{"id":2161,"date":"2022-03-13T13:04:02","date_gmt":"2022-03-13T13:04:02","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2161"},"modified":"2022-12-06T00:15:18","modified_gmt":"2022-12-06T00:15:18","slug":"4-2-3","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch4\/4-2-3\/","title":{"rendered":"4-2-3"},"content":{"rendered":"
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4.2<\/b> VELOCITY AND ACCELERATION ANALYSIS OF MECHANISMS-3<\/h1>\n

Example:<\/strong><\/p>\n

\"\"<\/p>\n

For the mechanism shown, the piston (2) has an upward velocity vA<\/sub> = v12<\/sub> = 180 mm\/s (linear velocity of link 2 with respect to link 1). We are to determine the velocity and acceleration of point D. In this example the loop closure, velocity and acceleration equations have been solved explicitly. It is left to the reader to derive the equations used.<\/p>\n

Define the fixed lengths:<\/p>\n

\"\" \u00a0 <\/b>\"\" \u00a0 <\/b>\"\" \u00a0<\/b>\"\" \u00a0<\/b>\"\" \u00a0<\/b>\"\" \u00a0<\/b>\"\"<\/b><\/h1>\n

Position Analysis<\/strong><\/p>\n

\"\" \u00a0<\/b>\"\" \u00a0\"\"<\/b><\/h1>\n

\"\"\u00a0<\/b> \"\"<\/b><\/h1>\n

\"\" \u00a0<\/b>\"\"<\/b><\/h1>\n

Path of point D (with A0<\/sub> as the origin and x-axis is the horizontal line for\u00a0 the coordinate frame used):<\/p>\n

\"\"<\/b><\/h1>\n

Velocity analysis<\/strong><\/p>\n

\"\"\u00a0<\/b> \"\"\u00a0<\/b> \"\"<\/b><\/h1>\n

\"\"<\/b><\/h1>\n

Acceleration analysis<\/strong><\/p>\n

\"\"<\/b><\/h1>\n

\"\"<\/b><\/h1>\n

\"\"<\/b><\/h1>\n

Some numerical results:<\/strong><\/p>\n

When: \"\";\u00a0<\/b> \"\";\u00a0 \"\";\u00a0 \"\"<\/b><\/p>\n

\"\"\u00a0 \"\"\u00a0 \u00a0\"\"\u00a0<\/b><\/p>\n

\"\" \"\" \"\" \"\"<\/b><\/p>\n

\"\" \u00a0\u00a0\"\" \"\" \"\"\"\"<\/b><\/p>\n

\"\" \"\"<\/b> <\/b><\/h1>\n

\"\"<\/b> \"\"<\/b><\/h1>\n

\"\"<\/b> \"\"<\/b><\/h1>\n

Polar plot of displacement, velocity and acceleration of point D is as shown in the figure below. The vectors are shown for s2\/1<\/sub>\u00a0= 60 mm.<\/p>\n

\"\"<\/p>\n

If we perform the velocity analysis graphically, we have to solve the velocity vector equation:<\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 v<\/strong>B4<\/sub> = v<\/strong>B3<\/sub> + v<\/strong>B4\/3<\/sub><\/p>\n

The velocity v<\/strong>B3<\/sub>\u00a0is equal to\u00a0v<\/strong>B2<\/sub>\u00a0= v<\/strong>12<\/sub>. The relative velocity\u00a0v<\/strong>B4\/3<\/sub>\u00a0must be parallel to the slider axis between links 3 and 4 since these two links can only translate along the slider axis relative to each other. v<\/strong>B4<\/sub>\u00a0is perpendicular to the line A<\/strong>0<\/sub>B<\/strong> since link 4 is in a fixed axis of rotation about A0<\/sub>. We first lay out v<\/strong>B2<\/sub>\u00a0with a\u00a0kv<\/sub> = 1 mm\/(mm\/s) and then draw a line parallel to the slider axis from the tip of\u00a0v<\/strong>B2<\/sub>. From the starting point we draw a line perpendicular to A<\/strong>0<\/sub>B<\/strong>. The intersection of these two lines gives us the magnitudes of v<\/strong>B4<\/sub> and\u00a0v<\/strong>B4\/3<\/sub>\u00a0(e.g. VB4<\/sub>\u00a0= 176.77mm\/kv<\/sub> = 176.77 mm\/s). When this velocity vector equation is solved, we can determine \u03c914<\/sub>\u00a0= VB4<\/sub>\/|A0<\/sub>B| (176.77\/239.54 = 0.738 s-1<\/sup>). The direction of this angular velocity must be clockwise in accordance to\u00a0v<\/strong>B4<\/sub>. \u03c914<\/sub> = \u03c913<\/sub>\u00a0since there is no relative rotation between links 3 and 4. Then we can solve the vector equation:<\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 v<\/strong>D<\/sub>\u00a0= v<\/strong>B3<\/sub>\u00a0+ v<\/strong>D\/B3<\/sub><\/p>\n

\"\"<\/p>\n

Velocity polygon for the example<\/p>\n

vD\/B3<\/sub>\u00a0= |DB|\u03c913<\/sub> (= 551.41\u00b70.738 = 406.91mm\/s) and the direction of v<\/strong>D\/B3<\/sub>\u00a0is perpendicular to DB<\/strong> in the sense of \u03c913<\/sub>. The sum of these two vectors gives us v<\/strong>D<\/sub> as shown in the figure. The result is in close agreement with the analytical solution.<\/p>\n

For the acceleration analysis we can write the acceleration vector equation:<\/p>\n

at<\/sup><\/strong>B4<\/sub>\u00a0+ an<\/sup><\/strong>B4<\/sub>\u00a0= a<\/strong>B3<\/sub>\u00a0+ at<\/sup><\/strong>B4\/3<\/sub>\u00a0+ ac<\/sup><\/strong>B4\/2<\/sub><\/p>\n

Considering B4<\/sub>, the tangential acceleration\u00a0at<\/sup><\/strong>B4<\/sub>\u00a0is of magnitude |BA0<\/sub>|a<\/span>14<\/sub>\u00a0(unknown\u00a0a<\/span>14<\/sub>) and its direction is perpendicular to BA<\/strong>0<\/sub>. The normal acceleration is of magnitude |BA0<\/sub>|\u03c914<\/sub>2<\/sup> = vB4<\/sub>2<\/sup>\/|BA0<\/sub>| (= 130.45 mm\/s2<\/sup>) and its direction is along BA<\/strong>0<\/sub>\u00a0towards A0<\/sub>, center of rotation. Since B3<\/sub>\u00a0and B2<\/sub>\u00a0are permanently coincident, they have the same velocity and acceleration. Assuming that the piston is moving at a constant speed,\u00a0\u00a0a<\/strong>B2<\/sub>\u00a0= a<\/strong>B3<\/sub>\u00a0= 0<\/strong>. The tangential relative acceleration\u00a0at<\/sup><\/strong>B4\/3<\/sub>\u00a0is along the relative path, which is the slider axis between 3 and 4. Its magnitude is an unknown. The Coriolis acceleration\u00a0ac<\/sup><\/strong>B4\/3<\/sub>\u00a0is of magnitude 2\u03c914<\/sub>vB4\/3<\/sub> (= 187.32 mm\/s2<\/sup>). Its direction is perpendicular to the slider axis and is determined by rotating the velocity vector v<\/strong>B4\/3<\/sub> by 90\u00ba\u00a0in the sense of \u03c914<\/sub>. Drawing these acceleration vectors by a certain scale\u00a0ka<\/sub> (= 1 mm\/(mm\/s2<\/sup>)), we determine the magnitudes of at<\/sup><\/strong>B4<\/sub>\u00a0and at<\/sup><\/strong>B4\/3<\/sub>. Next, for the acceleration of point D:<\/p>\n

a<\/strong>D<\/sub>\u00a0= a<\/strong>B3<\/sub>\u00a0+ an<\/sup><\/strong>D\/B<\/sub>\u00a0+ at<\/sup><\/strong>D\/B<\/sub><\/p>\n

Since we do not know the path of point D, we cannot separate\u00a0a<\/strong>D<\/sub>\u00a0in to its components.\u00a0a<\/strong>B3<\/sub>\u00a0= 0<\/strong>.\u00a0The normal relative acceleration\u00a0an<\/sup><\/strong>D\/B<\/sub> is of magnitude |BD|\u03c913<\/sub>2<\/sup>\u00a0(= 300.32 mm\/s2<\/sup>) along BD<\/strong> towards B. Note that \u03c913<\/sub>\u00a0= \u03c914<\/sub>, since there is no relative rotation between links 3 and 4. The relative tangential acceleration at<\/sup><\/strong>D\/B<\/sub> is of magnitude |BD|\u03b113<\/sub>. Since \u03b113<\/sub>\u00a0= \u03b114<\/sub>\u00a0= at<\/sup>B4<\/sub>\/|BA0<\/sub>| (= 0.640 s-2<\/sup>), at<\/sup><\/strong>D\/B<\/sub>\u00a0= 153.41 mm\/s2<\/sup>, perpendicular to DB<\/strong> in the sense of \u03b113<\/sub>. The sum of these known vectors will give us the acceleration of point D at the instant considered as shown below.<\/p>\n

\"\"<\/p>\n

Acceleration Polygon for the example<\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n

\"\"<\/a>\"\"<\/a>\"\"<\/a>\"\"<\/a>\"\" <\/p>\n","protected":false},"excerpt":{"rendered":"

4.2 VELOCITY AND ACCELERATION ANALYSIS OF MECHANISMS-3 Example: For the mechanism shown, the piston (2) has an upward velocity vA = v12 = 180 mm\/s (linear velocity of link 2 with respect to link 1). We are to determine the …<\/p>\n

4-2-3<\/span> Devam\u0131n\u0131 Oku »<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":1964,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"full-width-page.php","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-2161","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2161","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/users\/7747"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/comments?post=2161"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2161\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1964"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=2161"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}