SolSec 6.5 - 6- 37 Problems and Solutions Section 6.5 (6.40...

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6- 37 Problems and Solutions Section 6.5 (6.40 through 6.47) 6.40 Calculate the natural frequencies and mode shapes of a clamped-free beam. Express your solution in terms of E , I , ρ , and l . This is called the cantilevered beam problem. Solution: Clamped-free boundary conditions are wt w l tw l t x x xx xxx (,) 00 0 0 −= = = and assume E, I, ρ , l constant. The equation of motion is + = 2 2 4 4 0 w t EI A w x assume separation of variables ) ( ) ( ) , ( t q x t x w φ = to get EI A q q ω ′′′′ =− = ˙˙ 2 The spatial equation becomes ′′′′ − = ωφ A EI 2 0 define 0 that so 4 2 4 = = β EI A which has the solution: x C x C x C x C cosh sinh cos sin 4 3 2 1 + + + = Applying the boundary conditions: φφ () 0 0 = = ′′ = ′′′ = and ll yields that C 2 + C 4 = 0 C 1 + C 3 = 0
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6- 38 −− + + = −++ + = Cl C l C l C l C l C l ll l l c c c c 12 3 4 123 4 1 2 3 4 0 0 0101 1010 sin cos sinh cosh cos sin cosh sinh sin cos sinh cosh cos sin cosh sinh ββ β = 0 For a nonzero solution, the determinant must be zero to that (after expansion) = = sin sinh cos cosh cos cosh sin sinh ( sin sinh )(sin sinh ) ( cos cosh )( cos cosh ) ββββ l llll 0 Thus the frequency equation is cos l cosh l = -1 or l l n n cosh 1 cos = and frequencies are A EI n n ρ ω 4 = . The mode shapes are x C x C x C x C n n n n n n n n n φ cosh sinh cos sin 4 3 2 1 + + + = Using the boundary condition information that 1 3 2 4 and C C C C = = yields ) cosh (cos ) sinh (sin cosh sinh cos sin 2 1 2 1 2 1 l l C l l C l C l C l C l C + = + so that CC =− + + cos cosh sin sinh and the mode shapes can be expressed as: nn C xx + + + 2 cos cosh sin sinh sin cos + + + cos cosh sin sinh sinh cosh
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6- 39 6.41 Plot the first three mode shapes calculated in Problem 6.40. Next calculate the strain mode shape [i.e., ) ( x X ], and plot these next to the displacement mode shapes X ( x ). Where is the strain the largest? Solution: The following Mathcad session yields the plots using the values of β taken from Table 6.4. The strain is largest at the free end.
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6- 40 6.42 Derive the general solution to a fourth-order ordinary differential equation with constant coefficients of equation (6.100) given by equation (6.102).
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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 6.5 - 6- 37 Problems and Solutions Section 6.5 (6.40...

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