{"id":2222,"date":"2022-03-15T17:27:40","date_gmt":"2022-03-15T17:27:40","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2222"},"modified":"2022-08-03T18:05:51","modified_gmt":"2022-08-03T18:05:51","slug":"6-2-2","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch6\/6-2\/6-2-2\/","title":{"rendered":"6-2-2"},"content":{"rendered":"
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6.2.2. Principle of Superposition<\/strong>:<\/span><\/span><\/h1>\n

In the previous examples, there was only one known external force acting on one member of the mechanism and the system was brought to static equilibrium by an input or output force (or torque). The magnitude of this force or torque was an unknown. If there are two or more known external forces acting on one link, these forces can be combined into a single resultant and the problem reduces to the case we have already discussed. However, in real machinery there are several external forces acting on different links. For example, if we do not neglect the weight of the members, there will be at least one known external force on each link. In such a case, if we draw the free body diagrams of the links, no simplification will be possible and one has to write three equilibrium equations for each link. The resulting 3(l \u2212 <\/em>1) linear equations will include 3(l \u2212 <\/em>1) unknown joint force components and the input force (or torque). Usually simultaneous solution of the equilibrium equations would be required.. Another solution method the principal of superposition. This principal states that\u00a0the effect of the forces is the sum of the individual effects of the forces considered separately<\/strong><\/em><\/span>. In other words, if there are two or more external forces present, one can neglect all but one of the forces and determine the joint forces and the unknown reaction force that brings the system into equilibrium for this one external force. If the above procedure is carried out for each of the external forces, at each joint there will be different joint forces corresponding to each external force. The resultant joint force is the vectorial sum of all these forces. We shall explain this by solving the same problem by (a) without using the principal of superposition and (b) by using the Principal of superposition.<\/p>\n

Example 6.3.<\/span><\/strong><\/p>\n

\"\"<\/p>\n

For the mechanism shown |A0<\/sub>A| = a2<\/sub>\u00a0= 80, |AB| = a3<\/sub> = 100, |B0<\/sub>B| = a4<\/sub>\u00a0= 120, |A0<\/sub>B0<\/sub>| = a1<\/sub>\u00a0= 140, |AC| = b3<\/sub>\u00a0= 70, |BC| = 80 and |B0<\/sub>D| = b4 = 90 mm. When \u03b812<\/sub>\u00a0= 60\u00b0, from kinematic analysis \u03b813<\/sub> = 29,98\u00b0, \u03b814<\/sub>\u00a0= 96.40\u00b0. Two forces\u00a0F<\/strong>13<\/sub> = 50 N \u2220230\u00b0\u00a0and\u00a0F<\/strong>14<\/sub>\u00a0= 100 N \u2220200\u00b0\u00a0<\/sup>are acting on links 3 and 4 respectively.<\/p>\n

Solution without the principal of superposition<\/span><\/strong><\/p>\n

\"\"<\/p>\n

The free-body diagrams of the moving links are shown. The three equilibrium equations for link 4 are:<\/p>\n\n\n\n\n\n
S<\/span>Fx<\/sub> = F34x<\/sub>\u00a0+ G14x<\/sub> \u2212 F14<\/sub>\u00a0cos20\u00b0\u00a0= 0<\/td>\n(1)<\/td>\n<\/tr>\n
S<\/span>Fy<\/sub> = F34y<\/sub>\u00a0+ G14y<\/sub> \u2212 F14<\/sub>\u00a0sin20\u00b0\u00a0= 0<\/td>\n(2)<\/td>\n<\/tr>\n
S<\/span>MB0<\/sub> = \u2212a4 <\/sub>F34x<\/sub> sin\u03b814<\/sub>\u00a0+ a4 <\/sub>F34y<\/sub> cos\u03b814<\/sub> \u2212 b4 <\/sub>F14<\/sub> sin(20\u00b0 \u2212 \u03b814<\/sub>)\u00a0= 0<\/td>\n(3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

There are four unknowns in three equations, therefore the equations obtained from one free-body diagram is not enough to solve for the unknowns. Equations 1 and 2 can be used to solve for G14x<\/sub>\u00a0and G14y<\/sub>, only when F34xy<\/sub>\u00a0and F34y<\/sub>\u00a0are determined. The three equilibrium equations for link 3 must also be written (note that F<\/strong>34<\/sub>\u00a0and\u00a0F<\/strong>43<\/sub> have equal magnitude).<\/p>\n\n\n\n\n\n
S<\/span>Fx<\/sub> = F23x<\/sub> \u2212 F34x<\/sub> \u2212 F13<\/sub>cos50\u00b0 = 0<\/td>\n(4)<\/td>\n<\/tr>\n
S<\/span>Fy<\/sub> = F23y<\/sub> \u2212 F34y<\/sub> \u2212 F13<\/sub>sin50\u00b0\u00a0= 0<\/td>\n(5)<\/td>\n<\/tr>\n
S<\/span>MA<\/sub> = a3 <\/sub>F34x<\/sub> sin\u03b813<\/sub> \u2212 a3 <\/sub>F34y<\/sub> cos\u03b813<\/sub> \u2212 b3 <\/sub>F13<\/sub> sin(50\u00b0 \u2212 \u03b813<\/sub> \u2212 \u03b1) = 0<\/td>\n(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\n

where \u03b1\u00a0= 52.62\u00b0 (using the cosine theorem for triangle ABC). Equations 4 and 5 can be used to determine F23x<\/sub>\u00a0and F237x<\/sub>. Equations 3 and 6 must be used simultaneously to solve for F34x<\/sub>\u00a0and F34y<\/sub>. Substituting the known values into equations 3 and 6 results:<\/p>\n\n\n\n\n
\u2212119.25 F34x<\/sub>\u2212 13.38 F34y<\/sub> + 8748 = 0<\/td>\n(3)<\/td>\n<\/tr>\n
49.97 F34x<\/sub>\u00a0– 86.62 F34y<\/sub>\u00a0+ 1886 = 0<\/td>\n(6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Simultaneous solution of the two equations yield:<\/p>\n

F34x<\/sub>\u00a0= 66.60 N \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 F34y<\/sub> = 60.20 N\u00a0 \u00a0 \u00a0 \u00a0 \u00a0F<\/strong>34<\/sub> = 89.78 N \u222042.11\u00b0<\/p>\n

From equations 1 and 2:<\/p>\n

G14x<\/sub>\u00a0= 27.37 N \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 G14y<\/sub>\u00a0= 26.00 N \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 G<\/strong>14<\/sub> = 37.75 N \u2220\u221243.53\u00b0<\/p>\n

From Equations (4) and (5):<\/p>\n

F23x<\/sub>\u00a0= 98.74 N \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 F23y<\/sub>\u00a0= 98.50 N \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 F<\/strong>23<\/sub> = 139.46 N \u222044.93\u00b0<\/p>\n

Now, link 2 can be treated as a two force and a moment member (G<\/strong>12<\/sub> = \u2212F<\/strong>32<\/sub>\u00a0= F<\/strong>23<\/sub>). The moment equilibrium (about A0<\/sub>):<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0T12<\/sub> \u2212 a2 <\/sub>F23 <\/sub>sin(47.34\u00b0\u00a0\u2212 60\u00b0) = 0 \u00a0\u00a0\u00a0\u00a0or \u00a0\u00a0\u00a0\u00a0 T12<\/sub> = \u22122908 N\u00b7mm = 2.9 N\u00b7m (CW)<\/p>\n

Solution using the principal of superposition<\/span><\/strong><\/p>\n

\"\"<\/p>\n

The same problem will be solved usiing the principle of superposition. Let us consider two problems:<\/p>\n

\"\"<\/p>\n

In problem (a) the mechanism is under the action of F<\/strong>14<\/sub>\u00a0and in (b) F<\/strong>13<\/sub> only. If we denote the joint forces due to F<\/strong>14<\/sub>\u00a0by a single prime and the forces due to\u00a0F<\/strong>13<\/sub>\u00a0by double prime, we have the free-body diagrams due to\u00a0F<\/strong>14<\/sub>\u00a0as shown:<\/p>\n

\"\"<\/p>\n

Due to F<\/strong>14<\/sub>, the moment equilibrium for link 4 yields:<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 a4 <\/sub>F\u203234 <\/sub>sin(\u03b813<\/sub> \u2212 \u03b814<\/sub>) \u2212 b4 <\/sub>F14 <\/sub>sin(20\u00b0 \u2212 \u03b814<\/sub>) = 0<\/p>\n

from which F\u2032<\/strong>34<\/sub> = 79.54 N \u222029.98\u00b0 (= \u03b813<\/sub>).<\/p>\n

Since F\u2032<\/strong>34<\/sub> = \u2212F\u2032<\/strong>43<\/sub>\u00a0= F\u2032<\/strong>23<\/sub> = \u2212F\u2032<\/strong>32<\/sub>\u00a0= G\u2032<\/strong>12<\/sub> (links 3 and 2 are two-force and two force and a moment members respectively ; also action-reaction between bodies 2, 3, 4), we have to write the moment equilibrium for link 2 only:<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 T\u203212<\/sub>\u00a0\u2212 a2 <\/sub>F\u203232<\/sub> sin(\u03b813<\/sub> \u2212 \u03b812<\/sub>) = 0<\/p>\n

from which T\u203212<\/sub>\u00a0= \u22123184 N\u00b7mm = 3.18 N\u00b7m (CW)<\/p>\n

Due to F<\/strong>13<\/sub>\u00a0only:<\/p>\n

\"\"<\/p>\n

Link 4 is a two-force member, link 3 is a three force member while link 2 is a two-force and a moment member as before. The moment equilibrium equation for link 3 yields:<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 a3 <\/sub>F\u203343 <\/sub>sin(\u03b84<\/sub> \u2212 \u03b813<\/sub>) \u2212 b3 <\/sub>F13 <\/sub>sin(50\u00b0 \u2212 \u03b813<\/sub>) = 0<\/p>\n

from which F\u2033<\/strong>43<\/sub> = \u221220.57 N \u222096.40\u00b0 (= \u03b814<\/sub>) or F\u2033<\/strong>34<\/sub> = \u2212F\u2033<\/strong>43<\/sub>\u00a0= 20.57 N \u222096.40\u00b0<\/p>\n

F\u203323x<\/sub> and F\u203323y<\/sub> can be determined using the force equilibrium equations for link 3:<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 F\u203323x<\/sub>\u00a0= F13<\/sub> cos50\u00b0\u00a0+ F\u203343 <\/sub>cos(\u03b814<\/sub>) = 29.85 N<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 F\u203323y<\/sub> = F13<\/sub> sin50\u00b0\u00a0+ F\u203343 <\/sub>sin(\u03b814<\/sub>) = 58.74 N<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 F\u2033<\/strong>23<\/sub> = 65.89 N \u222063.06\u00b0<\/p>\n

Now, the moment equilibrium for link 2 yields:<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 T\u203312<\/sub> \u2212 a2 <\/sub>F\u203332 <\/sub>sin(63.06\u00b0 \u2212 \u03b812<\/sub>) = 0<\/p>\n

from which: T\u203312<\/sub> = 282 N\u00b7mm = 0.28 N\u00b7m (CCW)<\/p>\n

One can now superimpose the two solutions. For example, the torque T12<\/sub>\u00a0required for the original system will be:<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 T12<\/sub>\u00a0= T\u203212<\/sub>\u00a0+ T\u203312<\/sub> = \u22123184 + 282 = \u22122902 N\u00b7mm = 2.9 N\u00b7m (CW)<\/p>\n

Similarly:<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 F34x<\/sub>\u00a0= F\u203234x<\/sub>\u00a0+ F\u203334x<\/sub> = 79.54 cos(29.98\u00b0)+ 20.57 cos(96.40\u00b0) = 66.60 N<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 F34y<\/sub>\u00a0= F\u203234y<\/sub>\u00a0+ F\u203334y<\/sub> = 79.54 sin(29.98\u00b0)+ 20.57 sin(96.40\u00b0) = 60.19 N<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 F23x<\/sub>\u00a0= F\u203223x<\/sub>\u00a0+ F\u203323x<\/sub> = 79.54 cos(29.98\u00b0) + 29.85 = 98.75 N<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 F23y<\/sub>\u00a0= F\u203223y<\/sub>\u00a0+ F\u203323y<\/sub> = 79.54 sin(29.98\u00b0) + 58.74 = 98.49 N<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 F<\/strong>23<\/sub>\u00a0= G<\/strong>12<\/sub>\u00a0= 139.5 N \u222044.92\u00b0<\/p>\n

\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 F<\/strong>34<\/sub> = 89.77 N \u222042.11\u00b0<\/p>\n

The results are in good agreement with the results obtained previously (small differences are due to the round-off errors).<\/p>\n

Different formulations can be used depending on the type of the problem to be analysed, on the accuracy required, time to be spend and the available numerical computation facilities. For example, if a rough estimate is to be made for one particular position, a graphical computation may be the simplest (whether done by drafting instruments or by a drawing program). If the problem is to be repeated over and over for different positions and for different external load conditions (and maybe with different link lengths) a general formulation and writing a general computer program for the machine will be more feasible. If different types of structures are to be investigated in a development stage, use of a package program such as Working Model\u00ae or ADAMS\u00ae may be useful. Usually an engineer must check his results by performing the same computation in two or more different ways.<\/p>\n<\/div>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n

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

6.2.2. Principle of Superposition: In the previous examples, there was only one known external force acting on one member of the mechanism and the system was brought to static equilibrium by an input or output force (or torque). The magnitude …<\/p>\n

6-2-2<\/span> Devam\u0131n\u0131 Oku »<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":2217,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"full-width-page.php","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-2222","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2222","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=2222"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2222\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2217"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=2222"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}