SolSec 6.3

# SolSec 6.3 - 6 8 Problems and Solutions Section 6.3(6.8...

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6- 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 X ( x ) = α σιν σξ + β χοσ σξ The boundary conditions are . 0 ) ( ) 0 ( = = λ Ξ Ξ The expression for x b x a x X X σ σ σ σ σιν χοσ 29 ( ισ - = so at 0: 0 = σα α = 0 at l 0 = - σβ σιν σλ , β ≠ 0 so that σ l = n π or σ = n π / l where n starts a zero . Hence the mode shapes are of the form X n ( x ) = β ν χοσ νπξ λ for n = 1, 2, 3, … and for n = 0, X 0 ( x ) = β 0 χοσ 0 π λ ξ = β 0 a constant. The temporal solution is given by eq. (6.15) to be 2 2 ) ( ) ( σ - = τ Τ χ τ Τ ν ν & & so that the temporal solution of the first mode: && T 0 ( t ) + 0 χ 2 Τ 0 ( τ 29 = 0 = && Τ 0 ( τ 29 Τ 0 ( τ 29 = β + χτ 8

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6- 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 σ χοσ σιν 29 ( + = 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 σ 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 ϖ n = s n c = np l E r , n =1,2,3,... 6.10 It 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 =1878x2 π rad/s=11800 rad/s The first natural frequency is given computed in Example 6.3.1, Equation (6.63) as ϖ 1 = 2 π λ Ε ρ Ε ρ = ϖ 1 2 2 π 2 = (1180029 2 2 π 2 Ε ρ = 7.143×10 7 in Nm/kg. Examining the ratios from Table 2.1 for the values given yields that for Steel: E ρ = 2 ×10 11 2.8 ×10 3 = 7.143×10 7 Νμ /κγ Thus a steel bar with a length 4.5 meters will have a first natural frequency of 1878 Hz. This is something like a truck chassis. 9
6- 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 ϖ ν = (2 ν -1 29 π 2 λ Ε ρ Referring to values of r and E from table 1.2 yields (for ϖ 1 ): Steel π (229 (1 29 2.0 ×10 11 7.8 ×10 3 = 7,954 rad/s (1266Hz) Aluminum π (229 (1 29 7.1×10 10 2.7 ×10 3 = 8,055 rad/s (1282 Hz) Wood π (229 (1 29 5.4 ×10 9 6.0 ×10 2 = 4,712 rad/s (750 Hz) Carbon composite (student must hunt for E / ρ and guess a little) from Vinson and Sierakowski’s book on composites ρ / E = 3118 and 4897 ) 3118 ( 2 = π rad/s (780 Hz) 6.12 Derive the boundary conditions for a clamped-free bar with a solid lumped mass, of mass M attached to free end.

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