Design of Concrete Structures 15th Edition

Notes regarding reinforcement details. In the following problems, the reinforcement details are compressed to clarify the problem solutions. The reinforcement details are noted by the line: Reinforcement details Reinforcement is shown by bar size. Thus a No. 9 (No. 29) bar is As9. Expansion of the "Reinforcement details" line gives all bar sizes and

lg"lL C""nN,.,*:a .rePF^".",, Lr--r 7.b1 1-J r'r:e-O,. b(ops.pe lffio tblftL. rre lb/ftD+1 . +@ rbl+.AAJae lfil+ o,F TDqIFY4. \(an$.1,27 x l7o ' ze lLlq:iFtrl -s z.b? x l?D o :Zo lLf{tz4 No "c^Etblq &eNA\ rg*g^o.- So@ -z@ -vio"= i7P tblfr'Y -- Pq- =5.%rr i 5c.1 i.a*r.v laso{. = s"/af: (r-?)= -*%&- = o.5o,,*],r'.rSxbCC@r

1/8 1/8 1/8 16.1. A cantilever retaining wall is to be designed with geometry as indicated in Fig. P16.1 . Backfill material will be well-drained gravel having unit weight w = 120 pcf, internal friction angle ' = 33 , and friction factor against the concrete base f = 0.55. Backfill placed in front of

1/2 Problem 17.5 A column transfers a factored load of 700 kips to a 9-ft square footing shown in Figure P17.5, resulting in a factored uniform soil pressure of 8640 psf. Design the footing reinforcement using strut-and-tie methods. Material strengths are f'c = 4000 psi and fy = 60,000 psi. Because footings typically

tO.l /rtr*pe Joar 4eAZf 44L SeqH 4DDQlps>612 SAES 74,ed44 7rrlJol.try- TrlS ,s7fi<rF/s ovzoP r/67Lrr aqulen1ur ft4 Boztl 7or AilD Eofro?/ 6nES C*tecr. CLnanNces C-/crrnrt 16'*16e a trr(ase)tz'ty t6ttg6x4ex od:z)roPs$e{zr\Eorro\t2tt72z" 9laDqe3$r r (5.) ro,o a trr(ase)tz'ty t6ttg6x4ex od:z)roPs$e{zr\Eorro\t2tt72z" 9laDqe3$r r (5.) ro,o a trr(ase) tz'ty t6ttg6x4 ex od:z)roP s$e{zr\Eorro\ t2tt72z" 9laDqe 3$r r (5.) ro,o r*r Ctqj BorroH fe6.S -roPS?AEL f-.<oH <leDef,e L)L'Pr< 'TOP STEEL FOM AffiM

7KHVSHFLILHGFRQFUHWHVWUHQJWKfcIRUDQHZEXLOGLQJLVSVL&DOFXODWHWKHUHTXLUHG DYHUDJHfcr IRUWKHFRQFUHWHaLIWKHUHDUHQRSULRUWHVWUHVXOWVIRUFRQFUHWHZLWKDFRPSUHVVLYH VWUHQJWKZLWKLQSVLRIfcPDGHZLWKVLPLODUPDWHULDOVbLIWHVWUHVXOWVIRUFRQFUHWHZLWKfc SVLPDGHZLWKVLPLODUPDWHULDOVSURGXFHDVDPSOHVWDQGDUGGHYLDWLRQssRISVLDQGcLI WHVWVZLWKfc SVLPDGHZLWKVLPLODUPDWHULDOVSURGXFHDVDPSOHVWDQGDUGGHYLDWLRQssRI SVL 6ROXWLRQIF=5000psi D 1RSULRUUHVXOWVIFU IFSVL SVLE SULRUWHVWVIRUFRQFUHWHZLWKIFZLWKLQSVLRIIFRIWKHSURMHFWDQGVV SVL)URP7DEOHN DQGNVVLV SVL%HFDXVHIF SVLXVHHTVDQGDIFU IFNVV SVLIFU IFNVV SVL86(IFU SVLF SULRUWHVWVIRUFRQFUHWHZLWKIFZLWKLQSVLRIIFIRUWKHSURMHFWVV SVLDQGNLVIFU IFVV SVLIFU IFVV SVL86(IFU SVL&200(17LQFDVHVEDQGFWKHIFUZRXOGUHDVRQDEO\EHWDNHQDVSVL D 1RSULRUUHVXOWVIFU IFSVL SVLE SULRUWHVWVIRUFRQFUHWHZLWKIFZLWKLQSVLRIIFRIWKHSURMHFWDQGVV SVL)URP7DEOHN DQGNVVLV SVL%HFDXVHIF SVLXVHHTVDQGDIFU IFNVV SVLIFU IFNVV SVL86(IFU SVLF SULRUWHVWVIRUFRQFUHWHZLWKIFZLWKLQSVLRIIFIRUWKHSURMHFWVV SVLDQGNLVIFU IFVV SVLIFU IFVV SVL86(IFU SVL&200(17LQFDVHVEDQGFWKHIFUZRXOGUHDVRQDEO\EHWDNHQDVSVL D 1RSULRUUHVXOWV I FU IFSVL SVL E SULRUWHVWVIRUFRQFUHWHZLWKIFZLWKLQSVLRIIFRIWKHSURMHFWDQGVV SVL)URP7DEOH N DQGNVVLV

21.1 Determine the required embedment, hef, for a 5/8 in. diameter hex head bolt away from edges in 4000 psi cracked concrete with no supplemental reinforcement and carrying a sustained dead load of 6000 lbs, Figure P21.1. Given: fy= 36 ksi, futa = 58 ksi, ASe,N=0.226, and Abrg=0.454 in2. ( h ef =4

ZLr t l'(A.' %g \n' 3o' lPlL" iv+ rt = as.a if e 3(a) ?s.r f, s - Hi (t --'#\=*rr.r 6)*z= = -Zn1 Fi14" S -- r 4a1 ',8t ,tr' 1 68 ite-kl,rf,= 62.o gl4 ZLr t l'(A.' %g \n' 3o' lPlL" iv+ rt = as.a if e 3(a) ?s.r

'.i '11 4LI \,,IrTFl 4tnrr, - $ ilny }-tfzrcrr ol.! SLkr+ \ *Lh l+T rr.} tU,Jy B ?"fT- Uelrt lN l5 oTQaQte- ANb D-tQecftau -,, g Acc. {tn n . 1 Neecar GGN4, eF?e*at Fe* MFcr+ANr6M: RY \)rerOAL UazV, $ tr * StEc es($E a?tfuL+ FoG * E:QoATF t r)re ut.l ttt 1o Ftrfe2xt 0rL fOo 6: j3 l,ts), AeLEzT g frdrr)r; MJll ut igr e= Wi w"3 t

1/2 1/2 1/2 3.1. A 16 20 in. column is made of the same concrete and reinforced with the same six No. 9 (No. 29) bars as the column in Examples 3.1 and 3.2, except t hat a steel with yield strength f y = 40 ksi is used. The stress-strain diagram of this reinforcing

4.1Comparethecrackingmomentbasedonthegrosssectionpropertiesandthetransformedsectionpropertieswith4No.11(No.36)barsinFigureP4.1abasedonaconcretetensilecapacityof 4.1Comparethecrackingmomentbasedonthegrosssectionpropertiesandthetransformedsectionpropertieswith4No.11(No.36)barsinFigureP4.1abasedonaconcretetensilecapacityof 4.1Comparethecrackingmomentbasedonthegrosssectionpropertiesandthetransformed sectionpropertieswith4No.11(No.36)barsinFigureP4.1abasedonaconcretetensilecapacityof 7.5vfc. f'c4000psi Es29000000psi 24 36" 2Ec57000 f'cpsi3.605 106psib36in h24in Igbh3 2Ec57000 f'cpsi3.605 106psib36in h24in Igbh3 2 Ec57000 f'cpsi3.605 106 psi b36in h24in Ig bh 3 typ a)fc=4000psi 12 41472 in4 Mcr 7.5 f'cpsi Ig h 2 1639 in kip Reinforcement detailsCalculation of transformed sectionnEsEc As4As116.24 in2 nAs50.2 in2 ys2.5inybbh22nAsysbhnAs 11.48 inIgbh312 bhh2yb2 nAsybys2 45754 in4Mcrt7.5 f'cpsiIg1891 in kip Reinforcement detailsCalculation of transformed sectionnEsEc As4As116.24 in2 nAs50.2 in2 ys2.5inybbh22nAsysbhnAs 11.48 inIgbh312 bhh2yb2 nAsybys2

2/2 fs0.60fy fs36000 psi As bd 0.011kn() 22n n k 0.325j1 k 3 j 0.892 Moment due to concrete limits Moment due to steel limitMsc12fcbkddkd3 Msc 238 ft kip Mss Asfsjd Mss 255 ft kipThe maximum service moment is the minimum of the two values.Msmin Mss Msc Ms238 ft kip(d) Determine the nominal moment capacity of the

5.1 A rectangular beam secon is 10 in. wide and has a structural depth of 13.5 in. Given that the concrete strength is 4000 psi and there is no shear reinforcement in the secon, determine the nominal shear capacity allowed by the ACI Code. b10in d 13.5 in f'c4000 psi 1.0The

6.1 through 6.6 (based on the beam in Fig. 4.15, Example 4.13, fy = 60 ksi) For the No. 6 bars: ( ) 2.5 in. 1.5 0.5 0.75 2 2.375 in. 10 2 1.5 0.5 0.75 2 2 2.625

8.1. A rectangular beam is 16 in. wide and 29 in. deep. For c f= 4000 psi and y f= 60,000 psi, determine the required spacing of No. 4 (No. 13) closed stirrups for a factored shear of 80 kips and factored torsional moment of 50 ft-kips. The stirrup centroid is located 1.75

9.15 Using the column interaction diagrams in Appendix A ( c f= 4000 psi and y f= 60,000 psi) with t approximately equal to 2%, design square columns with equal reinforcement on all sides to carry each of the following loads and select the longitudinal and transverse reinforcement: (a) Pu = 2000 kips and Mu