FEMA P-751: Chapter 11: Wood Design

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11Wood DesignPeter W. Somers, P.E., S.E.Contents11.1 THREE-STORY WOOD APARTMENT BUILDING, SEATTLE, WASHINGTON . 311.1.1Building Description . 311.1.2Basic Requirements . 611.1.3Seismic Force Analysis. 911.1.4Basic Proportioning . 1111.2 WAREHOUSE WITH MASONRY WALLS AND WOOD ROOF, LOS ANGELES,CALIFORNIA . 3011.2.1Building Description . 3011.2.2Basic Requirements . 3111.2.3Seismic Force Analysis. 3311.2.4Basic Proportioning of Diaphragm Elements . 34

FEMA P752, NEHRP Recommended Provisions: Design ExamplesThis chapter examines the design of a variety of wood building elements. Section 11.1 features a threestory, wood-frame apartment building. Section 11.2 illustrates the design of the roof diaphragm and wallto-roof anchorage for the masonry building featured in Section 10.1. In both cases, only those portions ofthe designs necessary to illustrate specific points are included.Typically, the weak link in wood systems is the wood strength at the connections, but the desired ductilitymust be developed by means of these connections. Wood members have some ductility in compression(particularly perpendicular to grain) but little in tension. Nailed plywood shear panels developconsiderable ductility through yielding of nails and crushing of wood adjacent to nails. Because woodstructures are composed of many elements that must act as a whole, the connections must be consideredcarefully to ensure that the load path is complete. Tying the structure together is essential to goodearthquake-resistant construction.Wood elements often are used in low-rise masonry and concrete buildings. The same basic principlesapply to the design of wood elements, but certain aspects of the design (for example, wall-to-diaphragmanchorage) are more critical in mixed systems than in all-wood construction.Wood structural panel sheathing is referred to as “plywood” in this chapter. However, sheathing caninclude plywood and other products, such as oriented-strand board (OSB), that conform to the appropriatematerials standards.The calculations herein are intended to provide a reference for the direct application of the designrequirements presented in the 2009 NEHRP Recommended Provisions (hereafter, the Provisions) and itsprimary reference document, ASCE 7-05 Minimum Design Loads for Buildings and Other Structures(hereafter, the Standard) and to assist the reader in developing a better understanding of the principlesbehind the Provisions and the Standard.In addition to the Provisions, the documents below are referenced in this chapter. Although the Standardreferences the 2005 edition of the AF&PA SDPWS, this chapter utilizes the 2008 edition, which is themore recent, updated version. Note that the 2005 editions of the AF&PA NDS and AF&PA NDSSupplement are the latest versions.ACI 318American Concrete Institute. 2008. Building Code Requirements andCommentary for Structural Concrete.ACI 530American Concrete Institute. 2005. Building Code Requirements forMasonry Structures.ANSI/AITC A190.1American Institute of Timber Construction. 2002. Structural GluedLaminated Timber.ASCE 7American Society of Civil Engineers. 2005. Minimum Design Loads forBuildings and Other Structures.AF&PA GuidelineAmerican Forest & Paper Association 1996. Manual for EngineeredWood Construction (LRFD), Pre-Engineered Metal ConnectorsGuideline.AF&PA NDSAmerican Forest & Paper Association. 2005. National DesignSpecification.11-2

Chapter 11: Wood DesignAF&PA NDSAmerican Forest & Paper Association. 2005. National SupplementDesign Specification, Design Values for Wood Construction.AF&PA SDPWSAmerican Forest & Paper Association. 2008. Special Design Provisionsfor Wind and Seismic.WWPA RulesWestern Wood Products Association. 2005. Western Lumber GradingRules.11.1 THREE- is example features a wood-frame building with plywood diaphragms and shear walls.11.1.1 BuildingDescriptionThis three-story wood-frame apartment building has a double-loaded central corridor. The building istypical stick-frame construction consisting of wood joists and stud bearing walls supported by a concretefoundation wall and strip footing system. The seismic force-resisting system consists of plywood floorand roof diaphragms and plywood shear walls. Figure 11.1-1 shows a typical floor plan andFigure 11.1-2 shows a longitudinal section and elevation. The building is located in a residentialneighborhood a few miles north of downtown Seattle.The shear walls in the longitudinal direction are located on the exterior faces of the building and along thecorridor. The entire solid (non-glazed) area of the exterior walls has plywood sheathing, but only aportion of the corridor walls will require sheathing. In the transverse direction, the end walls and one lineof interior shear walls provide lateral resistance. It should be noted that while plywood sheathinggenerally is used at the exterior walls for reasons beyond just lateral load resistance, the interiorlongitudinal (corridor) and transverse shear walls could be designed using gypsum wallboard as permittedby AF&PA SDPWS Section 4.3.7.5. However, the corridor shear walls are not included in this exampleand the interior transverse walls are designed using plywood sheathing, largely due to the required shearcapacity.The floor and roof systems consist of wood joists supported on bearing walls at the perimeter of thebuilding, the corridor lines, plus one post-and-beam line running through each bank of apartments.Exterior walls are framed with 2 6 studs for the full height of the building to accommodate insulation.Interior bearing walls require 2 6 or 3 4 studs on the corridor line up to the second floor and 2 4 studsabove the second floor. Apartment party walls are not load-bearing; however, they are double walls andare constructed of staggered 2 4 studs at 16 inches on center. Surfaced, dry (seasoned) lumber is used forall framing to minimize shrinkage. Floor framing members are assumed to be composed of Douglas FirLarch material and wall framing is Hem-Fir No. 2, as graded by the WWPA Rules. The material andgrading of other framing members associated with the lateral design is as indicated in the example. Thelightweight concrete floor fill is for sound isolation and is interrupted by the party walls, corridor wallsand bearing walls.The building is founded on interior footing pads, continuous strip footings and concrete foundation walls(Figure 11.1-3). The depth of the footings and the height of the walls are sufficient to provide crawlspaceclearance beneath the first floor.11-3

FEMA P752, NEHRP Recommended Provisions: Design Examples148'-0"28'-0"25'-0"Post &beamlines25'-0"9'-0"56'-0"6'-0"9'-0"8'-0"11 2" Lightweight concreteover plywood deck onjoists at 16" o.c.Typical apartmentpartitionsFigure 11.1-1 Typical floor plan(1.0 ft 0.3048 gure 11.1-2 Longitudinal section and elevation(1.0 ft 0.3048 m)11-413'-0"

Chapter 11: Wood DesignPadfootingsContinuous footingand grade wallFigure 11.1-3 Foundation plan11.1.1.1 Scope. In this example, the structure is designed and detailed for forces acting in the transverseand longitudinal directions, including the following:§ Development of seismic loads using the Simplified Alternative Structural Design Criteria (hereinreferred to as the “simplified procedure”) contained in Standard Section 12.14.§ Design and detailing of transverse plywood walls for shear and overturning moment.§ Design and detailing of plywood floor and roof diaphragms.§ Design and detailing of wall and diaphragm chord members.§ Design and detailing of longitudinal plywood walls using the requirements for perforated shearwalls.The simplified procedure, new to the 2005 edition of the Standard, is permitted for relatively short,simple and regular structures utilizing shear walls or braced frames. The seismic analysis and designprocedure is much less involved than a building utilizing a seismic force resisting system listed inStandard Section 12.2 and analyzed using one of the procedures listed in Standard Section 12.6. SeeSection 11.1.2.2 for a more detailed discussion of what is and is not required for the seismic design. Inaccordance with Standard Section 12.14.1.1, the subject building qualifies for the simplified procedurebecause of the following attributes:§ Residential occupancy§ Three stories in height§ Bearing wall lateral system§ At least two lines of lateral force-resisting elements in both directions, at least one on each side ofthe center of mass§ No cantilevered diaphragms or structural irregularities11-5

FEMA P752, NEHRP Recommended Provisions: Design Examples11.1.2 BasicRequirements11.1.2.1 Seismic ParametersTable 11.1-1 Seismic ParametersDesign ParameterValueOccupancy Category (Standard Sec. 1.5.1)IIShort-Period Response, SS1.34Site Class (Standard Sec. 11.4.2)DSeismic Design Category (Standard Sec. 11.6)DSeismic Force-Resisting System(Standard Table 12.14-1)Wood Structural PanelShear WallsResponse Modification Coefficient, R6.511.1.2.2 Structural Design Criteria11.1.2.2.1 Ground Motion Parameter. Unlike the typical design procedures in Standard Chapter 12,the simplified procedure requires consideration of just one spectral response parameter, SDS. This isbecause the behavior of short, stiff buildings for which the simplified procedure is permitted will alwaysbe governed by short-period response. In accordance with Standard Section 12.14.8.1:SDS 2/3FaSSThe site coefficient, Fa, can be determined using Standard Section 12.14.8.1 with simple default valuesbased on soil type or using Standard Table 11.4-1 if the site class is known. Since Standard Table 11.4-1generally will result in more favorable value, that method is used for this example. Using SS 1.34 andSite Class D, Standard Table 11.4-1 lists a short-period site coefficient, Fa, of 1.0. Therefore, inaccordance with Standard Equation:SDS 2/3(1.0)(1.34) 0.8911.1.2.2.2 Seismic Design Category (Standard Sec. 11.6). Where the simplified procedure is used,Standard Section 11.6 permits the Seismic Design Category to be determined based on StandardTable 11.6-1 only. Based on the Occupancy Category and the design spectral response accelerationparameter, the subject building is assigned to Seismic Design Category D.11.1.2.2.3 Seismic Force-Resisting Systems (Standard Sec. 12.14.4). See Figure 11.1-4. For bothdirections, the load path for seismic loading consists of plywood floor and roof diaphragms and plywoodshear walls. Because the lightweight concrete floor topping is discontinuous at each partition and wall, itis not considered to be a structural diaphragm. In accordance with Standard Table 12.14-1, building has abearing wall system comprised of light-framed walls sheathed with wood structural panels. The responsemodification factor, R, is 6.5 for both directions.11-6

Chapter 11: Wood tedexteriorwall56'-0"25'-0"Some corridorwalls are usedas shear ll30'-0"Figure 11.1-4 Load path and shear walls(1.0 ft 0.3048 m)11.1.2.2.4 Diaphragm Flexibility (Standard Sec. 12.14.5). Standard Section 12.14.5 defines adiaphragm comprised of wood structural panels as flexible. Because the lightweight concrete floortopping is discontinuous at each partition and wall, it is not considered to be a structural diaphragm.11.1.2.2.5 Application of Loading (Standard Sec. 12.14.6). For the simplified procedure, seismic loadsare permitted to be applied independently in two orthogonal directions.11.1.2.2.6 Design and Detailing Requirements (Standard Sec. 12.14.7). The plywood diaphragms aredesigned for the forces prescribed in Standard Section 12.14.7.4. The design of foundations is perStandard Section 12.13 and wood design requirements are based on Standard Section 14.4 as discussed ingreater detail below. This example does not require any collector elements (Standard Sec. 12.14.7.3).11.1.2.2.7 Analysis Procedure (Standard Sec. 12.14.8). For the simplified procedure, only one analysisprocedure is specified and it is described in greater detail in Section 11.1.3.1 below.11.1.2.2.8 Drift Limits (Standard Sec. 12.14.8.5). Where the simplified procedure is used, there are notany specific drift limitations because the types of structures for which the simplified procedure isapplicable are generally not drift-sensitive. As specified in Standard Section 12.14.8.5, if a determinationof expected drift is required (for the design of cladding for example), then drift is permitted to becomputed as 1 percent of the building height unless a more detailed analysis is performed.11.1.2.2.9 Combination of Load Effects (Standard Sec. 12.14.3). The basic design load combinationsare as stipulated in Standard Chapter 2 as modified by the Standard Sec. 12.14.3.1.3. Seismic loadeffects according to the Standard Equations 12.14-5 and 12.14-6 are as follows:E QE 0.2SDSDE QE - 0.2SDSD11-7

FEMA P752, NEHRP Recommended Provisions: Design Exampleswhere seismic and gravity are additive and counteractive, respectively.For SDS 0.89, the design load combinations are as follows:(1.2 0.2SDS)D 1.0QE 0.5L 0.2S 1.38D 1.0QE 0.5L 0.2S(0.9 - 0.2SDS)D - 1.0QE 0.72D - 1.0QENote that there is no redundancy factor for the simplified procedure.11.1.2.3 Basic Gravity Loads§ Roof:Table 11.1-2 Roof Gravity Loads§ Load TypeValueLive/Snow Load(in Seattle, snow load governs over roof live load;in other areas this may not be the case)25 psfDead Load(including roofing, sheathing, joists, insulation and gypsum ceiling)15 psfFloor:Table 11.1-3 Floor Gravity Loads11-8Load TypeValueLive Load40 psfDead Load(1-1/2-in. lightweight concrete, sheathing, joists and gypsum ceiling.At first floor, omit ceiling but add insulation.)20 psfInterior Partitions and Corridor Walls(8 ft high at 11 psf)7 psf distributedfloor loadExterior Frame Walls(wood siding, plywood sheathing, 2 6 studs, batt insulation and5/8-in. gypsum wallboard)15 psf of wallsurfaceExterior Double Glazed Window Wall9 psf of wallsurfaceParty Walls(double-stud sound barrier)15 psf of wallsurfaceStairways20 psf

Chapter 11: Wood DesignTypical Footing (10 in. by 1 ft-6 in.) andStem Wall (10 in. by 4 ft-0 in.)690 plfApplicable Seismic Weights at Each LevelWroof Area (roof dead load interior partitions party walls) End Walls Longitudinal Walls182.8 kipsW3 W2 Area (floor dead load interior partitions party walls) End Walls Longitudinal Walls284.2 kipsEffective Total Building Weight, W751 kipsFor modeling the structure, the first floor is assumed to be the seismic base, because the short crawlspacewith concrete foundation walls is stiff compared to the superstructure.11.1.3 SeismicForceAnalysisThe analysis is performed manually following a step-by-step procedure for determining the base shear(Standard Sec. 12.14.8.1), vertical distribution of forces (Standard Sec. 12.14.8.2) and horizontaldistribution of forces (Standard Sec. 12.14.8.3). For a building with flexible diaphragms, StandardSection 12.14.8.3.1 allows the horizontal distribution of forces to be based on tributary areas andaccidental torsion need not be considered for the simplified procedure.11.1.3.1 Base Shear Determination. According to Standard Equation 12.14-11:V FS DSWRWhere F 1.2 for a three-story building, R 6.5 and W 751 kips as determined previously. Therefore,the base shear is computed as follows:V (1.2)(0.89)(751) 123.4 kips (both directions)6.511.1.3.2 Vertical Distribution of Forces. Forces are distributed as shown in Figure 11.1-5, where thestory forces are calculated according to Standard Equation 12.14-12 as follows:Fx wxVWThis results in a uniform vertical distribution of forces, where the story force is based on the relativeseismic weight of the story with all stories at the same seismic acceleration (as opposed to the triangularor parabolic vertical distribution used in the Equivalent Lateral Force procedure of Standard Sec. 12.8)11-9

FEMA P752, NEHRP Recommended Provisions: Design ExamplesFroofF3rdF2nd9'-0" 9'-0" 9'-0"56'-0"Roof3rdfloor2ndfloorGround floorFigure 11.1-5 Vertical shear distribution(1.0 ft 0.3048 m)The story force at each floor is computed as:Froof [182.8/751](123.4)F3rd [284.2/751](123.4)F2nd [284.2/751](123.4)Σ 30.0 kips 46.7 kips 46.7 kips 123.4 kips11.1.3.3 Horizontal Distribution of Shear Forces to Walls. Since the diaphragms are defined asflexible by Standard Section 12.14.5, the horizontal distribution of forces is based on tributary area to theindividual shear walls in accordance with Standard Section 12.14.8.3.1. For this example, forces aredistributed as described below.11.1.3.3.1 Longitudinal Direction. In this direction, there are four lines of resistance, but only theexterior walls are considered in this example. The total story force tributary to the exterior wall isdetermined as follows:(25/2)/56Fx 0.223FxThe distribution to each individual shear wall segment along this exterior line is discussed inSection 11.1.4.7 below.11.1.3.3.2 Transverse Direction. Again, based on the flexible diaphragm assumption, force is to bedistributed based on tributary area. As shown in Figure 11.1-4, there are three sets of two shear walls,each offset in plan by 8 feet. For the purposes of this example, each set of walls is assumed to be inalignment, resisting the same tributary width. The result is that the building is modeled with a diaphragmconsisting of two simple spans, which provides a more reasonable horizontal distribution of force than apure tributary area distribution.For a two-span, flexible diaphragm, the central walls will resist one-half of the total load, or 0.50Fx. Theother walls resist story forces in proportion to the width of diaphragm between them and the central walls.The left set of walls in Figure 11.1-4 resists (60/2)/148Fx 0.203Fx and the right set resists(88/2)/148Fx 0.297Fx, where 60 feet and 88 feet represent the dimension from the ends of the buildingto the centroid of the two central walls. Note that this does not exactly match the existing diaphragmspans, but is a reasonable simplification to account for the three sets of offset shear walls at the ends andmiddle of the building.11-10

Chapter 11: Wood Design11.1.3.4 Diaphragm Design Forces. As specified in Standard Section 12.14.7.4, the design forces forfloor and roof diaphragms are the same forces as computed for the vertical distribution in Section 11.1.3.2above plus any force due to offset walls (not applicable for this example).The weight tributary to the diaphragm, wpx, need not include the weight of walls parallel to the force. Forthis example, however, since the shear walls in both directions are relatively light compared to the totaltributary diaphragm weight, the diaphragm force is computed based on the total story weight, forconvenience. Therefore, the diaphragm

§ Design and detailing of wall and diaphragm chord members. § Design and detailing of longitudinal plywood walls using the requirements for perforated shear walls. The simplified procedure, new to the 2005 edition of the Standard, is permitted for relatively short, simple and regular structures utilizing shear walls or braced frames.

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