Understanding the concepts on Degrees of Freedom

AIM:- 

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

            To calculate the effective 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:

Mode:-1

M_n = ɸ_n^Tmɸ_n

The first mode of vibration (T₁) = 2 Seconds

ɸ_{1 =}[0.3340.640.8951.0781.173]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.640.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.3340.640.8951.0781.173]

M₁ = ɸ₁^Tmɸ₁

      _{= [0.3340.640.8951.0781.173]">[0.3340.640.8951.0781.173][0.3340.640.8951.0781.173][M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[�00000�00000�00000�00000�][0.3340.640.8951.0781.173]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.640.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.3340.640.8951.0781.173]}

M₁ = 3.861 M

Mode:-2

The second mode of vibration (T₂) = 0.6852 Seconds 

ɸ₂ =[-0.895-1.173-0.640.3341.078]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.640.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-0.895-1.173-0.640.3341.078]

M₂ = ɸ₂^Tmɸ₂

      _{= [-0.895-1.173-0.6410.3341.078]">[−0.895−1.173−0.6410.3341.078][-0.895-1.173-0.6410.3341.078][M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[�00000�00000�00000�00000�][-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 = 3.861 M

Mode:-3

The third mode of vibration (T₃) = 0.4346 Seconds

ɸ_{3 =}[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 = ɸ₃^Tmɸ₃

      _{= [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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[�00000�00000�00000�00000�][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 = 3.85 M

Mode:-4

The fourth mode of vibration (T₄) = 0.3383 Seconds

ɸ₄ =[-1.0780.8950.334-1.1730.641]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-1.0780.8950.334-1.1730.641]

M₄ = ɸ₄^Tmɸ₄

      _{= [-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]}

M₄ = 3.861 M

Mode:-5

The fifth mode of vibration (T₅) = 0.2966 Seconds

ɸ₅ =[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 = ɸ₅^Tmɸ₅

      _{= [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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[�00000�00000�00000�00000�][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 = 3.861 M

Step:-2

L_n = ɸ_n^Tmi

Mode:-1

The first mode of vibration (T₁) = 2 Seconds

ɸ_{1 =}[0.3340.640.8951.0781.173]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.640.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.3340.640.8951.0781.173]

L₁ = ɸ₁^TmI

      _{= [0.3340.640.8951.0781.173]">[0.3340.640.8951.0781.173][0.3340.640.8951.0781.173][M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[�00000�00000�00000�00000�][11111]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]}

L₁ = 4.121 M

Mode:-2

The second mode of vibration (T₂) = 0.6852 Seconds 

ɸ₂ =[-0.895-1.173-0.640.3341.078]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.640.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-0.895-1.173-0.640.3341.078]

L₂ = ɸ₂^TmI

      _{= [-0.895-1.173-0.6410.3341.078]">[−0.895−1.173−0.6410.3341.078][-0.895-1.173-0.6410.3341.078][M00000M00000M00000M00000M]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[�00000�00000�00000�00000�][11111]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]}

L₂ = -1.298 M

Mode:-3

The third mode of vibration (T₃) = 0.4346 Seconds

ɸ_{3 =}[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 = ɸ₃^TmI

      _{= [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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[�00000�00000�00000�00000�][11111]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]}

L₃ = 0.687 M

Mode:-4

The fourth mode of vibration (T₄) = 0.3383 Seconds

ɸ₄ =[-1.0780.8950.334-1.1730.641]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-1.0780.8950.334-1.1730.641]

L₄ = ɸ₄^TmI

      _{= [-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�][11111]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]}

L₄ = -0.381 M

Mode:-5

The fifth mode of vibration (T₅) = 0.2966 Seconds

ɸ₅ =[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 = ɸ₅^TmI

      _{= [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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[�00000�00000�00000�00000�][11111]">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]}

L₅ = 0.175 M

Step:-3

Γ_n =lnMn">lnMnln��

Mode:-1

Γ₁ =L1M1">L1M1�1�1

    = 4.121M3.861M">4.121M3.861M4.121�3.861�

    = 1.067

Mode:-2

Γ₂ =L2M2">L2M2�2�2

    = -1.298M3.861M">−1.298M3.861M-1.298�3.861�

    = -0.33

Mode:-3

Γ₃ =L3M3">L3M3�3�3

    = 0.687M3.85M">0.687M3.85M0.687�3.85�

    = 0.178

Mode:-4

Γ₄ =L4M4">L4M4�4�4

    = -0.381M3.861M">−0.381M3.861M-0.381�3.861�

    = -0.09

Mode:-5

Γ₅ =L5M5">L5M5�5�5

    = 0.175M3.861M">0.175M3.861M0.175�3.861�

    = 0.04

Step:-4

M_n^* =ln2Mn">ln2Mnln2��

Mode:-1

_M1^*  =L12M1">L12M1�12�1

      = (4.121M)23.861M">(4.121M)23.861M(4.121�)23.861�

      =  4.398M

Mode:-2

_M2^*  =L22M2">L22M2�22�2

       = (-1.298M)23.861M">(−1.298M)23.861M(-1.298�)23.861�

       = 0.436M

Mode:-3

_M3^*  =L32M3">L32M3�32�3

nbsp;   =">nbsp;   =����;   =frac{(0.687M)^2}{3.85M}`

       = 0.12M

Mode:-4

_M4^*  =L42M4">L42M4�42�4

       = (-0.381M)23.861M">(−0.381M)23.861M(-0.381�)23.861�

       = 0.03M

Mode:-5

_M5^*  =L52M5">L52M5�52�5

       = (0.175M)23.861M">(0.175M)23.861M(0.175�)23.861�

       = 0.007M

Step:-5

Modal mass participating ratio = M_n^* /M_j

  M_n^*    = Effective modal mass.

       M_j = Total mass = M+M+M+M+M = 5M

         

Mode:-1

• M₁^* /M_j = 0.88 = 88%

Mode:-2

• M₂^* /M_j = 0.086 = 8.6%

Mode:-3

• M₃^* /M_j = 0.024 = 2.4%

Mode:-4

• M₄^* /M_j = 0.006 = 0.6%

Mode:-5

• M₅^* /M_j = 0.001 = 0.1%

Step:-6

Now we need to obtain the base shear for each of the 5 modes of vibration.

V_bn =ln2Mn">ln2Mnln2�� X A_n

(T₁ ) = 2 seconds,  A₁= 0.27g

(T₂ ) = 0.6852 seconds, A₂ = 0.76g

(T₃ ) = 0.4346 seconds, A₃ = 1.03g

(T₄ ) = 0.3383 seconds, A₄ = 1.03g

(T₅ ) = 0.2966 seconds, A₅ = 0.76g

Mode:-1

•  V_b1= (M₁^* x A₁) = (4.39M x 0.27g) = 2.613 kips

Mode:-2

• V_b2 = (M₂^* x A₂) = (0.43M x 0.76g) = 0.719 kips

Mode:-3

• V_b3 = (M₃^* x A₃) = (0.12M x 1.03g) = 0.272 kips

Mode:-4

• V_b4 = (M₄^* x A₄) = (0.03M x 1.03g) = 0.067 kips

Mode:-5

• V_b5 = (M₅^* x A₅) = (0.007M x 0.76g) = 0.011 kips

Step:-7

Combined base shear as per SRSS rule:-

               =√((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.
•  = 88% + 8.6% = 96.6%

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.