Challenge

Understanding the concepts on Degrees of Freedom

AAmit Semwal17 Feb 2023

HTML

AIM:- To compute L_n , M_n and Γ_n for each of the five modes from the given equations.

To calculate the effectie modal mass M_n* and the modal mass participating ratio for the 5 modes.

To obtain the base shear for each of the 5 modes of vibration and the combined base shear as per SRSS rule.

PROCEDURE:-

Step:-1

Mn=ϕnTmϕn">Mn=ϕnTmϕn

Step:-1

Mn">Mn = ΦnTmΦn">ΦnTmΦnΦ

Mode:-1

The first mode o[0.3340.640.8951.0781.173]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.640.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.3340.640.8951.0781.173]

M1">M11 = Φ1TmΦ1">Φ1TmΦ1Φ1Φ1

= [0.3340.6410.8951.0781.173]">[0.3340.6410.8951.0781.173][0.3340.6410.8951.0781.173] [M00000M00000M00000M00000M]">⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0000x000000x00000x00000] [0.3340.640.8951.0781.173]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.640.8951.0781.173⎤⎥⎥⎦[0.3340.640.8951.0781.173]

By using the scientific calculater we calculate M1">M1x1 and we got, M1">M1x1 = 3.861M
M1">M1x1 = 3.861M

Mode:-2

The second mode of vibration (T2">T2x2) = 0.6852 seconds

Φ2">Φ2Φ2 = [-0.895-1.173-0.6410.3341.078]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.6410.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-0.895-1.173-0.6410.3341.078]
M2">M2x2 = Φ2TmΦ2">Φ2TmΦ2Φ2xΦ2

= [-0.895-1.173-0.641-0.3341.078]">[−0.895−1.173−0.641−0.3341.078][-0.895-1.173-0.641-0.3341.078] [M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[00000x0000x00000x00000] [-0.895-1.173-0.6410.3341.078]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.6410.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-0.895-1.173-0.6410.3341.078]

By using the scientific calculater we calculate M2">M2x2 and we got, M2">M2x2 = 3.86M
M2">M2x2 = 3.861M

Mode:-3

The third mode of vibration (T3">T3x3) = 0.4346 seconds

Φ3">Φ3Φ3 = [1.1730.334-1.078-0.6410.895]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[1.1730.334-1.078-0.6410.895]
M3">M3x3 = Φ3TmΦ3">Φ3TmΦ3Φ3xxΦ3

= [1.1730.334-1.078-0.6410.895]">[1.1730.334−1.078−0.6410.895][1.1730.334-1.078-0.6410.895] [M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[x00000x00000x00000x00000] [1.1730.334-1.078-0.6410.895]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[1.1730.334-1.078-0.6410.895]

By using the scientific calculater we calculate M3">M3x3 and we got, M3">M3 = 3.85M
M3">M3x3 = 3.85M

Mode:-4

The fourth mode of vibration (T4">T4x4) = 0.3383 seconds

Φ4">Φ4Φ4 = [-1.0780.8950.334-1.1730.641]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-1.0780.8950.334-1.1730.641]
M4">M4x4 = Φ4TmΦ4">Φ4TmΦ4Φ4Φ4

= [-1.0780.8950.334-1.1730.641]">[−1.0780.8950.334−1.1730.641][-1.0780.8950.334-1.1730.641] [M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[�00000�00000�00000�00000�] [-1.0780.8950.334-1.1730.641]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-1.0780.8950.334-1.1730.641]

By using the scientific calculater we calculate M4">M4x4 and we got, M4">M4x4 = 3.861M
M4">M4x4 = 3.861M

Mode:-5

The fifth mode of vibration (T5">T5�5) = 0.2966 seconds

Φ5">Φ5Φ5 = [0.641-1.0781.173-0.8950.334]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.641-1.0781.173-0.8950.334]
M5">M5 = Φ5TmΦ5">Φ5TmΦ5Φ5Φ5

= [0.641-1.0781.173-0.8950.334]">[0.641−1.0781.173−0.8950.334][0.641-1.0781.173-0.8950.334] [M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0000000000] [0.641-1.0781.173-0.8950.334]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.641-1.0781.173-0.8950.334]

By using the scientific calculater we calculate M5">M5 and we got, M5">M55 = 3.861M
M5">M5 = 3.861M

Step:-2

Ln">Ln = ΦnT">ΦnTΦm ι">m ι

Mode:-1

The first mode of vibration (T1">T1�1) = 2 seconds

Φ1">Φ1Φ1 = [0.3340.640.8951.0781.173]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.640.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.3340.640.8951.0781.173]
L1">L1�1 = Φ1Tm ι">Φ1Tm ιΦ1

= [0.3340.6410.8951.0781.173]">[0.3340.6410.8951.0781.173][0.3340.6410.8951.0781.173] [M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[00000x00000x00000x00000] [11111]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

By using the scientific calculater we calculate L1">L1 and we got, L1">L1 = 4.121M
L1">L1�1 = 4.121M

Mode:-2

The second mode of vibration (T2">T2) = 0.6852 seconds

Φ2">Φ2Φ2 = [-0.895-1.173-0.6410.3341.078]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.6410.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-0.895-1.173-0.6410.3341.078]
L2">L2 = Φ2Tm ι">Φ2Tm ιΦ2

= [-0.895-1.173-0.641-0.3341.078]">[−0.895−1.173−0.641−0.3341.078][-0.895-1.173-0.641-0.3341.078] [M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥000x00000x00000x00000] [11111]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

By using the scientific calculater we calculate L2">L2�2 and we got, L2">L2�2 = -1.298M
L2">L2�2 = -1.298M

Mode:-3

The third mode of vibration (T3">T3�3) = 0.4346 seconds

Φ3">Φ3Φ3 = [1.1730.334-1.078-0.6410.895]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[1.1730.334-1.078-0.6410.895]
L3">L3�3 = Φ3Tm ι">Φ3Tm ιΦ3��

= [1.1730.334-1.078-0.6410.895]">[1.1730.334−1.078−0.6410.895][1.1730.334-1.078-0.6410.895] [M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[00000x00000x00000x00000] [11111]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

By using the scientific calculater we calculate L3">L3x3 and we got, L3">L3x3 = 0.687M
L3">L3x3 = 0.687M

Mode:-4

The fourth mode of vibration (T4">T4x4) = 0.3383 seconds

Φ4">Φ4Φ4 = [-1.0780.8950.334-1.1730.641]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-1.0780.8950.334-1.1730.641]
L4">L4x4 = Φ4Tm ι">Φ4Tm ιΦ

= [-1.0780.8950.334-1.1730.641]">[−1.0780.8950.334−1.1730.641][-1.0780.8950.334-1.1730.641] [M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[�] [11111]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

By using the scientific calculater we calculate L4">L4x4 and we got, L4">L4x4 = -0.381M
L4">L4x4 = -0.381M

Mode:-5

The fifth mode of vibration (T5">T5x5) = 0.2966 seconds

Φ5">Φ5Φ5 = [0.641-1.0781.173-0.8950.334]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.641-1.0781.173-0.8950.334]
L5">L5x5 = Φ5Tm ι">Φ5Tm ι

= [0.641-1.0781.173-0.8950.334]">[0.641−1.0781.173−0.8950.334][0.641-1.0781.173-0.8950.334] [M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[00000x00000x00000x00000] [11111]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

By using the scientific calculater we calculate L5">L5 and we got, L5">L5 = 0.175M
L5">L5 = 0.175M

Step:-3

Γn">ΓnΓ = LnMn">LnMn

Mode:-1

Γ1">Γ1Γ1 = L1M1">L1M1x1x1 = 4.121M3.861M">4.121M3.861M4.121x3.861 = 1.067

Mode:-2

Γ2">Γ2Γ2 = L2M2">L2M2x2x2 = -1.298M3.861M">−1.298M3.861M-1.298.3.861= -0.33

Mode:-3

Γ3">Γ3Γ3 = L3M3">L3M3x3x3 = 0.687M3.85M">0.687M3.85M0.687x3.85 = 0.178

Mode:-4

Γ4">Γ4Γ4 = L4M4">L4M4x4x4 = -0.381M3.861M">−0.381M3.861M-0.381x3.861 = -0.09

Mode:-5

Γ5">Γ5Γ5 = L5M5">L5M5/5/5 = 0.175M3.861M">0.175M3.861M0.1753.861 = 0.04

Step:-4

Mn∗">Mn∗ = Ln2Mn">Ln2Mnx2

Mode:-1

M1∗">M1 = L12M1">L12M1x12x1
L1">L1 = 4.121M, M1">M1x1 = 3.861M
M1∗">M1∗ = (4.121M)23.861M">(4.121M)23.861M(4.121)23.861
M1∗">M1∗ = 4.39M

Mode:-2

M2∗">M2 = L22M2">L22M2x22x2
L2">L2= -0.298M, M2">M2x2 = 3.861M
M2∗">M2 = (-0.298M)23.861M">(−0.298M)23.861M(-0.298)23.861
M2∗">M2 = 0.43M

Mode:-3

M3∗">M3∗ = L32M3">L32M3x32x3
L3">L3 = 0.687M, M3">M3x3 = 3.85M
M3∗">M3∗ = (0.687M)23.85M">(0.687M)23.85M(0.687)23.85
M3∗">M3∗ = 0.12M

Mode:-4

M4∗">M4∗4∗ = L42M4">L42M4x42x4
L4">L44 = -0.381M, M4">M4x4 = 3.861M
M4∗">M4∗4∗ = (-0.381M)23.861M">(−0.381M)23.861M(-0.381)23.861
M4∗">M4∗4∗ = 0.03M

Mode:-5

M5∗">M5∗x5∗ = L52M5">L52M5/52/5
L5">L5x5 = 0.175M, M5">M55 = 3.861M
M5∗">M5∗x5∗ = (0.175M)23.861M">(0.175M)23.861M(0.175)23.861
M5∗">M5∗x5∗ = 0.007M

Step:-5

Modal mass participating ratio = Mn∗Mj">Mn∗Mj

Mn∗">Mn∗ = Effective modal mass.

Mj">Mj = Total mass = M+M+M+M+M = 5M

Mode:-1

M1∗Mj">M1∗Mj/1∗= 4.39M5M">4.39M5M4.39/5 = 0.88 = 88%

Mode:-2

M2∗Mj">M2∗Mj.2 = 0.43M5M">0.43M5M0.43/5 = 0.086 = 8.6%

Mode:-3

M3∗Mj">M3∗Mj3∗ = 0.12M5M">0.12M5M0.12/5= 0.024 = 2.4%

Mode:-4

M4∗Mj">M4∗Mj/4 = 0.03M5M">0.03M5M0.03x5= 0.006 = 0.6%

Mode:-5

M5∗Mj">M5∗Mj/5∗ = 0.007M5M">0.007M5M0.007/5 = 0.001 = 0.1%

Step:-6

Now we need to obtain the base shear for each of the 5 modes of vibration, as shown below.

Vbn">Vbn = Ln2Mn">Ln2Mn2×">××An">An = Mn∗">Mn∗×">××An">Ans

(T1">T1) = 2 seconds, A1">A1 = 0.27g

(T2">T2) = 0.6852 seconds, A2">A2 = 0.76g

(T3">T3) = 0.4346 seconds, A3">A3 = 1.03g

(T4">T4) = 0.3383 seconds, A4">A4 = 1.03g

(T5">T5) = 0.2966 seconds, A5">A5 = 0.76g

Mode:-1

Vb1">Vb1 = (M1∗">M1∗1∗x A1">A11) = (4.39M x 0.27g) = 2.619 kips

Mode:-2

Vb2">Vb2/2 = (M2∗">M2∗/2∗x A2">A2/2) = (0.43M x 0.76g) = 0.719 kips

Mode:-3

Vb3">Vb3/3 = (M3∗">M3∗/3∗x A3">A3/3) = (0.12M x 1.03g) = 0.272 kips

Mode:-4

Vb4">Vb4/4 = (M4∗">M4∗/4∗x A4">A4/4) = (0.03M x 1.03g) = 0.067 kips

Mode:-5

Vb5">Vb5/5 = (M5∗">M5∗/5∗x A5">A5/5) = (0.007M x 0.76g) = 0.011 kips

Step:-7

Combined base shear as per SRSS rule:-

Vb">Vb = Vb12+Vb22+Vb32+Vb42+Vb52">√Vb12+Vb22+Vb32+Vb42+Vb5212+22+32+42+52

= (2.613)2+(0.719)2+(0.272)2+(0.067)2+(0.011)2">√(2.613)2+(0.719)2+(0.272)2+(0.067)2+(0.011)2(2.613)2+(0.719)2+(0.272)2+(0.067)2+(0.011)2

= 3.20 kips

Step:-8

The two modes are to be considered such that a modal mass participating ratio of 90% is obtained

Mode:-

RESULT:-

- Calculated L_n , M_n and Γ_n for each of the five modes from the given equations.

- Calculated the effective modal mass M_n* and the modal mass participating ratio for the 5 modes.

- The base shear for each of the 5 modes of vibration are obtained.

- As per SRSS rule the combined base shear is obtained.

- The two modes are to be considered such that a modal mass participating ratio of 90% is obtained.

Understanding the concepts on Degrees of Freedom

Ship work like this
in your own course.

Talk to a career counsellor

Ship work like thisin your own course.

Talk to a career counsellor

Ship work like this
in your own course.