SEAU 5th Annual Education Conference 2-21-17
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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
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Design Example
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Design Examples
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IBC & SDPWS
2012 IBC
USE STANDARD
SDPWS-08
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IBC vs. SDPWS
USE STANDARD
2015 IBC SDPWS-15
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Shear Wall HoldownsFailures & Splitting
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Wood Shear Wall Design
Holdown Slip
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Multi Story Tie Down Systems
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Chord/Boundary Member Failure
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Chord/Boundary Member Failure
Chord failed in tension!
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Typical Failure Mode
Nail pull through, Edge tear, Nail yield
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Nails Working in Lumber
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Nails Resistance to Wall Racking
F
b
d
h
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Nails Yielding
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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
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NDS Yield Limit Equations
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Splitting happens because wood is relatively weak perpendicular to grain
Nails too close
(act like a wedge)
Material Properties of Wood
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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
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Minimum Edges Distances
3/8” MIN
Fastener Spacing
Fra
min
g W
idth
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Staggered Nailing
Fastener Spacing
Fra
min
g W
idth
3/8” MIN
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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
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Wood Shear Wall Design
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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
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Design Example – Type I SegmentedH
= 9
.0’
L = 10.0’
3.0’ 3.0’
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4.0’
4.0’
2.33
’2.
67’
V = 4159 lb
Both walls length are the same therefore no need to consider load distribution
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Segmented Wall Design
H =
9.0
’
L = 10.0’
3.0’ 3.0’
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4.0’
4.0’
2.33
’2.
67’
Wall PierHeight
Wall PierWidth
Design Example – Type I Segmented
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H =
9.0
’
3.0’
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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
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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
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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’
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Type I Segmented Summary
Use 1 side½” Struct I w/ 10d @ 3” o/c
4x4 with holdowns good for 6,685 lb(strength)
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Design Example – Type III PerforatedH
= 9
.0’
L = 10.0’
3.0’ 3.0’
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4.0’
4.0’
2.33
’2.
67’
V = 4159 lb
Entire wall sheathed. No straps or blocking as window head and sill locations.
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Design Example – Type III Perforated
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Perforated Shear Walls
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Determine Tension & Compression Chord Forces
H =
9.0
’
3.0’Li
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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’
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4.0’
4.0’
2.33
’2.
67’
Maximum opening height:
Percent of Full Height Sheathing:
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Capacity Adjustment Factor Co
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SDPWS Figure 4C
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Determine Height to Width Ratio
H =
9.0
’3.0’Li
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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
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Determine Required Wall NailingNominal shear capacity required :V strength = 1255 plv v ASD = 1255 x 0.7 = 875 plf
Where:
ASD Reduction Factor = 2.0
Use two side of sheathing:
V per side = 1755/2 = 875 plf
Use 2 sides½” Struct I w/ 10d @ 4” o/c
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Design Example – Type III Perforated
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Type III Perforated SummaryUse 2 sides½” Struct I w/ 10d @ 4” o/c
4x6 with holdowns good for 11,275 lb(strength)
Holdown good for 1670 lb each full height stud (strength)
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Multi-Story Shear Wall Design
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Wood Shear Wall Design
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Continuous Tie-Down Shear Walls
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Wall Edge-Modified Balloon Frame
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Top Plate/Chord Top Plate/Chord
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Wall Edge-Modified Balloon Frame
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Ribbon/Chord Ribbon/Chord
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Wall Edge-Platform Framed
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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
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Shear Wall Overturning Forces
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Fx
Fx
MOT
MOT
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Vwall above
Vfloor
h1 h2
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Shear Wall Overturning Forces
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h2Point of application of load to shear wall
Point of application of load to chord/collector
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Shear Wall Overturning Forces
Shear Wall Overturning Forces
Fx
Fx
MOT
MOT
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Vwall above
Vfloor
h1 h2
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Shear Wall Overturning Forces
h2Point of application of load to shear wall
Point of application of load to chord/collector
Framing Clips
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Shear Wall Overturning Forces
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Vwall above
Vfloor
h1 h2
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Shear Wall Overturning Forces
h3
So Which height to Use?
F
h
b
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How Much Load On Boundary Posts?
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w
TRIB
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Δa
Δa
Δa
Δa
δ
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Design shear wall boundary posts
Shear Wall Compression Forces
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Offset Posts
4-4x6
4-4x4
19.17’d’1.63’
d’’0.71’
Centroid of Posts
Centroid of Rod
d
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Concentric Posts
4-4x6
4-4x4
19.17’d’1.17’
d’’1.17’
Centroid of Posts
Centroid of Rod
d
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Design DetailMETAL PLATE CONNECTED TRUSSES @ 24”
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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
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Determine Chord Members
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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
’
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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
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Δa
Δa
Δa
Δa
δ
70
Design Tie down rods and bearing plates.
Shear Wall Uplift Forces
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Shear Wall Uplift Forces
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
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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
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Shear Wall Uplift Forces
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
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
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Bearing Plate Capacities
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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
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Differential Load:17.8-11.4k = 6.4k
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Bearing on Plates
Allowable Bearing Perpendicular to Grain Fc┴
Douglas Fir Larch 625
Hemlock Fir 405
LVL 480
LSL 435
ASD
LRFD
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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
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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
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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”
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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”
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Shear Wall Uplift Forces
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
83
10.0
’
19.17’
15.71 k 11.4 k
24.30 k 17.8 k
84
Differential Load:17.8-11.4k = 6.4k
Boundary Member Nailing
Vertical Shear Force:6.4k /10.0 = 640 plf
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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
Boundary Member Nailing
Sp
acin
g
4 POSTS
Edge Nail at Wall Ends
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Vertical Shear Force:6.4k /10.0 = 640 plf
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Boundary Member Nailing
Sp
acin
g
2 POSTS EACH SIDE 3 OR MORE POSTS EACH SIDE
Sp
acin
g
Edge Nail Spacing87
Boundary Member Nailing
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”
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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
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Wood Shear Wall Design
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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
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.
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2015 IBC‐Connectors in Treated Wood
Table from Simpson Strong-Tie Catalog
Fasteners in Treated Wood
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.
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Fasteners in Treated Wood
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.
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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.
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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.
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ACQ-C
CA-B
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Corrosion Resistant Fasteners
Hot Dipped Galvanized Toe Nails to Sill Plate
99
Corrosion Resistant Fasteners
Hot Dipped Galvanized Edge Nailing to Sill Plate
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101
Typical Wall Anchors
101
Sill Plate Anchorage
Thickness
102
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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.
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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?
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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
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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
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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
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Sill Plate Resisting Rotation
113
DRAFT
Early Physical Testing/Research
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DRAFT
DRAFT
Post-test documentation: (after +/- 2” displacement)Bolt stretching. Shallow concrete spall. Bolt did not pull out.
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• 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
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118
19.17’d’1.63’ d
Anchorage to Podium
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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
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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
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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
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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)
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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
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-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
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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
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Anchor Reinforcement
Courtesy of Simpson Strong-Tie131
Actual element thickness prevents adequate lengthof anchor reinforcement
Also called “ladder bars” in other publications
TU
Anchor Reinforcement
132
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Anchor Reinforcement
Courtesy of Simpson Strong-Tie133
Courtesy of Simpson Strong-Tie
Breakout
134
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Simpson Strongtie Website
135
Design Examples for Wind & Seismic loadingsDesign Tables for Wind and Seismic Loadings
Stiffness of Shear Walls
136
Wood Shear Wall Design
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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
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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
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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
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Wall Stiffness Graph
143143
System Stretch
144
Wood Shear Wall Design
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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
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Δ
Rod System Elongation
147
Bearing Plate Crushing
148
Differential Load:17.8-11.4k = 6.4k
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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
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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
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Design Detail
PREMANUFACTURED I JOISTS @ 24”
METAL PLATE CONNECTED TRUSSES @ 24”
153
154
Crushing Zone
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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
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157
Crushing Zone
158
Crushing Zone
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Crushing + Buckling
159
Crushing + Buckling
Squash Block
160
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Crushing + Buckling
Squash Block
161
162
Wood Crushing
A
B
C
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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
PREMANUFACTURED I JOISTS @ 24”
METAL PLATE CONNECTED TRUSSES @ 24”
164
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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
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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
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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.
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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
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Shear Wall Deflections
173
Wood Shear Wall Design
F4ROOF
F34TH FLR
F2 3RD FLR
F12ND FLR
ΔT
Strength or ASD Deflections?
174
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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
ΔT = ΔROOF + Δ4TH + Δ3RD + Δ2ND
Multi‐Story Deflections
176
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F4ROOF
F34TH FLR
F2 3RD FLR
F12ND FLR
ΔT = ΔROOF + Δ4TH + Δ3RD + Δ2ND
Δ4TH
Δ3RD
Δ2ND
Multi‐Story Deflections
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
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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
Multi‐Story Deflections
179
Eq. 4.2 -1
Eq. C4.3.2-1
180
Shear Wall Deflection Equation
F
h
b
da
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Eq. 4.2 -1
Eq. C4.3.2-1
181
Shear Wall Deflection Equation
F
h
b
da
What is the height “h” ???
182
Modified Balloon Frame
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hPoint of application of load to shear wall
Point of application of load to chord/collector
183
Point of Application of Load
Shear Wall Height “h”
184
Fx
Fx
h
b
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Fx
Fx
Δ
h
b
185
Shear Wall Height “h”
186
Shear Wall Deflections
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hPoint of application of load to shear wall
Point of application of load to chord/collector
Framing Clips
187
Platform Frame
Shear Wall Height “h”
Fx
Fx
h
b
188
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Fx
Fx
Δ
h
b
189
Shear Wall Height “h”
Eq. 4.2 -1
Eq. C4.3.2-1
190
Shear Wall Deflection Equation
F
h
b
da
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Ga Apparent Shear Stiffness
SDPWS-15 Table 4.3A
191
Shear Stiffness Gvtv
SDPWS-15 Table C4.2.2A
192
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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
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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
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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
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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
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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
Wood Shear Wall Design
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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
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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
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207
208
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Distribution of Shear to Shear Walls in a Line
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
Distribution of Shear to Shear Walls in a Line
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
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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
Distribution of Shear to Shear Walls in a Line
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
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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
Distribution of Shear to Shear Walls in a Line
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
214
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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
Aspect Ratio Factor (Strength)
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
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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)
Aspect Ratio Factor (Strength)
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
217
Comparison
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
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
Always controls?
218
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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
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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)
ASD unit shear capacity for seismic
Vsw1 = 520/2.0x1.0 = 260 plf
8.0’
8.0’
SW 1
Example 1‐Deflection Calculation Approach
222
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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
ASD unit shear capacity for seismic
Vsw2 = 520/2.0x0.81 = 210 plf
2.3’
8.0’ SW 2
Example 1‐Deflection Calculation Approach
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
Example 1‐Deflection Calculation Approach
224
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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
Example 1‐Deflection Calculation Approach
225
Testing of Holdowns
226
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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
Example 1‐Deflection Calculation Approach
228
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Part 1 Shear Wall 1 Uplift and Tiedown Elongation:
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
Example 1‐Deflection Calculation Approach
229
Bearing of Boundary Members
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:
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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
Part 2 Shear Wall 2 Deflection
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
Example 1‐Deflection Calculation Approach
232
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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
Tension and Compression Force
(neglecting vertical loads on wall):
2,594/2.05 = 1,266 lb
Elongation at ASD applied load:
(1266/4565) x 0.114 in = 0.03 in
2.3’
8.0’ SW 2
Example 1‐Deflection Calculation Approach
233
Bearing of Boundary Members
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:
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Part 2 Shear Wall 2 Deflection
2.3’
8.0’ SW 2
da
F
Example 1‐Deflection Calculation Approach
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
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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)
ASD unit shear capacity for seismic
vsw1 = 520 plf/2 x 1.0 = 260 plf
Example 2 – 2bs/h Adjustment Method
8.0’
8.0’
SW 1
NOT ACCUMULATIVE SMALLER VALUECONTROLS
238
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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)
ASD unit shear capacity for seismic
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
Example 2 – 2bs/h Adjustment Method
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
Example 2 – 2bs/h Adjustment Method
8.0’
8.0’ 2.3’
F
SW 1 SW 2
Compared to 2404 lb from Example 1
240
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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
Part 1 Shear Wall 1 Uplift and Tiedown Elongation:
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
Tension and Compression Force
(neglecting vertical loads on wall):
15,410/7.75 = 1,988 lb
Elongation at ASD applied load:
(1988/4565) x 0.114 in = 0.05 in
Example 3 ‐ Relative Rigidity Approach
242
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Part 1 Shear Wall 1 Deflection
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
Example 3 ‐ Relative Rigidity Approach
Part 2 Shear Wall 2 Deflection
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
Example 3 ‐ Relative Rigidity Approach
244
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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
Example 3 ‐ Relative Rigidity Approach
245
Part 1 Shear Wall 1 Deflection
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
Example 3 ‐ Relative Rigidity Approach
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Part 2 Shear Wall 2 Deflection
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
Example 3 ‐ Relative Rigidity Approach
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
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Force Transfer Around Openings
249
Wood Shear Wall Design
Force Transfer Around Openings
250
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Force Transfer Around Openings
251
252
Force Transfer Around Openings
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Force Transfer Around Openings
253
Force Transfer Around Openings
254
SDPWS §4.3.5.2
254
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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
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Example Detailing of F.T.A.O.
257
Location of Shear Walls
258
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Force Transfer Around Openings
259
Force Transfer Around Openings
260
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Segmented Shear Walls
h
b b
261
FTAO Shear Walls
h
L
b b
262
H
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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
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Drag Strut method
265
265
Cantilever Beam method
266
266
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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
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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
Force Transfer Around Openings
270
SDPWS §4.3.5.2
270
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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
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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
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EQ
. E
Q.
EQ. EQ.275
SEAOC/Thompson method
V = 4159 lb
T = 3,743 lb C = 3,743 lb
276
SEAOC/Thompson method
∑ Forces Vertical = 0
v
v
T = 3,743 lb
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277
SEAOC/Thompson method
1943#
1801#
T = 3,743 lb
4.33
’4.
67’
4.33
’
SEAOC/Thompson method
3.0’ 3.0’
2,080 2,080
278
V = 4159 lb
2,080 2,080
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279
SEAOC/Thompson method
∑ Forces Horizontal = 0
v
V = 4159 lb
280
SEAOC/Thompson method
2,080 lb
2,080 lb
10.0’
5.0’V = 4159 lb
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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
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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
Sum Moments about base
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
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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#
Sum Moments about base
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1943#
2.0 ft
2.54 ft
1392#
Freebody at Pier L1
287
Determine Tie Force:
Sum Moments about base
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
Sum Moments about base
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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
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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:
ASD Reduction Factor = 2.0
Use 1 side½” Struct I w/ 10d @ 3” o/c
292
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Type II Force Transfer SummaryUse 1 side½” Struct I w/ 10d @ 3” o/c
4x4 with holdowns good for 3,743 lb(strength)
Blocking and strap good for 1550 lb(strength)
293
Type I Segmented Summary
Use 1 side½” Struct I w/ 10d @ 3” o/c
4x4 with holdowns good for 6,685 lb(strength)
294
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Type III Perforated SummaryUse 2 sides½” Struct I w/ 10d @ 4” o/c
4x6 with holdowns good for 11,275 lb(strength)
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?
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Inflection Points
hb
ha
EQ
.
EQ
. aEQ.
bEQ.
Fin.Flr.
Fin.Flr.
H
297
Force Transfer Around Openings
298
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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
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SEAOC/Thompson method
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’
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Force Transfer Around Openings
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
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Determine V1, V2 & V3
305
H3
H2
H1
H
Ha
Hb
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
306
Determine a, b, c, d, e & f
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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
a b c d e fV1 V2 V3 V4
V4a V5a V6a
V4b V5b V6b
L4
308
Determine La, Lb, Lc & Ld
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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’
a b c d e fV1 V2 V3 V4
V4a V5a V6a
V4b V5b V6b
L412.75’ 11.5’9.5’ 4.0’ 4.0’ 11.5’
Dimensions Shown
310
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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
Sum Moments about base
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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
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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
Determine Tie Force:
U1
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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
Sum Moments about base
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
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a
592#
3.66 ft
2.54 ft
1392#
Freebody at Pier L1
319
Determine Tie Force:
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
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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
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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
Sum Moments about base
2.54 ft
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2218#
b c
440#1220#
531#2.54 ft531#
5.84 ft 2.10 ft
Freebody at Pier L2
209
174
2218#
209
325
Sum Moments about base
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
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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
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U1 U2
L1 L2
329
Freebody Segments
Wall Shears-SEAOC/Thompson Method
23367
174
23367
174
20944
174
209
20944
174
209
330
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853 lb 1220 lb
853 lb
440 lb
1220 lb 440 lb
331
Tie Forces-SEAOC/Thompson Method
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.
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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’
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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
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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
4. Find total Boundary Forces: O2L1 Lo1 L2 Lo2 L3 Lo3 L4
h
O2 = va x (Lo2) = 242 plf x 4.0’ = 968 lb
O2
338
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hb
ho
ha
5. Find total Boundary Forces: O3L1 Lo1 L2 Lo2 L3 Lo3 L4
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
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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
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hb
ho
ha
6. Calculate Corner Forces:
L1 Lo1 L2 Lo2 L3 Lo3 L4h
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
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hb
ho
ha
7. Calculate Tributary Lengths:
L1 Lo1 L2 Lo2 L3 Lo3 L4h
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 = L2 x Lo2/(L2+L3) = 12.75 x 4.0/(12.75+11.5) = 2.10’T4 = L3 x Lo2/(L2+L3) = 11.5 x 4.0/(12.75+11.5) = 1.90’
T3 T4
346
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hb
ho
ha
7. Calculate Tributary Lengths:L1 Lo1 L2 Lo2 L3 Lo3 L4
h
T5 = L3 x Lo3/(L3+L4) = 11.5 x 4.0/(11.5+11.5) = 2.0’T6 = L4 x Lo3/(L3+L4) = 9.5 x 4.0/(11.5+11.5) = 2.0’
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
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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
L1 Lo1 L2 Lo2 L3 Lo3 L4
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
SEAU 5th Annual Education Conference 2-21-17
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L
hb
ho
ha
8. Unit shears at sides of openings:
351
L1 Lo1 L2 Lo2 L3 Lo3 L4h
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
8. Unit shears at sides of openings:
352
L1 Lo1 L2 Lo2 L3 Lo3 L4
h
V4 = (V/L)(T6+L4)/L4V4 =(7,612/61.25)(2.0+11.5)/11.5 = 146 plf
V
V4T6
SEAU 5th Annual Education Conference 2-21-17
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L
hb
ho
ha
9. Check sum of shears in piers:
353
L1 Lo1 L2 Lo2 L3 Lo3 L4h
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
L1 Lo1 L2 Lo2 L3 Lo3 L4
h
Sum forcesV1+V2+V3+V4 = 1448 + 2576 + 1909 + 1679 = 7612 lb
V1
V
V2 V3 V4
SEAU 5th Annual Education Conference 2-21-17
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242 242 242
242 242 242
181 202 166 146
355
Wall Shears-Diekmann Method
Max. Shear
Wall Shears-SEAOC/Thompson Method
356
233
174
233
174
209
174
209
209
174
209
Max. Shear
356
SEAU 5th Annual Education Conference 2-21-17
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
Tie Forces-SEAOC/Thompson Method
Max. Tie Force
SEAU 5th Annual Education Conference 2-21-17
180
Questions?
SEAU
359