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Unformatted text preview: CHAPTER 3 LONGITUDINAL WAVES IN RODS 3.1 EQUATION OF MOTION We consider the longitudinaﬂ wave motion in an elastic body whose E
length is much greater than its cross—sectional dimensions, i.e., a rod.
We neglect any {arenaﬂ coninaciion 0d expanéion of the rod, and hence
this theory is an approximate one based on a Athengih 06 matehiaﬁA theony.
The approximation is very good if the wavelengths of the motion are long
compared with the rod's cross—sectional dimensions. In addition, it is
also assumed that plane cross—sections of the rod remain plane and parallel
during the deformation associated with the wave motion. Consider a section of an elastic rod, A+dA Let 0 be the normal stress in the Xdirection and u be the longitudinal displacement of any cross—section in the X~direction. Both 0 and u depend I only on the variables x and t.
Summing forces in the x—direction gives i 2
(O+d0)(A+dA) — (G)A = OK dx §—% , (3.1)
at where p is the mass density of the material, and K'is the average cross— sectional area. Equation (3.1) reduces to 2
odA+dOA= pA dXB—l‘zl ,
at
or
odA + doA = 6* azu
dX 8t2 Taking the limit as dx + 0 yields 2
3 _ 3 u
SE(Ao) — pA 3;? . (32) For the case A = constant, we would have swat. , (3.3) 2
u
31:2 Now from strength of materials theory with small deformations O = E8
(3.4)
8:22
3X ’ where 8 is the normal strain in the x—direction and E is modulus of elasticity. Combining (3.3) and (3.4) gives 32u azu 9
3X St
or (3.5)
=__l_u
uxx 2 tt
CL Where CL = XE , is the speed of propagation of the longitudinal disturbance.
* We then see that the dtépﬁaccmcnt u(x,t) is governed by the one—dimensional wave equation. *
It can also be shown that the Aineéé o(x,t), must also satisfy the one— dimensional wave equation (see homework exercise). ’1
1
l
i
l
<
E
l
l 3—3 Common boundary conditions which are used in conjunction with (3.5) are at a ﬁﬂee and, G = O e> EB = 0, while at a ﬁixed and, u = 0.
3x 32 TRAVELING WAVE SOLUTION RESULTS Recall the traveling wave solution to (3.5) was
u(X,t) = f(x—ct) + g(x+ct) , where c = c .
L Consider the case of an impact of the rod with velocity v to produce a wave traveling in the +x—direction. We therefore would have i u(X,t) = f(x~ct) . i
l
I
Note that g
E
Bu , E
=——'=— i
V at cf i
and
399.
8X _ E _ f '
Consequently,
,9 = _ X
E c
or E
O = —pvc . (3.6) I
Note for a wave traveling in the —x—direction we would find
I
i O = pvc . Equation (3.6) then gives the magnitude of the impact stress in terms of i the impact velocity. One can show the various reflection phenomena using the traveling wave solution. The results are that i
g
i
‘ 3—4 a u $+F€$S
i.) at a ﬁnea end, 0 = 0, —= O; a reflected wave has the same
3x A magnitude as the incident wave except opposite in sign, and at the end the displacement is doubled, sirgse ‘
ii.) at a ﬁéxed and, u = O; a reflected wave has the same magnitude /\ and Sign as the incident wave, and at the end the stress is doubled. 3.3 §§3ARAT10N OF VARIABLES SOLUTION Consider the free vibrations of an unrestrained rod of length R with a prescribed initial displacement rm“ ‘; —~ m % The problem is formulated as O<X<5Z, u =~l——u
XX 2 tt ’ — —
c O(O,t) = o(£,t) = O
ut(x,0) = 0 I
i
i
l
E
E
(3.7) 2
u(x,0) = f(X)
To separate the variables use the solution form
l
u(x,t) = X(X)T(t) g
i
! This produces the forms X = A sinkx + B coskx
l H
H C sinkct + D coslct . Now from conditions (3.7)2 3u(0,t) = sumo = 3x 3x 0 ’ O(0,t) = o(5L,t) = 0 => 3—5 so A = 0 sinAR = O é> A = ——'; n = l, 2, Note that the n = 0 case corresponds to the A=O case, which gives XT = constant. Relation (3.7)3 implies that Using superposition our solution may then be written as 8 a o nﬁx nﬁct
u(x,t) = Er‘+ 1 an cos ~Efcos g . (3.8) “M 11 Finally applying the last condition (3.7)4 yields that a 00 n'ITX
f(x) = 352 + i an cos 7Z , (3.9) n=l
which is the Fourier cosine series for f(x); hence 2
an = %f0f(£) cos 3%; dE . (3.10) Equations (3.8) and (3.10) then solve the problem for an arbitrary function f. 3'4 MW Consider now the longitudinal wave motion along a composite rod composed
of two semi—infinite rods of different material and cross—section. Assume a perfectly rigid bonding of the two rods. 3—6 Working in terms of the stress wave motion we write oi = incident stress pulse
or = reflected stress pulse
0t = transmitted stress pulse Now at the discontinuity of crosssection we must have the following matching conditions holding, (oi + or)Al = o A t 2
(3.11)
vi + vr = vt
where vi r are the incident, reflected or transmitted velocities, respectively.
! ’
Using (3.6) in (3.11)2 yields
_ 0i + gr _ Ot
_' " a
plcl plcl pzcz
or
O =.l (g _ 0 ) (3.12)
t u i r ’
p C .  .
Where u = l l . prodigch P4; 13 m/fccl wafer/cl Impedance pzcz Solving (3.11)l and (3.12) for or and o in terms of oi produces t
A
1—14—;
A2
0 = U.
r Al 1
l+u K—
2
(3.13)
A1
2 _
A2
0t Al Oi .
l+Ll K“
2 Note it follows from (3.13) that at a fixed end 3—7
while at a free end A + o :> 0r = —o. 3 o§;:c3 2 1 g 35 MW We now want to consider the case of variable cross—section where A = A(X). The rod or wave guide will now affect the wave amplitude because of the geometrical variation. x i
l
5
Recall the governing equation (3.2) read
2
gym) =pA8—%
8t
Using (3.4) in the above yields
2
3 Bu 3 u
—~ [EA ——] = pA ———
3X 3X atZ
Assuming E = constant, we get
11mg =_1_82u
5
A 8X 3x C2 at2
or (3.14)
Bzu + g;_gy£ = g;_ 82u
3x2 A 3X c2 Btz Consider next the case of a harmonic time dependent solution form, i.e., let u(x,t) = U(x)ei‘*’t . (3.15) Using (3.15) in (3.14) gives 2
A' w
n _ v ____ =
U + A U + 2 U 0 . (3.16)
c
For the case where A = constant (i.e., A' = 0), we get
(1)
cos —x
c
U: . w . (3.17)
Sln'— x c
Now assume A' is small, so that we are investigating slightly varying rods. In addition try the solution form U(X) = f(X) sin g(X) . (3.18) Substituting form (3.18) in (3.16) produces 2 A' w2
[f" — f(g') + Kf~f' +*§ f] sin g(X)
C (3.19)
I
+ [Zf'g' + fg” + %— fg'] cos g(X) = O
For (3.19) to be zero for all x we must have
2
2 A' w
n_ l _____ v __ =
f f(g) +A f +c2f o
A! 2f'g' + fg" + K—~fg' = O Rewriting equation (3.20)2 gives 1 u I
2.1+g_f.+é._=0 f g A Integrating the above yields 2 1n f + 1n g' + 1n A = Cl , where C1 = arbitrary constant. or 1n[f2Ag'] = Cl C
L szg' = e l = C2 , say . (3.21) I Now in (3.20)l, assuming A' is small the term %— f' is negligible in comparison to the other terms. In addition, we take f" << 1, so the term
f" is also negligible. Therefore (3.20)l becomes approximately 2 —f(g')2+%f = 0 .
C Using (3.21) in the above yields Solving for f gives cC
f=V ——2=KA_l/2 , (3.22) (DA /CzcI
where K = —~—. U) Hence, if we have a gradual variation in the cross—sectional area, then we
will have a small variation in the amplitude f. Using (3.21) and (3.22) gives §=C2=9
sz C Integrating the above g=gx+c3, 02% where C “ constant. 3 l Combining results (3.22) and (3.23) in (3.18) gives U(x) ='E— sin(%'x + C3) . (3.24) JR Note the comparison of (3.24) with the A' = 0 case (3.17). 3—10 36 VIBRATION OF RODS WITH RIGID END MASSES Consider the case of the free vibration of a finite uniform rod with a rigid end mass. The problem is formulated as _ L 1
uxx — 2 utt , O f X_: £
C 1
u(0,t) = 0 (3.25) :
AB R — W i
uX( $t) “ _ g utt( at) Separating the variables yields u(x,t) = X(X)T(t)
X(x) = Cl cosXx + C2 sinAx (3.26) T(t) II C cosACt + C4 Sinkct 3 Condition (3.25)2 gives while (3.25)3 implies that '  W.— 2 2 ‘ o
AE(C3 cosAct + C4 Slnkct)CZX coslﬂ — g A c C2 Sinl£
' (C3 coskct + C4 sinAct)
The previous relation may be written as
AEg = tanAI . (3.27) Wk c2 3—ll Letting
CL = AEgSL Q m mrmm‘s of rod Wcz MM ) M: was: of rigid wag/174 B = Xi ,
relation (3.27) then becomes
a = B tanB . (3.28) Relation (3.28) is a thanccndentaﬂ cquaiion for the eigenvalues X. It is sometimes called the ﬁhcqucncy equaiion of the system. The solution to (3.28) can be graphically illustrated by the following figure. g
i
3
E
i
K
.
I
E Hence from the figure for a given a, there will be k—roots (k = l, 2, ...) to equation (3.28). Call the roots Bk. Hence we write ka Bkct Bkct
uk(x,t) = sin —E—~(C3k cos % + 04k sin z ) , (3.29) which represents the motion for the k—th phincépaﬁ mode, The complete solution is the superposition of all modes, i.e., 3—12 ka Bkct . Bkct
3k cos + C Sln u(X,t) = f sin ——— (C
k=l £ z 4k 2 ) ’ (3'30) where 03k and C4k would be determined from the initial conditions. 37 L_A_TERAL INERTIA EFFECTS (Lame Thea/3]) We mentioned in the derivation of the governing equation (3.5), that
the effects of lateral contractionior expansion would be neglected. We now
re—examine this point and reformulate the analysis to include this effect. This modified analysis still follows a strength of materials approach II and initially assumes that 0y = O 0. So from Hooke's law, the transverse Z strains are given by _ .2 _ _ _ _ .Ee
€y—€Z—EOX vex—vax, 7 (3.31) where v is POM/son’xs We. The lateral displacements in the y and 2 directions are taken to be distributed linearly and so we write v=€y=\)y82 y 3X (3.32)
w=€z=—\)ziq. 2 8X With all the variables now expressed solely in terms of u(X,t), the governing equation of motion can be derived by using the eneagy method. 3—13 This method incorporates Hamiﬂton’é pntncépﬁc, and states that a system
moves in such a way as to make the total kinetic energy and system work an
* exiaemum. The details of this process are involved , so we give here only the end result 82u vZEZ 34u _ l 32u
——+ ——————————— , (3.33)
3x2 c 2 3x28t2 2 3t2
L CL
where k is the poﬁan aadtué 06 gynaiion of the cross—section, i.e., Ip = sz with IP being the poﬁad moment 06 anatia of the cross—sectional area. This
particular longitudinal wave theory is sometimes called Love's theory or
Rayleigh's theory. Equation (3.33) is not the one—dimensional wave equation, and so we
can expect significantly different propagational characteristics for this
case. Attempting a harmonic traveling wave solution form produces the fact
that, when lateral inertia is included, we have a ﬁaequencg dependent phaée ucﬂocity, i.e. dibpehAion (see homework exercise). 7':
For details see Graff or Love. E
g
E
3
§
s ...
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