Lecture 4 5 Urm Shear Walls

Classnotes for ROSE School Course in:
Classnotes for ROSE School Course in:
Masonry Structures
Masonry Structures

Lessons 4 and 5: Lateral Strength and Behavior of URM Shear Walls
flexural strength, shear strength, stiffness, perforated shear walls

Notes Prepared by:
Daniel P. Abrams
Willett Professor of Civil Engineering
University of Illinois at Urbana-Champaign
October 7, 2004

Masonry Structures, slide 1

Existing URM Buildings


Damage to Parapets

1996 Urbana Summer

1994 Northridge Earthquake, Filmore


Damage Can Be Selective

1886 Charleston, South Carolina


Damage to Corners

1994 Northridge Earthquake, LA


Damage to In-Plane Walls

1994 Northridge Earthquake, Hollywood

URM cracked pier, Hollywood


Damage to Out-of-Plane Walls
1886 Charleston, South Carolina

1996 Yunnan Province Earthquake, Lijiang Masonry Structures, slide 7

Likely Consequences

St. Louis Firehouse

1999 Armenia, Colombia Earthquake


2001 Bhuj Earthquake


Lateral Strength of URM Shear Walls


URM Shear Walls
P3
H3
n
Pi Pb = ∑ Pi
i =1
Hi
n
Vb = ∑ H i
P1 i =1
H1 diagonal n
hi tension crack M b = ∑ H i hi
i=1
flexural flexural
tension compression
crack cracks

Mb Vb
Pb
Ref: BIA Tech. Note 24C The Contemporary Bearing Wall - Introduction to Shear Wall Design
NCMA TEK 14-7 Concrete Masonry Shear Walls

URM Shear Walls
Design Criteria
(a) allowable flexural tensile stress: -fa + fb < Ft
Ft given in UBC 2107.3.5 (Table 21 - I); Ft = 0 per MSJC Sec. 2.2.3.2
pg. cc-35 of MSJC Commentary reads: Note, no values for allowable tensile stress
are given in the Code for in-plane bending because flexural tension in walls should be
carried by reinforcement for in-plane bending.

(b) allowable axial and flexural compressive stress:
fa fb
MSJC Sec. 2.2.3.1 and UBC 2107.3.4 unity formula: + < 1.0
Fa Fb

where:
Fa = allowable axial compressive stress (UBC 2107.3.2 or MSJC 2.2.3)
Fb = allowable flexural compressive stress = 0.33 f´m (UBC 2107.3.3 or MSJC 2.2.3)


Allowable Tensile Stresses, Ft
MSJC Table 2.2.3.2 and UBC Table 21-I

Mortar Type
Direction of Tension
and Portland Cement/Lime
or Mortar Cement Masonry Cement/Lime
Type of Masonry
all units are (psi) M or S N M or S N
tension normal
to bed joints
solid units 40 30 24 15
hollow units 25 19 15 9
fully grouted units 68* 58* 41* 26*

tension parallel
to bed joints
solid units 80 60 48 30
hollow units 50 38 30 19
fully grouted units 80* 60* 48* 29*

* grouted masonry is addressed only by MSJC

URM Shear Walls
Design Criteria
(c) allowable shear stresses:

UBC Sec. 2107.3.7 shear stress, unreinforced masonry:
clay units:
Fv = 0.3 (f’m)1/2 < 80 psi (7-44)
concrete units:
with M or S mortar Fv = 34 psi
with N mortar Fv = 23 psi

allowable shear stress may be increased by 0.2 fmd where fmd is
compressive stress due to dead load
V
Per UBC Sec. 2107.3.12 shear stress is average shear stress, fv =
Ae


URM Shear Walls
Design Criteria

MSJC Sec. 2.2.5.2: shear stress, unreinforced masonry:
Fv shall not exceed the lesser of:
(a) 1.5 (f’m)1/2
(b) 120 psi
(c) v + 0.45 Nv/An where v = 37 psi for running bond, w/o solid grout
37 psi for stack bond and solid grout
60 psi for running bond and solid grout
(d) 15 psi for masonry in other than running bond
VQ
Note: Per MSJC Sec. 2.2.5.1, shear stress is maximum stress, f v =
Ib


URM Shear Walls
Design Criteria

fvmax V 2
f v avg = = fv
Anet 3 max
fvavg 3 V
f v max =
2 Anet
for rectangular section


URM Shear Walls
Possible shear cracking modes.
strong mortar low vertical weak mortar
weak units compressive stress strong units

through masonry units sliding along bed joints stair step through bed
and head joints

Associated NCMA TEK Note
#66A: Design for Shear Resistance of Concrete Masonry Walls (1982)


Example: URM Shear Walls
Determine the maximum base shear per UBC and MSJC.
5000 lb. DL
8” CMU’s with face shell bedding
H block strength = 2800 psi
Type N Portland cement lime mortar
special inspection provided during construction
9’- 4”

Net section with face shell bedding:
H 80”
1.25”
9’- 4”

Anet = 2.5 (80) = 200 in 2
S net = 2.5 (80)2 / 6 = 2667 in 3
h'
r = 2.84quot; h’ = 112quot; = 39.4
6’ - 8” r


Example
Forces and Stresses:
Vb = 2 H
M b = Vb (1.5 x 9.33 ) x 12 = 168 Vb
f a dead = 2 (5000 lbs) / 200 in 2 = 50 psi (neglectin g self weight)
f b = M/S = 168 Vb /2667 = 0.0630 Vb

Maximum base shear capacity per UBC
shear stress
Fv = 23 + 0.2 f md = 23 + 0.2 (50) = 33 psi (Sec. 2107.3.7)
Vmax = (33 psi x 1.33) (200 in 2 ) = 7980 lbs. (does not govern)

flexural tensile stress
Ft = 19 psi x 1.33 = 25.3 psi (Ft per Sec. 2107.3.5 )
- f a + f b = Ft - 50 + 0.0630Vb = 25.3 V b = 1194 lbs. (governs)


Example
Maximum base shear capacity per UBC
flexural compressive stress
fa fb h 2 h'
+ < 1.33 Fa = 0.25f´ m [1 - ( ) ] = 0.23 f´ m for ≤ 99
Fa Fb 140r r
f´ m = 1850 psi per Table 21 - D for f´ u = 2800 psi and Type N mortar
Fa = 426 psi Fb = 0.33 (1850) = 616 psi

50 0 .0630 Vb
+ = 1.33 Vb = 11,857 lbs. (does not govern)
426 616

Maximum base shear capacity per MSJC
In lieu of prism tests, a lower bound compressive strength of 1861 psi will be used
based on the 2800 psi unit strength and Type N mortar per MSJC Spec. Table 2.
shear stress Fv = 1.5 (f´ m )1/2 = 1.5 (1861)1/2 = 65 psi or
Fv = 37 psi + 0.45 (50 psi) = 60 psi (governs)
2 2
Vmax = Fv Anet = (60 psi x 1.33) (200 in 2 )
3 3
= 10,629 lbs. (does not govern)


Example
Maximum base shear capacity per MSJC
flexural tensile stress
per MSJC Sec. 2.2.3.2 : Ft = 0 psi
- fa + fb = 0
- 50 + 0.0630 Vb = 0 V b = 794 lbs. (governs)
flexural compressive stress

fa fb Fa = 0.25f´ m [1 - (h/140r) 2 ] = 0.23 f´ m = 426 psi for h/r = 39.4
+ < 1.33
Fa Fb Fb = 0.33 (1861) = 620 psi
50 0.0630 Vb
+ = 1.33 Vb = 11,934 lbs. (does not govern)
426 620

Summary: Vb max
axial and flexural stress
shear stress
tension compression
UBC 7890 1194 11,857
MSJC 10,629 794 11,934


URM Shear Walls
Post-Cracked Behavior
σ v = fa P =σvA
Hh
H equilibrium : Hh = Pe ; e =
P
2P [1]
fm = = compressive edge stress ≤ Fa or Fb
bη
h η L ⎛L ⎞
= −e η = 3⎜ − e ⎟ [2]
3 2 ⎝2 ⎠
toe combining above equations :
heel 2P 2P 1
fm< Fa fm = =
bη ⎛L ⎞ L2 [3]
3b⎜ − e ⎟
P ⎝2 ⎠ L2
width = b η
e 3
4P
L/2 fm =
⎛ e ⎞
η 3 bL ⎜ 1 - 2 ⎟
⎝ L⎠


URM Shear Walls
Post-Cracked Behavior

Lateral Load, H
resultant load, P, shifts toward toe
toe crushing

cracking load
2 to 3 times
first flexural cracking

MSJC/UBC
assumed behavior
Lateral Deflection at Top of Wall

Note: shear strength should be checked considering effects of flexural cracking


Perforated URM Shear Walls


Lateral Stiffness of Shear Walls
Cantilevered shear wall
flexure shear :

∆ Hh 3 H 1
∆= + h
3E m I Av G

H common assumption s : Av = 5/6 Ag G = 0.4 E m
Hh 3 1.2 Hh
∆= +
3E m I Ag (0.4 E m ) bL3
Ig =
12
h 4 Hh 3 3 Hh Ag = bL
∆= +
E m bL3 bLE m
H ⎛ h ⎞⎡ ⎛ h ⎞ ⎤
2
∆= ⎜ ⎟ ⎢4⎜ ⎟ + 3⎥
bE m ⎝ L ⎠ ⎢ ⎝ L ⎠
⎣ ⎥
⎦
bE m
k = lateral stiffness = H/∆ =
L ⎛ h⎞ ⎡ ⎛ h ⎞2 ⎤
⎜ ⎟ ⎢4⎜ ⎟ + 3⎥
L ⎠⎣ ⎝ L ⎠
⎝ ⎢ ⎥
⎦


Pier between openings
flexure shear
3
Hh H 1
∆= + h
H 12E m I Av G
Hh 3 1.2 Hh
∆= +
E m bL3 bL(0.4 E m )
h
H ⎛ h ⎞ ⎡⎛ h ⎞ ⎤
2
∆= ⎜ ⎟ ⎢⎜ ⎟ + 3 ⎥
bE m ⎝ L ⎠ ⎢⎝ L ⎠
⎣ ⎥
⎦
H

L
bE m
k = lateral stiffness = H / ∆=
⎛ h ⎞ ⎡⎛ h ⎞ ⎤
2
⎜ ⎟ ⎢⎜ ⎟ + 3⎥
⎝ L ⎠ ⎢⎝ L ⎠
⎣ ⎥
⎦



k 1.6
bE m 1.4
cantilever
1
1.2 k=
⎛ h ⎞⎡ ⎛ h ⎞ ⎤
2

⎜ ⎟⎢ 4 ⎜ ⎟ + 3⎥
1.0 ⎝ L ⎠⎢ ⎝ L ⎠
⎣ ⎥
⎦
0.8 fixed pier
1
0.6 k=
⎛ h ⎞ ⎡⎛ h ⎞ ⎤
2
⎜ ⎟ ⎢⎜ ⎟ + 3 ⎥
0.4 ⎝ L ⎠ ⎢⎝ L ⎠
⎣ ⎥
⎦
0.2
h
0
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 L


References

Associated NCMA TEK Note:
61A Concrete Masonry Load Bearing Walls
- Lateral Load Distribution (1981)

Associated BIA Technical Note:
24C The Contemporary Bearing Wall
- Introduction to Shear Wall Design
24D The Contemporary Bearing Wall
- Example of Shear Wall Design
24I Earthquake Analysis of Engineered Brick Masonry Structures


Example: Lateral-Force Distribution
Determine the distribution of the lateral force, H, to walls A, B and C.
A B C

H 10’-0” 18’-0” 6’-0”
type of masonry and
wall thickness is the
same for each wall
h=15’

wall L h/L ki * ki / ∑ ki *based on cantilever action
bEm
A 10’ 1.50 0.0556 bEm 0.20 ki =
h ⎞⎡ ⎛ h ⎞ ⎤
2
⎛
B 18’ 0.83 0.2077 bEm 0.75 ⎜ ⎟ ⎢ 4 ⎜ ⎟ + 3⎥
⎝ L ⎠⎢ ⎝ L ⎠
⎣ ⎥
⎦
C 6’ 2.50 0.0143 bEm 0.05

Σki = 0.2776 bEm

Lateral-Force Distribution to Piers
Perforated Shear Walls equilibrium:
H = V1 + V2 + V3

∆ overall story stiffness:
H
H H
V3 H = K∆ or ∆ = =
K ∑ ki
V1 V2
where K = lateral story stiffness
h3 = ∑ k i = k1 + k 2 + k 3
h2
h1

bi Emi
ki = for fixed − fixed pier
⎛ hi ⎞ ⎡⎛ hi ⎞ ⎤
2
L2
⎜ ⎟ ⎢⎜ ⎟ + 3 ⎥
⎜L ⎟ ⎜L ⎟
L1 L2 L3 ⎝ i ⎠ ⎢⎝ i ⎠
⎣ ⎥
⎦

shear force attracted to single pier:
Vi = k i ∆i = k i ∆ = k i Η / ∑ k i
Vi = (k i / ∑ k i ) Η
k i / ∑ k i = distributi on factor


Example: Lateral Force Distribution to Piers
Determine the distribution of story shear, H, to each pier.
40” 40” 32” 8”grouted
H concrete block

24”

Va Vb Vc
112”

64”

A A
24”

Elevation
7.63”
56”

a b c

Section A-A


Example: Lateral Force Distribution to Piers
7.63 x48 = 366 x 7.63/2 = 1397
7501
piers a and c 7.63 x 40 = 305 x 20 = 6104 y= = 11.2quot;
671
7.63” 671 7501
I gross= 366(11.2 3.81) + 305(20− 11.2) + 7.63(40.0) /12= 84,273 4
− 2 2 3
in
y Av = (5/6) Aweb = (5/6) (7.63quot; )(40.0) = 254 in 2
48”

1 1
7.63”

ka = k c = = = 1.12 E
h3 h ( 64 )3 64
+ +
12EI GAv 12E(84,273) ( 0.4 E )( 254 )
40”
bEm 7.63 E
kb = = = 1.91 E
h ⎞ ⎡⎛ h ⎞ ⎤ 1 [1 + 3]
2
⎛
⎜ ⎟ ⎢⎜ ⎟ + 3 ⎥
pier b ⎝ L ⎠ ⎢⎝ L ⎠
⎣ ⎥
⎦
∑ ki = ( 1.12 + 1.91 + 1.12 ) Ε = 4.16 Ε
7.63”

64” DFa = DFc = 1 .12 / 4 .16 = 0 .27 DFb = 1 .91 / 4 .16 = 0 .46


Perforated Shear Walls
Axial Force due to Overturning
equilibrium of pier axial forces:
∑ pi = 0 for zero gravity force [1]
∑ pi = 0 = ∑ f ai Ai [2]
∑ Ai ≠ 0 thus, centroid of pier [3]
areas = neutral axis
y=
∑ Ai yi
[4]
∑ Ai
equilibrium of moments:
M = ∑ pi yi = ∑ f ai Ai yi [5]

y1 y2 from similar triangles:
y3
f ai = f max y i / c [6]
y1 y2 y3
y substituting in [5]:
fai= ave. axial ⎛ y ⎞ [7]
fmax M = ∑ ⎜ f max i ⎟ Ai yi
p2 p3
stress across ⎝ c ⎠
p1 pier “i”
f max f
c
M
M=
c
∑ Ai yi 2 M = max I
c
[8]


Axial Force due to Overturning

solving for fmax:

Mc
f max = [10]
I

substituting in [6]:
⎛ Mc ⎞ yi Myi
f ai = ⎜ ⎟ = [11]
⎝ I ⎠c I
MAi yi
pi = Ai f i =
I [12]
Ai yi
pi = M [13]
∑ Ai yi2

distribution factor
for overturning moment


Design Criteria for Piers between Openings
flexure: unreinforced piers
P fa f b
compressio n + < 1.0
Fa Fb
M
V tension - f a + f b < Ft

flexure: reinforced piers
h
neglecting axial force :
2M
compressio n f b = < Fb
V jkbd 2
M=Vih/2 M
tension f s = < Fs
As jd
P
P = Pdead + Plive + Plateral considering axial force :
(a) unity equation (UBC only)
(b) load - moment interaction diagram
P
(c) modify As by (only when tension controls)
bdFs


Design Criteria for Piers between Openings
P shear: unreinforced piers
VQ
M MSJC fv = < Fv
I net b
V
V
UBC fv = < Fv
Ae
h
shear: reinforced piers
V
V UBC fv = < Fv
M=Vih/2 bjd
V
P MSJC fv = < Fv
bd
Loading Combinations
UBC MSJC Sec. 2.1.1 Effect
D+L D+L Pmax for small lateral load
0.75(D+L+W/E) D+L+W/E Pmax and Mmax for large lateral
0.9D-0.75E 0.9D+E Pmin for smallest moment
capacity D+W


Example: Perforated Shear Wall
Check stress per the UBC for the structure shown below.
Design pier reinforcement if necessary.

14.9 kip
Special inspection is provided

10’-0”
Earthquake
f’m = 2500 psi
Loads
fully grouted but unreinforced
14.9 kip Grade 60 reinforcement
Type N mortar with Portland Cement

10’-0”
7.4 kip Gravity Loads
Level Dead Live
9’-8” 3 50 kip 80 kip
2 60 kip 80 kip
1 60 kip 80 kip

18’-8” total 170 kip 240 kip


Pier Dimensions 18’-8”

3’-4” 5’-4” 3’-4” 3’-4” 3’-4”
2’-8” 4’-0” 2’-8”

8” grouted
9’-4”

concrete block

7.63”
3’-4”

a b c

40” 32”


Stiffness of Pier “a”
y = [32 (7.63)2 / 2 + 7.63 (40 2 /2)] / 549 = 12.82quot;

a I g = (7.63)3 (32)/12 + 7.63(32)(12.82 - 7.63/2)2
7.63” + 7.63(40)3 /12 + 7.63(40)(20 - 12.82)2 = 76 ,470 in 4
7.63”
32”

1
ka = 3
h h
y +
12Em I (5/6 Aweb )(0.4 E m )
40”
1
ka = 3
48 48
+
12Em 76 ,740 (5/6 × 7.63 × 40)(0.4 E m )
1
ka = = 1.69 E m
0.120 0.472
+
Em Em

Stiffness of Pier “b”

Ig =7.63 (40) 3 / 12=40,693 in4
7.63”

b 1
kb =
h3 h
+
40” 12E m I (5/6 Aweb )(0.4 E m )

1
kb =
48 3 48
+
12E m 40 ,693 (5/6 × 7.63 × 40)(0.4 E m )
1
kb = = 1.43 Em
0.226 0.472
+
Em Em


Stiffness of Pier “c” 1
kc =
h3 h
+
32”

c 12E m I (5/6 Aweb )(0.4 E m )
1
kc =
112 3
7.63”

112
40” +
12E m 76 ,470 (5/6 × 7.63 × 40)(0.4 E m )
1
I g = 76 ,470 in 4 kc = = 0.38 E m
1.526 1.101
(same as Pier a) +
Em Em
Distribution of Story Shear to Piers

Vbase = 14.9 k + 14.9 k + 7.4 k = 37.2 k pier ki DF i Vi
a 1.69 Em 0.483 18.0
b 1.43 Em 0.409 15.2
c 0.38 Em 0.109 4.0
Σ k = 3.50 Em 1.000 37.2 k


Distribute Overturning Moments to Piers
y
7.63”
40.0”

a b c

124.0”
12.82”
211.2”
y 2 = 9 . 38 quot;
y 1 = 101 .8quot; y 3 = 96 .58 quot;

pier Ai yi Aiyi
a 549 12.8” 7038
b 305 124.0” 37,820
c 549 211.2” 115,949
∑ Ai =1403 ∑ Aiyi=160,807

y = 160 ,807 / 1403 = 114.6quot;

Distribute Overturning Moments to Piers

total story moment = M1 (@top of window opening, first story)
= 14.9k x 23.0’ + 14.9k x 13.0’ + 7.4k x 3.0’ = 558k-ft

2 Ai yi
pier Ai y Ai y i I A i y i2 + I Ai yi Pi = M1
In
(in2) (in) (1000 in4) (1000 in4) (1000 in4) (1000 in3) (kips)

a 549 101.8” 5689 76 5765 55.9 33.9

b 305 -9.38 27 41 68 -2.9 -1.8

c 549 -96.58” 5120 76 5196 -53.0 -32.1

∑A y
2
∑ Ai = 1403 i i + I = 11,029


Summary of Pier Forces

pier % gravity* Pd Pl Peq Veq Meq=Veq(h/2)
(kips) (kips) (kips) (kips) (kip-in)

a 0.361 61.4 86.6 33.9 18.0 432

b 0.319 54.2 76.6 -1.8 15.2 365

c 0.319 54.2 76.6 -32.1 4.0 224

*based on tributary wall length: pier a: (32” + 40” + 32”)/288 = 0.361
(assuming that floor loads are pier b: (32” + 40” + 20”)/288 = 0.319
applied uniformly to all walls) pier c: (20” + 40” + 32”)/288 = 0.319


Loading Combinations

axial compressive force, P moment, M shear, V d
case 1 case 2 case 3
pier D+L 0.75(D+L+E) 0.9D-0.75E 0.75Meq 0.75Veqx1.5*
(kips) (kips) (kips) (kip-in) (kips) (in.)

a 148.0 136.4 29.8 327 20.3 36

b 130.8 99.5 47.4 274 17.1 36

c 130.8 122.2 24.7 168 4.5 36

*UBC 2107.1.7 for Seismic Zones 3 and 4


Axial and Flexural Stresses, Load Case 1 = D + L

a b c

y y

pier PD+L fa Fa* fa/Fa
(kips) (psi) (psi)

a 148.0 270 543 0.497 < 1.0 ok

b 130.8 430 543 0.792< 1.0 ok

c 130.8 239 543 0.440< 1.0 ok

*Fa = 0.25f’m[1-(h/140r)2] r = 0.289t = 0.289 × 7.63 = 2.2quot; h = 9' −4quot; = 112quot;
Note that conservative assumption is used for Fa calculation, r is the lowest and h is the full height.

Axial and Flexural Stresses, Load Case 2: 0.75 (D + L + E)

a b c

y y

pier 0.75(PD+L+EQ) fa=P/A Fa fa/Fa 0.75Me Sg fb fb/Fb** fa/Fa+fb /Fb
(kips) (psi) (psi) (kip-in) (in3) (psi)

a 136.4 249 543 0.459 327 2813* 116 0.139 0.598 < 1.0 ok

b 99.5 326 543 0.600 274 2035 135 0.162 0.762 < 1.0 ok

c 122.2 223 543 0.411 168 2813* 60 0.072 0.483 < 1.0 ok

* minimum Sg is taken to give maximum fb for either direction of building sway
** Fb= 0.33f’m = 833 psi


Axial and Flexural Stresses, Load Case 3: 0.9D - 0.75Peq

a b c

y y

minimum axial compression: check tensile stress with Ft = 30 UBC Sec 2107.3.5
pier (0.9PD-0.75PEQ) fa=P/A 0.75Meq Sg fb - fa+fb
(kips) (psi) (kip-in) (in3) (psi) (psi)**

a 29.8 54 327 2813* 116 62 > 30 psi provide reinf.

b 47.4 155 274 2035 135 -20 < 30 psi ok

c 24.7 45 168 2813* 60 15 < 30 psi ok

* minimum Sg is taken to give maximum fb for either direction of building sway
**tensile stresses


Pier Shear Stress, Load Case 4 : 0.75E

a b c

y y

pier V=0.75Veq x 1.5 fv = V/Aweb fao = P/A* Fv = 23 + 0.2fao**
(kips) (psi) (psi) (psi)

a 20.3 67 54 34 < 67 provide shear reinf.

b 17.1 56 155 54 < 56 provide shear reinf.
c 4.5 15 45 32 > 15 ok

* from Case 3 0.9Pd-0.75Peq
** UBC 2107.3.7


Case Study: Large-Scale Test


Georgia Tech Large-Scale Test

24’

photo from Roberto Leon


Final Crack Pattern
Load Direction

slide from Roberto Leon


Final Crack Pattern

Load Direction


Results- Global Behavior

80
Wall 1 Force-Displacement Response
60

40
Base Shear (kip)

20

0
-0.30 -0.20 -0.10 0.00 0.10 0.20 0.30
-20

-40

-60

-80

Roof Displacement (in)


Overturning Effect (Vertical Stress)

Base strains recorded during loading in the push and pull direction

USA CERL Shaking Table Tests

12’

photos from S. Sweeney


Damage on North Wall
Final Cracking Pattern

Permanent offsets of 0.25” – 0.35”
due to rocking of pier.

slide from S. Sweeney


Peak Force vs. Deflection
North-South Uni-directional Motion
30
Base Shear (kip)

25
20 PGA = 0.33 g
PGA = 0.75 g
15
PGA = 0.98 g
10
PGA = 1.08 g
5
0
0 0.025 0.05 0.075 0.1 0.125 0.15 0.175
Average First Floor Deflection (in)

East-West Uni-directional Motion
25
Base Shear (kip)

20
PGA = 0.30 g
15
PGA = 0.75 g
10 PGA = 1.09 g
5 PGA = 1.40 g
0
0 0.1 0.2 0.3 0.4 0.5 0.6
Average First Floor Deflection (in)

slide from S. Sweeney

End of Lessons 4 & 5


Lecture 4 5 Urm Shear Walls

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Lecture 4 5 Urm Shear Walls