mid-rise shear walls & diaphragms
TRANSCRIPT
By: R. Terry Malone, PE, SE
Senior Technical Director
Architectural & Engineering Solutions
928-775-9119
Mid-rise Shear Walls
& Diaphragms
Avalon Bay Communities
Photo credit: Arden Photography
120 Union, San Diego, CATogawa Smith Martin
Copyright McGraw-Hill, ICC
Portions Based On:
Course Description
This presentation will provide an in-depth
look at shear wall and diaphragm designs
for low-rise and mid-rise buildings,
including those with rectangular, offset
and open-front plans. This presentation is
intended for structural engineers.
Mid-rise Codes
and Standards
Learning Objectives
Learning Objective 1
Discuss the different types of shear walls commonly used in low-rise
and mid-rise structures.
Learning Objective 2
Discuss complete load paths and the detailing required to maintain
lateral load paths to shear walls.
Learning Objective 3
Review shear wall code requirements.
Learning Objective 4
Discuss the methods available to analyze tall shear walls and discover
their effects on mid-rise structures.
Holes in diaphragm
(web discontinuity)Interconnection
ties?Diaphragm boundary offset
from shear wall (cantilever).
Large cantilevered
diaphragm
Irregular shaped shear wall
with opening (discontinuous)
Soft story
Continuous Load Paths?????
A quick note on residential projects
Presentation Topics
• Shear Wall Types
• Shear wall Anchorage
• SDPWS Code Requirements
• Complete Load Paths
• Offset Shear Walls and Relevant Code
Requirements
• Mid-rise Diaphragms and Tall Shear Walls
Shear Wall Configuration Options
Perforated WallsForce Transfer Around
Openings Walls
Segmented or
braced Walls
(IBC/IRC)
(No openings)
Wall
DL
V
v(plf)
Sh
ear
wall c
ho
rd
Sh
ear
wall c
ho
rd
h
Resist. mom. Arm 1C.L.
bearingC
C.L.
A.B.
T
b
Sheathing
joint as
occurs
2x blocking
as required
at joints.
DLhdr2DLhdr1
Anchor bolts
or nailing Hold down
anchor
V
v(plf)
WDL plf
tstuds
Sh
ear
wall c
ho
rd
Sh
ear
wall c
ho
rd
h1
Resist. mom. Arm 1
C
C.L.
A.B.
T b
Tstud + CLC.L.
bearing
Chord or
Strut
Chord
or Strut h2Wall
DL
DL hdr. DL hdr.
412
C.L.
SW
v
diaphragm shear
0M
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)
0M
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
4.3.4.2For 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
Wall
DL
V or F
Sh
ear
wall c
ho
rd
Sh
ear
wall c
ho
rd
h
C
T b
WDL plf
Segmented Shear Wall Deflection and Stiffness∆𝑺𝑾
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
∆𝒔𝒘 =𝟖
𝒗
𝑪𝒖𝒃𝒉𝟑
𝑬𝑨𝑩+
𝒗
𝑪𝒖𝒃𝒉
𝟏𝟎𝟎𝟎𝑮𝒂+
𝒉
𝒃∆𝒂
ATS
Wall stiffness k = 𝑭
∆
ATS
CX
∆𝑺𝑾=𝟖𝒗𝒉𝟑
𝑬𝑨𝒃+
𝒗𝒉
𝟏𝟎𝟎𝟎𝑮𝒂+ 𝒉∆𝒂
𝒃
Vertical elongation
• Device elongation
• Rod elongation
Bending
Apparent shear stiffness
• Nail slip
• Panel shear deformation
SDPWS 3 term deflection equation
4.3-1
∆𝑺𝑾=𝟖𝒗𝒉𝟑
𝑬𝑨𝒃+
𝒗𝒉
𝑮𝒗𝒕𝒗+𝟎. 𝟕𝟓𝒉𝒆𝒏+
𝒉∆𝒂
𝒃
Bending
Shear
Traditional 4 term deflection equation
C4.3.2-1
Nail slip
SDPWS combines
Rod elongation
(Wall rotation)
Segmented Shear Wall Deflection
SDPWS linear
3 term equation
Identical at 1.4 ASD
ASD unit shear
Traditional
4 term equation
Displacement, inches
Lo
ad
, p
lf
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 + uplift
Table 4.4.2-uplift only
VASD=Vn/2, Vstr=0.65Vn
Edge
nailing
Field nailing
12” 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
. . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Single row of
Fasteners
Double row of
Fasteners
Splices at Rim Joist or
Common Horizontal Member
See Table 4.4.1.6
for maximum anchor
bolt spacing
0.229x3x3 sq. plate
washer required
Uplift clips or anchorage
shall be on same side as
sheathing
. .
.
. .
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
.
Spacing ≥ 6”≥ 3”
3/4”
1/2”
1/2”
Increase
nailing for
uplift ea.
side
No Horizontal Sheathing Joint Over Studs
Nail spacing at common horizontal
framing > 3″ single row or > 6″ double
row (4.4.1.7 (1) and Fig. 4H)
. .
. .
. .
.
.
. .
.
. . . . . .
. . . . . .
Single row of Fasteners
Splices at Studs
w/ Blocking
Section
See Table 4.4.1.6
for maximum anchor
bolt spacing
0.229x3x3 sq. plate
washer required
Uplift clips or anchorage
shall be on same side as
sheathingSpacing ≥ 3”
Increase
nailing for
uplift ea.
side
3/4”
2x blocking
Sheathing splice
plate, same thk.
and face grain
orientation as
sheathing
Horizontal Sheathing Joint Over Studs
• Blocking to resist
shear, stud nailing
to resist uplift
• Sheathing Tension
splice over Studs
If resisting both shear
and uplift
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
vmax
Tint
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
wall
seg
men
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
wall
seg
men
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.
Sheathing
area 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.
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. o
pen
ing
heig
ht
vmax
Tint
V
Wall
segment
Wall
segment
Wall
segment
h m
ax
=20 f
t
Li (typ) Li (typ)Li (typ)
Vallow = (v Tabular ) x (CO) x ∑Li
Where:
Per SDPWS
Table 4.3.3.5𝑪𝟎 =
𝒓
𝟑 − 𝟐𝒓
𝑳𝒕𝒐𝒕𝚺𝑳𝒊
𝒓 =𝟏
𝟏 +𝑨𝟎𝒉𝚺𝑳𝒊
A0 = total area of openings
Hold down at ends
𝒗𝒎𝒂𝒙 = 𝑽
𝑪𝒐 ∑𝑳𝒊
𝒗𝒎𝒂𝒙 = 𝑻𝒊𝒏𝒕 =𝑽
𝑪𝟎𝚺𝑳𝒊
𝑻 = 𝑪 =𝑽𝒉
𝑪𝟎𝚺𝑳𝑰
At full-hgt. segments
CCC
C
C
AWC Section 4.3.6.1.3
Each 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.
∆𝑺𝑾=𝟖𝒗𝒉𝟑
𝑬𝑨𝒃+
𝒗𝒉
𝟏𝟎𝟎𝟎𝑮𝒂+ 𝒉∆𝒂
𝒃
vmax ∑Li
Open’g
Hdr
Max. o
pen
ing
heig
ht
vmax
V
Wall
segment
Wall
segment
h m
ax
=20 f
t
Li (typ)Li (typ)
∆𝑺𝑾
3’ 6’ 5.5’
14.5’
4’
3’
2’
1 2
A
B
3
C
D
4
9’
4500 lb
w=200 plf
(See recent Testing-APA Form M410 and SR-105)
Tie straps full
length of wall
per SDPWS
section 4.3.5.2
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 panels
or blocking
(Diekmann)-Vierendeel Truss/Frame
(b) Force Transfer Around Opening
Wall
pier
Wall
pier
Wall
pier
Wall
pier Wall
pier
Wall P
ier
heig
ht
Wall p
ier
heig
ht
Wall
pier
width
Cle
ar
heig
ht
Wall p
ier
heig
ht
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.
• 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 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:
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
C
A B
D
F
E
HG
I
L
J
K
T
V
V
M
M
MM
V
V
F=0 lb
F=0 lbF=0 lb
M
F=0
MF=0
Force Transfer Methodology (Diekmann)-Vierendeel Truss/Frame
F=0 lb
F
F
Gravity loads
to wall
1343.1 4243.1
w=200 plf
N.A.
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 plf
0M
0M
B.5
B.5
2691.38 lb
F4B.5
3912.92 lb
VB.5
BA
C
A
I
G
B
H
C
V2.53’
3’
Free-body of Upper Half and Upper Left Section
Point of
inflection
I
H
B.5
A
C
D
B
F
587.08
VB=587.08
VB.5=587.08
(587.08)
Vc=587.08
587.08
V2
.5
15
51
.7
V2
(1343.1)
3’
3’
F2b(V)
391.39
F2C(v)
391.399
91
.38
V2
39
1.3
9
(208.62)
F2A
F2B(H)
1381
1037.1
1160.3
F2C(H)
F1B
600
F1C
182.77
E
(0)
(0)
573.23
2’
2’
2’
1 2
0M
0M0M
0M
0M
0M
Units are in lb
Corner tie
strap force
Corner tie
strap force
HG
J
I
KL
15
91
.38
(99
1.3
8)
V3
268.8
1937.1
15
51
.7
15
51
.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
F4B
1268.5
F4C
4114.3F3C(H)
F3B(H)
F3A A
B
3
C
D
4
0M
0M
0M
3’ 3’
3’ 3’
2.5
B.5B.5
2124.9
1551.7
(4500)
(4500)
F2.5A
1274.9
(xxx) Shears and forces determined
in previous step.
0M
0M 0M
200 plf 200 plf 200 plf 200 plf
F2.5C
Resultant Forces on Wall Segments
0V
0H
0V
0V
0V0V
0H
0V
0H
1551.7
1706.9931931931
(0)
Model was based on anchor
bolts at 2’ o.c. and hold downs
at each end.
Notice inflection point and
frame like action
Axial Force DiagramShear Stresses
FEA Results
(See recent Testing-APA Form M410)
Advancements in FTAO Shear Wall Analysis
Basis of APA System Report
SR-105 (in development)
SEAOC Convention 2015
Proceedings
Advancements in Force Transfer Around
Openings for Wood Framed Shear Walls
APA System Report SR-105
(Report pending)
Refine 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.
20’-0”
L
5’-0”
L1
4’-0”
L22’-6”
L3
6’-0”
LO1
2’-6”
LO2
2’-
8”
ho
8’-
0”
H
V=4000 lbs.
1’-
4”
ha
4’-
0”
hb
1600
1600
Calculate O/T forces:
𝑭𝑶/𝑻=𝟒𝟎𝟎𝟎(𝟖)
𝟐𝟎= 1600 lbs.
Pier 1 Pier 2 Pier 3
Example Problem
(Note: No dead loads)
V=4000
20’-0”
5’-0” 4’-0” 2’-6”
6’-0” 2’-6”
2’-
8”
1600
1600
1’-
4”
4’-
0”
Calculate:
• Aspect ratios of pier sections
𝑷𝒊𝒆𝒓 𝟏 =𝒉
𝒅=
𝟐.𝟔𝟕
𝟓= 0.534 < 2:1
𝑷𝒊𝒆𝒓 𝟐 =𝟐.𝟔𝟕
𝟒= 0.668 < 2:1
𝑷𝒊𝒆𝒓 𝟑 =𝟐.𝟔𝟕
𝟐.𝟓= 1.07 < 2:1
𝑯𝒆𝒂𝒅𝒆𝒓 𝟏 =𝟔
𝟏.𝟑𝟑= 4.51 > 3.5:1
𝑯𝒆𝒂𝒅𝒆𝒓 𝟐 =𝟐.𝟓
𝟏.𝟑𝟑= 1.88 < 2:1
𝑺𝒊𝒍𝒍 𝟏 =𝟔
𝟒= 1.5 < 2:1
𝑺𝒊𝒍𝒍 𝟐 =𝟐.𝟓
𝟒= 0.625 < 2:1
.534:1 .668:1 1.07:1
4.51:1 1.88:1
.625:11.5:1
Pier 1 Pier 2 Pier 3
V=4000
20’-0”
5’-0” 4’-0” 2’-6”6’-0” 2’-6”
2’-
8”
8’-
0”
1600
1600
1’-
4”
4’-
0”
300 p
lf
400
300 p
lf
1200
Calculate:
• Aspect ratios of pier sections
• Unit shear above and below
opening
𝑽𝒉𝒅𝒓𝟏 =𝑭𝒐/𝒕(𝒉𝒂)
(𝒉𝒂+𝒉𝒃)=
𝟏𝟔𝟎𝟎(𝟏.𝟑𝟑𝟑)
𝟓.𝟑𝟑
= 400 lbs.
vhdr2 = 𝟒𝟎𝟎
𝟏.𝟑𝟑= 300 plf
𝑽𝒔𝒊𝒍𝒍𝟏 =𝟏𝟔𝟎𝟎(𝟒)
𝟓.𝟑𝟑= 1200 lbs.
vsill2 = 𝟏𝟐𝟎𝟎
𝟒= 300 plf
Method assumes the unit shear above
and below opening are equivalent
V=4000
FLo1=1800
5’-0” 4’-0” 2’-6”6’-0” 2’-6”
2’-
8”
8’-
0”
1600
1600
1’-
4”
4’-
0”
Calculate:
• Aspect ratios of pier sections
• Unit shear above and below
opening
• Total boundary force above
and below opening
Vhdr = 300 plf
FLo2=750
Vhdr=300
20’-0”
The total boundary force above and below the
opening = the unit shear above and below the
opening multiplied by the opening length
Bottom force
same
FLo1 = 300(6) = 1800 lbs.
FLo2 = 300(2.5) = 750 lbs.
1800 750
V=4000
20’-0”
5’-0” 4’-0” 2’-6”
6’-0” 2’-6”
2’-
8”
8’-
0”
1600
1600
1’-
4”
4’-
0”
Calculate:
• Aspect ratios of pier sections
• Unit shear above and below
opening
• Total boundary force above
and below opening
• Corner forces
𝑭𝟏 =𝑭𝑳𝒐𝟏(𝑳𝟏)
(𝑳𝟏+𝑳𝟐)=
𝟏𝟖𝟎𝟎(𝟓)
(𝟓+𝟒)= 1000 lbs.
𝑭𝟐 =𝟏𝟖𝟎𝟎(𝟒)
(𝟓+𝟒)= 800 lbs.
𝑭𝟑 =𝟕𝟓𝟎(𝟒)
(𝟒+𝟐.𝟓)= 461.50 lbs.
𝑭𝟒 =𝟕𝟓𝟎(𝟐.𝟓)
(𝟒+𝟐.𝟓)= 288.5 lbs.
F4=288.5F1=1000 F3=461.5F2=800
Corner forces are based on the boundary forces
above and below the opening and only the pier
sections adjacent to that opening
Bottom force
same
FLo1=1800 FLo2=750
F4=288.5F1=1000 F3=461.5F2=800
V=4000
20’-0”
L1=5’-0” L2=4’-0” L3=2’-6”
Lo1=6’-0” Lo2=2’-6”
2’-
8”
8’-
0”
1600
1600
1’-
4”
4’-
0”
Calculate:
• Aspect ratios of pier sections
• Unit shear above and below
opening
• Total boundary force above
and below opening
• Corner forces
• Tributary length of openings
𝑻𝟏 =𝑳𝟏(𝑳𝒐𝟏)
(𝑳𝟏+𝑳𝟐)=
𝟓(𝟔)
(𝟓+𝟒)= 3.33’
𝑻𝟐 =𝟒(𝟔)
(𝟓+𝟒)= 2.67’
𝑻𝟑 =𝟒(𝟐.𝟓)
(𝟒+𝟐.𝟓)= 1.54’
𝑻𝟒 =𝟐.𝟓(𝟐.𝟓)
(𝟒+𝟐.𝟓)= 0.96’
0.96’1.54’2.67’3.33’
The tributary length of the opening is the ratio
of the pier length multiplied by the opening
length divided by the sum of the lengths of the
piers on each side of the opening
T4T3T2T1
0.96’1.54’2.67’3.33’
V=4000
20’-0”
5’-0” 4’-0” 2’-6”
6’-0” 2’-6”
2’-
8”
8’-
0”
1600
1600
1’-
4”
4’-
0”
Vtop=200 plf
Calculate:
• Aspect ratios of pier sections
• Unit shear above and below
opening
• Total boundary force above
and below opening
• Corner forces
• Tributary length of openings
• Unit shear in piers
𝑽𝟏 =𝒗𝒕𝒐𝒑(𝑳𝟏+𝑻𝟏)
𝑳𝟏=
𝟐𝟎𝟎(𝟓+𝟑.𝟑𝟑)
(𝟓)
= 333.2 plf
𝑽𝟐 =𝟐𝟎𝟎(𝟐.𝟔𝟕+𝟒+𝟏.𝟓𝟒)
(𝟒)= 410.5 plf
𝑽𝟑 =𝟐𝟎𝟎(𝟎.𝟗𝟔+𝟐.𝟓)
(𝟐.𝟓)= 276.8 plf
V3=276.8 plfV2=410.5 plfV1=333.2 plf
The pier shear is equal to the unit shear at the
top of the wall multiplied by the pier length +
the tributary length adjacent to the pier divided
by the length of the pier
0.96’1.54’2.67’3.33’
T4T3T2T1
276.8410.5333.2
𝑽𝒕𝒐𝒑 =𝑽
𝑳=
𝟒𝟎𝟎𝟎
𝟐𝟎= 200 plf
V=4000
20’-0”
5’-0” 4’-0” 2’-6”
6’-0” 2’-6” 8’-
0”
1600
1600
200 plf
Calculate:
• Aspect ratios of pier sections
• Unit shear above and below
opening
• Total boundary force above
and below opening
• Corner forces
• Tributary length of openings
• Unit shear in piers
• Resisting forces
R1=v1L1=333.2(5)=1666 lbs.
4000 lbs.
R3=692R2=1642R1=1666
The resisting forces of the piers equal to the
pier unit shear multiplied by the pier length
69216421666
R2=v2L2=410.5(4)=1642 lbs.
R3=v3L3=276.8(2.5)=692 lbs.
V=4000
20’-0”
5’-0” 4’-0” 2’-6”
6’-0” 2’-6” 8’-
0”
1600
1600
200 plf
Calculate:
• Aspect ratios of pier sections
• Unit shear above and below
opening
• Total boundary force above
and below opening
• Corner forces
• Tributary length of openings
• Unit shear in piers
• Resistance forces
• Pier collector forces
F4=288.5F1=1000 F2+F3=800+461.5
R3=692
(403.5)
R2=1642
(380.5)
R1=1666
(Net=666)
R1-F1=1666-1000=666 lbs.
The net collector force at the corners of the
Opening is equal to the pier resisting force
minus the opening corner force
288.51000 1261.5(403.5)(380.5)(Net=666)
69216421666
R2-F2=1642-(800+461.5)
=380.5 lbs.
R3-F4=692-288.5=403.5 lbs.
V=4000
20’-0”
5’-0” 4’-0” 2’-6”
6’-0” 2’-6”
2’-
8”
8’-
0”
1600
1600
1’-
4”
4’-
0”
200 plf
Calculate:
• Aspect ratios of pier sections
• Unit shear above and below
opening
• Total boundary force above
and below opening
• Corner forces
• Tributary length of openings
• Unit shear in piers
• Resistance forces
• Pier collector forces
• Unit shears at corner zones
Va2=(R2-F2-F3)/L2=380.5/4 = 95.1 plf
161.4 plf95.1 plf133.2 plf
The corner zone unit shear is equal to the net
collector force divided by the pier length
161.495.1133.2
Va3=(R3-F4)/L3=403.5/2.5 =161.4 plf
Va1=(R1-F1)/L1=666/5 = 133.2 plf
V=403.5 lbs.
netV=380.5 lbs.
net
V=666 lbs.
net
V=4000
20’-0”
5’-0” 4’-0” 2’-6”6’-0” 2’-6”
2’-
8”
8’-
0”
1600
1600
1’-
4”
4’-
0”
300
300
200 plf
Calculate:
• Aspect ratios of pier sections
• Unit shear above and below
opening
• Total boundary force above
and below opening
• Corner forces
• Tributary length of openings
• Unit shear in piers
• Resistance forces
• Pier collector forces
• Unit shears at corner zones
• Verify design
∑H=V, 333.2(5)+410.5(4)+276.8(2.5) =4000 lbs. o.k.
133.2
333.2
Grid 1=va1(ha+hb)+v1(ho)=Fo/t, 133.2(4+1.33)+333.2(2.67)=1600 o.k.
Grid 2=va(ha+hb)-va1(ha+hb)-v1(ho)=0, 300(4+1.33)-133.2(4+1.33)-333.2(2.67)=0 o.k.
Grid 3=va2(ha+hb)+v2(ho)-va(ha+hb)=0, 95.1(5.33)+410.5(2.67)-300(5.33)=0 o.k.
Grid 4=va2(ha+hb)+v2(ho)-va(ha+hb)=0, -95.1(5.33)-410.5(2.67)+300(5.33)=0 o.k.
Grid 5=va(ha+hb)-va3(ha+hb)-v3(ho)=0, -300(5.33)+161.4(5.33)+276.8(2.67)=0 o.k.
Grid 6=va3(ha+hb)+v3(ho)=Fo/t, 161.4(5.33)+276.8(2.67)=1600 o.k.
654321
133.2
133.2
333.2
133.2
300
300
161.495.1
276.8
95.1 161.4
410.5410.5
300
300
300
300
95.1
95.1
𝜮𝑽 = 𝟎 𝒂𝒕 𝒂𝒍𝒍 𝒈𝒓𝒊𝒅 𝒍𝒊𝒏𝒆𝒔
20’-0”
L
5’-0”
L1
4’-0”
L2
2’-6”
L36’-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”
hb
1
TD
Multi-story walls?
Tall shear walls?
• 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
∆𝑨𝒗𝒆𝒓.= (∆𝒑𝒊𝒆𝒓 𝟏+∆𝒑𝒊𝒆𝒓 𝟐)+(∆𝒑𝒊𝒆𝒓 𝟏+∆𝒑𝒊𝒆𝒓 𝟐)
𝟒
• The remainder of the terms are identical to
the traditional equation.
• Deflections for a wall with multiple
openings is similar.
∆𝑨𝒗𝒆𝒓.=(∆𝒑𝒊𝒆𝒓 𝟏+∆𝒑𝒊𝒆𝒓 𝟐+∆𝒑𝒊𝒆𝒓 𝟑)+(∆𝒑𝒊𝒆𝒓 𝟏+∆𝒑𝒊𝒆𝒓 𝟐+∆𝒑𝒊𝒆𝒓 𝟑)
𝟔...
Hdr
Opening
Wall
Pier 2
Wall
Pier 1
Sill
∆𝑺𝑾=𝟖𝒗𝒉𝟑
𝑬𝑨𝒃+
𝒗𝒉
𝑮𝒗𝒕𝒗+𝟎. 𝟕𝟓𝒉𝒆𝒏+
𝒉∆𝒂
𝒃
Traditional 4 term deflection equation
∆𝑷𝒊𝒆𝒓 𝟏 ∆𝑷𝒊𝒆𝒓 𝟐
Reference APA SR-105
Loads Loads
∆𝑷𝒊𝒆𝒓 𝟏 ∆𝑷𝒊𝒆𝒓 𝟐
SW1 SW4
W ( plf)1 2
A
B
Diaphragm
support
Diaphragm
support
Stable
SW3
SW2
SW1 SW4
W ( plf)1 2
A
B
Diaphragm
support
Diaphragm
support
SW3
SW2
Unstable
Acts like open front diaphragm
No
support
Shear walls are parallel
to applied loads
Not all shear walls are
parallel to applied loads
Angled Shear Walls
24’
SW1 SW4
C.L.
W ( plf)
1
A
B
Diaphragm
support
Diaphragm
support
Diaphragm with Angled Shear Wall Layout (Base Model)
2
θ
9’ SW 3
9’ SW2
Analogous to a simple span beam with a pinned end and spring supported end
30’
100’
Θ=𝟏𝟎𝒐
Θ=𝟐𝟎𝒐
Θ=𝟑𝟎𝒐
Θ=𝟒𝟓𝒐
FEA Model
2x6 shear wall with ½”
wood structural panels
and 6x6 end posts (typ.)
½” wood structural
panels over G.L
roof beams
6x6 posts (typ.)
Structure is stable
P-delta problem.
Wall deflects too much
out-of-plane, which causes
rotation in the diaphragm
(Acts like open front diaphragm-shift
forces to side walls)
Large
deflection
Wall warps
and 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)Diaphragm
SW2
Structural Behavior
SW1 SW4
W ( plf)
1
A
B
Diaphragm
support
Diaphragm
support
Example- Circular Shear Wall Layout
2
SW3
SW2
C.R.
Circular Shear Walls
SW1
1
A
B
Diaphragm
support
C.R.
A
B
Diaphragm
support
SW1
Hold downHold down
Effective
area
SW1
TD1
TD1
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 WSP
w / GWB
w / GWB
Prescriptive / Proprietary Portal Frames
Prescriptive Code Portal Frames Proprietary
Portal Frames
IBC 2308.9.3.2
Source: strongtie.com
Pre-engineered Proprietary Shear Walls
Considerations:
Large Hold-down forces
Larger deflections
Larger foundations
Pre-engineered Narrow Wall Section
Proprietary
Pre-engineered Proprietary Truss Walls
Truss Walls
Truss Frames
Hybrid Wood/Steel Proprietary Systems
Source: hardyframe.com
Presentation Topics
• Shear Wall Types
• Shear wall Anchorage
• SDPWS Code Requirements
• Complete Load Paths
• Offset Shear Walls and Relevant Code
Requirements
• Mid-rise Diaphragms and Tall Shear Walls
120 Union, San Diego, CA
Togawa Smith Martin
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” m
ax.
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. See
schedule
1
12
𝟓𝒐 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 Issues
Courtesy 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 𝟗𝟎𝒐 at corner.
3” to 5”
Le
½” min. at
foundation
corner
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 shrinkage
occurs.
Continuous Rod Tie Downs with Shrinkage Compensation Devices
Source: strongtie.com
Automatic Tensioning Systems/Devises
ATS isolator nut
take-up device
and bearing plate
Coupler
Steel rod
Modified
balloon 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)
Legend
xx kips Tied off at each floor
(xx kips) Tied off at roof only
Tied off at each floor
Tied off at roof only
Skipped floor
Presentation Topics
• Shear Wall Types
• Shear wall Anchorage
• SDPWS Code Requirements
• Complete Load Paths
• Offset Shear Walls and Relevant Code
Requirements
• Mid-rise Diaphragms and Tall Shear Walls
120 Union, San Diego, CATogawa Smith Martin
2015 SDPWS
Shear Wall Capacity and Load Distribution
Allowable values shown in tables
are nominal shear values
Design values:
LRFD (Strength) = 0.8 x Vn
ASD = 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 Board
Use Table 4.3C for Gypsum and Portland Cement Plaster
Use 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
Unblocked
Table 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 0.4
6 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, construction
and same capacities
Dissimilar materials
(Not cumulative)
Dissimilar materials
Vallow = largest
Vallow=2x
Vallow =2x smaller or
largest, greater of.
VS1
VS2Same material, construction on both
sides, different capacities-Seismic
Side 1
Vsc
Side 2
𝐆𝐚𝐜 = 𝐆𝐚𝟏 + 𝐆𝐚𝟐
𝐯𝐒𝐂 = 𝐊𝐦𝐢𝐧𝐆𝐚𝐜 Combined apparent shear wall shear
capacity (seismic)
𝐊𝐦𝐢𝐧 =𝐯𝐒𝟏
𝐆𝐚𝟏𝐨𝐫
𝐯𝐒𝟐
𝐆𝐚𝟐Minimum 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 seismic
Vwc = combined nominal unit shear capacity wind
vS1=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
Segmented
SW
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 be
reduced 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 4.3.3.4
does not apply
Equal wall deflection∆𝒘𝒂𝒍𝒍
2. Where Vn x 0.1+0.9b/h of all structural
fiberboard shear walls with A/R>1:1, shear
distribution to individual full-height wall
segments shall be permitted to be taken as
proportional to design shear capacities.
(Traditional Method-by length)
Where multiplied by Vn x 0.1+0.9b/h, the
nominal 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 Approach
Traditional 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 𝟐𝒃
𝒉(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
𝟐𝑏
𝒉
A/R>2:1 A/R≤2:1h
Distribution of lateral forces to In-line Shear Walls
Method 2-Equal Deflection
∆𝑺𝑾𝟏 ∆𝑺𝑾𝟐
b1 b2
SW1 SW2h
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)
𝒗𝒔𝒘𝟏 =∆𝒔𝒘𝟐
𝟖𝒉𝟑
𝑬𝑨𝒃𝒔𝒘𝟏+
𝒉
𝟏𝟎𝟎𝟎𝑮𝒂+
𝒉𝟐
𝒌𝒃𝒔𝒘𝟏
• 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
Lo
ad
lb
s.
500
Deflection, in.
0.3
5
0.3
0.2
5
0.2
0.1
5
0.1
0.0
5
ASD strength and
deflection of shear
wall line
ASD strength and
deflection of SW2
ASD strength and
deflection of SW1
at SW2 defl. limit
ASD strength
of SW1 Per
SDPWS 4.3.4.2
Reserve
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 𝛀𝒐 = 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 stud
2x studs
Optional joint
Square plate washers prevent
cross 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 if
standard 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
Presentation Topics
• Shear Wall Types
• Shear wall Anchorage
• SDPWS Code Requirements
• Complete Load Paths
• Offset Shear Walls and Relevant Code
Requirements
• Mid-rise Diaphragms and Tall Shear Walls
120 Union, San Diego, CA
Togawa Smith Martin
Detailing for Continuous Load Paths
Karuna I
Holst Architecture
Photo: Terry MaloneContinuous Stacked In-line Load Paths
Simple Load Paths
Detailing for Continuous Load Paths
Karuna I
Holst Architecture
Photo: Terry MaloneDis-continuous 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
Co
llecto
r
C D
Alt. Config.
Sect. B
Anchor bolts
or nails
Shear
transfer
Conn.
Tie
strap
Nail shtg to
each 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 to
each 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
joist
Analyze this Section
as a transfer diaph.
or transfer wall
Header/collector
Section C
Section B Section A
B C
Co
llecto
r
C D
Alt. Config.
Sect. B
Anchor bolts
or nails
Shear
transfer
Conn.
Tie
strap
Nail shtg to
each 2x stud
Type 4 vert. irreg.
SDC B-FType 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 joints
in boundary element
Opening
Column
Possible perforated
or FTAO shear wallSegmented
shear wall
Header
Examples of Drag Struts, Collectors and chords at Exterior Boundaries
Sect.
Optional
load path
If balcony
Blocking
(typ.)
Header
Direction of Load
T
Shear Wall
(transfer wall)
Hdr.
Collector Chord
Chord
h1
Diaphragm sht’g.
elevation
Diaphragm sht’g.
elevation Parapet (typ.)
Vert
. S
tep
in d
iap
h.
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 Options
A
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 blocking
w/ vent holes
Venting Options B
B
Insulation
2x vertical
nailer E.S.
Truss
Baffle as
required
Insulation
Boundary
nailing
Screened
vent holes
Typical Solid Blocking
2x blocking
A
Insulation
2x rafter
A
Diaphragm 1 Diaphragm 2
Diaphragm 2
Boundary (typical)
Chord
Chord
Co
llecto
r
Str
ut
Str
ut
Chord
Str
ut
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 and Interior Shear Walls
SW
1
SW
2
SW
3
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
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
Diaphragm 1 Diaphragm 2
Diaphragm 2
Boundary (typical)
Chord
Chord
Co
llecto
r
Str
ut
Str
ut
Chord
Str
ut
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 and Interior Shear Walls
SW
1
SW
2
SW
3
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
Let’s take a look at
load paths to the shear
wall at grid line 2.
Special sheathing
nailing required
Strut/truss
(Call out force)
End nailed w/2-16d.
This connection
often has less
capacity than the
shears applied
(e. g. nail capacity
failure problem)
Cont. 2x plate w/
16d at calculated
Spacing (cross-grain
or end nail failure
problem)
16d at calculated
spacing (truss to
flat blk’g.)
Shear clips
Strut/truss
(multiple if
required)
Special sheathing
nailing required,
usually 8d or 10d
at 6” o.c. or 4” o.c.
Configuration A
Configuration CConfiguration B
Cross-grain
If truss deflects
Optional
Shth’g.
Strut/
truss
Typical Collector Framing and Connection Parallel to Shear
Wall
Lateral distribution
a b
V
V
L/2L/2
2x flat cross blk’g.
at 24” o.c. w/2 or
4-16d to plate
Horizontal
and vertical
distribution
Detail elevation
on drawings
Collector
SW
Typical interior
Shear wall or
braced wall
Roof trusses
@ 24” o.c.
Roof diaphragm
force to wall
No shear panels
installed or detailed.
GWB ceiling
buckling Trusses rotate because there
is nothing present to resist the
lateral forces, and the lateral
load is not transferred into the
wall.
Collector / strut
is missing
Typical Collector Framing and Connections- Roof Framing Perpendicular to Shear Wall
Corridor Shear Wall Line
Corridor Shear Wall with Shear Panels and Collector Added
Shear panels vary in detailing from
designer to designer (mini-shear walls).
Add member at end of wall as required.
Trusses
also brace
wall
Continuous drag strut
or collector is required
Tie straps at
end of wall
as required.
Ceiling
Sh
ear
wall h
eig
ht
Typical interior
Shear wall or
braced wall
Note:
When designing
the shear wall,
the forces from the
shear panels
above must be
transposed to the
shear wall below.
Framing Perpendicular to Wall
Wood shear panels
between trusses
Does not appear to have vertical blocking at truss
(no shear transfer for vertical shear force).
Shear panel ratios
3:1
The shear panels shown are 24” wide by 6’-0” high.
The framing could easily be 16” wide by 15’-0” high
or greater.
Wall studs
at 8” o.c.
If v=200 plf
400
400
1200
1200
2’
6’
Photo-Typical Shear PanelsCourtesy of Willdan Engineering
If 3:1 A/R
Side members nailed to
truss chords by side grain
nailing, full height
Typical Shear Panel Detailing
Side members nailed
to truss chords by
side grain nailing-2 16d
Top and bottom
members nailed
to truss chords by
end grain nailing-2 16d
Option 1 Option 2 Option 3
Blk’g. top and
bottom only
Side frm’g.
members
addedPanel edge
buckling
Add vertical
nailing member
in truss
Roof sheathing
Roof trusses
Steel strap
Shear panels
per detail 3
Hold down
straps as
requiredShear wall
2x, 3x, or 4x flat
blocking
Collector Framing Option 1
Continuous
ledger at
corridor
2x flat blk’g
(tight fit)
Potential
gaps
Typical Section
Note: If open web joists, continuous 2x
members can be nailed to blocking to
take compression forces.
Use Z clip to keep
blocking level
Bearing perp.
to grain? Alt. 4x blk’g.
Tie strapWSP
sht’g.
Joists
Example of Partial Strut/Collector
Collector force distribution
Strut/collector force diagram
Diaphragm unit
shear (plf) transfers
into blocking
For compression, blocking
acts as mini strut/collector
and 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
Collector/strutShear panel or truss
rotation blocking
Corridor Wall Line?
Fully sheathed?
These connections
are part of the complete
load path
Section A
Roof sheathing
Roof trusses
Splice strap
1 or 2 sides
Blocking at
tie strap
Shear panels
per detail 3
Hold down
straps as
requiredShear wall
2x blocking w/ A35
Clips (if req’d)
Single or multiple
continuous drag
members
Collector Framing Option 2(Assuming the collector does not fall at a truss joint)
Special strut
nailing full length
Connections for
shear transfer
Blocking
A
Edge nailing
each 2x block
Blocking
Section A
Roof sheathing Roof trusses
Shear panels
per detail 3
Hold down straps
as required
Shear wall
2x blocking w/
shear clips
(as req’d.)
Shear clips as required
Special strut nailing
full length
Connections for
shear transfer
Blocking with nails
and/or shear clips
(as req’d.)
Edge nailing
each 2x
Horizontal strap
is required across
joint if 2x members
can not be cont.
A
Single or multiple
continuous drag
members
Collector Framing Option 3(Assuming the collector does not fall at a truss joint)
Roof sheathing
Roof
trusses
Shear panels
per detail 3
Shear wall
2x blk’g.
Cont. drag
members
Shear clips
as requiredEdge nailing
each 2x
Horizontal strap is
required across joint
if 2x members can
not be continuous
A
Shear wall
Drag strut or collector Tru
sses (
typ
.)
A B
Section B
Special
nailing
Shear
panels
Special transfer area
nailing (drag shear plus
basic diaphragm shear)
Plan View
B
V
V F
F
Special nailing
required
Collector Framing Option 4(If the collector falls at a truss joint)
Presentation Topics
• Shear Wall Types
• Shear wall Anchorage
• SDPWS Code Requirements
• Complete Load Paths
• Offset Shear Walls and Relevant Code
Requirements
• Mid-rise Diaphragms and Tall Shear Walls
120 Union, San Diego, CA
Togawa Smith Martin
Offset Shear Walls – Code Requirements
SW 1
SW 2
Co
llecto
r
Collector
Collector
Out-of-plane Offsets In-plane Offsets
Dis
co
nti
nu
ou
s c
ho
rds
Tra
ns
ve
rse
Cant.
Mid-rise Multi-family
SWSWSWSW
SWSWSWSWSW
SW
Lds. Discontinuous struts
Longitudinal
Lds.
No exterior
Shear walls
Flexible, semi-rigid, or rigid???
TD3
TD1
TD2
SW4
SW3
SW5
SW2
SW1
Diaphragm stiffness changes
Multi Story, Multi-family
Wood Structure
1
2
3
Loads
I1 I2 I3
Higher shears
and nailing
requirements
ASCE7-10
1.3.5 - Cont. ld. Paths
12.1.3 - Cont. ld. paths-inter-conn. Ties
12.10.1-Openings, re-entrant. –transfer of dis-cont. forces
combined with other forces
12.10.2-Collector elements
I1 I2
SW4
SW3
SW5
SW1
1
2
3Transfer Area
Higher shears
and nailing
Reqmts.
Collector
(typ.)
Higher shears
and nailing
Reqmts.
SW2
Co
llecto
r
Str
ut/
ch
ord
Be
am
TD2
TD1
Str
ut
Main diaphragm
becomes TD3
Optional Framing
Layouts
Collector
(typ.)
Out-of-Plane Offset Shear WallsAssumed to act in the Same Line of Resistance
Loads
Co
llecto
r
Collector
Collector
TD1
SW1
SW2
Transfer
area
Offset
Discont.
drag strut
Dra
g
str
ut
Dra
g
str
ut
SW3
Co
llecto
r
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 irregularity
Re-entrant corner irregularity
SW
SW
(Typ.)
SW
SW (Typ.)
Transverse
walls typical
Longitudinal
exterior walls
typical
Corridor walls
typical
Transfer Area
Tying Shear Walls Across the Corridor
Typical Rectangular plans
collector
Co
lle
cto
r
collector
Co
lle
cto
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
Loads
Co
lle
cto
r
collectors
Corridor
walls
Jst. Jst.
2x6/2x8Tie straps at
all Joints (typ.)
Typical Corridor Section
Mech.
Co
lle
cto
r
Co
lle
cto
r
Co
lle
cto
r
Layout 1 Layout 3Layout 2
Section
Special nailing of the sheathing to the
collector is required the full length of the
collector (typ.)
Common Transverse Wall Layouts
Corridor
wallsCo
lle
cto
r
SW
SW
SWSW
SW
Layout 4
Co
lle
cto
r
Co
lle
cto
r
SW
SW
Layout 5
Transfer
Area (typ.)
Co
lle
cto
r
Co
lle
cto
r
SW
SW
Layout 6
Co
lle
cto
r
Tying Shear Walls Across the Corridor
3. Collectors for shear transfer
to individual full-height wall
segments shall be provided.
SDPWS 4.3.5.1
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 plfRL=RR=200(122/2)=12200 #
vR=12200/50=244 plf
Sum=40000 #
=200 plf(200’)
vSW=454.55 plf
vNet=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
11
2 p
lf8
8 p
lf
88
plf
11
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 Area
RL=RR=24(4/2)=48 #
vL=vR=48/6=8 plf
Sum=40000 #
=200 plf(200’)
5368 #
6608 #
Case 1-Full length offset walls
Example 1-Offset Shear Walls -Layout 6
SW
1
SW
2
SW3
2
B
3
C
A
D
SW4
156
156
164 236
244
244
164
8
160
160240
240
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
1224984
1464
8 236
Summing V=0
Summing V=0
All forces in lb., all shears in plf
O.K.
Case 1-Smaller resulting forces at corridor
Diaphragm 1 Diaphragm 2S
W1
SW
2
SW3
1 2
B
3
C
A
D
4
SW4
11
2 p
lf8
8 p
lf
88
plf
11
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 Area
RL=RR=24(4/2)=48 #
vL=vR=48/6=8 plf
Sum=40000 #
=200 plf(200’)
5368 #
6608 #
Case 2-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
2125.493188.2
4
3188.2
4
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 plf
vnet SW2=589.65-164-236=189.65 plf Checks, they
should be equal
SW1
F2AB=189.65(22)=4172.24 #
F2BC=(156+8)6=984 #
F2C=4172.24-984=3188.24 #
SW2
FSW2 =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
Case 2-Larger resulting forces at corridor
O.K.
22’
10’
12’
6’
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 plf
vnet SW2=911.27-164-236=511.27 plf Checks, they
should be equal
SW1
FSW1=511.27(10)=5112.7 #
FSW1 to 2B=(156+244)12=4800 #
F2BC=(156+8)6=984 #
F2C=5112.7-4800-984=-671.3 #
SW2
FSW2=511.27(12)=6135.24 #
FSW2 to 3C=(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
Case 3-Smaller resulting forces at corridor
O.K.
10’
12’
6’
447.5
447.5671.2
4
671.3
5112.7
6135.24
12’S
W1
10’
671.3
Potential buckling
problem w/ supporting
columns and beams
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
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 LFRS
load pathASCE 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
A.25
Diaphragm 1
Shear wall is not continuous to
foundation. Type 4 horizontal
and vertical irregularity
Chord
Co
llecto
r
Str
ut
Str
ut
Chord
Str
ut
Design supporting collector and
columns for over-strength factor
per ASCE 7-10 Section 12.3.3.3
SDC B-F and 12.10.2.1 SDC C-F.
Type 4 Horizontal Offset Irregularity-Seismic
SW
1
SW
2
SW
3
Type 4 horizontal irregularity-Out-of-plane offset irregularity occurs where there is a discontinuity in
lateral force resistance path. Out-of-plane offset of at least one of the vertical lateral force resisting
elements.
ASCE 7-10 Section 12.3.3.3 (SDC B-F)-Elements supporting discontinuous walls or frames. Type 4
horizontal or vertical irregularity.
• Beams, columns, slabs, walls or trusses.
• Requires over-strength factor of Section 12.4.3
1
2
B
3
C
A
Type 2 and Type 4 Irregularity SDC D-F(Interior collector similar)
Blocking
Continuous member
Alt. Collector/ strut
Connection
ASCE 7-10 Section 12.3.3.3Type 4 Irregularity (SDC B-F)
Elements supporting discont.
Walls:
•Beams, columns, slabs,
walls or trusses.
•Requires over-strength
factor of Section 12.4.3
(also see 12.10.2.1 SDC C-F
for collector requirements).
Diaphragm 1
Shear wall is not continuous to
foundation. Type 4 horizontal
and vertical irregularity
Chord
Co
llecto
r
Str
ut
Str
ut
Chord
Str
ut
Type 4 Horizontal Offset Irregularity-Seismic
SW
1
SW
2
SW
3
ASCE 7-10 Section 12.3.3.4 (SDC D-F) -Type 4 Horizontal irregularity or 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.
1
2
B
3
C
A
Design diaphragm connections to
SW and struts using 25% increase
per ASCE 7-10 Section 12.3.3.4.
(Grid lines 1, 2 and 3)
• Diaphragm shears
are not required to
be increased 25%.
• The transfer force
(SW) in SDC D-F
must be increased
by rho, per 12.10.1.1.
See 12.10.2 & 12.10.2.1
for collectors
Diaphragm 1
Shear wall is not continuous to
foundation. Type 4 horizontal
and vertical irregularity
Chord
Co
llecto
r
Str
ut
Str
ut
Chord
Str
ut
Type 4 Horizontal Offset Irregularity-Seismic
SW
1
SW
2
SW
3
ASCE 7-10 Section 12.3.3.4 (SDC D-F) -Type 4 Horizontal irregularity or 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:
• Collectors and their connections to vertical elements.
1
2
B
3
C
A
Collectors and their connections
to vertical elements per ASCE 7-10
Section 12.3.3.4. (Grid lines 1, 2 and 3)
Exception: Forces using the seismic load effects including the over-strength factor
of Section 12.4.3 need not be increased.
Type 2 and Type 4 Irregularity SDC D-F(Interior collector similar)
ASCE 7-10 Section 12.3.3.4 (SDC D-F)
Horizontal Type 2 Irreg.-Re-entrant corner,
horizontal and vertical Type 4 irregularity:
Requires a 25% increase in the diaphragm
design forces determined from 12.10.1.1 (Fpx)
for the following elements:
Exception: Forces using the seismic load
effects including the over-strength factor of
Section 12.4.3 need not be increased.
Blocking
Continuous member
Alt. Collector/
strut Connection
• Connections of diaphragm to
vertical elements and collectors.
CollectorShear Wall
• Collectors and their connections,
including their connections to
vertical elements.
Ftg.
Strut/Collector
Strut/Collector
SW1
Vr
V2
Dhr
Dh2
Ftg.
Su
pp
ort
co
lum
n
A BA.25
SW2
Shear wall system at grid line 1
In-plane Offset of Wall
Su
pp
ort
co
lum
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 anchorage is
designed for standard load
combinations of Sections
2.3 or 2.4.
Co
llecto
r
Assuming no
hold down
See struts and collectors
Connections
are designed for
standard load
combinations 2.3
or 2.4.
Ftg.
Support beam/collector
Collector
SW3
Vr
V2
Su
pp
ort
co
lum
n
Ftg.
Su
pp
ort
co
lum
n
A BA.33
Shear wall system at grid line 2
Out-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 or
2.4.
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
Footings and connections
are not required to be
designed for over-strength.
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
Struts and Collectors
Possible struts:
• Truss or rim joist
over wall
• Double top plate
• Beam/header
12.10.2 SDC B - Collectors can be designed w/o over-strength
but 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:
𝛀𝒐𝑭𝒙 - Forces determined by ELF Section 12.8 or Modal Response Spectrum Analysis
procedure 12.9
𝛀𝒐 𝑭𝒑𝒙 - Forces determined by Diaphragm Design Forces (Fpx), Eq. 12.10-1 or
𝑭𝒑𝒙 𝒎𝒊𝒏= 𝟎. 𝟐𝑺𝑫𝑺𝑰𝒆𝒘𝒑𝒙 - Lower bound seismic diaphragm design forces determined by
Eq. 12.10-2 (Fpxmin) using the Seismic Load Combinations of section
12.4.2.3 (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, 𝑭𝒑𝒙).
Struts and Collectors-Seismic
𝑭𝒑𝒙 𝒎𝒂𝒙= 𝟎. 𝟒𝑺𝑫𝑺𝑰𝒆𝒘𝒑𝒙- Upper bound seismic diaphragm design forces determined by
Eq. 12.10-2 (Fpxmax) using the Seismic Load Combinations of section
12.4.2.3 (w/o over-strength)-do not require the over-strength factor.
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
Sections do 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 5-In-plane Offset Segmented Shear Wall -with Gravity Loads
1’
Co
llecto
r VHdr=960 lb
VHdr=450 lb
A B C
ASCE 7 Table 12.3-2-Type 4
vertical irreg.- in-plane
discontinuity in the LFRS
if no hold down at B.
12.3.3.3 & 12.3.3.4
SDC B-F SDC D-F
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 below
with 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 lb
1’
12’
416.67 lb416.67 lb
416.67416.67
3370
1080
Upper Shear Wall
Rim joist
135 365
Co
llecto
r
1080 lb
330 plf (incl. wall DL)
Lower Shear Wall
TD shears-lbs. (plf)
w=230 plf (incl. wall DL)
+-8’4’
SW2
SW1
+250 p
lf
+416.6
7 p
lf
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 lb
3370 lb
Basic
Sh
ear
+
T
T
++
+
2000 lb
3000 lb
1’
8’
8’
8’4’
C
+Pos. direction -
Vertical Collector Forces
2nd floor Rim joist
++ SW1
SW2D
ep
th T
D1 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
Presentation Topics
• Shear Wall Types
• Shear wall Anchorage
• SDPWS Code Requirements
• Complete Load Paths
• Offset Shear Walls and Relevant Code
Requirements
• Mid-rise Diaphragms and Tall Shear Walls
120 Union, San Diego, CA
Togawa Smith Martin
Mid-rise Diaphragms and Shear Walls
Crescent Terminus building- credit: Richard Lubrant.
The analysis techniques provided in this presentation are intended to demonstrate one
method of analysis, but not the only means of analysis. The techniques and examples
shown here are provided as information for designers to consider to refine their own
techniques.
Design Considerations
Method of analysis for open-front or tall shear wall mid-rise structures is currently
evolving as more complex building geometries are becoming more prevalent. Building
shapes and footprints are driving the design procedures for all lateral resisting systems
and materials.
Design requirements:
• IBC1604.4- Analysis:
o Method of analysis shall take into account equilibrium, general stability,
geometric compatibility, and both short-and long term material properties.
o Shall be based on rational analysis in accordance with well established principals
of mechanics. Analysis shall provide complete load paths.
Considerations:
• Make sure that you have complete continuous lateral load paths.
• Address vertical / horizontal irregularities.
• Investigate diaphragm flexibility. Note: Possible changes in diaphragm
flexibility can occur from floor to floor: non-open front (flexible) vs. non-open
front or open front (rigid or semi-rigid).
• Check shear wall stiffnesses and the effects of tall shear walls.
• Check the effects of offset diaphragms and shear walls.
• Verify the horizontal distribution of forces within the diaphragm and to the
shear walls.
• Check story drift
76’
35’
6’
35’
Open Front & Non-open Front
Floor Plan w/ and w/o offsets
7654321
Typical
Unit
F
C
B
A
D
E
Open
Front
Non-Open
Front
8
SW
Typical SW
Typical
• ASCE 7-10 Section 12.3.1.1- (c), Light framed construction, meeting all conditions:
Longitudinal-Typically permitted to be idealized as flexible, provided exterior
shear walls exist. However, diaphragm can be semi-rigid or open front, even if
exterior walls exist.
Transverse-Traditionally assumed to be flexible diaphragm.
If token or questionable wall stiffness at exterior, use semi-rigid analysis using
the envelope method-(conservative), or semi-rigid, or idealize as rigid.
Non-open front
Open Front Structures:
• SDPWS 4.2.5.2 :
Loading parallel to open side - Model as semi-rigid (min.), shall include shear
and bending deformation of the diaphragm, or idealized as rigid.
Loading perpendicular to open side-traditionally assumed to be flexible.
Drift at edges of the structure ≤ the ASCE 7 allowable story drift when
subject to seismic design forces including torsion, and accidental torsion
(strength, multiplied by Cd). No set drift requirements required for wind (Drift
can be tolerated).
Includes flexible and rigid analysis
Strut/chord
Open
3
4
5
21
F
E
D
C
B
6 9 107 8
Strut/chord
Str
ut
Str
ut
(typ
.)
Strut
chord
Strut chord
Strut /chord
Str
ut
Strut/chord
Strut/chord
SW
1
SW5
SW2
SW3
SW6
SW4
Str
ut
MR
F1
Multiple
offset
diaphragm
Offset
strut
Support Support
Co
llecto
r
Collector
Collector
(typ.)
Collector
(typ.)
Collector
(typ.)
Co
llecto
r (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?
Complete Load Paths
Strut/chord
Open
3
4
5
21
F
E
D
C
B
6 9 107 8
Strut/chord
Str
ut
Str
ut
(typ
.)
Strut
chord
Strut chord
Strut /chord
Str
ut
Strut/chord
Strut/chord
SW
1
SW5
SW2
SW3
SW6
SW4
Str
ut
MR
F1
Support Support
Co
llecto
r
Collector
Collector
(typ.)
Collector
(typ.)
Collector
(typ.)
Discontinuous
diaphragm
chord/strut
Discontinuous
diaphragm
chord
Discont.
diaph.
chord
Discont.
diaphragm
chord
Discont.
diaphragm
chord
Co
llecto
r (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
F
E
D
C
6 9 107 8
Strut/chord
Str
ut
Str
ut
(typ
.)
Strut
chord
Strut chord
Strut /chord
Str
ut
Strut/chord
Strut/chord
SW
1
SW5
SW2SW3
SW6
SW4
Str
ut
MR
F1
Offset shear walls
and struts
Support Support
Co
llecto
r
Collector
Collector
(typ.)
Collector
(typ.)
Collector
(typ.)
Offset shear
walls
Co
llecto
r (t
yp
.)
A
Complete Continuous Lateral Load Paths
ASCE7-10 Section 1.4-Complete load paths are required
including members and their splice connections
B
Strut/chord
Open
3
4
5
21
F
E
D
C
B
6 9 107 8
Strut/chord
Str
ut
Str
ut
(typ
.)
Strut
chord
Strut chord
Strut /chord
Str
ut
Strut/chord
Strut/chord
SW
1
SW5
SW2
SW3
SW6
SW4
Str
ut
MR
F1
Opening
in diaph.
Support Support
Co
llecto
r
Collector
Collector
(typ.)
Collector
(typ.)
Collector
(typ.)
Co
llecto
r (t
yp
.)
Vertical
offset in
diaphragm
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.
Average drift
of walls
Is maximum diaphragm
deflection (MDD) >2x average
story drift of vertical elements,
using the Equivalent Force
Procedure of Section 12.8?
Maximum diaphragm
deflection
Semi-rigid (Envelope Method)
Is diaphragm
untopped steel
decking or Wood
Structural Panels
Idealize
as
flexible
Is span to depth ratio ≤ 3
and having no horizontal
irregularities ?
Is diaphragm concrete
slab or concrete filled
steel deck ?
Structural analysis must
explicitly include consideration
of the stiffness of the diaphragm
(i.e. semi-rigid modeling).
Is any of the following true?
1 & 2 family Vertical elements one Light framed construction
Dwelling of the following : where all of the following are
met:
1. Steel braced frames 1. Topping of concrete or
2. Composite steel and similar material is not
concrete braced frames placed over wood structural
3. Concrete, masonry, steel SW panel diaphragms except
or composite concrete and for non-structural topping
steel shear walls. not greater than 1 ½” thick.
2. Each line of vertical
elements of the seismic
force-resisting system
complies with the allowable
story drift of Table 12.12-1.
Idealize
as rigid
Idealize as
flexible
Yes
YesYes
No
No
No
No
No
Sta
rtASCE7-10 Section 12.3 Diaphragm Flexibility Seismic
Yes
Yes
Section 12.3.1- The structural analysis shall consider the relative stiffnesses of diaphragms and the
vertical elements of the lateral force resisting system.
(Could apply to CLT or Heavy timber diaphragms)
Is diaphragm untopped steel
decking , concrete filled steel
decks or concrete slabs, each
having a span-to-depth ratio
of two or less?
Yes
Yes
No
Diaphragm can
be idealized as
flexible
Is diaphragm of
Wood Structural
Panels ?
Diaphragm can
be idealized as
rigid
No
Is diaphragm untopped steel
decking , concrete filled steel
decks or concrete slabs, each
having a span-to-depth ratio
greater than two ?
Yes Calculate as
flexible or
semi-rigid
ASCE7-10, Section 26.2 Diaphragm Flexibility Wind
Start
Tall Shear Walls
Marselle Condominiums 5 stories of wood over 6 stories concrete Structural
Engineer engineer: Yu & Trochalakis, PLLC (podium) 2 above grade
Photographer: Matt Todd Photographer
Tall FTAO walls
or open front
structure???
∆𝑺𝑾=𝟖𝒗𝒉𝟑
𝑬𝑨𝒃+
𝒗𝒉
𝟏𝟎𝟎𝟎𝑮𝒂+ 𝒉∆𝒂
𝒃
A/R
h/d ≤ 2:1
Tall Shear WallsCurrently not addressed or required by code:
Engineering preference and/or judgement
+
+
Traditional Shear Walls: Max. A/R=3.5:1
• Traditional floor to floor method is good for
3 stories or less (w/ exception)
VR
V3
V2T
rad
itio
na
l
Me
tho
d
Ta
ll W
all
Me
tho
d∆𝑻= ∆𝒔𝒘𝟏+∆𝒔𝒘𝟐 + ∆𝒔𝒘𝟑
Testing shows that the traditional deflection
equation is less accurate for walls with
aspect ratios higher than 2:1.(Dolan)
Walls assumed
to be pinned at
top and bottom
at each floor, rigid
wall sections and
rigid out-of-plane
diaphragm stiffness
Tall Shear Wall
Floor to floor A/R’s and Stiffness of Shear WallsDoesn’t account for multi-story shear wall effects
Tall Shear Walls
• Tall walls greater than 3 stories should consider
flexure and wall rotation.
o Rotation and moment from walls above
and wall rotation effects from walls below.
𝐒𝐖
Tall 𝐒𝐖
∆𝐀𝐥𝐥𝐨𝐰 𝐬𝐭𝐨𝐫𝐲
𝐝𝐫𝐢𝐟𝐭
• Allowable story drift for traditional
and tall shear walls is checked floor
to floor.
Total displ.
of Tall Wall.
More flexible.
A/R=3.5:1
flr.-flr.
Total displ.
Traditional
walls
A/R=2:1
flr.-flr.
Rim joist
Should consider as
flexible because it is
unknown where rim
joist splices will occur
Platform framed
Semi-balloon framed
(Very flexible)
Flexibility does not
allow dead load to
aid in resisting wall
rotation
Stiffness might allow dead
load to aid in resisting wall
rotation, provided joints are
strategically placed.
(justify by calculation)
If diaphragm out-of-plane
stiffness=Flexible
Analyze entire wall as a
tall wall
If diaphragm out-of-plane
stiffness=Rigid (steel beam,
conc. beam) Analyze entire
wall as traditional floor to
floor
Compression
blocking
V
CT
M
Diaphragm
out-of-plane
Flexibility
Current Examples of Mid-rise Analysis
• APEGBC Technical & Practice Bulletin Revised April 8, 2015 “5 and 6 Storey Wood Frame Residential Building Projects (Mid-Rise)”-Based on FPInnovations
Mechanics Based Approach
• FPInnovations-Website ”A Mechanics-Based Approach for Determining Deflections of Stacked Multi-Storey
Wood-Based Shear Walls”, Newfield
• Shiotani/Hohbach Method-Woodworks Slide archive
http://www.woodworks.org/education/online-seminars/
• Thompson Method-Woodworks Website
http://www.woodworks.org/wp-content/uploads/HOHBACH-Mid-Rise-Shear-Wall-and-Diaphragm-Design-WSF-151209.pdf
Design Example: ”Design of Stacked Multi-Storey Wood-Based Shear Walls Using a Mechanics-Based Approach ”, Canadian Wood Council
Webinar
http://www.woodworks.org/wp-content/uploads/5-over-1-Design-Example.pdf
Paper
Current Examples of Tall Shear Wall Analysis
Deflections of Stacked Multi-story Shear Walls-U.S. StandardsBased on FPInnovations ”A Mechanics-Based Approach for Determining Deflections of Stacked Multi-Storey Wood-
Based Shear Walls”, Newfield-(Metric)
∆𝒊
Lwall
Lc
xtr
∆𝒏Where
∆𝒃,𝒊= 𝑽𝒊𝑯𝒊
𝟑
𝟑 𝑬𝑰 𝒊+
𝑴𝒊𝑯𝒊𝟐
𝟐 𝑬𝑰 𝒊
∆𝒔,𝒊= 𝑽𝒊𝑯𝒊
𝑮𝒗,𝒊𝒕𝒗,𝒊
∆𝒏,𝒊= 0.75𝑯𝒊𝒆𝒏,𝒊
∆𝒂,𝒊= 𝑯𝒊
𝑳𝒊𝒅𝒂,𝒊 (Includes affects of rod
elongation and crushing)
∆𝒓,𝒊 =
𝒋−𝟏
𝒊−𝟏
∆𝒃,𝒊+∆𝒓,𝒊 𝒘𝒂𝒍𝒍𝒔 𝒃𝒆𝒍𝒐𝒘
∆𝒂,𝒊
Rotational deflection
∆𝒃,𝒊
𝜃𝒊
𝒋=𝟏
𝒊−𝟏
𝜽𝒋
𝒊
𝒋
𝒊−𝟏
α𝒋
𝑯𝒊𝜶𝟏 + 𝜶𝟐 +…𝜶𝒊−𝟏
𝑯𝒊 𝜽𝟏 + 𝜽𝟐 +⋯𝜽𝒊−𝟏
𝑯𝒊
Level i
Level i-1
𝑰𝒕𝒓=𝑨𝒕,𝒕𝒓(𝒅𝟏𝟐) + 𝑨𝒄 𝒅𝟐
𝟐
Transformed composite section
ATS - cont. hold down used
𝑴𝒊𝑯𝒊𝟐
𝟐 𝑬𝑰 𝒊𝑽𝒊𝑯𝒊𝟑
𝟑 𝑬𝑰 𝒊
C.L.Brg
∆𝒊=∑𝑴𝒊𝑯𝒊
𝟐
𝟐 𝑬𝑰 𝒊+ ∑𝑽𝒊 𝑯
𝟑
𝟑 𝑬𝑰 𝒊+
𝑽𝒊𝑯𝒊
𝑮𝒗,𝒊𝒕𝒗,𝒊+ 0.75𝑯𝒊𝒆𝒏,𝒊 +
𝑯𝒊
𝑳𝒊𝒅𝒂,𝒊 + ∑𝒋−𝟏
𝒊−𝟏 ∆𝒃,𝒊+∆𝒓,𝒊
Shear
Defl.Nail Slip Rod
Elong.
crushing
Bending deflection Sum
rotation
below
∆𝒃,𝒊
𝒅𝒂,𝒊
Alt.
∆𝒔,𝒊= 𝑽𝒊𝑯𝒊
𝟏𝟎𝟎𝟎𝑮𝒂,𝒊
∆𝒔,𝒊 ∆𝒏,𝒊 ∆𝒂,𝒊 ∆𝒓,𝒊
∑𝑴𝒊𝑯𝒊𝟐
𝟐 𝑬𝑰 𝒊+∑𝑽𝒊 𝑯
𝟑
𝟑 𝑬𝑰 𝒊
𝐔. 𝐒. 𝐜𝐮𝐬𝐭𝐨𝐦𝐚𝐫𝐲 𝐮𝐧𝐢𝐭𝐬
Tall Wall Deflection
∑V5
VR
∑V4
∑V2
∑V3
𝑴𝟏=510.5’k
𝑴𝟐 = 𝟑𝟐𝟏. 𝟒𝟖′𝒌
𝑴𝟑 = 𝟏𝟔𝟑. 𝟖𝟕′𝒌
𝑴𝟒 = 𝟓𝟏. 𝟗𝟓′𝒌𝑯𝟓
𝑯𝟒
𝑯𝟑
𝑯𝟐
𝑯𝟏
5.195
11.192
15.761
18.902
20.659
α1
θ𝟏
∆1
α2
θ2
∆2
+
θ3
∆3
+
α3
θ4
∆4
+α4
∆5
+
𝜽𝟏 𝑯𝟐 + 𝑯𝟑
𝜽𝟏 𝑯𝟐
𝜽𝟏 𝑯𝟐 +𝑯𝟑 + 𝑯𝟒
𝜽𝟏 𝑯𝟐 +𝑯𝟑 + 𝑯𝟒 +𝑯𝟓
+α𝟏(𝑯𝟏 +𝑯𝟐)
𝑳𝒊
α1
α𝟏𝑯𝟏
𝑳𝒊
α𝟏(𝑯𝟏 + 𝑯𝟐)
𝑳𝒊
Deflection-Bending Deflection-Wall rotation)
∆𝑻𝒐𝒕𝒂𝒍= ∆𝟏 + 𝜽𝟏(𝑯𝟐+𝑯𝟑 ) +α𝟏(𝑯𝟏+𝑯𝟐+𝑯𝟑)
𝑳𝒊+ ∆𝟐 +α𝟐
(𝑯𝟐+𝑯𝟑)
𝑳𝒊+ 𝜽𝟐𝑯𝟑 + ∆𝟑Ex. 3rd flr.
𝐈𝐧𝐜𝐥𝐮𝐝𝐞𝐝 𝐢𝐧 ∆𝟏
translates to top translates to top
(Wall rotation)
+ +
+Rotation
+α𝟏(𝑯𝟏 +𝑯𝟐+𝑯𝟑)
𝑳𝒊
+α𝟏(𝑯𝟏 + 𝑯𝟐+𝑯𝟑 + 𝑯𝟒)
𝑳𝒊
+α𝟏(𝑯𝟏 + 𝑯𝟐+𝑯𝟑 + 𝑯𝟒 +𝑯𝟓)
𝑳𝒊
Distribution of shear to vertical resisting elements shall be
based on:
• Analysis where the diaphragm is modeled as :o Idealized as flexible-based on tributary area.
o Idealized as rigid-Distribution based on relative lateral
stiffnesses of vertical-resisting elements of the story below.
o Modelled as semi-rigid.
Not idealized as rigid or flexible
Distributed to the vertical resisting elements based on the relative stiffnesses of the
diaphragm and the vertical resisting elements accounting for both shear and flexural
deformations.
In lieu of a semi-rigid diaphragm analysis, it shall be permitted to use an enveloped
analysis (Minimum analysis).
Average drift of
walls
Maximum diaphragm deflection
(MDD) >2x average story drift of
vertical elements, using the ELF
Procedure of Section 12.8?
Maximum
diaphragm
deflection
Calculated as Flexible
• More conservatively distributes lateral forces
to corridor, exterior and party walls
• Allows easier determination of building drift
• Can over-estimate torsional drift
• Can also inaccurately estimate diaphragm
shear forces
• Can under-estimate forces distributed to the corridor walls
(long walls) and over-estimate forces distributed to the
exterior walls (short walls)
• Can inaccurately estimate diaphragm shear forces
Note:
Offsets in diaphragms can also
affect the distribution of shear
in the diaphragm due to changes
in the diaphragm stiffness.
2015 SDPWS 4.2.5 Horizontal Distribution of Shear (Revised)
∆=𝟓𝐯𝑳𝟑
𝟖𝑬𝑨𝒃+
𝒗𝑳
𝟒𝑮𝒕+ 𝟎. 𝟏𝟖𝟖𝐋𝒆𝒏 +
𝚺 𝜟𝑪𝑿
𝟐𝒃𝐈𝐁𝐂 𝐄𝐪. 𝟐𝟑 − 𝟏
I1 I2 I3Rigid or spring
Support ??
Exte
rio
r
sh
ear
walls
Typical Multi-residential Mid-rise unit
Rigid support or
Partial support
Support Support
TD1
TD2 No
Support
TD1
TD2
I1I2I3
No support
(Full cantilever)
Adjust terms of the equation for support
condition, stiffness and loading conditions
Wind Loads as applicableSeismic Loads
Support Support
Full support
(SW rigid)
Partial support
(Decreasing
SW stiffness)
No SW
support
If flexible
diaphragm
Corridor only SW
Condition B
Condition ACondition C
• Flexible
• Semi-rigid
• Rigid
• Semi-rigid
• Rigid
Effects of Tall Shear Walls
Note that possible changes in diaphragm flexibility can occur from floor to
floor: non-open front (flexible) vs. non-open front or open front (rigid or
semi-rigid).
open front?
Semi-rigid?
Flexible?
Effects of
openings?
Longitudinal Loading
No exterior
Shear walls ??
Drift by semi-rigid or
rigid analysis only
Calculation of drift at Edges is
required for all open front or
cantilever diaphragms (seismic only)
including torsion and accidental
torsion (Deflection-strength level
amplified by Cd. )
76’
156’
35’
6’
35’
Case Studies-Open-front & Non-Open Front w/ Offsets
26’ typ.
Typical
Unit
7654321
F
C
B
A
1.33 1.67
D
E
A.5
E.5
SW5SW4
SW3SW2
SW1
11.5’ 23’
27’ 35’
Open Front
Non-Open Front
The following information is from an on-going example calculation and is subject to
further revisions and validation. The information provided is project specific, and is for
informational purposes only. It is not intended to serve as recommendations or as the only
method of analysis available. As such, most of the information is based on the authors
opinions.
76’
156’
35’
6’
35’
Case Study-Non-Open Front Floor Plan With Offsets
26’ typ.
Typical
Unit
7654321
F
C
B
A
1.33 1.67
D
E
A.5
E.5
SW5SW4
SW3SW2
SW1
11.5’ 23’
27’ 35’
Varies’
• Flexible analysis
• Rigid analysis
• Semi-rigid (Envelope)
• ASCE 7-10 Section 12.3.1.1- (c), Light framed construction, meeting all conditions:
All Light framed construction
Non-structural concrete topping ≤ 1 ½”
Each elements of the seismic line of vertical force-resisting system complies with the
allowable story drift of Table 12.12-1
Longitudinal-Allowed to be idealized as flexible, provided exterior shear walls exist and
is in compliance with ASCE 7-10 Section 12.3.1.1 (c). If calculations show that the drift at
a line of lateral force resistance exceeds the allowable limit, flexible diaphragm behavior
cannot be used.
Open Front
Non-Open Front
ATS isolator nut
take-up device
and bearing plate
Coupler
Steel rod
Semi-balloon
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
Overt
urn
ing
Resis
tan
ce
26’
Co
rrid
or
∆𝒔𝒘𝒄𝒐𝒓𝒓.
6’
8’
12’
4’
4’
27’
11.5
’3’
11.5
’
Spring
Support
(SW stiff.)
I1I2I3
Computer Model
Tall or
narrow
SW
Full depth
M
V
35’
156’
Sym.
C.L.
3’23’
Spring
Support
(SW stiff.)K1 K2
Seismic
Moderately
stiff. support
.Positive
moment
∑ Kdiaph.
Shear Deflection per
USDA Research Note
FPL-0210
∆𝑫𝒊𝒂𝒑𝒉
∆ 𝑺𝑾𝑨𝒗𝒆𝒓.
∆𝟒∆𝟑∆𝟐∆𝟏
∆𝟒∆𝟑∆𝟐∆𝟏
9’ 9’ 9’ 4’ 4’
Co
rrid
or Ext.
SW
∆ 𝒔𝒘𝒄𝒐𝒓𝒓.
∆𝒔𝒘𝒆𝒙𝒕
∆𝒔𝒘 𝒆𝒙𝒕
Acc. torsion
Bending
Check Total deflection-Diaphragm flexibility
Check exterior SW stiffness effects
Off
se
t
Off
se
t
∑ Vsw∑ Vsw
Shear DeflectionNail slip deflection
Determination of Non-Open Front Diaphragm Flexibility
∆𝑨𝒗𝒆𝒓.𝒔𝒘
Bending increases
at start of offset
Roof Plot- Longitudinal Loads- A/R=1.5:1 (8” wall)
Base Line=0
Base Line=0
Base Line=0
En
d
Off
set
Off
set
51.13 k
46.4 k
43.21 k
4.477.66
80.48
0.674
8.95
414.2’k
526.55’k
692.1’k
Full
cantilever
A/R 1.5:1 wallA/R 2:1 wall
A/R 3.5:1 wall
Drift by semi-rigid
or rigid analysis
only
Spring support
If Shear walls
+
Same
as
=
Check Story Drift-3D model or 2D plane Grid
+Same
as
=
Collectors
Story Drift Check Longitudinal + Torsion (8’ ext. walls)
Significant shear
distributed to corridor
walls (cantilever action)
Notice curvature
(bending) of
offset sections
Maximum drift
(open front)
C.M.
ASCE 7-10 12.8.6
Story drift check
(Non-open front)
Story Drift Check Transverse + Torsion
Maximum drift
(Open front)Maximum drift (Open front)
Torsional irregularity????
C.M.
ASCE 7-10 12.8.6
Story drift check
(Non-open front)
76’
156’
35’
6’
35’
Case Study-Open Front Floor Plan With Offsets
26’ typ.
Typical
Unit
7654321
F
C
B
A
1.33 1.67
D
E
A.5
E.5
SW5SW4
SW3SW2
SW1
11.5’ 23’
27’ 35’
• Rigid analysis
• Semi-rigid (Envelope)
Open Front
Non-Open Front
Corridor walls typical
Shear panels
are required
over corridor
walls
Transverse
shear walls
typical
No exterior
shear walls
typical
Relevant 2015 SPDWS Sections-Open Front Structures
New definitions added:
• Open front structures
• Notation for L’ and W’ for
cantilever Diaphragms
• Collectors
Relevant Revised sections:
• 4.2.5- Horizontal Distribution
of Shears
• 4.2.5.1-Torsional Irregularity
• 4.2.5.2- Open Front Structures
(a)
L’
W’
Force
Cantilever Diaphragm
Plan
Open front
SW
SW
L’
Cantilever
DiaphragmPlan
W’
Open front
SW
SW
SW
ForceL’
W’
Open front
SW
SWForce
Cantilever
Diaphragm
Figure 4A Examples of Open Front Structures
(d)
L’
W’
Open front
SW
SW
Force
Cantilever
Diaphragm
SW
SWL’
Cantilever
Diaphragm
(c)(b)
Open front
4.2.5.2 Open Front Structures: (Figure 4A)
For resistance to seismic loads, wood-frame diaphragms in open front structures shall comply with
all of the following requirements:
1. The diaphragm conforms to:
a. WSP-L’/W’ ratio ≤ 1.5:1 4.2.7.1
b. Single layer-Diag. sht. Lumber- L’/W’ ratio ≤ 1:1 4.2.7.2
c. Double layer-Diag. sht. Lumber- L’/W’ ratio ≤ 1:1 4.2.7.3
2. The drift at edges shall not exceed the ASCE 7 allowable story drift when subject to seismic
design forces including torsion, and accidental torsion (Deflection-strength level amplified
by Cd. ).
3. For open-front-structures that are also torsionally irregular as defined in 4.2.5.1, the L’/W’
ratio shall not exceed 0.67:1 for structures over one story in height, and 1:1 for structures
one story in height.
4. For loading parallel to open side:
a. Model as semi-rigid (min.), shall include shear and bending deformation of the diaphragm,
or idealized as rigid.
5. The diaphragm length, L’, (normal to the open side) does not exceed 35 feet.
Exception:
Where the diaph. edge cantilevers no more than 6’ beyond the nearest line of vertical
elements of the LRFS.
4.2.5.2.1 For open front structures of 1 story, where L’≤25’ and L’/W’ ≤1:1, the diaphragm shall be
permitted to be idealized as rigid for the purposes of distributing shear forces through torsion.
W’
L’
Limit= 35’
Example-Open Front Diaphragms
Side wall
End wall
Side wall S
tru
t
Strut
Strut Strut
Str
ut
SW1
SW2
Str
ut/
ch
ord
W’
Side wall
Side wall
End wall or
Corridor wall
Str
ut
Strut
Strut Strut
Str
ut
SW1
SW2
Location of strut/chord
depends on framing
orientation
L’
Limit= 35’
Wind loading
Seismic
loading
4.2.6.1 Framing requirements:
• Diaphragm boundary elements shall be provided to transmit design
tension, compression and shear forces.
• Diaphragm Sheathing shall not be used to splice boundary elements.
Co
rrid
or
Corridor only shear walls
..
Check drift at edges
(extreme corners)
including torsion
plus accidental
torsion (Deflection-
strength level x Cd. )
w
∆𝒔𝒘
∆𝑫𝒊𝒂𝒑𝒉
Determination of Open Front Diaphragm Flexibility
26’
Co
rrid
or
Corridor only shear walls
.
∆𝒔𝒘
∆𝑫𝒊𝒂𝒑𝒉
I1I2I3
Computer Model
6’
8’
12’
4’
4’
27’
11.5
’3’
M
V
23’
156’
Sym.
C.L.
3’
Spring
Support
(SW stiff.) K1
Full depth
Seismic
Bending ∆
∑ Kdiaph.
Shear Deflection per
USDA Research Note
FPL-0210
Acc. torsion
∑ Vsw
∆𝟑
Shear DeflectionNail slip deflection
L’
Limit= 35’
• Check drift at edges (extreme corners) including torsion plus
accidental torsion (Deflection-strength level x Cd. )
• Seismic-meet ASCE7 drift
• Wind-deflections can be tolerated.
Final Diaphragm (Total) Deflections
∆𝟒∆𝟑∆𝟐∆𝟏
9.5’ 9.5’ 8’ 4’ 4’
Co
rrid
or
∆ 𝒔𝒘𝒄𝒐𝒓𝒓.
Roof
5th Floor
4th Floor
Co
rrid
or
. ∆𝑫𝒊𝒂𝒑𝒉
6’
8’
12’
4’
4’
27’
11.5
’3’
23’
∆ 𝒔𝒘𝒄𝒐𝒓𝒓. 35’
∆𝟒∆𝟑∆𝟐∆𝟏
9.5’ 9.5’ 8’ 4’ 4’
Co
rrid
or
∆ 𝒔𝒘𝒄𝒐𝒓𝒓.
∆𝟒∆𝟑∆𝟐∆𝟏
9.5’ 9.5’ 8’ 4’ 4’
Co
rrid
or
∆ 𝒔𝒘𝒄𝒐𝒓𝒓.I1I2I3
Seismic∑ Vsw
Check for
diaphragm
flexibility
Verify Diaphragm Flexibility:• If flexible, add blocking
• If still flexible, decrease nail spacing
to stiffen up
In summary:
• There are a number of shear wall framing systems available that
can be used to resisting lateral forces.
• Simple calculation methods and software are often available to
analyze and design most shear wall framing types.
• Boundary framing elements and their connections are required to
maintain complete load paths to the shear walls and must be fully
detailed on the drawings.
• Mid-rise structures require the consideration of stiffnesses of the
diaphragms and shear walls to determine the horizontal
distribution of shears within the diaphragm and to the walls.
• Wood Solution Paper
• Slide Archive-Workshop-Advanced Diaphragm Analysis
• Slide Archive-Offset Diaphragms and Shear Walls
• Webinar Archive- Offset Diaphragms -Part 1
Information on Website
Horizontally Offset Diaphragms
Example Offset Diaphragms
• Slide Archive-Workshop-Advanced Diaphragm Analysis
• Slide Archive-Offset Diaphragms and Shear Walls
• Webinar Archive- Offset Shear Walls-Part 2
Information on Website
Horizontally Offset Shear Walls
Example Offset Shear Walls
Additional Reference Materials
• The Analysis of Irregular Shaped
Structures: Diaphragms and Shear
Walls-Malone, Rice - 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
• Paper-The Analysis of Irregular Shaped
Diaphragms-Complete Example with narrative
and calculations
http://www.woodworks.org/wp-content/uploads/
Irregular-Diaphragms_Paper1.pdf
QUESTIONS?
This concludes Our Workshop Presentation on
Advanced Diaphragm Analysis
R. Terry Malone, P.E., S.E.
Senior Technical Director
WoodWorks.org
Contact Information:
928-775-9119