Seventh Edition VECTOR MECHANICS FOR ENGINEERS: CHAPTER �� DYNAMICS Ferdinand P. Beer Mechanical Vibrations E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University � ������������������������������������������������������� ��
Edition Seventh Vector Mechanics for Engineers: Dynamics Contents Introduction Sample Problem 19.4 Free Vibrations of Particles. Simple Forced Vibrations Harmonic Motion Sample Problem 19.5 Simple Pendulum (Approximate Solution) Damped Free Vibrations Simple Pendulum (Exact Solution) Damped Forced Vibrations Sample Problem 19.1 Electrical Analogues Free Vibrations of Rigid Bodies Sample Problem 19.2 Sample Problem 19.3 Principle of Conservation of Energy � ������������������������������������������������������� �� 19 - 2
Edition Seventh Vector Mechanics for Engineers: Dynamics Introduction • Mechanical vibration is the motion of a particle or body which oscillates about a position of equilibrium. Most vibrations in machines and structures are undesirable due to increased stresses and energy losses. • Time interval required for a system to complete a full cycle of the motion is the period of the vibration. • Number of cycles per unit time defines the frequency of the vibrations. • Maximum displacement of the system from the equilibrium position is the amplitude of the vibration. • When the motion is maintained by the restoring forces only, the vibration is described as free vibration . When a periodic force is applied to the system, the motion is described as forced vibration . • When the frictional dissipation of energy is neglected, the motion is said to be undamped . Actually, all vibrations are damped to some degree. � ������������������������������������������������������� �� 19 - 3
Edition Seventh Vector Mechanics for Engineers: Dynamics Free Vibrations of Particles. Simple Harmonic Motion • If a particle is displaced through a distance x m from its equilibrium position and released with no velocity, the particle will undergo simple harmonic motion , ( ) = = − δ + = − ma F W k x kx st � � + = m x kx 0 • General solution is the sum of two particular solutions , � � � � k k � � � � = + x C sin t C cos t � � � � 1 2 � � � � m m ( ) ( ) = ω + ω sin cos C t C t 1 n 2 n • x is a periodic function and ω n is the natural circular frequency of the motion. • C 1 and C 2 are determined by the initial conditions: ( ) ( ) = ω + ω = x C sin t C cos t C x 1 n 2 n 2 0 ( ) ( ) = � = ω ω − ω ω 1 = ω v x C cos t C sin t C v 1 n n 2 n n 0 n � ������������������������������������������������������� �� 19 - 4
Edition Seventh Vector Mechanics for Engineers: Dynamics Free Vibrations of Particles. Simple Harmonic Motion v 0 = ω C 1 n = C x 2 0 � � + • Displacement is equivalent to the x component of the sum of two vectors C C 1 2 which rotate with constant angular velocity ω . n 2 2 ( ) ( ) = ω + φ = ω + = x x m sin t x v x amplitude m 0 n 0 n − 1 ( ) = φ = ω tan v x phase angle 0 0 n π 2 τ = = period n ω n ω 1 n = = = f natural frequency n τ π 2 n � ������������������������������������������������������� �� 19 - 5
Edition Seventh Vector Mechanics for Engineers: Dynamics Free Vibrations of Particles. Simple Harmonic Motion • Velocity-time and acceleration-time curves can be represented by sine curves of the same period as the displacement-time curve but different phase angles. ( ) = ω + φ x x m sin t n � = v x ( ) = ω ω + φ x cos t m n n ( ) = ω ω + φ + π x sin t 2 m n n � � = a x 2 ( ) = − ω ω + φ sin x t m n n 2 ( ) = ω ω + φ + π x sin t m n n � ������������������������������������������������������� �� 19 - 6
Edition Seventh Vector Mechanics for Engineers: Dynamics Simple Pendulum (Approximate Solution) • Results obtained for the spring-mass system can be applied whenever the resultant force on a particle is proportional to the displacement and directed towards the equilibrium position. • Consider tangential components of acceleration and force for a simple pendulum, � � � − θ = θ sin F = W ml ma : t t g � � θ + θ = sin 0 l for small angles, g � � θ + θ = 0 l ( ) θ = θ ω + φ sin t m n π 2 l τ = = π 2 n ω g n � ������������������������������������������������������� �� 19 - 7
Edition Seventh Vector Mechanics for Engineers: Dynamics Simple Pendulum (Exact Solution) g � � An exact solution for θ + θ = sin 0 l π 2 φ l d � τ = 4 leads to n g 2 2 ( ) − θ φ 1 sin 2 sin 0 m which requires numerical solution. � � 2 K l � � τ = π 2 � � n � � π g � ������������������������������������������������������� �� 19 - 8
Edition Seventh Vector Mechanics for Engineers: Dynamics Sample Problem 19.1 SOLUTION: • For each spring arrangement, determine the spring constant for a single equivalent spring. • Apply the approximate relations for the harmonic motion of a spring-mass system. A 50-kg block moves between vertical guides as shown. The block is pulled 40mm down from its equilibrium position and released. For each spring arrangement, determine a ) the period of the vibration, b ) the maximum velocity of the block, and c ) the maximum acceleration of the block. � ������������������������������������������������������� �� 19 - 9
Edition Seventh Vector Mechanics for Engineers: Dynamics Sample Problem 19.1 SOLUTION: = = k 4 kN m k 6 kN m 1 2 • Springs in parallel: - determine the spring constant for equivalent spring - apply the approximate relations for the harmonic motion of a spring-mass system 10 4 k N/m ω = = = 14 . 14 rad s n m 20 kg π 2 τ = τ = 0 . 444 s n n ω n = δ + δ = ω P k k v x 1 2 m m n ( )( ) = P = v 0 . 566 m s 0 . 040 m 1 4.14 rad s m = = + k k k 1 2 δ 2 = 4 a x a = = 10 kN m 10 N m m m n 2 ) 2 ( )( = a 8 . 00 m s = 0 . 040 m 1 4.14 rad s m � ������������������������������������������������������� �� 19 - 10
Edition Seventh Vector Mechanics for Engineers: Dynamics Sample Problem 19.1 • Springs in series: = = k 4 kN m k 6 kN m 1 2 - determine the spring constant for equivalent spring - apply the approximate relations for the harmonic motion of a spring-mass system k 2 400N/m ω = = = 6 . 93 rad s n m 20 kg π 2 τ = τ = 0 . 907 s n n ω n = ω v x m m n ( )( ) = δ + δ = = P k k 0 . 040 m 6 .93 rad s v 0 . 277 m s 1 2 m P = = + k k k 2 1 2 = a x a δ m m n 2 ) 2 4 ( )( = a 1 . 920 m s = = = 0 . 040 m 6 .93 rad s 10 kN m 10 N m m � ������������������������������������������������������� �� 19 - 11
Edition Seventh Vector Mechanics for Engineers: Dynamics Free Vibrations of Rigid Bodies • If an equation of motion takes the form � � 2 2 � � + ω = θ + ω θ = x n x 0 or 0 n the corresponding motion may be considered as simple harmonic motion. • Analysis objective is to determine ω n . • Consider the oscillations of a square plate � � � � ( ) ( ) − θ = θ + θ W b sin mb I [ ] 2 2 2 ( ) ( ) 1 2 = + = = but I m 2 b 2 b mb , W mg 12 3 3 3 g g � � � � θ + θ ≅ θ + θ = sin 0 5 b 5 b π 3 g 2 5 b ω = τ = = π then , 2 n n ω 5 b 3 g n • For an equivalent simple pendulum, l = 5 b 3 � ������������������������������������������������������� �� 19 - 12
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