{"id":1484,"date":"2021-09-11T18:42:06","date_gmt":"2021-09-11T18:42:06","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=1484"},"modified":"2021-09-27T09:12:28","modified_gmt":"2021-09-27T09:12:28","slug":"3-4_ornek","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch3\/3-4\/3-4_ornek\/","title":{"rendered":"3-4_ornek"},"content":{"rendered":"
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\u00d6rnek: Alt\u0131 Uzuvlu Bir Mekanizmadaki Devreler<\/strong><\/h1>\n

\u015eekil 1-A da 6 uzuvlu bir mekanizma g\u00f6r\u00fclmektedir. Mekanizmada hi\u00e7 bir kapal\u0131 devre kalmamas\u0131 i\u00e7in bu sefer iki mafsal\u0131n s\u00f6k\u00fclmesi gereklidir. S\u00f6k\u00fclmesi gereken mafsal ki\u015fiden ki\u015fiye farkl\u0131 olabilir. Ancak devre kalmayacak \u015fekilde s\u00f6k\u00fclmesi gereken minimum mafsal say\u0131s\u0131 bir mekanizma i\u00e7in sabittir. Ayr\u0131ca mafsallar s\u00f6k\u00fcl\u00fcr iken uzuvlar\u0131n zincirden ayr\u0131lmamas\u0131 gereklidir. C ve B noktalar\u0131nda bulunan mafsallar s\u00f6k\u00fcld\u00fc\u011f\u00fcnde (\u015eekil 1-B), mekanizmada devre b\u0131rak\u0131lmam\u0131\u015f, a\u00e7\u0131k zincirler olu\u015fturulmu\u015ftur (istenir ise D ve E veya D ve A gibi iki farkl\u0131 mafsal s\u00f6k\u00fclebilir. Ancak \u00f6rne\u011fin B ve D mafsal\u0131 s\u00f6k\u00fclemez, \u00e7\u00fcnk\u00fc bu durumda bir uzuv zincirden kopar\u0131lm\u0131\u015f olur). C noktas\u0131n\u0131n konumunu 1, 2, 3 ve 1, 5, 4 uzuvlar\u0131ndan olu\u015fan a\u00e7\u0131k kinematik zincirleri kullanarak yazmak istedi\u011fimizde (B mafsal\u0131 s\u00f6k\u00fclm\u00fc\u015f durumda), gerekli konum parametreleri tan\u0131mlayarak elde etti\u011fimiz vekt\u00f6r devre denklemi:<\/p>\n

Ao<\/sub>A<\/strong> +\u00a0AC<\/strong> = Ao<\/sub>Do<\/sub><\/strong>\u00a0+ Do<\/sub>E<\/strong> +\u00a0EC<\/strong><\/p>\n

olur. Karma\u015f\u0131k say\u0131larla vekt\u00f6r devre denklemi yaz\u0131ld\u0131\u011f\u0131nda:<\/p>\n

a2<\/sub>ei\u03b812<\/sub><\/sup> + a3<\/sub>ei\u03b813<\/sub><\/sup>\u00a0= a1<\/sub> + a5<\/sub>ei\u03b815<\/sub><\/sup>\u00a0+ a4<\/sub>ei\u03b814<\/sub><\/sup><\/p>\n

\"\"<\/p>\n

(a1<\/sub> = |Ao<\/sub>Do<\/sub>|, a2<\/sub> = |Ao<\/sub>A|, a3<\/sub> = |AC|, b1<\/sub> = |AB|, c3<\/sub> = |CB|, a4<\/sub> = |EC|, a5<\/sub> = |Do<\/sub>E|, b5<\/sub> = |Do<\/sub>D|, c5<\/sub> = |ED|, a6<\/sub> = |BD|)<\/p>\n

B noktas\u0131n\u0131n konumu vekt\u00f6r\u00fc, benzer bir \u015fekilde 1, 2, 3 ve 1, 5, 6 uzuvlar\u0131ndan olu\u015fan a\u00e7\u0131k zincirleri kullanarak (C mafsal\u0131 a\u00e7\u0131k oldu\u011fu varsay\u0131larak), yaz\u0131ld\u0131\u011f\u0131nda elde edilen vekt\u00f6r devre denklemi ise<\/p>\n

A0<\/sub>A<\/strong> +\u00a0AB<\/strong> = A0<\/sub>D0<\/sub><\/strong>\u00a0+ D0<\/sub>D<\/strong> +\u00a0DB<\/strong><\/p>\n

olacakt\u0131r. Karma\u015f\u0131k say\u0131lar kullan\u0131larak yaz\u0131ld\u0131\u011f\u0131nda:<\/p>\n

a2<\/sub>ei\u03b812<\/sub><\/sup> + b3<\/sub>ei(\u03b813<\/sub> + \u03b13<\/sub>)<\/sup>\u00a0= a1<\/sub> + b5<\/sub>ei(\u03b815 <\/sub>+\u00a0<\/sub>\u03b15<\/sub>)<\/sup>\u00a0+ a6<\/sub>ei\u03b816<\/sub><\/sup><\/p>\n

Dikkat edilir ise, 3 ve 5 uzuvlar\u0131n\u0131n konumlar\u0131 birinci ve ikinci denklemlerde farkl\u0131 vekt\u00f6rlerle belirlenmektedir ( 3 uzvu i\u00e7in\u00a0AC<\/strong>\u00a0ve\u00a0AB<\/strong>\u00a0vekt\u00f6rleri, 5 uzvu i\u00e7in\u00a0D0<\/sub>E<\/strong>\u00a0ve\u00a0D0<\/sub>D<\/strong> vekt\u00f6rleri). Ancak bu vekt\u00f6rlerin \u015fiddetleri yine uzuv boyutlar\u0131 olup ayn\u0131 uzuv \u00fczerinde bulunan iki vekt\u00f6r\u00fcn birbirleri aras\u0131nda sabit bir a\u00e7\u0131 (\u03b13<\/sub>\u00a0ve \u03b15<\/sub>) bulunmaktad\u0131r. Bu a\u00e7\u0131lar o uzuvlar\u0131n boyutlar\u0131 ile tan\u0131ml\u0131d\u0131r. Bu nedenle her bir uzuv i\u00e7in sadece bir a\u00e7\u0131sal konum parametresi (\u03b813<\/sub> ve \u03b815<\/sub>) tan\u0131mlanm\u0131\u015ft\u0131r.<\/p>\n

Yazm\u0131\u015f oldu\u011fumuz iki devre kapal\u0131l\u0131k denkleminde be\u015f konum de\u011fi\u015fkeni bulunmaktad\u0131r (\u03b812<\/sub>, \u03b813<\/sub>, \u03b814<\/sub>, \u03b815<\/sub> ve \u03b816<\/sub>). Bu konum de\u011fi\u015fkenlerinden birisi ba\u011f\u0131ms\u0131z de\u011fi\u015fken olarak belirlenir ise, iki vekt\u00f6r devre denkleminden kalan di\u011fer d\u00f6rt de\u011fi\u015fkenin de\u011feri bulunabilir.<\/p>\n

E\u011fer B mafsal\u0131n\u0131n ba\u011flanmas\u0131n\u0131 C mafsal\u0131 ba\u011fland\u0131ktan sonra yapar isek bu durumda B mafsal\u0131 ile kapataca\u011f\u0131m\u0131z devre ECBDE devresi olacakt\u0131r (4-3 ve 5-6 a\u00e7\u0131k zincirleri) (\u00f6nceki durumda B mafsal\u0131 a\u00e7\u0131k oldu\u011fundan B mafsal\u0131 ile A0<\/sub>ABDD0<\/sub>A0<\/sub> devresi olu\u015fmu\u015ftu). Bu yeni durumda bu yeni devre i\u00e7in devre kapal\u0131l\u0131k denklemi<\/p>\n

EC<\/strong> +\u00a0CB<\/strong> = ED<\/strong> + DB<\/strong><\/p>\n

olacakt\u0131r. Karma\u015f\u0131k say\u0131lar ile bu denklem:<\/p>\n

a4<\/sub>ei\u03b814<\/sub><\/sup> + c3<\/sub>ei(\u03b813<\/sub> + \u03b23<\/sub>)<\/sup> = c5<\/sub>ei(\u03b815 <\/sub>+\u00a0<\/sub>\u03b25<\/sub>)<\/sup>\u00a0+ a6<\/sub>ei\u03b816<\/sub><\/sup><\/p>\n

yaz\u0131labilir. Her her \u00fc\u00e7 devre kapal\u0131l\u0131k denklemi ele al\u0131n\u0131r ise, elde edilen skaler denklem say\u0131s\u0131 bilinmeyen parametre say\u0131s\u0131ndan fazla olacakt\u0131r. Bu denklemleri i\u00e7in \u00e7\u00f6z\u00fcm ancak bilinmeyen parametre say\u0131s\u0131 kadar ba\u011f\u0131ms\u0131z skaler denklem olmas\u0131 ile m\u00fcmk\u00fcnd\u00fcr. Nitekim bu \u00f6rnek i\u00e7in elde edilmi\u015f olan birinci ve \u00fc\u00e7\u00fcnc\u00fc vekt\u00f6r devre denklemlerini taraf tarafa toplad\u0131\u011f\u0131m\u0131zda:<\/p>\n

A0<\/sub>A <\/strong>+ AC <\/strong>+ EC <\/strong>+ CB <\/strong>= D0<\/sub>E<\/strong> + EC <\/strong>+ ED <\/strong>+ DB<\/strong><\/p>\n

denklemi \u00e7\u0131kacakt\u0131r. Her iki tarafta bulunan EC<\/strong> vekt\u00f6r\u00fc gider. \u015eekil 2 de g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi ayn\u0131 uzuv \u00fczerinde bulunan \u00fc\u00e7 vekt\u00f6r uzvun \u03b8 ile belirlenen konumuna ba\u011f\u0131ml\u0131 kalmaks\u0131z\u0131n aralar\u0131nda sabit bir ili\u015fkiyi korurlar. A, B, C noktalar\u0131 ayn\u0131 uzuv \u00fczerinde oldu\u011funda:<\/p>\n

\"\"<\/p>\n

Ayn\u0131 uzuv \u00fczerinde bulunan ABC Noktalar\u0131<\/span><\/p>\n

b = |AB| , a = |AC|, c = |CB|<\/span><\/p>\n

AB<\/b> = bei(\u03b8 + \u03b1)<\/sup> , AC<\/b> = aei\u03b8<\/sup> , CB<\/b> = cei(\u03b8 + \u03b2)<\/sup><\/span><\/p>\n

AC <\/span><\/strong>+ CB<\/strong> = AB<\/strong> (Daima sa\u011flanacak olan vekt\u00f6r denklemi)<\/span><\/p>\n

Karma\u015f\u0131k say\u0131lar ile yazd\u0131\u011f\u0131m\u0131zda:<\/span><\/p>\n

aei\u03b8<\/sup>\u00a0<\/span>+ cei(\u03b8 + \u03b2)<\/sup> = (a + cei\u03b2<\/sup>)ei\u03b8<\/sup> = bei(\u03b8 + \u03b1)<\/sup> = bei\u03b1<\/sup>ei\u03b8<\/sup><\/span><\/p>\n

\u21d2\u00a0 a + cei\u03b2<\/sup> = bei\u03b1<\/sup><\/span><\/p>\n

Dikkat edilirse, denklemde de\u011fi\u015fken a\u00e7\u0131 olan ve uzvun konumunu belirleyen \u03b8 a\u00e7\u0131s\u0131 de\u011feri ne olursa olsun denklem sa\u011flanmaktad\u0131r.\u00a0<\/span><\/p>\n

\u00d6yle ise bu de\u011ferlendirme, \u00f6rne\u011fimizde 3 uzvu i\u00e7in\u00a0AC<\/strong> + CB<\/strong> =\u00a0AB<\/strong>\u00a0ve 5 uzvu i\u00e7in\u00a0D0<\/sub>E <\/strong>+\u00a0ED <\/strong>=\u00a0D0<\/sub>D<\/strong>\u00a0sonucunu verecektir. B\u00f6ylece birinci ve \u00fc\u00e7\u00fcnc\u00fc vekt\u00f6r devre denklemleri topland\u0131\u011f\u0131nda ikinci vekt\u00f6r devre denklemi elde edilmi\u015ftir. \u00d6yle ise bu \u00fc\u00e7 vekt\u00f6r devre denkleminden birisi di\u011fer iki denkleme ba\u011f\u0131ml\u0131d\u0131r. Bu nedenle elde edilmi\u015f olan \u00fc\u00e7 devre denkleminden sadece ikisi konum de\u011fi\u015fkenlerinin \u00e7\u00f6z\u00fcm\u00fc i\u00e7in kullan\u0131labilir.<\/p>\n<\/div>\n<\/div><\/div><\/div><\/div>

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A0<\/sub>A<\/strong> + AC<\/strong> + CE<\/strong> = A0<\/sub>D0<\/sub><\/strong>\u00a0+ D0<\/sub>E<\/strong><\/p>\n

A0<\/sub>A<\/strong> + AB<\/strong> + BD<\/strong> = A0<\/sub>D0<\/sub><\/strong>\u00a0+ D0<\/sub>D<\/strong><\/p>\n

a2<\/sub>ei\u03b812<\/sub><\/sup> + a3<\/sub>ei\u03b813<\/sub><\/sup> + a4<\/sub>ei\u03b814<\/sub><\/sup> = a1<\/sub> + a5<\/sub>ei\u03b815<\/sub><\/sup><\/p>\n

a2<\/sub>ei\u03b812<\/sub><\/sup> + b3<\/sub>ei(\u03b813 <\/sub>+\u00a0<\/sub>\u03b13<\/sub>)<\/sup> + a6<\/sub>ei\u03b816<\/sub><\/sup> = a1<\/sub> + b5<\/sub>ei(\u03b815 <\/sub>+\u00a0<\/sub>\u03b15<\/sub>)<\/sup><\/p>\n

 <\/p>\n<\/div>\n<\/div><\/div><\/div>

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\n\t\t\"\"\n\t<\/div>\n\n<\/div><\/div><\/div><\/div>
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A ve E gibi farkl\u0131 mafsallar s\u00f6k\u00fclm\u00fc\u015f kabul edilebilir (ve sonra birle\u015ftirilerek devre denklemi elde edilir). Mafsallar s\u00f6k\u00fcl\u00fcrken hi\u00e7bir uzvun zincirden kopmad\u0131\u011f\u0131na (en az bir mafsal ile bir ba\u015fka uzva ba\u011fl\u0131 oldu\u011funa) dikkat edilmelidir. Farkl\u0131 mafsallar s\u00f6k\u00fcld\u00fc\u011f\u00fcnde farkl\u0131 parametreler tan\u0131mlanabilir. Ancak her \u015fartta denklem say\u0131s\u0131 ve konum parametresi say\u0131s\u0131 bir mekanizma i\u00e7in sabit kalacakt\u0131r. Farkl\u0131 tan\u0131mlar kullan\u0131ld\u0131\u011f\u0131nda konum parametreleri aras\u0131nda sabit a\u00e7\u0131 farklar\u0131 olu\u015fabilir. Bu \u00f6rnekte B ve C mafsallar\u0131 yerine D ve E mafsallar\u0131 s\u00f6k\u00fcld\u00fc\u011f\u00fcnde \u03b814<\/sub> ve \u03b816<\/sub> a\u00e7\u0131 tan\u0131mlar\u0131 \u03c0 kadar fark etmi\u015ftir.<\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n

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

\u00d6rnek: Alt\u0131 Uzuvlu Bir Mekanizmadaki Devreler \u015eekil 1-A da 6 uzuvlu bir mekanizma g\u00f6r\u00fclmektedir. Mekanizmada hi\u00e7 bir kapal\u0131 devre kalmamas\u0131 i\u00e7in bu sefer iki mafsal\u0131n s\u00f6k\u00fclmesi gereklidir. S\u00f6k\u00fclmesi gereken mafsal ki\u015fiden ki\u015fiye farkl\u0131 olabilir. Ancak devre kalmayacak \u015fekilde s\u00f6k\u00fclmesi gereken …<\/p>\n

3-4_ornek<\/span> Devam\u0131n\u0131 Oku »<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":826,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-1484","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1484","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=1484"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1484\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/826"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=1484"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}