Design of Combined Footings

Size: px

Start display at page:

Download "Design of Combined Footings"

Darcy Watts
6 years ago
Views:

1 7 Design of Combine Footings Summary of combine footing esign is shown in the following steps. 1- Select a trial footing epth. - Establish the require base area of the footing: Uniform soil pressure is achieve by making the resultant of the service column loas coincie with the centroi of the footing base. 3- Evaluate the net factore soil pressure. 4- Check footing thickness for punching shear. 5- Draw S.F.D an B.M.D for footing: The shear force an bening moment iagrams for the footing are to be rawn, consiering the footing as an inverte beam subjecte to istribute loa. Column loas may be consiere as concentrate loas to simplify the analysis. 6- Check footing thickness for beam shear. Beam shear capacity is checke against critical factore shear evaluate from S.F.D. Depth of footing may be increase to satisfy beam shear requirement. 7- Compute the area of flexural reinforcement: Flexural reinforcement require in the longituinal irection is evaluate. Furthermore, the areas of flexural reinforcement uner columns in the transverse irection are compute. In computing this reinforcement, it will be assume that each column loa is uniformly istribute over a ban centere on the column an having a with extening a istance / from its faces. This area of reinforcement is to be uniformly istribute across the ban with (C +), an the rest of the footing is to be provie with shrinkage an temperature reinforcement. 8- Check bearing strength of column an footing concrete. 9- Check for anchorage of the reinforcement. 10- Prepare neat esign rawings showing footing imensions an provie reinforcement. Example (11.4): Design an appropriate footing/footings to support two columns A an B space at istance.1 m center-to-center, as shown in Figure a is 0 cm 30 cm an carries a ea loa of 0 tons an a live loa of 10 tons. Column B is 0 cm 40 cm in cross section but carries a ea loa of 30 tons an a live loa of 15 tons. With of footing is not to excee 1.0 m, an there is no property line restriction. 7

2 8 Use 300 kg / cm, f c D f. 0 m. q all gross.0 kg / cm, f y 400 kg / cm, ( ) γ soil t / m, an Figure a: Footing imensions Solution: 1- Select a trial footing epth: Assume that the footing is 50 cm thick. - Establish the require base area of the footing: ( net) ( 1.7) 0.5(.5) 16. t / m q all If isolate footings are to be use, PA 30 A1 req 1. 85m q 16. all ( net) B 1 m, an L 1.85 m. PB 45 A req. 78 m q 16. all ( net) B 1 m, an L.80 m. Overlapping of the two footings occurs, ruling out isolate footings as appropriate of footing types. A combine footing will be use instea. A req P q A all + P B ( net) A req 4.63m 16. To locate the resultant of the column forces, replace the column loa system with a resultant force system, or 8

3 9 P B (.1) R ( x ), or 45 (.1) ( )( x ) 1 an x 1. 6 m 1 1 x l + +, or 1 x1 l x m l l x x 0. 4 m (a) 1 1 Let B 1 m, length of footing l + l m an L 4. 63m, taken as 4.65 m 1.0 l + l. 55 m (b) 1 Aing (a) an (b) gives, l m an l m 1 3- Evaluate the net factore soil pressure: ( 0) ( 10) 40 tons P Au 1.0 ( 30) ( 15) 60 tons P Bu 1.0 P ( ) Au + PBu q u net 1.51 t / m LB 4.65 ( 1.0) 4- Check footing thickness for punching shear: Column B: Effective epth cm (lower layer) The factore shear force ( 1.51)( 0.817)( 0.617) tons V u 60 ( ) 86.8 cm b o Φ V c is the smallest of: 0.53 Φ f' c 1 + λ bo β + 40 / ( 0.75) ( 86.8)( 41.7) / tons λ Φ f' c bo ( 86.8) ( 41.7) / tons α V 0.7 s Φ c Φ + λ f' c bo b o.7 ( 0.75) ( 41.7) ( 86.8)( 41.7) / tons 9

4 30 Φ V c tons > tons Column A: The factore shear force ( 1.51)( 0.717)( 0.617) tons V u 40 ( ) 66.8 cm b o Φ V c is the smallest of: 0.53 Φ f' c 1 + λ bo β 0.53 ( 0.75) ( 66.8)( 41.7) / tons 30 / 0 λ Φ f' c bo ( 66.8) ( 41.7) / tons α V 0.7 s Φ c Φ + λ f' c bo b o.7 ( 0.75) ( 41.7) ( 66.8)( 41.7) / tons Φ V c tons > tons 5- Draw S.F.D an B.M.D for footing: S.F.D an B.M.D are shown in Figures an e. 30

5 31 (b) (c) () (e) Figure 11.13: (continue); (b) Footing imensions; (c) factore soil pressure; () shearing force iagram; (e) bening moment iagram 6- Check footing thickness for beam shear: ( 0.53) 300 ( 100)( 41.7) / tons Φ V c 0.75 Maximum factore shear force V u is locate at istance from the exterior face of column B, V u tons < 8.71 tons O.K. 7- Compute the area of flexural reinforcement: a. Bottom longituinal reinforcement: 0.85 ρ ( 300).353( 10) ( 3.7) 0.9 ( 100)( 41.7) ( 300) ( 100)( 41.7) 15.43cm , use 8 φ 16 mm b. Transverse reinforcement: Effective epth cm (upper layer) 31

6 3 Uner Column A: 40 ( 0.701) ( 0.701)( 1.0) Figure f: Transverse reinforcement (with of strips) 0. MuA 3. t.m 0.85 ρ ( 300).353 ( 10) ( 3.) ( 70.1)( 40.1) ( 300) so use ρ ρ min ( 70.1)( 50) cm, use 5 φ 14 mm Uner Column B: 60 ( 0.801) ( 0.801)( 1.0) 0. MuB 4.80 t.m ρ ( 300).353( 10) ( 4.8) 0.9 ( 80.1)( 40.1) ( 300) so use ρ ρ 0. min 0018 ( 80.1)( 50) cm, use 5 φ 14 mm Shrinkage Reinforcement in the short irection (bottom sie): ( )( 50) cm, use 6 φ 1 mm 8- Check bearing strength of column an footing concrete: ( 0.85)( 300)( 0)( 30) / tons 40 tons Φ P An 0.65 > Use minimum owel reinforcement, 0.005( 0)( 30) 3.0 cm ( 0.85)( 300)( 0)( 40) / tons 60 tons Φ P Bn 0.65 > Use minimum owel reinforcement, 0.005( 0)( 40) 4.0 cm 9- Check for anchorage of the reinforcement: Bottom longituinal reinforcement ( φ 16 mm ):: ψ t ψ e λ 1 an ψ s

7 33 c b is the smaller of: cm, or cm 8 ( ) cb + Ktr >.5, b 1.6 ( 0.8)( 400) (.5) l cm 3.5 take it equal to.5 Available length cm > cm, i.e., c b 5.1 cm Similarly, bottom transverse reinforcement is to be checke for anchorage, an will be foun satisfactory. 10- Prepare neat esign rawings showing footing imensions an provie reinforcement: Detaile esign rawings are shown in Figure g. Figure g: Design rawings 33

8 34 Example (11.5): Design a combine footing, to support two columns A an B space at istance 6.0 m center-tocenter as shown in Figure a. Column A is 40 cm 40 cm an carries a ea loa of 50 tons an a live loa of 30 tons. Column B is also 40 cm 40 cm in cross section but carries a ea loa of 70 tons an a live loa of 50 tons. Use 50 kg / cm, f c f y 400 kg / cm, an ( net) 1.50 kg / cm q all. Figure a: Footing imensions Solution: 1- Select a trial footing epth: Let footing thickness h 80 cm - Establish the require base area of the footing: A req P q A all + P B ( net) A req 13.33m 15.0 To locate the resultant of the column forces, ( 6 ) ( x), or 10 ( 6) ( )( x) P B R an x 3. 60m Length of footing L ( ) 7. 60m With of footing B 1. 75m, taken as 1.80 m Evaluate the net factore soil pressure: ( 50) ( 30) 108 tons P Au

9 35 ( 70) ( 50) 164 tons P Bu 1.0 P ( ) Au + PBu q u net t / m LB 7.60 ( 1.80) 4- Check footing thickness for punching shear: Effective epth cm (lower layer) Column A: Figure b: Critical sections for punching shear The factore shear force V u 108 ( 19.88)( )( 0.758) tons ( ) cm b o Φ V c is the smallest of: 0.53 Φ f' c 1 + λ bo β 0.53 ( 0.75) ( 63.)( 71.6) / tons 40 / 40 λ Φ f' c bo ( 63.) ( 71.6 )/ tons α V 0.7 s Φ c Φ + λ b o.7 ( 0.75) f' c bo ( 71.6 ) Φ V c 3.48 tons > tons Column B: ( 63.)( 71.6) / tons u The factore shear force V 164 ( 19.88)( ) tons ( ) cm b o 4 Φ V c is the smallest of: 35

10 Φ f' c 1 + λ bo β 0.53 ( 0.75) ( 446.4)( 71.6) / tons 40 / 40 λ Φ f' c bo ( 446.4) ( 71.6 )/ tons α V 0.7 s Φ c Φ + λ f' c bo b o.7 ( 0.75) ( 71.6) ( 446.4)( 71.6) / tons Φ V c 379.0tons > tons 5- Draw S.F.D an B.M.D for footing: S.F.D an B.M.D are shown in Figures e an f. (c) () (e) 36 Figure 11.14: (continue); (c) Factore soil pressure; () shearing force iagram (not to scale); (e) bening moment iagram (not to scale) 6- Check footing thickness for beam shear: Effective epth cm (lower layer) ( 0.53) 50 ( 180)( 71.60) / tons Φ V c 0.75

11 37 Maximum factore shear force V u is locate at istance from the left face of column B, V u tons 81.0 tons 7- Compute the areas of flexural reinforcement: ρ a- Top longituinal reinforcement: ( 50).353( 10) ( ) ( 0.9) 180( 71.6) ( 50) ( 180)( 71.6 ) 54.5cm , use 18 φ 0 mm b- Bottom longituinal reinforcement: ρ < ρ, so use ρ ρ 0. min 0018 min ( 180)( 80) cm, use 13 φ 16 mm c- Short irection: Uner Column A: Effective epth cm (upper layer) 108 ( ) ( )( 1.80) Figure f: Transverse reinforcement (with of strip) M Au t.m 0.85 ρ ( 50).353( 10) ( 14.70) 1 ( 75.05)( 0.9)( 70.1) ( 50) Use ρ ρ 0. min ( 75.05)( 80) 10.81cm, use 6 φ 16 mm Uner Column B: 164 ( 1.101) ( 1.101)( 1.80) MBu.3 t.m 37

12 38 ρ ( 50).353( 10) (.3) 0.9 ( 110.1)( 70.1) ( 50) min ρ < ρ, so use ρ ρ 0. min 0018 ( 110.1)( 80) cm, use 8 φ 16 mm Shrinkage Reinforcement in the short irection (bottom sie): ( )( 80) cm, use 4 φ 16 mm Shrinkage Reinforcement in the short irection (top sie): ( 760 )( 80) cm, use 55 φ 16 mm 8- Check bearing strength of column an footing concrete: ( 0.85)( 50)( 40)( 40) / tons 164 tons Φ P n 0.65 > i.e. use minimum owel reinforcement, 0.005( 40)( 40) 8.0 cm 9- Check for anchorage of the reinforcement: a- Top reinforcement ( φ 0mm ): ψ t 1.3 since more than 30 cm of fresh concrete is cast below the reinforcement, ψ e λ 1 an ψ s 0.8 c b is the smaller of: cm, or cb + Ktr b ( 0.8)( 1.3) ( 400) ( 3.5) cm 19 ( ).0 l cm.145, i.e., c b 4.9 cm Available evelopment length cm > cm b- Bottom reinforcement( φ 16mm ): ψ t ψ e λ 1 an ψ s 0. 8 c b is the smaller of: cm, or cm 1 ( ) cb + Ktr >.5, take it equal to.5 b 1.6 ( 0.8)( 400) ( 3.5) l cm.5, i.e., c b 6.81 cm 38

13 39 Available length cm > cm 10- Prepare neat esign rawings showing footing imensions an provie reinforcement: Detaile esign rawing is shown in Figure g. Figure g: Design rawing 39

Figure 1: Representative strip. = = 3.70 m. min. per unit length of the selected strip: Own weight of slab = = 0.

Figure 1: Representative strip. = = 3.70 m. min. per unit length of the selected strip: Own weight of slab = = 0. Example (8.1): Using the ACI Code approximate structural analysis, design for a warehouse, a continuous one-way solid slab supported on beams 4.0 m apart as shown in Figure 1. Assume that the beam webs