SolSec 6.3 - 6-Problems and Solutions Section 6.3 (6.8...

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Unformatted text preview: 6-Problems and Solutions Section 6.3 (6.8 through 6.29)6.8Calculate the natural frequencies and mode shapes for a free-free 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=+86-6.9Calculate the natural frequencies and mode shapes of a clamped-clamped bar.Solution: The calculation of the natural frequencies and mode shapes of a clamped-clamped bar is identical to that of the fixed-fixed 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, clamped-free 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. 96-6.11Compare the natural frequencies of a clamped-free 1-m aluminum bar to that of a 1-m bar made of steel, a carbon composite, and a piece of wood.Solution:For a clamped-free 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 clamped-free 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.

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SolSec 6.3 - 6-Problems and Solutions Section 6.3 (6.8...

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