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EXAMPLE 8.1 Dynamic Response of a Plucked String If a string of length I, ﬁxed at both ends, is plucked at its midpoint as shown in Fig. 8.5 and then
released, determine its subsequent motion. Solution: The solution is given by Eq. (8.30) with C" and D" given by Eqs. (8.33) and (8.34),
respectively. Since there is no initial velocity, w0(x) = 0, and so D" = 0. Thus the solution of Eq. (8.30)
reduces to 00 t
w(x, t) = 2 C” sin mm cos new (E.1)
n=1 1 l
where
2 1 mm
C" = —/ w0(x) sin — dx (E2)
1 0 l
The initial deﬂection w0(x) is given by
2h 1
—)£ forO s x S 5
W006) = 2M] _ x) 1 (E3)
1 fore S x S I
By substituting Eq. (E3) into Eq. (E2), C" can be evaluated:
2 "2 2h ’2h
Cn = / x sin mm dx i / (l x) sin mm dx
1 0 l l [,2 l 1
8h sinn—W forn=1,3,5,...
772n2
= (E4)
0 forn=2,4,6,...
“'0 (xi 0)
x 0%é+é—>\ 1 FIGURE 8.5 string. Initial deﬂection of the 596 CHAPTERS CONTINUOUS SYSTEMS By using the relation sin? = (—1)(”_1)’2, n = 1,3,5, (E5) the desired solution can be expressed as t 1
w(x, t) —— 8}:{sin 77 cos W; 9 sin 371” cos 37;“ + ~~ (E.6) 7T In this case, no even harmonics are excited. End Conditions Frequency
of Beam Equation Mode Shape (Normal Function) Value of ﬁn]
Pinnedpinned Sin 3”] = O Wn(x) = Cn[sianx] ﬁll : 7T
* i 321 : 2’7T
[331 = 377
B4] = 477'
:Free'free cos [3"] .cosh ﬁn] = 1 Wn(x) = Cn[sin[3,,x + sinh an [311 = 4.730041 Fixed—fixed 3:8 I Fixedfree a: Fixedpinned 9:1 Pinnedfree *:I + (1,, (cos an + cosh an)] where .ﬂ=< cos [3"] cosh Bnl : 1 sin 3”] — sinh is"! cosh B”! — cos [3,1]) Wn(x)n= Cn[sinh an — sin an + a”(cosh an ~ cos an)] where .”=( cos [3”] .cosh [3n] = —1 sinh [3n] — sin [3"] cos [3"] — cosh ﬁn!)
Wn(x) = Cn[sin an — sinh B"); — ,, (cos an — cosh [3,110] where 0% tan 3"] — tanh [3n] = 0 sin [3"] + sinh Bnl Wn(x) = Cn[sin an — sinh an > cos [3"] + cosh [3n] + a” (cosh an — cos an)] where an=< tan 3"] — tanh [3n] = 0
where any sin ﬁn] — sinh [3"] cos Bnl — cosh B"! Wn(x) = Cn[sin[3nx +01” sinthx] sin [3,,1
sinh Bnl ) 321 = 7.853205
B3l = 10.995608
B41: 14.137165
(Bl = 0 for rigid
body mode) 011: 4.730041
1321: 7.853205
B31: 10.995608
B41: 14.137165 [31]: 1.875104
[32] = 4.694091
[33]: 7.854757
[34] = 10.995541 Bll = 3.926602
[32] = 7.068583
[33] = 10.210176
B4] = 13.351768 [311 = 3.926602
[321 = 7.068583
B3l = 10.210176
B4] = 13.351768
(/31 = 0 for rigid
body mode) FIGURE 8.15 Common boundary conditions for the transverse Vibration of a beam. —
EXAMPLE 8.7 Natural Frequencies of a FixedPinned Beam Determine the natural frequencies of Vibration of a uniform beam ﬁxed at x = 0 and simply sup
ported at x = 1. Solution: The boundary conditions can be stated as W(O) : 0 (El)
d—W(0) = 0 (E2)
dx
W(l) = 0 (E3)
d2W de
El dxz (l) = 0 or dxz (l) = 0 (E4)
Condition (E. 1) leads to
C. + C3 = 0 (E5)
in Eq. (8.91), While Eqs. (E2) and (8.91) give
dW . .
E = B[— C1 sm Bx + C2 cos Bx + C3 s1nh Bx + C4 cosh Bx]x=0 = 0
x=0
or
Thus the solution, Eq. (8.91), becomes
W(x) = C1(cos Bx — cosh Bx) + C2(sin Bx — sinh Bx) (E.7)
Applying conditions (E3) and (E4) to Eq. (E.7) yields
. . _ (E.8)
C1(cos Bl  cosh Bl) + C2(s1n Bl — smh Bl) — 0
— C1(cos Bl + cosh Bl) — C2(sin Bl + sinh B1) = 0 (E9) For a nontrivial solution of C1 and C2, the determinant of their coefﬁcients must be zero—that is, (cos Bl — cosh Bl)
— (cos Bl + cosh Bl) (sin Bl — sinh Bl) _ —(sin [31 + sinh Bl) ' 0 (Em) 8.5 LATERAL VIBRATION OF BEAMS 617 Expanding the determinant gives the frequency equation
cos Bl sinh ,8] — sin ,Bl cosh 31 = 0
or tan Bl = tanh ﬁl (EU) The roots of this equation, [3"], give the natural frequencies of vibration to =(131)2 El “2 n=12 (E12)
n n s 9 9 . where the values of Bnl, n = l, 2, . . . satisfying Eq. (E.ll) are given in Fig. 8.15. If the value of C2
corresponding to B" is denoted as C2”, it can be expressed in terms of C1" from Eq. (E.8) as C _ _C cos Bnl — cosh Bnl E 13
2" _ 1" sin Bnl — sinh Bnl ( ' ) Hence Eq. (E.7) can be written as cos Bnl — cosh Bnl sin [3”] — sinh 3n] >(5in an — sinh Em] (13.14) Wn(x) = C1n:(cos an — cosh ﬁnx) — < The normal modes of vibration can be obtained by the use of Eq. (8.81)
wn(x, t) = W,,(x)(A,, cos cunt + B” sin wnt) (E.lS) with Wn(x) given by Eq. (E.l4). The general or total solution of the ﬁxedsimply supported beam
can be expressed by the sum of the normal modes: w(x, t) = 2 wn(x, t) (E.l6)
n=l 8.5 LATERAL VIBRATION 0F BEAMS — Forced Vibration of 21 Simply Supported Beam EXAMPLE 8.8 619 Find the steadystate response of a pinned—pinned beam subject to a harmonic force f(x, t) = f0 sin cot applied at x = a, as shown in Fig. 8.17. Solution
Approach: Mode superposition method. The normal mode functions of a pinnedpinned beam are given by (see Fig. 8.15; also Problem 8.31) Wn(x) = sin [3,, x = sin ml”
where
[3"] = I177
The generalized force Qn(l), given by Eq. (8.115), becomes
1 mm
Qn(t) = / f(x, 1‘) sin B" x dx = f0 sinT sin wt
0 The steadystate response of the beam is given by Eq. (8.117) 1 t
pAbwn/O Qn(7') Sln w” (t — ’7') d7 (111(1) = where l l
l
b = /W,2,(x)dx = /sin2 andx = —
0 o 2
The solution of Eq. (E.4) can be expressed as 2f0 Sin? .
ch“) = s1n wt f0 sin wt PM I 4 FIGURE 8.17 Pinned—pinned beam under harmonic force. (El) (E2) (E3) (E4) (E.5) (E.6) 620 CHAPTER 8 CONTINUOUS SYSTEMS Thus the response of the beam is given by Eq. (8.110): Zfo w 1 mm mm
,1: _ ' ' ' t E.7
w(x ) p Al "21 a)?! _ (02 sm 1 sm 1 sm to ( ) ...
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This note was uploaded on 09/08/2010 for the course MAE 315 taught by Professor Wu during the Summer '08 term at N.C. State.
 Summer '08
 WU

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