Challenge

Understanding the concepts on Degrees of Freedom

General:

Forming the mass matrix (m) as mij = 0 where i≠j & the story masses at the main diagonal where m11 for the first story & m55 for the 5th story.

Mass Matrix (m) =

Applying the following equations of the modal analysis:

Studying the 5 mode shapes of the structure brings the following results:

For Mode 1

* The modal transformed mass matrix (M1) =

= 3.861 m

* The modal property (L1) =

= 4.121 m

* The modal participation factor (ᴦ1) = L1/M1 = 3.861m/4.121m = 1.067

* The modal effective mass (M*1) = (L1)2/M1 = (3.861m)2/4.121m = 4.398 m

* The modal mass participation ratio = M*1/∑m = 4.398m/5m = 0.88

* Enter the response spectrum curve with T1 = 2 sec. to get the pseudo acceleration (An) = 0.27g where g is the gravitational acceleration.

* The base shear (Vb1) = M*1*An = 4.398m*0.27g = 1.187 mg

Using the same manner to calculate the same parameters for the rest of the mode shapes:

For Mode 2

* The modal transformed mass matrix (M2) =

= 3.861 m

* The modal property (L2) =

= -1.297 m

* The modal participation factor (ᴦ2) = L2/M2 = -0.335

* The modal effective mass (M*2) = (L2)2/M2 = 0.436 m

* The modal mass participation ratio = M*2/∑m = 0.09

* Enter the response spectrum curve with T2 = 0.6852 sec. to get pseudo acceleration (An) = 0.76g

* The base shear (Vb2) = 0.327 mg

For Mode 3

* The modal transformed mass matrix (M3) =

= 3.861 m

* The modal property (L3) =

= 0.683 m

* The modal participation factor (ᴦ3) = L3/M3 =0.177

* The modal effective mass (M*3) = (L3)2/M3 = 0.121 m

* The modal mass participation ratio = M*3/∑m = 0.02

* Enter the response spectrum curve with T3 = 0.4346 sec. to get pseudo acceleration (An) = 1.03g

* The base shear (Vb3) = 0.126 mg

For Mode 4

* The modal transformed mass matrix (M4) =

= 3.861 m

* The modal property (L4) =

= -0.381 m

* The modal participation factor (ᴦ4) = L4/M4 =-0.098

* The modal effective mass (M*4) = (L4)2/M4 = 0.038 m

* The modal mass participation ratio = M*4/∑m = 0.01

* Enter the response spectrum curve with T4 = 0.3383 sec. to get pseudo acceleration (An) = 1.03g

* The base shear (Vb4) = 0.039 mg

For Mode 5

* The modal transformed mass matrix (M5) =

= 3.861 m

* The modal property (L5) =

= 0.175 m

* The modal participation factor (ᴦ5) = L5/M5 =0.045

* The modal effective mass (M*5) = (L5)2/M5 = 0.008 m

* The modal mass participation ratio = M*5/∑m = 0.002

* Enter the response spectrum curve with T5 = 0.2966 sec. to get pseudo acceleration (An) = 1.03g

* The base shear (Vb5) = 0.008 mg

Final Results:

* Only the first 2 modes of shapes achieve 0.88 + 0.09 = 0.97 mass participation ratio which is greater than the minimum of 90% as per the code.

* The combined base with the SRSS method (Square Root of the Sum of Squares) =

sqrt [(1.187mg)2+(0.327 mg)2+(0.126 mg)2+(0.039 mg)2+(0.008 mg)2] = 1.15 mg

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