Elastic and inelastic relations..... mx+cx+Q(x)= -ma x Q x Q Q=kx elasticinelastic.

Elastic and inelastic relations

.. . ..mx+cx+Q(x)= -ma

x

Q

x

Q

Q=kx

elastic inelastic

Exercise 1 (A hysteretic energy dissipation index Eh)

A hysteretic energy dissipation index Eh corresponds to equivalent viscous damping factor h. Derive the equation (3.1) by calculating energy dissipation ΔU by viscous damping per cycle under the ‘resonant steady-state’ and putting ΔU = ΔW.

Eh = ΔW/2πFmDm (3.1)

ptay cos(

2

2cos1sin 2

c: damping coefficient(=2hm ), m: mass of the system : natural circular frequency(= ) k=Fm/Dm

y: amplitude, a: maximum amplitude(=Dm)p: input frequency, : phase difference ‘Resonant steady-state’ means p= .

m

k

　dtycdtdt

dy

dt

dycdy

dt

dycdyycU p

o 2

2

Example 2 (A hysteretic energy dissipation index Eh

in the case of elasto-plastic model)

Derive the equation shown below by calculating energy dissipation ΔW under the ‘resonant steady-state’ in the case of elasto-plastic model.

)1(2

hE

Excercise 2 (A hysteretic energy dissipation index Eh

in the case of Clough model)

Derive the equation (3.8) by calculating energy dissipation ΔW under the ‘resonant steady-state’ in the case of modified Clough model.

(3.8)

/1(1{1

hE

Simple method to get elastic periodof SDOF system

)(2 sk

mT

　　g

Aw

g

Wm f

3

12

nh

IEk c

)/(36(60

3350000

)/(36(20

2100000

23

1

2

mmNFF

mmNFF

E

cc

cc

c

　

　12

3bDI

w: a unit weight(=12000N/m2), Σ Ａ f: sum

of whole floor area of the building(m2) ,g: gravity(cm/s2), Fc: compressive strength of concrete (N/mm2) , b: width of a column(cm), D: depth of a olumn(cm), h: story height(cm), n: number of story

stiffness of concrete (N/mm2)

moment of inertia of a column (cm4)

Simple method to get base shear coefficientof SDOF system

　　

f

ccy

Aw

AC

τc: ultimate shear strength of columns(N/mm2=Fc/15), Ａ c: sum of 1st

story column section area(cm2) , w: a unit weight(=12000N/m2), ΣＡ f: sum of whole floor area of the building(m2)

Example 3 (Simple method to get fundamental parameters of SDOF system )

3-story building:column size: 60cm x 60cmFc= 24(N/mm2)

plan elevation

stor

y h

eigh

t: 3

.6m

⇔longitudinal direction

Exercise 3 (Simple method to get fundamental parameters of SDOF system )

12-story building:column size: 95cm x 95cmFc= 48(N/mm2)

plan

stor

y h

eigh

t: 3

.5m

elevation

⇔longitudinal direction

Newmark’s design criteria

・ property of energy conservation: For short period systems (T<0.5s) the energy dissipation is constant.

・ property of displacement conservation: For long period systems (T>0.5s) the response displacement is constant.

Newmark’s design criteria was used to decide strength of the system for each input strong ground motion.

Property of energy conservation

For short period systems (T<0.5s)The area of trapezoid OBCE= the one of OAD△

(μδy+(μ-1) δy)Qy/2=δL*QL/2

δy = Qy/k, δL = QL/k

12μ

QQ L

y

Property of displacement conservation

For long period systems (T>0.5s)inelastic response displacement is the same as elastic response displacement.

μδy= QL/k

δy = Qy/k

L

y

QQ

Example 5 (Newmark’s design criteria)

Calculate response displacement in the case that 1) elastic period=0.3 sec., base shear coefficient= 0.4, elastic response acceleration=0.8g at 0.3 sec.2) elastic period=1.0 sec., base shear coefficient= 0.1, elastic response acceleration=0.2g at 1.0 sec.

Exercise 5 (Newmark’s design criteria)

Calculate response displacement under the input of El-Centro NSusing elastic spectra (damping factor=0.05, Fig.4.3,(b), p.139) and Newmark’s design criteria in the case that 1) 7-story reinforced concrete building (base shear coefficient= 0.3, story height=2.8(m))2) 20-story steel building (base shear coefficient= 0.05, story height=3.5(m))

T=0.02H (T: period of the building (s), H: height of the building(m)) for reinforced concrete buildingT=0.03H (T: period of the building (s), H: height of the building(m)) for steel building

Tripartite response spectra

Response spectra which show response acceleration, velocity and displacement simultaneouslyusing the relations:

SV=ωSD

SA=ωSV=ω2SD

SD: response displacementSV: response pseudo-velocity SA: response pseudo-acceleration

Ductility factor

Ductility factor μ is defined as the ratio of the maximum response displacement x to the yielding displacement xy.

Ductility factor is used as the index of representing the damage level.

Ductility factor of short and long period system

Short period system long period system

Q

d dy dm

Qy

0

Q

d dy dm

Qy

0

μ=dm/dy=5 μ=dm/dy=2

Response ductility factor spectra

Ductility factors were calculated in the case of bilinear Takeda model under the input of El-Centro NS and Fukiai (Kobe EQ.) changing the base shear coefficient.


0.0

5.0

10.0

15.0

20.0

0 0.5 1 1.5 2 2.5 3

C=0.1C=0.2C=0.4C=0.8

Response ductility factor

Period(sec.)

Input: El-Centro NSBilinear Takeda model


0.0

5.0

10.0

15.0

20.0

0 0.5 1 1.5 2 2.5 3

C=0.1C=0.2C=0.4C=0.8


Period(sec.)

Input: Fukiai (Kobe EQ.)Bilinear Takeda model

Comparison of response ductility factor spectra

0.0

5.0

10.0

15.0

20.0

0 0.5 1 1.5 2 2.5 3

C=0.1C=0.2C=0.4C=0.8


Period(sec.)


0.0

5.0

10.0

15.0

20.0

0 0.5 1 1.5 2 2.5 3

C=0.1C=0.2C=0.4C=0.8


Period(sec.)


Required strength

Required strength which give maximum response ductility factors within constant values is important, because we need design base shear coefficient in the structural design.

↑Allowable ductility factor μa

Required strength spectra

0.0

0.5

1.0

1.5

2.0

0 0.5 1 1.5 2 2.5 3

μ a=1μ a=2μ a=4μ a=8

Base shaer coefficient

Period(sec.)


Required strength spectra

0.0

0.5

1.0

1.5

2.0

0 0.5 1 1.5 2 2.5 3



Period(sec.)


Comparison of required strength spectra

0.0

0.5

1.0

1.5

2.0

0 0.5 1 1.5 2 2.5 3



Period(sec.)

0.0

0.5

1.0

1.5

2.0

0 0.5 1 1.5 2 2.5 3



Period(sec.)



Example 8 (Calculation of required strength spectrum from response ductility factor spectra)

Calculate required strength (base shear coefficient) spectrum from response ductility factor spectra in the case that allowable ductility factor=4, model: bilinear Takeda model (α=0.5, β=0.01）under the input of El-Centro NS.

Response ductility factor spectra changing base shear coefficient of the system input motion: El-Centro NS T(s) Cy=0.005 0.01 0.02 0.05 0.1 0.2 0.3 0.4 0.5 0.100 99.999 99.999 99.999 99.999 99.999 38.270 8.309 1.826 1.174 0.200 99.999 99.999 99.999 99.999 65.070 19.300 6.894 2.350 1.557 0.500 99.999 99.999 99.999 45.490 13.500 6.573 3.564 2.734 1.987 1.000 99.999 73.020 24.880 13.080 4.469 2.071 1.474 1.285 1.030 1.500 74.490 33.440 13.380 5.099 1.954 0.947 0.632 0.474 0.379 2.000 47.700 21.080 7.728 3.817 1.831 0.887 0.592 0.444 0.355 3.000 21.190 9.574 3.388 1.865 1.114 0.570 0.380 0.285 0.228 99.999: greater than 100

Required strength spectrum (allowable ductility factor=4 ) by El-Centro NS

0.0

0.1

0.2

0.3

0.4

0 0.5 1 1.5 2 2.5 3

El-Centro_NS


Period(sec.)

requiredT(s) strength (base shear coefficient) 0.1 0.366 0.2 0.364 0.5 0.286 1.0 0.120 1.5 0.067 2.0 0.049 3.0 0.019

Exercise 8 (Calculation of required strength spectrum from response ductility factor spectra)

Calculate required strength (base shear coefficient) spectrum from response ductility factor spectra in the case that allowable ductility factor=4, model: bilinear Takeda model (α=0.5, β=0.01）under the input of Taft EW and compare it with El-Centro NS.

Response ductility factor spectra changing base shear coefficient of the system input motion: Taft EW T(s) Cy=0.005 0.01 0.02 0.05 0.1 0.2 0.3 0.4 0.5 0.100 99.999 99.999 99.999 99.999 84.430 1.109 0.720 0.540 0.432 0.200 99.999 99.999 99.999 74.450 23.240 4.561 1.577 1.079 0.859 0.500 99.999 99.999 32.930 13.910 5.297 1.396 1.168 0.864 0.691 1.000 91.840 37.000 10.040 3.978 1.734 0.793 0.529 0.396 0.317 1.500 51.110 18.410 7.052 2.825 1.389 0.655 0.437 0.327 0.262 2.000 31.780 11.420 4.222 1.352 0.855 0.428 0.285 0.214 0.171 3.000 15.710 6.619 2.097 0.957 0.479 0.239 0.160 0.120 0.096 99.999: greater than 100

Required strength spectra (allowable ductility factor=4 ) by El-Centro NS and Taft EW

0.0

0.1

0.2

0.3

0.4

0 0.5 1 1.5 2 2.5 3

El-Centro_NSTaft_EW


Period(sec.)

required strength (base shearT(s) coefficient) El-Centro Taft NS EW 0.1 0.366 0.197 0.2 0.364 0.219 0.5 0.286 0.133 1.0 0.120 0.050 1.5 0.067 0.042 2.0 0.049 0.022 3.0 0.019 0.016

Equivalent linear system

Inelastic responses can be estimated by equivalent linear systems with equivalent period and equivalent viscous damping.

x

Q

x

Q

inelastic

with equivalent viscous damping

elastic

equivalent period

Method to calculate inelastic responsesby equivalent linear systems

How to decide equivalent period and damping

equivalent period: period corresponding maximum displacement

equivalent damping: equivalent viscous damping

Example 9 (Required strength spectrum by equivalent linear system)

Calculate required strength (base shear coefficient) spectrum(T=0.1, 0.2, 0.5, 1.0, 1.5, 2.0 and 3.0 s)by equivalent linear system in the case that allowable ductility factor=4, model: bilinear Takeda model (α=0.5, β=0.01 ）under the input of El-Centro NSfrom elastic response spectra (h=0.05)→using the damping reduction factor inEquation shown below andcompare it with required strength spectrum by inelastic systemand Newmark’s design criteria.

hF h 101

5.1

Fh: reduction factor from h=0.05, h: damping factor

responseT(s) acceleration (cm/s2)0.1 560.70.2 640.70.3 695.9 0.4 601.80.5 820.5 1.0 507.41.5 186.7 2.0 174.9 3.0 112.3 4.0 45.4 6.0 31.0

Required strength spectrum (allowable ductility factor=4 ) by El-Centro NS using equivalent

linear system comparing with actual spectrum

0.0

0.1

0.2

0.3

0.4

0 0.5 1 1.5 2 2.5 3

InelasticEquivalent_linearNewmark's_criteria


Period(sec.)

Required strength (base shear coefficient) Inelastic Equivalent Newmark’s T(s) system linear design system criteria 0.1 0.366 0.386 0.216 0.2 0.364 0.363 0.217 0.5 0.286 0.306 0.316, 0.209 1.0 0.120 0.105 0.129 1.5 0.067 0.068 0.048 2.0 0.049 0.027 0.045 3.0 0.019 0.019 0.029

Required strength spectrum (allowable ductility factor=4 ) by El-Centro NS using equivalent

linear system comparing with actual spectrum

0.0

0.1

0.2

0.3

0.4

0 0.5 1 1.5 2 2.5 3

InelasticEquivalent_linear


Period(sec.)

Required strength (base shear coefficient)

Inelastic EquivalentT(s) system linear system 0.1 0.366 0.386 0.2 0.364 0.363 0.5 0.286 0.306 1.0 0.120 0.105 1.5 0.067 0.068 2.0 0.049 0.027 3.0 0.019 0.019

Exercise 9 (Required strength spectrum by equivalent linear system)

Calculate required strength (base shear coefficient) spectrum(T=0.1, 0.2, 0.5, 1.0, 1.5, 2.0 and 3.0 s)by equivalent linear system in the case that allowable ductility factor=4, model: bilinear Takeda model (α=0.5, β=0.01 ）under the input of Taft EWfrom elastic response spectra (h=0.05)→using the damping reduction factor inEquation shown below andcompare it with required strength spectrum by inelastic systemand Newmark’s design criteria.

hF h 101

5.1

Fh: reduction factor from h=0.05, h: damping factor

responseT(s) acceleration (cm/s2)0.1 211.90.2 422.50.3 398.40.4 387.40.5 340.31.0 156.41.5 129.32.0 84.93.0 47.24.0 27.16.0 22.7

Required strength spectrum (allowable ductility factor=4 ) by Taft EW using equivalent linear

system comparing with actual spectrum

0.0

0.1

0.2

0.3

0.4

0 0.5 1 1.5 2 2.5 3

InelasticEquivalent_linear


Period(sec.)

Required strength (base shear coefficient)

Inelastic EquivalentT(s) system linear system 0.1 0.197 0.255 0.2 0.219 0.233 0.5 0.133 0.094 1.0 0.050 0.051 1.5 0.042 0.028 2.0 0.022 0.016 3.0 0.016 0.014

Required strength spectrum (allowable ductility factor=4 ) by Taft EW using equivalent linear

system comparing with actual spectrum

Required strength (base shear coefficient) Inelastic Equivalent Newmark’s T(s) system linear design system criteria 0.1 0.197 0.255 0.082 0.2 0.219 0.233 0.163 0.5 0.133 0.094 0.131, 0.087 1.0 0.050 0.051 0.040 1.5 0.042 0.028 0.033 2.0 0.022 0.016 0.022 3.0 0.016 0.014 0.012

0.0

0.1

0.2

0.3

0.4

0 0.5 1 1.5 2 2.5 3

InelasticEquivalent_linearNewmark's_criteria


Period(sec.)

Example & Excersise10 (Prediction of response ductility factor from elastic response using equivalent linear syste

m and comparison with actual structural damage)

Calculate response ductility factor in the case of longitudinal direction of Kuoshing National Elementary School Building B under the input of recorded strong ground motion of 1999 Chi-chi, Taiwan earthquake using equivalent linear system and compare with actual structural damage.

Building A (Shikang Natl. Elem. Sch.)

Elastic response spectrum at Shikang National Elementary School Building A

0 0.5 1 1.5 2 2.5 30

500

1000

1500

2000Elastic response

acceleration(cm/s

2 )

Period(s)

Building B (Kuoshing Natl. Elem. Sch.)

0 0.5 1 1.5 2 2.5 30

500

1000

1500

2000Elastic response

acceleration(cm/s

2 )

Period(s)

Elastic response spectrum at Kuoshing National Elementary School Building B

Shikang

N.E.S.

Kuoshing

N.E.S.

Building A Building B

unit of weight 1.2 1.2 tonf/m2

Fc 154.7 216.5 kgf/cm2

No. of story 3 3

column1 depth(longitudinal) 33.0 50.0 cm

column1 depth(transvers) 46.8 58.0 cm





span length1(transvers) 783.2 500.0 cm

span length2(transvers) 261.3 cm

span length(longitudinal) 300.0 400.0 cm

story height 343.0 360.0 cm

No. of span(longitudinal) 16 5

No. of span(transvers) 2 5

column1 transvers

column2

column3

span length (longitudinal)

span length1 (transvers)

span length2 (transvers)

longitudinal

Information of the buildings

Simple method to get elastic periodof SDOF system

)(2 sk

mT

　　g

Aw

g

Wm f

3

12

nh

IEk c

)/(36(60

3350000

)/(36(20

2100000

23

1

2

mmNFF

mmNFF

E

cc

cc

c

　

　12

3bDI

w: a unit weight(=12000N/m2), Σ Ａ f: sum

of whole floor area of the building(m2) ,g: gravity(cm/s2), Fc: compressive strength of concrete (N/mm2) , b: width of a column(cm), D: depth of a olumn(cm), h: story height(cm), n: number of story

stiffness of concrete (N/mm2)

moment of inertia of a column (cm4)

equivalent period T’

factorductilityTT :,*2' 　

0.3Cymg

k 0.25k

0

k

0.01k

0

k

0

Cymg

Shear force

Displacement

Simple method to get base shear coefficientof SDOF system

　　

f

ccy

Aw

AC

τc: ultimate shear strength of columns(N/mm2=Fc/15), Ａ c: sum of 1st

story column section area(cm2) , w: a unit weight(=12000N/m2), ΣＡ f: sum of whole floor area of the building(m2)

hh E

F101

5.1

/1(1{1

hE

Equivalent viscous damping and damping reduction factor

Decide μso that response acceleration is equal to base shear coefficient at equivalent period T’ and equivalent viscous damping factor Eh from elastic response spectrum of damping factor 0.05.

Elastic and inelastic relations..... mx+cx+Q(x)= -ma x Q x Q Q=kx elasticinelastic.

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