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# Design Of Building Components And Systems CE 47900

Purdue

GPA 3.53

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This 37 page Class Notes was uploaded by Mrs. Brianne Jaskolski on Saturday September 19, 2015. The Class Notes belongs to CE 47900 at Purdue University taught by Julio Ramirez in Fall. Since its upload, it has received 81 views. For similar materials see /class/208027/ce-47900-purdue-university in Civil Engineering at Purdue University.

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Date Created: 09/19/15

CE 479 Wood Design Lecture Notes JAR Introduction zes 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 Fare Widths inches Minimum er ed Minimum dressed W Item Nominal Dry Green Nominul Dry Green Boards 34 58 i 34 714 I 1712 144 SillHEdOHd NOILOEIS Dimension Lumber 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 2quot4quot limitsz All species exoept Southern Pine see Table 4B Tabulated design values are for normal load duration and dry service conditions See NDS 43 for a comprehensive description of desagn value adjustment factors USE WITH TABLE 4A ADJUSTMENT FACTORS DESlgi39l values In pounds par aquare inch psri Tension Shear Compression Compression Modulus parallel parallel perpendicular parallel oi Grading Specres and Size Bending in gram 0 grain 13 grain to grain Elasticily I Rules cor r rmersial grade classilrcalrcn U F FC E agency Salsci Structural 375 500 20 255 l r25 1100000 No 625 375 120 265 j 600 1100000 No 8 Wider EDD 50 120 265 450 1000000 NELMA U No3 2330 900 120 365 275 900000 NSLB m Slud 2quot amp wider 475 275 120 265 300 V 900000 r JWPA Construction 700 400 V i 1211 265 gs 900000 Standard 2 4 wide 375 225 39 2 39 12o 265 39 475 99mm Uiility39 39 39 39 I 39 l5 100 39 120 39 265 39 SOD 39 80000 E l 39 39 Select Slruclural l z 1450 850 195 715 120i 1 7300000 No1 1050 600 195 T15 93930 EC 00UO m Nil 239 81 3939ILler 1000 500 13935 7 1 5 750 LEDODUD No3 575 350 195 l5 425 1300000 i JELI AA 2 8r wider 775 450 195 715 175 300000 Conmruclion 115i 675 195 715 1000 1400000 39 Siandard 2quot4 wide 39 650 375 195 39 715 39 7 75 LSOOJJGIJ U lily39 300 175 39 195 715 500 1200 ODD Allowable StressesDesi 11 Values NDS Tabulated values in the NDS Su lement 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 HEAKTWOOD SAPWOOD LA TEWOUD EA KAY W000 A N N UA L RNG Figure 44 Cross section of a log F l F x adjustment factors Design value 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 fl 3 Fl Design Value Tabulated Stress Allowable adjusted stress Stress Bending Fb Fb Tension parallel to grain F1 F1 Shear parallel to grain FV Fv Compression perpendicular Fcl Fclj to grain Compression parallel to Fe Fe grain Modulus of elasticity E 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 23 2etc 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 Adjustment Factors Sec 43 NDS 01 and Sunlement 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 SGrn 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 ofSGRN or SDry EMC exceeds 19for an extended period of time table Values should be multiplied by CM see Page 30 and others for Values in Tab e 4 NDSSupp 01 Wet Service Factor When dimension lumber ls used where moisture co 7 lelll wull exceed 9 an extended llme period design values shall be multiplied by me appropriale Wet service to mm the followmg fol r en llc 50p 6 4llcnl 7lcgs waging l0 B Load Duration Factor CD Wood an handle higher stresses ifloads are applied for a shortperiod oftime All tabulated Values apply to ormal 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 des39g uenlly Used Load i Table 232 Freq Durat on Factors CD1 Lulu Dunninn l V pica Des gn Loads Pol mallclll o 9 Dead Loud Ten yum I 0 Occupancy lee Loall l wo momlu l is s w Ll Sc Hillyx 125 onstructllmlmllll Ten minule l 5 m r llquakc Load ln pncl39 a lmpdul Load I Lon mum mm dwell m Apply m modulus ol clmvmh 1 nm in mwmwn pclTlellLlllJr ll glam 4W mlum r based 0 l dcfanm mm mm 2 l m llmlmn lac ml CA W lam Lbcmlml tolmccllnm or gr cr um I o 5 4mm m lllc mm lCl l low nlmcmbels t licl cvcmc m m rc CE 479 Wood Design Lecture Notes 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 39 to 4 thick shall be multiplied by the following size factors Size Factors CF Grades Width depth Select Structural amp tr No No2 No3 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 Cu ing esign values F for dimension lumber lquot to 4 thick shall be multiplied by the repetitive mcmhci factor C 2 HS when such members are used asjmstx truss chords rafters studs planks docking on similar members which are in Contact or spliced not more ilnm 14quot on centers um not less than 3 in number and u cjoineil by floor roofor other load distributing elements adequuic to support the design loa 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 eWise or load applied to narrow face If however load is a applied to the Wide face 7 the stresses may be increased by C V Flat Use Factor Cm cndmg design dlmb adjusted by size mum arc bu 39c on edge ISC c loud Ippllcd Lo mln39uw tun limclhirm lumber is med lliiiwlw Hum applied in wu u lime ihe bending design uluc Fquot hnll ulsn be mullir Plltd by he fullm nu Hm um I urmrs Flul Use Fuclnr mm V Tlucknei dcplhl t T 4quot T H i I l I L15 quot 115 llquot 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 l 39 AXIS m the Cross secuon at al beam is lht line on which hem l5 ncuher tension no compression mess Figure 1A Dimensions for Rectangular Cross Section Glued Liimmmcd Tunbm v Slum 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 no al 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 Table 233 Temperature Factor c ln Sm Ck Design Muislun mins Cnmlilmm El Dr NW n l11111bL lm l 111 11 Kim 1r 1111mm 1 1111t 1111111l1n1ur 1 4s14nnnu13resneerirsii 1 1l 11nd 1r senire L umlllhm 1 menunmneil 1mm 11 111 1 n syccil icd in 1111 pun G Form Factor Cf The purpose of this factor is to adjust tabulated bending stress F1 for nonrectangular sections see Section 334 inN S 0 334 Form Factor cf Tabulaied bending design values 11 101 bending members will1 enher a eirenim c1 ess seenen or 11 smmre cross section loaded in the plane ofthc dmgonal diamond sectlan 5111111 be multiplied by me follmving renn rise rers c1 Table 334 Form Factors c Ci ound Seerion 1 11 Diamond Sermon 1414 These fumi factors msnre ler n circular or dmmcnd sbrned bending member has me Same mumth capociry as a square bending member having he same crasssec area 1m erren1nr member is mperen 11 shall be nsidcrcd a beam oh ariablc cross 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 352 39 Inluea for Visualy 39 39 a 4quot thickquot2 Cont All species except Southern Fine see Table as Tabulated deetgn values are for normal load 39 39 p ND 4 IU adjustment factors 7 Us wm TAELE 4A ADQU TMENT FAcroR quot 7 7 7 Commasm cummm moan ts p cu ar pawuel m Guiding sooto m a a x 5 st Emmy mmcmzl g39aaa cummam E 9 2 in 3mm H 2 a My 2 s was 2w Wide Tabulated value Fb 850 psi Tab 4A Supp NDS 01 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 E p o l E o Equot 0 p D 139 0quot H mm 539 x a 111 n N O O O U z p O 0 H1 11 lt ll 39139 lt X D Z quotO I I I O O r n quot11 n 39 i II o 139 II 139 X 002000 00000 I D 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 x1xxl3x1105 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 xl3 683 psi Shear II to Grain FV FV 150 psi Factors NDS 43 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 psi Factors NDS 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 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 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 fb M SFbV with actualCF Sacl 2 Check shear Sec 34 NDS 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 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 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 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 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 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 pa 0quot as F For Green Lumber MC gt 19 ATOTAL 20 Along term Ashort term lt L180 ALive lt L240 For Seasoned Lumber gMC 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 compression muss ptl39pemllclllilr In grain shall be bilth on the not bcul39lng lr cud lllc hlhlwllhh Cmnplcssmn dc hh llllll per pendicular U L l C ml chklllan area a the and of bending members In allowance shall be made for the fact llmt us Ill member bends ju39usgtll39c upun the irmer edge or lhe beam male than al lhc member end 3104 Bearing Area Factor Cb cltltl0n deslgn values pmpcndlctllm cndsof l at l L 11 membel am L lcn olhcr localloh l ol39beanllgs less llmn hquot in nearer llmn 1quot m the and old mumbcL llic ldbulalcd com presslollr Value perpendicular lo gmlh E1 shall be pennincd m be mllllipllcrl by lhc lolluwmg bc nug am 39 lDL Ch la 10 2 where lquot hcahng lenglh measured parallel to glam m Tlus equation gives the following benlllig mm fucr tors 1V for lhc m lcalcd bedrng length on such small areas as plalci ml wusllels39 Table 3104 Bearing Area Factors ch 3 05 lquot 15quot g c L75 138 125 M9 113 no 397 3 iiquot or more I 00 CE 479 Wood Design Lecture Notes JAR Exam le Sawn Beam Desi n imension 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 12515 Fb 1587 79 m3 Try 2X6 S 756 in3 From Table 1B Supp From NDS Supplement Table 4A CF 13 F 1587x 1719psi 1255 3 7732m lt756 39ok W 1719 CE 479 Wood Design Lecture Notes JAR Check for Shear NDS Sec 34 3 V 7 7 Rect 2 bd CM Ct 10 A 825 in Table 1B Supp Conservative to use R Vm 460 13505 311 lbs v 7311565 psi 2 825 FV FVCDCMCIC139 180115207 psi gt565psz39 39ok Check De ections E39ECM CT CCE 17000 Ksi Table 4A Supp NDSOI 5wLLquot 5270135 1728 2057 L 384E 1 384 1700000208 L 135 x 12 2 240 240 067 gt 057 0k Use 2X6 N0 1 DFL MC S 19 Bearing Stress Check Sec 310 NDSOl Fem FcL CMCC17 625 Psi i amp ReqdA 05m2 FOL 625 Reqd 17 gt 9 033quot lt 6quot 0k b 15 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 SFt a F FI CD CM CW Ft Tabulated value Supp 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 i i a i El PROJECTED 1H AREA OF BOLT HOLE MEMBER WTH BOLT PROJECTED ARE A 0F 80L T HOLE PROJECTED AREA OF Grow6 FOR JFLT ME OR SHEAR PLATE MEMBER WITH CONVECTOR IN ONE 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 J I76 LBFT L 1 1 J l 1 J Paaz x P5132 066 Ema In N A a A Lower amen 1 EA 5 264k KB2 64K L L 4 15 s 3039 I 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 reguired 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 TIJ J 9 Lz I I j L J LOAD 7390 aorrom amen ar32 Lap D l 4 Q 75 30 J 1 II 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 3H 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 0 90 496 Free body diagram of joint A 39 444 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 m 1 a m 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 l3l4in3 thicknessl5 wide7l quot 1 Axial tension first check tension at net section midspanbolt location Because of bolted connection M0 at this section 5quot 0113 ml a l Am 725 15 15 18 075 956in2 444 fr 956 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 0464 ksi 464 psi reqd CD 115 Use that of shortestduration load in the combination Snow CM 10 Mcs 19 CE 479 Wood Design Lecture Notes JAR Ct 10 Ts 100 F CF 10 Table 4BAdjust Factors NDS 01 Supp Ft 825 115 949 pxi 949 gt 464 psi 2 Axial Tension Bending Ten 39 n TL 39 39 39 formula NDS Sec 391 page 21 7 tensiontension Eq 391 391 Bending and Axial Tension Membem subjected lo a combination ol bcmlmu lllKl mu icmlou soc Flgm39c 3H shall be so propomonud lull lquot 10 3971 3972 herb tabulated bending 195ng value mulupnea byall apuhcame acjuetmenl lactors except c tabulated nenmng design value multiplied bv all applicable admstmem factors except cv Fgtabulated design quot quot 39rquot by all Hquot L except CL This is because buckling is not an issue in tension CL beam stability factor 10 CE 479 Wood Design Lecture Notes JAR Thus 39 Tension 39 f 4440 725x15 RV 949 Psi same as pure tension Bending 39 fb M W S 813I4 408 psi at point of mdxbending stress no hole 822 psi F 1500 CD C39 CD 115 mead load included bending C39 10 Table 4B Adj factS 48 gt 24 Page 36 Fb 1500115 1725 psi Li S 10 F 5 g 093 lt 10 0k 908 I 725 Axial tension plus 39 due to bending net bending 39 l 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 015479 WaadDeaquecuxe Nmes JAR Cnmhinerl Bending and Cnmpressinn NDS 01 Sec 39 l 71 3 9 2 calls ND5 01 Eq 3 973 forthe case of bendmg about one orboth pnnclpal axls Column Bucklmg Lateral Torslonal Bucklmg of Beams aerolumn Interacuon P M PA l r n r 3 s 3 363 Strength in Compression Parallel to Grain m zlclunl Complcsslon sllcss or force palzlllel m x ccd llm ullmvablc compl39csslon dalgn 01 l all be bllscd 0 the no 11ml ulczl Sec II 21L llhl l cdll00d SCCHOD occurs ln the c incle part ol lllc column lenglh llm IS most sublcul lo polcnlml buckllllg thll lhc reduced scale 1 does nol occm m we cnm pa unlu cnlllmll lcnglll Ihal is mosl 51mm to polcnlilll bncllllng cum illluns org shall be ba Cd an gross sccnon arc llldtllumL 5 based on nul section men shall nol exceed lhc lablllalcil cmnprcssiun dcsl n lame pamllcl lo glam mllllipllcd by all appllr xccpl lhc column slnblllly Fatima C S FlCn hlllil CHM Wood Design Lecture Notes JAR 39 Combined Bending and Axial Loading 391 Bending and Axial Tension when K E Member subjcctcd ma combnmmn m bending and f F ywmmnana m maxralbendmg Mile lemon sec Flgm39c 3H shall bcso pmpomommm l M d l and 3971 K 39 Tldz r or blaxial bending PM E 392 RB iormxxal bendlng where r actual eagemse bending stress lbenclmg load applied to narrow face or membel tabulated mending dosvgn value multiplled by all f actual Hamse bendmg stress hendl g load appllcable adjustment lactors except 2 applled to We face of member tabulated hemlng desrgnvalue multiplied by all a 7 was lace mmenslon see Flgure 3n n h applicable amusunem factors except c 12 nanow face drmenslonlsee Figure 3n 392 Bending and Axial n Effeclnc column lengths 9 ml slmll be deter ompressm F hall mlncdumccorilance W113 12 i Fm and be detcrmmcd in accordance wit 7 and a shall be determined in 33 3 and 3 7 F n 2 Memberssubicclcdmacamhmauon ofbmdmgaboul ulancc with 2 3 and no orboih prinmpzll a 2m axial compression sec Flgr urc 31 shall be so nl opoltloncd mm 2 3 3 Eccentric Compression L i in Loading F n 1 ilkF in l I See 154 fox members subieclcd to combined bend u g and axial compression iluc lo ecccnmc loaillng or 393 1 ccccnmc loading in combination Wllll other loads 152 I 7710 Fm we N no hr 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 SNOILVI IOS GNV SNOlSlAOHd NQISEG 015479 WaadDeaquecuxe Nmes JAR Figure 31 Combined Bending and Axial Compression strong axis 0mg cross section In this case Equation 3 93 reduces to CE 479 Wood Design Lecture Notes JAR Example of Application J39 BENDMG LOAD T l l l l l l P P AXIAL quot392 L quot COMPRESSION A 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 E 479 Wood Design Lecture Notes JAR CE 37 Solid 39 371 Column Stab ity FactorY 0 I When 11 Lamplc n Icnglh In In x L I lun lncmbcr ls39 sllmmrlctl ul lateral displ mum m 5 lhruuglmul 1 J 72 rlm c cclilc uvlnmll lcnglh 1 for I mlul column man he deman m accordanccwnll pnm lplc of cnulnccnm mechanle one mulde rm dmnmmng umzcllvc cnlumn lenglll when cndrl39lxlly wmlllmns mo lule l is to multiply scum column length by the appro pn39alc effective length facmr speci ed in Appendlx o 6 41ml 1 7 1 Fl mm Ilw gt1Cl llCmCSST39 nl lhu mum 15de l ween mama w w solid columlh ul Iccmngularurms cc mm u u mu be lakcnzh Ihu lzllvm ltuc rcEGHUun m We App upmllc buckling lmlgm unl lrom Appcnlllx r39 714 llmlcndemm mm rm 0 d mlumm 1 d slmll um meccd 50 exccpl Ilml liming consll ucllml U41 mnu cocl 51m nul uxcccd 73 3 1 5 HM column Sklbllily quotmm 111 be calculated as follows u here Mam WEEFE Fl c vquot 24 39 c 3771 r lanulalen compresslon nnslgnvalue mulnplled by all anphcahle adjusvmenl laclors extent cp See 23 mo LBSQKCOVEJ V 0310 vlsual y graded Hl lbel 0354 mrmazmne evalualed lumber MEL 7 04 la la producls vnlllcuvr 1011 see 9 a 035 la mum nmuer poles and rules 19 for gluedlammamn umbev arstmctural composlle lunbev 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 t t I i r r r I I I I I l I I l I l Buckling modes l I i I I l l I I i I I I I i I II 1 i I o 1 1 Theoretical K value 05 07 10 10 20 20 Rocommondod design K when ideal conditions 065 080 12 10 210 24 approximated W Rotation fixed translation fixed End condition code Rotation free translation fixed Q Rotation fixed translation free 9 Rotation free translation free 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 222 496 Free body diagram of joint A 090 4 44 u T 39 4 44K 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 tvo 7390 rap cacao U 176 LBFf 03 L1 J TIFJ B A D L i r f L J LOAD 7 0 BOTTOM cyaeo ar g 32 4557 0 L 4 Q 75 3039 Jquot l 2539 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 ins 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 Q gt1lt z 2839122139 d 1 725 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 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 M wL2176752 12 8 81000 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 I fb1 s 10 F M 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 FCE 2645 psi 2 E 0884lt10 OK Use 2 x 8 No 1 Southern Fine 1502 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 Data Pu k 3 e in 125 fm ksi 15 h 1667 Pe at roof 375 ink Pe at 8 180 t in 763 Solicitations due to wind Section Pro 8 in block A in2 30 face shell bedding At 867 it below roof Mmax 768 ink l in4 309 face shell bedding Just under roof level M 133 ink A in2 415 fully bedded I in4 334 fully bedded Total Moments weight 0048 k ungrouted under roof M ink 508 lg in4 44419 grouted A in2 9156 at Max Mw M ink 948 r in weight 006 k grouted a Check stresses just below roof line fa s39 0105 Calculate stiffness param eters r in 234 Fa ksi 0280 Eq 212 hr 7051 Unity Eq 0502 less than 10 Ok Em ksi 1350 Note that axial stress exceeds bending fb ksi 0062719 Fb ksi 05 thus there is no net tension b Check stresses at maximum moment section due to wind fa ksi 0119 fb ksi Unity Eq 0660 less than 10 Ok There is no net tension since bending stress does not exceed axial compression fa 0002 ksi with type M PCL mortar from Table 2232 in MSJC 53002 Ft 25 psi for ungrouted masonry c Stability check Plt 025 Pe assiming a fully bedded masonry wall Note that e is the ratio of moment at the section not including wind effects and the axial load at the same section Pe k 4855188 025 Pe 1214 greater than Pmax 42 Kips e in Thus Ok

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