Lecture 22 –
Lecture 22 – De
Dead Loads, Live Loads
ad Loads, Live Loads & Lo
& Lo ad C
ad Comb
omb ination
ination s
s
DeDead Loadsad Loads
Dead loads include the
Dead loads include the weight of the weight of the physicphysic al al strucstruc tureture and the non-movable and the non-movable materials and objects att
materials and objects attached to the structure. ached to the structure. They are considered to beThey are considered to be permanent loads.
permanent loads. Building codes are of Building codes are of little use in little use in the determination of the determination of deaddead loads – they
loads – they must be hand-calculated. must be hand-calculated. Many references exist Many references exist tabulating thetabulating the typical weights of building materials, such as the Architectural Graphic
typical weights of building materials, such as the Architectural Graphic Standards, AISC Manual, et
Standards, AISC Manual, etc. c. Typically, dead loads are determTypically, dead loads are determined on ained on a “pounds per square foot” basis.
“pounds per square foot” basis. Ma
Materials: terials: WeWeight ight (lb. (lb. per per sq. sq. ft.)ft.) Ceilings:
Ceilings:
Channel
Channel suspended suspended acoustical acoustical 1.51.5 ½”
½” gypsum gypsum drywall drywall 22 Plaster
Plaster & & lath lath 88
Flooring: Flooring:
Concrete,
Concrete, normal normal weight weight per per 1” 1” thickness thickness 12.512.5 Precast
Precast concrete, concrete, 6” 6” plank, plank, no no topping topping 4040 ¾”
¾” plywood plywood subfloor subfloor 2.52.5 Steel
Steel decking, decking, 1½” 1½” 2.52.5
Walls & Partitions: (per height of wall) Walls & Partitions: (per height of wall)
4”
4” brick brick 4040
8”
8” concrete concrete block block CMU CMU 5555 12”
12” concrete concrete block block CMU CMU 8585 2x4
2x4 wood wood stud stud w/ w/ ½” ½” GWB GWB both both sides sides 88 4”
4” metal metal stud stud w/ w/ ½” ½” GWB GWB both both sides sides 66 4”
4” lightweight lightweight CMU CMU block block w/ w/ ½” ½” GWB GWB both both sides sides 2626
Roofing Materials: Roofing Materials:
Built-up
Built-up EPDM EPDM 6.56.5
Concrete
Concrete roof roof tile tile 9.59.5
Copper 2
Copper 2
Shingles,
Shingles, asphalt asphalt 2.82.8
Shingles,
Shingles, wood wood 2.52.5
Tile,
Tile, clay clay 16 16 - - 2020
Tile,
Tile, cement cement ribbed ribbed 1616 Slate, Slate, 3/16” 3/16” – – ¼” ¼” 7 7 - - 9.59.5 Slate, Slate, 3/8” 3/8” – – ½” ½” 14 14 - - 1818 Finish Materials: Finish Materials: Gypsum
Gypsum wallboard, wallboard, ½” ½” 22 Tile,
Tile, glazed glazed wall wall 3/8” 3/8” 33 Quarry
Quarry tile, tile, ¾” ¾” 99
Hardwood
Hardwood flooring, flooring, 25/32” 25/32” 44 Vinyl
Vinyl tile, tile, 1/8” 1/8” 1.51.5 Terrazzo,
Terrazzo, 1”, 1”, 2” 2” in in stone stone conc. conc. 2525
Insulation & Wate
Insulation & Waterproofinrproofin g:g:
Batt,
Batt, blankets blankets per per 1” 1” thickness thickness 0.30.3 Rigid
Example 1
GIVEN: The steel-framed floor structure as shown below, to be used as an office building. The construction is indicated and dead loads can be found from the table above. The floor-to-floor height = 12’-0”
REQUIRED: Determine the total dead load of the floor construction on a pounds-per-square foot basis.
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……….………. = 20 psf (per 1607.5 of the IBC) Acoustical hung ceiling…..………….= 1.5 psf
Mechanical/Electrical……..…………= 5 psf Sub-total = 88 psf
Determine the dead load of struct ural steel beams 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 psf
Ad ded to get her , th e to tal Dead L oad = 88 ps f + 13.9 p sf = 101.9 ps f
W 2 4 x 9 4 W18x35 W18x35 W18x35 W18x35 W 2 4 x 9 4 32’-0” 3 @ 7 ’ - 0 ” = 2 1 ’ - 0 ” 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
6” dia. std. wt. pipe col (typ.)
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 live loads based upon 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. ) 15 25 . 0 ( 0 T LL A K L L= + when KLL AT > 400 ft2 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 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 ft2
KLL AT = (2)(224 ft2) = 448 ft2 > 400 ft2 → live load reduction allowed. ) 15 25 . 0 ( 0 T LL A K L L= + ) 448 15 25 . 0 ( 50 + = psf L = 47.9 ps f
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 ft2
KLL AT = (2)(336 ft2) = 672 ft2 > 400 ft2 → live load reduction allowed. ) 15 25 . 0 ( 0 T LL A K L L= + ) 672 15 25 . 0 ( 50 + = psf L = 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. slab Trib. Area AT = ¼(32’)(21’) = 168 ft2
KLL AT = (4)(168 ft2) = 672 ft2 > 400 ft2 → live load reduction allowed. ) 15 25 . 0 ( 0 T LL A K L L= + ) 672 15 25 . 0 ( 50 + = psf L = 41.4 psf Handrail Loads
Section 1607.7 dictates loads on handrails, guards, grab bars and vehicle barriers. These loads must be 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
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:
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 ft2 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 framing plan from the previous examples. Assume the building is to be located in southern Florida.
REQUIRED: Determine the minimum design roof live load, Lr for the interior W18x35 filler beam, the exterior girder and the corner columns.
a) From IBC Table 1607.1→ use roof live load Lo = 20 psf
b) 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
c) Exterior girder:
Tributary area , AT = ½(32’)(21’) = 336 ft2 → R1 = 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 = ¼(32’)(21’) = 168 ft2 < 200 ft2 → R1 = 1.0 Lr = LoR1R2 = 20psf(1)(1) Lr = 20 psf
Load Combinations
Buildings and other structures and portions thereof shall be designed to resist the most crit ical 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: Load & Resistance Facto r Desig n Method :
D 1.4D D + L 1.2D + 1.6L + 0.5(Lr or S or R) D + L + (Lr or S or R) 1.2D + 1.6(Lr or S or R) + (f 1L or 0.8W) D + (W or 0.7E) + L + (Lr or S or R) 1.2D + 1.6W + f 1L + 0.5(Lr or S or R) 0.6D + W 1.2D + 1.0E + f 1L + f 2S 0.6D + 0.7E 0.9D + (1.0E or 1.6W) where: D = Dead loads L = Live loads Lr = Roof live loads S = Snow loads
E = Earthquake (seismic) loads W = Wind loads
R = Rain loads
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
Example 4
GIVEN: 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 Stress Design” methodology. Do not consider reduction in 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) W16x26 W16x26 W14x22 W 2 4 x 6 2 W14x22 W 2 4 x 6 2 25’-0” 3 @ 6 ’ - 0 ” = 1 8 ’ - 0 ” 6” dia. std. wt. pipe col (typ.)
Make a Table as shown:
Load Comb. Unifor m Load (PLF):
D 128 plf
D + L 128 plf + 0 = 128 plf
D + L + (Lr or S or R) 128 plf + 0 + (228 plf) = 356 plf