Dead Load Weight of Roof Trusses. Dead Load Weight of Floor Trusses

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2 Dead Load Weight of Roof Trusses Top Chord Bottom Chord Webs Weight per Lineal Foot (plf) 2x4 2x4 2x x4 2x8 2x x6 2x4 2x x6 2x6 2x x6 2x6 2x x6 2x8 2x x6 2x10 2x x6 2x10 2x x8 2x8 2x x8 2x8 2x x8 2x10 2x8 8.4 Dead Load Weight of Floor Trusses Depth in Inches Top Chord Bottom Chord Weight per Lineal Foot (plf) 12 Single Single Double Single Double Double Single Single Double Single Double Double Single Single Double Single Double Double 7.0

3 Problem 1 Given: Roof system consisting of the following: Asphalt shingles 30 lb. Felt 7/16 in. OSB sheathing 30 ft Fink trusses, 6/12 pitch and 2x4 chords and webs, 2' o.c. 6 in. of fiberglass insulation at ceiling (use.045 psf per inch) 5/8 in. gypsum board ceiling 30 psf snow load a) Determine the weights of the materials b) Identify the top chord and bottom chord dead load and live loads. c) Sketch the loading diagram for this truss (TCLL, TCDL, BCLL, BCDL). d) Calculate the end reactions for this truss assuming no overhangs.

4 Problem 2 Given: A floor joist is to be designed to meet a live load deflection criteria of L/360 and a total load deflection of L/240. The joist specified is a 2x10 DF/L #1. The span is 15 feet. The Uniform Design Live Load is 78 plf and the Uniform Design Dead Load is 30 plf. a) What is the actual live load deflection of the member and does it meet the design criteria? b) What is the total load deflection of the member and does it meet the design criteria?

5 Problem 3 Given: 20 ft. long, 18 in. deep floor trusses at 24" o.c. with single 4x2 chords designed to support the following loads: Floor live load = 40 psf Floor dead load = 20 psf During construction, a 20-piece stack of 5/8 in. by 4'x 12' gypsum board is stored near the mid-span of two of the trusses with the long dimension parallel to the trusses. At the time, the trusses are sheathed with ¾ in. T&G plywood and support approximately 1.5 psf of mechanicals and ductwork. a) Which is the greater load on the trusses, construction load or design load? b) If these truss materials have values of E= 1,700,000 psi and I = in 4 what is the expected maximum deflection of a typical truss under each load condition?

6 Problem 4 Given: An air conditioning unit weighing 2500 lbs. is to be supported by 5 trusses in the roof of a restaurant. The AC unit will rest on curbing, which runs perpendicular to the trusses and is spaced 4 ft apart. a) Based on the provisions of Section of ASCE 7, how much should the weight of the AC unit be increased? b) Assuming that the AC unit is supported equally by each truss, how much load should each truss be designed to support?

7 Problem 5 Given: A building is to be designed using the following loads: D = Dead Load = 20 psf LDF = L = Live Load = 40 psf L r = Roof Live Load = 20 psf S = Snow Load = 40 psf W = Wind Load = 22 psf E = Earthquake Load = 20 psf LDF = LDF = LDF = LDF = LDF = a) Fill in the Load Duration Factor (LDF) for each of these loads. b) Calculate the load for these roof load combinations: D + L r = D + S = D + W = D + E = c) Which is the load duration factor for the highest load combination in part b? LDF =

8 Bonus Problem 6 (not presented in course material, answer on following page) Given: A series of 30-ft, 2 by 4 Fink roof trusses spaced at 2 ft on center with 6/12 pitch are to be designed for use in a residence. The construction documents also provide the following information: 30 psf design roof snow load heavy clay tile roofing, no mortar 30 lb. roofing paper ⅝ inch plywood decking under tile roofing. 12 in. of fiberglass insulation at the ceiling (use psf/in) ½ inch drywall ceiling ½ psf ceiling load for A/C ducts, plumbing, and wiring The attic space is considered uninhabitable, but will be used for storage. a) The top and bottom chord dead load. b) The top chord live (include snow load as a live load) and bottom chord live load. c) Determine the total load and proper load duration. d) Sketch the loading diagram and calculate the reactions

9 Problem 6 - Answer a) The top and bottom chord dead loads: TCDL = Tile = 20.0 psf Felt =.30 psf Ply = 2.0 psf Total = 22.3psf x slope factor of = ½ truss = 1.1 psf Total = psf x tw of 2' = plf use 52 plf BCDL = ½ truss = 1.1 psf Insul =.54 psf Gyp = 2.2 psf Mech = 0.5 psf Total = 4.34 psf x tw of 2' = 8.68 plf use 9 plf b) Top and bottom chord Live Loads: TCLL = 30 psf (snow load controls over roof live load (L r ) Per ASCE 7, Section 4.9 when evaluating the Roof Live Load (L r ) it might be able to be reduced. L r = 20R 1 R 2 where 12 L r 20 R 1 = 1 since A t = 2'x30' = 60 ft2 < 200 R 2 = (6) = 0.90 L r = 20(1)(.90) = 18 psf BCLL = 20 psf c) Total Load & proper load duration Per ASCE 7, Section 2.4, the basic load combinations that apply are: DL DL + LL + SL DL = TCDL + BCDL = 26 psf psf = 30.5 psf Load duration factor = 0.9 (permanent) DL + LL + SL = = 80.5 psf Load duration factor = 1.15 (Snow) TCLL = 60 plf (snow load) TCDL = 52 plf Total TC = 112 plf BCLL = 40 plf BCDL = 9 plf Total BC = 49 plf R1 = R2 = = 2415 lb 2 Total = 161 plf

10 Bonus Problem 7 (not presented in course material, answer on following page) Given: An 18 inch deep truss 22 feet long is used in the floor of an office project. The floor assembly consists of linoleum tile over a hydronic floor heating system with 1-½ inch light weight concrete (assume hydronic heating system = weight of the lightweight concrete it replaces) and ¾ inch plywood subflooring. The ceiling attached to the bottom chord is ½ inch drywall and there is 6 inches of fiberglass insulation above the ceiling (use psf for insulation). Assume that the truss is spaced at 2 foot centers and has double 4x2 top and bottom chords. Assume a 1.0 pounds per square foot ceiling load for A/C ducts, plumbing, and wiring. Assume a 20 psf floor load for partitions. a) The top chord and bottom chord dead load. b) The top chord live and bottom chord live load. c) Draw the loading diagram and calculate the reactions. d) What would the live load be if this truss supported a floor in office lobby?

11 Bonus Problem 7 Answer a) The top chord and bottom chord dead load: TCDL = ½ truss = 1.65 psf (6.6 plf/2 = 3.3 psf /2) linoleum = 1.00 psf per ¼" 1½" concrete = 12.0 psf (8 psf/inch = 1.5x8) ply = 2.40 psf (0.4/1/8") Total = psf TCDL = 17 psf BCDL = ½ truss = 1.65 psf (6.6 plf/2 = 3.3 psf /2) ½" gypsum = 2.00 psf (0.55/1/8") 6" fiberglass = 0.27 psf (.045psf/inch) mechanicals = 1.00 psf Total = 4.92 psf` BCDL = 5 psf b) The top chord live and bottom chord live load: TCLL = Office = 50.0 psf (per ASCE 7, Table 4-1) Partitions = 20.0 psf (per ASCE 7, Sec , using given value) Total = 70.0 psf TCLL = 70 psf Note: Safe load per Table 4-1 specifies a 2000 lb over a 2.5ft x 2.5ft area. For trusses at 2 ft o.c. this translates into a uniform load of 640 plf across 2.5 ft of the span. The truss is analyzed for each possible location of the 640 plf load along the span. BCLL = 0 psf No storage available in floor trusses. BCLL = 0 psf c) Draw the loading diagram and calculate the reactions. Case 1 LL + DL 140 plf TCLL 34 plf TCDL 10 plf BCDL d) What would the live load be if this truss supported a floor in office lobby? Case 2 DL + Concentrated Load 2000# concentrated load randomly applied psf TCDL 4.92 psf BCDL Deflection criteria depends upon building designer, minimum = L/360 Office lobby floor live load = 100 psf per ASCE 7 Table 4-1