View
229
Download
0
Category
Preview:
Citation preview
7/28/2019 Solved MDOF Example
1/8
Earthquake Engineering Ahmed ElgamalMichael Fraser
Solved MDOF and Modal Analysis Example
1
3m
m
u2
u1
h1
h2
Ic
Ic
Elevation View
Use:
Esteel = 29,000 ksiIc = 164.8 in
4
h1 = 15 ft
h2 = 12 ft
in
skip
in
slb
sin
ftft
ft
lb
g
Wm
22
2209317.017.93
4.386
304030
=
=
==
Plan View
30 ft
40 ft
(a) For the two-story building shown above, define the two-degree-of-freedom free vibration matrix
equation in terms of k and m. Using this matrix equation, determine the natural frequencies 1 and2(in terms of k and m). Using these expressions for1 and2,plug in numerical values and determine and2 in radians. For each natural frequency, define and sketch the corresponding mode shape.
(b) Verify that the modes are orthogonal as expected.
(c) Normalize the first mode such that 1Tm1 = 1.0(d) Use the normalized first mode (from above) to verify that 1Tk1 = 12(e) Use the El Centro Response Spectrum and a damping ratio of 5% to estimate the maximum base shear
and moment.
(f) Find 0a and 1a in kmc 10 aa += for a viscous damping of 5% in modes 1 and 2.
3i
c
3i
c
floor3i
c
column h
48EI
h
EI
124kh
12EI
k =
=
= i
7/28/2019 Solved MDOF Example
2/8
Earthquake Engineering Ahmed ElgamalMichael Fraser
Solved MDOF and Modal Analysis Example
2
Write the equation of motion in matrix form:
=
+
+
0
0
0
03
2
1
22
221
2
1
u
u
kk
kkk
u
u
m
m
&&
&&
substitute in u2=&& and rearrange to get
=
+0
03
2
1
2
22
2
2
21
u
u
mkk
kmkk
Take the determinant and set equal to zero
032
22
2
2
21 =+
mkkkmkk
04321
2
2
2
1
42 =+ kkmkmkm
solving for2 yields
m
kkkkkk
6
1644 22212
1212
1
++=
m
kkkkkk
6
1644 22212
1212
2
+++=
( )( )( )
inkipin
inksi
h
EIk c 335.39
180
8.1642900048483
4
3
1
1 ===
( )( )
( )inkip
in
inksi
h
EIk c 826.76
144
8.1642900048483
4
3
2
2 ===
in
skipm
2
09317.0
=
substituting in k1, k2, and m:
7/28/2019 Solved MDOF Example
3/8
Earthquake Engineering Ahmed ElgamalMichael Fraser
Solved MDOF and Modal Analysis Example
3
2
22
1 95.101s
rad=
2
22
2 1138
s
rad=
s
rad097.101 =
s
rad734.332 =
Hzf 607.12
11 ==
Sec
fT 6223.0
1
1
1 ==
Hzf 369.5
2
22 ==
Sec
f
T 1863.01
2
2 ==
Determining 1st
Mode Shape:
Plug k1, k2, m, and1 = 10.097 rad/sec into
=
+
0
03
21
11
2
122
2
2
121
mkk
kmkk
( )( )( )( )
=
+0
0
09317.0097.10826.76826.76
826.7609317.03097.10826.76335.39
21
11
2
2
=
0
0
327.67826.76
826.76665.87
21
11
0826.76665.872111
=
let 121 = , then
( ) 01826.76665.87 11 =
876.0665.87
826.7611 ==
=
=000.1
876.0
21
11
1
7/28/2019 Solved MDOF Example
4/8
Earthquake Engineering Ahmed ElgamalMichael Fraser
Solved MDOF and Modal Analysis Example
4
Determining 2nd
Mode Shape:
Plug k1, k2, m, and2 = 33.734 rad/sec into
=
+
0
03
22
12
2
222
2
2
221
mkk
kmkk
( )( )( )( )
=
+
0
0
09317.0734.33826.76826.76
826.7609317.03734.33826.76335.39
22
12
2
2
=
0
0
200.29826.76
826.76917.201
22
12
0826.76917.201 2212 =
let 122 = , then
( ) 01826.76917.201 12 =
380.0917.201
826.7612
=
=
=
=
000.1
380.0
22
12
2
0
5
10
15
20
25
30
0 0.5 1 1.5
h
eight(ft)
11 = 0.876
21 = 1.000
0
5
10
15
20
25
30
-0.5 0 0.5 1 1.5
height(ft)
22 = 1.000
12 = -0.380
1
stMode Shape 2
ndMode Shape
7/28/2019 Solved MDOF Example
5/8
Earthquake Engineering Ahmed ElgamalMichael Fraser
Solved MDOF and Modal Analysis Example
5
(b) Verify that the modes are orthogonal as expected.
To verify that the modes are orthogonal, need to show that 0== jk TiTi , for ij
[ ]( ) [ ] 0
1
380.010317.9024485.0
1
380.0
09317.00
009317.031876.0 21 =
=
= xT 2
[ ] [ ] 01
380.05264.99310.24
1
380.0
826.76826.76
826.76826.76335.391876.01 =
=
+=T
Can also show 022 == 1k TT
(c) Normalize the first mode such that 1Tm1 = 1.0
[ ]( )
30766.01
876.0
09317.00
009317.031876.01 =
=T
Divide 1 by 30766.0 , therefore
=
=8029.1
5793.1
21
11
1
Check: [ ]( )
00.18029.1
5793.1
09317.00
009317.038029.15793.11 =
=T ok
(d) Use the normalized first mode (from above) to verify that 1Tk1 = 12
[ ] 212
2
2
1 097.10950.1018029.1
5793.1
826.76826.76
826.76826.76335.398029.15793.1 =
==
+=
s
rad
s
radT 2
7/28/2019 Solved MDOF Example
6/8
Earthquake Engineering Ahmed ElgamalMichael Fraser
Solved MDOF and Modal Analysis Example
6
(e) Use the El Centro Response Spectrum and a damping ratio of 5% to estimate the maximum base shear and
moment.
T1
T2
From Response Spectrum:
For T1 = 0.622 sec and = 5%, D1 = 2.8 in. and A1 = 0.8g = 309.12 in/s2
For T2 = 0.0.186 sec and = 5%, D2 = 0.31 in. and A2 = 0.9g = 347.76 in/s2
jnjn
n
njn mA
M
Lf =
7/28/2019 Solved MDOF Example
7/8
Earthquake Engineering Ahmed ElgamalMichael Fraser
Solved MDOF and Modal Analysis Example
7
=
==2
1
2
2
1
j
jij
j
jij
i
i
m
m
M
L
( )( ) ( )( )( )( ) ( )( )
0987.1109317.0876.027951.0
109317.0876.027951.0222
212
2
111
212111
2
1
2
1
2
1
1
1
1 =+
+=
++
==
=
=
mm
mm
m
m
M
L
j
jj
j
jj
( )( ) ( )( )( )( ) ( )( )
0977.0109317.0380.027951.0
109317.0380.027951.0222
222
2
121
222121
2
1
2
2
2
1
2
2
2 =++
=++
==
=
=
mm
mm
m
m
M
L
j
jj
j
jj
( ) ( ) kipsin
skip
s
inmA
M
Lf 159.83876.027951.012.3090987.1
2
21111
1
111 =
==
( ) ( ) kipsin
skip
s
inmA
M
Lf 609.3380.027951.076.3470977.0
2
21212
2
212 =
==
( ) ( ) kipsin
skip
s
inmA
M
Lf 643.31109317.012.3090987.1
2
22121
1
121 =
==
( ) ( ) kipsin
skip
s
inmA
M
Lf 166.3109317.076.3470977.0
2
22222
2
222 =
==
Base Shear
=
=N
j
jnn fV1
0
kipskipskipsfffVj
j 80.114643.31159.832111
2
1
101 =+=+== =
kipskipskipsfffVj
j 443.0166.3609.32212
2
1
202 ==+== =
Calculate Maximum Base Shear, V0 max
7/28/2019 Solved MDOF Example
8/8
Earthquake Engineering Ahmed ElgamalMichael Fraser
Solved MDOF and Modal Analysis Example
8
( ) ( ) ( ) ( ) kipskipskipsVVV 80.114443.080.114 222022
01max0 =+=+=
Base Moment
j
N
j
jnn dfM =
=1
0
( ) ( ) inkipsinkipsinkipsdfdfdfM jj
j =+=+== =
0.25221324643.31180159.83221111
2
1
101
( ) ( ) inkipsinkipsinkipsdfdfdfM jj
j ==+== =
16.376324166.3180609.3222112
2
1
202
Calculate Maximum Base Moment, M0 max
( ) ( ) ( ) ( ) ftkipsinkipsinkipsinkipsMMM ==+=+= 0.21028.2522316.3760.25221 222022
01max0
(f) Find 0a and 1a in kmc 10 aa += for a viscous damping of 5% in modes 1 and 2.
1 = 2 = = 0.05
( )( )( )
sradsradsrad
sradsrada
ji
ji777.0
734.33097.10
734.33097.10205.0
20 =+
=+
=
( )sradsradsrad
aji
100228.0
734.33097.10
205.0
21 =+
=+
=
srada 777.00 = & radsa 00228.01 =
Recommended