Understanding the concepts on Degrees of Freedom

Step:

Mode:-1

M_n = ɸ_n^Tmɸ_n

The first mode of vibration (T₁) = 2 Seconds

M₁ = ɸ₁^Tmɸ₁

_{M1 = 3.861 M}

Mode:-2

The second mode of vibration (T₂) = 0.6852 Seconds

M₂ = ɸ₂^Tmɸ₂

M₂ = 3.861 M

Mode:-3

The third mode of vibration (T₃) = 0.4346 Seconds

M₃ = ɸ₃^Tmɸ₃

_{M3 = 3.85 M}

Mode:-4

The fourth mode of vibration (T₄) = 0.3383 Seconds

M₄ = ɸ₄^Tmɸ₄

M₄ = 3.861 M

Mode:-5

The fifth mode of vibration (T₅) = 0.2966 Seconds

M₅ = ɸ₅^Tmɸ₅

_{M5 = 3.861 M}

Step:-2

L_n = ɸ_n^Tmi

Mode:-1

The first mode of vibration (T₁) = 2 Seconds

L₁ = ɸ₁^TmI

L₁ = 4.121 M

Mode:-2

The second mode of vibration (T₂) = 0.6852 Seconds

L₂ = ɸ₂^TmI

L₂ = -1.298 M

Mode:-3

The third mode of vibration (T₃) = 0.4346 Seconds

L₃ = ɸ₃^TmI

L₃ = 0.687 M

Mode:-4

The fourth mode of vibration (T₄) = 0.3383 Seconds

L₄ = ɸ₄^TmI

L₄ = -0.381 M

Mode:-5

The fifth mode of vibration (T₅) = 0.2966 Second

L₅ = ɸ₅^TmI

L₅ = 0.175 M

Step:-3

Γ_n =lnMnlnMn

Mode:-1

Γ₁ =L1M1L1M1

= 4.121M3.861M4.121M3.861M

= 1.067

Mode:-2

Γ₂ =L2M2L2M2

= −1.298M3.861M-1.298M3.861M

= -0.33

Mode:-3

Γ₃ =L3M3L3M3

= 0.687M3.85M0.687M3.85M

= 0.178

Mode:-4

Γ₄ =L4M4L4M4

= −0.381M3.861M-0.381M3.861M

= -0.09

Mode:-5

Γ₅ =L5M5L5M5

= 0.175M3.861M0.175M3.861M

= 0.04

Step:-4

M_n^* =ln2Mnln2Mn

Mode:-1

_M1^* =L12M1L12M1

= (4.121M)23.861M(4.121M)23.861M

= 4.398M

Mode:-2

_M2^* =L22M2L22M2

= (−1.298M)23.861M(-1.298M)23.861M

= 0.436M

Mode:-3

_M3^* =L32M3L32M3

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

= 0.12M

Mode:-4

_M4^* =L42M4L42M4

= (−0.381M)23.861M(-0.381M)23.861M

= 0.03M

Mode:-5

_M5^* =L52M5L52M5

= (0.175M)23.861M(0.175M)23.861M

= 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 =ln2Mnln2Mn 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.