Understanding the concepts on Degrees of Freedom

given that :- Different mode shaope values for different Time periods

Need to find:-Mn,LN,`Gamman`,Mn*,Base shear

`phi1=[[``0.334],[0.641],[0.895],[1.078],[1.173]]`

`M1=phi nTm Q`

`=[[0334,0.641,0.895,1.078,1.173]]*m[[0.034],[0.641],[0.895],[1.078],[1.173]]`

`=[154459/4000]m`

M1=3.8641m

similarly

WE KNOW THAT:-m

Five storeys of building:-

Mode	M values
Mode 1	M2=3.85
mode 2	M3=3.8613
mode 3	M4=3.8611
mode 4	M5=3.8612

Now we need to calculate Ln

`Ln=phi nTmL`

`L1=[[0.334,0.641,0.895,1.078,1.173]]*m[[1],[1],[1],[1],[1]]`

L1=4121/1000

l1=4.121

similarly

L2=-1.29

L3=0.68

L4=0.381

L5=0.175

NOW WE NEED TO CALCULATE Mn

```Mnast=(Ln^2/ M^n)`

M1=(4.121^2/3.8614)

M1=4.398

similarly

M*2=0.432

M*3=0.119

M*4=0.0357

`M*5=0.0079`

`Gamman=Ln/M^n`

`Gamma1=L^1/M^1`

=L1/M1=4.121/3.8614

`Gamma1=1.067`

SIMILARLY

`Gamma2=-0.336`

`Gamma3=0.1761`

`Gamma4=0.0986`

`Gamma5=0.045`

`OW WE NEED TO FIND THE BASE SHEAR Vbn=Ln^2* An/Mn`

we know that Pseudo Acceleration,An value from the graph

for mode 1 force vector `f1=(m1)(phi1)(Gamma1)(A1)`

`f1=m[[0.334],[0.641],[0.895],[1.078],[1.173]]*1.067*0.27`

`[[0.096],[0.187],[0.257],[0.3105],[0.335]]`

there fore Vb1=1.121

similarly

Vb2=0.31

Vb3=0.121

Vb4=0.035

Vb5=0.001