Based on:

Part 2-Offset Shear Walls

Presentation updated to 2012 IBC, ASCE 7-10Copyright McGraw-Hill, ICC

Presented by:

By: R. Terry Malone, PE, SESenior Technical DirectorArchitectural & EngineeringArchitectural & EngineeringSolutions

terrym@woodworks.org

“The Wood Products Council” isa Registered Provider with The American Institute of Architects

This course is registered with AIA CES for continuing professional education. As

Continuing Education Systems (AIA/CES), Provider #G516.

psuch, it does not include content that may be deemed or construed to be an approval or endorsement by the AIA of

Credit(s) earned on completion of this course will be reported to AIA CES for AIA members

yany material of construction or any method or manner ofhandling, using, distributing, or dealing in any material or AIA CES for AIA members.

Certificates of Completion for both AIA members and non-AIA members are available upon request

g yproduct.__________________________________

Questions related to specific materials, methods and services will be addressedrequest. methods, and services will be addressedat the conclusion of this presentation.

Course Description

A continuation from Part 1 this session will cover how to conductA continuation from Part 1, this session will cover how to conducta 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. Examplesassure that complete load paths are being established. Exampleswill be provided illustrating how to analyze in-plane and out-of-plane offset shear walls that are typically created by these diaphragms.

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Learning Objectives

• Segmentation of a Complex Diaphragm

Discuss methods of breaking down and analyzing complex diaphragms into manageable segments.

• In plane and Out of plane Offset Shear Walls• In-plane and Out-of-plane Offset Shear Walls

Discuss the various types of offset shear wall conditions.

• Out-of-plane Offset Shear WallsOut of plane Offset Shear WallsExamine the method of analyzing a diaphragm with offset shear walls for loading in the longitudinal direction.

• In plane Offset Shear Walls• In-plane Offset Shear WallsExamine a two story offset shear wall with varying width.

Presentation AssumptionsFlexible wood sheathed or un-topped steel deck diaphragms

(Also applies to semi rigid diaphragms)(Also applies to semi-rigid diaphragms)

• Loads to diaphragms and shear wallsp g• Strength level or allowable stress design• Wind or seismic forces (UNO).

• The loads are already factored for the appropriate load combination.y

Code References: • ASCE 7 10 “Minimum Design Loads for Buildings and Other Structures”• ASCE 7-10 “Minimum Design Loads for Buildings and Other Structures”• 2012 IBC

Design references:

• The Analysis of Irregular Shaped Structures: Diaphragms and Shear Walls-Malone, Rice

• Design of Wood Structures- Breyer, Fridley, Pollock, Cobeen• SEAOC Seismic Design Manual, Volume 2

W d E i i d C t ti H db k F h t Willi• Wood Engineering and Construction Handbook-Faherty, Williamson• Guide to the Design of Diaphragms, Chords and Collectors-NCSEA

Complete Example with narrative and calculations

http://www.woodworks.org/education-publications/research-papers/#

1 6 7

A Quick Note on Segmenting and analyzing Complex Diaphragms-Ch.8

3 4

2

1

5

6 7

L1 L3L2 L4

Diaphragm 1 Cantilever

2

A

5

Www W

w1

W

w2 w3

C

Open

w6w4 w5

W

d1

OM

RF

SW

TD1 TD2

Ww

w

WLw

Col

lect

or

Col

lect

or

2

Collector Collector

B

w7 w8

W d2

SW

Chord Collector Collector

ctor

Chord

C C

Dia

phra

gm

CollectorSupport

D

C

WLw

W d3

SW

SW

Col

lecChord

Offs

et

shea

r w

alls

w9

Collector

L5 L7L6Support

Www=123 plfW=200 plf (wind) W=200 plf W=200 plf (wind)

3 41 5 6 72Designed as diaph.

with opening

ecto

r

ctor

ctor

ecto

r Collector Collector Chord Chord Chord Chord

fram

e

A

Chords force transferred to TD2

Sub-Chord

TD1 W=200 plf (wind)Collector

Col

le

Col

lec

Collector

Col

lec

Col

l

Chord

Rig

id

Chord Collector

B

to TD2

Sub-Chord

TD2

Cant.Diaph.

Support

Collector

Stru

t

Chord

olle

ctor

Chord Chord C

WLw=77 plf

Chord

Co

Chord Stru

t

D

Support

Segmentation of the Diaphragm for Transverse Loading

3 41 5 6 72

S t

Diaph 1

Diaph.2

=123

plf

ASupport

ctor

ctor

ctor

ctor

Collector Collector Strut

Strut Strut

Cho

rd SW SW

fram

e

=77

plf

Strut force Diaph.1

Diaph 4

Ww

w=

lf

BCollector

Col

lec

Col

lec

Collector

Col

le

Col

lec

Strut Strut Strut

Collector

Rig

id

WLw

= transferred to TD2

TD1Diaph.3

Diaph.4

Diaph.5lf

W=2

00 p

l

C

TD2

Collector

Cho

rd

Strut

Cho

rd

SW

Cho

rd

W=2

00 p

DSupport

SWCho

rd

SW

C

Segmentation of the Diaphragm for Longitudinal Loading

Offset Shear Walls

SW 1

Collector

Col

lect

or

SW 2

Collector

SW 2

Out-of-plane Offsets In-plane Offsets

Out-of-Plane Offset Shear WallsAssumed to act in the Same Line of ResistanceAssumed to act in the Same Line of Resistance

SWTransfer area

• Offset walls are typically assumed to act in the same

olle

ctor

Collector

Dra

g st

rut

• Offset walls are typically assumed to act in the same line of lateral-force-resistance.

• Calculations are rarely provided showing how the wallsare interconnected to act as a unit, or to verify that a

Loads

Co

Collector

SWOffset

Discont.drag strut

D s complete lateral load path has been provided.

• Collectors are often not installed to transfer the disrupted forces across the offsets.

SWOffset

Dra

g st

rut

Collector

Discont.drag strut

ASCE 7-10 Section 14.5.2Where offset walls occur in the wall line, portions of the shear wall on each side of the offset shall be considered as separate shear walls unless provisions for force

SW

Col

lect

or

Collector

Collector as separate shear walls unless provisions for force transfer around the offset are provided.

Typical mid-rise multi-family structure at exterior wall line

Longitudinal Loading

Offset transverse walls In-line transverse walls

Cant.

No exteriorShear walls ??

Flexible, semi-rigid, or rigid???

SW3Loads

TD1

SW3

SW1

1

Loads

TD22

3

SW4

TD3SW2 Higher shears and nailing requirements

SW5

Diaphragm stiffnessM lti St M lti f il Diaphragm stiffness changesMulti Story, Multi-family

Wood StructureI1 I2 I3

Diaph.C L

Example 2-Diaphragm with Horizontal End OffsetLongitudinal Loading

C.L.

A10’5’

SW 2Drag strut

40 plf 160 plf

5’

200

plf

r and

rd

sTransfer

diaphragmTD1

35’

Diaphragm 1

p p

and

rds

2

Cho

rd

Col

lect

orTD

cho

r

50’Diaphragm 2Diaphragm 1

hordC

olle

ctor

TD

cho

r

15’

CollectorDiscontinuousDrag strutB

0 pl

f

Ch

200 plfSW 1

25’ 20’

1

80’

C15’

20 Drag strutDiscontinuousDrag strut

1 2 3 4+Pos. direction -

VA

SW 2A

vA -10’

Vnet=vsw-vdiaph

10’

2m 1

35’

Dia

ph.

C.L

.

(Net shear)

Neg

.

Dia

phra

gm

Dia

phra

gm

B

(Net shear)

VB

D

VBvB

15’

CSW 1

2

+15’Po

s.

VBVnet=vsw-vdiaph

vC

(Net shear)

4

BTransfer area

12 3

+Pos. direction -

Transfer Diaphragm and Net Diaphragm Shear

10’A

Vnet @ SWF2A

5’ 5’

Fend

10 SW 2Fstart

35’50’F2B

Trans

B

F3BF2BLong

F2BLong

15’

B

C15’ SW 1

F3B

F

Vnet @ SW

25’ 20’ 80’

CHECK: FstrutFchord

+Pos. direction -

F2BTrans

1 2 3 4

Longitudinal and Transverse Collector/Strut Force Diagrams

SW 2 Drag strutDrag strut

Example 3-Diaphragm with Horizontal End OffsetLongitudinal Loading-Offset Shear Walls

SupportSW 2A

10’25’ 5’Drag strut

5’ 80’Drag strut

Assumptions:1. Assume shear walls at grid lines B and C

f f

Support

200

plf

s35’

TD1s

Cho

rd

act along the same line of lateral-force-resistance.

2. Assume the total load distributed to grid lines A and B/C= wL/2 .

Cho

rd

colle

ctor

s

hord

3550’

Cho

rd

colle

ctor

Offset SW

SW 1

15’

Drag strut

Ch

B

200

plf

SW 3 SW 4

8’Collector

12’

Drag strut

Drag strut

SW

Offset SW

Drag strut

Drag strut is discontinuous

C15’

2

Pos. direction + -

8’12

45’Drag strut

Support

25’ 20’ 80’

1 2 3 4

Vsw2=wL/2 plf Vsw1, sw3, sw4=wL/2, vsw= V1,3,4/(Lsw1+Lsw3+Lsw4) plf

Total Shear to Shear Walls (Assumed)

SW 2A

10’

40 lf 160 lf

VA VA VA

35’ Pos.

40 plf 160 plf

200

plf

35

50’

P

FendF2B

B8’

SW 300 p

lf

SW 1 F2B

VB

Neg

.

200 plf17’

SW 4

F2B VB

1 2 3

C8’

2

SW 320Vnet=Vsw-Vdiaph

N SW 4

15’VC VC

Pos. direction + -

Basic Diaphragm Shears and Transfer Diaphragm Shear

Determine Force transferred Into Transfer Diaphragm

ASW 2

Vnet sw VA VA

V=Basic shear- TD shear plf

(Net shear)

35’50’

Pos.

(Net shear)

B8’ SW 1 VB

15’

C

8 SW 1

15’SW 3 SW 4Neg

.

8’Vnet sw

Vnet sw Vnet swV=Basic shear

VC VC

25’ 20’ 80’Pos. direction + -

V Basic shear- TD shear plf

(Net shear)

25 20 80

1 2 3No net change Net change

in TDNo net change

Net Diaphragm Shears4

-

ASW 2

Net shear Net shear (TD tension chord and Diaph.2 compression Chord)

F2B Special nailing F3B

Diaphragm 2

B8’ SW 1

F3B

15’

F2B

1

C

8 SW 18’ SW3 15’ SW 4

Net shearvB 2 3-vB

x1’

x2’

+F

-F

Transverse Collector Force Diagrams Pos. direction + -

So far, so good+vC

x1’+F

ASW 2

F2AFend

F3A

F4A=+xxx lb (Error)

5’ 5’10’

80’

Fstart

F3A10

Vnet sw

F2BNote: Neither force diagram closes to zero, therefore error.Notice that they do not close

F= F2B

B

FendNotice that they do not close by the same amount.

15’

B

CSW 415’

8’

12’ 45’

SW 1

SW 3

Vnet sw

Vnet swVnet sw

25’ 20’ 80’

F3C

F4C = -xxx lb (Error)

Vnet sw

FstartFend

Fstart

1 2 3

Longitudinal Strut Force Diagrams Pos. direction + -

8’

4600 lbA

SW 2

Calculated forces

Revisedforces

SW 2

Line needs to move in this direction

The shear wall shearsneeds to be lower in

Load distribution needs to

The shear wall shears

needs to be lower in order to move the force diagram in this direction

needs to increase towards line B/C. Increase the load to B/C by the

BSW 1

need to be higher in order to move the force diagram in this direction

Line needs to move

yamount off +/-.

15’

B

CSW 3 SW 4

5400 lb

Line needs to move in this direction

25’ 20’ 80’Pos. direction + -

5400 lb

1 2 3

Adjusted Longitudinal Strut Force Diagrams (8% increase to B/C)[Amount shifted to B/C depends on the offset to span ratio of the transfer diaphragm]

In-plane Offset Shear Walls

A 1

Transfer diaphragm grid line 1 to 3

ASCE 7 Section12.3.2.1-T 4 h i t l i i

ASCE 7 Section 12.3.2.2-Type 4 vertical irregularity out-of-plane discontinuity in the LFRS

B 3

2 Type 4 horizontal irreg. in-plane discontinuity in the LFRS

Potential buckling

A.75A.25

The deflection eq ation m st Elements requiring problem w/ supporting columns and beams

The deflection equation must be adjusted to account for the Uniformly distributed load plus the transfer force.

q gover-strength load combinations

ASCE 7 Section 12.3.3.4-Type 4 vertical or horizontal irregularity in SDC D, E, F design forces determined from Section 12.10.1.1 (Fpx) shall be increased 25% (See section for details)

ASCE 7 section 12.3.3.3Elements supporting discontinuous walls and frames (Rho required if SDC D thru F)

Example 4-In-plane Offset Segmented Shear Wall-with Gravity Loads

VHdr=450 lb

Hdr2000 lb

DL=150 plf

Transfer diaph. is required on

Wd Section does not comply with the required aspect

ti f f t d

8’ SW1(TD)

is required on the left side of the opening if sill section is added.

VHdr=960 lb

Sill

Blk’g. or rim joist

ratio for a perforated or FTO shear wall.

1’3000 lb

Hdr/collector

DL=250 plf

Nail shtg. To each 2x stud

8’

WdHold down location if Cantilever Method

SW2ctor

(TD)

Sill

if Cantilever Methodused.

Col

lec

A B C D

No hold down(option 1)Hold-down(option 2)

4’ 8’

12’ 6’ 1’

Increased nailing may be required at grid line 1

Ends of wall panels do not line up. Requires special nailing of sheathing into stud below.

Requires same number of studs above and belowwith boundary nailing each stud Solid blockingnailing each stud Solid blocking

required

Hold down

Nailing found in field was 12” o.c.

Photo-In-plane Offset Segmented Shear Walls

No hold-down belowHold down

5000 lb1’

12’416.67 lb416.67 lb

Rim joist

1080 lbw=230 plf (incl wall DL)VHdr=450 lb

++

2000 lb1080 lb

330 plf (incl. wall DL) w=230 plf (incl. wall DL)

+- 8’4’1020

1260 lb(-157 5)

1620 lb(+202)

8’135 365

SW1

+250

plf

1260(-157.5)

1020(-127.5)

60(+7 5)

1620(+202)

(-157.5) (+202)- shears

Aver.=250 plf +

2920 lb1080 lb

8’416.67416.67

33701080

TD shears-lbs. (plf)(+7.5) + shears

5000 lb2920 lb

Upper Shear Wall

ecto

r

lf

++259 2

VHdr=960 lb+450 lb3370 lb

ear

8’

+Pos. direction -

Col

le SW2

+416

.67

pl

+ ++

1 2

+259.2

Bas

ic S

he

Wall and Transfer Diaphragm Shears8’4’

Lower Shear Wall 2490 lb 9700 lb

+424.1 +618.7+289.2Sign Convention

2000 lb Roof 2000 lb Roof

8’++ SW1 ++ SW1

C

C

3000 lb1’

8

8’2nd floor Rim joist

SW1135 365

3000 lb1’

8’

8’2nd floor Rim joist

SW1135 365

T 416.67

365(8)+450=3370

VHdr=990 lb

416.67

T+

1’ 2 d floor Rim joist

SW2h

TD+259.2

C+

1’2nd floor Rim joist

SW2

h TD+259.2

VHdr 990 lb

+T

++C

Dep

th1 2

+424.2 +618.7+289.2+ ++

8’

Dep

th1 2

+424.2 +618.7+289.2

8’

8’4’

Vertical Collector Forces

T 8’4’ C

Horizontal Collector Forces

2490 lb 2490 lb9700 lb 9700 lb

+Pos. direction -

Vertical Collector Forces Horizontal Collector Forces

Sign ConventionCollector Force Diagrams

A

200’

SW 1

SW 2• No calculations

SW 2

Horizontal offset in chord/strutF=20 kips

• No collector

100’B

dow

wal

l

p

Strut/chord

1

C

Win

d

Strut/chordC.L. diaphragm

100’

2 3

4

diaphragm

Actual Project

v=304 plfv=328 plf

v=244 plfv=504 plf

v=997 plf v=588 plfv=390 plf v=997 plf

(d i d)

95 ft.

F=25278 lb @c l diaphragm

p (designed)

DiaphragmShears

F=25230 lb

c.l. diaphragm(calc’d=22500 lb)

F=24905 lb

Steel decking

Transfer diaphragms and collectors are required

45 ft.

F=7626 lb F=6991 lbF=7454 lb F=7512 lb4 ft offset

NOTE:v max=1864 plf

v max=1748 plfCriticalconnections

4 ft. offset

• Diaphragm designed as a simple rectangular diaphragm, no offset, using only a spreadsheet.

• Checked 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.

Actual Project

?QUESTIONS?This concludes The American Institute of ArchitectsContinuing Education Systems CoursePart 2 Offset Shear Walls

R. Terry Malone, P.E., S.E.Senior Technical DirectorWoodWorks.orgWoodWorks.org

Contact Information:terrym@woodworks.org

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Paper:Paper:http://www.woodworks.org/education-publications/research-papers/#