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# 59 Class Note for CE 47900 with Professor Ramirez at Purdue

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Date Created: 02/06/15

CE 479 Wood Design Lecture Notes JAR Introduction Sizes of Structural Lumber and Use Design Approach The design of structural wood is carried on the basis of allowable stresses at service load levels Structural calculations are based on the standard net size of a piece of lumber Most structural lumber is dressed lumber l Dressed Lumber Lumber that has been surfaced to the standard net size Which is less than the nominal size stated ie 8 X12 member nominal size 8 X 12 in actually is 7 12 X 11 12 in Standard net size NDS Table 1A Sec 3 Supplement 2001 Lumber is dressed on a planning machine for the purpose of obtaining smooth surfaces and uniform sizes Typically lumber Will be S4S surfaced on 4 sides Table 1A Nominal and Minimum Dressed Sizes of Sawn Lumber Thickness Inches Face wmms inches Minimum dressed Minimum dressed llem Nominal Dry l Glenn Nominul Dry Green Boards 34 58 Him 2 111 17010 1 34 2532 3 212 279I6 1714 1 17132 4 312 37916 1712 114 17932 5 442 458 g a 512 5758 0 7 612 6758 3 x 714 7712 g 9 8l4 812 1 10 914 91 g 11 101A 1042 1 12 11 14 11712 g 14 13 14 1371 I 10 1514 13712 m Dimension 2 112 1916 2 112 17916 Lumber 2712 2 2416 3 242 2916 3 2712 1916 4 312 3915 3712 3 3116 5 412 4758 4 3712 3916 6 512 553 471 4 4116 8 714 7712 D 10 914 9712 12 1114 11712 14 13 14 13712 16 1514 15712 2 Rough Sawn Large timbers are usually rough sawn to dimensions that are close to standard net sizes roughly 18 larger than the standard dressed size Rough surface is usually ordered specially for architectural purposes in smaller sizes 3 Full Sawn In this case a rough surface is obtained With actual size equal to the nominal size CE 479 Wood Design Lecture Notes JAR Wood Rating The majority of sawn lumber is graded by visual inspection and material graded in this way visually is known as visually graded structural lumber As the lumber comes out of the mill a person familiar with lumber grading rules examines each piece and assigns and stamps a grade There are two broad size classi cations of sawn lumber 0 Dimension Lumber smaller thinner sizes of structural lumber Dimension lumber usually ranges in the size from 2X2 through 4X16 In other words dimension lumber is any material with a thickness smaller dimension of a piece of wood and width is the larger dimension of 2 to 4 inches 0 Timbers are the larger pieces and have a minimum nominal dimension of 5 inches Thus the smallest practical size timber is a 6X6 inch The design properties given in the NDS supplement are based on two different sets of ASTM Standards 0 Ingrade procedures applied to Dimension lumber 0 Clear wood procedures applied to timbers The lumber grading rules which establish the allowable stresses for use in structural design have been developed over the years The relative size of the wood was used as a guide in anticipating its use Although most lumber is visually graded a small of lumber is MACHINE STRESS RATED by subjecting each piece of wood to a non distructive test This process is highly automated As lumber comes out of the mill it passes through a series of rollers In this process a bending load is applied about the minor axis of the cross section and the modulus of elasticity of each piece measured In addition the piece is visually inspected The material graded using MSR is limited to a thickness of 2 or less MSR has less variability in mechanical properties than visually graded lumber Consequently is often used to fabricate engineered wood products 0 Glulam beams 0 Wood Ijoists and light frame However stress rated boards are not commonly used for structural framing because they are very thin So we will focus on dimension lumber It must be remarked that the allowable stress depends on the species and on the size of the member Species A large number of species can be used to produce structural lumber The 2001 NDS supplement Sec 4 Page 29 contains allowable stresses for a large number of species The choice of species for use in design is a matter of economics typically For a given location only a few species groups may be available and it is prudent to check with local CE 479 Wood Design Lecture Notes JAR distributors as well as a wood products agency The species of tress used for structural lumber are classified as hardwoods and softwoods owing not necessarily to a description of the wood properties For example evergreens aka conifers are a large majority of the structural lumber This will be either DouglasFir or Southern Pine Table 4A Base Design Values for Visually Graded Dimension Lumber 2quot 4quot thickP39Z All species except Southern Pine see Table 4B Tabulated design values areer normal load duration and dry sewice conditions See NOS 43 fora comprehensive description of design value adjustment factors USE WITH TABLE 4A ADJUSTMENT FACTORS 0251er values in pounds per Smrare inch psri 1 Compression quot Tension Shear Compression Modulus parallel parallel perpendicular parallel of Grading Specres and Size Bendlng to gram 10 grain to grain to grain Elasllcily Rules commercial grade 03985511183110 FL 13 FE E Rgency Suleci Structural 375 500 120 265 1 25 1100000 No1 625 375 120 265 i300 1100000 No 2 amp wrder 600 350 120 265 450 1000090 NELMA No3 330 200 120 365 275 900000 NSLB Slud 2quot 8 wider 475 275 120 265 300 900000 WWPA Construction 39 mo 4013 120 265 325 1 900000 Standard 2394 wide 375 225 39 39 120 265 475 990nm Utility39 39 quot 39 175 100 39 120 39 265 300 800000 Select 51ruclura 1150 850 195 715 1200 1330000 No1 1050 500 195 T15 950 1500000 No2 239 amp under 1000 500 195 7 15 750 1500000 No3 575 350 195 15 425 1300000 iJELl39w39st Stud 239 8 wider 775 450 195 715 a175 1300000 Conslmc on 1150 675 195 715 1000 1400000 39 39 Slandard 2quot4quot wide 650 375 195 39 7 15 39 WE 1300000 quot U lil 300 1T5 195 715 500 1200000 39 SSHTVA N91531 Allowable StressesDesign Values NDS Tabulated values in the NDS Supplement Q Are determined by multiplying the tabulated stresses by the appropriate adjustment factors Thus becoming allowable design value F For example for tension parallel to the grain SAPWOOD Figure 44 Cross section of a log F l F x adjustment factors Design value A N N UA L RNG HEAKTWOOD EA KAY W000 LA TEWOUD CE 479 Wood Design Lecture Notes JAR For an acceptable design the axial tensile stress due to loads should not exceed the allowable adjusted stress ftSFI Design Value stress Tabulated Stress Allowable adjusted Stress Bending Tension parallel to grain Shear parallel to grain Compression perpendicular to grain Compression parallel to grain Modulus of elasticity Fb Ft Fv 1cl Fc E F12 Ft Fv 1cI F E Adjustment Factors Some decrease other increase tabulated values Examples CD load duration factor CM web service factor moisture content CF size factor Cfu at use factor Cf form factor Stresses and adjustment factors Stresses due to known loads NDS 2001 Section 3 P M 7 177 1 f f S 60 A CE 479 Wood Design Lecture Notes JAR Tabulated Values Stresses Tabulated design values listed in the NDS Supplement 2001 ED These values include reduction for safety F and are for normal load duration under the speci ed moisture service condition Modulus of elasticity E does not include reduction for safety and represent average values Dimension Lumber Page 29 NDS Supp 2001 Table 4A page 30 31 Base design value for visually graded dimension lumber except southern pine Table 4B page 3637 Base design value for visually graded southern pine Table 4C page 3942 Design values for mechanically graded dimension lumber MSR Timbers 5X5 and larger Table 4D page 4349 Design values for visually graded timbers all species Adiustment Factors Sec 43 NDS 01 and Suplement t0 NDS Tables A Wet Serviced Factor CM EMC Equilibrium moisture content the average moisture content that lumber assumes in service Moisture designation in grade stamp SGm surface green MC 19 in service SDry surfaced dry MC 15 in service These values can vary depending on environmental conditions in most buildings ranges from 7l4 EMC Special conditions must be analyzed individually CE 479 Wood Design mm Notes JAR Tabulated Values in NDS supplement apply to members with EMC of 19 or less regardless of SGRN or SDry IfEMC exceeds 19for an extended period of time table Values should be multiplied by CM see Page 30 and others for Values in Table 4 NDSSupp 01 Wet Service Factor C When dimensxon lumber ls med where moisture Cour tenl will exceed 9 fol an extended lime period design values shall be mulliplied by the appropriate Wet service fllctoxs from the followmg able Wet Service Factnrs cu FCL l 0 7 0 67 new 50p c v lllcnFCt lS mpli CM l 0 Fl B Load Duration Factor CD Wood can handle higher stresses ifloads are applied for a short period oftime All tabulated Values apply to normal duration loading 10 years The term duration of load refers to the total accumulated length oftime that a load is applied during the life of a structure Table 232 in NDS 01 provides CD to be used in the one associated with the shortest duration of time Whichever combination of loads together with the appropriate load duration factor produces the largest member size is the one that must be used in design Table 232 Frequently Used Load quot Duration Factors CD1 Lulld Dunnlnu 3 Typical Design Loads Pct mmcm o 9 Dead Load Ten yam LO Occupancy lee Loml lwo months l 15 Snow Loud chcll dllys l 25 Conslnlction Loud Ten mmum l 5 mum llquakc Load lmpacll 2 a lmpucl Load I Load dilmllon rman shall m Apply to mudulus ol clmmu l mu m comprcsslan pcrmellCulJr l0 gm 4ng lth r bvscd at l dclanm um lmm 2 l LCM umllon mars gr mlm l a l prmnl kid will 4mm pm mum clxcmlcal lllc lmplcl ml 0 mumum incl apply lo nl ulcmbcl M 1m m 1m or rm non muur shall not We in CE 479 Wood Design Lecture Notes JAR C Size Factor CF The size ofthe member has an effect on its unit stress See Supplement 01 Tables 4A 4B 4C 4D and 4E Size Factor C Tabulated bending tension and compression parallel to grain design values for dimenston lumber Z to 4 thick shall be multiplied by the follcwmg size factors Size Factors CF Fl Ft Ft Thickness breadth Grades Width depth 2 8L 3quot 4quot 2quot3 amp4quot 15 15 15 115 Select 5 14 14 14 ll Structural 6 13 11 l 3 11 N013 Btu 8quot 2 13 l 2 105 NoNu2 10quot I I 12 1 1 10 No3 12quot 0 11 10 10 IA St wider 0 9 10 09 09 Z 3quotamp4 Ll ll Ll 105 Stud 5quot amp 6 10 10 10 10 8 amp Wider Use No3 Grade tabulated design values and site actors quot 39 r 0 m 39 10 l u D Repetitive Member Factor C only FMf The system performance of a series of small closely spaced wood members Where failure of one member is not fatal see Supplement 01 Table 4A Adjustment Factors Repetitive Member Factor Cr Bending de en values F for dimension lumber lquot to 4 thick shall be multiplied by the repetitive mcmbci factor C r 115 when such members are used asjulstx truss chords rafters studs planks decking or similar members which are in contact or spuch not more than 24quot on centers are not less than 3 in number and nrcjuincrl by floor roof or other load distributing elements adequate to support the design load CE 479 Wood Design Lecture Notes JAR E Flat use Factor Cf Except for decking tabulated stress for dimension lumber apply to wood members that are stressed in exure about the strong axis 7 edgewise or load applied to narrow face If however load is a applied to the Wide face 7 the stresses may be increased by C Flat Use Factor Cm Bonding design dlucs idjnxicd by leL i39uciurs are based on edgcwiscmc loud lpplmd Lo ummw ldLL When lllmclhlml lumber is med ilziiwlw 110qu applied in Wide Fure illE bending design uiuc Fh lliill also be lnillli39 Plltd by he i39olimving lint mm l ncim39s Flul Use Facinr mm W Main dcpllil aquot Li i l 5quot i I 6quot L13 3quot 115 m amp wider 2 Tabulated bending stresses also for timber Beams amp stringers apply for bending about X axis NDS does not provide C for these cases 313 Definitions NEUTRAL AXIS m ihc Cl USS secrion oi a beam is the line on which lien is UCliliel39 tension nm compression mess Figure 1A Dimensions for Rectangular Cross Section Glued Liimmutcd Tunbei Snmi Lumbci CE 479 WoodDesign Lecture Notes JAR F Temperature Factor C The strength of the wood in service is increased as the temperature cools below the normal temp in most buildings On the other hand the strength decreases as temperatures are increased The factor Cl is the multiplier that is used to reduce tabulated stresses if higher than normal temperatures are encountered in a design situation Values of Cl are given in NDS Sec 234 for T gt 100 F Important to note that strength Will be regained When temperature returns to normal values Thus this factor applies for sustained conditions Table 233 Temperature Factor ct Design villus Ceuiluiimr rlim39l 101 Wu l l xiiriiiimin unisii lm il iiiuber iiiimsue iilier imi iiiimli me i 4 st 4uulu uresrieeiu i 6 mm ilr senire eeuiiiuiiii iiir pieiziiirieiiieii iinnii 39 iuiul pull me siiecil izdiul I 4 5 i G Form Factor Cf The purpose of this factor is to adjust tabulated bending stress F1 for nonrectangular sections see Section 334 inNDS 01 334 Form Factor cf Tilblllaled bending design values 11 ror bending members wiili eiilier a eiieulm eiess ieeiieu ei u sqmlm crosssectlon loaded in iiie plane orilie diagonal unimeuil seciiau shall be multiplied by ilie following renu rise lots Ci Table 334 Form Factors cf Ci Round Scerion l 18 Dlamond Section 1414 These foml factors lnslll e mm a circular or diamond sbspeii bending member luis ilie srime memeni capacity as a square bending member having iiie same crasssec tional area lfa Circular member is tapered ii shall be considered a beam on ariablc eress seeiien CE 479 Wood Design Lecture Notes JAR Example Determine the tabulated and allowable design values for the following member and loading condition 0 No 2 HemFir bending about strong axis 0 Floor beams 4x6 in 4 on centers Loads are DL Highhumidity conditions exist and moisture content may exceed 19 Stresses Bending NDS Supp 01 Table 4A Base Design Values for Visually Graded Dimension Lumber 2quot 4quot thick Cont All species except Southern Fine see Table 43 Tabuiated deeign vaiues are for normai ioad duration and dry service conditions See NDS 43 foi39 a comprehensive description of design value adiustment factors usa wni TAELE 47A ADJUSTMENT FAcroRs in pounds per summ mm psi Tension CampiEssioii Camp39essicn men is paraliei pawliei it aiming Species and SUE atiimiiu u glam ln grain Eiasiimly Rules comm1mm grade cussilinalioil FD F r E AgMC i i i I r smeci svucmmi man 925 150 05 iauo isnnuon Nu i s an iinn 725 150 495 i350 tsunami Na I 975 S95 ism 405 i350 1500000 N02 2 am sac 525 ISO 405 iano Laonuno WCLiE No a sac auo 150 405 v5 immune wwVA Stud 2 3 Wide 575 due I 50 4175 we 1209000 Cnnslmcllnn 975 mm iso 405 1550 iannimu smnaaia 2w Widry 550 325 150 405 me 1200500 r 250 we we me am 1iimaon Tabulated value Fb 850 psi Tab 4A Supp NDS 01 10 CE 479 Wood Design Lecture Notes JAR Factors NDS 01 Sec 43 Table 431 NDS 01 Page 27 Table 431 Applicability of Adjustment Factors for Sawn Lumber Load Duration Factor Wet Serv1ce Factor Temperature Factor Beam Stability Factor Size Factor Flat Use Factor Incising Factor Repetitive Member Factor Form Factor Column Stability Factor Buckling Stiffness Factor Bearing Area Factor U 0 Z 0 0 l E 0 Equot 0 0 7 139 a II 11 11 U39 x 3 VO I O 0 H1 111 l gt4 000 a 11 lt ll 39139 lt X D Z quotO I I I n quot11 139 II 139 N 00000 00000 I 9 K 111 ll 1 1 X z 000 O H n In many practical situations a number of adjustment factors may have a value of 10 A comprehensive summary of the modification factors for wood members is given in NDS Table 431 C D 2 load duration factors Sec 232 NDS C D 10 Table 232 controlled by live load CM 2 Wet service Sec 433 NDS 01 Supp Ch 4 Table 4A CM 2085 sinceMC gt 19 orl0 if Fb CF 1150 psi CF 2 size factor Sec 43 NDS 01 and Table 4A CF 13 sincerxCF 850x131105 psilt1150 psi CM 210 C Temperature factor Sec434NDS TSlOOOF Cl 11 CE 479 Wood Design Lecture Notes JAR CL 2 Beam Stability Factor Sec 435 and 333 NSD 01 Sec 4412 d b g 15 lt 20 no lateral support is required Sec 3332 CL 10 Also 3333 could be invokedif needed C I 0 Element is not loaded on its at side C Incising Factor Sec 438 done to increase treatment penetrations C 10 C Repetitive Member Factor Sec 4 3 9 NDS C 10 C Formfactor Sec 4310 amp334 C 10 Finally calculate allowable stress for bending Fb39 850 x1xx13x1105 psi Tension II to Grain FT Tension parallelto grain Table 4A Supp FT 525 psi Factors NDS 01 Sec 43 amp Table 431 CD 10 CM 10 Table 4A Supp Adj Factors CI 10 CF 13 C1 10 F 525 x13 683 psi Shear II to Grain FV FV 150 psi Factors NDS 43 12 CE 479 Wood Design Lecture Notes JAR CD 10 CM 097 Table 4A Supp Adj Factor CI 10 CZ 10 F 150 psix097146 psi Compression Jto grain FCi 405 psi MNDS 43 Table 431 Sec 433 and Table 4A CM 067 CI 10 CI 10 Cb Bearing area factor Sec 4313 Assume 1 2 6quot Cb 10 FCi 405 x 067 271 psi Compression II to grain Fc 1300 psi Factors NDS 43 Table 431 CD 10 Table 4A Supp CM 080r10 whenFcCFS750 psi CF 11 Table 4A Supp 39CM 08 since 1300X11gt750 CI 10 C1 10 Cp 10 This isabeam Fcy 1300x08x11x101144psi 13 CE 479 Wood Design Lecture Notes JAR Modulus of Elasticity MOE From Table 4A E 1300000 psi CM 2 09 Factors in Table 4A Supp Ct 10 CI 10 CT Buckling Sti hess factor for wood tresses 442 NA E 1300000 x 09 1170000 psi Ct 2 temperature factor C repetitive factor 14 CE 479 Wood Design Lecture Notes JAR Design Summa Beams 1 Determine trial beam size based on bending stress considerations long Bending stress 11 to grain 7 see Fig 61a For sawn lumber loadededgewise only are given tabulated values M Sreqd iv Fb Select trial member with use Table for dressed S4S Spr0v 2 Sreqd recheck for appropriate size factor C F since initially is unknown beam size so that M Sacl 2 Check shear Sec 34 NDS fb S F I with actual C F fV 15 E S F supp with app factors In this calculation a reduced shear d away from support face d overall depth can be used V Sec 3431a NDS 01 v V39 fV 15Z If this check shows the beam size selected to be inadequate the size is revised to provide sufficient A 15 CE 479 Wood Design Lecture Notes JAR De ection Criteria IBC 2003 Sec 35 NDS 01 Limits are established for de ections for beams trusses and similar members that are not to be exceeded under certain gravity loads Table 16043 in the IBC 2003 gives the necessary limits and other information necessary to ensure user comfort and to prevent excessive cracking of plaster ceilings TABLE 16043 DEFLECTION LIMITS quotquotquotquot39I I CONSTRUCTION L s or w39 D Ld399 Roof memberse Supporting plaster ceiling 1360 1360 1240 Supporting nonplaster ceiling l240 1240 1 180 Not supporting ceiling 1 180 1 180 1 120 Floor members 1360 1240 Exterior walls and interior partitions With brittle nishes 1240 With exible nishes 1120 Farm buildings 1180 Greenhouses 1 120 For SI 1 foot 3048 mm a For structural roo ng and siding made of formed metal sheets the total load de ection shall not exceed 160 For secondary roof structural members sup porting formed metal roo ng the live load de ection shall not exceed 1 l 50 For secondary wall members supporting formed metal siding the design wind load de ection shall not exceed 190 For roofs this exception only ap plies when the metal sheets have no roof covering b Interior partitions not exceeding 6 feet in height and exible folding and portable partitions are not governed by the provisions of this section The de ection criterion for interior partitions is based on the horizontal load de fined in Section l607l3 See Section 2403 for glass supports d For wood structural members having a moisture content of less than 16 per cent at time of installation and used under dry conditions the de ection re sulting from L 050 is permitted to be substituted for the de ection resulting from L D e The above de ections do not ensure against ponding Roofs that do not have suf cient slope or camber to assure adequate drainage shall be investigated for pending See Section 1611 for rain and pending requirements and Sec tion 15034 for roof drainage requirements 0 For Green Lumber MC gt 19 ATOTAL 20 Along term Ashort term lt L180 ALive lt L240 For Seasoned Lumber MC lt 19 16 CE 479 Wood Design Lecture Notes JAR ATOTAL 15ALong Term Asmm Termlt L180 ALive lt L240 Where ALong Tm immediate de ection due to the long term portion of the design load usually dead load Ashm Tm immediate de ection due to short term component of the design load usually live load Bearing Sec 310 NDS 01 3102 Bearing Perpendicular to Grain The actual ConlplL SSlOll 39 ptl39pelkllclllur ttt grain sllull ht based on the nut bcnl39lng ill I lltld shall non cued the ltlluwllhlc L39UIHPIL SSIOH dc lll llltlt pur pemliculttr lo gr HLlC S Vt llcnClllculmlllglmurlng area a the ends of bending membuls lttl allowance shall be made for the lad that us the member bends IJI39L SMIN upon the inner edge ol the buamtg is gleam thun al the member end 3104 Bearing Area Factor Cb Tablllzltcll Compl cltltloll leslgn tuhtcs pL lpclullCulhl nppl to beal mgs uruuy length at Itc ends at u membeL and to all llcmlngs 6quot nt more tn length at my other locallon lot hearings l 5 than uquot lu lullglll untl nut neum tlutu l tn the Cntl ol39u tnutuhct the ldbuldlctl Com tu gnun pressl n lcslgll value palpcndlclll r 10 gram Fri hull be pellllitlcd tn be lmllllpllctl by he following heanug am l39tlclm Cu lit to 2 where l heaung length measured parallel to grant m Tllls cquuttou gut the following bcallllg mm rue tots 1 rat the lndlcalt d bean length tm such small meas as plates nuu wusllets 5 4 Table 3104 Bearing Area Factors ch f t 05 lquot l5quot 2 3 4quot 6quotorllol e C L75 138 125 Ml 113 HO 100 1 l7 CE 479 Wood Design Lecture Notes JAR Example Sawn Beam Design Dimension Lumber Beams are spaced 16 inches on center roof beam Buckling of the compression zone is prevented by the plywood roof sheathing Material is No 1 Douglas Fir 7 larch Simple span is 13 ft 6 in Loads wD 19 lbft Dead load per lineal ft wL 27 lbft Live load per lineal ft TOTAL 46 lb ft Required load combination Sect 232 NDS 01 and Table 232 and Duration Factors Dalane 3 CD 09 DL CD 115 snowload Determine trial size based on bending and then check other criteria Sec 439 NDS 01 and Table 4A ofthe supplement Spacing lt 24 Cr 115 Table 4A Supp to NDS01 CF 120 MC lt 19 normal temperature conditions compression edge of bending member supported throughout in accordance with 4412 and no incision CMCTCL andCi10 Fb 1000 115 12 1151587 psi M Requ max w79m3 Fb 1587 Try 2X6 S 756 in3 From Table 1B Supp From NDS Supplement Table 4A CF 13 Fb39 1587x 1719psi 1255 3 7732m lt756 ok reqd 1719 18 CE 479 Wood Design Lecture Notes JAR Check for Shear NDS Sec 34 3K V2111 CM Ct 10 A 825 in Table 1B Supp Reel Conservative to use R Vm 460 13505 311 lbs 2 311 2 825 FV FVCDCMCIC139 180115207 psi gt565psz39 39ok fv 565 psi Check De ections E39ECMCTCCE 17000 Ksi Table 4A Supp NDSOI A 5wLLquot 5270135quotI728 L 384E I 384 I700000208 I L 135x12 240 H 067 gt 057 0k Use 2X6 N0 1 DFL MC S 19 Bearing Stress Check Sec 310 NDSOl Fem FCL CMCC17 625 Psi 033quot lt 6quot 0k WM 0 39u 19 CE 479 Wood Design Lecture Notes JAR Design of Tension Members Sec 38 NDS 01 Wood members are stressed in tension in a number of structured applications ie trusses Tension II to grain Table 431 NDS ft SF a F FI CD CM CW Ft Tabulated value Supp T where ft net lgt Am NDS Sec 312 The cross sectional area to be used in the tension stress calculation is the net area of the member This area is calculated by subtracting the projected area of any bolt holes from the grosscrosssectional area of the members 6 PROJECTED 1 6 7 AREA OF BOLT HOLE MEMBER WTH BOLT PROJECTED ARE A 0F 80L T HOLE PROJECTED AREA OF Grow6 FOR JFLT ME OR SHEAR PLATE MEMQQR WITH CONNECTOR w QAE FACE Figure 72 Netsection through two wood members One member is shown cut at a bolt hole The other is at a joint with a split ring or shear plate connector in one face plus the projected area of a bolt The bolt is required to hold the entire assembly wood members and con nectors together Photographs of split ring and shear plate connectors are included in Chap 13 Fig 132311 and b The projected areas for fasteners to be deducted from the gross area are as follows Nail holes disregarded Bolt holes computed as the hole diameter times the width of the wood member The hole diameter is between 132 and 116 in larger than the bolt diameter NDS Sec 812 In this book the bolt hole for strength calculation purposes is conser vatively taken as the bolt diameter plus 18 in Lag bolt holes a function of the connection details See NDS Appendix L for lag bolt dimensions Drill diameters for lead holes and shank holes are given in NDS Sec 912 Split ring and shear plate connectors a function of the connection details See NDS Appendix K for the projected areas of split rings and shear plates If more than one fastener is used the sum of the projected areas of all the fasteners at the critical section is subtracted from the gross area For staggered fastener pattern see NDS Sec 312 20 CE 479 Wood Design Lecture Notes JAR Example 72 Text for Spruce Pine Fir south No 1 Determine the required size of the lower tension chord in the truss shown below The loads are DL snow and the effects of roof slope on the magnitude of snow load have already been taken into account Joints are assumed pinned Connections will be made with a single row of 3A diameter bolts Trusses are 4 ftOin on centers Use No 2 southern pine surfaced d MC lt 019 Use NDSOl 4 I76 LBFT L l l Jj l 1 j Paaz x anz quot7 066 Ema In N 4 5 JL 3 i9 Lower amen EA 5 264k KB2 64K L L 4 15 s 3039 j 1 1 Figure 73a Uniform load on top chord converted to concentrated joint loads Total load horizontal plane 2 14DL 30SNOW 44 psf X 4 truss spacing 176 plf Truss analysis load to joint P 0176 x 75 32 laps Force in lower chord TAC 264 066 X 2 396 Kips service loads 21 CE 479 Wood Design Lecture Notes JAR Determine required size of tension member Assume chord will be a dimensional lumber 112 thick since MC S 19 CM 10 Table 431 NDS 01 FT39zFT CDCM CICFCI CD gt Table 232 NDS Dead Snow Load Combination CD gt 115 snow shorter duration CT 10 Sec 234 NDS T3100 F CF 2 10 see note Page 36 Supp NDS 01 thickness 2 1 12 thus already incorporated in table value E Tension parallel to the grain for No 2 southern pine 15 thick and assume 6 wide 2 725 psi Table 43 NDS 01 Supplement FT39 725 115 1 12834 psi Reqd AM 2 474 in2 E 834 Re qd Gross Area accounting for bolt hole AW 2 474 15 075 18 6053 A x 825 in2 gt 6053 Table IB Supp gm Use 2x6 No 2 Southern Pine surfaced dry 22 CE 479 Wood Design Lecture Notes JAR Example of Combined Bending and Tension LOAD r0 70 CHORD U I76 48 ms L1 J Ti J J b N F l 1 J j T L j LOAD 7390 BOTTOM amen r 32 Lap D L 4 Q 75 30 Combined Bending and Tension Sec 39 NDS 01 Let s take the truss that we ve designed the lower chord for tension only and place an additional distributed load of 32 lbft DL applied at the lower chord This load represents the weight of a ceiling supported by the bottom chord of the truss Figure 3quot Combined Bending and Axial Tension 23 CE 479 Wood Design Lecture Notes JAR Desigp Example Determine the size of the lower chord of the truss Use No 2 Southern pine surface dry MC S 19 Connections will be made with a single row of 34 diameter bolts also Connections are then assumed to be pinned Lateral buckling is prevented by ceiling Trusses are 4 on center i Determine Force in lower chord Resolve distributed load into joint loads 222 496 Free body diagram of joint A 03990 4f 4 44 u T 39 4 44 K 1 Estimate trial size of member from previous example a 2X6 was needed with the additional load in the bottom chord try 2X8 Calculate force in member Load diagram for lower truss chord taking advantage of symmetry 45 32 LdFr F l l t l r J raw44x 154 caquot r 1 5 l 1 Note that due to load application chord will be subjected to combined bending and tension 24 CE 479 Wood Design Lecture Notes JAR ii Member Design Try 2x8 from NDS Ol supp For No l 7 Southern Pine surface dry Table 4B Fl 1500 psi FT 825 psi SM l3l4l39n3 thicknessl5 wide7l quot 1 Axial tension first check tension at net section midspanbolt location Because of bolted connection M0 at this section 5quot plf izj N Am 725 15 15 18 075 956ln2 f 0464 ksl 464 psi reqd F 1 C D CM C C F C allowable The duration load factor used for the independent tension check is the CD of DL snow CD 115 Use that of shortestduration load in the combination Snow CM 10 Mcs 19 25 CE 479 Wood Design Lecture Notes JAR Ct 10 Ts 100 F CF 10 Table 4BAdjust Factors NDS 01 Supp Ft 825 115 949 pai 949 gt 464 psi 2 Axial Tension Bending Ten 39 n The combined tensile stresses are analyzed using a straightline interaction formula NDS Sec 391 page 21 a tensiontension Eq 391 391 Bending and Axial Tension Mcmbem subjccicd m a combmaiion ut ucuumg and axial cmmu taco Flgm39c 3H shall beso proportioned hm 3971 3972 tabulated neuutng destgn value mutttpneu byall apultcable amustmenl iactars except c tabulated nenutng aestgn value multiplied by all appttceme amustment factors except CL Fg tabulated design Value bending multiplied by all applicable adjustment factors except CL This is because buckling is not an issue in tension CL beam stability factor 10 26 CE 479 Wood Design Lecture Notes JAR Thus 39 Tension 39 f 4440 725x15 RV 949 psi same as pure tension M 32152x12 Bending 39 fi F 31314 408 psi at point of mambending stress no hole 822 psi 17 1500 CD C39 CD 115 mead load included bending C39 10 Table 4B Adj factS 48 gt 24 Page 36 F 1500 115 1725 psi LLiSL0 I b 408 g 093 lt 10 ok 1725 Axial tension plus compression due to bending net bending compression Check bending compression stress without axial tension remember axial tension in the truss chord is caused by snow Also selfweight of truss would reduce compression thus do not include Hence the combination of least axial tension with compression due to dead load on the chord causing bending is not critical Fb 1500x09 1350 psi 1350 gt 822psicompression Lateral buckling is prevented Use 2 x 8 No l southern pine check for shear and de ection due to chord bending must be carried out as in the beam example fbfl Fb Thus no need for check Eq 39 2 27 ca 479 WaadDeaglLecuxe Nuns JAR Cnmhined Bending and Cnmpressinn NDS 01 See 39 These members are referredto as beamrcolumns The basll stxalght hne mteracnon for bendlng and axlal tenslon Eq 3 971 ND5 01 has been modlfled as shown m Seehon 3 9 2 ofthe ND5 01 Eq 3 973 forthe ease of bendlng about one orboth pnnerpal axls and axlal compresslon Thls equahon ls lntendedto represent the followlng condluons Column Buckllng Lateral Torslonal Buckllng of Beams Beamrcolumn Interactlon P M The unlaxlal eornpressrve stressf PA whereA represents the net sectlonal area as per 3 o 3 363 Strength In Compression Parallel to Gra n The acuml Complcssion sllcss or ll e pulzlllcl lo glam shall nor cxcccd lhe lllmmblc compl39csslml deslgn mluc Ctllclllallolls 01 l sllzlll be bllscd on the Ilcl seer llml ulca 50 k 2 th ll llll lcdllcod scclloll occur lll the c illcal part ol lllc Column length lhal IS most subjccl ll polcnllul buckllllg Wllcll lllc reduced scolloll does not occnl m the annual pall ol lhe enlarnn Isllglll lhdl is mosl salnecl l0 polcmial hncllllng clllulllilllulls af shall be based on gross sccllon arcs In lldtllllonr r5 brlscd on nal scclmll men shall nol exceed lhc lablllalcd compression design nIllc parallel lo glam Illllltipllcd by all applr cable adjnsuncnl l aclors cxncpl lllc column slablllly factol S FrCDClC CMCJ 28 CE 479 Wood Design Lecture Notes JAR 39 Combined Bending and Axial Loading 391 Bending and Axial Tension Membem subjected lo a coinbiuniioii ol39lnouding nud non cnslon sec Flglll39c 3H slinll bcso piouoiuouodiuni r l T 1 0 Fl FD 3 971 and f 10 Ft 392 where R tabulated bending aegign value mulllDllEd by all applicable adluslmen factors except 2 l inoulaled bending deslgnvalue multlplled by all appllcable adluslmenl factors except cl 392 Bending and Axial Compression Monitorsnihiociodioncoinluununnorbondingnooni ono orbolh pnuoipnl a lino axial Compression sec Fl39gr uio 3n slum us so plopoltloncd ilini lei lo Y M1 nitFool 393 whom K E f in Hym lolerlhelunaxlalol blaxlalbemlng nun K E flltF 7 lorblaxialbendlng niiil ME Rm VDrblaXlal bentng r acliial edgewlse bendlng stress looming load applled lo nanoui face ol meinoei l aclual flalwlse bending stress oenoing load applled to Wide face of member a z Wlde lace dimension see Flgure 3n nanow face dlmenslon see Figul39e 3h Ellooiuo column lengths c and 39 shall be down milledlllnecordmlcelwlh3712 lg lg nndr 71 shall be determined in accordance wit 7 z nod 37 Edam nud Fm slmll oo dolomnnod in idaiioo Willi 2 3 and 33 3 393 Eccentric Compression Loading See 15 4 to members subjected lo cmnbmed bendr mg and axlal compression lluc io occeuino loading oi ccccnmc loading in ooiuouioiioii Wlili nilioi loads The combination of bending and axial compression is more critical due to the PA effect The bending produced by the transverse loading causes a de ection A The application of the axial load P then results in an additional moment WA this is also know as second order effect because the added bending stress is not calculated directly Instead the common practice in design speci cations is to include it by increasing ampli cation factor the computed bending stress in the interaction equation 29 SNOLLVI IOE UNV SNOISIAOHd NQISEG 015479 WananengnLeetrxe Netes JAR Figure 31 Combined Bending and Axial Campressian The most common case rnvolves axial eompressron eombrned wrtn bendrng abouttne strong axis ofthe cross seenon In tan case Equanon 3 93 reduees to and the amplrfreanon factor rs a number greater than 1 0 grven by the enpressron 1 Ampli cation factor for51 30 CE 479 Wood Design Lecture Notes JAR Example of Application J39 BENDHG LOAD T l l l l l l P P AXIAL a COMPRESSION quot2 7b toquot A DEFLECTOIV DUE 727 BENDMG LOAD b H m MPA ux 39 Figure 717s De ected shape of beam showing P A moment The computed bending stress 1 is based on the moment M from the moment diagram The moment diagram considers the e 39ects of the transverse load w but does not include the secondary moment P X A The PA effect is taken into account by amplifying the computed bending stress 13 The general interaction formula reduces to 1 pt Fe am Where f actual compressive stress PA F C allowable compressive stress parallel to the grain F c CDCMCCFCPC Note tlwitF C includes the Cp adj39usunentfactorfor stability Sec 3 7 NDS 01 to be considered in lengths of the column subject to buckling 31 CE 479 Wood Design Lecture Notes JAR Cp 377 Solid Columns 371 Column Stab ity FactorY OF 37 I I thn u Lamplchmn mcmbcr h39 summrlctl hmuglmm n5 lc 1 In plm cul Wanml displnccmcm m Illdircc unc 10 J IL Hm cclixc column laugh L for a mlul cuhmm mu m demnmmu m accordance an munmics of enmnccnng mechanm one mulde rm dumnnmg uchcnvc cnhnnn length when cndr xny mndiuuns mc known is to multiply actual column length by he uppro prime effective length facmr speci ed in Appendix J E new 1 7 Fm Snlid columm r1llllccmngularurmi we um Ilu Handcmcssmlm Z 39Ll dullbcmkcnaxlhu Izuvcr nl39llu mum l39 d 0H mm x v been adjusted x me Jppl opnule buckling lmgm cncl um K nom Appendix r 37 L4 11mlt1cndernus unmfm umlculumns I39M 51ml um owned 50 excel m dming conslmclmn M 51m not ucccd 75 32 7 15 I39lvu column swbxlily mum 1111 be calcnlmcd AS follows 39 g 24 c 145 x 39 3771 n hm tabulated compressmn desng value mumpned by a apphcahie adjustment amors extent CF See 231 K E r mm mayhemE1 V 03 or wsua y graded mme 3381 mrmazmne evaluamd umher MEL 04 YB or producls WYWCOV 011Sec Apomd 2 c 081039 sewn umber o 35 Vol vouw nmbev poles and piles 09 for gmsd Iammaled nmhev or structurm composne unber CE 479 Wood Design Lecture Notes JAR The actual unsupported column length multiplied by the appropriate length factor in Appendix G of the NDS 01 yields the effective column length lg i t I i l i r r 39139 I 1 I I I l I I l I l Buckling modes I39 i I l I I l l 39 l I I I I I t I II f 1 I o l 9 Theoretical K value 05 07 10 10 20 20 Rocommondod design K when ideal conditions 065 080 12 10 210 24 approximated 4 Rotation fixed translation fixed End condition 0060 x Rotation free translation fixed Q Rotation fixed translation free 9 Rotation free translation free 33 CE 479 Wood Design Lecture Notes JAR Design Problem The top chord of the truss analyzed in the case of tension and bending of the lower chord is considered for the case of combined compression and bending The bending of the top chord is due to dead plus snow being applied along the length of the member The truss analysis provides the forces in the member from AD 222 0 90 496 Free body diagram of joint A 39 444 u T 39 4 44 K 1st The length of the member from AD is 839 feet Although two load combinations D only and D S must be considered it has been determined that the D S combination controls the design and only those calculations are included herein rvo ro rap cameo U 176 LBFf 03 L1 J Ti F 75 F L 1 J I l L J LOAD 7390 earram cyaep n ur39 a 32 45 D L 4 a 75 3039 Let s try a 2 X 8 Southern Pine No 1 Table 4B NDS Supplement 01 FB 1500 psi FC 1650 psi parallel to the grain E 1700000 psi The tabulated values in 4B are size speci c thus CF comp II to the grain and CF bending are equal to 10 34 CE 479 Wood Design Lecture Notes I AR Section properties for 2X8 are A 1088 in2 and S 1314 i113 Axial check 1 Stability Column buckling can occur away from the truss joints Use gross are to calculate 2 PA 49601088 456 psi lateral support is provided by the roof diaphragm thus as the member is used edgewise buckling is prevented about the weak axis d2 direction Figure 31 Combined Bending and Axial Compression E gt1lt z 2839122139 d1 725 E E CACt 1700000 11 1700000 psi For Visually graded sawn lumber Section 3715 K53 03 c 08 FEE ch kE 9led12 03170000001392 2645 psi FEM FCCDCMCtCFC 165011510101010 1898 psi Substituting in Eq 371 results in C p 0792 Fig F Cy 1898 0792 1502 psi gt 456 psi OK Note that at the connections the reduced area should be used to check compression but there is no possibility of buckling braced location Assuming a bolt diameter plus 18 for the opening diameter of 0875 the net area is 15725875 f 956 in2 Thus the acting axial stress is 4960956 518 psi The design stress is F5 F C 1898 gt 518 OK 35 CE 479 Wood Design Lecture Notes JAR Bending check Assume pinned ends for the chord member and take load and span on the horizontal plane wL217675212 8 81000 M 149 ink MS 149 10001314 1130 psi The truss chord acting as a beam has full lateral support and the lateral stability factor C 10 ande 1500 115 1725psz39 gt 1130psi OK Combined Stress check There is no bending about the weak axis and the axial load is concentric Thus Equation 393 reduces to Z l F 51 I 7 1 F11 CE The load duration factor that controls is snow CD 115 Therefore the previously determined values of F c and F lb are still valid for use in the interaction formula The elastic buckling factor depends on the slendemess ratio about the strong axis of 1392 It must be noted that the buckling stress FC for the axial check is based on the lgdmax whereas for the PA effect is based on the axis about which the bending moment occurs strong axis based on 611 In this example it is a coincidence that the two values of the buckling stress FC are the same f F f s 10 f FCE 2645 psi 1130 Z 0884lt10 OK Use 2 x 8 No 1 Southern Fine 1502 1 456 1725 2645 It is appropriate in an actual design situation to check the Dload only case with a load duration factor CD of 09 36

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