Ch1-1(1-2)e - MAE351 Mechanical Vibrations Lecture 1 (Chap....

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1 MAE351 Mechanical Vibrations ecture 1 Lecture 1 hap 1 1 2) (Chap. 1.1-1.2) 1
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2 asic Mechanical Elements Basic Mechanical Elements of Vibrations m x c k m = mass k = stiffness amping 2 c = damping (Courtesy of Dr. D. Russell)
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3 Stiffness • From strength of materials, recall: ooke’sLaw: x 0 x 1 x 2 x () k f kx t   Hooke s Law: g 3 10 3 N f k f k nonlinear (or F) 10 N yield point quasi-linear 3 20 mm 0 x 0 20 mm x linear (w/o spring offset) (or x)
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4 ree ody diagram and equations Free-body diagram and equations of motion • Newton’s Law: k x y c m Friction-free surface x f ck mx f f  mg k f c 4 N
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5 tiffness and Mass Stiffness and Mass Vibration is cause by the interaction of two different forces one related to position (stiffness) and one related to acceleration (mass). roportional to displacement Displacement Stiffness ( k ) Proportional to displacement () f kx t   m k x Mass ( m ) statics k Mass Spring dynamics m f ma t mx t  5 Proportional to acceleration
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6 quation of Motion Equation of Motion From Newton’s Law for this simple mass-spring system, e two forces must be equal i e f the two forces must be equal i.e. f m = f k . isplacement ) ( ) ( t kx t x m k x Displacement 0 ) ( ) ( t kx t x m or m Mass Spring This is a 2nd order, ordinary differential equation (ODE) and all phenomena that have differential equations of this type for their quation of motion will exhibit 6 equation of motion will exhibit oscillatory behavior .
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7 xamples of Single egree f Examples of Single-Degree-of- Freedom Systems Pendulum Shaft and Disk l =length Torsional Stiffness k oment m Gravity g Moment of inertia J 2 n m l m g l  g sin o Mm  7 0 ) ( ) ( t l t 0 ) ( ) ( t k t J
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8 efinition of Vibration Vibration = “Mechanical oscillation about a reference position, Definition of Vibration which is a result of dynamic forces of that system” X otal ibratory Component, (t) Total Deflection Vibratory Component, x(t) Static Deflection, x st X total (t) = x st + x(t) t Time
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9 Solution to 2nd order DEs Lets assume a solution: x(t) () () 0 mx t kx t  ) t A t x n sin( ) ( Differentiating twice gives: t ) ( sin( ) ( cos( ) ( t x t A t x t A t x n n 2 2 - ) ) ( n n n Substituting back into the equation of motion gives: k t kA t A m n n n r ) ) 0 2 2 sin( sin( Natural 9 m k m n n or 0 frequency
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10 Summary of simple harmonic motion x(t) Period ) t A t x n sin( ) ( n T 2 Amplitude A t 0 x Slope here is v 0 Maximum Velocity n A n z cycles rad/s n n n 10 Hz s rad/cycle 2 2 2 n f
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Ch1-1(1-2)e - MAE351 Mechanical Vibrations Lecture 1 (Chap....

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