Fig. 1.1 Rectangular foundation plan.

Transcription

1 Footings Example 1 Design of a square spread footing of a seven-story uilding Design and detail a typial square spread footing of a six ay y five ay seven-story uilding, founded on stiff soil, supporting a 4 in. square olumn. The uilding has a 10 ft high asement. The ottom of the footing is 13 ft elow finished grade. The uilding is assigned to Seismi Design Category (SDC) B. Given: Column load Servie dead load D = 541 kip Servie live load L = 194 kip Seismi load E = ±18 kip (Column fore due to the uilding frame resisting the seismi load) Material properties Conrete ompressive strength f = 4 ksi Steel yield strength fy = 60 ksi Normalweight onrete = 1 Density of onrete = 150 l/ft 3 Fig. 1.1 Retangular foundation plan. Allowale soil-earing pressures D only: qall.d = 4000 psf D + L: qall,d+l = 5600 psf D + L + E: qall.lat = 6000 psf ACI Proedure Computation Step 1: Foundation type This ottom footing is 3 ft elow asement sla. Therefore, it is onsidered a shallow foundation. Step : Material requirements The mixture proportion must satisfy the duraility requirements of Chapter 19 (318) and strutural strength requirements. The designer determines the duraility lasses. Please see Chapter of SP-17 for an in-depth disussion of the ategories and lasses. ACI 301 is a referene speifiation that is in syn with ACI 318. ACI enourages By speifying that the onrete mixture shall e in aordane with ACI 301 and providing the exposure lasses, Chapter 19 requirements are satisfied. Based on duraility and strength requirements, and experiene with loal mixtures, the ompressive strength of onrete is speified at 8 days to e at least 4000 psi.

2 referening 301 into jo speifiations. There are several mixture options within ACI 301, suh as admixtures and pozzolans, whih the designer an require, permit, or review if suggested y the ontrator. Example 1 provides a more detailed reakdown on determining the onrete ompressive strength and exposure ategories and lasses. Step 3: Determine footing dimensions To alulate the footing ase area, divide the servie load y the allowale soil pressure. Area of footing = total servie load ( P) allowale soil pressure, q a Assuming a square footing. The unit weights of onrete and soil are 150 pf and 10 pf; lose. Therefore, footing self-weight will e ignored: D qall., D 541 k 135 ft (ontrols) 4 ksf D L 541 k+ 194 k 131 ft qall, DL 5.6 ksf D + L +E 541 k k + 18 k 16 ft q all., Lat 6 ksf l 135 ft 11.6 ft Therefore, provide 1 x 1 ft square footing. Step 4: Soil pressure The footing thikness is alulated in Step 5, footing design. Footing staility Beause there is no overturning moment, overall footing staility is assumed. Calulate fatored soil pressure P q u u Area

3 5.3.1(a) 5.3.1() 5.3.1(d) 5.3.1(e) Calulate the soil pressures resulting from the applied fatored loads. Load Case I: U = 1.4D Load Case II: U = 1.D + 1.6L Load Case IV: U = 1.D + E + L Load Case IV: U = 0.9D + E The load ominations inludes the seismi uplift fore. In this example, uplift does not our. U = 1.4D = 1.4(541 kip) = 757 kip 757 kip q u = 5.3 ksf 144 ft U = 1.D + 1.6L = 1.(541 kip) + 1.6(194 kip) = 960kip 960 kip q = 6.7 ksf 144 ft u (ontrols) U = 1.D + 1.0E + 1.0L = 1.(541 kip) + 18 kip + 1.0(194 kip) = 861 kip 861 kip q u = 6.0 ksf 144 ft U = 0.9D + 1.0E = 0.9(541 kip)+ 18 kip = 505 kip 505 kip q u = 3.5 ksf 144 ft Beause the footing has equal dimension in plan, it will e designed in one diretion and symmetry is assumed. Step 5: One-way shear design Fig. 1. One-way shear in longitudinal diretion.

4 1..1() Shear redution fator: Vn Vu Vn V Vs Therefore: V fwd V And satisfying: u V The ritial setion for one-way shear is at a distane d from the fae of the olumn (Fig. 1.). The engineer ould either assume a value for d that satisfies the strength Eq. (.5.5.1) y iteration or solve Eq. ( ). In this example, the first approah is followed: Assume that the footing is 30 in. thik. The over requirement is 3 in. to ottom of reinforement. Assume that No. 8 ars are used in the oth diretions and design for the more ritial ase (upper layer). Therefore, the effetive depth d: d = 30 in. 3 in. 1 in. 1 in./ = 5.5 in. ( l Vn Vu d) qu shear = 0.75 Assume Vs = 0 (no shear reinforement) V n = V (1 ft) 4 in. 5.5 in. Vu ( ) in. in. (1 ) 1 ft ft (1 ft)(6.7 ksf) = 31 kip V 0.75() 4000 psi (1 ft)(5.5 in.) (1 in./ft) = 348 kip V = 348 kip > Vu = 31 kip OK Therefore, assumed depth is adequate: h = 30 in.

5 Step 6: Two-way shear design The foundation will not e reinfored with shear reinforement for two-way ation. Therefore, the nominal shear strength for two-way foundation without shear reinforement is equal to the onrete shear strength: vn = v Under punhing shear theory, inlined raks are assumed to originate and propagate at 45 degrees away and down from the olumn orners. The punh area is alulated at an average distane of d/ from olumn fae on all sides (Fig. 1.3). o = 4( + d) ACI 318 permits the engineer to take the average of the effetive depth in the two orthogonal diretions when designing the footing, ut in this example the smaller effetive depth will e used. The two-way shear strength equations for nonprestressed memers must e satisfied and the least alulated value of (a), (), and () ontrols: v 4 f (a) 4 v ( ) f () where is ratio of the long side to short side of olumn; = 1 v ( sd ) f () o Fig. 1.3 Two-way shear. o = 4( ) = 198 in. v 4(1.0)( 4000 psi) = 53 psi 4 v ( )(1.0)( 4000 psi) = psi 1 (40)(5.5 in.) v ( )( 4000 psi) = 45 psi 198 in. Equation (a) ontrols; v = 53 psi

6 () s = 40, onsidered interior olumn V 4 fod Use a redution fator of 0.75: V (0.75)4 fod Vu qu [(a) ( d ) ] Chek if design strength exeeds required strength: V Vu? V = 0.75 (53 psi)(198 in.)(5.5 in.) = 177 kip 1000 l/kip V = 0.75(177 kip) = 958 kip 4 in.+5.5 in. Vu [(1 ft)(1 ft) ( ) ](6.7 ksf) 1 in./ft 851 kip V = 958 kip > Vu = 851 kip Two-way shear strength is adequate. OK Step 7: Flexure design The ritial setion is permitted to e at the fae of the olumn (Fig. 1.4). Fig. 1.4 Flexure in the longitudinal diretion. l Mu qu( ) ( ) / 4 in. 1 ft 1 in./ft M u (6.7 ksf)( ) (1 ft) /

7 ..1.1 Set ompression fore equal to tension fore at the olumn fae: C = T Mu = 1005 ft-kip (a) (a) (a) 1..1(a) C = 0.85f a and T = Asfy Af s y a and 0.85 f a Mn As f y ( d ) Sustitute for a in the equation aove. Use redution fator from Tale Setting Mn Mu = 1005 ft-k and solving for As: Chek the minimum reinforement ratio: l = Chek if the tension ontrolled assumption and the use of = 0.9 is orret. To answer the question, the alulated tensile strain in reinforement is ompared to the values in Tale 1... The strain in reinforement is alulated from similar triangles (refer to Fig. 1.5): t = ( d ) a where: = β and a = 0.8As 1 A s(60 ksi) a 0.1As 0.85(5 ksi)(1 ft) = 0.9 (0.1)A M (0.9) (60 ksi)(5.5 in. s n As ) As = 8.91 in. As,min = (1 ft)(1 in./ft)(30 in.) = 7.8 in. < As,req d = 8.91 in. OK Use 13 No. 8 ars distriuted uniformly aross the entire 1 ft width of footing. Note: Although not required y ode, some pratitioner distrute half the required ars in the mid third of the footing and distriute the remaining ars in the equally on oth sides. 0.1(13)(0.79 in. ) = 1.1 in t = (5.5 in. 1.1 in.) in. t = 0.06 > 0.005

8 Setion is tension ontrolled and = 0.9 Fig. 1.5 Strain distriution aross footing. Step 8: Transfer of olumn fores to the ase (a).8.3.() Fatored fores are transferred to the foundation at the ase of the olumn y earing on onrete and the reinforement dowels. The foundation is wider on all sides than the loaded area. Therefore, the nominal earing strength, Bn, is the smaller of the two equations. A Bn (0.85 fa1 ) A and 1 Bn (0.85 fa1 ) A Chek if.0 A where 1 A1 is the earing area of the olumn and A is the area of the part of the supporting footing that is geometrially similar to and onentri with the loaded area. A [(1 ft)(1 in./ft)] 6 A 1 (4 in.) Therefore, Eq. (.8.3.()) ontrols.

9 1..1(d) The redution fator for earing is 0.65: earing = Column fatored fores are transferred to the foundation y earing and through reinforement dowels. Provide dowel reinforement area of at least 0.005Ag and at least four ars. Bars are in ompression for all load ominations. Therefore, the ars must extend into the footing a ompression development length, ld, the larger of the two: l d f y r d 50 f ( fy rd) The footing depth must satisfy the following inequality so that the vertial reinforement an e developed within the provided depth: h l r d d 3 in. where Step 9: Footing details d, dwl, ars r = radius of No. 6 ent = 6d Development length B n (0.65)()(0.85)(4000 psi)(4 in.) Bn = 546 kip > 960 kip (Step 4) As,dowel = 0.005(4 in.) =.88 in. Use eight No. 6 ars 0.0(60,000 psi) ld (0.75 in.) 14.3 in psi l (60,000 psi)(0.75 in.) 13.5 in. d ld = 14.3 in. (ontrols) hreq d = 14.3 in. + 6(0.75 in.) in. + (0.75 in.) + 3 in. = 4.05 in. hreq d = 4.1 in. < hprov. = 30 in. OK OK Reinforement development is alulated at the maximum fatored moment, whih ours at the olumn fae. Bars must extend

10 a tension development length eyond the ritial setion. 3 f y { t e l s d } d 40 K f tr d where tar loation; not more than 1 in. of fresh onrete elow horizontal reinforement eoating fator; unoated sar size fator; No. 8 and larger = spaing or over dimension to enter of ar, whihever is smaller tr = transverse reinforement index It is permitted to use Ktr = 0. Ktr But the expression: must d not e taken greater than.5. l d 3 60,000 psi (1.0)(1.0)(1.0) { } d 40 (1.0) 4000 psi.5 No. 6: = 8.5d d K tr = 3.5 in in. No.8 ars: 8.5(1.0 in.) = 8.5 in. > 1 in. OK ld in the longitudinal diretion: ld,prov. = ((1 ft)(1 in./ft) 4 in.)/ 3 in. ld,prov. = 57 in. > ld,req d = 8.5 in. OK use straight No. 8 ars in oth diretions.

11 Step 10: Detailing