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Structural Analysis, 10th Edition Solution Manual - Document preview page 1

Structural Analysis, 10th Edition Solution Manual - Page 1

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Structural Analysis, 10th Edition Solution Manual

Gain confidence in solving textbook exercises with Structural Analysis, 10th Edition Solution Manual, a comprehensive guide filled with answers.

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Structural Analysis, 10th Edition Solution Manual - Page 1 preview image1–1.The floor of a heavy storage warehouse building is madeof 6-in.-thick stone concrete. If the floor is a slab having alength of 15 ft and width of 10 ft, determine the resultant forcecaused by the dead load and the live load.S0LUTI0NFrom Table 1-3,DL=[12lb/ft2.in.(6in.)](15ft)(10ft)=10,800lbFrom Table 1-4,LL=(250lb/ft2)(15ft)(10ft)=37,500lbTotal load:F=48,300lb=48.3kAns.Ans.F=48.3k1
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Structural Analysis, 10th Edition Solution Manual - Page 2 preview image
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Structural Analysis, 10th Edition Solution Manual - Page 3 preview image21–2.The wall is 15 ft high and consists of 2×4 in. studs,plastered on one side. On the other side there is 4-in. clay brick.Determine the average load in lb/ft of length of wall that thewall exerts on the floor.S0LUTI0NUsing the data tabulated in Table 1–3,4@in.claybrick:(39lb/ft2)(15ft)=585lb/ft2×4@in.studsplasteredon one side:(12lb/ft2)(15ft)=180lb/ftwD=765lb/ftAns.Ans.wD=765lb/ft
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Structural Analysis, 10th Edition Solution Manual - Page 4 preview image31–3.A building wall consists of 12-in. clay brick and12-in.fiberboard on one side. If the wall is 10 ft high, determine theload in pounds per foot that it exerts on the floor.SOLUTIONFrom Table 1–3,12@in.claybrick:(115lb/ft2)(10ft)=1150lb/ft1/2@in.fiberboard:(0.75lb/ft2)(10ft)=7.5lb/ftTotal:1157.5lb/ft=1.16k/ftAns.Ans.w=1.16k/ft
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Structural Analysis, 10th Edition Solution Manual - Page 5 preview image4*1–4.The “New Jersey” barrier is commonly used duringhighway construction. Determine its weight per foot of lengthif it is made from plain stone concrete.12 in.4 in.24 in.6 in.55◦75◦SOLUTIONCross@sectionalarea=6(24)+(12)(24+7.1950)(12)+(12)(4+7.1950)(5.9620)=364.54in2Use Table 1–2.w=144lb/ft3(364.54in2)(lft2144in2)=365lb/ftAns.Ans.w=365lb/ft
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Structural Analysis, 10th Edition Solution Manual - Page 6 preview image51–5.The precast floor beam is made from concrete having aspecific weight of23.6 kN/m3. If it is to be used for a floor of anoffice building, calculate its dead and live loadings per footlength of beam.1.5 m0.15 m0.3 m0.15 mSOLUTIONThe dead load is caused by the self-weight of the beam.wD=[(1.5m)(0.15m)+(0.15m)(0.3m)](23.6kN/m3)=6.372kN/m=6.37kN/mAns.For the office, the recommended line load for design inTable 1–4 is2.4kN/m2. Thus,wL=(2.40kN/m2)(1.5m)=3.60kN/mAns.Ans.wD=6.37kN/mwL=3.60kN/m
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Structural Analysis, 10th Edition Solution Manual - Page 7 preview image61–6.The floor of a light storage warehouse is made of150-mm-thick lightweight plain concrete. If the floor is a slabhaving a length of 7 m and width of 3 m, determine the resultantforce caused by the dead load and the live load.SOLUTIONFrom Table 1–3,DL=[0.015kN/m2.mm(150mm)](7m)(3m)=47.25kNFrom Table 1–4,LL=(6.00kN/m2)(7m)(3m)=126kNTotal Load:F=126kN+47.25kN=173kNAns.Ans.F=173kN
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Structural Analysis, 10th Edition Solution Manual - Page 8 preview image71–7.The precast inverted T-beam has the cross sectionshown. Determine its weight per foot of length if it is madefromreinforced stone concrete and twelve34-in.-diametercold-formed steel reinforcing rods.9 in.12 in.36 in.48 in.SOLUTIONFrom Table 1–2, the specific weight of reinforced stone concrete and the cold-formed steel areγC=150lb/ft3andγH=492lb/ft3, respectively.Reinforcedstoneconcrete:| (4812 ft) (1212 ft)+(912 ft) (3612 ft)−12(π4) (0.7512ft)2|(150lb/ft)=931.98lb/ftCold@formedsteel:|12(π4) (0.7512ft)2|(492lb/ft3)=18.11lb/ft950.09lb/ftwD=(950.09lb/ft)(1k1000lb)=0.950k/ftAns.Ans.wD=0.950k/ft
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Structural Analysis, 10th Edition Solution Manual - Page 9 preview image8*1–8.Thehollowcorepanelismadefromplainstoneconcrete. Determine the dead weight of the panel. The holeseach have a diameter of 4 in.7 in.12 ft12 in.12 in.12 in.12 in.12 in.12 in.SOLUTIONFrom Table 1–2,W=(144lb/ft3)[(12ft)(6ft)(712ft)−5(12ft)(π)(212ft)2]=5.29kAns.Ans.W=5.29k
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Structural Analysis, 10th Edition Solution Manual - Page 10 preview image91–9.The floor of a light storage warehouse is made of6-in.-thick cinder concrete. If the floor is a slab having a lengthof 10 ft and width of 8 ft, determine the resultant force causedby the dead load and that caused by the live load.SOLUTIONFrom Table 1–3,DL=(6in.)(9lb/ft2.in.)(8ft)(10ft)=4.32kAns.From Table 1–4,LL=(125lb/ft2)(8ft)(10ft)=10.0kAns.Ans.DL=4.32kLL=10.0k
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Structural Analysis, 10th Edition Solution Manual - Page 11 preview image101–10.The interior wall of a building is made from 2×4wood studs, plastered on two sides. If the wall is 12 ft high,determine the load in lb/ft of length of wall that it exerts onthe floor.SOLUTIONFrom Table 1–3,w=(20lb/ft2)(12ft)=240lb/ftAns.Ans.w=240lb/ft
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Structural Analysis, 10th Edition Solution Manual - Page 12 preview image111–11.The second floor of a light manufacturing building isconstructed from a 5-in.-thick stone concrete slab with anadded 4-in. cinder concrete fill as shown. If the suspendedceiling of the first floor consists of metal lath and gypsumplaster, determine the dead load for design in pounds persquare foot of floor area.4 in. cinderfill5 in. concrete slabceilingSOLUTIONFrom Table 1–3,5−in.concreteslab=(12)(5)=60.04−in.cinderfill=(9)(4)=36.0metallath & plaster=10.0Totaldeadload=106.0lb/ft2Ans.Ans.DL=106lb/ft2
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Structural Analysis, 10th Edition Solution Manual - Page 13 preview image12*1–12.A two-story hotel has interior columns for the roomsthat are spaced 6 m apart in two perpendicular directions.Determine the reduced live load supported by a typical interiorcolumn on the first floor under the public rooms.SOLUTIONTable 1–4:Lo=4.79kN/m2AT=(6m)(6m)=36m2KLL=4KLLAT=4(36)=144m2>37.2m2From Eq. 1-1,LL=Lo(0.25+4.571KLLTA)LL=4.79(0.25+4.5714(36))LL=3.02kN/m2Ans.3.02kN/m2>0.4Lo=1.916kN/m2OKAns.LL=3.02kN/m2
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Structural Analysis, 10th Edition Solution Manual - Page 14 preview image131–13.Afour-storyofficebuildinghasinteriorcolumnsspaced 30 ft apart in two perpendicular directions. If theflat-roof live loading is estimated to be30 lb/ft2, determine thereduced live load supported by a typical interior columnlocated at ground level.SOLUTIONFrom Table 1–4,Lo=50psfAT=(30)(30)=900ft2KLLAT=4(900)=3600ft2>400ft2From Eq. 1-1,L=La(0.25−151KLLAT)L=50(0.25−1514(900))=25psf%reduction=2550=50%>40%(OK)F=3[(25psf)(30 ft)(30ft)]+30psf(30ft)(30ft)=94.5kAns.Ans.LL=94.5k
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Structural Analysis, 10th Edition Solution Manual - Page 15 preview image141–14.The office building has interior columns spaced 5 mapart in perpendicular directions. Determine the reduced liveload supported by a typical interior column located on the firstfloor under the offices.SOLUTIONFrom Table 1–4,Lo=2.40kN/m2AT=(5m)(5m)=25m2KLL=4L=Lo(0.25+4.571KLLAT)L=2.40(0.25+4.5714(25))L=1.70kN/m2Ans.1.70kN/m2>0.4Lo=0.96kN/m2OKAns.LL=1.70kN/m2
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Structural Analysis, 10th Edition Solution Manual - Page 16 preview image151–15.A hospital located in Chicago, Illinois, has a flat roof,where the ground snow load is25 lb/ft2. Determine the designsnow load on the roof of the hospital.SOLUTIONCe=1.2Ct=1.0I=1.2pf=0.7CeCtIpgpf=0.7(1.2)(1.0)(1.2)(25)=25.2lb/ft2Ans.Ans.pf=25.2lb/ft2
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Civil Engineering
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Structural Analysis by Russell C. H...

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