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Introduction Springs are one of the most important components in vibration analysis because they are mechanical link between two particles in a mechanical system. In general a spring is a continuous system. As usual the spring has inertia but when we compare a vale of spring inertia to the other elements in the mechanical system we can see that this value is small so we can neglect it. The length of a spring when it is not subject to external forces is called the un-stretched length. Since the spring is made of a flexible material, the applied force F that must be applied to the spring to change its length by x is some continuous function of x. The appropriate form of the f(x) is determined by using the constitutive equation for the spring’s material. Since f(x) id infinitely differentiable at x = 0, it can be expanded by a Taylor series about x = 0. This procedure is called MacLaurin expansion: Since x is the spring’s change in length from its unscratched length, when x=0, F=0. Thus k0=0. When x is positive, the spring is in tension. When x is negative, the spring is negative, the spring is in compression. Many materials have the same properties in tension and compression. That is, if a tensile force F is required to lengthen the spring by, then a compressive force of the same magnitude F is required to shorten the spring by. For these materials, f(x) must be an odd function and cannot contain even powers. In reality all the springs are nonlinear. However in many situations x is small enough that the nonlinear terms of previous equation are small compared to the first variable that is k1*x. A linear spring obeys a force displacement law of where k is called the spring stiffness or spring constant and has dimensions of force pre length. The force in a spring whose force displacement law is given by (1.3) is conservative that is the work done by the spring force between two arbitrary displacement is independent of the path taken between the two displacement. Thus a potential energy function exists and for a linear spring is determined A torsional spring is a link in a mechanical system where application of a torque leads to an angular displacement between the ends of the torsional spring. A linear torsional spring has a relationship between an applied moment M and the angular displacement fi of where a torsional stiffness kt has dimensions of force times length. The potential energy function for a torsional spring is Example 1.1 The numerical values for the parameters in the system are: Determine the potential energy in the spring when the system is in the equilibrium. Develop an expression for the potential energy in the spring when the pulley is rotated clockwise from the equilibrium position Figure 1.1 – Free body diagram of static equilibrium position Let's start solving the example by defining A as statitc deflection of the spring, its change in length from its unstreched length when the system is in equilibrium. The static deflection is determined by applying hte equations of equilibrium to the free body diagram. The free body diagram show that it was assumed that the spring is compressed when the system is in equilibrium. The negative sign indiates that the spring is actually stretched. Its potential energy when the system is in equilibirim is: When the pulley is rotated through a clockwise angle θ from the system's equilibrium position, the spring is shortened in lenght 2rθ from its length when the system is in equilibrium. Thus the total change in lenght of the spring is and the potential energy in the spring is: Helical coil springs The helical coil spring is used in applications such as industrial machines and vehicle suspension systems. Consider a spring manufacture from a rod of circular cross section of diameter D. The shear modulus of the rod is G. The rod is formed into coil of N turns of radius r. It is assumed that the coil radius is much larger than the radius of the rod and that the normal to the plane of one coil nearly coincides with the axis of the spring. Figure 1.2. Free body diagram of cut coil spring exposes resultant shear force and resultant torque. Consider a helical coil spring when subject to an axis load F. Imagine cutting the rod with a knife at an arbitrary location in a coil, slicing the spring in two sections. The cut exposes an internal shear force F and an internal resisting torque Fr, as illustrated in previous figure. Assuming elastic behavior, the shear stress due to the resisting torque varies linearly with distance from the center of the rod to a maximum of Where J = Pi*D^4/32 is the polar moment of inertia of the rod. The shear stress due to the shear force varies nonlinearly with distance from the neutral axis. For r/D>>1 the maximum shear stress due to the resisting torque, and its effect is neglected. Principles of mechanics of materials can be used to show that the total change in length of the spring due to an applied force F Comparing Eq.(1.9) with Eq.(1.4) leads to the conclusion that the under the assumptions stated a helical coil spring can be modeled as a linear spring of stiffness. Example 1.2 A tightly wound spring is made from a 20 mm diameter bar of 0.2% C hardened steel (G=80e9 N/m^2). The coil diameter is 20 cm. the spring has 30 coils. What is the largest force that can be applied such that the elastic strength in shear of 220 e6 N/m^2 is not exceeded? What is the change in length of the spring when this force is applied? Assuming the shear force is negligible, the maximum shear stress in the spring when a force F is applied is: Thus the maximum allowable force is: The stiffness of this spring is calculated by: The total change in length of the spring due to application of the maximum allowable force is: Elastic elements as spring Application of a force F to the block of mass m of figure 1.3, results in a displacement x. The block is attached to a uniform thin rod of elastic modulus E, un-stretched length L, and cross sectional area Application of the force results in uniform normal strain in the rod of: The total strain energy developed due to the work of the force in stretching the rod is: If the force is suddenly removed, the block will oscillate about its equilibrium position. The initial strain energy is converted to kinetic energy and vice versa, a process which continues indefinitely. If the mass of the rod is small compared to the mass of the block, then inertia of the rod is negligible and the rod behaves as a discrete spring. From strength of materials, the force F required to change the length of the rod by x is: A comparison of Eq. (1.4) implies that the stiffness of the rod is: Figure1.3 – Longitudinal vibrations of mass attached to end of uniform thin rod can be modeled as a linear mass-spring system with k=AE/L The motion of a particle attached to an elastic element can be modeled as a particle attached to a linear spring, provided the mass of the element is small compared to the mass of the particle and a linear relationship between force and displacement exists for the element. In figure 1.4 A particle of mass m is attached to the midspan of a simply supported beam of length L, elastic modulus E, and cross-sectional moment of inertia I. The transverse displacement of the midspan of the beam due to applied static load F is Thus a linear relationship exists between transverse displacement and static load. Hence if the mass of the beam is small, the vibrations of the particle can be modeled as the vertical motion of a particle attached to a spring of stiffness. In general the transverse vibrations of a particle attached to a beam can be modeled as those of a particle attached to a linear spring. Let w(z) represent the displacement function of the beam due to a concentrated unit load applied at z = a. then the displacement at z=a due to a load F applied at z =a is: Then the spring stiffness for a particle placed at z = a is Figure 1.4 – a) The transverse vibrations of a block attached to a simply supported beam, b) are modeled by the mas spring system, provided the mas of the block beam is small compared to the mass of the block Figure 1.5 – Torsional stiffness of the shaft is JG/L Torsional oscillations occur in the system of Fig 1.5. A thin disk of mass moment of inertia I is attached to a circular shaft of length L, shear modulus G and polar moment of inertia J. When the disk is rotated through an angle from its equilibrium position, a moment develops between the dis and the shaft. Thus if the polar mass moment of inertia of the shaft is small compared with I, then the shaft acts as a torsional spring of stiffness: Example 1.3 A 200 kg machine is attached to the end of a cantilever beam of length L =2.5 m. Elastic modulus E=200e9 N/m^2, and cross sectional moment of inertia 1.8e-6m^4. Assuming the mass of the beam is small compared to the mass of the machine, what is the stiffness of the beam? Spring in combination Most of vibrational problems include more than one spring. That means they are in some combination. In this section will talk about what combinations of springs exist and how to solve them. The solution is very simple. You need to replace this combination of springs with one which has equivalent stiffness. The equivalent stiffness is determined such that the system with a combination of springs has the same displacement x as the equivalent system when both systems are subject to the same force F. A model of one degree of freedom system consisting of a block attached to a spring of an equivalent stiffness is illustrated in next figure. The resultant force acting on the block is : The springs in the system are parallel. The displacement of each spring in the system is the same, but the resultant force acting on the block is the sum of the forces developed in the parallel springs. If x is the displacement of the block, then the force developed in the i-th spring is ki*x and the resultant is Equating the forces form leads to The springs shown in next figure are in series. The force developed in each spring is the same and equal to the force acting on the block. The displacement of the block is the sum of the changes in length of the springs in the series combination. If xi is the change in length of the i-th spring then Since the force is the same in each spring, xi=F/ki and eq becomes Since the series combinations is to be replaced by a spring of an equivalent stiffness is used leading to We can see similarity with electrical components because electrical circuit component can also be placed in series and parallel and effect of the combination replaced by a single component with and equivalent value. The equivalent capacitance of capacitors in parallel or series is calculated like that of springs in parallel or series. The equivalent resistance of resistors in series is the sum of the resistances, whereas the equivalent resistance of resistors in parallel is calculated by using an equation similar to Eq.(1.26).
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1.7.Opruge Opruga je fleksibilna mehanička veza dviju čestica u mehaničkom sustavu. U stvarnosti opruga je kontinuirani sustav. Međutim, inercija opruge je obično mala uspoređujući njenu veličinu inercije s ostalim elementima u mehaničkom sustavu pa se iz tog razloga zanemaruje. Duljina opruge je ona duljina kada opruga nije potdvrgnuta opterećenju i naziva se nerastegnuta/početna duljina. U slučaju da je opruga izrađena od elstičnog materijal, sila koja dijeluje na oprugu F izaziva promjenu duljine opruge za x je kontinuirana funkcija od x. Ispravna formulacija određuje se koristeći konstitutivne jednadžbe za materijal od kojeg je opruga napravljena. Razvojem funkcije f(x) u Taylorov red dobivamo: Kao što smo prije naveli x je razlika između početne duljine opruge i njene konačne vrijednosti. Kada je x = 0, tada je i F=0 tada je i k0 = 0. Ako je x pozitivan, tada se opruga rasteže, a kada je x negativan tada se opruga tlači. Većina materijala ima iste značajke kod kompresije i rastezanja. To znači da vrijednost sile F koja je potrebna za rastezanje opruge mora biti ista i za njeno stezanje. Za takve materijale funkcija f(x) je neparna te slijedi: Sve opruge su inherentno nelinearne. Međutim u većini situacija x je jako mali tako da se neparan nelinearni sustav može svesti na samo jedan član. Linearna opruga se ponaša prema zakonu sile-pomaka: Sila koja djeluje na oprugu je konzervativna, a rad koji se vrši na oprugu je neovisan o putu između početne i krajnje duljine opruge. Funkcija potencijalne energije postoji i za lineranu oprugu iznosi:gdje je k krutost opruge ili konstanta opruge i njena je jedinica N/m. Torzijska opruga je veza u mehaničkom sustavu kod kojeg tozija dovodi do kutne deformacija krajeva opruge. Krutost linearne torzijske opruge proporcionalna je momentu M koji dijeluje na oprugu a obrnuto proporcionalna kutnoj deformaciji θ. Funkcija potencijalne energije za torzijsku oprugu definira se sljedećim izrazom: Zavojne torzijske opruge Zavojna torzijska opruga koristi se kod industrijskih strojeva i amortizacije vozila. Zamislimo orugu koja je izrađena od štapa čiji je poprečni presjeka promjera D, modula smicanja G. Štap se formira u oprugu s N koraka radiusa r. Slika 1.7 – Izometrijski prikaz zavojne,torzijske opruge Ako sada zavojnu torzijsku oprugu opteretimo aksijalnom silom F. Zamislimo da oprugu presječemo s nožem na proizvoljnoj poziciji. Sada kad smo podijeli oprugu na dva dijela na jednom djelu pojaviti će se unutarnje posmične sile F i unutarnji moment otpora S pretpostavkom da se radi o elastičnom odzivu opruge, tangencijalno će naprezanje zbog momenta otpora linearno varirati uodnosu na udaljenost od centra štapa prema maksimalnoj vrijednosti koja iznosi: U prethodnoj formuli je polarni moment inercije štapa. Tangencijalno naprezanje zbog posmične sile varira nelinearno s udaljenosti od neutralne osi. Za r/D>>1 maksimalno tangencijalno naprezanje koje se javlja zbog unutranje posmične sile puno je manja nego maksimalna vrijednost posmičnog napreznaja koje se javlja zbog momenta otpora i zbog toga se zanemaruje. Principi mehanike materijala mogu se koristi da bi pokazali ukupnu promjenu duljine opruge kada na oprugu dijeluje sila F i ona glasi: Uspoređujući jednadžbu 1.35 sa jednadžbom 1.30 dovodi do zaključka da se zavojna torzijska opruga može modelirati kao linearna opruga krutosti: Elastični elementi Dijelovanjem sile F na blok mase m rezultira pomakom x. Blok je spojen s tankim štapom modula elastičnosti E, početne duljine L, i poprečnog presjeka A. Dijelovanjem sile na sustav pojaviti će se naprezanja i deformacija u štapu. Ukupna energija deformacije nastaje zbog rada koji sila vrši na produljenje štapa. Ako sila prestane dijelovati na tijelo, blok će početi oscilirati oko ravnotežnog položaja. Inicijalna energija deformacije pretvara se u kinetičku energiju i obrnuto. Ako je masa štapa mala uspoređujući je s masom bloga, tada će inercija štapa biti zanemariva i štap će se ponašati kao diskretna opruga. Usporedbom varijabli 1.39 s jednadžbom 1.30 dobivamo jednadžbu krutosti štapa: Gibanje čestica sustava koje čine elastični element mogu se modelirati kao čestice linearne opruge, osiguravajući masu elementa malom u odnosu na masu čestice. Slika 1.8 – Sustav s elastičnim elementom i tijelom mase m Razmotrimo sustav u kojem tijelo mase m pričvrščeno na gredu duljine L, modula elastičnosti E, i momenta inercije poprečnog presjeka I. Poprečni pomak središnjeg djela grede pod utjecajem dijelovanje statičnog opterećenja F iznosi: Linearni odnos postoji između poprečnog pomaka i statičkog opterećenja. Ako je masa grede mala, tada se vibracije čestica grede mogu modelirati kao vertikalno gibanje u opruzi krutosti: Općenito poprečne vibracije čestica grede mogu se modelirati kao čestice linearne opruge. Neka predstavlja funkciju pomaka grede zbog opterećenja koje dijeluje na udaljenosti z = a. Tada će pomak z = a zbog sile F koja djeluje na toj udaljenosti biti: U sustavu koj se sastoji od diska mase m, momenta inercije I pričvrščen je za vratilo duljine L, modula smicanja G, i polarnog momenta inercije J. Kada disk pomakne iz ravnotežnog položaja za kut θ, pojavljuje se moment između diska i štapa: Ako je polarni moment inercije ptapa mali u usporedi s I, tada se vratilo ponaša kao torzijska opruga krutosti: 4.Kombinacija opruga Često se u praksi pojavljuju sustavi s oprugama u kombinaciji. Za analizu i proračun takvih vibracija potrebno je zamijeniti sustav opruga s jednom oprugom ekvivalentne krutosti Ekvivalentna krutost određuje se na temelju pretpostavke da ekvivalentna opruga mora imati isti pomak x kao i sustav opruga kada su podvrgnuti sili F. Model s jednim stupnjem slobode gibanja sastoji se od bloka mase m koji je pričvrščen za oprugu ekvivalentne krutosti. Rezultantna sila koja dijeluje na blok mase m: Opruge na sljedećoj slici su paralelnom spoju. Pomak svake opruge u sustavu je isiti, ali je rezultantna sila koja dijeluje na blok mase m suma svih sila koje se razvijaju u oprugama. Ako je x pomak bloka, tada je sila u i-toj opruzi kix ,a rezultanta: Ekvivalentnu krutost paralelong spoja opruga možemo dobiti : Osim paralenog spoja opruga postoji i serijski spoj. Sila koja se razvija u svakoj opruzi je jednaka onoj ukupnoj sili koja djeluje na blok. Pomak bloka je suma svih pomaka opruga u serijskom spoju. Ako je xi promjena duljine i-te opruge tada je: Budući da je sila za sve opruge jednaka vrijed: Budući da nas za analizu nekog sustava s više opruga u serijskom spoju zanima ekvivalentna krutost nju ćemo dobiti: