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

ASD–Problem 10Preliminary Design (continued from page 14)may have to be revisited if the beam size later changes. A similar process applies if shear is used to determinea trial section. There is no one right way to do this and the choice of method often depends on the experienceof the designer. A trial section of 5 in. by 9-5/8 in. was chosen to start the solution.Adjustment Factor CalculationsCommentsNote, C V and C L arenotusedinthesamecalculationfor moment capacity. The lower of the two is used.Volume FactorC V := 21112020⎛ ⎞ 12 5125 .⎜ ⎟ ⋅ ⎛ ⎝ 21.34⎠⎝ ⎜ in ⎞⎟ ⋅ ⎛ in⎞⎜ ⎟Depth⎠⎝ Width ⎠C V = 1.012Beam Stability FactorL u := LengthL u:= 29.922Depthl e := 1.63 . L u +3 . Depthl e = 498.3 in(l e ⋅ Depth)R B :=2WidthR B = 13.851COV E := 0.10E′ min := E ymin. CM . Ct120NDS 5.3.6Eq. 4.1: 21.34 ft is the distance between pointsof zero momentThis value cannot exceed 1, therefore C V =1.NDS 3.3.3The unsupported length is taken as the spanbetween the supportsEffective length chosen from Table 3.3.3Eq. 3.3-5R B is less than 50 as requiredAppendix F, Table F1E′ min := 7.3 × 10 5 psiF b_star := F . b_tension CD . CM . Ct . Cfu . CcF b_star = 2.300 × 10 3 psiE′ minF bE := 1.20⋅2R Bα :=F bEFb_ starα = 1.985⎛C L := 1 + α⎞⎛ 1 + α⎞α⎜ ⎟ − ⎜ ⎟ −⎝ 19 . ⎠ ⎝ 19 . ⎠ 095 .C L = 0.9562F b_star isthesameasF * b in NDS 3.3.3All factors except for C L or C V multiplied bythe NDS reference design value.Eq 3.3-6C L controls, rather than C V16 WOOD DESIGN FOCUS