Lecture 22 – Dead Dead Loads, Live Loads & Lo ad Comb Comb ination s Dead Dead Loads
Dead loads include the weight of the physic al struc ture and the non-movable materials and objects attached attached to the structure. They are considered to be permanent loads. Building codes are of little use in the determination of dead loads – they must be hand-calculated. Many references exist tabulating the typical weights of building materials, such as the Architectural Graphic Standards, AISC Manual, etc. etc. Typically, dead loads are determined determined on a “pounds per square foot” basis. Materials: Materials: Ceilings: Channel suspended acoustical ½” gypsum drywall Plaster & lath Flooring: Concrete, normal weight per 1” thickness Precast concrete, 6” plank, no topping ¾” plywood subfloor Steel decking, 1½” Walls & Partitions: (per height of wall) 4” brick 8” concrete block CMU 12” concrete block CMU 2x4 wood stud w/ ½” GWB both sides 4” metal stud w/ ½” GWB both sides 4” lightweight CMU block w/ ½” GWB both sides Roofing Materials: Built-up EPDM Concrete roof tile Copper Shingles, asphalt Shingles, wood Tile, clay Tile, cement ribbed Slate, 3/16” – ¼” Slate, 3/8” – ½” Finish Materials: Gypsum wallboard, ½” Tile, glazed wall 3/8” Quarry tile, ¾” Hardwood flooring, 25/32” Vinyl tile, 1/8” Terrazzo, 1”, 2” in stone conc. Insulation & Waterproofin Waterproofin g: Batt, blankets per 1” thickness Rigid insulation Lecture 22 - Page 1 of 9
Weight Weight (lb. per sq. ft.)
1.5 2 8 12.5 40 2.5 2.5 40 55 85 8 6 26 6.5 9.5 2 2.8 2.5 16 - 20 16 7 - 9.5 14 - 18 2 3 9 4 1.5 25 0.3 1.5
Example 1 GIVEN: The steel-framed floor structure as shown below, to be used as an office office building. The construction is indicated and dead loads can be found from from the table above. The floor-to-floor height = 12’-0” REQUIRED: Determine the total dead load of the floor floor construction on a pounds-per-square foot basis.
32’-0”
6” dia. std. wt. pipe col (typ.)
W18x35
4 9 x 4 2 W
W18x35 W18x35
4 9 x 4 2 W
W18x35
” 0 ’ 1 2 = ” 0 ’ 7 @ 3
Floor Floor construction: 4” conc. over 1½” metal deck ¾” quarry tile floor finish Partitions - 2x4 metal stud walls w/ ½” GWB both sides Acoustical hung ceiling below beams Mechanical/Electrical allowance = 5 sf
Add – up all superimposed dead loads as follows: 4” concrete slab 4” @ 12.5 psf/inch = 50 psf 1½” metal deck……….…….………..= 2.5 psf ¾” quarry tile ………….……………..= 9 psf Partitions……………….……………. Partitions……………….… …………. = 20 psf (per 1607.5 of the IBC) Acoustical hung ceiling…..………….= 1.5 1.5 psf Mechanical/Electrical……..…………= Mechanical/Electrical……..………… = 5 psf Sub-total = 88 psf Determine the dead load of struct ural steel steel beams and and col umns : 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 13.9 ps f
→
Ad ded to get her , th e to tal Dead L oad = 88 ps f + 13.9 p sf = 101.9 ps f
Lecture 22 - Page 2 of 9
Floor Live Loads
From the IBC, a Live Load is defined as “Those loads produced by the use and occupancy of the building or other structure and do not include construction or environmental 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, moveable equipment, and anything else that does not remain permanently stationary. IBC Section 1607 specifies prescribed minimum live loads. Table 1607.1 lists these prescribed minimum minimum live loads based upon anticipated anticipated occupancy. The architect or engineer-of-record is free to INCREASE these loads as he/she deems necessary – however these loads CANNOT BE DECREASED except under Section 1607.9 where a formula is given that may be used to reduce the live loads. L = L0 (0.25 +
15 K LL A T
)
when KLL AT > 400 ft 2 where: L = Reduced design live load per square foot of area supported by the member L0 = Unreduced design live load per square foot per Table 1607.1 < 100 psf KLL = Live load element factor per Table 1607.9.1 AT = Tributary area in square feet Example 2 GIVEN: The same floor floor system as shown in Example 1. REQUIRED: Determine the reduced live load, L, (if applicable) for the design of the 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 beams Trib. Area AT = (7’)(32’) = 224 ft 2 KLL AT = (2)(224 ft 2) = 448 ft 2 > 400 ft 2 → live load reduction allowed. L = L0 (0.25 +
L = 50 psf (0.25 +
15 K LL A T
15 448
)
) = 47.9 47.9 ps f
Lecture 22 - Page 3 of 9
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. slab Trib. Area AT = ½(32’)(21’) = 336 ft 2 KLL AT = (2)(336 ft 2) = 672 ft 2 > 400 ft 2 → live load reduction allowed. 15
L = L0 (0.25 +
L = 50 psf (0.25 +
K LL A T
15 672
)
) = 41.4 41.4 ps f
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. slab Trib. Area AT = ¼(32’)(21’) = 168 ft2 KLL AT = (4)(168 ft 2) = 672 ft 2 > 400 ft 2 → live load reduction allowed. 15
L = L0 (0.25 +
L = 50 psf (0.25 +
K LL A T
15 672
)
) = 41.4 41.4 ps f
Handrail Loads
Section 1607.7 dictates loads on handrails, guards, grab bars and vehicle barriers. These loads must be carried carried throughout the entire assembly and into the supporting structure. In particular, the minimum design loads on handrails (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
Lecture 22 - Page 4 of 9
Roof Live Loads
In general, design loading on roofs comes from snow. However, in areas where snow is not extreme, the minimum prescribed roof live load is: L r = L o R1R2 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 A T < 200 ft 2 = 1.2 – 0.001(AT) for 200 ft 2 < 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 3 GIVEN: The flat-roof flat-roof framing framing plan from the the previous examples. Assume the building is to be located in southern Florida. REQUIRED: Determine the minimum design roof live load, Lr for for the interior W18x35 filler beam, the exterior girder and the corner columns.
a) From IBC Table 1607.1 → use roof live load L o = 20 psf b) Interior filler beam: The tributary area, A T = 7’(32’) = 224 ft 2
R2 = 1
R1 = 1.2 – 0.001A T = 1.2 – 0.001(224 ft 2) = 0.976
→
since the roof is flat, F < 4
→
Lr = LoR1R2 = 20psf(0.976)(1) L r = 19.5 19.5 psf
Lecture 22 - Page 5 of 9
c) Exterior girder: Tributary area , AT = ½(32’)(21’) = 336 ft 2 → R1 = 1.2 – 0.001A T = 1.2 – 0.001(336 ft 2) = 0.864 R2 = 1
since the roof is flat, F < 4
→
Lr = LoR1R2 = 20psf(0.864)(1) L r = = 17.3 17.3 psf
d) Corner column: Tributary area , AT = ¼(32’)(21’) = 168 ft 2 < 200 ft 2 → R1 = 1.0 Lr = = LoR1R2 = 20psf(1)(1) L r = 20 psf ps f
Lecture 22 - Page 6 of 9
Load Combinations
Buildings and other structures and portions thereof shall be designed to resist the most crit ical effects effects of combinations of loads in accordance with either the Allowable Stress Design Method or the Load and Resistance Factor Design (LRFD) Method (or referred to as the “Strength” method) as prescribed in Section 1605.1: Al lo wab le St res s Des ig n Met ho d: D D+L D + L + (L r or S or R) D + (W or 0.7E) + L + (L r or S or R) 0.6D + W 0.6D + 0.7E
where: D = Dead loads L = Live loads Lr = = Roof live loads S = Snow loads E = Earthquake (seismic) loads W = Wind loads R = Rain loads
Load & Resistance Facto Facto r Desig Desig n Method Method :
1.4D 1.2D + 1.6L + 0.5(Lr or or S or R) 1.2D + 1.6(Lr or S or R) + (f 1L or 0.8W) 1.2D + 1.6W + f 1L + 0.5(Lr or or S or R) 1.2D + 1.0E + f 1L + f 2S 0.9D + (1.0E or 1.6W) f 1 = 1.0 for floors in places of public assembly, for live loads > 100 psf, and for parking garage live load f 1 = 0.5 for other live loads f 2 = 0.7 for roof configurations that do not shed snow off the structure f 2 = 0.2 for other roof configurations
Lecture 22 - Page 7 of 9
Example 4 GIVEN: The roof framing plan plan as shown below. Loads are indicated indicated as follows:
25’-0”
6” dia. std. wt. pipe col (typ.)
W14x22
2 6 x 4 2 W
W16x26
2 6 x 4 2 W
W16x26 W14x22
• • • • • •
” 0 ’ 8 1 = ” 0 ’ 6 @ 3
Superimposed roof roof dead load “D” “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 Stress Design” methodology. Do not consider reduction in live live load. 2) Determine the maximum moment on the W16x26 beam using the maximum 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 Lecture 22 - Page 8 of 9
Make a Table as shown: Load Comb. D D+L D + L + (L r or S or R) D + (W or 0.7E) + L + (L r or S or R)
Unifor m Load (PLF): (PLF): 128 plf 128 plf + 0 = 128 plf 128 plf + 0 + (228 plf) = 356 plf
Lecture 22 - Page 9 of 9
