Understanding the concepts on Degrees of Freedom

1. Mass matrix (m) =

2. Modified mass matrix (M) =

For the first mode (n=1) =

Similarly M2 (n=2)

Similarly M3 (n=3)

3. Modal property (L) = Ø_n ^T.m.l

For the first mode (n=1) =

Similarly

For (n=2)

For (N=2)

= -0.895-1.173-0.641+0.334+1.078

= Ans – (-1.297)

For (n=3)

= 1.173+0.334-1.078-0.641+0.895

= Ans – (0.683).

For (n=4)

= -1.078+0.895+0.334-1.173+0.641

= Ans – (-0.381).

For (n=5)

= 0.641-1.078+1.173-0.895+0.334

= Ans – (0.175).

4. 4. Modal participation factor ( Γ ) = L/M

For the first mode, Γ = L1/M1

Γ1 = 4.121/3.864

Γ1 = 1.067

Similarly

Γ2 = -1.297/3.86

Γ2 = -0.336

Γ3 = 0.683/3.862

Γ3 = 0.177

Γ4 = -0.381/3.86

Γ4 = -0.098

Γ5 = 0.175/3.86

Γ5 = 0.045

5. Effective modal mass (M*) =L_n²/M_n

For the first mode (n=1), M1* =L₁²/M₁

_{M1* =}(4.121m)² /3.864

_{M1* =4.395m}

_Similarly

For (n=2)

M2*= (-1.297m)² /3.86

M2* = 0.436m

For (n=3)

M3* = (0.683m)² /3.862

M3* = 0.121m

For (n=4)

M4* = (-0.381m)² /3.862

M4* = 0.038m

For (n=5)

M5*= (0.175m)² /3.86

M5* = 0.008m

6. Modal mass participation ratio = M₁ /mj

mj = m+m+m+m+m = 5m

Modal mass participation ratio of first mode (n=1) = 4.395/5 = 0.88

Similarly for Second mode (n=2) = 0.436/ 5 = 0.09

Third mode (n=3) = 0.121/ 5 = 0.024

Fourth mode (n=4) = 0.038/ 5 = 0.01

Fifth mode (n=5) = 0.008/ 5 = 0.002

Adding first and second mode ie = 0.88+0.09 = 0.97

In order to achieve minimum 90% modal mass participation ration , First and second modes are sufficient.

7. Base shear (V_{b ) =} M_n *. Aj

For the first mode, VB1 = 4.395mx 0.27g

(n=1) = 1.187 mg

Similarly for VB2 = 0.436mx0.75g

= 0.327mg

Similarly for VB3 = 0.121mx 1.04g

= 0.126mg

Similarly for VB4 = 0.038 mx 1.01g

= 0.039mg

Similarly for VB5 = 0.008mx 1.00g

= 0.008mg

8. Average base shear is calculated by the method of SRSS

√(Vb1² + Vb2² + Vb3² + Vb4² + Vb5² )

√(1.187² + 0.327² + 0.126² + 0.039² + 0.008² )

√(1.409 + 0.107 + 0.016 + 0.002 + 0.0001 )

√(1.5341)

Vb = 1.239mg