ASE 365 - HW 6 Solutions

ASE 365 - HW 6 Solutions - Solutions Homework Set 6 1 Draw...

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Unformatted text preview: Solutions: Homework Set 6 1. Draw FBD of m . Horizontal spring force: k (¯ x- x ). Vertical spring force: k (¯ y- y ). Force in inclined spring: k [0 . 6(¯ x- x ) + 0 . 8(¯ y- y )]. Use Newton’s second law in both x and y directions and rearrange to get EOMs: bracketleftbigg m m bracketrightbiggbraceleftbigg ¨ x ¨ y bracerightbigg + k bracketleftbigg 1 . 36 . 48 . 48 1 . 64 bracketrightbiggbraceleftbigg x y bracerightbigg = k bracketleftbigg 1 . 36 . 48 . 48 1 . 64 bracketrightbiggbraceleftbigg ¯ x ¯ y bracerightbigg 2. Setting det( K- ω 2 M ) = 0 gives a quadratic characteristic equation whose two roots give the natural frequencies ω 2 1 = k/m and ω 2 2 = 2 k/m . From ( K- ω 2 1 M ) u 1 = and ( K- ω 2 2 M ) u 2 = , obtain the eigenvectors u 1 = braceleftbigg- . 8 . 6 bracerightbigg , u 2 = braceleftbigg . 6 . 8 bracerightbigg The first eigenvector represents the first mode of vibration, which is simply vibration in the direction perpendicular to the inclined spring. The second mode is vibration aligned with the inclined spring. 3. Representing the displacement vector as braceleftbigg x ( t ) y ( t ) bracerightbigg = 2 summationdisplay i =1 u i η i ( t ) = [ u 1 u 2 ] braceleftbigg η 1 ( t ) η 2 ( t ) bracerightbigg = U η ( t ) and the acceleration vector in terms of the modal matrix U as well, and multiplying the EOMs by U T , gives, after forming the matrix products, the uncoupled modal EOMs bracketleftbigg m m bracketrightbiggbraceleftbigg ¨ η 1 ¨ η 2 bracerightbigg + bracketleftbigg k 2 k bracketrightbiggbraceleftbigg η 1 η 2 bracerightbigg = k bracketleftbigg- . 8 . 6 1 . 2 1 . 6 bracketrightbiggbraceleftbigg ¯ x ¯ y bracerightbigg (Note that having a fully populated matrix on the right-hand side of the equation does...
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