution of Soil PressureThedistributionofsoilpressureunderafootingisafunctionofthe typeofsoil,therelativerigidityofthesoilandthefooting, andthe depth of foundation at level of contact between footing and soilPLPLPLFooting on sandFooting on clay Equivalent uniform distribution Page 11Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaConcentrically loaded FootingsIftheresultantoftheloadsactingatthebaseofthefooting coincideswiththecentroid ofthefootingarea,thefootingis concentricallyloadedandauniformdistributionofsoilpressureis assumed in design, as shown in FigureCentroidal axisPLP/ABL Page 12Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaEccentrically Loaded FootingsFootings are often designed for both axial load and moment. Moment maybecausedbylateralforcesduetowindorearthquake,andby lateral soil pressures. FootingiseccentricallyLoadedifthesupportedcolumnisnot concentricwiththefootingareaorifthecolumntransmitsatits juncture with the footing not onlya verticalload butalsoa bending moment.MPLP/AMy/IyCentroidal axisPLP/AMy/IyeCentroidal axis Page 13Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaEccentrically Loaded FootingsThepressuredistributiononthebaseoffootingthatsupport combined concentric load(P) and moment (M) is given byPLP/AMy/Iy = = = =L 6e 1AP /12 BL (L/2) PeAPIy PeAPIMyAPq3minmaxeCentroidal axisWhereeTheeccentricityoftheloadrelativeto centroidal axis of area.IMomentofinertiaaboutcentroidalaxisythe distance from the centroidal axis to pointwherethestressesarebeing calculated Page 14Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaEccentrically Loaded FootingsPLLarge eccentricities cause tensile stresses on part of the base area of the footing. Since soil cannot resist tensile stressesFor rectangular footing, this occurs when the eccentricity exceedsek = L/6This is referred to as Kern distanceLoads applied within the kern, then pressuredistribution will be compression over the entirearea of the footingVarious pressure distribution for rectangularfooting are shown in the following pageP/AMy/IyBLKerneCentroidal axis L/6 L/6 B/6 Page 15Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaEccentrically Loaded FootingsPq=P/A=qavg a) Concentric load, e =0Pq>P/Ab) e < ekekeq2qavgc) e > ekekeResultant of soil pressureResultant of load on footing Page EX1- 1Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen Alshorafa Example # 1 Design an isolated footing to support an interior column 0.5m 0.3mPD= 1000 kN andPL= 600 kNMD= 120 kN.m andML= 70 kN.mUse fc= 25 MPa , fy= 420 MPa, andqall net = 200 kpaPM 0.5 0.3 Page EX1- 2Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 1- Select a trial footing depth Assume that the footing is 0.6 m thick 2- Establish the required base area of the footingCheck stress) 8.93m I and 8.75m (A m 2.5 3.5 Try 6.75m200 600) (1000qPA4 12 2.5 3.5 2 2all(net)sg 3= = = =+= =OK N q kPa 220 8.9 190 8.75 1600Iy MAPq 0 kPa 146 8.93 190 8.75 1600IC MAPqkern) the (within 0.583 0.119600) (1000 70) (120PMenet all 2 3.5s smax 2 3.5s smin 6 3.5 6L.,> =+ = + => =- = - == = < =++= = Page Ex1- 3Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 3- Evaluate the net factored soil pressure) 13.33m I and 10m (A m 2.5 4 Try 412 2.5 423= = = OK q kPa 1 13.33 190 10 1600Iy MAPq 0 kPa 132 13.33 190 10 1600IC MAPqkern) the (within 0 0.119600) (1000 70) (120PMenet all 2 4.0s smax 2 4.0s smin 6 4 6L, 89 67 .< =+ = + => =- = - == = < =++= =( )( )kPa 1 13.33 256 10 2160IC MAPqkPa 254.4 13.33 256 10 2160IC MAPqkN.m 256 1.6(70) 120 1.2 MkN 2 1.6(600) 1000 1.2 P2 4u umin u, 2 4u umax u,uu 6 . 77 160=- = - ==+ = + == + == + = Page Ex1- 4Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionPM 254.4 kPa 254.4 kPa 177.6 kPaL 177.6 kPa 254.4 kPaB 254.4 kPa Page Ex1- 5Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 4- Check footing thickness for punching shear 4.0 m 2.5mC 2+dC 1+d( ) ( )kN 2160 V kN 2465 V kN 3558 1000 530 3720 3720 530 40 2 25 12 1 0.75 d bbd 2 ' f 12 1kN 3697 1000 530 3720 0.5/0.3 2 1 25 6 1 0.75 d b 2 1 ' f 6 1kN 2465 1000 530 3720 25 3 1 0.75 d b ' f 3 1of smallest the is V kN 2160 2.5 4 2 177.6 254.4Vmm 3720 ] 300 530 500 530 2[ bmm 530 20 - 50 - 600 dmax u, Cooscocco cCmax u,oavg= > == + =+= + =+=== +== + + + == = Page Ex1- 6Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 5- Check footing thickness for beam shear In short direction (sec 1- 1) In long direction (sec 2- 2)u c cuavgV kN 1000 530 2 25 6 1 0.75 d b ' f 6 1 VkN Vmm 530 20 - 50 - 600 d> == == - - +== =828 500 740 5 . 2 53 . 0 2 5 . 0 4 2 4 . 254 231 4.0 m 2.5mdd 1 1 2 2d 177.6 kPa 255.4 kPa 231 kPau c cuavgV kN 1000 530 25 6 1 0.75 d b ' f 6 1 VkN Vmm 530 20 - 50 - 600 d> == == - - == =1325 4000 493 0 . 4 53 . 0 2 3 . 0 5 . 2 216d 216 kPa Page Ex1- 7Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 6- Compute the area of flexural reinforcement in each direction In long direction (Sec 1- 1) 4.0 m 2.5m 1 1 177.6 254.4 221 kPa 237.6 x 2.5 1.75( )( )mm 18 1 mm 4 Amm 2 00 2 600 0.0018 Amm 4717 00 2 530 0.00356 A 0.00219 2500 * 530 * 25 0.85 0.9 910 * 10 2- 1 - 1 420 25 0.85 530mm d & 00mm 2 bkN.m 910 2 1.75 2.5 237.6 2l wMface column at critical M2req s, 2min s, 2s 2 6 2 2uuuF == == = = == === =9 717 700 5 5 5 Page Ex1- 8Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 6- Compute the area of flexural reinforcement in each direction In short direction (Sec 2- 2) 4.0 m 2.5m 2 2( )( ) 16mm 1 3323mm 4320 1.6 1 2ent reinforcem band Central 1.6 2.5 4footing of dimension side Shortfooting of dimension side Long , 1 2ratio band Central 4320mm 600 4000 0.0018 A A 0.00125 4000 * 530 * 25 0.85 0.9 523 * 10 2- 1 - 1 420 25 0.85kN.m 523 2 1.1 4 216 2l wMface column at critical M2min s, req s,min 2 6 2 2uuuF = +== = =+== = => = === =7 216x4 Page Ex1- 9Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 7- Check for anchorage of the reinforcementBottom longitudinal reinforcement ( 18mm) =1.0 for bottom bars, =1.0 for uncoated bars =1.0 450 mmmm 4 18 2.5 (0.8)(1.0) (1.0)(1.0) 25 420 10 9l 2.5dK Cuse i.e., 2.5 3.6718 0 66.17dK Cdbtrbtr 35 ===+ > =+=+ Page Ex1- 10Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 7- Check for anchorage of the reinforcementBottom longitudinal reinforcement ( 16mm) =1.0 for bottom bars, =1.0 for uncoated bars =1.0 400 mmmm 388 16 2.5 (0.8)(1.0) (1.0)(1.0) 25 420 10 9l 2.5dK Cuse i.e., 2.5 4.1316 0 66dK Cdbtrbtr===+ > =+=+ Page ex1- 11Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 8- Prepareneatdesigndrawingsshowingfootingdimensionsand provided reinforcement 17 16 3 14 3 14 19 18 4.00 m 2.50 m 0.75m 0.75m 2.50 m 0.60 m Page EX2- 1Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen Alshorafa Example # 2Design a combined footing,to supporttwo columns A and Bspaced atdistance6.0mcenter-to-centerasshowninFigureColumnAis 0.4mx0.4mandcarriesadeadloadof500kNsandaliveloadof 300kNsColumnBisalso0.4mx0.4mincrosssectionbutcarriesa dead load of 700 kNs and a live load of 500 kNs.Use fc= 25 MPa , fy= 420 MPa, andqall net = 150 kpaA B 0.4 0.4 0.4 0.4Propertylimit 6 m ?? m?? mPD =500kNPL =300kNPD =750kNPL =450kN Page EX2- 2Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 1- Establish the required base area of the footingTo locate the resultant of the column forces M@start =0.0 800(0.2)+1200 (6) =2000 (x) x= 3.8 m 2all(net)B Agm 1150 qP PA 33 . 3450 750 300 500=+ + +=+=Ps =800kNPs =1200kNPs =2000kN 6 m 0.2 m X=3.8 mABPs =800kNPs =1200kNPs =2000kN X=3.8 mAB X=3.8 m Page EX2- 3Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 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( )( )kPa 197.4 1.8 7.6 1620 1080AP PqkN.m 1620 1.6(450) 750 1.2 PkN 1080 1.6(300) 500 1.2 Pu B, u A,uuB,uA,=+=+== + == + = Page EX2- 4Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 4- Check footing thickness for punching shearColumn AThe factored shear force Vu =1080 - 197.4(1.115)(0.758) =914 kN b =2(400+715/2)+400+715=2630 mmVc = 2350 kN > Vu = 914OKkN 5968 1000 715 2630 2630 715 30 2 25 12 1 0.75 d bbd 2 ' f 12 1control kN 2350 1000 715 2630 25 3 1 0.75 d b ' f 3 1of smallest the is V oosco cC=+ =+== 7.6 m 1.8 m 758 1115 1115 1115AB Page EX2- 5Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 4- Check footing thickness for punching shear [contd.]Column BThe factored shear force Vu =1620 - 197.4(1.115) 2 =1375 kN b =4(400+715)=4460 mmVc = 3986 kN > Vu = 1375OKkN 8383 1000 715 4460 4460 715 40 2 25 12 1 0.75 d bbd 2 ' f 12 1control kN 3986 1000 715 4460 25 3 1 0.75 d b ' f 3 1of smallest the is V oosco cC=+ =+== 7.6 m 1.8 m 758 1115 1115 1115AB Page EX2- 6Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 5- Draw S.F.D and B.M.D for footing 1080kN 1620 kN 197.37 x1.8= 355.26 kN/m 71 1009 1122.6 497.4 2.84 m 7.10 1425.7 348 0.915 798 Page EX2- 7Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 6- Check footing thickness for beam shear Effective depth d= 800 75 10=715 mm (lower layer)MaximumfactoredshearforceVuislocatedatdistancedfromthe face of column B, Vu,critical =798 kN 7- Compute the areas of flexural reinforcementa) Top longitudinal reinforcementu c cV kN 8041000 715 1800256 10.75 d b ' f6 1 V > == =( ) 20mm 18 Use , mm 5508 715 1800 0.00428 A 0.00428 1800 * 715 * 25 0.85 0.9 1425.7 * 10 2- 1 - 1 420 25 0.85 2req s, 2 6F = == = Page EX2- 8Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 7- Compute the areas of flexural reinforcement [contd.]b) Bottom longitudinal reinforcementc) Short Direction : Effective depth d= 800 75 16- 8=700 mm (Upper layer)( ) 16 13 Use , mm 2592 800 1800 0.0018 A A 0.00101 1800 * 715 * 25 0.85 0.9 348 * 10 2- 1 - 1 420 25 0.85 2min s, req s,min 2 6F = = =< = = 7.6 m 1.8 m 400+700 400+350 1100 750 0.4 0.4 0.4 0.4 Page EX2- 9Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionUnder Column A 7.6 m 1.8 m 400+700 400+350 1100 750 0.4 0.4 0.4 0.4( ) 16 Use6 . mm 1080 750 800 0.0018 A 750 * 700 * 25 0.85 0.9 147 * 10 2- 1 - 1 420 25* 0.85 mm 700 d mm, 750 bkN.m 147 2 0.4 1.82 0.750.75) * (1.8 1080 M 2min s,min 2 6 2u A,F = = 387 mm 10- Prepare neat design drawings showing footing dimensions andprovided reinforcementmm 3 1 2.5 (0.8)(1.0) (1.0)(1.0) 25 420 10 9l 2.5 to equal it take 2.5, 416 0 6dK Cdbtr 87 6 26 .1 . 8==> =+=+ Page EX2- 13Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 14@100 13 16 B 0.75m 1.10 m 1.80 m 0.80 m 8 16 B 6 16 B 14@100 18 20 T 7.60 m 0.4m 0.4m 5.6 m 1.2 m 6 16 8 16 14@100 13 16 18 20 14@100 14@100 14@100 Page EX2- 1Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen Alshorafa Example # 3Design a combined footing,to supporttwo columns A and Bspaced atdistance4.0mcenter-to-centerasshowninFigureColumnAis 0.4mx0.4m andcarriesadeadloadof800kNs andaliveloadof 400kNsColumnBisalso0.3mx0.3m incrosssectionbutcarriesa dead load of 500 kNs and a live load of 250 kNs.Use fc= 28 MPa , fy= 420 MPa, andqall net = 180 kpaA B 0.4 0.4 0.3 0.3Propertylimit 4 m?? mPD =800kNPL =400kNPD =500kNPL =250kN Page EX2- 2Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 1- Establish the required base area of the footingTo locate the resultant of the column forces M@start =0.0 750(4) =1950 (x) x =1.538 m 2all(net)B Agm 10.833180 250 500 400 800qP PA =+ + +=+=Ps =1200kN Ps =750kNPs =1950kN 4 m 0.2 mA BPs =1950kNABPs =1200kN Ps =750kN X=1.54 m Page EX2- 3Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionTry a rectangular combined footing Length of footing L=2(1.54+ 0.20)=3.48mi.e. this type of footing is not possible if uniform soil pressure is to be maintained below the footing.Try a trapezoidal combined footingAssume no projection of footing beyond column B. L=0.2+4+0.15=4.35m()2 2 1m 10.833 L2B B=+( )m 4.98 B Bor , m 10.833 4.352B B2 1 2 2 1= +=+ Page EX2- 3Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 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) 4.35 mA B 0.4 0.4 0.3 0.3 4.0 m 1.0 mc.gXbar =1.74 mm 3.99 0.99 - 4.98 B & m, 0 B 3 4.35 4.98B 4.98 1.74m 4.98 B B but , 3LB B 2B Bx 1 2 2 2 1 2 1 2 1bar= = =+== +++=99 .( )( )kPa 240 10.875 1000 1600AP PqkN.m 1000 1.6(250) 500 1.2 PkN 1600 1.6(400) 800 1.2 Pu B, u A,uuB,uA,=+=+== + == + = A=10.55 m 2 Page EX2- 4Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 4- Check footing thickness for punching shear Column AThe factored shear force,Vu =1600 - 240(1.065)(0.733)=1412.8 kNbo =2(400+665/2)+400+665 =2530 mmVc = 2225 kN > Vu = 1413OKkN 5500 1000 665 2530 2530 665 30 2 28 12 1 0.75 d bbd 2 ' f 12 1control kN 2225 1000 665 2530 28 3 1 0.75 d b ' f 3 1kN 3338 1000 665 2530 1 2 1 28 6 1 0.75 d b 2 1 ' f 6 1of smallest the is V oosco coccC=+ =+=== + =+A 0.4 0.4 0.3 0.3B 0.633 0.733 0.965 1.065 Page EX2- 4Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 4- Check footing thickness for punching shear [contd.]Column BThe factored shear force,Vu =1000 - 240(0.965)(0.633)=854 kNbo =2(300+665/2)+300+665 =2230 mmVc = 1962 kN > Vu = 854 OKkN 5369 1000 665 2230 2230 665 30 2 28 12 1 0.75 d bbd 2 ' f 12 1control kN 1962 1000 665 2230 28 3 1 0.75 d b ' f 3 1kN 2943 1000 665 2230 1 2 1 28 6 1 0.75 d b 2 1 ' f 6 1of smallest the is V oosco coccC=+ =+=== + =+A 0.4 0.4 0.3 0.3B 0.633 0.733 0.965 1.065 Page EX2- 6Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 5- Draw S.F.D and B.M.D for footingq 1 = 240 x 1.0= 240 kN/m & q 2 = 240 x 4= 960 kN/mTheintensityofsoilpressureatdistancexfromtheleftsideof footing is qux =960- 165.5xThe shear force is given asVux =960x- 82.76x 2+C 1,Where C 1 is the constant of integration accounting for shear due to column loadsThebendingmomentisgivenbyMux =480x 2- 27.59x 3+C 2 , Where C 2istheconstantofintegrationaccountingforbendingdueto column loads.The section of maximum bending moment corresponds to the sectionof zero shear force, or 960x- 82.76x 2- 1600=0.0Solving this equation gives x 1 = 2.02 m, and x 2 =9.71m (rejected).Substituting x 1 =2.05m in moment equation gives Mu,max =1187kN.m Page EX2- 6Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 1600 kN 1000 kN 240 960xqu(x) 188 1412 962 38 18.9 1187 2.8 2.02 0.865 0.815 671 701kN 671 V kN 1433 1000 665 3257286 10.75 d b ' f6 1 Vuc c= > == = 6- Check footing thickness for beam shearSection A-A Effective depth d=750 75 10= 665mmCritical section for beam shear is located at distance 665+200= 865mm to theright side of center of column A. X=0.2+0.865=1.065 mVu =960x- 82.76x 2- 1600=671 kNThe width of footing at @ x=1.065 is equal to 3266 mm Page EX2- 6Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 1600 kN 1000 kN 240 960xqu(x) 188 1412 962 38 18.9 1187 2.8 2.02 0.865 0.815 671 701kN 701 V kN 732.8 1000 665 1666286 10.75 d b ' f6 1 Vuc c= > == = 6- Check footing thickness for beam shearSection B-B Effective depth d=750 75 10= 665mmCritical section for beam shear is located at distance 665+150= 815mm to theleft side of center of column B. X = 4.2 - 0.815 = 3.385 mVu =960x- 82.76x 2- 1600=701 kNThe width of footing at @ x=3.385 is equal to 1666 mm Page EX2- 7Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 7- Compute the areas of flexural reinforcementa) Top longitudinal reinforcement The width of footing at @ x=2.02 is equal to 2600 mmb) Bottom longitudinal reinforcement( ) 20@160mm) ( or 20mm 16 Use, mm 4842 665 2600 0.0028 A 0.0028 2600 * 665 * 28 0.85 0.9 1187 * 10 2- 1 - 1 420 25 0.85 2req s, 2 6F F= == = 0.0018 used so min min= = < ,mm 14@100 Use /m, mm 1350 750 1000 0.0018 A A2min s, req s,= = = Page EX2- 9Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionUnder Column A( )mm 2 7 Use . mm 2560 733 651 0.00537 A 0.00537 733 * 651 * 28 0.85 0.9 600 * 10 2- 1 - 1 420 28* 0.85 mm 651 d mm, 733 bkN.m 2 2 0.4 - 3.75 3.75 1600M 3.75m 2 3.5 4l 651mm 10 14 75 750layer upper for depth Effective2req s, 2 6 2u A,avg 2 600F = ==== == ==+== - - - =A 0.4 0.4 0.3 0.3B 0.733 0.633 4.0 m 3.5 m 1.44 m 1.0m Page EX2- 9Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionUnder Column B( ) 14mm 6 Use . mm 855 633 750 0.0018 A A 0.00087 633 * 651 * 28 0.85 0.9 87 * 10 2- 1 - 1 420 28* 0.85 mm 651 d mm, 633 bkN.m 87 2 2 0.3 - 1.22 1.22 1000M 1.22m 2 1.0 1.44l 651mm 10 14 75 750layer upper for depth Effective2min s, req s,min 2 6 2u A,avg= = =< === == ==+== - - - = A 0.4 0.4 0.3 0.3B 0.733 0.633 4.0 m 3.5 m 1.44 m 1.0m Page EX2- 11Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionShrinkage Reinforcement in the short direction (top and bottom sides) 9- Check for anchorage of the reinforcementa) Top longitudinal reinforcement ( 20mm) =1.3 for top bars, =1.0 for uncoated bars =0.8 for 20mm, =1.0 for normal weight concrete C the smallest of 75+10=85 mm 50 mm i.e., C is taken as 50 mm Available length =2020- 75=1945 > 720 mm 14@100mm Use . 1350mm 1000 750 0.0018 A2min s,= = 2.5 2.2722 0 50dK Cbtr< =+=+mm 720 222.27 (0.8)(1.0) (1.3)(1.0)28 42010 9ld== Page EX2- 12Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionb) Bottom longitudinal reinforcement ( 14mm)= =1.0 =0.8 for 14mm, =1.0 for normal weight concrete C the smallest of 75+7=82 mm 50mm i.e., C is taken as 50 mm Available length =1400- 75=1325 > 387 mmmm 320 14 2.5 (0.8)(1.0) 28 420 10 9l 2.5 to equal it take 2.5, 3.5714 0 50dK Cdbtr==> =+=+ Page EX2- 12Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionc) Bottom transverse reinforcement ( 14mm) = = = 1.0 =0.8 for 14mm, C the smallest of 75+7=82 mm [(633- 75- 7)/(2x5)]= 55.1 mm i.e., C is taken as 55.1 mm Available length =350- 75=275 < 320 mmBottom longitudinal bars need to be hooked at their ends to provide additional anchorage.mm 320 14 2.5 (1.0)(0.8) 28 420 10 9l 2.5 to equal it take 2.5, 3.9414 0 5dK Cdbtr==> =+=+ 1 . 5 Page EX2- 12Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutiond) Bottom transverse reinforcement ( 22mm) = = = 1.0 =1.0 for 22mm, C the smallest of 75+11=86 mm [(733- 75- 11)/(2x6)]= 53.9 mm i.e., C is taken as 53.9 mm Available length =1550- 75=1475 > 642 mm 10- Prepareneatdesigndrawingsshowingfootingdimensionsand provided reinforcementmm 642 22 2.45 (1.0) 28 420 10 9l 2.5 2.4522 0 53.9dK Cdbtr==> =+=+ Page EX2- 13Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 0.733m 0.633 m 0.75 m 0.4m 0.3m 3.65 m 7 22 6 14 14@100 20@160 14@100B 1.0 m 4.0 m A 7 22 B 6 14 B 14@100 T&B 14@100 B 7.60 m 20@160 T Page EX2- 1Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen Alshorafa Example # 4Design a combined footing, to supportthree columnsA, B, and C as shown in Figure. ColumnAis 0.4mx0.4m and carries a deadload of 500kNsandaliveloadof300kNs,columnBisalso0.4mx0.4min crosssectionbutcarriesadeadloadof800kNs andaliveloadof 480 kNs , and column C is also 0.4mx0.4m in cross section but carries a dead load of 600 kNs and a live load of 360 kNs.Use fc= 28 MPa , fy= 420 MPa, andqall net = 180 kpaA 0.4 0.4 4.5 m?? mPD =500kNPL =300kNPD =800kNPL =480kNB 0.4 0.4C 0.4 0.4PD =600kNPL =360kN 5.5 mL 1L 2 Page EX2- 2Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 1- Establish the required base area of the footingTo locate the resultant of the column forces M@A =0.0 1280(4.5)+960(10)=3040 (x) x =5.05 m 2all(net)C B Agm 16.89180 qP P PA =+ +=+ +=960 1280 800 800 kN 1280 kN R =3040 kN 4.5 mA B 960 kNC 5.5 m 800 kN 1280 kN R =3040 kNAB 960 kNCxbar =5.05 m Page EX2- 3Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionLength of footing Ltotal =2(5.05+L 1) ,assume L 1 = 0.6 mLtotal =2(5.05+0.6)=11.3 m & L 1 +5.05 = (10- 5.05)+L 2 L 2 =0.7 mL x B = Area 11.3 x B =16.89 B=1.495 m, taken as 1.5m 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 (lower layer)( )( )( )kPa 242.12 1.5 11.3 1296 1728 1080AP PqkN.m 1296 1.6(360) 600 1.2 PkN.m 1728 1.6(480) 800 1.2 PkN 1080 1.6(300) 500 1.2 Pu B, u A,uuC,uB,uA,= + +=+== + == + == + = Page EX2- 4Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 4- Check footing thickness for punching shearColumn BThe factored shear force Vu =1728 - 242.12(1.115)(1.115) =1427 kN b =4(400+715) =4460 mm= + =+= + =+> ==kN 8872 1000 715 4460 4460 715 40 2 8 2 12 1 0.75 d bbd 2 ' f 12 1kN 6328 1000 715 4460 1 2 1 28 6 1 0.75 d b 2 1 ' f 6 1OK V control kN 4219 1000 715 446028 3 10.75 d b ' f 3 1Vooscoccu o cc 4.5 m 1.5 m 5.5 m 0.6 0.7A B C 1115 1115B d/2 d/2 Page EX2- 6Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 5- Draw S.F.D and B.M.D for footing 1080 kN 1728 kN 242.12 x1.5= 363.18 kN/m 217.9 862.1 1041.7 2.37 m 66 958 0.915 772.2 1296 kN 955.7 254.2Vuc =709.4 2.87 m 1405 142 89 Page EX2- 6Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 6- Check footing thickness for beam shear Effective depth d=800 75 10= 715mm (lower layer)MaximumfactoredshearforceVuislocatedatdistance 715+200=915mm to the left side of center of column C.Vu =1041.7- 0.915x363.18=709.4 kN 7- Compute the areas of flexural reinforcementa) Top longitudinal reinforcementOK kN 7 V kN 71000 715 1500286 10.75 d b ' f6 1 Vu c c 4 . 09 4 . 09 = = == =( ) 20mm 18 Use , mm 5443 715 1500 0.005074 A 0.005074 1500 * 715 * 28 0.85 0.9 1405 * 10 2- 1 - 1 420 28 0.85 2req s, 2 6F = == = Page EX2- 8Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 7- Compute the areas of flexural reinforcement [contd.]b) Bottom longitudinal reinforcementc) Short Direction : Effective depth d= 800 75 16- 8=700 mm (Upper layer) 14@100 Use , mm 2160 800 1500 0.0018 A A 0.0018 use so , 2min s, req s,min minF = = == = < 1.5 mA B C 1.1 m 1.1 m 1.1 m Page EX2- 9Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionUnder Column AUnder Column B( ) 16) or(8 14@100 Use . mm 1584 1100 800 0.0018 A A 0.000538 1100 * 700 * 28 0.85 0.9 109 * 10 2- 1 - 1 420 28* 0.85 kN.m 109 2 2 0.4 - 1.51.5 1080M 2min s, req s, 2 6 2u A,= = ==== =( ) 16) (8 or 14@100 Use . mm 1584 1100 800 0.0018 A A 0.000866 1100 * 700 * 28 0.85 0.9 175 * 10 2- 1 - 1 420 28* 0.85 kN.m 175 2 2 0.4 - 1.51.5 1728M 2min s, req s, 2 6 2u B,= = ==== = Page EX2- 9Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionUnder Column CShrinkage Reinforcement in the short direction (top and bottom)( ) 16) (8 or 14@100 Use . mm 1584 1100 800 0.0018 A A 0.000647 1100 * 700 * 28 0.85 0.9 131 * 10 2- 1 - 1 420 28* 0.85 kN.m 131 2 2 0.4 - 1.51.5 1296M 2min s, req s, 2 6 2u C,= = ==== = 14@100mm Use . 1440mm 1000 800 0.0018 A2min s,= = Page EX2- 11Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 9- Check for anchorage of the reinforcementa) Top longitudinal reinforcement ( 20mm) =1.3 for top bars, =1.0 for uncoated bars =0.8 for 20mm, =1.0 for normal weight concrete C the smallest of 75+10=85 mm [1500- 2(75)- 20]/(2x17)=39.1 mm i.e., C is taken as 39.1 mm Available length =2370+600- 75=2895 > 762 mm 2.5 120 0 39.1dK Cbtr< =+=+95 .mm 762 201.95 (1.0) (1.3)(0.8)28 42010 9ld== Page EX2- 11Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 9- Check for anchorage of the reinforcementb) Bottom reinforcement ( 14mm)= = = 1.0and =0.8 C the smallest of 75+10=85 mm and 100/2 =50 mm i.e., C is taken as 50 mmFor longitudinalreinforcement. Available length =600- 75=525mm > 320 mmFor transversereinforcement. Available length =550- 75=475mm > 320 mm 2.5 to equal it take 2.514 0 5dK Cbtr, 57 . 30> =+=+mm 320 142.5 (1.0)(0.8)28 42010 9ld== Page EX2- 4Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 4.5 m 1.5 m 5.5 m 0.6 0.7A B C 1.10 m 0.80 m 8 16 14@100 18 20 14@100 8 16 1.10 m 8 16 1.10 m 14@100 11.30 m 14@100 8 16 B 8 16 B 8 16 B 14@100 14@100 14@100 B 18 20T Page EX2- 1Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen Alshorafa Example # 5Designacombinedfooting,tosupportthreecolumnsA,B,andC usinga0.5mwidespinebeamasshowninFigure.ColumnAis 0.4mx0.4mandcarriesadeadloadof500kNsandaliveloadof 300kNs,columnBisalso0.4mx0.4mincrosssectionbutcarriesa deadloadof800kNs andaliveloadof480kNs,andcolumnCis also 0.4mx0.4min cross section but carries adeadloadof600kNs and a live load of 360 kNs.Use fc= 28 MPa , fy= 420 MPa, andqall net = 180 kpaA 0.4 0.4PD =500kNPL =300kNPD =800kNPL =480kNB 0.4 0.4C 0.4 0.4PD =600kNPL =360kNA B C 5.5 m 4.5 m Page EX2- 2Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 1- Establish the required base area of the footingTo locate the resultant of the column forces M@A =0.0 1280(4.5)+960(10)=3040 (x) x =5.05 m 2all(net)C B Agm 16.89180 qP P PA =+ +=+ +=960 1280 800 800 kN 1280 kN R =3040 kN 4.5 mA B 960 kNC 5.5 m 800 kN 1280 kN R =3040 kNAB 960 kNCxbar =5.05 m Page EX2- 3Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionLength of footing Ltotal =2(5.05+L 1) ,assume L 1 = 0.6 mLtotal =2(5.05+0.6)=11.3 m & L 1 +5.05 = (10- 5.05)+L 2 L 2 =0.7 mL x B = Area 11.3 x B =16.89 B=1.495 m, taken as 1.5m 2- Evaluate the net factored soil pressure 3- Select a trial footing depth Assume that the footing is 0.3 m thick Effective depth d = 300 75 7 = 218 mm (lower layer)( )( )( )kPa 242.12 1.5 11.3 1296 1728 1080AP PqkN.m 1296 1.6(360) 600 1.2 PkN.m 1728 1.6(480) 800 1.2 PkN 1080 1.6(300) 500 1.2 Pu B, u A,uuC,uB,uA,= + +=+== + == + == + = Page EX2- 6Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 4- Check footing thickness for beam shear Effective depth d=300 75 7= 218 mm For a strip 1-m wideMaximumfactoredshearforceVuislocatedatdistancedfromthe face of beamVu =242.12x1.0x0.282 =68.3 kN < Vc =144 kNkN 1441000 218 1000286 10.75 d b ' f6 1 Vc c== = 1 m 0.5 m d=0.218 0.5- 0.218=0.282 1.5 m Page EX2- 6Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolution 5- Compute the area of flexural reinforcement for the slab a- Transverse direction: For a strip 1-m wideMaximum factored moment is located at the face of beamMu =242.12(1.0)(0.5) 2 =30.3 kN.m 2 1 m 0.5 m 0.5 m 1.5 m( ) 12@200mm Use , mm 540 300 1000 0.0018 A A 0.0017 1000 * 218 * 28 0.85 0.9 30.3 * 10 2- 1 - 1 420 28 0.85 2min s, req s,min 2 6F = = =< = = Page EX2- 11Design of Concrete Structure IIUniversity of PalestineFootingInstructor:Eng. Mazen AlshorafaSolutionb- Longitudinal direction:For shrinkage reinforcement in the Longitudinal direction of the slab 5- Draw S.F.D and B.M.D for footing 12@200mm Use . 540mm 1000 300 0.0018 A2min s,= = 1080 kN 1728 kN 242.12 x1.5= 363.18 kN/m 217.9 862.1 1041.7 2.37 m 66 958 0.915 772.2 1296 kN 955.7 254.2Vuc =709.4 2.87 m 1405 142 89