Type:
Solution Manual
Resource:
Structural Wood Design
Edition:
2nd Edition
Author(s):
Abi Aghayere Jason Vigil
Chapter 1
1-10
Solutions
S1-1
Determine the total shrinkage across the width and thickness of a green triple 2x4 Douglas Fir Larch top plates loaded perpendicular to grain as the moisture content decreases from an initial value of 30% to a final value of 12%.
For a 2 x 4 sawn lumber, the actual thickness = 1.5 in. For a 2 x 4 sawn lumber, the actual width, d1 = 3.5 in. M1 = 30 and M2 = 12 (a) Shrinkage across the width of the 2 x 4 continuous blocking: The shrinkage parameters from Table 1-3 for shrinkage across the width of the 2x4 are a = 6.031, and b = 0.215 The final width d2 is given as, 6.031- 0.215 (12) 1 100 d2 = 3.5 6.031- 0.215(30) 1 100
= 3.365 in.
Thus, the total shrinkage across the width of the (triple) 2x4 is d1 – d2 = 3.5 in. – 3.365 in. = 0.135 in. (b) Shrinkage across the thickness of the (triple) 2x 4 top plates: The shrinkage parameters from Table 1-3 for shrinkage across the thickness of the 2x4 plates are a = 5.062, and b = 0.181
The final thickness d2 of each plate is given as, 5.062 - 0.181(12) 1 100 d2 =1.5 5.062 - 0.181(30) 1 100
= 1.451 in.
The total shrinkage across the thickness of the triple top plates is the sum of the shrinkage in each of the individual wood member calculated as 3 plates x (d1 – d2) = 3(1.5 in. – 1.451 in.) = 0.147 in.
Chapter 1
1-11
Solutions
S1-2
Determine the total shrinkage over the height of a 2-story building with the exterior wall cross-section shown below as the moisture content decreases from an initial value of 25% to a final value of 12%.
For a 2 x 6 sawn lumber, the actual thickness = 1.5 in. For a 2 x 12 sawn lumber, the actual width, d1 = 11.25 in. M1 = 25 and M2 = 12 (a) Shrinkage across the width of the 2 x 12 header joist: The shrinkage parameters from Table 1-3 for shrinkage across the width of the 2x12 are a = 6.031, and b = 0.215 The final width d2 is given as, 6.031- 0.215 (12) 1 100 d2 =11.25 6.031- 0.215(25) 1 100
= 10.93 in.
Thus, the total shrinkage across the width of the 2- 2x12 header joist is 2(d1 – d2)= 2(11.25 in. – 10.93 in.) = 0.64 in. (b) Shrinkage across the thickness of a 2x 6 top plate: The shrinkage parameters from Table 1-3 for shrinkage across the thickness of the 2x6 plates are a = 5.062, and b = 0.181
The final thickness d2 of each plate is given as, 5.062 - 0.181(12) 1 100 d2 =1.5 5.062 - 0.181(25) 1 100
= 1.465 in.
In the section shown in Figure 1.24, there are a total of 7- 2x6 top and sole or sill plates, and 2 2x12 header joists. The total shrinkage of the building cross-section will be the sum of the shrinkage across the thickness of the 2x6 top and sole/sill plates plus the shrinkage across the width of the 2x12 header joists. That is,
Chapter 1
Solutions
S1-3
7, 2x6 plates x (d1 – d2) = 7(1.5 in. – 1.465 in.) = 0.245 in.
The longitudinal shrinkage or shrinkage parallel to grain in the 2x6 wall studs is negligible. Therefore, the total shrinkage across over the height of the two-story building, which is the sum of the shrinkage of all the wood members at the floor level, is 0.245 in. + 0.64 in. = 0.89 in.
1-12
How many board feet (bf) are there in a 4 x 16 x 36 ft long wood member? How many Mbf are in this wood member? Determine how many pieces of this wood member would amount to 4.84 Mbf or 4840 bf?
# of bf = 4x16x(36’x12)/144 = 192 bf = 192/1000 Mbf = 0.192 Mbf # of pieces = 4840 bf/192 bf = 25.2 or 25 pieces
Chapter 2
2-1.
Solutions
S2-1
Calculate the total uniformly distributed roof dead load in psf of horizontal plan area for a sloped roof with the design parameters given below. • • • • • •
2x8 rafters at 24” on centers Asphalt shingles on ½” plywood sheathing 6” insulation (fiberglass) Suspended Ceiling Roof slope: 6-in-12 Mechanical & Electrical (i.e. ducts, plumbing etc) = 5 psf
Solution: 2x8 rafters at 24” on-centers = 1.2 psf Asphalt shingles (assume ¼” shingles) = 2.0 psf ½” plywood sheathing = (4 x 0.4 psf/1/8” plywood) = 1.6 psf 6” insulation (fiberglass) = 6 x 1.1 psf/in. = 6.6 psf Suspended Ceiling = 2.0 psf Mechanical & Electrical (i.e. ducts, plumbing etc) = 5.0 psf Total roof dead load, D (psf of sloped roof area) = 18.4 psf The total dead load in psf of horizontal plan area will be: 2 2 w DL = D 6 +12 , psf of horizontal plan area 12
=18.4 psf (1.118 ) = 20.6 psf of horizontal plan area
2-2.
Given the following design parameters for a sloped roof, calculate the uniform total load and the maximum shear and moment on the rafter. Calculate the horizontal thrust on the exterior wall if rafters are used. • • • • •
Roof dead load, D= 20 psf (of sloped roof area) Roof snow load, S = 40 psf (of horizontal plan area) Horizontal projected length of rafter, L2 =14 ft Roof slope: 4-in-12 Rafter or Truss spacing = 4’ 0
Solutions: 2 2 Sloped length of rafter, L1 = 4 +12 (14 ft ) =14.8 ft 12