SolSec 6.3

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

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6- 8 Problems and Solutions Section 6.3 (6.8 through 6.29) 6.8 Calculate 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 be Xx a x b x ( ) sin cos =+ σσ The boundary conditions are . 0 ) ( ) 0 ( = = l X X The expression for x b x a x X X σ sin cos ) ( is = so at 0: 00 =⇒ = aa at l =− bl b sin , so that σ l = n π or σ = n π / l . Hence the mode shapes are of the form l x n b n π cos The temporal solution is given by eq. (6.15) to be 2 2 ) ( ) ( = t T c t T n n & & so that ˙˙ ( ) ( ) ( ) sin cos . Tt l cT b T t a c l tb c l t 1 2 11 12 0 += = + ππ and

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6- 9 6.9 Calculate 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., x b x a x X σ cos sin ) ( + = where a and b are constants to be determined. The clamped boundary conditions require that X (0) = X ( l ) = or 0 = b or X = a sin x 0 = a sin σ l or σ = n π / l Hence the mode shapes will be of the form X n = a n sin σ n x Where σ ν = n π / l . The frequencies are determined from the temporal solution and become ωσ π ρ nn c n l E n == = , , , ,. .. 123 6.10 It is desired to design a 12.82 m, clamped-free bar such that the first natural f requency is 100 Hz. Of what material should it be made? Solution: Note that in the first printing of the book the frequency should have been 1000 Hz or to get a precise answer use a 12.82 meter bar. The point eventually is for the students to realize that longitudinal frequencies are much higher then bending frequencies introduced in the next section. Proceeding with a 100 Hz and a 12.82 meter bar, first change the frequency into radians: 100 Hz =100x2 π rad/s The first natural frequency is given computed in Example 6.3.1, Equation (6.63) as ω ρρ 11 2 2 2 2 2 2 52 7 24 100 2 4 1 6 10 12 82 2 63 10 =⇒ = = ⇒= × = × l ll EE E () .( . ) . in Nm/kg. Examining the ratios from Table 2.1 for the values given yields that for Aluminum: E = × × 71 10 27 10 263 10 10 3 7 . . . Nm/kg Thus an Aluminum bar with a length of 12.28 meters will have a first natural frequency of 100 Hz. This is very long! Thus one would expect that normal size bars would not have such low natural frequencies.
6- 10 6.11 Compare 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 ρ π ω E l n n 2 ) 1 1 ( = Referring to values of r and E from table 1.2 yields (for ω 1 ): Steel () .

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## This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

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

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