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Unformatted text preview: 6Problems and Solutions Section 6.3 (6.8 through 6.29)6.8Calculate the natural frequencies and mode shapes for a freefree bar. Calculate the temporal solution of the first mode.Solution:Following example 6.31 (with different B.C.s), the spatial response of the bar will beX(x)=+The boundary conditions are .)()(==The expression for xbxaxXX29(=so at 0:== 0at l= , 0so that l= nor = n/lwhere nstarts a zero. Hence the mode shapes are of the formXn(x)= for n= 1, 2, 3, and for n= 0,X(x)==a constant.The temporal solution is given by eq. (6.15) to be22)()(=&&so that the temporal solution of the first mode:&&T(t)+2(29= 0 =&&(29(29=+866.9Calculate the natural frequencies and mode shapes of a clampedclamped bar.Solution: The calculation of the natural frequencies and mode shapes of a clampedclamped bar is identical to that of the fixedfixed string since the equations of motion are mathematically the same. The solution of this problem is thus given at the beginning of section 6.2, but is repeated here: Applying separation of variable to eq. (6.56) yields that the spatial variable must satisfy eq. (6.59) of example 6.3.1, i.e., xbxaxX29(+=where aand bare constants to be determined. The clamped boundary conditions require that X(0) = X(l) = or0 = bor X= asinx0 = asinlor = n/lHence the mode shapes will be of the formXn= ansinnxWhere = n/l. The frequencies are determined from the temporal solution and becomen=snc=nplEr, n=1,2,3,...6.10It is desired to design a 4.5 m, clampedfree bar such that the first natural frequency is 1878 Hz. Of what material should it be made?Solution:First change the frequency into radians:1878 Hz =1878x2rad/s=11800 rad/sThe first natural frequency is given computed in Example 6.3.1, Equation (6.63) as1=2=12422= (11800292422= 7.143 107in Nm/kg. Examining the ratios from Table 2.1 for the values given yields that for Steel:E=2 10112.8 103= 7.143 107 /Thus asteelbar with a length 4.5 meters will have a first natural frequency of 1878 Hz. This is something like a truck chassis. 966.11Compare the natural frequencies of a clampedfree 1m aluminum bar to that of a 1m bar made of steel, a carbon composite, and a piece of wood.Solution:For a clampedfree bar the natural frequencies are given by eq. (6.6.3) as=(2 1292Referring to values of r and E from table 1.2 yields (for 1):Steel(229 (1292.0 10117.8 103= 7,954rad/s (1266Hz)Aluminum(229 (1297.1 10102.7 103= 8,055rad/s (1282 Hz)Wood(229 (1295.4 1096.0 102= 4,712rad/s (750 Hz)Carbon composite (student must hunt for E/and guess a little) from Vinson and Sierakowskis book on composites /E= 3118 and4897)3118(2=rad/s (780 Hz)6.12Derive the boundary conditions for a clampedfree bar with a solid lumped mass,...
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This note was uploaded on 11/28/2010 for the course ME 4440 taught by Professor Hill during the Winter '09 term at Detroit Mercy.
 Winter '09
 Hill
 Natural Frequency

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