Five-story Building And Load Data Of Example - 1375 Words

Five-story Building And Load Data Of Example (Math Problem Sample) Content: LOAD ANALYSIS IN BUILDINGSNameName Course No.Professors NameInstitutions NameLocationDateProblem#1: (30 points)Given the five-story building and load data of Example 3.1 (Text Book), determine the total axial dead and live loads at the base of column B2 assuming that the storage occupancy is on level 4 and that all other floors are typical office occupancy.Given * 5 story building * Beams: 22 by 20.5 * Columns: 22 by 22 * Joist: 16+4.5 by 7+53 * Roof: Ordinary flat (slope of on 12) from IBC: LL = 20 psf * Level 4: LL = 125 psf * All floors: offices from IBC: LL = 50 psf * Superimposed DL: 10 psf * Normal weight concrete: w = 150 pcfRequired: Total axial DL and LL at base of Column B2Dead loads: The axial dead load supported by column B2 consists of the weight of the structural members and the superimposed dead load that are tributary to this column: * Weight of joists = 841000*25+222*20=39.5 kips * Weight of beams = 22*20.5144*20*1501000=9.4 kips * Weight of column (typical story) = 22*22144*10*1501000=5.0 kips * Weight of column (first story) = 22*22144*12*1501000=6.1 kips * Superimposed dead load = 101000*25+222*20=4.7 kipsLive load: IBC table 1607.1 is used to determine the nominal live loads on the basis of the given occupancies. Live load reductions are taken wherever applicable.Roof: Lo = 20 psfReduced live load is determined by IBC eq. (16.26):Lr=LoR1R2The Tributary area, At=25+222*20=470 ft^2Because 200 ft2 600 ft2, R1 is determined by IBC Eq. (16.28):R1=1.2-0.001At=1.2-0.001*470=0.73A roof slope of on 12 means that F = 1/2; because F 4, R2 = 1,Thus, Lr = 20 * 0.73 * 1 = 15 psf 12 psfAxial Live load=151000*470=7.1 kipsLevel 4: Because level 4 is storage with a live load of 125 psf, which exceeds 100 psf, the live load is not permitted to be reduced (IBC 1607.10.1.2)Axial live load=1251000*470=58.8 kipsTypical Floor: * Reducible. From IBC Table 1607.1, reducible nominal live load for an office occupancy = 50 psf (lobby and corridor loads are neglected per the problem statement). Reduced live load L is determined by IBC Eq. (16.23):L=Lo*o.25+15KllAt=50*0.25+15KllAt0.50Lo for members supporting one floor0.40Lo for members supporting two or more floorsThe live load element factor KLL = 4 for an interior column (IBC table 1607.10.1), and the tributary area AT at a particular floor level is equal to the sum of the tributary areas for that floor and all the floors above it where the live load can be reduced. Thus,L=50*(0.25+154*470) = 29.8 kipsTherefore, the total axial dead load at base of column B2:No. of floors = 5No. of typical floors = 4Total Weight of Joists = 39.5*5 = 197.5 kipsTotal Weight of beams = 9.4*5 = 47.0 kipsTotal super-imposed dead load = 4.7*5 = 23.5 kipsTotal weight of column (on typical story) = 5.0*4 =20.0 kipsTotal weight of column (on first floor) = 6.1 kipsTotal Axial dead load at base of Column B2 = 294.1 kipsAnd the total axial live load at base of column B2:Roof (Level 6) = 7.1 kipsLevel 5 (typical floor) = 29.8 kipsLevel 4 = 58.8 kipsLevel 3 (typical floor) = 29.8 kipsLevel 2 (typical floor) = 29.8 kipsTotal Axial Live load at base of Column B2 = 155.3 kipsProblem#2: (30 points)Given the five-story building and load data of Example 3.1(Text Book), determine the dead and live loads along the span of the beam on column line A between 1 and 2 on (a) a typical floor with office occupancy, (b) the floor with storage occupancy, and (c) the roof.Given: * 5 story building * Beams: 22 by 20.5 * Columns: 22 by 22 * Joist: 16+4.5 by 7+53 * Roof: Ordinary flat (slope of on 12) from IBC: LL = 20 psf * Level 4: LL = 125 psf * All floors: offices from IBC: LL = 50 psf * Superimposed DL: 10 psf Normal weight concrete: w = 150 pcfRequired: The dead and live loads along the span of the beam on typical floor, level 4 and roofArea loading (superimposed dead load and live load) is assumed to be applied on all floor areaThe beam is in X-direction (perpendicular to slab strip) and joist s are in Y-direction (parallel to slab strip).The panel ratio is 25/5 and is greater than 2 (one way action), hence the slab is one-way supported.*ppf is pounds per foot * RoofDead LoadWeight of joists = 841000*5= 0.42 ppfWeight of beam rib = 22*(20.5-4.5)/ (144)*150/1000 = 0.37 ppfSuperimposed dead load = 10/1000*5= 0.05 ppfTotal Dead Load = 0.84 ppfLive LoadFrom problem #1 above, uniform Live load = 15/1000*5 = 0.075ppfUltimate Uniform load on joist, wu = 1.4wD+1.7wL = 1.4*0.84+1.7*0.075 = 1.30 ppfThe joists in turn exert a point load on the beam = 1.3*25/ 2 = 16.25 kipsHence on the roof level, the beam will have 16.25 kips point loads where the joists rest on the beam. * Level 4Dead LoadWeight of joists = 841000*5= 0.42 ppfWeight of beam rib = 22*(20.5-4.5)/ (144)*150/1000 = 0.37 ppfSuperimposed dead load = 10/1000*5= 0.05 ppfTotal Dead Load = 0.84 ppfLive LoadFrom problem #1 above, uniform Live load = 125/1000*5 = 0.625ppfUltimate Uniform load on joist, wu = 1.4wD+1.7wL = 1.4*0. 84+1.7*0.625 = 2.24 ppfThe joists in turn exert a point load on the beam = 2.24*25/ 2 = 30.0 kipsHence on the roof level, the beam will have 30 kips point loads where the joists rest on the beam....