J. Marcondes, in
Encyclopedia of Vibration, 2001 Products are tested, without any packaging, to determine their inherent ability to sustain shock, vibration and
compression forces. In many cases, due to the lack of prototypes for testing, the information may be estimated, based on the properties of the components of the product. However, due to the complexity of most products, it is best to perform product-testing in order to obtain the most accurate data possible. This is important because any inaccurate information used about the product will produce a nonoptimal design, which will result either in product damage or overpackaging (often never
detected). Typically, one determines the ability of the product to sustain impacts without damage. This level, in acceleration units, g, is the ‘product shock fragility’ (see SHOCK).
To obtain vibration data about the product, a prototype is placed on a vibration tester and a test is conducted to identify the resonant frequency of its components, in each of the orthogonal directions (Figure 4). There are basically two methods for this; resonance search using a sine sweep test, and monitored component random vibration. In the sine sweep test, components of the product are observed for resonance (visually or audibly). In the random vibration method, components
need to be monitored by mounting a small accelerometer on the components, and determining the vibration transmission between product and component. The added mass of the accelerometer needs to be taken into account, using: Figure 4. Product fragility – vibration sensitivity. [5]ωRc=ωRca(mc+ma/mc)0.5 Obtaining information about the product compression resistanceA compression test may be used to determine the product's ability to sustain compression forces. This is done by applying orthogonal compression forces until product failure or some other predetermined limit is achieved. Read full chapter URL: https://www.sciencedirect.com/science/article/pii/B0122270851001612 Denny–Brown StabilizerW. Burger M.Sc. Extra Master, A.G. Corbet Extra Master, in Ship Stabilizers, 1966 Principles of OperationA schematic diagram is shown in Fig. 7.8.  FIG. 7.8. Multra control system. The three main sensors detect roll angle, roll velocity and roll acceleration. The precession of the roll and the velocity gyro are converted into electrical signals via linvars. The acceleration unit is a servo unit and derives its signals by differentiating the movements of the roll velocity gyroscope. The transmitter CX2 is operated by the helm. It is normal that small movements of helm of up to say 5° either side do not affect the control but when helm is increased beyond this point, linear signals are derived which may provide maximum stabilizing power for full helm. Each linvar and also the output from the acceleration amplifier (Fig. 6.19) and the helm transmitter (if fitted) are connected to the tappings of a potentiometer. The potentiometers are connected in series to create a summation circuit where the various signals are added (or subtracted) according to the sign of the function each represents. The output from the summation circuit is applied to an amplifier which energizes the output transformer of the control equipment. The output transformer is connected to the operating circuit of the fin. The operating circuit of one fin is shown in Fig. 7.8—the operating circuit(s) of the other fin(s), being, of course, identical. The signal is passed to a phase-conscious rectifier and the output from this is connected to the operating armature coil of a differential electromechanical relay (see inset). This relay is polarized with d.c. and excitation given to the armature by the signal will unbalance the flux through the airgap. A force is so produced and the armature is deflected proportionally both in sense and magnitude to the d.c. current applied. This movement will operate the pilot valve of the hydraulic relay. The output arm of this relay actuates the rotary servo valve of the tilting pump (p. 94). The fin feedback system employs the synchro CX1. Mechanical movement of the rotor arm is converted into an electrical reset signal which is fed back to the hydraulic relay. The input controlling the hydraulic relay at any moment is the difference between the control and the resetting (feedback) signals. The desired cut-off ratio can be arranged by making the sensitivity of the initial stages of amplification such that full control of the hydraulic relay is obtained with say one-quarter of the maximum signal. This represents a 4:1 cut-off ratio. Read full chapter URL: https://www.sciencedirect.com/science/article/pii/B9780080115047500137 A Coupling Procedure of Finite Element and Scaled Boundary Finite Element Methods for Soil–Structure Interaction in the Time Domain1Junyi Yan, ... Feng Jin, in Seismic Safety Evaluation of Concrete Dams, 2013 AbstractA coupling procedure of finite element and scaled boundary finite element (SBFE) methods is presented for three-dimensional dynamic analysis of unbounded soil–structure interaction in the time domain. Based on linear system theory, the acceleration unit-impulse response matrix of the unbounded soil calculated by the SBFE method is converted into time-independent matrices; thus, the time history of the unit-impulse response function of the unbounded medium can be replaced by a state-variable description. Interaction forces between unbounded soil and structure are evaluated by linear equations instead of time-consuming convolution integrals. Since only partial information on the unit-impulse response function is needed to capture the character of the unbounded medium, and the history of the response function can be truncated, the computational effort spent in the SBFE method can be reduced. The accuracy of the procedure can be controlled with prescribed parameters. Numerical examples of foundation responses agree with previous results and are more efficient than the direct convolution integral method. Read full chapter URL: https://www.sciencedirect.com/science/article/pii/B9780124080836000064 Fluid DynamicsNICOLAE CRACIUNOIU, BOGDAN O. CIOCIRLAN, in Mechanical Engineer's Handbook, 2001 1.2.2 BRITISH ENGINEERING SYSTEM OF UNITSThe fundamental units in this system (called the FPS system) are foot (ft), pound (lb), and second (sec), corresponding to the length, force, and time fundamental mechanical dimensions. Other units are ft3 for the unit volume, ft/sec2 for unit acceleration, ft-lb for unit work, and lb/ft2 for unit pressure. The unit for mass is called the slug. The slug can be derived from Newton's second law applied for freely falling mass, namely weight (lb) = mass (slugs) ×g (32.2 ft/sec2. then, mass (slugs) = weight (lb)/g (32.2 ft/sec2). Thus, 1 slug = lb-sec2/ft. The temperature unit is the degree Fahrenheit (F) or, on the absolute scale, the degree Rankine (R). Read full chapter URL: https://www.sciencedirect.com/science/article/pii/B9780124713703500097 Sensors for ControlClarence W. de Silva, in Encyclopedia of Physical Science and Technology (Third Edition), 2003 IX Piezoelectric TransducersSome substances, such as barium titanate and single-crystal quartz, can generate an electrical charge and an associated potential difference when they are subjected to mechanical stress or strain. This piezoelectric effect is used in piezoelectric transducers. Direct application of the piezoelectric effect is found in pressure and strain measuring devices, and many indirect applications also exist. They include piezoelectric accelerometers and velocity sensors and piezoelectric torque sensors and force sensors. It is also interesting to note that piezoelectric materials deform when subjected to a potential difference (or charge). Some delicate test equipment (e.g., in vibration testing) use piezoelectric actuating elements (reverse piezoelectric action) to create fine motions. Also, piezoelectric valves (e.g., flapper valves), directly actuated using voltage signals, are used in pneumatic and hydraulic control applications and in ink-jet printers. Miniature stepper motors based on the reverse piezoelectric action are available. Consider a piezoelectric crystal in the form of a disc with two electrodes plated on the two opposite faces. Since the crystal is a dielectric medium, this device is essentially a capacitor, which may be modeled by a capacitance C. Accordingly, a piezoelectric sensor may be represented as a charge source with a series capacitive impedance, in an equivalent circuit. The impedance from the capacitor is given by (26)Z=1jωC. As is clear from Eq. (26), the output impedance of piezoelectric sensors is very high, particularly at low frequencies. For example, a quartz crystal may present an impedance of several megohms at 100 Hz, increasing hyperbolically with decreasing frequencies. This is one reason why piezoelectric sensors have a limitation on the useful lower frequency. The other reason is the charge leakage. The sensitivity of a piezoelectric crystal may be represented either by its charge sensitivity or by its voltage sensitivity. Charge sensitivity is defined as (27) Sq=∂q∂F=1A∂q∂p. Here q denotes the generated charge and F denotes the applied force, for a crystal with surface area A, where p is the stress (normal or shear) or pressure applied to the crystal surface. Voltage sensitivity Sv is given by the change in voltage due to a unit increment in pressure (or stress) per unit thickness of the crystal. Thus, in the limit, we have (28) Sv=1d∂v∂p, where d denotes the crystal thickness. The sensitivity of a piezolectric element is dependent on the direction of loading. This is because the sensitivity depends on the crystal axis. Sensitivities of several piezoelectric material along their most sensitive crystal axis are listed in Table III. TABLE III. Sensitivities of Several Piezoelectric Material
IX.A Piezoelectric AccelerometerNow we will discuss a piezoelectric motion transducer—the piezoelectric accelerometer—in more detail. A piezoelectric velocity transducer is simply a piezoelectric accelerometer with a built-in integrating amplifier in the form of a miniature integrated circuit. Accelerometers are acceleration-measuring devices. It is known from Newton's second law that a force (f) is necessary to accelerate a mass (or inertia element), and its magnitude is given by the product of mass (M) and acceleration (a). This product (Ma) is commonly termed inertia force. The rationale for this terminology is that if a force of magnitude Ma were applied to the accelerating mass in the direction opposing the acceleration, then the system could be analyzed using static equilibrium considerations. This is known as d'Alembert's principle. The force that causes acceleration is itself a measure of the acceleration (mass is kept constant). Accordingly, mass can serve as a front-end element to convert acceleration into a force. This is the principle of operation of common accelerometers. There are many different types of accelerometers, ranging from strain gage devices to those that use electromagnetic induction. For example, force that causes acceleration may be converted into a proportional displacement using a spring element, and this displacement may be measured using a convenient displacement sensor. Examples of this type are differential-transformer accelerometers, potentiometer accelerometers, and variable-capacitance accelerometers. Alternatively, the strain, at a suitable location of a member, that was deflected due to inertia force may be determined using a strain gage. This method is used in strain gage accelerometers. Vibrating-wire accelerometers use the accelerating force to tension a wire. The force is measured by detecting the natural frequency of vibration of the wire (which is proportional to the square root of tension). In servo force-balance (or null-balance) accelerometers, the inertia element is restrained from accelerating by detecting its motion and feeding back a force (or torque) to exactly cancel out the accelerating force (torque). This feedback force is determined, for instance, by knowing the motor current, and it is a measure of the acceleration. The advantages of piezoelectric accelerometers (also known as crystal accelerometers) over other types of accelerometers are their light weight and high-frequency response (up to about 1 MHz). However, piezoelectric transducers are inherently high-output-impedance devices that generate small voltages (on the order of 1 mV). For this reason, special impedance-transforming amplifiers (e.g., charge amplifiers) have to be employed to condition the output signal and to reduce loading error. A schematic diagram for a compression-type piezoelectric accelerometer is shown in Fig. 25. The crystal and the inertia mass are restrained by a spring of very high stiffness. Consequently, the fundamental natural frequency or resonant frequency of the device becomes high (typically 20 kHz). This gives a reasonably wide useful range (typically up to 5 kHz). The lower limit of the useful range (typically 1 Hz) is set by factors such as the limitations of the signal-conditioning systems, the mounting methods, the charge leakage in the piezoelectric element, the time constant of the charge-generating dynamics, and the signal-to-noise ratio. A typical frequency response curve for a piezoelectric accelerometer is shown in Fig. 26. FIGURE 25. A compression-type piezoelectric accelerometer. FIGURE 26. A typical frequency response curve for a piezoelectric accelerometer. In compression-type crystal accelerometers, the inertia force is sensed as a compressive normal stress in the piezoelectric element. There are also piezoelectric accelerometers that sense inertia force as a shear strain or tensile strain. For an accelerometer, acceleration is the signal that is being measured (the measurand). Hence, accelerometer sensitivity is commonly expressed in terms of electrical charge per unit acceleration or voltage per unit acceleration. Acceleration is measured in units of acceleration due to gravity (g), and charge is measured in picocoulombs (pC), which are units of 10−12 coulombs (C). Typical accelerometer sensitivities are 10 (pC)/g and 5 mV/g. Sensitivity depends on the piezoelectric properties and on the mass of the inertia element. If a large mass is used, the reaction inertia force on the crystal will be large for a given acceleration, thus generating a relatively large output signal. Large accelerometer mass results in several disadvantages, however. In particular, 1.The accelerometer mass distorts the measured motion variable (mechanical loading effect). 2.A heavy accelerometer has a lower resonant frequency and, hence, a lower useful frequency range (Fig. 26). For a given accelerometer size, improved sensitivity can be obtained by using the shear-strain configuration. In this configuration, several shear layers can be used (e.g., in a delta arrangement) within the accelerometer housing, thereby increasing the effective shear area and, hence, the sensitivity in proportion to the shear area. Another factor that should be considered in selecting an accelerometer is its cross-sensitivity or transverse sensitivity. Cross-sensitivity primarily results from manufacturing irregularities of the piezoelectric element, such as material unevenness and incorrect orientation of the sensing element. Cross-sensitivity should be less than the maximum error (percentage) that is allowed for the device (typically 1%). The technique employed to mount the accelerometer to an object can significantly affect the useful frequency range of the accelerometer. Some common mounting techniques are 1.Screw-in base 2.Glue, cement, or wax 3.Magnetic base 4.Spring-base mount 5.Hand-held probe Drilling holes in the object can be avoided by using the second through fifth methods, but the useful range can decrease significantly when spring-base mounts or hand-held probes are used (typical upper limit of 500 Hz). The first two methods usually maintain the full useful range, whereas the magnetic attachment method reduces the upper frequency limit to some extent (typically 1.5 kHz). Piezoelectric signals cannot be read using low-impedance devices. The two primary reasons for this are 1.High output impedance in the sensor results in small output signal levels and large loading errors. 2.The charge can quickly leak out through the load. The charge amplifier is the commonly used signal-conditioning device for piezoelectric sensors. This device was discussed previously. Because of impedance transformation, the impedance at the charge amplifier output becomes much smaller than the output impedance of the piezoelectric sensor. This virtually eliminates loading error. Also, by using a charge amplifier circuit with a large time constant, charge leakage speed can be decreased. Read full chapter URL: https://www.sciencedirect.com/science/article/pii/B0122274105001393 Embedded Software for Networking ApplicationsSrinivasa Addepalli, in Software Engineering for Embedded Systems, 2013 AcceleratorsPEHs also include several acceleration blocks to take the load away from the cores on routine and computationally intensive algorithmic tasks. There are three types of acceleration engines in PEHs – look-aside accelerators, ingress accelerators, and egress accelerators. Look-aside accelerators are the ones used by the software running in cores while processing the packets. Cryptography, pattern matching, compression/decompression, de-duplication and timer management are some examples of acceleration units that are normally included in multicore SoCs. Ingress acceleration is done by PEHs on the incoming packets before they are handed over to software running on the cores, in order to save core cycles in certain tasks. Parsing of headers and making extracted values of header fields available along with the packets to software is one type of ingress acceleration. Similar to this, some other popular ingress accelerations in PEHs include packet header integrity checks, reassembly of IP fragments into full IP packets, aggregation of consecutive TCP segments and de-tunneling. Egress acceleration is done by PEMs before transmitting packets to the interfaces. Cores, after processing the packets, send the packets to the PEH. The PEH then applies actions before sending them out. Quality of service shaping, link aggregation and tunneling are some of the acceleration functions that are part of PEHs. Read full chapter URL: https://www.sciencedirect.com/science/article/pii/B9780124159174000244 Using the Luenberger Observer in Motion ControlGeorge Ellis, in Control System Design Guide (Fourth Edition), 2012 18.3.2 Experiment 18E: Using Observed Acceleration FeedbackExperiment 18E models the acceleration-feedback system of Figure 18.21 (see Figure 18.22). The velocity loop uses the observed velocity feedback to close the loop. A single term, KTEst/JTEst, is formed at the bottom center of the model to convert the current-loop output to acceleration units; to convert the observed acceleration to current units, the term 1 + KAFB is formed via a summing block (at bottom center) and used to scale the control law output, consistent with Figure 18.20. An explicit clamp is used to ensure that the maximum commanded current is always within the ability of the power stage. The Live Scope shows the system response to a torque disturbance (the command generator defaults to zero). The observer from Experiment 18A is used without modification. Figure 18.22. Experiment 18E: Using observed acceleration feedback. The improvement of acceleration feedback is evident in the step-disturbance response as shown in Figure 18.23. Recall that without acceleration feedback, the control law gains were set as high as was practical; the non-acceleration-feedback system took full advantage of the reduced phase lag of the observed velocity signal. Even with these high control law gains, acceleration feedback produces a substantial benefit as can be seen by comparing Figures 18.23a and 18.23c. Figure 18.23. Response to a 5-Nm step disturbance without and with acceleration feedback: (a) without acceleration feedback (KAFB = 0.0); (b) with minimal acceleration feedback (KAFB = 1.0); (c) with more acceleration feedback (KAFB = 10.0). The disturbance response Bode plot of the system with and without acceleration feedback is shown in Figure 18.24. The acceleration feedback system provides substantial benefits at all frequencies below about 500 Hz. Figure 18.24 shows two levels of acceleration feedback: KAFB = 1.0 and KAFB = 5. This demonstrates that as the KAFB increases, disturbance response improves, especially in the lower frequencies, where the idealized model of Figure 18.19 is accurate. Note that acceleration feedback tests the stability limits of the observer. Using the observer as configured in Experiment 18E, KAFB cannot be much above 1.0 without generating instability in the observer. The problem is cured by reducing the sample time of the observer through changing the Live Constant “TSAMPLE” to 0.0001. This allows KAFB to be raised to about 15 without generating instability. It should be pointed out that changing the sample time of a model is easy but may be quite difficult in a working system. See Question 18.5 for more discussion of this topic. Figure 18.24. Bode plot of velocity loop disturbance response without (KAFB = 0) and with (KAFB = 1.0, 10.0) acceleration feedback. Read full chapter URL: https://www.sciencedirect.com/science/article/pii/B9780123859204000187 Motion and InertiaK.M. SMITH C.ENG., M.I.E.E., P. HOLROYD C.ENG., M.I.MECH.E., A.M.I.STRUCT.E., in Engineering Principles for Electrical Technicians, 1968 5.6 Momentum, inertia, mass, and weightIn Chapter 1 we considered the equilibrium of a set of forces, the forces being either stationary or moving at a constant speed. Let us now investigate the action of a body when a gradually increasing force is applied to it. Take a 1-lb mass and place it on a smooth table (Fig. 5.5). Attach a piece of string to the mass and let the other end of the string hang over the side of the table with a weight hanger fastened to it. Apply weights to the weight hanger until the 1-lb mass just moves at a constant speed along the table. The system is now in equilibrium, the weights on the weight hanger being sufficient to overcome the frictional resistance of the 1-lb mass on the table. FIG. 5.5. Now add an extra weight on to the hanger and notice the response of the 1-lb mass; it begins to move faster and faster, that is, it accelerates. The extra force on the string has produced an acceleration on the 1-lb mass. Another weight added to the hanger will increase the acceleration of the 1-lb mass. Sir Isaac Newton stated a second law relating the applied force and the resulting acceleration. “The rate of change of momentum is proportional to the impressed force.” To be able to apply this law requires certain terms to be defined and understood. MOMENTUMMomentum is defined as the quantity of motion possessed by a body and is measured by the product of the mass of the body and its velocity. INERTIAInertia is a measure of the resistance a body has to change of motion. A tanker full of petrol requires a larger force to accelerate it than does an empty tanker. Similarly, to stop a full tanker requires a larger braking force than to stop the empty one, i.e. the full tanker has more inertia than the empty one. MASSThe mass of a body is the numerical measure of the inertia of the body. The only way to alter the mass of the body is to cut a portion away or to add more matter to the body. WEIGHTThe weight of a body is a measure of the earth's attraction on the body or the gravitational force exerted on the body. It was stated in Chapter 1 that the gravitational attraction decreases as the bodies move away from each other; therefore it follows that the weight of a body will alter depending on where the body is weighed. There will be a slight difference in the weight of an object at the top of a high mountain to the weight in a deep pit. Newton's second law states Force∝ rate of change of momentum. Let υ1=initial velocity(ft/s ),υ2=final velocity(ft/s),m=mass (lb),t=time to change from υ1 to υ2(sec), then first momentum = mv1 second momentum = mv2. Change in momentum = m( υ2−υ1). Rate of change of momentum =m(υ2− υ1)t, but acceleration is defined as υ2−υ1t=f. Therefore the rate of change of momentum = mass × acceleration. From Newton's second law; force ∝ rate of change of momentum. (12)Force∝mass×acceleration. (13)Force=a constant×mass×acceleration. 5.7 Systems of unitsTo be able to use expression (13) requires the value of the constant to be found, or alternatively the constant can equal unity if a force is chosen which will give a unit mass a unit acceleration. This forms the basis of the different systems of units when dealing with problems involving forces and accelerations. A force of 1 lbf is the gravitational attraction on a mass of 1 lb. If we take a mass of 1 lb and let it fall freely it will have an acceleration of g ft/s2 and therefore we can write 1 lbf (force) = k (constant) × 1 lb (mass) × g ft/s2 (acceleration) ∴k=1(lbf) 1(lb)×g(ft/s2)=1lbf-s2glb-ft. The following systems take the value of k = 1 and derive new units for either force or mass. FOOT-POUND-SECOND (F.P.S.) SYSTEMS1. Absolute system of units. New unit of force. Unit of force is the poundal (pdl). Unit of mass is the pound (lb). Unit of acceleration is the foot per second per second (ft/s2). Definition. The poundal is that force which, when acting on a mass of one pound, produces an acceleration of one foot per second per second. Force=mass×acceleration1 (pdl)=l (lb)×l (ft/s2).Hencegpdl=1 lbf. 2.Gravitational or Engineer's system of units. New unit of mass. Unit of force is the pound-force (lbf). Unit of mass is the slug. Unit of acceleration is the foot per second per second (ft/s2). Definition. The slug is that mass which, when acted on by a force of one pound-force, has an acceleration of one foot per second per second. Force=mass×acceleration1 (lbf)=l (slug)×l (ft/s2).Hencegslug=glb or a body weighing W lbf has a mass of W lb or W g slugs.METRE-KILOGRAMME-SECOND (M.K.S.) SYSTEM3. Absolute system of units. New unit of force. Unit of force is the newton (N). See also Section 10.8. Unit of mass is the kilogramme (kg). Unit of acceleration is the metre per second per second (m/s2). Definition. The snewton is that force which, when acting on a mass of one kilogramme, produces an acceleration of one metre per second per second. Force=mass×acceleration1(N)=1(kg)×1(m/s2). Other systems have been used in the past such as the centimetre-gramme-second (c.g.s.) system, but the m.k.s. system has replaced these. The International System of units (SI) is to be adopted for use in this country and a gradual change-over will take place in the next few years. Read full chapter URL: https://www.sciencedirect.com/science/article/pii/B9780080129853500108 Materials: The Stuff That Surrounds UsMichael Ashby, Kara Johnson, in Materials and Design (Third Edition), 2014 Noise ManagementSound is caused by vibration; its pitch depends on its frequency. The (youthful) human ear responds to frequencies from about 20 to about 20,000 hz, corresponding to wavelengths of 17 m to 17 mm. The bottom note on a piano is 28 hz, the top note 4186 hz. The most important range, from the point of view of acoustic design, is roughly 500–4000 hz. Sound pressure is measured in Pascals (pa), but because audible sound pressure has a range of about 106 it is more convenient to use a logarithmic scale with units of decibels (db). Figure 4.10 shows levels, measured in db. Sound levels above 50 db can impair concentration. Sustained sound levels above 90 db cause damage to hearing. 4.10. How Loud Is Loud?
Vibration, like noise, becomes damaging above a critical threshold. The level of vibration is characterized by the acceleration (units, m/sec2) associated with it; safe practice requires levels below 1 m/sec2 (0.1 g where g is the acceleration due to gravity). The frequency, too, is important: less than 1 hz gives sea-sickness; 1–100 hz gives breathing difficulties and headaches; 10–1000 hz causes loss of sense of touch or “white finger”. Duration, too, is important – the longer the duration, the lower the level should be. The mechanism for reducing sound and vibration levels within a given enclosed space (a room, for instance) depends on where it comes from. If it is generated within the room, one seeks to absorb the sound. If it comes from outside, one seeks to insulate the space to keep the sound out. And if it is transmitted through the frame of the structure itself (deriving from a machine tool, or from traffic), one seeks to isolate the structure from the source of vibration. In product design it is usually necessary to absorb sound. Soft porous materials absorb incident, airborne sound waves, converting them into heat. (Sound power, even for a very loud noise, is small, so the temperature rise is negligible.) Porous or highly flexible materials such as low density polymer and ceramic foams, plaster, and fiberglass absorb well; so do woven polymers like carpets and curtains. The proportion of sound-absorbed by a surface is called the “sound-absorption coefficient”. A material with a coefficient of 0.8 absorbs 80% of the sound that hits it, reflecting 20% back; one with a coefficient of 0.03, absorbs only 3% of the sound, reflecting 97%. Figure 4.11 shows sound-absorption coefficients for a number of materials. 4.11. Soaking up Sound
Vibration is damped by isolating the source of vibration from the rest of the structure with high-damping rubber mounts, foams, elastomer grips, etc. A composite of rubber filled with cork particles is a good choice here. The low shear modulus of the rubber isolates shear waves, and the compressibility of the cork adds a high impedance to compressive waves. Read full chapter URL: https://www.sciencedirect.com/science/article/pii/B9780080982052000044 Damping considerations in tall buildingsAlberto Lago, ... Antony Wood, in Damping Technologies for Tall Buildings, 2019 3.1.2 Multiple Degrees-of-Freedom SystemWhen more than one independent coordinates (i.e., more than one displacement modes) are required to express the overall motion of a system, an MDOF system is needed to idealize the structure (Fig. 3.6). Figure 3.6. An idealized three DOF system. 3.1.2.1 Equations of motionSimilar to the SDOF system, the equation of motion of an MDOF system can be developed on the basis of the equilibrium of effective forces corresponding to each degree of freedom (DOF): (3.24)FIi(t)+ Fdi(t)+FSi(t)=Fi(t) where FIi(t) is the inertial force associated with the ith DOF, Fdi( t) the damping force associated with the ith DOF, FSi(t) the spring force associated with the ith DOF, and Fi(t) the applied dynamic force associated with the ith DOF (due to wind or earthquake). Assuming that the MDOF system has N degrees of freedom, the inertial force, damping force, and spring force associated with the ith DOF are expressed as follows, respectively: (3.25)FIi(t)=mi1·ü1(t)+mi2·ü2(t )+…+miN·üN(t) (3.26)Fdi(t)= ci1·u̇1(t)+ci2·u̇2(t)+…+ciN·u̇ N(t) (3.27)FSi(t)=ki1·u1(t)+ki2·u2(t )+…+kiN·uN(t ) where i =1 to N (i.e., a set of N equations is obtained for each of the above forces) and mij mass associated with the ith DOF due to unit acceleration along the jth DOF (i.e., mass influence coefficient) cij damping constant associated with the ith DOF due to unit velocity along the jth DOF (i.e., damping influence coefficient) kij stiffness associated with the ith DOF due to unit displacement along the jth DOF (i.e., stiffness influence coefficient) Substituting Eqs. (3.25)–(3.27) in Eq. (3.24), the equation of motion in matrix form can be expressed as: (3.28)Mü(t)+CU˙(t)+KU(t)=F (t) where M is the mass matrix (N × N), C the damping matrix (N × N), K the stiffness matrix (N × N), F(t) the force vector (N × 1), Ü(t) the acceleration vector (N × 1), U̇(t) the velocity vector (N × 1), and U(t) the displacement vector (N × 1). The format of above matrices and vectors is as follows, respectively: (3.29)M=[ m11⋯m1N⋮⋱⋮mN1⋯m NN] (3.30)C=[c11⋯c1N⋮⋱⋮cN1⋯cNN] (3.31)K=[k11⋯ k1N⋮⋱⋮k N1⋯kNN] (3.32)F(t)=[F1(t)⋮FN(t) ] (3.33)Ü(t)=[ü1(t)⋮ü N(t)] (3.34)U̇(t)=[u̇1(t)⋮u̇N(t) ] (3.35)U(t)=[u1(t)⋮uN( t)] 3.1.2.2 Free vibration propertiesFor an MDOF system with N DOF, a series of N independent (displacement) vectors, ϕ, can be identified. These are called natural modes of vibration or mode shapes. Fig. 3.7 illustrates the mode shapes of an idealized two-story building structure with two DOF (Only horizontal displacements are considered). Figure 3.7. Natural vibration modes of an idealized two DOF building system. Consider the undamped, free-vibration case, for example, the damping matrix is zero (i.e., C=0), and the external force vector is a zero vector (i.e., F(t)=0) in Eq. (3.28), undamped, free vibration of the MDOF system can be expressed by: (3.36) MÜ(t)+KU( t)=0 The response of the structure can be modeled by the multiplication of two independent variables: the set of mode shape vectors ϕm(x) that control the shape of the deformation at location x, and a scalar function of time qm(t) that controls the amplitude of the motion. Mathematically, this can be written as: (3.37)U(t)=ϕq(t) where ϕ is a matrix containing all the mode shapes, and q(t) is a vector containing all the individual functions qm(t). It can be shown (Chopra, 2002) that the solution of Eq. (3.36) requires qm( t) to be harmonic; therefore, substituting a harmonic function of circular frequency ωm for the time-dependent displacement [qm(t)=Amcosωmt+ Bmsinωmt] in Eq. (3.37), and then the resulting expression in Eq. (3.36), the following equation is obtained after mathematical manipulations: (3.38)[−ωm2Mϕm +Kϕm]qm(t)=0 The constants Am and Bm can be obtained using the initial conditions of displacement and velocity. Since qm(t)=0 is its trivial solution, the solution of following real eigenvalue problem gives the vibration properties of the MDOF system: (3.39)[K−ωm2M]ϕ m=0 Neglecting the trivial solution (ϕm=0), the nontrivial solution of the above equation reads: (3.40) det|K−ωm2M|=0 which results in N real and positive frequencies ωm (m=1,2,…,N) with the arrangement ordered as ω1<ω2<…<ωN. Having determined the frequencies ωm, the solution of Eq. (3.39) gives the corresponding vectors ϕm (mode shapes). Note that the solution of Eq. (3.39) for ϕm represents the shape of a vector with relative values (not absolute values) of N displacements (degrees of freedom): (3.41)ϕm={ϕ1mϕ2m ϕ3m⋮ϕNm} Assembling all the N frequencies and corresponding Nm modes into a compact matrix form, the spectral (diagonal) matrix of eigenvalue problem Ω2 and the modal matrix Φ are expressed, respectively, by: (3.42)Ω2=[ω120⋮0 0ω22⋮0……⋱… 00⋮ωN2] (3.43)Φ=[ϕ 11ϕ21⋮ϕN1 ϕ12ϕ22⋮ ϕN2……⋱…ϕ1Nm ϕ2Nm⋮ϕ NNm] An important property of the natural modes is the orthogonality of different modes, that is, ωm ≠ωr, as expressed below: (3.44)ϕmTK ϕr=0ϕmTMϕr=0 where ϕmT is the transpose of the mth mode shape vector. The orthogonality condition demonstrates that the following matrices are diagonal: (3.45)K¯=ΦTKΦM¯=ΦTMΦ where the diagonal elements (generalized, modal, stiffness, and mass) read: (3.46)K¯m=ϕmTKϕm=ωm2 M¯mM¯m=ϕmTMϕm The normalization of natural modes is an important process to standardize the elements associated with different DOF. There are several approaches to normalize each mode: •Normalize by the amplitude of the DOF with the largest modal amplitude •Normalize by the modal amplitude at the roof •Normalize to achieve M ¯m=ϕmTMϕm= 1 (computer programs commonly apply this approach) In addition, for earthquake engineering purposes, it is useful to define the mode participation factor as follows: (3.47)Γm=ϕmTMrM¯m where r is the influence vector which takes into account how the ground acceleration is transmitted to the different DOF. If the ground motion is purely horizontal (no rotational input at the base), and only horizontal DOF are considered, then r becomes a vector of 1s and induces a rigid body motion in all modes. It can be shown that the product of the mode participation factor and the mode shape of the mth mode, Γmϕm, is independent of how the modes are normalized. Therefore it can be a better tool to understanding the behavior of the building than the mode shape alone as it gives insight into its lateral deflected shape. Fig. 3.8 shows the product Γmϕm for a generic shear building. It can be seen, for example, how the first and second modes have opposite signs at the roof, and therefore, their sum diminishes the total response of the building at that level. This product is sometimes referred to as the effective mode shape of the building, and it is primarily affected by the lateral load-resisting system of the building and the distribution of stiffness along its height (Miranda and Akkar, 2005). Figure 3.8. Effective mode shapes for the first four modes of a generic shear building. Adapted from Cruz, C., 2017. Evaluation of Damping Ratios Inferred from the Seismic Response of Buildings. PhD Thesis Stanford University, California.For a damped MDOF system, free vibration leads to the following equation of motion [by substituting F(t)=0 into Eq. (3.28)]: (3.48)MÜ(t)+CU̇(t)+KU(t)=0 Solutions to the above equation may differ depending on whether the format of the damping matrix is classical or nonclassical. The damping matrix is classical if the following condition is met (Caughey and O’Kelly, 1965): (3.49)CM−1K=KM −1C In this case, all the natural modes of the system are identical to those of the undamped system. For a classically damped MDOF system, all the modes are uncoupled, and thus premultiplying Eq. (3.48) by ΦT , the following equation of motion is obtained: (3.50)M¯q¨(t)+C¯q̇(t)+K¯q(t)=0 where M ¯ and K¯ are defined in Eq. (3.45), and the damping matrix is diagonal: (3.51)C¯ =ΦTCΦ Eq. (3.50) gives N uncoupled differential equations in modal coordinates qm(t) for a classically damped system, as follows: (3.52)M¯mq¨m(t)+C¯mq̇m(t)+K ¯mqm(t)=0 where M¯m and K¯m are expressed in Eq. (3.46) and the generalized (modal) damping is given by: (3.53) C¯m=ϕmTC ϕm Classical modal analysis is valid for a classically damped MDOF system. Indeed, for each mode (e.g., mth mode), a damping ratio can be derived in the way similar to the case of the SDOF system in Section (3.1.1), which is expressed as: (3.54)ζm =C¯m2M¯mωm Dividing Eq. (3.52) by M¯m, the following equation in a modal form is found: (3.55)q¨m(t)+2ζmωmq̇m(t) +ωm2qm(t)=0 The solution of the above equation is identical to that of Eq. (3.6) of an SDOF system and reads: (3.56)qm(t)=e−ζmωmt[q m(0)cosωmDt+ q̇m(0)+ζmωmqm(0)ωmDsinωmDt] where the damped frequency corresponding to the mth mode is defined by: (3.57) ωDm=ωm1−ζm 2 If the damping matrix does not satisfy Eq. (3.49), then the system is considered to have nonclassical damping. In this case, the natural modes are complex (nonreal values), and the damping matrix C is not diagonal. To solve the equation of motion for a nonclassically damped system, other procedures, for complex eigenvalue analysis, are employed. This is beyond the scope of this book, and interested readers should refer to Chopra (2012). 3.1.2.3 Forced vibration propertiesThe equation of motion of a damped MDOF system (Eq. (3.28)) can be solved in the similar way as an SDOF system mentioned above. If the MDOF system with N DOF is excited by both an external force to the main body, F(t), and a ground acceleration force to the base, −M1u¨g(t), the equation of motion of the system reads: (3.58)MU¨(t)+CU̇(t)+KU(t)=F(t)−M1u¨g(t) where 1 is called the effective vector of order N with elements equal to unity. Note that this equation is valid only if the building has a fixed condition at the base, that is, the flexibility of the soil is neglected (typical assumption). The above matrix-form equation contains N differential equations, each of which corresponds to a particular DOF. Therefore the relative displacement associated with each DOF can be computed using numerical methods, similar to those applied for the SDOF system in Section 3.1.1.3. Note that the total acceleration of the system is represented by the combination of relative acceleration of the main body (building stories) and the ground acceleration u¨g(t) as follows (Fig. 3.9): Figure 3.9. Schematic accelerations in a (A) building frame and (B) tower-like system. (3.59)U¨t(t)=u¨g(t)1+U¨(t) Read full chapter URL: https://www.sciencedirect.com/science/article/pii/B9780128159637000038 Which of the following is the acceleration?The rate of change of velocity is known as acceleration. The rate of change of displacement is known as velocity.
What are 3 examples of acceleration?Examples. An object was moving north at 10 meters per second. ... . An apple is falling down. ... . Jane is walking east at 3 kilometers per hour. ... . Tom was walking east at 3 kilometers per hour. ... . Sally was walking east at 3 kilometers per hour. ... . Acceleration due to gravity.. Which of the following is a unit of acceleration?Unit of acceleration is the metre per second per second (m/s2).
What is the acceleration answer?Acceleration is defined as. The rate of change of velocity with respect to time. Acceleration is a vector quantity as it has both magnitude and direction. It is also the second derivative of position with respect to time or it is the first derivative of velocity with respect to time.
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Which of the following is the of acceleration?
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