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 = 480 lbPLr = 1,120 lbLoad combination:D + LrSpan:L = 9 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 volume factor, Cv, for a 3.125” x 21” Douglas Fir glulam that is 21 feet long and simply supported.

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 = 280 lb/ftwLr = 150 lb/ftLoad combination:D + LrSpan:L = 12 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 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:PD = 440 lbPLr = 800 lbLoad combination:D + LrSpan:L = 6 ftMember size:4 x 8Stress grade and species:No. 1 & Better 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 maximum axial stress in the member. Assume normal temperatures, no incising, and that all loads act in the directions shown. Ignore the weight of the member.Load:PD = 5,000 lbPL = 0 lbPLr = 0 lbPS = 0 lbPR = 0 lbPW = 8,000 lbPE = 8,500 lbQD = 2,500 lbQL = 2,000 lbQLr = 4,500 lbQS = 3,000 lbQR = 2,500 lbQW = 0 lbQE = 0 lbSpan:L = 12 ft Member size:4 x 12 Stress grade and species:No. 1 & Better Douglas Fir-Larch Unbraced length:lu = L/2 = 6 ft Moisture content:MC < 19 percent 

What is the wet service factor, CM, for the bending strength of softwood glulams?

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 = 100 lb/ftwLr = 450 lb/ftLoad combination:D + LrSpan:L = 11 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 bending Fbx+ reference design value for 26F-1.9E softwood glulams.

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 = 360 lbPLr = 800 lbLoad combination:D + LrSpan:L = 13 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

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 = 4,000 lbPL = 4,000 lbPLr = 1,500 lbPS = 3,000 lbPR = 2,000 lbPW = 3,500 lbPE = 0 lbLength:L = 11 ftMember size:4 x 10Stress grade and species:No. 2 Douglas Fir-LarchUnbraced length:lu = L = 11 ftMoisture content:MC < 19 percent