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Find two natrial frequencies of transverse vibration of a beam fixed at 60th axis. Use lumped mass matrix. Assume length of beam as 1 unit, FI = 106 unit, ζA=106 unit

Find two natrial frequencies of transverse vibration of a beam fixed at 60th axis. Use lumped mass matrix. Assume length of beam as 1 unit, FI = 106 unit, ζA=106 unit and Use consistent mass matrix.


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a) Consistent Mass Matrix
Element Matrix B given by,

Element matrix equation is given by
EIh3e[63he63he3he2h2e3heh2e63he63he3heh2e3he2h2e]w2δAhe420[15622he5413he22he4h2e13he3h2e5413he15622he13he3h2e22he4h2e]{V1V2V3V4}={Q1Q2Q3Q4}

Take he=0.5andEI=δA=106

16[61.561.51.50.51.50.2561.561.51.50.251.50.5]w2840[15611546.51116.57.5546.5156116.57.5111]{V1V2V3V4}={Q1Q2Q3Q4}

[9624962424824496249624244248]w2[0.1860.0130.06450.00750.0130.00120.00750.0090.06450.00750.1860.0130.00750.0090.0130.0012]{V1V2V3V4}={Q1Q2Q3Q4}

Global Matrix equation is given by,

106[962496240024824400962419209624244016244009624962400244248]w2[0.1860.0130.06450.0075000.0130.00120.00750.009000.06450.00750.37200.06450.00750.00750.00900.00240.00750.009000.06450.00750.1860.013000.00750.0090.0130.0012]{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,[1920.372w200160.0024w2]=[00]

For non-trival solution,
|1920.372w200160.0024w2|=0
(1920.372w2)(160.0024w2)=0=>0.00884w46.4w2+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[962496240024824400962419209624244016244009624962400244248]w2[0.250000000.00160000000.50000000.00320000000.250000000.0016]{V1V2V3V4V5V6}={Q1Q2Q3Q4Q5Q6}

Imposing boundary conditions,
V1=V2=V4=V6=0,andQ3=Q4=0forbalancingmatrixreducesto,
i.e,[1920.5w200160.0032w2]=[00]

For non-trival solution,
|1920.5w200160.0032w2|=0

(1920.5w2)(160.0032w2)=0=>0.0016w48.164w2+3072=0
By Solving above equation we get,
w2=4693.42and409.084
w1=68.5andw2=20.22

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