Essay Timber Design Solution

5184 WordsOct 21, 201221 Pages
Chapter 2 Solutions Page 1 of 19 Problem 2.1 a) See Appendix A and Appendix B for weights of roofing, sheathing, framing, insulation, and gypsum wallboard. Asphalt shingles 3/8 in. plywood sheathing (3/8 in.) (3.0 psf/in) 2x6 @ 16 in. o.c. Fiberglass loose insulation (5.5 in.) (0.5 psf/in) Gypsum wallboard (1/2 in.) (5.0 psf/in) Roof Dead Load (D) along roof slope Convert D to load on a horizontal plane: = 2.0 psf = 1.1 psf = 1.4 psf = 2.75 psf = 2.5 psf = 9.75 psf Roof slope = 3:12 Hypotenuse = (9 + 144)½ = 12.37 Don horizontal plane = (9.75 psf) (12.37/12) = 10.1 psf [NOTE that this does not include an allowance for weight of re-roofing over existing roof. For each additional layer of shingles add 2.0 psf…show more content…
= (density)(cross-sectional area)/(width of tributary area) Glulam girder s.w. (33)(6.75/12)(33/12)/(20) = 2.55 psf 4x14 purlin s.w. (33)(3.5/12)(13.25/12)/(8) = 1.33 psf Built-up roof (5 ply w/o gravel) = 2.50 psf 15/32 in. plywood sheathing (15/32 in.) (3.0 psf/in) = 1.41 psf 2x4 @ 24 in. o.c. = 0.6 psf Average Dead Load of entire Roof = 8.4 psf (NOTE that this does not include an allowance for weight of re-roofing over existing roof.) b) Subpurlin Dead Load: (2.50 + 1.41 + 0.6 psf) (2 ft) = 9.0 lb/ft c) Purlin Dead Load: (2.50 + 1.41 + 0.6 + 1.33 psf) (8 ft) = 46.7 lb/ft d) Girder Dead Load: (2.50 + 1.41 + 0.6 + 1.33 + 2.56 psf) (20 ft) = 168 lb/ft e) Column Dead Load: (2.50 + 1.41 + 0.6 +1.33 + 2.56 psf) (20 ft) (50 ft) = 8400 lb f) R1 = 1.0 since not considering tributary area R2 = 1.0 for slope less than 4 in 12 Basic Roof Live Load: Lr = 20 R1 R2 = 20 psf g) R = 5.2(ds + dh) = 5.2 (5 in. + 0.5 in.) = 28.6 psf ______________________________________________________________________ Chapter 2 Solutions Page 5 of 19 Problem 2.5 a) Tributary area (on a horizontal plane): AT = (20 ft)(13 ft) = 260 ft2 > 200 ft2 b) R1 = 1.2 – 0.001 AT = 0.94 R2 = 1.0 for slope less than 4 in 12 Roof Live Load: Lr = 20 R1 R2 = 18.8 psf wLr
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