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Lecture 22 - Page 1 of 9

Lecture 22 Dead Loads, Live Loads & Load Combinations

Dead Loads

Dead loads include the weight of the physical structureand the non-movablematerials and objects attached to the structure. They are considered to bepermanent loads. Building codes are of little use in the determination of deadloads they must be hand-calculated. Many references exist tabulating thetypical weights of building materials, such as the Architectural GraphicStandards, AISC Manual, etc. Typically, dead loads are determined on apounds per square foot basis.

Materials: Weight (lb. per sq. ft.)

Ceilings:

Channel suspended acoustical 1.5

gypsum drywall 2

Plaster & lath 8

Flooring:Concrete, normal weight per 1 thickness 12.5Precast concrete, 6 plank, no topping 40

plywood subfloor 2.5

Steel decking, 1 2.5

Walls & Partitions: (per height of wall)

4 brick 40

8 concrete block CMU 55

12 concrete block CMU 85

2x4 wood stud w/ GWB both sides 8

4 metal stud w/ GWB both sides 6

4 lightweight CMU block w/ GWB both sides 26Roofing Materials:

Built-up EPDM 6.5

Concrete roof tile 9.5

Copper 2

Shingles, asphalt 2.8

Shingles, wood 2.5

Tile, clay 16 - 20

Tile, cement ribbed 16Slate, 3/16 7 - 9.5

Slate, 3/8 14 - 18Finish Materials:

Gypsum wallboard, 2Tile, glazed wall 3/8 3

Quarry tile, 9

Hardwood flooring, 25/32 4

Vinyl tile, 1/8 1.5

Terrazzo, 1, 2 in stone conc. 25

Insulation & Waterproofing:

Batt, blankets per 1 thickness 0.3Rigid insulation 1.5


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Example 1GIVEN: The steel-framed floor structure as shown below, to be used as an officebuilding. The construction is indicated and dead loads can be found from thetable above. The floor-to-floor height = 12-0REQUIRED: Determine the total dead load of the floor construction on apounds-per-square foot basis.

Add up all superimposeddead loads as follows:

4 concrete slab 4 @ 12.5 psf/inch = 50 psf1 metal deck....= 2.5 psf quarry tile ...= 9 psfPartitions.. = 20 psf (per 1607.5 of the IBC)

Acoustical hung ceiling...= 1.5 psfMechanical/Electrical..= 5 psf

Sub-total = 88 psf

Determine the dead load of structural steel beams and columns :

4 W18x35 x 32-0 long = 4480 lbs.

2 W24x94 x 21-0 long = 3948 lbs.4 6 std. wt. stl. cols @ 18.97 plf x 12-0 long .= 911 lbs.

Sub-total = 9339 lbs.

Taking this weight and dividing by the area 9339 lbs/(32 x 21)= 13.9 psf

Added together , the total Dead Load = 88 psf + 13.9 psf = 101.9 psf

W24x94

W18x35

W18x35

W18x35

W18x35

W24x94

32-0

3@7-0=21-0

Floor construction:4 conc. over 1 metal deck quarry tile floor finishPartitions - 2x4 metal stud walls w/ GWB both sides

Acoustical hung ceiling below beamsMechanical/Electrical allowance = 5 sf

6 dia. std. wt.pipe col (typ.)


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Floor Live Loads

From the IBC, a Live Load is defined as Those loads produced by the use andoccupancy of the building or other structure and do not include construction orenvironmental loads such as wind load, snow load, rain load, earthquake load,flood load or dead load.

Examples of things contributing to live loads include people, furniture, moveableequipment, and anything else that does not remain permanently stationary.

IBC Section 1607 specifies prescribed minimumlive loads. Table 1607.1 liststhese prescribed minimum live loads based upon anticipated occupancy. Thearchitect or engineer-of-record is free to INCREASE these loads as he/shedeems necessary however these loads CANNOT BE DECREASED exceptunder Section 1607.9 where a formula is given that may be used to reduce thelive loads.

)1525.0(0TLL

AKLL +=

when KLLAT> 400 ft2

where:L = Reduced design live load per square foot of area supported by the memberL0= Unreduced design live load per square foot per Table 1607.1 < 100 psfKLL= Live load element factor per Table 1607.9.1

AT= Tributary area in square feet

Example 2GIVEN: The same floor system as shown in Example 1.REQUIRED: Determine the reduced live load, L, (if applicable) for the design ofthe interior W18x35 filler beams, the W24x94 girder and the corner columns.

a) Interior filler beams:

From Table 1607.1 use L0= 50 psf (office)From Table 1607.9.1, use KLL= 2 for interior beamsTrib. Area AT= (7)(32) = 224 ft

2

KLLAT= (2)(224 ft

2

) = 448 ft

2

> 400 ft

2

live load reduction allowed.

)15

25.0(0TLL

AKLL +=

)448

1525.0(50 += psfL = 47.9 psf


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b) Edge girders:

From Table 1607.1 use L0= 50 psf (office)From Table 1607.9.1, use KLL= 2 for edge beam w/o cant. slabTrib. Area AT= (32)(21) = 336 ft

2

KLLAT= (2)(336 ft2) = 672 ft2> 400 ft2live load reduction allowed.

)15

25.0(0TLL

AKLL +=

)672

1525.0(50 += psfL = 41.4 psf

c) Corner columns:

From Table 1607.1 use L0= 50 psf (office)

From Table 1607.9.1, use KLL= 4 for ext. column w/o cant. slabTrib. Area AT= (32)(21) = 168 ft

2

KLLAT= (4)(168 ft2) = 672 ft2> 400 ft2live load reduction allowed.

)15

25.0(0TLL

AKLL +=

)672

1525.0(50 += psfL = 41.4 psf

Handrail Loads

Section 1607.7 dictates loads on handrails, guards, grab bars and vehiclebarriers. These loads must be carried throughout the entire assembly andinto the supporting structure. In particular, the minimum design loads onhandrails (excluding vehicle barriers) is:

Uniform load = 50 PLF acting at the top applied from ANY direction

or

Point load = 200 lbs acting at the top applied from ANY direction


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Roof Live Loads

In general, design loading on roofs comes from snow. However, in areaswhere snow is not extreme, the minimum prescribed roof live load is:

Lr= LoR1R2 in units of PSF

where:Lr= roof live load in pounds per square foot of horizontal projection

Lo= unreduced roof live load per Table 1607.1

R1= 1.0 for AT< 200 ft2

= 1.2 0.001(AT) for 200 ft2< AT< 600 ft

2

R2= 1 if F < 4

= 1.2 0.05F if 4 < F < 12= 0.6 if F > 12

F = the number of inches rise per foot slope on sloped roof= rise-to-span ratio multiplied by 32 for arch or dome roof

Example 3GIVEN: The flat-roof framing plan from the previous examples. Assume thebuilding is to be located in southern Florida.REQUIRED: Determine the minimum design roof live load, Lrfor the interiorW18x35 filler beam, the exterior girder and the corner columns.

a) From IBC Table 1607.1 use roof live load Lo= 20 psfb) Interior filler beam:

The tributary area, AT= 7(32) = 224 ft2 R1= 1.2 0.001AT

= 1.2 0.001(224 ft2)= 0.976

R2 = 1 since the roof is flat, F < 4

Lr= LoR1R2

= 20psf(0.976)(1)

Lr= 19.5 psf


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c) Exterior girder:

Tributary area , AT= (32)(21) = 336 ft2R1= 1.2 0.001AT

= 1.2 0.001(336 ft2)= 0.864

R2 = 1 since the roof is flat, F < 4

Lr= LoR1R2= 20psf(0.864)(1)

Lr= 17.3 psf

d) Corner column:

Tributary area , AT= (32)(21) = 168 ft2< 200 ft2R1= 1.0

Lr= LoR1R2= 20psf(1)(1)

Lr= 20 psf


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Load Combinations

Buildings and other structures and portions thereof shall be designed toresist the most crit ical effects of combinations of loadsin accordancewith either the Allowable Stress Design Method or the Load andResistance Factor Design (LRFD) Method (or referred to as the Strength

method) as prescribed in Section 1605.1:

Al lowable Stress Design Method: Load & Resistance Factor Design Method :

D 1.4D

D + L 1.2D + 1.6L + 0.5(Lror S or R)

D + L + (Lror S or R) 1.2D + 1.6(Lror S or R) + (f1L or 0.8W)

D + (W or 0.7E) + L + (Lror S or R) 1.2D + 1.6W + f1L + 0.5(Lror S or R)

0.6D + W 1.2D + 1.0E + f1L + f2S

0.6D + 0.7E 0.9D + (1.0E or 1.6W)

where:D = Dead loadsL = Live loadsLr= Roof live loadsS = Snow loadsE = Earthquake (seismic) loadsW = Wind loadsR = Rain loads

f1= 1.0 for floors in places of publicassembly, for live loads > 100 psf, andfor parking garage live load

f1= 0.5 for other live loads

f2= 0.7 for roof configurations that donot shed snow off the structure

f2= 0.2 for other roof configurations


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Example 4GIVEN: The roof framing plan as shown below. Loads are indicated as follows:

Superimposed roof dead load D (not including beam weight) = 17 psf

Roof live load Lr = 20 psf

Roof snow load S = 38 psf Roof earthquake load E = N/A

Roof wind load W = -11 psf (NOTE a negative number indicates uplift)

Roof rain load R = 31 psf (NOTE this is 6 of water @ unit wt. = 62.4 pcf)

REQUIRED:1) Determine the maximum uniform load on the W16x26 steel beam

considering the 6 load combinations above assuming Allowable StressDesign methodology. Do not consider reduction in live load.

2) Determine the maximum moment on the W16x26 beam using themaximum uniform load obtained in Part 1.

D = 6(17 psf) + 26 plf= 128 plf

L = 0 since L is considered a floor live load

Lr= 6(20 psf)= 120 plf

S = 6(38 psf)= 228 plf

E = 0 since building is located in a non-seismic zone

W = 6(-11 psf)= -66 plf

R = 6(31 psf)= 186 plf

W16x26

W16x26

W14x22

W24x62

W14x22

W24x62

25-0

3@6-0=18-0

6 dia. std. wt.pipe col (typ.)


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Make a Table as shown:

Load Comb. Uniform Load (PLF):D 128 plf

D + L 128 plf + 0 = 128 plf

D + L + (Lror S or R) 128 plf + 0 + (228 plf) = 356 plfD + (W or 0.7E) + L + (Lror S or R)

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