Homework #3 ( Due September 24)

David Wagner 2007/09/23 12:53

Solve problems 4.3, 4.4, 4.8, 4.12, 4.14, 4.16, 4.17, 4.20, and 4.23 in your textbook.

Problem 4.3

  • a. Moisture Content: The percentage of water to oven-dry weight, MC={{moist weight - oven dry weight}/{oven dry weight}}*100 percent
  • b. Fiber Saturation Point: This is the moisture content with all of the free water removed and all of the bound water remaining.
  • c. Equilibrium Moisture Content: the average moisture content the lumber has in service.

Problem 4.4

Moisture Content Ranges

  • a. Dry Lumber: 19% or less,
  • b. Green Lumber: over 19%

Problem 4.8

Short Axis X-XLong Axis Y-Y
Dimension (inxin)DressedArea (in²)Section Modulus (in³)ISection Modulus (in4)I
a. 2×4 1½x3½ 5.25 3.063 5.359 1.313 0.984
b. 8×8 7½x7½ 56.25 70.31 263.7 70.31 263.7
c. 4×10 3½x9¼ 32.38 49.91 230.8 18.89 33.05
d. 6×16 5½x15½ 85.25 220.2 1707 78.15 214.9

Design Values for Wood Construction, 2005 Table 1B

Problem 4.12

DimensionClassNDS Supplement Table
a. 10×12 P&T 4D
b. 14×14 P&T 4D
c. 4×8 Dimension Lumber 4A
d. 4×4 Dimension Lumber 4A
e. 2×12 Dimension Lumber 4A
f. 6×12 B&S 4D
g. 8×12 B&S 4D
h. 8×10 P&T 4D

Problem 4.14

Grades for Visually Graded Hem-Fir

  • a. Dimension Lumber: Select Structural, No.1 & Better, No.1, No.2, No.3, Stud, Construction, Standard, Utility;Table 4A
  • b. Beams and Strings (B&S): Select Structural, No.1, No.2; NDS Supplement Table 4D
  • c. Posts and Timbers (P&T): Select Structural, No.1, No.2; NDS Supplement Table 4D

Problem 4.16

Reference design values for No.1 DF-L (psi)

SizeFbFtFvFc⊥FcEEminRulesTable
a. 10×10 1,200 825 170 625 1,000 1,600,000 580,000WCLIB,WWPA4D
b. 12×14 1,200 825 170 625 1,000 1,600,000 580,000WCLIB,WWPA4D
c. 4×16 1,000 675 180 625 1,500 1,700,000 620,000WCLIB,WWPA4A
d. 4×4 1,000 675 180 625 1,500 1,700,000 620,000WCLIB,WWPA4A
e. 2×10 1,000 675 180 625 1,500 1,700,000 620,000WCLIB,WWPA4A
f. 6×12 1,350 675 170 625 925 1.600,000 580,000WCLIB,WWPA4D
g. 6×8 1,350 675 170 625 925 1.600,000 580,000WCLIB,WWPA4D
h. 10×14 1,350 675 170 625 925 1.600,000 580,000WCLIB,WWPA4D

Problem 4.17

f_Xn = K_F F_x

Size FbnFtnFvnFc⊥nFcnEnEmin-n
KF2.16/φ2.16/φ2.16/φ1.875/φ2.16/φ1.5/φ
φ0.850.800.750.900.900.85
a. 10×10 2203 1426 275 1055 1944 1600000 739500
b. 12×14 2203 1426 275 1055 1944 1600000 739500
c. 4×16 1836 1166 292 1055 2916 1700000 790500
d. 4×4 1836 1166 292 1055 2916 1700000 790500
e. 2×10 1836 1166 292 1055 2916 1700000 790500
f. 6×12 2479 1166 275 1055 1798 1600 739500
g. 6×8 2479 1166 275 1055 1798 1600 739500
h. 10×14 2479 1166 275 1055 1798 1600 739500

Problem 4.20

The load duration factor CD is the ASD factor used to account for the effects of load durations differing from those used to find the NDS table values.

while the time effect factor λ is used much as φ is used, to adjust the LRFD design resistance.

Problem 4.23

The reference design values Fc⊥ and E are not subject to adjustment for duration of loading by ASD.

The LRFD time effect Factor λ applies to load combinations and not to reference design values.


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