Understanding the concepts on Degrees of Freedom

• L_n , M_n , M_n* and Γ_n for each of the five modes are to be computed from Equations below.

(1) calculation of M_n

_{mass matrix(m)=}

_{modifined mass matrix (Mn)=(ϕn)T.m.ϕn}

_{For first mode,n=1}

_{M1=[0.334 0.641 0.895 1.078 1.173]}

_{=m[0.334x0.334 0.641x0.641 0.895x0.895 1.078x1.078 1.173x1.173]}

_{= 3.861m}

_{similary for 2nd mode , n=2 , M2=3.861m}

_{similary for 3nd mode , n=3 , M3=3.861m}

_{similary for 4th mode , n=4 , M4=3.861m}

_{similary for 5th mode , n=5 , M5=3.861m}

(2) calculation of L_n

_{Modal property (Ln)=(ϕn)T.m.l , where l is influnce vector and it is assumed here 100% utilization of seismic factor. It is unit matrix.}

_{For first mode,n=1}

_{L1=[0.334 0.641 0.895 1.078 1.173]}

_{=m[0.334 + 0.641 + 0.895 + 1.078 + 1.173]}

_{= 4.121m}

_{similary for 2nd mode , n=2 , L2= -1.297m}

_{similary for 3nd mode , n=3 , L3= 0.683m}

_{similary for 4th mode , n=4 , L4= -0.381m}

_{similary for 5th mode , n=5 , L5= 0.175m}

(3) calculation of Γ_n

_{Modal participation factor (Γn)= Ln/Mn}

_{For first mode,n=1}

_{Γ1= 4.121m/ 3.861m}

_=1.067

_{similary for 2nd mode , n=2 , Γ2= -0.335}

_{similary for 3nd mode , n=3 , Γ3= 0.177}

_{similary for 4th mode , n=4 , Γ4= -0.098}

_{similary for 5th mode , n=5 , Γ5= 0.045}

(3) calculation of M_n*

_{Effective modal mass (Mn*)= Ln^2/Mn}

_{For first mode,n=1}

_{M1*= (4.121m)^2/ 3.861m}

_=4.398m

_{similary for 2nd mode , n=2 , M2*= 0.436m}

_{similary for 3nd mode , n=3 , M3*= 0.121m}

_{similary for 4th mode , n=4 , M4*= 0.038m}

_{similary for 5th mode , n=5 , M5*= 0.008m}

_{Now modal mass participation ratio = Mn*/mj}

_{where , mj= total mass of building= m+m+m+m+m=5m}

_{For first mode,n=1}

_{∴ modal mass participation ratio = M1*/mj}

_{= 4.398m/5m}

_{= 0.88}

_{similary for 2nd mode , n=2 , = 0.09}

_{similary for 3nd mode , n=3 , = 0.02}

_{similary for 4th mode , n=4 , = 0.01}

_{similary for 5th mode , n=5 , = 0.02}

• _{Adding first two mode mass participation ratio,we get (0.88+0.09)=0.97. so achive minimum 90% of modal mass participation ratio, first two modes are enough.}

• calculation of base shear (Vb)

As we know base shear= M_n*.An , An is calculated from graph

for the first mode , n=1, T=2 sec Vb1= 4.398mx0.27g

=1.187mg

_{similary for 2nd mode , n=2 , T=0.6852 s , Vb2= 0.327mg}

_{similary for 3nd mode , n=3 , T=0.4346 s , Vb2= 0.126mg}

_{similary for 4th mode , n=4 , T=0.3383s, Vb2= 0.039mg}

_{similary for 5th mode , n=5 , T=0.2966 s , Vb2= 0.008mg}

_{Now, Avg. shear is calculated by the method of square root of sum of square=square root[(VB1)^2+(VB2)^2+(VB3)^2+(VB4)^2+(VB5)^2]}

_{= square root[(0.327mg)^2+(0.126mg)^2+(0.039mg)^2+(1.187mg)^2+(0.008mg )^2]}

_{= 1.15 mg}

_{THANK YOU}