Mechanical: New Applications
http://www.maplesoft.com/applications/category.aspx?cid=199
en-us2014 Maplesoft, A Division of Waterloo Maple Inc.Maplesoft Document SystemThu, 24 Apr 2014 00:55:46 GMTThu, 24 Apr 2014 00:55:46 GMTNew applications in the Mechanical categoryhttp://www.mapleprimes.com/images/mapleapps.gifMechanical: New Applications
http://www.maplesoft.com/applications/category.aspx?cid=199
Descartes & Mme La Marquise du Chatelet And The Elastic Collision of Two Bodies
http://www.maplesoft.com/applications/view.aspx?SID=153515&ref=Feed
<p><strong><em> ABSTRACT<br /> <br /> The Marquise</em></strong> <strong><em>du Chatelet in her book " Les Institutions Physiques" published in 1740, stated on page 36, that Descartes, when formulating his laws of motion in an elastic collision of two bodies B & C (B being more massive than C) <span >having the same speed v</span>, said that t<span >he smaller one C will reverse its course </span>while <span >the more massive body B will continue its course in the same direction as before</span> and <span >both will have again the same speed v.<br /> <br /> </span>Mme du Chatelet, basing her judgment on theoretical considerations using <span >the principle of continuity</span> , declared that Descartes was <span >wrong</span> in his statement. For Mme du Chatelet the larger mass B should reverse its course and move in the opposite direction. She mentioned nothing about both bodies B & C as <span >having the same velocity after collision as Descartes did</span>.<br /> <br /> At the time of Descartes, some 300 years ago, the concept of kinetic energy & momentum as we know today was not yet well defined, let alone considered in any physical problem.<br /> <br /> Actually both Descartes & Mme du Chatelet may have been right in some special cases but not in general as the discussion that follows will show.</em></strong></p><img src="/view.aspx?si=153515/Elastic_Collision_image1.jpg" alt="Descartes & Mme La Marquise du Chatelet And The Elastic Collision of Two Bodies" align="left"/><p><strong><em> ABSTRACT<br /> <br /> The Marquise</em></strong> <strong><em>du Chatelet in her book " Les Institutions Physiques" published in 1740, stated on page 36, that Descartes, when formulating his laws of motion in an elastic collision of two bodies B & C (B being more massive than C) <span >having the same speed v</span>, said that t<span >he smaller one C will reverse its course </span>while <span >the more massive body B will continue its course in the same direction as before</span> and <span >both will have again the same speed v.<br /> <br /> </span>Mme du Chatelet, basing her judgment on theoretical considerations using <span >the principle of continuity</span> , declared that Descartes was <span >wrong</span> in his statement. For Mme du Chatelet the larger mass B should reverse its course and move in the opposite direction. She mentioned nothing about both bodies B & C as <span >having the same velocity after collision as Descartes did</span>.<br /> <br /> At the time of Descartes, some 300 years ago, the concept of kinetic energy & momentum as we know today was not yet well defined, let alone considered in any physical problem.<br /> <br /> Actually both Descartes & Mme du Chatelet may have been right in some special cases but not in general as the discussion that follows will show.</em></strong></p>153515Fri, 07 Mar 2014 05:00:00 ZDr. Ahmed BaroudyDr. Ahmed BaroudyHohmann Elliptic Transfer Orbit with Animation
http://www.maplesoft.com/applications/view.aspx?SID=151351&ref=Feed
<p>Abstract<br /><br />The main purpose of this article is to show how to use Hohmann elliptic transfer in two situations:<br />a- When one manned spaceship is trying to catch up with an other one <br />on the same circular orbit around Earth.<br />b- When delivering a payload from Earth to a space station on a circular <br />orbit around Earth using 2-stage rocket .<br /><br />The way we set up the problem is as follows:<br />Consider two manned spaceships with astronauts Sally & Igor , the latter<br />lagging behind Sally by a given angle = 4.5 degrees while both are on the same<br />circular orbit C2 about Earth. A 2d lower circular orbit C1 is given. <br />Find the Hohmann elliptic orbit that is tangent to both orbits which allows<br />Sally to maneuver on C1 then to get back to the circular orbit C2 alongside Igor.<br /><br />Though the math was correct , however the final result we found was not !! <br />It was somehow tricky to find the culprit!<br />We have to restate the problem to get the correct answer. <br />The animation was then set up using the correct data. <br />The animation is a good teaching help for two reasons:<br />1- it gives a 'hand on' experience for anyone who wants to fully understand it,<br />2- it is a good lesson in Maple programming with many loops of the type 'if..then'.<br /><br />Warning<br /><br />This particular animation is a hog for the CPU memory since data accumulated <br />for plotting reached 20 MB! This is the size of this article when animation is <br />executed. For this reason and to be able to upload it I left the animation <br />procedure non executed which drops the size of the article to 300KB.<br /><br />Conclusion<br /><br />If I can get someone interested in the subject of this article in such away that he or <br />she would seek further information for learning from other sources, my efforts<br />would be well rewarded.</p><img src="/view.aspx?si=151351/Elliptic_image1.jpg" alt="Hohmann Elliptic Transfer Orbit with Animation" align="left"/><p>Abstract<br /><br />The main purpose of this article is to show how to use Hohmann elliptic transfer in two situations:<br />a- When one manned spaceship is trying to catch up with an other one <br />on the same circular orbit around Earth.<br />b- When delivering a payload from Earth to a space station on a circular <br />orbit around Earth using 2-stage rocket .<br /><br />The way we set up the problem is as follows:<br />Consider two manned spaceships with astronauts Sally & Igor , the latter<br />lagging behind Sally by a given angle = 4.5 degrees while both are on the same<br />circular orbit C2 about Earth. A 2d lower circular orbit C1 is given. <br />Find the Hohmann elliptic orbit that is tangent to both orbits which allows<br />Sally to maneuver on C1 then to get back to the circular orbit C2 alongside Igor.<br /><br />Though the math was correct , however the final result we found was not !! <br />It was somehow tricky to find the culprit!<br />We have to restate the problem to get the correct answer. <br />The animation was then set up using the correct data. <br />The animation is a good teaching help for two reasons:<br />1- it gives a 'hand on' experience for anyone who wants to fully understand it,<br />2- it is a good lesson in Maple programming with many loops of the type 'if..then'.<br /><br />Warning<br /><br />This particular animation is a hog for the CPU memory since data accumulated <br />for plotting reached 20 MB! This is the size of this article when animation is <br />executed. For this reason and to be able to upload it I left the animation <br />procedure non executed which drops the size of the article to 300KB.<br /><br />Conclusion<br /><br />If I can get someone interested in the subject of this article in such away that he or <br />she would seek further information for learning from other sources, my efforts<br />would be well rewarded.</p>151351Wed, 04 Sep 2013 04:00:00 ZDr. Ahmed BaroudyDr. Ahmed BaroudyEquilibrium Configurations of Cantilever Beam under Terminal Load
http://www.maplesoft.com/applications/view.aspx?SID=142643&ref=Feed
<p>This worksheet implements the calculation of the equilibrium shapes of<br /> initially straight inextensible and unshearable elastic cantilever beam.Three types of load conditions are implemented:<br />1) Follower load problem.<br />2) Unknown load parameter problem<br />3) Conservative load problem</p>
<p>For each of these problems animation of beam deformation are given.</p><img src="/view.aspx?si=142643/5a35294a40cc0022a2745d42d103edbb.gif" alt="Equilibrium Configurations of Cantilever Beam under Terminal Load" align="left"/><p>This worksheet implements the calculation of the equilibrium shapes of<br /> initially straight inextensible and unshearable elastic cantilever beam.Three types of load conditions are implemented:<br />1) Follower load problem.<br />2) Unknown load parameter problem<br />3) Conservative load problem</p>
<p>For each of these problems animation of beam deformation are given.</p>142643Sat, 26 Jan 2013 05:00:00 ZDr. milan batistaDr. milan batistaHardening of Aluminium Alloy AA 7075 T 7351
http://www.maplesoft.com/applications/view.aspx?SID=140361&ref=Feed
<p><span id="ctl00_mainContent__documentViewer"><span><span class="body summary">This worksheet is concerned with the hardening of aluminium alloy, the behaviour of which can be expressed by a simple power law with two hardening parameters. Based upon experimental data these parameters have been determined by both a linear regrssion and the nonlinear <em>MARQUARDT-LEVENBERG algorithm.</em></span></span></span></p><img src="/applications/images/app_image_blank_lg.jpg" alt="Hardening of Aluminium Alloy AA 7075 T 7351" align="left"/><p><span id="ctl00_mainContent__documentViewer"><span><span class="body summary">This worksheet is concerned with the hardening of aluminium alloy, the behaviour of which can be expressed by a simple power law with two hardening parameters. Based upon experimental data these parameters have been determined by both a linear regrssion and the nonlinear <em>MARQUARDT-LEVENBERG algorithm.</em></span></span></span></p>140361Wed, 14 Nov 2012 05:00:00 ZJosef BettenJosef BettenClassroom Tips and Techniques: Fourier Series and an Orthogonal Expansions Package
http://www.maplesoft.com/applications/view.aspx?SID=134198&ref=Feed
The OrthogonalExpansions package contributed to the Maple Application Center by Dr. Sergey Moiseev is considered as a tool for generating a Fourier series and its partial sums. This package provides commands for expansions in 17 other bases of orthogonal functions. In addition to looking at the Fourier series option, this article also considers the Bessel series expansion.<img src="/view.aspx?si=134198/thumb.jpg" alt="Classroom Tips and Techniques: Fourier Series and an Orthogonal Expansions Package" align="left"/>The OrthogonalExpansions package contributed to the Maple Application Center by Dr. Sergey Moiseev is considered as a tool for generating a Fourier series and its partial sums. This package provides commands for expansions in 17 other bases of orthogonal functions. In addition to looking at the Fourier series option, this article also considers the Bessel series expansion.134198Mon, 14 May 2012 04:00:00 ZDr. Robert LopezDr. Robert LopezSpherical Pendulum with Animation
http://www.maplesoft.com/applications/view.aspx?SID=132143&ref=Feed
<p>Some years ago I have written a Maple document ( already on Maple's online) on the subject of animating a simple pendulum for large angles of oscillation. This gave me the chance to test Maple command JacobiSN(time, k). I was very much pleased to see Maple do a wonderful job in getting these Jacobi's elliptic functions without a glitch.<br />Today I am back to these same functions for a similar purpose though much more sophisticated than the previous one.<br />The idea is:<br />1- to get the differential equations of motion for the Spherical Pendulum (SP),<br />2- to solve them,<br />3- to use Maple for finding the inverse of these Elliptic Integrals i.e. finding the displacement z as function of time,<br />4- to get a set of coordinates [x, y, z] for the positions of the bob at different times for plotting,<br />5- finally to work out the necessary steps for the purpose of animation.<br />It turns out that even with only 3 oscillations where each is defined with only 20 positions of the bob for a total of 60 points on the graph, the animation is so overwhelming that Maple reports:<br /> " the length of the output exceeds 1 million".<br />Not withstanding this warning, Maple did a perfect job by getting the animation to my satisfaction. <br />Note that with only 60 positions of the bob, the present article length is equal to 11.3 MB! To be able to upload it, I have to save it without running the last command related to the animation. Doing so I reduced it to a mere 570 KB.<br /><br />It was tiring to get through a jumble of formulas, calculations and programming so I wonder why I have to go through all this trouble to get this animation and yet one can get the same thing with much better animation from the internet. I think the reason is the challenge to be able to do things that others have done before and secondly the idea of creating something form nothing then to see it working as expected, gives (at least to me) a great deal of pleasure and satisfaction.<br />This is beside the fact that, to my knowledge, no such animation for (SP) has been published on Maple online with detailed calculations & programming as I did.<br /><br /></p><img src="/view.aspx?si=132143/433082\Spherical_Pendulum_p.jpg" alt="Spherical Pendulum with Animation" align="left"/><p>Some years ago I have written a Maple document ( already on Maple's online) on the subject of animating a simple pendulum for large angles of oscillation. This gave me the chance to test Maple command JacobiSN(time, k). I was very much pleased to see Maple do a wonderful job in getting these Jacobi's elliptic functions without a glitch.<br />Today I am back to these same functions for a similar purpose though much more sophisticated than the previous one.<br />The idea is:<br />1- to get the differential equations of motion for the Spherical Pendulum (SP),<br />2- to solve them,<br />3- to use Maple for finding the inverse of these Elliptic Integrals i.e. finding the displacement z as function of time,<br />4- to get a set of coordinates [x, y, z] for the positions of the bob at different times for plotting,<br />5- finally to work out the necessary steps for the purpose of animation.<br />It turns out that even with only 3 oscillations where each is defined with only 20 positions of the bob for a total of 60 points on the graph, the animation is so overwhelming that Maple reports:<br /> " the length of the output exceeds 1 million".<br />Not withstanding this warning, Maple did a perfect job by getting the animation to my satisfaction. <br />Note that with only 60 positions of the bob, the present article length is equal to 11.3 MB! To be able to upload it, I have to save it without running the last command related to the animation. Doing so I reduced it to a mere 570 KB.<br /><br />It was tiring to get through a jumble of formulas, calculations and programming so I wonder why I have to go through all this trouble to get this animation and yet one can get the same thing with much better animation from the internet. I think the reason is the challenge to be able to do things that others have done before and secondly the idea of creating something form nothing then to see it working as expected, gives (at least to me) a great deal of pleasure and satisfaction.<br />This is beside the fact that, to my knowledge, no such animation for (SP) has been published on Maple online with detailed calculations & programming as I did.<br /><br /></p>132143Mon, 26 Mar 2012 04:00:00 ZDr. Ahmed BaroudyDr. Ahmed BaroudyClassroom Tips and Techniques: An Undamped Coupled Oscillator
http://www.maplesoft.com/applications/view.aspx?SID=129521&ref=Feed
<p>Even for just three degrees of freedom, an undamped coupled oscillator modeled by the ODE system <em>M</em> ü + <em>K</em> u = 0 is difficult to solve analytically because, ultimately, a cubic characteristic equation has to be solve exactly. Instead, we simultaneously diagonalize <em>M</em> and <em>K</em>, the mass and stiffness matrices, thereby uncoupling the equations, and obtaining an explicit solution.</p><img src="/view.aspx?si=129521/thumb.jpg" alt="Classroom Tips and Techniques: An Undamped Coupled Oscillator" align="left"/><p>Even for just three degrees of freedom, an undamped coupled oscillator modeled by the ODE system <em>M</em> ü + <em>K</em> u = 0 is difficult to solve analytically because, ultimately, a cubic characteristic equation has to be solve exactly. Instead, we simultaneously diagonalize <em>M</em> and <em>K</em>, the mass and stiffness matrices, thereby uncoupling the equations, and obtaining an explicit solution.</p>129521Tue, 10 Jan 2012 05:00:00 ZDr. Robert LopezDr. Robert LopezDamage Effective Stress Concepts
http://www.maplesoft.com/applications/view.aspx?SID=129456&ref=Feed
<p>During the last two or three decades many scientists have devoted much effort to the stress analysis in a damaged material, and the notation <em>damage effective stress </em>has been introduced. </p>
<p>In the following some various <em>damage effective stress concepts </em>should be reviewed.</p><img src="/view.aspx?si=129456/428320\12aff582b7c2a4ef667abfccba992187.gif" alt="Damage Effective Stress Concepts" align="left"/><p>During the last two or three decades many scientists have devoted much effort to the stress analysis in a damaged material, and the notation <em>damage effective stress </em>has been introduced. </p>
<p>In the following some various <em>damage effective stress concepts </em>should be reviewed.</p>129456Sun, 08 Jan 2012 05:00:00 ZProf. Josef BettenProf. Josef BettenClassroom Tips and Techniques: Simultaneous Diagonalization and the Generalized Eigenvalue Problem
http://www.maplesoft.com/applications/view.aspx?SID=128444&ref=Feed
<p>This article explores the connections between the generalized eigenvalue problem and the problem of simultaneously diagonalizing a pair of <em>n × n</em> matrices.</p>
<p>Given the <em>n × n</em> matrices <em>A</em> and <em>B</em>, the <em>generalized eigenvalue problem</em> seeks the eigenpairs <em>(lambda<sub>k</sub>, x<sub>k</sub>)</em>, solutions of the equation <em>Ax = lambda Bx</em>, or <em>(A - lambda B) x = 0</em>. If <em>B</em> is nonsingular, the eigenpairs of <em>B<sup>-1</sup> A</em> are solutions. If a matrix <em>S</em> exists for which<em> S<sup>T</sup> A S = Lambda</em>, and <em>S<sup>T</sup> B S = I</em>, where <em>Lambda</em> is a diagonal matrix and <em>I</em> is the <em>n × n</em> identity, then <em>A</em> and <em>B</em> are said to be <em>diagonalized simultaneously</em>, in which case the diagonal entries of <em>Lambda</em> are the generalized eigenvalues for <em>A</em> and <em>B</em>. Such a matrix <em>S</em> exists if <em>A</em> is symmetric and <em>B</em> is positive definite. (Our definition of positive definite includes symmetry.)</p><img src="/view.aspx?si=128444/thumb.jpg" alt="Classroom Tips and Techniques: Simultaneous Diagonalization and the Generalized Eigenvalue Problem" align="left"/><p>This article explores the connections between the generalized eigenvalue problem and the problem of simultaneously diagonalizing a pair of <em>n × n</em> matrices.</p>
<p>Given the <em>n × n</em> matrices <em>A</em> and <em>B</em>, the <em>generalized eigenvalue problem</em> seeks the eigenpairs <em>(lambda<sub>k</sub>, x<sub>k</sub>)</em>, solutions of the equation <em>Ax = lambda Bx</em>, or <em>(A - lambda B) x = 0</em>. If <em>B</em> is nonsingular, the eigenpairs of <em>B<sup>-1</sup> A</em> are solutions. If a matrix <em>S</em> exists for which<em> S<sup>T</sup> A S = Lambda</em>, and <em>S<sup>T</sup> B S = I</em>, where <em>Lambda</em> is a diagonal matrix and <em>I</em> is the <em>n × n</em> identity, then <em>A</em> and <em>B</em> are said to be <em>diagonalized simultaneously</em>, in which case the diagonal entries of <em>Lambda</em> are the generalized eigenvalues for <em>A</em> and <em>B</em>. Such a matrix <em>S</em> exists if <em>A</em> is symmetric and <em>B</em> is positive definite. (Our definition of positive definite includes symmetry.)</p>128444Tue, 06 Dec 2011 05:00:00 ZDr. Robert LopezDr. Robert LopezRelaxation von Brain Tissue
http://www.maplesoft.com/applications/view.aspx?SID=125272&ref=Feed
<p>In seiner Dissertation hat <em>AIMEDIEU </em>(2004) nichtlineare Relaxation von <em>Brain Tissue </em>untersucht. Er fand gute Übereinstimmungen zwischen dem Wurzel-t-Gesetz (<em>BETTEN, </em>2008) und seinen eigenen Experimenten.</p>
<p> Im Folgenden wird gezeigt, dass <em>PRONY</em>-Reihen weniger geeignet sind, den experimentellen Befund zu beschreiben.</p><img src="/applications/images/app_image_blank_lg.jpg" alt="Relaxation von Brain Tissue" align="left"/><p>In seiner Dissertation hat <em>AIMEDIEU </em>(2004) nichtlineare Relaxation von <em>Brain Tissue </em>untersucht. Er fand gute Übereinstimmungen zwischen dem Wurzel-t-Gesetz (<em>BETTEN, </em>2008) und seinen eigenen Experimenten.</p>
<p> Im Folgenden wird gezeigt, dass <em>PRONY</em>-Reihen weniger geeignet sind, den experimentellen Befund zu beschreiben.</p>125272Tue, 30 Aug 2011 04:00:00 ZProf. Josef BettenProf. Josef BettenPlastic method of structural analysis
http://www.maplesoft.com/applications/view.aspx?SID=102534&ref=Feed
<p>This worksheet contains a step-by-step method for the analysis of 2D frames with all kind of boundary conditions or joints between elements.</p>
<p>In each step, determined by the creation of a new plastic hinge, numeric information (displacements, reactions and stresses) and graphic information (beam diagrams and deformed shape) are obtained.</p>
<p>Finally, accumulated beam diagrams and accumulated deformed shapes are also obtained.<br /><br /></p><img src="/view.aspx?si=102534/Image1.PNG" alt="Plastic method of structural analysis" align="left"/><p>This worksheet contains a step-by-step method for the analysis of 2D frames with all kind of boundary conditions or joints between elements.</p>
<p>In each step, determined by the creation of a new plastic hinge, numeric information (displacements, reactions and stresses) and graphic information (beam diagrams and deformed shape) are obtained.</p>
<p>Finally, accumulated beam diagrams and accumulated deformed shapes are also obtained.<br /><br /></p>102534Mon, 14 Mar 2011 04:00:00 ZAntolín Lorenzana IbánAntolín Lorenzana IbánRelaxation due to the Sqrt(t)-Law in Comparison with the MAXWELL-Fluid
http://www.maplesoft.com/applications/view.aspx?SID=100763&ref=Feed
<p>This document is concerned with the relaxation functions (11.72a,b) taken</p>
<p>from BETTEN's book: Creep Mechanics, Third Edition, Springer-Verlag 2008,</p>
<p>in more detail. The parameters in the MAXWELL-model (11.72b) and the</p>
<p>sqrt(t)-law (11.72a) are denoted by b and B, respectively. A relation between</p>
<p>b and B has been found, so that the simple MAXWELL-model can be regarded</p>
<p>as the best approximation to the sqrt(t)-law.</p><img src="/view.aspx?si=100763/maple_icon.jpg" alt="Relaxation due to the Sqrt(t)-Law in Comparison with the MAXWELL-Fluid" align="left"/><p>This document is concerned with the relaxation functions (11.72a,b) taken</p>
<p>from BETTEN's book: Creep Mechanics, Third Edition, Springer-Verlag 2008,</p>
<p>in more detail. The parameters in the MAXWELL-model (11.72b) and the</p>
<p>sqrt(t)-law (11.72a) are denoted by b and B, respectively. A relation between</p>
<p>b and B has been found, so that the simple MAXWELL-model can be regarded</p>
<p>as the best approximation to the sqrt(t)-law.</p>100763Sun, 09 Jan 2011 05:00:00 ZProf. Josef BettenProf. Josef BettenElastic-Plastic Transition # Appendix.mws
http://www.maplesoft.com/applications/view.aspx?SID=99879&ref=Feed
<p><strong>Abstract </strong></p>
<p><strong><br /></strong>In addition to the recently submitted worksheet, dated by November 21, 2010,</p>
<p>uni-axial stress-strain relations have been examined, if they are suitable to</p>
<p>describe the elastic-plastic transition.</p>
<p><em>Keywords: </em>Envelope; modified tanh approach; modified <em>GAUSS </em>error function;</p>
<p> modified rational function;</p><img src="/view.aspx?si=99879/maple_icon.jpg" alt="Elastic-Plastic Transition # Appendix.mws" align="left"/><p><strong>Abstract </strong></p>
<p><strong><br /></strong>In addition to the recently submitted worksheet, dated by November 21, 2010,</p>
<p>uni-axial stress-strain relations have been examined, if they are suitable to</p>
<p>describe the elastic-plastic transition.</p>
<p><em>Keywords: </em>Envelope; modified tanh approach; modified <em>GAUSS </em>error function;</p>
<p> modified rational function;</p>99879Sun, 05 Dec 2010 05:00:00 ZProf. Josef BettenProf. Josef BettenTwo Bodies Revolving Around Their Center of Mass with ANIMATION
http://www.maplesoft.com/applications/view.aspx?SID=99587&ref=Feed
<p>For any isolated system of two bodies revolving around each other by virtue of the gravitational attraction that each one exerts on the other, the general motion is best described by using a frame of reference attached to their common Center of Mass (CM). The reason is that their motion is in fact around their CM as we shall see. <br />For an isolated system the momentum remains constant so that the CM is either moving along a straight line or is at rest.<br />For an Earth's satellite we can always take the motion of the satellite relative to Earth using a geocentric frame of reference. <br />The reason is that:<br /> the mass of the satellite being insignificant compared to Earth's <br /> mass, the revolving satellite doesn't affect Earth at all so<br /> that the CM of Earth-satellite system is still the center of the Earth.<br /> Hence we use the center of the Earth as the origin of a rectangular<br /> coordinates system.<br /> <br />In this article we use Maple powerful animation routines to study the motion of two bodies having comparable masses revolving about each other by showing: <br />1- their combined motion as seen from their common Center of Mass,<br />2- their relative motion as if one of them is fixed and the other one is moving. <br />In this last instance the frame of reference is attached to the the body that is supposed to be at rest.<br /><br /></p><img src="/view.aspx?si=99587/thumb.jpg" alt="Two Bodies Revolving Around Their Center of Mass with ANIMATION" align="left"/><p>For any isolated system of two bodies revolving around each other by virtue of the gravitational attraction that each one exerts on the other, the general motion is best described by using a frame of reference attached to their common Center of Mass (CM). The reason is that their motion is in fact around their CM as we shall see. <br />For an isolated system the momentum remains constant so that the CM is either moving along a straight line or is at rest.<br />For an Earth's satellite we can always take the motion of the satellite relative to Earth using a geocentric frame of reference. <br />The reason is that:<br /> the mass of the satellite being insignificant compared to Earth's <br /> mass, the revolving satellite doesn't affect Earth at all so<br /> that the CM of Earth-satellite system is still the center of the Earth.<br /> Hence we use the center of the Earth as the origin of a rectangular<br /> coordinates system.<br /> <br />In this article we use Maple powerful animation routines to study the motion of two bodies having comparable masses revolving about each other by showing: <br />1- their combined motion as seen from their common Center of Mass,<br />2- their relative motion as if one of them is fixed and the other one is moving. <br />In this last instance the frame of reference is attached to the the body that is supposed to be at rest.<br /><br /></p>99587Mon, 29 Nov 2010 05:00:00 ZDr. Ahmed BaroudyDr. Ahmed BaroudyElastic-Plastic Transition
http://www.maplesoft.com/applications/view.aspx?SID=99226&ref=Feed
<p>This worksheet is concerned with the generalization of nonlinear material laws found in experiments to multi-axial states of stress. For engineering applications it is very important to generalize uni-axial relations to tensorial constitutive equations valid for multi-axial states of stress. This can be achieved by utilizing interpolation methods for tensor functions developed by the Author. It can be shown that the scalar coefficients in tensorial constitutive equations are functions of the set of irreducible invariants (<em>integrity basis) </em>and experimental data. In the following uni-axial relations describing the elastic-plastic transition in uni-axial tests have been generalized to multi-axial states of stress.</p>
<p><em>Keywords: </em>Elastic-plastic transitions under uni-axial load; tensorial generalization; irreducible invariants; tensorial interpolation; example;</p><img src="/view.aspx?si=99226/maple_icon.jpg" alt="Elastic-Plastic Transition" align="left"/><p>This worksheet is concerned with the generalization of nonlinear material laws found in experiments to multi-axial states of stress. For engineering applications it is very important to generalize uni-axial relations to tensorial constitutive equations valid for multi-axial states of stress. This can be achieved by utilizing interpolation methods for tensor functions developed by the Author. It can be shown that the scalar coefficients in tensorial constitutive equations are functions of the set of irreducible invariants (<em>integrity basis) </em>and experimental data. In the following uni-axial relations describing the elastic-plastic transition in uni-axial tests have been generalized to multi-axial states of stress.</p>
<p><em>Keywords: </em>Elastic-plastic transitions under uni-axial load; tensorial generalization; irreducible invariants; tensorial interpolation; example;</p>99226Sun, 21 Nov 2010 05:00:00 ZProf. Josef BettenProf. Josef BettenCross Contraction Numbers
http://www.maplesoft.com/applications/view.aspx?SID=97899&ref=Feed
<p>This worksheet is concerned with the formulation of the ratio between <em>transverse </em>and <em>longitudinal </em>strains of isotropic and anisotropic materials, called <em>cross contraction numbers. </em>Some examples for practical use have been discussed in more detail, where <em>elastic, elastic-plastic, </em>and <em>plastic deformations </em>should be taken into account.</p>
<p><em>Keywords: POISSON</em>'s ratio; Isotropic and Anisotropic Materials; Elastic, Elastic-Plastic, and Plastic Deformations;</p><img src="/view.aspx?si=97899/maple_icon.jpg" alt="Cross Contraction Numbers" align="left"/><p>This worksheet is concerned with the formulation of the ratio between <em>transverse </em>and <em>longitudinal </em>strains of isotropic and anisotropic materials, called <em>cross contraction numbers. </em>Some examples for practical use have been discussed in more detail, where <em>elastic, elastic-plastic, </em>and <em>plastic deformations </em>should be taken into account.</p>
<p><em>Keywords: POISSON</em>'s ratio; Isotropic and Anisotropic Materials; Elastic, Elastic-Plastic, and Plastic Deformations;</p>97899Sun, 17 Oct 2010 04:00:00 ZProf. Josef BettenProf. Josef BettenPassive Differential Gear
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<p>The Differential Gear component represents a differential gear that distributes rotational motion about the longitudinal axis (driveshaft) to two lateral axes.<br />The longitudinal motion is divided by the user defined gear ratio, and then split between the lateral shafts.</p><img src="/view.aspx?si=97280/custom_component_s.jpg" alt="Passive Differential Gear" align="left"/><p>The Differential Gear component represents a differential gear that distributes rotational motion about the longitudinal axis (driveshaft) to two lateral axes.<br />The longitudinal motion is divided by the user defined gear ratio, and then split between the lateral shafts.</p>97280Wed, 29 Sep 2010 04:00:00 ZMaplesoftMaplesoftRavigneaux Planetary Gear Set
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<p>The Ravigneaux block is a double planetary gear set constructed from two custom component gear pairs, a Planet-Planet and a Ring-Planet. The custom component documentation can be found in the attachments folder.</p><img src="/view.aspx?si=97254/custom_component_s.jpg" alt="Ravigneaux Planetary Gear Set" align="left"/><p>The Ravigneaux block is a double planetary gear set constructed from two custom component gear pairs, a Planet-Planet and a Ring-Planet. The custom component documentation can be found in the attachments folder.</p>97254Tue, 28 Sep 2010 04:00:00 ZMaplesoftMaplesoftRack and Pinion
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<p>A rack and pinion is a type of linear actuator that is used for the conversion of rotary motion to linear, or vice versa. The circular “pinion” engages teeth on a linear gear bar or “rack.” An example of its use is in rack-and-pinion steering in automobiles. The pinion is attached to the bottom end of the steering column and turns with the steering wheel. The rack meshes with the pinion and is free to move left and right in response to the angular input at the steering wheel.</p><img src="/view.aspx?si=97252/sim_icon.jpg" alt="Rack and Pinion" align="left"/><p>A rack and pinion is a type of linear actuator that is used for the conversion of rotary motion to linear, or vice versa. The circular “pinion” engages teeth on a linear gear bar or “rack.” An example of its use is in rack-and-pinion steering in automobiles. The pinion is attached to the bottom end of the steering column and turns with the steering wheel. The rack meshes with the pinion and is free to move left and right in response to the angular input at the steering wheel.</p>97252Tue, 28 Sep 2010 04:00:00 ZMaplesoftMaplesoftWorm Gear
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<p>A worm drive is an arrangement of gears consisting of a worm and worm wheel. The worm has only one tooth wrapped continuously around its circumference, much like a thread on a screw. The worm is meshed with a worm-gear or worm-wheel, whose axis is perpendicular to that of the worm. Worm drives are a simple and compact way to achieve a high torque, low speed gear ratio.</p><img src="/view.aspx?si=97251/thumb.jpg" alt="Worm Gear" align="left"/><p>A worm drive is an arrangement of gears consisting of a worm and worm wheel. The worm has only one tooth wrapped continuously around its circumference, much like a thread on a screw. The worm is meshed with a worm-gear or worm-wheel, whose axis is perpendicular to that of the worm. Worm drives are a simple and compact way to achieve a high torque, low speed gear ratio.</p>97251Tue, 28 Sep 2010 04:00:00 ZMaplesoftMaplesoft