{"id":2738,"date":"2022-03-25T13:17:27","date_gmt":"2022-03-25T13:17:27","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2738"},"modified":"2022-09-02T16:47:13","modified_gmt":"2022-09-02T16:47:13","slug":"example-2","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch6\/6-3\/example-2\/","title":{"rendered":"Example 2"},"content":{"rendered":"
\n
\n\t
Example 2<\/span><\/b><\/h6>\n

In this second example the same centric slider-crank mechanism of the previous example is considered. However, in addition to\u00a0F<\/strong>14<\/sub> two additional forces,\u00a0F<\/strong>13<\/sub>\u00a0and\u00a0F<\/strong>12<\/sub>, act on the links. F12<\/sub> = 100 N , F13<\/sub> = 150 N and F14<\/sub> = 100 N.<\/p>\n

\"\"<\/p>\n

The free body diagrams of the links are shown below. The pin radii and the friction coefficient are that of the previous example.<\/p>\n

\"\"<\/p>\n

The force equilibrium equations for link 4:<\/p>\n\n\n\n\n\n
S<\/span>Fx<\/sub>\u00a0= F34x<\/sub>\u00a0\u2212 F14<\/sub>\u00a0\u2212 \u03bcG14<\/sub>=0\u00a0 \u00a0 \u00a0 \u00a0or\u00a0 \u00a0 \u00a0 \u00a0F34x<\/sub> = 100 + 0.1G14\u00a0<\/sub><\/td>\n(i)<\/td>\n<\/tr>\n
S<\/span>Fy<\/sub> = F34y<\/sub>\u00a0\u2212 G14<\/sub>\u00a0= 0\u00a0 \u00a0 \u00a0 \u00a0or\u00a0 \u00a0 \u00a0 \u00a0G14<\/sub>\u00a0= F34y<\/sub><\/td>\n(ii)<\/td>\n<\/tr>\n
S<\/span>MB<\/sub>\u00a0= \u2212sG14<\/sub> \u2212 5G14<\/sub>\u00a0+ 5F34<\/sub>\u00a0= 0 \u00a0 \u00a0 \u00a0or\u00a0 \u00a0 \u00a0 \u00a0s = 5(F34<\/sub>\u00a0\u2212 G14<\/sub>)\/G14<\/sub><\/td>\n(iii)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

If s is within the physical boundaries of the two contacting surfaces, the above equations are valid (contact Mode I or II will exist). Also:<\/p>\n\n\n\n
\\displaystyle {\\text{F}}_{\\text{34}}=\\sqrt{{{{\\text{F}}_{\\text{34x}}}^2+{{\\text{F}}_{\\text{34y}}}^2}}<\/span><\/span><\/td>\n(iv)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

For link 3:<\/p>\n\n\n\n\n\n
S<\/span>Fx<\/sub>\u00a0= \u2212F34x <\/sub> + F13<\/sub>cos(210\u00b0)\u00a0+ F23x<\/sub> = 0\u00a0 \u00a0 \u00a0 \u00a0or\u00a0 \u00a0 \u00a0 \u00a0F23x<\/sub> = F34x <\/sub>+ 129.904<\/td>\n(v)<\/td>\n<\/tr>\n
S<\/span>Fy<\/sub> = F34y<\/sub> + F13<\/sub>sin(210\u00b0) + F23y<\/sub> = 0 \u00a0 \u00a0 \u00a0 or\u00a0 \u00a0 \u00a0 F23y<\/sub>\u00a0= 75 \u2212 F34y<\/sub><\/td>\n(vi)<\/td>\n<\/tr>\n
S<\/span>MB<\/sub> = \u2212a3<\/sub>F34y<\/sub>cos\u03b813<\/sub> \u2212 a3<\/sub>F34x<\/sub>sin\u03b813<\/sub> \u2212 25F23<\/sub> \u2212 5F34<\/sub> \u2212 400F13<\/sub>sin(210\u00b0 \u2212 \u03b813<\/sub>) = 0 \u00a0 \u00a0 \u00a0or
\n783.045F34y<\/sub>\u00a0\u2212 163.829F34x<\/sub> \u2212 25F23<\/sub> \u2212 5F34<\/sub> \u2212 40005.218 = 0<\/td>\n		(vii)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

and<\/p>\n\n\n\n
\\displaystyle {\\text{F}}_{\\text{23}}=\\sqrt{{{{\\text{F}}_{\\text{23x}}}^2+{{\\text{F}}_{\\text{23y}}}^2}}<\/span><\/span><\/td>\n(viii)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

For link 2:<\/p>\n\n\n\n\n\n
S<\/span>Fx<\/sub> = \u2212F23x<\/sub>\u00a0+ G12x<\/sub>\u00a0+ F12<\/sub>cos(\u221260\u00b0) = 0\u00a0 \u00a0 \u00a0 \u00a0or\u00a0 \u00a0 \u00a0 \u00a0G12x<\/sub> = F23x<\/sub>\u00a0\u2212 50<\/td>\n(ix)<\/td>\n<\/tr>\n
S<\/span>Fy<\/sub> = F23y<\/sub> + G12y<\/sub>\u00a0+ F12<\/sub>sin(\u221260\u00b0) = 0 \u00a0 \u00a0 \u00a0 or\u00a0 \u00a0 \u00a0 G12y<\/sub>\u00a0= F23x<\/sub> + 86.603<\/td>\n(x)<\/td>\n<\/tr>\n
S<\/span>MA<\/sub>0<\/sub> = T12<\/sub> + 100F12<\/sub>sin(\u221260\u00b0 \u2212 \u03b812<\/sub>) + 200F23y<\/sub>sin(\u03b812<\/sub>) \u2212 200F23x<\/sub>cos(\u03b812<\/sub>) \u2212 25F23<\/sub> \u2212 5G12<\/sub>\u00a0= 0 \u00a0 \u00a0 \u00a0or
\nT12<\/sub> = 9063.078 \u2212 163.830F23x<\/sub> + 114.715F23y<\/sub> +25F23<\/sub> + 5G12<\/sub><\/td>\n		(xi)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

and<\/p>\n\n\n\n
\\displaystyle {\\text{G}}_{\\text{12}}=\\sqrt{{{{\\text{G}}_{\\text{12x}}}^2+{{\\text{G}}_{\\text{12y}}}^2}}<\/span><\/span><\/td>\n(xii)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

In the equations the forces are in N (Newton) and moments are in N\u00b7mm.\u00a0There are 12 equations in 12 unknowns (F34x<\/sub>\u00a0, F34y<\/sub>\u00a0, F34<\/sub>, F23x<\/sub>, F23y<\/sub>, F23<\/sub>, G12x<\/sub>\u00a0, G12y<\/sub>, G12<\/sub>\u00a0, G14<\/sub>\u00a0, s and T12<\/sub>).\u00a0\u00a0 Equations (iv, viii, xii) are non-linear, whereas the remaining equations are linear. Direct solution of these equations will not be attempted. An iterative procedure is more suitable in such cases.<\/p>\n

Substituting Equations (i) and (ii) into equation (vii) and solving for G14<\/sub>\u00a0yields:<\/p>\n\n\n\n
G14<\/sub> = 265.872 + 0.033F23<\/sub>\u00a0+ 0.007F34<\/sub><\/td>\n(xiii)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The last two terms of equation (xiii) are due to friction at the joints and it is logical that these terms will be small in comparison to the first term, which is the value of G14\u00a0<\/sub>when \u03bc = 0. As the first guess for G14<\/sub>, if we neglect the effect of friction, the value of G14<\/sub>\u00a0can be easily determined (G14<\/sub> = 265.872 N). We can than solve for F34x<\/sub> using equation (i), F34y<\/sub>\u00a0and F34<\/sub> from equation (ii), F34y<\/sub>\u00a0= G14<\/sub> then use equation (iv). Next using equations (v, vi, viii)\u00a0 we can solve for F23x\u00a0<\/sub>, F23y\u00a0<\/sub>, F23\u00a0<\/sub>. The values of F23<\/sub>\u00a0and F34<\/sub>\u00a0thus found can now be used to refine the value of G14<\/sub>. The procedure is repeated until there is no significant change in G14<\/sub>. The iteration steps for this example are shown in the table below. The procedure in general rapidly converges, since the friction effect in mechanisms is quite small (if the mechanism is not at a critical position). Also one can see the effect of friction on the solution.<\/p>\n

Force\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a02. Step\u00a0 \u00a0 \u00a0 \u00a03. Step\u00a0 \u00a0 \u00a0 4. Step \u00a0 \u00a0 \u00a0 \u00a0 5. Step<\/p>\n

G14<\/sub>(eq. xiii) (G14<\/sub>0<\/sup>\u00a0= F34y<\/sub>)\u00a0 \u00a0 265.872\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0278.484\u00a0\u00a0\u00a0\u00a0\u00a0278.854\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0278.865<\/p>\n

F34x<\/sub> (eq. iv)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 126.587\u00a0 \u00a0 \u00a0 127.848\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0127.885\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0127.886<\/p>\n

F34y<\/sub> (eq. iv)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0294.469\u00a0\u00a0\u00a0\u00a0\u00a0306.429\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0306.780\u00a0 \u00a0 \u00a0 \u00a0 306.791<\/p>\n

F23x<\/sub> (eq. v)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 256.491\u00a0\u00a0\u00a0\u00a0\u00a0257.752\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0257.789\u00a0 \u00a0 \u00a0 \u00a0 257.790<\/p>\n

F23y<\/sub>\u00a0(eq. vi)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-190.872\u00a0\u00a0 \u00a0-203.484\u00a0\u00a0 \u00a0\u00a0-203.854\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0-203.865<\/p>\n

F23\u00a0<\/sub> (eq. vii)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 319.718\u00a0\u00a0\u00a0\u00a0\u00a0 328.393\u00a0\u00a0\u00a0\u00a0\u00a0 328.652\u00a0 \u00a0 \u00a0 \u00a0328.659<\/p>\n

(All values in Newton)<\/p>\n

After this iterative method of finding\u00a0F<\/strong>34<\/sub>, G<\/strong>14<\/sub>\u00a0and\u00a0F<\/strong>23<\/sub>, one can than solve for the other unknowns from the remaining equations:<\/p>\n

s = 0.96 mm\u00a0 (eq. iii),\u00a0\u00a0\u00a0\u00a0\u00a0G12x<\/sub> = 207.790 N\u00a0 (eq. ix),\u00a0\u00a0\u00a0\u00a0\u00a0G12y<\/sub>\u00a0= \u2212117.262 N (eq. x)\u00a0 \u00a0 \u00a0and\u00a0\u00a0\u00a0\u00a0\u00a0G12<\/sub> = 238.594 N (eq. xii)<\/p>\n

T12<\/sub>\u00a0= 47.148 N\u00b7m (CW)<\/p>\n

Since the friction effect is usually small, the above iterative method will converge to a solution in very few steps.<\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n

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

Example 2 In this second example the same centric slider-crank mechanism of the previous example is considered. However, in addition to\u00a0F14 two additional forces,\u00a0F13\u00a0and\u00a0F12, act on the links. F12 = 100 N , F13 = 150 N and F14 = …<\/p>\n

Example 2<\/span> Devam\u0131n\u0131 Oku »<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":2224,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"full-width-page.php","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-2738","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2738","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=2738"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2738\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2224"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=2738"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}