Determine the ASD adjusted design shear strength, Fv’, for t…

Determine the ASD adjusted design shear strength, Fv’, for the following beam. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:wD = 160 lb/ftwLr = 420 lb/ftLoad combination:D + LrSpan:L = 11 ftMember size:4 x 10Stress grade and species:No. 1 & Better Douglas Fir-LarchUnbraced length:lu = 0Moisture content:MC < 19 percentLive load deflection limit:Allow. Δ ≤ L/360

Determine the maximum actual bending stress in the following…

Determine the maximum actual bending stress in the following beam. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:wD = 160 lb/ftwLr = 300 lb/ftLoad combination:D + LrSpan:L = 6 ftMember size:4 x 12Stress grade and species:Select Structural Douglas Fir-LarchUnbraced length:lu = 0Moisture content:MC > 19 percentLive load deflection limit:Allow. Δ ≤ L/360

Determine the ASD adjusted design compression strength perpe…

Determine the ASD adjusted design compression strength perpendicular 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 = 260 lb/ftwLr = 240 lb/ftLoad combination:D + LrSpan:L = 10 ftMember size:4 x 6Stress 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 modulus of elasticity, E’, for th…

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:wD = 140 lb/ftwLr = 390 lb/ftLoad combination:D + LrSpan:L = 12 ftMember size:4 x 6Stress grade and species:No. 1 & Better 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 b…

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 = 520 lbPLr = 1,120 lbLoad combination:D + LrSpan:L = 10 ftMember size:4 x 14Stress grade and species:Select Structural Douglas Fir-LarchUnbraced length:lu = 0Moisture content:MC > 19 percentLive load deflection limit:Allow. Δ ≤ L/360

A wood column is pin connected at its top and bottom. Determ…

A wood column is pin connected at its top and bottom. Determine the controlling slenderness ratio, le/d, for the column. Assume normal temperatures, no incising, and that all loads are downward. Ignore the weight of the member.Load:PD = 2,000 lbPL = 2,000 lbPLr = 0 lbPS = 3,000 lbPR = 4,000 lbPW = 2,000 lbPE = 0 lbLength:L = 14 ftMember size:4 x 8Stress grade and species:No. 1 Douglas Fir-LarchUnbraced length:lu = L = 14 ftMoisture content:MC < 19 percent

Determine the ASD adjusted design compression strength paral…

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:PD = 560 lbPLr = 960 lbLoad combination:D + LrSpan:L = 8 ftMember size:4 x 12Stress 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…

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 = 200 lbPLr = 1,120 lbLoad combination:D + LrSpan:L = 7 ftMember size:4 x 8Stress grade and species:Select Structural Douglas Fir-LarchUnbraced length:lu = 0Moisture content:MC < 19 percentLive load deflection limit:Allow. Δ ≤ L/360

Determine the ASD adjusted design tension strength, Ft’, for…

Determine the ASD adjusted design tension strength, Ft’, for the following beam. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:PD = 560 lbPLr = 1,440 lbLoad combination:D + LrSpan:L = 7 ftMember size:4 x 14Stress grade and species:No. 1 & Better 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…

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 = 440 lbPLr = 1,280 lbLoad combination:D + LrSpan:L = 7 ftMember size:4 x 6Stress grade and species:No. 1 Douglas Fir-LarchUnbraced length:lu = 0Moisture content:MC > 19 percentLive load deflection limit:Allow. Δ ≤ L/360