The Wood Products Council The AIA/CES High Load … · Diaphragm Design Example Rigid vs. Flexible...

High Load Diaphragm Designf P li d R ffor Panelized Roofs

A Cost Effective Solution for Large Low Slope RoofsLisa Podesto, PE

Technical Director

“The Wood Products Council” is a Registered Provider with The American Institute of Architects Continuing Education Systems (AIA/CES). Credit(s) earned on completion of this program will be ( ) ( ) p p greported to AIA/CES for AIA members. Certificates of Completion for both AIA members and non-AIA members are available upon request.

This program is registered with AIA/CES for continuing professional education. As such, it does not include content that may be deemed or construed to be an approval or endorsement by the AIA of any

t i l f t ti th d f h dli imaterial of construction or any method or manner of handling, using, distributing, or dealing in any material or product.

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

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This presentation is protected by US and International Copyright laws ReproductionInternational Copyright laws. Reproduction,

distribution, display and use of the presentation without written permission of the speaker is

prohibitedprohibited.

© The Wood Products Council 2010© The Wood Products Council 2010

Learning Objectives

At the end of this program, participants will be able to:

1 Hi hli ht i t t hi h l d di h d ti li1. Highlight important high load diaphragm detialing

2. Explore sub diaphragm design techniques

3. Introduce the collective chord modification theoryy

4. Discover high load effects on diaphragm deflection

Overview

C l l ti M th d Calculation Methods Rigid vs. FlexibleHigh Load Diaphragm TableHigh Load Diaphragm TableSub-diaphragmsCollective Chord designCollective Chord designDiaphragm Deflection

Diaphragm Design Example

Rigid vs. Flexible Wood Diaphragms

Flexible, Rigid and Semi-Rigid Diaphragms

Flexible Diaphragm load is distributed to shear walls by

tributary area Rigid Diaphragm load is distributed to shear walls by wall

stiffness and torsionstiffness and torsion Semi-rigid Between flexible and rigid dependent on stiffness Between flexible and rigid, dependent on stiffness

Diaphragm (Plan View)

w

L/2 L/2

Flexible Diaphragm

w

sw

.25wL .25wL.50wLFlexible

di

L/2 L/2L/2 L/2

Rigid – All walls Identical

w

sw

.333wL .333wL.333wLRigid (noTorsion)Torsion)

L/2 L/2

Flexible vs. Rigid

w

2K 2KKStiffness

.25wL .25wL.50wLFlexible

Rigid (noTorsion)

.40wL .40wL.20wL

L/2 L/2L/2 L/2

Flexible, Rigid or Semi-Rigid

Which do you have? Prescribed flexible Calculated flexible Prescribed rigid Prescribed rigid Else, “semi-rigid”

Prescribed Flexible Diaphragm

In many cases wood diaphragms are permittedIn many cases wood diaphragms are permitted to be idealized as flexible

ASCE 7-05 Sec. 12.3.1.1 exempts one- and two-family dwellings from rigid diaphragm analysisdwellings from rigid diaphragm analysis.

CBC 2007 Sec 1613.6.1 adds following text to the ASCE provisions.


CBC 2007 Sec. 1613.6.1 Diaphragms constructed of wood structural panels….

Shall also be permitted to be idealized as flexible,Shall also be permitted to be idealized as flexible,provided all of the following conditions are met:


1 C t t i i t t l d i l th 1 51. Concrete topping is non-structural and is less than 1.5 in.

2 Each line of vertical elements of LFRS complies with2. Each line of vertical elements of LFRS complies with allowable story drift of ASCE7-05 Table 12.12-1

3 Vertical elements of LFRS are light framed walls3. Vertical elements of LFRS are light framed walls sheathed with wood structural panels or steel sheets

4. Cantilever portions of the diaphragm designed in4. Cantilever portions of the diaphragm designed in accordance with Sec. 2305.2.5

Calculated Flexible Diaphragm

ASCE 7-05 Sec. 12.3.1.3Diaphragms are permitted to be idealized as fl ibl hflexible when: The diaphragm deflection is more than two times the

average story drift of adjoining shear wallsaverage story drift of adjoining shear walls

2 DIAPHRAGM 2 x SHEARWALLS

Calculated Flexible Diaphragm

SHEARWALLS

(Average Deflection) SHEARWALLS

DIAPHRAGM

The longer the diaphragm the morediaphragm the more likely it is to calculate as flexible

Prescribed Rigid Wood Diaphragms (CBC 2305.2.5)

Open front Cantilevered diaphragmsg

Semi-Rigid Diaphragm

Semi-rigid results in force distribution somewhere between rigid and flexible Thus, an envelope approach can be used

where the both rigid and flexible models are used and the highest forces from each are selected

Deflections (4-term eqn’s)

Shear Wall (IBC §2305.3.2)

adbhhe

tGvh

EAbvh

n 75.08 3

Diaphragm (IBC §2305.2.2)

vv btGEAb

bX

LetG

vLEAbvL c

n

2)(

188.048

5 3

btGEAb vv 248

APA L350 (www.apawood.org) has comprehensive li ti f i t t d llisting of input parameters and examples

Deflections (4-term equations)

Diaphragm (CBC 2305.2.2)

bending shear nail slip chord connection slipTotal bb vv nn cc

bX

LetG

vLEAbvL c

n

2)(

188.048

5 3

SDPWS –

btGEAb vv 248

0.25 v L 1000Ga

SDPWS unblocked and bl k d1000Ga blocked

Deflections (4-term equations)

Shear Wall (CBC 2305.3.2)Total bb vv nn aaTotal bb vv nn aa

dhhevhvh 7508 3

bending shear nail slip anchorage slip

a

vv

db

hetGEAb

n 75.0

v h 1000Ga

SPDWS

APA L350 (www.apawood.org) has comprehensive listing of input parameters and examples

1000Ga

g p p p

Deflection (3-term eqn.)

Diaphragm (SDPWS §4.2.2)Diaphragm (SDPWS §4.2.2)

XvLvL c )(2505 3 W

XGvL

EAWvL c

a 2)(

100025.0

85

Ga values for blocked and unblocked diaphragms

Diaphragms and Shear Walls

Deflection of Unblocked Diaphragms is 2.5 times the deflection of blocked diaphragm.p g

If framing members are spaced more than 24”o.c., testing indicates further deflection increase of about 20% or 3 times the deflection offurther deflection increase of about 20%, or 3 times the deflection of a comparable blocked diaphragm. (This is based on limited testing of the diaphragm by APA)

Large & High Load Diaphragms How to design for lateral loads

Table 4.2B in SDPWS referenced in 2009 IBC

High Load Diaphragm Design Table 4.2B in SDPWS referenced in 2009 IBC Based on APA full scale testing APA report 138 ES 1952 now incorporated in code

3x normal diaphragm shear values 1800 plf ASD for seismic 1800 plf – ASD for seismic 2520 plf - ASD for wind

40% increase for wind loads All edges are blocked 8’-10’ panel width with purlins at each end

Utilizes multiple rows of nails

Fastener Pattern – Figure 4C in 2008 SDPWS for use with High-Load Diaphragm Table

4” nominal three lines

3” nominal two lines

Avoid Nail Splitting

2X4 3X4Slide provided by John Lawson, S.E., Kramer and Lawson

4.2.7.1 notesHigh-Load Diaphragm Table

Loads were limited by lumber splitting

High Load Diaphragm Table

Loads were limited by lumber splitting.

2 x 42 x 4

Clarification to High Load Diaphragm Table

Intermediate NailingMaximum spacing 12” o.c.Exception: 6” o c for spans greater thanException: 6 o.c for spans greater than

32” o.c.

Intermediate Member Size2x framing allowed at intermediate framing members where fasteners are 12” or 6” o cmembers where fasteners are 12 or 6 o.c.

Notes to High Load Diaphragm Table

Th h l i th t bl f 1 d The shear values in the table are for cases 1 and 2

The shear values are applicable to cases 3 4 5 The shear values are applicable to cases 3,4,5 and 6 provided fasteners at all continuous edges are spaced in accordance with the boundary fastener spacing

Diaphragm Layout Cases

Load Perpendicular to Cont. Edge

Load Parallel to Cont. Edge


Boundary, edge and intermediate nailing(case 1 and 2)

Continuous Panel Edge Nailing (Panel Edge Nailing- for case 1, 2Boundary Nailing for caseBoundary Nailing - for case3,4,5 and 6)

Boundary N ili

Intermediate Nailing

Nailing

Note: Framing omitted for clarityPanel Edge

Nailingg


B d d d i t di t iliBoundary, edge and intermediate nailing(case 3,4,5,6)

Boundary Nailing

Edge Nailing

(use boundary nailing at ti d tNailing

Field Nailing

continuous edge per note d.)

Nailing

Note: Framing omitted for clarity

Seismic Diaphragm-to-Wall Anchorage Forces

Sub-diaphragm Concept

Sub-diaphragm is a portion of a larger wood diaphragm designed to anchor and transfer localdiaphragm designed to anchor and transfer local forces to primary diaphragm struts and main diaphragm (2006 IBC 2302.1)p g ( )

Advantages

Eliminates the need for long-span design of walls forEliminates the need for long span design of walls for out-of-plane bending

Transfers anchorage forces to main members, thus reducing the number of connections required to fulfill continuous cross tie requirements.

Members used as cross-ties are typically better Members used as cross-ties are typically better suited for accommodating the necessary connections

Reduces cost – the larger the roof the greater are the savings provided by the use of sub-diaphragms.

How to design for lateral loads

Normal Diaphragm Design

Connections required foreach line of sub-purlins

Lateral Load

1800= 1800connections


Sub-Diaphragm Design

T i l l d t fTypical load transfer

Lateral Load = 102= 102

connections

Aspect ratio 2.5:1 max.


Normal Diaphragm Design

Connections required forh li f li

Lateral Load

each line of purlins

Typical Load Transfer

Subdiaphragm (Typical) – max aspect ratio = 2 5:1ratio = 2.5:1

Subdiaphragm is designed the same as a diaphragm

Sub-diaphragm Summary

f h h bd h f Use of the the subdiaphragm concept often reduces number of connectionsR d t f d f Reduces cost of wood roofs

APA document (Z350) provides connection details and has tables to aid the designerdetails and has tables to aid the designer

Reference

Examples: Sub-diaphragmsContinuous cross-tiesAnchorage details

APA Publication Z350

Reference

Examples: Diaphragms DesignSub-diaphragm DesignDeflection Calculations

APA Publication L350


W k ll ll d d t i b ildi

Traditional Chord Design

Works well on small and moderate size buildings

Lateral Load


M i l l b ildi

Collective Chord Design

More economical on large buildings Realistic way to model chord action

Lateral Load


Traditional Chord vs. Collective ChordBased on 8’ oc tie spacing

X Y Traditional Collective

6 2 kipsX 120’ 160’ 19 kips 6.2 kips

max

400’ 400’ 40 kips 4.5 kips max

Y

max

750’ 1100’ 211 kips 9.0 kips max

Results of Example done by Kramer and Lawson


Multiple Nailing Zones

E i t i l d ti Economizes on material and time Less nails Less nailing time

1

2

Less nailing time

2

34

2

1


T E ti t h f

Diaphragm Deflection Calculations

Two Equations to choose from 2006 IBC – traditional equation 2005 AF&PA NDS – simplified equation **suggested you 2005 AF&PA NDS simplified equation suggested you

use this equation** Collective Chord Modification Reduces diaphragm deflection calculations Complicates equation for moment of inertia See John Lawson’s paper for resulting equation See John Lawson s paper for resulting equation

Multiple nailing zonesMore accurate deflections when taken into account Using virtual work method, equation is derived for you in

John Lawson’s paper


Calculation Methods Resources/Examples

Hi h L d Di h CBC bl 2306 3 2 &High Load Diaphragm CBC table 2306.3.2 &Diaphragms and Shear Walls Design/Construction Guide -APA form L350A

Sub-Diaphragm Diaphragms and Shear Walls Design/Construction Guide -APA form L350ALateral Load Connections for Low-Slope R f Di h APA F N Z350ARoof Diaphragms – APA Form No. Z350A

Collective Chord “Thinking Outside the Box: New approaches to very large flexible diaphragm” by John LLawson

Diaphragm Deflection “Thinking Outside the Box: New approaches to very large flexible diaphragm” by John Lawson

High Load Diaphragm Design Example

Design Criteria

192’ x 120’ tilt-up building Panelized Roof System Panelized Roof System 8” – 30 ft high wall with 4 ft parapet Check for seismic load only Check for seismic load only Importance Factor – 1.0 Seismic Category – D (SS = 1 68 S1 = 0 6)Seismic Category D (SS 1.68, S1 0.6)

NOTE: The example is simplified to illustrate specific points and d t i l d ll l d bi ti d ll d i h kdoes not include all load combinations and all design checks otherwise required.

Design Process

Part A. Diaphragm Design

Diaphragm Loads (Seismic only)

Diaphragm Analysis (Transverse)


Structural Panel and Fastener Pattern S l ti (T )

Selection (Transverse)

Diaphragm Analysis (Longitudinal) Structural Panel and Fastener Pattern

Selection (Longitudinal)

( g )

Diaphragm Loads

Vertical LoadsDL Roof = 10 psfLL Roof = 30 psf

W ROOF= 192 x 120 x 10 = 230,400 lbs (16.3%)W WALL = (30/2+4) x 100 x 2 x (192 +120) = 1,185,600 lbs (83.7%)W TOTAL = 230,400 + 1,185,600 = 1,416,000 lbsp

DL Wall = 100 psfTOTAL , , , , ,

Seismic LoadsCS = SDS/(R/I)

V TOTAL = 0.28 x 1,416,000 = 396,480 lbs

V TRANSVERSE = (120 x 10 + 19 x 100 x 2) x 0.28 = 1,400 lbs (plf)CS SDS/(R/I) CS max = SD1/T(R/I)CS min = 0.5S1/T(R/I)SDS = 1.12

V LONGITUDINAL = (192 x 10 + 19 x 100 x 2) x 0.28 = 1,600 lbs (plf)

SDS 1.12SD1 = 0.6R = 4, I = 1SDC – Category DSDC Category D

CS =1.12/(4/1) = 0.28 > 0.01 CS max =0.6/.3(4/1) = 0.50>0.28 V = CS W=0.28WS max ( )CS min =0.5x0.6/(4/1) = 0.075<0.28

V CS W 0.28W

Diaphragm Loads (Transverse)

w = 1,400 plfcase 4

2

EA B C D192’

48’ 48’ 48’48’

case

148 48 4848

Purlin

Sub-purlin

2

3Girder

N

4

N

Diaphragm Layout Cases

CASE 4

SE

2C

A

(Case 4)


Load w:

w = 1400 plf(Case 4)

192’x

Shear V:

V 1 400 192 /2 134 400 lb v = 1 120 plfV max = 1,400 x 192 /2 = 134,400 lbsv max = 134,400 / 120’ = 1,120 plfv 40 = (134,400 – 40x1,400)/120 = 653 plfv 72 = (134,400 – 72x1,400)/120 = 280 plf

v max = 1,120 plf

v 40 = 653 plf

72 ( , , ) pv 72 = 280 plf

1,120 plf 653 plf @ 40’ 280 plf @ 72’

(Case 4)

A B C D E

(Case 4)

1

A B C D192’

48’ 48’ 48’48’

E

1

2

A D C AD

3N

4

1 120 lf ( 4)

High Load Diaphragm Table

v = 1,120plf (case 4)

2640/2=1320

653 lf ( 4)

High Load Diaphragm Table (2306.3.2)

v = 653 plf (case 4)

1340/2 = 670

The table gives shear values for Case 1 and 2. For cases 3,4,5,6 values are applicable providing fasteners at all continuous edges are spaced in accordance with boundary fastening spacing.

Panel and Nailing Pattern Selection

v max = 1,120 plf < 1320 plfv max = 134,400 / 120’ = 1,120 plf

19/32” R t d Sh thi E 1

A

case 2 and 4 19/32” Rated Sheathing Exposure 14x Framing3 rows of 10d Common Nails@ 4”, 4”, 12”

case 2 and 4

v 40 = 653 plf > 670 plfv 40 = (134,400 – 40x1,400)/120 = 653 plf

19/32” Rated Sheathing Exposure 1

B

case 2

3x Framing2 rows of 10d Common Nails@ 4”, 6”, 12” (adjusted 4”,4”,12” )

adjust edge spacing to 4” o.c.D

case 4

v 72 = (134,400 – 72x1,400)/120 = 280 plf v 72 = 280 plf < 320 plf

19/32” Rated Sheathing Exposure 12x Framing

Ccase 2 and 4

2x Framing1 row of 10d Common Nails@ 6”, 6”, 12”

Panel and Nailing Pattern Selection (Transverse)

A B C D192’

48’ 48’ 48’48’

E

1

A D C AD

2m

on N

ails

”, 4”

, 12”

12”

mon

Nai

ls

”, 4”

, 12”

3

min

gof

10d

Com

m4”

, 12”

amin

gs

of 1

0d @

4”

g 0d @

6”,

6”,

Nmin

gof

10d

Com

m4”

, 12”

amin

gs

of 1

0d @

4”

4

4x F

ram

3 ro

ws

@ 4

”, 4

3x F

ra2

row

s

2x F

ram

ing

1 ro

ws

of 1 N

4x F

ram

3 ro

ws

@ 4

”, 4

3x F

ra2

row

sCapacity 1,290 plf 650 plf 320 plf 1,290 plf650 plf

N

Diaphragm Loading (Longitudinal)

A B C D192’

E

N

case

2

1

48’ 48’ 48’48’ W= 1,600 plf

2

3

44

Diaphragm Analysis (Longitudinal)

Load w:

w = 1600 plf

120’x

Shear V:

V max = 1,600 x 120 /2 = 96,000 lbsv max = 96,000 / 192’ = 500 plf

(96 000 32 1 600)/192 233 lf

v max = 500 plf

v = 233 plfv 32 = (96,000 – 32x1,600)/192 = 233 plf v 32 = 233 plf

Panel & Fastener Pattern Selection

v max = 500 < 650 < 1,290 plfv max = 96,000 / 192’ = 500 plf

19/32” Rated Sheathing Exposure 1

B

19/32” Rated Sheathing Exposure 13x Framing2 rows of 10d Common Nails@ 4”, 6”, 12”

v 72 = 233 plf < 320 plfv 32 = (96,000 – 32x1,600)/192 = 233 plf

19/32” Rated Sheathing Exposure 12 F i

C

2x Framing1 row of 10d Common Nails@ 6”, 6”, 12”

Panel & Fastener Pattern Selection

A B C D192’

E

1

48’ 48’ 48’48’

3 F i

B

2C

3x Framing2 rows of 10d @ 4”, 6”, 12”

3


N4


B4

650 plf320 plfCapacity

Panel & Fastener Pattern Selection (Combined)

A B C D192’

E

1

192’48’ 48’ 48’48’

2


n N

ails

Nai

ls

12”

CD D


g 10d

Com

mon

2”v

g 0d C

omm

on2”

v

g 0d @

4”,

4”,

N

3

AA

3x Framing2 rows of 10d @ 4”, 6”, 12” 4x

Fra

min

g3

row

s of

1@

4”,

4”, 1

4x F

ram

ing

3 ro

ws

of 1

@ 4

”, 4”

, 12

3x F

ram

ing

2 ro

ws

of 1

B

4

AA B

High Load Diaphragm Fastener Pattern

Boundaries

Intermediate

Other edges

High Load Diaphragm Fastener Pattern

BD

4” or 6”

3” nominal two lines

4 or 6

A

4”

4” nominal three lines

Design Example (Continued)

Part B. Wall to Diaphragm Anchorage

Anchorage Forces (Seismic only)

Sub diaphragm Analysis and Design (E

Sub-diaphragm Analysis and Design (E-

W)

W ll h t S b li (E W)

Wall anchorage to Sub-purlin (E-W)

Cross-tie Load Transfer (E-W) Cross-tie Load Transfer (N-S)

Wall Anchorage Force

FP= 0.8 I SDS wp (ASCE 7-05 equation 12.11-1 )

FP= 0.80 x 1.0 x 1.12 x wp = .90FP= 0.90 x (100 x 19) = 1,734 plf > 400x1.12 > 280 plf

wp = 100 x 34 x 17/30 = 1,927plf

L SUB E-W = 1,734 x 20/1290 = 27 ft

Sub-diaphragm Depth

L SUB E-W = 1,734 x 4/1290= 5.4 ftL SUB N-S = 1,734 x 8/650 = 22 ft < 40 ft (girder spacing)

L SUB E-W = 1,734 x 8/1290= 11 ft

E-W USE: 32 ft wide sub-diaphragm

F TIE = 1,734x 4 = 6,940 lbs F TIE = 1,734 x 8 = 13,872 lbs

F TIE = 1,734 x 40 = 69,360 lbs

TIE , ,

w = 1 734 plf

Sub-diaphragm Design (E-W)w 1,734 plf

1w = 1,734 plf

1

22

3

4

192’

48’ 48’ 48’48’ NEA B C D

192’

Sub-diaphragm Design (E-W)

ll l d f h f lf Wall load for anchorage force = 1, 734 plf

Length-to-width = 40/32 = 1.25 < 2½ (o.k.)Length to width 40/32 1.25 < 2½ (o.k.)

Subdiaphragm Shear 1,084 plf < 1,290 plf

(v=(wl/2)/width =1,734x20/32 = 1,084) main diaphragm sheathing/nailing is adequate for subdiaphragm

Maximum chord force = 10,834 lb

(T = wl2/8x32 = 1,734 x 402/(8x32), important to check combined tension-bending)

Wall Subpurlin Anchorage (E-W)

b l ’ h b l Subpurlins @ 2’ oc, use every other subpurlin to transfer wall forces into the sub-diaphragm (wall

d t b h k d f b di b tneed not be checked for bending between anchors)

6,940 lb per subpurlin anchorF TIE = 1,734x 4 = 6,940 lbs

Continuous Cross Ties

1

2

Large number of connections are

3

connections are required for just one line of sub-purlins

4

Fewer connections are required for one line of purlins.

4

192’

48’ 48’ 48’48’

EA B C D

192’

Sub-diaphragm Load Transfer (E-W)

Continuity Ties

Typical Sub-diaphragm

192’

48’ 48’ 48’48’

EA B C D

192’


6,940 lbs

lf1,734x18/32=976 plf

Subdiaphragm(Case 2 E W direction)

40'

1,73

4 pl

(Case 2 E-W direction)1

6,940 lbs

32'


Ties at 4’-0” o.c.

Sub-purlin to Wall

Sub-purlin to Sub-purlin connection

Connection

Wall-to-Subpurlin Connection (Design for 7,000 lbs)

APA wood structuralTack weld hanger orpanel sheathing Tack weld hanger orprovide Pneutek pins.

Subpurlin

Diaphragm to wall anchorage

Add steel box to hangerfor compressive stress

cret

e or

wal

l

Diaphragm to wall anchorage using embedded straps shall be attached to or hooked around the reinforcing steel or terminated so as to directly transfer force to the

Steel channel Con

cC

MUas to directly transfer force to the

reinforcing steel. (ASCE 7-05 12.11.2.2.5)

Anchorage Element Design

Strength design forces for steel elements of the wall anchorage system shall be 1.4 times the force otherwise required by this section

(ASCE 7-05 12.11.2.2.2)(ASCE 7 05 12.11.2.2.2)

Subpurlin-to-Subpurlin Continuity Tie Connection

PurlinStrap installed oversheathing (not shown)

Subpurlin

Plan

Anchorage (E-W) Wall-to-Girder

4’ tributary area, same force as wall-to-subpurlin connection

6,940 lb per subpurlin anchor

Design for 7 000 lbsWall-to-Girder Connection

Ledger/diaphragm chord (shown behind)

APA wood structural

Design for 7,000 lbs

APA wood structural panel sheathing

Concrete orCMU llCMU wall

Girder(glulam shown)

Design for 7 000 lbsWall-to-Girder Connection

APA Wood Structural


APA Wood Structural Panel Sheathing

GirderT t d Girder (glulam shown)

Top mounted hanger

Continuous Girder Ties (E-W)

Continuity Ties

Typical Sub-diaphragm

192’

48’ 48’ 48’48’

EA B C D

192’


976 plf6,940 lbs

69,400 lbs

f

1,734x18/24=976 plf

Subdiaphragm

40'

,734

plf

(Case 2 E-W direction)1

69,400 lbs

6,940 lbs 976 plf

32'


A l d i t f d i t th i d f th As load is transferred into the girder from the subdiaphragm, the axial load in the girder increase from 6,940 lb to 69,400 lb, ,

The girder-to-girder connection must resist 69,400 lb

976 x 32 = 27,360,

976 x 32 = 27,360

69,400 lb6,940 lb

9 6 3 ,360

6,940 lb69,400 lb

,

Girder-to-Girder Connection


Girder (glulam beam shown)

Wood structural panel sheathing not shown for clarity

Hanger

Tension ties on both sides of girder

Hanger

50,000 lbs. Use (10) 3/4" diameter bolts75,000 lbs. Use (12) 1" diameter bolts

Continuous Purlin Ties (N-S)

8 ‘ typical N-S continuity ties located at each purlin line (Typ.)

13,900 lb 1,734x8 =13,872 lb

Wall-to-Purlin Connection


Inserts to 6" wide tension tieembossed to go over hanger

APA wood structural

panel sheathingprovide approx. 12K

over hanger panel sheathing

Top-mounthanger Glulam purling

Full length steel channel

p

Elevation

Purlin-to-Purlin Continuity Tie Connection

APA wood structural panel sheathing

Purlin (Typ.)Purlin (Typ.)

Elevation

Wood structural panel sheathing not shown for clarity

Plan

Course Evaluations

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We’re here to answer your yquestions.

This concludes The AmericanThis concludes The American Institute of Architects Continuing

Education Systems Course

WoodWorks!Lisa Podesto, P.E.

Office: 530.596.4031Cell: 530.520.7966

[email protected] woodworks orgwww.woodworks.org

The Wood Products Council The AIA/CES High Load … · Diaphragm Design Example Rigid vs. Flexible...

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