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
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 = 320 lbPLr = 1,120 lbLoad combination:D + LrSpan:L = 12 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
Determine the maximum actual deflection of the following bea…
Determine the maximum actual deflection of 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 = 180 lb/ftLoad combination:D + LrSpan:L = 13 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 maximum bending moment in the following beam….
Determine the maximum bending moment in the following beam. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:wD = 120 lb/ftwLr = 240 lb/ftLoad combination:D + LrSpan:L = 10 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 maximum bending moment in the following beam….
Determine the maximum bending moment in the following beam. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:PD = 200 lbPLr = 2,240 lbLoad combination:D + LrSpan:L = 6 ftMember size:4 x 10Stress 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 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:PD = 400 lbPLr = 1,920 lbLoad combination:D + LrSpan:L = 7 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