Offset Diaphragms and Shear Walls

Transcription

1 Excerpts From: Wood Products ouncil Offset s and Shear Walls Presentation updated to 2012 I, SE 710 opyright McGrawHill, I Presented by: y: R. erry Malone, PE, SE Senior echnical Director rchitectural & Engineering Solutions terrym@woodworks.org Resources For You No ost Project echnical Education ase Studies Design ools and more at woodworks.org Schools Midrise/multifamily ommercial orporate Franchise Retail Institutional Recreational Healthcare Marselle ondominium rchitect: P rchitects Photo: Matt odd

2 Evaluations How did we do? Have we helped? Potentialase Studies I/ES redits he Wood Products ouncil is a Registered Provider with he merican Institute of rchitects ontinuing Education Systems (I/ES). redit(s) earned on completion of this program will be reported to I/ES for I members. ertificates of ompletion for both I members and noni members are available upon request. his program is registered with I/ES for continuing professional education. s such, it does not include content that may be deemed or construed to be an approval or endorsement by the I of any material of construction or any method or manner of handling, using, distributing, ib ti or dealing in any material or product. Questions related to specific materials, methods, and services will be addressed at the conclusion of this presentation. opyright Materials his presentation is protected by US and International opyright laws. Reproduction, distribution, display and use of the presentation without written permission of the speaker is prohibited. he Wood Products ouncil 2012 Outline / Learning Objectives Develop a better understanding of code requirements regarding the distribution of discontinuous member forces within the. Learn how to recognize areas within the that require special analysis and detailing. Discuss how to solve complex s to allow for more interesting and flexible layouts. Understand how to analyze offset shear walls to act as a single line of lateralforceresistance resistance.

3 t Presentation ssumptions ll s discussed are flexible wood sheathed or untopped steel deck s. he method of analysis is also relevant to internal load paths within semirigid s. ode and asic Information oundary Elements omplete Load Paths he analysis and design of simple rectangular s is already understood. ode References: SE 710 Minimum Design Loads for uildings and Other Structures 2012 I Design references: he nalysis of Irregular Shaped Structures: s and Shear WallsMalone, Rice Design of Wood Structures reyer, Fridley, Pollock, obeen SEO Seismic Design Manual, Volume 2 Wood Engineering and onstruction HandbookFaherty, Williamson Guide to the Design of s, hords and ollectorsnse, Mays oundary Elements oundary Elements: location where shear is transferred into or out of the. ransfer is either to a boundary element or to another force resisting element. oundary members include s and drag struts at and shear wall perimeters, interior openings, discontinuities, and reentrant corners. ollector / Drag strut: horizontal element parallel and in line with the applied force that collects and transfers shear forces to the vertical elements of the lateralforceresistingsystem and/or distributes forces within the. ransfer (Sub): portion of a larger wood or untopped steel deck designed to anchor and transfer local forces to primary struts and the main. Fundamental Principles: shear wall is a location where forces are resisted (supported), and therefore defines a boundary location. Note: ll edges of a shall be supported by a drag strut, collector,, shear wall, or other lateral resisting element. 1 hord boundary (typical) 1 hord 2 and shear wall sheathing shall not be used to splice boundary elements. s shall be designed for both shear and bending stresses resulting from design forces. t discontinuities, such as openings or reentrant corners, the design shall assure that the dissipation or transfer of edge () forces combined with other forces in the is within shear and tension capacity of the. What does this mean? Framing members, blocking, and connections shall extend into the a sufficient distance to develop the force transferred into the. What does this mean? ollector elements shall be provided that are capable of transferring forces originating in other portions of the structure to the element providing resistance to those forces. Required for all SD and wind Stru 1 1 boundary (typical) W2 hord 2 boundary (typical) 1 2 Note: Interior shear walls without a collector or a complete alternate load path are NO LLOWED! hord hord r ollector oundary Elements 3

4 t t 1 2 hord tor hord Deflection if tie oundary oundary (Longitudinal loading) ollector Reentrant corner 1 boundary hord ollect Deflection if no tie Loads hord ut Stru 4 earing will occur if collectors 3 are not installed at reentrant corner. Note: ll edges of a shall be supported by a drag strut,, shear wall or other lateral resisting element. Deflected curve if proper tie Deflected curve if no tie omplete ontinuous Lateral Load Paths 2 ny system or method of construction shall be based upon a rational analysis in accordance with wellestablished principals of mechanics. Such analysis shall result in a system that provides a complete load path capable of transferring loads from their point of origin to the load resisting elements and down into the foundation. (includes members and their connections and splices) 3 Design Requirements Openings in shear panels that materially effect their strength shall be fully detailed on the plans and shall have their edges adequately reinforced to transfer all shear stresses. What does this mean? Design of coldformed steel shear walls and s shall be in accordance with ISILateral. Steel deck s shall be in compliance with SE7 Section (discontinuities, collectors, struts). [ not a complete quote] oundary Elements L Shaped uildingsransverse Loading Offset shear 1 walls / Vertical / 5 offset ollector (typ.) Open Opening in / ollector ollector Discont. (typ.) (typ.) diaph. ollector / Multiple l 2 offset D / E 3 4 Intermediate 4 Offset shear walls offset and struts /strut 1 r (typ.) ollector (typ.) 7 recommended installing collectors at reentrant corners to provide continuity. SE 710 section requires interconnection ties. SE 710 section t discontinuities such as openings and reentrant corners, the design shall assure that the dissipation or transfer of edge () forces combined with other forces in the is within shear and tension capacity of the. I section requires continuous load pathsloading transverse and longitudinally. ypical ommercial s Discontinuities r (typ.) ollecto MRF F1 Method of nalysis he Visual Shear ransfer Method FY M FX Positive i Direction Lds. ransverse Direction Shears pplied to Sheathing hi Elements Unit shear transferred from the sheathing element intothe the boundary element (plf) Symbol for 1 ft x 1 ft square piece of sheathing in static equilibrium (typ.) Unit shear acting on sheathing element (plf) Shears ransferred Into oundary Elements

5 1 2.L. hord Force Diagram w=uniformly load Resisting 1 Produces compression wall () into boundary element () in () shear transferred tension into boundary element (typ.) 3 in ompr. Resisting () wall () 2 Forces Positive sign convention Positive diaph. shear elements Pos. 1ft ft. x1ft ft. squaresheathing sheathing element symbol at any location in the. Produces tension into boundary element Maximum moment asic Shear Diagram hord Force Diagram () Neg. in ompr. Resisting wall () in tension Negative diaphr. shear elements Forces ll edges of a shall be supported by a boundary element (, strut, collector) or other vertical lateral force resisting element (shear wall, frame). Shear Distribution Into a Simple he Visual Shear ransfer Method Introduction to ransfer s and ransfer reas support W ( plf) (Longitudinal member) Longitudinal ollector (ypically a tie strap and flat blocking are called out) ransfer area (Longitudinal member).l. ransfer Members and Elements ransverse drag strut support 4 Partial length collectors do not constitute a complete load path. his force must be transferred out to the main s. complete load path is required. Disrupted Righting (resisting) forces ollector Rotation of section Righting (resisting) forces ransfer 1 area 2 3 Disrupted ransfer area without transverse collectors ransfer Mechanism hord Main llector hord/o D1 ransfer ( eam) ollector Full lldepth llector hord/ol Main 2 3 NOE: ollector must extend the full depth of the transfer ransfer using beam concept hord ll ollector Sect. Details are seldom cut PlanPartial length collector he length of the collector is typically determined by dividing the collector force by the nailing capacity. (Wrong!!) he collector is typically checked for tension only. ompression rarely checked. ypical callout MS14 tie strap x 10 0 with (xx) 10d nails over 2x flat blocking. Lap 28 onto wall. ransfer rea Higher. unit shear (plf) transfers into blocking F1 F2 locking acts as ministrut/collector and transfers (accumulates) forces into the next block. /collector force diagram F3 F4 ollector force distribution Example of Partial /ollector earing perp. to grain? ont. tie strap over F(total)= F1F2F3F4 No gaps allowed. sheathing is not allowed to transfer strut/collector tension or compression forces.

6 1 2 hord vnet=300(75)= 225 plf he ransfer is affected by the basic shear and the transfer shear hord force at notch ( ) discontinuity Disrupted vnet=300(250)= 550 plf No outside force is changing the basic shear in this area Legend 300 plf asic shear (250 plf) ransfer shear from discontinuous =xxx plf Net (basic shear / D shear) ollector o 500 plf 2 hord (D support) vnet =225 (75) = 150 plf D1 vnet =225 (250) = 475 plf ollector 3 Main ollector o hord (D support) ransfer depth DD 300 plf asic Shear Diagram at transfer ngth r len LD ransfer Main plf plf asic Procedure Method by Edward F. Diekmann V= (a), Shear = V LD DD 75 plf Subtract from basic b dd to basic a Disrupted 250 plf No outside force is changing the V= (b), Shear = V LD DD basic shear in this area ransfer Shears nalogous to a beam with a concentrated Load. vnet= 300 (250) = 550 plf asic shear ransfer LD Net (basic shear / D) 1 ransfer depth F( a) R hord, strut L or shear wall 2 ransfer D1 b R 1 ransfer depth F( a) R hord, strut b or shear wall ransfer Discont. ll ollector F hord, strut t F hord / strut or shear wall hord, strut or shear wall a F( b) R L L aph. length ransfer Dia Simple Span ransfer Discont. F R hord / strut nalogous to a simple span beam with a concentrated load D1 2 F( L) R b b a L h nsfer length ran F R R nalogous to a propped cantilever beam with a concentrated load Propped antilever ransfer Simple Span and Propped antilever ransfer s Example 1 with Horizontal End Offset s with Horizontal Offsets ransverse Loading w=200 plf.l. support plf D1 F2 M=0 1 /R=2.5: ctor and hords ollec D ch ctor and hords ollec D ch M2= ft.lb F2 = lb lb F2 ollector Free body for F lb V= lb support V= lb Sign onvention lb

7 ransfer and Net Shear plf plf 150 plf asic shear diagram plf.l lb v= (15) 50(20) = plf lb lb 37.1.L. 1 v=150(107.1)=42.9 plf (Net resulting shear) lb v=70(107.1)=37.1 plf (Net resulting shear) Neg. gram ransfer d shear diag lb v= (35) = 250 plf 50(20) 250 Pos lb plf lb v=150(250)= 400 plf v=70(250)= 320 plf (Net resulting shear) (Net resulting shear) Sign onvention lb No net change Nt Net change occurs in No net change 1 2 transfer 3 Legend plf asic shear (240 plf) ransfer =xxx plf Net (basic shear / transfer diaph. ) lb F=6000 lb oundary (this is not an of transfer insignificant area force.) plf 0 plf allout all nailing on drawings: Standard nailing oundary nailing ollector nailing hord, & ollector Force Diagrams Sign onvention 2 4 Example 2 with Horizontal End Offset Longitudinal Loading plf net plf net 1 Offset Shear Walls plf w plf net 10.5 plf net 50 ollector plf net ollector plf net ollector Pos. direction Outofplane Offsets Inplane Offsets

8 OutofPlane Offset Shear Walls ssumed to act in the Same Line of Resistance Example 4 with Horizontal End Offset Longitudinal LoadingOffset Shear Walls Drag strut t 2 Drag strut t strut Drag ctor ollec ransfer area ollector ollector Whenever there are offset walls, they are typically assumed to act in the same line of lateralforceresistance. alculations are rarely provided showing how the walls are interconnected to act as a unit, or to verify that a complete lateral load path has been provided. ollectors are rarely installed to transfer the disrupted forces across the offsets. 200 plf hord hord collectors D1 hord collectors ssumptions: 1. ssume shear walls at grid lines and act along the same line of lateralforceresistance. 2. ssume the total load distributed to grid lines and /= wl/2. hord 50 drag strut drag strut Offset rag rut Dr str ollector Loads ollector ollector ypical Midrise multifamily structure at exterior wall line 1 8 Offset Drag strut 200 plf strut is discontinuous ollector Drag strut 12 Offset 3 Drag strut Pos. direction otal Shear to Shear Walls (ssumed) Vsw2=wL/2=200(50)/2=5000 lb, vsw2=5000/10=500 plf Vsw1, sw3, sw4=wl/2=200(50)/2=5000 lb, vsw=5000/(8815)=161.3 plf lb 700 lb 2 (8.85 plf) (28 plf) lb (40.9 plf) F=387 lb (Error) plf 160 plf 200 plf F= lb 1 agram 1 asic shear dia lb 8 (28 plf) vsw=161.3 plf vnet= plf lb F=590.3 lb lb 00 plf plf lb 8 3 Pos. Neg lb (20.66 plf) 1510 lb (15.1 plf) shear diagram 2 asic s 4510 lb (45.1 plf) F=590.3 lb plf net vsw=161.3 lb vnet=116.2 plf Note: Neither force diagram closes to zero, therefore error. Notice that they do not close by the same amount F = lb (Error) Determine Force transferred Into ransfer asic Shears and ransfer Shear Pos. direction Longitudinal Force Diagrams Pos. direction

9 2 Revised forces alculated forces 4600 lb with Intermediate Horizontal Offset ransverse and Longitudinal Loading he shear wall Line needs to move needs to be lower in F=36 lb in this direction order to move the force diagram in this direction Load distribution needs to increase towards line /. Increase the load to / by the amount off /. 3 he shear wall need to be higher in order to move the force diagram in this direction Line needs to move in this direction djusted Longitudinal Force Diagrams (8% increase to /) [mount shifted to / depends on the offset to span ratio of the transfer ] F=18 lb lb Pos. direction wplf w4 plf hord w1 plf w2 plf w3 plf D1 ollector r diagram asic shear r ollecto Fstrut/2 gram asic shear diag a Fstrut/2 w5 plf s or struts r ollector ollector D2 Pos. direction r ollector w5 plf 4.L. ar diagram asic shea hord Inplane Offset Shear Walls Hdr Nail shtg to Section oundary nailing should be installed at each 2x stud at hold down and each plate lk g. or rim joist Nail shtg to each 2x stud (option 2) nalyze this section as a transfer diaph. Section ector oll Wd Sill each 2x stud Section ompression locks required at all H.D. locations Shear transfer onn. Header/collector Window ie strap Section does not comply with the required aspect ratio for a segmented shear wall. ie strap Section D nchor bolts or nails Need hold down (option 1) Hold down Increased nailing may be required (ypical) due to additional shear from upper hold down (add vertical from hold down to basic horizontal shear). lt. onfig. Inplane Offset Segmented Shear Walls

10 Ends of wall panels do not line up. Requires special nailing of sheathing into stud below. Questions? his concludes he merican Institute of rchitects ontinuing Education Systems ourse Requires same number of studs above and below with boundary nailing each stud Solid blocking required R. erry Malone, P.E., S.E. Prescott Valley, rizona ontact Information: Nailing found No holddown below in field was 12 o.c. PhotoInplane Offset Segmented Shear Walls