Rectangular Footings Analysis and Design

Transcription

1 Rectangular Footings Analysis an Design ORGAN STATE UNIVERSITY SCHOOL OF ARCHITECTURE AND PLANNING LECTURE VII Dr. Jason E. Caralambies = = Wat Does a Footing Do?! Te main function of a footing is to isperse te loa of te structure in a wier space so tat te soil reacts to it an a minimal settlement will be acieve.

2 Types of Footings

3 How Footings Fail! q_max>q_allowable (i.e. soil failure)! Column loa compression failure.! Footing fails in flexure: How Footings Fail! Footing fails in one way sear. ΦV c 1way =Φ L f ' c

4 elt=cllqfkp=c^fi! Footing fails in two way (puncing) sear. ΦV c1 =Φ[ 4 β c ] f ' c β 0 ΦV c =Φ[ α s β c ΦV c3 =4Φ f ' c β 0 ] f ' c β 0 Were αs is a value pertaining to te location of te point loa, an β c =b/ Note: Usually Φ vc3 governs In Class Example! A reinfoce concrete footing nees to be esigne to support column A3 in te inicate rawing. Accoring to te preliminary structural analysis tat was conucte, te column is subjecte to bot axial loas as well as moments. For reasons pertaining to uctwork, plumbing an oter utilities tat run uner te builing, te lengt (L) of te footing nees to be restricte to te imension specifie.

5 FOOTING UNDER AXIAL LOAD AND OENT (R. Klingner) A reinfoce concrete footing nees to be esigne to support column A3 in te inicate rawing. Accoring to te preliminary structural analysis tat was conucte, te column is subjecte to bot axial loas as well as moments. For reasons pertaining to uctwork, plumbing an oter utilities tat run uner te builing, te lengt (L) of te footing nees to be restricte to te imension specifie. L 1in γ conc := 150 lbf ft 3 ASP := 5 b := 30in f' c := 5.ksi := 30in f y := 60ksi P Dini := 100 P D := P Dini 1. P D = 10 P Lini := 30 P L := P Lini 1.6 P L = 368 P u := P D P L P u = 488 D1ini := 0k' D1 := D1ini 1. D1 = 0k' Dini := 5k' D := Dini 1. D = 30 k' L1ini := 600k' L1 := L1ini 1.6 L1 = 960 k' Lini := 100k' L := Lini 1.6 L = 160 k' u1 := D1 L1 u1 = 960 k' u := D L u = 190 k' 1 Estimate size of footing (ASP usually governs): P Dini P Lini q serv = A g q ft_ini C 1 D1ini L1ini I 1 C Dini Lini I Assuming a value for B: Here we can try a series of values until te ASP conition is met.try a value for "B" an Solving oment of Inertia an istance "c" along bot irections: B := 14ft LB 3 I 1 := I 1 1 = ft 4 BL 3 I := I 1 = ft 4 A g := LB A g = B L C 1 := C 1 = 7ft C := C = 5.08 ft Here we may make an initial assumption for te slab tickness, giving te appropriate cover tickness, an a generous 1.3" for iameter of rebars running along te long axis : lbf Try: := 14in Cover clear := 3in t := Cover clear 1.3in t = 18.3 in q ft_ini := t 150 ft 3 q ft_ini = 0.88 ( P Dini P Lini ) ( D1ini L1ini ) C 1 q serv q ( Dini Lini ) C A ft_ini := g I 1 I ASP conition := if ( q serv < ASP, "OK", "Reesign" ) q serv = 4.87 ASP conition = "OK"

6 Verify tat soil is in compression trougout te area of te footing: P Dini P Lini q min := A g q min = 0.3 q ft_ini C 1 C D1ini L1ini Dini Lini I 1 I q min_conition := if q min > 0, "OK", "Reesign" q min_conition = "OK" Assume tat tickness is controlle by puncing sear: Note: Puncing sear epens on axial loas only We will solve by trial an error: a) Assume a ept (one alreay above for te 1st roun) b) Calculate β0 c) Verify tat ΦVc>Vn require an accoring to ACI Note: Vc>Vn oes not apply to slabs an footings ( ) ) revise an re-iterate It is generally suggeste tat a goo starting point for te ept of a footing to use a rule of tumb measurement of 1.5*(Column wit or iameter). However, wit a relatively low ASP, we may want to minimize te loa we sall apply to te soil. We can initially aress te conition of Vu<ΦVc an iterate te process wit a series of values for. Anoter process tat we can follow is te alreay given conition above were we establise an initial q value for te footing. Tat can yiel a maximum acceptable value for te footing's tickness: See formula below. After iteration we assume a minimum acceptable value for knowing tat we can not surpass te value given as t_max. ASP q serv q ft_ini t max := t γ max = 8.53 in q ft := γ conc t conc q ft = 0.3 P u q net := q A net = 3.43 β 0 := ( b) β 0 = ft g A ext_col := ( b ) A ext_col = Vu way := q net A g A ext_col Vu way = b Given te following: Φ :=.75 α s := 40 β c := β c = 1 4 f' ΦVc 1 Φ c := ( 1ksi) β β c psi 0 ΦVc = α s f' c ΦVc := Φ ( 1 ksi) β β 0 psi 0 ΦVc 1000 = f' c ΦVc 3 := Φ ( 4) ( 1ksi) β psi 0 ΦVc = 5.69 ΦVc way := min ΦVc 1, ΦVc, ΦVc 3 ΦVc way = 5.69 Punc_Sear conition := if Vu way < ΦVc way, "Approve", "Not Approve" Punc_Sear conition = "Approve"

7 3 Verify tickness cosen is goo for 1-way sear: We ceck for te moments applie on eac irection separately: a) First on te major axis, b) Ten on te seconary axis. ( q' q max ) Vu 1way L ( B ) = We start wit te main axis: P D P ( L ) u1 q' := q' 4.43 ( P D P L ) u1 C 1 = BL I 1 q max := q BL I max = q' q max V u1way L ( B ) := V u1way = If iger tan Vu, ΦV c1way = Φ f' c L ΦV it is approve. c1way = OneWaySear conition := if V u1way < ΦV c1way, "Approve", "Not Approve" OneWaySear conition = "Not Approve" Assuming a new value for ""...Try: := 1in t := Cover clear 1.3in t = 5.3 in P D P ( L ) u1 q' := q' 4.67 ( P D P L ) u1 C 1 = BL I 1 q max := q BL I max = q' q max V u1way L ( B ) := V u1way = If iger tan Vu, ΦV c1way = Φ f' c L ΦV it is approve. c1way = OneWaySear conition := if V u1way < ΦV c1way, "Approve", "Not Approve" OneWaySear conition = "Approve" Now we ceck te beavior about te seconary axis: P D P ( L ) u q' := q' 3.89 ( P D P L ) u C = BL I q max := q BL I max = 4. ( q' q max ) V u1way B ( L ) := V u1way = ΦV c1way = Φ f' c B If iger tan Vu, it is approve. ΦV c1way = OneWaySear conition := if V u1way < ΦV c1way, "Approve", "Not Approve" OneWaySear conition = "Approve"

8 4 Ceck Flexure - in eac irection: Critical section occurs at column face. In eac irection, only axial loa an moment in tat irection will contribute. Starting wit te main axis: Φ :=.9 P D P ( L ) u1 q'' := q'' 3.94 ( P D P L ) u1 C 1 = q BL I max := q 1 BL I max = momentarm istance along footing B B B u := L q'' ( q max q'' ) 3 u = k' 15 u A s_ini := A f y 1 s_ini = in To be more precise we can follow te stanar formula for flexure an apply it on tis slab as te footing is essentially a slab. Please note tat te value "b" refers to te geometric base of te element we are analyzing, tus "b" is te lengt L in tis case. Also te value of "" is roug as it refers to te centroi of flexural rebars running along ifferent irections. See more precise imensions on bottom of page: Φ n = Φω b f' c ( ω) u u Given tat n = k' Φ n 0.9 Φ u We can assume te minimum value for n u n := Φ n = k' OR n = k'' Tis formula is erive from u = As*fy*(-a/) Were As = p*b* an a = As*fy/(0.85*f'c*b) We en up wit a quaratic equation If we solve for ω. Te value "b" is cange to "L" from ere on. Also, we consier tat"' sall be te istance of extreme top fiber to centroi of longituinal rebars along te longest imension (wic in tis case is "B": ω 0.59ω = n L f' c n L f' c = L f' c 5 L f' c 59 L f' c n ω := ω = L f' c Continuing wit te process we apply te following formula to etermine te area of steel neee. Please note tat te term "b" in te formula, again soul replace by our geometric base of "L". not te "b" of te column sie. f' c f' c A s = ω b Replacing te "b" to "L": A f s := ω L A y f s = in y ACI etermines maximum spacing. 18" is te max we can use. Ten #9 bars will not suffice. Take ten #10 bars at 1" O.C. For te sort irection, for reasons of simplification at tis stage, we can make an initial assumption tat #10 bars will also be use. Determine te total ept "t" of te footing b := 1.7in Note: te aition of 0.5" was to prevent a possible t roun 1.5 b Cover clear :=.5 truncation (rouning own) an tus guarantee tat in in in in t = 6 in te minimum necessary tickness is at least met. Tus we can precisely etermine te actual "" values for te long an te sort irections b long := t Cover clear long =.37 in sort := t Cover clear 1.5 b sort = 1.1 in

9 Now we can solve for te seconary axis: P D P ( L ) u q' := q' 3.6 ( P D P L ) u C = BL I q max := q BL I max = 4. momentarm istance along footing L L L u := B q' ( q max q' ) 3 u = k' u n := Φ n = k' OR n = k'' We en up wit a quaratic equation If we solve for ω. Te value "b" is cange to "B" from ere on: ω 0.59ω = n B sort f' c n B sort f' c = ω := 5 10 B f' c sort 5 B f' c sort 59 B f' c n 59 B f' c ω = f' c A s := ω B f sort A s = in y Tis value owever is very low. Refer to ACI (9.6.1.) an ( ) Terefore: A s961 max 00 3 f' c =, B t A f y f s961 = in A s9613 := 1.3 A s A s9613 = in y A s_conition := if ( A s9613 < A s961, A s9613, A s961 ) A s_conition = 5.74 in Dimensions coul be reajuste at tis point to minimize use of steel rebars along te sort axis. For minor axis of a rectangular beam we ave to establis a ban witin wic most steel soul be place. Accoring to ACI a ban wit equal to te span of te sort irection receives a certain minimum percentage of te reinforcement. Te formula is te following Reinforcement_in_Ban_Wit = ( Total_Reinforcement_in_sort_irection) β 1 L β = B tat is consiering tat "L" is te lengt of te long irection an "B" te span of te sort, but in te case te terms ave been reverse. We can set a efault formula wit te terms "min" an "max" applie so tat te process is automate to give a value beta equal or iger tan 1. max( L, B) β := β = Terefore, witin a ban as wie as te sort sie of te footing te area of steel min( B, L) soul be as follows: A s A s_central_ban := A β 1 s_central_ban = in A s_rest := A s A s_central_ban A s_rest = 0.7 in 5 Verify soil pressure maximum is not exceee: w:= t L B γ conc w = C 1 P Dini P Lini w D1ini L1ini Dini Lini q soil := q A g I 1 I soil = C ASP conition := if q soil < ASP, "OK", "Reesign" ASP conition = "OK"