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 = 5,500 lbQD = 4,000 lbQL = 1,500 lbQLr = 500 lbQS = 4,500 lbQR = 0 lbQW = 0 lbQE = 0 lbSpan:L = 9 ft Member size:4 x 6 Stress grade and species:No. 1 Douglas Fir-Larch Unbraced length:lu = L/2 = 4.5 ft Moisture content:MC < 19 percent 

A wood member is loaded as shown. Using ASD, determine the adjusted bending strength, Fb’. Assume normal temperatures, no incising, and that all loads act in the directions shown. Ignore the weight of the member.Load:PD = 4,500 lbPL = 0 lbPLr = 0 lbPS = 0 lbPR = 0 lbPW = 8,500 lbPE = 7,000 lbQD = 1,500 lbQL = 3,000 lbQLr = 4,000 lbQS = 1,000 lbQR = 0 lbQW = 0 lbQE = 0 lbSpan:L = 12 ft Member size:4 x 6 Stress grade and species:Select Structural Douglas Fir-Larch Unbraced length:lu = 0 ft Moisture content:MC < 19 percent 

A wood member is loaded as shown. Using ASD, determine the adjusted tension strength, Ft’. Assume normal temperatures, no incising, and that all loads act in the directions shown. Ignore the weight of the member.Load:PD = 3,000 lbPL = 0 lbPLr = 0 lbPS = 0 lbPR = 0 lbPW = 6,000 lbPE = 5,000 lbQD = 2,000 lbQL = 5,000 lbQLr = 0 lbQS = 4,000 lbQR = 4,000 lbQW = 0 lbQE = 0 lbSpan:L = 13 ft Member size:4 x 6 Stress grade and species:Select Structural Douglas Fir-Larch Unbraced length:lu = L/2 = 6.5 ft Moisture content:MC > 19 percent 

Use the AWC Special Design Provisions for Wind and Seismic, 2015 Edition and ASD procedures to answer the following question.A building with a non-structural interior wall is exposed to a transverse load of w = 235 lb/ft. The roof is sheathed with single-layer straight lumber sheathing. Is the diaphragm satisfactory in regard to the maximum diaphragm aspect ratio? Let L = 68 ft, a = 50 ft, b = 18 ft, be = 12 ft, and bi = 4 ft.

If a column is loaded to failure, buckling will occur about the axis that has the largest slenderness ratio.

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 = 2,500 lbPL = 1,000 lbPLr = 500 lbPS = 5,000 lbPR = 3,500 lbPW = 4,500 lbPE = 0 lbLength:L = 12 ftMember size:4 x 6Stress grade and species:Select Structural Douglas Fir-LarchUnbraced length:lu = L = 12 ftMoisture content:MC < 19 percent

Use the AWC Special Design Provisions for Wind and Seismic, 2015 Edition and ASD procedures to answer the following question.A single-story wood-frame building is exposed to a wind load of wU = 295 lb/ft. For simplicity, assume this load was determined from the load combinations, i.e. do not apply the load combinations to it, and that the building has no windows or doors. Let L = 28 ft, b = 20 ft, and the height of the walls be 8.5 ft. Ignoring the effects of gravity loads, what is the net maximum uplift force at the base of the shear walls?

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 = 3,000 lbPL = 1,500 lbPLr = 0 lbPS = 4,500 lbPR = 1,000 lbPW = 3,000 lbPE = 0 lbLength:L = 9 ftMember size:4 x 14Stress grade and species:No. 1 & Better Douglas Fir-LarchUnbraced length:lu = L = 9 ftMoisture content:MC < 19 percent

A wood member is loaded as shown. Using ASD, determine the adjusted tension strength, Ft’. Assume normal temperatures, no incising, and that all loads act in the directions shown. Ignore the weight of the member.Load:PD = 2,500 lbPL = 0 lbPLr = 0 lbPS = 0 lbPR = 0 lbPW = 5,500 lbPE = 7,000 lbQD = 0 lbQL = 500 lbQLr = 5,000 lbQS = 3,000 lbQR = 4,500 lbQW = 0 lbQE = 0 lbSpan:L = 14 ft Member size:4 x 8 Stress grade and species:No. 1 Douglas Fir-Larch Unbraced length:lu = L/2 = 7 ft Moisture content:MC < 19 percent 

A wood member is loaded as shown. Using ASD, determine the adjusted bending strength, Fb’. Assume normal temperatures, no incising, and that all loads act in the directions shown. Ignore the weight of the member.Load:PD = 2,500 lbPL = 0 lbPLr = 0 lbPS = 0 lbPR = 0 lbPW = 9,000 lbPE = 5,000 lbQD = 3,000 lbQL = 4,000 lbQLr = 5,000 lbQS = 1,000 lbQR = 1,000 lbQW = 0 lbQE = 0 lbSpan:L = 14 ft Member size:4 x 10 Stress grade and species:Select Structural Douglas Fir-Larch Unbraced length:lu = 0 ft Moisture content:MC < 19 percent