Flow Induced Vibration

FLOW INDUCED VIBRATIONS IN PIPES, A FINITE ELEMENT APPROACH IVAN GRANT Bachelor of Science in Mechanical Engineering Nagpur University Nagpur, India June, 2006 submitted in partial ful? llment of requirements for the degree MASTERS OF SCIENCE IN MECHANICAL ENGINEERING at the CLEVELAND STATE UNIVERSITY May, 2010 This thesis has been approved for the department of MECHANICAL ENGINEERING and the College of Graduate Studies by: Thesis Chairperson, Majid Rashidi, Ph. D. Department & 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 essential support and assistance throughout my graduate career, and also for their guidance which immensely contributed towards the completion of this thesis. This thesis would not have been realized without their support. I would also like to thank Dr. Asuquo. B. Ebiana and Dr. Rama. S. Gorla for being in my thesis committe e. Thanks are also due to my parents,my brother and friends who have encouraged, supported and inspired me. FLOW INDUCED VIBRATIONS IN PIPES, A FINITE ELEMENT APPROACH IVAN GRANT ABSTRACT Flow induced vibrations of pipes with internal ? uid ? ow is studied in this work.Finite Element Analysis methodology is used to determine the critical ? uid velocity that induces the threshold of pipe instability. The partial di? erential equation of motion governing the lateral vibrations of the pipe is employed to develop the sti? ness and inertia matrices corresponding to two of the terms of the equations of motion. The Equation of motion further includes a mixed-derivative term that was treated as a source for a dissipative function. The corresponding matrix with this dissipative function was developed and recognized as the potentially destabilizing factor for the lateral vibrations of the ? id carrying pipe. Two types of boundary conditions, namely simply-supported and cantilevered were consi dered for the pipe. The appropriate mass, sti? ness, and dissipative matrices were developed at an elemental level for the ? uid carrying pipe. These matrices were then assembled to form the overall mass, sti? ness, and dissipative matrices of the entire system. Employing the ? nite element model developed in this work two series of parametric studies were conducted. First, a pipe with a constant wall thickness of 1 mm was analyzed. Then, the parametric studies were extended to a pipe with variable wall thickness.In this case, the wall thickness of the pipe was modeled to taper down from 2. 54 mm to 0. 01 mm. This study shows that the critical velocity of a pipe carrying ? uid can be increased by a factor of six as the result of tapering the wall thickness. iv TABLE OF CONTENTS ABSTRACT LIST OF FIGURES LIST OF TABLES I INTRODUCTION 1. 1 1. 2 1. 3 1. 4 II Overview of Internal Flow Induced Vibrations in Pipes . . . . . . Literature Review . . . . . . . . . . . . . . . . . . . . . . . . . . Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Composition of Thesis . . . . . . . . . . . . . . . . . . . . . . . iv vii ix 1 1 2 2 3 FLOW INDUCED VIBRATIONS IN PIPES, A FINITE ELEMENT APPROACH 2. 1 Mathematical Modelling . . . . . . . . . . . . . . . . . . . . . . . 2. 1. 1 2. 2 Equations of Motion . . . . . . . . . . . . . . . . . . . 4 4 4 12 12 Finite Element Model . . . . . . . . . . . . . . . . . . . . . . . . 2. 2. 1 2. 2. 2 2. 2. 3 Shape Functions . . . . . . . . . . . . . . . . . . . . . Formulating the Sti? ness Matrix for a Pipe Carrying Fluid 14 Forming the Matrix for the Force that conforms the Fluid to the Pipe . . . . . . . . . . . . . . . . . . . . . 21 2. 2. 4 2. 2. 5Dissipation Matrix Formulation for a Pipe carrying Fluid 26 Inertia Matrix Formulation for a Pipe carrying Fluid . 28 III FLOW INDUCED VIBRATIONS IN PIPES, A FINITE ELEMENT APPROACH 31 v 3. 1 Forming Global Sti? ness Matrix from Elemental Sti? ness Matrices . . . . . . . . . . . . . . . . . . . . 31 3. 2 Applying Boundary Conditions to Global Sti? ness Matrix for simply supported pipe with ? uid ? ow . . . . 33 3. 3 Applying Boundary Conditions to Global Sti? ness Matrix for a cantilever pipe with ? uid ? ow . . . . . . . 34 3. 4 MATLAB Programs for Assembling Global Matrices for Simply Supported and Cantilever pipe carrying ? uid . . . . . . . . . . 35 35 36 3. 5 3. 6 MATLAB program for a simply supported pipe carrying ? uid . . MATLAB program for a cantilever pipe carrying ? uid . . . . . . IV FLOW INDUCED VIBRATIONS IN PIPES, A FINITE ELEMENT APPROACH 4. 1 V Parametric Study . . . . . . . . . . . . . . . . . . . . . . . . . . 37 37 FLOW INDUCED VIBRATIONS IN PIPES, A FINITE ELEMENT APPROACH 5. 1 Tapered Pipe Carrying Fluid . . . . . . . . . . . . . . . . . . . . 42 42 47 50 50 51 54 MATLAB program for Simply Supported Pipe Carrying Fluid . . MATLAB Program for Cantilever Pipe Carrying Fluid . . . . . . MATLAB Program for Tapered Pipe Carrying Flu id . . . . . . 54 61 68 VI RESULTS AND DISCUSSIONS 6. 1 6. 2 Contribution of the Thesis . . . . . . . . . . . . . . . . . . . . . Future Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . BIBLIOGRAPHY Appendices 0. 1 0. 2 0. 3 vi LIST OF FIGURES 2. 1 2. 2 Pinned-Pinned Pipe Carrying Fluid * . . . . . . . . . . . . . . Pipe Carrying Fluid, Forces and Moments acting on Elements (a) Fluid (b) Pipe ** . . . . . . . . . . . . . . . . . . . . . . . . . 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 Force due to Bending . . . . . . . . . . . . . . . . . . . . . . . . .Force that Conforms Fluid to the Curvature of Pipe . . . . . Coriolis Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Inertia Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Pipe Carrying Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . Beam Element Model . . . . . . . . . . . . . . . . . . . . . . . . . Relationship between Stress and Stra in, Hooks Law . . . . . . 2. 10 Plain sections remain plane . . . . . . . . . . . . . . . . . . . . . 2. 11 Moment of Inertia for an Element in the Beam . . . . . . . . . 2. 12 Pipe Carrying Fluid Model . . . . . . . . . . . . . . . . . . . . . 3. 1 3. 2 3. 4. 1 Representation of Simply Supported Pipe Carrying Fluid . . Representation of Cantilever Pipe Carrying Fluid . . . . . . . Pinned-Free Pipe Carrying Fluid* . . . . . . . . . . . . . . . . . Reduction of Fundamental Frequency for a Pinned-Pinned Pipe with increasing Flow Velocity . . . . . . . . . . . . . . . . 4. 2 Shape Function Plot for a Cantilever Pipe with increasing Flow Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4. 3 Reduction of Fundamental Frequency for a Cantilever Pipe with increasing Flow Velocity . . . . . . . . . . . . . . . . . . . . 5. 1 Representation of Tapered Pipe Carrying Fluid . . . . . . . 39 40 41 42 vii 5. 2 6. 1 Introducing a Taper in the Pipe Carrying Fluid . . . . . . . . Representation of Pipe Carrying Fluid and Tapered Pipe Carrying Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 47 viii LIST OF TABLES 4. 1 Reduction of Fundamental Frequency for a Pinned-Pinned Pipe with increasing Flow Velocity . . . . . . . . . . . . . . . . 38 4. 2 Reduction of Fundamental Frequency for a Pinned-Free Pipe with increasing Flow Velocity . . . . . . . . . . . . . . . . . . . . 40 5. 1 Reduction of Fundamental Frequency for a Tapered pipe with increasing Flow Velocity . . . . . . . . . . . . . . . . . . . . . . 46 6. 1 Reduction of Fundamental Frequency for a Tapered Pipe with increasing Flow Velocity . . . . . . . . . . . . . . . . . . . . . . . 48 6. 2 Reduction of Fundamental Frequency for a Pinned-Pinned Pipe with increasing Flow Velocity . . . . . . . . . . . . . . . . 49 ix CHAPTER I INTRODUCTION 1. 1 Overview of Internal Flow Induced Vibrations in Pipes The ? ow of a ? uid through a pipe can impose pressures on the walls of the pipe c ausing it to de? ect under certain ? ow conditions. This de? ection of the pipe may lead to structural instability of the pipe.The fundamental natural frequency of a pipe generally decreases with increasing velocity of ? uid ? ow. There are certain cases where decrease in this natural frequency can be very important, such as very high velocity ? ows through ? exible thin-walled pipes such as those used in feed lines to rocket motors and water turbines. The pipe becomes susceptible to resonance or fatigue failure if its natural frequency falls below certain limits. With large ? uid velocities the pipe may become unstable. The most familiar form of this instability is the whipping of an unrestricted garden hose.The study of dynamic response of a ? uid conveying pipe in conjunction with the transient vibration of ruptured pipes reveals that if a pipe ruptures through its cross section, then a ? exible length of unsupported pipe is left spewing out ? uid and is free to whip about and im pact other structures. In power plant 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 simply supported pipe containing ? id were carried out by Ashley and Haviland[2]. Subsequently,Housner[3] derived the equations of motion of a ? uid conveying pipe more completely and developed an equation relating the fundamental bending frequency of a simply supported pipe to the velocity of the internal ? ow of the ? uid. He also stated that at certain critical velocity, a statically unstable condition could exist. Long[4] presented an alternate solution to Housner’s[3] equation of motion for the simply supported end conditions and also treated the ? xed-free end conditions. He compared the analysis with experimental results to con? rm the mathematical model.His experi mental results were rather inconclusive since the maximum ? uid velocity available for the test was low and change in bending frequency was very small. Other e? orts to treat this subject were made by Benjamin, Niordson[6] 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 critical velocity pipes supported at both ends bow out and buckle[1]. Straight Cantilever pipes fall into ? ow induced vibrations and vibrate at a large amplitude when ? ow velocity exceeds critical velocity[8-11]. . 3 Objective The objective of this thesis is to implement numerical solutions method, more 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 internal ? uid ? ow has been formed and dis- 3 cussed. The governing equatio n of motion is a partial di? erential equation that is fourth order in spatial variable and second order in time. Parametric studies have been performed to examine the in? uence of mass distribution along the length of the pipe carrying ? id. 1. 4 Composition of Thesis This thesis is organized according to the following sequences. The equations of motions are derived in chapter(II)for pinned-pinned and ? xed-pinned pipe carrying ? uid. A ? nite element model is created to solve the equation of motion. Elemental matrices are formed for pinned-pinned and ? xed-pinned pipe carrying ? uid. Chapter(III)consists of MATLAB programs that are used to assemble global matrices for the above cases. Boundary conditions are applied and based on the user de? ned parameters fundamental natural frequency for free vibration is calculated for various pipe con? urations. Parametric studies are carried out in the following chapter and results are obtained and discussed. CHAPTER II FLOW INDUCED VIBRATION S IN PIPES, A FINITE ELEMENT APPROACH In this chapter,a mathematical model is formed by developing equations of a straight ? uid conveying pipe and these equations are later solved for the natural frequency and onset of instability of a cantilever and pinned-pinned pipe. 2. 1 2. 1. 1 Mathematical Modelling Equations of Motion Consider a pipe of length L, modulus of elasticity E, and its transverse area moment I. A ? uid ? ows through the pipe at pressure p and density ? t a constant velocity v through the internal pipe cross-section of area A. As the ? uid ? ows through the de? ecting pipe it is accelerated, because of the changing curvature of 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. Referring to ? gures (2. 1) and 4 5 Figure 2. 1: Pinned-Pinned Pipe Carrying Fluid * (2. 2),balancing the forces 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 length of the pipe is opposed by the shear stress of the ? uid friction against the tube walls. The sum of the forces parallel Figure 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 perimeter of the pipe, and ? s the shear 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 tension in the pipe. The forces on the element of the pipe normal to the pipe axis accelerate the pi pe 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 length of the empty pipe. The bending moment 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 above 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 pressure 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 given 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 length of the pi pe and the ? uid in the pipe is given by M = m + ? A. The next section describes the forces acting on the pipe carrying ? uid for each of the components of eq(2. 9) Y F1 X Z EI ? 4Y ? x4 Figure 2. 3: Force due to Bending Representation of the First Term in the Equation of Motion for a Pipe Carrying Fluid 8 The term EI ? Y is a force component 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 Representation of the Second Term in the Equation of Motion for a Pipe Carrying Fluid The term ? Av 2 ? Y is a force component acting on the pipe as a result of ? ow ? x2 around a curved pipe. In other words the momentum 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 t he Equation of Motion for a Pipe Carrying Fluid ? Y The term 2? Av ? x? t is the force required to rotate the ? id element as each point 2 in the span 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 Fourth 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 Element 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 elasticity of the pipe is E,and the area 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 area A. Flow of the ? uid through the de? ecting pipe is accelerated 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 approximate the unknown by an expression given as n w= i=1 Ni ai where Ni are the interpolating shape functions prescribed in terms of linear independent functions and ai are a set of unknown 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 Substituting the values of L1 and L2 we get (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 Matrix for a Pipe Carrying Fluid ?1 ?2 W1 W2 Figure 2. 8: Beam Element Model For a two dimensional beam 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. 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, ? 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 strain energy at point R is given by UR . The internal strain energy at point R can be expressed 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 Stress and Strain, Hooks Law Also; ? =E Rel ation between stress and strain for elastic material, Hooks Law Substituting 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: Plain sections remain plane Assuming plane sections remain same, = 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 integrate the internal strain energy at point R over the volume. The internal strain energy for the entire beam is given as: UR dv = U vol (2. 26) Substituting 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 above 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) Substituting 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 above 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 nothing 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 property of the transpose, yields (AB)T = B T AT (2. 35) We can express the Potential Energy expressed in equation(2. 34) in terms 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. 3 8) 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 substit uting the values of shape functions from equations (2. 42) to (2. 45) into equation(2. 41) and integrating 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 Carrying 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 causing 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 angle made by the pipe due to the ? uid ? ow with the neutral axis. ? and ? represent the unit vectors along the X i j ? nd Y axis and r and ? rep resent 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 + sin ? i j (2. 47) ? ? = ? sin + cos i j Expression for slope at point R is given by; tan? = dW dx (2. 48) (2. 49) 22 Since the pipe undergoes a small de? ection, hence ? 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? erentiate the above equation to get 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 ? + r[cos + sin + r? [? sin + cos i j] i j] Sin ce ? 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 cross-section of the pipe. ? ? Substituting the value of Ry from equation(2. 60) yields; 1 ? ? ? ? ? ? ? ? ? T = ? f A[W 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 A[W 2 + vf ? 2 + 2W vf ? ] 2 (2. 63) Now 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 integrating 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 to 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 determined 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 N4 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 integrating it over the length gives us the elemental 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 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 ? ? [D]e = ? ? 30 30 ? 6 30 6 ? ? ? ? ? 6 1 ? 6 0 [D]e represents the elemental 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, volume 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 vecto r 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) Kinetic 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 volume. Also,mass can be expressed as the product of density and volume 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 Lagr ange 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. 100) (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 Elemental Sti? ness Matrices Inorder to form a Global Matrix,we start with a 6Ãâ€"6 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 4Ãâ€"4 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 build 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 matrix ? ? 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 6Ãâ€"6 Global Sti? ess Matrix formulated in eq(3. 5) is mo di? ed to a 4Ãâ€"4 Global Sti? ness Matrix. It is as follows; Y 1 2 X L Figure 3. 1: Representation of Simply Supported 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. Hence we end up with the Sti? ness Matrix shown in eq(3. 6) 34 3. 3 Applying 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 6Ãâ€"6 Global Sti? ness Matrix formulated in eq(3. 5) is modi? ed to a 4Ãâ€"4 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 of 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 outside diameter 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 Appendix 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 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. 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 Appendix 2 for the complete MATLAB Program. 0 * Flow Induced Vibrat ions,Robert D.Blevins,Krieger. 1977,P 297 CHAPTER IV FLOW INDUCED VIBRATIONS IN PIPES, A FINITE ELEMENT APPROACH 4. 1 Parametric Study 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. ) diameter and a . 0001 m (0. 004 in. ) thick wall. The other parameters are: 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 MATLAB program for the simply supported pipe with ? uid ? ow is utilized 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 Table 4. 1: Reduction of Fundamental Frequency for a Pinned-Pinned Pipe with increasing Flow Velocity Velocity of Fluid(v) Ve locity 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. 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 carrying ? 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 Fr equency 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 curvature 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 input 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 outer 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 †¢ Outer 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 th e 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 = ID + 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 Solving 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 utilized 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. 5798 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 basic 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 elements 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 Frequency 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 approxim ately three times larger than the pinned-pinned pipe,thus making it more stable. 50 6. 1 Contribution of the Thesis Developed 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. †¢ Determined the e? ect of ? uid velocities and density on the vibrations of a thin walled Simply Supported and Cantilever pipe carrying ? uid. †¢ 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 turbulence developed in ? uid, whe reas the situation is clearly more complex in two-phase ? ow since the ? uctuation of the mixture itself is added to the inherent turbulence of each phase. †¢ Extend 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 Divergence of a Simply Supported Pipe†.National Aeronautics and Space Administration Report NASA TN D-2870 June(1965). [2] Ashley,H and G. Haviland †Bending Vibrations of a Pipe Line Containing Flowing Fluid†. J. Appl. Mech. 17,229-232(1950). [3] Housner,G. W †Bending Vibrations of a Pipe Line Containing Flowing Fluid†. J. Appl. Mech. 19,205-208(1952). [4] Long. R. H †Experimental and Theoretical 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 Industry 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 Stability 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 Theory†. Proc. Roy. Soc. (London). Ser. A 293,512-527(1966). [12] S. S. Rao †The Finite Element Method in Engineering†. Pergamon Press Inc. 245294(1982). [13] Michael. R. Hatch †Vibration Simulation Using Matlab and Ansys†. Chapman and 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 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 . lc ; n um elements =input ( ’ Input 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; 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 55 m =zeros (2? nu ) ;% C r e a t i n g G l o b a l Mass 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 conforms 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: nu m 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 ? ro hA? 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( d of4 , 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(: ,1:1)=[]; k ( ( 2 ? mnu? 2 ) : ( 2 ? mnu? 2 ) , : ) = [ ] ; k ( : , ( 2 ? mnu? 2 ) : ( 2 ? mnu? 2 ) ) = [ ] ; k x(1:1 ,:)=[]; x(: ,1:1)=[]; 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(1:1 ,:)=[]; d(: ,1:1)=[]; 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 nodes ( n ) i s % e q u a l t o num ber 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

Negative and Positive Effects of Peer Relationships

Possessing a functional or dysfunctional family is of much importance to a healthy development, helping children through peer pressure, acceptance, and the anxiety of belonging. Yet how important is the environment that a child is raised on, this being shared or non-shared? How difficult or easy can peer pressure be? Will peer pressure help or deter a child from being functional. How much do these factors affect development from childhood to adolescence? This paper will explain the different stages of childhood to adolescence, and how a child and adolescence copes with nature and nurture .Family is of great importance to having a functional or dysfunctional development; it will help or impede the child to have a support system in place. The key areas are the family structure, function, shared and non-shared environment. A functional family is a family bonds and works together toward achieving needs. In a difference, the dysfunctional family is the opposite; non-shared environment. In a shared environment, the children are by same parents in the same home and have a crucial role to the development of the Middle childhood and Adolescence period.Children raised by a functional family have some standard to behavior. Parents are first role model. In a functional family, the words that come out from the young child mouth are very selective. The parents built them with such image of respect for both in and out of their environment. Other than behavioral factors, there are conditions that impair the development of the middle childhood. Children living in a non-shares environment maybe malnourished poorly guided, as a result of dysfunctional family. The attitude of such children will most likely be very raw, and disrespectful.Regardless of the family structure, there are stress factors that affect the middle childhood and adolescence development such as separation from both parents if they are divorced, delayed puberty because of malnutrition, adaptation to new environm ent and peer pressure . Positive impacts of peers and peer groups could be moral development, close friendships, and stability. Negative impacts of peers and peer groups could range from rejection, to bullying, and to no sense of belonging. During the Middle Childhood stage, children tend to look for a sense of belonging.As changes occur within and around them, they develop somewhat of their own culture. This culture allows the child to involve himself with others and create a bond that can compromise, share, and defend one another as an equal (Burger, 2010). As these bonds develop, moral growth is also developed. The morals that a child develops during these stages, sets a foundation for his or her morals to continue throughout adulthood. This childhood culture many times allows the children to develop close friendships. Close friendships become like an extended family to some of these children.These friendships can also hold another positive effect on the child, by providing stabi lity. Many children are affected by family disasters, such as divorce, and single parents. The stability of a close friend developing during this time continues to impact the child positively in to Adolescence and adulthood. It is always easier to discuss the positive impacts that peers can and may have on children in the Middle Childhood age, but negative impacts play a large role in a person’s life. One of the largest negative impacts that peers and peer groups can have on a child between the ages of 7-11, is rejection.Rejection impacts the child from the time it begins and continues all throughout the development stages. Bullying is known to impact the child deeply through words and physical actions. Upon peers and peer groups, bullying and rejection can lead to of place. The negative impact of bullying and rejection can lead a child and adolescence into depression or even no sense of belonging. The effects that peers have on a young adolescent can determine how that child behaves and also how the child views the world around them. During adolescent years children often find themselves in scenarios that could harm their future wellbeing.More than ever in this society as children find the need to belong or fit in to the popular crowd in school for acceptance they often tend to mimic their friends behaviors. This is often a result of the individual child trying to find him or herself. The results however, are not always negative. In fact, there are some positive results that may occur as a result of copying their peers. Simply stated, â€Å"When teens surround themselves with people who make good decisions and who are involved with positive activities and choices, it makes the adolescent child want to be better† (Stock, 2010 pg. 2).Positive peers influence adolescents and can drive the child toward improved confidence, and improved grades in school. Inversely, the same can be said for the adolescent child who decides that he or she wants to be l ike his or her friends who have a negative influence. Children who fall into this category are those that are of the bandwagon philosophy. Those negatively impacted by peers often show signs of lower grades in school, increased distance from family. In fact, â€Å"peer pressure can lead to experimentation with drugs and alcohol, and various high risks behaviors† (Fact Sheets, 2009 pg. ). The changes in the adolescent child can have lasting effects depending on which type of peer influences that child may be surrounded by. The negative impact of peer pressure can be strong; however there are also positive influences. Healthy influences are important to have in our lives from birth until death. Additional pressures that adolescents face compared to middle childhood are mostly from the bodily maturation process. Puberty has an enormous impact on the attitude and character that an adolescent presents.In addition, adolescents face a period of identity confusion. According to Inter national Child and Youth Care Network (2001), â€Å"Identity formation arises from repudiation of childhood identifications and the assumption of new configuration with both internal and societal recognitions. † Understanding the effects of puberty and its effects on the child’s identity, for boys, growing facial hair, voice changes, and the onslaught of acne can impede the need to find the identity of his ego. Girls will face breast growth, menstrual cycles, and acne as well during adolescence.The pressures that arise out of coping with these changes are very different, important, and upsetting for an adolescent compared to a nine-year-old. Stages in a child’s life are diverse and will be expressed in the child’s behavior. When a child reaches middle childhood, he or she will be in what is called the latency stage. Latency stage is the time the child starts to make friends of the same sex, is subjective to specific sexual behaviors, and thinks in terms o f morality, intellectual, and social skills start to develop.As an adolescent, they would be in their final phase of the developmental stage which the child focuses more on a heterosexual relationship outside of the family. Given that adequate or appropriate adjustment to the environment or situations, a child can pass through preceding the preceding stages with the most favorable gratification. If not behaviors that inhibit a person's ability to adjust to particular situation can produce unfavorable results. Oswalt, (2008). Moral reasoning will portray the outcome of a child’s behavior in society .There are three levels to moral reasoning and they develop in stages. First is pre-conventional reasoning which there is no integral part of values and external rewards and punishments force reasoning. Second is conventional reasoning that can be characterized by various integral parts of values, usually these values are put in place by parents. The third level is post conventional reasoning, this is when morals are altogether incorporated and is not contingent on non-essential sources. The media's impact and communal outlook with the development with adolescent has been far-reaching.Today's children are pummeled with unachievable values on what is considered beautiful, exposed to violent images of sex and violence. Exposure to such expectations and combined with the physical and emotional changes, children are more likely to struggle with sexuality, be subjected to violence at school, and incur eating disorders. These early years in childhood development are times when children are most vulnerable and when psychological disorders like depression and other temperamental disorders start to appear during this stage of life.Having a functional or dysfunctional family is of great importance to a healthy development. The environment and peer pressure are also of great importance, helping or deterring a child’s function in society. Children must believe they are safe, protected helping the child through key stages of development .

Marketing Management †Caselet on Organic Foods Essay

Overall Understanding of the Case, 3 C’s Scenario and Analysis Of Complete Marketing Environment : Organic food refers to food items that are produced, processed and packaged without using chemicals. Organic food is increasingly becoming popular due to its perceived health benefits over conventional food. The industry is growing rapidly since the past five years and has caught the attention of farmers, manufacturers and, above all, consumers. The health benefits of organic food are more perceived than real. However, the public opinion that organic food is healthier than conventional food is quite strong and is the sole reason for about 30% growth in the organic food industry since the past 5-6 years In general, organic food consumers, manufacturers and farmers strongly believe in organic food having following benefits over non organic food: ? Better health: Since organic food is not prepared using chemical fertilizers and pesticides, it does not contain any traces of these strong chemicals and might not affect the human body. Better taste: People strongly believe that organic food tastes better than non organic food. The prominent reason for this belief is that it is produced using organic means of production. Further organic food is often sold locally resulting in availability of fresh produce in the market. Environment safety: As harmful chemicals are not used in organic farming, there is minimal soil, air and water pollution; thus ensuring a safe world for future generations to live in. Animal welfare: Animal welfare is an important aspect of producing organic milk, organic meat, organic poultry, and organic fish. People feel happy that the animals are not confined to a miserable caged life while eating organic animal products. ? ? ? Popular organic food items include organic tea, organic coffee, organic wine, organic meat, organic beef, organic milk, organic honey, organic vegetables, organic fruits, organic rice, organic corn, organic herbs, organic essential oils, organic coconut oil and organic olive oil. Indians prefer organic preserves, tea, honey, cashew butter and various flours. The organic concept is not limited to food items. Due to excessive usage of harmful chemicals in cosmetics, people are turning towards organic cosmetics, clothing etc. ACNielsen, survey suggests preference of consumers for functional foods – foods that have additional health benefits. India was among the top ten countries where health food, including organic food, was demanded by the consumers. Around the globe, Organic alternatives are purchased mainly for health reasons. Over two thirds of survey respondents think that organic foods are healthier for them and their children. This healthy perception is consistently strong across the Asia Pacific region. Online Indians top the list with 87 percent of consumers purchasing organic foods motivated by the perceived benefits they can derive for their children’s’ health. Health Benefit for kids is seen, hence even willing to pay more. Organic food market is still a developing market, especially in India. Most of the people who eat organic food don’t buy it regularly. Hence need to sell other health products too if one wishes to get a daily (or at least weekly) visit from their customers. Indian organic food consumer needs education. ? There are many consumers who are unaware of the difference between natural and organic food. Many people purchase products labeled as Natural thinking that they are Organic. Organic food refers to food items that are produced, manufactured and handled using organic means. Natural food, on the other hand, generally refers to food items that are not altered chemically or synthesized in any form. These are derived from plants and animals. Thus a natural food item is not necessarily organic and vice versa. Further, consumers are not aware of the certification system. Since certification is not compulsory for domestic retail in India, many fake organic products are available in the market. ? Q. 1 (a) Your organization , which already markets branded Indian spices and flours, is planning to enter the domestic market with organic foods in related products. Segment the market for organic foods, specify the target you as an organization will choose and develop a positioning platform. The Company is already into branded Indian spices and flours ( Consumer non durable category ) and has a robust Intensive Distribution channel & market share. They can use this strength in making the Marketing environment conducive to launching organic foods in the related products. Understand Consumer decision making process : Company needs to focus on the benefits and experiences ( Basic need of Hunger, Security need of being safer option and Health ) produced by these Organic foods over non organic ones & Question why their consumer will / is buying the Organic products. Concentrate on the Health need, influence consumers by educating them and providing adequate information on Organic foods. Allow them to evaluate alternatives and make firm and favourable purchase decisions. The company will expand market share basis this and increase customer satisfaction & resultant profits. Create pull with adequate & clear communication and use ultimate push at dealer / distributor end – i. e. modern retail format – in-store promotions In terms of opportunity will look at i) Product Development i. e. new organic products in present market and also ii) Related Product Diversification – New organic product in new market – from normal branded Spices & Flours to Organic ones. Can further expand the product line by adding packaged Organic Bread with added supplements / vitamins ; Whole grain, high fibre products ; Cereals and pulses. At a later stage also introduce other Organic food product lines like Vegetables, Fruits, Fruit juices etc. Marketing will determine which attributes are most important in influencing a consumer’s choice to purchase. Initially being a Consumer nondurable ( FMCG ) category of product, the involvement will be low. But with the increased awareness, emphasis on the emotional Health need, Credence attribute, consistent quality and value to customer, can look at moving our products to high involvement and Brand loyality. Therefore, creating awareness amongst consumers and within the distribution chain could be seen as an opportunity for marketers to find a positioning for Organic foods claiming health benefits, to reach out to a wider base. Product – Will prefer to choose Organic flours first then spices to begin with as flours are Base foods which are consumed in larger quantity as compared to spices which are Ingredient foods, consumed in lesser quantity. Also base foods have a larger reach and are considered more important as compared to ingredient foods. Segmentation : Name of Product : Potential customer needs : Organic Foods – Flours – Base Foods Basic Hunger need : Base Food Security need : Safe/No Harmful Chemicals / Clean Sub market : Determining variables : Consumer Characteristics Emotional need : Healthy Tasty and Nutritious Pure / Unadulterated High Fibre content Non Polluting / Environment Friendly Base Food : Flour Health & Nutrition B2B – Health Food Restaurants Health & nutrition value Availability Awareness Calorie / Nutrition info Children Health conscious Mothers Earning members of family Elders in family with special health needs. Further Segmentation Geographic – Metro – A class cities Demographic – Income – High income group families Age / Occupation – Children : 3-15 yrs / Students Earning members of family : 25 – 60 yrs /Professionals Elders : 60 yrs + / Retired Psychographic – Health and fitness conscious With special health needs – like Nutrition, BP, Diabetes, Tension, Obesity etc Cultural shifts: lifestyle, leisure Target chosen : Customer : Mother – working / homemaker Consumer : Children Will first target Children Health needs, then at later stages of acceptance can cover all other areas of consumers and also introduce more product lines and variants as discussed earlier. Position : Positioning by Benefits : Basis is the benefit consumer gets in using the product, Product Promise POSITIONING STATEMENT : IT IS A ORGANIC FLOUR, BETTER THAN THE NORMAL NON ORGANIC FLOURS BECAUSE IT IS HARMFUL CHEMICAL FREE AND HENCE MORE HEALTHY AND NUTRITIOUS FOR GROWING CHILDREN The company will talk about the Farm to Fork concept – ie from unpolluted Creation to Consumption Everything is Organic, in the purest form. There is nothing artificial about it For a. Healthier India Will educate existing consumers, create awareness wrt FAB of Organic Foods through all possible areas of Media like TV, Print, In store promotions, Outdoor, Sampling, Demos, Direct mailers, Schools info sessions, Nutrition talks / classes, School Canteens, Health food outlets, Health food restaurants, Exercise areas, Public chat rooms, online etc Q. 1(b) Based on the above question’s answer – Assume that the organic food category is in the introduction phase of PLC, accordingly what would be the major marketing decisions at this stage? Do you expect the adoption rate to be fast or slow? Justify. Major marketing decisions at this introduction phase of PLC As discussed above in the Overall Understanding of the Case, 3 C’s Scenario and Analysis Of Complete Marketing Environment and the Product STP, initially mass market the product through existing Intensive distribution network. This stage will see more of Product information through Promotional campaigns to increase Public awareness, which will then Stimulate demand. The later growth & maturity stages of PLC will see more of Brand Promotion and Market segmentation ( also see above ) Product chosen – Base food – flour , can move to whole grain organic bread / biscuits / pasta Also later Organic ingredient foods like spices, condiments Expand in the area of Organic pulses, vegetables, fruits, juices & preserves. Price – Premium / atleast 20% above normal non organic foods Place – Modern trade / also Health food Restaurants tie up like Subway etc which will exclusively use our organic flours for their whole wheat & multi grain breads. Promotion – In shop education, in store activation to educate consumer. Media like TV, Print, In store promotions, Outdoor, Sampling, Demos, Direct mailers, Schools info sessions, Nutrition talks / classes, School Canteens, Health food outlets, Health food restaurants, Exercise areas, Public chat rooms, online etc Initial Adoption rate will be slow. But with sustained awareness will wait for the tipping point in our favour. As new product category, greater awareness is required There is already a Movement towards natural foods from synthetic / chemical based foods : Beverages – Aerated drinks to juices, Tropicana, Real, no preservatives – LMN, Neebooz, Lemonez, Green tea. Base foods – white to brown – whole grain, multigrain breads, flour Cosmetics – study says 66% of what we apply is absorbed by body – harmful chemicals etc, Hence move towards organic & safe products eg Body Shop, Biotique etc Organic food market small now but Volume to come from growing trend of Health & Environment conscious campaigns and followers who will in turn be major influencers in buying decisions of people. Market will grow and prosper – by creating new market for safe, healthy, and environmentally friendly Organic Foods. Also with this the Company will maintain credibility and constantly improve quality and offerings to suit customer needs and hence hope to create a big brand with loyal customers over a sustained period of time.

3 works of art Essay Example | Topics and Well Written Essays - 1500 words

3 works of art - Essay Example The gown looks like the nuns’ uniform and she is lazily enveloped into the reverie or dream about the angel. This is an epitome of the spiritualism that an image can depict. Moreover, the environment that hooks up the angel and St.Teresa looks dreamy, implying the supernatural feeling that the paint is showing. Lastly, there is intense light that illuminates the shiny structure that St.Teresa is lying on and the gowns that both are wearing. This is a symbol of divine power in a scared place. In this paint, Caravaggio is showing a pretty realistic image of St.peter being crucified. He is keen on the posture of the people and the wood used to make the crucifix. It is a paint that shows the crucifixion of St.Peter as he asked people to do. He is crucified upside down as a symbol of not imitating his Lord Jesus Christ. The painting is so real that it shows the cloth that is wrapped around him to conceal his loins, the people lifting the cross and the obscurity of their faces in the shadows. The picture arouses feelings of sadness by looking at the way St.Peter has left his mouth open with his long beard and dully eyes, showing so much pain. The people lifting the crucifix are also showing fatigue since it is clear that they are straining to lift the crucifix to its intended right position (Brett & Kate, para8). This paint is a fascinating image that describes the evolution of the Catholic Church by Peter Paul Rubens. He is so artistic in the painting of women and men in their full physical features lying on each other in various postures. There is also the presence of leopards in shady color trying to roar at the infants on the ground. This is a description of the spread of Catholic Churches across the world and the numerous challenges it has faced. Behind the leopards, the adults and the infants is a dark blue sky that is not fully clear. It shows the beautiful sky, symbolizing the hope and feasibility of the

Question Essay Example | Topics and Well Written Essays - 250 words - 8

Question - Essay Example When the level of technology is low or out of date, the system level autonomy is limited due to lack of proper requirements while when the level of technology is good, the system level is less limited. This matters a lot. Low level of investment is likely to limit the level of the autonomy of the system-level whereby this will be due to lack of adequate resources while on the other hand if the investment is good the system-level may not be much limited. The surrogate hierarchy of decision making is based on the fact the in the absence of the competent advanced directives, decision still have to be made. In this case the surrogate hierarchy comes into play. The hierarchy of the surrogate decision making is therefore as follows: In case it reaches the decision making point when the suffering patient is out of reach the first people to be consulted in this case are the patients spouse with help in the final decision making. State registered domestic partners are also allowed to come in and play this role together with the sibling. This may come in place due to self selection and convenience selection. Self selection occurs when individuals allocate resources for themselves while convenience occur when occur when resources are easy to allocate This principle is composed of all the conditions that the society experiences and the positivity behind those conditions. All community must actively show concern towards improving the health awareness in order to make it a better place. This principle has three essential elements which are social welfare, peace and security and respect for persons. It is in line with the concept of human dignity, common good and human rights. It implies what the society at large can offer to individual person. This principle requires that every person within the society be able to access basic health care services that are necessary. This is a policy which its objective is to educate and create awareness to the

Challenges Facing the Gulf Cooperation Council Thesis Proposal

Challenges Facing the Gulf Cooperation Council - Thesis Proposal Example The council members criticized the world’s failure to take action on Syria’s current situation. The council condemns Iran’s interference in its internal affairs. Divergent views by GCC member states make it intricate for them to achieve a unified decision with regards to Syria. For instance, Qatar and Saudi Arabia have openly supported the opposition forces in their struggle against Bashar al-Assad’s regime (Ehteshami, 2013). All GCC member countries want better relations with Iran, yet condemn it for interfering with its internal affairs. Iran is suspected to support opposition protests in Bahrain (Rashwan, 2012). GCC neighbors are going through a difficult political transition such as Yemen. Other nations like Jordan are strategically located in relation to the Gulf region. As such, Jordan is included in the security calculations of the Gulf region. Nevertheless, Jordan is subject to pressure from Israeli. This leaves Jordan at the state of economic cris is (Masters, 2013). The diverse experiences that GCC neighbors are going through make it hard for them to achieve a unified decision.   The research seeks to explore aspects of the GCC council and its decision-making mechanisms. The research also seeks to examine the state of countries neighboring the GCC member states. These are such as Iraq, Syria, Yemen, Jordan   International relations study gained momentum in the 19th century following world war two. Three schools of thought emerged in a bid to explain the international system. These are liberalism, realism, and constructivism. However, realists and liberalists are the prominent schools of thoughts explaining international relations. Realism sees international relations as being founded on selfish motives where individual states seek power.

Report Essay Example | Topics and Well Written Essays - 7500 words

Report - Essay Example This difference is important to understand as it impacts not only businesses and investors but also governments in their policy formulation. This paper, therefore, aims to study whether different economic indicators influence large and small cap stocks by looking to answer the question: The key research question explored the relationship between economic indicators and stock prices small cap and as large cap companies in the US using available data for the last 10 years (from Jan 2001 to June 2011) based on 9 leading economic indicators and on two stock indices (one each for large and small cap stocks). The economic indicators were the independent variables while the stock price was the dependent variable in the multiple regression models developed for the two stock indices. Using public records, data were collected from publicly available financial statements of corporations. The proposed relationships were tested through regression analyses. Results indicated that the two stock cat egories are influenced by a different set of economic indicators except the Consumer Price Index, the Industrial Production Index (IPI), and the Consumer Confidence Index. The implication for economic change includes governments’ awareness of the need to monitor these factors as their Key Performance Indicators for measuring the impact and success of their past and future economic policy decisions on businesses. Table of Content Chapter 1Introduction 6 1.1Background and Context 6 1.2Objectives 6 1.3Achievements 7 1.4Overview of Dissertation 8 1.5Problem Description 8 Chapter 2Literature Review 10 Chapter 3Research Method 19 Chapter 4Data Analysis and Results 26 4.1Test of Variables 26 4.2Regression Analyses for the stock indices 26 4.2.1Regression analysis: S&P 500 27 4.2.2Regression analysis: S&P SmallCap 600 34 4.2.3Check for the predictability of regression models 43 Chapter 5Discussion, Conclusions, and Recommendations 46 5.1Summary 46 5.2Evaluation 49 5.3Future Work 51 R eferences 52 Appendix A: Normal distribution of dependent variables 55 Appendix B: Linearity of relation between economic indicators & S&P 500 56 Appendix C: Linearity of relation between economic indicators & S&P SmallCap 600 57 Appendix D: Clarification of concepts and terms 58 Qualifications & Experience 62 Abstract 1 Table of Contents 2 Chapter 1 Introduction 5 Chapter 2 Literature Review 9 Chapter 3 Research Method 18 Chapter 4 Data Analysis and Results 25 Chapter 5 Discussion, Conclusions, and Recommendations 43 References 49 Appendix A: Normal distribution of dependent variables 51 Appendix B: Linearity of relation between economic indicators & S&P 500 52 Appendix C: Linearity of relation between economic indicators & S&P SmallCap 600 53 Appendix D: Clarification of concepts and terms 54 Qualifications & Experience 58 List of Tables Table 1 9 Table 2 27 Table 3 28 Table 4 29 Table 5 29 Table 6 30 Table 7 34 Table 8 36 Table 9 36 Table 10 37 Table 11 38 Table 12 39 Table 13 39 Table 14 46 List of Figures Figure 1: Histogram of regression model residuals for S&P 500 32 Figure 2: Variance of residuals for S&P 500 regression model 32 Figure 3: Scatter plot of residuals for regression model for S&P

My Experiences As a Nursing Student Essay Example | Topics and Well Written Essays - 1750 words - 11

My Experiences As a Nursing Student - Essay Example However, any medical professional can execute the process as long as he or she follows the protocols of medical operations (Callara, L. 2008, 57). The criterion (protocol) that must be followed, starts with checking the absence of possible air leaks into the bottle of a chest drain. Air leaks are usually noted when a patient breathes out using force or coughing with vigor, unlike normal exhalation. Secondly, a check-up for the fluid volume into the chest drains to establish if it is normal or low follows. If it minimal, say, 10ml per hour, chest drain is usually present (Daly, J. Speedy, S. & Jackson, D. 2009, 68). The establishments of a respiratory difficulty also define that there is a chest drain problem. Similarly, an increased bleeding risk that is associated with a decrease in coagulation also shows that there is a chest drain. Having established the explained steps, evidence from radiology for the absence of air or accumulation of fluid in the chest will be required before th e removal of chest drain process begins (Jacob, A. & Sonali, J. 2007, 29). My work in the whole chest drain process was to prepare the equipment used for the procedure and care for the patient since before and after the procedure, the patient is put under a closed –chest underwater seal that drains air and fluids to enhance the expansion of lungs (Basford, L. & Slevin, O. 2003, 39). The equipment I prepared for the doctor to carry out the procedure was a sterile dressing pack, gloves, stitch cutter, a solution for the cleansing of the skin, a clinical waste bag, a sterile swab, clamps and a dressing that is non-adherent (Quinn, F. 1998, 45). The predominant procedure is ensuring that the patient lies in an upright position to guarantee the expansion of lungs so that there is easy optimal drainage of air and fluids enhanced by gravity (Fagin, C. 2000, 40).  Ã‚  

Principles of Marketing Financial Services Essay

Principles of Marketing Financial Services - Essay Example Around the world, financial services organizations are driven to increase revenues while decreasing costs. They need the ability to quickly go to market with new products, deepen customer relationships, increase revenue per customer, and improve the accuracy of their strategic decision making. At the same time, they need to control operational and compliance costs, ensure interoperability of existing applications and infrastructure, and provide seamless interactions with their customers. All within an increasingly complex compliance environment. High-performance financial institutions understand the link between operational performance and financial performance. The ability to uncover and turn vital insights into operational value levers—such as customer-facing activities, exposure to credit risk, and market share—and link them to financial value levers like operating margin, loan-loss ratios, and cost of assets in a timely and accurate fashion is key to optimizing over all performance and driving value creation. Abbey’s vision is to be the outstanding financial services company in the UK. It is the sixth largest bank in the United Kingdom and aims to achieve the largest position in the market. The purpose of Abbey is to achieve above average growth in share holder value over the long term by meeting stakeholders needs. A detailed and more specific description of the products and services offered by Abbey includes – Current Accounts, Credit Cards, Loans, Mortgages, Insurance, Savings, ISAs and Child Trust Fund. ... The ability to uncover and turn vital insights into operational value levers-such as customer- facing activities, exposure to credit risk, and market share-and link them to financial value levers like operating margin, loan-loss ratios, and cost of assets in a timely and accurate fashion is key to optimizing overall performance and driving value creation -> Abbey An Overview : Vision: Abbey's vision is to be the outstanding financial services company in theUK. It is the sixth largest bank in the United Kingdom and aims to achieve thelargest position in the market. Purpose: The purpose of Abbey is to achieve above average growth in shareholder value over the long term by meeting stake holders needs. Products and Services: A detailed and more specific description about theproducts and services offered by Abbey includes - Current Accounts, CreditCards, Loans, Mortgages, Insurance, Investments and Pensions, Savings, ISAs and Child Trust Fund. Strategies Implemented : The right mixture of strategies which suits the market place and customers are being implemented so that there would be an optimum utilization of resources and maximization of profits can be achieved. -> SWOT Analysis : Its main strength is its expert knowledge of finance. This meant that it should focus on this area. Its main weakness is its size. As only the sixth largest bank it could not offer thesame range of products as bigger banks. This meant that it should offer a simplerange. The main opportunity was to provide simpler products which customer wouldbetter understand. The main threat is from other banks, who might want to take over Abbey, so it

Time Management Essay Example for Free

Time Management Essay Study: â€Å"Working to live: Why university students balance full-time study and employment According to Valerie Holmes, within this group 83 per cent of students worked at some point during term-time of their degree programmed. In total 58 per cent of those students who worked did so to either cover or contribute to basic costs of living. While the majority of students felt they could balance work and study, half of all students questioned felt that working could have a negative impact on their degree classification. Valerie Holmes, (2008) Working to live: Why university students balance full-time study and employment, Education + Training, Vol. 50 Iss: 4, pp.305 – 314 http://www.emeraldinsight.com/journals.htm?articleid=1728331 | The work–study relationship: experiences of full†time university students undertaking part†time employment Journal of Education and Work Volume 23, Issue 5, 2010 Ralph Halla* Pages 439-449 Publishing models and article dates explained Received: 21 Apr 2010 Accepted: 14 Jul 2010 Version of record first published: 29 Nov 2010 Abstract Work and study commitments of full†time undergraduate students at the University of New South Wales were investigated in four surveys conducted in 1994, 1999, 2006 and 2009. Respondents to the surveys reported the amount of time they spent during term time in paid employment, studying outside of formal class hours and in leisure activities (1999 and 2006 only). Fifty full†time students in 2006 and 37 in 2009 who were identified through the survey as working in excess of 10 hours per week were interviewed about their work and study relationships. Findings are consistent with UK studies showing an increase in part†time work by full†time students. In addition, a steady decrease was found in hours of study outside normal class time and in time spent in leisure activities. Reasons for working offered by interviewees were predominantly financial although many reported that gaining work experience, even in areas not related to their studies, was an important consideration. While some of the students interviewed felt that the government should provide more support for full†time students, the majority thought that the university should cater more for the needs of working students by providing more online facilities for assignment submission and communication and more flexible timetables and submission requirements. In the absence of any likely moves by governments to provide financial support to students, universities need to recognize the increasing demands placed on full†time students by part†time work and to implement procedures to assist working students. http://www.tandfonline.com/loi/cjew20 Literature: More students balance school with jobs By Jacob Serebrin | January 25th, 2012 | More than half of full-time university students in Quebec work while attending school and more than 40 per cent of all undergraduates work more than 20 hours weekly says a new study by the Fà ©dà ©ration à ©tudiante universitaire du Quà ©bec, a provincial lobby group that wants lower tuition. On top of that, more than twice as many full-time students aged 20 to 24 in the province work part-time jobs than students did in the 1970s. The workloads are hurting their educations: 43 per cent of full-time undergraduates say that their jobs have negatively affected their studies and 30 per cent say their jobs mean they’ll take longer to finish. It’s worst for PhD students—six in 10 say work forced them to prolong their studies. It’s not just students in Quebec who are putting in long hours between classes. According to the 2011 Canadian University Survey Consortium study 56 per cent of undergraduates in Canada work. The average number of hours is 18 per week. Nearly a fifth (18 per cent) work more than 30 hours weekly. One third of working students report â€Å"a negative impact on their academic performance.† The latest research also builds on a November 2010 report put out by FÉUQ that said employment income accounts for more than 50 per cent of the average full-time student’s income in Quebec. Predictably, FÉUQ is using the results of both studies to argue against a tuition increase that will take effect this fall. The hike will see tuition for in-province student’s rise by $325 a year to $3,793 in 2016. It’s easy to dismiss FÉUQ’s concerns–the province has the lowest fees in the country. But the fact that so many students are working so much suggests many are already at the breaking point. It also rebuts the claim by Quebec politicians that the increase would return tuition to 1968-9 levels, adjusted for inflation, which is what finance minister Raymond Backhand told the National Assembly. The claim that today’s students are paying less than past students has also been a favorite of the Conference of Rectors and Principals of Quebec Universities, which represents administrators. Perhaps tuition was indeed more expensive in the 1968-9s. But in the 1970s, students could afford to work less in coffee shops and clothing stores—and more on their studies—than students of today. http://oncampus.macleans.ca/education/2012/01/25/more-students-balancing-school-and-part-time-jobs/ Vol. 1, Issue 1 spring 2005 The Effects of College Student Employment on Academic Achievement By: Lauren E Watanabe Mentor: Jana Jasinski Review of Literature As money and resources become more scarce for college students, jobs become more of a necessity rather than an after school activity. Any changes to students routines will lead to changes in academics, whether they are positive or negative. Employment among college students has been increasing rapidly. Its effect on the academic performance of students has been questioned by many researchers (Green, 1987). Some of the issues raised in the literature concern matters such as the number of hours worked, whether or not the students jobs pertain to their majors, and the students workloads. As more students are employed, they face having to balance their academic requirements, extracurricular activities, and employment responsibilities to maintain their lifestyles (Furr Elling, 2000). The literature reviewed below examines how employment has affected academic achievement. Much of the research indicating that employment negatively affects students academic achievement stated that an increase in the amount of hours worked was the most influential factor. In one study, more hours worked decreased the likelihood of being an A student (Pritchard, 1996). According to Furr and Elling (2000), 29% of the students working 30-39 hours per week and 39% of those students working full time indicated that work had a negative and frequent impact on their academic progress. Those who take on part-time jobs are less engaged in school before they enter the labor force, and part-time employment, especially for more than 20 hours weekly, further exacerbates this problem (Steinberg et al., 1993, p. 175). Furr and Elling (2000) also found that upperclassmen worked more hours than freshmen, indicating that the older students would be more likely to suffer in their academics. Therefore, w orking full time has an even greater impact on academics because, often times, working 40 or more hours further decreases a students college grade point average (GPA) and is negatively related to completion of a bachelors degree (Astin, 1993). The act of balancing school work with the labor market may also lead students to put forth less effort into both because they are spreading themselves too thin (Astin, 1993). According to these researchers, it is not the job itself that causes the problems, but the overload on the amount of time worked because students who work more hours each week spend less time on homework, [and] pay attention in class less often (Steinberg Dornbusch, 1991, p. 307). Not all of the research has shown negative GPA effects from the amount of hours a student is employed. Some findings indicated that employment had either a positive effect or none at all. A number of researchers, for example, found that hard work built stronger academic character because it taught the students time management skills, gave them experience outside of the classroom, and provided them with more satisfaction in college (Pennington, Zvonkovic, Wilson, 1989). Dallam and Hoyt (1981) suggested that a good balance between stu dents credit hours and working hours forced students to be more organized and to have better time management. They also found that students who worked between 1 and 15 hours per week showed a slightly higher GPA than those whose workloads were heavier and those who were not working at all (Dallam Hoyt, 1981; Li-Chen Wooster, 1979). Not only were higher GPAs found in students that maintained jobs, but Green (2001) also stated that they had gained job skills, experience, knowledge of a variety of jobs, a sense of accomplishment, a feeling of responsibility, and money for personal and school expenses (p. 329). Other researchers, when comparing high and low academic performance and the amount of hours students worked, found that the amount of hours employed did not have an adverse effect on their academics (Pinto, Parente, Palmer, 2001). Similarly, Watts (2002) analysis of 19 students at the University of Brighton found that 4 of 12 working undergraduates said that working did not affect their academics and 5 said that it actually had a positive impact. Although some of the previously mentioned studies used samples of high school students rather than undergraduates, their results were consistent. The fact that some contained samples of less than 50 students, however, may have accounted for some of the differences between the positive and negative academic results. Not accounting for the amount of time actually put into the job, researchers have found that the type of employment a student holds has an impact on academics. Dead-end jobs such as a cashier or fast food worker tend to have a negative effect (Li-Chen Wooster, 1979), whereas high-quality, part-time jobs that seemed to develop career-related skills may in effect contribute to increased levels of career maturity, and these types of jobs are more likely to be flexible and work with students schedules (Healy, OShea, Crook, 1985). These types of jobs allow for hands-on experience that cannot be gained in the classroom alone. For example, of the 600 full-time students at Lamar University surveyed, 91 out of 215 students whose jobs related to their majors had a mean GPA of 2.98, while those whose jobs were career unrelated had a mean GPA of 2.66 (Li-Chen Wooster, 1979). Also, student comments suggested that employment related to a potential career provided additional experience. For example, 10 out of 23 comments of a 120 nursing student survey at a university indicated that they were gaining more practical experience . . . and that as all [their] employment is in care areas, [they felt] it [had] extended [their] experience (Lee, 1999, p. 448). As money and resources become more scarce for college students, jobs become more of a necessity rather than an after school activity. Any changes to students routines will lead to changes in academics, whether they are positive or negative. Though the research results were not always consistent, it was a common theme that the more hours worked led to decreased academic performance, but that working in general did not necessarily have a negative effect on grades. When it came to students jobs as they applied to their majors, the effects were positive in that they provided experience beyond the classroom (Lee, Mawdsley, Rangeley, 1999). The following study will look at these variables as well as class standing, the amount of credit hours taken, and flexibility of the work schedule in order to determine the positive or negative relationship of working and academics. Other variables, such as demographic factors, will also be examined. http://www.urj.ucf.edu/vol1issue1/watanabe/literature.php