Masonry Structures, lesson 10 slide 1
Seismic design and assessment ofMasonry Structures
Seismic design and assessment ofMasonry Structures
Lesson 10October 2004
Masonry Structures, lesson 10 slide 2
Out-of-plane seismic response of urm walls
Parapet failure
(Newcastle Earthquake Study, The Institution of
Engineers, Australia, 1990) Out-of-plane damage(from the 1997 Umbria-Marche
earthquake, Italy, Blasi et al., 1999).
Masonry Structures, lesson 10 slide 3
Out-of-plane seismic response of urm walls
Global and partial overturning mechanisms of
façade walls
Masonry Structures, lesson 10 slide 4
Tests performed at Ismes, Bergamo, (Benedetti et al. 1996)
Out-of-plane collapse followed by global collapse
Masonry Structures, lesson 10 slide 5
Out-of-plane seismic response of urm walls
Earthquake Excitation atfootings
In-plane shear walls response filters theground motion andtransmits to floor diaphragms
Floor diaphragm response amplifies accelerations andtransmits load to out-of-plane walls
Out-of-plane wall
Shakingmotion
Parapet wall
Seismic Load Path for Unreinforced Masonry Buildings
(Doherty 2000,adapted after Priestley)
Masonry Structures, lesson 10 slide 6
Out-of-plane seismic response of urm walls
Important issues:-Strength of wall against out-of-plane forces and relevant mechanisms of resistance
-Out-of-plane displacement capacity of walls
-Evaluation of out-of-plane dynamic response
-Definition of seismic demand on walls considering filtering effect of building and diaphragms
Masonry Structures, lesson 10 slide 7
Out-of-plane strength of urm walls
∆c
Self weight above crack = W/2
Base reaction = O+W
(A) (B)OverburdenForce = O
One-way bending, before cracking
One-way bending, after cracking
Wall subjected to horizontal distributed loading (wind or inertia forces)
Masonry Structures, lesson 10 slide 8
Out-of-plane strength of urm walls, low vertical stress
Mid-height Displacement (∆)
Semi-rigid Non-linear F-∆Relationship
∆instability
Un-cracked Elastic Capacity
Displacement Control Loading
Force Control/Dynamic Loading (Explosive)
App
lied
late
ralF
orce
, w (k
N/m
)
Un-cracked Linear Elastic Behaviour
Res(1)
Relastic
(Doherty 2000)
Masonry Structures, lesson 10 slide 9
Out-of-plane strength of urm walls, high vertical stress
Mid-height Displacement (∆)
Semi-rigid nonlinear F-∆relationship
∆instability
Force/DynamicControl Loading
Un-cracked Linear Elastic Behaviour
Res(1)
Relastic
DisplacementControl Loading
App
lied
late
ralF
orce
, w (k
N/m
)
(Doherty 2000)
Masonry Structures, lesson 10 slide 10
Out-of-plane flexural strength of nonloadbering walls
Two fundamental resistance mechanisms:
tensile strength of masonry to resist one-way and two-way
bending arching action
Masonry Structures, lesson 10 slide 11
Flexural strength of nonloadbering walls
Vertical one-way flexure: already treatedClearly, when zero or very low vertical compression is present, tensile strength of bedjoints assume an importan role and should not be neglected.
Horizontal flexure:
Masonry Structures, lesson 10 slide 12
Horizontal flexure
Empirical relations for brick masonry (single-wythe):
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
jt
njttp f
fCf σ1'
for toothed failure, where fjt = tensile stregth normal to bedjoints, C=2 for stress in MPa;
jttbtp fff 55.045.0' +=
for failure in units, where ftb = tensile strength of unit (can be assumed as 0.1 times the compressive strength)
Masonry Structures, lesson 10 slide 13
Two-way flexure (single-wythe)
Most walls are supported on three or four sides. Evaluation is difficult because the walls are statically indeterminate and the material is anisotropic.Wall tend to show fracture lines whose pattern depends also on geometry of wall, degree of rotational restraint, presence of vertical compression, besides properties of materials.
Common design/assessment methods make use of •simplistic conservative approaches such as the crossed strip method or•yield line and fracture line approaches
Masonry Structures, lesson 10 slide 14
Arching action
Masonry Structures, lesson 10 slide 15
Arching action
tfC c )1( γ−=
)( 0∆−= tCM γResisting moment:
)(802 ∆−= t
hCp γ
According to EC6 and BS: (1-γ)=0.1
From which, neglecting deflection (ok for h/t < 25):2
72.0 ⎟⎠⎞
⎜⎝⎛=htfp c
Masonry Structures, lesson 10 slide 16
Arching action with gap
Ltg
2)(4 γ≤
gLtg g
−
∆=
22γ
From geometry of similar triangles:
This can be inserted in the equation of the previous slide:
tgL
tgLg
g γγ 44)(≅
−=∆
Arch action can be activated only if
)(82 gt
hCp ∆−= γ
Masonry Structures, lesson 10 slide 17
Influence of movement of supports on arching action
The thrust force C can produce some displacement of the supports.
The displacement can be treated as an equivalent gap. If the movement is a function of the thrust force, a number of trials or iterations may be required to determine the optimum stress level.
Axial shortening due to compression of the wall and corresponding deflection can be treated similarly as an equivalent gap.
Masonry Structures, lesson 10 slide 18
Post-cracking behaviour loadbearing walls in single bending
h
h/2
∆
∆/2
P
W/2
W/2
P+W
H=∆W/2h
H=∆W/2h
w h/2
P
H=∆W/2h
W/2
∆o
R
x
w = ma
Priestley, 1985
hWH⋅∆⋅
=2
From moment equilibrium about R: ∆⋅+
⋅=⋅
⋅∆⋅
+∆⋅+∆⋅+
⋅=⋅ Rhwh
hWPWhwxR
822228
22
From moment equilibrium about O:
( )∆−⋅⋅= xRh
w 2
8
Static instability occurs when ∆=x
Masonry Structures, lesson 10 slide 19
Post-cracking behaviour in single bending: statics
Calculation of force-deflection curve: first, calculate curvature at midheight section and the associated moment, then evaluate displacement at midheight assuming a given curvature distribution.
R
fcrack
x
t
Mcrack=Rt/6
fcrack=2R/t
φcrack=fcrack/Et
x=t/6
2
2 88 h
MwhwM crackcrack
crackcrack
⋅=⇒
⋅= JE
hwcrackcrack ⋅⋅
⋅⋅=∆
3845 4
Masonry Structures, lesson 10 slide 20
Post-cracking behaviour in single bending: statics
After cracking, calculation behaviour is non linear and the following relationship are used for a given value of deformation/stress at the right edge of the section, from which the corresponding value of x is found:
f
xx
t
R
M =Rx
f =2R/(3(t/2-x))
φ =f/(3E(t/2-x))
crackcrack
∆⋅=∆φφ
Conservatively, it can beassumed that displacement si proportional to curvature at midheight :
At ultimate, a rectangular stress block is assumed and the corresponding curvature is evaluated, from which the displacement at midheight.
Masonry Structures, lesson 10 slide 21
Post-cracking behaviour in single bending: statics
Having determined the displacement ∆ at midheight corresponding to a givenvalue of x, from the equation:
( )∆−⋅⋅= xRh
w 2
8
the corresponding value of distributed horizontal force is determined, and the nonlinear force-displacement curve is determined point by point.
Note: static instability may occur before the attainment of the ultimate stress. The behaviour is elastic nonlinear, i.e. the wall will load and unload alongthe same curve.
Masonry Structures, lesson 10 slide 22
Note: if the two halves of the cracked wall were considered as rigid blocks, the force-displacement curve would be given by the blue line, which can be obtained by simple equilibrium of rigid blocks.:
Post-cracking behaviour in single bending: rigid block behaviour
App
lied
Late
ral F
orce
F0 Bi-linear F- ∆Relationship
Real semi-rigid Non-linear F- ∆Relationship
∆u=∆instability
K0
'Semi-rigid thresholdresistance'
Masonry Structures, lesson 10 slide 23
∆
∆
Post-cracking behaviour in single bending: rigid block behaviour
Masonry Structures, lesson 10 slide 24
1. LVP(TWD)
2 LVP (MWD)
4. INSTRON
3. LVDT (BWD)
5. ACCEL(TA)
6. ACCEL(TTA)
7. ACCEL(TWA)
8. ACCEL(MWA)
9. ACCEL(BWA)
Stationary reference frame: Rigid connection to strong floor
Wall specimen
Timber wall catch
Tests performed at University of Adelaide, Australia (Doherty et al. 2000)
Out-of-plane dynamic experimental behaviour
Masonry Structures, lesson 10 slide 25
Braced steel frame (Representing URM shear wall action)
Cornice support
Table bearing support rails
10mm stiff rubber spacer
Wall Specimen
(after Doherty, 2000)
Masonry Structures, lesson 10 slide 26
300% NAHANNI DISPLACEMENTS
-10
-5
0
5
10
15
0.01
0.62
1.23
1.84
2.45
3.06
3.67
4.28
4.89 5.5 6.11
6.72
7.33
7.94
8.55
9.16
9.77
10.4
TIME (secs)
DISP
LACE
MENT
(mm)
300% NAHANNI ACCELERATIONS
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
00.5
61.1
1
1.67
2.22
2.78
3.33
3.89
4.44 5
5.55
6.11
6.66
7.22
7.77
8.33
8.88
9.44
9.99
10.5
TIME (secs)
ACCE
LERA
TION (
g)
300% NAHANNI VELOCITIES
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
0.01
0.57
1.13
1.69
2.25
2.81
3.37
3.93
4.49
5.05
5.61
6.17
6.73
7.29
7.85
8.41
8.97
9.53
10.1
10.6
TIME (secs)
VELO
CITY (
m/S)
PGD 14mm
PGV 207mm/sec
PGA 0.65g
300%
Nahanni
Out-of-plane dynamic experimental behaviour
Masonry Structures, lesson 10 slide 27
80% PACOIMA DAM DISPLACEMENTS
-50
-40
-30
-20
-10
0
10
20
30
40
0.02 0.6 1.18
1.76
2.34
2.92 3.5 4.08
4.66
5.24
5.82 6.4 6.98
7.56
8.14
8.72 9.3 9.88
10.5 11 11.6
12.2
12.8
13.4
TIME (secs)
DISPL
ACEM
ENT (
mm)
80% PACOIMA DAM ACCELERATIONS
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0
0.58
1.16
1.74
2.32 2.9 3.48
4.06
4.64
5.22 5.8 6.38
6.96
7.54
8.12 8.7 9.28
9.86
10.4 11 11.6
12.2
12.8
13.3
TIME (secs)
ACCE
LERA
TION (
g)
80% PACOIMA DAM VELOCITIES
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
0.02
0.62
1.22
1.82
2.42
3.02
3.62
4.22
4.82
5.42
6.02
6.62
7.22
7.82
8.42
9.02
9.62
10.2
10.8
11.4 12 12.6
13.2
TIME (secs)
VELO
CITY (
m/S)
PGD 43.2mm
PGV 245mm/sec
PGA 0.35g
80%
Pacoima Dam
Out-of-plane dynamic experimental behaviour
Masonry Structures, lesson 10 slide 28
Safety wih respect to collapse: force or displacement?
Clearly, any methodology should take into consideration the nonlinear nature of the response.
Earlier proposal by Priestley (1985): force based assessment based on equal energy equivalence;
Equivalent “resistance” of a linear response
Masonry Structures, lesson 10 slide 29
Design accelerations to be compared with
“resistance”
Masonry Structures, lesson 10 slide 30
(from Tena-Colunga & Abrams)
A more rigorous evaluation of the flexibility of diaphragms would be made by an adequate dynamic global model.
Masonry Structures, lesson 10 slide 31
Force-based assessment with q-factor
Eurocode 8 & Italian code approach
The effect of the seismic action may be evaluated considering a force system proportional to the masses (concentrated or distributed) of the element; the resultant force (Fa) on the element, is computed as:
Fa = Wa Sa / qa
Wa is the weight of the element.qa is the behaviour factor, to be considered equal to 1 for cantilever elements (for example chimneys and parapets fixed only at the base) equal to 2 for non-structural walls, equal to 3 (Italian code) for structural walls which do not exceed specified slenderness limits.
Sa is the seismic coefficient to apply at the structural or non-structural wall, that considers dynamic interaction with the building.
Masonry Structures, lesson 10 slide 32
Force-based assessment with q-factor
gSa
0.5)/TT(11
Z/H)3(1gSa
S g2
1a
ga ≥⎥
⎦
⎤⎢⎣
⎡−
−++
=
agS is the design ground accelerationZ is the height from the foundation of the centre of the mass of the elementH is the height of the structureg is the gravity accelerationTa is the first period of vibration of the wall element in the considered direction (out-of-plane), estimated also in an approximate way.T1 is the first period of vibration of the structure in the considered direction.
Masonry Structures, lesson 10 slide 33
Recent trends: displacement based assessment
Mid-height Displacement
App
lied
Lat
eral
For
ce
Rigid bodyCalculated curve
Fu
∆u∆2
F0
Griffith et al., 2000-today
F0
∆1
K0
FU
∆2 ∆U
K1
∆MAX
K2 KS
Dynamic response is estimated via an equivalent secant stiffness (K2) and an equivalent s.d.o.f. system with suitable effective damping (5% suitable for one-way vertical bending).
Masonry Structures, lesson 10 slide 34
Recent trends: displacement based assessment
Displacement demand must be evaluated using an elastic spectrum which takes into account the filtering effect of the structure, e.g.(Italian code)
( )( )
2s
s 1 d s g 2 2s 1
1 s1 s D d s g 2
1 DD s d s g 2
3 1 Z HTT < 1.5T ∆ (T ) = a S 0 54 1 1 T T
1 5T T Z1.5T T < T ∆ (T ) =a S 1 9 2 4H4
1 5T T ZT T ∆ (T ) =a S 1 9 2 4H4
.
. . .
. . .
⎛ ⎞+⎜ ⎟−⎜ ⎟π + −⎝ ⎠
⎛ ⎞≤ +⎜ ⎟π ⎝ ⎠
⎛ ⎞≤ +⎜ ⎟π ⎝ ⎠
agS is the design ground accelerationZ is the height from the foundation of the centre of the mass of the elementH is the height of the structureg is the gravity accelerationTa is the first period of vibration of the element in the considered direction, estimated also in an approximate way.T1 is the first period of vibration of the structure in the considered direction.