{"id":2162,"date":"2022-03-13T13:04:14","date_gmt":"2022-03-13T13:04:14","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2162"},"modified":"2022-12-06T02:40:13","modified_gmt":"2022-12-06T02:40:13","slug":"4-2-4","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch4\/4-2-4\/","title":{"rendered":"4-2-4"},"content":{"rendered":"
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4.2<\/b> VELOCITY AND ACCELERATION ANALYSIS OF<\/strong> MECHANISMS-4<\/h1>\n

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

M<\/span>echanism shown in the figure\u00a0<\/span>is used in a hay bailing machine. We are to determine the velocity and acceleration of point F on link 6 when the input link is rotating at a constant velocity of 4 rad\/s. The link lengths are as given on the figure.<\/p>\n

\"\"<\/p>\n

In this problem, for the position analysis stepwise solution and for the velocity and acceleration analysis matrix inversion method will be used.<\/p>\n

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

\u00a0\"\" \u00a0 \"\" \u00a0 \u03c912<\/sub> := 4\u00a0 \"\"<\/strong><\/span><\/p>\n

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

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

Generate the input crank angle for every 5\u00b0:<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

xBk<\/sub>, yBk<\/sub>\u00a0are the rectangular coordinates of B with respect to C0<\/sub> with x-axis along QC0<\/sub>.<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Convert rectangular coordinates to polar coordinates.<\/p>\n

\"\"\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Cosine theorem for angle BC0<\/sub>C.<\/p>\n

 <\/p>\n

\"\"\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Cosine theorem for the transmission angle BCC0<\/sub><\/p>\n

 <\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Note that \u03d5, \u03b2\u00a0and s are used as dummy variables.<\/p>\n

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

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

x numbers with A0<\/sub>\u00a0as the origin.<\/p>\n

Coordinates of F in complex numbers:<\/p>\n

\"\"<\/p>\n

In\u00a0f<\/span>igure below, the path of point F is shown (the x sign is the position of point F\u00a0when \u03b812<\/sub> = 90\u00b0).<\/p>\n

\"\"<\/p>\n

For velocity and acceleration analysis create the coefficient matrix:<\/p>\n

\"\"<\/p>\n

The angular velocities are found from the velocity loop equations.<\/p>\n

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

The angular acceleration of the links are found from the acceleration loop equations.<\/p>\n

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

Velocity and acceleration of point F:<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

The polar plots of the velocity and acceleration vectors of point F for a complete cycle are shown below (velocity values are measured in mm\/s).<\/p>\n

\"\"(Velocity of F at an instant is the vector from point 0 to a point on the curve; e.g. velocity of Point F when \u03b812<\/sub> = 90\u00b0\u00a0is the line drawn from 0 to the mark x)<\/p>\n

The polar plots of the velocity and acceleration vectors of point F for a complete cycle are shown below (acceleration values are measured in mm\/s2<\/sup>).<\/p>\n

\"\"
\n(Acceleration\u00a0<\/span>of F at an instant is the vector from point 0 to a point on the curve; e.g. acceleration\u00a0<\/span>of Point F when \u03b812<\/sub> = 90\u00b0 is the line drawn from o to the mark x)<\/p>\n

In case of graphical solution for the position given, link 2 is in a fixed axis of rotation with \u03c912<\/sub> = 4 rad\/s. Therefore vA<\/sub> = vB<\/sub> = 728 mm\/s and while v<\/strong>A<\/sub> is to the left, v<\/strong>B<\/sub>\u00a0is to the right. Using a scale factor kv<\/sub> =0.5 mm\/(mm\/s) we draw these known vectors (see figure). Considering the loop formed by links 1, 2, 3 and 4, the velocity equation is:<\/p>\n

v<\/strong>C<\/sub>\u00a0= v<\/strong>B<\/sub>\u00a0+ v<\/strong>C\/B<\/sub><\/p>\n

v<\/strong>C\/B<\/sub>\u00a0is perpendicular to CB<\/strong> and v<\/strong>C<\/sub>\u00a0is perpendicular to CC<\/strong>0<\/sub>. At the given position, since BA<\/strong>0<\/sub>\u00a0and CC<\/strong>0<\/sub> are parallel,\u00a0v<\/strong>C<\/sub> and\u00a0v<\/strong>B<\/sub> must be equal and v<\/strong>C\/B<\/sub> = 0<\/strong> and \u03c913<\/sub> = 0. In such a case v<\/strong>D<\/sub> = v<\/strong>C<\/sub> = v<\/strong>B<\/sub>. Next, consider point E\u00a0<\/strong>which is a permanently coincident point on links 5 and 6. For points D and E on link 5:<\/p>\n

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

Similarly, for points A and E on link 6:<\/p>\n

v<\/strong>E<\/sub>\u00a0= v<\/strong>A<\/sub>\u00a0+ v<\/strong>E\/A<\/sub><\/p>\n

v<\/strong>D<\/sub> and\u00a0v<\/strong>A<\/sub>are of known magnitude and direction. The relative velocities v<\/strong>E\/D<\/sub> and v<\/strong>E\/A<\/sub>\u00a0must be perpendicular to ED<\/strong> and EA<\/strong> respectively. Although these two equations cannot be solved separately, when they are equated to each other, the two relative velocity magnitudes are the unknowns. Therefore simultaneous solution is made. For the velocity of point F, we can apply Mehmke’s theorem or solve the two vector equations<\/p>\n

v<\/strong>F<\/sub>\u00a0= v<\/strong>E<\/sub>\u00a0+ v<\/strong>F\/E<\/sub><\/p>\n

v<\/strong>F<\/sub>\u00a0= v<\/strong>A<\/sub>\u00a0+ v<\/strong>F\/A<\/sub><\/p>\n

simultaneously.<\/p>\n

\"\"
\nV<\/span>elocity\u00a0P<\/span>olygon<\/strong><\/p>\n

\u00a0For the acceleration analysis we use the same points that have been used for the velocity analysis. However we have to write the acceleration of these points in terms of their components. Since link 2 is rotating at a constant speed, the tangential accelerations of points A and B are zero (a<\/strong>A<\/sub>\u00a0= an<\/sup><\/strong>A<\/sub>, a<\/strong>B<\/sub>\u00a0= an<\/sup><\/strong>B<\/sub>). aA<\/sub>\u00a0= aB<\/sub> = 2912 mm\/s2<\/sup>, both towards the center of rotation. Using a scale factor\u00a0ka<\/sub> = 0.2 mm\/(mm\/s2<\/sup>), we draw these vectors (see figure). Next we write the acceleration of point C as:<\/p>\n

an<\/sup><\/strong>C<\/sub>\u00a0+ at<\/sup><\/strong>C<\/sub>\u00a0= an<\/sup><\/strong>B<\/sub>\u00a0+ an<\/sup><\/strong>C\/B<\/sub>\u00a0+ at<\/sup><\/strong>C\/B<\/sub><\/p>\n

an<\/sup><\/strong>C<\/sub>\u00a0is a vector along CC<\/strong>0<\/sub>\u00a0towards C0<\/sub>\u00a0and is of magnitude |CC0<\/sub>|\u03c913<\/sub>2<\/sup> = vC<\/sub>2<\/sup>\/|CC0<\/sub>|= 1045 mm\/s2<\/sup>.\u00a0an<\/sup><\/strong>C\/B<\/sub>\u00a0= 0<\/strong> since \u03c913<\/sub>\u00a0= 0 for this position. at<\/sup><\/strong>C<\/sub>\u00a0and at<\/sup><\/strong>C\/B<\/sub>\u00a0are perpendicular to CC<\/strong>0<\/sub> and CB<\/strong>, respectively (unknown magnitudes). Note that while \u03c913<\/sub>\u00a0= 0, \u03b113<\/sub>\u00a0\u2260 0. at<\/sup><\/strong>D\/B<\/sub>\u00a0can now be determined since \u03b113<\/sub>\u00a0= at<\/sup>C\/B<\/sub>\u00a0\/|CB| (an<\/sup><\/strong>D\/B<\/sub> = 0<\/strong>). Hence a<\/strong>D<\/sub>\u00a0is known. For point E we can write:<\/p>\n

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

and<\/p>\n

a<\/strong>E<\/sub>\u00a0= a<\/strong>A<\/sub>\u00a0+ an<\/sup><\/strong>E\/A<\/sub>\u00a0+ at<\/sup><\/strong>E\/A<\/sub><\/p>\n

an<\/sup>E\/D<\/sub>\u00a0= |ED|\u03c915<\/sub>2<\/sup>\u00a0= vE\/D<\/sub>2<\/sup>\/|ED| = 1366 mm\/s2<\/sup>\u00a0 \u00a0(along ED<\/strong>, towards D)<\/p>\n

an<\/sup>E\/A<\/sub>\u00a0= |EA|\u03c916<\/sub>2<\/sup>\u00a0= vE\/A<\/sub>2<\/sup>\/|EA| = 1269 mm\/s2<\/sup>\u00a0 \u00a0(along EA<\/strong>, towards A)<\/p>\n

Tangential acceleration components\u00a0at<\/sup><\/strong>E\/D<\/sub>\u00a0and at<\/sup><\/strong>E\/A<\/sub>\u00a0are perpendicular to the lines ED<\/strong> and EA<\/strong> respectively (unknown magnitudes). The two equations can now be solved simultaneously to determine the velocity of point E. For the acceleration of point F, Mehmke’s theorem is applied (triangle aef on the acceleration polygon is similar to the triangle AEF of link 6). The result is shown in the figure below. Hence aF<\/sub>\u00a0= 1075.5\/0.2 =5378 mm\/s2<\/sup>\u00a0, downwards as shown.<\/p>\n

\"\"
\nAcceleration Polygon<\/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-4 Example: Mechanism shown in the figure\u00a0is used in a hay bailing machine. We are to determine the velocity and acceleration of point F on link 6 when the input link is rotating at …<\/p>\n

4-2-4<\/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-2162","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2162","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=2162"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2162\/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=2162"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}