Transcript of "Module 7 (RESPONSE OF INELASTIC S.D.O.F SYSTEMS TO EARTHQUAKE LOADING)"
1.
University of Engineering & Technology,
Peshawar, Pakistan
CE-409: Introduction to Structural Dynamics and
Earthquake Engineering
MODULE 7
RESPONSE OF INELASTIC SDOF SYSTEMS TO
EARTHQUAKE LOADING
Prof. Dr. Akhtar Naeem Khan &
drakhtarnaeem@nwfpuet.edu.pk
Prof. Dr. Mohammad Javed
mjaved@nwfpuet.edu.pk
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2.
BASIC ASPECTS OF SEISMIC DESIGN
Designing buildings to behave elastically during earthquakes without
damage may render the project economically unviable.
As a consequence, The design philosophy for earthquake resistant
design of structure is to allow damage and thereby dissipate the energy
input to it during the earthquake.
Therefore, the traditional earthquake-resistant design philosophy
requires that normal buildings should be able to resist:
(a)Minor and frequent shaking with no/un-notable damage to structural
and non-structural elements;
(b) Moderate shaking with minor to moderate damage (repairable) to
structural and non-structural elements; and
(c) Severe and infrequent shaking with damage to structural elements, but
with NO collapse (to save life and property inside/adjoining the building).
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3.
BASIC ASPECTS OF SEISMIC DESIGN
Earthquake-Resistant Design Philosophy for buildings:
(a)Minor (Frequent) Shaking – No/Hardly any damage,
(b) Moderate Shaking – Minor to moderate structural damage, and
(c) Severe (Infrequent) Shaking – Structural damage, but NO collapse
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4.
BASIC ASPECTS OF SEISMIC DESIGN
Buildings are designed only for a fraction of the force that they
would experience, if they were designed to remain elastic during
the expected strong ground shaking (see given below figure) , and
thereby permitting damage (inelastic range) see figure on next
slide.
Basic strategy of earthquake design: Calculate maximum elastic forces and
CE-409: MODULE 7 (Fall-2013)
reduce by a factor to obtain design forces.
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5.
BASIC ASPECTS OF SEISMIC DESIGN
Structures must have sufficient initial stiffness to ensure the nonoccurrence of structural damage under minor shaking. Thus, seismic
design balances reduced cost and acceptable damage, to make the
project viable.
For this reason, design against earthquake effects is called as
earthquake-resistant design and not earthquake-proof design.
Earthquake-Resistant and NOT Earthquake-Proof: Damage is expected
during an earthquake in normal constructions (a) undamaged building,
and (b) damagedCE-409: MODULE 7 (Fall-2013)
building.
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6.
BASIC ASPECTS OF SEISMIC DESIGN
Two aspects are worth consideration:
1.Force carrying ability under seismic demand
2. Ability to absorb energy under seismic demand
Note:
Compromise can be made on force carrying ability
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7.
BASIC ASPECTS OF SEISMIC DESIGN
The design for only a fraction of the elastic level of seismic
forces is possible, only if the building can stably withstand large
displacement demand through structural damage without collapse
and undue loss of strength. This property is called ductility (see
Figure on next slide).
It is relatively simple to design structures to possess certain
lateral strength and initial stiffness by
appropriately
proportioning the size and material of the members. But,
achieving sufficient ductility is more involved and requires
extensive laboratory tests on full-scale specimen to identify
preferable methods of detailing.
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8.
BASIC ASPECTS OF SEISMIC DESIGN
Sharp reduction in strength w/o
significant displacements after
peak strength
Ductility: Buildings are designed and detailed to develop favorable
failure mechanisms that possess:
1.specified lateral strength,
2.reasonable stiffness and, above all,
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3. good post-yield MODULE 7 (Fall-2013)
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9.
BASIC ASPECTS OF SEISMIC DESIGN
Peak base shear induced in a linearly elastic system by ground
motion is Vb = (A/g)w. where w is the weight of the system and A is
the pseudo acceleration corresponding to the natural vibration period
and damping of the system.
Most buildings (as already discussed) are designed, however, for
base shear smaller than the elastic base shear ,Vb = (A/g)w
This becomes clear from figure on next slide, where the base shear
coefficient A/g from the design spectrum of Fig. 6.9.5(chopra’s book),
scaled by 0.4 to correspond to peak ground acceleration of 0.4g, is
compared with the base shear coefficient specified in the 2000
International Building Code
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10.
Construction of Design Spectrum (firm soil)
Acceleration sensitive region
Velocity sensitive region
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Displacement
sensitive region
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11.
Design Spectrum for various values of ζ
2.71
Figure: Pseudo- acceleration design spectrum (84.1 th percentile) drawn on linear
scale for ground motions with go = 1g , u go = 48 in/sec, and u go = 36 in. ;
u
ζ = 1,2,5,10 and CE-409: MODULE 7 (Fall-2013)
20 %.
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11
12.
BASIC ASPECTS OF SEISMIC DESIGN
A/g= 0.4* (2.71g) =1.09
R, is detailed in the slides to
follow. This
is a factor
which primarily depend upon
material , structural system
and detailing and is used to
work out the design base
shear. In the literature, it is
generally referred to as
force reduction factor for a
structural system
R=1.5
R=8
Comparison of base shear coefficients
from elastic design spectrum and
International Building Code 2000.
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13.
Response of Elastoplastic SDOF system to
Earthquake loading
In this lecture, we will study the earthquake response of
elastic-perfectly plastic ( referred as elastoplastic systems) SDOF
systems to earthquake motions. Note that elastoplastic system is an
idealized response of a non-linear system
u
fs
Lateral force, fS
Lateral displacement, u
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14.
Elastoplastic idealization of a non-linear system
fy=
Lateral force at which
yielding
start in idealized
elastoplastic
system.
Also
known as yield strength
uy= Yield displacement
in
elastoplastic system. It is also
called yield deformation
um = Maximum displacement in
idealized elastoplastic system
Note That initial stiffness, k, of
both the systems must be same
k
Force-deformation curve : actual and
elastoplastic idealization based on equal
energy principle
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14
15.
Elastic system corresponding to a given elastoplastic
system
u
fs
Elastoplastic system
and
corresponding elastic system has
the same stiffness. Similarly
both systems have same mass
and damping. Consequently,
natural vibration period, Tn, of
elastoplastic
system
and
corresponding elastic system is
the same as long as u ≤ uy.
k, m and ζ are same
for the two systems
um = peak deformations in elastoplastic system; and, uo = peak deformation
in the corresponding linear elastic system when both are subjected to same
ground motion.
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16.
Normalized yield strength of an elastoplastic
system, f y
The normalized yield strength f y of
an elastoplastic system is defined as:
f
f =
f
y
y
o
Where fo and uo are the peak values of force and deformation,
respectively, in the linear elastic system corresponding to
elastoplastic system under the same ground motion.
For brevity the notation fo is used instead of fso (elastic resisting
force) as followed in previous lectures
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17.
Normalized yield strength of an elastoplastic
system, f y
fo can be interpreted as the strength required for the structure to
remain within its linear elastic limit during the ground motion.
If the normalized yield strength, f y = f y /f o of a system is less
than 1.0, the system will deform beyond its linearly elastic limit.
e.g., f y = 0.75 implies that the yield strength of the elastoplastic
system is 0.75 times the strength required for the system to remain
elastic during the ground motion.
f y = f y /f o = 0.75
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18.
Normalized yield strength of an elastoplastic
system, f y
f y ku y u y
f y can also be expressed as: f y =
=
=
f o kuo uo
Please note again that uy is the displacement at which yielding
start in the elastoplastic system .
Whereas, uo is the peak displacement in the corresponding
elastic system .
This uo must not be confused with the maximum displacement
in the elastoplastic system, um
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19.
Yield strength reduction factor, Ry
fy can also be related to fo through a yield strength reduction factor,
Ry as:
f o uo
Ry =
=
f y uy
In other words
f
1
1
R = =
=
f f
f
f
o
y
y
y
y
o
Ry is greater than 1 for a system that deforms into inelastic range.
Ry=2 implies that the yield strength of the elastoplastic system is
the strengthfrequired for the system to remain elastic divided by 2.
i.e .,
fy =
o
2
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20.
Displacement ductility factor
Ductility factor,μ, of an elastoplastic system is defined as the ratio of
peak (or absolute maximum) deformation to the yield deformation.
um
µ=
uy
fs
fy
uy
u
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um
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21.
Relation b/w μ and f y
fy
uy
fy
1
um um
fy = =
⇒
=
⇒
=
. fy
fo u o
uo u y
uo uy
um
µ
⇒
= µ. f y =
uo
Ry
This relationship couples the peak displacements of elastoplastic
(um) and corresponding elastic (uo) system
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22.
Elastoplastic system under cyclic loading
-fs
+fs
b
c
h
a
g
f
d
e
Elastoplastic force-deformation relation
a-b-c =+ve loading, c-d = unloading, d-e-f = -ve loading , f-g= unloading, g-h= +ve
loading
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23.
Elastoplastic system under cyclic loading
a
+fs
c
b
b
d
c
c
h
a
g
f
d
Loading
f
d-e-f
c-d
a-b-c
e
e
d
f g
- fs
Unloading
+fs
Unloading
Reloading
f-g
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g h
Reloading
g-h
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24.
Equation of motion for elastoplastic system
The EOM for an elastic SDOF system subjected to ground
motion is:
mu + cu + f s = − mu g (t)
Force fs corresponding to deformation u, in case of inelastic system,
is not single valued and depends upon the history of deformations
and on whether the deformation is increasing (positive velocity) or
decreasing (negative velocity) see Figure on slide 22. Thus the
resisting force fs in case of inelastic system can be expressed as:
f s = f s (u,u)
⇒ mu + cu + f s (u,u) = −mu g (t)
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25.
Equation of motion for elastoplastic system
c
1
⇒u + u + f s (u,u) = −u g (t)
m
m
Substituting
f s (u,u) ~
c = 2ζmωn and
= f s (u,u)
fy
2ζmωn
1 ~
u+
u + f s (u,u)f y = −u g (t)
m
m
since f y = k .u y = (ωn 2 m)u y
~
⇒ u + 2ζωn u + ωn u y f s (u,u) = −u g (t)
2
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26.
Minimum strength required for a
system to remain linear elastic
Consider an elastic SDOF system with
weight w, Tn=0.5 sec, and ζ=0. The
u
Tn=0.5 sec,
ζ=0
deformation response history of the system
subjected to El Centro ground motion is
ug , g
shown in the below given figure.
u go = 0.319g
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27.
Minimum strength required for a system to remain
linear elastic
Time variation of fs/w (i.e ratio of elastic resisting force to the weight
of system) for the system on previous slide is shown. For an undamped
f s /w = ku/w = mA / w = A /( w / m) = − t /g
u
system,
It can be seen that fo/w=1.37 or fo=1.37w . i.e., The minimum
strength required for the structure (Tn=5% and ζ=0) to remain
elastic (when subjected to 1940 El-centro earthquake) is 1.37w
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28.
Effect of yielding on deformation response history of
elastoplastic system
Now consider an elastoplastic SDOF system with same properties
(as given on previous slide i.e.,w, Tn=0.5 sec, and ζ=0) and with a
normalized yield strength of
f = 0.125 or f /f = 0.125
y
y
o
⇒ f = 0.125 f = 0.125 *1.37w = 0.171w
y
o
The deformation response history of the system (developed using
EOM for elastoplastic systems given at the end of slide 25)
subjected to El- Centro ground motion is shown on the next slide
for first 10 sec. The peak displacement also occur in first 10
seconds
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29.
Effect of yielding on deformation response history of
elastoplastic system
10 sec
Response of elastoplastic system with Tn = 0.5 sec, ζ = 0, and f y = 0.125 to ElCE-409: MODULE 7 (Fall-2013)
Centro ground motion: (a) deformation; (b) resisting force and acceleration; (c) time 29
30.
Effect of f y on deformation response history of
elastoplastic system
Now we examine how the response of elastoplastic system is affected
by its yield strength. Consider four SDOF systems all with identical
properties in their linear elastic range (i.e Tn=0.5 sec and ζ=5%) but with
different normalized yield strengths of f y = 1.0, 0.5, 0.25 and 0.125
To keep the discussion simple at this stage, it is assumed that the
elastoplastic systems considered in discussion can indefinitely yield in
f y range.
plastic = 1.0
implies a
linearly elastic system
f y = f y /f o
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31.
Effect of f y on deformation response history of
elastoplastic system
f = 1.0
u, in
y
f y = fo
h
h
a
a
g
f
c
b
c
b
g
d
e
f
d
e
um=maximum displacement in elastoplastic system subjected to El-centro 1940
ground motion, up = permanent/residual displacement in the elastoplastic system
(at the end of El-centro 1940 ground motion). up=0 in case of elastic system
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32.
Effect of f y on deformation response history of
elastoplastic system
u, in
f y = 0.5
fo
f y = 0 .5 f o
c
b
h
a
g
f
d
e
Typical +ve loading-unloading and
-ve loading-unloading for a single cycle
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33.
Effect of f y on deformation response history of
elastoplastic system
f y = 0.25
fo
f y = 0.25 f o
c
b
h
Typical +ve loading-unloading and
-ve loading-unloading for a single cycle
a
g
f
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d
e
33
34.
Effect of f y on deformation response history of
elastoplastic system
f y = 0.125
fo
Typical +ve loading-unloading and
-ve loading-unloading for a single cycle
f y = 0.125 f o
c
b
h
a
g
d
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f
e
34
35.
Effect of f y on ductility demand, μ, and residual
deformation, up , of elastoplastic system
um − in.
um 1
µ=
.
uo f y
1.00
2.25
1.00
0
0.50
1.62
1.44
0.17
0.25
1.75
3.11
1.1
0.125
2.07
7.36
1.13
fy
up-in
Tn=0.5 sec, ζ=5% and peak value of disp.
in elastic SDOF system=uo=2.25"
up
El-centro 1940
up is the residual displacement in the elastoplastic system at the end of
ground motion. up=0 in case of elastic system
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36.
Ductility demand, μD
The values of μ as calculated on previous slide are known as
ductility demand.
Thus the ductility demand imposed by El Centro ground motion
on inelastic systems having f y = 0.5,0.25,0.125 are 1.44, 3.11 and
7.36 respectively.
Ductility demand represents a requirement on the design in the
sense that the ductility capacity, previously defined as
displacement ductility factor (i.e., the ability to deform beyond the
elastic limit) should exceed the ductility demand.
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37.
Ductility demand, μD
The ductility demand for the system with
, T n=0.5 sec
and ζ =5% was found to be 3.11 when subjected to El-Centro ground
motion.
A system with above mentioned properties and having ductility
capacity greater than 3.11 will survive collapse when subjected to El
Centro 1940 ground motion. However, another system with same
properties but having a ductility capacity of 3 will collapse when
subjected to El-Centro 1940ground motion
It may be noted that μ is used for displacement ductility factor (i.e
ductility capacity) as well as ductility demand in the text book being
followed. However, we will follow μD for ductility demand and μC for
ductility capacity. Note, here μ will be taken as ductility factor
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38.
Effect of Tn on ductility demand, μD
μ = 8.0 = 1/f y = R y
μ = 4.0 = 1/f y = R y
μ = 2.0 = 1/f y = R y
μ = 1.0 = 1/f y = R y
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39.
Effect of Tn on ductility demand, μ
Following observations can be made from the figure given on
previous slide.
For systems with Tn in displacement sensitive region
(long structures) the ductility demand is independent of T n and
approximately equal to R (i.e. 1 / f y )
y
For systems with Tn in velocity sensitive region (intermediate to
long structures) the ductility demand may be larger or smaller than
fy
Ry; and the influence of , although small, is not negligible.
For systems with Tn in acceleration sensitive region
(short structures) the ductility demand much be much larger than
Ry, specially in case of very short structures
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40.
Construction of constant ductility response spectrum
uy corresponding to various values of f are determined by
y
u
f =
or u = u .f
u
the equation:
y
y
y
o
y
o
Constant ductility response spectrum for Dy is drawn using
= uy.
Dy
Design spectrums for Vy (Pseudo-velocity response spectrum)
and Ay (Pseudo-acceleration response spectrum) can be
constructed using the relations:
2
2π
2π
V = D .ω = D & A = D .ω = D
T
T
2
y
y
n
y
y
n
CE-409: MODULE 7 (Fall-2013)
y
n
y
n
40
41.
Relations b/w and yield strength ,fy , and base
shear coefficient for elasto plastic system, Ay/g
f = k.u = ( mω ) u = m( ω u ) = mA
2
y
⇒
2
n
y
n
y
y
y
A
w
f = .A =
.w
g
g
y
y
y
f
A
⇒ =
w
g
y
y
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42.
Inelastic pseudo-acceleration response spectrum for
constant ductility factors
Effect of Tn on fy/w (i.e. yielding
base shear coefficient is insignificant
when Tn≥ 1.5 sec
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44.
Inelastic Pseudo-velocity design spectrum
The very first step in the construction of inelastic design spectrum for
constant ductility is to develop the elastic design spectrum using
procedure explained in previous lecture
Once the elastic design spectrum is developed, the inelastic design
spectrum for constant ductility is obtained by dividing its various
branches by Ry (details given on next slide).
One of the proposal suggested by Newmark and Hall (Figure 7.11.3)
for correlating Ry with Tn is:
1
Tn < Ta
R y = 2μ −1
Tb < Tn < Tc'
μ
Tn > Tc
Where Ta,Tb,……… are mentioned on the inelastic design spectra given
on next slide. It must be noted that Ta=Ta′ ,Tb=Tb ′, Td=Td ′, Te=Te ′ and
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44
T =T ′.
45.
Inelastic Pseudo-velocity design spectrum (New-mark Hall)
Tc≠Tc ′ as V and A are divided
by different values value of Ry
CE-409: MODULE 7 (Fall-2013)
Td=Td ′ as V and D are divided
by same value of Ry i.e. μ
45
46.
Inelastic design spectra (Newmark-Hall) for firm
soil with PGA=1g
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47.
Inelastic Pseudo-acceleration design spectrum
(Newmark-Hall)-log scale
Once Vy is calculated by Vy=V/μ, then
Ay can be be easily calculated by:
2π
A y = Vyωn =
T Vy
n
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49.
Relation between um and Ay determined from inelastic
pseudo- acceleration design spectrum
u = μu
It is already known
m
( mA ) = A
f
u = =
k ( ω m) ω
y
Where
y
y
y
2
y
2
n
T
or u = A
2π
n
2
n
y
y
2
T
⇒ u = μu = μ A
2π
n
m
y
CE-409: MODULE 7 (Fall-2013)
y
49
50.
Inelastic peak deformation design spectrum
Hall)-log scale
(Newmark-
In order to draw inelastic deformation design spectrum, inelastic peak
deformations, um, is calculated using following relation (slide 49)
2
um
Tn
= µ
Ay
2π
where Ay is calculated
by equation mentioned
on slide 47 i.e
2π
A y = Vyωn = Vy
T
n
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51.
Application of the Inelastic design spectrum:
1. Structural design for allowable ductility
Consider a SDOF system having allowable ductility,μ, which is
decided on the ductility capacity of the material and design details
selected
It is desired to determine the design yield strength, fy, and the design
deformation, um, for the system.
For the known values of Tn, ζ , μ the value of Ay/g is determined of
from Figure 7.11.5 or 7.11.6 (Chopra’s book) given on slides 46 and 47.
e.g., Ay=0.49g for Tn=1 sec, ζ=5% and μ=4 as shown on next slide. The
required yield strength is determined from relation:
A
f =
.w
g
y
y
For above determined value of Ay, the
corresponding value of fy is 0.49gw/g =
0.49w
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52.
Application of the Inelastic design spectrum:
1. Structural design for allowable ductility
0.49g
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53.
Application of the Inelastic design spectrum:
1. Structural design for allowable ductility
The peak deformation, um, can be related to Ay as follows
( A/g.w ) = A
f
R = =
( A /g.w ) A
f
o
y
A
⇒ Ay =
Ry
y
y
y
Please recall that A is the elastic pseudoacceleration
2
T
We have already derived the relation u = μ
A
2π
n
m
y
2
μ T
or u =
A
R 2π
n
m
y
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54.
Application of the Inelastic design spectrum:
2. Evaluation of existing structures
Consider the simplest possible structures, SDF system, having
mass m, initial stiffness k at small displacement.
The yield strength fy of the structure are determined from its
properties: dimensions, member sizes, and design details
(reinforcement in R.C. structures, connections in steel structures).
fy can be determined from any suitable method from existing
analytical methods based on extensive laboratory works. Result of
a pushover analysis? for determining fy is shown on next slide.
Tn for small oscillation is computed from k and m, and the
damping ratio ζ from field tests
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55.
Application of the Inelastic design spectrum:
2. Evaluation of existing structures Draw a sketch of
frame with plastic
hinges
First plastic hinge
fs
Collapse
fs
fy
u
u
uy
um
Force-displacement curve of a building using Pushover analysis?
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56.
Application of the Inelastic design spectrum:
2. Evaluation of existing structures
For a system with known Tn and ζ, A is read from elastic design
spectrum
Ay for known value of fy and Ry can be determined using:
A
f
f =
.w or A =
g
w
g
and
R = A/A
y
y
y
y
y
y
With Tn already known, μ for calculated value of Ry can be
determined by using the applicable equation determined from
three 3 equations given on slide 44
μ Tn
peak deformation um, can be determined
by using eqn. derived on slide 52
CE-409: MODULE 7 (Fall-2013)
2
u =
A
R 2π
m
y
56
57.
Problem M7.1
Consider a one-story frame with lumped weight w, Tn = 0.25 sec, and fy
= 0.512w. Assume that ζ = 5% and elastoplastic force–deformation
behavior. Determine the lateral deformation for the design earthquake
has a peak acceleration of 0.5g and its elastic design spectrum is given by
Fig. 6.9.5 multiplied by 0.5
Solution:
For a system with Tn = 0.25 sec, A = (2.71g)0.5 = 1.355g from Fig.
6.9.5 (slide 10)
Slide 41
Slide 53
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58.
Problem M7.1 (contd…..)
Slide 54
2
Slide 56
μ T
u =
A
R 2π
n
m
y
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59.
Application of the Inelastic design spectrum:
3. Direct displacement based seismic design of structures
Displacement Based Seismic Design (DBSD) is defined broadly as
any seismic design in which displacement related quantities are used
directly to judge performance acceptability. This performance
acceptability for various limit states/performance levels in general is
referred to as Performance Based Seismic Design (PBD) among
earthquake engineering community
A simple DBSD approach could be to specify a drift limit
corresponding to a defined damage level, and then require that the
drift under the specified seismic loading does not exceed the specified
drift. This procedure is in contrast with Force Based Seismic Design
(FBSD) procedure in which the acceptability of structural
performance is judged on the basis of force –based quantities.
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60.
Application of the Inelastic design spectrum:
3. Direct displacement based seismic design of structures
A simple example of a force-based procedure is the familiar
requirement that the design base shear strength under seismic
loading shall not be less than some fraction of base shear calculated
assuming linear elastic structural response.
We followed FBSD process in previous slides of this module due
to the reason that currently seismic codes are based on FBSD
procedure because of their familiarity for design against other
loading such as gravity and wind.
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61.
Application of the Inelastic design spectrum:
3. Direct displacement based seismic design of structures
After earthquakes of 1994 Northridge, USA and 1995 Kobe,
Japan, earthquake engineering community is seriously making
effort to PBD (see figure on next slide) which is essentially based on
DBSD procedure.
It is worth mentioning that neither of the two procedures (i.e.,
FBSD and DBSD) can be totally taken independent of the decision
making parameters involved in the two procedures. Inherently, both
involves relevant parameters related to forces and displacements,
however, decisions are based on Forces in FBSD and
Displacements/Drifts in DBSD.
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62.
Application of the Inelastic design spectrum:
3. Direct displacement based seismic design of structures
It is a common practice in earthquake engineering to indicate structural damages
in terms of performance levels.
Performance levels as per FEMA 356?
Immediate occupancy
(O)
Life safety
Collapse prevention
(LS)
Operational
(CP)
(IO)
Moderate damages
Very light damages
(Building can be
occupied. No repair
work required)
Light damages
(Building can be
occupied but will
need repair work)
Severe damages
(Building can be
occupied after
subsequent repair)
(Building is far beyond
the economically
feasible repair)
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63.
Application of the Inelastic design spectrum:
3. Direct displacement based seismic design of structures
LS
IO
CP
O
CP performance level
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O performance level
IO performance level
LS performance level 63
64.
Application of the Inelastic design spectrum:
3. Direct displacement based seismic design of structures
The inelastic design spectrum is also useful for direct Displacementbased design of structures.
The goal is to determine the initial stiffness and yield strength of the
structure necessary to limit the deformation to some acceptable value.
Applied to an elastoplastic SDF system (Fig. 7.12.1), such a design
procedure may be implemented as a sequence of the following steps:
1. Estimate the yield deformation uy for the
system.
2. Determine acceptable plastic rotation θp of the
hinge at the base.
3. Determine the design displacement um from
um = uy + hθp
and design ductility factor from
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65.
Application of the Inelastic design spectrum:
3. Direct displacement based seismic design of structures
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66.
Application of the Inelastic design spectrum:
3. Direct displacement based seismic design of structures
Problem M 7.2: Consider a long reinforcedconcrete viaduct that is part of a freeway. The total
weight of the superstructure, 13 kips/ft, is supported
on identical bents 30 ft high, uniformly spaced at
130 ft. Each bent consists of a single circular
column 60 in. in diameter (Fig. E7.3a). Using the
displacement-based design procedure, design the
longitudinal reinforcement of the column for the
design earthquake has a peak acceleration of 0.5g and its
elastic design spectrum is given by Fig. 6.9.5 multiplied
by 0.5
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67.
Problem M 7.2 contd….
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68.
Problem M 7.2 contd….
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69.
Problem M 7.2 contd….
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70.
Problem M 7.2 contd….
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71.
Home Assignment No. 6
Solve problems 7.7 and 7.8 from chopra’s book
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