Understanding the concepts on Degrees of Freedom

Q] 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?

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)^T m 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)^T m 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)^T m 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)` = 3.86M
• `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)^T m 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)^T m 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)^T m 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)^T m 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(2)(L)` = `underset(2)(Phi)^T m 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)^T m 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)^T m 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)^T m 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

`underset(n)(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)``times``underset(n)(A)` = `underset(n)(M)^**``times``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)^**`x `underset(1)(A)`) = (4.39M x 0.27g) = 2.619 kips

Mode:-2

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

Mode:-3

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

Mode:-4

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

Mode:-5

• `underset(b5)(V)` = (`underset(5)(M)^**`x `underset(5)(A)`) = (0.007M x 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.