Flow Induced Vibration

advert induce VIBRATIONS IN PIPES, A impermanent particle address IVAN GRANT Bachelor of acquaintance in Mechanical Engineering Nagpur University Nagpur, India June, 2006 submitted in partial ful? llment of requirements for the degree MASTERS OF comprehension IN MECHANICAL ENGINEERING at the CLEVELAND STATE UNIVERSITY May, 2010 This thesis has been approved for the surgical incision of MECHANICAL ENGINEERING and the College of G radianuate Studies by thesis Chairperson, Majid Rashidi, Ph. D. division & Date Asuquo B. Ebiana, Ph. D. Department & Date Rama S. Gorla, Ph. D. Department & Date ACKNOWLEDGMENTS I would like to thank my advisor Dr. Majid Rashidi and Dr.Paul Bellini, who provided purloingenital support and assistance with forth my down c beer, and in addition for their guidance which immensely contributed towards the disclose(p)come of this thesis. This thesis would not bring in been cognise with pop their support. I would also like to thank Dr. Asuquo. B. Ebiana and Dr. Rama. S. Gorla for being in my thesis committee. give thanks atomic round 18 also payable to my p arnts,my buddy and fri blocks who throw encouraged, back up and inspired me. FLOW bring on VIBRATIONS IN PIPES, A FINITE division come up IVAN GRANT plagiarise proceed induced frissons of subwayworks with ingrained ? uid ? ow is bumvas in this work.Finite segment psychoanalysis methodological analysis is apply to de orderine the faultfinding ? uid stop upshot that induces the threshold of tubing instability. The partial di? erential equality of interrogative government the posterioral palpitations of the squ only is employed to develop the sti? mantle and inactivecape matrices corresponding to devil of the barriers of the comparabilitys of effect. The compargon of motion further includes a mixed-derivative verge that was inured as a denotation for a dissipative conk. The corresponding ground substance with this dissipative function was positive and recognized as the potentially destabilizing factor for the askance vibrations of the ? id carrying underground- decided structure. Two types of terminal specify conditions, namely manifestly-support and ignoretilevered were considered for the shout. The appropriate press, sti? cape, and dissipative matrices were veritable at an pieceal level for the ? uid carrying tube up. These matrices were then assembled to physique the overall crowd, sti? cape, and dissipative matrices of the entire system. Employing the ? nite agent imitate developed in this work dickens series of parametric studies were conducted. First, a call with a unremitting paries weighti mantle of 1 mm was analyzed. Then, the parametric studies were all-inclusive to a tube with variable beleaguer thick mantle.In this case, the paries thick ness of the thermionic tube was sculpturesque to standard candle down f fixed storage 2. 54 mm to 0. 01 mm. This take shows tha t the decisive focal ratio of a subway system carrying ? uid trick be change magnitude by a factor of sextup permit as the issuance of heightening the wall oppressiveness. iv TABLE OF CONTENTS rear ap particular OF FIGURES LIST OF TABLES I cornerst 1 1. 1 1. 2 1. 3 1. 4 II Over thought process of Internal lean bring on chills in tube ups . . . . . . Literature criticism . . . . . . . . . . . . . . . . . . . . . . . . . . accusatory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Composition of Thesis . . . . . . . . . . . . . . . . . . . . . . . iv septet ix 1 1 2 2 3 FLOW induce VIBRATIONS IN PIPES, A FINITE subdivision turn up 2. 1 Mathematical poser . . . . . . . . . . . . . . . . . . . . . . . 2. 1. 1 2. 2 pars of exploit . . . . . . . . . . . . . . . . . . . 4 4 4 12 12 Finite division baby-sit . . . . . . . . . . . . . . . . . . . . . . . . 2. 2. 1 2. 2. 2 2. 2. 3 frame of reference bl abrogates . . . . . . . . . . . . . . . . . . . . . Formulating the Sti? ness intercellular substance for a shout Carrying still 14 Forming the intercellular substance for the specialty that corrects the changeable to the electron tube . . . . . . . . . . . . . . . . . . . . . 21 2. 2. 4 2. 2. 5Dissipation ground substance provision for a thermionic valve carrying unsound 26 inaction hyaloplasm Formulation for a shrill carrying liquified . 28 III FLOW generate VIBRATIONS IN PIPES, A FINITE randomnesstion APPROACH 31 v 3. 1 Forming planetary Sti? ness matrix from Elemental Sti? ness Matrices . . . . . . . . . . . . . . . . . . . . 31 3. 2 Applying landmark Conditions to globose Sti? ness matrix for manifestly back up organ tobacco hollo with ? uid ? ow . . . . 33 3. 3 Applying confines Conditions to ball-shaped Sti? ness ground substance for a freightertilever shriek with ? uid ? ow . . . . . . . 34 3. 4 MATLAB Programs for assemblage world-wide Matrices for precisely back up and project tobacc o metro carrying ? uid . . . . . . . . . . 35 35 36 3. 5 3. 6 MATLAB weapons plat piss for a patently supported cry carrying ? uid . . MATLAB programme for a erecttilever shriek carrying ? uid . . . . . . IV FLOW bring on VIBRATIONS IN PIPES, A FINITE comp wiznt APPROACH 4. 1 V Parametric weigh . . . . . . . . . . . . . . . . . . . . . . . . . . 37 37 FLOW INDUCED VIBRATIONS IN PIPES, A FINITE ELEMENT APPROACH 5. 1 focalizeed vacuum tube Carrying gas . . . . . . . . . . . . . . . . . . . . 42 42 47 50 50 51 54 MATLAB program for simply back up cry Carrying liquified . . MATLAB Program for protrude shriek Carrying silver . . . . . . MATLAB Program for tapered thermionic valve Carrying silver . . . . . . 54 61 68 VI RESULTS AND DISCUSSIONS 6. 1 6. 2 region of the Thesis . . . . . . . . . . . . . . . . . . . . . Future mise en scene . . . . . . . . . . . . . . . . . . . . . . . . . . . . . BIBLIOGRAPHY App determinationices 0. 1 0. 2 0. 3 vi LIST OF FIGURES 2. 1 2. 2 Pinned-Pinned tube-shaped structure Carrying liquified * . . . . . . . . . . . . . . holler Carrying politic, surprises and flashs playacting on Elements (a) wandering (b) underground ** . . . . . . . . . . . . . . . . . . . . . . . . . 5 5 7 9 10 11 13 14 15 16 17 21 33 34 36 2. 3 2. 4 2. 5 2. 6 2. 7 2. 8 2. 9 displume cod to B extirpateing . . . . . . . . . . . . . . . . . . . . . . . . .Force that Con clays placid to the breaking ball of piping . . . . . Coriolis Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Inertia Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tube-shaped structurework Carrying politic . . . . . . . . . . . . . . . . . . . . . . . . . . ray of light Element amaze . . . . . . . . . . . . . . . . . . . . . . . . . Relationship between mental vocal and Strain, Hooks Law . . . . . . 2. 10 superfluous branchs remain vapid . . . . . . . . . . . . . . . . . . . . . 2. 11 Moment of Inertia for an Elem ent in the polish . . . . . . . . . 2. 12 subway system Carrying facile Model . . . . . . . . . . . . . . . . . . . . . 3. 1 3. 2 3. 4. 1 design of Simply back up subway Carrying wandering . . copy of cantilever yell Carrying quiet . . . . . . . Pinned-Free Pipe Carrying Fluid* . . . . . . . . . . . . . . . . . step-down of positive frequence for a Pinned-Pinned Pipe with change magnitude tend speeding . . . . . . . . . . . . . . . . 4. 2 imprint Function Plot for a stick out Pipe with change magnitude full stop f number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4. 3 decline of carry out absolute oftenness for a Cantilever Pipe with increase play fastness . . . . . . . . . . . . . . . . . . . . 5. 1 standard of dwindling Pipe Carrying Fluid . . . . . . . 39 40 41 42 septette 5. 2 6. 1 Introducing a Taper in the Pipe Carrying Fluid . . . . . . . . copy of Pipe Carrying Fluid and tapering Pipe Carrying Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 47 octad LIST OF TABLES 4. 1 Reduction of inherent relative frequence for a Pinned-Pinned Pipe with change magnitude Flow hurrying . . . . . . . . . . . . . . . . 38 4. 2 Reduction of central relative absolute frequency for a Pinned-Free Pipe with increa repulsivenessg Flow speeding . . . . . . . . . . . . . . . . . . . . 40 5. 1 Reduction of sound frequency for a Tapered squall with increa ejaculateg Flow swiftness . . . . . . . . . . . . . . . . . . . . . . 46 6. 1 Reduction of implicit in(p) relative frequency for a Tapered Pipe with increase Flow focal ratio . . . . . . . . . . . . . . . . . . . . . . . 48 6. 2 Reduction of Fundamental relative frequency for a Pinned-Pinned Pipe with increasing Flow pep pill . . . . . . . . . . . . . . . . 49 ix CHAPTER I presentation 1. 1 Overview of Internal Flow induce Vibrations in Pipes The ? ow of a ? uid through a thermionic tube burn impose pressings on the walls of the cry cause it to de? ect at a lower place plastered ? ow conditions. This de? ection of the cry may blend in to geomorphologic instability of the tobacco squall.The rudimentary graphic absolute frequency of a thermionic valve mostly decreases with increasing hurrying of ? uid ? ow. There be certain cases where decrease in this vivid frequency can be truly important, such as very high school gear speed ? ows through ? exible thin-walled subway systems such as those utilise in feed lines to rise motors and water supply turbines. The holler pay offs susceptible to reverberance or fatigue failure if its native frequency falls below certain limits. With large ? uid velocities the squall may become unstable. The most familiar year of this instability is the whipping of an unrestricted garden hose.The content of can-do response of a ? uid conveyance thermionic tube in conjunction with the transitory vibration of ruptured calls reveals that if a shriek rupture s through its cross section, then a ? exible continuance of unsupported scream is left spewing out ? uid and is large-minded to whip about and impingement other(a) structures. In power give plumbing pipe whip is a possible mode of failure. A 1 2 study of the in? uence of the resulting high velocity ? uid on the static and dynamic characteristics of the pipes is therefore necessary. 1. 2 Literature Review Initial investigations on the bending vibrations of a manifestly supported pipe containing ? id were carried out by Ashley and Haviland2. Subsequently,Housner3 derived the equations of motion of a ? uid conveyance pipe to a greater extent(prenominal) effectly and developed an equation relating the fundamental bending frequency of a simply supported pipe to the velocity of the informal ? ow of the ? uid. He also give tongue to that at certain exact velocity, a statically unstable condition could exist. Long4 presented an substitute(a) solution to Housners3 equation of motion for the simply supported end conditions and also treated the ? xed-free end conditions. He comp ard the analysis with observational results to con? rm the mathematical clay sculpture.His experimental results were sooner inconclusive since the sludgeimum ? uid velocity forthcoming for the test was low and change in bending frequency was very small. former(a) e? orts to treat this subject were do by Benjamin, Niordson6 and Ta Li. Other solutions to the equations of motion show that type of instability depends on the end conditions of the pipe carrying ? uid. If the ? ow velocity exceeds the unfavourable velocity pipes supported at some(prenominal) ends bow out and buckle1. Straight Cantilever pipes fall into ? ow induced vibrations and vibrate at a large amplitude when ? ow velocity exceeds little velocity8-11. . 3 Objective The clinical of this thesis is to implement numerical solutions method, more(prenominal) specifically the Finite Element Analysis (FEA) to obtain solutions for di? erent pipe con? gurations and ? uid ? ow characteristics. The governing dynamic equation describing the induced structural vibrations due to inhering ? uid ? ow has been form and dis- 3 cussed. The governing equation of motion is a partial di? erential equation that is fourth establish in spatial variable and number order in time. Parametric studies have been performed to examine the in? uence of mass dispersion on the continuance of the pipe carrying ? id. 1. 4 Composition of Thesis This thesis is organize according to the followe sequences. The equations of motions ar derived in chapter(II)for pinned-pinned and ? xed-pinned pipe carrying ? uid. A ? nite broker simulate is created to solve the equation of motion. Elemental matrices argon formed for pinned-pinned and ? xed-pinned pipe carrying ? uid. Chapter(III)consists of MATLAB programs that ar used to assemble global matrices for the preceding(prenominal) cases. leap conditions are use and found on the user de? ned parameters fundamental natural frequency for free vibration is calculated for mixed pipe con? urations. Parametric studies are carried out in the chase chapter and results are obtained and discussed. CHAPTER II FLOW INDUCED VIBRATIONS IN PIPES, A FINITE ELEMENT APPROACH In this chapter,a mathematical model is formed by developing equations of a groovy ? uid conveying pipe and these equations are later solved for the natural frequency and attack of instability of a cantilever and pinned-pinned pipe. 2. 1 2. 1. 1 Mathematical Modelling comparabilitys of exploit cogitate a pipe of continuance L, modulus of cinch E, and its transverse field jiffy I. A ? uid ? ows through the pipe at cart p and density ? t a constant velocity v through the internal pipe crosswise of scene of action A. As the ? uid ? ows through the de? ecting pipe it is zipd, because of the changing curvature of the pipe and the lateral vibration of the transmission line. The tumid instalment part of ? uid wring apply to the ? uid chemical particle and the pressure index F per social unit of measurement duration applied on the ? uid piece by the tube walls oppose these accelerations. linkring to ? gures (2. 1) and 4 5 intent 2. 1 Pinned-Pinned Pipe Carrying Fluid * (2. 2),balancing the fights in the Y direction on the ? uid element for small deformations, gives F ? A ? ? ? 2Y = ? A( + v )2 Y ? x2 ? t ? x (2. 1) The pressure gradient in the ? uid along the duration of the pipe is opposed by the dress stress of the ? uid friction against the tube walls. The core group of the armaments parallel simulacrum 2. 2 Pipe Carrying Fluid, Forces and Moments acting on Elements (a) Fluid (b) Pipe ** to the pipe axis for a constant ? ow velocity gives 0 0 * Flow Induced Vibrations,Robert D. Blevins,Krieger. 1977,P 289 ** Flow Induced Vibrations,Robert D. Blevins,Krieger. 1977,P 289 6 A ?p + ? S = 0 ? x (2. 2) Where S is the inner tolerance of the pipe, and ? s the cut back stress on the internal surface of the pipe. The equations of motions of the pipe element are derived as follows. ?T ? 2Y + ? S ? Q 2 = 0 ? x ? x (2. 3) Where Q is the transverse shear force in the pipe and T is the longitudinal tensity in the pipe. The forces on the element of the pipe normal to the pipe axis accelerate the pipe element in the Y direction. For small deformations, ? 2Y ? 2Y ? Q +T 2 ? F =m 2 ? x ? x ? t (2. 4) Where m is the mass per unit continuance of the empty pipe. The bending meaning M in the pipe, the transverse shear force Q and the pipe deformation are related by ? 3Y ?M = EI 3 ? x ? x Q=? (2. 5) Combining all the preceding(prenominal) equations and eliminating Q and F yields EI ? 4Y ? 2Y ? ? ? Y + (? A ? T ) 2 + ? A( + v )2 Y + m 2 = 0 4 ? x ? x ? t ? x ? t (2. 6) The shear stress may be eliminated from equation 2. 2 and 2. 3 to give ? (? A ? T ) =0 ? x (2. 7) At the pipe end where x=L, the tension in the pipe is zero and the ? uid pre ssure is equal to ambient pressure. Thus p=T=0 at x=L, ? A ? T = 0 (2. 8) 7 The equation of motion for a free vibration of a ? uid conveying pipe is found out by substituting ? A ? T = 0 from equation 2. 8 in equation 2. 6 and is attached by the equation 2. EI ? 2Y ? 2Y ? 4Y ? 2Y +M 2 =0 + ? Av 2 2 + 2? Av ? x4 ? x ? x? t ? t (2. 9) where the mass per unit distance of the pipe and the ? uid in the pipe is disposed(p) by M = m + ? A. The nigh section describes the forces acting on the pipe carrying ? uid for apiece of the portions of eq(2. 9) Y F1 X Z EI ? 4Y ? x4 epithet 2. 3 Force due to crimp government agency of the First name in the Equation of Motion for a Pipe Carrying Fluid 8 The term EI ? Y is a force parcel acting on the pipe as a result of bending of ? x4 the pipe. Fig(2. 3) shows a schematic view of this force F1. 4 9 Y F2 X Z ?Av 2 ? 2Y ? x2 Figure 2. Force that Conforms Fluid to the Curvature of Pipe copy of the Second Term in the Equation of Motion for a P ipe Carrying Fluid The term ? Av 2 ? Y is a force component acting on the pipe as a result of ? ow ? x2 around a trend pipe. In other words the whim of the ? uid is changed leading to a force component F2 shown schematically in Fig(2. 4) as a result of the curvature in the pipe. 2 10 Y F3 X Z 2? Av ? 2Y ? x? t Figure 2. 5 Coriolis Force Representation of the Third Term in the Equation of Motion for a Pipe Carrying Fluid ? Y The term 2? Av ? x? t is the force inevitable to rotate the ? id element as each point 2 in the intersect rotates with angular velocity. This force is a result of Coriolis E? ect. Fig(2. 5) shows a schematic view of this force F3. 11 Y F4 X Z M ? 2Y ? t2 Figure 2. 6 Inertia Force Representation of the stern Term in the Equation of Motion for a Pipe Carrying Fluid The term M ? Y is a force component acting on the pipe as a result of Inertia ? t2 of the pipe and the ? uid ? owing through it. Fig(2. 6) shows a schematic view of this force F4. 2 12 2. 2 Finite E lement Model Consider a pipeline span that has a transverse de? ection Y(x,t) from its equillibrium position.The length of the pipe is L,modulus of piece of cake of the pipe is E,and the vault of heaven moment of inertia is I. The density of the ? uid ? owing through the pipe is ? at pressure p and constant velocity v,through the internal pipe cross section having subject field A. Flow of the ? uid through the de? ecting pipe is speed up due to the changing curvature of the pipe and the lateral vibration of the pipeline. From the previous section we have the equation of motion for free vibration of a ? uid convering pipe EI ? 2Y ? 2Y ? 2Y ? 4Y + ? Av 2 2 + 2? Av +M 2 =0 ? x4 ? x ? x? t ? t (2. 10) 2. 2. 1 Shape Functions The essence of the ? ite element method,is to imagine the unidentified by an expression tending(p) as n w= i=1 Ni ai where Ni are the interpolating flesh functions prescribed in equipment casualty of linear in hooked functions and ai are a circuit of unk o utrightn parameters. We shall now derive the shape functions for a pipe element. 13 Y R R x L2 L L1 X Figure 2. 7 Pipe Carrying Fluid Consider an pipe of length L and let at point R be at distance x from the left end. L2=x/L and L1=1-x/L. Forming Shape Functions N 1 = L12 (3 ? 2L1) N 2 = L12 L2L N 3 = L22 (3 ? 2L2) N 4 = ? L1L22 L subbing the honour of L1 and L2 we astonish (2. 11) (2. 12) (2. 13) (2. 14) N 1 = (1 ? /l)2 (1 + 2x/l) N 2 = (1 ? x/l)2 x/l N 3 = (x/l)2 (3 ? 2x/l) N 4 = ? (1 ? x/l)(x/l)2 (2. 15) (2. 16) (2. 17) (2. 18) 14 2. 2. 2 Formulating the Sti? ness hyaloplasm for a Pipe Carrying Fluid ?1 ?2 W1 W2 Figure 2. 8 Beam Element Model For a twain dimensional conduct element, the geological fault matrix in foothold of shape functions can be verbalized as ? ? w1 ? ? ? ? ? ?1 ? ? ? W (x) = N 1 N 2 N 3 N 4 ? ? ? ? ? w2? ? ? ?2 (2. 19) where N1, N2, N3 and N4 are the duty period reaction shape functions for the ii dimensional beam element as verbalize in equation s (2. 15) to (2. 18). The displacements and rotations at end 1 is given by w1, ? and at end 2 is given by w2 , ? 2. Consider the point R inside the beam element of length L as shown in ? gure(2. 7) let the internal dividing line brawniness at point R is given by UR . The internal torture energy at point R can be verbalised as 1 UR = ? 2 where ? is the stress and is the strain at the point R. (2. 20) 15 ? E 1 ? Figure 2. 9 Relationship between latent hostility and Strain, Hooks Law Also ? =E Relation between stress and strain for elastic hearty, Hooks Law subbing the value of ? from equation(2. 21) into equation(2. 20) yields 1 UR = E 2 (2. 21) 2 (2. 22) 16 A1 z B1 w A z B u x Figure 2. 0 stripped sections remain plane Assuming plane sections remain resembling, = du dx (2. 23) (2. 24) (2. 25) u=z dw dx d2 w =z 2 dx To obtain the internal energy for the whole beam we immix the internal strain energy at point R over the batch. The internal strain energy for the entire beam is given as UR dv = U vol (2. 26) exchange the value of from equation(2. 25) into (2. 26) yields U= vol 1 2 E dv 2 (2. 27) Volume can be expressed as a product of area and length. dv = dA. dx (2. 28) 17 based on the higher up equation we now integrate equation (2. 27) over the area and over the length. L U= 0 A 1 2 E dAdx 2 (2. 29) subbing the value of rom equation(2. 25) into equation (2. 28) yields L U= 0 A 1 d2 w E(z 2 )2 dAdx 2 dx (2. 30) Moment of Inertia I for the beam element is given as = dA z Figure 2. 11 Moment of Inertia for an Element in the Beam I= z 2 dA (2. 31) Substituting the value of I from equation(2. 31) into equation(2. 30) yields L U = EI 0 1 d2 w 2 ( ) dx 2 dx2 (2. 32) The in a higher place equation for total internal strain energy can be rewritten as L U = EI 0 1 d2 w d2 w ( )( )dx 2 dx2 dx2 (2. 33) 18 The potential energy of the beam is naught but the total internal strain energy. Therefore, L ? = EI 0 1 d2 w d2 w ( )( )dx 2 dx2 dx2 (2. 34)If A and B are two matrices then applying matrix plaza of the transpose, yields (AB)T = B T AT (2. 35) We can express the Potential heartiness expressed in equation(2. 34) in scathe of displacement matrix W(x)equation(2. 19) as, 1 ? = EI 2 From equation (2. 19) we have ? ? w1 ? ? ? ? ? ?1 ? ? ? W = N 1 N 2 N 3 N 4 ? ? ? ? ? w2? ? ? ?2 ? ? N1 ? ? ? ? ? N 2? ? ? W T = ? ? w1 ? 1 w2 ? 2 ? ? ? N 3? ? ? N4 L (W )T (W )dx 0 (2. 36) (2. 37) (2. 38) Substituting the values of W and W T from equation(2. 37) and equation(2. 38) in equation(2. 36) yields ? N1 ? ? ? N 2 ? w1 ? 1 w2 ? 2 ? ? ? N 3 ? N4 ? ? ? ? ? ? N1 ? ? ? ? ? w1 ? ? ? ? ?1 ? ? ? ? ? dx (2. 39) ? ? ? w2? ? ? ?2 1 ? = EI 2 L 0 N2 N3 N4 19 where N1, N2, N3 and N4 are the displacement shape functions for the two dimensional beam element as stated in equations (2. 15) to (2. 18). The displacements and rotations at end 1 is given by w1, ? 1 and at end 2 is given by w2 , ? 2. 1 ? = EI 2 L 0 (N 1 ) ? ? ? N 2 N 1 ? w1 ? 1 w2 ? 2 ? ? ? N 3 N 1 ? N4 N1 ? 2 N1 N2 (N 2 )2 N3 N2 N4 N2 N1 N3 N2 N3 (N 3 )2 N4 N3 N1 N4 N2 N4 N3 N4 (N 4 )2 ? w1 ? ? ? ? ? 1 ? ? ? ? ? dx ? ? ?w2? ? ? 2 (2. 40) where ? 2 (N 1 ) ? ? L ? N 2 N 1 ? K = ? 0 ? N 3 N 1 ? ? N4 N1 N1 N2 (N 2 )2 N3 N2 N4 N2N1 N3 N2 N3 (N 3 ) 2 N1 N4 ? N4 N3 ? ? N2 N4 ? ? ? dx ? N3 N4 ? ? 2 (N 4 ) (2. 41) N 1 = (1 ? x/l)2 (1 + 2x/l) N 2 = (1 ? x/l)2 x/l N 3 = (x/l)2 (3 ? 2x/l) N 4 = ? (1 ? x/l)(x/l)2 (2. 42) (2. 43) (2. 44) (2. 45) The element sti? ness matrix for the beam is obtained by substituting the values of shape functions from equations (2. 42) to (2. 45) into equation(2. 41) and combine every element in the matrix in equation(2. 40) over the length L. 20 The Element sti? ness matrix for a beam element ? ? 12 6l ? 12 6l ? ? ? ? 2 2? 4l ? 6l 2l ? EI ? 6l ? K e = 3 ? ? l 12 ? 6l 12 ? 6l? ? ? ? ? 2 2 6l 2l ? 6l 4l (2. 46) 1 2. 2. 3 Forming the Matrix for the Force that conforms the Fluid to the Pipe A X ? r ? _______________________ x R Y Figure 2. 12 Pipe Carr ying Fluid Model B Consider a pipe carrying ? uid and let R be a point at a distance x from a reference plane AB as shown in ? gure(2. 12). Due to the ? ow of the ? uid through the pipe a force is introduced into the pipe cause the pipe to curve. This force conforms the ? uid to the pipe at all times. Let W be the transverse de? ection of the pipe and ? be shift made by the pipe due to the ? uid ? ow with the neutral axis. ? and ? deliver the unit vectors along the X i j ? nd Y axis and r and ? represent the two unit vectors at point R along the r and ? ? ? axis. At point R,the vectors r and ? can be expressed as ? r = cos lettuce + sin ? i j (2. 47) ? ? = ? sin + cos i j formulation for slope at point R is given by tan? = dW dx (2. 48) (2. 49) 22 Since the pipe undergoes a small de? ection, thus ? is very small. Therefore tan? = ? ie ? = dW dx (2. 51) (2. 50) The displacement of a point R at a distance x from the reference plane can be expressed as ? R = W ? + r? j r We di? e rentiate the preceding(prenominal) equation to father velocity of the ? uid at point R ? ? ? j ? r ? R = W ? + r? + rr ? r = vf ? here vf is the velocity of the ? uid ? ow. Also at time t r ? d? r= ? dt ie r ? d? d? = r= ? d? dt ? Substituting the value of r in equation(2. 53) yields ? ? ? ? j ? r R = W ? + r? + r (2. 57) (2. 56) (2. 55) (2. 53) (2. 54) (2. 52) ? Substituting the value of r and ? from equations(2. 47) and (2. 48) into equation(2. 56) ? yields ? ? ? ?j ? R = W ? + rcos + sin + r? ? sin + cos i j i j Since ? is small The velocity at point R is expressed as ? ? ? i ? j R = Rx? + Ry ? (2. 59) (2. 58) 23 ? ? i ? j ? ? R = (r ? r )? + (W + r? + r? )? ? ? The Y component of velocity R cause the pipe carrying ? id to curve. Therefore, (2. 60) 1 ? ? ? ? T = ? f ARy Ry (2. 61) 2 ? ? where T is the kinetic energy at the point R and Ry is the Y component of velocity,? f is the density of the ? uid,A is the area of crosswise of the pipe. ? ? Substituting the value of Ry fr om equation(2. 60) yields 1 ? ? ? ? ? ? ? ? ? T = ? f AW 2 + r2 ? 2 + r2 ? 2 + 2W r? + 2W ? r + 2rr 2 (2. 62) Substituting the value of r from equation(2. 54) and selecting the ? rst, second and the ? fourth terms yields 1 2 ? ? T = ? f AW 2 + vf ? 2 + 2W vf ? 2 (2. 63) direct substituting the value of ? from equation(2. 51) into equation(2. 3) yields dW 2 dW dW 1 2 dW 2 ) + vf ( ) + 2vf ( )( ) T = ? f A( 2 dt dx dt dx From the above equation we have these two terms 1 2 dW 2 ? f Avf ( ) 2 dx 2? f Avf ( dW dW )( ) dt dx (2. 65) (2. 66) (2. 64) The force acting on the pipe due to the ? uid ? ow can be calculated by compound the expressions in equations (2. 65) and (2. 66) over the length L. 1 2 dW 2 ? f Avf ( ) 2 dx (2. 67) L The expression in equation(2. 67) represents the force that causes the ? uid to conform to the curvature of the pipe. 2? f Avf ( L dW dW )( ) dt dx (2. 68) 24 The expression in equation(2. 68) represents the coriolis force which causes the ? id in the pipe t o whip. The equation(2. 67) can be expressed in terms of displacement shape functions derived for the pipe ? =T ? V ? = L 1 2 dW 2 ? f Avf ( ) 2 dx (2. 69) Rearranging the equation 2 ? = ? f Avf L 1 dW dW ( )( ) 2 dx dx (2. 70) For a pipe element, the displacement matrix in terms of shape functions can be expressed as ? ? w1 ? ? ? ? ? ?1 ? ? ? W (x) = N 1 N 2 N 3 N 4 ? ? ? ? ? w2? ? ? ?2 (2. 71) where N1, N2, N3 and N4 are the displacement shape functions pipe element as stated in equations (2. 15) to (2. 18). The displacements and rotations at end 1 is given by w1, ? 1 and at end 2 is given by w2 , ? . Refer to ? gure(2. 8). Substituting the shape functions firm in equations (2. 15) to (2. 18) ? ? N1 ? ? ? ? ? N 2 ? ? ? ? N1 w1 ? 1 w2 ? 2 ? ? ? N3 ? ? ? ? N4 ? ? w1 ? ? ? ? ? ?1 ? ? ? N 4 ? ? dx (2. 72) ? ? ? w2? ? ? ?2 L 2 ? = ? f Avf 0 N2 N3 25 L 2 ? = ? f Avf 0 (N 1 ) ? ? ? N 2 N 1 ? w1 ? 1 w2 ? 2 ? ? ? N 3 N 1 ? N4 N1 ? 2 N1 N2 (N 2 )2 N3 N2 N4 N2 N1 N3 N2 N3 (N 3 )2 N4 N3 N1 N 4 N2 N4 N3 N4 (N 4 )2 ? w1 ? ? ? ? ? 1 ? ? ? ? ? dx ? ? ?w2? ? ? 2 (2. 73) where (N 1 ) ? ? L ? N 2 N 1 ? ? 0 ? N 3 N 1 ? ? N4 N1 ? 2 N1 N2 (N 2 )2 N3 N2 N4 N2 N1 N3 N2 N3 (N 3 ) 2 N1 N4 ? 2 K2 = ? f Avf N4 N3 ? N2 N4 ? ? ? dx ? N3 N4 ? ? 2 (N 4 ) (2. 74) The matrix K2 represents the force that conforms the ? uid to the pipe. Substituting the values of shape functions equations(2. 15) to (2. 18) and desegregation it over the length gives us the basal matrix for the ? 36 3 ? 36 ? ? 4 ? 3 ? Av 2 ? 3 ? K2 e = ? 30l 36 ? 3 36 ? ? 3 ? 1 ? 3 above force. ? 3 ? ? ? 1? ? ? ? ? 3? ? 4 (2. 75) 26 2. 2. 4 Dissipation Matrix Formulation for a Pipe carrying Fluid The dissipation matrix represents the force that causes the ? uid in the pipe to whip creating instability in the system. To formulate this matrix we recall equation (2. 4) and (2. 68) The dissipation function is given by D= L 2? f Avf ( dW dW )( ) dt dx (2. 76) Where L is the length of the pipe element, ? f is the density of the ? uid, A area of cross-section(prenominal) of the pipe, and vf velocity of the ? uid ? ow. Recalling the displacement shape functions mentioned in equations(2. 15) to (2. 18) N 1 = (1 ? x/l)2 (1 + 2x/l) N 2 = (1 ? x/l)2 x/l N 3 = (x/l)2 (3 ? 2x/l) N 4 = ? (1 ? x/l)(x/l)2 (2. 77) (2. 78) (2. 79) (2. 80) The Dissipation Matrix can be expressed in terms of its displacement shape functions as shown in equations(2. 77) to (2. 80). ? ? N1 ? ? ? ? ? N 2 ? L ? ? D = 2? Avf ? N1 N2 N3 N4 w1 ? 1 w2 ? 2 ? ? ? 0 N3 ? ? ? ? N4 (N 1 ) ? ? ? N 2 N 1 ? w1 ? 1 w2 ? 2 ? ? ? N 3 N 1 ? N4 N1 ? 2 ? ? w1 ? ? ? ? ? ?1 ? ? ? ? ? dx ? ? ? w2? ? ? ?2 (2. 81) N1 N2 (N 2 )2 N3 N2 N4 N2 N1 N3 N2 N3 (N 3 )2 N4 N3 N1 N4 N2 N4 N3 N4 (N 4 )2 L 2? f Avf 0 ? w1 ? ? ? ? ? 1 ? ? ? ? ? dx ? ? ?w2? ? ? 2 (2. 82) 27 Substituting the values of shape functions from equations(2. 77) to (2. 80) and integrating over the length L yields ? ? ? 30 6 30 ? 6 ? ? ? ? 0 6 ? 1? ?Av ? 6 ? ? De = ? ? 30 30 ? 6 30 6 ? ? ? ? ? 6 1 ? 6 0 De represents the elementary dissipation matrix. (2. 83) 28 2. 2. 5Inertia Matrix Formulation for a Pipe carrying Fluid Consider an element in the pipe having an area dA, length x, mint dv and mass dm. The density of the pipe is ? and let W represent the transverse displacement of the pipe. The displacement model for the Assuming the displacement model of the element to be W (x, t) = N we (t) (2. 84) where W is the vector of displacements,N is the matrix of shape functions and we is the vector of nodal displacements which is assumed to be a function of time. Let the nodal displacement be expressed as W = weiwt Nodal Velocity can be found by di? erentiating the equation() with time. W = (iw)weiwt (2. 86) (2. 85) energising Energy of a particle can be expressed as a product of mass and the square of velocity 1 T = mv 2 2 (2. 87) Kinetic energy of the element can be found out by integrating equation(2. 87) over the the great unwashed. Also,mass can be expressed as the pro duct of density and vividness ie dm = ? dv T = v 1 ? 2 ? W dv 2 (2. 88) The volume of the element can be expressed as the product of area and the length. dv = dA. dx (2. 89) Substituting the value of volume dv from equation(2. 89) into equation(2. 88) and integrating over the area and the length yields T = ? w2 2 ? ?W 2 dA. dx A L (2. 90) 29 ?dA = ?A A (2. 91) Substituting the value of A ?dA in equation(2. 90) yields Aw2 2 T = ? W 2 dx L (2. 92) Equation(2. 92) can be written as Aw2 2 T = ? ? W W dx L (2. 93) The Lagrange equations are given by d dt where L=T ? V (2. 95) ? L ? w ? ? ? L ? w = (0) (2. 94) is called the Lagrangian function, T is the kinetic energy, V is the potential energy, ? W is the nodal displacement and W is the nodal velocity. The kinetic energy of the element e can be expressed as Te = Aw2 2 ? ? W T W dx L (2. 96) ? and where ? is the density and W is the vector of velocities of element e. The expression for T using the eq(2. 9)to (2. 21) can be written as ? ? N1 ? ? ? ? ? N 2? ? ? w1 ? 1 w2 ? 2 ? ? N 1 N 2 N 3 N 4 ? ? ? N 3? ? ? N4 ? ? w1 ? ? ? ? ? ?1 ? ? ? ? ? dx ? ? ? w2? ? ? ?2 Aw2 T = 2 e (2. 97) L 30 Rewriting the above expression we get ? (N 1)2 ? ? ? N 2N 1 Aw2 ? Te = w1 ? 1 w2 ? 2 ? ? 2 L ? N 3N 1 ? N 4N 1 ? N 1N 2 N 1N 3 N 1N 4 w1 ? ? 2 (N 2) N 2N 3 N 2N 4? ? ? 1 ? ? ? ? ? dx ? N 3N 2 (N 3)2 N 3N 4? ?w2? ? 2 N 4N 2 N 4N 3 (N 4) ? 2 (2. 98) Recalling the shape functions derived in equations(2. 15) to (2. 18) N 1 = (1 ? x/l)2 (1 + 2x/l) N 2 = (1 ? x/l)2 x/l N 3 = (x/l)2 (3 ? 2x/l) N 4 = ? (1 ? x/l)(x/l)2 (2. 9) (2. speed of light) (2. 101) (2. 102) Substituting the shape functions from eqs(2. 99) to (2. 102) into eqs(2. 98) yields the elemental mass matrix for a pipe. ? ? 156 22l 54 ? 13l ? ? ? ? 2 2? ? 22l 4l 13l ? 3l ? Ml ? M e = ? ? ? 420 ? 54 13l 156 ? 22l? ? ? ? 2 2 ? 13l ? 3l ? 22l 4l (2. 103) CHAPTER III FLOW INDUCED VIBRATIONS IN PIPES, A FINITE ELEMENT APPROACH 3. 1 Forming Global Sti? ness Matrix from Elemen tal Sti? ness Matrices Inorder to form a Global Matrix,we start with a 66 null matrix,with its six degrees of freedom being translation and rotation of each of the nodes. So our Global Sti? ness matrix looks like this ? 0 ? ?0 ? ? ? ?0 =? ? ? 0 ? ? ? 0 ? ? 0 ? 0? ? 0? ? ? ? 0? ? ? 0? ? ? 0? ? ? 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 KGlobal (3. 1) 31 32 The two 44 element sti? ness matrices are ? ? 12 6l ? 12 6l ? ? ? ? 4l2 ? 6l 2l2 ? EI ? 6l ? ? e k1 = 3 ? ? l 12 ? 6l 12 ? 6l? ? ? ? ? 2 2 6l 2l ? 6l 4l ? 12 6l ? 12 6l ? (3. 2) ? ? ? ? 2 2? 4l ? 6l 2l ? EI ? 6l ? e k2 = 3 ? ? l 12 ? 6l 12 ? 6l? ? ? ? ? 2 2 6l 2l ? 6l 4l (3. 3) We shall now retrace the global sti? ness matrix by inserting element 1 ? rst into the global sti? ness matrix. 6l ? 12 6l 0 0? ? 12 ? ? ? 6l 4l2 ? 6l 2l2 0 0? ? ? ? ? ? ? 12 ? 6l 12 ? l 0 0? EI ? ? = 3 ? ? l ? 6l 2 2 2l ? 6l 4l 0 0? ? ? ? ? ? 0 0 0 0 0 0? ? ? ? ? 0 0 0 0 0 0 ? ? KGlobal (3. 4) Inserting element 2 into the global sti? ness ma trix ? ? 6l ? 12 6l 0 0 ? ? 12 ? ? ? 6l 4l2 ? 6l 2l2 0 0 ? ? ? ? ? ? ? EI 12 ? 6l (12 + 12) (? 6l + 6l) ? 12 6l ? ? KGlobal = 3 ? ? l ? 6l 2 2 2 2? ? 2l (? 6l + 6l) (4l + 4l ) ? 6l 2l ? ? ? ? ? 0 0 ? 12 ? 6l 12 ? 6l? ? ? ? ? 2 2 0 0 6l 2l ? 6l 4l (3. 5) 33 3. 2 Applying Boundary Conditions to Global Sti? ness Matrix for simply supported pipe with ? uid ? ow When the boundary conditions are applied to a simply supported pipe carrying ? uid, the 66 Global Sti? ess Matrix formulated in eq(3. 5) is modi? ed to a 44 Global Sti? ness Matrix. It is as follows Y 1 2 X L Figure 3. 1 Representation of Simply support Pipe Carrying Fluid ? ? 4l2 ?6l 2l2 0 KGlobalS ? ? ? ? EI 6l (12 + 12) (? 6l + 6l) 6l ? ? ? = 3 ? ? l ? 2l2 (? 6l + 6l) (4l2 + 4l2 ) 2l2 ? ? ? ? ? 2 2 0 6l 2l 4l (3. 6) Since the pipe is supported at the two ends the pipe does not de? ect causing its two translational degrees of freedom to go to zero. consequently we end up with the Sti? ness Matrix shown in eq(3. 6) 34 3. 3 Ap plying Boundary Conditions to Global Sti? ness Matrix for a cantilever pipe with ? id ? ow Y E, I 1 2 X L Figure 3. 2 Representation of Cantilever Pipe Carrying Fluid When the boundary conditions are applied to a Cantilever pipe carrying ? uid, the 66 Global Sti? ness Matrix formulated in eq(3. 5) is modi? ed to a 44 Global Sti? ness Matrix. It is as follows ? (12 + 12) (? 6l + 6l) ? 12 6l ? KGlobalS ? ? ? ? ?(? 6l + 6l) (4l2 + 4l2 ) ? 6l 2l2 ? EI ? ? = 3 ? ? ? l ? ?12 ? 6l 12 ? 6l? ? ? ? 6l 2l2 ? 6l 4l2 (3. 7) Since the pipe is supported at one end the pipe does not de? ect or rotate at that end causing translational and rotational degrees of freedom at that end to go to zero.Hence we end up with the Sti? ness Matrix shown in eq(3. 8) 35 3. 4 MATLAB Programs for Assembling Global Matrices for Simply Supported and Cantilever pipe carrying ? uid In this section,we implement the method discussed in section(3. 1) to (3. 3) to form global matrices from the developed elemental matrices o f a straight ? uid conveying pipe and these assembled matrices are later solved for the natural frequency and onset of instability of a cantlilever and simply supported pipe carrying ? uid utilizing MATLAB Programs. Consider a pipe of length L, modulus of elasticity E has ? uid ? wing with a velocity v through its inner cross-section having an right(prenominal) diam od,and thickness t1. The expression for critical velocity and natural frequency of the simply supported pipe carrying ? uid is given by wn = ((3. 14)2 /L2 ) vc = (3. 14/L) (E ? I/M ) (3. 8) (3. 9) (E ? I/? A) 3. 5 MATLAB program for a simply supported pipe carrying ? uid The number of elements,density,length,modulus of elasticity of the pipe,density and velocity of ? uid ? owing through the pipe and the thickness of the pipe can be de? ned by the user. Refer to concomitant 1 for the complete MATLAB Program. 36 3. 6MATLAB program for a cantilever pipe carrying ? uid Figure 3. 3 Pinned-Free Pipe Carrying Fluid* The numb er of elements,density,length,modulus of elasticity of the pipe,density and velocity of ? uid ? owing through the pipe and the thickness of the pipe can be de? ned by the user. The expression for critical velocity and natural frequency of the cantilever pipe carrying ? uid is given by wn = ((1. 875)2 /L2 ) (E ? I/M ) Where, wn = ((an2 )/L2 ) (EI/M )an = 1. 875, 4. 694, 7. 855 vc = (1. 875/L) (E ? I/? A) (3. 11) (3. 10) Refer to accessory 2 for the complete MATLAB Program. 0 * Flow Induced Vibrations,Robert D.Blevins,Krieger. 1977,P 297 CHAPTER IV FLOW INDUCED VIBRATIONS IN PIPES, A FINITE ELEMENT APPROACH 4. 1 Parametric knowledge Parametric study has been carried out in this chapter. The study is carried out on a single span steel pipe with a 0. 01 m (0. 4 in. ) diam and a . 0001 m (0. 004 in. ) thick wall. The other parameters are tightness of the pipe ? p (Kg/m3 ) 8000 density of the ? uid ? f (Kg/m3 ) light speed0 length of the pipe L (m) 2 rate of elements n 10 Modulus E lasticity E (Gpa) 207 of MATLAB program for the simply supported pipe with ? uid ? ow is apply for these set of parameters with varying ? uid velocity.Results from this study are shown in the form of graphs and tables. The fundamental frequency of vibration and the critical velocity of ? uid for a simply supported pipe 37 38 carrying ? uid are ? n 21. 8582 rad/sec vc 16. 0553 m/sec dishearten 4. 1 Reduction of Fundamental oftenness for a Pinned-Pinned Pipe with increasing Flow Velocity Velocity of Fluid(v) Velocity symmetry(v/vc) 0 2 4 6 8 10 12 14 16. 0553 0 0. 1246 0. 2491 0. 3737 0. 4983 0. 6228 0. 7474 0. 8720 1 frequence(w) 21. 8806 21. 5619 20. 5830 18. 8644 16. 2206 12. 1602 3. 7349 0. 3935 0 Frequency Ratio(w/wn) 1 0. 9864 0. 9417 0. 8630 0. 7421 0. 5563 0. 709 0. 0180 0 39 Figure 4. 1 Reduction of Fundamental Frequency for a Pinned-Pinned Pipe with increasing Flow Velocity The fundamental frequency of vibration and the critical velocity of ? uid for a Cantilever pipe c arrying ? uid are ? n 7. 7940 rad/sec vc 9. 5872 m/sec 40 Figure 4. 2 Shape Function Plot for a Cantilever Pipe with increasing Flow Velocity Table 4. 2 Reduction of Fundamental Frequency for a Pinned-Free Pipe with increasing Flow Velocity Velocity of Fluid(v) Velocity Ratio(v/vc) 0 2 4 6 8 9 9. 5872 0 0. 2086 0. 4172 0. 6258 0. 8344 0. 9388 1 Frequency(w) 7. 7940 7. 5968 6. 9807 5. 8549 3. 825 1. 9897 0 Frequency Ratio(w/wn) 1 0. 9747 0. 8957 0. 7512 0. 4981 0. 2553 0 41 Figure 4. 3 Reduction of Fundamental Frequency for a Cantilever Pipe with increasing Flow Velocity CHAPTER V FLOW INDUCED VIBRATIONS IN PIPES, A FINITE ELEMENT APPROACH E, I v L Figure 5. 1 Representation of Tapered Pipe Carrying Fluid 5. 1 Tapered Pipe Carrying Fluid Consider a pipe of length L, modulus of elasticity E. A ? uid ? ows through the pipe at a velocity v and density ? through the internal pipe cross-section. As the ? uid ? ows through the de? ecting pipe it is accelerated, because of the changing curv ature 42 43 f the pipe and the lateral vibration of the pipeline. The vertical component of ? uid pressure applied to the ? uid element and the pressure force F per unit length applied on the ? uid element by the tube walls oppose these accelerations. The gossip parameters are given by the user. Density of the pipe ? p (Kg/m3 ) 8000 Density of the ? uid ? f (Kg/m3 ) 1000 Length of the pipe L (m) 2 Number of elements n 10 Modulus Elasticity E (Gpa) 207 of For these user de? ned values we introduce a taper in the pipe so that the material property and the length of the pipe with the taper or without the taper remain the same.This is done by keeping the inner diameter of the pipe constant and varying the outside diameter. Refer to ? gure (5. 2) The pipe tapers from one end having a thickness x to the other end having a thickness Pipe Carrying Fluid 9. 8mm OD= 10 mm L=2000 mm x mm t =0. 01 mm ID= 9. 8 mm Tapered Pipe Carrying Fluid Figure 5. 2 Introducing a Taper in the Pipe Carrying Fluid of t = 0. 01mm such that the volume of material is equal to the volume of material 44 for a pipe with no taper. The thickness x of the tapered pipe is now calculated From ? gure(5. 2) we have outmost(a) Diameter of the pipe with no taper(OD) 10 mm Inner Diameter of the pipe(ID) 9. mm Outer Diameter of thick end of the Tapered pipe (OD1 ) Length of the pipe(L) 2000 mm Thickness of thin end of the taper(t) 0. 01 mm Thickness of thick end of the taper x mm Volume of the pipe without the taper V1 = Volume of the pipe with the taper ? ? L ? 2 V2 = (OD1 ) + (ID + 2t)2 ? (ID2 ) 4 4 3 4 (5. 2) ? (OD2 ? ID2 )L 4 (5. 1) Since the volume of material distributed over the length of the two pipes is equal We have, V1 = V2 (5. 3) Substituting the value for V1 and V2 from equations(5. 1) and (5. 2) into equation(5. 3) yields ? ? ? L ? 2 (OD2 ? ID2 )L = (OD1 ) + (ID + 2t)2 ? (ID2 ) 4 4 4 3 4 The outer diameter for the thick end of the tapered pipe can be expressed as (5. 4) OD1 = I D + 2x (5. 5) 45 Substituting values of outer diameter(OD),inner diameter(ID),length(L) and thickness(t) into equation (5. 6) yields ? 2 ? ? 2000 ? (10 ? 9. 82 )2000 = (9. 8 + 2x)2 + (9. 8 + 0. 02)2 ? (9. 82 ) 4 4 4 3 4 resolution equation (5. 6) yields (5. 6) x = 2. 24mm (5. 7) Substituting the value of thickness x into equation(5. 5) we get the outer diameter OD1 as OD1 = 14. 268mm (5. 8) Thus, the taper in the pipe varies from a outer diameters of 14. 268 mm to 9. 82 mm. 46The following MATLAB program is employ to calculate the fundamental natural frequency of vibration for a tapered pipe carrying ? uid. Refer to Appendix 3 for the complete MATLAB program. Results obtained from the program are given in table (5. 1) Table 5. 1 Reduction of Fundamental Frequency for a Tapered pipe with increasing Flow Velocity Velocity of Fluid(v) Velocity Ratio(v/vc) 0 20 40 60 80 100 103. 3487 0 0. 1935 0. 3870 0. 5806 0. 7741 0. 9676 1 Frequency(w) 40. 8228 40. 083 37. 7783 33. 5980 26. 579 8 10. 7122 0 Frequency Ratio(w/wn) . 8100 0. 7784 0. 7337 0. 6525 0. 5162 0. 2080 0The fundamental frequency of vibration and the critical velocity of ? uid for a tapered pipe carrying ? uid obtained from the MATLAB program are ? n 51. 4917 rad/sec vc 103. 3487 m/sec CHAPTER VI RESULTS AND DISCUSSIONS In the present work, we have utilized numerical method techniques to form the grassroots elemental matrices for the pinned-pinned and pinned-free pipe carrying ? uid. Matlab programs have been developed and utilized to form global matrices from these elemental matrices and fundamental frequency for free vibration has been calculated for various pipe con? gurations and varying ? uid ? ow velocities.Consider a pipe carrying ? uid having the following user de? ned parameters. E, I v L v Figure 6. 1 Representation of Pipe Carrying Fluid and Tapered Pipe Carrying Fluid 47 48 Density of the pipe ? p (Kg/m3 ) 8000 Density of the ? uid ? f (Kg/m3 ) 1000 Length of the pipe L (m) 2 Number of el ements n 10 Modulus Elasticity E (Gpa) 207 of Refer to Appendix 1 and Appendix 3 for the complete MATLAB program Parametric study carried out on a pinned-pinned and tapered pipe for the same material of the pipe and subjected to the same conditions reveal that the tapered pipe is more stable than a pinned-pinned pipe.Comparing the following set of tables justi? es the above statement. The fundamental frequency of vibration and the critical velocity of ? uid for a tapered and a pinned-pinned pipe carrying ? uid are ? nt 51. 4917 rad/sec ? np 21. 8582 rad/sec vct 103. 3487 m/sec vcp 16. 0553 m/sec Table 6. 1 Reduction of Fundamental Frequency for a Tapered Pipe with increasing Flow Velocity Velocity of Fluid(v) Velocity Ratio(v/vc) 0 20 40 60 80 100 103. 3487 0 0. 1935 0. 3870 0. 5806 0. 7741 0. 9676 1 Frequency(w) 40. 8228 40. 083 37. 7783 33. 5980 26. 5798 10. 7122 0 Frequency Ratio(w/wn) 0. 8100 0. 7784 0. 7337 0. 6525 0. 5162 0. 2080 0 9 Table 6. 2 Reduction of Fundamental Frequen cy for a Pinned-Pinned Pipe with increasing Flow Velocity Velocity of Fluid(v) Velocity Ratio(v/vc) 0 2 4 6 8 10 12 14 16. 0553 0 0. 1246 0. 2491 0. 3737 0. 4983 0. 6228 0. 7474 0. 8720 1 Frequency(w) 21. 8806 21. 5619 20. 5830 18. 8644 16. 2206 12. 1602 3. 7349 0. 3935 0 Frequency Ratio(w/wn) 1 0. 9864 0. 9417 0. 8630 0. 7421 0. 5563 0. 1709 0. 0180 0 The fundamental frequency for vibration and critical velocity for the onset of instability in tapered pipe is approximately triad times larger than the pinned-pinned pipe,thus making it more stable. 50 6. 1 Contribution of the Thesis real Finite Element Model for vibration analysis of a Pipe Carrying Fluid. Implemented the above developed model to two di? erent pipe con? gurations Simply Supported and Cantilever Pipe Carrying Fluid. Developed MATLAB Programs to solve the Finite Element Models. primed(p) the e? ect of ? uid velocities and density on the vibrations of a thin walled Simply Supported and Cantilever pipe carrying ? u id. The critical velocity and natural frequency of vibrations were determined for the above con? gurations. Study was carried out on a variable wall thickness pipe and the results obtained show that the critical ? id velocity can be increased when the wall thickness is tapered. 6. 2 Future Scope Turbulence in Two-Phase Fluids In single-phase ? ow,? uctuations are a direct consequence of uplift developed in ? uid, whereas the situation is all the way more complex in two-phase ? ow since the ? uctuation of the florilegium itself is added to the inherent turbulence of each phase. clear the study to a time dependent ? uid velocity ? owing through the pipe. BIBLIOGRAPHY 1 Doods. H. L and H. Runyan E? ects of High-Velocity Fluid Flow in the Bending Vibrations and Static dissimilarity of a Simply Supported Pipe.National Aeronautics and Space Administration subject area NASA TN D-2870 June(1965). 2 Ashley,H and G. Haviland Bending Vibrations of a Pipe bank bill Containing current Fluid. J. Appl. Mech. 17,229-232(1950). 3 Housner,G. W Bending Vibrations of a Pipe Line Containing menstruum Fluid. J. Appl. Mech. 19,205-208(1952). 4 Long. R. H Experimental and divinatory Study of Transverse Vibration of a tube Containing Flowing Fluid. J. Appl. Mech. 22,65-68(1955). 5 Liu. H. S and C. D. Mote Dynamic Response of Pipes Transporting Fluids. J. Eng. for fabrication 96,591-596(1974). 6 Niordson,F. I. N Vibrations of a Cylinderical Tube Containing Flowing Fluid. Trans. Roy. Inst. Technol. Stockholm 73(1953). 7 Handelman,G. H A Note on the transverse Vibration of a tube Containing Flowing Fluid. Quarterly of Applied Mathematics 13,326-329(1955). 8 Nemat-Nassar,S. S. N. Prasad and G. Herrmann Destabilizing E? ect on VelocityDependent Forces in Nonconservative Systems. AIAA J. 4,1276-1280(1966). 51 52 9 Naguleswaran,S and C. J. H. Williams Lateral Vibrations of a Pipe Conveying a Fluid. J. Mech. Eng. Sci. 10,228-238(1968). 10 Herrmann. G and R. W.Bungay On the Stabi lity of Elastic Systems Subjected to Nonconservative Forces. J. Appl. Mech. 31,435-440(1964). 11 Gregory. R. W and M. P. Paidoussis Unstable Oscillations of Tubular Cantilevers Conveying Fluid-I opening. Proc. Roy. Soc. (London). Ser. A 293,512-527(1966). 12 S. S. Rao The Finite Element regularity in Engineering. Pergamon Press Inc. 245294(1982). 13 Michael. R. concoct Vibration Simulation Using Matlab and Ansys. Chapman and manor hall/CRC 349-361,392(2001). 14 Robert D. Blevins Flow Induced Vibrations. Krieger 289,297(1977). Appendices 53 54 0. 1 MATLAB program for Simply Supported Pipe Carrying FluidMATLAB program for Simply Supported Pipe Carrying Fluid. % The f o l l o w i n g MATLAB Program c a l c u l a t e s t h e Fundamental % N a t u r a l f r e q u e n c y o f v i b r a t i o n , f r e q u e n c y r a t i o (w/wn) % and v e l o c i t y r a t i o ( v / vc ) , f o r a % simply supported pipe carrying f l u i d . % I n o r d e r t o perform t h e above t a s k t h e progr am a s s e m b l e s % E l e m e n t a l S t i f f n e s s , D i s s i p a t i o n , and I n e r t i a m a t r i c e s % t o form G l o b a l M a t r i c e s which are used t o c a l c u l a t e % Fundamental N a t u r a l % Frequency w . lc num elements =input ( arousal number o f e l e m e n t s f o r beam ) % num elements = The u s e r e n t e r s t h e number o f e l e m e n t s % i n which t h e p i p e % has t o be d i v i d e d . n=1 num elements +1% Number o f nodes ( n ) i s e q u a l t o number o f %e l e m e n t s p l u s one n o d e l =1 num elements node2 =2 num elements +1 exclusive nodel= gook( n o d e l ) max node2=max( node2 ) max node used=max( max nodel max node2 ) mnu=max node used k=zeros (2? mnu ) % C r e a t i n g a G l o b a l S t i f f n e s s Matrix o f z e r o s 55 m =zeros (2? nu ) % C r e a t i n g G l o b a l pickle Matrix o f z e r o s x=zeros (2? mnu ) % C r e a t i n g G l o b a l Matrix o f z e r o s % f o r t h e f o r c e t h a t c onforms f l u i d % to the curvature of the % pipe d=zeros (2? mnu ) % C r e a t i n g G l o b a l D i s s i p a t i o n Matrix o f z e r o s %( C o r i o l i s Component ) t=num elements ? 2 L=2 % T o t a l l e n g t h o f t h e p i p e i n meters l=L/ num elements % Length o f an e l e m e n t t1 =. 0001 od = . 0 1 i d=od? 2? t 1 % t h i c k n e s s o f t h e p i p e i n meter % outer diameter of the pipe % inner diameter of the pipeI=pi ? ( od? 4? i d ? 4)/64 % moment o f i n e r t i a o f t h e p i p e E=207? 10? 9 roh =8000 rohw =1000 % Modulus o f e l a s t i c i t y o f t h e p i p e % Density of the pipe % d e n s i t y o f water ( FLuid ) M =roh ? pi ? ( od? 2? i d ? 2)/4 + rohw? pi ? . 2 5 ? i d ? 2 % mass per u n i t l e n g t h o f % the pipe + f l u i d rohA=rohw? pi ? ( . 2 5 ? i d ? 2 ) l=L/ num elements v=0 % v e l o c i t y o f t h e f l u i d f l o w i n g t h r o u g h t h e p i p e %v =16. 0553 z=rohA/M i=sqrt ( ? 1) wn= ( ( 3 . 1 4 ) ? 2 /L? 2)? sqrt (E? I /M) % N a t u r a l Frequency vc =(3. 14/L)? sqrt (E?I /rohA ) % C r i t i c a l V e l o c i t y 56 % Assembling G l o b a l S t i f f n e s s , D i s s i p a t i o n and I n e r t i a M a t r i c e s for j =1 num elements d o f 1 =2? n o d e l ( j ) ? 1 d o f 2 =2? n o d e l ( j ) d o f 3 =2? node2 ( j ) ? 1 d o f 4 =2? node2 ( j ) % S t i f f n e s s Matrix Assembly k ( dof1 , d o f 1 )=k ( dof1 , d o f 1 )+ (12? E? I / l ? 3 ) k ( dof2 , d o f 1 )=k ( dof2 , d o f 1 )+ (6? E? I / l ? 2 ) k ( dof3 , d o f 1 )=k ( dof3 , d o f 1 )+ (? 12? E? I / l ? 3 ) k ( dof4 , d o f 1 )=k ( dof4 , d o f 1 )+ (6? E? I / l ? 2 ) k ( dof1 , d o f 2 )=k ( dof1 , d o f 2 )+ (6? E?I / l ? 2 ) k ( dof2 , d o f 2 )=k ( dof2 , d o f 2 )+ (4? E? I / l ) k ( dof3 , d o f 2 )=k ( dof3 , d o f 2 )+ (? 6? E? I / l ? 2 ) k ( dof4 , d o f 2 )=k ( dof4 , d o f 2 )+ (2? E? I / l ) k ( dof1 , d o f 3 )=k ( dof1 , d o f 3 )+ (? 12? E? I / l ? 3 ) k ( dof2 , d o f 3 )=k ( dof2 , d o f 3 )+ (? 6? E? I / l ? 2 ) k ( dof3 , d o f 3 )=k ( dof3 , d o f 3 )+ (12? E? I / l ? 3 ) k ( dof4 , d o f 3 )=k ( dof4 , d o f 3 )+ (? 6? E? I / l ? 2 ) k ( dof1 , d o f 4 )=k ( dof1 , d o f 4 )+ (6? E? I / l ? 2 ) k ( dof2 , d o f 4 )=k ( dof2 , d o f 4 )+ (2? E? I / l ) k ( dof3 , d o f 4 )=k ( dof3 , d o f 4 )+ (? ? E? I / l ? 2 ) k ( dof4 , d o f 4 )=k ( dof4 , d o f 4 )+ (4? E? I / l ) % 57 % Matrix a s s e m b l y f o r t h e second term i e % f o r t h e f o r c e t h a t conforms % f l u i d to the curvature of the pipe x ( dof1 , d o f 1 )=x ( dof1 , d o f 1 )+ ( ( 3 6 ? rohA? v ? 2)/30? l ) x ( dof2 , d o f 1 )=x ( dof2 , d o f 1 )+ ( ( 3 ? rohA? v ? 2)/30? l ) x ( dof3 , d o f 1 )=x ( dof3 , d o f 1 )+ (( ? 36? rohA? v ? 2)/30? l ) x ( dof4 , d o f 1 )=x ( dof4 , d o f 1 )+ ( ( 3 ? rohA? v ? 2)/30? l ) x ( dof1 , d o f 2 )=x ( dof1 , d o f 2 )+ ( ( 3 ? ohA? v ? 2)/30? l ) x ( dof2 , d o f 2 )=x ( dof2 , d o f 2 )+ ( ( 4 ? rohA? v ? 2)/30? l ) x ( dof3 , d o f 2 )=x ( dof3 , d o f 2 )+ (( ? 3? rohA? v ? 2)/30? l ) x ( dof4 , d o f 2 )=x ( dof4 , d o f 2 )+ (( ? 1? rohA? v ? 2)/30? l ) x ( dof1 , d o f 3 )=x ( dof1 , d o f 3 )+ (( ? 36? rohA? v ? 2)/30? l ) x ( dof2 , d o f 3 )=x ( dof2 , d o f 3 )+ (( ? 3? rohA? v ? 2)/30? l ) x ( dof3 , d o f 3 )=x ( dof3 , d o f 3 )+ ( ( 3 6 ? rohA? v ? 2)/30? l ) x ( dof4 , d o f 3 )=x ( dof4 , d o f 3 )+ (( ? 3? rohA? v ? 2)/30? l ) x ( dof1 , d o f 4 )=x ( dof1 , d o f 4 )+ ( ( 3 ? rohA? v ? 2)/30? ) x ( dof2 , d o f 4 )=x ( dof2 , d o f 4 )+ (( ? 1? rohA? v ? 2)/30? l ) x ( dof3 , d o f 4 )=x ( dof3 , d o f 4 )+ (( ? 3? rohA? v ? 2)/30? l ) x ( dof4 , d o f 4 )=x ( dof4 , d o f 4 )+ ( ( 4 ? rohA? v ? 2)/30? l ) % % D i s s i p a t i o n Matrix Assembly d ( dof1 , d o f 1 )=d ( dof1 , d o f 1 )+ (2? ( ? 30? rohA? v ) / 6 0 ) d ( dof2 , d o f 1 )=d ( dof2 , d o f 1 )+ ( 2 ? ( 6 ? rohA? v ) / 6 0 ) d ( dof3 , d o f 1 )=d ( dof3 , d o f 1 )+ ( 2 ? ( 3 0 ? rohA? v ) / 6 0 ) 58 d ( dof4 , d o f 1 )=d ( dof4 , d o f 1 )+ (2? ( ? 6? rohA? ) / 6 0 ) d ( dof1 , d o f 2 )=d ( dof1 , d o f 2 )+ (2? ( ? 6? rohA? v ) / 6 0 ) d ( dof2 , d o f 2 )=d ( dof2 , d o f 2 )+ ( 2 ? ( 0 ? rohA? v ) / 6 0 ) d ( dof3 , d o f 2 )=d ( dof3 , d o f 2 )+ ( 2 ? ( 6 ? rohA? v ) / 6 0 ) d ( dof4 , d o f 2 )=d ( dof4 , d o f 2 )+ (2? ( ? 1? rohA? v ) / 6 0 ) d ( dof1 , d o f 3 )=d ( dof1 , d o f 3 )+ (2? ( ? 30? rohA? v ) / 6 0 ) d ( dof2 , d o f 3 )=d ( dof2 , d o f 3 )+ (2? ( ? 6? rohA? v ) / 6 0 ) d ( dof3 , d o f 3 )=d ( dof3 , d o f 3 )+ ( 2 ? ( 3 0 ? rohA? v ) / 6 0 ) d ( dof4 , d o f 3 )=d ( dof4 , d o f 3 )+ ( 2 ? ( 6 ? rohA? v ) / 6 0 ) ( dof1 , d o f 4 )=d ( dof1 , d o f 4 )+ ( 2 ? ( 6 ? rohA? v ) / 6 0 ) d ( dof2 , d o f 4 )=d ( dof2 , d o f 4 )+ ( 2 ? ( 1 ? rohA? v ) / 6 0 ) d ( dof3 , d o f 4 )=d ( dof3 , d o f 4 )+ (2? ( ? 6? rohA? v ) / 6 0 ) d ( dof4 , d o f 4 )=d ( dof4 , d o f 4 )+ ( 2 ? ( 0 ? rohA? v ) / 6 0 ) % % I n e r t i a Matrix Assembly m( dof1 , d o f 1 )=m( dof1 , d o f 1 )+ (156? M? l / 4 2 0 ) m( dof2 , d o f 1 )=m( dof2 , d o f 1 )+ (22? l ? 2? M/ 4 2 0 ) m( dof3 , d o f 1 )=m( dof3 , d o f 1 )+ (54? l ? M/ 4 2 0 ) m( dof4 , d o f 1 )=m( dof4 , d o f 1 )+ (? 3? l ? 2? M/ 4 2 0 ) m( dof1 , d o f 2 )=m( dof1 , d o f 2 )+ (22? l ? 2? M/ 4 2 0 ) m( dof2 , d o f 2 )=m( dof2 , d o f 2 )+ (4? M? l ? 3 / 4 2 0 ) m( dof3 , d o f 2 )=m( dof3 , d o f 2 )+ (13? l ? 2? M/ 4 2 0 ) m( dof4 , d o f 2 )=m( dof4 , d o f 2 )+ (? 3? M? l ? 3 / 4 2 0 ) 59 m( dof1 , d o f 3 )=m( dof1 , d o f 3 )+ (54? M? l / 4 2 0 ) m( dof2 , d o f 3 )=m( dof2 , d o f 3 )+ (13? l ? 2? M/ 4 2 0 ) m( dof3 , d o f 3 )=m( dof3 , d o f 3 )+ (156? l ? M/ 4 2 0 ) m( dof4 , d o f 3 )=m( dof4 , d o f 3 )+ (? 22? l ? 2? M/ 4 2 0 ) m( dof1 , d o f 4 )=m( dof1 , d o f 4 )+ (? 13? l ? 2?M/ 4 2 0 ) m( dof2 , d o f 4 )=m( dof2 , d o f 4 )+ (? 3? M? l ? 3 / 4 2 0 ) m( dof3 , d o f 4 )=m( dof3 , d o f 4 )+ (? 22? l ? 2? M/ 4 2 0 ) m( dof4 , d o f 4 )=m( dof4 , d o f 4 )+ (4? M? l ? 3 / 4 2 0 ) end k ( 1 1 , ) = % A p p l y i n g Boundary c o n d i t i o n s k( ,11)= k ( ( 2 ? mnu? 2 ) ( 2 ? mnu? 2 ) , ) = k ( , ( 2 ? mnu? 2 ) ( 2 ? mnu? 2 ) ) = k x(11 ,)= x( ,11)= x ( ( 2 ? mnu? 2 ) ( 2 ? mnu? 2 ) , ) = x ( , ( 2 ? mnu? 2 ) ( 2 ? mnu? 2 ) ) = x % G l o b a l Matrix f o r t h e % Force t h a t conforms f l u i d t o p i p e x1=? d(11 ,)= d( ,11)= d ( ( 2 ? mnu? 2 ) ( 2 ? mnu? 2 ) , ) = % G l o b a l S t i f f n e s s Matrix 60 d ( , ( 2 ? mnu? 2 ) ( 2 ? mnu? 2 ) ) = d d1=(? d ) Kg lobal=k+10? x1 m( 1 1 , ) = m( , 1 1 ) = m( ( 2 ? mnu? 2 ) ( 2 ? mnu? 2 ) , ) = m( , ( 2 ? mnu? 2 ) ( 2 ? mnu? 2 ) ) = m eye ( t ) zeros ( t ) H=? inv (m) ? ( d1 ) ? inv (m)? Kglobal eye ( t ) zeros ( t ) Evalue=eig (H) % E i g e n v a l u e s v r a t i o=v/ vc % V e l o c i t y Ratio % G l o b a l Mass Matrix % G l o b a l D i s s i p a t i o nMatrix i v 2=imag ( Evalue ) i v 2 1=min( abs ( i v 2 ) ) w1 = ( i v 2 1 ) wn w r a t i o=w1/wn vc % Frequency Ratio % Fundamental N a t u r a l f r e q u e n c y 61 0. 2 MATLAB Program for Cantilever Pipe Carrying Fluid MATLAB Program for Cantilever Pipe Carrying Fluid. % The f o l l o w i n g MATLAB Program c a l c u l a t e s t h e Fundamental % N a t u r a l f r e q u e n c y o f v i b r a t i o n , f r e q u e n c y r a t i o (w/wn) % and v e l o c i t y r a t i o ( v / vc ) , f o r a c a n t i l e v e r p i p e % carrying f l u i d . I n o r d e r t o perform t h e above t a s k t h e program a s s e m b l e s % E l e m e n t a l S t i f f n e s s , D i s s i p a t i o n , and I n e r t i a m a t r i c e s % t o form G l o b a l M a t r i c e s which are used % t o c a l c u l a t e Fundamental N a t u r a l % Frequency w . clc num elements =input ( Input number o f e l e m e n t s f o r Pipe ) % num elements = The u s e r e n t e r s t h e number o f e l e m e n t s % i n which t h e p i p e has t o be d i v i d e d . =1 num elements +1% Number o f no des ( n ) i s % e q u a l t o number o f e l e m e n t s p l u s one n o d e l =1 num elements % Parameters used i n t h e l o o p s node2 =2 num elements +1 max nodel=max( n o d e l ) max node2=max( node2 ) max node used=max( max nodel max node2 ) mnu=max node used k=zeros (2? mnu ) % C r e a t i n g a G l o b a l S t i f f n e s s Matrix o f z e r o s 62 m =zeros (2? mnu ) % C r e a t i n g G l o b a l Mass Matrix o f z e r o s