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CHAPTER ELEVENFOOTINGS10Example (11.1):Design an isolated footing to support an interior column 25 cm 60 cm in cross section thatcarries a dead load of 60 tons, a live load of 40 tons, a dead load moment of 15 t.m, and a liveload moment of 10 t.m (service loads and moments). Use f c′ 250 kg/cm 2 , f y 4200 kg/cm 2 ,q all ( gross ) 2.0 kg/cm 2 , γsoil 1.7 t/m3 , and D f 2.0 m .Note that there is no restriction on foundation projections on any side of the footing.Solution:1- Select a trial footing depth:Assume that the footing is 50 cm thick.2- Establish the required base area of the footing:q all (net ) 20 1.5 (1.7 ) 0.5 (2.5) 16.2 t / m 2Eccentricity e M 15 10 0.25 m P 60 40Figure 11.10.aApproximate footing area is given as:Areq Pq all (net ) 60 40 6.17 m 216.2Try 3.10 m 2.0 m 0.5 m footing

CHAPTER ELEVENFOOTINGS11In order to have uniform soil pressure under the footing, the footing is to be positioned in such away to balance the given moment through shifting the centroid of the footing 0.25 m away fromthe centroid of the column as shown in Figures 11.10.a and 11.10.b.3- Evaluate the net factored soil pressure:Pu 1.20 (60 ) 1.60 (40 ) 136 tonsP136qu (net ) u 21.94 t/m2 , as shown in Figure 11.10.b.L B (3.1)(2.0 )Figure 11.10.b4- Check footing thickness for punching shear:Average effective depth d 50 – 7.5 – 1.6 40.9 cmThe factored shear forceVu (21.94 )[(3.1)(2.0 ) 1.009 (0.659 )] 121.44 tonsbo 2 (60 25 40.9 40.9 ) 333.6 cmΦ Vc is the smallest of:Figure 11.10.c: Critical section for punching shear

CHAPTER ELEVENΦ Vc 0.53 ΦFOOTINGS2 f ' c 1 λ bo dβ 0.53 (0.75) 250 1 Φ Vc λ Φ122 (333.6 )(40.9 ) / 1000 157.22 tons60 / 25 f ' c bo d 0.75 250 (333.6 )(40.9 ) / 1000 161.80 tons α d Φ Vc 0.27 Φ s 2 λ bo f ' c bo d 0.27 (0.75 ) 40 (40.9 ) 2 Φ Vc 157.22 tons333.6 250 (333.6 ) (40.9 ) / 1000 301.61 tons 121.44 tonsi.e. footing thickness is adequate for resisting punching shear.5- Check footing thickness for beam shear in each direction:a- In the short direction:The critical section for beam shear is given in Figure 11.10.d.Φ Vc 0.75 (0.53 ) 250 (200 )(40.9 ) / 1000 51.41 tonsMaximum factored shear force Vuis located at distance d from faces of column,Vu (21.94 )(2.0 )(1.091) 47.87 tons 51.41 tonsb- In the long direction:The critical section for beam shear is given in Figure 11.10.e.Φ Vc 0.75 (0.53 ) 250 (310 )(40.9 ) / 1000 79.69 tonsMaximum factored shear force Vuis located at distance d from faces of column,Vu (21.94 )(3.1)(0.466 ) 31.69 tons 79.69 tons

CHAPTER ELEVENFOOTINGSFigure 11.10.d: Critical section for Figure 11.10.e: Critical section forbeam shear (short direction)beam shear (long direction)6- Compute the area of flexural reinforcement in each direction:a- Reinforcement in long direction:(1.5 )2M u (21.94 )(2.0 ) 49.37 t .m2ρ 0.85 (250 ) 2.353 (10 )5 (49.37 ) 1 1 0.00406724200 ()()()()0.9020040.9250 As 0.004067 (200 )(40.9 ) 33.27 cm2 , use 14 φ 18 mm in the long direction.b- Reinforcement in short direction:M u (24.52 )(3.1)ρ (0.875 )2 26.04 t .m20.85 (250 ) 2.353 (10 )5 (26.04 ) 1 1 0.00134624200 ()()()()0.9031040.9250 As 0.0018 (310 )(50 ) 27.9 cm2 2 Ascentral band reinforcement 1 β 2 (27.9 ) 21.88 cm2 1310/200 Use 15 φ 14 mm in the central bandFor the side bands, As (27.9 21.88) 6.02 cm 213

CHAPTER ELEVENFOOTINGSUse 2 φ 14 mm in the right side band, and 3 φ 14 mm in the left side band.7- Check for bearing strength of column and footing concrete:For column,Φ Pn 0.65 (0.85 )(250 )(25 )(60 ) / 1000 207.18 tons 136 tonsi.e. use minimum dowel reinforcement, As 0.005 (25)(60 ) 7.5 cm 2The column is to be designed for axial force plus bending moment.8- Check for anchorage of the reinforcement:a. Reinforcement in long direction ( φ 18 mm ):ψ t ψ e λ 1 and ψ s 0.8cb is the smaller of:7.5 0.9 8.4 cm , or200 15 1.8 6.54 cm , i.e., cb 6.54 cm14 (2)cb Ktr 6.54 0 3.63 2.5 , take it equal to 2.5db1.8ld 0.8(1.8 ) (4200 ) 43.71 cm3.5 (2.5 ) 250Available length 100.0 – 7.5 92.5 cm 43.71 cmb. Reinforcement in short direction ( φ 14 mm ):ψ t ψ e λ 1 and ψ s 0.8cb is the smaller of:7.5 0.7 8.2 cm , or200 15 1.4 6.56 cm , i.e., cb 6.56 cm14 (2)cb Ktr 6.56 0 4.69 2.5 , take it equal to 2.5db1.4ld 0.8(1.4 )(4200 ) 34.0 cm3.5 (2.5 ) 250Available length 87.5 – 7.5 80.0 cm 34.0 cm14

CHAPTER ELEVENFOOTINGS9- Prepare neat design drawings, showing footing dimensions and providedreinforcement:Design drawings are shown in Figure 11.10.f.Figure 11.10.f: Design drawings15

CHAPTER ELEVENFOOTINGS16Example (11.2):Design an isolated edge footing to support an edge column 70 cm 25 cm in cross section andcarries a dead load of 25 tons and a live load of 20 tons.Usef c′ 250 kg / cm 2 ,f y 4200 kg / cm 2 ,q all ( gross ) 2.5 kg / cm2 ,D f 1.5 m .Figure 11.11.a: Footing dimensionsSolution:1- Select a trial footing depth:Assume that the footing is 40 cm thick, shown in Figure 11.11.a.2- Establish the required base area of the footing:qall (net ) 25 1.1(1.7 ) 0.4 (2.5) 22.13 t/m2Try 1.3 m 3.0 m 0.4 m footingq all (net ) 25 20 6 (25 20 )(0.3) 21.3 (3)3 (1.3)γ soil 1.7 t / m3 , and

CHAPTER ELEVENqmax 11.54 15.97 27.51 t/m2FOOTINGS 22.13 t/m2 (compressive)qmin 11.54 15.97 4.43 t/m2 (tensile), as shown in Figure 11.11.b.Try 1.0 m 4.0 m 0.4 m footingq all (net ) 25 20 6 (25 20 )(0.15) 21.0 (4 )4 (1.0 )qmax 11 .25 10.125 21.375 t / m2 22.13 t / m2 (compressive)q min 11 .25 10.125 1.125 t / m 2 (compressive), as shown in Figure 11.11.c.Use 1.0 m 4.0 m 0.4 m footingFigure 11.11.b: Net soil pressureFigure 11.11.c: Net soil pressure3- Evaluate the net factored soil pressure:Pu 1.20 (25 ) 1.60 (20 ) 62 tonsqu (net ) 626 (62 )(0.15 ) 1.0 (4 )4 (1.0 )2qu max 15.50 13.95 29.45 t/m2qu min 15.50 13.95 1.55 t/m2, shown in Figure 11.11.d.17

CHAPTER ELEVENFOOTINGSFigure 11.11.d: Critical section for punching shear and factored soilpressure4- Check footing thickness for punching shear:Average effective depth d 40 – 7.5 – 1.6 30.9 cmThe factored shear force 5.61 29.45 Vu 62 (0.8545 )(0.559 ) 53.72 tons2 bo 2 (85.45) 25.0 30.9 226.8 cmΦ Vc is the smallest of:Φ Vc 0.53 Φ 2 f ' c 1 λ bo dβ 0.53 (0.75) 250 1 Φ Vc λ Φ2 (226.8 )(30.9 ) / 1000 75.51 tons70 / 25 f ' c bo d 0.75 250 (226.8 ) (30.9 ) / 1000 83.11 tons α d Φ Vc 0.27 Φ s 2 λ bo f ' c bo d 0.27 (0.75 ) 30 (30.9 ) 2 Φ Vc 75.51 tons226.8 53.72 tons 250 (226.8 ) (30.9 ) / 1000 136.59 tons18

CHAPTER ELEVENFOOTINGS19i.e. footing thickness is adequate for resisting punching shear.5- Check footing thickness for beam shear in each direction:Figure 11.11.e: Critical section for Figure 11.11.f: Critical section forbeam shear (short direction)beam shear (short direction)a- In the short direction:The critical section for beam shear is shown in Figure 11.11.e.Φ Vc 0.75 (0.53 ) 250 (100 )(30.9 ) / 1000 19.42 tonsMaximum factored shear force Vu is located at distance d from faces of column. The factoredshear force 1.55 29.45 Vu (1.0 )(1.566 ) 24.27 tons2 19.42 tonsIncrease footing thickness to 50 cm and check for beam shear. The critical section for beam shearis shown in Figure 11.11.f.Φ Vc 0.75 (0.53 ) 250 (100 )(40.9 ) / 1000 25.71 tonsThe factored shear force 1.55 29.45 Vu (1.0 )(1.466 ) 22.72 tons2 25.71 tonsi.e. footing thickness is adequate for resisting shear.6- Compute the area of flexural reinforcement in each direction:a- Reinforcement in short direction:

CHAPTER ELEVEN[FOOTINGS] 9.92 1.55 M u 4 (1.55 )(0.3 )2 / 2 4 (0.3 )(0.1) 0.78 t .m2 As 0.0018 (400)(50) 36.0 cm 2 2 AsCentral band reinforcement 1 β 2 (36) 14.40 cm2 1 400 / 100 Use 13 φ 12 mm in the central bandFor the side bands, As (36 14.40) 21.60 cm2Use 10 φ 12 mm in each of the side bandsb- Reinforcement in long direction: 1.55 29.45 2 M u (1.0 / 2 )(1.875 ) 27.25 t .m2 ρ 0.85 (250 ) 2.353 (10 ) 5 (27.25 ) 1 1 0.004510724200 ()()()0.910040.9250 As 0.0045107 (100 )(40.9 ) 18.45 cm2 , use 10 φ 16 mm in the long direction7- Check for bearing strength of column and footing concrete:For column,Φ Pn 0.65 (0.85 )(250 )(25 )(70 ) / 1000 241.72 tons 62 tonsi.e. use minimum dowel reinforcement, As 0.005 (25 )(70 ) 8.75 cm28- Check for anchorage of the reinforcement:a- Reinforcement in long direction ( φ 16 mm ):ψ t ψ e λ 1 and ψ s 0.8cb is the smaller of:7.5 0.8 8.3 cm , or100 15 1.6 4.17 cm , i.e., cb 4.17 cm10 (2 )20

CHAPTER ELEVENFOOTINGS21cb Ktr 4.17 0 2.61 2.5 , take it equal to 2.5db1.6ld 1.6 (0.8 ) (4200 ) 38.86 cm3.5 (2.5 ) 250Available length 187.5 – 7.5 180.0 cm 38.86 cmb- Reinforcement in short direction ( φ 12 mm ):ψ t ψ e λ 1 and ψ s 0.8cb is the smaller of:7.5 0.6 8.1 cm , or400 15 1.2 6.19 cm , i.e., cb 6.19 cm31 (2)cb Ktr 6.19 0 5.16 2.5 , take it equal to 2.5db1.2ld 1.2 (0.8 )(4200 ) 29.14 cm3.5 (2.5 ) 250Available length 30.0 – 7.5 22.5 cm 29.14 cmHook all bars at their ends to provide additional anchorage length.9- Prepare neat designreinforcement:drawings showing footing dimensions and providedDesign drawings are shown in Figure 11.11.g.As seen in this example, the design leads to a footing that is long and narrow. Either combined orstrap footing is a better solution for this eccentric case.

CHAPTER ELEVENFOOTINGSFigure 11.11.g: Design drawings22

CHAPTER ELEVENFOOTINGS23Design of Wall FootingsWhen a wall carries a uniformly distributed line load, wall sections along the length of the wallbehave equally. Consequently, the design of the footing can be based on a strip 1-m wide alongthe length of the wall. Design of wall footings is summarized in the following steps.1- Select a trial footing depth:According to ACI Code 15.7, depth of footing above reinforcement is not to be less than 15 cmfor footings on soil. Establish the required base width of the footing:Width of footing is established by dividing the total service load by the allowable net soilpressure.2- Evaluate the net factored soil pressure.3- Check footing thickness for beam shear in the transverse direction:The critical section for beam shear is located at distance d from the face of the wall.4- Compute the area of flexural reinforcement:If a footing carries a concrete wall, ACI Code 15.4.2 specifies that the critical section for momentbe taken at the face of the wall. Main reinforcement is provided in the short direction whileshrinkage reinforcement is provided in the long direction of the footing.5- Check for bearing strength of wall and footing concrete.6- Check for anchorage of the reinforcement:Both, flexural and dowel reinforcement lengths are checked for anchorage to prevent bond failureof the dowels in the footing and to prevent failure of the lap splice between the dowels and thewall reinforcing bars.7- Prepare neat designreinforcement.drawings showing footing dimensions and provided

CHAPTER ELEVENFOOTINGS24Example (11.3):Design a footing to support a reinforced concrete wall 20 cm thick as shown in ﺧﻄﺄ! ﻟﻢ ﯾﺘﻢ اﻟﻌﺜﻮر ﻋﻠﻰ . ﻣ ﺼﺪر اﻟﻤﺮﺟ ﻊ .a. The wall supports a service dead load of 10 t/m and a service live load of 7.5 t/m inaddition to its own weight.Use f c′ 300 kg / cm 2 , f y 4200 kg / cm 2 , q all ( gross ) 1.8 kg / cm 2 , γ soil 1.7 t / m3 , andD f 2.0 m .Figure 11.12.a: Footing dimensionsSolution:1- Select a trial footing depth:Assume that the footing is 30 cm thick.2- Establish the required base width of the footing:For a strip 1-m wide along the wall,q all (net ) 18 1.7 (1.7 ) 0 .3 (2.5 ) 14 .36 t/m2Total service load /m 10 7.5 0.20 (1.7 3) (2.5) 19.85 tonsAreq 19.85 1.38 m214.36Use B 1.4 m.3- Evaluate the net factored soil pressure:qu (net ) 1.2 (10 ) 1.6 (7 .5 ) 1.2 (0.2 )(4.7 )(2.5 ) 26.82 19.16 t/m21.4 (1.0 )1.4

CHAPTER ELEVENFOOTINGS4- Check footing thickness for beam shear:Effective depth d 30 – 7.5 – 0.70 21.8 cmΦ Vc 0.75 (0.53) 300 (100 )(21.8 ) / 1000 15.0 tonsMaximum factored shear force Vu is located at distance d from the face of wall, 1.4 0.2 Vu (19.16 )(1.0 ) 0.218 7.32 tons 15.0 tons2 i.e. provided footing thickness is adequate for resisting beam shear.5- Compute the area of flexural reinforcement:a- Flexural reinforcementM u 19.16ρ (0.6 )2 3.45 t .m20.85 (300 ) 2.353 (10 )5 (3.45 ) 1 1 0.001954200 0.9 (100 ) (21.8 )2 (300 ) As 0.00195 (100 ) 21.8 4.25 cm2As ,min 0.0018 (100 )30 5.40 cm2use φ 12 mm @ 205 cmb- Shrinkage reinforcement:As 0.0018 (140 )(25) 6.30 cm 2 , use 8φ 10 mm6- Check for bearing strength of wall and footing concrete:Φ Pn 0.65 (0.85 )(300 )(20 )(100 ) / 1000 331.5 tons 26.82 tonsi.e, use minimum dowel reinforcement, As 0.005 (20 )(100 ) 10.0 cm27- Check for anchorage of the reinforcement:a- Flexural reinforcement ( φ 12 mm ):ψ t ψ e λ 1 and ψ s 0.8C is the smaller of:7.5 0.6 8.1 cm , or15 7.5 cm , i.e., cb 7.5 cm225

CHAPTER ELEVENFOOTINGScb Ktr 7.5 0 6.25 2.5 ,db1.2ld take it equal to 2.51.2 (0.8 )(4200 ) 26.60 cm3.5 (2.5 ) 300Available development length 60.0 –7.5 52.5 cm 26.60 cmO.Kb- Dowel reinforcement ( φ 10 mm ):To calculate required development length for φ 10 mm bars,ld 0.075 (1.0 )(4200)300 18.18 cm ,orld 0.0044 (1.0 )(4200 ) 18.48 cm , orl d 20.0 cmAvailable length 30 - 7.5 - 1.2 -1 20.30 cm 20 cm O.Kc- Reinforcement splices:To calculate splice length for φ 10 mm bars,lsplice 0.0073 (1.0 )(4200) 30.66 cm8- Prepare neat design drawings, showing footing dimensions and providedreinforcement:De