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

Structural Wood Design Using ASD and LRFDBy John “Buddy” Showalter, P.E.IntroductionA new publication entitled Structural Wood Design UsingASD and LRFD is being developed as a companion designtool to the 2005 National Design Specification ® (NDS ® )forWood Construction. It will be available beginning in theSpring of 2005 through the American Forest & Paper Association(AF&PA). The authors are Dr. Dan L. Wheat, P.E. of theUniversity of Texas at Austin and Dr. Steven M. Cramer, P.E.of the University of Wisconsin–Madison. The book is intendedto aid instruction on structural design of wood structuresusing both allowable stress design (ASD) and load andresistance factor design (LRFD). It will allow direct comparisonof ASD and LRFD for wood structures on a problemby problem basis.BackgroundServing as the basis for the new document, a design aid,entitled LRFD Solved Example Problems for Wood Structures,which was published in 2000 as a companion to AF&PA’sLRFD Manual, has been updated to include parallel ASD solutionsto the 40 LRFD example problems previously developed.It was originally co-sponsored by the American WoodCouncil of AF&PA, Southern Pine Council (Southern ForestProducts Association and the Southeastern Lumber ManufacturersAssociation) and the Wood Truss Council of America,and published by the International Codes Council.The design examples in Structural Wood Design UsingASD and LRFD range from simple to complex and covermany design scenarios. This design aid is intended for useby practicing engineers, many of whom currently use ASD,but who may want to compare and contrast it with LRFD;and by academics, whose teaching objectives may vary.Some problems have been posed as stand-alone problems,but others belong to a set in which all examples are associatedwith one structure. The focus of this design tool is education;it is not to propose specific designs or details. It ishoped that the format, which has facing pages with ASDand LRFD solutions, will allow readers to verify the comparablework efforts and common design equations of ASDand LRFD. As well, the new design aid will include examplesthat incorporate such provisions as those for localstresses in fastener groups, namely row tear-out and grouptear-out, which were not a part of AF&PA’s LRFD Manual.Example ProblemTo illustrate the content of Structural Wood Design UsingASD and LRFD, an example problem has been reproducedon the following pages. It is a uniformly loaded beam withan overhang, and it is to be designed for the simultaneousapplication of dead and snow load. All problems will be inthis general format; that is, the ASD solution is shown onthe left page, with the LRFD solution shown on the facingpage. This will allow for an easy comparison of the two setsof methods.Solutions have been developed using Mathcad ® Professionalsoftware by MathSoft ® ,Inc.ThoseunfamiliarwithMathcad should note that the symbol “:=” is equivalent tothe traditional equals “=” sign.Problem 10. Cantilevered Glulam Beam DesignSelect the size of the glued laminated timber beam toresist an unfactored uniform dead load of 20 lb perfoot (includes beam self-weight) and 180 lb per foot ofsnow load. Full lateral support is provided only at thepin and roller supports. Normal temperature and dryconditions prevail. Use visually graded Southern Pinecombination 20F-V5. Only flexure and shear are to bechecked.Winter 2004 11

ASD Solution Problem 10Structural AnalysisASD–Problem 10P dead := 20 lbfftP snow := 180 lbfftw:=P dead +P snoww = 200 lbfftEquations of EquilibriumftReaction B := w⋅32ft⋅16 24 ftReaction B = 4.267 × 10 3 lbfReaction A := w⋅32ft − Reaction BReaction A = 2.133 × 10 3 lbfV A := Reaction AV B_left := V A −w⋅24ftV B_left = –2.667 × 10 3 lbfV B_right := Reaction B +V B_leftV B_right = 1.600 × 10 3 lbfM positive := V A ⋅ ⎛ ⎝ ⎜ V A ⎞ 1⎟ ⋅w ⎠ 2M positive = 1.365 × 10 5 lbf . inM negative := w⋅8ft⋅4ftM negative = 7.680 × 10 4 lbf . in12 WOOD DESIGN FOCUS

LRFD Solution Problem 10Structural AnalysisLRFD–Problem 10P dead := 20 lbfftP snow := 180 lbfftw:=1.2P dead +1.6P snoww = 312 lbfftEquations of EquilibriumftReaction B := w⋅32ft⋅16 24 ftReaction B = 6.656 × 10 3 lbfReaction A := w⋅32ft − Reaction BReaction A = 3.328 × 10 3 lbfV A := Reaction AV B_left := VA −w⋅24ftV B_left = –4.160 × 10 3 lbfV B_right := Reaction B +V B_leftV B_right = 2.496 × 10 3 lbfM positive := V A ⋅ ⎛ ⎝ ⎜ V A ⎞ 1⎟ ⋅w ⎠ 2M positive = 2.130 × 10 5 lbf . inM negative := w⋅8ft⋅4ftM negative = 1.198 × 10 5 lbf . inWinter 2004 13

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

LRFD–Problem 10Member InformationReference Design ValuesNDS Supplement Table 5ASection PropertiesThe following reference design values are tabulated Length := 24 ⋅12 in NDS 3.3.4.4in the NDS Supplement. They will be adjusted⎛for LRFD later in the solution when other wood Depth := 9 +5 ⎞⎜ ⎟ in⎝ 8⎠specific adjustments are appliedTrial sizeWidth := 5 inTrial sizeF b_tension := 2000 psiArea := Depth . WidthF b_compression := 2000 psi Area = 48.125 in 2F v := 300 psiE y :=1400000 psi3⎡( Width ⋅ Depth ) ⎤⎢⎥S xx :=⎣ 12 ⎦⎛ Depth⎞⎜ ⎟⎝ 2 ⎠E ymin := 730000 psi S xx = 77.201 in 3Adjustment FactorsNDS Table 5.3.1C M := 1.0 Wet service factor K F_b := 216 . Format conversion factor for bendingφ bC t := 1.0 Temperature factor K F_s := 15 .φ sC fu := 1.0 Flat use factor K F_s := 216Format conversion factorfor stability.φ vFormat conversion factor for shearC c := 1.0 Curvature factor The K F factors convert reference design valuesand moduli (for stability) to LRFD referenceλ := 0.8 Time effect factor resistances –see Table N1 Appendix Nφ b := 0.85Bending resistance factorφ v := 0.75 Shear resistance factorφ s := 0.85Stability resistance 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 controlling one.This is easily done by selecting which load combination gives the highest M/λ value [not the highest momentalone]. Once the critical load combination is determined, an estimated section modulus couldbecalculatedas(continued on page 17)Winter 2004 15

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

LRFD–Problem 10Preliminary Design (continued from page 15)M, leaving off for simplicity other adjustment factors of F b . Note that this estimate includes the conversion–bymeans of K F –of a normal duration allowable design value F b to a short-term reference resistance, which thenλKFφbFbis adjusted by λ and φ. However, there are at least four other options for getting section modulus. One is simply toguess a beam size. Another is to calculate M/F b . Yet another is to select a section modulus to satisfy deflection criteria–butnone is specified in this problem–which often controls. The last option is to calculate the required sectionmodulus using the adjusted design value, F′ b , that is, with all of the adjustment factors applied to F b , but realizingthat some adjustment factors, such as C v , may have to be revisited if the beam size later changes. A similar process appliesif shear is used to determine a trial section. There is no one right way to do this and the choice of method oftendepends on the experience of 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 := φ s. KF_s . Eymin . 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 := 1.095 × 10 6 psiF b_star := λ . φ .bKF_b . Fb_tension . CM . Ct . Cfu . CcF b_star = 3.456 × 10 3 psiE′ minF bE := 1.20⋅2R Bα :=F bEFb_ starα = 1.982⎛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 design value.Eq 3.3-6C L controls, rather than C VWinter 2004 17

ASD-Problem 10Solve for Adjusted StressBending DesignNDS 3.3F′ b := F b_tension. CL . CD . CM . Ct . Cfu . CcF′ b := 2198 psiM positive = 136533 lbf . inf b := M positiveS xxf b := 1769 psif b

LRFD–Problem 10Solve for Adjusted StressBending DesignNDS 3.3F′ b := λ . φ b. KF_b . Fb_tension . CL . CM . Ct . Cfu . CcF′ b := 3302 psiM positive = 212992 lbf . inf b := M positiveS xxf b := 2759 psif b