Understanding the concepts on Degrees of Freedom

Question 1:

A five storey structure with 5 translational DOF is shown here. Each storey has mass ‘m’. The eigenvalue problem has been solved and the 5 periods of vibration T_n and their corresponding mode shapes, φ_n has been derived.

The fundamental period of vibration (T₁ = 2 seconds) and its associated mode shape:

The second mode of vibration (T₂ = 0.6852 seconds) and its associated mode shape:

The third mode of vibration (T₃ = 0.4346 seconds) and its associated mode shape:

The fourth mode of vibration (T₄ = 0.3383 seconds) and its associated mode shape:

The fifth mode of vibration (T₄ = 0.2966 seconds) and its associated mode shape:

• L_n , M_n and Γ_n for each of the five modes are to be computed from Equations below.
• Also, calculate the effective modal mass M_n* and the modal mass participating ratio for the 5 modes. The mass matrix will be required to evaluate these parameters.

• Finally, obtain the base shear for each of the 5 modes of vibration and the combined base shear as per SRSS rule. The pseudo acceleration design spectrum to obtain the base shear for each mode is provided below.

• How many modes are to be considered such that a modal mass participating ratio of 90% is obtained?

Ans.

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:-1

`underset(n)(M)=underset(n)(phi)^Tm underset(n)(phi)`

Mode:-1

The first mode of vibration `(underset(1)(T))` = 2 seconds

`underset(1)(phi)=[[0.334],[0.64],[0.895],[1.078],[1.173]]`

`underset(1)(M)=underset(1)(phi)^Tm underset(1)(phi)`

`=[0.334][0.641][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]]`

By using the scientific calculater we calculate `underset(1)(M)` and we got,`underset(1)(M)` = 3.861M

`underset(1)(M)`= 3.861M

Mode:-2

The second mode of vibration (`underset(2)(T)`) = 0.6852 seconds

`underset(2)(phi)=[[-0.895],[-1.173],[-0.641],[0.334],[1.078]]`

`underset(2)(M)=underset(2)(phi)^Tm underset(2)(phi)`

`=[-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]]`

By using the scientific calculater we calculate `underset(2)(M)` and we got,`underset(2)(M)` = 3.861M

`underset(2)(M)`= 3.861M

Mode:-3

The third mode of vibration (`underset(3)(T)`) = 0.4346 seconds

`underset(3)(phi)=[[1.173],[0.334],[-1.078],[-0.641],[0.895]]`

`underset(3)(M)=underset(3)(phi)^Tm underset(3)(phi)`

`=[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]]`

By using the scientific calculater we calculate `underset(3)(M)` and we got,`underset(3)(M)` = 3.85M

`underset(3)(M)`= 3.85M

Mode:-4

The fourth mode of vibration (`underset(4)(T)`) = 0.3383 seconds

`underset(4)(phi)=[[-1.078],[0.895],[0.334],[-1.173],[0.641]]`

`underset(4)(M)=underset(4)(phi)^Tm underset(4)(phi)`

`=[-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]]`

By using the scientific calculater we calculate `underset(4)(M)` and we got,`underset(4)(M)` = 3.861M

`underset(4)(M)`= 3.861M

Mode:-5

The fifth mode of vibration (`underset(5)(T)`) = 0.2966 seconds

`underset(5)(phi)=[[0.641],[-1.078],[1.173],[-0.895],[0.334]]`

`underset(5)(M)=underset(5)(phi)^Tm underset(5)(phi)`

`=[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]]`

By using the scientific calculater we calculate `underset(5)(M)` and we got,`underset(5)(M)` = 3.861M

`underset(5)(M)`= 3.861M

Step:-2

`underset(n)(L)= underset(n)(phi)^T m iota`

Mode:-1

The first mode of vibration (`underset(1)(T)`) = 2 seconds

`underset(1)(phi)=[[0.334],[0.64],[0.895],[1.078],[1.173]]`

`underset(1)(L)=underset(1)(phi)^Tm iota`

`=[0.334][0.641][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]]`

By using the scientific calculater we calculate `underset(1)(L)` and we got,`underset(1)(L)` = 4.121M

`underset(1)(L)`= 4.121M

Mode:-2

The second mode of vibration (`underset(2)(T)`) = 0.6852 seconds

`underset(2)(phi)=[[-0.895],[-1.173],[-0.641],[0.334],[1.078]]`

`underset(1)(L)=underset(1)(phi)^Tm iota`

`=[-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]]`

By using the scientific calculater we calculate `underset(2)(L)` and we got,`underset(2)(L)` = 1.298M

`underset(2)(L)`= 1.298M

Mode:-3

The third mode of vibration (`underset(3)(T)`) = 0.4346 seconds

`underset(3)(phi)=[[1.173],[0.334],[-1.078],[-0.641],[0.895]]`

`underset(3)(L)=underset(3)(phi)^Tm iota`

`=[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]]`

By using the scientific calculater we calculate `underset(3)(L)` and we got,`underset(3)(L)` = 0.687M

`underset(3)(L)`= 0.687M

Mode:-4

The fourth mode of vibration (`underset(4)(T)`) = 0.3383 seconds

`underset(4)(phi)=[[-1.078],[0.895],[0.334],[-1.173],[0.641]]`

`underset(4)(L)=underset(4)(phi)^Tm iota`

`=[-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]]`

By using the scientific calculater we calculate `underset(4)(L)` and we got,`underset(4)(L)` = -0.381M

`underset(4)(L)`= -0.381M

Mode:-5

The fifth mode of vibration (`underset(5)(T)`) = 0.2966 seconds

`underset(5)(phi)=[[0.641],[-1.078],[1.173],[-0.895],[0.334]]`

`underset(5)(L)=underset(5)(phi)^Tm iota`

`=[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]]`

By using the scientific calculater we calculate `underset(5)(L)` and we got,`underset(5)(L)` = 0.175M

`underset(5)(L)`= 0.175M

Step:-3

`Gamma=underset(n)(L)/ underset(n)(M)`

Mode:-1

`underset(1)Gamma =underset(1)(L)/ underset(1)(M)`=`(4.121M)/(3.861M)`=1.067

Mode:-2

`underset(2)Gamma =underset(2)(L)/ underset(2)(M)`=`(-1.298M)/(3.861M)`= -0.33

Mode:-3

`underset(3)Gamma =underset(3)(L)/ underset(3)(M)`=`(0.687M)/(3.85M)`= 0.178

Mode:-4

`underset(4)Gamma =underset(4)(L)/ underset(4)(M)`=`(-0.381M)/(3.861M)`= -0.09

Mode:-5

`underset(5)Gamma =underset(5)(L)/ underset(5)(M)`=`(0.175M)/(3.861M)`= 0.04

Step:-4

`underset(n)(M)^*=(underset(n)(L)^2)/underset(n)(M)`

Mode:-1

`underset(1)(M)^*=(underset(1)(L)^2)/underset(1)(M)`

`underset(1)(L)`= 4.121M `underset(1)(M)`= 3.861M

`underset(1)(M)^*=(4.121M)^2/(3.861M)`

`underset(1)(M)^*`= 4.39M

Mode:-2

`underset(2)(M)^*=(underset(2)(L)^2)/underset(2)(M)`

`underset(2)(L)`= 0.298M `underset(2)(M)`= 3.861M

`underset(2)(M)^*=(-0.298M)^2/(3.861M)`

`underset(2)(M)^*`= 0.43M

Mode:-3

`underset(3)(M)^*=(underset(3)(L)^2)/underset(3)(M)`

`underset(3)(L)`= 0.687M `underset(3)(M)`= 3.85M

`underset(3)(M)^*=(0.687M)^2/(3.85M)`

`underset(3)(M)^*`= 0.12M

Mode:-4

`underset(4)(M)^*=(underset(4)(L)^2)/underset(4)(M)`

`underset(4)(L)`= 0.381M `underset(4)(M)`= 3.861M

`underset(4)(M)^*=(0.381M)^2/(3.861M)`

`underset(4)(M)^*`= 0.03M

Mode:-5

`underset(5)(M)^*=(underset(5)(L)^2)/underset(5)(M)`

`underset(5)(L)`= 0.175M `underset(5)(M)`= 3.861M

`underset(5)(M)^*=(0.175M)^2/(3.861M)`

`underset(5)(M)^*`= 0.007M

Step:-5

Modal mass participating ratio = `(underset(n)(M)^*)/underset(j)(M)`

`underset(n)(M)^*`= Effective modal mass.

`underset(j)(M)`= Total mass = M+M+M+M+M = 5M

Mode:-1

`(underset(1)(M)^*)/underset(j)(M)=(4.39M)/(5M)`= 0.88 = 88%

Mode:-2

`(underset(2)(M)^*)/underset(j)(M)=(0.43M)/(5M)`= 0.086 = 8.6%

Mode:-3

`(underset(3)(M)^*)/underset(j)(M)=(0.12M)/(5M)`= 0.024 = 2.4%

Mode:-4

`(underset(4)(M)^*)/underset(j)(M)=(0.03M)/(5M)`= 0.006 = 0.6%

Mode:-5

`(underset(5)(M)^*)/underset(j)(M)=(0.007M)/(5M)`= 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.

`underset(bn)(V)=(underset(n)(L)^2)/underset(n)(M) ast underset(n)(A)=underset(n)(M)^* ast underset(n)(A)`

`underset(1)(T)`= 2 seconds , `underset(1)(A)`= 0.27g

`underset(2)(T)`= 0.6852 seconds , `underset(2)(A)`= 0.76g

`underset(3)(T)`= 0.4346 seconds , `underset(3)(A)`= 1.03g

`underset(4)(T)`= 0.3383 seconds , `underset(4)(A)`= 1.03g

`underset(5)(T)`= 0.2966 seconds , `underset(5)(A)`= 0.76g

Mode:-1

`underset(b1)(V)=underset(1)(M)^* ast underset(1)(A)`= `4.39M ast 0.27g = 2.619`kips

Mode:-2

`underset(b2)(V)=underset(2)(M)^* ast underset(2)(A)`= `0.43M ast 0.76g = 0.719`kips

Mode:-3

`underset(b3)(V)=underset(3)(M)^* ast underset(3)(A)`= `0.12M ast 1.03g = 0.272`kips

Mode:-4

`underset(b4)(V)=underset(4)(M)^* ast underset(4)(A)`= `0.03M ast 1.03g = 0.067`kips

Mode:-5

`underset(b5)(V)=underset(5)(M)^* ast underset(5)(A)`= `0.007M ast 0.76g = 0.011`kips

Step:-7

Combined base shear as per SRSS rule:-

`underset(b)(V)=sqrt(underset(b1)(V)^2+underset(b2)(V)^2+underset(b3)(V)^2+underset(b4)(V)^2+underset(b5)(V)^2)`

`sqrt((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.

`(underset(1)(M)^*)/(underset(j)(M))+(underset(2)(M)^*)/underset(j)(M)`= 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.