{"id":1290,"date":"2021-09-09T22:11:12","date_gmt":"2021-09-09T22:11:12","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=1290"},"modified":"2021-09-27T14:52:14","modified_gmt":"2021-09-27T14:52:14","slug":"5-3","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch5\/5-3\/","title":{"rendered":"5-3"},"content":{"rendered":"
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5.3 Mekanik Avantaj<\/h1>\n

\u0130lkokul y\u0131llar\u0131ndan itibaren bildi\u011fimiz basit levye prensibinde, \u015fekilde g\u00f6sterilen sistemde oldu\u011fu gibi, sistem dengede ise, d\u00f6nme ekseninde moment toplam\u0131 s\u0131f\u0131r olmas\u0131 gereklili\u011finden:<\/p>\n

F1<\/sub>l<\/em>1<\/sub> = F2<\/sub>l<\/em>2<\/sub>\u00a0 \u00a0 \u00a0veya\u00a0 \u00a0 F2<\/sub>\/F1<\/sub> = l<\/em>1<\/sub>\/l<\/em>2<\/sub> = Mekanik Avantaj<\/p>\n

olarak tan\u0131mlanmaktad\u0131r. Bu y\u00f6ntemle y\u00fcksek F2<\/sub>\u00a0kuvvetini s\u0131n\u0131rl\u0131 bir F1<\/sub>\u00a0kuvveti ile elde etmek ister isek l<\/em>1<\/sub>\/l<\/em>2<\/sub> oran\u0131n\u0131 y\u00fcksek se\u00e7memiz gerekecektir. Buna bir farkl\u0131 yakla\u015f\u0131m ise, kayb\u0131n olmad\u0131\u011f\u0131n\u0131 kabul ederek, enerji sak\u0131n\u0131m\u0131ndan birim zamanda yap\u0131lan i\u015f ile elde edilen i\u015fin e\u015fit olmas\u0131 \u015fart\u0131ndan: F1<\/sub>v1<\/sub>\u00a0= F2<\/sub>v2\u00a0<\/sub>olacakt\u0131r. Burada v1<\/sub>\u00a0ve v2<\/sub>\u00a0kuvvetlerin etki etti\u011fi noktada h\u0131z\u0131d\u0131r.<\/p>\n

\"\"<\/p>\n

Ayn\u0131 kavram\u0131 mekanizmalarda uygular isek, \u00f6rne\u011fin bir d\u00f6rt-\u00e7ubuk mekanizmas\u0131 i\u00e7in giri\u015f momenti T12<\/sub>\u00a0ve \u00e7\u0131k\u0131\u015f kolu direnci T14<\/sub>\u00a0ise, enerji sak\u0131n\u0131m\u0131ndan dolay\u0131 (kay\u0131plar ihmal edilerek):<\/p>\n

Giri\u015f G\u00fcc\u00fc = \u00c7\u0131k\u0131\u015f G\u00fcc\u00fc<\/p>\n

T12<\/sub>\u03c912<\/sub> = T14<\/sub>\u03c914<\/sub>\u00a0 \u00a0 \u00a0veya\u00a0 \u00a0 Mekanik Avantaj = ma<\/sub> = T14<\/sub>\/T12<\/sub> = \u03c912<\/sub>\/\u03c914<\/sub> = |I14<\/sub>I24<\/sub>|\/|I12<\/sub>I24<\/sub>|<\/p>\n

olacakt\u0131r. Dikkat edilir ise mekanik avantaj h\u0131z oran\u0131n\u0131n tersidir. Ayr\u0131ca, e\u011fer giri\u015f ve \u00e7\u0131k\u0131\u015f kollar\u0131na belirli bir kol boyunda bu kuvvetler uygulan\u0131yor ise:<\/p>\n

Mekanik Avantaj = ma<\/sub> = F14<\/sub>\/F12<\/sub> = (T14<\/sub>\/r14<\/sub>)\/(T12<\/sub>\/r12<\/sub>) = (\u03c912<\/sub>r12<\/sub>)\/(\u03c914<\/sub>r14<\/sub>) = (|I14<\/sub>I24<\/sub>|r12<\/sub>)\/(|I12<\/sub>I24<\/sub>|r14<\/sub>)<\/p>\n

olacakt\u0131r. I14<\/sub>I24<\/sub>\u00a0= B0<\/sub>I24<\/sub>\u00a0uzunlu\u011fu s\u0131f\u0131r olur ise mekanik avantaj s\u0131f\u0131r olacak, I12<\/sub>I24<\/sub> = A0<\/sub>I24<\/sub> uzunlu\u011fu s\u0131f\u0131r ise, mekanik avantaj sonsuz olacakt\u0131r. Tabi ki sistem elastikli\u011finden ve bo\u015fluklardan dolay\u0131 sonu\u00e7ta mekanik avantaj sonsuz olamaz. Ancak mekanik avantaj \u00e7ok y\u00fcksek de\u011ferlere ula\u015facakt\u0131r. Bu \u015fekilde basit levyelerle elde edilemeyecek kadar b\u00fcy\u00fck mekanik avantaj elde edilebilecektir. Nitekim preslerde, s\u0131k\u0131\u015ft\u0131rma penslerinde, ta\u015f k\u0131rma gibi y\u00fcksek g\u00fc\u00e7 isteyen her i\u015fte y\u00fcksek mekanik avantaj sa\u011flayan mekanizmalar kullan\u0131lmaktad\u0131r (a\u015fa\u011f\u0131daki \u015fekle bak\u0131n\u0131z).<\/p>\n

Sonucun daha da genelle\u015ftirilmesi m\u00fcmk\u00fcnd\u00fcr. 1 uzvuna ba\u011fl\u0131 i ve j uzuvlar\u0131 oldu\u011funda (i giri\u015f, j ise \u00e7\u0131k\u0131\u015f uzvu olacakt\u0131r), bu iki uzvun h\u0131z\u0131 aras\u0131nda (\u00e7ak\u0131\u015fan Iij<\/sub> ani d\u00f6nme merkezinde ba\u011f\u0131l h\u0131z s\u0131f\u0131r oldu\u011fundan) \u03c91i<\/sub>|I1i<\/sub>Iij<\/sub>| = \u03c91j<\/sub>|I1j<\/sub>Iij<\/sub>| ili\u015fkisinin oldu\u011fu belirlenmi\u015fti. \u00d6yle ise, enerjinin sak\u0131n\u0131m\u0131 prensibi kullan\u0131ld\u0131\u011f\u0131nda<\/p>\n

Mekanik Avantaj = ma<\/sub> = T1j<\/sub>\/T1i<\/sub>\u00a0= \u03c91i<\/sub>\/\u03c91j<\/sub>\u00a0= |I1j<\/sub>Iij<\/sub>|\/|I1i<\/sub>Iij<\/sub>|<\/p>\n

olacakt\u0131r (uzuvlara kuvvet uyguland\u0131\u011f\u0131nda denklemde uygun de\u011fi\u015fiklikler yap\u0131l\u0131r). Bu \u015fekilde, mekanizmalar\u0131n mekanik avantajlar\u0131n\u0131n belirlenmesi h\u0131z oranlar\u0131n\u0131n belirlenmesi problemine indirgenmi\u015ftir. Dikkat edilir ise, y\u00fcksek mekanik avantaj elde edilmesi i\u00e7in mekanizman\u0131n \u00f6l\u00fc konumlara yak\u0131n olmas\u0131 gerekecektir.<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\u00d6rnek:<\/strong><\/p>\n

\u015eekilde g\u00f6sterilen per\u00e7in makinas\u0131 i\u00e7in \u03b812<\/sub>\u00a0= 110\u00b0\u00a0iken F16<\/sub>\/T12<\/sub>\u00a0oran\u0131n\u0131 bulun.<\/p>\n

\"\"<\/p>\n

Mekanizma \u03b812<\/sub>\u00a0= 110\u00b0\u00a0konumunda \u00e7izilmi\u015ftir. S\u00fcrt\u00fcnme ihmal edildi\u011finde, enerjinin sak\u0131n\u0131m\u0131ndan dolay\u0131 :<\/p>\n

T12<\/sub>\u03c912<\/sub> = F16<\/sub>v16<\/sub><\/p>\n

yazabiliriz. 1, 2 ve 6 uzuvlar\u0131 aras\u0131nda bulunan ani d\u00f6nme merkezlerinden yararlanarak:<\/p>\n

v16<\/sub>\u00a0= |I12<\/sub>I26<\/sub>|\u03c912<\/sub><\/p>\n

olacakt\u0131r. \u00d6yle ise:<\/p>\n

F16<\/sub>\/T12<\/sub> = \u03c912<\/sub>\/v16<\/sub> = \u03c912<\/sub>\/(|I12<\/sub>I26<\/sub>|\u03c912<\/sub>) = 1\/|I12<\/sub>I26<\/sub>|<\/p>\n

dir. I12<\/sub>, A0<\/sub>\u00a0merkezidir. Bu durumda problem I26<\/sub>\u00a0ani d\u00f6nme merkezini bulunmas\u0131 ile \u00e7\u00f6z\u00fclecektir. \u015eekilde g\u00f6sterildi\u011fi gibi, mafsal eksenleri d\u00f6nme merkezleri olarak belirlendi\u011finde I26<\/sub>\u00a0ani d\u00f6nme merkezini bulmak i\u00e7in ilk olarak I24<\/sub>\u00a0ve I46<\/sub>\u00a0ani d\u00f6nme merkezlerini bulmam\u0131z gerekir. I24<\/sub>, I12<\/sub>I14<\/sub>\u00a0ile I32<\/sub>I34<\/sub>\u00a0do\u011frular\u0131n\u0131n kesim noktas\u0131 ve I46<\/sub>\u00a0ise I14<\/sub>I16<\/sub>\u00a0ile I45<\/sub>I56<\/sub>\u00a0do\u011frular\u0131n\u0131n kesim noktas\u0131 olarak belirlenir (kullan\u0131lan ani d\u00f6nme merkezleri mafsal eksenleridir). Bu iki ani d\u00f6nme merkezi belirlendikten sonra I26<\/sub>, I12<\/sub>I16<\/sub>\u00a0ile I24<\/sub>I46<\/sub>\u00a0do\u011frular\u0131n\u0131n kesim noktas\u0131d\u0131r. Dikkat edilir ise, I16<\/sub>\u00a0sonsuzda oldu\u011fundan I12<\/sub>I16<\/sub>\u00a0do\u011frusu A0<\/sub>\u00a0dan \u00e7izilen 1, 6 uzuvlar\u0131 aras\u0131ndaki kayar mafsala dik bir do\u011frudur. I46<\/sub>\u00a0ise BC do\u011frusu (bu do\u011fru I45<\/sub>I56<\/sub>\u00a0do\u011frusudur) ile B0<\/sub>\u00a0dan kayar mafsal eksenine dik do\u011frudur (I45<\/sub>I56<\/sub>\u00a0do\u011frusu). BC ile B0<\/sub>C do\u011frular\u0131 \u00e7ak\u0131\u015ft\u0131\u011f\u0131nda B0<\/sub>\u00a0(I14<\/sub>) ile I26<\/sub>\u00a0\u00e7ak\u0131\u015facakt\u0131r. Bu durumda I26<\/sub>\u00a0ani d\u00f6nme merkezi ise, A0<\/sub>\u00a0(I12<\/sub>) ile \u00e7ak\u0131\u015f\u0131r. Ayn\u0131 durum A0<\/sub>A ile AB do\u011frular\u0131n\u0131n \u00e7ak\u0131\u015fmas\u0131 durumunda s\u00f6z konusudur. Bu durumda I24<\/sub>, A0<\/sub>\u00a0noktas\u0131 ile \u00e7ak\u0131\u015facakt\u0131r. Bu durumda per\u00e7inleme s\u0131ras\u0131nda I12<\/sub>I26<\/sub>\u00a0mesafesi \u00e7ok k\u0131sa olaca\u011f\u0131ndan F16<\/sub>\/T12<\/sub>\u00a0oran\u0131 b\u00fcy\u00fck de\u011ferler alacakt\u0131r.<\/p>\n

\"\"<\/p>\n

AutoCad k\u00fct\u00fc\u011f\u00fc i\u00e7in t\u0131klay\u0131n\u0131z: Mekanikpres.dwg<\/a>.<\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n

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

5.3 Mekanik Avantaj \u0130lkokul y\u0131llar\u0131ndan itibaren bildi\u011fimiz basit levye prensibinde, \u015fekilde g\u00f6sterilen sistemde oldu\u011fu gibi, sistem dengede ise, d\u00f6nme ekseninde moment toplam\u0131 s\u0131f\u0131r olmas\u0131 gereklili\u011finden: F1l1 = F2l2\u00a0 \u00a0 \u00a0veya\u00a0 \u00a0 F2\/F1 = l1\/l2 = Mekanik Avantaj olarak tan\u0131mlanmaktad\u0131r. Bu …<\/p>\n

5-3<\/span> Devam\u0131n\u0131 Oku »<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":1285,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-1290","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1290","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=1290"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1290\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1285"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=1290"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}