Understanding the concepts on Degrees of Freedom

`M_1=phi_1^Tmphi_1`

`M_1=[(0.334,0.641,0.895,1.078,1.173)][(m,0,0,0,0),(0,m,0,0,0),(0,0,m,0,0),(0,0,0,m,0),(0,0,0,0,m)][(0.334),(0.641),(0.895),(1.078),(1.173)]`

`M_1=[0.334m,0.641m,0.895m,1.078m,1.173m][(0.334),(0.641),(0.895),(1.078),(1.173)]`

`M_1=0.111556m+0.410881m+0.801025m+1.162084m+2.346m`

`M_1=3.86m`

`M_2=phi_2^Tmphi_2`

`M_2=[(-0.895,-1.173,-0.641,0.334,1.078)][(m,0,0,0,0),(0,m,0,0,0),(0,0,m,0,0),(0,0,0,m,0),(0,0,0,0,m)][(-0.895),(-1.173),(-0.641),(0.334),(1.078)]`

`M_2=[-0.895m,-1.173m,-0.641m,0.334m,1.078m][(-0.895),(-1.173),(-0.641),(0.334),(1.078)]`

`M_2=3.86m`

`M_3=phi_3^Tmphi_3`

`M_3=[(1.173,0.334,-1.078,-0.641,0.895)][(m,0,0,0,0),(0,m,0,0,0),(0,0,m,0,0),(0,0,0,m,0),(0,0,0,0,m)][(1.173),(0.334),(-1.078),(-0.641),(0.895)]`

`M_3=[1.173m,0.334m,-1.078m,-0.641m,0.895m][(1.173),(0.334),(-1.078),(-0.641),(0.895)]`

`M_3=3.86m`

`M_4=phi_4^Tmphi_4`

`M_4=[(-1.078,0.895,0.334,-1.173,0.641)][(m,0,0,0,0),(0,m,0,0,0),(0,0,m,0,0),(0,0,0,m,0),(0,0,0,0,m)][(-1.078),(0.895),(0.334),(-1.173),(0.641)]`

`M_4=3.86m`

`M_5=phi_5^Tmphi_5`

`M_5=[(0.641,-1.078,1.173,-0.895,0.334)][(m,0,0,0,0),(0,m,0,0,0),(0,0,m,0,0),(0,0,0,m,0),(0,0,0,0,m)][(0.641),(-1.078),(1.173),(-0.895),(0.334)]`

`M_5=3.86m`

`L_1=phi_1^Tmiota`

`iota=[(1),(1),(1),(1),(1)]`

`L_1=[(0.334,0.641,0.895,1.078,1.173)][(m,0,0,0,0),(0,m,0,0,0),(0,0,m,0,0),(0,0,0,m,0),(0,0,0,0,m)][(1),(1),(1),(1),(1)]`

`L_1=[0.334m,0.641m,0.895m,1.078m,1.173m][(1),(1),(1),(1),(1)]`

`L_1=0.334m+0.641m+0.895m+1.078m+1.173m`

`L_1=4.12m`

`L_2=phi_2^Tmiota`

`L_2=[(-0.895,-1.173,-0.641,0.334,1.078)][(m,0,0,0,0),(0,m,0,0,0),(0,0,m,0,0),(0,0,0,m,0),(0,0,0,0,m)][(1),(1),(1),(1),(1)]`

`L_2=[-0.895m,-1.173m,-0.641m,0.334m,1.078m][(1),(1),(1),(1),(1)]`

`L_2=-0.895m-1.173m-0.641m+0.334m+1.078m`

`L_2=-1.297m`

`L_3=phi_3^Tmiota`

`L_3=[(1.173,0.334,-1.078,-0.641,0.895)][(m,0,0,0,0),(0,m,0,0,0),(0,0,m,0,0),(0,0,0,m,0),(0,0,0,0,m)][(1),(1),(1),(1),(1)]`

`L_3=[1.173m,0.334m,-1.078m,-0.641m,0.895m][(1),(1),(1),(1),(1)]`

`L_3=1.173m+0.334m-1.078m-0.641m+0.895m`

`L_3=0.683m`

`L_4=phi_4^Tmiota`

`L_4=[(-1.078,0.895,0.334,-1.173,0.641)][(m,0,0,0,0),(0,m,0,0,0),(0,0,m,0,0),(0,0,0,m,0),(0,0,0,0,m)][(1),(1),(1),(1),(1)]`

`L_4=[-1.078m,0.895m,0.334m,-1.173m,0.641m][(1),(1),(1),(1),(1)]`

`L_4=-1.078m+0.895m+0.334m-1.173m+0.641m`

`L_4=-0.381m`

`L_5=phi_5^Tmiota`

`L_5=[(0.641,-1.078,1.173,-0.895,0.334)][(m,0,0,0,0),(0,m,0,0,0),(0,0,m,0,0),(0,0,0,m,0),(0,0,0,0,m)][(1),(1),(1),(1),(1)]`

`L_5=[0.641m,-1.078m,1.173m,-0.895m,0.334m][(1),(1),(1),(1),(1)]`

`L_5=0.641m-1.078m+1.173m-0.895m+0.334m`

`L_5=0.175m`

`tau_n=L_n/M_n`

`tau_1=L_1/M_1`

`tau_1=(4.12m)/(3.86m)`

`tau_1=1.067`

`tau_2=L_2/M_2`

`tau_2=(-1.297m)/(3.86m)`

`tau_2=-0.336`

`tau_3=L_3/M_3`

`tau_3=(0.683m)/(3.86m)`

`tau_3=0.177`

`tau_4=L_4/M_4`

`tau_4=(-0.381m)/(3.86m)`

`tau_4=-0.0987`

`tau_5=L_5/M_5`

`tau_5=(0.175m)/(3.86m)`

`tau_5=0.0453`

`M_n^X=L_n^2/M_n`

`M_1^X=L_1^2/M_1`

`M_1^X=(16.9744m^2)/(3.86m)`

`M_1^X=4.398m`

`M_2^X=L_2^2/M_2`

`M_2^X=(1.682209m^2)/(3.86m)`

`M_2^X=0.436m`

`M_3^X=L_3^2/M_3`

`M_3^X=(0.466489m^2)/(3.86m)`

`M_3^X=0.121m`

`M_4^X=L_4^2/M_4`

`M_4^X=(0.145161m^2)/(3.86m)`

`M_4^X=0.037m`

`M_5^X=L_5^2/M_5`

`M_5^X=(0.030625m^2)/(3.86m)`

`M_5^X=0.008m`

Modal mass participating ratio= `M_n^X/`5 model(5m)`

Modal mass participating ratio of Model 1= 4.398/5=0.8796

Modal mass participating ratio of Model 2= 0.436/5=0.0872

Modal mass participating ratio of Model 3= 0.121/5=0.0242

Modal mass participating ratio of Model 4= 0.037/5=0.0074

Modal mass participating ratio of Model 5= 0.008/5=0.0016

`V_(bn)=(L_n^2)/M_nA_n=M_n^XA_n`

`V_(b1)=M_1^XA_1`

For T1=2 A1=0.279 from graph

`V_(b1)=4.398m*0.279g`

`V_(b1)=1.227mg`

`V_(b2)=M_2^XA_2`

T2=0.6852 A2=0.769 from graph

`V_(b2)=0.436m*0.769g`

`V_(b2)=0.527mg`

`V_(b3)=M_3^XA_3`

T3=0.4346 A3=1.039 from graph

`V_(b3)=0.121m*1.039g`

`V_(b3)=0.126mg`

`V_(b4)=M_4^XA_4`

T4=0.3383 A4=1.039 from graph

`V_(b4)=0.037m*1.039g`

`V_(b4)=0.038mg`

`V_(b5)=M_5^XA_5`

T5=0.2966 A5=1.039 from graph

`V_(b5)=0.008m*1.039g`

`V_(b5)=0.0083mg`

Finally, obtaining the base shear for each of the 5 modes of vibration and the combined base shear as per SRSS rule

=`sqrt(V_(b1)^2+V_(b2)^2+V_(b3)^2+V_(b4)^2+V_(b5)^2`

=`sqrt(1.227^2+0.527^2+0.126^2+0.038^2+0.0083^2`

=1.342mg

How many modes are to be considered such that a modal mass participating ratio of 90% is obtained?

Modal mass participating ratio of Model 1 & 2= 0.8796+0.0872=0.9668/100=96.68% which exceeds 90% or

Modal mass participating ratio of Model 1 &3=0.8796+0.0242=0.9038/100=90.38% which exceeds 90%