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SEAU 5th Annual Education Conference 2-21-17

1

WOOD SHEAR WALL DESIGN

By: Doug Thompson S.E., SECB

STB Structural Engineers, Inc.1

Learning Objectives

Become Familiar with:

1. Shear wall holdowns, failures & splitting

2. Types of shear walls

3. Multi-story shear wall design

4. Shear wall anchorage

5. Stiffness of shear walls

6. System stretch

7. Distribution of shear to shear walls in a line

8. Force Transfer Around Openings

2


2

Design Example

3

Design Examples

4


3

5

IBC & SDPWS

2012 IBC

USE STANDARD

SDPWS-08

6

IBC vs. SDPWS

USE STANDARD

2015 IBC SDPWS-15


4

Shear Wall HoldownsFailures & Splitting

7

Wood Shear Wall Design

Holdown Slip

8


5

Multi Story Tie Down Systems

9

Chord/Boundary Member Failure

10


6

Chord/Boundary Member Failure

Chord failed in tension!

11

Typical Failure Mode

Nail pull through, Edge tear, Nail yield

12


7

Nails Working in Lumber

13

Nails Resistance to Wall Racking

F

b

d

h

14


8

Nails Yielding

15

MODE III– fastener yield in bending at one plastic hinge and bearing –dominated yield of wood fibersMODE IV

– fastener yield in bending at two plastic hinges and bearing –dominated yield of wood fibers

NDS Dowel Yield Equations

16


9

NDS Yield Limit Equations

17

Splitting happens because wood is relatively weak perpendicular to grain

Nails too close

(act like a wedge)

Material Properties of Wood

18


10

Splitting will not occur perpendicular to grain, no matter how close nails are

Splitting occurs parallel to grain

Staggering

Staggering a line of nails parallel to wood grain

minimizes splitting

Material Properties of Wood

20

Nail Splitting Avoidance

2x4 3x4


11

Minimum Edges Distances

3/8” MIN

Fastener Spacing

Fra

min

g W

idth

21

Staggered Nailing

Fastener Spacing

Fra

min

g W

idth

3/8” MIN

22


12

Framing at Adjoining Panel Edges

1/8” Gap is a requirement from the Wood Structural Panel manufacturer

3x at panel edge required:8d @ 2” o/c or less - staggered (410 plf ASD)10d @ 3” o/c or less - staggered (665 plf ASD)SDC D,E or F: 350 plf ASD, 560 plf LRFD

Types of Shear Walls

24



13

3 Types of Shear Walls

I. Segmented1. Aspect Ration

2:1 for Seismic2. Aspect Ratio up

to 3 ½ to 1 if allowable shear reduced by 2w/h

II. Force Transfer1. Code does not

provide guidance for this method

2. Different approaches using rational analysis can be used.

III. Perforated1. Specific code

requirements2. Wall capacity is

based upon empirical equations and tables

SDPWS §4.3.5.1 SDPWS §4.3.5.2 SDPWS §4.3.5.3

25

Design Example – Type I SegmentedH

= 9

.0’

L = 10.0’

3.0’ 3.0’

26

4.0’

4.0’

2.33

’2.

67’

V = 4159 lb

Both walls length are the same therefore no need to consider load distribution


14

Segmented Wall Design

H =

9.0

’

L = 10.0’

3.0’ 3.0’

27

4.0’

4.0’

2.33

’2.

67’

Wall PierHeight

Wall PierWidth

Design Example – Type I Segmented

28


15

H =

9.0

’

3.0’

29

4.0’

4.0’

2.33

’2.

67’

3.0’

V = 4159 lb

V/pier = V/2

Segmented Wall Design

Aspect Ratio Factor (WSP for Strength

(SDPWS 4.3.4.2)

ASD unit shear capacity for seismic

vsw1 = 520 plf/2 x 0.57 = 148 plf

h=9.0’

bs=3.0’

SW

2 bs /h Adjustment for Aspect Ratio

30


16

Determine Required Wall Nailing

Nominal shear capacity required :v ASD = 485 plf

Where:

ASD Reduction Factor = 2.0

Aspect Ratio Factor = 0.875

Use 1 side½” Struct I w/ 10d @ 3” o/c

31

Determine Uplift for at Holdown

Uplift force:

Mot = F x h

Mot = 2780 x 9.0

Mot = 25,020 ft-lb

Using no resisting loads:

Strength Uplift = Mot/d

Uplift = 25,020/2.58 =6,685 lbs

Check bearing on plate from 4x4 for ASD:

F = 2780 lbs

3.0’

9.0’

2.42’

32


17

Type I Segmented Summary


4x4 with holdowns good for 6,685 lb(strength)

33

Design Example – Type III PerforatedH

= 9

.0’

L = 10.0’

3.0’ 3.0’

34

4.0’

4.0’

2.33

’2.

67’

V = 4159 lb

Entire wall sheathed. No straps or blocking as window head and sill locations.


18

Design Example – Type III Perforated

35

Perforated Shear Walls

36


19

Determine Tension & Compression Chord Forces

H =

9.0

’

3.0’Li

37

3.0’Li

T = 11,275 lb C = 11,275 lb

Check bearing on plate

from 4x4 for ASD:

Need 4x6 Posts at Ends

Design Example – Type III PerforatedH

= 9

.0’

L = 10.0’

3.0’ 3.0’

38

4.0’

4.0’

2.33

’2.

67’

Maximum opening height:

Percent of Full Height Sheathing:


20

Capacity Adjustment Factor Co

39

SDPWS Figure 4C

40


21

Determine Height to Width Ratio

H =

9.0

’3.0’Li

41

SegmentHeight

SegmentWidth

3.0’Li

Adjusted Length ∑Li= 6.0 x 0.67 = 4.0 ft

Determine Maximum Induced Shear

SDPWS §4.3.5.3.3

42


22

Determine Required Wall NailingNominal shear capacity required :V strength = 1255 plv v ASD = 1255 x 0.7 = 875 plf

Where:


Use two side of sheathing:

V per side = 1755/2 = 875 plf

Use 2 sides½” Struct I w/ 10d @ 4” o/c

43

Design Example – Type III Perforated

44


23

Type III Perforated SummaryUse 2 sides½” Struct I w/ 10d @ 4” o/c


Holdown good for 1670 lb each full height stud (strength)

45

Multi-Story Shear Wall Design

46



24

Continuous Tie-Down Shear Walls

47

Wall Edge-Modified Balloon Frame

48

Top Plate/Chord Top Plate/Chord

48


25

Wall Edge-Modified Balloon Frame

49

Ribbon/Chord Ribbon/Chord

50

Wall Edge-Platform Framed


26

51

Wall Edge-Platform Framed

F4ROOF

F34TH FLR

H1+

H2+

H3

H1+

H2

H1

H1+

H2+

H3+

H4

STORY HEIGHT H1

STORY HEIGHT H2

STORY HEIGHT H3

STORY HEIGHT H4ΣMOT

LEVEL 4

ΣMOT

LEVEL 3

ΣMOT

LEVEL 2

ΣMOT

LEVEL 1

F2 3RD FLR

F12ND FLR

dT C

b52

d’

Shear Wall Overturning Forces


27


53

Fx

Fx

MOT

MOT

53

Vwall above

Vfloor

h1 h2

54



28

h2Point of application of load to shear wall

Point of application of load to chord/collector

55



Fx

Fx

MOT

MOT

56


29

Vwall above

Vfloor

h1 h2

57


h2Point of application of load to shear wall


Framing Clips

58



30

Vwall above

Vfloor

h1 h2

59


h3

So Which height to Use?

F

h

b

60


31

How Much Load On Boundary Posts?

61

w

TRIB

61

Δa

Δa

Δa

Δa

δ

62

Design shear wall boundary posts

Shear Wall Compression Forces


32

63

Offset Posts

4-4x6

4-4x4

19.17’d’1.63’

d’’0.71’

Centroid of Posts

Centroid of Rod

d

64

Concentric Posts

4-4x6

4-4x4

19.17’d’1.17’

d’’1.17’

Centroid of Posts

Centroid of Rod

d


33

Design DetailMETAL PLATE CONNECTED TRUSSES @ 24”

65

PREMANUFACTURED I JOISTS @ 24”

Shear Wall Chord Forces

Level

(ft-k)

ASD

(k)

d’

(ft)

d

(ft)

ASD DemandCompression

Strength Demand Compression

(k) (k)Drag Truss

-- -- -- -- 1.32 1.80

Roof 51.50.36

51.13 19.67 3.56 4.85

4th Floor 177.60.89

21.13 19.67 8.66 11.84

3rd Floor 348.31.63

01.29 19.50 15.71 21.80

2nd Floor 541.22.81

21.63 19.17 24.30 34.37

66


34

Determine Chord Members

67

LevelChordPosts

TotalArea

le

(ft)Cf Cp

Bearing

Cap.(kips)

ASDDeman

d(kips)

StabilityCapacity

(kips)

D/CRatio

RoofFour 3 x

435.0 7.82 1.15 0.258 21.88 3.56 24.93 0.16

4th FloorFour 3 x

435.0 8.84 1.15 0.192 21.88 8.66 18.54 0.47

3rd FloorFour 4 x

449.0 8.84 1.15 0.192 30.63 15.71 25.96 0.61

2nd FloorFour 4 x

677.0 8.84 1.1 0.200 48.13 24.30 40.70 0.60

Compression Members

4×4 PostCD = 1.6

Ke 1 = 1.0 d1 = 3.5”

68

Kl e

1=

8.9

4’

9.44

’


35

Bearing of Boundary Members

4 - 4×4 Posts

Chord compression force (D+L+E)

LRFD

P = 21,800 Ab = 49.0 in2 lb > 6 inches Cb = 1.0

69

Δa

Δa

Δa

Δa

δ

70

Design Tie down rods and bearing plates.

Shear Wall Uplift Forces


36


Level

(ft-k)

d

(ft)

StrengthUplift

ASD Uplift Differential load

Per Floor

(lbs)(ft-lb) (ft-lb) (lb)

Drag Truss -- -- -- -- 1.32 --

Roof 31.0 19.67 51.5 14.2 2.43 0

4th Floor 66.7 19.67 177 30.5 6.09 3,660

3rd Floor 102 19.50 348 46.7 11.4 5,332

2nd Floor 128 19.17 541 63.0 17.8 6,373

71

Rod Capacities & Elongations

LevelPlate

Height(ft)

ASDTensionDemand

(kips)

RodDia.

d(in)

Eff.Dia.de

(in)

Ag

(in2)Ae

(in2)

Fu

(ksi)

Fy

(ksi)

ASD Rod

Capacity

ASDRod

Elong.

(in)(kips)

Roof 8.21 2.430.62

50.52

70.307 0.226 58 36 6.68 0.037

4th

Floor9.44 6.09

0.625

0.527

0.307 0.226 58 36 6.68 0.105

3rd

Floor9.44 11.43

0.875

0.755

0.601 0.462 58 36 13.07 0.097

2nd

Floor9.44 17.80 1.00

0.865

0.785 0.606 120 105 35.33 0.11572


37


Level

(ft-k)

d

(ft)

StrengthUplift


Per Floor


Drag Truss -- -- -- -- 1.32 --

Roof 31.0 19.67 51.5 14.2 2.43 0

4th Floor 66.7 19.67 177 30.5 6.09 3,660

3rd Floor 102 19.50 348 46.7 11.4 5,332

2nd Floor 128 19.17 541 63.0 17.8 6,373

73

10.0

’

19.17’

15.71 k 11.4 k

24.30 k

Shear Wall Boundary Forces

17.8 k

Uplift forcefrom above

74

Differential Load:17.8-11.4k = 6.4k


38

Bearing Plate Capacities

75

Level

Bearing Plate Bearing

FactorCb

Bearing

Load(kips)

Allowable

Capacity(kips)

Width

(in)

Length

(in)

Thickness(in)

Hole dia.(in)

ABrg

(in2)

Roof 3.0 5.5 0.6 0.8125 15.98 1.07 2.433 10.69

4th

Floor3.0 3.5 0.4 0.8125 9.98 1.11 3.660 6.92

3rd

Floor3.0 5.5 0.6 1.0625 15.61 1.07 5.332 10.44

2nd

Floor3.0 5.5 0.6 1.1875 15.39 1.07 6.373 10.29

Bearing Plate Check

76

76



39

Bearing on Plates

Allowable Bearing Perpendicular to Grain Fc┴

Douglas Fir Larch 625

Hemlock Fir 405

LVL 480

LSL 435

ASD

LRFD

77

Bearing Plate Check

4.0”× 4.0” Bearing Plate

NDS 3.10.4

When lb < 6 inches

When lb > 6 inches

Differential Load = 6,373 lb

78

lb

Differential Load


40

Bearing Zone Through Framing

Framing at the floors resists both downward overturning forces and upward overturning forces.

As a general rule the downward loads increase as they go down the structure and the boundary posts stack and increase as they go down.

Engineer needs to consider how the loads are transferred through the framing for the differential loads.

79

80

Transfer Overturning Through Floor Framing

2nd Floor

3rd Floor

5,332 lb

6,373 lb

MOT


41

81

Bearing Zone

Bearing Check:

Differential load at 3rd floor

5,332 lb

Thickness of framing at floor:

(2 × 1.5) + (23/32) + 3.5 = 7.2”

Bearing plate width = 5.5”

Bearing width at bottom of 4 x 4 top plate:

(5.5 + 7.2 + 7.2) = 19.9 inches

7.2”

5.5”

5.5” 6.0”

19.9”

82

Bearing Zone

Neglecting the trimmer stud, there are 3 - 4 x 4 compression posts within the bearing area:

Bearing area:

(3.5 × 3.5) × 3 = 36.7 in2

Bearing stress:

7.2”

5.5” 6.0”

19.9”

5.5”


42


Level

(ft-k)

d

(ft)

StrengthUplift


Per Floor


Drag Truss -- -- -- -- 1.32 --

Roof 31.0 19.67 51.5 14.2 2.43 0

4th Floor 66.7 19.67 177 30.5 6.09 3,660

3rd Floor 102 19.50 348 46.7 11.4 5,332

2nd Floor 128 19.17 541 63.0 17.8 6,373

83

10.0

’

19.17’

15.71 k 11.4 k

24.30 k 17.8 k

84


Boundary Member Nailing

Vertical Shear Force:6.4k /10.0 = 640 plf


43

85

Nailing to Boundary Posts

4-4x6

4-4x4

19.17’d’1.63’

d’’0.71’

Centroid of Posts

Centroid of Rod

d

2nd Flr


Sp

acin

g

4 POSTS

Edge Nail at Wall Ends

86

Vertical Shear Force:6.4k /10.0 = 640 plf


44


Sp

acin

g

2 POSTS EACH SIDE 3 OR MORE POSTS EACH SIDE

Sp

acin

g

Edge Nail Spacing87


88

E.N. SPACING PER PLAN

EDGE NAIL SPACING TO EACH BOUNDARY POST

4 POSTS 6+ POSTS

2 4” 6”

3 6” 9”

4 8” 12”

6 12” 12”

88


45

89

Wall Level

ASD

Cumulative

OT Tension Forces

ft

ASD

Cumulative OT

Compression Forces

ft

Story Height

ft

Estimated Wood

Shrinkage &

Settlement per floor

C 4 2430 3560 8’-3” 0.25

C 3 6090 8660 9’-4” 0.25

C 3 11400 15710 9’-4” 0.25

C 1 17800 24300 9’-4” 0.25

Sample Drawing Specification:

Shear Wall Ancorage

90



46

Fasteners in Treated Wood

Connectors that are used in exterior applications and in contact with preservative-treated wood shall have coating types and weights in accordance with treated wood or connector manufacturer’s recommendations. In the absence of manufacturer’s recommendations, a minimum of ASTM A 653, type G185 zinc-coated galvanized steel, or equivalent shall be used.

91


Exception: Plain carbon steel fasteners, including nuts and washers in SBX/DOT and zinc borate preservative-treated wood in an interior, dry environment shall be permitted.

92


47

93

2015 IBC‐Connectors in Treated Wood

Table from Simpson Strong-Tie Catalog


Fasteners for fire-retardant-treated wood used in exterior applications or wet or damp locations shall be of hot-dipped zinc-coated galvanized steel, stainless steel, silicon bronze or copper. Fasteners other than nails, timber rivets, wood screws and lag screws shall be permitted to be mechanically deposited zinc-coated steel with coating weights in accordance with ASTM B 695. Class 55 minimum.

94


48


Fasteners for fire-retardant-treated wood used in interior locations shall be in accordance with the manufacturer’s recommendations. In the absence of manufacturer’s recommendations, Section 2304.9.5.3 shall apply.

95

2015 IBC‐Fasteners in Treated Wood

Exception: Plain carbon steel fasteners, including nuts and washers in SBX/DOT and zinc borate preservative-treated wood in an interior, dry environment shall be permitted.

96


49

SummaryThe Reason for Higher Fasteners Corrosion Rate

in ACQ compared to CCA is due to the:

Larger amount of copper in ACQ. Copper is “Noble” metal and causes electrochemical reaction with less Noble metals

Active ingredients in ACQ not corrosion inhibitors; chrome & arsenic in CCA serve as inhibitors

Quat in ACQ is surfactant - absorbs water. Chromate in CCA is water repellant, lowers moisture content.

97

ACQ-C

CA-B

98


50

Corrosion Resistant Fasteners

Hot Dipped Galvanized Toe Nails to Sill Plate

99

Corrosion Resistant Fasteners

Hot Dipped Galvanized Edge Nailing to Sill Plate

100


51

101

Typical Wall Anchors

101

Sill Plate Anchorage

Thickness

102


52

F

Δ

103

h

b

Shear Wall Rotation

Rotation

Holdown

Exception: Standard cut washers are permitted where anchor bolts are designed to resist shear only and meet the following requirements:

Wall is designed with provisions of SDPWS Section 4.3.5.1 (segmented conventional shear wall) neglecting dead load stabilizing moment.Shear wall aspect ratio h/b does not exceed 2:1The nominal unit shear capacity of the shear wall does not exceed 980 plf for seismic or 1370 plf for wind.

For ASD seismic: (980/2.0) = 490 plf

SPDWS‐15Sill Plate ‐ Washer Plate Reqmts.

104


53

Wall Width

Cle

ar H

eigh

t

Summary: Washer Plates can be omitted:h/b < 2.0Seismic ASD shear < 490 plfHoldowns designed for overturning force(s) neglecting dead loads.

105

SPDWS‐15Sill Plate ‐ Washer Plate Reqmts.

What’s Wrong With This Picture?

106


54

Washer plate shall extend to within ½ inch of the edge of the sill plate on the side(s) with sheathing.

Sill Plate ‐ Washer Plate Reqmts.

4” Studs107

Sill Plate Splitting Failure

108


55

109

Heavy Plate Washers w/ Slots

Standard Cut Washers are required at Heavy Plate Washers with Diagonal Slots

Requires taller projection from concrete.

½” MAX.

Sill Plate Requirements

110

3x Sill 2x Sill

3x Sill Required:8d @ 2” o/c or less - staggered (410 plf ASD)10d @ 3” o/c or less - staggered (665 plf ASD)SDC D,E or F: 350 plf ASD, 560 plf LRFD


56

Washer plate shall extend to within ½ inch of the edge of the sill plate on the side(s) with sheathing.

Sill Plate ‐ Washer Plate Reqmts.

4” Studs 6” Studs & Larger

Alternate to staggered anchor bolts is larger plates

111

F

h

b

Shear Wall Rotation

Shear Wall Rotation

112


57

Sill Plate Resisting Rotation

113

DRAFT

Early Physical Testing/Research

114


58

DRAFT

DRAFT

Post-test documentation: (after +/- 2” displacement)Bolt stretching. Shallow concrete spall. Bolt did not pull out.

116


59

• For sill pates of 2x or 3x nominal thickness, the allowable lateral design strength for shear parallel to grain of sill plate anchor bolts is permitted to be determined using the lateral design value for a bolt attaching a wood sill plate to concrete, as specified in AF&PA NDS Table 11E, provided the anchor bolts comply with all of the following:1. The maximum anchor nominal diameter is 5/8”;2. Anchors are embedded into concrete a minimum of 7”;3. Anchors are located a minimum of 1-3/4” from the edge of the

concrete parallel to the length of the wood sill plate; and4. Anchors are located a minimum of 15 anchor diameters from the

edge of the concrete perpendicular to the length of the wood sill plate.

IBC-2015, 2305.1.2 - Sill plate anchor bolts

117

118

19.17’d’1.63’ d

Anchorage to Podium


60

119

Anchorage to Podium

Discontinuous Systems Past codes exempted slabs supporting

light frame construction. Exemption has been removed. ASCE 7-10 §12.3.3.3 requires

amplification of seismic loads in the design of structural elements supporting discontinuous walls.

For Light-framed Shear WallsΩ0 = 3.0

If “flexible diaphragms” used, can reduce to Ω0 = 2.5 (footnote “g”)

TU

PU PU

120

Anchorage to Podium

Discontinuous Systems ASCE 7-10 §12.3.3.3 states that the

connections of the discontinuous wall to the supporting element need only be adequate to resist the forces for which the discontinuous wall was designed.

TU

PU PU


61

121

Anchorage to Podium

Discontinuous SystemsThe expanded commentary (3rd printing of ASCE 7-10) of §12.3.3.3 provides further explanation:

“For wood light-frame shear wall construction, the final sentence of §12.3.3.3 results in the shear and overturning connections at the base of a discontinued shear wall (i.e., shear fasteners and tie-downs) being designed using the load combinations of §2.3 or 2.4 rather than the load combinations with overstrength factor of §12.4.3.”

TU

PU PU

122

Anchorage to Podium

Discontinuous SystemsASCE 7-10 §12.4.3.1:

one possible route to reduce the calculated overstrength load occurs when it can be shown that yielding of other elements (tie down rod, shear wall, diaphragm, collector, etc.) will occur below the overstrength-level forces. When this is the case, the seismic load effects including overstrength can be reduced to a lower value.

TU

PU PU


62

123

Anchorage to Podium

Discontinuous SystemsASCE 7-10’s expanded commentary on §12.4.3 provides further explanation:

“The standard permits the seismic load effects, including overstrength factor, to be taken as less than the amount computed by applying Ω0 tothe design seismic forces where it can be determined that yielding of other elements in the structure limits the amount of load that can be delivered to the element and, therefore, the amount of force that can develop in the element.”

TU

PU PU

Rod Capacities

Level

ASDTensionDemand

(kips)

RodDia.

d(in)

Eff.Dia.de

(in)

Ag

(in2)Ae

(in2)

Fu

(ksi)

Fy

(ksi)

Roof 0.32 0.625 0.527 0.307 0.226 60 36 6.684th

Floor 4.48 0.625 0.527 0.307 0.226 60 36 6.683rd

Floor 10.10 0.875 0.755 0.601 0.462 120 105 13.822nd

Floor 16.42 0.875 0.755 0.601 0.462 120 105 27.05

124

ASDTensionDeman

d(kips)

ASD Rod

Capacity

(kips)


63

125

Anchorage to Podium

Discontinuous SystemsACI 318, does apply a factor similar to an overstrength factor to brittle concrete breakout failure modes if they govern the anchorage design. ACI 318, requires material overstrength to be considered with tension controlsNeed to consider the ratio of ultimate to yielding of the anchor:

ACI 318 has conditions for threaded rods.

TU

PU PU

DRAFT126


64

-Nonlinear modeling in SAP2000

Plastic Hinge Zones

127

Better Knowledge

• Recent Testing– Reinforced element (slab)– Building design; reinforcing in area of “cone”

• Unreduced calculated concrete capacity (CCD) aligns with test results for reinforced element

• Anchor reinforcement provides higher capacity for the anchor. Can different shapes or arrangements of reinforcement increase concrete capacity?

128


65

Anchor Reinforcement – ACI 318 Appendix D

Courtesy of Simpson Strong-Tie129

Actual element thickness prevents adequate lengthof anchor reinforcement

Also called “ladder bars” in other publications

TU

Anchor Reinforcement

130


66



Actual element thickness prevents adequate lengthof anchor reinforcement

Also called “ladder bars” in other publications

TU


132


67



Courtesy of Simpson Strong-Tie

Breakout

134


68

Simpson Strongtie Website

135

Design Examples for Wind & Seismic loadingsDesign Tables for Wind and Seismic Loadings

Stiffness of Shear Walls

136



69

F

h

b

da

Determination of Wall Rigidity

137

F

h

b

da

Determination of Wall Rigidity

Eq. 4.2 -1

Eq. C4.3.2-1

138


70

The da Tie down Displacement Factor

Why include crushing , shrinkage & elongation?Crushing can increase drifts 20 to 30 percent.

Tie down displacements can increase drifts 200 to 300 percent.

139

The da Tie down Displacement Factor

The codes are not clear on inclusion of crushing and shrinkage.

Seismic Design Manuals have included effects

IBC now has a revised definition of da:“Vertical elongation of overturning anchorage (including fastener slip, device elongation, anchor rod elongation, etc.) at the design shear load (v)”.

140140


71

Wood Shrinkage

Wood only shrinks perpendicular to grain.

The amount of shrinkage (or expansion) in wood is directly proportional to the change in moisture content.

The higher the moisture content at time of construction, the more shrinkage that can occur in the structure.

141141

The Indeterminant Process

Wall rigidities are based upon displacement under a given load.

Wall load is not known.

Wall rigidity largely effected by the type of tie down device.

Wall rigidities largely effected by moisture content of lumber at installation.

142142


72

Wall Stiffness Graph

143143

System Stretch

144



73

F

h

b

Tie DownDisplacement da

145

System Stretch

Δa

Δa

Δa

Δa

δ

146

Tension Side:

Rod system elongation and bearing plate crushing.

System Stretch

Compression Side:

Crushing of wood framing at boundary posts


74

Δ

Rod System Elongation

147

Bearing Plate Crushing

148



75

Bearing Plate Crushing

149

Level

ASDBearing

Load(kips)

StrengthBearing

Load(kips)

BearingPlateABrg

(in2)

fc┴

(ksi)0.73F’c┴

(ksi)

Crush(in)

Roof 2.433 3.475 15.98 0.217 0.456 0.010

4th Floor 3.660 5.229 9.98 0.524 0.456 0.028

3rd Floor 5.332 7.618 15.61 0.488 0.456 0.024

2nd Floor 6.373 9.105 15.39 0.592 0.456 0.036

LRFD Design:

10.0

’

19.17’

15.71k 11.4k

24.3 k 17.8 k

Accumulative Compression

Sill Plate Crushing

150


76

Sill Plate Crushing

151

Sill Plate Crushing

152

Level ChordPosts

ASDDemand

(kips)

StrengthDemand

(kips)

TotalArea(in2)

fc┴

(ksi)0.73F’c┴

(ksi)Crush

(in)

RoofFour3 x 4

3.56 4.85 35.0 0.139 0.456 0.030

4th FloorFour 3 x 4

8.66 11.84 35.0 0.338 0.456 0.074

3rd FloorFour 4 x 4

15.71 21.80 49.0 0.445 0.456 0.098

2nd FloorFour 4 x 6

24.30 34.37 77.0 0.446 0.456 0.098


77

Design Detail


METAL PLATE CONNECTED TRUSSES @ 24”

153

154

Crushing Zone


78

Sill Plate Crushing

F’c┴

0.73F’c┴

0.02 0.04Crushing (inches)

Bea

ring

Pre

ssur

e

155

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16

Deformation, inEq. 1.0 Eq. 2.0 Eq. 3.0

Fc┴ Load Deformation Curve

0.73 F’c┴ F’c┴

156


79

157

Crushing Zone

158

Crushing Zone


80

Crushing + Buckling

159

Crushing + Buckling

Squash Block

160


81

Crushing + Buckling

Squash Block

161

162

Wood Crushing

A

B

C


82

Deformation Adjustment Factors

Detail Bearing Condition Deformation adjustment

factor

C1. Wood-to-wood (both perpendicular to grain) 2.5

B2. Wood-to-wood (one parallel and one perpendicular to grain)

1.75

A3. Metal-to-wood (wood loaded perpendicular to grain)

1.0

163

Design Detail


METAL PLATE CONNECTED TRUSSES @ 24”

164


83

165

Combined Bearing/Crushing

A B

C

B

Tie‐Down Displacements

166

Level

StrengthRod

Elong.(in)

Shrinkage(Vertical

Displacement)

(in)

ChordCrushing

(in)

BearingPlate

Crushing(in)

Take-upDeflectionElongation

(in)

TotalDisplacement

da

(in)

Roof 0.051 0.03 0.030 0.010 0.030 0.1524th Floor 0.147 0.03 0.074 0.028 0.030 0.3113rd Floor 0.135 0.03 0.098 0.024 0.030 0.3182nd Floor 0.161 0.03 0.098 0.036 0.030 0.356

A + B + C


84

Wood Shrinkage

Wood only shrinks perpendicular to grain.

The amount of shrinkage (or expansion) in wood is directly proportional to the changein moisture content.

The higher the moisture content at time of construction, the more shrinkage that can occur in the structure.

167

Shrinkage Zone

168


85

Shrinkage Zone

169

LevelVertical Displacement Design

Displacement(in)

Per Floor Cumulative

4th Floor 0.216 0.648 3/43rd Floor 0.216 0.432 1/22nd Floor 0.216 0.216 1/4

Shrinkage Compensation

170

Shrinkage of 0.091 inch + settlement of 0.125 inch = 0.216 inch.


86

Δ

Shrinkage Compensation

171

Take‐up Devices

Devices are proprietary

Purpose is to compensate for building shrinkage and settlement.

Keep rotating the nut down or use a compression spring.

172


87

Shear Wall Deflections

173


F4ROOF

F34TH FLR

F2 3RD FLR

F12ND FLR

ΔT

Strength or ASD Deflections?

174


88

175

F4ROOF

F34TH FLR

F2 3RD FLR

F12ND FLR

ΔT

Δ4TH

Δ3RD

Δ2ND

ΔT = ΔROOF + Δ4TH + Δ3RD + Δ2ND

Story‐by‐Story Deflections

F4ROOF

F34TH FLR

F2 3RD FLR

F12ND FLR


Multi‐Story Deflections

176


89

F4ROOF

F34TH FLR

F2 3RD FLR

F12ND FLR


Δ4TH

Δ3RD

Δ2ND


177

F4 = 6,000 lbROOF

F3 = 7,000 lb4TH FLR

F2 = 5,000 lb3RD FLR

F1 = 2,500 lb2ND FLR

½” Structual I sheathing

10d @ 4” o/c E.N. spacing

4-4x6 Boundary Studs

Δa = 0.25 inch

b

30.0 ft

10.0

’10

.0’

10.0

’10

.0’

Design Example Multi‐Story Deflections

178


90

F4ROOF

F34TH FLR

F2 3RD FLR

F12ND FLR

Δ2ND = 0.04 in

Δ3RD = 0.15 in

Δ4TH = 0.48 in

ΔT = 0.89 + 0.48 + 0.15 + 0.04 = 1.56 in


179

Eq. 4.2 -1

Eq. C4.3.2-1

180

Shear Wall Deflection Equation

F

h

b

da


91

Eq. 4.2 -1

Eq. C4.3.2-1

181


F

h

b

da

What is the height “h” ???

182

Modified Balloon Frame


92

hPoint of application of load to shear wall


183

Point of Application of Load

Shear Wall Height “h”

184

Fx

Fx

h

b


93

Fx

Fx

Δ

h

b

185


186

Shear Wall Deflections


94

hPoint of application of load to shear wall


Framing Clips

187

Platform Frame


Fx

Fx

h

b

188


95

Fx

Fx

Δ

h

b

189


Eq. 4.2 -1

Eq. C4.3.2-1

190


F

h

b

da


96

Ga Apparent Shear Stiffness

SDPWS-15 Table 4.3A

191

Shear Stiffness Gvtv

SDPWS-15 Table C4.2.2A

192


97

Fastener Slip en

Where Vn is the load per nail in pounds

Fastener Size

Fastener Slip en

FabricatedGreen

Fabricated Dry

8d Common (Vn/857)1.869 (Vn/616)3.018

10d Common (Vn/977)1.894 (Vn/769)3.276

193

So Which Equation to Use?

194


98

So Which Equation to Use?

Quick calculation for deflection:3 Term EquationEasily done on calculator or software

Comprehensive analysis of structure:4 Term EquationSpreadsheet program automatically computes Gvtv and en from look-up

tables

195

L1 L2 L3 L4

Shear Wall Deflection

V

196


99

Shear Wall Deflection

SHEAR WALL DEFLECTION USING PERFORATED SHEAR WALL METHOD

SDPWS §4.3.2.1

197

SDPWS §4.3.2.1

r = sheathing area ratio

Ltot = total length of a perforated shear wall including the lengths of perforated shear wall segments

and the lengths of segments containing openings

A0 = total area of openings in the perforated shear wall individual opening areas calculated as the opening width times the clear opening height.

h = height of the perforated shear wall

ΣLi = sum of perforated shear wall segment lengths, feet

198


100

E.N. SPACING PER PLAN

EDGE NAIL SPACING TO EACH POST

4 POSTS 6+ POSTS

2 4” 6”

3 6” 9”

4 8” 12”

6 12” 12”

Deflection of Wall with Openings

199

Longitudinal Direction

Interior walls designed with continuous tie-downs

200

Exterior walls designed with FTAO


101

Shear Wall Comparisons

Wall v (plf)(ft) (ft)

NailSpace (in)

A0 r C0 vmax

(in)

1, 4 370 9.44 -- 6 112 0.84 0.96 386 0.15

1, 4 ΣLi = 64.5

2a,3a 478 9.44 18.06 -- -- -- -- 0.29

2b,3b 478 9.44 24.06 -- -- -- -- 0.28

2c,3c 478 9.44 18.06 -- -- -- -- 0.29

2, 3 Σ = 60.0

201

Distribution of Shear to Shear Walls in a Line

202



102

SDPWS §4.3.3.1

203

1. The Deflection Calculation Approach

2. The Adjustment Factor Approach

C4.3.3.4 The distribution of shear force to shear walls in a line is in proportion to the stiffness of each shear wall. In design, at a given deflection the force of each wall is determined by multiplying the wall stiffness times the deflection (e.g. commonly referred to as distribution based on relative stiffness or the equal deflection approach).

SDPWS §C4.3.3.4

SDPWS Commentary

3. Relative StiffnessApproach

204


103

205

1. The Deflection Calculation Approach

2. The Adjustment Factor Approach

3. The Relative Stiffness Approach

SDPWS §4.3.3.1

F

SW 1 SW 2

206


104

207

208


105


8.0’

8.0’ 2.3’

4.3.3.4.1 Deflection calculation approach1. Distribute shear to

provide same deflection in each shear wall

2. Account for distribution of shear based on stiffness of each shear wall. Distribution of shear is not directly proportional to shear wall length

209


8.0’

8.0’ 2.3’

4.3.3.4.1 Deflection calculation approach

Utilize shear wall deflection equations from SDPWS 4.3.2:

210


106

For wood structural panel shear walls, distribute shear in proportion to reduced strengths in accordance with 2bs/h factor

Accounts for reduced stiffness of high aspect ration shear walls


4.3.3.4.1 Exception, Adjustment factor approach (2bs/h) for WSP

4.3.3.4.1 Exception, Adjustment Factor (Stiffness)

AspectRatioh/bs

2:1 3:1 3 ½:1

2bs/h 1.00 0.67 0.57

211

Aspect Ratio Reduction Factor

212


107

Aspect Ratio vs. Deflection

b/h

2b/h

213

Accounts for reduced stiffness of high aspect ration shear walls

Not cumulative with aspect ratio factor for strength (4.3.4.2)

Applies for both wind and seismic


4.3.3.4.1 Exception, Adjustment factor approach (2bs/h) for WSP


AspectRatioh/bs

2:1 3:1 3 ½:1

2bs/h 1.00 0.67 0.57

214


108

Aspect Ratio Factor (Strength)

New factor for WSP shear walls that accounts for reduced strength of high aspect ratio shear walls

215

For wood structural panel shear walls with h/bs > 2:1, unit shear capacities are determined by multiplying by the Aspect Ratio Factor

1.25 – 0.125 h/bs

Applies to segmented and FTAO shear walls


4.3.4.2 – Shear Wall Aspect Ratio Factors (for strength)

4.3.4.2 Aspect Ratio Factor (Strength)

AspectRatioh/bs

2:1 3:1 3 ½:1

1.25-0.125h/bs 1.00 0.875 0.813

216


109

Applies for both wind and seismic

4.3.4.2 Aspect Ratio Factor (strength) is less sever than adjustment for stiffness (where 4.3.3.4.1 Exception is used)


4.3.4.2 – Shear Wall Aspect Ratio Factors (for strength)


AspectRatioh/bs

2:1 3:1 3 ½:1

1.25-0.125h/bs 1.00 0.875 0.813

217

Comparison


AspectRatioh/bs

2:1 3:1 3 ½:1

1.25-0.125h/bs 1.00 0.875 0.813


AspectRatioh/bs

2:1 3:1 3 ½:1

2bs/h 1.00 0.67 0.57

Always controls?

218


110

Design Example

Wall Construction:

15/32” OSB (WSP) Blocked

2x4 Douglas Fir framing

Common nails

8d @ 6” o/c E.N.

8d @ 12” o/c F.N.

End posts – Double 2x4’s

EA = 16,800,00 lb

da = 0.125 in at 3,500 lb uplift

219

Determine Apparent Shear Stiffness

Ga = 13 kips/in

220


111

Example 1‐Deflection Calculation Approach

SDPWS Example C.4.3.3.4.1-1

8.0’

8.0’ 2.3’

F

SW 1 SW 2

221

Shear Wall 1 (SW1)

Nominal shear capacity for seismic = 520 plf (SDPWS Table 4.3A)

SW1 Aspect Ratio (h/bs)

8.0/8.0 = 1.0

Aspect Ratio Factor (WSP) for strength = 1.0 (SDPWS 4.3.4.2)


Vsw1 = 520/2.0x1.0 = 260 plf

8.0’

8.0’

SW 1


222


112

Shear Wall 2 (SW2)

Nominal shear capacity for seismic = 520 plf (SDPWS Table 4.3A)

SW2 Aspect Ratio (h/bs)

8.0/2.3 = 3.5

Aspect Ratio Factor (WSP) for strength

1.25 -0.125 h/bs (SDPWS 4.3.4.2)

1.25-0.125x8/2.3 = 0.81


Vsw2 = 520/2.0x0.81 = 210 plf

2.3’

8.0’ SW 2


223

Maximum design unit shear wall segment based upon 4.3.4.2 Aspect Ratio Factor (Strength)

Next step – address distribution of shear based on 4.3.3.4.1 (i.e. distribution based on relative stiffness of SWI and SW2

8.0’

8.0’ 2.3’

F

SW 1 SW 2

Vsw1=260 plf Vsw2=210 plf


224


113

Part 1 Shear Wall 1 Deflection

8.0’

8.0’

SW 1Where:

v sw1 = 260 plf

h = 8.0 ft

EA = 16,800,000 lb

b sw = 8.0 ft

Ga = 13 k/in

da = 0.06 in

F

How do I calculate da?

da


225

Testing of Holdowns

226


114

Simpson Holdown Catalog

227

Part 1 Shear Wall 1 Uplift and Tiedown Elongation:

Anchorage Stiffness:

k = 4,565/0.114 in = 40,000 lbs/in


228


115


8.0’

8.0’

SW 1

F

da

Overturning Force:

260 plf * 8.0 ft = 2,080 lbs

Moment Overturning:

2,080 lb x 8.0 ft = 16,640 ft-lbs

Tension and Compression Force

(neglecting vertical loads on wall):

16,640/7.75 = 2,147 lb

Elongation at ASD applied load:

(2147/4565) x 0.114 in = 0.05 in


229


2 - 2×4 Posts

ASD

P = 2,147 lb Ab = 10.5 in2 lb < 3 inches from end of member Cb = 1.0

230

Crushing:


116

Simplification of 3rd Term

b

h

da

vsw

Solve for elongation in terms

of induced unit shear v:

By substitution:

Problem is we don’t know what da is

231


Where:

δ = 0.228

EA = 16,800,000 lb

b sw = 2.3 ft

Ga = 13 k/in

k = 45,000 lb/in

2.3’

8.0’ SW 2

da

F

Same as SW1

< 210 plf


232


117

Part 2 Shear Wall Uplift and Tiedown Elongation:

Determine daSW2 for unit shear force = 141 plf

Overturning Force:

141 plf * 2.3 ft = 324 lbs

Moment Overturning:

324 lb x 8.0 ft = 2,594 ft-lbs



2,594/2.05 = 1,266 lb


(1266/4565) x 0.114 in = 0.03 in

2.3’

8.0’ SW 2


233


2 - 2×4 Posts

ASD

P = 1,266 lb Ab = 10.5 in2 lb < 3 inches from end of member Cb = 1.0

234

Crushing:


118


2.3’

8.0’ SW 2

da

F


235

Example 1 – Deflection Calc Approach

Part 3 – Sum Design Strengths of SW1 and SW2

Vsw1 = 260 plf x 8.0 ft = 2080 lb

Vsw2 = 141 plf x 2.3 ft = 324 lb

V wall line = 2080 + 324 = 2404 lb

8.0’

8.0’ 2.3’

F

SW 1 SW 2

236


119

Example 2 – 2bs/h Adjustment Method

SDPWS Example C4.3.3.4.1-2

8.0’

8.0’ 2.3’

F

SW 1 SW 2

237

Shear Wall 1 (SW1)

Nominal shear capacity for seismic 520 plf (SDPWS Table 4.3A)

SW1 Aspect ratio (h/bs) = 1.0

Adjustment factor (based upon stiffness) =2bs/h = 1.0

(SDPWS 4.3.3.4.1 Exception 1)

Aspect Ratio Factor (WSP for Strength = 1.0

(SDPWS 4.3.4.2)


vsw1 = 520 plf/2 x 1.0 = 260 plf


8.0’

8.0’

SW 1

NOT ACCUMULATIVE SMALLER VALUECONTROLS

238


120

Shear Wall 2 (SW2)

Nominal shear capacity for seismic 520 plf (SDPWS Table 4.3A)

SW2 Aspect ratio (h/bs) = 3.5

Adjustment factor (based upon stiffness) =2bs/h = 0.57

(SDPWS 4.3.3.4.1 Exception 1)

Aspect Ratio Factor (WSP for Strength = 0.81

(SDPWS 4.3.4.2)


vsw1 = 520 plf/2 x 0.57 = 148 plf

8.0’

2.3’

SW 1

NOT ACCUMULATIVE SMALLER VALUECONTROLS

Compared to 141 plf from Example 1


239

Sum Design Strengths

Vsw1 = 260 plf x 8.0 ft = 2080 lb

Vsw2 = 148 plf x 2.3 ft = 340 lb

V wall line = 2080 + 340 = 2480 lb


8.0’

8.0’ 2.3’

F

SW 1 SW 2

Compared to 2404 lb from Example 1

240


121

Example 3 ‐ Relative Rigidity Approach

8.0’

8.0’ 2.3’

SW 1 SW 2

Arbitrarily use total design shear and distribute forces to the two walls based upon there tributary lengths (per foot basis)V wall line = 2480 lbs

V

241


8.0’

8.0’

SW 1

F

da

Overturning Force:

1,926 lbs

Moment Overturning:

1,926 lb x 8.0 ft = 15,410 ft-lbs



15,410/7.75 = 1,988 lb


(1988/4565) x 0.114 in = 0.05 in


242


122


8.0’

8.0’

SW 1Where:

v sw1 = 1926/8.0 = 241 plf

h = 8.0 ft

EA = 16,800,000 lb

b sw = 8.0 ft

Ga = 13 k/in

da = 0.07

in

F

da



2.3’

8.0’ SW 2

da

F

Where:

v sw1 = 554/2.3 = 241 plf

h = 8.0 ft

EA = 16,800,000 lb

b sw = 2.3 ft

Ga = 13 k/in

da = 0.05 in


244


123

F

b

da

Shear Wall

Flb

∆In

kk/in

SW1 1926 0.216 8.94 2134

SW2 554 0.382 1.45 346

∑ 2480 10.39 2480

Part 3 Shear Wall Rigidities


245


8.0’

8.0’

SW 1Where:

v sw1 = 2134/8.0 = 267 plf

h = 8.0 ft

EA = 16,800,000 lb

b sw = 2.3 ft

Ga = 13 k/in

da = 0.07

in

F

da



124


2.3’

8.0’ SW 2

da

F

Where:

v sw1 = 346/2.3 = 150 plf

h = 8.0 ft

EA = 16,800,000 lb

b sw = 2.3 ft

Ga = 13 k/in

da = 0.05 in


247

Side‐by‐Side Comparison

DesignExample

Design Method

SW1

(plf)

∆sw1

(in)

SW2

(plf)

∆sw2

(in)

Design Total Shear Force(lbs)

1Deflection

Calc.Approach

260 0.228 141 0.228 2404

22bs/h

Adjust.260 Not

Calculated148 Not

Calculated2480

3 RelativeStiffness

267 0.239 150 0.239 2480

248


125

Force Transfer Around Openings

249



250


126


251

252



127


253


254

SDPWS §4.3.5.2

254


128

Force Transfer Shear Walls

ADVANTAGESSystem is conventionalRequires less lumber at shear

wall jambsEliminates large holdown

devicesDISADVANTAGES Requires smaller windows, no

balconies. Usually used for hotel structures

but not apartment/condo structures.

255

Example Detailing of F.T.A.O.

256


129

Example Detailing of F.T.A.O.

257

Location of Shear Walls

258


130


259


260


131

Segmented Shear Walls

h

b b

261

FTAO Shear Walls

h

L

b b

262

H


132

263

Joint Research Report on FTAO

Joint Venture:

APA – Engineered Wood Association

University of British Columbia, Vancover, B.C.

USDA Forest Products Laboratory, Madison, WI

FTAO Testing

264


133

Drag Strut method

265

265

Cantilever Beam method

266

266


134

Measured vs. Predicted Strap Forces

267

267

Conclusions from testing:

Comparison of analytical methods with tested values for walls detailed as FTAOThe drag strut technique was consistently un-conservativeThe cantilever beam technique was consistently ultra-conservativeSEAOC/Thompson provides similar results as DiekmannSEAOC/Thompson & Diekmann techniques provided reasonable agreement with measured strap forces

268


135

Design Example

The height of a wall pier shall be defined as the clear height of the opening.

The width of a wall shall be defined as the sheathed width of the wall.

Overturning forces resisted by coupling beams above and/or below openings.

269


270

SDPWS §4.3.5.2

270


136

SEAOC/Thompson method

H =

9.0

’

L = 10.0’

3.0’ 3.0’

271

4.0’

4.0’

2.33

’2.

67’

Wall PierHeight

Wall PierWidth

Design Example

272


137

Design Example

H =

9.0

’

L = 10.0’

3.0’ 3.0’

273

4.0’

4.0’

2.33

’2.

67’

V = 4159 lb

Split wall in halfH

L

274

V = 4159 lb

T = 3,743 lb C = 3,743 lb


138

EQ

. E

Q.

EQ. EQ.275


V = 4159 lb

T = 3,743 lb C = 3,743 lb

276


∑ Forces Vertical = 0

v

v

T = 3,743 lb


139

277


1943#

1801#

T = 3,743 lb

4.33

’4.

67’

4.33

’


3.0’ 3.0’

2,080 2,080

278

V = 4159 lb

2,080 2,080


140

279


∑ Forces Horizontal = 0

v

V = 4159 lb

280


2,080 lb

2,080 lb

10.0’

5.0’V = 4159 lb


141

H1

4.33

’

L1

V/2

V

T

L10.0’

V

281

SEAOC/Thompson method5.0’L2 Tie

a

1801#

2.0 ft

2.33 ft

2080#1801#

1546#534#

5.0 ft

2.0 ft3.0 ft

Freebody at Pier U1

4.33 ft

2080#

282

Sum Moments about base


142

a

1801#

2.0 ft

2.33 ft

2080#1801#

1546#534#

5.0 ft

2.0 ft3.0 ft

Freebody at Pier U1

4.33 ft

2080#

283


Determine Tie Force:

a

1801#

2.0 ft

2.33 ft

2080#1801#

1546#534#

5.0 ft

2.0 ft3.0 ft

Freebody at Pier U1

4.33 ft

2080#

284

773178

693


143

H1

4.33

’ L1

V/2

V

T L10.0’

1801#

285

SEAOC/Thompson method5.0’L2

Tie

V

∑ Forces Vertical = 0

1943#

625#

2080#

3743#

1801#

11.66 ft

2.0 ft3.0 ft

2.0 ft

4.67 ft

2.67 ft

2080#

Freebody at Pier L1

286

1455#



144

1943#

2.0 ft

2.54 ft

1392#

Freebody at Pier L1

287



1943#

852#540#

2080#

202#

1801#

5.0 ft

2.0 ft3.0 ft

2.0’

4.67 ft

2.67 ft

1392#

Freebody at Pier L1

288



145

1943#

3.66 ft

2.54 ft

Freebody at Pier L1

289

180

2080#

852#540#

693

765

3.0 ft

2.0 ft

Wall Shears-SEAOC/Thompson Method

773178

693

765180

693

178

693

773

180

693

765

290


146

1546 lb 1546 lb

1530 lb 1530 lb

291

Tie Forces-SEAOC/Thompson Method

Determine Required Wall Nailing

Nominal shear capacity required :v ASD = 765 x 0.7 =535 plf

Where:



292


147

Type II Force Transfer SummaryUse 1 side½” Struct I w/ 10d @ 3” o/c


Blocking and strap good for 1550 lb(strength)

293

Type I Segmented Summary



294


148

Type III Perforated SummaryUse 2 sides½” Struct I w/ 10d @ 4” o/c


Holdown good for 1670 lb each full height stud (strength)

295

H

L

L1 L2

V1 V2

V

296

What happens when there are unequal wall piers lengths?


149

Inflection Points

hb

ha

EQ

.

EQ

. aEQ.

bEQ.

Fin.Flr.

Fin.Flr.

H

297


298


150

FTAO Shear Walls

SEAOC/Thompson methodShown in the SSDM’s 1997 UBC thru 2009 IBC editions. Assumed inflection points at mid opening. Had uniform jamb widths.

SEAOC/Thompson methodShown in the SSDM’s 2012 and 2015 IBC editions. Proportions inflection points. Uses different jamb widths.

Diekmann methodPresented at the 2015 SEAOC Convention. Uses different jamb widths.

299

Inflection Points

hb

ha

EQ

.

EQ

. aEQ.

bEQ.

Fin.Flr.

Fin.Flr.

H

300


151


301

THE METHOD ASSUMES THE FOLLOWING: •The unit shear at the sides of the opening is uniform.•The location of the inflection points above and below the openings is proportional to the wall pier lengths adjacent to that specific opening.•The location of the inflection points at the sides of the window openings is proportional to the wall lengths above and below that specific opening.•The vertical shear forces above and below the opening are determined by static free body equilibrium of the free body of that wall pier element.

Wall Dimensions

1.67

’2.

54’

4.0’

8.21

’

302

L1 L4 L2 L5 L3 L6 L48.0’ 12.75’ 11.5’9.5’ 4.0’ 4.0’ 9.5’


152


H3

H2

H1

H

Ha

Hb

1.67

’2.

54’

4.0’

8.21

’

hb

h

a

2.41

’ 1.

59’

303

H3

H2

H1

H

Ha

hb

h

a

La Lb Lc Ld

L1 L5 L2 L6 L3 L7

a b c d e fV1 V2 V3 V4

V4a V5a V6a

V4b V5b V6b

L4

304

Determine V1, V2 & V3


153

Determine V1, V2 & V3

305

H3

H2

H1

H

Ha

Hb

hb

h

a

La Lb Lc Ld

L1 L5 L2 L6 L3 L7


V4a V5a V6a

V4b V5b V6b

L4

306

Determine a, b, c, d, e & f


154

307

Determine a, b, c, d, e & f

H3

H2

H1

H

Ha

Hb

hb

h

a

La Lb Lc Ld

L1 L5 L2 L6 L3 L7


V4a V5a V6a

V4b V5b V6b

L4

308

Determine La, Lb, Lc & Ld


155

Determine La, Lb, Lc & Ld

L1 L4 L2 L5 L3 L6 L4

La11.66’

Lb20.69’

Lc15.59’

Ld11.6’

8.0’ 12.75’ 11.5’9.5’ 4.0’ 4.0’ 9.5’

a b c d e f

309

H3

H2

H1

H

Ha

Hb

1.67

’2.

54’

4.0’

8.21

’

hb

h

a

2.41

’ 1.

59’

La11.66’

Lb20.69’

Lc15.4’

Ld13.5’

L1 L4 L2 L5 L3 L68.0’


V4a V5a V6a

V4b V5b V6b

L412.75’ 11.5’9.5’ 4.0’ 4.0’ 11.5’

Dimensions Shown

310


156

3.26

’4.

95’

11.66’ 20.69’ 15.4’ 13.5’

U1 U2 U3 U4

L1 L2 L3 L4

H8.

21’

311

Label the Freebody Segments

V1=1392#

V1=1392#

3.26

’4.

95’

11.66’

U1

L1

H 8.21

’

V4a

V4b

V1

312

Freebodies at Piers U1 & L1



157

V1=1392#

V1=1392#

3.26

’4.

95’

11.66’

U1

L1

H 8.21

’

V4a

V4b

V1

313

Freebodies at Piers U1 & L1

a

389#

1.59 ft

1.67 ft

1392#

389#

853#540#

11.66 ft

3.66 ft8.0 ft

Freebody at Pier U1

3.26 ft

233

174

1392#

314 Sum Moments about base


158

a

389#

1.59 ft

1.67 ft

1392#

389#

852#540#

11.66 ft

3.66 ft8.0 ft

Freebody at Pier U1

3.26 ft

23367

174

1392#

315

a

V4a=389#

1.67 ft

H3

3.66 ft

Determine Tie Force at Pier U1

1392#

316


U1


159

a

592#

852#540#

1392#

202#

389#

11.66 ft

3.66 ft8.0 ft

1.81 ft

4.95 ft

2.54 ft233

174

1392#

Freebody at Pier L1

317


a

592#

852#540#

1392#

202#

389#

11.66 ft

3.66 ft8.0 ft

1.81 ft

4.95 ft

2.54 ft23367

174

1392#

Freebody at Pier L1

318


160

a

592#

3.66 ft

2.54 ft

1392#

Freebody at Pier L1

319


L1

c

3.26 ft

349#349#

1220#

5.84 ft

20.7 ft

2.10 ft12.75 ft

Freebody at Pier U2

1.67 ft

1.59 ft

209

174

2218#

2218#

209

320

bSum Moments about base


161

c

3.26 ft

349#349#

440#1220#

5.84 ft

20.7 ft

2.10 ft12.75 ft

Freebody at Pier U2

1.67 ft

1.59 ft

209

174

2218#

2218#

209

321

bSum Moments about base

c

349#349#

440#558#1220#

5.84 ft

20.7 ft

2.10 ft12.75 ft

Freebody at Pier U2

20944

174

2218#

2218#

209

322b


162

c

349#349#

5.84 ft 2.10 ft

Freebody at Pier U2

1.67 ft

323

b

1.67 ft

Determine Tie Force:Determine Tie Force:

U2

2218#

b c

1220#

531#2.54 ft

531#

5.84 ft 2.10 ft

Freebody at Pier L2

209

174

2218#

209

324


2.54 ft


163

2218#

b c

440#1220#

531#2.54 ft531#

5.84 ft 2.10 ft

Freebody at Pier L2

209

174

2218#

209

325


2218#

b c

440#1220#

531#2.54 ft531#

5.84 ft 2.10 ft

Freebody at Pier L2

209

174

2218#

209

326

558#

44

12.75 ft


164

b c

531# 2.54 ft531#

5.84 ft 2.10 ft

Freebody at Pier L2

327

L2

Determine Tie Force: Determine Tie Force:

U1 U2 U3 U4

L1 L2 L3 L4

328

Freebody Segments


165

U1 U2

L1 L2

329

Freebody Segments


23367

174

23367

174

20944

174

209

20944

174

209

330


166

853 lb 1220 lb

853 lb

440 lb

1220 lb 440 lb

331


Diekmann method

332

THE METHOD ASSUMES THE FOLLOWING: The unit shear above and below the opening is uniform.The corner forces are based on the shear above and below the openings and only the piers adjacent to that specific opening.The tributary length of the opening is the basis for calculating the shear to each pier. This tributary length is the ratio of the length of the pier multiplied by the length of the opening it is adjacent to, divided by the sum of the length of the pier and the length of the pier on the other side of the opening.


167

Diekmann method

333

For example, T1= (L1*Lo1)/ (L1+L2) The shear in each pier is the total shear divided by the L of the wall, multiplied by the sum of the length of the pier and its tributary length, divided by the length of the pier:(V/L)(L1+T1)/L1The unit shear of the corner zones is equal to subtracting the corner forces from the panel resistance, R. R is equal to the shear of the pier multiplied by the pier length:Va1 = (v1L1-F1)/L1

Wall Dimensions

1.67

’2.

54’

4.0’

8.21

’

334

8.0’ 12.75’ 11.5’9.5’ 4.0’ 4.0’ 9.5’


168

hb

ho

ha

1. Calculate the tiedown forces:L1 Lo1 L2 Lo2 L3 Lo3 L4

h

V

H = Vh/L = 7,612 lb x 8.21’/61.25’ = 1,020 lb

H H

O1 O2 O3

335

hb

ho

ha

2. Solve for shears above and belowL1 Lo1 L2 Lo2 L3 Lo3 L4

h

V

va = vb = H/(ha+hb) = 1,020 lb/(1.67’ + 2.54’) = 242 plf

H H

O1 O2 O3

336


169

hb

ho

ha

3. Find total Boundary Forces: O1L1 Lo1 L2 Lo2 L3 Lo3 L4

h

O1 = va x (Lo1) = 242 plf x 9.5’ = 2,299 lb

O1

337

hb

ho

ha


h

O2 = va x (Lo2) = 242 plf x 4.0’ = 968 lb

O2

338


170

hb

ho

ha


h

O3 = va x (Lo3) = 242 plf x 4.0’ = 968 lb

O3

339

hb

ho

ha

4. Determine Corner ForcesL1 Lo1 L2 Lo2 L3 Lo3 L4

h

The corner forces are based on the shear above and below the openings and only the piers adjacent to that unique opening.

O1 O2 O3

340


171

hb

ho

ha

5. Calculate Corner Forces:

L1 Lo1 L2 Lo2 L3 Lo3 L4h

F1 = O1(L1)/(L1+L2) = 2,299 lb x 8.0/(8.0 + 12.75) = 886 lbF2 = O1(L2)/(L1+L2) = 2,299 lb x12.75/(8.0 +12.75) = 1,413 lb

O1

F1 F2

F1 F2

O2

F3 F4

F3 F4

O3F5 F6

F5 F6

341

hb

ho

ha

6. Calculate Corner Forces:L1 Lo1 L2 Lo2 L3 Lo3 L4

h

F3 = O2(L2)/(L2+L3) = 968 lb x12.75/(12.75+11.5) = 509 lb F4 = O2(L3)/(L2+L3) = 968 lb x 11.5/(12.75+11.5) = 459 lb

O2

F3 F4

F3 F4

342


172

hb

ho

ha

6. Calculate Corner Forces:


F5 = O3(L3)/(L3+L4) = 968 lb x11.5/(11.5+9.5) = 484 lb F6 = O3(L4)/(L3+L4) = 968 lb x 9.5/(11.5+9.5) = 484 lb

O3F5 F6

F5 F6

343

hb

ho

ha

6. Calculate Tributary Lengths:

344

L1 Lo1 L2 Lo2 L3 Lo3 L4

h

Ratio of the length of the pier x length of the openings it is adjacent to, then/ (length of the pier + length of the pier on the other side of the opening)

T1 T2 T3 T4 T5 T6


173

hb

ho

ha

7. Calculate Tributary Lengths:


T1 = L1 x Lo1/(L1+L2) = 8.0 x 9.5/(8.0+12.75) = 3.66’T2 = L2 x Lo1/(L1+L2) = 12.75 x 9.5/(8.0+12.75) = 5.84’

T1 T2

345

hb

ho

ha

7. Calculate Tributary Lengths:L1 Lo1 L2 Lo2 L3 Lo3 L4

h


T3 T4

346


174

hb

ho

ha

7. Calculate Tributary Lengths:L1 Lo1 L2 Lo2 L3 Lo3 L4

h


T5 T6

347

L

hb

ho

ha

8. Unit shears at sides of openings:L1 Lo1 L2 Lo2 L3 Lo3 L4

h

The shear force in each pier = the total shear / the L of thewall x (L pier + its tributary length)/ by the L pier

V1

V

T1 T2 T3 T4 T5 T6V2 V3 V4

348


175

L

hb

ho

ha

8. Unit shears at sides of openings:L1 Lo1 L2 Lo2 L3 Lo3 L4

h

V1 = (V/L)(L1+T1)/L1 V1 = (7,612/61.25)(8.0+3.66)/8.0=181 plf

T1 V1

V

L

hb

ho

ha

8. Unit shears at sides of openings:

350


h

V2 = (V/L)(T2+L2+T3)/L2 V2 =(7,612/61.25)(5.84+12.75+2.10)/12.75 = 202 plf

T2

V

V2 T3


176

L

hb

ho

ha


351


V3 = (V/L)(T4+L3+T5)/L3V3 =(7,612/61.25)(1.90+11.5+2.0)/11.5 = 166 plf

T5

V

V3T4

L

hb

ho

ha


352


h

V4 = (V/L)(T6+L4)/L4V4 =(7,612/61.25)(2.0+11.5)/11.5 = 146 plf

V

V4T6


177

L

hb

ho

ha

9. Check sum of shears in piers:

353


V1xL1 = 181 x 8.0 = 1448 lb V2xL2 = 202 x 12.75 = 2576 lbV3xL3 = 166 x 11.5 = 1909 lb V4xL4 = 146 x 11.5 = 1679 lb

V1

V

V2 V3 V4

L

hb

ho

ha

9. Check sum of shears in piers:

354


h

Sum forcesV1+V2+V3+V4 = 1448 + 2576 + 1909 + 1679 = 7612 lb

V1

V

V2 V3 V4


178

242 242 242

242 242 242

181 202 166 146

355

Wall Shears-Diekmann Method

Max. Shear


356

233

174

233

174

209

174

209

209

174

209

Max. Shear

356


179

866 lb 1413 lb

866 lb

509 lb

1413 lb 509 lb

357

Tie Forces-Diekmann Method

Max. Tie Force

853 lb 1220 lb

853 lb

440 lb

1220 lb 440 lb

358


Max. Tie Force


180

Questions?

SEAU

359

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