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Reinforced Concrete Design, 9th Edition Solution Manual

Reinforced Concrete Design, 9th Edition Solution Manual makes solving textbook questions easier with expertly crafted solutions.

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Reinforced Concrete Design, 9th Edition Solution Manual - Page 1 preview imageInstructor’s Solutions ManualTo accompanyReinforced Concrete DesignNinthEditionAbi O. AghayereDrexel University
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Reinforced Concrete Design, 9th Edition Solution Manual - Page 8 preview imageProb 1-11Part (a):Dead LoadsSlab weight(8”/12)(150pcf)=100psf½” light wt. floor finish=4psfsusp. ceiling=2psfM/E (industrial)=20psfPartitions=20psf_______________________________________________DL= 146psf (with partitions)DL= 126psf (without partitions)Part (b):Live Load = 100psf1.4D = (1.4)(126) = 176.4psf1.2D + 1.6L = (1.2)(126) + (1.6)(100)=311psfcontrolsPart (c):Load on spandrel beamTrib. width = (18 ft/2) + 1.0 ft = 10 ftCladding:Block: (55psf)(10 ft.–2’-2”) = 431 plfBrick: (40psf)(10 ft.)= 400 plfBeam Stem:(8”/12)(12”/12)(150pcf) = 225 plfService Dead Load:(126psf)(10 ft) + 431plf + 400 + 225 plf = 2316 plf =2.32 k/ftService Live Load:(100psf)(10 ft) = 1000 plf =1.0 k/ftPart (d):Factored Load on spandrel beam:1.2D + 1.6L = (1.2)(2.32) + (1.6)(1.0)=4.4plf1-5
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Reinforced Concrete Design, 9th Edition Solution Manual - Page 9 preview imagePart (e):Load on interior beamTrib. width = (18 ft/2) + (18 ft/2) + 1.0 ft = 19 ftBeam Stem:(8”/12)(12”/12)(150pcf) = 225 plfService Dead Load:(126psf)(19 ft) + 225 plf = 2619 plf =2.62 k/ftService Live Load:(100psf)(19 ft) = 1000 plf =1.9 k/ftPart (f):Factored Load on interior beam:1.2D + 1.6L = (1.2)(2.62) + (1.6)(1.9)=6.2k/ftPart (g):Spandrel Beam:2)36)(4.4(2LwVuu===79.2 k8)36)(4.4(8LwM22uu===713 k-ft1-6
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Reinforced Concrete Design, 9th Edition Solution Manual - Page 10 preview imageProb1-12Roof LoadsTypical Floor LoadsDead Load:7” slab (7”/12 x 150 pcf)= 87.5 psf5 ply + gravel= 6.5 psfInsulation= 3.5 psfM & E= 10 psfSuspended ceiling= 2 psfTotal roof dead load,D= 109.5 psfRoof live load (snow)S= 35 psfDead Load:7” slab (7”/12 x 150 pcf)= 87.5 psf0.5” light wt floor finish= 4 psfPartitions= 20 psfM & E= 10 psfSuspended ceiling= 2 psfTotal roof dead load,D= 123.5 psfFloor live load (office)L= 50 psfA.Roof:wDLslab= 109.5 psf;wLLslab= 35 psfUsing the load combinations, we find the maximum factored load to be,wuslab= 1.2 (109.5psf) + 1.6 (35 psf) = 188 psfFloor:wDLslab= 123.5 psf;wLLslab= 50 psfUsing the load combinations, we find the maximum factored load to be,wuslab= 1.2 (123.5 psf) + 1.6 (50 psf) = 229 psfB.(i)TYPICAL INTERIORFLOORBEAMTributary width of beam = 16 ftBeam stem weight = (16”/12) [(26”–7”)/12] x 150 pcf = 317 Ib/ft = 0.32 kip/ftwDLbeam= 123.5 psf x 16 ft + 317 Ib/ft = 2293 Ib/ft = 2.3 kip/ftwLLbeam= 50 psf x 16 ft = 800 Ib/ft = 0.8 kip/ftwsbeam= wDLbeam+ wLLbeam=3.1 kip/ftwubeam= wuslabx beam trib width + 1.2 x beam stem weight= 229 psf x 16 ft + 1.2 (317 Ib/ft) = 4045 Ib/ft =4.1 kip/ftFor a beam subjected to a uniformly distributed load, the beam reactions, assuming simplesupports, are:RDL= wDLbeamx L/2 = 2.3 kip/ft x 32 ft /2 = 37 kipRLL= wLLbeamx L/2= 0.8 kip/ft x 32 ft /2 = 12.8 kipRu= wubeamx L/2or 1.2 (RDL) + 1.6 (RLL)= 4.1 kip/ft x 32 ft/2 =66 kipor 1.2 (37 kip) + 1.6 (12.8 kip) =66 kip (same!)1-7
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Reinforced Concrete Design, 9th Edition Solution Manual - Page 11 preview image(ii)TYPICAL INTERIORROOFBEAMTributary width of beam = 16 ftBeam stem weight = (16”/12) [(26”–7”)/12] x 150 pcf = 317 Ib/ft = 0.32 kip/ftwDLbeam= 109.5 psf x 16 ft + 317 Ib/ft = 2070 Ib/ft = 2.1 kip/ftwLLbeam= 35 psf x 16 ft = 560 Ib/ft = 0.56 kip/ftwsbeam= wDLbeam+ wLLbeam=2.7 kip/ftwubeam= wuslabx beam trib width + 1.2 x beam stem weight= 188 psf x 16 ft + 1.2 (317 Ib/ft) = 3389 Ib/ft =3.4 kip/ftFor a beamsubjected to a uniformly distributed load, the beam reactions, assuming simplesupports, are:RDL= wDLbeamx L/2 = 2.1 kip/ft x 32 ft /2 = 34 kipRLL= wLLbeamx L/2= 0.56 kip/ft x 32 ft /2 = 9 kipRu= wubeamx L/2or 1.2 (RDL) + 1.6 (RLL)= 3.4 kip/ft x 32 ft/2 =55 kipor 1.2 (34 kip) + 1.6 (9 kip) =55 kip (same!)(iii)TYPICALFLOORSPANDREL BEAMTributary width of beam = (16 ft / 2 ) + [(16”/12)/2] = 8.67 ftBeam stem weight = (16”/12) [(26”–7”)/12] x 150 pcf = 317 Ib/ft = 0.32 kip/ftExterior Cladding load = 55 psfblockx 7.83’height+ 40 psfbrickx 10’height= 0.83 kip/ftwDLbeam= 123.5 psf x 8.67 ft + 317 Ib/ft + 830 Ib/ft = 2218 Ib/ft = 2.2 kip/ftwLLbeam= 50 psf x 8.67 ft = 434 Ib/ft = 0.44 kip/ftwsbeam= wDLbeam+ wLLbeam=2.7 kip/ftwubeam= wuslabx beam trib width + 1.2 x beam stem weight= 229 psf x 8.67 ft + 1.2 (317 Ib/ft + 830 Ib/ft) = 3362 Ib/ft =3.4 kip/ftFor a beamsubjected to a uniformly distributed load, the beam reactions, assuming simplesupports, are:RDL= wDLbeamx L/2 = 2.2 kip/ft x 32 ft /2 = 36 kipRLL= wLLbeamx L/2= 0.44 kip/ft x 32 ft /2 = 7 kipRu= wubeamx L/2or 1.2 (RDL) + 1.6 (RLL)= 3.4 kip/ft x 32 ft/2 =55 kipor 1.2 (36 kip) + 1.6 (7 kip) =55 kip (same!)1-8
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Reinforced Concrete Design, 9th Edition Solution Manual - Page 12 preview image(iv)TYPICALROOFSPANDREL BEAMTributary width of beam = (16 ft / 2 ) + [(16”/12)/2] = 8.67 ftBeam stem weight = (16”/12) [(26”–7”)/12] x 150 pcf = 317 Ib/ft = 0.32 kip/ftOn the roof beam and girder, there is usually a one to two feet high parapet. However, if weassumeNO PARAPETin this example, the exterior cladding load on the roof spandrel beamwill be:Exterior Cladding load = 55 psfblockx 0’height+ 40 psfbrickx 0’height= 0 kip/ftwDLbeam= 109.5 psf x 8.67 ft + 317 Ib/ft + 0 Ib/ft = 1267 Ib/ft = 1.3 kip/ftwLLbeam= 35 psf x 8.67 ft = 303 Ib/ft = 0.3 kip/ftwsbeam= wDLbeam+ wLLbeam=1.6 kip/ftwubeam= wuslabx beam trib width + 1.2 x beam stem weight= 188 psf x 8.67 ft + 1.2 (317 Ib/ft + 0 Ib/ft) = 2010 Ib/ft =2.0 kip/ftFor a beam subjected to a uniformly distributed load, the beam reactions, assuming simplesupports, are:RDL= wDLbeamx L/2 = 1.3 kip/ft x 32 ft /2 = 21 kipRLL= wLLbeamx L/2= 0.3 kip/ft x 32 ft /2 = 5 kipRu= wubeamx L/2or 1.2 (RDL) + 1.6 (RLL)= 2.0 kip/ft x 32 ft/2 =32 kipor 1.2 (21 kip) + 1.6 (5 kip) =33 kip (same!)C.(i)TYPICAL INTERIORFLOORGIRDERGirder stem weight = (18”/12) [(26”–7”)/12] x 150 pcf = 357 Ib/ft = 0.36 kip/ftThis girder has TWO interior beams framing into it at the MIDSPA, in addition to a uniformlydistributed loading (UDL) due to the girder stem weight. Therefore, the total concentrated loadsacting on the girder are:PDL girder= 2 beams x RDL beam= 2 x 37 kip= 74 kip(acts at midspan)PLL girder= 2 x RLL beam= 2 x 12.8 kip= 26 kip (acts at midspan)Ps girder= PDL girder+ PLL girder= 74 + 26=100 kip(acts at midspan)Pu girder= 1.2 PDL girder+ 1.6 PLL girder= 1.2(74) + 1.6(26)=131 kip(acts at midspan)1-9
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Reinforced Concrete Design, 9th Edition Solution Manual - Page 13 preview imageThe uniformly distributed load (UDL) on this girder is due to the girder stem weight. Thus,wDLgirder= girder stem weight = 357 Ib/ftwLLgirder= 0The total UDL on the girder are:wsgirder=wDLgirder+wLLgirder= 357 Ib/ftwugirder= 1.2wDLgirder+ 1.6wLLgirder= 1.2 (357 Ib/ft) + 1.6 (0) = 430 Ib/ft =0.43 kip/ftThe REACTIONS for a GIRDER subjected to a uniformly distributed load, w, in addition toconcentrated loads at MIDSPAN (see plan) from the beam reactions, assuming simplesupports, are:RDL= wDLgirderx L/2 + PDLgirder/2 = (0.36 kip/ft x 32 ft /2) + (74 kip /2 ) = 43 kipRLL= wLLgirderx L/2 + PLLgirder/2 = (0 kip/ft x 32 ft /2) + (26 kip /2 ) = 13 kipRu= wugirderx L/2 + Pu girder/2 = (0.43 kip/ft x 32 ft /2) + (131 kip /2 ) =73 kip(ii)TYPICAL INTERIORROOFGIRDERGirder stem weight = (18”/12) [(26”–7”)/12] x 150 pcf = 357 Ib/ft = 0.36 kip/ftThis girder has TWO interior beams framing into it at the MIDSPAN, in addition to a uniformly distributed loading (UDL) due tothe girder stem weight. Therefore, the total concentrated loads acting on the girder are:PDL girder= 2 beams x RDL beam= 2 x 34 kip= 68 kip(acts at midspan)PLL girder= 2 x RLL beam= 2 x 9 kip= 18 kip (acts at midspan)Ps girder= PDL girder+ PLL girder= 68 + 18=86 kip(acts at midspan)Pu girder= 1.2 PDL girder+ 1.6 PLL girder= 1.2(68) + 1.6(18)=111 kip(acts at midspan)The uniformlydistributed load (UDL) on this girder is due to the girder stem weight. Thus,wDLgirder= girder stem weight = 357 Ib/ft;wLLgirder= 0The total UDL on the girder are:wsgirder=wDLgirder+wLLgirder= 357 Ib/ftwugirder= 1.2wDLgirder+ 1.6wLLgirder= 1.2 (357 Ib/ft) + 1.6 (0) = 430 Ib/ft =0.43 kip/ftThe REACTIONS for a GIRDER subjected to a uniformly distributed load, w, in addition toconcentrated loads at MIDSPAN (see plan) from the beam reactions, assuming simplesupports, are:1-10
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Reinforced Concrete Design, 9th Edition Solution Manual - Page 14 preview imageRDL= wDLgirderx L/2 + PDLgirder/2 = (0.36 kip/ft x 32 ft /2) + (68 kip /2 ) = 40 kipRLL= wLLgirderx L/2 + PLLgirder/2 = (0 kip/ft x 32 ft /2) + (18kip /2 ) = 9 kipRu= wugirderx L/2 + Pugirder/2 = (0.43 kip/ft x 32 ft /2) + (111 kip /2 ) =63 kip(iii)TYPICALFLOORSPANDREL GIRDERGirder stem weight = (18”/12) [(26”–7”)/12] x 150 pcf = 357 Ib/ft = 0.36 kip/ftExterior Cladding load = 55 psfblockx 7.83’height+ 40 psfbrickx 10’height= 0.83 kip/ftThis girder has ONE interior beam framing into it at the MIDSPAN, in addition to a uniformlydistributed loading (UDL) due to the girder stem weight, the girder edge distance, AND theexterior cladding. Therefore, the total concentrated loads acting on the girder are:PDL girder= 1 beam x RDL beam= 1x 37 kip= 37kip(acts at midspan)PLL girder= 1 x RLL beam= 1 x 12.8 kip= 13 kip (acts at midspan)Ps girder= PDL girder+ PLL girder= 37 + 13=50 kip(acts at midspan)Pu girder= 1.2 PDL girder+ 1.6 PLL girder= 1.2(37) + 1.6(13)=65 kip(acts at midspan)Uniform floor load on Girder edge distance:wDL edge distance= (9”/12) x 123.5 psf = 93 Ib/ftwLL edge distance= 0 (Because the WHOLE girder edge distance is occupied by the block wallcladding and thus there can be no live load in that edge distance)The uniformly distributed load (UDL) on this girder is due to the girder stem weight, theload on the girder edge distance, AND the exterior cladding. Thus,wDLgirder= girder stem weight + girder edge distance + exterior cladding= 357 Ib/ft + 93 + 830 Ib/ft= 1280 Ib/ft ;wLLgirder= 0The total uniformly distributed load (UDL) on the girder is:wsgirder=wDLgirder+wLLgirder= 1280 Ib/ftwugirder= 1.2wDLgirder+ 1.6wLLgirder= 1.2 (1280 Ib/ft) + 1.6 (0) = 1536 Ib/ft =1.54 kip/ftThe REACTIONS for a GIRDER subjected to a uniformly distributed load, w, in addition toconcentrated loads at MIDSPAN (see plan) from the beam reactions, assuming simplesupports, are:RDL= wDLgirderx L/2 + PDLgirder/2 = (1.28 kip/ft x 32 ft /2) + (37 kip /2 ) = 39 kipRLL= wLLgirderx L/2 + PLLgirder/2 = (0 kip/ft x 32 ft /2) + (13 kip /2 ) = 6.5 kipRu= wugirderx L/2 + Pugirder/2 = (1.54 kip/ft x 32 ft /2) + (65 kip /2 ) =57 kip1-11
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Reinforced Concrete Design, 9th Edition Solution Manual - Page 15 preview image(iv)TYPICALROOFSPANDREL GIRDERGirder stem weight = (18”/12) [(26”–7”)/12] x 150 pcf = 357 Ib/ft = 0.36 kip/ft(no parapet)This girder has ONE interior beam framing into it at the MIDSPAN, in addition to auniformlydistributed loading(UDL) due to the girder stem weight, the load on the girder edgedistance, AND the exterior cladding. Therefore, the total concentrated loads acting on thegirder are:PDL girder= 1 beam x RDL beam= 1x 34 kip= 34kip(acts at midspan)PLL girder= 1 x RLL beam= 1 x 9 kip= 9 kip (acts at midspan)Ps girder= PDL girder+ PLL girder= 34 + 9=43 kip(acts at midspan)Pu girder= 1.2 PDL girder+ 1.6 PLL girder= 1.2(34) + 1.6(9)=55 kip(acts at midspan)Uniform roof load on Girder edge distance:wDL edge distance= (9”/12) x 109.5 psf = 82 Ib/ftwLL edge distance= (9”/12) x 35 psf = 26 Ib/ftThe uniformly distributed load (UDL) on this girder is due to the girder stem weight, theload on the girder edge distance, AND the exterior cladding. Thus,wDLgirder= girder stem weight + girder edge distance + exterior cladding/parapet= 357 Ib/ft + 82 Ib/ft + 0 Ib/ft = 439 Ib/ft (assuming no parapet)wLLgirder= 26 Ib/ftThe total UDL on the girder are:wsgirder=wDLgirder+wLLgirder= 465 Ib/ftwugirder= 1.2wDLgirder+ 1.6wLLgirder= 1.2 (439 Ib/ft) + 1.6 (26 Ib/ft) = 568 Ib/ft =0.57 kip/ftThe REACTIONS for a GIRDER subjected to a uniformly distributed load, w, in addition toconcentrated loads at MIDSPAN (see plan) from the beam reactions, assuming simplesupports, are:RDL= wDLgirderx L/2 + PDLgirder/2 = (0.44 kip/ft x 32 ft /2) + (34 kip /2 ) = 24 kipRLL= wLLgirderx L/2 + PLLgirder/2 = (0.026 kip/ft x 32 ft /2) + (9 kip /2 ) = 4.9 kipRu= wugirderx L/2 + Pugirder/2 = (0.57 kip/ft x 32 ft /2) + (55 kip /2 ) =36.6 kip1-12
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Reinforced Concrete Design, 9th Edition Solution Manual - Page 31 preview imageProb 2-33fc’= 3000 psi; fy= 60,000 psi; span, L = 14 ft; bar diameter = 5/8 = 0.625”; slab thickness, h =7”(a)Selfweight= (7”/12)(150 pcf) = 87.5 psfEffective depth, d = 7”–1.5” cover–(0.625”/2) = 5.18”Asfor #5 @ 14” o.c. = 0.265 in2/ft()()()22(0.265)(60)0.52"0.85(0.85)(3)(12)0.520.90.265605.18225.89./.88(5.89)0.239(14)1.21.62391.2(5087.5)1.6,46.7( )sycnsyuuusA faf baMA fdftkft widthMwksfwDLpsfLTherefore LpsfbA====−=−=−====+=++=2,minmax0.0018(12)(7")0.15/55(7")35"18"# 4@15" . .# 4@14" . . ()inft widthshoro cUseo cmatches spacing of main rebar=====Note: Check c/dtand the assumed strength reduction factor,.1= 0.85 for fc’ = 3000 psic =(0.52”)/0.85 = 0.61”c/dt= 0.61/5.18 = 0.118 < 0.375, therefore,= 0.9 as assumed.Prob 2-34fc’= 4000 psi; fy= 60,000 psi2-16
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