Unit 2: Unit code: QCF Level: 4 Credit value: 5 Engineering Science L/60/404 OUTCOME 2 - TUTORIAL 3 FREE VIBRATIONS UNIT CONTENT OUTCOME 2 Be able to determine the behavioural characteristics of elements of dynamic engineering systems Uniform acceleration: linear and angular acceleration; Newton s laws of motion; mass moment of inertia and radius of gyration of rotating components; combined linear and angular motion; effects of friction Energy transfer: gravitational potential energy; linear and angular kinetic energy; strain energy; principle of conservation of energy; work-energy transfer in systems with combine linear and angular motion; effects of impact loading Oscillating mechanical systems: simple harmonic motion; linear and transverse systems; qualitative description of the effects of forcing and damping Students following the mechanical course will find this material is basically revision of work on basic dynamics from national level. Students who have not done basic mechanical dynamics might find it useful to study the modules on basic dynamics on the web site. You should judge your progress by completing the self assessment exercises. It is assumed that students doing this tutorial already understands BASIC DYNAMICS including velocity, acceleration, inertia, momentum and angular motion.most of this tutorial is revision of the material you should already have studied with some extra studies added to it. CONTENTS. Revise Basic Vibration Theory 2. Free Vibration Examples 3. Simple Harmonic Motion 4. Natural Vibrations 5. Simple Pendulum 6. Mass on Spring 7. Transverse Vibrations 8. Torsional Vibrations 9. Oscillation of a Floating Body 0. Forcing, Damping and Resonance
. REVISION OF BASIC VIBRATION THEORY. FREQUENCY measure the number of times something happens in a unit time normally second. Hence if something completes 20 cycles in second the frequency is 20 cycles/second. The cycle per second is called the Hertz. If something vibrates n times in t seconds the frequency is f = n/t Hertz. Calculate the frequency of a pendulum that oscillates 0 times in 8 seconds. PERIODIC TIME is the time taken for complete cycle and is denoted with T. If something oscillates n times in t seconds the periodic time is T = t/n = /f AMPLITUDE is the displacement of the varying quantity. This is best explained with a graph. In cases where the oscillation does not form a sine wave, the principle is the same. Figure SELF ASSESSMENT EXERCISE No. Calculate the periodic time of a pendulum that oscillates 0 times in 8 seconds. Mains electricity has a frequency of 50 Hz. What is the periodic time? What is the periodic time and amplitude shown on the graph?
2 FREE VIBRATIONS- EXAMPLES A free vibration is one that occurs naturally with no energy being added to the vibrating system. The vibration is started by some input of energy but the vibrations die away with time as the energy is dissipated. In each case, when the body is moved away from the rest position, there is a natural force that tries to return it to its rest position. Here are some examples of vibrations with one degree of freedom. Figure 2 Note that the mass on the spring could be made to swing like a pendulum as well as bouncing up and down and this would be a vibration with two degrees of freedom. The motion that all these examples perform is called SIMPLE HARMONIC MOTION (S.H.M.). This motion is characterised by the fact that when the displacement is plotted against time, the resulting graph is basically sinusoidal. Displacement can be linear (e.g. the distance moved by the mass on the spring) or angular (e.g. the angle moved by the simple pendulum). Although we are studying natural vibrations, it will help us understand S.H.M. if we study a forced vibration produced by a mechanism such as the Scotch Yoke.
3. SIMPLE HARMONIC MOTION The diagram shows a Scotch Yoke Mechanism which produces an up and down (reciprocating) motion of point P called simple harmonic motion. The wheel revolves at radians/sec and the pin forces the yoke to move up and down. The pin slides in the slot and Point P on the yoke oscillates up and down as it is constrained to move only in the vertical direction by the hole through which it slides. At any moment point P has a displacement x, velocity v and an acceleration a. Figure 3 The pin is located at radius R from the centre of the wheel. The vertical displacement of the pin from the horizontal centre line at any time is x. This is also the displacement of point P. The yoke reaches a maximum displacement equal to R when the pin is at the top and R when the pin is at the bottom. This is the amplitude of the oscillation. If the wheel rotates at radian/sec then after time t seconds the angle rotated is = t radians. Figure 4 From the right angle triangle we find x = R sin(t) Velocity is the rate of change of distance with time and in calculus form v = dx/dt. If we differentiate x we get v = dx/dt = R cos(t). The maximum velocity or amplitude is R and this occurs as the pin passes through the horizontal position and is plus on the way up and minus on the way down. This makes sense since the tangential velocity of a point moving in a circle is v = R and at the horizontal point they are the same thing. Acceleration is the rate of change of velocity with time and in calculus form a = dv/dt. Differentiating v we get a = dv/dt = - 2 R sin(t). Since R sin(t) = x then substituting x we find a = - 2 x This is the usual definition of S.H.M. The equation tells us that any body that performs sinusoidal motion must have an acceleration that is directly proportional to the displacement and is always directed to the point of zero displacement. The constant of proportionality is 2. Any vibrating body that has a motion that can be described in this way must vibrate with S.H.M. and have the same equations for displacement, velocity and acceleration. It was assumed that time started as the pin moved through the point x = 0 for the first time. We could start timing the motion at any point in the cycle so if the starting point occurred when the initial angle was φ the equations would be Displacement x = R sin(t + ). Velocity v = dx/dt = R cos(t + ) Acceleration a = dv/dt = - 2 R sin(t + )
Let s examine the graphs representing these equations. The graphs on the left show no initial starting angle and the graphs on the right show a starting angle of φ. Note how in each case, as we move from x to v and then to a, the graphs move 90 o to the left and acceleration is ½ cycle out of phase with the displacement. Figure 5
WORKED EXAMPLE No. The displacement of a body performing simple harmonic motion is described by the following equation x = A sin (t + ) where A is the amplitude, is the natural frequency and is the phase angle. Given A = 20 mm, = 50 rad/s and = /8 radian, calculate the following. i. The frequency. ii. The periodic time. iii. The displacement, velocity and acceleration when t = T/4. Sketch the raphs of x, v and a and confirm your answers. SOLUTION First deduce the frequency. f = /2 = 50/2 = 7.96 Hz. Next deduce the periodic time. T = /f = 0.26 s Next deduce the time t. t = T/4 = 0.034 s Next write out the equation for displacement and solve x at t = 0.034 s π x 20 sin (50 x 0.34) 8 π x 20sin.57 20sin.963 8.48 mm 8 Next write down the equations for v and a a - 20ω x 20sin ωt v 20ω cos 2 φ ωt φ 20 x 50 x cos.963-382.2 mm/s 2 2 φ - 20 x 50 sin(.963) - 46203 mm/s sin ωt The plots of x, v and a confirm these answers. Figure 6
4. NATURAL VIBRATIONS - EXPLANATION In the following work we will show how some simple cases of natural vibrations are examples of simple harmonic motion. Remember that one important point common to all of them is that there must be a natural force that makes the body move to the rest position. Another point common to all the following examples is that the body has mass (inertia) and that in order to accelerate, there must be an inertia force or torque present. A free vibration has no external energy added after it starts moving so it follows that all the forces and all the moment of force acting on the body must add up to zero. This is the basis of the analysis. 5. SIMPLE PENDULUM The restoring force in this case is gravity. When the pendulum is displaced through an angle, the weight tries to restore it to the rest position. This analysis is based on moment of force (torque). RESTORING TORQUE Figure 7 Weight = mg To find any moment we must always use the distance to the centre of rotation measured at 90 o (normal) to the direction of the force. In this case the distance is L sin Denote the torque as T g.this tries to return the mass to the rest position. T g = weight x mg (Lsin) INERTIA TORQUE Since the pendulum has angular acceleration as it slows down and speeds up, it requires an inertia torque to produce it. Denote this torque as T i. From Newton s second law for angular motion T i = I is the angular acceleration and I is the moment of inertia. The mass is assumed to be concentrated at radius L. (If it was not, the problem would be more complicated).the moment of inertia is then simply given as I = ml 2
BALANCE OF MOMENTS If there is no applied torque from any external source, then the total torque on the body must be zero. T g + T i = 0 mg Lsin + ml 2 = 0 mg Lsin = - ml 2 g Lsin = -L 2 The sine of small angles is very similar to the angle itself in radians (try it on your calculator). The smaller the angle, the truer this becomes. In such cases sin() = radians and so we may simplify the equation to g = -L = -(g/l) This meets the requirements for SHM since the acceleration is directly proportional to the displacement and the minus sign indicates that it is always accelerating towards the rest point. It follows that the constant of proportionality is so (g/l). 2 = g/l = (g/l) ½ If the displacement Note that the displacement in this example is angle and should not be confused with the angle on the Scotch Yoke. The frequency of oscillation is obtained from f = /2. Hence f 2 g L Note that the mass makes no difference to the frequency. On earth we can only change the frequency by altering the length L. If we took the pendulum to the moon, it would oscillate more slowly since gravity is smaller. In outer space where g is very close to zero, the pendulum would have no weight and would not swing at all if moved to the side. Remember also that the above equation is only true if the pendulum swings through a small angle. If the angle is large, the motion is not perfect S.H.M.
WORKED EXAMPLE No.2 A mass is suspended from a string 60 mm long. It is nudged so that it makes a small swinging oscillation. Determine the frequency and periodic time. SOLUTION f = (/2) (g/l) ½ f = (/2) (9.8/0.06) ½ = 2.035 Hz T = /f = 0.49 s
6 MASS ON SPRING Most natural oscillations occur because the restoring force is due to a spring. A spring is any elastic body which when stretched, compressed, bent or twisted, will produce a force or torque directly proportional to displacement. Examples range from the oscillation of a mass on the end of a spring to the motion of a tree swaying in the wind. Let's start with a simple mass suspended on a spring. Figure 8 Consider the mass is pulled down with a force F as shown. The spring is stretched a distance x o. This is called the initial displacement. When the mass is the released, it oscillates up and down with simple harmonic motion. Let s analyse the forces involved. F is the applied force in Newton. F s = Spring force that tries to return the mass to the rest position. From spring theory we know that F s = k x x is the displacement from the rest position at any time and k is the spring stiffness. Since the motion of the mass clearly has acceleration then there is an inertia force F i. From Newton s second law of motion we know that F i = mass x acceleration = M a Balancing forces gives F = F i + F s = M a + k x If the mass is disturbed and released so that it is oscillating, the applied force must be zero and this is the requirement for it to be a free natural oscillation. 0 = M a + k x Rearrange to make a the subject a = -(k/m) x This is the equation for SHM and tells us the acceleration is directly proportional to displacement and it is directed towards the rest position. If x was plotted against time, a sinusoidal graph would k result. The constant of proportionality is and this must be the angular frequency squared so M The frequency of oscillation is f ω ω k M k M Because this is a natural oscillation the frequencies are often denoted as n and f n. This equation is true for all elastic oscillations.
WORKED EXAMPLE No.3 A spring of stiffness 20 kn/m supports a mass of 4 kg. The mass is pulled down 8 mm and released to produce linear oscillations. Calculate the frequency and periodic time. Sketch the graphs of displacement, velocity and acceleration. Calculate the displacement, velocity and acceleration 0.05 s after being released. SOLUTION k 20000 ω 70.7 rad/s M 4 ω f.25 Hz T 0.0899 s f The oscillation starts at the bottom of the cycle so x o = -8 mm. The resulting graph of x against time will be a negative cosine curve with an amplitude of 8 mm. The equations describing the motion are as follows. x = x o cost When t = 0.05 seconds x = -8 cos(70.7 x 0.05) x = 7.387 mm. (Note angles are in radian) This is confirmed by the graph. If we differentiate once we get the equation for velocity. v = -x o sin t v = -x o sin t = -70.7 (-8)sin(70.7 x 0.05) v = -27 mm/s This is confirmed by the graph. Differentiate again to get the acceleration. a = - 2 x o cost and since x = x o cost a = - 2 x a = -70.7 2 x 7.387 = -36 934 mm/s 2 This is confirmed by the graph. Figure 9
SELF ASSESSMENT EXERCISE No.2. Calculate the frequency and periodic time for the oscillation produced by a mass spring system given that the mass is 0.5 kg and the spring stiffness is 3 N/mm. (2.3 Hz, 0.08 s). 2. A mass of 4 kg is suspended from a spring and oscillates up and down at 2 Hz. Determine the stiffness of the spring. (63.6 N/m). The amplitude of the oscillation is 5 mm. Determine the displacement, velocity and acceleration 0.02 s after the mass passes through the mean or rest position in an upwards direction. (.243 mm, 60.86 mm/s and -96.4 mm/s 2 ) 3. From recordings made of a simple harmonic motion, it is found that the frequency is 2 Hz and that at a certain point in the motion the velocity is 0.3 m/s and the displacement is 20 mm, both being positive downwards in direction. Determine the amplitude of the motion and the maximum velocity and acceleration. Write down the equations of motion. Note that the data given is at time t = 0. You will have to assume that x = x o cos(t + ) at time t=0 Ans. x= 0.03 cos(t - 50 o ) v = -0.394 sin(t - 50 o ) a = -57.9 x
7. TRANSVERSE VIBRATIONS A transverse oscillation is one in which the motion is sideways to the length. This occurs in shafts and beams. They may be solved as either angular or linear motion and linear is chosen here. CANTILEVER Figure 0 Consider a mass on the end of a cantilever beam. If the mass is moved sideways to the beam, the beam bends and it will always be found that the force is directly proportional to displacement. In other words, it is simple transverse spring. The stiffness of the beam may be found from beam theory and it can be shown that in the above case F 3EI k 3 x L k 3EI The theory is the same as a mass on the spring so f 3 M ML SIMPLY SUPPORTED BEAM Figure For a simply supported beam with a point load at the middle it can be shown that 48EI k and f 3 L STATIC DEFLECTION 48EI 3 ML The foregoing work can be simplified by simply measuring the static deflection produced by the addition of the weight. Suppose the mass deflects a distance x s under its own weight. The force is the weight so F = Mg. It follows that Mg k x s This formula works for any elastic system. Note that none of the forgoing work takes into account the mass of the beam or shaft and unless this is small in comparison to the concentrated mass, the formulae does not give accurate answers. f k M g x s
WORKED EXAMPLE No.4 A rod 20 mm diameter and.2 m long is rigidly fixed at one end and has a mass of 2 kg concentrated at the other end. Ignoring the weight of the rod, calculate the frequency of transverse vibrations. Take E = 200 GPa. SOLUTION First calculate the second moment of area. For a circular section : I πd 64 4 π x 0.02 64 4 7.854 x 0 9 m 4 f 3EI ML 3 9 3 x 200 x 0 x 7.854 x 0 3 2 x.2 9 5.88 Hz WORKED EXAMPLE No.5 A horizontal shaft may be treated as a simply supported beam and has a mass of 40 kg placed at the middle and it deflects mm under the weight. Ignore the mass of the shaft and calculate the frequency of transverse oscillations. SOLUTION f g x s 9.8 0.00 5.76 Hz SELF ASSESSMENT EXERCISE No.3. A simple cantilever has a mass placed on the end and it deflects.2 mm. Calculate the frequency of transverse vibrations that will occur. (4.39 Hz)
8. TORSIONAL OSCILLATIONS Note it is not clear if the syllabus intends this work to be covered so it may left at your discretion. A torsional oscillation occurs when a restoring torque acts on a body which is displaced by turning it about its axis. The restoring torque may be caused by twisting a shaft or by some other form of spring. A torsional spring produces a torque (T) directly proportional to the angle of twist (Q) such that T = k t. k t is the torsional spring constant and for a simple shaft this may be derived in terms of the material dimensions and properties. From the tutorial on torsion we found that: T Gθ T GJ from which the torsionalspring constant is k t J L θ L Remember that J is the polar second moment of area, L is the length and G is the modulus of rigidity for the material. For a solid round shaft J = D 4 /32. Consider a flywheel of moment of inertia I on the end of a vertical shaft. If a torque T is applied to the flywheel it must Figure 2 ) Overcome the inertia (inertia torque) by the law T = I 2) Twist the shaft (spring torque) by the law T = k t Hence T = I + K t When the flywheel is released, the torque is zero so 0 =I + K t k t Rearrange to make the acceleration () the subject α θ I This shows that the angular acceleration is directly proportional to displacement angle and always towards the rest position so the motion is S.H.M. The constant of proportionality 2 hence: ω 2 k t I f ω k I t
WORKED EXAMPLE No.6 A vertical shaft has a torsional stiffness of 57 Nm/rad. The shaft has a flywheel of moment of inertia 30 kg m 2 mounted on the end. Determine the frequency of the torsional oscillations. SOLUTION K t = 570.8 Nm/radian f = (/2)((K t /I)) /2 f = (/2)((570.8/30)) /2 =.52 Hz SELF ASSESSMENT EXERCISE No.4. A large rotating machine may be idealised as a shaft with a torsional stiffness of 3500 Nm/radian and moment of inertia 350 kg m 2. Calculate the frequency of natural torsional oscillations. (0.256 Hz)
9. OSCILLATION OF A FLOATING BODY You may have observed that some bodies floating in water bob up and down. This is another example of simple harmonic motion and the restoring force in this case is buoyancy. Consider a floating body of mass M kg. Initially it is at rest and all the forces acting on it add up to zero. Suppose a force F is applied to the top to push it down a distance x. The applied force F must overcome this buoyancy force and also overcome the inertia of the body. Figure 3 BUOYANCY FORCE The pressure on the bottom increases by p = g x. The buoyancy force pushing it up on the bottom is F b and this increases by p A. Substitute for p and F b = g x A INERTIA FORCE The inertia force acting on the body is F i = M a BALANCE OF FORCES The applied force must be F = F i + F b and this must be zero if the vibration is free. 0 Fi Fb Ma ρ g x A Make the acceleration the subject. ρag a - x M This shows that the acceleration is directly proportional to displacement and is always directed towards the rest position so the motion must be simple harmonic and the constant of proportionality must be the angular frequency squared. ρag ω 2 M ω f ω ρag M ρag M SELF ASSESSMENT EXERCISE No.5. A cylindrical rod is 80 mm diameter and has a mass of 5 kg. It floats vertically in water of density 036 kg/m 3. Calculate the frequency at which it bobs up and down. (Ans. 0.508 Hz)
0. FORCING, DAMPING AND RESONANCE FORCED OSCILLATIONS Machines and structures have natural frequencies. When something disturbs them, they tend to oscillate at their natural frequency. The disturbance may act in such a way as to make the oscillation build up. A good illustration of this is a child s swing. The swing is a pendulum with its own natural frequency. If pushed at the critical moment, the amplitude of the swing grows and grows. If the person stops pushing, the motion dies down as friction removes the energy. In engineering, rotating machinery that is not balanced will produce centrifugal force that makes the machine bounce about. In large machines such as turbines and generators, this could cause catastrophic damage. Such events occur when a disturbing force adds energy to the oscillation. Example of this are Wind induced vibrations in bridges, chimneys and overhead power lines. Vibrations from an engine making the deck of a ship vibrate. The wheel on a car getting out of balance. Unbalanced machinery. Musical instruments make use of these phenomena but generally forced vibrations are to be avoided as they cause damage. Friction can be our friend in this instance as it opposes the oscillation. Devices designed to produce friction and stop vibrations are called dampers or dashpots. They dissipate the energy as heat. An example is the shock absorber in a vehicle suspension. Without them, the vehicle will oscillate up and down on its suspension springs. DAMPING An ideal dashpot produces a force that opposes motion and this force is directly proportional to velocity v. The damping force is ideally given by F = c v where c is the damping coefficient and has units of N s/m The diagram shows how a damper might be fitted to damp the motion of a mass on a spring. (This could be an idealised car suspension system). Figure 4 A free natural vibration with a damper will produce a motion that decays in amplitude as the energy is dissipated. This is illustrated by the next diagram.
The diagram shows that depending on the damping, the motion will die out quickly or over a longer period of time. RESONANCE Figure 5 In an idealised forced oscillation we might think of an up and down force F applied to the mass. The diagram shows an unbalanced wheel rotating on a platform resting on a spring. The rotation will force the spring to move up and down. The system wants to vibrate at the natural frequency but the force acts at a frequency corresponding to the rotational frequency f. We will find that if the rotation is slow, the amplitude of the oscillation will be small. As we speed the wheel up the amplitude will grow and when we reach the natural frequency (f = fn) the oscillation will be large. If we increase the speed still further, the oscillation reduces and at very high speeds hardly any vibration at all will occur. This demonstrates resonance. When the forcing frequency is close to the natural frequency, we have resonance. The graph shows how the amplitude peaks at the resonant frequency. The result is affected by damping and the greater the damping, the smaller the peak. If large enough damping is used, the peak disappears altogether. Figure 6