Determine the ASD adjusted design compression strength parallel to grain, Fc’, for the following beam. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:wD = 240 lb/ftwLr = 210 lb/ftLoad combination:D + LrSpan:L = 13 ftMember size:4 x 8Stress grade and species:No. 2 Douglas Fir-LarchUnbraced length:lu = 0Moisture content:MC < 19 percentLive load deflection limit:Allow. Δ ≤ L/360

Determine the ASD adjusted design bending strength, Fb’, for the following beam. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:PD = 480 lbPLr = 2,080 lbLoad combination:D + LrSpan:L = 7 ftMember size:4 x 10Stress grade and species:No. 2 Douglas Fir-LarchUnbraced length:lu = 0Moisture content:MC < 19 percentLive load deflection limit:Allow. Δ ≤ L/360

A wood member is loaded as shown. Using ASD, determine the adjusted bending strength, Fb’. Assume normal temperatures, no incising, and that all loads act in the directions shown. Ignore the weight of the member.Load:PD = 1,500 lbPL = 0 lbPLr = 0 lbPS = 0 lbPR = 0 lbPW = 5,500 lbPE = 6,000 lbQD = 1,000 lbQL = 4,000 lbQLr = 4,000 lbQS = 500 lbQR = 1,000 lbQW = 0 lbQE = 0 lbSpan:L = 8 ft Member size:4 x 10 Stress grade and species:Select Structural Douglas Fir-Larch Unbraced length:lu = 0 ft Moisture content:MC < 19 percent 

Determine the maximum actual shear stress in the following beam. Do not reduce the shear based on NDS Section 3.4.3. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:wD = 200 lb/ftwLr = 270 lb/ftLoad combination:D + LrSpan:L = 10 ftMember size:4 x 12Stress grade and species:No. 1 Douglas Fir-LarchUnbraced length:lu = 0Moisture content:MC < 19 percentLive load deflection limit:Allow. Δ ≤ L/360

Determine the ASD adjusted minimum modulus of elasticity, Emin’, for the following beam. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:wD = 180 lb/ftwLr = 300 lb/ftLoad combination:D + LrSpan:L = 14 ftMember size:4 x 14Stress grade and species:No. 2 Douglas Fir-LarchUnbraced length:lu = 0Moisture content:MC > 19 percentLive load deflection limit:Allow. Δ ≤ L/360

Determine the maximum actual shear stress in the following beam. Do not reduce the shear based on NDS Section 3.4.3. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:PD = 480 lbPLr = 2,080 lbLoad combination:D + LrSpan:L = 8 ftMember size:4 x 12Stress grade and species:No. 1 Douglas Fir-LarchUnbraced length:lu = 0Moisture content:MC < 19 percentLive load deflection limit:Allow. Δ ≤ L/360

For glulams, the wet service factor would be used to determine which of the following adjusted design values? Select all that apply.

What is the wet service factor, CM, for the modulus of elasticity of softwood glulams?

A 3 x 8 rafter frames into a 2 x 8 stud wall. The load is a combination of PD = 120 lb and PS = 510 lb. Consider the D + S load combination. Determine the ASD actual bearing stress fc⊥ in the top plate of the wall.

Determine the ASD adjusted modulus of elasticity, E’, for the following beam. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:PD = 200 lbPLr = 2,080 lbLoad combination:D + LrSpan:L = 14 ftMember size:4 x 14Stress grade and species:No. 2 Douglas Fir-LarchUnbraced length:lu = 0Moisture content:MC > 19 percentLive load deflection limit:Allow. Δ ≤ L/360