Understanding the concepts on Degrees of Freedom

1.

Given

5 Storey Building with 5DOF and Mass m

The fundamental Period of Vibration T1 = 2 Sec

The fundamental Period of Vibration T2 = 0.685 Sec

The fundamental Period of Vibration T3 = 0.436 Sec

The fundamental Period of Vibration T4 = 0.338 Sec

The fundamental Period of Vibration T5 = 0.296 Sec

To Calculate

• Ln , Mn and n for each of five modes.
• Effective modal mass Mn* and the modal mass participating ration
• Base shear for each mode and combined bas shear.
• How may modes to be considered such that a modal mass participating ratio is 90%

Soln

Mass Matrix m =

Modified Mass matrix M = Ø^T m Ø

For First mode n=1 M =

On solving Modified matrix M = 3.860m

Similarly for mode n = 2 , Modified Matrix M2 = 3.85m

for mode n = 3 , Modified Matrix M3 = 3.85m

for mode n = 4 , Modified Matrix M4 = 3.85m

for mode n = 5 , Modified Matrix M5 = 3.85m

Modal Property

L = Ø^T m i

For first mode n=1

L1 =

L1 = 4.121 m

Similarly for mode n = 2 , Modal Property L2 = -1.297m

for mode n = 3 , Modal Property L3 = 0.683m

for mode n = 4 , Modal Property L4 = -0.381m

for mode n = 5 , Modal Property L5 = 0.175m

Effective Modal Mass M*

M* = Ln²/Mn

For mode n= 1

M*= (4.121m)² / 3.86m

= 4.399m

Similarly for mode n=2, M* = 0.436m

for mode n=3, M* = 0.121

for mode n=4, M* = 0.037

for mode n=5, M* = 0.007

Modal mass Participation Ratio

M = M*/m_j

Where mj is the total mass acing on the structure = m+m+m+m+m

= 5m

For First mode n=1, M = 4.399m/5m

= 0.88

For n=2 M = 0.08

For n=3 M = 0.02

For n=4 M = 0.07

For n=5 M = 0.001

Adding firs two modal mass participation ration e get, 0.88 + 0.08 = 0.96 i.e 96%

Which is higher than 90% mentioned in the question. Hence first two modes are enough.

Base Shear

Vb = Mn* An

For first mode A1 is 0.27g for time period of 2 sec value taken from graph.

Vb1 = 4.398m * 0.27 g

= 1.187mg

Similarly For Vb2 = 0.335mg

For Vb3 = 0.124mg

For Vb4 = 0.038mg

For Vb5 = 0.0072mg

Average Base shear is Calculated by

V_B = Sqroot (Vb1²+Vb2²+Vb3²+Vb4²+Vb5²)

V_B = 1.2mg