2001 ASD Supplements - unprotected PDF - American Wood Council

ASD STRUCTURAL LUMBER SUPPLEMENTL-6721 + αc ⎛ 1 + αc ⎞ αcCP= − ⎜ ⎟ −2c ⎝ 2c ⎠ c1 + 0.573=−16 .= 0.483( 0.483)( 5.25)( 975)P ′ == 2,470 lb.Example 3:Species: Douglas Fir-Larch2⎛ 1 + 0.573⎞0.573⎜ ⎟ −⎝ 16 . ⎠ 080 .Application — Simple columnColumn Capacity — y-axisFcE=KcEE′( le/dy)0.3x1,600,000=2( 144/5.5)= 700 psiFcEαcy=Fc*700=1,000= 0.7002Size: 6 x 8 (5.5 in. by 7.5 in.) by 12 ft.Grade: No. 1 (dry)F cN: 1,000 psiEN: 1,600,000 psiA: 41.25 in 2Column Capacity — x-axisFcE==K E′cE2( le/dx)( 0.3)( 1,600,000)2( 144/7.5)= 1,302 psiFcEαcx=Fc*1,302.08=1,000= 1.30221 + αcy ⎛ 1 + αcy ⎞ αcyCPy= −c⎜⎝ c⎟ −2 2 ⎠ c1 + 0.700=−16 .= 0.5592⎛ 1 + 0.700⎞0.700⎜ ⎟ −⎝ 16 . ⎠ 080 .Joist Capacity TablesThe general design equation for flexural bending is:M¢³Mwhere( 0.559)( 41.25)( 1,000)P′x== 23, 059 lb.M = moment due to design loadsM′ = allowable moment capacityAAPPENDIX21 + αcx ⎛ 1 + αcx ⎞ αcxCPx= − ⎜ ⎟ −2c ⎝ 2c ⎠ c1 + 1.302=−16 .= 0.774( 0.774)( 41.25)( 1,000)P′x== 31,928 lb.2⎛ 1 + 1.302⎞1.302⎜ ⎟ −⎝ 16 . ⎠ 080 .The general design equation for flexural shear is:V ¢³ Vwhere:Example 4:V = shear force due to design loadsV′ = allowable shear capacitySpecies: Douglas Fir-LarchApplication — floor joistSize: 2 x 6 (1.5 in. by 5.5 in.)Grade: No. 2C F: 1.3 (size factor)AMERICAN FOREST & PAPER ASSOCIATION