If there are 4 levels, m1 = 102.32 ton, m2 = 23.72 ton, m3 = 9.63 ton, m4 = 4.42 ton k1 = 4.31 kN/m, k2 = 19.7 kN/m, k3 = 30.19 kN/m, k4 = 36.44 kN/m a1 = 1 1. Determine the three modal frequencies and periods 2. Determine the normalized mode shape factors 3. Draw the mode shapes 4. Determine the participation factors
If there are 4 levels, m1 = 102.32 ton, m2 = 23.72 ton, m3 = 9.63 ton, m4 = 4.42 ton k1 = 4.31 kN/m, k2 = 19.7 kN/m, k3 = 30.19 kN/m, k4 = 36.44 kN/m a1 = 1 1. Determine the three modal frequencies and periods 2. Determine the normalized mode shape factors 3. Draw the mode shapes 4. Determine the participation factors
Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
5th Edition
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Saeed Moaveni
Chapter9: Mass And Mass-related Variables In Engineering
Section: Chapter Questions
Problem 3P: Determine the specific gravity of the following materials: gold ( = 1208 lb/ft3), platinum ( = 1340...
Related questions
Question
If there are 4 levels,
m1 = 102.32 ton, m2 = 23.72 ton, m3 = 9.63 ton, m4 = 4.42 ton
k1 = 4.31 kN/m, k2 = 19.7 kN/m, k3 = 30.19 kN/m, k4 = 36.44 kN/m
a1 = 1
1. Determine the three modal frequencies and periods
2. Determine the normalized mode shape factors
3. Draw the mode shapes
4. Determine the participation factors
reference: pictures attached
![m3 = 65.26 ton
m₂ = 138.68 ton
m₁ = 157.85 ton
m₁ = 157.85 ton
m2 = 138.68 ton
m3 = 65.26 ton
= 0.158
= 0.139
kN.s2
Equilibrium of mass 1:
-m₁ü1-K₁U1 + k2(U1 + U₂) = 0
m₁ü1 + (K1 + K2)U1 - K2U2 = 0
= 0.0653
mm
kN.s2
mm
kN.s2
mm
k3 = 71.034
k₂
= 89.931
k₁ = 57.081
ZEZE
kN
kN
m
kN
m
kN
mm
kN
mm
k₁= 57080.65
k2 = 89930.68
K3 = 71034.28 =71.034
kN
mm
Equilibrium of mass 3:
-M3Ü3-K3(U3 - U₂) = 0
MÇÜ3 + K3(U3 - U2) = 0
= 57.081
=89.931
KN
mm
KN
mm
KN
mm
3
K₂U₂
-K₂ U₁
U₁
ki
k₁ (U₂-U₁)
k₁U₁
det
[m] =
det
m₂
0
m₂Ü₁
Equilibrium of mass 2:
-m₂Ü2-K₂(U₂-U₁) + K3(U3-U2) = 0
m2Ü2 + k2u2 - k2u1 - K3U3+ k3U2 = 0
m2Ü2 + (k2 + K3)U2 - K2u1 - K3U3 = 0
0
-k₂
0
0
1. Mass and Stiffness Matrices
[m, 0 0 U₁ [k₁ + K₂
0 Ü₂} +
m3] (03
[0.158
0
0
[147.012
0
0
0
0.139
0 0.0653
-89.931
0
[k] =-89.931 160.965 -71.034
0
-71.034 71.034
2. Period and Modal Frequencies
[k₁+k₂ m₁Wn²
-k₂
0
k₂ (U₂-U₁)
U₂-U₁
-89.931
0
[147.012 0.158Wn2
K₂
Wn²2 = 995.06
Wn²3= 2060.32
k₂ (U₂ - U₁)
kN.s²
mm
-k₂
k₂ + k3 k3 U₂ =
-K3
-K₂
k₂ + k3 m₂Wn²
-K3
kz (Uz - U₂)
◄ M₂Ü₂
KN
mm
-89.931
kz (Uz - U₂)
k₂
rad
Uz - U₂
160.965 0.139Wn2
18-8
k₂ (U₂ - U₂)
0
-K3
k3 m3Wn²]
K₂ (Uz - U₂)
-71.034
71.034 0.0653Wn2]
[(147.012 -0.158Wn2) (160.865 -0.139Wn2) (71.034 -0.0653Wn2) + 0
+0]
- [(0) + (147.012 -0.158Wn2)(-71.034) (-71.034) + (71.034
-0.0653Wn2)] = 0
(0.001429Wn6 + 4.544Wn4-3473.084Wn² + 1680931.963)
- (1324.282Wn² + 1316291.575) = 0
-1.429x10-3(Wn²)³ + 4.544(Wn²)2-3473.084Wn² + 364640.388 = 0
Wn² = 124.47
Tn₁ = 0.5630 s
Wn₁ = 11.16
Wn2= 31.54 rad
S
Wn3 = 45.39 rad
0
-71.034
◄
Tn₂ = 0.1992 s
Tn30.1384 s
Mz üz
k₂ (Uz - U₂)
2
4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F268b5e58-a998-4863-bb36-1406cfdadf6e%2Fd0123d9a-ed67-4dbf-93da-593001ffffc2%2Fm08dqvv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:m3 = 65.26 ton
m₂ = 138.68 ton
m₁ = 157.85 ton
m₁ = 157.85 ton
m2 = 138.68 ton
m3 = 65.26 ton
= 0.158
= 0.139
kN.s2
Equilibrium of mass 1:
-m₁ü1-K₁U1 + k2(U1 + U₂) = 0
m₁ü1 + (K1 + K2)U1 - K2U2 = 0
= 0.0653
mm
kN.s2
mm
kN.s2
mm
k3 = 71.034
k₂
= 89.931
k₁ = 57.081
ZEZE
kN
kN
m
kN
m
kN
mm
kN
mm
k₁= 57080.65
k2 = 89930.68
K3 = 71034.28 =71.034
kN
mm
Equilibrium of mass 3:
-M3Ü3-K3(U3 - U₂) = 0
MÇÜ3 + K3(U3 - U2) = 0
= 57.081
=89.931
KN
mm
KN
mm
KN
mm
3
K₂U₂
-K₂ U₁
U₁
ki
k₁ (U₂-U₁)
k₁U₁
det
[m] =
det
m₂
0
m₂Ü₁
Equilibrium of mass 2:
-m₂Ü2-K₂(U₂-U₁) + K3(U3-U2) = 0
m2Ü2 + k2u2 - k2u1 - K3U3+ k3U2 = 0
m2Ü2 + (k2 + K3)U2 - K2u1 - K3U3 = 0
0
-k₂
0
0
1. Mass and Stiffness Matrices
[m, 0 0 U₁ [k₁ + K₂
0 Ü₂} +
m3] (03
[0.158
0
0
[147.012
0
0
0
0.139
0 0.0653
-89.931
0
[k] =-89.931 160.965 -71.034
0
-71.034 71.034
2. Period and Modal Frequencies
[k₁+k₂ m₁Wn²
-k₂
0
k₂ (U₂-U₁)
U₂-U₁
-89.931
0
[147.012 0.158Wn2
K₂
Wn²2 = 995.06
Wn²3= 2060.32
k₂ (U₂ - U₁)
kN.s²
mm
-k₂
k₂ + k3 k3 U₂ =
-K3
-K₂
k₂ + k3 m₂Wn²
-K3
kz (Uz - U₂)
◄ M₂Ü₂
KN
mm
-89.931
kz (Uz - U₂)
k₂
rad
Uz - U₂
160.965 0.139Wn2
18-8
k₂ (U₂ - U₂)
0
-K3
k3 m3Wn²]
K₂ (Uz - U₂)
-71.034
71.034 0.0653Wn2]
[(147.012 -0.158Wn2) (160.865 -0.139Wn2) (71.034 -0.0653Wn2) + 0
+0]
- [(0) + (147.012 -0.158Wn2)(-71.034) (-71.034) + (71.034
-0.0653Wn2)] = 0
(0.001429Wn6 + 4.544Wn4-3473.084Wn² + 1680931.963)
- (1324.282Wn² + 1316291.575) = 0
-1.429x10-3(Wn²)³ + 4.544(Wn²)2-3473.084Wn² + 364640.388 = 0
Wn² = 124.47
Tn₁ = 0.5630 s
Wn₁ = 11.16
Wn2= 31.54 rad
S
Wn3 = 45.39 rad
0
-71.034
◄
Tn₂ = 0.1992 s
Tn30.1384 s
Mz üz
k₂ (Uz - U₂)
2
4
![3. Normalized Mode Shape Factor; a₁ = 1
rad
Wn₁ 11.165
[127.352 -89.931
0
-89.931 143.693 -71.034
0
-71.034 62.906
127.352-89.931 (az) + 0 = 0
az 1.416
Wn₂ = 31.54 rad
s
[-10.013 -89.931
-89.931 23.010
0 -71.034
0
-71.034
6.115
-10.013-89.931(az) + 0 = 0
a2 = -0.111
Wn3 = 45.39 rad
a2 = -1.982
[M] = [+]¹[m][4] =
. Modal Matrix
1
1
1
[] 1.416 -0.111 -1.982
1.598 -1.302 2.215
-178.199 -89.931
a₁
-89.931 -124.751 -71.034a2 = (3) =a₂ = -1.982
0
2.215
-71.034 -63.418]
-178.199-89.931(a2) + 0 = 0
-89.931-124.751(a2) -71.034(a3) = 0
[M]
[K] = [+]¹[k][4] =
[₁][m]1)
[₁][m][₁]
10
Ts =
38-8-8-
Tn1 0.563
Ts 0.56
1.416
1.598 10.158 0
1
1
-0.111 -1.302 0 0.139 0.001416
1.416 -0.111 -1.982
-1.982 2.215
0
0
-1.302 2.215
6. Peak Displacement
Cv = 0.56; Ca = 0.4; 1 = 1.5
C₂ 0.56
= 0.56
2.5Ca 2.5(0.4)
}-{8}
=
{1}=a2=1.416
(a3 (1.598)
-89.931+ 143.693(az) -71.034(a3) = 0
a 1.598
A₁ 14637
D₁ =
124.47
Wn²₁
D₂=A₂ 14715
Wn²₂
995.06
14715
Wn² 2060.32
D3=A₁ =
0
1
1.416 1.598 1[147.012 -89.931
1
1
-0.111 -1.302-89.931 160.965 -71.034 1.416 -0.111 -1.982
11 -1.982 2.215
0 -71.034 71.034 1.598 -1.302 2.215
[74.997 0.120 -0.220
[K] = 0.120 268.845 -0.0499
-0.220 -0.0499 2108.027]
-89.931 +23.010(az) -71.034(a3) = 0
a3 = -1.302
Tn2
Ts
a3 = 2.215
0.603 2.9x10-4 -9.7x10-41
2.9x10-4
0.27 2.6x10-4
-9.7x104 2.6x10-4 1.024
[1 -1.982 2.215]
[0.158
[11.982 2.215] 0
0
[0.158
= 14.788 mm
0.1992
0.56
= 7.142 mm
0
0
1
0
0.139 0
0
0.06531
= 1.01 <1
0.56
0.563
A₁ = =19= (1.5) (9.81) 14.637 = 14637mm
s²
A₂ = 2.5Calg = 2.5(0.4)(1.5)(9.81)= 14.715 = 14715
= 0.356 < 1
• Maximum Floor Displacement
[uo] = [+][r][D]
0.139
0
-1.982
0 0.0653 2.215
A3 = 2.5Calg = 2.5(0.4)(1.5) (9.81)= 14.715=14715
• Maximum Spectral Design
= 117.595 mm
mm
kN²
0.111
Tn3 0.1384
Ts
0.56
0.027
1.024
=
mm
$²
s²
= 0.026
1
89.490
[u₁] = [₁][₁][D₂] = 1.416 [0.761][117.595] = 126.718 mm
1.598]
[143.005)
1
3.179
[u₂] = [₂][₂][D₂] = -0.111 [0.215][14.788] = -0.353 mm
-1.302]
1
-4.140.
0.186
[us] = [3][[₂][D₂] = -1.982 [0.026] [7.142] =-0.368 mm
2.215
0.411.
• Peak Displacement at each level
Uimax=√√(89.49)2 + (3.179)2 + (0.186)² = 89.547 mm
U2max=√√(126.718)² + (-0.353)² + (-0.368)² = 126.719 mm
U3max=√(143.005)² + (-4.14)² + (0.411)² = 143.066 mm
5
= 0.247 < 1
7
[0.158
0
0
4. Draw Mode Shape
Mode 1:
1
(₁) a2 1.416
(a3) (1.598)
0
0.139
0
Mode 2:
(2)=a2=-0.111
(a3)
Mode 3:
(3)=a2
(a3)
{₁}=
5. Participation Factor
[₁ = ₁³(m][1]
[₁]¹[m][:]
[₂=
.302)
-1.982
2.215
[₂] [m][1]
[₂] [m][₂]
=1.416 {$₂2) =
(1.598)
[1
1 1416 1.598] 0
[0.158
[1 1416 1.598] 0
0
-8--8--
1.302)
=
[0.158 0
0.139
0
0.139
0
{Vb} = {f}{1}
[1759.924 2192.375
{Vb} = 499.869 -48.813
[0.158
0
[1 -0.111 -1.302 0 0.139
0
0
1.598
0.158
-0.111-1.302] 0 0.139
0
0
1.416
0
0
0.0653 1
0
1
0
1.416
0.0653 1.598
2215
0.459
0.60 0.761
7. Maximum Equivalent Static Floor Forces
[f] = [m][4][r][A]
[f] =
0
0
0.065311
01
0
0 -0.111
0.0653-1.302
0.058
0.270
-1.982
2.215)
0
1 [0.761 0
1
0 1.416 -0.111 -1.982 0 0.215
0.0653] [1.598 -1.302 2.215 0
0
1₁4
0
0
14715]
= 0.215
[14637 0
0 14715
0
60.449
0
[1759.924
499.869
[f] =2192.375 -48.813 -105.403 kN
55.338
1152.14 -268.982
fimax = √√(1759.924)² + (499.869)² + (60.449)² = 1830.534 KN
f2max=√√(2192.375)² + (−48.813)² + (−105.403)² = 2195.45 KN
f3max = √√(1152.14)² + (−268.982)² + (55.338)² = 1184.416 KN
8. Maximum Base Shear
1152.14
-268.982
60.449 -105.403 55.338
18-1
[5104.799]
= 182.074 kN
10.384
Vomax=√√(5104.799)² + (182.074)² + (10.384)² = 5108.056 kN
6
0
0
0.026]
8](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F268b5e58-a998-4863-bb36-1406cfdadf6e%2Fd0123d9a-ed67-4dbf-93da-593001ffffc2%2Fsobefx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:3. Normalized Mode Shape Factor; a₁ = 1
rad
Wn₁ 11.165
[127.352 -89.931
0
-89.931 143.693 -71.034
0
-71.034 62.906
127.352-89.931 (az) + 0 = 0
az 1.416
Wn₂ = 31.54 rad
s
[-10.013 -89.931
-89.931 23.010
0 -71.034
0
-71.034
6.115
-10.013-89.931(az) + 0 = 0
a2 = -0.111
Wn3 = 45.39 rad
a2 = -1.982
[M] = [+]¹[m][4] =
. Modal Matrix
1
1
1
[] 1.416 -0.111 -1.982
1.598 -1.302 2.215
-178.199 -89.931
a₁
-89.931 -124.751 -71.034a2 = (3) =a₂ = -1.982
0
2.215
-71.034 -63.418]
-178.199-89.931(a2) + 0 = 0
-89.931-124.751(a2) -71.034(a3) = 0
[M]
[K] = [+]¹[k][4] =
[₁][m]1)
[₁][m][₁]
10
Ts =
38-8-8-
Tn1 0.563
Ts 0.56
1.416
1.598 10.158 0
1
1
-0.111 -1.302 0 0.139 0.001416
1.416 -0.111 -1.982
-1.982 2.215
0
0
-1.302 2.215
6. Peak Displacement
Cv = 0.56; Ca = 0.4; 1 = 1.5
C₂ 0.56
= 0.56
2.5Ca 2.5(0.4)
}-{8}
=
{1}=a2=1.416
(a3 (1.598)
-89.931+ 143.693(az) -71.034(a3) = 0
a 1.598
A₁ 14637
D₁ =
124.47
Wn²₁
D₂=A₂ 14715
Wn²₂
995.06
14715
Wn² 2060.32
D3=A₁ =
0
1
1.416 1.598 1[147.012 -89.931
1
1
-0.111 -1.302-89.931 160.965 -71.034 1.416 -0.111 -1.982
11 -1.982 2.215
0 -71.034 71.034 1.598 -1.302 2.215
[74.997 0.120 -0.220
[K] = 0.120 268.845 -0.0499
-0.220 -0.0499 2108.027]
-89.931 +23.010(az) -71.034(a3) = 0
a3 = -1.302
Tn2
Ts
a3 = 2.215
0.603 2.9x10-4 -9.7x10-41
2.9x10-4
0.27 2.6x10-4
-9.7x104 2.6x10-4 1.024
[1 -1.982 2.215]
[0.158
[11.982 2.215] 0
0
[0.158
= 14.788 mm
0.1992
0.56
= 7.142 mm
0
0
1
0
0.139 0
0
0.06531
= 1.01 <1
0.56
0.563
A₁ = =19= (1.5) (9.81) 14.637 = 14637mm
s²
A₂ = 2.5Calg = 2.5(0.4)(1.5)(9.81)= 14.715 = 14715
= 0.356 < 1
• Maximum Floor Displacement
[uo] = [+][r][D]
0.139
0
-1.982
0 0.0653 2.215
A3 = 2.5Calg = 2.5(0.4)(1.5) (9.81)= 14.715=14715
• Maximum Spectral Design
= 117.595 mm
mm
kN²
0.111
Tn3 0.1384
Ts
0.56
0.027
1.024
=
mm
$²
s²
= 0.026
1
89.490
[u₁] = [₁][₁][D₂] = 1.416 [0.761][117.595] = 126.718 mm
1.598]
[143.005)
1
3.179
[u₂] = [₂][₂][D₂] = -0.111 [0.215][14.788] = -0.353 mm
-1.302]
1
-4.140.
0.186
[us] = [3][[₂][D₂] = -1.982 [0.026] [7.142] =-0.368 mm
2.215
0.411.
• Peak Displacement at each level
Uimax=√√(89.49)2 + (3.179)2 + (0.186)² = 89.547 mm
U2max=√√(126.718)² + (-0.353)² + (-0.368)² = 126.719 mm
U3max=√(143.005)² + (-4.14)² + (0.411)² = 143.066 mm
5
= 0.247 < 1
7
[0.158
0
0
4. Draw Mode Shape
Mode 1:
1
(₁) a2 1.416
(a3) (1.598)
0
0.139
0
Mode 2:
(2)=a2=-0.111
(a3)
Mode 3:
(3)=a2
(a3)
{₁}=
5. Participation Factor
[₁ = ₁³(m][1]
[₁]¹[m][:]
[₂=
.302)
-1.982
2.215
[₂] [m][1]
[₂] [m][₂]
=1.416 {$₂2) =
(1.598)
[1
1 1416 1.598] 0
[0.158
[1 1416 1.598] 0
0
-8--8--
1.302)
=
[0.158 0
0.139
0
0.139
0
{Vb} = {f}{1}
[1759.924 2192.375
{Vb} = 499.869 -48.813
[0.158
0
[1 -0.111 -1.302 0 0.139
0
0
1.598
0.158
-0.111-1.302] 0 0.139
0
0
1.416
0
0
0.0653 1
0
1
0
1.416
0.0653 1.598
2215
0.459
0.60 0.761
7. Maximum Equivalent Static Floor Forces
[f] = [m][4][r][A]
[f] =
0
0
0.065311
01
0
0 -0.111
0.0653-1.302
0.058
0.270
-1.982
2.215)
0
1 [0.761 0
1
0 1.416 -0.111 -1.982 0 0.215
0.0653] [1.598 -1.302 2.215 0
0
1₁4
0
0
14715]
= 0.215
[14637 0
0 14715
0
60.449
0
[1759.924
499.869
[f] =2192.375 -48.813 -105.403 kN
55.338
1152.14 -268.982
fimax = √√(1759.924)² + (499.869)² + (60.449)² = 1830.534 KN
f2max=√√(2192.375)² + (−48.813)² + (−105.403)² = 2195.45 KN
f3max = √√(1152.14)² + (−268.982)² + (55.338)² = 1184.416 KN
8. Maximum Base Shear
1152.14
-268.982
60.449 -105.403 55.338
18-1
[5104.799]
= 182.074 kN
10.384
Vomax=√√(5104.799)² + (182.074)² + (10.384)² = 5108.056 kN
6
0
0
0.026]
8
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VIEWStep 2: Formulate mass matrix
VIEWStep 3: Formulate stiffness matrix
VIEWStep 4: Natural frequency of MDOF system
VIEWStep 5: Determine mode shape-1
VIEWStep 6: Determine mode shape-2
VIEWStep 7: Determine mode shape-3
VIEWStep 8: Plot the mode shapes
VIEWStep 9: Determine the participation factors
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