Part 2-Offset Shear Walls

Presentation Based On: Part 2-Offset Shear Walls Presentation updated to 2012 IB, ASE 7-10 2008 SDPWS opyright McGraw-Hill, I Presented by: By: R. Terry Malone, PE, SE Senior Technical Director Architectural & Engineering Solutions terrym@woodworks.org

The Wood Products ouncil is a Registered Provider with The American Institute of Architects ontinuing Education Systems (AIA/ES), Provider #G516. redit(s) earned on completion of this course will be reported to AIA ES for AIA members. ertificates of ompletion for both AIA members and non-aia members are available upon request. This course is registered with AIA ES for continuing professional education. As such, it does not include content that may be deemed or construed to be an approval or endorsement by the AIA of any material of construction or any method or manner of handling, using, distributing, or dealing in any material or product. Questions related to specific materials, methods, and services will be addressed at the conclusion of this presentation.

ourse Description A continuation from Part 1, this session will cover how to conduct a preliminary breakdown of a complex diaphragm to better understand the distribution of forces within the diaphragm and assure that complete load paths are being established. Examples will be provided illustrating how to analyze in-plane and out-ofplane offset shear walls that are typically created by these diaphragms.

Learning Objectives Segmentation of a omplex Diaphragm Discuss methods of breaking down and analyzing complex diaphragms into manageable segments. In-plane and Out-of-plane Offset Shear Walls Discuss the various types of offset shear wall conditions. Out-of-plane Offset Shear Walls Examine a method of analyzing a diaphragm with offset shear walls for loading in the longitudinal direction. In-plane Offset Shear Walls Examine a two story offset shear wall with varying width.

Presentation Assumptions Flexible wood sheathed or un-topped steel deck diaphragms (an also apply to semi-rigid and rigid diaphragms) Loads to diaphragms and shear walls Strength level or allowable stress design Wind or seismic forces (UNO). The loads are already factored for the appropriate load combination. ode References: ASE 7-10 Minimum Design Loads for Buildings and Other Structures 2012 IB Analysis and Design references: The Analysis of Irregular Shaped Structures: Diaphragms and Shear Walls- Malone, Rice WoodWorks- The Analysis of Irregular Shaped Diaphragms Design of Wood Structures- Breyer, Fridley, Pollock, obeen SEAO Seismic Design Manual, Volume 2 Wood Engineering and onstruction Handbook-Faherty, Williamson Guide to the Design of Diaphragms, s and s-nsea

omplete Example with narrative and calculations http://www.woodworks.org/education-publications/research-papers/#

A Quick Note on Segmenting and analyzing omplex Diaphragms-h.8 1 Diaphragm 1 3 4 6 7 antilever L1 L2 L3 L4 2 5 W Www W A w1 w2 w3 Diaphragm 2 D B Www SW Support W w4 w7 WLw w5 Open TD1 TD2 Offset shear walls SW W w6 SW w8 w9 OMRF WLw W d1 d2 d3 L5 L6 L7 Support

1 2 3 4 Designed as diaph. with opening W=200 plf (wind) W=200 plf Www=123 plf W=200 plf (wind) 5 6 7 A B Support Strut TD1 Sub- Sub- WLw=77 plf D TD2 Strut Rigid frame s force transferred to TD2 W=200 plf (wind) ant. Diaph. Support Segmentation of the Diaphragm for Transverse Loading

1 2 3 4 5 6 7 Www=123 plf Strut Diaph.1 Strut W=200 plf SW TD1 SW Diaph.2 Strut Strut Diaph.3 Strut D W=200 plf SW Strut Diaph.4 TD2 SW Rigid frame Support WLw=77 plf A Strut force transferred to TD2 Strut Diaph.5 SW B Support Segmentation of the Diaphragm for Longitudinal Loading

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

Out-of-Plane Offset Shear Walls Assumed to act in the Same Line of Resistance Drag strut SW 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 SW Loads s are required to be installed to transfer the disrupted forces across the offsets. Discont. drag strut SW Drag strut Typical mid-rise multi-family structure at exterior wall line ASE 7-10 Section 14.5.2 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. heck for Type 2 horizontal irregularity Re-entrant corner irregularity

Longitudinal Loading Offset transverse walls In-line transverse walls ant. No exterior Shear walls?? Flexible, semi-rigid, or rigid???

Loads 1 SW3 2 SW1 TD1 TD2 3 SW4 SW2 TD3 Higher shears and nailing requirements SW5 Multi Story, Multi-family Wood Structure I1 I2 Diaphragm stiffness changes I1 I2 I3

Example 2-Diaphragm with Horizontal End Offset Longitudinal Loading A 5 SW 2 10 Diaph..L. Drag strut 40 plf 160 plf 5 200 plf B 35 Diaphragm 1 Discontinuous Drag strut and TD chords Transfer diaphragm TD1 and TD chords Diaphragm 2 50 15 200 plf Discontinuous Drag strut 200 plf Drag strut SW 1 15 25 20 80 1 2 3 Pos. direction + - 4

A VA Vnet=vsw-vdiaph SW 2 10 va - $= &'/" 35 Diaphragm 1 (Net shear) Diaphragm 2 Neg. Diaph..L. B VB 15 VB Pos. vb + v Vnet=vsw-vdiaph 15 SW 1 (Net shear) 2 $= &'/" 1 2 3 B Transfer area 4 Transfer Diaphragm and Net Diaphragm Shear Pos. direction + -

A F2A Vnet @ SW 5 5 10 SW 2 Fend Fstart 35 F2B Long F2B Trans F2B Long F3B 50 B 15 F3B Vnet @ SW 15 SW 1 F2B Trans 25 20 80 HEK: Fstrut Fchord Pos. direction + - 1 2 3 4 Longitudinal and Transverse /Strut Force Diagrams

Example 3-Diaphragm with Horizontal End Offset Longitudinal Loading-Offset Shear Walls A 200 plf B 35 SW 1 Drag strut 15 Drag strut is discontinuous SW 2 25 5 10 5 Offset SW Drag strut 200 plf collectors TD1 Drag strut collectors 12 Drag strut 80 Assumptions: 1. Assume shear walls at grid lines B and act along the same line of lateral-forceresistance. 2. Assume the total load distributed to grid lines A and B/= wl/2. Offset SW 50 SW 3 Drag strut SW 4 45 15 Support Pos. direction + - Support 25 20 80 1 2 3 4

Total Shear to Shear Walls (Assumed) Vsw2=wL/2 plf Vsw1, sw3, sw4=wl/2, vsw= V1,3,4/(Lsw1+Lsw3+Lsw4) plf A VA SW 2 10 VA VA 40 plf 160 plf 200 plf 35 Pos. Fend F2B 50 B SW 1 VB 17 Vnet=Vsw-Vdiaph F2B 200 plf 200 plf F2B SW 3 Neg. V VB V SW 4 15 1 Determine Force transferred Into Transfer Diaphragm 2 2 3 Pos. direction + - Basic Diaphragm Shears and Transfer Diaphragm Shear

A Vnet sw SW 2 VA VA 35 V=Basic shear - TD shear plf (Net shear) Pos. 50 B SW 1 Vnet sw 15 V=Basic shear - TD shear plf (Net shear) 25 20 No net change Net change in TD 1 2 Net Diaphragm Shears 3 No net change Neg. SW 3 15 SW 4 Vnet sw V 80 VB V Pos. direction + - Vnet sw 4

A SW 2 - Net shear Net shear (TD tension chord and Diaph.2 compression ) F2B Special nailing F3B Diaphragm 2 F3B B SW 1 1 15 SW3 15 SW 4 F2B -vb 2 3 Net shear x2 -F +F +v x1 So far, so good Transverse Force Diagrams Pos. direction + -

A F2A Fend 5 SW 2 5 10 F3A 80 F4A=+xxx lb (Error) Fstart Vnet sw Fend F2B F= F2B Note: Neither force diagram closes to zero, therefore error. Notice that they do not close by the same amount. B SW 1 Vnet sw 15 12 Vnet sw SW 3 45 15 SW 4 Vnet sw F3 Fstart 25 20 Fend 80 Fstart F4 = -xxx lb (Error) 1 2 3 Longitudinal Strut Force Diagrams Pos. direction + -

A SW 2 Revised forces alculated forces 4600 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 need to be higher in order to move the force diagram in this direction Load distribution needs to increase towards line B/. Increase the load to B/ by the amount off +/-. B SW 1 Line needs to move in this direction 15 SW 3 SW 4 5400 lb 25 20 80 1 2 3 Pos. direction + - Adjusted Longitudinal Strut Force Diagrams (8% to 20% increase to B/) [Amount shifted to B/ depends on the offset to span ratio of the transfer diaphragm]

In-plane Offset Shear Walls

Diaphragm load at 2 nd floor Transfer diaphragm grid line 1 to 3 B Drag strut Drag strut F W A SW1 SW2 1 SW3 / beam ASE 7 Table 12.3-1-Type 4 horizontal out-of-plane offset irregularity in the LFRS ASE 7 Table 12.3-2-Type 2 4 vertical irreg.-in-plane discontinuity in the LFRS 3 SW4 Drag strut Drag strut SW5 A.75 A.25 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 ASE 7 section 12.3.3.3 Elements supporting discontinuous walls and frames (heck requirements for Rho)

Example 4-In-plane Offset Segmented Shear Wall -with Gravity Loads VHdr=450 lb DL=150 plf 2000 lb Hdr 1 3000 lb No hold down (option 1) Hold-down (option 2) 4 Blk g. or rim joist Nail shtg. To each 2x stud 12 SW1 DL=250 plf SW2 Wd VHdr=960 lb Sill Hdr/collector Wd Sill A B D 6 1 Section does not comply with the required aspect ratio for a perforated or FTO shear wall. ASE 7 Table 12.3-2-Type 4 vertical irreg.- in-plane discontinuity in the LFRS

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

2000 lb w=230 plf (incl. wall DL) VHdr=450 lb 5000 lb 416.67 lb Rim joist 12 1 416.67 lb 1080 lb 330 plf (incl. wall DL) Aver.=250 plf + + Pos. direction + - 5000 lb Wall and Transfer Diaphragm Shears + 135 365 SW1 1080 lb 2920 lb +450 lb 3370 lb Upper Shear Wall Sign onvention +250 plf 1260 lb (-157.5) 1260 (-157.5) - 1620 60 (+202) (+7.5) TD shears-lbs. (plf) + shears 3370 416.67 1080 416.67 + 4 4 +259.2 1020 (-127.5) 1 2 2490 lb Lower Shear Wall 9700 lb SW2 + + + +424.1 +289.2 + +618.7 1620 lb (+202) - shears VHdr=960 lb +416.67 plf Basic Shear

2000 lb Roof 2000 lb Roof 416.67 + SW1 + 135 365 365(8)+450=3370 + SW1 + 135 365 3000 lb 1 T 2 nd floor Rim joist 416.67 VHdr=990 lb 3000 lb 2 nd floor Rim joist 1 + T + +259.2 SW2 1 2 + + + Depth TD +259.2 SW2 1 2 + + + Depth TD T +424.2 +289.2 +618.7 +424.2 +289.2 +618.7 4 T 4 2490 lb 9700 lb 2490 lb 9700 lb Vertical Forces Horizontal Forces Force Diagrams Pos. direction + - Sign onvention

QUESTIONS?* This*concludes*The*American*Ins9tute*of*Architects* on9nuing*educa9on*systems*ourse* Part*2A*Offset*Shear*Walls* R. Terry Malone, P.E., S.E. Senior Technical Director WoodWorks.org ontact Information: terrym@woodworks.org Events/Presentation Archives (slide handouts)-free Paper: http://www.woodworks.org/education-publications/research-papers/#