Advanced Diaphragm Analysis-Workshop

Transcription

1 Presentation ased On: dvanced Diaphragm nalysis-workshop Presentation updated to 2015 I, SE SDPWS opyright McGraw-Hill, I Presented by: y: R. Terry Malone, PE, SE Senior Technical Director rchitectural & Engineering Solutions terrym@woodworks.org

2 omplete Example with narrative and calculations Download Process: WoodWorks.org website Publications-Media tab Wood Solutions Papers

3 ourse Description This presentation will provide a review of a method of analysis that can be used to address irregularities that commonly occur in todays structures. Topics will include: Diaphragms with horizontal offsets In-plane and out-of-plane offset shear walls Diaphragms with openings Tall shear walls

4 Learning Objectives asic Information Discuss boundary elements, complete lateral load paths, diaphragm flexibility and review an analytical method used for solving complex diaphragms and shear walls. Diaphragms With Horizontal Offset Learn how to analyze a diaphragm with a horizontal offset and how to transfer forces across areas of discontinuity. In-plane and Out-of-plane Offset Shear Walls Learn how to analyze in-plane and out-of plane offset shear walls. Diaphragms With Openings Learn how to analyze a diaphragm with an interior opening and opening at the support wall line, which creates a discontinuity in the diaphragm web.

5 Presentation ssumptions ssumptions: Loads to diaphragms and shear walls Strength level or allowable stress design Wind or seismic forces. The loads are already factored for the appropriate load combinations. ode and Standards: SE 7-10 Minimum Design Loads for uildings and Other Structures 2015 I 2015 SDPWS (Spid-wiz)- Special Design Provisions for Wind and Seismic nalysis and Design references: The nalysis of Irregular Shaped Structures: Diaphragms and Shear Walls- Malone, Rice Woodworks-The nalysis of Irregular Shaped Diaphragms (paper) 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 Design of Wood Structures- reyer, Fridley, Pollock, obeen Wood Engineering and onstruction Handbook-Faherty, Williamson Guide to the Design of Diaphragms, hords and s-nse

6 Evolution to omplex uildings Simple structures omplex structures The method of analysis is: 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 Today s presentation focuses on: Developing continuous load paths across areas of discontinuities. Flexible wood sheathed or un-topped steel deck diaphragms. an also be used for semi-rigid diaphragms. Wood diaphragms are well suited for these shapes as they can be easily adapted to the building shape and are cost effective.

7 Mid-rise Multi-family Marselle ondominiums Structural Engineer engineer: Yu & Trochalakis, PLL Photographer: Matt Todd Photographer 5 stories of wood over 6 stories concrete (podium) 2 above grade

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

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

10 asic Information oundary Elements omplete Load Paths Diaphragm Flexibility Method of nalysis

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

12 Strut Strut Strut 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) 1 2 Diaphragm 1 oundary (typical) hord Diaphragm 2 oundary (typical) Diaphragm 1 Diaphragm 2 hord 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) 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 ) 3 Note: Interior shear walls require a full depth collector unless a complete alternate load path is provided hord Required for Seismic and wind

13 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 Shear wall Shear wall Shear wall Shear Distribution if ontinuous

14 Strut 1 hord Deflection if tie Diaphragm 2 oundary hord Diaphragm 1 Diaphragm 2 Diaph. oundary (Longitudinal loading) Deflection if no tie hord Strut Strut 4 2 Diaphragm 1 boundary 3 Re-entrant corner Tearing will occur if collectors are not installed at re-entrant corner. 3 hord 1 2 Loads Deflected curve if proper tie Deflected curve if no tie oundary Elements L Shaped uildings-transverse Loading

15 asic Information oundary Elements omplete Load Paths Diaphragm Flexibility Method of nalysis

16 Strut (typ.) 1 Strut Strut Strut (typ.) MRF Strut/chord (typ.) 6 Strut/chord Open Strut /chord (typ.) (typ.) Strut chord Support Multiple offset diaphragm What does this mean? E 2 Strut chord Strut/chord F 4 Strut/chord Offset 3 strut 4 omplete ontinuous Lateral Load Paths D Support 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.

17 Strut (typ.) 1 Strut Strut Strut (typ.) MRF1 1 Strut/chord 2 5 (typ.) Discontinuous diaphragm 5 chord 3 Open Strut/chord Discont. diaphragm chord Strut /chord Discont. diaph. chord Support (typ.) 2 Strut chord E Strut/chord F Discont. diaphragm 4 chord Strut/chord 3 (typ.) 4 Strut chord D Support Discontinuous diaphragm chord/strut SE7-10 Section 1.4-omplete load paths are required including members and their splice connections omplete ontinuous Lateral Load Paths

18 Strut (typ.) 1 Strut Strut Strut (typ.) MRF1 1 Offset shear walls 2 Strut/chord 5 3 (typ.) Open Strut/chord Strut /chord (typ.) (typ.) Strut chord Support E 2 Strut chord Strut/chord F Strut/chord 3 4 D Support 4 Offset shear walls and struts SE7-10 Section 1.4-omplete load paths are required including members and their splice connections omplete ontinuous Lateral Load Paths

19 Strut (typ.) 1 Strut Strut Strut (typ.) MRF Strut/chord (typ.) Open 6 Vertical offset in diaphragm Strut/chord Strut /chord (typ.) Opening in diaph. (typ.) Strut chord Support E 2 Strut chord Strut/chord F Strut/chord 3 4 D Support 4 Design: 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. omplete ontinuous Lateral Load Paths

20 asic Information oundary Elements omplete Load Paths Diaphragm Flexibility Method of nalysis

21 Start SE7-10 Section 12.3 Diaphragm Flexibility Seismic Section The structural analysis shall consider the relative stiffnesses of diaphragms and the vertical elements of the lateral force resisting system. Is any of the following true? Yes Is diaphragm untopped steel decking or Wood Structural Panels 1 & 2 family Vertical elements one Light framed construction Dwelling of the following : where all of the following are met: 1. Steel braced frames 1. Topping of concrete or 2. omposite steel and similar material is not concrete braced frames placed over wood structural 3. oncrete, masonry, steel panel diaphragms except or composite concrete and steel shear walls. for non-structural topping not greater than 1 ½ thick. 2. Each line of vertical elements of the seismic force-resisting system complies with the allowable story drift of Table Yes Idealize as flexible No No Idealize as flexible Structural analysis must explicitly include consideration of the stiffness of the diaphragm (i.e. semi-rigid modeling). Semi-rigid (Envelope Method) Is diaphragm concrete slab or concrete filled steel deck? Yes No verage drift of walls Yes No Is span to depth ratio 3 and having no horizontal irregularities? Yes No (ould apply to LT or Heavy timber diaphragms) Maximum diaphragm deflection Is maximum diaphragm deflection (MDD) >2x average story drift of vertical elements, using the Equivalent Force Procedure of Section 12.8? Idealize as rigid

22 SE7-10, Section 26.2 Diaphragm Flexibility Wind Start Is diaphragm untopped steel decking, concrete filled steel decks or concrete slabs, each having a span-to-depth ratio of two or less? Yes Diaphragm can be idealized as rigid No Is diaphragm of Wood Structural Panels? No Yes Diaphragm can be idealized as flexible Is diaphragm untopped steel decking, concrete filled steel decks or concrete slabs, each having a span-to-depth ratio greater than two? Yes alculate as flexible or semi-rigid

23 asic Information oundary Elements omplete Load Paths Diaphragm stiffness/flexibility Method of nalysis

24 The Visual Shear Transfer Method How to visually show the distribution of shears through the diaphragm + Sheathing element symbol for 1 ft x 1 ft square piece of sheathing in static equilibrium (typ.) FY +M Lds. + - FX Positive Direction + - 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

25 + - T Resisting wall (+) shears Strut in tension Strut in comp. Strut Forces Resisting wall (+) shears Positive sign convention Support Positive diaph. shear elements Pos. + Diaphragm.L. w=uniform load 1 2 (-) (-) + 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 asic Shear Diagram (-) (-) 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).

26 Introduction to Transfer Diaphragms and Transfer reas Transfer Diaph. length 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) Maximum TD spect Ratio=4:1 (Similar to main diaph.) T (strap/blocking or beam/truss) hord TD1 Transfer rea (support) Transfer Diaph. depth T 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) 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<=30 o.k. to use strap/blocking, If > 30 use beam/truss Increase TD depth if shears are too high in transfer area

27 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

28 Drag strut Diaph..L. W ( plf) 4 Diaphragm chord Discontinuous diaphragm chord 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 Diaphragm support Transfer rea Diaphragm chord Longitudinal Typical callout Steel tie strap x ga. x width x length with (xx) 10d nails over 2x, 3x or 4x flat blocking. Lap x -y onto wall. Transfer Diaphragm Members and Elements Diaphragm support

29 Example of Partial Strut/ 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 F3 F4 + + ont. tie strap over F(total)= F1+F2+F3+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. force distribution

30 Strut Resisting forces Resisting forces hord/ hord/ Support 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 Rotation of section Transfer area without transverse collectors Transfer Mechanism 3 Disrupted chord Discontinuous diaphragm chord hord T Transfer Diaphragm ( eam) Support 2 3 NOTE: TD1 Full depth must extend the full depth of the transfer diaphragm Main chord Transfer using beam concept

31 L Transfer Diaph. length L Transfer Diaph. length 1 Transfer Diaph. depth hord, strut or shear wall 2 F( a) R L 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. hord / strut a b F Transfer Diaphragm TD1 hord, strut or shear wall Support F( L) R b a b R hord, strut or shear wall Support F( b) R L R Discont. F hord / strut 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

32 Transfer diaphragm length LTD LTD 1 Method of nalysis-method by Edward F. Diekmann hord 2 (TD support) 3 Main chord -75 plf - T(a), Shear = V V= LTD DTD vnet=+300-(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=+300+(250)= +550 plf No outside force is changing the basic diaphragm shear in this area T +500 plf TD1 vnet =+225 +(250) = +475 plf (TD support) TD depth DTD Main chord +300 plf plf plf asic Shear Diagram at transfer diaphragm 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 +250 plf T(b), Shear = V V= LTD DTD Transfer Diaphragm Shears nalogous to a beam with a concentrated Load.

33 Direction of shear transferred into collector Resulting net shear diagram acting on collector plf +150 plf plf +475 plf Net shears 325 plf 325 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= = +325 plf Shear right= =+325 plf force=area of shear diagram Fcollector=( )(Lcollector) 2 Force Diagram Shear Distribution Into The

34 Diaphragms with Horizontal Offsets

35 Strut Strut Strut Irregularity Requirements for Diaphragms with Horizontal End Offsets-Seismic Type 2 Horizontal Irregularity (Re-entrant corner) exists where both projections > 15% of plan dimension in given direction. SD D-F Triggers Section an also trigger a Type 3 Horizontal Irregularity- abrupt discontinuity or variation in stiffness in diaphragm SD D-F Support Diaph. chord Diaph..L. 1 TD1 2 Diaph. chord Discontinuous Elements and forces Support 1 2 Support Diaph. chord 3 Varying depth and stiffness Support 4

36 Strut Strut Strut Irregularity Requirements for Diaphragms with Horizontal End Offsets-Seismic SE 7-10 Section % increase is required in diaphragm (inertial) design forces (Fpx) for Type 2 or Type 3 horizontal irregularities located in (SD D-F) for the following elements: onnections of diaphragm to vertical elements and collectors (diaphragm supporting elements-td) s and their connections to vertical elements Exception: Forces using the seismic load effects including the over-strength factor of Section need not be increased. Diaph..L. Design collector and connections to using 25% increase per SE 7-10 Section (Grid lines 1 and 4) Design diaphragm connections to and struts using 25% increase per SE 7-10 Section (Grid lines 1 and 4) Diaphragm shears are not required to be increased 25%. See & for collectors 1 Support onnection design with 25% incr. Diaph. chord 1 2 Support TD1 Support Diaph. chord Design transfer diaphragm connection to collector using 25% increase per SE 7-10 Section Diaph. chord 2 3 Design transfer diaphragm connections to boundary elements (chords) at Transfer Support 4 Diaphragm using 25% increase per SE 7-10 Section (Grid line and )

37 7500 lb 3500 lb 15 TD chords TD chords Example 1-Diaphragm with Horizontal End Offset-Transverse Loading alcs w=200 plf Support Diaph..L. Diaph. chord 35 1 TD plf F2 M= M2 ft.-lb F2= lb /R=2.5: lb F2 Diaph. chord Free body for F2 Support Support RL =12500 lb Discontinuous diaphragm chord Diaph. chord + - Sign onvention 80 Support RR=12500 lb

38 15 Pos. Neg. TD shear diagram Transfer Diaphragm and Net Diaphragm Shear plf plf 150 plf asic shear diagram plf Diaph..L lb v= (15) 50(20) = plf alcs lb v=150-(107.1) =+42.9 plf (Net resulting shear) lb v=150+(250)= +400 plf (Net resulting shear) an be > 3x basic shear No net change Net change occurs in TD 3 35 v=70-(107.1) =-37.1 plf (Net resulting shear) lb 15 v=70+(250)= +320 plf (Net resulting shear) 80 v= lb (35) = +250 plf 50(20) plf Sign onvention lb No net change Legend plf asic diaphragm shear (240 plf) Transfer diaphragm shears =xxx plf Net shears (basic shear +/- TD shears)

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

40 15 Diaph..L. alcs plf net plf net T F=6000 lb Special nailing (sum of shears to collector or highest boundary nailing-greater of) F= lb plf net F=6000 lb (this is not an insignificant force.) F=3750 lb Transverse Force Diagrams Sign onvention

41 1 b 2 3 w1 Deflection of rectangular diaphragm Diaph. P.L. w2 Deflection of offset diaph. b Diaphragm Deflection Equations Equation variables for offset diaphragms Varying uniform loads oncentrated loads from discontinuous shear walls Varying moments of inertia, and sometimes Different support conditions T7 Modify the bending and shear portion of the standard rectangular deflection equation to fit the model: I1 TL = 5vL3 8Eb + vl Le 4G n + Σ Δ X t 2b ending deflection Shear deflection I2 SDPWS combines Nail slip djusted for nonuniform nailing (T-7/P) hord slip Standard deflection equation for simple span, rectangular, rigid supports, fully blocked, uniformly loaded, constant cross section ( at.l.) L annot use I Eq P equation Δ TL = Δ + Δ S Le n + Σ(Δ X) 2b where Δ = a b mm EI 1 dx + b c mm EI 2 dx, and Δ S = bt b wx 2G 2 a b b 2 dx + b c wxdx = 0 L vv Gt dx Shear Deflection -USD Research Note FPL-0210 Simplification of the conventional energy method. The integrations of the equations can be reduced to multiplying the total area of the shear diagram due to the general loading by the ordinate of the shear diagram due to a dummy load applied at the desired point of shear deflection. NOTE: Multiply deflection x 2.5 for unblocked diaphragm Multiply nail slip by 1.2 if not Structural I plywood

42 V Seismic Wind W1 W2 W V W3 W1 W2 W W3 Seismic Wind onfiguration 1 Plan Shear wall above onfiguration 2 Plan I1 I2 I3 I2 I1 I3 Diaphragm Stiffness Diaphragm Stiffness Diaphragm ending Deflection Modeling- Other Layouts

43 4/6/12 6/6/12 6/12 U ase I 6/12 U ase I 6/12 U ase I plf 320 plf 285 plf plf 42.9 plf 37.1 plf 70 plf x1 x2 x3 x4 Special nailing along collectors Sum of shears to collector or highest boundary nailinggreater of plf plf 150 plf asic shear diagram 70 plf Transfer diaphragm oundary (Typ.) allout all nailing on drawings: Standard diaphragm nailing oundary nailing nailing Diaphragm boundary heck the shear capacity of the nailing along the collector /6/12 Transfer area oundary (High shear area) oundary locations Diaphragm Nailing allouts

44 plf hord hord 200 plf and TD chords and TD chords Example 2-Diaphragm with Horizontal End Offset -Longitudinal Loading alcs Diaph..L. Drag strut 40 plf 160 plf 35 Diaphragm 1 Transfer diaphragm TD1 Diaphragm 2 50 Discontinuous Drag strut Discontinuous Drag strut 200 plf Drag strut Pos. direction + - Sign onvention 4

45 15 Pos. asic shear diagram Neg. Diaph..L. asic shear diagram Vsw=5000 lb vsw=500 plf plf net 700 lb v=-40.9-(10.5)=-51.4 plf 210 lb lb plf Vsw=5000 lb V = wl lb Pos. direction v=+45.1+(24.5) =+69.6 plf Legend v=+15.1-(10.5)=+4.6 plf (Net resulting shear) v=+15.1+(24.5) =+39.6 plf 700 lb lb Transfer Diaphragm and Net Diaphragm Shear plf 375 plf asic diaphragm shear =xxx plf Net diaphragm shears (basic shear +/- TD Shears) (240 plf) Transfer diaphragm shears lb Transfer area V = wl 2 Vsw=5000 lb W=200 plf W=40+160=200 plf alcs Sign onvention Vsw=5000 lb vsw=333.3 plf plf net 4

46 15 F=700 lb plf net plf 5 net F= -957 lb F=3529 lb plf net alcs 35 F=819 lb F=700 lb F= -819 lb F=700 lb plf net F=368 lb F=368 lb F= lb strut 4050 lb chord Pos. direction + - Sign onvention Longitudinal and Transverse /Strut Force Diagrams plf net 4

47 Window wall No calculations No collector Horizontal offset in chord/strut F=20 kips Strut/chord 1.L. diaphragm 100 Strut/chord Example-ctual Project

48 95 ft. v=304 plf v=504 plf v=328 plf v=997 plf v=244 plf v=588 plf v=390 plf v=997 plf (designed) 45 ft. Steel decking F=25230 lb F=22500 c.l. diaphragm (My calc=25278 lb) F=24905 lb Diaphragm Shears Transfer diaphragms and collectors are required F=7626 lb F=7454 lb 4 ft. offset F=6991 lb F=7512 lb NOTE: v max=1864 plf ritical connections v max=1748 plf Diaphragm designed as a simple rectangular diaphragm, no offset, using only a spreadsheet. hecked only diaphragm shear and chord force (maximum depth, not offset depth). No collectors, connection designs or details at re-entrant corners. Forces on trusses at collectors were not called out on drawing. Example-ctual Project

49 Diaphragm 2 Www Quick Note on Segmentation Offset shear walls W W OMRF WLw d1 d2 d3 1 Diaphragm antilever 3 4 L1 L2 L3 L4 2 5 W Www W w1 w2 w3 w4 w5 w6 TD1 Open TD2 W hord Support w7 hord w8 D WLw w9 L5 L6 L7 Support

50 Strut Rigid frame Designed as diaph. with opening W=200 plf (wind) W=200 plf Www=123 plf W=200 plf (wind) hord hord Sub-hord hord hord hords force transferred to TD2 hord TD1 Sub-hord W=200 plf (wind) hord Support TD2 Strut hord hord hord ant. Diaph. WLw=77 plf D hord hord Support Segmentation of the Diaphragm for Transverse Loading

51 W=200 plf hord hord W=200 plf Www=123 plf hord hord hord Rigid frame WLw=77 plf Strut Diaph.1 Diaph.2 Strut Strut Support Strut force transferred to TD2 Strut TD1 Strut Diaph.3 TD2 Diaph.4 Strut Strut Diaph.5 D Support Segmentation of the Diaphragm for Longitudinal Loading

52 V=4000 lb V=4000 lb Neg. 20 Pos. Pos. 20 Neg. 1 2 W=200 plf lb plf 100 plf 200 plf hord 8000 lb -200 asic Shear Diagram TD1-200 plf -100 plf -300 plf TD lb hord lb F=16000 lb lb Transfer Diagram Shear lb hord hord lb lb Transfer Diagram Shear Sign onvention + - nalysis Option 1-nalyze as Diaphragm with Intermediate Offset

53 20 W=100 plf 2000 lb 2000 lb 20 W=200 plf W=100 plf lb W=200 plf 2000 lb W=100 plf W=100 plf plf hord 300 plf 200 plf asic Shear Diagram plf -300 plf W=100 plf W=100 plf hord hord hord 40 F=16000 lb -500 plf hord hord asic Shear Diagram lb hord 40 ssumes small diaphragms are supported off of main diaphragm 120 asic Shear Diagram hord lb nalysis Option 2-nalyzing as separate diaphragms

54 W=200 plf 3 4 hord 600 plf 200 plf /R=6:1>4:1, N.G. Fchord=18000 lb 200 plf Options 1, 2 and 3 Values Options 1 and lb asic Shear Diagram 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 3-Ignoring lower diaphragm sections (Not recommended) 40

55 spect ratios all diaphragms 4:1 max Using partial sections???? (Not recommended) 4 ont. collector required (full depth of diaph). ant. Diaphragm 1 Transfer area Diaphragm 2 Diaphragm 3 No exterior Shear walls Unshaded areas ride off of main designated diaph. s Non-shear wall Shear wall

56 Offset Shear Walls 1 2 Out-of-plane Offsets In-plane Offsets

57 Relevant Irregularities Per SE 7-10 Horizontal Irregularities Table and Vertical Irregularities Table SE 7 Table Type 4 vertical irregularity: inplane offset discontinuity in the LFRS (if no H.D. at.25) Transfer diaphragm grid line 1 to 3 See Section SE 7 Table Type 4 horizontal irregularity: out-ofplane offset discontinuity in the LFRS load path 2 SE 7 Table Type 4 vertical irregularity: inplane offset discontinuity The deflection equation must be adjusted to account for the uniformly distributed load plus the transfer force. Potential buckling problem w/ supporting columns and beams Elements requiring over-strength load combinations

58 Strut Type 4 Horizontal & Vertical Offset Irregularity-Seismic Type 4 horizontal irregularity-out-of-plane offset irregularity occurs where there is a discontinuity in lateral force resistance load path. Out-of-plane offset of at least one of the vertical lateral force resisting elements. Type 4 vertical irregularity-in-plane discontinuity in vertical lateral force resisting element occurs where there is an offset of vertical seismic force resisting element resulting in overturning demands on a beam, column, truss, wall or slab. SE 7-10 Section (SD -F) Elements supporting discontinuous walls or frames. SE 7-10 Section (SD D-F) Increases in force due to irregularity 1 Diaphragm 2 3 Strut Strut hord Shear wall is not continuous to foundation. Type 4 horizontal and vertical irregularity hord

59 ontinuous member SE 7-10 Section (SD -F) Elements supporting discont. Walls or frames: locking or shear panels pplies to: eams, columns, slabs, walls or trusses. Requires over-strength factor of Section lso see SD -F for collector requirements. lt. / strut onnection Type 4 Vertical Irregularity SD D-F (Interior collector similar)

60 Diaphragm 1 Strut Strut Strut SE 7-10 Section (SD D-F) -Type 4 Horizontal and Type 4 vertical irregularity requires a 25% increase in the diaphragm (inertial) design forces determined from (Fpx) for the following elements: onnections of diaphragm to vertical elements and collectors. s and their connections to vertical elements. Exception: Forces using the seismic load effects including the over-strength factor of Section need not be increased Shear wall is not continuous to foundation. hord Design diaphragm connections to and struts (Grid lines 1, 2 and 3) Diaphragm shears are not required to be increased 25%. The transfer force (3) in SD D-F must be increased by rho, per hord See & for collectors s and their connections to vertical (Grid lines 1, 2 and 3) Type 4 Horizontal Irregularity-Seismic

61 ontinuous member SE 7-10 Section (SD D-F) Type 2 Horizontal Re-entrant corner Irregularity Type 3 Diaphragm discontinuity irregularity Type 4 horizontal or vertical irregularity: Requires a 25% increase in the diaphragm design forces (Fpx) determined from for the following elements: locking or shear panels onnections of diaphragm to vertical elements and collectors. Shear Wall lt. / strut onnection s and their connections, including their connections to vertical elements. Type 2 Horiz., Type 3 Horiz., and Type 4 Vert. & Horiz. Irregularity SD D-F (Interior collector similar)

62 Struts and s-seismic Struts / collectors and their connections shall be designed in accordance with SE 7-10 sections: SD - s can be designed w/o over-strength but not if they support discontinuous walls or frames SD thru F- s and their connections, including connections to the vertical resisting elements require the over-strength factor of Section , except as noted: Shall be the maximum of: Ω o F x - Forces determined by ELF Section 12.8 or Modal Response Spectrum nalysis procedure 12.9 Ω o F px - Forces determined by Diaphragm Design Forces (Fpx), Eq or F px min = 0. 2S DS I e w px - Lower bound seismic diaphragm design forces determined by Eq (Fpxmin) using the Seismic Load ombinations of section (w/o over-strength)-do not require the over-strength factor. Exception: F px max = 0. 4S DS I e w px - Upper bound seismic diaphragm design forces determined by Eq (Fpxmax) using the Seismic Load ombinations of section (w/o over-strength)-do not require the over-strength factor. 1. In structures (or portions of structures) braced entirely by light framed shear walls, collector elements and their connections, including connections to vertical elements need only be designed to resist forces using the standard seismic force load combinations of Section with forces determined in accordance with Section (Diaphragm inertial Design Forces, F px ).

63 Support column Support column Dhr.25 olumn elements are designed for standard load combinations 2.3 or 2.4. Strut/ Wall element is designed for standard load combinations of SE 7-10 Sections 2.3 or Vr Shear wall anchorage is designed for standard load combinations of Sections 2.3 or 2.4. Dh2 Strut/ V2 Shear wall hold downs are designed for standard load combinations of Sections 2.3 or Wall designed for over-strength per SE 7-10 Section (SD -F) and/or (SD D-F) Ftg. Ftg. SE 7 Table Type 4 vertical irregularity- In-plane offset discontinuity in the LFRS See struts and collectors ssuming no hold down Shear wall system at grid line 1 In-plane Offset of Wall

64 Support column Support column Dhr.33 Wall element is designed for standard load combinations of SE 7-10 Sections 2.3 or 2.4. Vr Shear wall hold downs and connections are designed for standard load combinations of SE 7-10 Sections 2.3 or 2.4. Footings are not required to be designed for over-strength. Ftg. 3 Dv Support beam/collector and columns shall be designed in accordance with SE 7-10 section: Elements supporting discontinuous walls or frames require over-strength Factor of Section Ftg. Dh2 V2 onnection is designed for standard load combinations of Sections 2.3 or 2.4. SE 7 Table Type 4 horizontal irregularity- out of-plane offset discontinuity in the LFRS See struts and collectors Shear wall system at grid line 2 Out-of-Plane Offset

65 Support column Support column Wall element, hold downs, and connections are designed for standard load combinations of SE 7-10 Sections 2.3 or Dhr 4.33 Dh2 Vr Wall element designed for standard load combinations of Sections 2.3 or 2.4. onnections are designed for standard load combinations of Sections 2.3 or 2.4. Footings are not required to be designed for overstrength. Ftg. Ftg. Support beam/collector Grade beam 5 Gr. m. DL V2 onnections are designed for standard load combinations of SE 7-10 Sections 2.3 or 2.4. SE 7 Table Type 4 vertical irregularity- In-plane offset discontinuity in the LFRS See struts and collectors and columns shall be designed in accordance with SE 7-10 section: Elements supporting discontinuous walls or frames require over-strength factor of Section Shear wall system at grid line 3-In-plane Offset

66 Drag strut Drag strut Out-of-Plane Offset Shear Walls ssumed to act in the Same Line of Resistance Transfer area 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. Discont. drag strut Offset Loads s are required to be installed to transfer the disrupted forces across the offsets. Discont. drag strut Typical mid-rise multi-family structure at exterior wall line SE 7-10 Section Where offset walls occur in the wall line, the shear walls on each side of the offset should be considered as separate shear walls unless provisions for force transfer around the offset are provided. In the plane of the diaphragm heck for Type 2 horizontal irregularity Re-entrant corner irregularity

67 Mid-rise Multi-family ant. orridor only shear walls Offset shear walls No exterior Shear walls Flexible, semi-rigid, or rigid???

68 Strut/chord Strut eam Loads Higher shears and nailing requirements TD1 TD2 3 (typ.) Transfer rea Higher shears and nailing Reqmts. TD1 TD TD3 Higher shears and nailing Reqmts. 2 Main diaphragm becomes TD3 Multi Story, Multi-family Wood Structure 5 (typ.) 5 I1 I2 I1 I2 I3 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 elements

69 plf hord hord collectors hord collectors 200 plf hord Example 3-Diaphragm with Horizontal End Offset Longitudinal Loading-Out-of-plane offset Shear Walls Drag strut Drag strut 80 Support TD1 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/ Offset 1 8 Drag strut Offset Drag strut 12 3 Drag strut Support Drag strut is discontinuous Pos. direction

70 200 plf Neg. 200 plf asic shear diagram 1 Pos. asic shear diagram 2 Total Shear to Shear Walls (ssumed) Vsw2=wL/2=200(50)/2=5000 lb, Vsw1, sw3, sw4=wl/2=200(50)/2=5000 lb, lb (-28 plf) vsw2=5000/10=500 plf vsw=5000/(8+8+15)=161.3 plf lb (+8.85 plf) alcs lb (-40.9 plf) 40 plf 160 plf 35 F= lb F=590.3 lb lb (+28 plf) vsw= plf vnet= plf Determine Force transferred Into Transfer Diaphragm lb plf lb lb (+15.1 plf) lb ( plf) asic Diaphragm Shears and Transfer Diaphragm Shear 4510 lb (+45.1 plf) Pos. direction + - Sign onvention

71 15 Neg. Pos. asic shear diagram vsw=500 lb vnet= plf lb (-40.9 plf) alcs v=-40.9+(8.85) = plf 35 v=+15.1+(8.85) = plf vsw=161.3 lb vnet=133.3 plf 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 lb (+15.1 plf) Pos. direction No net change Sign onvention Legend plf asic diaphragm shear =xxx plf Net diaphragm shears (basic shear +/- TD Shears) (240 plf ) Transfer diaphragm shears lb 4 vsw=161.3 lb (+45.1 plf) vsw=161.3 lb vnet=116.2 plf vnet=116.2 plf

72 alcs plf net 8.85 plf net Net shear (TD tension chord and Diaph.2 compression chord) F=141.7 lb F= lb F= lb F=141.6 lb F= F= plf net So far, so good Transverse Force Diagrams Pos. direction + - Sign onvention

73 15 F= lb F=-700 lb F=+3659 lb 80 F=+387 lb (Error) alcs F=-860 lb F= lb F=590.3 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 vsw=161.3 lb vnet=116.2 plf vsw=161.3 lb vnet=116.2 plf F= lb F=-1030 lb F = lb F = lb 80 F = lb (Error) Longitudinal Strut Force Diagrams Pos. direction + - Sign onvention

74 15 alcs F= lb 2 F=700 lb F=802.5 lb F= lb F=364 lb F=3236 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=36 lb Load distribution needs to increase towards line /. Increase the load to / by the amount off +/-. alculated forces 4600 lb 1 F=693.6 lb F=87.5 lb F=416.6 lb F=18 lb Line needs to move in this direction lb F=75 lb F=957.8 lb F= lb djusted Longitudinal Strut Force Diagrams (8% increase to /) [mount shifted to / depends on the offset to span ratio of the transfer diaphragm]

75 Tying Shear Walls cross the orridor?? Special nailing of the sheathing to the collector is required the full length of the collector (typ.) Transfer rea (typ.) orridor walls orridor walls Section collectors Layout 1 Layout 2 Layout 3 Layout 4 Layout 5 Layout 6 SDPWS s for shear transfer to individual full-height wall segments shall be provided. Loads Tie straps at all Joints (typ.) Jst. ommon Transverse Wall Layouts Mech. 2x6/2x8 Jst. Typical orridor Section

76 2 1 Layout 1-Full length walls aligned 1 2 W=200 plf v= plf vnet=54.55 plf 6 Diaphragm 1 Diaphragm RL=RR=200(78/2)=7800 # vr=7800/50=156 plf 1200 RL=RR=200(122/2)=12200 # vr=12200/50=244 plf D 7800 # Sum=20000 # # Sum=40000 # =200 plf(200 )

77 88 plf plf 24 plf 112 plf 1 88 plf Example 4-Offset Shear Walls cross the orridor-layout 6 Layout 6, ase1-full length offset walls W=200 plf # 3608 # Transfer rea RL=RR=24(4/2)=48 # 122 vl=vr=48/6=8 plf RL=RR=112(78/2)=4368 # vr=4368/28=156 plf Diaphragm Diaphragm RL=RR=88(82/2)=3608 # vr=3608/22=164 plf RL=RR=88(122/2)=5368 # vr=5368/28=244 plf RL=RR=112(118/2)=6608 # vr=6608/28=236 plf 5368 # # D 82 Sum=7976 # Sum=20048 # Sum=11976 # Sum=40000 # =200 plf(200 ) 118

78 ase 1-Smaller resulting forces at corridor Total load to grid lines 2 & 3 R23= =20048 # O.K. 2 3 Summing V=0 L=22+22=44 V=20048 # v=20048/44= plf vnet 1= =55.64 plf vnet 2= =55.64 plf hecks, they should be equal 1 F2=55.64(22)=1224 # F2=(156+8)6=984 # F2= =240 # D Summing V=0 Sum=20048 # 2 F3D=55.64(22)=1224 # F3=(236+8)6=1464 # F3= = -240 # F23=F23=240(4)/6=160 # O.K. ll forces in lb., all shears in plf Shear at transfer area=240/6+8=48 plf

79 88 plf plf 24 plf 112 plf 1 88 plf Layout 6, ase2-full length plus partial length offset shear walls W=200 plf # D # 78 Transfer rea RL=RR=24(4/2)=48 # vl=vr=48/6=8 plf RL=RR=112(78/2)=4368 # RL=RR=88(122/2)=5368 # vr=4368/28=156 plf vr=5368/28=244 plf Diaphragm Diaphragm RL=RR=112(118/2)=6608 # RL=RR=88(82/2)=3608 # vr=6608/28=236 plf vr=3608/22=164 plf Sum=7976 # Sum=20048 # Sum=11976 # Sum=40000 # =200 plf(200 ) # #

80 ase 2-Larger resulting forces at corridor Total load to grid lines 2 & 3 R23= =20048 # O.K Summing V= L=22+12=34 V=20048 # v=20048/34= plf vnet 1= = plf vnet 2= = plf hecks, they should be equal 1 F2=189.65(22)= # F2=(156+8)6=984 # F2= = # D Summing V=0 Sum=20048 # 2 F2 =189.65(12)= # F2 to 3=(236+8)6+( )10=5464 # F3= = # F23=F23= (4)/6= # O.K. ll forces in lb., all shears in plf Shear at transfer area= /6+8= plf

81 88 plf plf 24 plf 112 plf plf Layout 6, ase3 - Partial length offset shear walls W=200 plf # D # 78 Transfer rea RL=RR=24(4/2)=48 # vl=vr=48/6=8 plf RL=RR=112(78/2)=4368 # RL=RR=88(122/2)=5368 # 12 vr=4368/28=156 plf vr=5368/28=244 plf Diaphragm Diaphragm RL=RR=88(82/2)=3608 # RL=RR=112(118/2)=6608 # vr=3608/22=164 plf 12 vr=6608/28=236 plf Sum=7976 # Sum=20048 # Sum=11976 # Sum=40000 # =200 plf(200 ) # #

82 ase 3-Smaller resulting forces at corridor Total load to grid lines 2 & R23= =20048 # O.K. 156 Summing V=0 244 L=12+10=22 V=20048 # v=20048/22= plf D Summing V=0 Sum=20048 # vnet 1= = plf vnet 2= = plf hecks, they should be equal 1 F1=511.27(10)= # F1 to 2=( )12=4800 # F2=(156+8)6=984 # F2= = # 2 F2=511.27(12)= # F2 to 3=(236+8)6+( )10=5464 # F3= = # O.K. ll forces in lb., all shears in plf F23=F23=671.3(4)/6=120 # Shear at transfer area=671.3/6+8=119.9 plf

83 In-plane Offset Shear Walls

84 Example 5-In-plane Offset Segmented Shear Wall -with Gravity Loads lb VHdr=450 lb DL=150 plf Hdr lb 1 DL=250 plf lk g. or rim joist Nail shtg. To each 2x stud Wd VHdr=960 lb Sill Hdr/collector Wd Sections do not comply with the required aspect ratio for a perforated or FTO shear wall. 8 2 No hold down (option 1) Hold-down (option 2) Sill 6 1 SE 7 Table Type 4 vertical irreg.- in-plane discontinuity in the LFRS if no hold down at & SD -F SD D-F

85 Ends of wall panels do not line up. Requires special nailing of sheathing into stud below. Requires same number of studs above and below with boundary nailing each stud Solid blocking required Hold down Nailing found in field was 12 o.c. No hold-down below Photo-In-plane Offset Segmented Shear Walls Hold down

86 plf asic Shear +250 plf 2000 lb w=230 plf (incl. wall DL) VHdr=450 lb 5000 lb lb Rim joist lb 1080 lb 330 plf (incl. wall DL) ver.=250 plf lb 8 Pos. direction lb +450 lb 3370 lb 5000 lb Wall and Transfer Diaphragm Shears Upper Shear Wall Sign onvention lb (-157.5) 1260 (-157.5) (+7.5) TD shears-lbs. (plf) (-127.5) Lower Shear Wall (+202) + shears lb 9700 lb 1620 lb (+202) - shears 8 VHdr=960 lb

87 Depth TD Depth TD 2000 lb Roof 2000 lb Roof (8)+450= lb 1 T 8 2 nd floor Rim joist VHdr=990 lb 3000 lb 8 2 nd floor Rim joist 1 + T T T lb 9700 lb 2490 lb lb Vertical Forces Horizontal Forces Force Diagrams Pos. direction + - Sign onvention

88 Diaphragms With Large Openings Interior and End Openings 1 2 collector 3 4 Diaph. 5.L. w plf collector SE 7-10 Section (SD D-F) - Horizontal irregularity Type 3 requires a 25% increase in the diaphragm design forces determined from (Fpx) for the following elements: onnections of diaphragm to vertical elements and collectors (diaphragm supporting elements). s and their connections to vertical elements. D collector collector Diaphragm shears are not required to be increased 25%. Exception: Forces using the seismic load effects including the over-strength factor of Section need not be increased. Use of over-strength forces is not commonly considered to be triggered for boundary elements at diaphragm openings. However, the 25% increase does apply. Type 3 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.

89 Roof pop-up section with opening below. Skylight or atrium opening lerestory windows End opening Stairwell access to roof ommon Openings In Diaphragms

90 Harrington Recovery enter Structural engineer: Pujara Wirth Torke, Inc. Photographer: urtis Walz Openings in diaphragm

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

92 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 3 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

93 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 Fv = v4 H 2 H H L Opening L Steel straps and blocking v1 v v Upper ound Fh = v2l for left collector Lower ound v2 Fh = v2 L 2 for right collector v4 v4 Lv H v3 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

94 plf External loads not included in check, plf Pattern nailing has to increase-n.g. 318 plf Nail capacity has to increase n.g plf R. = =1.2: lb lb lb plf plf 318 plf plf 318 plf L. opening v=150 plf.l. opening v=300 plf v=150 plf F V = 150(10) 2 L V = = 750 lb F H = = 1080 lb = < 10 o.k L H = = < o.k. Opening is not a factor 12 L x 10 H Opening Located at low shear Lower bound F V = 300(10) 2 L V = = 1500 lb = 12.5 > 10 n.g. F H = = 2160 lb Lower bound 2160 L H = = 120 > 12 n.g. Opening is a factor 12 L x 10 H Opening Located at high shear F V = 150(35) 2 = 2625 lb heck If Size and Location of Opening In Diaphragm Is Factor L V = = < 35 ut > depth Lower bound F H = = 2160 lb L H = = > 12 n.g. Opening is a factor 12 L x 35 H Opening Located at low shear

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

96 Shear distribution follows analysis Shear Distribution in Diaphragm

97 /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

98 D 1 2 w plf TD1 F2 F3 F2 F3 F4 R M 2 0 M 0 M 4 0 V1 V1 Element II Element IV V2 3 3 Inflection point. F=0 Opening F=0 Element I Element III V3 4 F4 TD2 V4 asic shear diagram without openings V2L V2R V3 V4L V4R asic Shear Diaphragm With Opening 5 V5 V5 Diaph..L. Typical method of analysis (P Report 138), T-7, and FPInnovations 1. alculate the chord forces at grid lines 2, 3, and 4 using FD s. 2. Determine the basic diaphragm shears without an opening. 3. 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. 3. alculate the chord force at grid line 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

99 1600 lb V=25120 V=21120 V=19520 V=13920 V=6720 V=8320 V= lb 1600 lb Example 6-Pop-up Roof Section /R main diaphragm and upper section=3.33:1 /R TD1=TD2=3.0:1 o.k. Wind Loads (SD) Main W=200 plf t opening Ww=123 plf Lw=77 plf t pop-up (20 psf) Ww=50 plf Lw=30 plf Diaph. 6.L. W=200 plf W=200 plf W=123 plf TD1 W=50 plf Open to below W=30 plf TD W=200 plf W=123 plf W=50 plf F3 Sub-hord D RL= W=77 plf RR=21280 Sub-hord TD1 W=30 plf v1 v2r v2l v3 v4l 8 ΣM = 0 D W=77 plf RL=25120 F3D v4r asic Shear Diaphragm With Opening (plf) v5 v6 Open to below

100 V4L=8320 lb V4R=6720 lb 1600 lb External loads not used in rough calc., except reaction Upper limit 8320 lb FPInnovations: Hgt.> 0.15 ddiaph. 28 Width>0.15 Ldiaph. End dist.< 3x width Detailed analysis required L. opening Wwind 318 plf lb plf 318 plf Shears w/ opening 318 plf 60 Unit shear right = Unit shear left = F V = 112(20) 2 L V = = 1120 lb = 112 plf = 208 plf = < 20 o. k. F H = = 4160 lb Lower bound for right collector 4160 L H = = < 40 o. k. 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= 352 plf, nail cap.= 358 plf Use 6/6/12 locked, H.F. (includes 1600 lb.) lso < 12 header depth, o.k. v= 112 plf 318 plf all nail cap.= 318 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

101 (352 plf) lb lb (360.9 plf) (284.4 plf) 7964 lb 7964 lb (284.4 plf) (208 plf) 5824 lb (352 plf) 9416 lb 9416 lb (784.6 plf) (496.3plf) 5956 lb 5956 lb (496.3 plf) (208 plf) 2496 lb 200 plf 123 plf 123 plf F lb F4 470 lb F lb 200 plf Diaph..L lb TD1 F2 Element II Element I + + F lb Sub-hord 7076 lb 50 plf 50 plf ΣM = 0 ΣM = lb 2496 lb (112) plf) (112) plf) 1600 lb Start here ΣV = 0 TD2 28 D 20 F2 Sub-hord F lb lb F2D 6827 lb lb + F4D lb RL=25120 lb 77 plf 77 plf V1 V2 V3 V4L=8320 lb V4R=6720 lb V plf Element IV ΣM = 0 Lb ΣM = 0 V Lb 30 plf V (60) Lb Free-body of hord Forces and Segment Forces F3D Element III 20 5 The sum of the section shears must match the basic diaphragm shear Values without an opening, V=0.

102 w lb = = T.D lb = lb = lb 6453 lb = lb Pos. Neg. Neg = D lb Transfer diaphragm shears Sign convention asic Shear Diagram Net Shears-Left Transfer Diaphragm

103 3363 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

104 1600 lb F=12800 lb w F=9524 lb F=6453 lb TD T T F=1476 lb D Sign convention Force Diagrams-Left Side

105 lb w F=7076 lb T T F=3170 lb F=2018 lb TD2 F=4384 lb T F=4924 lb F=1696 lb D Force Diagrams-Right Side + - Sign convention

106 lb lb ver.=640.5 ver.=352.2 ver.= loses to 13.6 zero F=486.7 lb 470 F=13295 lb F=20340 lb F=18550 lb F=18858 lb F=6829 lb 6827 F=18206 lb F=13282 lb F=18568 lb F=18876 lb D ver.=341.5 ver.=352.2 ver.= T lb lb hord Force Diagrams loses to zero

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

108 sum w1 w2 b b b w3 I1 I2 P L I1 Diaph..L. w1 Deflection of diaph. w/ opening D b Diaphragm Deflection Equations Equation variables for offset diaphragms Varying uniform loads oncentrated loads from discontinuous shear walls Varying moments of inertia T7 Modify the bending and shear portion of the standard rectangular deflection equation to fit the model, where: Δ TL = Δ + Δ S Le n + Σ(Δ X) 2b where Δ = a b mm EI 1 dx + b c mm EI 2 dx + c d mm EI 1 dx, and SDPWS combines TL = 5vL3 8Eb + vl Le 4G n + Σ Δ X t 2b annot use I Eq Δ S = bt 2G 2 a b wxdx + b c wx b b 2 dx + ending deflection Shear deflection Nail slip djusted for nonuniform nailing (T-7/P) hord slip Standard deflection equation for simple span, rectangular, rigid supports, fully blocked, uniformly loaded, constant cross section ( at.l.) Shear Deflection -USD Research Note FPL-0210 Simplification of the conventional energy method The integrations of the equations can be reduced to multiplying the total area of the shear diagram due to the general loading by the ordinate of the shear diagram due to a dummy load applied at the desired point of shear deflection. NOTE: Multiply deflection x 2.5 for unblocked diaphragm Multiply nail slip by 1.2 if not Structural I plywood

109 Opening (typ.) s (typ.) TD TD Skylight, atrium, mechanical openings, stairwell openings, etc. Multiple Opening Issues

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

111 Strut Example 7- Intermediate Horizontal Offset at End Wall With Strut w1 plf (WW) w plf 4 D Section 10 hord 18 Skylight (Enclosed area) hord + 16 Section 16 + TD1 /R= Uniform shear in walls and in diaph. at grid line 1 D w1 1 2 V3 F2 Section Sum Shears w2 Section R1 V3 F2 hord hord 3 56 w2 plf (LW) Varies Diaph..L. asic Shear Diagram -

112 vdiaph 1 2 Neg. No shear transferred Strut Pos. Neg. 1 3 Net Shear Strut Net Shear 1 T 2 T T T F2 + V F D Net shear diagram Support VD Transfer diaphragm shears Support Resulting Strut, and hord Force Diagrams if Strut

113 plf plf plf V=11800 lb V=8800 lb Example 8 -Intermediate notch at End Wall Without Strut W=123 plf W=200 plf 4 56 D lb if strut (-38%) lb if strut (+38%) hord R1=8800 lb (v=400 plf) 16 Section hord Open area Section TD1 /R= hord 8800 lb lb lb Shear in diaphragm at grid line 2 Is based on depth hord V= lb v=310.5 plf F2= lb R1=8800 lb D W= Section W= R1D=6200 lb Section F2=5584 lb V= lb v=310.5 plf R1D=6200 lb (V=387.5 plf) W=77 plf Sum Shears Shear Varies Diaph..L. Forces in red are from previous example asic Shear Diagram R=15000 lb R=15000 lb -

114 Neg. Pos. Neg v=210.7-(123.7) = +87 plf v=157.1-(123.7) = plf R1 F2= D 2 v=210.7+(255.24) = plf R1D v=210.7-(117) = plf varies + Transfer diaphragm net shears plf plf plf v=157.1+(255.24) = plf v= (117) = +40 plf F2= Transfer diaphragm shears Transfer Forces to s asic shear diagram Transfer diaphragm and net diaphragm shears Sign onvention

115 F=5684 lb 10 R1 400 F=5684 lb F=5584 lb F=5584 lb D R1D Support Longitudinal hord Force Diagrams

116 -400 plf plf 1 1 Vsw=8800 lb 2 3 vsw= plf vnet= = plf F=4000 F= F= Net shear diagram Vsw=6200 lb vsw=387.5 plf vnet=0 plf 2 D 400 R1=8800 F= F= F= F= lb R1D=6200 lb Traverse Strut/ Force Diagrams

117 Relevant 2015 SPDWS Sections (Diaphragm Stiffness, Distribution of Shears and Open Front / antilever Diaphragms) (a) (b) (c) Force Open front L Force Force L W Open front L Open front antilever Diaphragm Plan W Plan antilever Diaphragm W antilever Diaphragm Figure 4 Examples of Open Front Structures New definitions added: (d) Open front antilever Diaphragm Open front structures Notation for L and W for cantilever Diaphragms s Force L Relevant Revised sections: Horizontal Distribution of Shears Torsional Irregularity Open Front Structures Open front W L antilever Diaphragm