Understanding the concepts on Degrees of Freedom

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 (T5 = 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 complete Ln, Mn & γn for each of thr five modes from the given equations.

To calculate the effective modal mass Mn* and the modal mass participating ratio for the 5 nodes.

To obtain the base shear for each of the 5 modes of the vibration and the combined base shear as per SRSS rule.

PROCEDURE:-

Step:-1

Mn = Φ_n^T x m x Φ_n

Model-1:

The first mode of vibration (T₁) = 2 seconds

Φ₁ =`[[0.344],[0.641],[0.895],[1.078],[1.173]]`

M_{1 =} Φ₁^T x m x Φ₁

= [ 0.344 0.641 0.895 1.078 1.173] x `[[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]]` x `[[0.344],[0.641],[0.895],[1.078],[1.173]]`

By using the scientific calculator, we calculate M₁ & we got, M_{1 =} 3.861 m

Model-2:

The first mode of vibration (T₂) = 0.6852 seconds

Φ₂ = `[[- 0.895],[- 1.173],[- 0.641],[0.334],[1.078]]`

M_{2 =} Φ₂^T x m x Φ₂

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

By using the scientific calculator, we calculate M₂ & we got, M_{2 =} 3.86 m

Model-3:

The first mode of vibration (T₃) = 0.4346 seconds

Φ₃ = `[[1.173],[0.334],[- 1.078],[- 0.641],[0.895]]`

M_{3 =} Φ₃^T x m x Φ₃

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

By using the scientific calculator, we calculate M₃ & we got, M_{3 =} 3.85 m

Model-4:

The first mode of vibration (T₄) = 0.3383 seconds

_Φ4 = `[[- 1.078],[0.895],[0.334],[- 1.173],[0.641]]`

M_{4 =} Φ₄^T x m x Φ₄

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

By using the scientific calculator, we calculate M₄ & we got, M_{4 =} 3.861 m

Model-5:

The first mode of vibration (T₅) = 0.2966 seconds

Φ₅ = `[[0.641],[- 1.078],[1.173],[- 0.895],[0.334]]`

M_{5 =} Φ₅^T x m x Φ₅

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

By using the scientific calculator, we calculate M₅ & we got, M_{5 =} 3.861 m

Step:-2

Ln = Φ_n^T x m x l

Model-1:

The first mode of vibration (T₁) = 2 seconds

Φ₁ = `[[0.344],[0.641],[0.895],[1.078],[1.173]]`

L_{1 =} Φ₁^T x m x l

= [ 0.344 0.641 0.895 1.078 1.173] x `[[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]]` x `[[1],[1],[1],[1],[1]]`

By using the scientific calculator, we calculate L₁ & we got, L_{1 =} 4.121 m

Model-2:

The first mode of vibration (T₂) = 0.6852 seconds

Φ₂ = `[[- 0.895],[- 1.173],[- 0.641],[0.334],[1.078]]`

L_{2 =} Φ₂^T x m x l

= [ -0.895 -1.173 -0.641 0.334 1.078] x `[[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]]` x `[[1],[1],[1],[1],[1]]`

By using the scientific calculator, we calculate L₂ & we got, L_{2 =} -1.298 m

Model-3:

The first mode of vibration (T₃) = 0.4346 seconds

Φ₃ = `[[1.173],[0.334],[- 1.078],[- 0.641],[0.895]]`

L_{3 =} Φ₃^T x m x l

= [ 1.173 0.334 -1.078 -0.641 0.895] x`[[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]]` x `[[1],[1],[1],[1],[1]]`

By using the scientific calculator, we calculate L₃ & we got, L_{3 =} 0.687 m

Model-4:

The first mode of vibration (T₄) = 0.3383 seconds

Φ₄ = `[[- 1.078],[0.895],[0.334],[- 1.173],[0.641]]`

L_{4 =} Φ₄^T x m x l

= [ -1.078 0.895 0.334 -1.173 0.641]x`[[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]]` x `[[1],[1],[1],[1],[1]]`

By using the scientific calculator, we calculate L₄ & we got, L_{4 =} -0.381 m

Model-5:

The first mode of vibration (T₅) = 0.2966 seconds

Φ₅ = `[[0.641],[- 1.078],[1.173],[- 0.895],[0.334]]`

L_{5 =} Φ₅^T x m x l

= [ 0.641 - 1.078 1.173 -0.895 0.334] x `[[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]]` x `[[1],[1],[1],[1],[1]]`

By using the scientific calculator, we calculate L₅ & we got, L_{5 =} 0.175 m

Step:-3

T_n = L_n / M_n

Model-1:

• T₁ = L₁ / M₁

= 4.121 m / 3.861 m

= 1.067

Model-2:

• T₂ = L₂ / M₂

= -1.298 m / 3.861 m

= - 0.33

Model-3:

• T₃ = L₃ / M₃

= 0.687 m / 3.861 m

= - 0.178

Model-4:

• T₄ = L₄ / M₄

= - 0.381 m / 3.861 m

= - 0.09

Model-5:

• T₅ = L₅ / M₅

= 0.175 m / 3.861 m

= - 0.04

Step:-4

M_n = L²_n / M_n

Model-1:

• M*₁ = L²₁ / M₁

= L₁ = 4.121m, M₁ = 3.861m

M*₁= (4.121m) ² / 3.861m

M*₁= 4.39m

Model-2:

• M*₂ = L²₂ / M₂

= L₂ = - 0.298m, M₂ = 3.861m

M*₂= (- 0.298m) ² / 3.861m

M*₂= 0.43m

Model-3:

• M*₃ = L²₃ / M₃

= L₃ = 0.687m, M₃ = 3.85m

M*₃= (0.687m) ² / 3.85m

M*₃= 0.12m

Model-4:

• M*₄ = L²₄ / M₄

= L₄ = 0.687m, M₄ = 3.85m

M*₄= (4.121m) ² / 3.85m

M*₄= 0.03m

Model-5:

• M*₅ = L²₅ / M₅

= L₅ = 0.175m, M₅ = 3.861m

M*₅= (0.175m) ² / 3.861m

M*₅= 0.007m

Step:-5

• Modal mass participating ratio = M*n / Mj
• M*_n = Efficitive modal mass.
• M_{j = Total mass = M+M+M+M+M = 5M}

• M*₁ / Mj = 4.39m / 5m = 0.88 = 88%
• M*₂ / Mj = 0.43m / 5m = 0.086 = 8.6%
• M*₃ / Mj = 0.12m / 5m = 0.024 = 2.4%
• M*₄ / Mj = 0.03m / 5m = 0.006 = 0.6%
• M*₅ / Mj = 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.

V_bn = L²_{n /} M_n x A_n = M*_n 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

• V_{b1 =} M*₁ x A₁ = (4.39m x 0.27g) = 2.619 kips
• V_{b2 =} M*₂ x A₂ = (0.43m x 0.76g) = 0.719 kips
• V_{b3 =} M*₃ x A₃ = (0.12m x 1.03g) = 0.272 kips
• V_{b4 =} M*₄ x A₄ = (0.03m x 1.03g) = 0.06 kips
• V_{b5 =} M*₅ x A₅ = (0.007m x 0.76g) = 0.011 kips

Step:-7

Combined base shear as per SRSS rule:

V_{b = `sqrt(V²_"b1" + V²_"b2" + V²_"b3" + V²_"b4" + V²_"b5"`}

_{= `sqrt(2.613² + 0.719² + 0.272² + 0.067² + 0.011²"`}

= 3.20 kips

Step:-8

• The Two modes are to be considered such that a modal mass participating ratio of 90% is obtained.
• (M*₁ / M_j ) + (M*₂ / M_j)
• 88% + 8.6%
• 96.6%

RESULT:

• Calculated Ln, Mn & for each of the five modes from the given equations.
• Calculated the effective modal mass Mn* 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.