Understanding the concepts on Degrees of Freedom

AIM:- To obtain Ln, Mn and Γn for each of the five modes associated wih the model, calculate the effective modal mass M*n and the modal mass participating ratios for the five modes and also obtain the base shear as per SRSS rule.

INTRODUCTION- All the solutions are going to be explained with step by step procedure.

SOLUTION:- 

1.1 Calculation of Ln, Mn and Γn for each modes, effective modal mass and modal mass participating ratios

a) For first mode

^ΦT_{1 = [ 0.334  0.641  0.895 1.078 1.173]}

_{Mass matrix , m = [ [ m , 0 , 0,  0,  0 ] , [ 0 , m , 0 , 0, 0], [0,0,m,0,0], [ 0, 0, 0,0,m ,0 ], [ 0 , 0, 0, 0, m] ]}

 M₁ = Φ^T₁ * m * Φ₁ = 3.861 m

L₁ = Φ^T₁ * m * I = [ 0.334, 0.641, 0.895, 1.078, 1.173 ] * m * [ [1], [1], [1], [1], [1] ] = 4. 121 m

Γ₁ = L₁ / M₁ = 4.121 m / 3.861 m = 1.067

M^*₁ = L²₁/ M_{1 = 4.397 m}

Modal mass participation ratios for first mode is given by = (4.397 m/ 5 m) * 100 = 87.94%

b) For Second mode 

_{Mass matrix , m = [ [ m , 0 , 0,  0,  0 ] , [ 0 , m , 0 , 0, 0], [0,0,m,0,0], [ 0, 0, 0,0,m ,0 ], [ 0 , 0, 0, 0, m] ]}

 M₂ = Φ^T₂ * m * Φ₂ = 3.861 m

 L₂ = Φ^T₂ * m * I = - 1.3 m

Γ₂ = L₂ / M₂ = -1.3 m/ 3.861 m = -0.336 m

M^*₂ = L²₂/ M₂ = (1.3m)^2/3.861 m = 0.437 m

Modal mass participation ratios for second mode is given by = (0.437 m/ 5 m) * 100 = 8.74 %

c) For third mode 

_{Mass matrix , m = [ [ m , 0 , 0,  0,  0 ] , [ 0 , m , 0 , 0, 0], [0,0,m,0,0], [ 0, 0, 0,0,m ,0 ], [ 0 , 0, 0, 0, m] ]}

M₃ = Φ^T₃ * m * Φ₃ = 3.861 m

L₃ = Φ^T₃ * m * I = [ 1.173, 0,334 , -1.078, -0.641, 0.895 ] * m * [ [1], [1], [1], [1], [1] ] = 0.683 m

Γ₃ = L₃ / M₃ = 0.683 m/ 3.861 m = 0.177 m

M^{* 3} = L²₃/ M₃ = (0.683m)^2/3.861 m = 0.12 m

Modal mass participation ratios for third mode is given by = (0.12 m/ 5 m) * 100 = 2.4 %

d) For fourth mode 

_{Mass matrix , m = [ [ m , 0 , 0,  0,  0 ] , [ 0 , m , 0 , 0, 0], [0,0,m,0,0], [ 0, 0, 0,0,m ,0 ], [ 0 , 0, 0, 0, m] ]}

M₄ = Φ^T₄ * m * Φ₄ = 3.864 m

L₄ = Φ^T₄ * m * I = [-1.078, 0.895, 0.334, -1.173, 0.641] * m * [ [1], [1], [1], [1], [1] ] = -0.381 m

Γ₄ = L₄ / M₄ = -0.381 m/ 3.864 m = -0.098  m

M^*₄ = L²₄/ M₄ = (0.381m)^2/3.861 m = 0.037 m

Modal mass participation ratios for fourth mode is given by = (0.037 m/ 5 m) * 100 = 0.74 %

e) For fifth mode 

_{Mass matrix , m = [ [ m , 0 , 0,  0,  0 ] , [ 0 , m , 0 , 0, 0], [0,0,m,0,0], [ 0, 0, 0,0,m ,0 ], [ 0 , 0, 0, 0, m] ]}

 M₅ = Φ^T₅ * m * Φ₅ = 3.861 m

L₅ = Φ^T₅ * m * I = [0.641, -1.078, 1.173, -0.895, 0.334] * m * [ [1], [1], [1], [1], [1] ] = 0.175 m

Γ₅ = L₅ / M₅ = 0.175  m/ 3.864 m =  0.045  m

M^*₅ = L²₅/ M₅ = (0.175 m)^2/3.861 m =  7.931 * 10^-3  m

Modal mass participation ratios for fifth mode is given by = ( 7.931 * 10^-3  m/ 5 m) * 100 = 0.15 %

1.2 Calculation of base shear forces

 V_b¹ = M^*₁ * A₁ = 4.397 m * 0.27g ( T₁ = 2s) = 1.187 m*g =2.658 kips

 V_b2 = M^*₂ * A₂ = 0.437 m * 0.7g ( T₂ = 0.6852 s) = 0.3059 m*g =0.685 kips

 V_b3 = M^*₃* A₃ = 0.12 m * 1.039 g ( T₃ = 0.4346 s) = 0.279 kips

 V_b4 = M^*₄* A₄ = 0.037 m * 1.039 g ( T4 = 0.3383 s) = 0.085 kips

V_b5 = M^*₅* A₅ = 7.931 * 10^-3 m * 0.76 g ( T₅ = 0.2966 s) = 0.011 kips

1.3 Combined base shear force as per SRSS rule

V_b = Sq rt ( 2.658^2 + 0.685^2+ 0.279^2 + 0.085 ^2 + 0.011^2 ) = 3 kips

1.4 Modal mass participating ratios

Modal mass ratio is obtained in 2 model = (8.74+ 87.94 ) % = 96.68%

RESULT:-  a) Ln, Mn and Γn for each of the five modes associated wih the model have been illustrated properly.

               b) Effective modal mass have been illustrated properly 

               c)  modal mass participating ratios for the five modes are also been obtained the base shear as per SRSS rule.