The Analysis of Offset Diaphragms and Shear...
47
By: R. Terry Malone, PE, SE Senior Technical Director Architectural & Engineering Solutions [email protected] Presentation updated to 2012 IBC, ASCE 7-10 Copyright McGraw-Hill, ICC and the Wood Products Council Presentation based on: Webinar Presentation, June 5, 2014: The Analysis of Offset Diaphragms and Shear Walls
Transcript of The Analysis of Offset Diaphragms and Shear...
By: R. Terry Malone, PE, SESenior Technical DirectorArchitectural & Engineering [email protected]
Presentation updated to 2012 IBC, ASCE 7-10Copyright McGraw-Hill, ICCand the Wood Products Council
Presentation based on:Webinar Presentation, June 5, 2014:
The Analysis of Offset Diaphragms and Shear Walls
Outline / Learning Objectives
• Basic InformationDiscuss boundary elements, complete lateral resisting load path requirements, related code sections, and how to establish complete lateral load paths across areas of discontinuity.
• Discuss an analytical method used for the analysisDiscuss an analytical method used for solving complex diaphragms and shear walls using “Transfer Diaphragms” and the “Visual Shear Transfer Method.”
• Horizontally offset Diaphragms Review the analysis of flexible offset diaphragms consisting of structural wood panel sheathing or un-topped steel decking.
• In-plane and out-of-plane offset shear walls Review the analysis and effects of horizontally out-of-plane and in-plane offset shear walls.
http://www.woodworks.org/education-publications/research-papers/#
Complete Example with narrative and calculations
Marselle Condominiums 5 stories of wood over 6 stories concrete Structural Engineer engineer: Yu & Trochalakis, PLLC (podium) 2 above gradePhotographer: Matt Todd Photographer
Mid-rise Multi-family
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
Presentation AssumptionsFlexible wood sheathed or un-topped steel deck diaphragms
(Can also apply to semi-rigid and rigid diaphragms)
• 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 combination.
Code and Standards: • ASCE 7-10 “Minimum Design Loads for Buildings and Other Structures”• 2012 IBC• 2008 SDPWS-”Special Design Provisions for Wind and Seismic”
Analysis and Design references:
• The Analysis of Irregular Shaped Structures: Diaphragms and Shear Walls-Malone, Rice
• Woodworks-The Analysis of Irregular Shaped Diaphragms• Design of Wood Structures- Breyer, Fridley, Pollock, Cobeen• SEAOC Seismic Design Manual, Volume 2• Wood Engineering and Construction Handbook-Faherty, Williamson• Guide to the Design of Diaphragms, Chords and Collectors-NCSEA
• Boundary Elements • Complete Load Paths• Method of Analysis
Basic Information
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 without a collector or a complete alternate load path are NOT ALLOWED!
Diaphragm Boundary Elements
SW1
SW2
SW3
Note: All edges of a diaphragm shall be supported by a boundary element.
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 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.
Required for Seismic and wind
1 2
B
3
C
A
Strut/chord
Open
3
4
5
21
FE
D
C
B
6 9 107 8
Strut/chord
Stru
t
Stru
t (ty
p.)
Strutchord
Strut chord
Strut /chord
Stru
t
Strut/chord
Strut/chord
SW1
SW5
SW2SW3
SW6
SW4
Stru
t
MR
F1
Multiple offset diaphragm
Offset strut
Support Support
Col
lect
or
Collector
Collector(typ.)
Collector(typ.)
Collector (typ.)
Col
lect
or (t
yp.)
A
Complete Continuous Lateral Load Paths
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-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?
Strut/chord
Open
3
4
5
21
FE
D
C
B
6 9 107 8
Strut/chord
Stru
t
Stru
t (ty
p.)
Strutchord
Strut chord
Strut /chord
Stru
t
Strut/chord
Strut/chord
SW1
SW5
SW2SW3
SW6
SW4
Stru
t
MR
F1
Support Support
Col
lect
or
Collector
Collector(typ.)
Collector(typ.)
Collector (typ.)
Discontinuousdiaphragm chord/strut
Discontinuousdiaphragm chord
Discont.diaph. chord
Discont.diaphragm chord
Discont.diaphragm chord
Col
lect
or (t
yp.)
A
Complete Continuous Lateral Load Paths
ASCE7-10 Section 1.4-Complete load paths are required including members and their splice connections
Strut/chord
Open
3
4
5
21
FE
D
C
B
6 9 107 8
Strut/chord
Stru
t
Stru
t (ty
p.)
Strutchord
Strut chord
Strut /chord
Stru
t
Strut/chord
Strut/chord
SW1
SW5
SW2SW3
SW6
SW4
Stru
t
MR
F1
Offset shear wallsand struts
Support Support
Col
lect
or
Collector
Collector(typ.)
Collector(typ.)
Collector (typ.)
Offset shear walls
Col
lect
or (t
yp.)
A
Complete Continuous Lateral Load Paths
ASCE7-10 Section 1.4-Complete load paths are required including member and their splice connections
Strut/chord
Open
3
4
5
21
FE
D
C
B
6 9 107 8
Strut/chord
Stru
t
Stru
t (ty
p.)
Strutchord
Strut chord
Strut /chord
Stru
t
Strut/chord
Strut/chord
SW1
SW5
SW2SW3
SW6
SW4
Stru
t
MR
F1
Openingin diaph.
Support Support
Col
lect
or
Collector
Collector(typ.)
Collector(typ.)
Collector (typ.)
Col
lect
or (t
yp.)
Vertical offset indiaphragm
A
Complete Continuous Lateral Load Paths
Design: • 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.
• Boundary Elements • Complete Load Paths• Method of Analysis
Basic Information
Method of AnalysisThe Visual Shear Transfer Method
+ -
+
Positive Direction
+ -
Transverse Direction (shown)
Lds.
Shears Applied to Sheathing Elements
FY
FX
+M
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)
+
-
+-
+ -
-+
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
(support)
(support)
Chord
Chord
TD1TD Ratio=4:1
Maximum 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.
Transfer Diaphragm
• sub-diaphragm• Transfers local forces out to primary
chords/struts of the main diaphragm.
What doesthis mean?
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.(Wrong!)
• The collector is often checked for tension only. (Wrong!) Compression is rarely checked.
Typical calloutCMST14 tie strap x 10’-0” with (xx) 10d nails over 2x flat blocking. Lap 2-8” onto wall.
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.
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
SW
Col
lect
or
T 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
Basic ProcedureMethod by Edward F. Diekmann
+500plf
Main chord
Main chord
Disruptedchord
Basic diaph.shear TD shears
vnet= 300 + (250) = 550 plfNet shear
The transfer diaph.aspect Ratio should be similar to the main diaphragm.
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 on collector
325 plf 325 plf
Net directionof shears actingon 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 forceon collector
B
2 3
Net shear
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
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
Diaphragms with Horizontal Offsets
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
RL lb
200 plf
25’
F2A
Free body for F2B
1 2
A
B
∑M=035’
M2B ft.-lbF2B lb
Support RL lb
RR lb
Example 1-Diaphragm with Horizontal End OffsetTransverse Loading
V2 lb
V3 lb
A/R=2.5:1
Support
SW 1
SW 2
Diaph.C.L.
25’ 20’
15’
80’
V=Basic shear- TD shear plf
F2Bv= DTD
Vc plf
VA lb
VC lbv= Net shear> 2x
Basic shear
v2R plf v2L plf
v1 plf
v3 plf
1 2
A
B
3
C
Basic shear diagram
Net TD shears (basic shear +/- transfer diaph. shears)
No net change Net change occurs in TD
No net change
35’
Transfer Diaphragm and Net Diaphragm Shear
+ -
Sign Convention
4
15’
35’
RL lb
RR lb
(Net shear)
Pos.
Neg
.
v4 plf
TD s
hear
dia
gram
F2B
v= DTDVA plf
SW 1SW 2
Diaph.C.L.
25’ 20’
15’
F3C
FCL
C
T F2B
T
F2A
1 2
A
B
3
C
F2B
+v2
-v3x1’
x2’+F lb-F lb
Longitudinal Chord Force Diagrams
0 plf
+ -
Sign Convention4
17.5’
v1 v2L
Support
Support
0 plf
FCL
F2B
F3C
SW 1
25’ 20’
15’
F=6000 lb
F=6000 lb(this is not an insignificant force.)
F3B
F3B
T C
1 2
A
B
3
C
Transverse Collector Force Diagrams
+ -
Sign Convention
Net shear Net shear
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
TD135’
50’
DiscontinuousDrag strut
1 2
A
B
3
C
4
10’5’
15’
200
plf
Diaphragm 2Diaphragm 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 OffsetLongitudinal 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
VB
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
Offset Shear Walls
SW 1
SW 2
Col
lect
or
Collector
Collector
Out-of-plane Offsets In-plane Offsets
Out-of-Plane Offset Shear WallsAssumed to act in the Same Line of Resistance
Loads
Col
lect
or
Collector
Collector
SW
SW
Transfer area
Offset
Discont.drag strut
Dra
g st
rut
Dra
g st
rut
SW
Col
lect
or
Collector
Collector
• 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 unless provisions for force transfer around the offset are provided.
Check for Type 2 horizontal irregularityRe-entrant corner irregularity
Cant.
Longitudinal Loading
Offset transverse walls In-line transverse walls
No exteriorShear walls ??
Flexible, semi-rigid, or rigid???
TD3
TD1
TD2
SW4
SW3
SW5
SW2
SW1
Diaphragm stiffness changes
Higher shears and nailing requirements
Multi Story, Multi-family Wood Structure
1
2
3
Loads
I1 I2 I3
SW 1
SW 2
25’ 20’
15’
200
plf
Drag strut
Cho
rd
colle
ctor
s
Cho
rd
80’
35’50’
Drag strut is discontinuous
1 2
A
B
3
C
4
10’25’ 5’
15’
200
plf
TD1C
hord
co
llect
ors
Pos. direction + -
SW 3 SW 4
8’
8’
Drag strut
Collector
5’
12’45’
80’
Drag strut
Drag strutC
hord
Drag strut
Example 3-Diaphragm with Horizontal End OffsetLongitudinal Loading-Offset Shear Walls
Assumptions:1. Assume shear walls at grid lines B and C
act along the same line of lateral-force-resistance.
2. Assume the total load distributed to grid lines A and B/C= wL/2 .
Offset SW
Offset SW
Support
Support
SW 2
35’
50’
1 2
A
B
3
C
10’
8’
Pos. direction + -
8’
2
SW 3
Pos.
40 plf 160 plf
200
plf
200
plf
SW 1
Vnet=Vsw-Vdiaph
FendF2B
F2B
VA
VB
Basic Diaphragm Shears and Transfer Diaphragm ShearN
eg.
200 plf
Vsw2=wL/2 plf∑ Vsw1, sw3, sw4=wL/2, vsw=∑V1,3,4/(Lsw1+Lsw3+Lsw4) plf
Determine Force transferred Into Transfer Diaphragm
Total Shear to Shear Walls (Assumed)
17’
SW 4
15’
F2B
VA
VC
VA
VC
VB
25’ 20’
15’
80’
1 2
A
B
3
C
No net change Net change in TD
No net change
35’
Net Diaphragm Shears
8’ SW 1
15’SW 3 SW 4
SW 2
Pos. direction + -
50’
Vnet sw
Pos.
Neg
.
8’
4
V=Basic shear- TD shear plf
(Net shear)
Vnet sw
Vnet sw Vnet swV=Basic shear- TD shear plf
(Net shear)
VA
VC
VA
VC
VB
15’
-
F2B
F2B
1
2
A
B
3
C
Transverse Collector Force Diagrams
8’ SW 18’ SW3 15’ SW 4
Pos. direction + -
SW 2
Net shear
Special nailing
So far, so good
Net shear (TD tension chord and Diaph.2 compression Chord)
Net shear-vB
+vC
x1’
x2’
+F
-F
F3B
F3B
Diaphragm 2
25’ 20’
15’
80’
F= F2B
Fstart
F3C
1 2
A
B
3
C
Longitudinal Strut Force Diagrams
SW 4
Pos. direction + -
SW 2
15’
Fend
F2B
F4C = -xxx lb (Error)
F2AFend
F3A
F4A=+xxx lb (Error)
Note: Neither force diagram closes to zero, therefore error.Notice that they do not close by the same amount.
8’
12’
8’
45’
5’ 5’10’
80’
SW 1
SW 3
Vnet sw
Vnet swVnet sw
Vnet sw
FstartFend
Fstart
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
Pos. direction + -
SW 2
5400 lb
Line needs to move in this direction
The shear wall shearsneed 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
In-plane Offset Shear Walls
Wd
Hdr
Sill
Nail shtg. To each 2x stud
No hold down(option 1)Hold-down(option 2)
Blk’g. or rim joist
Section does not comply with the required aspect ratio for a perforated or FTO shear wall.
4’
8’
8’
8’
12’
1’
2000 lb
3000 lb
Hdr/collector
Wd
Sill
DL=150 plf
DL=250 plf
SW2
SW1
6’
Example 4-In-plane Offset Segmented Shear Wall-with Gravity Loads
1’
Col
lect
or
VHdr=960 lb
VHdr=450 lb
A B C D
ASCE 7 Table 12.3-2-Type 4 vertical irreg.- in-plane discontinuity in the LFRS
Ends of wall panels do not line up. Requires special nailing of sheathing into stud below.
Nailing found in field was 12” o.c.
Requires same number of studs above and belowwith boundary nailing each stud Solid blocking
required
Photo-In-plane Offset Segmented Shear Walls
No hold-down below
Hold down
Hold down
Wall and Transfer Diaphragm Shears
5000 lb
8’
8’4’
2920 lb1080 lb
++
2000 lb
8’
8’
+Pos. direction -
5000 lb1’
12’416.67 lb416.67 lb
416.67416.673370
1080
Upper Shear Wall
Rim joist
135 365
Col
lect
or
1080 lb330 plf (incl. wall DL)
Lower Shear Wall
TD shears-lbs. (plf)
w=230 plf (incl. wall DL)
+- 8’4’
SW2
SW1
+250
plf
+416
.67
plf
1260(-157.5)
1020(-127.5)
60(+7.5)
1620(+202)
2490 lb 9700 lb
+ ++
+
1 2
+424.1
+259.2
+618.7+289.2
VHdr=450 lb
VHdr=960 lb
1260 lb(-157.5)
1620 lb(+202)
Sign Convention
+ shears
- shears
Aver.=250 plf +
+450 lb3370 lb
Bas
ic S
hear
+T
T
++
+
2000 lb
3000 lb1’
8’
8’
8’4’
C
+Pos. direction -
Vertical Collector Forces
2nd floor Rim joist
++ SW1
SW2D
epth
TD
1 2
+424.2
+259.2
+618.7
135 365
+289.2
Roof
+
T
C
++
+
2000 lb
3000 lb
8’
1’
8’
8’
8’4’ C
2nd floor Rim joist
++ SW1
SW2
Dep
th T
D
1 2
+424.2
+259.2
+618.7
135 365
+289.2
Roof C
Horizontal Collector Forces
T
C
2490 lb 2490 lb
8’
416.67
365(8)+450=3370
9700 lb
VHdr=990 lb
416.67
Sign ConventionCollector Force Diagrams
9700 lb
Questions?This concludes
The National Council of Structural Engineers Association Webinar Presentation-June 5, 2014
The Analysis of Offset Diaphragms and Shear Walls
R. Terry Malone, P.E., S.E.Senior Technical DirectorWoodWorks.orgPrescott Valley, Arizona
Contact Information:Technical or design assistance - [email protected] Book related questions - [email protected]
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