Diaphragm And Shear Walls With Offsets and Openings

Transcription

1 Presentation ased On: Diaphragm nd Shear Walls With Offsets and Openings Presentation updated to 2015 I, SE SDPWS opyright McGraw-Hill, I y: R. Terry Malone, PE, SE Presented by: Senior Technical Director rchitectural & Engineering Solutions terrym@woodworks.org

2 ourse Description The structural configurations of modern buildings often require complex lateral load paths that incorporate multiple horizontal and vertical offsets, multiple irregularities and fewer lateral-resisting elements. This presentation will provide a review of methods of analysis that can be used to address openings and offsets that now commonly occur in wood-frame diaphragms and shear walls. Topics will include: Diaphragms with horizontal offsets Diaphragms with large openings Shear walls with offsets and openings Reference odes and Standards

3 Learning Objectives asic Information Discuss diaphragm boundary elements, complete lateral load paths, and review analytical methods used to solve complex diaphragms and shear walls. Diaphragms With Horizontal Offsets Learn how to analyze a diaphragm with a horizontal offset and how to transfer forces across areas of discontinuity. Diaphragms With Large Openings Demonstrate methods of analyzing a diaphragm with an interior opening and opening at the support wall line. Shear Walls with Offsets and Openings Review how to analyze offset shear walls and shear walls with openings.

4 Interconnection ties? Irregular shaped shear wall with opening (discontinuous) Large cantilevered diaphragm Soft story ontinuous Load Paths????? Holes in diaphragm (web discontinuity) Diaphragm boundary offset from shear wall (cantilever). quick note on residential projects

5 asic Information oundary Elements Method of nalysis

6 W ( plf) hord SW SW Strut ollector hord Diaphragm support Diaphragm support Strut- receives shears from one side only*. ollector- receives shears from both sides. *[Drag struts and collectors are synonymous in SE7] Struts, ollectors, and hords- (my) Terminology

7 Diaphragm oundary Elements Fundamental Principles: shear wall is a location where diaphragm forces are resisted (supported), and therefore defines a diaphragm boundary location. Note: ll edges of a diaphragm shall be supported by a boundary element. (SE 7-10 Section 11.2) SDPWS ollectors for shear transfer to individual full-height wall segments shall be provided. 1 2 Strut SW1 Strut Diaphragm 1 oundary (typical) hord Diaphragm 2 oundary (typical) Diaphragm 1 Diaphragm 2 Note: Interior shear walls require a full depth collector unless a complete alternate load path is provided hord SW ollector Diaphragm oundary Elements: hords, drag struts, collectors, Shear walls, frames oundary member locations: Diaphragm and shear wall perimeters Interior openings reas of discontinuity Re-entrant corners. Diaphragm and shear wall sheathing shall not be used to splice boundary elements. (SDPWS 4.1.4) 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. (SE 7-10 Section ) hord SW2 Strut Required for Seismic and wind

8 Shear failure Not enough shear capacity Splitting failure ottom section not supported Failure Modes Would you do this? lternate load path lternate load path

9 Note: Diaphragm sections act as notched beams (shear distribution-diagonal tension or compression) High stress concentrations at end of the wall Shear wall Shear wall Shear wall ode does not allow the sheathing to be used to splice or act as boundary elements Shear Distribution if No ollector ollector Shear wall Shear wall Shear wall Shear Distribution if ontinuous ollector

10 Example assumes uniform distribution of shears thru depth of diaphragm, not parabolic distribution Net= Net= Loc. -20 Loc. Sheathing element with shears acting on all sides Net= 0 Will act as three separate simple span diaphragms with supports as shown Net=0 Must have full length horizontal chord/collector here to collect unbalanced shears-will still result in tearing at end of shear wall. Loc. 1 2 Loc. Strut SW1 Strut Loc. Loc. hord hord SW Loc. hord Loc. hord hord SW2 Strut hord

11 SW1 Strut hord Deflection if tie Diaphragm 1 Diaphragm 2 Diaph. oundary (Longitudinal loading) ollector ollector Diaphragm 2 oundary Deflection if no tie hord hord Strut Strut SW4 SW2 Diaphragm 1 boundary SW Re-entrant corner Tearing will occur if collectors are not installed at re-entrant corner. hord 1 2 Loads Deflected curve if proper tie Deflected curve if no tie oundary Elements L Shaped uildings-transverse Loading

12 The Method of nalysis The Visual Shear Transfer Method How to visually show the distribution of shears through the diaphragm The method of analysis: an be used for all construction types. Straight forward and simple to use. rational method of analysis based on simple statics! Well Documented over several decades FY +M FX Positive Direction Lds Sheathing element symbol for 1 ft x 1 ft square piece of sheathing in static equilibrium (typ.) Transverse Direction (shown) Shears pplied to Sheathing Elements Transfer shears Unit shear acting on sheathing element (plf) Unit shear transferred from the sheathing element into the boundary element (plf) Shears Transferred Into oundary Elements

13 + - T Resisting wall (+) shears Strut in comp. Strut Forces Strut in tension Resisting wall (+) shears Positive sign convention Support Positive diaph. shear elements SW 1 + SW 2 Pos. + Diaphragm.L. w=uniform load 1 2 (-) (-) SW + Diaphragm shear transferred into boundary element (typ.) + - Maximum moment 1 ft. x 1 ft. square sheathing element symbol at any location in the diaphragm. T SW asic Shear Diagram - SW - - (-) (-) Neg. Shear Distribution Into a Simple Diaphragm The Visual Shear Transfer Method Strut in ompr. Resisting wall shears (+) Strut in tension Support T Strut Forces Negative diaph. shear elements ll edges of a diaphragm shall be supported by a boundary element (chord, strut, collector) or other vertical lateral force resisting element (shear wall, frame).

14 Introduction to Transfer Diaphragms and Transfer reas Transfer Diaphragm hord (support) Sub-diaphragm-don t confuse w/ sub-diaphragms supporting conc./masonry walls Transfers local forces out to primary chords/struts of the main diaphragm. ( ased on method, SE 7 Section 1.4 and SDPWS 4.1.1) T ollector TD1 ollector (strap/blocking or beam/truss) ollector Transfer rea hord (support) Transfer Diaph. depth Transfer Diaph. length T Maximum TD spect Ratio=4:1 (Similar to main diaph.) Framing members, blocking, and connections shall extend into the diaphragm a sufficient distance to develop the force transferred into the diaphragm. (SDPWS 4.2.1) ollector Length?-My rule of thumb: What does this mean? heck length by dividing discontinuous force by the nailing capacity (other issues need to be considered) Length=full depth of transfer the diaphragm, set by /R If L<=0 o.k. to use strap/blocking, If > 0 use beam/truss Increase TD depth if shears are too high in transfer area

15 2 TD aspect ratio 8:1 4:1 maximum allowed 16 Warping/racking Over-stressed. (Notice deformation in transfer diaphragm) TD spect Ratio Too High

16 1 2 Diaph..L. W ( plf) 4 Diaphragm chord SW Discontinuous diaphragm chord Diaphragm chord Diaphragm support Transfer rea SW The length of the collector is often determined by dividing the collector force by the diaphragm nailing capacity. (aution-other issues need to be considered!) The collector is often checked for tension only. (Wrong!) ompression forces occur when the loads reverse direction. Diaphragm chord Longitudinal ollector Typical callout Steel tie strap x ga. x width x length with (xx) 10d nails over 2x, x or 4x flat blocking. Lap x -y onto wall. Transfer Diaphragm Members and Elements Drag strut Diaphragm support

17 Example of Partial Strut/ollector 2x flat blk g (tight fit) WSP sht g. earing perp. to grain? Tie strap Potential gaps lt. 4x blk g. Joists Use Z clip to keep blocking level Note: If open web joists, continuous 2x members can be nailed to blocking to take compression forces. Typical Section F1 + Diaphragm unit shear (plf) transfers into blocking Strut/collector force diagram F2 F F4 + + ont. tie strap over F(total)= F1+F2+F+F4 For compression, blocking acts as mini strut/collector and transfers (accumulates) forces into the next block. No gaps allowed. Diaphragm sheathing is not allowed to transfer strut/collector tension or compression forces. ollector force distribution

18 SW Support SW Main chord This force must be transferred out to the main chords. complete load path is required. (SE 7 Section 1.4 and SDPWS 4.1.1) Disrupted chord Transfer area 1 2 Resisting forces ollector Rotation of section Resisting forces Transfer area without transverse collectors Transfer Mechanism Disrupted chord Discontinuous diaphragm chord hord T Strut hord/ollector Transfer Diaphragm ( eam) Support 2 NOTE: TD1 ollector Full depth SW hord/ollector ollector must extend the full depth of the transfer diaphragm Main chord Transfer using beam concept

19 1 2 Transfer Diaph. depth F( a) R = hord, strut L or shear wall Support R 1 Transfer Diaph. depth Discont. F hord / strut hord, strut or shear wall 2 F( a) R = b Support R F Transfer Diaphragm TD1 Discont. ollector hord / strut hord, strut or shear wall Support F( b) R = L a b L Transfer Diaph. length F R Discont. F hord / strut Transfer Diaphragm TD1 hord, strut or shear wall a b Support F( L) R = b L Transfer Diaph. length R F Simple Span Transfer Diaphragm nalogous to a simple span beam with a concentrated load Propped antilever Transfer Diaphragm nalogous to a propped cantilever beam with a concentrated load Simple Span and Propped antilever Transfer Diaphragms

20 1 Method of nalysis-method by Edward F. Diekmann hord 2 (TD support) Main chord -75 plf - T(a), Shear = V V= LTD DTD SW vnet=+00-(75)= +225 plf vnet =+225 (75)= +150 plf The transfer diaph. spect Ratio should be similar to the main diaphragm. hord force at discontinuity Disrupted chord vnet=+00+(250)= +550 plf No outside force is changing the basic diaphragm shear in this area T ollector +500 plf TD1 vnet =+225 +(250) = +475 plf ollector (TD support) TD depth DTD ollector Main chord +00 plf plf plf Transfer diaphragm length asic Shear Diagram at transfer diaphragm LTD T T No outside force is changing the basic diaphragm shear in this area + Subtract from basic shears a b dd to basic diaphragm shears LTD +250 plf T(b), Shear = V V= LTD DTD Transfer Diaphragm Shears nalogous to a beam with a concentrated Load.

21 Direction of shear transferred into collector Resulting net shear diagram acting on collector plf +150 plf + + ollector plf +475 plf Net shears 25 plf 25 plf Lcollector Net direction of shears acting on collector Note: The net shears will not always be equal. Dir. of force on collector Place the net diaphragm shear on each side of the collector Place the transfer shears on each side of the collector Sum shears on collector (based upon direction of shears transferred onto collector). Shear left= = +25 plf Shear right= =+25 plf ollector force=area of shear diagram Fcollector=(25+25)(Lcollector) 2 ollector Force Diagram Shear Distribution Into The ollector

22 Diaphragms with Horizontal Offsets Strut SW1 Support 1 Strut/chord Strut /chord Discont. diaph. chord Multiple offset Diaphragm E ollector Strut (typ.) 2 Discontinuous diaphragm chord SW5 ollector (typ.) Strut ollector (typ.) SW2 F Discont. diaphragm chord Open Strut chord 4 5 Strut Strut/chord SW6 SW ollector (typ.) ollector ollector (typ.) Strut/chord Offset strut Strut/chord SW4 Strut chord nalysis: SE7-10 Sections: Design shall be based on a rational analysis t diaphragm discontinuities such as openings and re-entrant corners, the design shall assure that the dissipation or transfer of edge (chord) forces combined with other forces in the diaphragm is within shear and tension capacity of the diaphragm. D MRF1 Discont. diaphragm chord Support Discontinuous diaphragm chord/strut

23 Vertically offset Diaphragms? Openings in diaphragm Offsets in the diaphragm and walls Harrington Recovery enter Structural engineer: Pujara Wirth Torke, Inc. Photographer: urtis Walz

24 Example 1-Diaphragm with Horizontal End Offset-Transverse Loading w=200 plf Support Diaph..L. Diaph. chord 5 SW 1 ollector TD chords TD1 ollector TD chords plf F2 M=0 5 SW 2 50 M2 ft.-lb F2= lb Diaph. chord /R=2.5:1 ollector 25 F lb Free body for F2 15 Support RL =12500 lb Discontinuous diaphragm chord 7500 lb Support lb Diaph. chord + - Sign onvention 80 Support RR=12500 lb 1 2 4

25 Transfer Diaphragm and Net Diaphragm Shear 57.1 plf plf 150 plf asic shear diagram plf Diaph..L lb v= (15) = plf 50(20) alcs SW lb v=150-(107.1) =+42.9 plf (Net resulting shear) lb v=150+(250)= +400 plf (Net resulting shear) an be > x basic shear No net change Net change occurs in TD 5 v=70-(107.1) =-7.1 plf (Net resulting shear) lb Pos. Neg. TD shear diagram 15 v=70+(250)= +20 plf (Net resulting shear) 80 v= lb (5) = +250 plf 50(20) plf Sign onvention SW lb No net change Legend 4 75 plf asic diaphragm shear (240 plf) Transfer diaphragm shears =xxx plf Net shears (basic shear +/- TD shears)

26 +20 lb lb lb -7.1 Diaph..L. alcs F=7200 lb F= lb SW 1 F= lb F= lb T plf F= lb SW 2 Support lb plf net 70 plf T 0 plf -250 F =7200 lb Sign onvention 1 2 Longitudinal hord Force Diagrams Support 4

27 Diaph..L. alcs SW plf net plf net T F=6000 lb Special nailing (sum of shears to collector or highest boundary nailing-greater of) F=748.5 lb SW plf net F=6000 lb (this is not an insignificant force.) F=750 lb Transverse ollector Force Diagrams Sign onvention

28 57.2 plf x1 x2 x x4 Special nailing along collectors 20 plf 285 plf 214. plf 42.9 plf 7.1 plf 70 plf Sum of shears to collector or highest boundary nailinggreater of 57.1 plf 214. plf 150 plf asic shear diagram 70 plf Transfer diaphragm oundary (Typ.) 4/6/12 6/6/12 6/12 U ase I 6/12 U ase I 6/12 U ase I allout all nailing on drawings: Standard diaphragm nailing oundary nailing ollector nailing Diaphragm boundary heck the shear capacity of the nailing along the collector 1 2 4/6/12 Transfer area oundary (High shear area) oundary locations Diaphragm Nailing allouts

29 Example 2-Diaphragm with Horizontal End Offset -Longitudinal Loading 5 SW Diaph..L. Drag strut 40 plf 160 plf 200 plf 5 hord Diaphragm 1 Discontinuous Drag strut ollector and TD chords Transfer diaphragm TD1 ollector ollector and TD chords Diaphragm 2 50 hord plf Discontinuous Drag strut 200 plf Drag strut SW Pos. direction + - 4

30 + 5 Pos. direction 700 lb SW v=-40.9-(10.5)=-51.4 plf 700 lb - 15 asic shear diagram +28 Vsw=5000 lb vsw=500 plf plf net v=+45.1+(24.5) 1 =+69.6 plf 2 Sign onvention Legend 210 lb lb Diaph..L. asic shear diagram v=+15.1-(10.5)=+4.6 plf (Net resulting shear) v=+15.1+(24.5) =+9.6 plf 700 lb plf plf asic diaphragm shear Transfer Diaphragm and Net Diaphragm Shear Pos. Neg. (240 plf) Transfer diaphragm shears =xxx plf Net diaphragm shears (basic shear +/- TD Shears) 4510 lb Transfer area lb plf SW 1 Vsw=5000 lb vsw=. plf 288. plf net 4

31 F=700 lb -28 F=529 lb plf 5 net SW alcs 2.4 plf net F= -957 lb 10.5 plf net 5 F=819 lb F=700 lb F= -819 lb F=700 lb plf +9.6 net F=68 lb F=68 lb F=42.5 lb strut 4050 lb chord Pos. direction Longitudinal and Transverse ollector/strut Force Diagrams + 15 SW 1 Sign onvention 288. plf net - 4

32 Quick Note on Segmentation 8000 lb W=200 plf hord 00 plf 100 plf 200 plf ollector lb -200 Pos TD1 +00 ollector asic Shear Diagram -100 plf -200 plf -00 plf TD Neg lb ollector hord ollector lb F = lb Neg. 20 Pos lb Transfer Diagram Shear lb hord V=4000 lb V=4000 lb hord lb lb Transfer Diagram Shear nalysis Option 1 nalyze as Diaphragm with Intermediate Offset Sign onvention + -

33 lb W=200 plf 2000 lb W=100 plf W=100 plf plf +500 W=100 plf W=100 plf hord hord hord ollector asic Shear Diagram lb plf 200 plf hord asic Shear Diagram lb W=200 plf hord 40 F=16000 lb 2000 lb ssumes small diaphragms are supported off of main diaphragm 120 ollector -200 plf -00 plf W=100 plf hord plf hord hord W=100 plf asic Shear Diagram lb nalysis Option 2-nalyzing as separate diaphragms

34 1 2 W=200 plf 4 hord Values Options 1 and plf lb 20 asic Shear Diagram /R=6:1>4:1, N.G. 200 plf Fchord=18000 lb 200 plf Options 1, 2 and plf 200 plf hord 40 F=16000 lb Still must make a connection ssumes main diaphragm takes all of the load. Lower diaphragms are ignored plf lb nalysis Option -Ignoring lower diaphragm sections (Not recommended) 40

35 Offset Shear Walls 1 Strut SW1 Strut/chord E ollector Strut /chord 2 Offset shear walls Strut (typ.) SW5 ollector (typ.) Strut ollector (typ.) SW2 F Open Strut chord 4 5 Strut Strut/chord SW6 SW ollector (typ.) ollector ollector (typ.) Strut/chord Offset shear walls and struts Strut/chord SW4 Strut chord D MRF1 Support Support Support omplete ontinuous Lateral Load Paths

36 Out-of-Plane Offset Shear Walls ssumed to act in the Same Line of Resistance Discont. drag strut Drag strut SW1 Offset ollector SW2 Transfer area ollector TD1 ollector Loads Offset walls are often assumed to act in the same line of lateral-force-resistance. alculations are seldom 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 required to be installed to transfer the disrupted forces across the offsets. Discont. drag strut SW Drag strut ollector ollector TD2 ollector Typical mid-rise multi-family structure at exterior wall line Where offset walls occur in the wall line, the shear walls on each side of the offset should be considered as separate shear walls and should be tied together by force transfer around the offset (in the plane of the diaphragm). heck for Type 2 horizontal irregularity Re-entrant corner irregularity

37 Mid-rise Multi-family Lds. Discontinuous struts Longitudinal ant. SW SW SW Lds. Discontinuous chords Transverse SW SW SW SW SW SW SW No exterior Shear walls Flexible, semi-rigid, or rigid???

38 Loads 1 2 SW1 Higher shears and nailing requirements TD1 TD2 SW ollector (typ.) SW4 1 2 SW1 Transfer rea Higher shears and nailing Reqmts. eam Strut TD1 TD2 SW SW4 SW2 TD Higher shears and nailing Reqmts. SW2 ollector Strut/chord Main diaphragm becomes TD Multi Story, Multi-family Wood Structure SW5 I1 I2 I1 I2 I ollector (typ.) SW5 Optional Framing Layouts SE7-10 Diaphragm stiffness changes ont. ld. Paths ont. ld. paths-inter-conn. Ties Openings, re-entrant. transfer of dis-cont. forces combined with other forces ollector elements

39 Example -Diaphragm with Horizontal End Offset Longitudinal Loading-Out-of-plane offset Shear Walls 200 plf hord 5 SW 1 8 Drag strut 15 Drag strut is discontinuous SW Offset SW Drag strut 200 plf hord collectors TD1 ollector Drag strut hord collectors 12 Drag strut 80 ssumptions: 1. ssume shear walls at grid lines and act along the same line of lateral-forceresistance. 2. ssume the total load distributed to grid lines and /= wl/2. Offset SW 50 hord SW Drag strut SW Support Pos. direction + - Support

40 Total Shear to Shear Walls (ssumed) Vsw2=wL/2=200(50)/2=5000 lb, vsw2=5000/10=500 plf Vsw1, sw, sw4=wl/2=200(50)/2=5000 lb, vsw=5000/(8+8+15)=161. plf lb (-28 plf) SW lb (+8.85 plf) alcs lb (-40.9 plf) 200 plf 1 5 SW plf 160 plf asic shear diagram lb (+28 plf) vsw=+161. plf vnet=+1.29 plf F=1066. lb Determine Force transferred Into Transfer Diaphragm 2 F=590. lb 590. lb 200 plf plf 590. lb 8 SW Pos lb (+15.1 plf) Pos. direction asic Diaphragm Shears and Transfer Diaphragm Shear Neg lb ( plf) asic shear diagram lb (+45.1 plf) Sign onvention

41 vsw=500 lb vnet= plf SW lb (-40.9 plf) alcs v=-40.9+(8.85) =-2.05 plf SW 1 vsw=161. lb vnet=1. plf v=+15.1+(8.85) =+2.95 plf 15 v=+15.1-(20.66)=-5.56 plf v=+45.1-(20.66)=+24.4 plf (Net resulting shear) No net change Net change In TD Net Diaphragm Shears 8 Pos. Neg lb (+15.1 plf) Pos. direction No net change Sign onvention Legend 4 75 plf asic diaphragm shear =xxx plf Net diaphragm shears (basic shear +/- TD Shears) (240 plf ) Transfer diaphragm shears asic shear diagram SW 4510 lb SW 4 vsw=161. lb (+45.1 plf) vsw=161. lb vnet=116.2 plf vnet=116.2 plf

42 SW 2 alcs plf net 8.85 plf net F=141.7 lb F= 09.8 lb F= 09.9 lb 1 8 SW 1 F=141.6 lb 15 F= F= plf net 8 SW 15 SW 4 So far, so good Transverse ollector Force Diagrams Pos. direction + - Sign onvention

43 F=+819. lb F=-700 lb SW F=+659 lb 80 F=+87 lb (Error) alcs F=-860 lb F=1066. lb F=590. lb vsw=500 lb vnet= plf F= lb Note: Neither force diagram closes to zero, therefore error. Notice that they do not close by the same amount SW SW vsw=161. lb vnet=116.2 plf vsw=161. lb vnet=116.2 plf SW F= lb F=-100 lb F = lb F = lb 80 F =-86.9 lb (Error) 1 2 Longitudinal Strut Force Diagrams Pos. direction + - Sign onvention

44 alcs F=88.5 lb SW 2 F=700 lb F=802.5 lb F= lb F=64 lb F=26 lb Line needs to move in this direction The shear wall shears needs to be lower in order to move the force diagram in this direction The shear wall shears needs to be higher in order to move the force diagram in this direction Revised forces F=6 lb Load distribution needs to increase towards line /. Increase the load to / by the amount off +/-. alculated forces 4600 lb SW 1 15 F=69.6 lb F=87.5 lb F=416.6 lb F=18 lb Line needs to move in this direction SW SW lb F=75 lb F=957.8 lb F= lb 1 2 djusted Longitudinal Strut Force Diagrams (8% increase to /) [mount shifted to / depends on the offset to span ratio of the transfer diaphragm]

45 Tying Shear Walls cross the orridor or a Large Diaphragm exterior walls SDPWS ollectors for shear transfer to individual full-height wall segments shall be provided. ollector Transverse walls orridor walls SW SW collector SW collector SW (Typ.) ollector SW Transfer rea Where offset walls occur in the wall line, the shear walls on each side of the offset should be considered as separate shear walls and should be tied together by force transfer around the offset (in the plane of the diaphragm). SW (Typ.) Tying Shear Walls cross a Large Diaphragm Must follow engineering mechanics and all force diagrams must close to zero or be resolved by other methods.

46 Example Total load to grid lines 2 & R2= =20048 # O.K. D SW SW Summing V= SW SW Summing V=0 Sum=20048 # LSW=22+12=4 VSW=20048 # vsw=20048/4= plf vnet SW1= = plf vnet SW2= = plf hecks, they should be equal SW1 F2=189.65(22)= # F2=(156+8)6=984 # F2= = # SW2 FSW2 =189.65(12)= # FSW2 to =(26+8)6+(26+164)10=5464 # F= = # F2=F2=188.24(4)/6= # O.K. ll forces in lb., all shears in plf Shear at transfer area=188.24/6+8=59.7 plf

47 Diaphragms With Large Openings Interior and End Openings Use of over-strength forces is not commonly considered to be triggered for boundary elements at diaphragm openings. 1 2 Strut SW1 Support Strut/chord Strut /chord E ollector Strut (typ.) SW5 ollector (typ.) TD1 SW2 Strut ollector (typ.) Strut chord F Open 5 Strut Strut/chord SW6 ollector (typ.) ollector SW ollector (typ.) Strut/chord Strut/chord ollector SW4 Open Design: 4 I 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. Type Horizontal Irregularity-SD D-F-Diaphragm Discontinuity Irregularity. Diaphragm discontinuity irregularity exists where there is an abrupt discontinuity or variation in stiffness, including a cut-out or open area greater than 50% gross enclosed diaphragm area, or a change in effective diaphragm stiffness of more than 50% from one story to the next. TD2 TD Strut chord MRF1 Strut chord Support D

48 Harrington Recovery enter Structural engineer: Pujara Wirth Torke, Inc. Photographer: urtis Walz ommon openings in diaphragm2 Roof pop-up section with opening below. Skylight or atrium opening lerestory windows End openings

49 Interior Openings Diaph. 5.L. w plf D

50 I 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. Local shears higher Local shears lower asic Shear Diagram FPInnovations Design example: Designing for openings in wood diaphragm It is strongly recommended that analysis for a diaphragm with an opening should be carried out except where all four of the following items are satisfied: a. Depth no greater than 15% of diaphragm depth; b. Length no greater than 15% of diaphragm length; c. Distance from diaphragm edge to the nearest opening edge is a minimum of times the larger opening dimension; d. The diaphragm portion between opening and diaphragm edge satisfies the maximum aspect ratio requirement. (all sides of the opening) Location and Magnitude of Shear Stairwells Elevators Most openings of any significant size should be checked. Size of opening ffect of Size and location in Diaphragm

51 Opening size is not a factor if: 1. alculated strap length does not exceed L or H 2. Shear does not increase beyond nailing capacity L External loads are not included in the check. The unit shears in upper section are constant vert. and horiz. across upper section Lh!$ = $' ( % H H L Opening L Steel straps and blocking ollector v1 v v Upper ound!" = $%& for left collector Lower ound v2!" = $% & % for right collector v4 ollector ollector v4 Lv H v I/Diekmann Minimum recommended steel strap lengths if opening size is not a factor I Section : If opening > 4 ft. use Figure (1) 16 ga. X 1 ½ x L or H strap w/ (16)16d or engineered 2. Diekmann, T-7: Rule of thumb as shown above. Diekmann Method Of hecking If Opening Size Is Factor

52 plf External loads not included in check, plf Pattern nailing has to increase-n.g. 18 plf Nail capacity has to increase n.g plf ). +. =,%,- =1.2: lb lb lb plf plf 18 plf plf 18 plf L. opening v=150 plf.l. opening v=00 plf v=150 plf!. =,/-(,-) % &. = 2/- 6,7,7- = 2/- 45! ( =,7- ; =,-7-45 = /. '6 <,- o.k. Lower bound 12 L x 10 H Opening Located at low shear!. = 6--(,-) % &. = =,/-- 45,/-- '7-6;- = 12.5 > 10 n.g.! ( = 6;- ; = %,;- 45 Lower bound %,;- & ( = 6,7 6-- =,%-< >,% n.g. Opening is a factor 12 L x 10 H Opening Located at high shear!. =,/-(6/) % = %;%/ 45 heck If Size and Location of Opening In Diaphragm Is Factor &. = %;%/ '7-6;- = %,. 77/< < 6/ ut > depth Lower bound! ( = 6;- ; = %,;- 45,-7- & ( = = ;. '% <,% 6,7,/- o.k. Opening is not a factor %,;- 6,7,/- & ( = =,%. 7/ >,% n.g. Opening is a factor 12 L x 5 H Opening Located at low shear

53 M V M M=0 D R Local forces 1 2 T T hord forces are assumed to be zero at these locations due to contraflexure (inflection points). M=0. V V I.P.. V V I.P..L. opening Some examples apply load to one side of the diaphragm only W 4 T T Vierendeel truss action Displacement and Local Forces 5 R

54 Shear distribution follows analysis Shear Distribution in Diaphragm

55 /R Opening /R /R Opening /R Easy to visualize if header section is replaced by a wire. nalyze by envelope method: Diaph. with opening Diaphragm w/ interior offset /R TD /R Opening /R /R TD Transfer Diaphragm (TD) Transfer diaphragms are required if the opening size does affect the shear or tension capacity of the diaphragm. T 7, Diekmann, FPInnovations If the sections above, below or on each side of the opening does not meet code aspect ratio limits it should be ignored (not stiff enough). ll sections must meet ode required aspect ratios. spect Ratio Issues

56 D 1 2 w plf R TD1 F2 F2 F F=0 Opening F ΣM 2 = 0 ΣM = 0 ΣM 4 = 0 V1 V1 Element II Element IV V2 Inflection point. F=0 Element I Element III V 4 F4 F4 TD2 V4 asic shear diagram without openings V2L V2R V V4L V4R asic Shear Diaphragm With Opening 5 V5 V5 Diaph..L. Typical method of analysis (P Report 18), T-7, and FPInnovations 1. alculate the chord forces at grid lines 2,, and 4 using FD s. 2. Determine the basic diaphragm shears without an opening.. Determine the diaphragm shears with an opening. 4. reak the sections above and below the opening into elements as shown. 5. Determine the local forces at each corner of each segment by FD s. 6. Determine the net resulting shears and forces (+/-) by combing the shears with and without an opening using a table. Using the visual shear transfer method 1. Determine shear (V4) at grid line reak the sections above and below the opening into elements as shown.. alculate the chord force at grid line. 4. Starting at grid line 4 and moving to the left, sum forces at each corner of each segment to determine the local forces, by FD s. 5. alculate all chord, collector forces, and transfer diaphragm shears and forces using the visual shear transfer method. Opening nalysis-diekmann method

57 Example 4-Pop-up Roof Section /R main diaphragm and upper section=.:1 /R TD1=TD2=.0:1 o.k. Wind Loads (SD) Main W=200 plf t opening Ww=12 plf Lw=77 plf t pop-up (20 psf) Ww=50 plf Lw=0 plf 1 2 W=200 plf 1600 lb TD1 4 5 Diaph..L. 6 W=12 plf W=200 plf W=50 plf Open to below W=0 plf 1600 lb TD W=200 plf 1600 lb TD1 W=12 plf 1280 W=50 plf F Sub-hord Sub-hord W=0 plf D RL= V=25120 V=21120 V=19520 v2r v1 v2l W=77 plf 40 V=1920 V=820 V=6720 v v4l 20 V= RR=21280 SW >? = - D W=77 plf RL=25120 FD 1280 v4r asic Shear Diaphragm With Opening (plf) v5 v6 Open to below

58 External loads not used in rough calc., except SW reaction Upper limit 820 lb FPInnovations: Hgt.> 0.15 ddiaph. 28 Width>0.15 Ldiaph. End dist.< x width Detailed analysis required L. opening Wwind 18 plf lb 1600 lb plf 18 plf 208 V4L=820 lb V4R=6720 lb 112 Shears w/ opening 18 plf 60 DEFG HIJKL LFMIG = ;2%- ;- DEFG HIJKL 4JOG= 76%- '-!. =,,%(%-) % &. =,,%- 6,7 %-7 =,,%- 45 =,,% N4O = %-7 N4O =,-. % < %-..! ( = %-7 %- = ',;- 45 Lower bound for right collector ',;- & ( = 6,7,,% = %-. % < DTD Detailed analysis is not required However, a check for the upper bound (left side of opening will produce : Increased nailing Vertical and horizontal collector lengths > opening width and height Therefore, a detailed analysis is required v= 419 plf, nail cap.= 480 plf Use 4/6/12 locked, H.F. v= 52 plf, nail cap.= 58 plf Use 6/6/12 locked, H.F. (includes 1600 lb.) lso < 12 header depth, o.k. v= 112 plf 18 plf all nail cap.= 18 plf Use 6/12 Unblocked, H.F. 40 W x 20 H Opening asic Shear Diaphragm Without Opening (plf) Nailing Pattern (SD Values) heck If Size and location of Opening in Diaphragm is ritical

59 12 SW D 200 plf TD1 20 (52 plf) 1600 lb (52 plf) 9416 lb F lb F4 470 lb F 2040 lb F lb lb Sub-hord 7076 lb F2 Sub-hord F lb lb lb F2D 6827 lb lb (784.6 plf) (60.9 plf) 12 plf Element II + (496.plf) (284.4 plf) 5956 lb 7964 lb 5956 lb 7964 lb 1280 lb (496. plf) (284.4 plf) 12 plf + + (208 plf) (208 plf) 2496 lb 5824 lb F4 F4D 5824 lb 2496 lb Start here >. = lb 200 plf RL=25120 lb 77 plf 77 plf V1 V2 V V4L=820 lb V4R=6720 lb ΣV2 = = Lb 50 plf 50 plf >? = - >? = - 0 plf Element IV >? = - >? = - ΣV = = 1920 Lb 0 plf ΣV4 = = 112(60) = 820 Lb 1600 lb Free-body of hord Forces and Segment Forces FD Element I Element III (112) plf) (112) plf) TD Diaph..L. The sum of the section shears must match the basic diaphragm shear Values without an opening, V=0.

60 w lb = =+69.7 T.D lb 52-61= lb = lb 645 lb 52-44= lb Pos. Neg. Neg = D lb Transfer diaphragm shears Sign convention asic Shear Diagram Net Shears-Left Transfer Diaphragm

61 6 lb vnet= =-56.2 plf 4 w 5 Pos. Neg lb lb vnet= = plf 1600 lb TD vnet= = plf vnet= = plf + Sign convention - Neg lb lb Transfer diaphragm shears vnet= =+51.45plf Net Shears-Right Transfer Diaphragm 4924 lb asic Shear Diagram vnet= = plf D

62 F=12800 lb w F=9524 lb F=645 lb TD lb 60.7 T T F=1476 lb D Sign convention ollector Force Diagrams-Left Side

63 w F=7076 lb T T F=170 lb F=2018 lb lb F=484 lb TD T F=4924 lb F=1696 lb D ollector Force Diagrams-Right Side + - Sign convention

64 lb lb ver.=640.5 ver.=52.2 ver.= loses to 1.6 zero F=486.7 lb 470 F=1295 lb 1280 F=2040 lb 2040 F=18550 lb F=18858 lb F=6829 lb 6827 F=18206 lb F=1282 lb F=18568 lb 1280 F=18876 lb D ver.=41.5 ver.=52.2 ver.=22.65 T lb lb hord Force Diagrams loses to zero

65 T T SW 1 TD1 TD2 SW 2 T T T D Final Strut/hord Force Diagrams

66 End Openings Diaph. 5.L. w plf Does not meet /R (Envelope) D D

67 Example 5- Intermediate Horizontal Offset at End Wall With Strut 1 2 w1 plf (WW) w plf 4 D SW lb Strut SW lb 12 Section 10 hord 18 Skylight (Enclosed area) hord + 16 Section 16 + TD1 /R=.7 ollector ollector ollector ollector 15 Uniform shear in walls and in diaph. at grid line 1 D w1 1 2 V F2 Section Sum Shears w2 Section R1 V F2 hord hord SW 56 R=15000 lb w2 plf (LW) Varies Diaph..L. R=15000 lb asic Shear Diagram -

68 1 1 2 Net Shear Strut Net Shear vdiaph No shear transferred SW 1 T Strut SW 2 T T T F2 Pos. Neg. Neg. + V F2 - - SW D Net shear diagram Support VD Transfer diaphragm shears Support Resulting Strut, ollector and hord Force Diagrams if Strut

69 Example 6 -Intermediate notch at End Wall Without Strut 1 2 W=12 plf W=200 plf 4 56 D SW lb if strut (-8%) SW lb if strut (+8%) hord R1=8800 lb (v=400 plf) 16 R1D=6200 lb (V=87.5 plf) Section hord Open area Section W=77 plf Sum Shears Shear Varies TD1 /R=.7 ollector ollector ollector V=11800 lb 15 ollector V=8800 lb hord 8800 lb lb lb Shear in diaphragm at grid line 2 Is based on depth hord 150 Diaph..L. V=681.6 lb 22 v=10.5 plf 16 F2= lb R1=8800 lb D W= Section W= R1D=6200 lb Section F2=5584 lb V= lb v=10.5 plf 119 Forces in red are from previous example asic Shear Diagram SW R=15000 lb R=15000 lb plf plf plf -

70 1 2 2 D SW 1 SW 2 R1 v=210.7+(255.24) = plf R1D v=210.7-(12.7) = +87 plf v=210.7-(117) = +9.7 plf varies + Transfer diaphragm net shears plf plf plf v=157.1-(12.7) = +.4 plf v=157.1+(255.24) = plf v= (117) = +40 plf F2= Pos. Neg. Neg F2= Transfer diaphragm shears Transfer Forces to ollectors asic shear diagram Transfer diaphragm and net diaphragm shears Sign onvention

71 SW 1 F=5684 lb 10 R F=5684 lb 18 F=5584 lb F=5584 lb SW 16 SW 2 D R1D Support Longitudinal hord Force Diagrams

72 +. plf -400 plf Net shear diagram SW 1 Vsw=6200 lb vsw=87.5 plf vnet=0 plf SW 2 D 1 Vsw=8800 lb 2 vsw=7. plf vnet=7.-400=. plf 400 R1=8800 F=4000 F=4917 F= F= F= F= F= lb SW R1D=6200 lb Traverse Strut/ollector Force Diagrams

73 Shear Walls with Openings Perforated Walls FTO Shear Walls antilever Shear Walls Shear Walls with Small Openings

74 Example 7- FTO Shear Walls 2 (Diekmann)-Vierendeel Truss/Frame Shear panels or blocking 4500 lb Strut/collector 1 2 (Typical boundary Member) w=200 plf (See recent Testing-P Form M410 and SR-105) Many examples ignore gravity loads 4 ont. Rim joist min. per SDPWS Section (2008 requirement) Typical boundary member D nchor bolts or nails Tie straps full length of wall per SDPWS section

75 Optional pplication of Loads lb Strut/collector lb 1000 lb 1500 lb Strut/collector lb 69 plf 1500 lb Strut/collector Hdr/Strut/collector 1 2 W=200 plf 10. plf 4

76 Wall Pier height Wall pier height Wall pier Wall pier Dr. ollector (typ.) Wd. oundary members Wall pier Note: Not shown as having to comply w/ /R Foundation wall Wd. Overall width Wall pier Wall pier width Wall pier Wall pier width W SDPWS Figure 4E (b) Force Transfer round Opening lear height Wall pier height Limitations: The aspect ratio limitations of Table 4..4 shall apply to the overall wall and the pier sections on each side of the openings Sections exceeding.5:1 aspect ratio shall not be considered a part of the wall. Minimum pier width=2-0. full height pier section shall be located at each end of the wall. Where a horizontal offset occurs, portions on each side of the offset shall be considered as separate FTO walls. ollectors for shear transfer shall be provided through the full length of the wall. llowable Shear Wall spect Ratios For FTO Shear Walls

77 Force Transfer Distribution ased on stiffness (FE analysis) Example results single opening with shallow headers lb Strut/collector /R=2.6<.5 O.K /R=2.62<. 5 O.K. D lb /R=10.5>.5 N.G lb /R=<.5 O.K. This slide shows force distribution and location of I.P s due to sill section not meeting /R requirements Not to scale

78 Force Transfer Methodology ased on stiffness Example results double opening with shallow headers This slide shows force distribution and location of I.P s if header section not meeting /R requirements R 7.17 F=0 lb F=0 lb F=0 lb T F=0 lb F=0 lb F=0 lb H.5 Q F=0 lb S F=0 lb I Tie strap/blocking full width N P O N M L K J 25.5 D Not to scale T

79 Point of inflection is assumed to occur at mid-length (Typ.) 1 2 M V w=200 plf V M Gravity loads to wall M G H M V F=0 lb I V F=0 F=0 lb F=0 lb M V D F=0 lb J V M locking E F F M Tie strap/blocking full width V F V L M N.. K M D T 14.1 F=0 M Force Transfer Methodology (Diekmann)-Vierendeel Truss/Frame

80 lb plf 14.5 G F lb 0 lb 0 lb H I 4 Point of inflection M = 0 F lb plf 4500 lb M = 0 2 H V lb I V.5 V lb lb lb 0 lb Free-body of Upper Half and Upper Left Section

81 1 200 plf plf plf 200 plf 4 H = 0 (4500) F2 F2.5 F M = M = V = 0 G H V = 0 2 M = 0 M = 0 M = F F2(H) (0) H = 0 F(H) V= V = 0 F4 600 F2b(V) F(V) V= orner tie strap force 2 (208.62) M = 0 (587.08) V.5= V2 (0) (991.8) Units are in lb (xxx) Shears and forces determined in previous step. V (0) I (912.92) V.5= (2691.8) 2 F V = 0 D M = 0 Vc= F2(v) 91.9 F2(H) 57.2 orner tie strap force F (0) F(V) F(H) J M = 0 V= V = 0 F M = 0 E V2 V = M = 0 F V L M = 0 V K M = 0 (4500) D H = 0 (14.1) Resultant Forces on Wall Segments (424.1)

82 Example 8 - locking and Strapping Partial Width (with uniform load) w=200 plf lbs. Support Support T.D.1 Transfer diaphragm sections T.D.2 D Support Support

83 Pier Section Forces and Loads Support w=200 plf Pos. Neg. Neg. F F T.D.2 Support Transfer diaphragm shears asic Shear Diagram From Gravity Loads D lbs. + Sign convention - v v.6 v4 Summing

84 v= = plf vnet= =+24.9 plf lbs lbs lbs. vnet= = plf w=200 plf vnet= =+0.27 plf v= 1656 = plf 2.67 vnet= = plf T.D lbs. vnet= = plf lbs vnet= = plf v= = plf lbs. Transfer diaphragm shears + Sign convention Transfer Diaphragm Shears and Net Shears lbs. (+49.2 plf) asic Shear Diagram lbs. ( plf) D lbs lbs. ( plf) Summing

85 1975.8! =,P2/. 7 45H ! =,P62., 45H. T.D.2! = 6;2;. ; 45H T Sign convention D Horizontal ollector Forces lbs.

86 F=0 lbs F=0 lbs T F= lbs. F=110.5 lbs T.D F= lbs. F=2270. lbs. ollector T F=424.1 lbs. Sign convention Summing D Vertical ollector Forces lbs.

87 dvancements in FTO Shear Wall nalysis (See recent Testing-P Form M410) asis of P System Report SR-105 (in development) dvancements in Force Transfer round Openings for Wood Framed Shear Walls P System Report SR-105 (Report pending) Refine rational design methodologies to match test results Used test results from full-scale wall configurations nalytical results from a computer model llows asymmetric piers and multiple openings. SEO onvention 2015 Proceedings

88 Point of inflection is assumed to occur at mid-length (Typ.) 1 2 V M /R=1.96<.5 O.K. FTO Shear Wall-Double Opening-No Gravity Loads (Diekmann method)-vierendeel Truss/Frame M V M V V /R=2.7<.5 M M V M V O.K Tie strap/blocking full width N.. P O N M D E F R Q 7 F=0 lb F=0 lb F=0 lb /R=1.08<.5 O.K. F=0 lb T S L=24 F M V M V F M V /R=2.11<.5 M V T O.K. M F=0 M Not to scale Force Transfer Methodology (Diekmann)-Vierendeel Truss/Frame 4 F=0 lb F=0 lb F=0 lb /R=1.94<. F=0 5 lb L O.K. K 5 6 J G H I V M M V D H=9

89 D E F R F=0 lb T F=0 lb G H 4 T 750 F=0 lb F=0 lb H = 0 L=24 F=0 lb F=0 lb M = V = 0 M = V = 0 F G 2 R H T H = Force Transfer Methodology (Diekmann)-Vierendeel Truss/Frame

90 1 2 (4500) M = G M = 0 M = 0 (0) plf 2 (750) M = 0 (1218.8) (0) plf (750) plf (0) M = ( ) (0) M = 0 0 (75 0) 75 0 (0) 75 plf 2 75 G 281. plf (0) H I 1125 (1125 ) (750) D M = plf (0) M = 0 D plf (0) J 2 M = E F L L F L K (1687.5) M = Not to scale (4500) D H = 0 (1687.5) Resultant Forces on Wall Segments

91 Wall Design with Shallow Headers Force Transfer Methodology ased on stiffness Example results double opening with shallow headers This slide shows force distribution and location of I.P s if header section not meeting /R requirements R 7.17 F=0 lb F=0 lb F=0 lb T F=0 lb F=0 lb F=0 lb H.5 Q F=0 lb S F=0 lb I Tie strap/blocking full width N P O N M L K J 25.5 D Not to scale T

92 Example 9 - antilever Wall Method- (Diekmann) V1-2 1 Sections do not comply with allowable aspect ratios. (Imagine if section is replaced by wire) T only,, M, V Typically, the tie strap would be 6 required to extend the full width of the pier section if a TD is not used V-4 V T.D.1 T.D.2 T.D. 5 D locking Tie strap/blocking full width (Typ.) Fo/t 4 4 V Fo/t E

93 V1-2 V-4 V5-6 T.D.1 T.D.2 T.D. v-d 5 M = 0 F2 4 F v- 2 v-d M = 0 F4 D 6 5 F5 M = 0 8 vd-e M = 0 Fo/t. V2 V2 V = 0 F2D v=579.7 plf M = 0 V2- H = 0 V V V = 0 FD vd-e M = 0 4 V4 V4 V = 0 E v-e M = 0 V4-5 H = 0 V5 V5 V = 0 F5D 6 Fo/t Wall Section Forces

94 V + V5-6 V5-6 6 Pos. T.D. T.D. F5 F5 F5 TD net shear 5 8 Neg. V5 - or Pier basic shear V5 - VE + + E - + Fo/t Summing Sign convention Transfer Diaphragm TD Design

95 When Does an Opening in a Shear Wall Require a Detailed nalysis? (Mainly mechanical, electrical and plumbing openings) I Openings in shear panels (Diaphragms or shear walls) that materially effect their strength shall be fully detailed on the plans and shall have their edges adequately reinforced to transfer all shear stresses. FPInnovations: (Designing for openings in wood diaphragms) Note: Recommendations should be similar for shear walls as stated below: It is strongly recommended that analysis for a shear wall with an opening should be carried out except where all five of the following items are satisfied: a. Depth no greater than 15% of shear wall height; b. Length no greater than 15% of shear wall length; c. Distance from shear wall edge to the nearest opening edge is a minimum of times the opening dimension in the given direction; d. The shear wall portion between opening and diaphragm edge satisfies the maximum aspect ratio requirement. (all sides of the opening) e. The shear does not exceed the nailing capacity of the wall without an opening. H L Wall-Small Opening llowable /R=.5 Most opening sizes should be checked for effects. Suggestions: Use as rough guidance for MEP openings If opening is window, use perforated wall design If shear stresses are too high, use FTO method

96 V=4500 lbs. riteria: Opening depth 0.15H Opening length 0.15L Distance from edge x the opening dimension in the given direction; The shear wall portions around opening satisfies the maximum /R. The shear does not exceed the nailing capacity Of the wall w/o an opening. Opening: 0.15H= 0.15(9)= L= 0.15(14.5)=2.17 >2 Edges: Vert.= (1.5)=4.05 >.65 Horiz.= (2)=6 Violates vertical edge distance T 1 2 (01.6 plf) E (62.1 plf) G (62.1 plf) I (00.8 plf) (62.7) J (67.8) (62.7) 2 Wx1.5 H opening K (67.8) D (02.) F (58) H (58) L (01.5) D Nailing capacity of wall w/o opening 4 /12 =80 plf SD (seismic) 15/2 sheathing y inspection, wall meets shear criteria Wall Potentially in ompliance with ll riteria

97 V=4500 lbs. riteria: Opening depth 0.15H Opening length 0.15L Distance from edge x the opening dimension in the given direction; The shear wall portions around opening satisfies the maximum /R. The shear does not exceed the nailing capacity Of the wall w/o an opening. Opening: 0.15H= 0.15(9)=1.5 < L= 0.15(14.5)=2.17 >2 Violates opening height Edges: Vert.= (2)=6 > Horiz.= (2)=6 > Violates edge distance T Nailing capacity of wall w/o opening 4 /12 =80 plf SD (seismic) 15/2 sheathing 1 2 (275.8 plf) E (41.8 plf) G (41.8 plf) I (284.4 plf) (41.8) 2 Wx2 H opening J (87.9) (41.8) K (87.9) D (299.5) F (4) H (4) L (02.1) y inspection, portions of wall are 9% over-stressed in shear Wall Not in ompliance with ll riteria D

98 V=4500 lbs. riteria: Opening depth 0.15H Opening length 0.15L Distance from edge x the opening dimension in the given direction; The shear wall portions around opening satisfies the maximum /R. The shear does not exceed the nailing capacity Of the wall w/o an opening. 1 2 (258.6 plf) E (41.8 plf) G (41.8 plf) I (289.6 plf) (465.5) 2 Wx2 H opening J (72.) (465.5) K (72.) D (294) F (4) H (4) L (0.7) Opening: 0.15H= 0.15(9)=1.5 <2 0.15L= 0.15(14.5)=2.17 >2 Violates opening height Edges: Vert.= (1.5)=4.05 >2 Horiz.= (2)=6 > Violates edge distance Nailing capacity of wall w/o opening 4 /12 =80 plf SD (seismic) 15/2 sheathing T y inspection, portions of wall are 2% over-stressed in shear Wall Not in ompliance with ll riteria D

99 In summary: omplex structural layouts require a more detailed analysis than conventional layouts due to the irregularities and discontinuities encountered. Simple calculation methods and software are available to solve the most difficult problems. The framing and connections are required to maintain complete load paths and must be fully detailed on the drawings.

100 Reference Materials The nalysis of Irregular Shaped Structures: Diaphragms and Shear Walls-Malone, Rice-ook published by McGraw-Hill, I Woodworks Presentation Slide rchives-workshop-dvanced Diaphragm nalysis NEHRP (NIST) Seismic Design Technical rief No. 10-Seismic Design of Wood Light-Frame Structural Diaphragm Systems: Guide for Practicing Engineers SEO Seismic Design Manual, Volume 2 Woodworks-The nalysis of Irregular Shaped Diaphragms (paper). omplete Example with narrative and calculations. Diaphragms_Paper1.pdf

101 Method of nalysis References Example Offset Diaphragms and Shear Walls Offset Diaphragms Information on Website Webinar rchive- Offset Diaphragms -Part 1 Webinar rchive- Offset Shear Walls-Part 2 Slide rchive-workshop-dvanced Diaphragm nalysis Slide rchive-offset Diaphragms and Shear Walls