# A hysteretic energy dissipation index Eh

Urban and Civil

Nov 25, 2013 (4 years and 7 months ago)

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Elastic and inelastic relations

..

.

..

mx+cx+Q(x)=
-
ma

x

Q

x

Q

Q=kx

elastic

inelastic

m
x
Q
x
c
p
x
)
(

k
t
c
t
m
x
t
x
c
x
x
t
m
p
x

3
6
2
3
3
6
2
0
0
0
0

0
0
2
3
3
x
t
x
x
t
x

Linear acceleration method

Exercise 1

(A hysteretic energy dissipation index E
h
)

A hysteretic energy dissipation index E
h

corresponds to equivalent
viscous damping factor h. Derive the equation (3.1) by calculating
energy dissipation
ΔU

by viscous damping per cycle under the
-
state’ and putting
ΔU

=
ΔW.

E
h

= ΔW/2πF
m
D
m

(3.1)

pt
a
y
cos(
2
2
cos
1
sin
2

c
: damping coefficient(=
2hm
),

m
: mass of the system

: natural circular frequency(= )

k=F
m
/D
m

y
: amplitude
, a
:

maximum amplitude(=
D
m
)

p
:

input frequency,

: phase difference

-
state’ means

p= .

m
k

dt
y
c
dt
dt
dy
dt
dy
c
dy
dt
dy
c
dy
y
c
U
p
o

2
2

Example 2 (A hysteretic energy dissipation index E
h

in the case of elasto
-
plastic model)

Derive the equation shown below by calculating energy dissipation
ΔW

-
state’ in the case of elasto
-
plastic
model
.



)
1
(
2

h
E
Excercise 2 (A hysteretic energy dissipation index E
h

in the case of Clough model)

Derive the equation (3.8) by calculating energy dissipation
ΔW

-
state’ in the case of modified Clough
model
.

(3.8)



/
1
(
1
{
1
h
E
Simple method to get elastic period

of SDOF system

)
(
2
s
k
m
T

g
A
w
g
W
m
f

3
12
nh
I
E
k
c

)
/
(
36
(
60
3350000
)
/
(
36
(
20
2100000
2
3
1
2
mm
N
F
F
mm
N
F
F
E
c
c
c
c
c

12
3
bD
I

w
: a unit weight(=12000N/m
2
),
Σ

f
: sum
of whole floor area of the building(m
2
) ,

g: gravity(cm/s
2
),
F
c
: compressive strength
of concrete (N/mm
2
)

,
b
: width of a
column(cm),
D
: depth of a olumn(cm),
h
: story height(cm),
n
: number of story

stiffness of concrete (N/mm
2
)

moment of inertia
of a column (cm
4
)

Simple method to get base shear coefficient

of SDOF system

f
c
c
y
A
w
A
C

τ
c
: ultimate shear strength of columns(N/mm
2
=F
c
/15),

c
:
sum of 1st
story column section area(cm
2
) ,
w
:
a unit weight(=12000N/m
2
),

Σ

f
:
sum of whole floor area of the building(m
2
)

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

)

3
-
story building:

column size: 60cm x 60cm

Fc= 21(N/mm2)

plan

elevation

story height: 3.6m

longitudinal direction

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

)

12
-
story building:

column size: 95cm x 95cm

Fc= 48(N/mm
2
)

plan

story height: 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.

††
䙯F

††

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

δ
y
+(μ
-
1)
δ
y
)Q
y
/2=
δ
L*
Q
L
/2

δ
y
=

Q
y
/k,
δ
L
=

Q
L
/k

1

Q
Q
L
y

Property of displacement conservation

For
long

period systems (T>0.5s)

inelastic response displacement is
the same as elastic response
displacement.

μ
δ
y
= Q
L
/k

δ
y
=

Q
y
/k

L
y
Q
Q

Example 4 (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 4 (Newmark’s design criteria)

Calculate response displacement under the input of El
-
Centro NS

using 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 building

T=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 simultaneously

using the relations:

S
V
=ωS
D

S
A
=ωS
V

2
S
D

S
D
: response displacement

S
V
: response pseudo
-
velocity

S
A
: response pseudo
-
acceleration

Ductility factor

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

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

d
y

d
m

Qy

0

Q

d

d
y

d
m

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.

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=0.1
C=0.2
C=0.4
C=0.8
Response ductility factor
Period(sec.)
Input: El
-
Centro NS

Bilinear Takeda model

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=0.1
C=0.2
C=0.4
C=0.8
Response ductility factor
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=0.1
C=0.2
C=0.4
C=0.8
Response ductility factor
Period(sec.)
Input: Fukiai (Kobe EQ.)

Bilinear Takeda model

0.0
5.0
10.0
15.0
20.0
0
0.5
1
1.5
2
2.5
3
C=0.1
C=0.2
C=0.4
C=0.8
Response ductility factor
Period(sec.)
Input: El
-
Centro NS

Bilinear Takeda model

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
μa=1
μa=2
μa=4
μa=8
Base shaer coefficient
Period(sec.)
Input: El
-
Centro NS

Bilinear Takeda model

Required strength spectra

0.0
0.5
1.0
1.5
2.0
0
0.5
1
1.5
2
2.5
3
μa=1
μa=2
μa=4
μa=8
Base shaer coefficient
Period(sec.)
Input: Fukiai (Kobe EQ.)

Bilinear Takeda model

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
μa=1
μa=2
μa=4
μa=8
Base shaer coefficient
Period(sec.)
0.0
0.5
1.0
1.5
2.0
0
0.5
1
1.5
2
2.5
3
μa=1
μa=2
μa=4
μa=8
Base shaer coefficient
Period(sec.)
Input: El
-
Centro NS

Bilinear Takeda model

Input: Fukiai (Kobe EQ.)

Bilinear Takeda model

Example 5 (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

Example 5 (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
Base shaer coefficient
Period(sec.)

required

T(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 5 (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

Exercise 5 (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_NS
Taft_EW
Base shaer coefficient
Period(sec.)

required strength

(base shear

T(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 responses

by equivalent linear systems

How to decide equivalent period and damping

equivalent period:

period corresponding
maximum displacement

equivalent damping:

equivalent viscous damping

Example 6 (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 NS

from elastic response spectra (h=0.05)→

using the damping reduction factor in

Equation shown below and

compare it with required strength spectrum by inelastic system

and Newmark’s design criteria.

h
F
h
10
1
5
.
1

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

response

T(s) acceleration

(cm/s
2
)

0.1 560.7

0.2 640.7

0.3 695.9

0.4 601.8

0.5 820.5

1.0 507.4

1.5 186.7

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
Inelastic
Equivalent_linear
Newmark's_criteria
Base shaer coefficient
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
Inelastic
Equivalent_linear
Base shaer coefficient
Period(sec.)

Required strength

(base shear coefficient)

Inelastic Equivalent

T(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 6 (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 EW

from elastic response spectra (h=0.05)→

using the damping reduction factor in

Equation shown below and

compare it with required strength spectrum by inelastic system

and Newmark’s design criteria.

h
F
h
10
1
5
.
1

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

response

T(s) acceleration

(cm/s
2
)

0.1 211.9

0.2 422.5

0.3 398.4

0.4 387.4

0.5 340.3

1.0 156.4

1.5 129.3

2.0 84.9

3.0 47.2

4.0 27.1

6.0 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
Inelastic
Equivalent_linear
Base shaer coefficient
Period(sec.)

Required strength

(base shear coefficient)

Inelastic Equivalent

T(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
Inelastic
Equivalent_linear
Newmark's_criteria
Base shaer coefficient
Period(sec.)
Example 7 (Prediction of response ductility factor from
elastic response using equivalent linear system 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
3
0
500
1000
1500
2000
Elastic response
acceleration(cm/s
2
)
Period(s)

Building B (Kuoshing Natl. Elem. Sch.)

0
0.5
1
1.5
2
2.5
3
0
500
1000
1500
2000
Elastic 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

column2 depth(longitudinal)

52.3

50.0

cm

column2 depth(transvers)

52.4

58.0

cm

column3 depth(longitudinal)

40.0

50.0

cm

column3 depth(transvers)

65.0

58.0

cm

span length1(transvers)

783.2

500.0

cm

span length2(transvers)

261.3

cm

span l

ength(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 period

of SDOF system

)
(
2
s
k
m
T

g
A
w
g
W
m
f

3
12
nh
I
E
k
c

)
/
(
36
(
60
3350000
)
/
(
36
(
20
2100000
2
3
1
2
mm
N
F
F
mm
N
F
F
E
c
c
c
c
c

12
3
bD
I

w
: a unit weight(=12000N/m
2
),
Σ

f
: sum
of whole floor area of the building(m
2
) ,

g: gravity(cm/s
2
),
F
c
: compressive strength
of concrete (N/mm
2
)

,
b
: width of a
column(cm),
D
: depth of a olumn(cm),
h
: story height(cm),
n
: number of story

stiffness of concrete (N/mm
2
)

moment of inertia
of a column (cm
4
)

equivalent period
T’

factor
ductility
T
T
:
,
*
2
'

0.3Cymg
k
0.25k
0
k
0.01k
0
k
0
Cymg
Shear force
Displacement
Simple method to get base shear coefficient

of SDOF system

f
c
c
y
A
w
A
C

τ
c
: ultimate shear strength of columns(N/mm
2
=F
c
/15),

c
:
sum of 1st
story column section area(cm
2
) ,
w
:
a unit weight(=12000N/m
2
),

Σ

f
:
sum of whole floor area of the building(m
2
)

h
h
E
F
10
1
5
.
1



/
1
(
1
{
1
h
E
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 E
h
from elastic response spectrum of damping
factor 0.05.