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Shear Walls 1
Shear Walls
• Load Distribution to Shear Walls• Shear wall stiffness• Shear
walls with openings• Diaphragm types
• Types of Masonry Shear Walls• Maximum Reinforcement
Requirements• Shear Strength• Example: simple building
Shear Walls 2
Shear Walls: Stiffness
_____ stiffness predominates
_______ stiffness predominates
Both shear and bending stiffness
are important
d
h
h/d < 0.25 0.25 < h/d < 4.0 h/d >4.0
__________
___________
Lateral Force Resisting System
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Shear Walls 3
Shear Wall Stiffness9.1.5.2 Deflection calculations shall be
based on cracked section properties. Assumed properties shall not
exceed half of gross section properties, unless a cracked-section
analysis is performed.
Cantilever wall
cant
h = height of wallAv =shear area; (5/6)A for
a rectangleEv = G = shear modulus
(modulus of rigidity); given as 0.4Em(4.2.2.2.2)
Ev = ___________, where ν is Poisson’s ratio
t = thickness of wallL = length of wall
342
Lh
Lh
tEk mcant
Fixed wall (fixed against rotation at top)
fixed
32
Lh
Lh
tEk mfixed
Real wall is probably between two cases; diaphragm provides some
rotational restraint, but not full fixity.
Shear Walls 5
T- or L- Shaped Shear WallsSection 5.1.1 Wall intersections
designed either to:a) ____________________: b)
____________________
Connection that transfers shear: (must be in running bond)a)
Fifty percent of masonry units interlockb) Steel connectors at max
4ft.c) Intersecting bond beams at max 4 ft. Reinforcing of at least
0.1in2 per foot of wall
2-#4’sMetal lath or wire screen to support grout
1/4in. x 11/2in. x 28in. with 2in. long 90 deg bends at each end
to form U or Z shape
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Shear Walls 6
Effective Flange Width (5.1.1.2.3)
Effective flange width on either side of web shall be smaller of
actual flange width, distance to a movement joint, or:
• Flange in compression: 6t• Flange in tension:
• Unreinforced masonry: 6t• Reinforced masonry: 0.75 times
floor-to-floor wall
height
Shear Walls 7
Example: Flanged Shear Wall
40in.
112i
n.6t
=48i
n.56
in.
Elevation
Plan
Given: Fully grouted shear wallRequired: Stiffness of
wallSolution: Determine stiffness from basic principles.
A
Find centroid from outer flange surface in
inininininininy 16.11
671204062.781.34862.7
2
I
vA
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Shear Walls 9
Example: Flanged Shear Wall
m
mmvvnm
Ein
Einin
inEin
EAPh
IEPh
PPk 157.0
4.0305112
860003112
1
3 2433
Shear Walls 10
Shear Walls with Openings
Qamaruddin, M., Al-Oraimi, S., and Hago, A.W. (1996).
“Mathematical model for lateral stiffness of shear walls with
openings.” Proceedings, Seventh North American Masonry Conference,
605-617.
1. Divide wall into piers.2. Find flexibility of each pier3.
Stiffness is reciprocal of flexibility4. Distribute load according
to stiffness
1
512
2322
6
3
EAh
kkkkkkkk
EIhf
bbtt
bbtt
bbtt
btt kkkk
kkVhM21
bbtt
tbb kkkk
kkVhM21
Bottom spandrel
Top spandrel
h bh t
tt h
hk
bb h
hk
If hb = 0
1
512
122
6
3
EAh
kk
EIhf
t
t
12 t
tt k
kVhM
121
t
tb k
kVhMAs the top spandrel decreases in height, the top approaches
a fixed condition against rotation.
h
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Shear Walls 11
Example - Shear Walls with Openings
2ft 2ft 2ft4ft 6ft
4ft
4ft
8ft
Given:
A B C
Required:A. Stiffness of wallB. Forces in each pier under
10 kip horizontal load
Shear Walls 12
Example - Shear Walls with OpeningsSample calculations: Pier Bh
= 4ft ht = 4ft hb = 8ft kt = h/ht = 4ft/4ft = .0 kb = h/hb =
4ft/8ft = 0.5
12
3tLI tfttLA 2
1
512
2322
6
3
EAh
kkkkkkkk
EIhf
bbtt
bbtt
EtEtf
k 0210.0/6.47
11
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Shear Walls 14
Example - Shear Walls with OpeningsPier h (ft) ht (ft) hb (ft)
kt kb I (ft3) A (ft) f s f k
A 12 4 0 3 inf 0.667t 2t 308.4 18 326.4/Et 0.0031EtB 4 4 8 1 0.5
0.667t 2t 41.6 6 47.6/Et 0.0210EtC 4 4 8 1 0.5 0.667t 2t 41.6 6
47.6/Et 0.0210Et
f is flexural deformation; s is shear deformation
Total stiffness is 0.0451EtAverage stiffness from finite element
analysis 0.0440Et Solid wall stiffness 0.1428Et (free at top);
0.25Et (fixed at top)
Shear Walls 15
Example - Shear Walls with OpeningsSample calculations: Pier B
kipsk
EtEtV
kkV bb 66.4100451.0
0210.0
bbtt
btt kkkk
kkVhM21
bbtt
tbb kkkk
kkVhM21
Pier V (kips) Mt (k-ft) Mb (k-ft)
A 0.68 3.50 4.66B 4.66 11.19 7.46C 4.66 11.19 7.46
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Shear Walls 17
Example - Shear Walls with Openings
2ft 2ft 2ft4ft 6ft
4ft
To find axial force in piers, look at FBD of top 4ft of
wall.10k
4.66k 4.66k0.68k3.50k-ft 11.19k-ft 11.19k-ft
Moment direction on piers
This 65.9k-ft moment is carried by axial forces in the piers.
Axial forces are determined from Mc/I, where each pier is
considered a concentrated area.Find centroid from left end.
i
ii
AAy
y
Shear Walls 19
Example - Shear Walls with Openings
The axial force in pier A is:
FBD’s of piers:
A
B C
0.68k4.66k-ft4.45k
4.66k7.46k-ft0.45k
4.66k7.46k-ft4.90k
4.66k11.19k-ft
4.90k
4.66k11.19k-ft
0.45k
0.68k3.50k-ft
4.45k
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Shear Walls 21
Example - Shear Walls with OpeningsFBD’s of entire wall to
obtain forces in bottom right spandrel:
0.68k4.66k-ft4.45k
A B C
10k
Fx=9.32kMz=110.8k-ft
Fy=4.45k
xy
kFxFxkkFx
32.9068.010
kFyFykFy
45.4
045.4
ftkMzMzftkftk
ftkftkM z
8.1100)11(45.4)1(45.4)66.4()16(10
Z
Shear Walls 22
Shear Walls with Openings
• Requirements only in strength design• Length ≥ 3 x thickness;
width ≤ 6 x thickness • Height < 5 x length • Factored axial
compression ≤ 0.3Anf’m (9.3.4.3.1)• One bar in each end cell
(9.3.4.3.2)• Minimum reinforcement is 0.0007bd (9.3.4.3.2)
Shear walls with openings will be composed of solid wall
portions, and portions with piers between openings.
Piers:
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Shear Walls 23
Shear Walls: Building Layout1. All elements either need to be
isolated, or will participate in carrying
the load2. Elements that participate in carrying the load need
to be properly
detailed for seismic requirements 3. Most shear walls will have
openings4. Can design only a portion to carry shear load, but need
to detail rest of
structure
Shear Walls 24
Coupled Shear Walls
________ Shear Wall ___________ Shear Wall
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Shear Walls 25
Diaphragms
• Diaphragm: _____________ system that transmits ____________
forces to the vertical elements of the lateral load resisting
system.
• Diaphragm classification:• ______________: distribution of
shear force is based on
tributary ________ (wind) or tributary _______ (earthquake)•
____________: distribution of shear force is based on
relative ______________.
__________
___________
Lateral Force Resisting System
Typical classifications:__________: Precast planks without
topping, metal deck without concrete, plywood sheathing_________:
Cast-in-place concrete, precast concrete with concrete topping,
metal deck with concrete
Shear Walls 26
Rigid Diaphragms
Direct Shear:
i
iv RR
RRVF
Torsional Shear: 2ii
iitt dRR
dRRVeF
V = total shear forceRRi = relative rigidity of lateral force
resisting element idi = distance from center of stiffnesse =
eccentricity of load from center of stiffness
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Shear Walls 27
Diaphragms: Example
Given: The structure shown is subjected to a 0.2 kip/ft
horizontal force. Relative rigidities are given, where the relative
rigidity is a normalized stiffness.Required: Distribution of force
assuming:• flexible diaphragm• rigid diaphragm.
RR
= 4
RR
= 5
RR
= 1
50 ft 50 ft
0.2 kip/ft
PLAN VIEW
Shear Walls 28
Flexible Diaphragms: Example
Solution: Flexible diaphragm – windDistribute based on tributary
area
For seismic, the diaphragm load would be distributed the same
(assuming a uniform mass distribution), but when wall weights were
added in, the forces could be different.
RR
= 4
RR
= 5
RR
= 1
50 ft 50 ft
PLAN VIEW
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Shear Walls 29
Rigid Diaphragms: Example
Solution: Rigid diaphragm
RR
= 4
RR
= 5
RR
= 1
50 ft 50 ft
Wal
l 1
Wal
l 2
Wal
l 3
x
Wall x RR x(RR)123
050
100
451
0250100
Total 10 350
Center of stiffness = 350/10 = 35 ft
Wall RR d (ft) RR(d) RR(d2) Fv Ft Ftotal123
451
-351565
-1407565
490011254225
8102
-4.12.21.9
3.912.23.9
Total 10 350 10250 20 0.0 20
Shear Walls 30
Diaphragms Design ForcesSolution: Forces are shown for a rigid
diaphragm.
The moment is generally taken through chord forces, which are
simply the moment divided by the width of the diaphragm. In masonry
structures, the chord forces are often take by bond beams.
50 ft 50 ft
0.2 kip/ft
3.9 k 12.2 k 3.9 k
V (k)
M (k-ft)
3.9
-3.9
6.1
-6.1
38
19.5 ft
38
-55
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Shear Walls 31
Drag Struts and Collectors• Shear forces: generally
considered
to be uniformly distributed across the width of the
diaphragm.
• Drag struts and collectors: transfer load from the diaphragm
to the lateral force resisting system.
w
PLAN VIEW
v
vL/2 vL/2
WEST WALL ELEVATION
v
vL
EAST WALL ELEVATION
L/3 L/3 L/2
Shear Walls 32
Diaphragm BehaviorThree lateral force resisting systems:
Length=20 ft; Height=14 ft
W24x68
W14
x68
Moment Resisting Framek = 83.2 kip/in
W16x40
W14
x68
Braced Framek = 296 kip/in
L 4x4x5/16
W16x40
Masonry Shear Wallk = 1470 kip/in
L 4x4x5/16
E = 1800 ksiFace shell bedding
End cells fully grouted
Lateral force resisting systems at 24 ft o.c. Diaphragm assumed
to be concrete slab, E=3120ksi, ν=0.17, variable thickness, load of
1 kip/ft.
Diaphragm
Lateral Force Resisting System
24 ft 24 ft
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Shear Walls 33
Diaphragm Behavior
Shear Walls 34
Shear Walls_____________ (unreinforced) shear wall (7.3.2.2):
Unreinforced wall
_____________ (unreinforced) shear wall (7.3.2.3): Unreinforced
wall with prescriptive reinforcement.
40db or 24 in.
Joint reinforcement at 16 in. o.c. or bond beams at 10 ft.
Reinforcement not required at openings smaller than 16 in. in
either vertical or horizontal direction
≤ 16 in.
Control joint
≤ 8 in.
≤ 10 ft.
≤ 8 in.Corners and end of walls
Within 16 in. of top of wall Structurally connected floor and
roof levels
Reinforcement of at least 0.2 in2
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Shear Walls 35
Shear Walls_____________ reinforced shear wall (7.3.2.4):
Reinforced wall with prescriptive reinforcement of detailed plain
shear wall.
______________ reinforced shear wall (7.3.2.5): Reinforced wall
with prescriptive reinforcement of detailed plain shear wall.
Spacing of vertical reinforcement reduced to 48 inches.
___________ reinforced shear wall (7.3.2.6):1. Maximum spacing
of vertical and horizontal reinforcement is min{1/3 length of
wall, 1/3 height of wall, 48 in. [24 in. for masonry in other
than running bond]}.2. Minimum area of vertical reinforcement is
1/3 area of shear reinforcement3. Shear reinforcement anchored
around vertical reinforcing with standard hook4. Sum of area of
vertical and horizontal reinforcement shall be 0.002 times
gross
cross-sectional area of wall5. Minimum area of reinforcement in
either direction shall be 0.0007 times gross
cross-sectional area of wall [0.0015 for horizontal
reinforcement for masonry in other than running bond].
Shear Walls 36
Minimum Reinforcement of Special Shear Walls
Use specified dimensions, e.g. 7.625 in. for 8 in. CMU
walls.
Reinforce-ment Ratio
8 in. CMU wall 12 in. CMU wallAs (in2/ft) Possibilities As
(in2/ft) Possibilities
0.0007 0.064#4@32#5@56
0.098#4@24#5@32#6@48
0.0010 0.092#4@24#5@32#6@40
0.140#4@16#5@24#6@32
0.0013 0.119#4@16#5@32#6@40
0.181#5@16#6@24
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Shear Walls 37
Special Shear Walls: Shear Capacity
Minimum shear strength (7.3.2.6.1.1):• Design shear strength,
Vn, greater than shear
corresponding to 1.25 times nominal flexural strength, Mn
• Except Vn need not be greater than 2.5Vu.
Normal design: φVn has to be greater than Vu. Thus, Vn has to be
greater than Vu/φ = Vu/0.8 = 1.25Vu. Thus, the requirement
approximately doubles the shear.
Shear Walls 38
Seismic Design CategorySeismic Design Category Allowed Shear
Walls
A or BC
D and higher
Response modification factor:Seismic design force divided by
response modification factor, which accounts for ductility and
energy absorption.
Shear Wall ROrdinary plain 1.5Detailed plain 2Ordinary
reinforced 2Intermediate reinforced 3.5Special reinforced 5
Knoxville, TennesseeCategory C for Use Group I and IICategory D
for Use Group III (essential
facilities
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Shear Walls 39
Maximum reinforcing (9.3.3.5)No limits on maximum reinforcing
for following case (9.3.3.5.4):
5.1 and 1 RdV
M
vu
u Squat walls, not designed for ductility
In other cases, can design by either providing boundary elements
or limiting reinforcement.
Boundary element design (9.3.6.5):More difficult with masonry
than concreteBoundary elements not required if:
gA1.0 mu fP geometrically symmetrical sections
gA05.0 mu fP geometrically unsymmetrical sections
AND
1wu
u
lVM OR 3
wu
u
lVM
mnu fAV 3 AND
Shear Walls 40
Maximum reinforcing (9.3.3.5)
Reinforcement limits: Calculated usingMaximum stress in steel of
fyAxial forces taken from load combination
D+0.75L+0.525QECompression reinforcement, with or without lateral
ties, permitted to be
included for calculation of maximum flexural tensile
reinforcement
Uniformly distributed reinforcement
Compression steel with area equal to tension steel
ymu
muyy
vymu
mum
v
s
f
dPbf
dA
64.0
syymumuyymu
mum
s
Eddf
bdPbf
bdA
,min
64.0
= 1.5 ordinary walls = 3 intermediate walls = 4 special
walls
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Shear Walls 41
Maximum reinforcingConsider a wall with uniformly distributed
steel:
Strain mu
y
Stress
Steel in tension
fy
Steel in compression
0.8f’m
PTCC ssm
bdfC vymu
mumm
8.08.0
y
y
y
yy
ymu
ysys AfT
21
y
As taken as total steeldv is actual depth of masonry
Portion of steel in tension
Yielded steel
Elastic steel
ymu
ymusy
mu
ymu
ymu
musy
mu
y
mu
ymu
ymu
musys
Af
Af
AfC
5.0
5.0
21
Shear Walls 42
Maximum reinforcingPCCT mss
ymu
muysy
ymu
ymusy
ymu
yysyss AfAfAfCT
5.05.0
PCPbdfAfCT mvymu
mum
ymu
muysyss
8.08.0
ymu
muyy
vymu
mum
v
s
f
dPfb
dA
64.0
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Shear Walls 43
Maximum reinforcing, εs = 4εy
Shear Walls 44
Maximum reinforcing, εs = 3εy
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Shear Walls 45
Maximum reinforcing, εs = 1.5εy
Shear Walls 46
Shear Strength (9.3.4.1.2)
umnvvu
um PfAdV
MV 25.075.10.4
Maximum Vn is: gmnvn fAV 6 25.0)/( vuu dVM
vyv
s dfsAV
5.0
Vertical reinforcement shall not be less than one-third
horizontal reinforcement; reinforcement shall be uniformly
distributed, max spacing of 8 ft (9.3.6.2)
gnsnmn VVV 8.0
Mu/Vudv need not be taken > 1.0Pu = axial load
gmnvn fAV 4 0.1)/( vuu dVMInterpolate for values of Mu/Vudv
between 0.25 and 1.0
gmnvvu
un fAdV
MV
25
34
γg = 0.75 for partially grouted shear walls and 1.0
otherwise
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2013 Draft Code Changes 47
Partially Grouted Walls
MethodVexp / Vn
Mean St. Dev.
Partially Grouted Walls (Minaie et al, 2010; 60 tests)
2008 Provisions 0.90 0.26Multiply shear strengths by Ag/An 1.53
0.43
Using just face shells 1.77 0.78
Fully Grouted Walls (Davis et al, 2010; 56 tests)
2008 Provisions 1.16 0.17
0.90/1.16 = 0.776; rounded to 0.75
Strength Design 48
ExampleGiven: 10ft. high x 16 ft long 8 in. CMU shear wall;
Grade 60 steel, f’m=2000psi; 2-#5 each end; #5 at 48in. (one space
will actually be 40 in.); superimposed dead load of 1 kip/ft. SDS =
0.4Required: Maximum seismic load based on flexural capacity; shear
steel required to achieve this capacity. Solution: Check capacity
in flexure using 0.9D+1.0E load combination.
Pu = (0.9-0.2SDS)D = (0.9-0.2*0.4)(1k/ft+0.38k/ft) =
0.82(1.38k/ft) = 1.13 k/ft
Total = 1.13k/ft(16ft) = 18.1 kips
Weight of wall: 38 psf(10ft) = 0.38 k/ft(Wall weight from
lightweight units, grout at 48 in. o.c.)
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Shear Walls 49
ExampleSolve for stresses, forces and moment in terms of c,
depth to neutral axis.
8 inc
52 in92 in
140 in188 in
T1 T2 T3 T4 C1C2
Strength Design 51
Example• Use trial and error to find c such that Pn = 18.1
kips/0.9 = 20.1 kips
• Pn = C1 + C2 – T1 – T2 – T3 – T4• c = 14.85 inches
• C1 = 97.6 kips; C2 = 15.5 kips
• T1 = 37.2 kips; T2 = 18.6 kips; T3 = 18.6 kips T4 = 18.6 kips
(all yielded)
• Sum moments about middle of wall (8 ft from end) to find M
• Mn = 37.2(188-96) + 18.6(140-96) + 18.6(92-96) + 18.6(52-96) +
97.6[96-8/2] + 15.5[96-8-(0.8*14.85-8)/2]
• Mn = 13660 kip-in = 1140 kip-ft; Mu = 0.9(1140 kip-ft) = 1025
kip-ft
• Seismic = Mu/h = (1025 kip-ft)/(10ft) = 102.5 kips
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Strength Design 52
ExampleCalculate net shear area, Anv, including grouted cells.
Net area
Net area: 2685)5.262.7)(8(51925.2 ininininininAnv
Maximum Vn
625.016/10// ftftdVhVdVM vuuvuu
kipskipsVV nu 9.919.1148.08.0 Maximum Vu
Maximum shear controls; seismic = 91.9 kips
Strength Design 54
Example
kkVVVg
nmgnns 2.6475.0
0.8975.09.114
kipskipspsiinVnm 0.891.1325.02000685625.075.10.4 2
Top of wall (critical location for shear):Pu = (0.9-0.2SDS)D =
0.82(1k/ft) = 0.82 k/ftTotal: 0.82k/ft(16ft) = 13.1 kips
ns
vyvvy
vns V
dfAsdf
sAV
5.0 5.0
Use #5 bars in bond beams. Determine spacing.
Use __ bond beams, spaced at ____ in. o.c. vertically.
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Example: Maximum ReinforcingCheck with boundary elements:
• Geometrically symmetrical section:
• 0.1fʹmAn = 0.1(2.0ksi)(685in2) = 137 kips
• Check with Pu = (1.2+0.2SDS)D = 1.28(1.38k/ft) = 1.77 k/ft
• Total = 1.77k/ft(16ft) = 28.3 kips OK
• Since Mu(Vudv) = 0.625 ≤ 1 No boundary elements required
• Maximum reinforcement provisions satisfied
If we needed to check maximum reinforcing, there are at least
two ways. These are shown on the following slides.A live load of 1
k/ft is assumed for maximum reinforcement calculations.
Strength Design 56
Strength Design 57
Example: Maximum Reinforcing
Check on axial load:a) Set steel strain to limit and find cb)
Find axial load for this cc) If applied axial load is less than
axial load, OK
Steel strain = 1.5y = 0.00310c = 0.0025/(0.0025+0.00310)*192 =
85.7 in.From spreadsheet, Pu = 258 kips
Pu = D+0.75L+0.525QE = (1.38k/ft + 0.75(1k/ft))16ft = 34.1 kips
OK
Note: Did not include compression steel which is allowed when
checking maximum reinforcement.If compression steel had been
included, Pu = 299 kips
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Strength Design 58
Example: Maximum Reinforcing
Find steel strain for given axial load:a) Through trial and
error (or solver), find c for given axial loadb) Determine steel
strainc) If steel strain greater than limit, OK
P = 34.1 kips (from D+0.75L+0.525QE)c = 20.4 in. (from
spreadsheet and solver)s = (192-20.4)/20.4*0.0025 = 0.0210 = 10.2y
OK
Note: Did not include compression steel which is allowed when
checking maximum reinforcing. If compression steel had been
included: c = 11.3 in. s = 0.0400 = 19.3y
Strength Design 59
Example: Special Shear Wall
If this were a special shear wall:• Vn ≥ shear corresponding to
1.25Mn (7.3.2.6.1.1)
• 1.25Mn = 1.25(1140 k-ft) = 1425 k-ft
• Vn = Mn/h = 1425k-ft/10ft = 142.5 kips
• Vn = 142.5k/0.8 = 178.1kips
• But, maximum Vn is 114.9 kips (64% of required).
Cannot build wall. Need to fully grout to get greater shear
capacity.
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Strength Design 60
Example: Special Shear WallFully grouted:
• Use spreadsheet to find Mn corresponding to Pn = 20.1 kips
• Mn = 1156 kip-ft; Mu = 1041 kip-ft a 1.4% increase
• Seismic = 104.1 kip-ft
• 1.25Mn = 1.25(1156 k-ft) = 1445 k-ft
• Vn = Mn/h = 1445k-ft/10ft = 144.5 kips
• Vn = 144.5k/0.8 = 180.6kips
21494.196.625.7 inininAnv
kipskipspsiinVnm 2.1941.1325.020001494625.075.10.4 2 No shear
steel required; choose horizontal reinforcement to satisfy
prescriptive requirements.
Strength Design 61
Example: Special Shear Wall
Prescriptive reinforcement:• Minimum reinforcement in either
direction: 0.0007A
• Vertical: 0.0007(1494 in2) = 1.05 in2
• 7-#5’s provided: 7(0.31 in2) = 2.17 in2 OK• Spacing is min{1/3
length of wall, 1/3 height of wall, 48 in.} = {192
in./3, 120 in./3, 48 in.} = min{64, 40, 48 in.}• Need to
decrease spacing of vertical reinforcement to 40 inches
• Use spacings of 40, 32, 40, 32, 40 inch to match wall
coursing.• This will have a small impact on moment capacity,
increasing
Mu from 1041 kip-ft to 1166 kip-ft, a 12% increase• This
increases the required nominal shear strength to 202.4
kips. This is slightly greater than the nominal shear strength
due to the masonry (194.2 kips), but with the prescriptive
reinforcement, everything will be OK.
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Strength Design 62
Example: Special Shear Wall
Prescriptive reinforcement:• Minimum reinforcement in either
direction: 0.0007A
• Horizontal: 0.0007(120 in)(7.625 in) = 0.64 in2
• Use #5’s at 40 in (required spacing); total is 3 bars• 3(0.31
in2) = 0.92 in2 OK
• Total reinforcement: 0.002A• Vertical: 2.79in2/1494in2 =
0.00187A• Horizontal: 0.92in2/915in2 = 0.00100A• Total = 0.00187A +
0.00100A = 0.00287A OK
Shear Walls 63
ExampleGiven: Wall system constructed with 8 in. grouted CMU
(Type S mortar).Required: Determine the shear distribution to each
wall
40in 40in64in
112i
n
Elevation
Plan
6t=4
8in
56in
Solution: Use gross properties (will not make any difference for
shear distribution).
ba c
40in 40in
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Shear Walls 64
ExamplePier b inE
inin
inin
inE
Lh
Lh
Etkb 286.03
641124
64112
62.7
3422
Piers a and c Previously determined stiffness from basic
principles.
inEk ca 157.0,
Portion of shear load = stiffness of pier over the sum of the
stiffnesses.
VVinEinEinE
inEVk
kV aa 26.0157.0286.0157.0157.0
Vb = 0.48V
Deflection calculations:k = (0.600in.)E = (0.600in.)(1800ksi) =
1080 kip/inReduce by a factor of 2 to account for cracking: k = 540
kip/in
Shear Walls 65
ExampleRequired: Design the system for a horizontal earthquake
load of 35 kips, a dead load of 1200 lb/ft length of building, a
live load of 1000 lb/ft length of building, and a vertical
earthquake force of 0.2 times the dead load. Use a special
reinforced shear wall (R>1.5).
Solution: Use strength design. Check deflection.
inink
kkP 065.0
/54035
000579.0112065.0
inin
hor
1728h
-
Shear Walls 66
Example
Use load combination 0.9D+EMu = 16.8k(112in)(1ft/12in) =
156.8k-ft Pu = 0.9(14.1k) - 0.2(14.1k) = 9.87k
Use prescriptive reinforcement of (0.002-0.0007)Ag = 0.0013Ag in
vertical directionAs = 0.0013(64in)(7.625in) = 0.63in2
Wall b: Vub = 0.48Vu = 0.48(35k) = 16.8kD =
1.2k/ft(_____________)(1ft/12in) +
0.075ksf(112in)(64in)(1ft2/144in2) = 14.1kips
Tributary width
Minimum prescriptive reinforcement for special shear walls:Sum
of area of vertical and horizontal reinforcement shall be 0.002
times
gross cross-sectional area of wallMinimum area of reinforcement
in either direction shall be 0.0007 times gross
cross-sectional area of wall
Shear Walls 67
Example: Flexural SteelUse load combination 0.9D+E Mu =
156.8k-ft Pu = 9.87k
Try 2-#4 each end; 1-#4 middle: As=1.00in2
Middle bar is at 28 in. (8 cells, cannot place bar right in
center)From spreadsheet: Mn=152.7k-ft NG
Note: used solver to find value
Try 1-#6 each end; 1-#6 middle: As=1.32in2 Mn=194.1k-ft OKTry
2-#5 each end; 1-#5 middle: As=1.55in2 Mn=219.1k-ft OK
Use 2-#5 each end; 1-#5 middle: As=1.59in2
-
Shear Walls 68
Example: Maximum ReinforcingCheck maximum reinforcing:Axial
force from load combination D+0.75L+0.525QE
L=1k/ft(104in)(1ft/12in) = 8.67kipD+0.75L+0.525QE =
14.1k+0.75(8.67k)+0.525(0) = 20.60kip
For maximum reinforcing calculations, compression steel can be
included even if not laterally supportedFor #5 bars and P =
20.60kip, c = 6.35in.. (from solver on spreadsheet)
yymus ininine
kdkdd 421.100211.00025.0
35.635.660
OK
Shear Walls 69
Example: Shear
kipsklbkpsiinin
PfAdVMV umnvuunm9.5028.725.01000/12000)64)(625.7(25.2
25.0/75.10.4
175.164112
inVinV
dVM
u
u
vu
u Use Mu/Vudv = 1.0
kipsVn 9.50Ignore any shear steel:
kipskipskipsVn 8.167.409.508.0 OK
Find Pu neglecting wall weightPu = 0.9(10.4k) - 0.2(10.4k) =
7.28k
-
Shear Walls 70
Example: Shear
Mn = Mn/ = 219.1k-ft/0.9 = 243.4k-ftV corresponding to Mn:
[243.3k-ft/(112in)](12in/ft) = 26.1kips1.25 times V: 1.25(26.1k) =
32.6kips < Vn=40.7kips OK
Wall b: Vertical bars: 2-#5 each end; 1-#5 middleHorizontal
bars: #4 at 24 in. o.c. (Total 5-#4)
Minimum prescriptive horizontal reinforcement: 0.0007AgAs =
0.0007(64in)(7.625in) = 0.34in2
Maximum spacing is {64in./3, 48in.} = 21.3 in.Use bond beams at
16 in. o.c. (I would be fine with 24 in.)
Shear Walls 71
Example
Use load combination 0.9D+EMu = 9.1k(112in)(1ft/12in) = 84.9k-ft
Pu = 0.9(11.1k) - 0.2(11.1k) = 7.77k
Minimum prescriptive reinforcement: 0.0013AgAs =
0.0013(88in)(7.625in) = 0.87in2
Wall a: Vua = 0.26Vu = 0.26(35k) = 9.1kD =
1.2k/ft(20in+40in)(1ft/12in) +
0.075ksf(112in)(40in+48in)(1ft2/144in2) = 11.1kips
Design issues:Flange in tension: easy to get steel; problems
with maxFlange in compression: hard to get steel; max is easySteel
in both flange and web affect maxNeed to also consider loads in
perpendicular directionNeed to check shear at interface of flange
and web
-
Shear Walls 72
ExampleUse load combination 0.9D+E Mu = 84.9k-ft Pu = 7.77k
To be consistent with other wall, try #5 barsTry 3 in the flange
(either distributed as shown, or one in
the corner, and two in the jambTry 2 in the web jamb
Flange in tension: Mn=150.8k-ftMax reinforcing check:
s=0.0127=6.14y
Note: with 2-#5 in flange, Mn=106.9k-ft, but might need 3 bars
for perpendicular direction or out-of-plane
Axial force for max from load combination
D+0.75L+0.525QEL=1k/ft(60in)(1ft/12in) = 5.00kipD+0.75L+0.525QE =
11.1k+0.75(5.00k)+0.525(0) = 14.85kip
Flange in compression: Mn=125.8k-ftMax reinforcing check:
s=0.0564=27.3y
OK
OK
Shear Walls 73
Example: Shear
Mn = Mn/ = 150.8k-ft/0.9 = 167.6k-ftV corresponding to Mn:
[167.6k-ft/(112in)](12in/ft) = 18.0kips1.25 times V: 1.25(18.0k) =
22.4kips < Vn=25.4kips
Walls a,c: Vertical bars: 5-#5 as shown in previous
slideHorizontal bars: #4 at 24 in. o.c. (Total 5-#4)
Minimum prescriptive horizontal reinforcement: 0.0007AgUse 5
bond beams, 24in. o.c. (top spacing is only 32in.)
kipsklbkpsiininVm 7.312.425.01000/12000)40)(625.7(25.2
18.240112
inVinV
dVM
u
u
vu
u Use Mu/Vudv = 1.0
kipsVn 7.31Ignore any shear steel: kipskipsVn 4.257.318.0 OK
OK
-
Shear Walls 74
ExampleSection 5.1.1.2.4 requires shear to be checked at
interface. Check intersection of flange and web in walls a and
c.
kipsksiinfAV ysu 8.556031.03 2 Approximate shear force is
tension force in flexural steel.
kipskipsVn 7.689.858.0 OK
kipslbkpsiinin
PfAVdMV umnvm9.851000/12000)112)(625.7(0.175.10.4
25.0/75.10.4
Conservatively take Mu/Vudv=1.0, and Pu=0.
Shear Walls 75
Example: Drag StrutsDistributed shear force: ftk
ftin
ink
LVu /875.1
112
22435
Look at top of wall:
3.33ft 3.33ft5.33ft3.33ft 3.33ft
Drag strut Drag strut9.1k 9.1k16.8k
1.88k/ft 1.88k/ft 1.88k/ft 1.88k/ft1.88k/ft
2.8k 3.4k 2.8k3.4k