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 =}`[[0.334],[0.64],[0.895],[1.078],[1.173]]`

M₁ = ɸ₁^Tmɸ₁

_{= `[[0.334,0.64,0.895,1.078,1.173]]``[[M,0,0,0,0],[0,M,0,0,0],[0,0,M,0,0],[0,0,0,M,0],[0,0,0,0,M]]``[[0.334],[0.64],[0.895],[1.078],[1.173]]`}

M₁ = 3.861 M

Mode:-2

The second mode of vibration (T₂) = 0.6852 Seconds

ɸ₂ =`[[-0.895],[-1.173],[-0.64],[0.334],[1.078]]`

M₂ = ɸ₂^Tmɸ₂

_{= `[[-0.895,-1.173,-0.641,0.334,1.078]]``[[M,0,0,0,0],[0,M,0,0,0],[0,0,M,0,0],[0,0,0,M,0],[0,0,0,0,M]]``[[-0.895],[-1.173],[-0.641],[0.334],[1.078]]`}

M₂ = 3.861 M

Mode:-3

The third mode of vibration (T₃) = 0.4346 Seconds

ɸ_{3 =}`[[1.173],[0.334],[-1.078],[-0.641],[0.895]]`

M₃ = ɸ₃^Tmɸ₃

_{= `[[1.173,0.334,-1.078,-0.641,0.895]]``[[M,0,0,0,0],[0,M,0,0,0],[0,0,M,0,0],[0,0,0,M,0],[0,0,0,0,M]]``[[1.173],[0.334],[-1.078],[-0.641],[0.895]]`}

M₃ = 3.85 M

Mode:-4

The fourth mode of vibration (T₄) = 0.3383 Seconds

ɸ₄ =`[[-1.078],[0.895],[0.334],[-1.173],[0.641]]`

M₄ = ɸ₄^Tmɸ₄

_{= `[[-1.078,0.895,0.334,-1.173,0.641]]``[[M,0,0,0,0],[0,M,0,0,0],[0,0,M,0,0],[0,0,0,M,0],[0,0,0,0,M]]``[[-1.078],[0.895],[0.334],[-1.173],[0.641]]`}

M₄ = 3.861 M

Mode:-5

The fifth mode of vibration (T₅) = 0.2966 Seconds

ɸ₅ =`[[0.641],[-1.078],[1.173],[-0.895],[0.334]]`

M₅ = ɸ₅^Tmɸ₅

_{= `[[0.641,-1.078,1.173,-0.895,0.334]]``[[M,0,0,0,0],[0,M,0,0,0],[0,0,M,0,0],[0,0,0,M,0],[0,0,0,0,M]]``[[0.641],[-1.078],[1.173],[-0.895],[0.334]]`}

M₅ = 3.861 M

Step:-2

L_n = ɸ_n^Tmi

Mode:-1

The first mode of vibration (T₁) = 2 Seconds

ɸ_{1 =}`[[0.334],[0.64],[0.895],[1.078],[1.173]]`

L₁ = ɸ₁^TmI

_{= `[[0.334,0.64,0.895,1.078,1.173]]``[[M,0,0,0,0],[0,M,0,0,0],[0,0,M,0,0],[0,0,0,M,0],[0,0,0,0,M]]``[[1],[1],[1],[1],[1]]`}

L₁ = 4.121 M

Mode:-2

The second mode of vibration (T₂) = 0.6852 Seconds

ɸ₂ =`[[-0.895],[-1.173],[-0.64],[0.334],[1.078]]`

L₂ = ɸ₂^TmI

_{= `[[-0.895,-1.173,-0.641,0.334,1.078]]``[[M,0,0,0,0],[0,M,0,0,0],[0,0,M,0,0],[0,0,0,M,0],[0,0,0,0,M]]``[[1],[1],[1],[1],[1]]`}

L₂ = -1.298 M

Mode:-3

The third mode of vibration (T₃) = 0.4346 Seconds

ɸ_{3 =}`[[1.173],[0.334],[-1.078],[-0.641],[0.895]]`

L₃ = ɸ₃^TmI

_{= `[[1.173,0.334,-1.078,-0.641,0.895]]``[[M,0,0,0,0],[0,M,0,0,0],[0,0,M,0,0],[0,0,0,M,0],[0,0,0,0,M]]``[[1],[1],[1],[1],[1]]`}

L₃ = 0.687 M

Mode:-4

The fourth mode of vibration (T₄) = 0.3383 Seconds

ɸ₄ =`[[-1.078],[0.895],[0.334],[-1.173],[0.641]]`

L₄ = ɸ₄^TmI

_{= `[[-1.078,0.895,0.334,-1.173,0.641]]``[[M,0,0,0,0],[0,M,0,0,0],[0,0,M,0,0],[0,0,0,M,0],[0,0,0,0,M]]``[[1],[1],[1],[1],[1]]`}

L₄ = -0.381 M

Mode:-5

The fifth mode of vibration (T₅) = 0.2966 Seconds

ɸ₅ =`[[0.641],[-1.078],[1.173],[-0.895],[0.334]]`

L₅ = ɸ₅^TmI

_{= `[[0.641,-1.078,1.173,-0.895,0.334]]``[[M,0,0,0,0],[0,M,0,0,0],[0,0,M,0,0],[0,0,0,M,0],[0,0,0,0,M]]``[[1],[1],[1],[1],[1]]`}

L₅ = 0.175 M

Step:-3

Γ_n =`frac{Ln}{Mn}`

Mode:-1

Γ₁ =`frac{L1}{M1}`

= `frac{4.121M}{3.861M}`

= 1.067

Mode:-2

Γ₂ =`frac{L2}{M2}`

= `frac{-1.298M}{3.861M}`

= -0.33

Mode:-3

Γ₃ =`frac{L3}{M3}`

= `frac{0.687M}{3.85M}`

= 0.178

Mode:-4

Γ₄ =`frac{L4}{M4}`

= `frac{-0.381M}{3.861M}`

= -0.09

Mode:-5

Γ₅ =`frac{L5}{M5}`

= `frac{0.175M}{3.861M}`

= 0.04

Step:-4

M_n^* =`frac{Ln^2}{Mn}`

Mode:-1

_M1^* =`frac{L1^2}{M1}`

= `frac{(4.121M)^2}{3.861M}`

= 4.398M

Mode:-2

_M2^* =`frac{L2^2}{M2}`

= `frac{(-1.298M)^2}{3.861M}`

= 0.436M

Mode:-3

_M3^* =`frac{L3^2}{M3}`

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

= 0.12M

Mode:-4

_M4^* =`frac{L4^2}{M4}`

= `frac{(-0.381M)^2}{3.861M}`

= 0.03M

Mode:-5

_M5^* =`frac{L5^2}{M5}`

= `frac{(0.175M)^2}{3.861M}`

= 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 =`frac{Ln^2}{Mn}` 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.