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Lesson9 2nd Part
1. Seismic design and assessment of
Seismic design and assessment of
Masonry Structures
Masonry Structures
Lesson 9, continued
October 2004
Masonry Structures, lesson 9 part 2 slide 1
Limitations of the storey mechanism approach
To perform a separate analysis for each storey, it is necessary to make assumptions on the
boundary conditions of the piers, i.e. on their rotational restraints: fixed-fixed, or fixed-
free, or other.
These assumptions are strongly affected by the strength and stiffness of the coupling
horizontal structural elements: plain unreinforced masonry spandrel beams, or r.c. slabs, or
r.c. ring beams, which may or may not crack or fail as horizontal loads increase.
The state of stress of these elements cannot be determined accurately on the basis of a
separate analysis for each storey, but only from a global analysis of the whole multi-storey
structure. In principle, only by knowing how much the coupling element are stressed can
the engineer judge if cracking or failure can be expected, and, as a consequence, what kind
of boundary conditions can be assumed for the piers.
A variation in the axial force of the piers may take place under the overturning effect of the
horizontal loads, affecting the flexural and shear strength of the individual piers. This effect
may not be of relevance in low-rise squat buildings, but it can be in a more general context.
Again, an evaluation of this effect can be made only very approximately with a separate
storey-by-storey analysis.
Masonry Structures, lesson 9 part 2 slide 2
2. Limitations of the storey mechanism approach
The storey-mechanism approach must therefore always be applied with a clear
understanding of its meaning and limitations, otherwise it can lead, in some cases, to
unrealistic and unconservative results.
The engineer can improve to some extent the results with a proper choice of boundary
conditions (end rotation) for the piers, but still some structural configurations of multi-
storey walls or buildings cannot be analysed properly with such method.
Masonry Structures, lesson 9 part 2 slide 3
URM MASONRY
SPANDREL
BEAMS UNDER at first cracking
SEISMIC
ACTION
Crack patterns from
an experimental
cyclic test on a full- at ultimate
scale masonry
building prototype
(University of Pavia,
1994)
Masonry Structures, lesson 9 part 2 slide 4
3. Strength of urm spandrel beams
Very little information is available on the behaviour of urm spandrel beams
subjected to cyclic shear. A proposal for strength evaluation which could be
suitable for applications is as follows.
Unreinforced masonry spandrels can be considered as structurally effective
only if they are regularly bonded to the adjoining walls and resting on a floor
tie beam or on an effective lintel.
The verification of unreinforced masonry coupling beams, in presence of a
known axial horizontal force, is carried out in analogy of the vertical walls.
If the axial load is not known from the model (for instance, when the analysis
is carried out with the hypothesis of in-plane infinitely rigid floors), but
horizontal elements with tensile strength (such as steel ties or r.c. ring beams)
are present in proximity of the masonry beam, the resisting values may be
assumed not greater than the following values associated to the shear and
flexural failure mechanisms.
Masonry Structures, lesson 9 part 2 slide 5
Strength of urm spandrel beams
The shear strength Vt of an unreinforced masonry coupling
beam, connected to a r.c. ring beam or a lintel and effectively
bonded at the ends, may be computed in a simplified way as
follows:
Vt = h t fv0
where: h is the section height of the masonry beam;
t is the width (thickness) of the beam
fv0 = is the shear strength in absence of compression.
Masonry Structures, lesson 9 part 2 slide 6
4. Strength of urm spandrel beams
The maximum resisting moment, associated to the flexural mechanism, always in
presence of horizontal elements resisting to tension actions in order to balance the
horizontal compression in masonry beams, may be evaluated as follows:
[
M u = H p h / 2 1 ā H p /(0.85 f hu ht ) ]
where: Hp is the minimum between the tension strength of the element in tension
placed horizontally and the value 0.4fhuht
fhu= is the compression strength of masonry in the horizontal direction (in the plane of
the wall).
The shear strength, associated to this mechanism, may be computed as:
V p = 2M u / l
where l is the clear span of the masonry beam.
The value of shear strength for the unreinforced masonry beam element shall be
assumed as the minimum between Vt and Vp.
Masonry Structures, lesson 9 part 2 slide 7
Non linear static modelling: beyond the storey mechanism approach
āStorey mechanismā Refined finite
element
Ok up to 2
(3?) storeys
Gambarotta & Lagomarsino, Papa
Macro-element modelling & Nappi., LourenƧo,ā¦
TomaževiÄ, Braga & Dolce
fascia
maschio
nodo
MAS3D (Braga, PEFV (DāAsdia & SAM (Magenes, Della TREMURI (Lagomarsino,
Liberatore, Spera) Viskovic) Fontana, Bolognini) Penna & Galasco)
Masonry Structures, lesson 9 part 2 slide 8
5. Requirements for non linear models
ā¢ Low or moderate computational burden to allow the modeling of whole
buildings:
ā¢ discretization of the structure with macro-elements: the elements have
dimensions comparable to the inter-storey height or with the size of openings
(doors, windows), to reduce the number of degrees of freedom of the model.
ā¢ Reliability of results:
ā¢ all the fundamental failure mechanisms should be accounted for with suitable
failure criteria;
ā¢ the model should give a good estimate of the overall deformational behaviour
under horizontal loads.
Masonry Structures, lesson 9 part 2 slide 9
Overview of some macroelement models for urm
EQUIVALENT TRUSS
APPROACH
(Pagano et al., 1984-1990)
Masonry Structures, lesson 9 part 2 slide 10
6. Overview of some macroelement models for urm
MULTI-FAN MODEL, MAS3D (Braga,
Liberatore, Spera, 1990-2000)
No-tension stress field simulated as a set of
āradialā stress fields for which an
analytical formulation in closed form exists.
Masonry Structures, lesson 9 part 2 slide 11
Overview of some macroelement models for urm
Pier or spandrel elem.
āJointā element
PEFV (DāAsdia & Viskovic 1990-today)
Linear elastic finite elements with variable
(adaptive) geometry. Masonry Structures, lesson 9 part 2 slide 12
7. Overview of some macroelement models for urm
TREMURI
(Lagomarsino, Penna,
Galasco 1997- today)
Beam-columns-type
elements with
internal degrees of
freedom and
coupling of
rotation/axial
displacement to
simulate rocking.
Allows dynamic
analysis also.
Masonry Structures, lesson 9 part 2 slide 13
Overview of some macroelement models for urm
SAM (Magenes, Della Fontana, Bolognini 1998- today)
Equivalent 3ād frame model
ā¢Simplified strength criteria for all elements, including r.c. ring beams, easily adaptable
to code-like formulations.
ā¢Simplified multi-linear constitutive rules are used (extension of concepts already
present in early storey-mechanism formulations)
ā¢Flexural (ārockingā) failure:a plastic hinge is introduced at the end of the effective
length where Mu is attained
ā¢Shear failure: plastic shear deformation Ī³ occurs when Vu is attained
ā¢Suitable for both urm and reinforced masonry.
ā¢Crude idealization but effective results especially for prediction of behaviour at
ultimate
Masonry Structures, lesson 9 part 2 slide 14
8. Nonlinear equivalent frame
rigid i
offset
H1 Īø = chord rotation
i'
Ļ = flexural deform.
effective
Ī³ = shear deformation
length Heff
j'
rigid H2
offset
j
V
V
Spandrel
Shear force- Pier element element
V
shear V u
u
deformation
behaviour in
the case of Ī±V
u
shear failure
mechanism Ī³
Ī³ Ī³ Ī³
1 2
Ī³ = Īøuā Ļ
Masonry Structures, lesson 9 part 2 slide 15
Nonlinear equivalent frame
80
70 F.E.M.
SAM (w. brittle spandrels)
Total base shear (kN)
60
50
40
30
20
10
0
0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040
Total displacement at 3rd floor (m)
URM wall with weak spandrels: Damage pattern predicted by
No storey mechanism refined nonlinear f.e.m. analysis
Masonry Structures, lesson 9 part 2 slide 16
10. 5-storey urm wall with r.c. ring beams: equivalent frame model
Masonry Structures, lesson 9 part 2 slide 19
5-storey urm wall: nonliner equivalent frame pushover analysis
Global angular deformation (%)
0.000 0.078 0.156 0.234 0.312 0.390 0.468
1400 0.42
Pushover analysis with
Analysis A
first-mode (linear) force 1200 0.36
distribution.
Base shear coefficient
Total base shear (kN)
1000 0.30
Analysis B
R.c. beams: elasto-plastic
800 0.24
beam elements (w. flexural Analysis C
hinging). 600 0.18
Analysis G
The analyses from A to G 400
No r.c. ring
0.12
show the effect of beams
200 0.06
decreasing strength and
stiffness of the r.c. beams 0 0.00
on the response of the wall. 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
Roof displacement (m)
Masonry Structures, lesson 9 part 2 slide 20
11. 5-storey urm wall: nonliner equivalent frame pushover analysis
20 5th FLOOR
Coupling elements (masonry soft storey
spandrels and r.c. beams) can
16 4th FLOOR
affect not only the strength,
global overturning
but also the overall deformed of cantilever walls
shape and collapse mechanism 12
3rd FLOOR
Height (m)
2nd FLOOR
8
1st FLOOR
4
Analysis A
Analysis C
soft storey Analysis G
0
0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08
Horizontal displacement (m)
Masonry Structures, lesson 9 part 2 slide 21
Nonlinear equivalent frame
Comparison of a 3-d storey ā 1000
Forza alla base-Spostamento
mechanism analysis and a 3-d 900
storey mechanism
nonlinear frame analysis: two POR
SAM
800
storey urm building with rigid
700
floor diaphragms and r.c. ring
600
beams. SAM
Forza [KN]
500
The flexural and shear strength 400
criteria of masonry walls are 300
kept the same for both methods 200
100
0
0 0.01 0.02 0.03
Spostame nto [m]
Masonry Structures, lesson 9 part 2 slide 22
12. Use of nonlinear static analysis in seismic design/assessment
The non linear static analysis is based on the application of gravity loads and of a
horizontal force system that, keeping constant the relative ratio between the acting
horizontal forces, is scaled in order to monotonically increase the horizontal displacement
of a control point on the structure (for example, the centre of the mass of the roof), up to
the achievement of the ultimate conditions.
A suitable distribution of lateral loads should be applied to the building. At least two
different distributions must be applied:
-a āmodalā pattern, based on lateral forces that are proportional to mass multiplied by the
displacement associated to the first mode shape
- a āuniformā pattern, based on lateral forces that are proportional to mass regardless of
elevation (uniform response acceleration).
Lateral loads shall be applied at the location of the masses in the model, taking into
account accidental eccentricity.
Masonry Structures, lesson 9 part 2 slide 23
Use of nonlinear static analysis in seismic design/assessment
The relation between base shear force and the control displacement (the ācapacity
curveā) should be determined by pushover analysis for values of the control
displacement ranging between zero and a sufficiently large value, which must exceed by
a suitable margin the displacement demand which will be estimated under the design
earthquake (target displacement) .
The target displacement is calculated as the seismic demand derived from the design
response spectrum by converting the capacity curve into an idealized force-displacement
curve of an equivalent single-degree-of-freedom system.
For the evaluation of the displacement demand of the equivalent s.d.o.f. system,
different procedures can be followed, depending on:
ā¢ how the seismic input is represented (acceleration spectra, displacement spectra,
composite A-D spectra);
ā¢ how the inelastic and hysteretic behaviour of the structure is accounted for (equivalent
viscous damping, ductility demand, energy dissipation demand).
Masonry Structures, lesson 9 part 2 slide 24
13. Use of nonlinear static analysis in seismic design/assessment
An example of procedure (e.g. as adopted by EC8 and Italian code):
Forza alla base-Spostamento
800
T ET T O
705
Step 1: carry out 700
the pushover 600 DLS
analysis with the 564 ULS
Base shear (kN)
chosen force 500
Forza [KN]
distribution. Plot 400
capacity curve
300
and determine the
performance 200
limit states of
100
interest
0
0 0.01 0.0146 0.02
Spostamento [m]
Roof displacement (m)
Masonry Structures, lesson 9 part 2 slide 25
Use of nonlinear static analysis in seismic design/assessment
Ī = ām iĪ¦ 2i
Ī¦vibration of the structure,mass displacementdirection, normalized
array that represents the
in the considered
in the first mode of
mĪ¦
to the unit value of the relative component of the control point.
ā i i
2000 Fb
Fb
1800
F* =
1600 Ī
Base shear [kN]
1400
1200 Step 2: determine an
1000 equivalent bilinear
800
600
s.d.o.f. system
400
200 dc
0
dc
0 5 10 15 20 25 30 d* =
Roof displacement [cm] Ī
N m*
m* = ā mi Ī¦ i T * = 2Ļ
i =1 k*
Masonry Structures, lesson 9 part 2 slide 26
14. Use of nonlinear static analysis in seismic design/assessment
Forza alla base-Spostamento
900
800
Capacity
F*max curve
700
F*y
600
0.8F*max
0.7F*max
500 Sistema equivalente SDOF
TETTO
Forza [KN]
Equivalent
Base shear (kN)
400 Bilineare
bilinear
SDOF
300
200
100
0
0 d*y 0.01 d*max 0.02
Displacement
Spostamento [m] (m)
Masonry Structures, lesson 9 part 2 slide 27
Use of nonlinear static analysis in seismic design/assessment
Elastic displacement spectrum
Step 3: using the elastic
response spectrum, calculate
ā
the displacement demand on if T*ā„TC d max = d e , max = S De ( T *)
the sdof system
if T*<TC
d e , max ā” TC ā¤
ā¢1 + (q * ā 1 ) T * ā„ ā„ d e , max
ā
d max =
q* ā£ ā¦
elastic acceleration
m* S e (T * ) spectrum
q = *
d* =
dc Fy*
ā
d max Ī
N m*
m* = ā mi Ī¦ i T * = 2Ļ
i =1
k*
Masonry Structures, lesson 9 part 2 slide 28
15. Use of nonlinear static analysis in seismic design/assessment
Step 4: convert the displacement demand on the
equivalent sdof into the control displacement
and find target point on capacity curve and Īd max = d c ,max
*
compare with displacement capacity.
2000
Stato Limite DS
1800
1600
Taglio alla base [kN]
1400
1200
1000
800
600
400
ā 200
d max d c , max
0
0 5 10 15 20 25 30
Spostamento copertura [cm]
Masonry Structures, lesson 9 part 2 slide 29
Use of nonlinear static analysis in seismic design/assessment
Available on ftp site:
Relevant chapters of new Italian seismic code
(English translation available! Thanks Paolo)
Relevant chapters of FEMA 356
Eurocode 8 (see Annex B)
Masonry Structures, lesson 9 part 2 slide 30
16. When and how to use storey-mechanism method
Eurocode 8: āFor low-rise masonry buildings, in which structural wall
behaviour is dominated by shear, each storey may be analyzed
independently. Such requirements are deemed to be satisfied if the number of
storey is 3 or less and if the average aspect ratio of structural walls is less
than 1.0.
ā¦.
New Italian seismic code: āFor buildings with number of storeys greater
than two, the structural model should take into account the effects due to the
variation of the vertical forces due to the seismic action and should guarantee
the local and global equilibrium. ā
Masonry Structures, lesson 9 part 2 slide 31
Earlier use of storey-mechanism method (TomaževiÄ)
du Ī¦u
Āµu = = Ultimate ductility
de Ī¦e
q2 +1 q behaviour factor
Āµu ā„
2 (force reduction factor),
specified by code (e.g.
1.5-2.0 for urm)
Ī¦ = d/h storey drift
a S ā Ī²0
H du , j ā„ Vdesign , j = Ļ j ā Wtot ā S d (T ; q ) = k j ā Wtot ā
g q
Masonry Structures, lesson 9 part 2 slide 32
17. Use of storey-mechanism method with present EC8 procedure
ā¢Evaluate elastic period of building T1 , e.g. using approximate formulae.
ā¢Estimate elastic base shear from elastic acceleration spectrum:
Fel,base = Se(T1) Wtot /g = Se(T1) Mtot
ā¢Evaluate ratio between interstorey shear Vj of the storey j being considered and the
total base shear: N
Ļ j = V j / Fbase Vj = āF
i= j
i
where Fi is the seismic force at the i-th floor.
ā¢The equivalent sdof is defined by putting F* = Vj and d*= interstorey displacement
ā¢Evaluate q* = Ļ j Fel,base /F*y
ā¢Calculate d*max= d*y [1+(q*-1)Tc/T1] (not greater than q d*y ) and check d*maxā¤ du
Masonry Structures, lesson 9 part 2 slide 33