lecture 4 5 urm shear walls
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Masonry Structures, slide 1
Classnotes for ROSE School Course in:Masonry Structures
Classnotes for ROSE School Course in:Masonry Structures
Notes Prepared by:Daniel P. Abrams
Willett Professor of Civil EngineeringUniversity of Illinois at Urbana-Champaign
October 7, 2004
Lessons 4 and 5: Lateral Strength and Behavior of URM Shear Wallsflexural strength, shear strength, stiffness, perforated shear walls
Masonry Structures, slide 2
Existing URM Buildings
Masonry Structures, slide 3
Damage to Parapets
1994 Northridge Earthquake, Filmore
1996 Urbana Summer
Masonry Structures, slide 4
Damage Can Be Selective
1886 Charleston, South Carolina
Masonry Structures, slide 5
Damage to Corners
1994 Northridge Earthquake, LA
Masonry Structures, slide 6
Damage to In-Plane Walls
1994 Northridge Earthquake, Hollywood
URM cracked pier, Hollywood
Masonry Structures, slide 7
Damage to Out-of-Plane Walls
1886 Charleston, South Carolina
1996 Yunnan Province Earthquake, Lijiang
Masonry Structures, slide 8
Likely Consequences
St. Louis Firehouse
1999 Armenia, Colombia Earthquake
Masonry Structures, slide 9
2001 Bhuj Earthquake
Masonry Structures, slide 10
Lateral Strength of URM Shear Walls
Masonry Structures, slide 11
URM Shear Walls
∑
∑
∑
n
1=iiib
n
1=iib
n
1=iib
h H= M
H= V
P= P
Ref: BIA Tech. Note 24C The Contemporary Bearing Wall - Introduction to Shear Wall DesignNCMA TEK 14-7 Concrete Masonry Shear Walls
P3
Pb
hi
H3
Hi
H1
Pi
P1
flexural tension crack
flexural compression cracks
VbMb
diagonaltension crack
Masonry Structures, slide 12
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.
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)
0.1Ff
Ff
b
b
a
a <+
(b) allowable axial and flexural compressive stress:
MSJC Sec. 2.2.3.1 and UBC 2107.3.4 unity formula:
Masonry Structures, slide 13
Allowable Tensile Stresses, FtMSJC Table 2.2.3.2 and UBC Table 21-I
Direction of Tensionand
Type of Masonry
Mortar TypePortland Cement/Lime
or Mortar Cement Masonry Cement/Lime
M or S M or SN Ntension normalto bed joints
solid unitshollow unitsfully grouted units
tension parallelto bed joints
solid unitshollow unitsfully grouted units
40 25 68*
80 50 80*
30 19 58*
60 38 60*
24 15 41*
48 30 48*
15 9
26*
30 19 29*
* grouted masonry is addressed only by MSJC
all units are (psi)
Masonry Structures, slide 14
URM Shear WallsDesign 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 psiwith N mortar Fv = 23 psi
ev A
Vf =Per UBC Sec. 2107.3.12 shear stress is average shear stress,
allowable shear stress may be increased by 0.2 fmd where fmd is compressive stress due to dead load
Masonry Structures, slide 15
URM Shear WallsDesign Criteria
IbVQfv =Note: Per MSJC Sec. 2.2.5.1, shear stress is maximum stress,
(c) allowable shear stresses:
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 grout60 psi for running bond and solid grout
(d) 15 psi for masonry in other than running bond
Masonry Structures, slide 16
URM Shear WallsDesign Criteria(c) allowable shear stresses:
fvavg
for rectangular section
fvmax
netmaxv
maxvnet
avgv
AV
23f
f32
AVf
=
==
Masonry Structures, slide 17
URM Shear Walls
Possible shear cracking modes.
strong mortarweak units
through masonry units
Associated NCMA TEK Note#66A: Design for Shear Resistance of Concrete Masonry Walls (1982)
low vertical compressive stress
sliding along bed joints
weak mortarstrong units
stair step through bed and head joints
Masonry Structures, slide 18
Example: URM Shear WallsDetermine the maximum base shear per UBC and MSJC.
5000 lb. DL
H
H
9’-4
”9’
-4”
6’ - 8”
8” CMU’s with face shell beddingblock strength = 2800 psiType N Portland cement lime mortar special inspection provided during construction
2net in 200 (80) 2.5 A ==
32net in 2667 6 / (80) 2.5 S ==
4.39r'h 112" h’ 2.84" r ===
Net section with face shell bedding:80”
1.25
”
Masonry Structures, slide 19
Example
bbb
2deada
bbb
b
V0.0630 /2667V 168 M/S f weight)selfg (neglectin psi50 in 200 / lbs)(5000 2 f
V 168 12x ) 9.33x (1.5 V MH 2 V
===
==
==
=Forces and Stresses:
Maximum base shear capacity per UBC
govern) not (does lbs.7980 )in(200 1.33)x psi (33 V
2107.3.7)(Sec. psi 33 (50) 0.2 23 f 0.2 23 F2
max
mdv
==
=+=+=
shear stress
flexural tensile stress
(governs) lbs.1194 Vb = 25.3 0.0630V 50 - F f f-) 2107.3.5Sec. per(F psi 25.3 1.33x psi 19 F
btba
tt
=+=+
==
Masonry Structures, slide 20
flexural compressive stress
33.1Ff
Ff
b
b
a
a <+ 99 rh' for f´ 0.23 ])
140rh( -[1 0.25f´ F m
2ma ≤==
33.1616
V0630.042650 b =+ govern) not (does lbs. 11,857 Vb =
ExampleMaximum base shear capacity per UBC
Maximum base shear capacity per MSJCIn 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
govern) not (does lbs. 10,629
)in(200 1.33)x psi(60 32 A F
32 V
(governs) psi60 psi)(50 0.45 psi 37 F or psi65 (1861)1.5 )(f´1.5 F
2netvmax
v
1/21/2mv
=
==
=+=
===
psi 616 (1850) 0.33 Fpsi 426 Fmortar NType and psi 2800 f´ for D-Table 21 per psi1850 f´
ba
um===
==
Masonry Structures, slide 21
ExampleMaximum base shear capacity per MSJC
flexural tensile stress
0= V0.0630 +50 -0= f + f-
psi0 = F: 2.2.3.2Sec. MSJC per
b
ba
t
(governs) lbs. 794 Vb =
flexural compressive stress
govern) not (does lbs.11,934 = V 33.1620
V 0630.042650
33.1Ff
Ff
bb
b
b
a
a
=+
<+psi620 (1861) 0.33 F
39.4 h/r for psi 426 f´ 0.23 ](h/140r) -[10.25f´ F
b
m2
ma
==
====
shear stress tension compressionaxial and flexural stress
UBCMSJC
7890 11,8571194
Vb maxSummary:
10,629 11,934794
Masonry Structures, slide 22
URM Shear Walls
⎟⎠⎞
⎜⎝⎛
Le 2 -1 bL 3
4P= f m
Post-Cracked Behavior
h
av f=σ
toe
fm< Fa
3η
e
ηL/2
width = b
heel
H
P
PHhe;PeHh:mequilibriu
AP v
==
= σ
[1]bam F or Fstress edge ecompressivb
P2f ≤==η
[2]⎟⎠⎞
⎜⎝⎛ −=−= e
2L3 e
2L
3ηη
[3]22
LL1
e2Lb3
P2b
P2f
:equationsabove combining
m⎟⎠⎞
⎜⎝⎛ −
==η
Masonry Structures, slide 23
Note: shear strength should be checked considering effects of flexural cracking
URM Shear WallsPost-Cracked Behavior
Lat
eral
Loa
d, H
Lateral Deflection at Top of Wall
first flexural cracking
resultant load, P, shifts toward toetoe crushing
2 to
3 ti
mes
cr
acki
ng lo
ad
MSJC/UBC assumed behavior
Masonry Structures, slide 24
Perforated URM Shear Walls
Masonry Structures, slide 25
Lateral Stiffness of Shear WallsCantilevered shear wall
h G1
AH
I3EHh=
:shearflexure
vm
3+∆∆
H
L
h bL= A12bL= I
g
3
g)E(0.4 A
Hh 1.2 + I3E
Hh=
E0.4 = G A 5/6= A :sassumption common
mgm
3
mgv
∆
m3
m
3
bLEHh 3 +
bLEHh4 = ∆
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛ 3
Lh4
Lh
bEH=
2
m∆
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛ 3
Lh4
Lh
bE= H/= stiffness lateral= k2
m∆
Masonry Structures, slide 26
Lateral Stiffness of Shear WallsPier between openings
hG1
AH
I12EHh=
shearflexure
vm
3+∆
H
H
h
L
)EbL(0.4Hh 1.2 +
bLEHh=
m3
m
3∆
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛ 3
Lh
Lh
bEH=
2
m∆
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
===
32
Lh
Lh
bE/H stiffnesslateralk m∆
Masonry Structures, slide 27
Lateral Stiffness of Shear Walls
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.20 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
mbEk
Lh
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
=
3Lh4
Lh
1k2
cantilever
fixed pier
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
=
3Lh
Lh
1k2
Masonry Structures, slide 28
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
Masonry Structures, slide 29
Example: Lateral-Force DistributionDetermine the distribution of the lateral force, H, to walls A, B and C.
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
=
3Lh4
Lh
bEk2
mi
*based on cantilever action
type of masonry and wall thickness is the same for each wall
iii k/k*k h/LLwall ∑
A 10’ 1.50 0.0556 bEm 0.20
10’-0”
h=15’
A
H 18’-0”
B
B 18’ 0.83 0.2077 bEm 0.75
C
6’-0”
C 6’ 2.50 0.0143 bEm 0.05
Σki = 0.2776 bEm
Masonry Structures, slide 30
321 V V V H ++=equilibrium:
factorondistributik / kk / (k V
kkkk V
ii
iii
iiiiii
= ∑
)∑=
∑ / = = =
ΗΗ∆∆
shear force attracted to single pier:
Lateral-Force Distribution to PiersPerforated Shear Walls
pierfixedfixedfor
3Lh
Lh
Ebk2
i
i
i
i
miii −
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
overall story stiffness:
321i
i
kkkk stiffnessstory lateral Kwhere
kH
KHorK H
++ = ∑==
∑ = = = ∆∆
h3
L1
L2
H ∆
V1 V2
L3L2
V3
h 1 h 2
Masonry Structures, slide 31
A
H
56” a
40”11
2”
24”
64”
24”
7.63”
Va
Example: Lateral Force Distribution to PiersDetermine the distribution of story shear, H, to each pier.
Section A-A
Elevation
b
40” 32”
Vb
A
8”groutedconcrete block
c
Vc
Masonry Structures, slide 32
piers a and c
4322gross in84,273/127.63(40.0)11.2)305(203.81)366(11.2I =+−+−=
"2.11671
7501y ==610420x30540x7.631397 7.63/2x 366 x48 7.63
====
671 7501
40”
7.63”
Example: Lateral Force Distribution to Piers
y
48”
7.63
”
2webv in 254 )(40.0)(7.63" (5/6) A (5/6) A ===
E12.1
)254)(E4.0(64
)12E(84,273)64(
1
GAh
12EIh
1= k= k 3
v
3ca =+
=+
pier b
64”
7.63
”
E91.1 3]+ [1 1E 7.63
3Lh
Lh
bE= k2
mb ==
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
Ε164Ε121911121ki . = ). + . + .( = ∑
46.0 16.4/91.1 DF27.0 16.4/12.1 DF DF bca =====
Masonry Structures, slide 33
Perforated Shear WallsAxial Force due to Overturning
fmaxfai= ave. axial stress acrosspier “i”
c
y1 y2 y3
y1y2 y3
y
M
p1p2 p3
[1]forcegravity zero forp i 0 = ∑equilibrium of pier axial forces:
[5]iiaiii yAfyp M ∑=∑=
equilibrium of moments:
[6]c/yff imaxai =from similar triangles:
substituting in [5]:[7]
iii
max yAcyfM ∑ ⎟
⎠⎞
⎜⎝⎛=
[8]Ic
fM yAc
fM max2ii
max == ∑
[2]iaii Afp ∑= = ∑ 0
[3]axis neutral areas
pier of centroid thus,A i
=
≠∑ 0
[4]∑
∑=i
iiA
yAy
Masonry Structures, slide 34
Perforated Shear WallsAxial Force due to Overturning
[10]
solving for fmax:
IMc= f max
substituting in [6]:
IyM=
cy
IMc= f ii
ai ⎟⎠⎞
⎜⎝⎛ [11]
[12]IyMA= fA= p ii
iii
[13]
distribution factor for overturning moment
2ii
iii yA
yA M= p∑
Masonry Structures, slide 35
Perforated Shear WallsDesign Criteria for Piers between Openings
b2b F jkbd2M fncompressio <=
:forceaxialneglectingflexure: reinforced piers
flexure: unreinforced piers
1.0 Ff
Ffncompressio
bb
aa <+
P
P = Pdead + Plive + Plateral
V
V
h
M
P
M=Vih/2
tba < F f +f-tension
ss
s FjdA
M ftension <=
only) (UBC equation unity (a):forceaxialgconsiderin
diagramninteractiomoment -load (b)
controls) tension (only when bdF
PAs
s by modify (c)
Masonry Structures, slide 36
ve
v
vnet
v
FAV= f
FbI
VQ= f
<
<
UBC
MSJC
shear: unreinforced piers
shear: reinforced piers
vv FbjdV= f < UBC
vv FbdV= f < MSJC
Perforated Shear WallsDesign Criteria for Piers between Openings
P
V
V
h
M
P
D+L D+L Pmax for small lateral loadUBC MSJC Sec. 2.1.1 EffectLoading Combinations
M=Vih/2
0.75(D+L+W/E) D+L+W/E Pmax and Mmax for large lateral0.9D-0.75E 0.9D+E Pmin for smallest moment capacity D+W
Masonry Structures, slide 37
Example: Perforated Shear WallCheck stress per the UBC for the structure shown below. Design pier reinforcement if necessary.
Gravity LoadsLevel Dead Live3 50 kip 80 kip
Special inspection is providedf’m = 2500 psifully grouted but unreinforcedGrade 60 reinforcementType N mortar with Portland Cement
Earthquake Loads
14.9 kip
7.4 kip
10’-
0”10
’-0”
9’-8
”
14.9 kip14.9 kip
18’-8”
2 60 kip 80 kip1 60 kip 80 kip
total 170 kip 240 kip
Masonry Structures, slide 38
Example: Perforated Shear Wall18’-8”Pier Dimensions
9’-4
” 8” groutedconcrete block
3’-4
”
40” 32”
3’-4” 5’-4” 3’-4” 3’-4”3’-4”
2’-8
”4’
-0”
2’-8
”
7.63”
a b c
Masonry Structures, slide 39
Example: Perforated Shear Wall
12.82" 549 / /2)](40 7.63 2/ (7.63) [32 y 22 =+=
)E)(0.4 A (5/6
h I12E
h1k
mwebm
3a
+=
Stiffness of Pier “a”
7.63”
32”
7.63
”
40”y
a
423
23g
in470,76 12.82) 0 -7.63(40)(2 /12 7.63(40)
7.63/2) -2.827.63(32)(1 (32)/12(7.63) I
=++
+=
)E40)(0.4 7.63 (5/6
48 740,6712E
841k
mm
3a
××+
=
m
mm
a E 69.1
E472.0
E120.0
1k =+
=
Masonry Structures, slide 40
Example: Perforated Shear Wall
43g in 40,693 12 / (40) 7.63I ==
)E)(0.4 A (5/6
h I12E
h1k
mwebm
3b
+=
Stiffness of Pier “b”
7.63
”
40”
b
)E40)(0.4 7.63 (5/6
48 693,4012E
841k
mm
3b
××+
=
m
mm
b E 43.1
E472.0
E226.0
1k =+
=
Masonry Structures, slide 41
Example: Perforated Shear Wall
)E)(0.4 A (5/6
h I12E
h1k
mwebm
3c
+=
32”
7.63
”
40”
Stiffness of Pier “c”
4g in 470,76I =
(same as Pier a)
b 1.43 Em 0.409 15.2 c 0.38 Em 0.109 4.0
Σk = 3.50 Em 1.000 37.2 k
k 2.37k4.7k 9.14k 9.14Vbase =++= pier ki DF i V i
a 1.69 Em 0.483 18.0
Distribution of Story Shear to Piers
c
)E40)(0.4 7.63 (5/6
112 470,7612E
1121k
mm
3c
××+
=
m
mm
c E 38.0
E101.1
E526.1
1k =+
=
Masonry Structures, slide 42
7.63”
40.0
”
12.82”
Distribute Overturning Moments to PiersExample: Perforated Shear Wall
"6.1141403/807,160y ==
pier Ai yi Aiyi
a 549 12.8” 7038
a
"8.101y 1 =
y
124.0”
"38.9y 2 =
b
b 305 124.0” 37,820
"58.96y 3 =
c
211.2”
c 549 211.2” 115,949
∑ Ai =1403 ∑ Aiyi=160,807
Masonry Structures, slide 43
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
Distribute Overturning Moments to PiersExample: Perforated Shear Wall
11,029IyA 1403A 2iii =+=∑ ∑
b 305 -9.38 27 41 68 -2.9 -1.8
c 549 -96.58” 5120 76 5196 -53.0 -32.1
pier
(in2)
Ai y
(in)
2
ii yA(1000 in4) (1000 in4)
IyA 2ii +
(1000 in4)
1n
iii M
IyA = P
(kips)
iiyA(1000 in3)
I
a 549 101.8” 5689 76 5765 55.9 33.9
Masonry Structures, slide 44
*based on tributary wall length: pier a: (32” + 40” + 32”)/288 = 0.361(assuming that floor loads are pier b: (32” + 40” + 20”)/288 = 0.319applied uniformly to all walls) pier c: (20” + 40” + 32”)/288 = 0.319
Summary of Pier ForcesExample: Perforated Shear Wall
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
Masonry Structures, slide 45
Loading CombinationsExample: Perforated Shear Wall
*UBC 2107.1.7 for Seismic Zones 3 and 4
axial compressive force, P moment, M shear, V dcase 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
Masonry Structures, slide 46
*Fa = 0.25f’m[1-(h/140r)2] 112"4"9'h 2.2"7.630.2890.289tr =−==×==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 1 = D + LExample: Perforated Shear Wall
pier PD+L fa Fa* fa/Fa
(kips) (psi) (psi)
a
y
a 148.0 270 543 0.497 < 1.0 ok
b
b 130.8 430 543 0.792< 1.0 ok
y
c
c 130.8 239 543 0.440< 1.0 ok
Masonry Structures, slide 47
Example: Perforated Shear WallAxial and Flexural Stresses, Load Case 2: 0.75 (D + L + E)
* minimum Sg is taken to give maximum fb for either direction of building sway** Fb= 0.33f’m = 833 psi
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
y
a 136.4 249 543 0.459 327 2813* 116 0.139 0.598 < 1.0 ok
b
b 99.5 326 543 0.600 274 2035 135 0.162 0.762 < 1.0 ok
y
c
c 122.2 223 543 0.411 168 2813* 60 0.072 0.483 < 1.0 ok
Masonry Structures, slide 48
minimum axial compression: check tensile stress with Ft = 30 UBC Sec 2107.3.5
Example: Perforated Shear WallAxial and Flexural Stresses, Load Case 3: 0.9D - 0.75Peq
* minimum Sg is taken to give maximum fb for either direction of building sway**tensile stresses
pier (0.9PD-0.75PEQ) fa=P/A 0.75Meq Sg fb - fa+fb(kips) (psi) (kip-in) (in3) (psi) (psi)**
a
y
a 29.8 54 327 2813* 116 62 > 30 psi provide reinf.
b
b 47.4 155 274 2035 135 -20 < 30 psi ok
y
c
c 24.7 45 168 2813* 60 15 < 30 psi ok
Masonry Structures, slide 49
Pier Shear Stress, Load Case 4 : 0.75EExample: Perforated Shear Wall
* from Case 3 0.9Pd-0.75Peq** UBC 2107.3.7
pier V=0.75Veq x 1.5 fv = V/Aweb fao = P/A* Fv = 23 + 0.2fao**(kips) (psi) (psi) (psi)
a
y
a 20.3 67 54 34 < 67 provide shear reinf.
b
b 17.1 56 155 54 < 56 provide shear reinf.
y
c
c 4.5 15 45 32 > 15 ok
Masonry Structures, slide 50
Case Study: Large-Scale Test
Masonry Structures, slide 51
Georgia Tech Large-Scale Test
24’
photo from Roberto Leon
Masonry Structures, slide 52
Final Crack Pattern
Load Direction
slide from Roberto Leon
Masonry Structures, slide 53
Final Crack Pattern
Load Direction
slide from Roberto Leon
Masonry Structures, slide 54
Results- Global Behavior
Wall 1 Force-Displacement Response
-80
-60
-40
-20
0
20
40
60
80
-0.30 -0.20 -0.10 0.00 0.10 0.20 0.30
Roof Displacement (in)
Bas
e Sh
ear (
kip)
Masonry Structures, slide 55
Overturning Effect (Vertical Stress)
Base strains recorded during loading in the push and pull direction
slide from Roberto Leon
Masonry Structures, slide 56
USA CERL Shaking Table Tests
12’
photos from S. Sweeney
Masonry Structures, slide 57
Damage on North Wall
Permanent offsets of 0.25” – 0.35” due to rocking of pier.
Final Cracking Pattern
slide from S. Sweeney
Masonry Structures, slide 58
Peak Force vs. DeflectionNorth-South Uni-directional Motion
0
510
15
2025
30
0 0.025 0.05 0.075 0.1 0.125 0.15 0.175 Average First Floor Deflection (in)
Bas
e Sh
ear
(kip
)
PGA = 0.33 g
PGA = 0.75 gPGA = 0.98 gPGA = 1.08 g
East-West Uni-directional Motion
0
5
10
15
20
25
0 0.1 0.2 0.3 0.4 0.5 0.6
Average First Floor Deflection (in)
Base
She
ar (k
ip)
PGA = 0.30 g
PGA = 0.75 g
PGA = 1.09 gPGA = 1.40 g
slide from S. Sweeney
Masonry Structures, slide 59
End of Lessons 4 & 5
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