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Five-story Building And Load Data Of Example (Math Problem Sample)

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Problem#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.
Problem#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.
Problem#3: (10 points) Determine the modulus of elasticity Ec for a lightweight concrete mixture with a unit weight of 1,760 kg/m3 and Images equal to (a) 21, (b) 28, and

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LOAD ANALYSIS IN BUILDINGS
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Problem#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 ½ 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 pcf
Required: Total axial DL and LL at base of Column B2
Dead 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 kips
Live 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 psf
Reduced live load is determined by IBC eq. (16.26):
Lr=LoR1R2
The Tributary area, At=25+222*20=470 ft^2
Because 200 ft2 < 600 ft2, R1 is determined by IBC Eq. (16.28):
R1=1.2-0.001At=1.2-0.001*470=0.73
A roof slope of ½ on 12 means that F = 1/2; because F < 4, R2 = 1,
Thus, Lr = 20 * 0.73 * 1 = 15 psf > 12 psf
Axial Live load=151000*470=7.1 kips
Level 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 kips
Typical 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+15KllAt
≥0.50Lo for members supporting one floor
≥0.40Lo for members supporting two or more floors
The 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 kips
Therefore, the total axial dead load at base of column B2:
No. of floors = 5
No. of typical floors = 4
Total Weight of Joists = 39.5*5 = 197.5 kips
Total Weight of beams = 9.4*5 = 47.0 kips
Total super-imposed dead load = 4.7*5 = 23.5 kips
Total weight of column (on typical story) = 5.0*4 =20.0 kips
Total weight of column (on first floor) = 6.1 kips
Total Axial dead load at base of Column B2 = 294.1 kips
And the total axial live load at base of column B2:
Roof (Level 6) = 7.1 kips
Level 5 (typical floor) = 29.8 kips
Level 4 = 58.8 kips
Level 3 (typical floor) = 29.8 kips
Level 2 (typical floor) = 29.8 kips
Total Axial Live load at base of Column B2 = 155.3 kips
Problem#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 ½ 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 pcf
Required: The dead and live loads along the span of the beam on typical floor, level 4 and roof
Area loading (superimposed dead load and live load) is assumed to be applied on all floor area
The beam is in X-direction (perpendicular to slab strip) and joists 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
* Roof
Dead Load
Weight of joists = 841000*5= 0.42 ppf
Weight of beam rib = 22*(20.5-4.5)/ (144)*150/1000 = 0.37 ppf
Superimposed dead load = 10/1000*5= 0.05 ppf
Total Dead Load = 0.84 ppf
Live Load
From problem #1 above, uniform Live load = 15/1000*5 = 0.075ppf
Ultimate Uniform load on joist, wu = 1.4wD+1.7wL = 1.4*0.84+1.7*0.075 = 1.30 ppf
The joists in turn exert a point load on the beam = 1.3*25/ 2 = 16.25 kips
Hence on the roof level, the beam will have 16.25 kips point loads where the joists rest on the beam.
* Level 4
Dead Load
Weight of joists = 841000*5= 0.42 ppf
Weight of beam rib = 22*(20.5-4.5)/ (144)*150/1000 = 0.37 ppf
Superimposed dead load = 10/1000*5= 0.05 ppf
Total Dead Load = 0.84 ppf
Live Load
From problem #1 above, uniform Live load = 125/1000*5 = 0.625ppf
Ultimate Uniform load on joist, wu = 1.4wD+1.7wL = 1.4*0.84+1.7*0.625 = 2.24 ppf
The joists in turn exert a point load on the beam = 2.24*25/ 2 = 30.0 kips
Hence on the roof level, the beam will have 30 kips point loads where the joists rest on the beam.
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