# Timber Design Solution

Topics: Dead and live loads, Flat roof, Beam Pages: 32 (5182 words) Published: October 10, 2012
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)
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 along roof slope, or (2.0 psf)(12.37/12)=2.1 psf on horizontal plane.]

2x4 @ 16 in. o.c.
Gypsum wallboard (½ in.)

= 10.0
= 0.9
= 2.5
= 13.4

psf
psf
psf
psf

D = (13.4 psf) (8 ft) = 107.2 lb/ft
d) R1 = 1.0 since not considering tributary area
R2 = 1.0 for slope less than 4:12
Basic Roof Live Load: Lr = 20 R1 R2 = 20 psf
e) R1 = 1.0 since not considering tributary area
R2 = 1.0 for slope of 4:12
Basic Roof Live Load: Lr = 20 R1 R2 = 20 psf
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Chapter 2 Solutions
Page 2 of 19

Problem 2.2
a) See Appendix A and Appendix B for weights of concrete roof tiles, lumber sheathing, and framing.
Concrete tile
(950 lb / 100 ft2) = 9.5
15/32 in. structural panel (plywood) sheathing (15/32) (3.0 psf/in.) = 1.4 2x8 @ 16 in. o.c.
= 1.9
= 12.8
Convert D to load on a horizontal plane:

psf
psf
psf
psf

Roof slope = 6:12
Hypotenuse = (36 + 144)½ = 13.42
Don horizontal plane = (12.8 psf) (13.42/12) = 14.3 psf

b) See Appendix A and Appendix B for weights of framing, insulation, gypsum lath and plaster. Fiberglass loose insulation
2x6 @ 16 in. o.c.
Gypsum wallboard

(10 in.) (0.5 psf/in)
(½ in.) (5 psf/in)

=
=
=
=

5.0
1.4
2.5
8.9

psf
psf
psf
psf

c) roof slope = 6:12 (F = 6)
R1 = 1.0 since not considering tributary area
R2 = 1.2 – 0.05 F = 1.2 – (0.05)(6) = 0.9
Basic Roof Live Load: Lr = 20 R1 R2 = 18 psf
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Chapter 2 Solutions
Page 3 of 19

Problem 2.3
a) See Appendix A and Appendix B for weights of roofing, sheathing, and suspended ceiling. Built-up roof (5 ply w/ gravel)
½ in. plywood sheathing
(½ in.) (3.0 psf/in)
Roof trusses @ 24 in. o.c.
(9 lb/ft)÷(2 ft)
Suspended acoustic ceiling: Acoustical fiber tile
Suspended acoustic ceiling: Channel-suspended system

= 6.5
= 1.5
= 4.5
= 1.0
= 1.0
= 14.5

psf
psf
psf
psf
psf
psf

b) See Appendix A and Appendix B for weights of framing, sheathing, and suspended ceiling. Concrete
(150 lb/ft3) (0.125 ft)
2x10 @ 16 in. o.c.
5/8 in. plywood sheathing
(5/8 in.) (3.0 psf/in)
Air duct
Suspended acoustic ceiling: Acoustical fiber tile
Suspended acoustic ceiling: Channel-suspended system

= 18.8
= 2.4
= 1.9
= 0.5
= 1.0
= 1.0
= 25.6

psf
psf
psf
psf
psf
psf
psf

c) roof slope = 0.25:12
R1 = 1.0 since not considering tributary area
R2 = 1.0 for roof slope less than 4 in 12
Basic Roof Live Load: Lr = 20 R1 R2 = 20 psf
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Chapter 2 Solutions
Page 4 of 19

Problem 2.4
a) See Appendix A and Appendix B for weights of roofing, sheathing, and subpurlins. Assume Douglas-Fir/Larch (G = 0.5) at 12% m.c. for 4x14 purlin and 6.75x33 glulam girder. Using density formula from NDS Supplement:

⎤ ⎡ m.c. ⎤
G
3
density = 62.4⎢
⎥ ⎢1 + 100 ⎥ = 33 lb/ft

⎣1 + G (0.009 )(m.c.) ⎦ ⎣
[Note that 33 lb/ft3 is a reasonable (and typically conservative)...