
a) Consistent Mass Matrix
Element Matrix B given by,
Element matrix equation is given by
EIh3e[6−3he−6−3he−3he2h2e3heh2e−63he63he−3heh2e3he2h2e]−w2δAhe420[156−22he5413he−22he4h2e−13he−3h2e54−13he15622he13he−3h2e22he4h2e]{V1V2V3V4}={Q1Q2Q3Q4}
Take he=0.5andEI=δA=106
16[6−1.5−6−1.5−1.50.51.50.25−61.561.5−1.50.251.50.5]−w2840[156−11546.5−111−6.5−7.554−6.5156116.5−7.5111]{V1V2V3V4}={Q1Q2Q3Q4}
∴[96−2496−24−248244−96249624−244248]−w2[0.186−0.0130.06450.0075−0.0130.0012−0.0075−0.0090.0645−0.00750.1860.0130.0075−0.0090.0130.0012]{V1V2V3V4}={Q1Q2Q3Q4}
Global Matrix equation is given by,
∴106[96−24−96−2400−24824400−96241920−9624−244016−24400−9624962400−244248]−w2[0.186−0.0130.06450.007500−0.0130.0012−0.0075−0.009000.0645−0.00750.37200.06450.00750.0075−0.00900.0024−0.0075−0.009000.0645−0.00750.1860.013000.0075−0.0090.0130.0012]{V1V2V3V4V5V6}={Q1Q2Q3Q4Q5Q6}
Imposing global boundary condition.
V1=0,V2=0,V5=0,V6=0
For Balancing, Q3=0,Q4=0, matrix reduces to,
106{[1920016]−w2[0.372000.0024]}{V3V4}={00}
i.e,[192−0.372w20016−0.0024w2]=[00]
For non-trival solution,
|192−0.372w20016−0.0024w2|=0
∴(192−0.372w2)(16−0.0024w2)=0=>0.00884w4−6.4w2+3072=0
By Solving above equation we get,
w2=6722.91and516.91
∴w1=82andw2=22.74
b) Lump Mass matrix
Lumped mass Element matrix is given by
[M]e=δAhe[120000(178)h2e0000120000(178)h2e]
Now δA=106he=0.5
[M]e=106[0.2500000.001600000.2500000.0016]
One assembly of two element global matrix will be,
M=106[0.250000000.00160000000.50000000.00320000000.250000000.0016]
Now global matrix equation will be,
∴106[96−24−96−2400−24824400−96241920−9624−244016−24400−9624962400−244248]−w2[0.250000000.00160000000.50000000.00320000000.250000000.0016]{V1V2V3V4V5V6}={Q1Q2Q3Q4Q5Q6}
Imposing boundary conditions,
V1=V2=V4=V6=0,andQ3=Q4=0forbalancingmatrixreducesto,
i.e,[192−0.5w20016−0.0032w2]=[00]
For non-trival solution,
|192−0.5w20016−0.0032w2|=0
∴(192−0.5w2)(16−0.0032w2)=0=>0.0016w4−8.164w2+3072=0
By Solving above equation we get,
w2=4693.42and409.084
∴w1=68.5andw2=20.22