By: R. Terry Malone, PE, SESenior Technical DirectorArchitectural & Engineering Solutionsterrym@woodworks.org928-775-9119

Diaphragm and Shear Wall Design

Avalon Bay CommunitiesPhoto credit: Arden Photography

120 Union, San Diego, CATogawa Smith Martin

Copyright McGraw-Hill, ICC

Portions Based On:

Course Description

Intended for structural engineers who would like a deeper understanding of advanced shear wall and diaphragm principles, this presentation will provide an in-depth look at wood-frame shear wall and diaphragm designs commonly used for non-residential and multi-family low-rise and mid-rise buildings, including those with rectangular, offset and open-front plans. Topics will include:

• Basic information and method of analysis

• Diaphragms with horizontal offsets

• Diaphragms with large openings

• Shear wall types and anchorage

• SDPWS and ASCE 7 shear wall code requirements

• Offset shear walls and code requirements

Presentation Assumptions

Assumptions:

• 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.

Code and Standards:

Evolution to Complex Buildings

• Simple structures Complex structures

• The method of analysis:§ Can be used for all construction types.

§ Is Straight forward and simple to use. “A 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.

Can also be used for semi-rigid or rigid diaphragms.

Wood diaphragms are well suited for these shapes as they can be

easily adapted to the building shape and are cost effective.

Dis

cont

inuo

us c

hord

sTr

ansv

erse

Cant.

Mid-rise Multi-family

SWSWSWSW

SWSWSWSWSW

SW

Lds. Discontinuous strutsLongitudinal

Lds.

No exteriorShear walls

Flexible, semi-rigid, or rigid???

Harrington Recovery CenterStructural engineer: Pujara Wirth Torke, Inc.Photographer: Curtis Walz

Offsets in the diaphragm and walls

Vertically offset Diaphragms?

Openings in diaphragm

• Boundary Elements • Complete Load Paths• Method of Analysis

Basic Information

SW

Struts, Collectors, and Chords- (my) Terminology

W ( plf)

Chord

Stru

t

Diaphragm support

Col

lect

or

Strut- receives shears from one side only*. Collector- receives shears from both sides.

*[ Drag struts and collectors are synonymous in ASCE7]

SW

Diaphragm support

Chord

W ( plf)

Diaphragm 1 Diaphragm 2

Diaphragm 2 Boundary (typical)

Chord

Chord

Col

lect

or

Stru

tSt

rut

Chord

Stru

t

Fundamental Principles:A shear wall is a location where diaphragm forces are resisted(supported), and therefore defines a diaphragm boundary location.

Note: Interior shear walls require a full depth collector unless a complete alternate load path is provided

Diaphragm Boundary Elements

SW1

SW2

SW3

Note: All edges of a diaphragm shall be supported by a boundary element. (ASCE 7-10 Section 11.2)

Diaphragm 1 Boundary (typical)

• Diaphragm Boundary Elements:• Chords, drag struts, collectors, Shear walls,

frames• Boundary member locations:

• Diaphragm and shear wall perimeters• Interior openings• Areas of discontinuity• Re-entrant corners.

• Diaphragm and shear wall sheathing shall not be used to act as or splice boundary elements. (SDPWS 4.1.4)

• Collector 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. (ASCE 7-10 Section 12.10.2)

Required for Seismic and wind

1 2

B

3

C

A

3. Collectors for shear transfer to individual full-height wall segments shall be provided.

SDPWS 4.3.5.1

Would you do this?Failure Modes

Shear failureNot enough shear capacity

Splitting failureBottom section not supported

Alternate load path

Alternate load path

Shear wall

High stress concentrations at end of the wall

Shear Distribution if No Collector

Note:Diaphragm sections act as notched beams(shear distribution-diagonal tension or compression)

Collector

Shear Distribution if Continuous Collector

Code does not allow the sheathing to be used to splice or act as boundary elements

Shear wall

Shear wall

Shear wall

Shear wall

Shear wall

Loads

Col

lect

or

Collector

Diaphragm 1 Diaphragm 2S

trut

Chord Chord

Str

utS

trutDiaphragm 2

Boundary

Diaphragm 1boundary

Re-entrant cornerTearing will occur if collectors are not installed at re-entrant corner.

1 2

A

B

3

C

Deflection if no tie

Deflected curve if proper tieDeflected curve if no tie

SW1

SW2 SW3

SW4

Chord

Chord

Boundary Elements “L” Shaped Buildings-Transverse Loading

Deflection if tie

Diaph.Boundary(Longitudinal loading)

• Boundary Elements • Complete Load Paths• Method of Analysis

Basic Information

Strut/chord

Open

3

4

4

21

F

E

D

C

B

5 8 96 7

Strut/chord

Str

ut (t

yp.)

Strutchord

Strut chord

Strut /chord

Strut/chord

Strut/chord

SW

1

SW2

SW7SW5

SW6

Str

ut

MR

F1

Multiple offsets

Offset struts

Support Support

Col

lect

or

Collector

Collector(typ.)

Collector(typ.)

Collector (typ.)

Col

lect

or (t

yp.)

Analysis: ASCE7-10 Sections:• 1.3.1.3.1-Design shall be based on a rational analysis• 12.10.1-At diaphragm discontinuities such as openings and re-entrantcorners, 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.

Complete Load PathsDiscontinuousdiaphragm chords

Offset shear wallsand struts

ASCE7-10 Section 1.4-Complete load paths are required including members and their splice connections

Openingin diaph.

Vertical offset indiaphragm

SW3

10

A

SW4

Strut/chord

Open

3

4

5

21

F

E

D

C

B

6 9 107 8

Strut/chord

Str

ut (t

yp.)

Strutchord

Strut chord

Strut /chord

Strut/chord

Strut/chord

SW

1

SW2

SW7SW5

SW

9

SW6

Str

ut

MR

F1

Support Support

Col

lect

or

Collector

Collector(typ.)

Collector(typ.)

Collector (typ.)

Col

lect

or (t

yp.)

A

Complete Load Paths

SW

8

SW3

Transfer area

Collectors for shear transfer to individual full-height wall segments shall be provided.

SDPWS 4.3.5.1 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).

4

SW4

• Boundary Elements • Complete Load Paths• Method of Analysis

Basic Information

The Visual Shear Transfer MethodHow to visually show the distribution of shears through the diaphragm

+ -

+

Positive Direction

+ -

Transverse Direction (shown)

Lds.

Shears Applied to Sheathing Elements

FY

FX

+M

Sheathing element symbol for 1 ft x 1 ft square piece of sheathing in static equilibrium (typ.)

+

+ -

Shears Transferred Into Boundary Elements

Unit shear transferred from the sheathing element into the boundary element (plf)

Unit shear acting on sheathing element (plf)

Transfer shears

+

-

+-

+ -

-+

Basic Shear DiagramPositive diaph.shear elements

Pos.

Neg.

Diaphragm shear transferred into boundary element (typ.)

Strut intension

Resisting wallshears

Resisting wallshears

Resisting wallshears

Strut inCompr.

Strut incomp.

Strut intension

SW 2

SW 1

SW 3

DiaphragmC.L.

Strut Forces Strut ForcesT

C

T

T

C

C

1 2

A

B

Negative diaph.shear elements

(-)

(+)

(+)

(+)

(-)

(-)

(-)

+ -

Positive sign convention

Maximum moment

1 ft. x 1 ft. square sheathingelement symbol at any location in the diaphragm.

Shear Distribution Into a Simple DiaphragmThe Visual Shear Transfer Method

Support Support

SW

SW

All edges of a diaphragm shall be supported by a boundary element (chord, strut, collector) or other vertical lateral force resisting element (shear wall, frame).

w=uniform load

Introduction to Transfer Diaphragmsand Transfer Areas

Col

lect

or

T Collector(strap/blocking or beam/truss)

(support)

(support)

Chord

Chord

TD1

Col

lect

or

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)

Transfer Diaphragm • Sub-diaphragm-don’t confuse w/ sub-diaphragms

supporting conc./masonry walls

• Transfers local forces out to primary chords/struts of the main diaphragm. ( Based on method, ASCE 7 Section 1.4 and SDPWS 4.1.1)

• Maximum TD Aspect Ratio=4:1 (Similar to main diaph.)

What doesthis mean?

Collector Length?-My rule of thumb:

• Check length by dividing discontinuous force by the nailing capacity (other issues need to be considered)

• Length=full depth of transfer the diaphragm, set by A/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

Transfer Area

Transfer Diaph. depth

Tran

sfer

Dia

ph. l

engt

h

2’

16’

TD aspect ratio 8:14:1 maximum allowed

Over-stressed.(Notice deformationin transfer diaphragm)

TD Aspect Ratio Too High

Warping/racking

SW

SW

Diaph.C.L.

W ( plf)

Longitudinal Collector

Diaphragm chord

Discontinuousdiaphragm chord

1 2

A

B

3

CDiaphragm chord

Dra

g st

rut

Diaphragm support

Diaphragmsupport

Transfer Diaphragm Members and Elements

4

Transfer Area

Diaphragm chord

• The length of the collector is oftendetermined by dividing the collector force by the diaphragm nailing capacity.(Caution-other issues need to be considered!)

• The collector is often checked for tension only. (Wrong!) Compression forcesoccur when the loads reverse direction.

Typical calloutSteel tie strap x ga. x width x length with (xx) 10d nails over 2x, 3x or 4x flat blocking. Lap x’-y” onto wall.

2x flat blk’g(tight fit)

Potential gaps

Typical Section

Use Z clip to keep blocking level

Bearing perp. to grain? Alt. 4x blk’g.Tie strap

WSP sht’g.

Joists

Example of Partial Strut/Collector

Collector Compression force distribution

Strut/collector force diagram

Diaphragm unit shear (plf) transfers into blocking

For compression, blocking acts as mini strut/collectorand transfers (accumulates) forces into the next block.

+ ++

No gaps allowed. Diaphragm sheathing is not allowed to transfer strut/collector tension or compression forces.

F4F3F2F1

Cont. tie strap over

F(total)=∑F1+F2+F3+F4

Transfer Mechanism

1 2

B

3

C

T C

Main chord

Main chord

B

C

A

2 3

Disruptedchord

Res

istin

g fo

rces

Transfer area without transverse collectors

Transfer using beam concept

Transfer Diaphragm

( Beam)

Transferarea

Disruptedchord

Support

Support

Collector

Rotationof section

This force must be transferred out to the main chords. A complete load path is required.(ASCE 7 Section 1.4 and SDPWS 4.1.1)

Collector must extend the full depth of the transfer diaphragm

NOTE:

SW

CollectorFull depthChord

Cho

rd/C

olle

ctor

Stru

t

TD1

Cho

rd/C

olle

ctor

SW

Discontinuousdiaphragm chord

SW

Res

istin

g fo

rces

TD1

Support

Support

Discont. Chord / strutF

ab

A

C

B

21Transfer Diaph. depth

Tran

sfer

Dia

ph. l

engt

h

LaFRA)(

=

L

LbFRC)(

=

TransferDiaphragm

Analogous to a simple span beam with a concentrated load

TD1

Support

Support

Discont. Chord / strutF

A

C

B

21Transfer Diaph. depth

Tran

sfer

Dia

ph. l

engt

h

bLFRB)(

=

L

baFRA)(

=

TransferDiaphragm

Analogous to a propped cantilever beam with a concentrated load

Simple Span Transfer Diaphragm

Collector Chord, strutor shear wall

Chord, strut or shear wall

Chord, strut or shear wall

Chord, strut or shear wall

F

F

RB

RA

RC

RA

Propped Cantilever Transfer Diaphragm

ab

Simple Span and Propped Cantilever Transfer Diaphragms

Discont. Chord / strutF

SW

Col

lect

orT T

+

-

Analogous to a beam with a concentrated Load.

Chord force at discontinuity

Subtract from basic shears

Add to basic diaphragmshears

1

A

B

2

C

Collector

(TD support)

(TD support)

Chord

TD1

Basic Shear Diagram at transfer diaphragm

-75 plf

+250 plf

+300plf +225 +225

plf plf

vnet=+300+(250)= +550 plf

vnet =+225–(75)= +150 plf

3

TD depthTr

ansf

er d

iaph

ragm

leng

th

+

, Shear =VC

DTD DTD

, Shear = VADTD

vnet=+300-(75)= +225 plf

vnet =+225 +(250)= +475 plf

Transfer Diaphragm Shears

ab

VA=

VC=

LTD

T(b)LTD

T(a)LTD

LTD

Method of Analysis-Method by Edward F. Diekmann

+500plf

Main chord

Main chord

Disruptedchord

No outside force is changing the basic diaphragmshear in this area

No outside force is changing the basic diaphragm shear in this area

T

C

Col

lect

or

+ +

+ +

+225 plf +150 plf

+550 plf +475 plf

Resulting net

shear diagram

acting on collector

325 plf 325 plf

Net direction

of shears acting

on collector

Shear Distribution Into The Collector

Direction of shear

transferred into

collector

Collector

• Collector force=area of shear diagram

Shear left=+550-225= +325 plf

2

• Place the net diaphragm shear

on each side of the collector

• Sum shears on collector (based upon

direction of shears transferred onto

collector).

Fcollector=(325+325)(Lcollector)Dir. of force

on collector

B

2 3

Net shears

Note: The net shears

will not always be

equal.

Lcollector

• Place the transfer shears on each side

of the collector

Shear right=+475-150=+325 plf

Collector Force

Diagram

Diaphragms

• Horizontal Offsets• End offsets• Large interior Openings

Diaphragm Design

SW1SW2

Diaph.C.L.

Diaph. chord CollectorC

olle

ctor

Diaph. chord

1 2

A

B

3

C

Col

lect

or

Support

Support

TD1

4

Diaph. chordSupport

Irregularity Requirements for Diaphragms with Horizontal End Offsets-Seismic

Support

A Type 2 Horizontal Irregularity (Re-entrant corner) exists where both projections > 15% of plan dimension in given direction. SDC D-F

• Triggers Section 12.3.3.4• Can also trigger a Type 3 Horizontal Irregularity- abrupt discontinuity or

variation in stiffness in diaphragm SDC D-F

Stru

tStru

tSt

rut

DiscontinuousElements and forces

Varying depth and stiffness

SW1

SW2

Diaph.C.L.

Diaph. chord CollectorC

olle

ctor

Diaph. chord

1 2

A

B

3

C

Col

lect

or

Support

Support

TD1

4

Diaph. chord

Connection design with 25% incr.

Support

Irregularity Requirements for Diaphragms with Horizontal End Offsets-Seismic

Support

ASCE 7-10 Section 12.3.3.4 –A 25% increase is required in diaphragm (inertial) design forces (Fpx) for Type 2 or Type 3 horizontal irregularities located in (SDC D-F) for the following elements:

• Connections of diaphragm to vertical elements and collectors (diaphragm supporting elements-TD)• Collectors and their connections to vertical elements

Design diaphragm connections to SW and struts using 25% increase per ASCE 7-10 Section12.3.3.4. (Grid lines 1 and 4) S

trut

Design transfer diaphragm connections to boundary elements (chords) at Transfer Diaphragm using 25% increase per ASCE 7-10 Section 12.3.3.4. (Grid line C and A)

Str

utS

trut

Design transfer diaphragm connection to collector using 25% increase per ASCE 7-10 Section 12.3.3.4.

Exception: Forces using the seismic load effects including the over-strength factor of Section 12.4.3 need not be increased.

Design collector andconnections to SW using 25% increase per ASCE 7-10 Section 12.3.3.4. (Grid lines 1 and 4)

See 12.10.2 & 12.10.2.1 for collectors

• Diaphragm shears are not required to be increased 25%.

SW 1SW 2

Diaph.C.L.

25’ 20’

15’

w=200 plf

Diaph. chord Collector

Col

lect

or

TD c

hord

s

Diaph. chord

80’

35’50’

Discontinuousdiaphragm chord

1 2

A

B

3

C

Col

lect

or

TD c

hord

s

Support

Support

TD1

+ -

Sign Convention

4

Diaph. chord

F2B

12500 lb

200 plf

25’

F2A

Free body for F2B

1 2

A

B

∑M=035’

M2B ft.-lbF2B=7142.9 lb

Support RL =12500 lb

RR=12500 lb

Example 1-Diaphragm with Horizontal End Offset-Transverse Loading

7500

lb

3500

lb

A/R=2.5:1

Support

Calcs

SW 1

SW 2

Diaph.C.L.

25’ 20’

15’

80’

+214.3 +42.9 -37.1

+400 +70

+70

+320

v=150-(107.1)=+42.9 plf

v=70-(107.1)=-37.1 plf

7142.9 lb

v= 50(20)

7142.9(35)= +250 plf

7142.9 lb

2142.9 lb

5000 lb

v= 7142.9(15)

50(20)= -107.1 plf

v=150+(250)= +400 plfv=70+(250)= +320 plf

TD

sh

ea

r d

iag

ram

(Net resulting shear)Can be > 3x basic shear

150 plf 214.3 plf

357.1 plf

70 plf

+250

-107.1

1 2

A

B

3

C

Basic shear diagram

375 plf Basic diaphragm shear

=xxx plf Net shears (basic shear +/- TD shears)(240 plf) Transfer diaphragm shears

No net change Net change occurs in TD

No net changeLegend

35’

+ -

Sign Convention

+357.1 +214.3 +42.9 -37.1 +70

4

-250

+70

+357.1

-250

15’

35’

12500 lb

12500 lb(Net resulting shear)

(Net resulting shear) (Net resulting shear)

Po

s.

Neg

.

-250 plf

Transfer Diaphragm and Net Diaphragm Shear Calcs

SW 1SW 2

Diaph.C.L.

25’ 20’

15’

+42.9 -37.1

+400

F=7142.5 lb

F =7200 lb

F=7200 lb

F=7812.5 lb

C

T

+70

70 plf+320

7142.9 lb

T

7142.9 lb

1 2

A

B

3

C

F=7142.9 lb

+42.9

-37.110.72’

9.27’+230 lb-172.1 lb

Longitudinal Chord Force Diagrams

0 plf

+ -

Sign Convention

+357.1 +214.3 +42.9 -37.1

4

-250

-250

17.5’

+357.1 +214.3

Support

Support

F=7812.5 lb

357.1 plf net

0 plf

Calcs

SW 1SW 2

Diaph.C.L.

25’ 20’

15’

80’

+357.15

+214.3 +42.9 -37.1

+400

+70

+320

F=6000 lb

F=6000 lb(this is not an insignificant force.)

F=3748.5 lb

F=3750 lb

+70

T

Special nailing

(sum of shears to collector or highest boundary nailing-greater of)C

1 2

A

B

3

C

Transverse Collector Force Diagrams

+ -

Sign Convention

+357.15 +214.3 +42.9 -37.1 +70

4

171.4 plf net 107.1 plf net

250 plf net

Calcs

Diaph.C.L.

w1

A B

1

2

3

Deflection of rectangular diaphragm Deflection of

offset diaph.

C

w2Diaphragm Deflection EquationsEquation variables for offset diaphragms• Varying uniform loads • Concentrated loads from discontinuous shear walls• Varying moments of inertia, and sometimes• Different support conditions

ATC7• Modify the bending and shear portion of the

standard rectangular deflection equation to fit the model:

∆"#=%&#'

()*++&#-./

+ 0. 2((#34 +5 6789+

Bending deflection Shear

deflection

Nail slip Adjusted for non-uniform nailing (ATC-7/APA)

Chordslip

NOTE: Multiply deflection x 2.5 for unblocked diaphragmMultiply nail slip by 1.2 if not Structural I plywood

P

Shear Deflection -USDA Research Note FPL-0210• Simplified energy method (Virtual Work).• The integrations of the equations can be reduced to

multiplying the total area of the shear diagramdue to the general loading by the ordinate of the shear diagram due to a dummy load applied at the desired point of shear deflection.

I1 I2

SDPWS combines

Standard deflection equation for simple span, rectangular, rigid supports, fully blocked, uniformly loaded, constant cross section (∆ at C.L.)

6"# = 6: + 6; + 0. 2((<34 +=(67?)

9+

ABCDC

6: = ∫F+ GH

IJ2KL + ∫+

M GH

IJ9KL, OPK

6; =+/

9Q*9∫F+ RS

+T

+

9 KL + ∫+MRSUS = ∫0

# &V

./KL

WXY IZ. 9' − 2APA equation

Cannot use

bb’

L All terms relate to simple span uniformly

loaded

Diaphragm Nailing Callouts

10d

@ 4

/6/1

2 B

1

2 3 Transfer area Boundary (High shear area)

Transfer diaphragm Boundary (Typ.)

Boundary locations

Diaphragm boundary

357.

2 pl

f

320

plf

285

plf

214.

3 pl

f 42

.9 p

lf

37.1

plf

70 p

lf

214.3 plf 357.1 plf

70 plf Basic shear diagram150 plf

10d

@ 6

/6/1

2 B

10d

@ 6

/12

UB

Cas

e I

10d @ 4/6/12 B

10d

@ 6

/12

UB

Cas

e I

10d

@ 6

/12

UB

Cas

e I

Check the shear capacity of the nailing along the collector

Callout all nailing on drawings:• Standard diaphragm nailing• Boundary nailing• Collector nailing

x4x3x1 x2 Special nailing along collectors Sum of shears to collector or highest boundary nailing-greater of

SW 1

Diaph.C.L.

25’ 20’

15’

200

plf

Cho

rd

Collector

Col

lect

or a

nd

TD c

hord

s

80’

Transfer diaphragm

TD1

35’

50’

DiscontinuousDrag strut

1 2

A

B

3

C

4

10’5’

15’

200

plf

Diaphragm 2

Diaphragm 1

SW 2Drag strut

40 plf 160 plf

Cho

rd

5’

+Pos. direction -

200 plfDrag strut

Col

lect

or a

nd

TD c

hord

s

Example 2-Diaphragm with Horizontal End Offset -Longitudinal Loading

DiscontinuousDrag strut

15’

35’

VA SW 2A

B

C

4

SW 1

+Pos. direction -

Dia

ph.

C.L

.

vA

2

BTransfer area

Transfer Diaphragm and Net Diaphragm Shear

(Net shear)

+

-10’

15’

12 3

Vnet=vsw-vdiaph

Neg

.Po

s.

VB

Dia

phra

gm 2

Dia

phra

gm 1

FB

Vnet=vsw-vdiaph

!" = $%&

!" = $%&

vB

vC

(Net shear)

25’ 20’

15’

80’

35’50’

CHECK: FstrutFchord

10’ SW 2

1 2

A

B

3

C

4

15’ SW 1

Vnet @ SW

+Pos. direction -

Longitudinal and Transverse Collector/Strut Force Diagrams

F2A5’ 5’

F3B

F3B

F2BLong

F2BTrans

Fend

Fstart

F2BTrans

F2BLong

Vnet @ SW

Let’s Take a Break

• Horizontal Offsets• End offsets• Large interior Openings

Diaphragm Design

End Openings

w plfDiaph.C.L.

1 2

A

B

3

CD

4 51 2

A

B

3

CD

4

Does not meet A/R

(Envelope)

SW 1

SW 3

SW 2

Basic Shear Diagram

Diaph.C.L.

1 2

A

B

3

C

4

D

TD1A/R=3.73

Skylight(Enclosed area)

w1 plf (WW)

w2 plf (LW)

Varies+

-

Stru

t

w plf

Chord

Chord

Chord

Chord

Collector

Collector

Col

lect

or

Col

lect

or

Section A

Section B

F2CC

D

R1

V3B

Section B

F2B

A

B

1 2

V3A

Section A +

+

w1

w2

Example 3- Intermediate Horizontal Offset at End Wall With Strut

Sum Shears

Uniform shear in walls and in diaph. at grid line 1

10’

12’

18’

16’

16’

15’

56’

150’

SW 1

SW 3

SW 2

Resulting Strut, Collector and Chord Force Diagrams if Strut

1 2

A

B

3

C

D

Support Support

+

-

-

F2C

F2B

VA

VD

vdia

phN

et S

hear

Stru

t N

et S

hear

Net shear diagram

Stru

t

C

T

C

T

C TC

T

C

No

shea

rtr

ansf

erre

d

Pos.

Neg

. N

eg.

Transfer diaphragm shears

1

SW 1

SW 3

Example 4 -Intermediate notch at End Wall Without Strut

SW 2

Basic Shear Diagram

Diaph.C.L.

1 2

A

B

3

C

4

D

Open area

10’

12’

18’

16’

16’

119’

150’

56’

W=123 plf

W=77 plf

Shear Varies

+210

.7 p

lf

+157

.1 p

lf

15’

+

-

W=200 plf

Chord

Chord

Chord

Chord Collector

Collector

Col

lect

or

Col

lect

or

Section A

Section B

16’

F2C=5584 lb

16’C

D

1 2

R1D=6200 lb

V=4968.4 lbv=310.5 plf

Section B

R1B=8800 lb

16’F2B=5684.4 lb

22’A

B

1 2

V=6831.6 lbv=310.5 plf

Section A

+

+

W=123

W=77

Sum Shears

Shear in diaphragm at grid line 2 Is based on depth

R=15000 lb

R=15000 lb V=11

800

lb

8800 lb+6200 lb15000 lb

6428 lbif strut(-38%)

8571 lbif strut(+38%)

V=88

00 lb

+310

.5 p

lfR1B=8800 lb

R1D=6200 lb(V=387.5 plf)

(v=400 plf)

TD1A/R=3.73

Forces in red are from previous example

SW 1

+255.2

-123.7

Neg

. P

os.

-117

Neg

.

SW 2

v=157.1+(255.24)= +412.3 plf

+ -

Sign Convention

Transfer diaphragm and net diaphragm shears

+310.5 plf

varies

+210.7 plf+157.1 plf

v=210.7+(255.24)= +465.9 plf

v=210.7-(117)= +93.7 plf

v=210.7-(123.7)= +87 plf

v=157.1-(123.7)= +33.4 plf

v=157.19-(117)= +40 plf

1 2

A

B

3

C

D

1855.8

1755.4

+

Basic shear diagram

2 3

+157.1

+157.1

+310.5

+310.5

F2C=5584

F2B=5684.4

R1B

R1D

87

33.4

412.3

465.9

40

93.7

Transfer diaphragm shears

Transfer diaphragm net shears

Transfer Forces to Collectors

SW 1

SW 3

SW 2

Longitudinal Chord Force Diagrams

F=5584 lb

F=5584 lb

33.4

412.3

465.9 412.3

40

F=5684 lb

F=5684 lb

87

465.9

93.7

1 2

A

B

3

C

D

Support

R1B

R1D

18’

10’

12’

16’

16’

15’

400

310.5387.5

310.5

SW 3

SW 2

Traverse Strut/Collector Force Diagrams

F=3469.2

F=4917

Vsw=6200 lbvsw=387.5 plfvnet=0 plf

-400 p

lf+333.3

3 p

lf

F=4000

F=3469.2

F=2271.4

F=1872.2

F=1872.2 lb

33.4

412.3

93.7

40

87

465.9

93.7

40

465.9

412.3

33.487

1 2

A

B

3

C

D

SW 1

Vsw=8800 lbvsw=733.33 plfvnet=733.3-400=333.33 plf

R1B=8800

R1D=6200 lb

157.1

157.1

310.5387.5

310.5400

157.1

157.1

Net shear diagram

• Horizontal Offsets• End offsets• Large interior Openings

Diaphragm Design

ASCE 7-10 Section 12.3.3.4 (SDC D-F) -Horizontal irregularity Type 3 requires a 25% increase in the diaphragm design forces determined from 12.10.1.1 (Fpx) for the following elements:• Connections of diaphragm to vertical

elements and collectors (diaphragm supporting elements).

• Collectors and their connections to vertical elements.

• Use of over-strength forces is not commonly considered to be triggered for boundary elements at diaphragm openings. However, the 25% increase does apply.

Diaphragms With Large Interior Openings

w plf

Diaph.C.L.

1 2

A

B

3

C

D

4 5

Type 3 Horizontal Irregularity-SDC 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.

Diaphragm shears are not required to be increased 25%.

Exception: Forces using the seismic load effects including the over-strength factor of Section 12.4.3 need not be increased.

collector

collector

collector

collector

Roof pop-up section with opening below.

Common Openings In Diaphragms

Skylight or atrium opening

Clerestory windows

Stairwell access to roof

End opening

Optional strut

Harrington Recovery CenterStructural engineer: Pujara Wirth Torke, Inc.Photographer: Curtis Walz

Openings in diaphragm

Affect of Size and location in Diaphragm

Basic Shear Diagram

Location and Magnitude of Shear

Size of opening

Local shears lower

Local shears higher

StairwellsElevators

IBC 2305.1.1 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.

Most openings of any significant size should be checked.

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 largeropening dimension;

d. The diaphragm portion between opening and diaphragm edge satisfies the maximum aspect ratio requirement. (all sides of the opening)

FPInnovationsDesign example: Designing for openings in wood diaphragm

C.L. opening

I.P.

I.P.

1 2

A

B

3

C

D

54

Displacement and Local Forces

Vierendeel trussaction

V

V V

V

T

T

C

C

W

T

T

C

C

Local forces

.

.

R R

Some examples apply load to one side of the diaphragm only

Chord forces are assumedto be zero at these locationsdue to contraflexure (inflection points). M=0

Sub-Chord

Sub-Chord

M

V M

M=0

Shear Distribution in Diaphragm

Shear distribution follows analysis

A/R

A/RA/R A/R

A/RA/R

ATC 7, Diekmann, FPInnovationsIf 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).

All sections must meet Code required aspect ratios.

TransferDiaphragm

(TD)

TD

Aspect Ratio Issues

Opening

A/RA/ROpening

Easy to visualize if header section is replaced by a wire.

Transfer diaphragms are required if the opening size affects the shear or tension capacity of the diaphragm.

Opening

TD

R

w plf

F2

F2 F3

F4

F4

02 =SM 03 =SM 04 =SM

V4

Diaph.C.L.

TD2 TD1 Opening

F3

1 2

A

B

3

C

D

4

Element III

Element II

Element IV

Element I

Inflection point.

V2V3

V1

Basic shear diagram without openings

F=0

5

V5

Basic Shear Diaphragm With Opening

V4R

V2RV3V1

V5

Opening Analysis-Diekmann method

Typical method of analysis (APA Report 138), ATC-7, and FPInnovations

1. Calculate the chord forces at grid lines 2, 3, and 4 using FBD’s.

2. Determine the basic diaphragm shears without an opening.

3. Determine the diaphragm shears with an opening.

4. Break the sections above and below the opening into elements as shown.

5. Determine the local forces at each corner of each segment by FBD’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 4. 2. Break the sections above and below the

opening into elements as shown.3. Calculate the chord force at grid line 3.4. Starting at grid line 4 and moving to the left,

sum forces at each corner of each segmentto determine the local forces, by FBD’s.

5. Calculate all chord, collector forces, andtransfer diaphragm shears and forcesusing the visual shear transfer method.

V4LV2L

F=0

Example 5-Pop-up Roof Section

W=200 plf

Diaph.C.L.

1 2

A

B

3

C

D

4 6

40’20’

60’

200’

28’

20’

12’

20’

5

W=50 plf

W=123 plfW=200 plf

W=77 plf

W=30 plf

Basic Shear Diaphragm With Opening (plf)

v4R

v2Rv3v1

v5

v4Lv2L

v6

TD1 TD2

A/R TD1=TD2=3.0:1 o.k.

A/R main diaphragm and upper section=3.33:1

Wind Loads (ASD)MainW=200 plf

At openingWw=123 plfLw=77 plf

At pop-up (20 psf)Ww=50 plfLw=30 plf

V=2

720

V=6

720

V=8

320

V=1

3920

V=1

9520

V=2

1120

RR=21280RL=25120

V=2

5120

1600

lb

1600

lb

1 2

A

B

3

C

D

W=123 plfW=200 plf

W=77 plf

W=30 plfTD1

RL=25120

1600

lb

W=50 plf13280

13280

!" = $ 40’20’

8’

Open to below

W=50 plf

W=30 plf

W=80 plf

SW1600 lb

1600 lb

Open tobelow

F3A

F3D

Sub-Chord

Sub-Chord

FPInnovations:Hgt.> 0.15 dDiaph.Width>0.15 Ldiaph.End dist.< 3x widthDetailed analysis required

SW 1

Diaph.C.L.

0 0

0 0

+

+

+

+10104 l

b

7964 l

b

7964 l

b

5824 l

b

5824 l

b2496 l

b

(360.9

plf

)

(284.4

plf

)

(284.4

plf

)

(208 p

lf)

2496 l

b

5956 l

b

5956 l

b

9416 l

b

(784.6

plf

)

(496.3

plf

)

(208 p

lf)

(496.3

plf

)

F2C

F4A

F4C

F4BF2B

F2A

F4DF2D

13280 lb

13280 lb

12810 lb 7076 lb

18204 lb6827 lb

6453 lb 4924 lb

200 plf

1 2

A

B

3

C

D

54

Start here

470 lb 20340 lb

200 plf123 plf123 plf

77 plf 77 plf

28’

12’

20’

20’

20’ 20’

20’

2112016009416101042 =++=SV Lb

13920595679643 =+=SV Lb83201600)60(112

249658244

=+=+=SV

Lb

TD1 TD2

Free-body of Chord Forces and Segment Forces

!" = $ !" = $

!" = $!" = $

The sum of the section shears must match thebasic diaphragm shear Values without an opening, ∑V=0.

Sub-Chord

Sub-Chord

V1 V2 V3 V4L=8320 lb V4R=6720 lb

50 plf50 plf

30 plf30 plf

RL=25120 lb

1600 lb1600 lb9416 l

b10104 l

b

(112)

plf

)(1

12)

plf

)

(352 p

lf)

(352 p

lf)

!% = $

F3A

F3D

Element III

Element II

Element IV

Element I

Net Shears-Left Transfer Diaphragm

20’

+ -

A

B

1

C

D

3

T.D.1

Sign convention

Neg.

Pos.

Neg.

352418.7

Basic Shear Diagram

w

Transfer diaphragm shears

2

6453 lb

879.6 lb

12810 lb

20’

7236.6 lb

418.7-361=+57.7

418.7+278.7=+639.7

418.7-44=+374.7

+374.7

360.

978

4.6

360.

9

-361

+278.7

-44

352-361=-9

-

352+278.7= +630.7

352-44=+308

1600 lb

6453 lb

12810 lb

20’

-+

A

B

4

C

D

TD2

Sign convention

Transfer diaphragm shears

vnet=112-168.2=-56.2 plf

vnet=112+185.7=+297.7 plf

vnet=112-60.55=+51.45plf

5

+45.33+112

Basic Shear Diagram

vnet=45.33-168.2=-122.8 plf

vnet=45.33+185.7=+231.02 plf

vnet=45.33-60.55 =-15.22 plf

45.33

45.33

45.33

w

45.33

45.33

4924 lb

1211 lb

7076 lb

3363 lb

-168.2

+185.7

-60.55

1600 lb

4924 lb

7076 lb

Net Shears-Right Transfer Diaphragm

Neg.

Neg.

Pos.

Collector Force Diagrams-Left Side

-+

A

B

1

C

D

3

TD1

Sign convention

w

2

F=12800 lb F=9524 lb

1600

lb

F=1476 lb

F=6453 lb

57.7

697.3

9

630.7

697.3

347.7

630.7

308

360.9308

784.6

630.7

308

9

630.7

360.9

T

C

T

C

Collector Force Diagrams-Right Side

2.67’

-+

A

B

3

C

D

4

TD2

Sign convention

2.83’

w

3.6

F=3170 lb F=2018 lb

1600

lb

F=1696 lb

F=4384 lb

56.2

297.7

122.8

231.03

297.7

51.45

231.03

15.22

15.22

45.33

231.03

15.22

122.8

231.03

297.7

5824

51.45

56.15

297.7

2496

51.45

F=7076 lb

F=4924 lb

45.33

45.33

45.33

45.33

TC

T C

T

C

A57.67

D20’ 20’40’

T

Chord Force Diagrams

1 2 3 54

09 784.6 208 56.2 122.8 45.33

51.45 15.22 045.33

496.3 496.3

374.9 308 360.9 208284.4 284.4

C

F=486.7 lbF=13295 lb

F=20340 lbF=18858 lb

F=18876 lb

F=18550 lb

Closes to zero

F=6829 lb

F=18206 lb

Aver.=640.5

Aver.=352.2

F=13282 lb F=18568 lb

+397 lb-34.75 lb

Aver.=322.65

Aver.=352.2

Aver.=89.5

13.6’ 106.4’

+498.9 lb-12.15 lb

18868

2034013280

470

Aver.=341.5

1886818204

13280

6827

Closes to zero

SW 1 SW 2

Final Strut/Chord Force Diagrams

1 2

A

B

3

C

D

54

TD1 TD2

T T

C

C

T

C T

T

C

C

Shear Walls

Shear Wall Design

• Shear Wall Types

• Shear wall Anchorage

• SDPWS Code Requirements

• Complete Load Paths

• ASCE 7 Offset Shear Wall Requirements

• Offset Shear Walls

Shear Wall Configurations

Perforated Walls Force Transfer Around Openings Walls

Segmented or braced Walls

(IBC/IRC)(No openings)

WallDL

V1

v(plf)

Sh

ea

r w

all

ch

ord

Sh

ea

r w

all

ch

ord

h

Resist. mom. Arm 1C.L. bearingC

C.L. A.B.

T

b

Sheathing joint as occurs

2x blockingas requiredat joints.

DLhdr2DLhdr1

Anchor bolts or nailing Hold down anchor

V2

v(plf)

WDL plf

tstuds

Sh

ea

r w

all

ch

ord

Sh

ea

r w

all

ch

ord

h1

Resist. mom. Arm 1

C

C.L. A.B.

T b

Tstud + CLC.L. bearing

Chord orStrut

Chordor Strut

Fr1

Fr2

h2WallDL

DL hdr. DL hdr.

412

C.L.SW

v diaphragm shear

0=åM

Fv

Fchord/strut

Fchord/strut

Fv

Vvert

Vhoriz

Segmented Shear Walls

Segmented Wall Types- Standard & Sloped

L bearing

Hdr / strut Hdr / strut

Resist. mom. Arm 2

Resist. mom. Arm 2

v(plf)

0=åM

Dead Load Distribution

b/2

b

This section Supported by the first floor bearing width

This section in bearing

Preference & Engineering Judgement

Rigid Body

Semi-Rigid Body

Rigid Body

45 60

Short Walls

Long Walls

b/2

b/2

Wood structural panels –

Unblocked

Wood structural panels –

Blocked

Particleboard – Blocked

Diagonal sheathing,

conventional

Gypsum board

Portland Cement Plaster

Structural Fiberboard

Type Maximum height-width ratios

AWC SDPWS Table 4.3.4-Maximum shear wall dimension ratios-Wind and seismic

Footnotes

1. Walls having aspect ratios exceeding 1.5:1 shall be blocked shear walls.

Allowable Aspect Ratios & Adjustment Factors

4.3.4.2

For wood structural panel shear walls with aspect ratios (h/b) greater than 2:1, the nominal shear

capacity shall be multiplied by the Aspect Ratio Factor (WSP) =1.25-0.125h/b.

For structural fiberboard shear walls with aspect ratios (h/b) greater than 1:1, the nominal shear

capacity shall be multiplied by the Aspect Ratio Factor (fiberboard) =1.09-0.09 h/b.

C4.3.4.2

2: 1

3.5:1

2:1

2:1

2:1(1)

2:1 (1)

3.5:1

WallDL

V or F

Shea

r wal

l cho

rd

Shea

r wal

l cho

rd

h

CT b

WDL plf

Segmented Shear Wall Deflection and Stiffness∆"#

Deflection of unblocked segmented shear wall• Use Eq.4.3-1 with v/Cub per 4.3.2.2 and Table 4.3.3.2

• Max. height unblocked=16 feet

v v v

∆$% =' (

)*+,-

./0 +()*+

,233345

+ ,+ ∆5

ATS

Wall stiffness k = 6∆

ATS

CX

∆"#= '(,-./+ +

(,233345

+ ,∆5+

Vertical elongation• Device elongation• Rod elongation

Bending

Apparent shear stiffness• Nail slip• Panel shear deformation

SDPWS 3 term deflection equation

4.3-1

∆"#= '(,-./+ +

(,4(7(

+3. 9:,;<+ ,∆5+

Bending

Shear

Traditional 4 term deflection equation

C4.3.2-1

Nail slip

SDPWS combines

Rod elongation(Wall rotation)

Segmented Shear Wall Deflection

SDPWS linear3 term equation

Identical at 1.4 ASD

ASD unit shear

Traditional4 term equation

Displacement, inches

Load

, plf

Perforated Shear Walls- Empirical Design

Hdr

Hold downs at Ends per section 4.3.6.4.2

Intermediate uplift anchorage is required at each full height panel locations… ”in addition to…” per section 4.3.6.4.2.1.

Collector per SDPWS section 4.3.5.3(Full length of wall)

Top and bottom of wall cannot be Stepped or sloped

Opening

Hdr

Opening

Use other methods

Typical boundary member

All full hgt. sections must meet the aspect ratio requirements of section 4.3.4.1.

Common sheathing joint locations

vmaxTint

Li (typ)

V

Wall segment

Wall segment

Wall segment

Header sections do not have to comply with aspect ratios.

Openings are allowed at end of wall but cannot be part of wall

Inter. uplift anchorage

Inter. uplift anchorage

Inter. uplift anchorage

• APA Diaph-and-SW Construction Guide• AWC-Perforated Shear Wall Design

Reference examples

F

(a) Perforated Shear Wall

AWC SDPWS Figure 4C

Bottom

of roof

diaph.

framing

Floor

diaph.

Floor

diaph.

Bottom

of diaph.

framing

Bottom

of diaph.

framing

Segment

Width, b

Sto

ry a

nd

wa

ll

se

gm

en

t h

eig

ht,

h

Wd.

Wd.

Dr.

Wall

segment

Segment

Width, b

Wall

segment

Segment

Width, b

Wall

segment

Segment

Width, b

Wall

segment

Sto

ry a

nd

wa

ll

se

gm

en

t h

eig

ht,

h

Sheathing is

provided above

and below all

openings

Figure 4D for

segmented walls

is the same as this

figure, except this

section is added.

Foundation wall

Allowable Perforated Shear Wall Aspect Ratios

• Sections exceeding 3.5:1 aspect

ratio shall not be considered a

part of the wall.

• The aspect ratio limitations of

Table 4.3.4 shall apply.

• Vn ≤ 1740 plf seismic-WSP 1 side

• Vn ≤ 2435 plf wind-WSP 1 side

• Vn ≤ 2435 plf wind-WSP 2 sides

• A 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 perforated walls.

• Collectors for shear transfer shall

be provided through the full

length of the wall.

• Uniform top-of-wall and bottom-of

wall plate lines. Other conditions

require other methods.

• Maximum wall height ≤ 20’.

Limitations:

Maximum Shear @ full hgt. sect.

Sheathingarea ratio

Empirically determined shear resistance reduction factors, C0 – based on maximum opening height and percentage of full height wall segments.

Sugiyama Method-Reduced shear capacity is multiplied by the full length of the wall.

APA, IBC, SDPWS Method- Reduced shear capacity is multiplied by the sum of the lengths of the full height sections (slightly more conservative).

Hdr

Opening

Hdr

Max

. ope

ning

hei

ght

vmax

Tint

V

Wall segment

Wall segment

Wall segment

h m

ax=2

0 ft

Li (typ) Li (typ)Li (typ)

Vallow = (v Tabular ) x (CO) x ∑LiWhere:

Per SDPWS Table 4.3.3.5!" =

$% − '$

()*)+(,

$ = -- + /"

0+(,A0 = total area of openings

Hold down at ends

1234 = 5

!* ∑(,

1234 = 7,8) = 5!"+(,

7 = ! = 50!"+(9

At full-hgt. segments

CCC

C

C

AWC Section 4.3.6.1.3Each end of each segment shall be designed for the compression force C.

Table 4.3.3.5 Shear Capacity Adjustment Factor, CO

In the design of perforated shear walls, the length of each perforated shear wall segment with an aspect ratio greater than 2:1 shall be multiplied by 2b/h for the purposes of determining Li and ∑Li. The provisions of Section 4.3.4.2 and the exception to Section 4.3.3.4.1 (1.25-0.125h/b) shall not apply to perforated shear wall segments. Where perforated shear walls have WSP on 1 side and GWB on the opposite side, the combined shear capacity shall be in accordance with the provisions of Section 4.3.3.3.2.

4.3.4.3

Perforated Shear Wall Deflection

Vertical elongation• Strap nail slip• Device elongation• Rod elongation

Bending

Apparent shear stiffness• Nail slip• Panel shear deformation

Deflection of perforated shear wall

4.3-1

4.3.2.1 Deflection of Perforated Shear Walls: The deflection of a perforated shear wall shall be calculated in accordance with 4.3.2, where v in equation 4.3-1 is equal to vmax obtained in equation 4.3-9 and b is taken as ∑Li.

∆"#= %&'()*+ +

&'-.../0

+ '∆0+

vmax ∑Li

Open’g

Hdr

Max

. ope

ning

hei

ght

vmax

V

Wall segment

Wall segment

h m

ax=2

0 ft

Li (typ)Li (typ)

∆"#

3’ 6’ 5.5’14.5’

4’

3’

2’

1 2

A

B

3

C

D

4

9’

4500 lbw=200 plf

(See recent Testing-APA Form M410 and T555)

Tie straps as required to develop the corner forces into the wall Anchor bolts

or nails

(Typical boundary Member)

Typical boundary member

2’ min. per SDPWS Section 4.3.5.2(2008 requirement)

Many examples ignore gravity loads

FTAO Shear Walls

Cont. Rim joist

Strut/collector

Shear panelsor blocking

(Diekmann)-Vierendeel Truss/Frame

https://www2.strongtie.com/webapps/sitebuiltshearwalldesigner/https://www.apawood.org/ftao

TDAlt.ld.path

A/R

A/R

A/R

A/R

(b) Force Transfer Around Opening

Wall

pier

Wall

pier

Wall

pier

Wall

pier Wall

pier

Wa

ll P

ier

he

igh

t

Wa

ll p

ier

he

igh

t

Wall

pier

width

Cle

ar

he

igh

t

Wa

ll p

ier

he

igh

t

Overall width

AWC SDPWS Figure 4E

Dr.

Wall

pier

width

Boundary

members

Collector (typ.)

Foundation wall

Allowable Shear Wall Aspect Ratios For FTAO Shear Walls

Note: Not

shown as

having to

comply w/

A/R

Wd.

Wd.

• The aspect ratio limitations of

Table 4.3.4 shall apply to the

overall wall and the pier sections

on each side of the openings

• Minimum pier width=2’-0”.

• A 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 FTAO walls.

• Collectors for shear transfer shall

be provided through the full

length of the wall.

Limitations:

ATC 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).

2.67’ 5.25’ 2.33’

10.25’

7’

0.5’

2’

1 2

A

B

3

CD

4

9.5’

V A/R=2.63<3.5 O.K.

A/R=2.62<3.5 O.K.

A/R=10.5>3.5 N.G.

A/R=3<3.5 O.K.

Fo/t

Force Transfer Methodology –Based on stiffness (FEA analysis)

This slide shows force distribution and location of I.P’s due to sill section not meeting A/R requirements

Not to scale

A/C unit

WD

Strut/collector

Fo/t

V

C

1 2

A

3

D

4

T

F=0 lb

F=0 lbF=0 lb

Force Transfer Methodology –Based on stiffness

F=0 lb

5 6

6’6.5’6’3.5’ 3.5’

25.5’

1’

3.5’

3.5’

2.5’

F=0 lb

F=0 lbF=0 lb

F=0 lb

This slide shows force distribution and location of I.P’s if header section not meeting A/R requirements

Example results – double opening with shallow headers

Not to scale

Tie strap/blocking full width

N.A.

4500

C

Tie strap/blocking full width

Blocking

Point of inflection is assumed to occur at mid-length (Typ.)

MM

V

V

V

V

M

M

1 2

A

B

3

C

D

4

CA B D

F

E

H GI

L

J

K

T

V

V

M

M

F=0 lb

F=0 lbF=0 lb

M

F=0

MF=0 Force Transfer Methodology (Diekmann)-Vierendeel Truss/Frame

F=0 lb

FF

Gravity loads to wall

1343.1 4243.1

w=200 plf

N.A.

Example Cont.

MM

V

V

4500 lb

4’

14.5’

208.62 lb

208.62 lb

991.38 lb

587.08 lb

F1B.5

VB.5

1 2

A

B

3

A

B

4

2’

2’

4500 lb

200 plf

200 plf0=åM

0=åMB.5

B.5

2691.38 lbF4B.5

3912.92 lbVB.5

BA C

A D

E

B

D

E

V2.53’

3’

Free-body of Upper Half and Upper Left Section

Point of inflection B.5

0 lb 0 lb

0 lb

L

A

L

K

B

I

587.08

VB=587.08

VB.5=587.08 (587.08)

Vc=587.08

587.08

V2.

5

1551

.7

V2

(1343.1)

3’

3’

F2b(V)391.39

F2C(v)391.39

991.

38

V2

391.

39

(208.62)

F2A

F2B(H)

1381

1037.1

1160.3

F2C(H)

F1B600

F1C182.77

J

(0)

(0)

573.23

2’

2’

2’

1 2

0=åM

0=åM0=åM

0=åM

0=åM

0=åM

Units are in lb

Corner tie strap force

Corner tie strap force

DC

F

E

GH

1591

.38

(991

.38)

V3

268.8

1937.1

1551

.7

1551

.7

236.17

3676.6

V3

(4243.1)

(2691.38)VB.5=3912.92

VB=3912.92

3912.92

(3912.92)

VC=3912.92

3912.92

(0)

5.5’

3’

2’

2’

2’

F3(V)1422.88

F3B(V)1422.9

F4B1268.5

F4C4114.3

F3C(H)

F3B(H)

F3A A

B

3

C

D

4

0=åM

0=åM

0=åM

3’ 3’

3’ 3’

2.5

B.5 B.5

2124.9

1551.7

(4500)

(4500)

F2.5A1274.9

(xxx) Shears and forces determined in previous step.

0=åM

0=åM 0=åM

200 plf 200 plf 200 plf 200 plf

F2.5C

Resultant Forces on Wall Segments

0=åV

0=åH

0=åV

0=åV

0=åV0=åV

0=åH

0=åV

0=åH

1551.7

1706.9931931931

(0)

Unit shear

196

150 496 796

196

386 517 517 43

359

711

711

Transfer diaphragmsections

3’ 6’ 5.5’

14.5’

4’

3’

2’

1 2

A

B

3

C

D

4

T.D.2T.D.1

2.67’ 2.67’

9’

3.6

Example 2 - Blocking and Strapping Partial Width (with uniform load)

4500 lbs.

Support

Support

Support

Support

1.2w=200 plf

Lap as required

A/R

A/R

A/R

A/R

Partial length straps/anchorsConnected to 1st bay blocking only

APA Wall tests 8 and 9 – M410

Blocking at corners

Anchor location

Partial length straps/anchorsConnected to 1st bay blocking only

APA Wall tests 8 and 9 – M410

Photo credit APA:

Wall 8 – IMG_1297

Wall 9 – IMG_1667100_0018100_0021

4243.1 lbs.

Pier Section Forces and Loads 2.67’

-+

A

B

3

C

D

4

T.D.2

Sign convention

1937.1

1975.8

2.83’

Support

Support

F

F

3676.6

Neg.

Pos.

Neg. 1551

.715

91.4

v3.6 v4v3

Basic Shear DiagramFrom Gravity Loads

Summing

w=200 plf

Transfer diaphragm shears

3.6

3676.6 lbs.

1551

.7

4243.1 lbs.

1937.1 lbs.

Transfer Diaphragm Shears and Net Shears

2.67’

-+

1937.1

3676.6 lbs.

281.1 lbs.

2020.6 lbs.

v= 281.12.67

=- 105.3 plf

v= 1656

2.67= +620.2 plf

v= 2020.6

2.67= -756.8 plf

-105.3

-756.8

+620.2

+471.5+408.6

A

B

3

C

D

4

T.D.2

Sign convention

Transfer diaphragm shears

1591

.4

vnet=349.23-105.38=+243.93 plf

vnet=349.23+620.2=+969.45 plf

vnet=349.23-756.8=-407.55 plf

1975.8 lbs.

3.6

+303.3+243.9

+1028.8+969.5

+1028.8+969.5

-348.2-407.6

-348.2-407.6 +471.5+408.6

3677.1 lbs.(+408.57 plf)

4243.1 lbs.(+471.46 plf)

3143.1 lbs.(+349.23 plf)

Basic Shear Diagram

Summing

vnet=408.57-105.38=+303.27 plf

vnet=408.6+620.2=+1028.8 plf

vnet=408.57-756.8=-348.21 plf

408.6

408.6

408.6

471.5

471.5

471.5

w=200 plf

4243.1 lbs.Horizontal Collector Forces

2.67’

-+

A

B

3

C

D

4

T.D.2

Sign convention

C

T969.45

969.5

407.55

1028.81937.1

3676.6

243.9 303.3

348.2

1028.8

243.9 303.3 408.6 471.5

! = #$%&. # ()*.

! = %+&+. + ()*.

! = #$&,. - ()*.

1975.8 2.83’

4243.1 lbs.Vertical Collector Forces

2.67’

-+

408.6

471.5

A

B

3

C

D

4

T.D.2

Sign convention

F=1103.5 lbs.

F=2774.4 lbs.

T

348.2

1591.4

407.61551.7

1028.8

969.5

969.5

303.3

F=0 lbs.

Summing

F=2270.3 lbs.

F=210.54 lbs.

F=0 lbs.

F=4243.1 lbs.

480.6

C

C

T

Collector

243.93

480.6

471.5

C

(See recent Testing-APA Form M410)

Advancements in FTAO Shear Wall Analysis

T555Refine rational design methodologies to match test results

§ Used test results from full-scale wall configurations

§ Analytical results from a computer model

§ Allows asymmetric piers and multiple openings.

https://www.apawood.org/ftao

20’-0”L

5’-0”L1

4’-0”L2

2’-6”L3

6’-0”LO1

2’-6”LO2

4’-8

”ho

2

8’-0

”H

4000 lbs.

V

1’-4

”ha

2’-0

”hb

2

1600

1600

2’-8

”ho

14’

-0”

hb1

TD

Multi-story walls?Tall shear walls?

What happens when different depths?

• The deflection of the wall is the average of the deflection of the piers as shown (acting both ways combined) using the 4 term eq.

Single opening

∆"#$%.= (∆()$% *+∆()$% ,)+(∆()$% *+∆()$% ,).

• The remainder of the terms are identical to the traditional equation.

• Deflections for a wall with multiple openings is similar.

∆"#$%.=(∆()$% *+∆()$% ,+∆()$% /)+(∆()$% *+∆()$% ,+∆()$% /)

0 ...

Hdr

Opening Wall Pier 2

Wall Pier 1

Sill

∆12= 4#5/6"7 +

#59#:#

+;. <=5$>+ 5∆?7

Traditional 4 term deflection equation

∆@)$% * ∆@)$% ,

Reference APA T555

Loads Loads

∆@)$% * ∆@)$% ,

Let’s Take a 10 Minute Break

Combined Shear and Uplift Shear Walls-WSP-APA SR-101C

. . . . . . .

. . . . . . .

. . . . . . . .

. . . . . . . .

Typical Panel Fastening

Single row of Fasteners Double row of Fasteners

.

.

.

.

.

.

.

.

.

.

. . . . . . . . .

. .

. .

.

• Min. panel thickness=7/16”

• Sheathing can be applied vertically or horiz.

• Min. fastener spacing=3” single row, 6” Dbl. row

• All horiz. edges shall be blocked

• Applies to all shear wall types

• Capacity: Table 4.4.1-shear + upliftTable 4.4.2-uplift only

VASD=Vn/2, Vstr=0.65Vn

Edge nailing

Field nailing12” o.c.

Spacing

≥ 6”

≥ 3”

3/4” 1/2”

1/2”

1/2”

1/2” 3/4”

Straps are required at edges of an opening to transfer uplift forces

SW1 SW4

W ( plf)1 2

A

B

Diaphragm support

Diaphragmsupport

Stable

SW3

SW2

SW1 SW4

W ( plf)1 2

A

B

Diaphragm support

Diaphragmsupport

SW3

SW2

Unstable

Acts like open front diaphragm(SDPWS 4.2.5.2)

No support

Shear walls are parallel to applied loads

Not all shear walls are parallel to applied loads

Angled/Skewed Shear WallsNo Angled/Skewed Shear Wall Testing

Structure is stable

P-delta problem. Wall deflects too much out-of-plane, which causes rotation in the diaphragm(Acts like open front diaphragm-shifts forces to side walls)

Large deflection

Wall warpsand translates

Concentric steel braced frame or shear walls both sides of the building-resist rotational forces.

Diaphragm rotates

Diaphragm with Angle Shear Wall-Horizontal Type 5 Irregularity Transverse Loading

Wall stiffness(SW1 & longit. walls)

DiaphragmSW2

Structural Model

Top of wall braces

24’SW1

C.L.

1

A

B

SW support

θ

θθθθ =!"#

=$"#=%"#=&'#

SW1

Effe

ctiv

e w

idth

1

A

BDiaphragm support

C.R.

A

BDiaphragm support

Hold down

Hold down

Effective width

SW1

TD1

SW1

Effe

ctiv

e w

idth Cantilever

section TD1

No Circular Shear Wall Testing

WSP thickness depends on radius

Circular Shear Walls

Source: nees.org

Mid-Ply Shear Walls

¡ Test Results:

• Mid-ply walls 2.4 to 2.8 x stiffer than standard shear walls monotonic loading.

• Mid-ply walls 1.8 to 2.2 x stiffer than standard shear walls cyclic loading.

• Better energy dissipaters, 3 to 5x better than standard shear walls

¡ Mid-ply walls:• Carry high shear demand

• Reduce torsional effects

WSP

WSP

WSP

WSP

GWB

GWB

ATS system

Typical Mid-Ply Shear Walls

Fasteners are in double shear

16” o.c. typ.

3 Layer WSP

1 Layer WSPw / GWB

w / GWB

Prescriptive / Proprietary Portal Frames

Prescriptive Code Portal Frames Proprietary Portal

Frames IBC 2308.9.3.2

Source: strongtie.com

Truss Walls

Truss Frames

Hybrid Wood/Steel Proprietary Systems

Shear Wall Design

• Shear Wall Types

• Shear wall Anchorage

• SDPWS Code Requirements

• Complete Load Paths

• ASCE 7 Offset Shear Wall Requirements

• Offset Shear Walls

Hold Down Anchors Systems

Standard Hold Down (Bucket Style)

Strap Hold Down

…………

………

Continuous RodAutomatic Tensioning System

ATS

Typical Hold Down Details at Foundation

1

12

18”

max

.

Bolts, nails or screws per manufacturer

Self positioning hold down. Install per manufacturer’s instruction

Raise 3” for every ¼” offset from H.D. centerline(Manufacturers instructions)for misplaced anchors

C.L.

Install anchor per manufacturer’s instruction

Number of studs as required by design.

1

12!" max.1 ½” max.

Top of brg. plate

Approved coupler Preservative treated barrier may be required

Minimum wood member thickness

3” to 5”

Standard (discrete) Hold Downs

Bucket Style Hold Downs

Opt. supplied manufacturers anchor

Multiple Hold Downs at Corners

Source: DartDesignInc.com

Source: strongtie.com

Hold downs installed at an angle

Field Installation Issues

Large knots

Courtesy of Willdan Engineering

Hold down misplaced.Sheathing is removed.

Field Installation IssuesCourtesy of Willdan Engineering

Strap to be installed plumb and tight to studs. Fill all holes.

#4 rebar. Extend 2 x Le each way beyond strap.Bend !"# at corner.

3” to 5”

Le

½” min. at foundationcorner

Possible foundation corner location

Strap Hold Down Installations

…………

…………

Loose anchor strap-stabbed into foundation-no vibration

Too close to vent opening

Field Installation Issues

Slack in hold down strap (bent)Often stabbed into foundation without vibration.

Field Installation Issues

Slack in strap (bent) due to vertical shrinkage of framing.

Field Installation Issues

Typical Hold Down Details at FloorStandard (discrete) Hold Downs

Loose bolted connection due to vertical shrinkage of framing. Tighten after shrinkage

Nail after shrinkageoccurs.

Continuous Rod Tie Downs with Shrinkage Compensation Devices

Source: strongtie.com

Automatic Tensioning Systems/Devises

ATS isolator nut take-up deviceand bearing plate

Coupler

Steel rod

Modifiedballoon framing

Platform framing

Automatic Tensioning Systems/Devises

• Restraints are required at roof and each floor level to get best results

• Software programs are available for design

• Must be installed per manufacturers recommendations

Number and size of studs as required by design

Anchorage into foundation as required by calculation and per manufacturers recommendations

4 kips

Automatic Tensioning Systems/Devises

27 kips

14 kips(27 kips)Full hgt.

2 kips2 kips

7 kips 7 kips

13.5 kips 13.5 kips

Tied off at roof only• Lack of redundancy• Increased costs• Increased drift• Single device must

accommodate shrinkage at all floors

Tied off at roof and all floors• Accommodates

shrinkage at each floor• Greater redundancy• Reduced drift• Lower costs (Can be

subjective)

Legendxx kips Tied off at each floor(xx kips) Tied off at roof only

Tied off at each floorTied off at roof only

Skipped floor

Shear Wall Design

• Shear Wall Types

• Shear wall Anchorage

• SDPWS Code Requirements

• Complete Load Paths

• ASCE 7 Offset Shear Wall Requirements

• Offset Shear Walls

2015 SDPWS

SDPWS Shear Wall Code Requirements

Allowable values shown in tables are nominal shear values

Design values:LRFD (Strength) = 0.8 x VnASD = Vn/2

SheathingMaterial

MinimumNominal

PanelThickness

(in.)

MinimumFastener

PenetrationIn Framing

Member or Blocking

(in.)

FastenerType & Size

ASeismic

Panel Edge Fastener Spacing (in.)

Table 4.3A Nominal Unit Shear Capacities for Wood-Framed Shear Walls

6 4 3 2 6 4 3 2 Vs Ga Vs Ga Vs Ga Vs Ga Vw Vw Vw Vw

1,3,6,7

5/16 1-1/4 6d 400 13 10 600 18 13 780 23 16 1020 35 22 560 840 1090 1430

BWind

Panel Edge Fastener Spacing (in.)

Wood Based Panels4

(plf) (kips/in.) (plf) (kips/in.)(plf) (kips/in.)(plf) (kips/in.) (plf) (plf) (plf) (plf)

Nail (common orGalvanized box)

Wood Structural Panels-Structural 1

Use Table 4.3B for WSP over ½” or 5/8” GWB or Gyp-sheathing BoardUse Table 4.3C for Gypsum and Portland Cement PlasterUse Table 4.3D for Lumber Shear Walls

OSB PLY OSB PLY OSB PLY OSB PLY4,5

3/8 460 19 14 720 24 17 920 30 20 1220 43 24 645 1010 1290 17107/16 1-3/8 8d 510 16 13 790 21 16 1010 27 19 1340 40 24 715 1105 1415 1875 15/32 560 14 11 860 18 14 1100 24 17 1460 37 23 785 1205 1540 2045 15/32 1-1/2 10d 680 22 16 1020 29 20 1330 36 22 1740 51 28 950 1430 1860 2435

Increased 40%*

Blocked

Shear Wall Capacity-Wood Based Panels

*40% Increase based on reducing the load factor from 2.8, which was originally used to develop shear capacities down to 2.0 for wind resistance. (2006 IBC commentary)

Supported Edges

UnblockedTable 4.3.3.3.2 Unblocked Shear Wall Adjustment Factor, Cub

Intermediate Framing

Stud Spacing (in.)12 16 20 24

6 12 0.8 0.6 0.5 0.46 6 1.0 0.8 0.6 0.5

vub= vb Cub Eq. 4.3-2

vub = Nominal unit shear value for unblocked shear wall

vb = Use nominal shear values from Table 4.3A for blocked WSP shear walls with stud spacing at 24” o.c. and 6” o.c nailing

Shear Wall Capacity-Wood Based Panels

Summing Shear Capacities-4.3.3.3

Same material, constructionand same capacities(can double Ga)

Dissimilar materials(Not cumulative)

Dissimilar materials

Vallow = largest

Vallow=2x

Vallow =2x smaller orlargest, greater of.

VS1

VS2Same material, construction on both sides, different capacities-Seismic

Side 1

Vsc

Side 2

!"# = !"% + !"'()* = +,-.!"# Combined apparent shear wall shear

capacity (seismic)

+,-. = ()%!"%

/0 ()'!"'

Minimum of

Where:

(Exception-for wind,add capacities)

Gac= Combined apparent shear wall shear stiffness

Ga1= Apparent shear wall shear stiffness side 1(Ga2 side 2)-from SDPW Tables 4A-4D

vS1=Nominal unit shear capacity of side 1 (vS2 side 2)

VS1

VS2

VS1

Vsc or Vwc

Vsc or Vwc

Vsc or Vwc

Vsc = combined nominal unit shear capacity seismicVwc = combined nominal unit shear capacity windvS1=Nominal unit shear capacity of side 1 (vS2 side 2)

The shear wall deflection shall be calculated using the combined apparent shear wall shear stiffness, Gac and the combined nominal unit shear capacity, vsc, using the following equations:

2 layers WSP 1 side of wall

When two layers of WSP sheathing are applied to the same side of a conventional wood-frame wall, the shear capacity for a single-sided shear wall of the same sheathing-type, thickness, and attachment may be doubled provided that the wall assembly meets all of the following requirements:

• Panel joints between layers shall be staggered

• Framing members located where two panels abut shall be a minimum of 3 x framing.

• Special sheathing attachment requirements

• Special retrofit construction requirements

• Double 2x end studs, if used, shall be stitch-nailed together based on the uplift capacity of the double-sheathed shear wall.

Shear Wall A/R Adjustment Factors-SDPWS Section 4.3.4

• WSP’s with A/R > 2:1 multiply Vn x (1.25-0.125h/b)per Section 4.3.4.2.

• Struct. Fiberboard with A/R > 1:1 multiply Vn x (1.09-0.09h/b) per Section 4.3.4.2.

• Accounts for reduced unit shear capacity in high aspect ratio walls due to loss of stiffness as A/R increases.

Capacity Adjustments-Wind and seismic

SegmentedSW

PSW

• Segments with an aspect ratio > 2:1 shall be multiplied by 2b/h for the purposes of determining Li and ∑Li. The provisions of Section 4.3.4.2 and the exceptions to Section 4.3.3.4.1 (equal deflection) shall not apply to perforated shear wall segments.

Justification-Based on tests:

Distribution of lateral forces to In-line Shear Walls

• All walls are of same materials and construction

1. Where Vn of all WSP shear walls having A/R>2:1 are multiplied by 2b/h, shear distribution to individual full-height wall segments is permitted to be taken as proportional to design shear capacities of individual full height wall segments.(Traditional Method-by length)

Where multiplied by 2b/h the Vn need not bereduced additionally by Aspect Ratio Factor(WSP)= 1.25-0.125h/b per Section 4.3.4.2.

Exception:

Section 4.3.4.2-Limitation of section

Same materials in same wall line

Summing is allowed

Dissimilar materials in same wall line

Section 4.3.3.4does not apply

Equal wall deflection∆"#$$

2. Where Vn x 0.1+0.9b/h of all structural fiberboard shear walls with A/R>1:1, sheardistribution to individual full-height wall segments shall be permitted to be taken asproportional to design shear capacities. (Traditional Method-by length)

Where multiplied by Vn x 0.1+0.9b/h, thenominal shear capacities need not be reduced additionally by Section 4.3.4.2

• Shear distribution to individual SW’s shall provide same calculated deflection in each shear wall (Default Method)

per 2015 SDPWS)

Distribution of lateral forces to In-line Shear Walls

• All walls are of same materials and construction

Section 4.3.3.4-Limitation of section

Method 1-Simplified ApproachTraditional Method

b1 b2

SW1 SW2

h

Example Calculation per Commentary(assumes same unit shear capacity)

SW1 and SW2• Find nominal shear capacity• Check aspect ratio-adjust per SDPWS 4.3.4.1 exception

• Aspect ratios > 2:1 use reduction factor !"# (further reductions in 4.3.4.2 are not

required)• Convert to ASD unit shear capacity• Sum design strengths = vsw1 x b1 + vsw2 x b2 for line strength• Will produce similar results to equal deflection method

!$#

A/R>2:1 A/R≤2:1h

Distribution of lateral forces to In-line Shear Walls

Method 2-Equal Deflection

∆"#$ ∆"#%

b1 b2

SW1

SW2

h

Method of Analysis

SW1 and SW2• Find nominal unit shear capacity• Check aspect ratio-adjust per SDPWS 4.3.4.2• Aspect ratios > 2:1 use reduction factor

1.25-0.125h/b per 4.3.4.2• Determine Ga and EA values• Determine ASD unit shear capacity• Calculate deflection of larger SW (SW2)

&'($ = ∆'(%*+,

-./'($0 +$111230

+%4/'($

• Sum design strengths = vsw1 x b1 + vsw2 x b2 for line strength

Example Calculation per Commentary(assumes same unit shear capacity)

• Determine unit shear in smaller SW (SW1) that will produce same deflection

3000

2500

2000

1500

1000

Load

lbs.

500

SW L

ine

SW2

Deflection, in.

0.350.3

0.250.2

0.150.

1

0.05

ASD strength and deflection of shear wall line

ASD strength and deflection of SW2

ASD strength and deflection of SW1at SW2 defl. limit

ASD strength of SW1 Per SDPWS 4.3.4.2

SW1Reserve capacity

Construction Requirements

3x’s at adjoining panels required when:

• Nails are spaced 2” o.c.• 10d nails are spaced 3” o.c. and have penetration >1.5”• Nominal unit shear capacity on either side of shear wall > 700 plf

(SDC D-F)

4.3.7.1 - 3x Stud: Requirements:

4.3.6.3.1 Adhesives: Adhesive attachment of shear wall sheathing shall not be used alone, or in combination with mechanical fasteners.

Exception: Approved adhesive attachment systems shall be permitted for wind and seismic design in Seismic Design Categories A, B, and C where R = 1.5 and !" = 2.5, unless other values are approved. Not permitted in SDC D.

Footnote:6. Where panels are applied on both faces of a shear wall and nail spacing is less than

6" on center on either side, panel joints shall be offset to fall on different framing members. Alternatively, the width of the nailed face of framing members shall be 3" nominal or greater at adjoining panel edges and nails at all panel edges shall be staggered.

Panel jnts.staggered

Adjoining Panel jnts.

Adjoining Panel jnts.

Adjoining Panel jnts.

Panel joints Staggered Panel joints Aligned

3x stud

2x stud2x studs

Optional joint

Square plate washers preventcross grain bending-splitting of sill plate

Sill plate

Cross grain bending (tension perp. to grain stresses)

Wood structural panel 1/2” Max

4.3.6.4.3 Anchor Bolts: • Foundation anchor bolts shall have a steel plate washer under each nut. • Minimum size-0.229”x3”x3” in. • The hole in the plate washer - Diagonally slotted, width of up to 3/16” larger

than the bolt diameter, and a slot length not to exceed 1-3/4” is permitted ifstandard cut washer is provided between the nut and the plate.

• The plate washer shall extend to within 1/2" of the edge of the bottom plate on the side(s) with sheathing.

• Required where sheathing nominal unit shear capacity is greater than 400 plf for wind or seismic. (i.e. 200 plf ASD, 320 plf LRFD)

Square Plate Washers

Shear Wall Anchorage

Standard cut washer if slotted hole

Standard cut washers • Permitted to be used where anchor bolts are designed to resist shear only and

the following requirements are met: a) The shear wall is designed segmented wall with required uplift anchorage at

shear wall ends sized to resist overturning neglecting DL stabilizing moment.

b) Shear wall aspect ratio, h:b, does not exceed 2:1. c) The nominal unit shear capacity of the shear wall does not exceed 980 plf

for seismic or 1370 plf for wind.

Standard cut plate washers

Sill plate

Cross grain bending (tension perp. to grain stresses)

Wood structural panel

Cut Washers

Lunch

Information Overload

Deer in Headlight

Shear Wall Design

• Shear Wall Types

• Shear wall Anchorage

• SDPWS Code Requirements

• Complete Load Paths

• ASCE 7 Offset Shear Wall Requirements

• Offset Shear Walls

Detailing for Continuous Load Paths

Karuna IHolst Architecture

Photo: Terry MaloneContinuous Stacked In-line Load Paths

Simple Load Paths

Detailing for Continuous Load Paths

Karuna IHolst Architecture

Photo: Terry MaloneDiscontinuous Load Paths

Complicated Load Paths

In-plane Offset Segmented Shear Walls

Hold Down(option 1)No irreg.

Wd

Hdr

Sill

Wd

Hold down(Typical)

Tie strap

Boundary nailing should be installed at each 2x stud at hold down and each plate

A

Compr. blocks required at all H.D. locations

Blk’g. or rim joist

Header/collector

Section C

Section B Section A

B C

Col

lect

or

C D

Alt. Config.

Sect. B

Anchor boltsor nails

Shear transfer Conn.

Tie strap

Nail shtg toeach 2x stud

Type 4 vert. irreg.SDC B-F

• Type 4 Vertical Irregularity, in-plane offset • ASCE 7-10 12.3.3.3 Elements supporting discontinuous walls SDC B-F• ASCE 7-10 12.3.3.4 25% increase in Fpx SDC D-F (connections)

In-plane Offset Segmented Shear Walls

Wd

Hdr

Sill

Nail shtg toeach 2x stud

(option 2) Wd

Hold down(Typical)

Tie strap

Boundary nailing should be installed at each 2x stud at hold down and each plate

A

Compr. blocks required at all H.D. locations

Blk’g. or rim joistAnalyze this Section as a transfer diaph.or transfer wall

Header/collector

Section C

Section B Section A

B C

Col

lect

or

C D

Alt. Config.

Sect. B

Anchor boltsor nails

Shear transfer Conn.

Tie strap

Nail shtg toeach 2x stud

Type 4 vert. irreg.SDC B-F

Type 4 vert. irreg.SDC B-F

• Type 4 Vertical Irregularity, in-plane offset • ASCE 7-10 12.3.3.3 Elements supporting discontinuous walls SDC B-F• ASCE 7-10 12.3.3.4 25% increase in Fpx SDC D-F (connections)

Hold down(Option 2 -only)

Wd. Wd. Wd.

Opening

Floor or roof sheathing

Blocking or continuous rim joist

Continuous rim joist, beam or special truss can be used as strut / collector or chord.

Double top plate can be used as strut / collector or chord.

Splice at all jointsin boundary element

Opening

Column

Possible perforated or FTAO shear wall

Segmentedshear wall

Header

Examples of Drag Struts, Collectors and chords at Exterior Boundaries

Sect.

Header

Direction of Load

Roof sheathing

Shear panel

Double top plate used as strut / collector or chord.

Examples of Drag Struts, Collectors and chords at Exterior Boundaries

Flat strap and blocking used as strut / collector or chord.

All shear is transferred into top plates, then into SW

All shear is transferred flat strap and blocking, then into SW

SW

SW

Non-SWNon-SW

Non-SWNon-SW

Non-SW

Non-SW

Non-Strut/collector

Strut/collector

Strut/collector

T

Shear Wall(transfer wall)

Hdr.

Collector Chord

Chord

h1

Diaphragm sht’g. elevation

Diaphragm sht’g. elevation

Parapet (typ.)

Vert

. Ste

pin

dia

ph.

Transfer area

Complete Load Path to Foundation Roof at Different Elevations-Chord Forces

Soil pressure

Chord force

Fo/tFo/t

Foundation

Parapet (typ.)

Tie strap

TT

If strut action

Shear Wall

T

h2

No shear wall perpendicular to this wall at step

Blocking not full height.No diaph. Shr. Transfer (boundary nailing?).Truss top chords in cross-grain bending.

Cross grain bending at gang-nail plate

Assumed bearing of block against truss chord

Incomplete Load Path-Blocking Issues

NEW: See 2015 IBC Figures 2308.6.7.2(1) and 2308.6.7.2(2) for possible solutions

2015 IBC Figure 2308.6.7.2(1)

2015 IBC Figure 2308.6.7.2(2)

Insulation

Screened vent

Typical shear panel

Boundary nailing

Middle 3rd available for vent holes-No continuous vents

2x blocking

Venting OptionsA

A

2x vertical nailer

Truss

2x blocking

2x framing all 4 sides

Insulation

Insulation

Insulation

Screened vent holes

Typical shear panel

2x framing all 4 sides

Boundary nailing

Outriggers per manuf. instructions

Baffle

2x blockingw/ vent holes

Venting Options B

B

Insulation

2x vertical nailer E.S.

Truss

Baffle asrequired

Insulation

Boundary nailing

Screened vent holes

Typical Solid Blocking

2x blocking

A

Insulation

2x rafter

A

Shear Wall Design

• Shear Wall Types

• Shear wall Anchorage

• SDPWS Code Requirements

• Complete Load Paths

• ASCE 7 Offset Shear Wall Requirements

• Offset Shear Walls

ASCE 7 Offset Shear Wall Requirements

SW 1

SW 2

Col

lect

or

Collector

Collector

Out-of-plane Offsets In-plane Offsets

Potential bucklingproblem w/ supporting columns and beams

F (Rho = structure)SDC D-F Diaphragm load at 2nd floor

SW1

SW3

SW4

SW5Collector / beam

Collector

Drag

strut

Drag

strut

Drag strut

Drag

strut

A

B3

2

1

A.75

A.33

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

Elements requiring over-strength load combinations

Transfer diaphragm grid line 1 to 3 See Section 12.10.1.1(transfer forces)

ASCE 7 Table 12.3-2 Type 4 vertical irregularity: in-plane offset discontinuity

ASCE 7 Table 12.3-1 Type 4 horizontal irregularity: out-of-plane offset discontinuity in the LFRSload path

W (Fpx w/ Rho=1)

ASCE 7 Table 12.3-2 Type 4 vertical irregularity: in-plane offset discontinuity in the LFRS (if no H.D. at A.25)

Relevant Irregularities Per ASCE 7-10 Horizontal Irregularities Table 12.3-1 and Vertical Irregularities Table 12.3-2

SW2

H.D.

H.D.

A.25

Diaphragm

Shear wall is not

continuous to

foundation.

Type 4 horizontal

and vertical

irregularity

Chord

Co

lle

cto

r

Str

ut

Str

ut

Chord

Str

ut

Type 4 Horizontal & Vertical Offset Irregularity-SeismicS

W1

SW

2

SW

3

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.

• ASCE 7-10 Section 12.3.3.3 (SDC B-F)

Elements supporting discontinuous walls or frames.

• ASCE 7-10 Section 12.3.3.4 (SDC D-F)

Increases in force due to irregularity1

2

B

3

C

A

Co

lle

cto

r S

W

SW

ab

ov

e

or

No

SW

SW

SW

SW

SW

SW

Co

lle

cto

r

Collector

1 2

Diaphragm

Str

ut Cantilever?

Type 4 Vertical Irregularity SDC D-F

SW3

Strut

. . .

. .

Blockingor shear panels

Continuous member

Also see 12.10.2.1 SDC C-F for collector requirements.

ASCE 7-10 Section 12.3.3.3 (SDC B-F)Elements supporting discont. Walls or frames:

Applies to:•Beams, columns, slabs, walls ortrusses.•Requires over-strength factor ofSection 12.4.3

2

. . .

. .

Diaphragm 1

Shear wall is not continuous to foundation.

Chord C

olle

ctor

Str

ut

Chord

Str

utType 4 Horizontal Irregularity-Seismic

SW

1S

W2

SW

3

ASCE 7-10 Section 12.3.3.4 (SDC D-F) -Type 4 Horizontal and Type 4 vertical irregularity requires a 25% increase in the diaphragm (inertial) design forces determined from 12.10.1.1 (Fpx) for the following elements:

• Connections of diaphragm to vertical elements and collectors.• Collectors and their connections to vertical elements.

1

2

B

3

C

A

Collectors and their connections to vertical (Grid lines 1, 2 and 3)

• Diaphragm shears are not required to be increased 25%.

• The transfer force (SW3) in SDC D-F must be increased by rho, per 12.10.1.1.

See 12.10.2 & 12.10.2.1 for collectors

Exception: Forces using the seismic load effects including the over-strength factor of Section 12.4.3 need not be increased.

Design diaphragm connections to SW and struts (Grid lines 1, 2 and 3)

SW

4S

W5

Type 2 Horiz., Type 3 Horiz., and Type 4 Vert. & Horiz. Irregularity SDC D-F(Interior collector similar)

SW

SW

Strut

ASCE 7-10 Section 12.3.3.4 (SDC 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 diaphragmdesign forces (Fpx) determined from 12.10.1.1for the following elements:

. . .

. .

Blockingor shear panels

Continuous member

Strut/collector

Alt. Collector/ strut Connection

• Connections of diaphragm to vertical elements and collectors.

3

Type 2 Horiz., Type 3 Horiz., and Type 4 Vert. & Horiz. Irregularity SDC D-F(Interior collector similar)

SW

SW

Strut

ASCE 7-10 Section 12.3.3.4 (SDC 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 diaphragmdesign forces (Fpx) determined from 12.10.1.1for the following elements:

. . .

. .

Blockingor shear panels

Continuous member

Strut/collector

Alt. Collector/ strut Connection

• Connections of diaphragm to vertical elements and collectors.

Collector

• Collectors and their connections, including their connections to vertical elements.

Shear Wall3

Ftg.

Strut/Collector

Strut/Collector

SW1

Vr

V2

Dhr

Dh2

Ftg.

Supp

ort c

olum

n

A BA.25

SW2

Shear wall system at grid line 1In-plane Offset of Wall

Supp

ort c

olum

n

ASCE 7 Table 12.3-2-Type 4 vertical irregularity- In-plane offset discontinuity in the LFRS

Wall element is designed for standard load combinations of ASCE 7-10 Sections 2.3 or 2.4.

Shear wall anchorage is designed for standard load combinations of Sections 2.3 or 2.4.

Column elements are designed for standard load combinations 2.3 or 2.4.

Wall designed for over-strength per ASCE 7-10 Section 12.3.3.3 (SDC B-F) and/or 12.3.3.4 (SDC D-F)

Shear wall hold downs are designed for standard load combinations of Sections 2.3 or 2.4.

Col

lect

or

Assuming no hold down

See struts and collectors

Connectionsare designed for standard load combinations 2.3 or 2.4.

Ftg.

Support beam/collector

Collector

SW3

Vr

V2

Sup

port

col

umn

Ftg.

Sup

port

col

umn

A BA.33

Shear wall system at grid line 2Out-of-Plane Offset

Footings and connections are not required to be designed for over-strength.

Dhr

Dh2

Dv

ASCE 7 Table 12.3-1-Type 4 horizontal irregularity- out of-plane offset discontinuity in the LFRS

Wall element is designed for standard load combinations of ASCE 7-10 Sections 2.3 or2.4.

Shear wall hold downs and connections are designed for standard load combinations of ASCE 7-10 Sections 2.3 or 2.4.

Elements supporting discontinuous walls or frames require over-strength Factor of Section 12.4.3.

Collector and columns shall be designed in accordance with ASCE 7-10 section: 12.3.3.3

Connection is designed for standard load combinations of Sections 2.3 or 2.4.

See struts and collectors

Grade beamFtg.

Support beam/collector

Collector

SW4

SW5

Vr

V2S

up

po

rt c

olu

mn

Gr. Bm.

DL

Ftg.

Su

pp

ort

co

lum

n

AB A.75

A.33

Shear wall system at grid line 3-In-plane Offset

ASCE 7 Table 12.3-2-Type 4 vertical

irregularity- In-plane offset discontinuity

in the LFRS

Wall element, hold downs,

and connections are

designed for standard

load combinations of

ASCE 7-10 Sections 2.3

or 2.4.

Elements supporting discontinuous walls or

frames require over-strength factor of Section 12.4.3.

Collector and columns shall be designed in

accordance with ASCE 7-10 section: 12.3.3.3

Connections are

designed for standard

load combinations of

Sections 2.3 or 2.4.

Dhr

Dh2

Connections are

designed for standard

load combinations of

ASCE 7-10 Sections

2.3 or 2.4.

Wall element

designed for

standard load

combinations of

Sections 2.3 or 2.4.

See struts and collectors

Footings and connections

are not required to be

designed for over-strength.

Concrete Podium slab

Support beam/collector

Collector

SW4

SW5

Vr

V2Su

ppor

t col

umn

Gr. Bm. DLSu

ppor

t col

umn

AB A.75 A.33

Discontinuous Shear wall system at Podium Slab

Wall element, hold downs, and connections are designed for standard load combinations of ASCE 7-10 Sections 2.3 or 2.4.

Connections are designed for standard load combinations of Sections 2.3 or 2.4.

Dhr

Dh2

Connections are designed for standard load combinations of ASCE 7-10 Sections 2.3 or 2.4.

Wall element designed for standard load combinations of Sections 2.3 or 2.4.

Collector, columns, Podium slabs and beam shall be designed using the over-strength factor of ASCE 7-10 section 12.3.3.3

Elements supporting discontinuous walls or frames require over-strength factor of Section 12.4.3.

Concrete beam

All connections to the members above slab are designed for standard load combinations of ASCE 7-10 Sections 2.3 or 2.4.

Conc. wall

Seismic reactions from the flexible upper portion shall be amplified by the ratio

!"#$$%& $'&()'*!"+',%& $'&()'*

> 1.0

for connections embedded into slab (does not apply to gravity reactions).

12.10.2 SDC B - Collectors can be designed w/o over-strengthbut not if they support discontinuous walls or frames.

12.10.2.1 SDC C thru F- Collectors and their connections, including connections to the vertical resisting elements require the over-strength factor of Section 12.4.3, except as noted:

Shall be the maximum of:!"#$ - Forces determined by ELF Section 12.8 or Modal Response Spectrum Analysis

procedure 12.9 !" #%$ - Forces determined by Diaphragm Design Forces (Fpx), Eq. 12.10-1 or

#%$&'(= *. ,-.-/01%$ - Lower bound seismic diaphragm design forces determined byEq. 12.10-2 (Fpxmin) using the Seismic Load Combinations of section12.4.2.3 (w/o over-strength)-do not require the over-strength factor.

Struts / collectors and their connections shall be designed in accordance with ASCE 7-10 sections:

Exception:

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 12.4.2.3 with forces determined in accordance with Section 12.10.1.1 (Diaphragm inertial Design Forces, #%$).

Struts and Collectors-Seismic

#%$&2$= *. 3-.-/01%$- Upper bound seismic diaphragm design forces determined byEq. 12.10-2 (Fpxmax) using the Seismic Load Combinations of section12.4.2.3 (w/o over-strength)-do not require the over-strength factor.

Shear Wall Design

• Shear Wall Types

• Shear wall Anchorage

• SDPWS Code Requirements

• Complete Load Paths

• ASCE 7 Offset Shear Wall Requirements

• Offset Shear Walls

Out-of-Plane Offset Shear Walls

Loads

Col

lect

or

Collector

Collector

TD1

SW1

SW2

Transfer area

Offset

Discont.drag strut

Dra

g st

rut

Dra

g st

rut

SW3

Col

lect

or

Collector

Collector

TD2

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

• Calculations 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.

• Collectors 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

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).

Check for Type 2 horizontal irregularityRe-entrant corner irregularity

SW 1

SW 2

25’ 20’

15’

200

plf

Drag strut

Ch

ord

co

llect

ors

Ch

ord

80’

35’

50’

Drag strut is discontinuous

1 2

A

B

3

C

4

10’25’ 5’

15’

200

plf

TD1C

ho

rd

colle

cto

rs

Pos. direction + -

SW 3 SW 4

8’

8’

Drag strut

Collector

5’

12’45’

80’

Drag strut

Drag strutC

ho

rd

Drag strut

Diaphragm with Horizontal End OffsetLongitudinal Loading-Out-of-plane offset Shear Walls

Offset SW

Offset SW

Support

Support

SW 2

35’

50’

1 2

A

B

3

C

10’

8’

Pos. direction

+8.85

+ -

Ba

sic

sh

ea

r d

iag

ram

2

8’

2

SW 3

Po

s.

40 plf 160 plf

20

0 p

lf

20

0 p

lf

SW 1

Ba

sic

sh

ea

r d

iag

ram

1

vsw=+161.3 plfvnet=+133.29 plf

F=1066.3 lb

F=590.3 lb

590.3 lb590.3 lb

-700 lb(-28 plf)

+700 lb (+28 plf)

+177.1 lb(+8.85 plf)

4510 lb(+45.1 plf)

1510 lb(+15.1 plf)

-4090 lb (-40.9 plf)

Basic Diaphragm Shears and Transfer Diaphragm Shear

-20.66 Ne

g.

-413.2 lb(-20.66 plf)

+28

200 plf

-28

Vsw2=wL/2=200(50)/2=5000 lb, vsw2=5000/10=500 plf

Vsw1, sw3, sw4=wL/2=200(50)/2=5000 lb, vsw=5000/(8+8+15)=161.3 plf

Determine Force transferred Into Transfer Diaphragm

Total Shear to Shear Walls (Assumed)

17’

Calcs

Sign Convention

4600 lb

25’ 20’

15’

80’

1 2

A

B

3

C

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]

SW 1

SW 3 SW 4

SW 2

15’ 5400 lb

F=1954.5 lb

F=75 lb

F=957.8 lb

F=416.6 lb

F=802.5 lb

F=693.6 lb

F=1169.6 lb

F=700 lb F=36 lb

F=18 lb

F=364 lb

F=87.5 lb

F=3388.5 lbF=3236 lb

Line needs to move in this direction

The shear wall shearsneeds to be higher in order to move the force diagram in this direction

Line needs to move in this direction

The shear wall shearsneeds to be lower in order to move the force diagram in this direction

Load distribution needs to increase towards line B/C. Increase the load to B/C by the amount off +/-.

Calculated forces

Revisedforces

200

plf

200

plf

SW

SW

(Typ.)

SW

SW (Typ.)

Tra

ns

ve

rs

e

wa

lls

exterior walls

Corridor walls

Transfer Area

collector

Co

llecto

r

collector

Co

llecto

r

Tying Shear Walls Across the Corridor or a Large Diaphragm

Tying Shear Walls Across a Large Diaphragm

SW

SW

3. Collectors for shear transfer

to individual full-height wall

segments shall be provided.

SDPWS 4.3.5.1

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).

Must follow engineering mechanics and all

force diagrams must close to zero or be

resolved by other methods.

Diaphragm 1 Diaphragm 2

SW

1S

W2

1 2

B

3

C

A

D

W=200 plf

78’ 122’

6’

22’

22’

50’

7800 # Sum=20000 # 12200 #

156

156 244

244

RL=RR=200(78/2)=7800 #vR=7800/50=156 plf

RL=RR=200(122/2)=12200 #vR=12200/50=244 plf

Sum=40000 #=200 plf(200’)

vSW=454.55 plfvNet=54.55 plf

1200

1200

Layout 1-Full length walls aligned

Diaphragm 1 Diaphragm 2

SW

1

SW

2

SW3

1 2

B

3

C

A

D

4

SW4

112

plf

88 p

lf

88 p

lf11

2 p

lf

24 p

lf

W=200 plf

78’ 122’

82’

4’

118’

6’

22’

22’

50’

Sum=7976 # Sum=20048 # Sum=11976 #

156

156

164 236

236

244

244

164

8 8

RL=RR=112(78/2)=4368 #vR=4368/28=156 plf

4368 #

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

3608 #

Transfer AreaRL=RR=24(4/2)=48 #vL=vR=48/6=8 plf

Sum=40000 #=200 plf(200’)

5368 #

6608 #

Layout 2-Full length offset walls

SW

1

SW

2

SW3

2

B

3

C

A

D

SW4

156

156

164 236

244

244

164

8

160

16024

0

24

0

Total load to grid lines 2 & 3

R23=4368+3608+48+5368+6608+48=20048 # O.K.

LSW=22+22=44’

VSW=20048 #

vSW=20048/44=455.64 plf

vnet SW1=455.64-156-244=55.64 plf

vnet SW2=455.64-164-236=55.64 plf Checks, they

should be equal

SW1

F2AB=55.64(22)=1224 #

F2BC=(156+8)6=984 #

F2C=1224-984=240 #

SW2

F3CD=55.64(22)=1224 #

F3CB=(236+8)6=1464 #

F3B=1224-1464= -240 #

FB23=FC23=240(4)/6=160 #

Shear at transfer area=240/6+8=48 plf

Sum=20048 #

1224

1224240

240

8 236

Summing V=0

Summing V=0

All forces in lb., all shears in plf

O.K.

Diaphragm 1 Diaphragm 2S

W1

SW

2

SW3

1 2

B

3

C

A

D

4

SW4

112

plf

88 p

lf

88 p

lf11

2 p

lf

24 p

lf

W=200 plf

78’ 122’

82’

4’

118’

6’

22’

22’

50’

Sum=7976 # Sum=20048 # Sum=11976 #

156

156

164 236

236

244

244

164

8 8

RL=RR=112(78/2)=4368 #vR=4368/28=156 plf

4368 #

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

3608 #

Transfer AreaRL=RR=24(4/2)=48 #vL=vR=48/6=8 plf

Sum=40000 #=200 plf(200’)

5368 #

6608 #

Layout 3-Full length plus partial length offset shear walls

10’

12’

SW

1

SW

2

SW3

2

B

3

C

A

D

SW4

156

156

164 236

244

244

164

8

2125.49

2125.4931

88.2

4

3188

.24

Total load to grid lines 2 & 3

R23=4368+3608+48+5368+6608+48=20048 # O.K.

LSW=22+12=34’VSW=20048 #vSW=20048/34=589.65 plf

vnet SW1=589.65-156-244=189.65 plfvnet SW2=589.65-164-236=189.65 plf Checks, they

should be equal

SW1F2AB=189.65(22)=4172.24 #F2BC=(156+8)6=984 #F2C=4172.24-984=3188.24 #

SW2FSW2 =189.65(12)=2275.76 #FSW2 to 3B=(236+8)6+(236+164)10=5464 #F3B=2275.76-5464= -3188.24 #

FB23=FC23=3188.24(4)/6=2125.49 #

Shear at transfer area=3188.24/6+8=539.37 plf

Sum=20048 #

4172.24

2275.76

3188.24

3188.24

8

236

Summing V=0

Summing V=0

All forces in lb., all shears in plf

O.K.

22’

10’

12’

6’

Diaphragm 1 Diaphragm 2S

W2

SW3

1 2

B

3

C

A

D

4

SW4

112

plf

88 p

lf

88 p

lf11

2 p

lf

24 p

lf

W=200 plf

78’ 122’

82’

4’

118’

6’

22’

22’

50’

Sum=7976 # Sum=20048 # Sum=11976 #

156

156

164 236

236

244

244

164

8 8

RL=RR=112(78/2)=4368 #vR=4368/28=156 plf

4368 #

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

3608 #

Transfer AreaRL=RR=24(4/2)=48 #vL=vR=48/6=8 plf

Sum=40000 #=200 plf(200’)

5368 #

6608 #

Layout 4 - Partial length offset shear walls

10’

12’

SW

110

’

12’

Total load to grid lines 2 & 3

R23=4368+3608+48+5368+6608+48=20048 # O.K.

LSW=12+10=22’VSW=20048 #vSW=20048/22=911.27 plf

vnet SW1=911.27-156-244=511.27 plfvnet SW2=911.27-164-236=511.27 plf Checks, they

should be equal

SW1FSW1=511.27(10)=5112.7 #FSW1 to 2B=(156+244)12=4800 #F2BC=(156+8)6=984 #F2C=5112.7-4800-984=-671.3 #

SW2FSW2=511.27(12)=6135.24 #FSW2 to 3B=(236+8)6+(236+164)10=5464 #F3B=6135.24-5464= 671.24 #

FB23=FC23=671.3(4)/6=120 #

Shear at transfer area=671.3/6+8=119.9 plf

SW

2

SW3

2

B

3

C

A

D

SW4

156

156

164 236

244

244

164

8

Sum=20048 #

671.24

8

236

Summing V=0

Summing V=0

All forces in lb., all shears in plf

O.K.

10’

12’

6’

120

12067

1.24

671.

3

5112.7

6135.24

12’S

W1

10’

671.3

120 Union, San Diego, CATogawa Smith Martin

Carbon 12, Portland, ORPATH Architecture

DraftR. Terry Malone, P.E., S.E., Senior Technical Director

•Overview•Relevant ASCE 7-16 and 2015 SDPWS Code Requirements

1.Diaphragm Flexibility-Seismic2.Open Front Diaphragms3.Torsional Irregularities4.Redundancy5.Diaphragm Design Forces

•Basic Information•Lateral Load Calculations

1.Calculate Seismic Forces2.Rigid Diaphragm Analysis3.Shear Wall Deflections

Seismic Design Example of a Cantilever Wood Diaphragm

Scott Breneman, PHD, PE, SE, Senior Technical Director

Reference Materials

http://www.woodworks.org/wp-content/uploads/Irregular-Diaphragms_Paper1.pdf

• The Analysis of Irregular Shaped Structures: Diaphragms and Shear Walls-Malone, Rice-Book published by McGraw-Hill, ICC

• Woodworks Presentation Slide Archives-Workshop-Advanced Diaphragm Analysis

• NEHRP (NIST) Seismic Design Technical Brief No. 10-Seismic Design of Wood Light-Frame Structural Diaphragm Systems: A Guide for Practicing Engineers

• SEAOC Seismic Design Manual, Volume 2

• Woodworks-The Analysis of Irregular Shaped Diaphragms (paper). Complete Example with narrative and calculations.

• Webinar Archive- Offset Diaphragms -Part 1• Webinar Archive- Offset Shear Walls-Part 2• Slide Archive-Workshop-Advanced Diaphragm Analysis • Slide Archive-Offset Diaphragms and Shear Walls

• Webinar Archive- Lateral Design Considerations for Mid-rise Structures-Stacked multi-story shear walls (MBA)

Information on Website

Method of Analysis and Webinar References

Diaphragms OpeningsOffset Shear WallsOffset Diaphragms

Mid-rise Design Considerations

Presentation Slide Archives, Workshops, White papers, research reportsInformation on Website

Questions?This concludes Our Workshop Presentation on Advanced Diaphragm Analysis

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

Contact Information:terrym@woodworks.org928-775-9119