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Unformatted text preview: CHAPTER 6 VIBRATION OF MEMBRANES 61 EQUATION OF MOTION Consider now the transverse vibrations of an elastic membrane extended in two dimensions. The membrane equilibrium position lies in the x—y
plane. The membrane has negligible bending resistance, and gravitational
body forces are to be neglected. It is assumed that constant tension forces are applied uniformly to the membrane in all directions. i
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‘v
x
w
l
l
l
» :2 ‘ ' W  ' V y 7 I i
Z The transverse deflection measure w(x,y,t) is in the z—direction. Isolating a dX by dy element of membrane, and viewing along the y—axis shows L 6—2 A similar view occurs along the Xaxis. Using Newton's Law in the z—direction gives 2 2 2
Tdy g“Eld}: + de §—% dy = pdxdyaig .
3X 8y 3t Dividing by dxdy, we can rewrite the above as 2 2
3 3 w .gjg (6 l) 2w l
“'2‘ + _ = ‘2‘ ’ 3x By C at
where c = 6. Note that T = tenSion/length and p = mass/area. From
equation (6.1), we see that the transverse motion of a membrane is governed
by the Modémemiomﬂ wave equation. A The usual boundary condition specifies zero deflection at the supporting points of the membrane. 62 yiBRATIONS OF A RECTANGULAR MEMBRANE We look now at a specific problem of a vibrating membrane of rectangular shape released from rest from some prescribed initial deflection. (61,0)
The problem is formulated in Cartesian coordinates as, W + w = l—w
xx yy C2 tt W(O,Y:t) = W(a9Y9t) = O
(6.2)
w(X,O,t) = w(X,b,t) = O
W(X9Y9O) = f(X9Y) = 0 o i
l
i Using separation of variables, we look for solution forms w(X,y,t) = X(X)Y(y)T(t)  Using this form in (6.2)l yields ll X” + aZX 0 Y” + PZY = o (6.3) T + (a2 + 82)c2 T = 0 ,
where a and B are separation constants. The solutions to equations(6.3) are i
l
J
E X = A sindx + B cosux Y = H sinBy + D cosBy (6.4) T = E sin cVaZ + 82 t + F cos cVaz + 82 t , with A, B, H, D, E and F being constants. Using the boundary conditions (6.2)2 3 5 gives
’ 9 i
i B = D = 0 a = 23 ; m = 1,2, ... J
5 INT
= ~—  = 1 2 .. B b 9n 9 3 [
w E = O . Consequently applying superposition our solution becomes w w 2 2
w(x,y,t) = Z 2 Am cos[ﬁct EE+ ~133]sin EEE sin 2%1 . (6.5) m=l n=l n a b a Finally using condition (6.2)4 yields 00 00 f(x,y) = Z Z A sin Eﬂ§ sin EEK , m=l n=l mm a b which is the doubﬂc Foucicn Aime AehLeA for f(x,y); From the theory of Fourier series the coefficients may be found by the usual orthogonality properties 6—4 giving
b
4 ‘3 ,mﬂ€.mm
Amn =‘gg g g f(E,n) Sln‘“;“ Sln —E— dadn  (6'6) The frequencies of vibration for this case are given by 2 2 w _.e e_ 2_
55‘" 2 2 + (6.7)
a b f =
Note the degenenatc cube where a = b; this implies a non—uniqueness of the
frequency associated with each mode shape, i.e., wmn = wnm (see Morse [25]). The shapes of the modal patterns are shown in the following figure. Note that zeta dcﬁﬁeciion lines are called nodaﬁ Zineb. / /
mu; rcunv%%%§g%%;‘
.Aiglugﬁﬁiﬂn‘? —Shapes of the} ﬁrst four normal modes of a rectangular membrane. Arrows
point to the nodal lines. 63 AXISYMMETRIC VIBRATION OFAN INFINITE MEMBRANE Consider next an axisymmetric problem of a membrane of infinite extent
with some prescribed initial displacement. The problem is now formulated
in polar coordinates with the deflection being independent of 6. The governing equation (6.1) becomes for this case
+.__———=————————— , W=W(r,t) . The initial conditions are to be w(r,0) = f(r)
(6.9)
wt(r,0) = 0 Applying the zero order Hankel transform (see equation (1.82)) over the variable r in equations (6.8) and (6.9) yields 62% dt2 + 62x26 = 0 (6.10) w(A,O) = f(A) (6.11)
wt(x,0) = 0 ,
where w = w(X,t).
The solution to (6.10) subject to conditions (6.11) is
6(x,t) = ?(A) cos cAt . (6.12) Finally using (6.12) with the inversion integral formula (1.87), we find that 00 w(r,t) = f AE(A) cos ext JO(Xr)dA . (6.13)
0 As a special case, the initial condition is taken to be —;§ , r < a w(r,0) = f(r) = Ha (6.14)
0 , r > a
For this case, the limit is to be taken as a + 0. Using the general solution
form (6.13) we get (see Graff [13] for details) 1 ct
 —*'~——————————— 0 < r < ct
2 9
2W r1/2 c t2 _ r2)3/2 O , ct < r < w W(r,t) = (6.15) This specific solution (with a +‘0) would correspond to a plucked membrane. The following figure compares this type of solution with the corresponding string problem. 3
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§ ; ——Comparison between the behavior of a plucked string and a. plucked
membrane. The ﬁrst sketches show the initial shapes; the lower ones the shapes at
successive instants‘ later. One quarter of the membrane has been cut away to show the shape of the cross section. ...
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This note was uploaded on 10/03/2011 for the course MCE 565 taught by Professor Staff during the Spring '11 term at Rhode Island.
 Spring '11
 Staff

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