Excerpts From:

Part I-Offset Diaphragms Copyright Materials

Presentation updated to 2012 IBC, ASCE 7-10

Copyright McGraw-Hill, ICC

This presentation is protected by US and International Copyright laws. laws Reproduction, Reproduction distribution, display and use of the presentation without written permission of the speaker is prohibited. prohibited

By: R. Terry Malone, PE, SE Presented by:

Senior Technical Director Architectural & Engineering Solutions

© The Wood Products Council 2014

[email protected]

Course Description “The Wood Products Council” is a Registered Provider with The American Institute of Architects Continuing Education Systems (AIA/CES), Provider #G516.

Credit(s) earned on completion of this course will be reported to AIA CES for AIA members members. Certificates of Completion for both AIA members and non-AIA members are available upon request request.

This course is registered with AIA CES for continuing pprofessional education. As such, it does not include content that may be deemed or construed to be an approval or endorsement byy the AIA of any material of construction or any method or manner of handling, using, distributing, y material or or dealingg in any product. __________________________________ _________

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

Lateral force resisting systems in today’s today s structures are much more complex than they were several decades ago, incorporating multiple horizontal and vertical offsets in the diaphragms, multiple irregularities, and fewer lateral resisting elements. This two part presentation will provide a brief review of the method used to analyzed these complex structures. In part 1, topics will include code requirements, how to recognize diaphragm irregularities and discontinuities, how shears are distributed through complex diaphragms, the method of analysis used to solve the transfer of forces across areas of discontinuity, and the analysis of flexible wood sheathed or untopped steel decking diaphragms with horizontal offsets.

Transverse Loading Semi-Rigid

Learning Objectives •

Offset walls

In-line walls Not effected

Not effected

Basic Information Discuss boundary elements, complete lateral resisting load path th requirements i t and d related l t d code d sections. ti

•

Cant.

Examine Common Types of Discontinuities

Discuss the Analytical Method of Analysis Review an analytical method used for solving complex diaphragms and shear walls using “Transfer Diaphragms” and the “Visual Shear Transfer Method.”

•

Lc

•

Fu ull cantileverr

Examine common types of discontinuities and irregularities and discuss how to establish complete lateral load paths across areas of discontinuity.

Offset Diaphragms-Examples Review the analysis of flexible offset diaphragms for loading in the transverse and longitudinal directions.

No exterior Shear walls (Not recommended)

W’ Wc

Presentation Assumptions Flexible wood sheathed or un-topped steel deck diaphragms The method of analysis is also relevant to internal load path analysis within semi semi-rigid rigid diaphragms diaphragms.

•

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

•

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

Code References: • •

ASCE 7-10 7 10 “Mi “Minimum i D Design i L Loads d ffor B Buildings ildi and d Oth Other Structures” St t ” 2012 IBC

Design references: •

The Analysis of Irregular Shaped Structures: Diaphragms and Shear WallsMalone, Rice

• • • •

Design of Wood Structures- Breyer, Fridley, Pollock, Cobeen SEAOC Seismic Design Manual, Volume 2 Wood Engineering and Construction Handbook-Faherty, Williamson Guide to the Design of Diaphragms, Chords and Collectors-NCSEA, Mays

Basic Information • Boundary Elements • Complete Load Paths • Method of Analysis

Diaphragm Boundary Elements

Chord

Fundamental Principles:

C

SW

Diaphragm Boundary Elements: • Chords, drag struts, collectors, Shear walls, frames • Boundary member locations: • Diaphragm and shear wall perimeters • Interior openings • Areas of discontinuities • Re-entrant corners.

•

Diaphragm and shear wall sheathing shall not be used to splice boundary elements.

•

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.

Chord

SW1

Deflection if no tie Chord

3 Re-entrant corner Tearing g will occur if collectors are not installed at re-entrant corner.

Diaphragm 1 SW3 boundary

Loads

Chord

C

1

2

Diaphragm 1

Chord

Diaphragm 2 boundary (typical)

N t Note: All edges of a diaphragm shall be supported by a drag strut, chord, shear wall or other lateral resisting element. Deflected curve if proper tie Deflected curve if no tie

Boundary Elements “L” Shaped Buildings-Transverse Loading

SW3 S

Diaphragm 1 boundary (typical)

Diaphragm 2 Note: Interior shear walls without a collector or a complete alternate load path are NOT ALLOWED! Chord

Basic Information • Boundary Elements • Complete Load Paths • Method of Analysis

SW4

B

Collector

SW2

Stru ut

Diaphragm 2

Collec ctor

Strut

Diaph. Boundary (Longitudinal loading)

A

Strut

Diaphragm Di h 2 Boundary

SW1 Deflection if tie

C

Chord

Chord

Chord

Chord

Collector

SW

Required for all SDC and wind

A

B

Strut, Collector, and Chord- (my) Terminology

Diaphragm 1

TD1

3

2

Strut

Strut- receives shears from one side only. Collector- receives shears from both sides. Discontinuous Chord-perpendicular to the applied load and diaphragm receives axial tension and compression chord forces forces. Chord

1

Strutt

Chord/Collector

Diaphragm support

SW

Strut

Chord Diaphragm support

Chord

Chord/Collector

Strut

SW

shall be supported by a boundary element.

Strut S

Collector

Note: All edges of a diaphragm T

SW2 S

SW

•

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

Collec ctor

W ( plf)

Strrut (typ.)

Analysis: • • •

1

3

6

7

8

A

D

Support

Discontinuous diaphragm chord/strut

Design shall be based on a rational analysis Complete load path from point of origin to lateral force resisting element (includes members and their connections and splices) p ) 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.

C

Open Opening In diaph.

10

Vertical offset in diaphragm

Discont. diaphragm chord MRF1

SW5

Collector (t p ) (typ.)

Collector ((typ.)

Strut chord h d

Strut/chord

Strut

Discont. diaphragm chord

Strut

B

9

Strut/chord

SW6

Vertical offset in diaphragm

Collector ll t C ll t Collector Strut /chord C (typ.) (typ.) Discont. diaph. chord Strut/chord Support C ll t Collector Offset SW2 Strut chord Multiple E strut SW3 SW4 offset Strut/chord F diaphragm Discont. Offset shear walls 4 diaphragm and struts chord

Offset shear walls 2

Discontinuous diaphragm 5 chord

C ll t Collector Strut /chord Collector (typ.) (typ.) Discont. diaph. chord CollectorStrut/chord Support Offset SW2 Strut chord Multiple E strut SW4 SW3 offset Strut/chord F diaphragm Discont. Offset shear walls 4 di h diaph. and struts chord Strrut (typ.)

Opening in diaph.

Strut

Open

Collector (typ.)

C

SW5 Collector (t p ) (typ.) Strut

Collector

Strut SW1

Strut/chord

10

Strut/chord

SW6 B

9

Strut

A

8

Collector ((typ.)

7

SW1

3

6

MRF1

1

Offset shear walls 2

Discontinuous diaphragm 5 chord

Strut chord h d

D

Discont. diaph. diaph chord/strut

• ASCE 7-10 section 12.10.1-At diaphragm g discontinuities such as openings g and re-entrant corners, the design shall assure that the dissipation or transfer of edge (chord) forces combined with other forces in the diaphragm is within shear and tension capacity of the diaphragm. What does this mean?

Complete Continuous Lateral Load Paths

Method of Analysis

Basic Information • Boundary Elements • Complete Load Paths • Method of Analysis

Support

The Visual Shear Transfer Method Symbol for 1 ft x 1 ft square piece of sheathing in static equilibrium (typ.)

+

Lds.

FY +M FX

Positive P ii Direction +

-

+

-

Transverse Direction (shown)

Sh Shears Applied A li d to t Sheathing Sh thi Elements El t + +

-

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

Shears Transferred Into Boundary Elements

1

2

w=uniform load

+ SW 1

(-)

(-)

Diaphragm shear transferred into boundary element (typ.)

SW 3

+ -

Strut in comp.

C

Resisting wall (+) shears h

B

1 ft ft. x 1 ft ft. square sheathing element symbol at any location in the diaphragm. (-)

+

SW 2

Resisting wall shears ((+))

T

10’

SW

Support

Maximum moment -

Basic Shear Diagram

+

B

Support

Pos.

Positive diaph. shear elements

Strut Forces Negative N ti diaph. di h shear elements

Neg.

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).

20’ C

Shear Distribution Into a Simple Diaphragm

Colllector

Colllector

• T

Collector

Chord (support)

A

-433

+542 Diaphragm

Shear

Collector force 2

F=975(10) =9750 lb

Support 16250 lb

Visual Shear Transfer Method (method used in later examples)

F=487(20) =9740 lb

-433

(+/- round off)

C

+542

Vsw=+1462 vnet=+487 Support 29250 lb

Positive sign convention

Calculating Collector Forces

1

2

3

Diaph. C.L.

4

W ( plf) Diaphragm chord

Transfer Diaphragm (sub-diaphragm):

SW

A portion of a larger diaphragm designed to anchor and transfer local forces to primary diaphragm chords/struts of the main diaphragm. •

TD Ratio=4:1 Maximum

+ +

Collector force is equal to the area of the shear diagram

A

TD1

SW 3

100’ 542 542 plf plf + 433 plf Basic Shear Diagram -

Traditional Method

Introduction to Transfer Diaphragms and Transfer Areas •

Diaphragm p g shear distribution into the collector

B

The Visual Shear Transfer Method

Chord (support)

433 plf

Support 13000 lb

Net shears 2

Direct shears 2

Diaph. 2

C 80’

A

10’

SW 2 20’

Add shears together for net shears

T

-

Strut Forces Positive sign convention

Tributary width to center wall

Strut in tension

(-)

SW 1

B

Tributary width to left and right wall

C

Diaph.1

T

At discontinuities, such as openings or re-entrant corners, the design shall assure that the dissipation or transfer of edge (chord) forces combined with other forces in the diaphragm is within shear and tension capacity of the diaphragm. What does this mean? Framing members, blocking, and connections shall extend into the diaphragm a sufficient distance to develop the force transferred into the diaphragm. What does this mean?

SW

Discontinuous diaphragm chord

Longitudinal Collector (Typically a tie strap and flat blocking are called out)

Drag sttrut

Strut in tension

Strut in Compr.

-

+

Unit shear acting on diaphragm sheathing element

-975 p plf

T

C

SW

+487 p plf

Resisting wall (+) shears

3

W=325 plf

A

Unit shear transferred into boundary element

vsw = +1462 plf vdiaph h1 = - 433 plf vdiaph h2 = - 542 plf

A

2

Legend Collectorr

Diaphragm C.L.

1

B

Diaphragm chord C

Diaphragm support

Diaphragm chord Transfer area

Transfer Diaphragm Members and Elements

Diaphragm support

A

Partial length collectors do not constitute a complete load path.

2

Chord Collector

TD1 C

SW 2

3

NOTE: Collector must extend the full depth p of the transfer diaphragm

Transfer area without transverse collectors

Transfer Mechanism

• The length of the collector is typically F2 F1 F3 determined by dividing the collector force by the diaphragm nailing capacity. (Wrong!!) • The collector is typically checked for tension only. Compression rarely checked. (Wrong!!) Blocking acts as ministrut/collector and transfers (accumulates) forces into the next block.

+

2

A

Chord

3

(TD support)

Main chord

-75 plf

C

Disrupted h d chord

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

T

+

b

B

LTD

Co ollector

3 +225 plf

Add to basic diaphragm shears

T

Direction of shear transferred into collector

+150 plf

•

Place the net diaphragm shear on each side of the collector

•

Place the transfer shears on each side of the collector

•

Sum shears on collector (based upon direction of shears transferred onto collector).

+ Net direction of shears acting on collector

Collector +

+

+550 plf +475 plf N t shear Net h

Shear left=+550-225= +325 plf 325 plf

325 plf

Note: The net shears will not always y be equal.

Shear right=+475-150=+325 plf •

Collector force=area of shear diagram

a

B

TD1

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

T(a) VA= , Shear = VA LTD DTD

Subtract from basic shears

Transfer diaphragm llength LTD

Chord force at discontinuity

Co ollector

SW

F(total)=™F1+F2+F3+F4

Example of Partial Strut/Collector

2

The Transfer Diaph Diaph. Aspect Ratio should be similar to the main diaphragm.

+

-

vnet=+300-(75)= +225 plf vnet =+225–(75)= +150 plf

SW

F4

+

Collector force distribution

Transfer using beam concept

1

Bearing perp. grain? to g Cont. tie strap over

Strut/collector force diagram

Plan-Partial length collector

C

3

Diaphragm unit shear (plf) transfer into blocking

Details are seldom cutt

Main chord

Support

Transfer Area Higher shears.

Sect.

Collector Full depth

Strut

Disrupted chord

Typical callout CMST14 tie strap x 10’-0” 10 0 with (xx) 10d nails over 2x flat blocking. Lap 2-8” onto wall.

Main chord

Chord d/Collector

Chord d/Collector

Resis sting force es

Chord

Rotation of section

Transfer T f area

Discontinuous diaphragm chord B

C

1

T

Collector

Disrupted chord

B

SW

Transfer Diaphragm ( Beam)

Resis sting force es

This force must be transferred out to the main chords. A complete load path is required.

Support

SW

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

(TD support)

C

Main chord

TD depth

+

T

+250 plf

VC No outside force VC= T(b), Shear = No outside force is changing the DTD LTD DTD is changing the +500 +300 basic diaphragm Transfer Diaphragm Shears basic diaphragm plf plf +225 +225 shear in this area Analogous to a beam with a + shear in this area plf plf concentrated Load. vnet= 300 + (250) = 550 plf Basic Shear Diagram at transfer diaphragm Net shear Basic diaph. shear TD shears

Basic Procedure

Method by Edward F. Diekmann

Lcollector

Dir. of force on collector

Fcollector=(325+325)(Lcollector)

Resulting net shear diagram on collector

Shear Distribution Into The Collector

2

Transfer Di h depth Diaph. d th

1

2 F (a) L

RA

Chord, strut or shear wall

2 Transfer Di h depth Diaph. d th

RA

A

Chord, strut or shear wall

Diaphragms with Horizontal Offsets

F (a) b

A

Discont. Collector Chord / strut

B

b L

TD1 Support

Chord, strut or shear wall

F

RB

F(L) b

B

RB

a

F

a

TD1

Transfer Diaphragm

L Trans sfer Diaph. le ength

Transfer Diaphragm

RA

Support

b

Support

RA

Tra ansfer Diaph h. length

1

Support

Chord, strut or shear wall

C

RC

F (b) L

RC

F

Discont. Chord / strut

Simple Span Transfer Diaphragm

C

F

Propped Cantilever Transfer Diaphragm

Analogous to a simple span beam with a concentrated load

Analogous to a propped cantilever beam with a concentrated load

Simple Span and Propped Cantilever Transfer Diaphragms

Transfer Diaphragm and Net Diaphragm Shear

w=200 p plf

357.1 plf +357.1

Diaph. CL C.L.

TD1

Colllector TD c chords

Collector chords TD c

M2=250 ft.-k F2 =7142.9 lb

Collector

+42.9

v= 7142.9(15) = -107.1 plf 50(20)

70 plf

-37.1

+70

35’

35’ v=150-(107.1) =+42 9 plf =+42.9 (Net shear)

SW 2 50’

F2

25’

+357.1

+214.3

v=70-(107.1) ( ) = -37.1 plf (Net shear) +42.9

-37.1

+70

7142.9 lb

B

12500 lb

Diaph. chord

B

35’

+214.3

SW 1

F2 ™M=0 ™M 0

A

B

A/R=2.5:1

2

200 plf

15’

SW 1 35’

150 plf

Basic shear diagram 1

-107.1

+70

214.3 plf

Diaphr. chord

Support

2142.9 lb

Free body for F2

7142.9 lb

+400

+320

+70

15’

Pos.

A

Diaph. C.L.

A

Neg.

Transverse Loading

TD shea ar diagram

Example 1-Diaphragm with Horizontal End Offset

v=

-250

-250 plf

SW 2 7142.9(35) 50(20)

= +250 plf -250

+250

C

15’

12500 lb Support

1

V=75 500 lb

Support 12500 lb Discontinuous diaphragm chord 25’ 2

Diaph. chord V=35 500 lb

C

20’ 3

5000 lb v=70+(250)= +320 plf Net shear

v=150+(250) = +400 plf Net shear

+

-

Sign Convention +

80’

-

Support Sign Convention 12500 lb 4

25’ N nett change No h 1

2

20’ N t change Net h occurs in TD

12500 lb

80’ N nett change No h 3

Legend 375 plf (240 plf) =xxx plf

4

Basic diaphragm shear Transfer diaphragm shears Net TD shears (basic shear +/- transfer diaph. shears)

+42.9 +230 lb

9.275’ -172.1 lb -37.1

10.725’

7142.9 lb

A

A +357.2 +357.1

C

+214.3 +42.9

171.4 plf net

107.1 plf net

+214.3

-37.1

-250 +42.9

Special nailing along collectors

+70

-37.1

F=7200 F 7200 lb F=7812.5 lb

Sum of shears to collector or highest boundary nailinggreater of

0 plf SW 1 SW 1

F=3748 F 3748.5 5 lb

SW 2 F=7142.9 lb

F=6000 lb

C

F=7142.5 lb F=7812.5 lb +214.3 T +42.9

+357.1

T

-37.1

+357.2

+214.3

+42.9

-37.1

+70

+70

15’

B

7142.9 lb

+400

357.1 plf net T 70 plf 0 plf

+320

C

15’

B

-250

Diaph. C.L.

+ +

25’

20’

-

250 plf net F=3750 lb

20’

1

4

3

25’

Sign Convention

17 5’ 17.5’

2

+70

-

Support

Sign Convention 1

+320

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

Support F =7200 lb

+400

C

2

3

Transverse Collector Force Diagrams

Longitudinal Chord Force Diagrams

Example 2-Diaphragm with Horizontal End Offset Diaph. CL C.L.

SW 2

A

5’

10’

5’

Diaphragm 1

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

Diaphragm boundary

Discontinuous Drag strut

B

Check the shear capacity of the nailing along the collector

15’ 1

10d @ 4/6/12 B

1

C 2

3

Transfer area boundary

Boundary locations

TD1

Diaphragm 2

200 plf Discontinuous Drag strut 20’ 2

SW 1

Drag strut

15’ 80’

3

4 Pos. direction

Diaphragm Nailing Callouts

50’

Collector

25’ 1

Transfer diaphragm

Collector and TD chorrds

10d @ 6/12 UB Case I C

10d @ 6/12 UB Case I C

10d @ 6/12 UB Case I

10d @ 6/6/12 B

10d @ 4/6/12 B

Basic shear diagram

35’

Collectorr and TD chorrds

70 plf

200 plf

150 plf

160 p plf

Chord

214.3 plf

Transfer diaphragm Boundary (Typ.)

200 plf 2

40 p plf 357.1 plf

Drag strut

Ch hord

37.1 plf 70 plf

285 plf

Longitudinal Loading

x4

x3 214.3 plf 42.9 plf

x2

320 plf

357.2 pllf

x1

+

-

Vsw=5000 lb vsw=500 plf 448.6 plf net

F=700 lb

Diaph. C.L. +1 15.1 plf

700 lb +15.1

+39.6

+69.6

+69.6

+45.1

C

2 2

1

3

+24.5

490 lb

+

+45.1

4510 lb

B

Legend 375 p plf (240 plf)

Basic diaphragm p g shear Transfer diaphragm shears

=xxx plf

Net TD shears (basic shear +/-TD Shears)

23.4 plf net

+39.6

-40.9

2

+40 9 +40.9

3

10.5 plf A net

F= -957 lb

51.4-40.9=10.5 plf

51 4 28=23 4 plf 51.4-28=23.4 28-4.6=23.4 plf

35’ F=819 lb F=700 lb

+28

+28

+4.6

Vsw=5000 lb

15’ SW 1

+4.6

C

+39.6

+39.6

+69.6

+69.6

39.6 p plf

39.6-15.1=24.5 39.6 15.1 24.5 plf

69.6 plf

69.6-45.1=24.5 plf

F=368 lb

35 plf net

B

+15 1 +15.1 +45.1

15.1-4.6=10.5 plf

35 plf

B

50’

Vsw=5000 lb vsw=333.3 plf 288.3 plf net

Transfer area

-51.4

15’

+39.6

-28

Vsw=5000 lb

W=20 00 plf

15’

Neg.

+4.6

+45.1 plf

+4.6

F=3529 lb 5’

-51 4 -51.4

-40.9 40 9 plf

-

F=700 lb

Pos.

+28

700 lb B

v=+15.1+(24.5)= =+15 1+(24 5)=+39.6 +39 6 plf v=+45.1+(24.5)=+69.6 plf

Dia aphragm 2

Diiaphragm 1

v=+15.1-(10.5)= 15.1 (10.5) +4.6 4.6 plf (Net shear)

10’ SW 2 448.6 plf net

A

-10.5

v=-40.9-(10.5)= -51.4 plf

35’

5’

W=40+160=20 W 00 plf

-51.4 51 4 SW 2 10’

-28

-40.9 plf

A

700 lb

4090 lb

210 lb

50’

Resulting direction of shear in collectors and chords

F=368 lb

+45.1

15’ SW 1 288 3 plf net 288.3

F= -819 lb F=4323.5 lb strut 4050 lb chord

+39.6

20’

25’

80’

Pos. direction

+

-

4 +69.6

+69.6

1 Pos. direction

Transfer Diaphragm and Net Diaphragm Shear

Questions? This concludes AIA Presentation Part 11- on Offset Diaphragms R. Terry Malone, P.E., S.E. Senior Technical Director WoodWorks.org Prescott Valley, Arizona Contact Information: [email protected] WoodWorks woodworks.orgg Events/Presentation Archives (slide handouts)-free

+

-

2

3

Longitudinal and Transverse Collector/Strut Force Diagrams

4