2001 ASD Supplements - unprotected PDF - American Wood Council

L-66 APPENDIXAppendix:Example of Chapter 5 Capacity TableDevelopment 1Tension Capacity TablesThe general design equation for tension members is:T¢³TwhereExample 1:T = tension force due to design loadsT′ = allowable tension capacitySpecies: Hem-FirApplication — tension memberSize: 2 x 6 (1.5 in. by 5.5 in.)Grade: 1650f-1.5E MSRF t′: 1,020 psiA: 8.25 in. 2Tension CapacityT′ = FA ′t== 8,415 lbs( 1, 020)( 8.25)Stud Wall Capacity Tables/Column Capacity TablesThe general design equation is:P ¢ ³ Pwhere:P = compressive force due to design loadsP′ = allowable compression capacityAxial CapacityP¢= C AF¢where:CPPc1 + αc ⎛ 1 + αc ⎞ αc= − ⎜ ⎟ −2c ⎝ 2c ⎠ cFcEα c =Fc*2andFcE=KcEE′l e / d2( )F* c= tabulated compression design value multipliedA = areaC Pby all applicable adjustment factors except C p= column stability factorF c′ = allowable parallel-to-grain compression designvalueE′ = allowable modulus of elasticityK cE= 0.3 for visually graded lumber= 0.384 for MEL= 0.418 for MSRc = 0.8 for solid sawn membersExample 2: Application — wall stud laterallysupported in the weak direction and axiallyloaded (prismatic column)Species: Southern PineFSize: 2 x 4 (1.5 in. by 3.5 in.) by 8'Grade: Stud (dry)cE=F c′ : 975 psiE′ : 1.4 x 10 6 psiA: 5.25 in. 2C D= 1.0K e= 1.0K E′cE( le/ d )20.3 × 1,400, 000=( 96/3.5)= 558 psiFcEα c =Fc*558=975= 0.57321Due to rounding, the calculated and tabulated values may differ.AMERICAN WOOD COUNCIL