Solution Manual For Structural Analysis, 8th Edition

Preview (16 of 558 Pages)

100%

Solution Manual For Structural Analysis, 8th Edition - Page 1 preview image

Loading page ...

11–1.The floor of a heavy storage warehouse building ismade of 6-in.-thick stone concrete. If the floor is a slabhaving a length of 15 ft and width of 10 ft, determine theresultant force caused by the dead load and the live load.From Table 1–3DL=[12 lbft2#in.(6 in.)] (15 ft)(10 ft)=10,800 lbFrom Table 1–4LL=(250 lbft2)(15 ft)(10 ft)=37,500 lbTotal LoadF=48,300 lb=48.3 kAns.1–2.The floor of the office building is made of 4-in.-thicklightweight concrete. If the office floor is a slab having alength of 20 ft and width of 15 ft, determine the resultantforce caused by the dead load and the live load.From Table 1–3DL=[8 lbft2#in. (4 in.)] (20 ft)(15 ft)=9600 lbFrom Table 1–4LL=(50 lbft2)(20 ft)(15 ft)=15,000 lbTotal LoadF=24,600 lb=24.6 kAns.1–3.The T-beam is made from concrete having a specificweight of 150 lbft3. Determine the dead load per foot lengthof beam. Neglect the weight of the steel reinforcement.w=(150 lbft3) [(40 in.)(8 in.)+(18 in.) (10 in.)]w=521 lbftAns.a1 ft2144 in2b26 in.40 in.8 in.10 in.

Solution Manual For Structural Analysis, 8th Edition - Page 2 preview image

Loading page ...

Solution Manual For Structural Analysis, 8th Edition - Page 3 preview image

Loading page ...

2*1–4.The “New Jersey” barrier is commonly used duringhighway construction. Determine its weight per foot oflength if it is made from plain stone concrete.1–5.The floor of a light storage warehouse is made of150-mm-thick lightweight plain concrete. If the floor is aslab having a length of 7 m and width of 3 m, determine theresultant force caused by the dead load and the live load.From Table 1–3DL=[0.015 kNm2#mm (150 mm)] (7 m) (3 m)=47.25 kNFrom Table 1–4LL=(6.00 kNm2) (7 m) (3 m)=126 kNTotal LoadF=126 kN+47.25 kN=173 kNAns.12 in.4 in.24 in.6 in.55°75°Cross-sectional area=6(24)+(24+7.1950)(12)+(4+7.1950)(5.9620)=364.54 in2Use Table 1–2.w=144 lbft3(364.54 in2)=365 lbftAns.a1 ft2144 in2ba12ba12b

Solution Manual For Structural Analysis, 8th Edition - Page 4 preview image

Loading page ...

31–7.The wall is 2.5 m high and consists of 51 mm102 mmstuds plastered on one side. On the other side is 13 mmfiberboard, and 102 mm clay brick. Determine the averageload in kNm of length of wall that the wall exerts on the floor.8 in.8 in.4 in.4 in.6 in.6 in.6 in.20 in.Use Table 1–3.For studsWeight=0.57 kNm2(2.5 m)=1.425 kNmFor fiberboardWeight=0.04 kNm2(2.5 m)=0.1 kNmFor clay brickWeight=1.87 kNm2(2.5 m)=4.675 kNmTotal weight=6.20 kNmAns.1–6.The prestressed concrete girder is made from plainstone concrete and four-in. cold form steel reinforcingrods. Determine the dead weight of the girder per foot of itslength.34Area of concrete=48(6)+4(14+8)(4)-4()=462.23 in2Area of steel=4()=1.767 in2From Table 1–2,w=(144 lbft3)(462.23 in2)+492 lbft3(1.767 in2)=468 lbfta1 ft2144 in2ba1 ft2144 in2ba38b2a38b2d12c2.5 mAns.

Solution Manual For Structural Analysis, 8th Edition - Page 5 preview image

Loading page ...

41–10.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. cinder fill5 in. concrete slabceilingFrom Table 1–3,5-in. concrete slab=(12)(5)=60.04-in. cinder fill=(9)(4)=36.0metal lath & plaster=10.0Total dead load=106.0 lbft2Ans.1–9.The interior wall of a building is made from 24wood studs, plastered on two sides. If the wall is 12 ft high,determine the load in lbft of length of wall that it exerts onthe floor.From Table 1–3w=(20 lbft2)(12 ft)=240 lbftAns.*1–8.A building wall consists of exterior stud walls withbrick veneer and 13 mm fiberboard on one side. If the wallis 4 m high, determine the load in kNm that it exerts on thefloor.For stud wall with brick veneer.w=(2.30 kNm2)(4 m)=9.20 kNmFor Fiber boardw=(0.04 kNm2)(4 m)=0.16 kNmTotal weight=9.2+0.16=9.36 kNmAns.

Solution Manual For Structural Analysis, 8th Edition - Page 6 preview image

Loading page ...

51–11.A four-story office building has interior columnsspaced 30 ft apart in two perpendicular directions. If theflat-roof live loading is estimated to be 30 lbft2, determinethe reduced live load supported by a typical interior columnlocated at ground level.Floor load:Lo=50 psfAt=(30)(30)=900 ft27400 ft2L=Lo(0.25+)L=50(0.25+)=25 psf% reduction==50%740% (OK)Fs=3[(25 psf)(30 ft)(30 ft)]+30 psf(30 ft)(30 ft)=94.5 kAns.25501524(900)152KLLAT*1–12.A two-story light storage warehouse has interiorcolumns that are spaced 12 ft apart in two perpendiculardirections. If the live loading on the roof is estimated to be25 lbft2, determine the reduced live load supported bya typical interior column at (a) the ground-floor level, and(b) the second-floor level.At=(12)(12)=144 ft2FR=(25)(144)=3600 lb=3.6 kSinceAt=4(144) ft27400 ft2L=12.5(0.25 +)=109.375 lbft2(a) For ground floor columnL=109 psf70.5Lo=62.5 psfOKFF=(109.375)(144)=15.75 kF=FF+FR=15.75 k+3.6 k=19.4 kAns.(b) For second floor columnF=FR=3.60 kAns.152(4)(144)

Solution Manual For Structural Analysis, 8th Edition - Page 7 preview image

Loading page ...

61–14.Atwo-storyhotelhasinteriorcolumnsfortherooms that are spaced 6 m apart in two perpendiculardirections. Determine the reduced live load supported by atypical interior column on the first floor under the publicrooms.Table 1–4Lo=4.79 kNm2AT=(6 m)(6 m)=36 m2KLL=4L=Lo(0.25+)L=4.79(0.25+)L=3.02 kNm2Ans.3.02 kNm270.4Lo=1.916 kNm2OK4.5724(36)4.572KLLAT1–13.The office building has interior columns spaced 5 mapart in perpendicular directions. Determine the reducedlive load supported by a typical interior column located onthe first floor under the offices.From Table 1–4Lo=2.40 kNm2AT=(5 m)(5 m)=25 m2KLL=4L=Lo(0.25+)L=2.40(0.25+)L=1.70 kNm2Ans.1.70 kNm270.4Lo=0.96 kNm2OK4.5724(25)4.572KLLAT

Solution Manual For Structural Analysis, 8th Edition - Page 8 preview image

Loading page ...

7*1–16.Wind blows on the side of the fully enclosedhospital located on open flat terrain in Arizona. Determinethe external pressure acting on the leeward wall, which hasa length of 200 ft and a height of 30 ft.V=120 mihKzt=1.0Kd=1.0qh=0.00256KzKztKdV2=0.00256Kz(1.0)(1.0)(120)2=36.86Kz1–15.Wind blows on the side of a fully enclosed hospitallocated on open flat terrain in Arizona. Determine theexternal pressure acting over the windward wall, which hasa height of 30 ft. The roof is flat.V=120 mihKzt=1.0Kd=1.0qz=0.00256KzKztKdV2=0.00256Kz(1.0)(1.0)(120)2=36.86KzFrom Table 1–5,zKzqz0–150.8531.33200.9033.18250.9434.65300.9836.13Thus,p=q G Cp–qh(G Cpi)=q(0.85)(0.8)-36.13 (;0.18)=0.68q<6.503p0–15=0.68(31.33)<6.503=14.8 psf or 27.8 psfAns.p20=0.68(33.18)<6.503=16.1 psf or 29.1 psfAns.p25=0.68(34.65)<6.503=17.1 psf or 30.1 psfAns.p30=0.68(36.13)<6.503=18.1 psf or 31.1 psfAns.

Solution Manual For Structural Analysis, 8th Edition - Page 9 preview image

Loading page ...

81–17.A closed storage building is located on open flatterrain in central Ohio. If the side wall of the building is20 ft high, determine the external wind pressure acting onthe windward and leeward walls. Each wall is 60 ft long.Assume the roof is essentially flat.From Table 1–5, forz=h=30 ft,Kz=0.98qh=36.86(0.98)=36.13From the text==1 so thatCp= -0.5p=q GCp-qh(GCp2)p=36.13(0.85)(-0.5)-36.13(;0.18)p= -21.9 psf or-8.85 psfAns.200200LoBV=105 mihKzt=1.0Kd=1.0q=0.00256KzKztKdV2=0.00256Kz(1.0)(1.0) (105)2=28.22KzFrom Table 1–5zKzqz0–150.8523.99200.9025.40Thus, for windward wallp=qGCp–qh(GCpi)=q(0.85)(0.8) – 25.40(;0.18)=0.68q<4.572p0 – 15=0.68 (23.99)<4.572=11.7 psf or 20.9 psfAns.p20=0.68 (25.40)<4.572=12.7 psf or 21.8 psfAns.Leeward wall==1 so thatCp= -0.5p=q GCp-qh(GCpi)p=25.40(0.85)(-0.5)-25.40 (;0.18)p= -15.4 psf or-6.22 psfAns.6060LB1–16.Continued

Solution Manual For Structural Analysis, 8th Edition - Page 10 preview image

Loading page ...

91–18.The light metal storage building is on open flatterrain in central Oklahoma. If the side wall of the buildingis 14 ft high, what are the two values of the external windpressure acting on this wall when the wind blows on theback of the building? The roof is essentially flat and thebuilding is fully enclosed.V=105 mihKzt=1.0Kd=1.0qz=0.00256KzKztKdV2=0.00256Kz(1.0)(1.0)(105)2=28.22KzFrom Table 1–5For 0…z…15 ftKz=0.85Thus,qz=28.22(0.85)=23.99p=q GCp-qh(GCpi)p=(23.99)(0.85)(0.7)-(23.99)(0.18)p= -9.96 psf orp= -18.6 psfAns.;1–19.Determine the resultant force acting perpendicularto the face of the billboard and through its center if it islocated in Michigan on open flat terrain. The sign is rigidand has a width of 12 m and a height of 3 m. Its top side is15 m from the ground.qh=0.613KzKztKdV2Sincez=h=15 mKz=1.09Kzt=1.0Kd=1.0V=47 msqh=0.613(1.09)(1.0)(1.0)(47)2=1476.0 Nm2Bs==4, sh==0.2From Table 1–6Cf=1.80F=qhGCfAs=(1476.0)(0.85)(1.80)(12)(3)=81.3 kNAns.31512 m3 m

Solution Manual For Structural Analysis, 8th Edition - Page 11 preview image

Loading page ...

101–21.The school building has a flat roof. It is located in anopen area where the ground snow load is 0.68 kNm2.Determine the snow load that is required to design the roof.pf=0.7CcCtIspgpf=0.7(0.8)(1.0)(1.20)(0.68)=0.457 kNm2Alsopf=pf=Ispg=(1.20)(0.68)=0.816 kNm2usepf=0.816 kNm2Ans.1–22.The hospital is located in an open area and has aflat roof and the ground snow load is 30 lbft2. Determinethe design snow load for the roof.Sincepq=30 lbft2720 lbft2thenpf=Ispg=1.20(30)=36 lbft2Ans.*1–20.A hospital located in central Illinois has a flat roof.Determine the snow load in kNm2that is required todesign the roof.pf=0.7CcCtIspgpf=0.7(0.8)(1.0)(1.20)(0.96)=0.6451 kNm2Alsopf=Ispg=(1.20)(0.96)=1.152 kNm2usepf=1.15 kNm2Ans.

Solution Manual For Structural Analysis, 8th Edition - Page 12 preview image

Loading page ...

112–1.Thesteelframeworkisusedtosupportthereinforced stone concrete slab that is used for an office. Theslab is 200 mm thick. Sketch the loading that acts alongmembersBEandFED. Take,.Hint: SeeTables 1–2 and 1–4.b=5 ma=2 m200 mm thick reinforced stone concrete slab:(23.6 kN>m3)(0.2 m)(2 m)=9.44 kN>mLive load for office: (2.40 kN>m2)(2 m)=Ans.Due to symmetry the vertical reaction atBandEareBy=Ey=(14.24 kN>m)(5)>2=35.6 kNThe loading diagram for beamBEis shown in Fig.b.480kN>m14.24kN>mBeamFED.The only load this beam supports is the vertical reaction of beamBEatEwhich isEy=35.6 kN. The loading diagram for this beam is shown in Fig.c.BeamBE.Sincethe concrete slab will behave as a one way slab.Thus, the tributary area for this beam is rectangular shown in Fig.aand the intensityof the uniform distributed load isba=5 m2 m=2.5,ABCDEFbaa

Solution Manual For Structural Analysis, 8th Edition - Page 13 preview image

Loading page ...

122–2.Solve Prob. 2–1 with,.b=4 ma=3 mBeamBE.Since, the concrete slab will behave as a two way slab. Thus,the tributary area for this beam is the hexagonal area shown in Fig.aand themaximum intensity of the distributed load is200 mm thick reinforced stone concrete slab: (23.6 kN>m3)(0.2 m)(3 m)=14.16 kN>mLive load for office: (2.40 kN>m2)(3 m)=Ans.Due to symmetry, the vertical reactions atBandEare=26.70 kNThe loading diagram for BeamBEis shown in Fig.b.BeamFED.The loadings that are supported by this beam are the vertical reactionof beamBEatEwhich isEy=26.70 kN and the triangular distributed load of whichits tributary area is the triangular area shown in Fig.a. Its maximum intensity is200 mm thick reinforced stone concrete slab: (23.6 kN>m3)(0.2 m)(1.5 m)=7.08 kN>mLive load for office: (2.40 kN>m2)(1.5 m)=Ans.The loading diagram for BeamFEDis shown in Fig.c.3.60kN>m10.68kN>mBy=Ey=2c12 (21.36 kN>m)(1.5 m)d+(21.36 kN>m)(1 m)2720kN>m21.36kN>mba=4362ABCDEFbaa

Solution Manual For Structural Analysis, 8th Edition - Page 14 preview image

Loading page ...

132–3.The floor system used in a school classroom consistsof a 4-in. reinforced stone concrete slab. Sketch the loadingthat acts along the joistBFand side girderABCDE. Set,.Hint: See Tables 1–2 and 1–4.b=30 fta=10 ftAEbaaaaBCDFJoistBF.Since, the concrete slab will behave as a one way slab.Thus, the tributary area for this joist is the rectangular area shown in Fig.aand theintensity of the uniform distributed load is4 in thick reinforced stone concrete slab: (0.15 k>ft3)(10 ft)=0.5 k>ftLive load for classroom: (0.04 k>ft2)(10 ft)=Ans.Due to symmetry, the vertical reactions atBandFareBy=Fy=(0.9 k>ft)(30 ft)>2=13.5 kAns.The loading diagram for joistBFis shown in Fig.b.GirderABCDE.The loads that act on this girder are the vertical reactions of thejoists atB,C, andD, which areBy=Cy=Dy=13.5 k. The loading diagram forthis girder is shown in Fig.c.0.4k>ft0.9k>fta412 ftbba=30 ft10 ft=3

Solution Manual For Structural Analysis, 8th Edition - Page 15 preview image

Loading page ...

14*2–4.Solve Prob. 2–3 with,.b=15 fta=10 ftAEbaaaaBCDFJoistBF.Since, the concrete slab will behave as a two wayslab. Thus, thetributaryareaforthejoististhehexagonalareaasshownin Fig.aand the maximum intensity of the distributed load is4 in thick reinforced stone concrete slab: (0.15 k>ft3)(10 ft)=0.5 k>ftLive load for classroom: (0.04 k>ft2)(10 ft)=Ans.Due to symmetry, the vertical reactions atBandGareAns.The loading diagram for beamBFis shown in Fig.b.GirderABCDE.The loadings that are supported by this girder are the verticalreactions of the joist atB,CandDwhich areBy=Cy=Dy=4.50 k and thetriangular distributed load shown in Fig.a. Its maximum intensity is4 in thick reinforced stone concrete slab:(0.15 k>ft3)(5 ft)=0.25 k>ftLive load for classroom: (0.04 k>ft2)(5 ft)=Ans.The loading diagram for the girderABCDEis shown in Fig.c.0.20kft0.45kfta412ftbBy=Fy=2c12 (0.9k>ft)(5ft)d+(0.9k>ft)(5ft)2=4.50k0.4k>ft0.9k>fta412ftbba=15 ft10 ft=1.5 < 2

Solution Manual For Structural Analysis, 8th Edition - Page 16 preview image

Loading page ...

152–5.Solve Prob. 2–3 with,.b=20 fta=7.5 ftAEbaaaaBCDFBeamBF.Since, the concrete slab will behave as a one wayslab. Thus, the tributary area for this beam is a rectangle shown in Fig.aand theintensity of the distributed load is4 in thick reinforced stone concrete slab: (0.15 k>ft3)(7.5 ft)=0.375 k>ftLive load from classroom: (0.04 k>ft2)(7.5 ft)=Ans.Due to symmetry, the vertical reactions atBandFareAns.The loading diagram for beamBFis shown in Fig.b.BeamABCD.The loading diagram for this beam is shown in Fig.c.By=Fy=(0.675 k>ft)(20 ft)2=6.75 k0.300k>ft0.675k>fta412ftbba=20 ft7.5 ft=2.772

Preview Mode

This document has 558 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

Structural Analysis, 10th Edition Solution Manual

Solution Manual For Elementary Surveying: An Introduction To Geomatics, 13th Edition

Elementary Surveying: An Introduction to Geomatics, 15th Edition Solution Manual

Solution Manual for Reliability Engineering, 1st Edition

Solution Manual For Steel Structures: Design And Behavior, 5th Edition

Dynamics of Structures, 5th Edition Solution Manual

Solution Manual for Design Concepts for Engineers, 5th Edition

Solution Manual For Civil And Environmental Systems Engineering, 2nd Edition

Solution Manual For Dynamics of Structures, 4th Edition

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition

Company

Explore

Study Tools