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:PD = 240 lbPLr = 1,280 lbLoad combination:D + LrSpan:L = 11 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. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:PD = 360 lbPLr = 1,600 lbLoad combination:D + LrSpan:L = 13 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

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 = 100 lb/ftwLr = 450 lb/ftLoad combination:D + LrSpan:L = 13 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 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:wD = 120 lb/ftwLr = 360 lb/ftLoad combination:D + LrSpan:L = 6 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 volume factor, Cv, for a 3.125” x 21” Douglas Fir glulam that is 28 feet long and simply supported.

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 = 120 lb/ftwLr = 360 lb/ftLoad combination:D + LrSpan:L = 12 ftMember size:4 x 8Stress 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 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 = 240 lbPLr = 1,600 lbLoad combination:D + LrSpan:L = 7 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 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 = 1,120 lbLoad combination:D + LrSpan:L = 6 ftMember size:4 x 12Stress 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 beam. Assume normal temperatures, bending about the strong axis, and no incising. Ignore the weight of the beam.Load:wD = 100 lb/ftwLr = 180 lb/ftLoad combination:D + LrSpan:L = 14 ftMember size:4 x 10Stress grade and species:Select Structural Douglas Fir-LarchUnbraced length:lu = 0Moisture content:MC > 19 percentLive load deflection limit:Allow. Δ ≤ L/360

For deflection calculations, the adjusted modulus of elasticity Ex’ for a hardwood glulam beam 16F-V3 with ASD load combination (D + 0.6W), under 16% moisture content, and a constant temperature of 130°F is _______.