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 = 440 lbPLr = 1,280 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

A pin-connected wood truss is loaded and supported as shown. Determine the axial force produced in member FG if PB = 800 lb, PC = 1,800 lb, PD = 3,200 lb, L = 6 ft, and h = 8 ft. Positive choices indicate tension. Negative choices indicate compression.

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 = 220 lb/ftwLr = 150 lb/ftLoad combination:D + LrSpan:L = 10 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 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 = 180 lb/ftwLr = 420 lb/ftLoad combination:D + LrSpan:L = 13 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 client has come to the family planning clinic to discuss the use of contraceptives. The nurse should do which of the following to facilitate a productive discussion?  

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 = 240 lbPLr = 1,440 lbLoad combination:D + LrSpan:L = 12 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 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 = 14 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 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 = 240 lb/ftwLr = 330 lb/ftLoad combination:D + LrSpan:L = 8 ftMember size:4 x 6Stress 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 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 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:PD = 280 lbPLr = 1,280 lbLoad combination:D + LrSpan:L = 13 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