Understanding the concepts on Degrees of Freedom
ANS:MASS MATRIX:(M);(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)
modified mass matrix (m) on t.mn.on
for the first mode (0.334 o.641 0.895 1.078 1.173){
(0.334 o.641 0.895 1.078 1.173) { m 0 0 0 0 (0.334
0 m 0 0 0 0.641
0 0 m 0 0 0.895
0 0 0 m 0 1.078
0 0 0 0 m} 1.173)
M1: 3.861m
modal properties(L)=On t.m .l
where l= influence vector ( assuming 100% utilization of seismic force)= { 1
1
1
1
1}
for first mode =
(0.334 o.641 0.895 1.078 1.173 ) { m 0 0 0 0 1
0 m 0 0 0 1
0 0 m 0 0 1
0 0 0 m 0 1
0 0 0 0 m} 1
L1 =4.121m
modal participation factor L/M
n=1
modal participation factor = 4.21m/3.861m
modal participation factor =1.067
similarly = -0.335
0.177
-0.098
0.045
modal participation factor Ln2/Mn
n=1
modal participation factor = 4.21m2/3.861m
modal participation factor =1.067
similarly = 0.436
0.121
0.038
0.008
modal mass participation ratioMn *mj
where mj = total mass of building =m+m+m+m+m=5m
modal mass participation ratio of first mode (n=1) 4.398m/5m=0.88
second mode (n=2)=0.09
third mode (n=3)= 0.02
fourth mode (n=4)= 0.01
fifth mode (n=5) =o.o2
adding first two mass participating ratios we get o.88+0.09=0.97
to achieve mim 90% of modal mass participation ratio first two modes are enough
base shear (Vb) Mn*An
for the first mode Vb1 = 4.398 *0.27g
(n=1)
Vb1=1.187mg
similarly; Vb2=0.327mg
Vb3=0.126mg
Vb4=0.039mg
Vb5=0.008mg
avg base shear is calculated by the method of square root of sum sqaue
vVb1+Vb2+Vb3+Vb4+VB5
VB =1.15MG
RESULT : SO GET 90% OF RESULT IN 2 MODES
_