Structural Wood Design Using ASD and LRFD - American Wood ...

ASD–Problem 10Member InformationReference Design ValuesNDS Supplement Table 5ASection PropertiesF b_tension := 2000 psi Length := 24 ⋅12in NDS 3.3.3.4F b_compression := 2000 psi⎛Depth := 9 +5 ⎞⎜ ⎟ in⎝ 8⎠Trial sizeF v := 300 psi Width := 5 in Trial sizeE y :=1400000 psiArea := Depth . WidthE ymin := 730000 psi Area = 48.125 in 23⎡( Width ⋅ Depth ) ⎤⎢⎥S xx :=⎣ 12 ⎦⎛ Depth⎞⎜ ⎟⎝ 2 ⎠S xx = 77.201 in 3Adjustment FactorsNDS Table 5.3.1C D := 1.15C M := 1.0C t := 1.0C fu := 1.0C c := 1.0Load duration factorWet service factorTemperature factorFlat use factorCurvature factorDesign CalculationsPreliminary DesignThe preliminary sizing of members is done in a variety of ways by designers. In general, a beam must be examinedfor several load combinations and the designer is responsible for determining the critical or controllingone. This can be done by selecting which load combination gives the highest M/C D value [not the highest momentalone]. Once the critical load combination is determined, an estimated section modulus could be calculatedas , leaving off for simplicity other adjustment factors of F b . However, there are at least four otherMCDFboptions for getting trial member sizes. One is simply to guess a beam size. Another is to calculate M/F b .Yetanotheris to select a section modulus to satisfy deflection criteria–but none is specified in this problem–which oftencontrols. The last option is to calculate the required section modulus using the adjusted design value, F′ b ,that is, with all of the adjustment factors applied to F b , but realizing that some adjustment factors, such as C v ,(continued on page 16)14 WOOD DESIGN FOCUS