Plasma Vessel Vertical Supports. Pendulum solution: Equivalent Friction Factor calculation

Similar documents
Connections between Plasma and Outer Vessels Ports: Bellows FE Models

Finite Element Modeling for Transient Thermal- Structural Coupled Field Analysis of a Pipe Joint

2012 MECHANICS OF SOLIDS

Grid supports design for dual-cooled fuel rods

D : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown.

Stresses Analysis of Petroleum Pipe Finite Element under Internal Pressure

Equilibrium of a Particle

ROTATING RING. Volume of small element = Rdθbt if weight density of ring = ρ weight of small element = ρrbtdθ. Figure 1 Rotating ring

Due Monday, September 14 th, 12:00 midnight

2007 Problem Topic Comment 1 Kinematics Position-time equation Kinematics 7 2 Kinematics Velocity-time graph Dynamics 6 3 Kinematics Average velocity

Multi Linear Elastic and Plastic Link in SAP2000

Physics for Scientists and Engineers 4th Edition, 2017

Downloaded from Downloaded from / 1

Boundary Nonlinear Dynamic Analysis

The... of a particle is defined as its change in position in some time interval.

Stress Analysis Lecture 3 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

N = Shear stress / Shear strain

CHAPTER 10 ROTATION OF A RIGID OBJECT ABOUT A FIXED AXIS WEN-BIN JIAN ( 簡紋濱 ) DEPARTMENT OF ELECTROPHYSICS NATIONAL CHIAO TUNG UNIVERSITY

Mechanics of Solids. Mechanics Of Solids. Suraj kr. Ray Department of Civil Engineering

Mechanical Engineering Ph.D. Preliminary Qualifying Examination Solid Mechanics February 25, 2002

Chapter 3. Load and Stress Analysis. Lecture Slides

Rotation. PHYS 101 Previous Exam Problems CHAPTER

Measurement and modelling of contact stiffness

Chapter 11. Displacement Method of Analysis Slope Deflection Method

PHYS 1114, Lecture 33, April 10 Contents:

STATICS. FE Review. Statics, Fourteenth Edition R.C. Hibbeler. Copyright 2016 by Pearson Education, Inc. All rights reserved.

DHANALAKSHMI COLLEGE OF ENGINEERING, CHENNAI DEPARTMENT OF MECHANICAL ENGINEERING ME 6603 FINITE ELEMENT ANALYSIS PART A (2 MARKS)

Two Tier projects for students in ME 160 class

1. Replace the given system of forces acting on a body as shown in figure 1 by a single force and couple acting at the point A.

PERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR - VALLAM THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK

Institute of Structural Engineering Page 1. Method of Finite Elements I. Chapter 2. The Direct Stiffness Method. Method of Finite Elements I

LOWELL, MICHIGAN, MAY 28, Every Patriotic American Salutes His Nation's Flag

Translational Motion Rotational Motion Equations Sheet

Methods of Analysis. Force or Flexibility Method

Q9.1. A. t = 1 s B. t = 2 s C. t = 3 s D. t = 4 s E. t = 5 s Pearson Education, Inc.

The University of Melbourne Engineering Mechanics

MODELLING OF TRIPLE FRICTION PENDULUM BEARING IN SAP2000

Institute of Structural Engineering Page 1. Method of Finite Elements I. Chapter 2. The Direct Stiffness Method. Method of Finite Elements I

STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS

Example 4: Design of a Rigid Column Bracket (Bolted)

Homework #2 MEAM Automotive Body Structures 1998 Winter

NORMAL STRESS. The simplest form of stress is normal stress/direct stress, which is the stress perpendicular to the surface on which it acts.

Advanced Structural Analysis EGF Cylinders Under Pressure

Title. Author(s)DONG, Q.; OKAZAKI, T.; MIDORIKAWA, M.; RYAN, K.; SAT. Issue Date Doc URL. Type. Note. File Information BEARINGS

T k b p M r will so ordered by Ike one who quits squuv. fe2m per year, or year, jo ad vaoce. Pleaie and THE ALTO SOLO

NON-LINEAR VISCOELASTIC MODEL OF STRUCTURAL POUNDING

R13. II B. Tech I Semester Regular Examinations, Jan MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) PART-A

1. A pure shear deformation is shown. The volume is unchanged. What is the strain tensor.

Example of Calculating the Nominal Life

UNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation.

Stress Analysis of Radial and Non- Radial Nozzle Connections in Ellipsoidal Head Pressure Vessel

Prediction of dynamic behavior of workpieces in ultrasonic plastic welding

Rotational & Rigid-Body Mechanics. Lectures 3+4

A 3D-ball bearing model for simulation of axial load variations

Shear stresses around circular cylindrical openings

THE EFFECT OF GEOMETRY ON FATIGUE LIFE FOR BELLOWS

Arbitrary Normal and Tangential Loading Sequences for Circular Hertzian Contact

TECHNICAL REPORT BALLTEC LTD. FINITE ELEMENT ANAYLSIS AND FATIGUE ASSESSMENT OF KN ANCHOR CONNECTOR REPORT NO REVISION NO.

CHAPTER 7 FINITE ELEMENT ANALYSIS OF DEEP GROOVE BALL BEARING

Solution: T, A1, A2, A3, L1, L2, L3, E1, E2, E3, P are known Five equations in five unknowns, F1, F2, F3, ua and va

Advanced Higher Physics. Rotational motion

MEI STRUCTURED MATHEMATICS 4763

QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS

AP Physics. Harmonic Motion. Multiple Choice. Test E

Finite Element Method in Geotechnical Engineering

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK. Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV

QUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A

Answers to questions in each section should be tied together and handed in separately.

LOWELL. MICHIGAN, THURSDAY, MAY 23, Schools Close. method of distribution. t o t h e b o y s of '98 a n d '18. C o m e out a n d see t h e m get

Example-3. Title. Description. Cylindrical Hole in an Infinite Mohr-Coulomb Medium

Numerical Modeling of Interface Between Soil and Pile to Account for Loss of Contact during Seismic Excitation


TutorBreeze.com 7. ROTATIONAL MOTION. 3. If the angular velocity of a spinning body points out of the page, then describe how is the body spinning?

Lecture 8: Flexibility Method. Example

Mechanics of Materials II. Chapter III. A review of the fundamental formulation of stress, strain, and deflection

A Finite Element Study of Elastic-Plastic Hemispherical Contact Behavior against a Rigid Flat under Varying Modulus of Elasticity and Sphere Radius

THE BEHAVIOUR OF REINFORCED CONCRETE AS DEPICTED IN FINITE ELEMENT ANALYSIS.

Cone-shaped socket connections for cylindrical members

Physics 201 Midterm Exam 3

Solution of Differential Equation by Finite Difference Method

Preliminary Design Review: Loads, Structures, and Mechanisms. Michael Cunningham, Shimon Gewirtz, Rajesh Yalamanchili, Thomas Noyes

Samantha Ramirez, MSE. Stress. The intensity of the internal force acting on a specific plane (area) passing through a point. F 2

Phys 2210 S18 Practice Exam 3: Ch 8 10

Massachusetts Institute of Technology 22.68J/2.64J Superconducting Magnets. February 27, Lecture #4 Magnetic Forces and Stresses

STRESS, STRAIN AND DEFORMATION OF SOLIDS

2. (a) Explain different types of wing structures. (b) Explain the advantages and disadvantages of different materials used for aircraft

BI-DIRECTIONAL SEISMIC ANALYSIS AND DESIGN OF BRIDGE STEEL TRUSS PIERS ALLOWING A CONTROLLED ROCKING RESPONSE

Final Exam Name: Box# Physics

Final Exam Spring 2014 May 05, 2014

4) Vector = and vector = What is vector = +? A) B) C) D) E)

STATICS. Friction VECTOR MECHANICS FOR ENGINEERS: Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

Drop Test Simulation of a BGA Package: Methods & Experimental Comparison

Distributed: Wednesday, March 17, 2004

Rotational Kinematics and Dynamics. UCVTS AIT Physics

Quantitative Skills in AP Physics 1

GATE SOLUTIONS E N G I N E E R I N G

PHY 001 (Physics I) Lecture 7

FINITE GRID SOLUTION FOR NON-RECTANGULAR PLATES

Strength of Material. Shear Strain. Dr. Attaullah Shah

Princeton/MuMu/99-24 May 20, 1999 ANSYS Coupled-Field Analysis in the Simulation of Liquid Metal Moving in the Magnetic Field

Transcription:

WENDELSTEIN 7- X Andrea Capriccioli Plasma Vessel vertical supports. Pendulum solution: Equivalent Friction Factor calculation KKS.-Nr.: 1- xxx Max-Planck- Institut für Plasmaphysik Dok.-Kennz.: -x000xx.0 Plasma Vessel Vertical Supports. Pendulum solution: Equivalent Friction Factor calculation 0 A. Capriccioli Rev. Datum erstellt geprüft genehmigt Bemerkungen A. Capriccioli Page 1 of 11

I. Introduction and Description of the Model... 3 II. Results... 5 III. Conclusions... 8 ANNEX 1: PV sliding Stiffness... 9 ANNEX 2: FE Model Validation... 10 ANNEX 3: friction coeff. versus Pendulum length... 11 A. Capriccioli Page 2 of 11

I. INTRODUCTION and DESCRIPTION OF THE MODEL A schematic ANSYS 3-D FE model was created to perform a parametric analysis of the friction coefficient of the Plasma Vessel (PV) vertical supports. The actual design takes into consideration the Pendulum solution for the PV vertical supports. This solution is made up of two ball-and-socket joints connected each other with a long tube. The diameter of the ball-and-socket is 80 mm and the length of the tube varies from ~680 mm (outer PV support AEA) to ~1300-1950 mm (inner PV supports AFF-AEX). The elements used in the model are SOLID45 (3-D Structural Solid) and CONTAC52 (3-D Pointto-Point Contact) for a total of about 15.500 nodes and elements (see Fig.1). In Fig.2 it s possible to see the ball-socket joint and the Figures 3 and 4 show the non linear elements between the components ball and the socket. The gap between the two spherical surfaces (null gap) is set by means of the Real Constant value (GAP). The Fig. 5 shows the constraints on the base of the lower and upper sockets. Fig.1 View of the opened ball-and-socket joint. Fig.2 A. Capriccioli Page 3 of 11

The gap direction is set by means of the real constants NX, NY, and NZ (gap direction vector). The material Property MU defines the coefficient of friction μ (=0.2 in the Ref. case). Fig.3 The nodes on the top of the upper socket are coupled in the vertical direction (Y); they have identical displacements in horizontal (X) direction (1 mm see Fig.6) and null displacement in the other horizontal (Z) direction. Uniform Vertical Forces act on all the nodes. Fig.4 The nodes on the base of the lower socket are fully fixed. Fig.5 A. Capriccioli Page 4 of 11

II. RESULTS The Fig.6 show the upper ball-and-socket joint (the socket is attached to the PV legs) shifted in X (horizontal) direction. The lower and upper socket bases remain horizontal while the internal spherical parts rotate with the fixed friction coefficient. In Fig.7 it s possible to see the vertical (Y) displacements of the lower joint. Fig.6 Fig.7 A. Capriccioli Page 5 of 11

Upper and lower joints. Axial sections (in XY plane): Views of the Vertical (Y) displ. In the present parametric analysis the parameters are the diameter of the spherical surface of contact (ball-socket) and the length of the connection element that joins the two rotation centers. When one of these parameter changes, the reaction force changes too and the ratio between the reaction forces and the total vertical load (applied to the upper nodes) gives a sort of global Equivalent Plane Sliding friction factor (EPSff). A. Capriccioli Page 6 of 11

Three values for the Pendulum Length have been selected (500, 1000 and 2000 mm): per each length there are 4 different diameters (70, 80, 100 and 120 mm). The Table 1 and the Graph 1 show the ratio between the reaction force in X direction and the vertical load: the values represent the EPSlff (the real friction coeff. assigned to the surfaces in contact is ff=0.2). In Graph 5 ANNEX 3 the EPSlff values are reported versus the Pendulum length. Spheres Diam. [mm] Pendulum Length 500 mm Pendulum Length 1000 mm Pendulum Length 2000 mm 70 0.0335 0.0167 0.0083 80 0.0409 0.0195 0.0098 100 0.0464 0.0234 0.0117 120 0.0557 0.0276 0.0139 Table 1 6.0E-02 Equivalent Plane Sliding friction factor 5.0E-02 4.0E-02 3.0E-02 2.0E-02 1.0E-02 0.0E+00 60 70 80 90 100 110 120 130 Graph 1 Sphere Diameter [mm] Pendulum Length L=500 mm Pendulum Length L=1000 mm Pendulum Length L=2000 mm For the actual vertical supports length (with ball-and-socket Diam.=100 mm and ff=0.2), the results are: - AEA port (L~0.680 m): EPSff ~ 0.0343 - AFF port (L~1.300 m) : EPSff ~ 0.018 - AEX port (L~1.950 m): EPSff ~ 0.012 With an average value = 0.0214 A. Capriccioli Page 7 of 11

III. CONCLUSIONS The curves shown in Graph 1 can be summarized, in first approximation, with the function [1f]: in the Graph 2 the comparison between the ANSYS and the formula results is reported (only for ff=0.2). It is possible to note that the differences increase when the Pendulum length decreases. Equivalent Plane Sliding friction factor (EPSff) EPSff ff K (d / 80) (1000 / L) [1f] Where: ff is the friction factor of the surfaces in contact K = 0.095 (constant) d is the diameter [mm] of the spheres in the pendulum joints L is the distance [mm] between the two spheres centers ( Pendulum Length) Equivalent Plane Sliding friction factor 6.0E-02 5.0E-02 4.0E-02 3.0E-02 2.0E-02 1.0E-02 0.0E+00 60 70 80 90 100 110 120 130 Graph 2 Sphere Diameter [mm] Pendulum Length L=500 mm Pendulum Length L=1000 mm Pendulum Length L=2000 mm Formula (L=500) Formula (L=1000) Formula (L=2000) In ANNEX 1 is reported the analysis of the PV sliding stiffness: the equivalent friction coefficient used in the calculations was 0.05. In the light of the present analysis, it is necessary decrease the constant components due to the friction. The average of the new equivalent friction factors is 0.0214 and in first approximation the constant component 13.5 kn should be divided by 2.34 (=0.05/0.0214); the other constant 11.5 kn by 1.46 (=0.05/0.0343) while the last 8.3 kn by 3.33 (0.05/0.015). A. Capriccioli Page 8 of 11

Short summary of Plasma Vessel sliding stiffness (PV as Rigid Body ) All the values are referred to a single AEU port. ANNEX 1 X Y PV 72 Young s modulus = 10 3 * E 1 kn UX=1 mm RX=16.4 kn UX=1 mm RX=2.9 kn 2.4 kn/mm + 13.5 kn 1340 mm 8.93 kn/mm 11960 kn/rad UY=1 mm RY=11.7 kn 6638 mm 3.4 kn/mm + 8.3 kn (min) 0.058 kn/mm 0.48 kn/mm (+ 11.5 kn max) AEU Port ~387 kn/mm UY=1 mm RY=3.4 kn Translation along X (CSYS=20) and results in RSYS=1: NO Friction NODE FX [N] FY [N] Ftot [N] UX [mm] UY [mm] UX in csys 20 [mm] UY in csys 20 [mm] 24646-2595.5 1.68E-05 2596-1 -2.05E-04 1 0 24653-1027.5 3162.3 3325-0.30922 0.95099 0.096 0.904 121636 2127.4 1545.7 2630 0.8089 0.58794 0.654 0.346 218493 2035.4-1478.8 2516 0.80914-0.58762 0.655 0.345 315350-1048.8-3228 3394-0.30883-0.95112 0.095 0.905 Ftot [N] 14460 U tot 2.500 2.500 Kr [N/mm]= 2429 UX = 1 mm projection of U tang. in X dir (CSYS=20) U tang U rad Radial Dir Tangential Dir Rotation around a Z and results in RSYS=1: NO Friction 24646 1.96E-05 3353.7 3354 9.96E-07 1 24653 2.75E-05 3354.4 3354 9.63E-07 1 121636 2.80E-05 3357.7 3358 9.97E-07 1 218493 4.10E-13 3351.5 3352 9.70E-07 1 315350 2.92E-05 3358.5 3359 9.78E-07 1 Kt [N/mm]= 3355 1 72 projection of U rad. in X dir (CSYS=20) 1kN

ANNEX 2: FE Model Validation The Fig.8 shows a simplified 2D model of the ball-and-socket joint. In this case the contact is between a cylinder and three points on the socket. In this case it is easy to know the contact forces and to verify the model. With the same parameters (friction coeff., diameter of the ball/cylinder and pendulum length) the results are very close to the 3D Model (and practically identical to the formula [1]). In the Graph 3 the comparison between the 2 Models is reported (only for ff=0.2). 6.0E-02 5.0E-02 4.0E-02 3.0E-02 2.0E-02 1.0E-02 0.0E+00 Graph 3 60 70 80 90 100 110 120 130 Sphere Diameter [mm] Pendulum Length L=500 mm Pendulum Length L=1000 mm Pendulum Length L=2000 mm 2D Model (L=500) 2D Model (L=1000) 2D Model (L=2000) Equivalent Plane Sliding friction factor A. Capriccioli Page 10 of 11

ANNEX 3 EPSlff values versus Pendulum length. Equivalent Plane Sliding friction factor 6.0E-02 5.5E-02 5.0E-02 4.5E-02 4.0E-02 3.5E-02 3.0E-02 2.5E-02 2.0E-02 1.5E-02 1.0E-02 5.0E-03 0.0E+00 500 Graph 5 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 Pendulum Length [mm] Pendulum Sphere d=70 mm Pendulum Sphere d=100 mm Pendulum Sphere d=80 mm Pendulum Sphere d=120 mm A. Capriccioli Page 11 of 11