Lesson10

Seismic design and assessment of
Seismic design and assessment of
Masonry Structures
Masonry Structures

Lesson 10
October 2004

Masonry Structures, lesson 10 slide 1

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).



Global and partial
overturning mechanisms of
façade walls


Out-of-plane collapse followed by global collapse

Tests performed
at Ismes,
Bergamo,
(Benedetti et al.
1996)



Seismic Load Path for Unreinforced Masonry Buildings
Floor diaphragm response Shaking
amplifies accelerations and motion
transmits load to out-of-
plane walls
Parapet wall

In-plane shear walls
response filters the
ground motion and
transmits to floor Earthquake Excitation at
diaphragms footings

Out-of-plane wall
(Doherty 2000,adapted after Priestley)



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


Out-of-plane strength of urm walls

Wall subjected to horizontal distributed loading (wind or inertia forces)

(A) (B)
Overburden
Force = O

Self weight
above crack =
W/2

∆c

One-way bending, One-way bending,
before cracking after cracking

Base reaction =
O+W


Out-of-plane strength of urm walls, low vertical stress

Relastic
Force Control/Dynamic Loading (Explosive)
Un-cracked Elastic Capacity

Displacement Control Loading
Applied lateral Force, w (kN/m)

Semi-rigid Non-linear F-∆
Res(1) Relationship

Un-cracked Linear Elastic
Behaviour

Mid-height Displacement (∆) ∆instability
(Doherty 2000)


Out-of-plane strength of urm walls, high vertical stress

Res(1) Semi-rigid nonlinear F-∆
Applied lateral Force, w (kN/m)

relationship
Relastic
Force/Dynamic
Control Loading
Displacement
Control Loading
Un-cracked Linear Elastic
Behaviour
Mid-height Displacement (∆) ∆instability

(Doherty 2000)


Out-of-plane flexural strength of nonloadbering walls

Two fundamental resistance mechanisms:

tensile strength of masonry to
resist one-way and two-way
arching action
bending


Flexural strength of nonloadbering walls

Vertical one-way flexure: already treated
Clearly, when zero or very low vertical compression is present, tensile
strength of bedjoints assume an importan role and should not be neglected.

Horizontal flexure:


Horizontal flexure

Empirical relations for brick
masonry (single-wythe):
⎛ σ ⎞
f tp ' = C f jt ⎜1 + n ⎟
⎜ f jt ⎟
⎝ ⎠

for toothed failure, where fjt = tensile
stregth normal to bedjoints, C=2 for stress in
MPa;

f tp ' = 0.45 f tb + 0.55 f jt

for failure in units, where ftb = tensile
strength of unit (can be assumed as 0.1 times
the compressive strength)


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


Arching action


Arching action

C = fc (1−γ )t

Resisting moment: M = C(γt − ∆0 )

8C
p= (γt − ∆0 )
h2
According to EC6 and BS: (1-γ)=0.1
2
⎛t ⎞
From which, neglecting deflection (ok for h/t < 25): p = 0.72fc ⎜ ⎟
⎝ h⎠

Arching action with gap

Arch action can be 4(γt)2
g≤
activated only if L

From geometry of
similar triangles:

g 2∆g
=
2γt L − g

g(L − g) gL
∆g = ≅
4γt 4γt

8C
This can be inserted in the equation of the previous slide: p= (γt − ∆g )
h2

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.


Post-cracking behaviour loadbearing walls in single bending

Priestley, 1985 From moment equilibrium about O:

From moment W ⋅∆ w ⋅ h2 W ∆ W ⋅ ∆ h w ⋅ h2
equilibrium about R: H = R⋅x = + ⋅ + P⋅∆ + ⋅ = + R⋅∆
2⋅h 8 2 2 2⋅h 2 8
8
P P w= ⋅ R ⋅ (x − ∆ )
h2
H=∆W/2h H=∆W/2h

∆/2
h/2

W/2
h/2

∆ w w = ma
h

W/2
Static instability
∆
W/2 occurs when ∆=x
o
x
H=∆W/2h
P+W R


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.

Mcrack=Rt/6
fcrack
fcrack=2R/t
R
φcrack=fcrack/Et
x
x=t/6
t

w ⋅ h2 8 ⋅ M crack 5 ⋅ wcrack ⋅ h 4
M crack = crack ⇒ wcrack = ∆ crack =
8 h2 384 ⋅ E ⋅ J



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:

M =Rx Conservatively, it can be
assumed that displacement si
f f =2R/(3(t/2-x)) proportional to curvature at
midheight :
R φ =f/(3E(t/2-x))
φ
x ∆= ⋅ ∆ crack
x φcrack
t

At ultimate, a rectangular stress block is assumed and the corresponding
curvature is evaluated, from which the displacement at midheight.



Having determined the displacement ∆ at midheight corresponding to a given
value of x, from the equation:
8
w= ⋅ R ⋅ (x − ∆ )
h2
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 along
the same curve.


Post-cracking behaviour in single bending: rigid block behaviour

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.:

F0 Bi-linear F- ∆
'Semi-rigid threshold Relationship
resistance' Real semi-rigid
Non-linear F- ∆
Relationship
Applied Lateral Force

K0

∆u=∆instability


Post-cracking behaviour in single bending: rigid block behaviour

∆

∆


Out-of-plane dynamic experimental behaviour

Wall specimen 5. ACCEL
(TA)
Timber wall
1. LVP
catch
(TWD)

7. ACCEL
(TWA)

2 LVP 8. ACCEL
(MWD) (MWA)

9. ACCEL
(BWA)
3. LVDT
Stationary
(BWD)
4. INSTRON 6. ACCEL reference frame:
(TTA) Rigid connection
to strong floor

Tests performed at University of Adelaide, Australia (Doherty et al. 2000)


Braced steel Cornice
frame support
(Representin
g URM
shear wall
action)
Wall
Specimen

Table 10mm stiff rubber spacer
bearing
support rails (after Doherty, 2000)


300% NAHANNI DISPLACEMENTS
15
DISPLACEMENT (mm)

10

5
PGD 14mm
0
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

-5

-10
TIME (secs)

300% NAHANNI VELOCITIES
0.25
0.2

300% PGV
VELOCITY (m/S)

0.15
0.1

207mm/sec
0.05

Nahanni 0
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

-0.05
-0.1
-0.15
TIME (secs)

300% NAHANNI ACCELERATIONS
0.8

0.6
ACCELERATION (g)

PGA
0.4

0.2

0.65g
0
0
0.56
1.11
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

-0.2

-0.4

-0.6
TIME (secs)


80% PACOIMA DAM DISPLACEMENTS
40

30

20
PGD
DISPLACEMENT (mm)
10

43.2mm
0

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
-10

-20

-30

-40

-50
TIME (s ecs )

80% PACOIMA DAM VELOCITIES
0.3

80% 0.25

PGV
0.2
VELOCITY (m/S)

0.15

0.1

Pacoima
245mm/sec
0.05

0

Dam
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
-0.05

-0.1

-0.15
TIME (s ecs )

80% PACOIMA DAM ACCELERATIONS
0.4

0.3

PGA
0.2
ACCELERATION (g)

0.1

0

0.35g
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
-0.1

-0.2

-0.3

-0.4
TIME (s ecs )


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


Design accelerations
to be compared with
“resistance”


A more rigorous evaluation of the flexibility of diaphragms would be made
by an adequate dynamic global model.

(from Tena-Colunga & Abrams)

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.


Force-based assessment with q-factor

a g S ⎡ 3(1 + Z/H) ⎤ a gS
Sa = ⎢ − 0.5⎥ ≥
g ⎣1 + (1 − Ta /T1 ) 2
⎦ g

agS is the design ground acceleration
Z is the height from the foundation of the centre of the mass of the element
H is the height of the structure
g is the gravity acceleration
Ta 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.


Recent trends: displacement based assessment

Griffith et al., 2000-today

F0
F0 Rigid body
Applied Lateral Force

Calculated curve

Fu FU

K1

K2 KS K0

∆2 ∆u ∆1 ∆2 ∆MAX ∆U

Mid-height Displacement

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).


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)

Ts2 ⎛ 3 (1 + Z H ) ⎞
Ts < 1.5T1 ∆ d (Ts ) = a gS ⎜ − 0.5 ⎟
4π ⎜ 1 + (1 − Ts T1 )
2 2
⎟
⎝ ⎠
1.5T1Ts ⎛ Z⎞
1.5T1 ≤ Ts < TD ∆ d (Ts ) =a g S ⎜1.9 + 2.4 ⎟
4π 2 ⎝ H⎠
1.5T1TD ⎛ Z⎞
TD ≤ Ts ∆ d (Ts ) =a gS ⎜ 1.9 + 2.4 ⎟
4π 2 ⎝ H⎠

agS is the design ground acceleration
Z is the height from the foundation of the centre of the mass of the element
H is the height of the structure
g is the gravity acceleration
Ta 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.


Lesson10

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