mid-rise shear walls & diaphragms - woodworks topics • shear wall types • shear wall...

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

Presentation updated to 2015 IBC, ASCE 7-10, 2015 SDPWS

Copyright McGraw-Hill, ICC

Portions Based On:

mailto:[email protected]

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

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

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=𝟖𝒗𝒉𝟑

𝑬𝑨𝒃+

𝒗𝒉

𝟏𝟎𝟎𝟎𝑮𝒂+ 𝒉∆𝒂

𝒃

∆𝑺𝑾

Vertical elongation

• Device elongation

• Rod elongation

Bending

Apparent shear stiffness

• Nail slip

• Panel shear deformation

Deflection of standard segmented shear wall

Tests were based on Aspect Ratio (A/R) = 2:1

Deflection permitted to be calculated based

on rational analysis where apparent shear

stiffness accounts for:


• Non-linear nail slip

4.3-1

Deflection of unblocked segmented shear wall

• Use Eq.4.3-1 with v/Cub per 4.3.2.2

v v v

∆𝒔𝒘 =𝟖𝒗

𝑪𝒖𝒃𝒉𝟑

𝑬𝑨𝑩+

𝒗

𝑪𝒖𝒃𝒉

𝟏𝟎𝟎𝟎𝑮𝒂+𝒉

𝒃∆𝒂

ATS

Wall stiffness k = 𝑭

∆

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

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 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 M

ax. 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

Ao = total area of openings

Hold down at ends

𝒗𝒎𝒂𝒙 = 𝑽

𝑪𝒐 ∑𝑳𝒊

𝒗𝒎𝒂𝒙 = 𝑻𝒊𝒏𝒕 =𝑽

𝑪𝟎𝚺𝑳𝒊

𝑻 = 𝑪 =𝑽𝒉

𝑪𝟎𝚺𝑳𝑰

At full-hgt. segments

Perforated Shear Wall Deflection

Vertical elongation

• Nail slip

• Device elongation

• Rod elongation

Bending

Apparent shear stiffness

• Nail slip


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

(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

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)

(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”

L22’-6”

L3

6’-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

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

Side 1

Vsc

Side 2

𝐆𝐚𝐜 = 𝐆𝐚𝟏 + 𝐆𝐚𝟐

𝐯𝐒𝐂 = 𝐊𝐦𝐢𝐧𝐆𝐚𝐜 Combined apparent shear wall shear

capacity (seismic)

𝐊𝐦𝐢𝐧 =𝐯𝐒𝟏

𝐆𝐚𝟏𝐨𝐫𝐯𝐒𝟐

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 table

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

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.

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)

Where multiplied by 2b/h the Vn need not be

reduced 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 4.3.4.2

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.

Where Vn x 0.1+0.9b/h, the nominal shear

capacities need not be reduced by 1.09-0.09h/b

• 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 SW2h

Example Calculation per Commentary

(assumes same unit shear capacity)

SW1 and SW2

• Determine ASD unit shear capacity

• Check aspect ratio-adjust per SDPWS 4.3.4.1 (and/or exception), Aspect ratios

> 2:1 use reduction factor 𝟐𝒃𝒔

𝒉(further reductions in 4.3.4.2 are not required)

• Sum design strengths = vsw1 x b1 + vsw2 x b2

for line strength

• Will produce similar results to equal deflection method


Method 2-Equal Deflection

∆𝑺𝑾𝟏 ∆𝑺𝑾𝟐

b1 b2

SW1 SW2h

Method of Analysis

SW1 and SW2

• Determine ASD unit shear capacity

• Check aspect ratio-adjust per SDPWS

4.3.4.2

• Determine Ga and EA values

• Calculate deflection of larger SW (SW2)

𝒗𝒔𝒘𝟏 =∆𝒔𝒘𝟐−

𝒉∆𝒂𝒃

𝟖𝒉𝟑

𝑬𝑨𝒃+𝒉

𝟏𝟎𝟎𝟎𝑮𝒂

• 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

Presentation Topics






Requirements


Typical Floor Plan

Typical

Unit

Transverse walls

typical

Longitudinal

exterior walls

typicalCorridor walls

typical

26’

StrutSW1

SW2

35’ maximum if open front

(i.e. No ext. wall)

Wind loading

Seismic

loading

Co

rrid

or

6’

8’

12’

4’

4’

27’

11.5’

3’

23’

Unit 1 of 6

3’

11.5’

SW

3

Typical Offset Exterior Walls - Plan Layout

Let’s look at

exterior offset

walls

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

I1 I2

SW

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.)

Framing and Transfer Diaphragm Layout Options

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 or Diaphragm??

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


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’

120

120671.2

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 discontinuity12.3.3.3 Discont. Walls SDC B-F

12.3.3.4 25% incr. SDC D-F

ASCE 7 Table 12.3-1

Type 4 horizontal irregularity-out-of-

plane offset discontinuity in the LFRS12.3.3.3 Discont. Walls SDC B-F

12.3.3.4 25% incr. SDC D-FASCE 7 Table 12.3-2

Type 4 vertical irregularity- in-

plane offset discontinuity in

the LFRS (if no H.D. at A.25) 12.3.3.3 Discont. Walls SDC B-F

12.3.3.4 25% incr. SDC D-F

Relevant Irregularities Per ASCE 7-10 Tables 12.3-1 and 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.

• 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)

• 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

Exception: Forces using the seismic load effects including the over-strength factor

of Section 12.4.3 need not be increased.

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)

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

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 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



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 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.

Presentation Topics






Requirements


Mid-rise Diaphragms and Shear Walls

Crescent Terminus building- credit: Richard Lubrant.

Important Design considerations: Continuous load paths

Vertical/horizontal irregularities

Diaphragm stiffnesses and flexibility

Shear wall stiffness and the effects of tall walls

Horizontal distribution of forces within the diaphragm and

shear walls

The effects of offset diaphragms and shear walls

• ASCE 7-10 Section 12.3.1.1- (c), Light framed construction, meeting all conditions:

Longitudinal-Allowed 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

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 ratio when

subject to seismic design forces including torsion, and accidental torsion

(strength, multiplied by Cd). Drift check not required for wind.

Includes flexible and rigid analysis

Typical Floor Plan

Typical

Unit

Open Front / Cantilever Diaphragm-Corridor Shear Walls Only

Transverse

shear walls

typical

No exterior

shear walls

typicalCorridor walls

typical

26’

StrutSW1

SW2

35’ Maximum

(2015 SDPWS)

Wind loading

Seismic

loading

Co

rrid

or 6’

8’

12’

4’

4’

27’

11.5’

3’

23’

Unit 1 of 6

3’

11.5’

Shear panels are

required over

corridor walls

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

W’

L’

Shear

Defl.Nail

sliprotation End shear

wall lateral

translation

Example-Simple Open Front Diaphragm Deflection (APA)

End wall

Side wall

∆𝑨=𝑽𝒎𝒂𝒙 𝑳

𝟐𝑮𝒕+ 0.376L’𝒆𝒏 +

𝟐∆𝒔𝒘𝑳

𝒃+ ∆𝒆𝒘

∆𝒔𝒘𝟏

∆𝒔𝒘𝟐

∆𝑨

∆𝒔𝒊𝒅𝒆𝒘𝒂𝒍𝒍

𝜽

𝜽

Note:

If there are splices in the struts, chord slip must

be added to the deflection equation

Str

ut

Strut

Strut Strut

Str

ut

SW1

SW2

Str

ut

/ ch

ord

Bending term required if

chords cont. across corridor

Rotation term needs to be

adjusted if different end

wall lengths

∆𝒔𝒘

∆𝒔𝒘∆𝑨

𝜽

∆𝑬𝒘

b

L

En

d w

all

Side wall

APA Figure

Open

end

w

Boundary element not

restrained for bending

e

V

F

F

Side wall

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 Horizontal Distribution of Shear (Revised)

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.

Can be idealize as rigid when the computed maximum in-plane deflection of the

diaphragm is less than or equal to two times the average deflection of adjoining

vertical elements of the LFRS under equivalent tributary lateral load.

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.

It shall be permitted to use an enveloped analysis -Larger of the shear forces resulting

from analyses where idealized as flexible and idealized as rigid (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 accurately distributes lateral forces

to corridor, exterior and party walls

• Allows better determination of building drift

• 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.

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 ratio 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.

b. Drift at edges of the structure ≤ the ASCE 7 allowable story drift ratio when

subject to seismic design forces including torsion, and accidental torsion. Torsion

(Deflection-strength level amplified by Cd. )

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.

• Idealized as Flexible

• Idealized as Rigid

ASCE7-10 Section 12.3-Diaphragm Flexibility

or…..

12.3.1 Diaphragm Flexibility- The structural Analysis shall consider the relative

stiffnesses of the diaphragm and the vertical elements of the seismic force-

resisting system. Unless a diaphragm can be idealized as either flexible or rigid,

the structural analysis shall explicitly include consideration of the stiffness of the

diaphragm (i.e. semi-rigid modeling assumption).

• You can model it as Semi-rigid

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

Methods Used:

• Shiotani/Hohbach Method

• Thompson Method

• FPInnovations Method

• Computer models (Cashew)

A/R

h/d ≤ 3.5:1

Tall Shear Walls

+

+

Traditional Method

• Traditional Method good for 3 stories or less

(w/ exception)

• Tall walls greater than 3 stories require

consideration of flexure and wall rotation.

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 cannot be used for walls with

aspect ratios higher than 2:1, because

deflections can not be adequately predicted.(Dolan)

Walls assumed

to be pinned at

top and bottom

at each floor-Rigid

wall sections

∆𝟑.𝟓:𝟏∆𝟐:𝟏

Tall Wall

A/R, Stiffness of Shear Walls

Diaphragm

out-of-plane

stiffness=Flexible

Analyze as

tall wall

Diaphragm

out-of-plane

stiffness=Rigid

Analyze as

floor to floor

Doesn’t account for multi-story shear wall effects

FPInnovations-Deflections of Stacked Multi-story Shear Walls”A Mechanics-Based Approach for Determining Deflections of Stacked Multi-Storey Wood-Based Shear

Walls”, Newfield (metric units)

• Mechanics based approach

• Uses 5 part deflection equation

∆𝒊= ∆𝒃,𝒊 + ∆𝒔,𝒊 + ∆𝒏,𝒊 + ∆𝒂,𝒊 + ∆𝒓,𝒊

𝑴𝒐 =

𝒊

𝒏

𝑽𝒊𝑯𝒊

𝑽𝒐 =

𝒊

𝒏

𝑽𝒊

Ls

Lc

ytr

∆𝒏

Where

∆𝒃,𝒊= 𝑽𝒊𝑯𝒊𝟑

𝟑 𝑬𝑰 𝒊+ 𝑴𝒊𝑯𝒊𝟐

𝟐 𝑬𝑰 𝒊

∆𝒔,𝒊= 𝑽𝒊𝑯𝒊

𝑳𝒊𝑩𝒗,𝒊

∆𝒏,𝒊= 0.0025𝑯𝒊𝒆𝒏,𝒊

∆𝒂,𝒊= 𝑯𝒊

𝑳𝒊𝒅𝒂,𝒊 (Includes affects of rod elongation

and crushing)

∆𝒓,𝒊= 𝑯𝒊 ∑𝒋−𝟏𝒊−𝟏 𝑽𝒋𝑯𝒋

𝟐

𝟐 𝑬𝑰 𝒋+𝑴𝒋𝑯𝒋

𝑬𝑰 𝒋+𝑯𝒊 ∑𝒋−𝟏

𝒊−𝟏 𝒅𝒂,𝒋

𝑳𝒋

∆𝒂,𝒊

Bending

Panel

Shear

Nail slip

Anchor

Elongation

Wall

Rotation

5 part deflection equation

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 . 𝒚𝒕𝒓𝟐 +𝑨𝒄 . 𝑳𝒄 − 𝒚𝒕𝒓

𝟐

Uses transformed composite section

(transformed bending stiffness)

where continuous tie down (ATS)

∆𝒃,𝒊

𝑰=𝑨 . 𝑳𝟐/2 if discrete hold down

(wood chord members used)

(cont. hold down used)

∆=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

Longitudinal Loading

No exterior

Shear walls ??

Drift by semi-rigid or

rigid analysis only

Calculation of drift at corners is

required for all open front or

cantilever diaphragms (seismic only)

including torsion and accidental

torsion (Deflection-strength level

amplified by Cd. )

Determination of Open Front Diaphragm Flexibility

26’

StrutSW1

SW2

35’

Wind loading

Seismic

loading

Areas of non-clarity:• Determination of diaphragm flexibility if open front

• Effects/requirements of blocking

• Wind requirements for open front diaphragms

• Exterior wall contribution, ignore? Model as non-resisting?

Corr

idor

Corridor only shear walls

.

∆𝒔𝒘

∆𝑫𝒊𝒂𝒑𝒉 I1I2I3

Computer Model

6’

8’

12’

4’

4’

27’

11.5’

3’

M

V

23’

Unit 1 of 6

156’

Sym.

C.L.

3’

11.5’

Spring

Support

(SW stiff.) K1

Full depth

Wind

Seismic

Bending ∆

Does not

include

rotational

effects

∑ Kdiaph.

Shear Deflection per

USDA Research Note

FPL-0210

Location of strut/chord

depends on framing

orientation

26’

StrutSW1

SW2

Wind loading

Seismic

loading

Co

rrid

or

∆𝒔𝒘𝒄𝒐𝒓𝒓.

6’

8’

12’

4’

4’

27’

11.5’

3’

11.5’

Unit 1 of 6

Determination of Diaphragm Flexibility and Shear Distribution With

Exterior Shear Walls

Narrow Exterior shear wallsFlr. / Flr. hgt.= 10’-0”

∆𝒔𝒘 𝒆𝒙𝒕

Spring

Support

(SW stiff.)

I1I2I3

Computer Model

SW

3

Tall or

narrow

SW

Full depth

M

V

35’

156’

Sym.

C.L.

3’

23’

Spring

Support

(SW stiff.)K1 K2

Not considered open front because of ext. SW-However, ?????

1K

2.5K

5K

Wind

Seismic

10KNo stiff. or

support

Low stiff.

Moderately

stiff.

Rigid

support

. .Positive

moments

100% 97.5%95.2%91% 0%

2.5%4.8%9%

Place holders

∑ Kdiaph.

.Shear Deflection per

USDA Research Note

FPL-0210

∆𝑫𝒊𝒂𝒑𝒉

∆ 𝑺𝑾𝑨𝒗𝒆𝒓.

2D Computer Model-Longitudinal Loading

Spring support

if Shear walls = SW stiffness

Drift by semi-rigid or

rigid analysis only

+Same

as

=23’11.5’

Calculation of drift at corners is required for all open front or

cantilever diaphragms (seismic only) including torsion and

accidental torsion (Deflection-strength level amplified by Cd. )

Drift by semi-rigid

or rigid analysis

only

Spring support

If Shear walls

+

Same as

=

2D Computer Model-Transverse Loading

Calculation of drift at

corners is required for all

open front or cantilever

diaphragms (seismic only)

including torsion and

accidental torsion

(Deflection-strength level

amplified by Cd. )

Model asModel as

Note: Rigid diaphragm

analysis can cause

torsional irregularity even

if uniform distribution of

transverse walls

Reference Materials

http://www.woodworks.org/publications-media/solution-papers/

http://www.woodworks.org/wp-content/uploads/Irregular-

Diaphragms_Paper1.pdf

• The Analysis of Irregular Shaped Structures: Diaphragms and Shear Walls-Malone,

Rice-Book published by McGraw-Hill, ICC

• Woodworks-The Analysis of Irregular Shaped Diaphragms (paper)

• Woodworks Webinar-Offset Diaphragms-Part 1

• Woodworks Presentation Slide Archives-Workshop-Advanced Diaphragm Analysis

• Woodworks Webinar-Offset Shear Walls- Part 2-Coming on December 16

• 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

Woodworks-The Analysis of Irregular Shaped

Diaphragms (paper). Complete Example with narrative and

calculations.

mid-rise shear walls & diaphragms - woodworks topics • shear wall types • shear wall...

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