PART I. Basic Concepts

Transcription

1 PART I. Basic Concepts. Introduction. Basic Terminology of Structural Vibration.. Common Vibration Sources.. Forms of Vibration.3 Structural Vibration

2 . Basic Terminology of Structural Vibration The term vibration describes repetitive motion that can be measured and observed in a structure Unwanted vibration can cause fatigue or degrade the performance of the structure. Therefore it is desirable to eliminate or reduce the effects of vibration In other cases, vibration is unavoidable or even desirable. In this case, the goal may be to understand the effect on the structure, or to control or modify the vibration, or to isolate it from the structure and minimize struc-tural response.

3 ... Common Vibration Sources Industrial operations such as pressing or forging are common sources of ground-borne vibration It can be critically important to ensure that the natural frequencies of the structural system do not match the operating frequencies of the equipment The dynamic amplification which can occur if the frequencies do coincide can lead to hazardous structural fatigue situations, or shorten the life of the equipment.

4 ... Common Vibration Sources

5 .. Forms of Vibration Free vibration is the natural response of a structure to some impact or displacement Forced vibration is the response of a structure to a repetitive forcing function that causes the structure to vibrate at the frequency of the ecitation Sinusoidal vibration is a special class of vibration. The structure is ecited by a forcing function that is a pure tone with a single frequency Random vibration is very common in nature. The vibration you feel when driving a car result from a comple combination of the rough road surface, engine vibration, wind buffeting the car's eterior, etc

6 Figure..Sinusoidal vibration (top) and random vibration (bottom)

7 .3 Structural Vibration K C M The simplest vibration model is the single-degree-offreedom, or mass-spring-damper model It consists of a sim-ple mass (M) that is suspended by an ideal spring with a known stiffness (K) and a dashpot damper from a fied support If you displace the mass by pulling it down and releasing it, the mass will respond with motion similar to Fig.. 4

8 Figure.4. Free Vibration of Mass-Spring-Damper Model

9 .3 Structural Vibration Frequency is measured in cycles per second with units of Hertz (Hz). Assuming the damping is small, then the mathematical relationship is given by () ω = n K M A larger stiffness will result in a higher n, and a larger mass will result in a lower n. Usually damping is described in terms of the damping ratio--. That ratio is related to C by () ζ = C Mωn

10 Modal shapes Beam st Mode shape nd Mode shape Pinned Ends 3rd Mode shape Fig.. 5 A beam fied at both ends. Each mode shape corresponds to a peak frequency at which all points move in unison The higher modes are harder to detect and have less effect on the overall vibration of the structure.

11 PART II. Vibration of Plates. General Theory. Transverse Vibration of a rectangular plate.

12 . General Theory In three-dimensional elasticity theory the stress at a point is specified by the si quantities: σ σ σ the components of direct stress;,, z y σe, σyz, σz the components of shear stress.

13 Z J dy H z D C Y y F B 0 d A X Fig... Stress components on the face ABCD of an element

14 The components of direct stress and strain are related by Hooke's law, etended to include Poisson's ratio effects, ε = E σ v( σy + σz) (3) with analogous epressions for ε ε, The y and, z components of shear stress and strain are related by ε y = σ G y, etc (4) In these equationse,g andv are the elastic constants, Young's modulus, shear modulus (or modulus of rigidity) and Poisson's ratio, respectively. E= G( + v) (5)

15 ε = Increase in length Initial length of = [ u + ( u / ) d ] d d = u (6) Similarly, y v y ε z = w z ε = and

16 Y u D C D A B v A B y X Fig... Displacement in the plane OXY

17 The deformed shape of the element is the parallelogram A B C D and the shear strain is (a + ). Thus for small angles v u εy = α+ β = + y (7) Similarly, ε yz = w + v y z and ε u w z = + z

18 For an element of sides d, dy and dz (Fig. 5.) the force due to the stress σ is σ dydz ; if the corresponding strain is ε, the etension in the X-direction is ε d and the work done on the element is ( σ dydz)( ε d) = σ ε dv (8) Using, ε y = α + β, the work done on the element is σ y ε y dv Generalizing for a three-dimensional state of stress, the strain energy in an elastic body δ = ( σ ε + σ yε y + σ zε z + σ yε y + σ yzε yz + σ zε z ) dv V (9)

19 . Transverse Vibrations of Rectangular Plate Consider the plate element shown below Fig.(.5 )..The plate is thin and of uniform thickness h; thus the free surfaces of the plate are the planes z=± h.the direct stress in the transverse direction σ zero. This stress must be zero at the free surfaces, and provided that the plate is thin, it is reasonable to assume that it is zero at any section z. z is

20 Stresses in the middle plane of the plate (membrane stresses) are neglected, i.e. transverse forces are supported by bending stresses, as in fleure of a beam. Plane sections that are initially normal to the middle plane remain plane and normal to it. Thus with this assumption the shear strains σ z and σ yzare zero. Only the transverse displacement w (in the Z- direction) has to be considered.

21 Z B C Z B C O A z (a) R O (b) Fig..3. Element of plate. (a) Unstrained. (b) Strained A M y M Z O S M y B d y d Y S y M y My + dy y M y S y pyft (,)() S + y Sy dy M y My + dy y + S d A C M M + d M y My + d Fig..5 Forces and moments on an element of a plate h X If OA represents the middle surface of the plate, then OA = from the third assumption; also = Rdθ, where R is the radius of curvature of the deformed middle surface. Thus the strain in BC, at a distance z from the middle surface,

22 Thus the strain in BC, at a distance z from the middle surface, ε BC BC R + z R z R = = = BC R The relation between the curvature and the displacement of the middle surface, w, is: R = w Thus ε z w = (0) Similarly ε y = z w y

23 Thus the strain in BC, at a distance z from the middle surface, ε ( ) B C BC R + z R z BC R R = = = The relation between the curvature and the displacement of the middle surface, w, is: R w = Thus ε = z w (0) Similarly ε y = z w y

24 v u + y, where u and v are the displacements at depth z in the X- and Y- directions, respectively. Using the assumption hat sections normal to the middle plane remain normal to it, u w = z. Similarly v w = z y Thus ε y w = z y () (In equation () the term w y is the twist of the surface.)

25 From the second assumption and equations (3) and (4), the stress-strain relations for a thin plate are σ E = v ε + ε ( v ) y A Z u= z w σ E = v ε + ε σ ( v ) y y Also y = E v ( ) ε y () Z w X Fig. Relation between u and w

26 Also σ y E = ( v) ε y Substituting from equations (0), () and () in the strain energy epression (9), ab h/ E δ= ε + ε y+ vεεy+ ( v ) ε εy dzdyd 00 h/ ( v ) ab D w w w w w = v ( v) dyd y y 00 (3) after integrating with respect to z, where Eh D= 3 ( v )

27 If is the density of the plate, the kinetic energy a b h / w I= ρ t 0 0 h / dzdyd ab w = ρh t 00 dyd (4) after integrating with respect to z. Epressions (3) and (4) will be used with the finite element methods.

28 the moments shown in Fig..5, we have M h / = h / σ zdz M y h / = h / σ y zdz ; and h / M = M = σ zdz y y y h / (5)

29 M The equilibrium equations, obtained by resolving in the Z-direction and taking moments about the Y- and X-aes, are, after dividing by d dy: S S y w + + py (, ) f( t) = ρh y t M + y S = y 0 and M M y y + + S y = y 0 Eliminating S and S y M My M y w py (, ) f( t) = ρh y y t (6)

30 Substituting from equations (0) to () in equations (5) and integrating with respect to z. ; w w M = D + v y, and w M y = D + v y w My = D( v) w y (7) Substituting from equations (7) in equation (6) gives the equilibrium equation for an element of the plate in terms of w and its derivatives, w w w D + + ρh w py (, ) f() t = y y t (8)

31 PART III. Application of FEM 3. Finite Element Method (FEM) 3. Element Stiffness and Mass Matrices 3.. Aial Element 3.3 FEM for In-plane Vibration 3.4 FEM for Transverse Vibration.

32 3. Finite Element Method In static analysis. The finite element method solves equations of the form [ K]{ U} = { P} (9) where [K] is a global stiffness matri, {U} is a vector of nodal displacement, and {P} is a vector of nodal forces. The dynamic analysis usually starts with an equation of the form ([ K] ω [ M] ){ U} = { P} (0) where in this equation denotes the frequencies of vibration and [M] is the mass matri

33 For a free vibration, we have {P} = 0 and Eq. 0. reduces to the following epression: ([ K] ω [ M] ){ U} = { 0} () For a nontrivial solution, we have [ K] = ω [ M] () which represents the generalized Eigenvalue problems.

34 u u f Figure (.6). Truss f element. u( ) = b + b (3) where b and b are constants. In matri form Equation (3) may be written as follows: u ( ) { } b = b (4)

35 By using Eq. (3) and applying the boundary conditions u(0) = u and u(l) = u, where L is the length of the elements, we find u 0 = b + (0) b ( ) ( ) = + u L b Lb Equations (5) and (6) are written in matri form as follows u 0 b = u L b The same two equations may also be written in a matri form that relates b and b and u and u in the following way b L 0 u = b u By substituting Eq. (8) into Eq. (4), we find u L u (5) (6) (7) (8) or, in general u u ( ) { H H } = u u () = L (9) where H, in Eq. (30) are referred to as shape functions or interpolation functions (30)

36 The aial strain ε ε u = L L u in Fig.7 is the rate change of u( ) with respect to. Therefore, by differentiating Eq. (9) with respect to, we find (30) By applying Hooke's law, the normal stress at cross sections along the length of the element is given by the epression σ = Eε (3) where is Young's modulus of elasticity and is given by Eq. (3).

37 0 u f u f d d du d du EA T L = { } { } = 0 u u f f d u u L L L L u EA u L = L L d 0 The total energy stored in the element is defined by where A is the cross-sectional area of the member. By using Eq. (30), the epression given by Eq. (33) may be written as follows: When A is constant, we have (35) (33) (34)

38 and equation 34. yields EAL L L = L L u u u u { u u } { f f } AE u u = { u u} { f f} L u u (36) For a stationary condition, we have = u i On this basis, use of Eq. (37) yields AE u f = L u f 0 (37) (38) By comparing Eq. (38) with Eq. (9), we conclude that the stiffness matri [K] of the element is (39) AE [ K ] = L

39 A plate 3.3 Finite Element Method for Inplane Vibration of Plate Y Y V 3 V 4 U 3 U (5) (6) (9) V l y l V U U Fig..7 Rectangular plate for inplane deformation X 3 (3) (4) (8) () () (7) 3 Fig..8 Plate consisting of four inplane deformation X

40 l l y δ = h ( σ ε + σ ε + σ ε ) ddy δ e y y y y 0 0 e the strain energy of the element the strain energy can be epressed in terms of displacements, using the stress-strain relations () and the strain-displacement relations (7) and (8), ( v) (40) ly l Åh u v u v u v = v ddy v y y y 0 0 u ly l u v u v v =,, D ddy + y y y 0 0 (4) u v + y

41 The assumed displacement functions are u= a + a+ ay+ ay 3 4 v= a + a+ ay+ ay u g 0 a v = 0 g and or a where a, a,... a 8 (4) g = [ y y] is a vector containing the coefficients

42 u e u u u u 3 = 4 y y 4 Na v = = a 5 v v a l a 0 ly a 3 l l ll a l 0 0 a l 0 a 3 y 7 v l 4 ly ll y a 8 Thus where a= Bu e B = N and As (43) v y = a + a y 7 8 u = a + a y 4, u v + = a3 + a4+ a6 + ay 8 y

43 u v y u v + y = Ga where 0 0 y G = (44) y Substituting from equation (44) in equation (4) l y l ly l T T δ e = T T T agdgaddy = u eb G DGddyBu T = u ekeue where the element stiffness matri l y l T T Ke = B G DGddy B (45) 0 0 e

44 The kinetic energy of the element l y l u v I e = ρ h + t t 0 0 e u ly l u v t = ρh dyd t t v 0 0 t = u& T M u& e e e where the element mass matri ly l g T g 0 = T ρ 0 g T g 0 0 M B h dydb (46) dyd (47)

45 3.4 Finite Element Method for Transverse Vibrations of Plates The assumed displacement function is w= a + a+ ay+ a + ay+ a y + a + a y+ ay + a y + a y+ a y = ga Y Eqt.(48) ( w3, φ3, ψ3) 3 4 ( w3, φ3, ψ3) ( w, φψ, ) ( w, φ, ψ ) X Fig..9 Rectangular plate for fleure of a plate

46 Substituting nodal values of and y in equation (48), e w w w Na φ φ ψ ψ = = e e a N w Bw = = 00,, y l l e w w w w w C ddy y y y w y δ = From equation (3) the strain energy epression for the element can be written (50) (49)

47 where C v 0 = Dv ( v) w w y w y From Equation (48) y 0 0 6y 0 = y 0 6y a= Ga y 0 6 6y Using Eqs. (49) and (5) in equation (50) δ = w K w T e e e e (49)

48 from Eqs. (48 where the stiffness matri ly l T T Ke = B GCGddyB 00 The kinetic energy of the element l l y w I e = ρh t (48) and (49) 00 dyd w = gbwe & t T Thus I = wmw & & where the element mass matri e e e e ly l T T Me = B ρh g dydb 00 (53)

49 For forced vibration the matri equation && (54) Mw + Kw = p Considering the con-tribution to p from a particular element, p e, and supposing that this element is subjected to a transverse applied force per unit are a of py (, ) f( t),application of the principle of virtual work gives. l l y T p δw pyft (, )() δwydyd (, ) e e = 00

50 δ w e δ wy (, ) list the virtual increments in the element nodal values and is the virtual displacement ay point (,y). Using the Eqs. (48) and (49), ly l T eδ e = () (, ) δ e 0 0 p w f t pygb wdyd and ly l T p e = f() t p( ygdyd, ) B 0 0 (55)

51 PART IV. References Demeter G. Fertis Mechanical and Structural Vibration. J. F. Hall - Finite Element Analysis in Earthquake Engineering to the international handbook of Earthquake Engineering and Engineering Seismology part B, 003. G. D.Manolis and D.E. Beskos - Boundary Element Method in Elastodynamics, 988.