315 CH_8 - — EXAMPLE 8.1 Dynamic Response of a Plucked...

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Unformatted text preview: — 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] Pinned-pinned 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 Fixed-free a: Fixed-pinned 9:1 Pinned-free *: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 Fixed-Pinned 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 fixed-simply 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 steady-state 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 pinned-pinned 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 steady-state 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.

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315 CH_8 - — EXAMPLE 8.1 Dynamic Response of a Plucked...

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