Mid-rise Shear Walls & Diaphragms

By: R. Terry Malone, PE, SE

Senior Technical Director

Architectural & Engineering Solutions

[email protected]

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:

mailto:[email protected]

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.


Discuss complete load paths and the detailing required to maintain

lateral load paths to shear walls.


Review shear wall code requirements.


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


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.


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:


• 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:



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:



opening


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:



opening


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:



opening


and below opening

• Corner forces


• 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:



opening


and below opening

• Corner forces



• 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:



opening


and below opening

• Corner forces



• 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:



opening


and below opening

• Corner 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:



opening


and below opening

• Corner 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


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






Requirements


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


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


Slack in hold down strap (bent)

Often stabbed into foundation

without vibration.


Slack in strap (bent) due

to vertical shrinkage of

framing.


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


Automatic Tensioning Systems/Devises

ATS isolator nut

take-up device

and bearing plate

Coupler

Steel rod

Modified

balloon framing

Platform framing


• 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


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






Requirements


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)


• 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


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






Requirements



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 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






Requirements



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


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


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


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


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


Case 2-Larger resulting forces at corridor

O.K.

22’

10’

12’

6’


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


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


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




Chord

Co

llecto

r

Str

ut

Str

ut

Chord

Str

ut


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




Chord

Co

llecto

r

Str

ut

Str

ut

Chord

Str

ut


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


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



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



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




Dhr

Dh2

Connections are



ASCE 7-10 Sections

2.3 or 2.4.

Wall element

designed for

standard load

combinations of



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


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






Requirements



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


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


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


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???


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

http://www.woodworks.org/education/online-seminars/

http://www.woodworks.org/wp-content/uploads/HOHBACH-Mid-Rise-Shear-Wall-and-Diaphragm-Design-WSF-151209.pdf

http://www.woodworks.org/wp-content/uploads/5-over-1-Design-Example.pdf

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


• 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


Determination of Open Front Diaphragm Flexibility

26’

Co

rrid

or

Corridor only shear walls

.

∆𝒔𝒘


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


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


∆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:

[email protected]

928-775-9119

Mid-rise Shear Walls & Diaphragms

Documents

Transcript of Mid-rise Shear Walls & Diaphragms