{"id":2133,"date":"2022-03-12T23:28:54","date_gmt":"2022-03-12T23:28:54","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2133"},"modified":"2022-12-06T01:08:10","modified_gmt":"2022-12-06T01:08:10","slug":"3-5","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch3\/3-5\/","title":{"rendered":"3-5"},"content":{"rendered":"
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3.5<\/b> Graphical Solution of Loop Closure Equations<\/h1>\n

In performing the motion analysis the links are represented as straight lines that join the kinematic elements of the link. Since the tips of these lines are labelled, they are directed lines of certain magnitude and direction. Hence, in effect once we label the joints the diagram we thus obtain is the graphical representation of the loop closure equation (this was why we were able to write the loop closure equations, anyhow). For the graphical solution, this diagram must be drawn on paper by a certain scale, ks, which is generally called as the space scale. It is the ratio of the magnitudes drawn on the sheet to the original magnitude. If you are to draw the mechanism on the computer, you may use the exact dimensions, no matter how big or how small the mechanism is in its actual size.<\/p>\n

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

Consider the four-bar mechanism in F(a). If four links of different dimensions were given, you would have several different possible ways of joining these links. However, if we know an assembled position, the mechanism and its motion must be unique. Let us determine the positions of the links when the input link is rotated to some other angular position \u03b812<\/sub>\u2032.<\/p>\n

The new position of the vector\u00a0A0<\/sub>A<\/strong>\u00a0(A0<\/sub>A\u2032<\/strong>) is completely known when is \u03b812<\/sub>\u2032 given. The vector loop equation:<\/p>\n

A0<\/sub>A<\/strong> + AB<\/strong> = A0<\/sub>B0<\/sub><\/strong>\u00a0+ B0<\/sub>B<\/strong><\/p>\n

must also be true when\u00a0<\/strong>A is replaced by A\u2032 (known point) and B is replaced by B\u2032 (unknown), since the mechanism will form a closed loop in this second position:<\/p>\n

A0<\/sub>A\u2032<\/strong>+ AB\u2032<\/strong>\u00a0= A0<\/sub>B0<\/sub><\/strong>\u00a0+ B0<\/sub>B\u2032<\/strong><\/p>\n

This equation can also be written in the form:<\/p>\n

A\u2032B\u2032<\/strong>\u00a0= A0<\/sub>B0<\/sub><\/strong> \u2212 A0<\/sub>A\u2032<\/strong>\u00a0+ B0<\/sub>B\u2032<\/strong><\/p>\n

or:<\/p>\n

A\u2032B\u2032<\/strong>\u00a0= A\u2032A0<\/sub><\/strong>\u00a0+ A0<\/sub>B<\/strong>0<\/strong>\u00a0<\/sub>+ B0<\/sub>B\u2032<\/strong><\/p>\n

\"\"<\/p>\n

The sum of the vectors\u00a0A\u2032A0<\/sub><\/strong>\u00a0+ A0<\/sub>B0<\/sub><\/strong>\u00a0is the vector\u00a0A\u2032B0<\/sub><\/strong>. We know the magnitude and direction of the vectors A\u2032A0<\/sub><\/strong>\u00a0and\u00a0A0<\/sub>B0<\/sub><\/strong>. We know the magnitudes of the vectors\u00a0A\u2032B\u2032<\/strong>\u00a0and\u00a0B0<\/sub>B\u2032<\/strong>. Since the starting positions A\u2032 and B0<\/sub>\u00a0for the vectors\u00a0A\u2032B\u2032<\/strong>\u00a0and\u00a0B0<\/sub>B\u2032<\/strong>\u00a0are known, in order to determine the new position of B (B\u2032), one need only construct circles with centres at A\u2032 and B0<\/sub>\u00a0and radii equal to the corresponding link lengths (AB and B0<\/sub>B, respectively) as shown in Figure 2.21b. The two circles will have two intersections B\u2032 and B\u2033;. However, one of these solutions is discarded since the mechanism cannot jump into one of these positions without disconnecting the joints (see figure). If the two circles drawn cannot intersect at all, then the mechanism cannot be assembled for that crank angle \u03b812<\/sub>\u2032.<\/p>\n

The above procedure is nothing but the solution of a triangle whose three sides are known. For great many mechanisms the position analysis using graphical technique will reduce to the solution of a triangle. As is well known, a triangle is fully determined if we know the three sides (SSS), two angles and one side (ASA) or the two sides and one angle (SAS or SSA). All these three cases can be encountered in the position analysis.<\/p>\n

Example 2.1.<\/em><\/strong><\/p>\n

\"\"<\/p>\n

Consider an inverted slider-crank mechanism shown in Figure above. We are asked to determine the positions of the links when the input link, 2, is rotated to some other angular position \u03b812<\/sub>\u2032. The loop closure equation is:<\/p>\n

A0<\/sub>A<\/strong> = A0<\/sub>B0<\/sub><\/strong>\u00a0+ B0<\/sub>C<\/strong> + CA<\/strong><\/p>\n

The vector\u00a0A0<\/sub>B0<\/sub><\/strong>\u00a0is fixed. The vector\u00a0A0<\/sub>A\u00a0<\/strong>has a fixed magnitude and its angular orientation is defined by the given angle \u03b812<\/sub>.\u00a0B0<\/sub>C<\/strong>\u00a0has a fixed magnitude but an unknown angular orientation whereas for the vector\u00a0CA<\/strong>\u00a0both the magnitude and the direction can change. However, the vectors\u00a0B0C<\/strong>\u00a0and\u00a0CA<\/strong>\u00a0are perpendicular to each other and the axis of the slider must always pass through point A. Points AB0C will always form a right angled triangle. In order to construct the triangle AB0<\/sub>C for the given points A and B0<\/sub>\u00a0(defined by the vectors\u00a0A0<\/sub>A\u00a0<\/strong>and\u00a0A0<\/sub>B0<\/sub><\/strong>), we draw a circle with centre at B0<\/sub>\u00a0and radius B0<\/sub>C. If we now draw a line passing through A and tangent to this circle, the new positions for links 3 and 4 will be found.<\/p>\n

\"\"<\/p>\n

In a complete motion analysis it is necessary to draw the mechanism in various phases of its cycle (for all possible input positions). Through such an analysis one can determine the limiting positions (positions at which there is the displacement of the link involved is at an extreme position, e.g. at this position there is a reversal of motion).<\/p>\n