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 = 280 lbPLr = 2,240 lbLoad combination:D + LrSpan:L = 12 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 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 = 160 lb/ftwLr = 390 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 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 = 120 lb/ftwLr = 360 lb/ftLoad combination:D + LrSpan:L = 8 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 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 = 440 lbPLr = 1,280 lbLoad combination:D + LrSpan:L = 13 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 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 = 100 lb/ftwLr = 420 lb/ftLoad combination:D + LrSpan:L = 8 ftMember size:4 x 14Stress 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 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 = 120 lb/ftwLr = 150 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 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 = 520 lbPLr = 960 lbLoad combination:D + LrSpan:L = 9 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

A 4 x 12 rafter frames into a 3 x 4 stud wall. The load is a combination of PD = 170 lb and PS = 650 lb. Consider the D + S load combination. Determine the ASD actual bearing stress fc⊥ in the top plate of the wall.

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 = 270 lb/ftLoad combination:D + LrSpan:L = 6 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 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 = 280 lb/ftwLr = 210 lb/ftLoad combination:D + LrSpan:L = 6 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