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

Instructions:

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

Content:

LOAD ANALYSIS IN BUILDINGS

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Date

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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