Lumped and Consistent Masses, Finite element analysis

1. ff:(; For a uniform cross-section bar as shown in figure 6.6 of length L = I m made up of
Jmaterial having E = 2 x 10
11
N/m
2
and p = 7800 kg/m
3
estimate the natura~ lfo
f
frequencies of axial vibrations of the bar using both consistent and lumped mass
matrices. Use a two element mesh. If the exact solution is given by the relation
CO; = ~~ {¾ ,i =1, J, 5, ••••• oo
compare your answer and give your comments, A = 30 x 10-6
m2
•
1 L 2 3
A
~ Fig. 6.6 [May 2011, Dec. 2012, May 2016, May 2018, Dec. 2018, M.U ,fl
. •v
Solution:
L 2
(i) Natural frequencies roi using Consistent Mass Matrix
The element matrix equation is given by
3
L = 1 m
E = 2 x 1011
N/m2
p = 7800 kg/m3
A = 30 X 10-6
m2
1~[~l -/]{~:} = ro
2
p:h, [~ ~]{~:}
we take two elements of equal length he = 0.5 m
Cancelling A on either side of the equation, we get
2 x0t 1
[ ~
1
-/ ]{ ~:} = 002 780x 0.5 [ ~ ~]{ ~:}
i.e. 108[_: -4
4
]{~:} = 002 [ :.:
5
o/:] {~J
L

2. 'o,,.-dd{Jr a ,,,,.iforn• cro.,·s-section bar of length L made up of a material whose Young'iY ·
{Si ~ d 1 ~ a,id ,lc11,,·ity are Kive11 by .E and p. Estimate the natural 'requenc/es oif axial
"'" " ,,.. . J'
ti n of t/ie bar ,,.,.t,,g bot/, con.,·lstent and lumped ma,w; matricet,.
,,/bra o ~-
4
__..______
L
______-'j
l?lg. 6.5 (D~c. 2016, M.V.J
S
olution: Since actual values of£, L and pare not given, we shaJJ consider only a single
element to find the frequency.
(I) Natural frequency using Consistent Mass Matrix
The element matrix equation is given by
~~[}l ~1][~:} = ro
2
p:h, [ ~ ; ](iJ
·: we take one element he = L
Cancelling A on either side of the equation and simplifying, we get
E ro2p L - E - w2p L U1
L 3 L 6 _ O
_ E _ w 2
p L E _ w
2
p L
J L 6 L 3
rnpobe b h t '
o und a ry co nditi o ns i. e . U 1 = 0, write t e equ a wn
E w2p L = 0
L 3
') JE
:. w· = ~
Ill) N pl
lt lu r~1 f
1 rt:qu e n cy Uij lng Lumped Mass Matrix
-ll rnPect ,n . . . . A h r l O I
i:l t,;:,; mu tr1x is g ive n h y p , " " ,
w = fl_ , ff
L VP
-

3. . (7) Find the natural frequencies ·of axial vibration Jo~ a bar shown in figure 6.6 usin ·,
I. I b . . d l d t · . g one
inear e ement y using consistent an umpe mass ma rices.
Take,
E = 2 X 1011 N/m2
and p = 7800 kg/mJ
Fig. 6.7
~
[May 2014, M.U~ f
Solution :
(i) 'D f '
o md ron using consistent mass matrix using 1 element
L = 1 m
E =2 X 10
11
N/m2
p = 7800 kg/m3
A= 30 mm2

4. ~J f tnd the natural Jrequ~ncy of axial vibration, of a fixed Jree bar r>f uniform crob•i
ectirm of 5(J mm
2
and Ungth of J metu using consivtent and lumped man matrix and
rompare the natural Jrequ~,rcit.s with exact frequencies. Take E = 200 GPa and den1i1,
= 7!ffi (J kg!m1
. Ta~ t-.·o lin ra r ~l~ments.
tD«. 2014, MX.1
A = 50 mm2
=50 x HJ--f, m2
n N 2
E = 200 GPa = 2 x HJ 3
/m
3
p = 7860 kg/m
~
<
,lu t i<, n : 1 CD ~
l
,1 •
2 3
1 m ~
◄
Fig. 6JJ

5. 11111
taAM IHI
492
(91 Find rhc natural frequencin of longitudinal vibratio11.~ of tlie constr • d
· Ulnt Sit ·
of arrru A and 2A and of equal lengths (L), a.1t s'1own ;,, figure 6 9 Co PPed Jha/t
· · mpare ti,
nhfo;ned using lumped mass matrix approacl, , e results
ill Lt
A
L
Fig. 6.9 [May 2015 (old), Dec. 2019, M.u.1
Solution : To fi nd natura l frequencies
IA] Lumped mass matri.x approach
(i ) The element matrix equation for this problem is
olpA he
2
(ii ) Now, we bave two elements
:. Element matrix equation for element 1 is
21E [_ ~l][~J = w2AL [~ ~][~J
:. Element matrix equation for element 2 is
A/[_
~
I][~;]'= ro
2
p/L[~ ~] [~:)
(W) Global matrix equation is given by
- 2 0 ]{U
1
)
3 - 1 U2
- 1 1 U3
w
2
pAL [
2
= 2 0
0
Impose global boundary condi tions i.e. U1
= 0, we get
[ p2f, [}]~]]-.,z [~ ~]][~:) = (~)
i.e.
H
}]~l ] 0)2 [ ~ ~ ]] [ ~ : ) = ( ~) where k =
0
3
0
2 E
P Lz
g

6. i 1 -, rri! 1/;r nn,11rnl f'r·,0
Qno1f'_
' n.f 11 rfol d h rnrimts nf a bor of tat(fr,rm c rn.fls section of
in ,,,m ... n'1d n_f ltnKth , m . 1ala n =-= l >" JO ~ N!mm ;, and () == 8000 kg/m
1
. Take two
[May 2012. Uec. 2015 , M.U .j
~olntinn : n,, 1dc the I hcde <.1l,m:t in of th e problem into two e lements.
.,.
CV
,,_~
~
I Q)
G) gj •
~
ill &
Fig. 6.1
The element matrix equation for the pr9blem is obtained from equation (6.7) viz.
oipA he [2 1] {U1}
6 1 _
2 U2
· · we Lake two elements of equal length he = 0.5 m
Car1celling A on either side of the equation, we get
E [ 1 -IJ{U1) _
he- -1 1 . V2
w
2
phe [2 1] {U1}
6 1 2 U2
2x JO
11
[ 1-lJ{U1]
o.s - 1 1 .u2
(J} 2 8000x0.5 [2 l]fU1}
6 _1 2 lU2
1.c. 1OR [
4
-
4
1
{U1)·
- 4 4 U2
w2 [l.3= 0.671 { U1}
_0.6 7 l.33j U2
:e that.in these proble.ms, units of length are to be taken in m only.

7. °";J:Y
{f.f,?.
i c?::,f,~
find the natural frequency of axial vibrations of a bar of uniform cross-section of 20 mm .,9
z
fl
..1nd l.ength 1 m. Take E = 2 X 105
N/mm2
and p = 8000 kg/m3• Ta.ke two linear elements.
Compare the natural frequencies with exact frequencies.
/ ~
Uune 2017, M.U.1 -~,
.·
.ution : Refer above solution for Consistent Mass Matrix.
act Solution is given by
l - 1, 3, 5, .....
7t 2 X 1011
For i =1 0)1 - -2 8000
- 7854 rad/s
31t 2 X 1.011
For i :::: 2 ())2 - 2 8000
=== 23562 rcd/s
:~bu tati ng the r es ults

8. 6..3 CONShTfNl ANO lUMl'fO tiA'-~ MAHUCf~
M1M1 ~P H)D mm.11 b, w no~lrnr ttlld l11Mr~d l'JMH 1t111trit=I-! :' IJ, rfr~ "'" sam, for linear
,Cr r1(!;mml. lM•1101!, Jaot- 2017. M.U.j
lff., 1 :"W fC tf'.t lU>n f' 4
ml lTI in cqu,1 1,on th.7). we hove conqidcrcd masG of the
f"h:Dh rH at r. f "'. 1h J,,1 nl"Uh'ti I h «"m-: 'h,u t th e c kmc n t A lsn. we have u~cd th e sa rne
!,l Hlfi" luvCU1J,t1 fe r , "'....,,r11tti _ b(~th ma~" iHH.l stiffness rn at riccs. Hence. these mass
mtHrh'. ' tiH ,-.al k d a !- , on~hrent metrftes.
, '-- r.i ai~l."l l F ·· t-iuh' the Cf"mrlctc mc_, s~ of the cle me nt eq u a lly at the two nodes as
1':bJ'-'·n i1n lflE'-' ..C (J:.
ffi
pA h~
2
Fig, 6~
2
r.n1e m3'-'- matni fo rmed in :this v.-ay is ca lled as Lumped mass matrix and is given as
l-fJ =
p A he [ 1
2 0
~] for a bar element
~d, aot.:.gt ~ of L-umped ~fa ss ~fa trices
, J > Tht.> ~umpre: m.::.~~ ma trix is a diago nal matrix.
l I f gcn, ~!uc- p: Gbkm s are so.
lved by iteration methods an d dynamic respunse
:~k..1btwo ~re uftt:n ma de by taking incremental time steps. Hence. computations
...~c :,me ccr1~u~rng. D iagon al ma tr ix fo rm eases :i nd re duces co mputations.
rl I f i nd tlt1: ,.,,, o nowrof fnqut'neies of t rt.UIS )1erse n'hrations of a bram fix~d at ht.1th tlt/U as
~hu • n 1.11 fr,;Urt· 6. J . L :) t C,nn i ~,, W Ma 't' ":. Mal r ix.
I nn it
- - .
---.-.--.._
-
-
_____
--~_··---d
- ·------ ,t
l·I~. td
E l
) ·•
J.. ."1
= I 06
units
_ J0° units
.J
~l