![\"neden](\"https:\/\/cn-hawe.com\/wp-content\/uploads\/2026\/03\/why-do-your-parts-consistently-measure-oversized_w1200.jpg\")
<\/figure>\n\n\n\nO 1.000\u2033 V kal\u0131b\u0131 ve 10 numara \u00e7eli\u011fe d\u00f6nelim. Ka\u011f\u0131t \u00fczerinde, kal\u0131nl\u0131\u011f\u0131 \u00e7\u0131kar, ikiye b\u00f6l ve yar\u0131\u00e7ap\u0131n\u0131 elde et. Tertemiz. Tatmin edici.<\/p>\n\n\n\n
\u015eimdi havada b\u00fck.<\/p>\n\n\n\n
Z\u0131mba, sac\u0131 kal\u0131b\u0131n omuzlar\u0131 aras\u0131na iter ama malzemeyi asla tamamen V i\u00e7ine zorlamaz. \u0130\u00e7 yar\u0131\u00e7ap, malzeme z\u0131mba burnunun \u00fczerinden gerilirken ve o omuzlar aras\u0131nda s\u00fcz\u00fcl\u00fcrken olu\u015fur. Son a\u00e7\u0131 tamamen z\u0131mban\u0131n ne kadar derine indi\u011fine ba\u011fl\u0131d\u0131r. Birka\u00e7 binde bir fazla strok mu? Daha dar a\u00e7\u0131. Birka\u00e7 binde bir eksik mi? Daha a\u00e7\u0131k.<\/p>\n\n\n\n
Bu, her ayarda temas geometrisinin de\u011fi\u015fti\u011fi anlam\u0131na gelir.<\/p>\n\n\n\n
Form\u00fcl, yar\u0131\u00e7ap\u0131 kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131n\u0131n belirledi\u011fini varsayar. Hava b\u00fck\u00fcmde yar\u0131\u00e7ap\u0131 makinenin strok konumu belirler. Bunlar ayn\u0131 model de\u011fildir. Biri geometridir. Di\u011feri ise harekettir.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: E\u011fer pres b\u00fck\u00fcm makinen strok tekrarlamas\u0131n\u0131 birka\u00e7 binde bir dahilinde tutam\u0131yorsa, statik bir yar\u0131\u00e7ap denklemi seni tam olarak nas\u0131l kurtarabilir?<\/p>\n\n\n\n
Peki strok kontroll\u00fc bir s\u00fcreci kal\u0131p kontroll\u00fcym\u00fc\u015f gibi ele ald\u0131\u011f\u0131m\u0131zda ne olur?<\/p>\n\n\n\n
Hava b\u00fck\u00fcm s\u00fcrecini oturtmal\u0131 b\u00fck\u00fcm olarak ele alman\u0131n tehlikesi<\/h3>\n\n\n\n![\"Hava](\"https:\/\/cn-hawe.com\/wp-content\/uploads\/2026\/03\/The-danger-of-treating-an-air-bending-process-like-a-bottoming-operation_w1200.jpg\")
<\/figure>\n\n\n\nOturtmal\u0131 b\u00fck\u00fcm, kaba kuvvettir. Malzemeyi kal\u0131ba sokars\u0131n, ta ki tam uyana kadar. Geri yaylanma, hava b\u00fck\u00fcmden \u00fc\u00e7 ila be\u015f kat fazla tonaj uygulad\u0131\u011f\u0131n i\u00e7in ezilerek yok olur. A\u00e7\u0131y\u0131 kal\u0131p belirler. Yar\u0131\u00e7ap fiilen damgalanm\u0131\u015f olur.<\/p>\n\n\n\n
Form\u00fcl\u00fcn ait oldu\u011fu d\u00fcnya bu.<\/p>\n\n\n\n
Ama bir hava b\u00fckme kurulumunda \u201cmatemati\u011fi do\u011fru yapmaya\u201d \u00e7al\u0131\u015f\u0131rken pres tonaj\u0131n\u0131 art\u0131r\u0131rsan\u0131z, kal\u0131plar\u0131 \u00e7atlatabilecek veya ko\u00e7u zorlayabilecek a\u015f\u0131r\u0131 tonajla fl\u00f6rt ediyorsunuz demektir. Operat\u00f6rlerin, bir bijon somununu s\u0131kar gibi yaylanmay\u0131 dengelemek i\u00e7in bas\u0131n\u00e7 ekledi\u011fini g\u00f6rd\u00fcm. Par\u00e7a do\u011fru a\u00e7\u0131y\u0131 tutturabilir. Ancak tak\u0131m, bunun bedelini daha sonra \u00f6der.<\/p>\n\n\n\n
Hava b\u00fckme, o \u015fiddetten ka\u00e7\u0131nmak i\u00e7in tasarlanm\u0131\u015ft\u0131r. Kaba kesinli\u011fi esneklik ve daha d\u00fc\u015f\u00fck tonajla de\u011fi\u015ftirir. Sadece strok derinli\u011fini de\u011fi\u015ftirerek ayn\u0131 kal\u0131pta birden fazla a\u00e7\u0131 \u00e7al\u0131\u015fabilirsiniz. Bu \u00e7ok y\u00f6nl\u00fcl\u00fck, \u00e7o\u011fu at\u00f6lyenin neredeyse her \u015fey i\u00e7in onu kullanmas\u0131n\u0131n nedenidir.<\/p>\n\n\n\n
Yine de, art\u0131k kullanmad\u0131\u011f\u0131m\u0131z bir i\u015flem i\u00e7in tasarlanm\u0131\u015f bir form\u00fcle h\u00e2l\u00e2 tutunuyoruz.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Vuru derinli\u011fini dinamik olarak ayarlamak m\u0131 daha ucuz, yoksa kitab\u00ee ger\u00e7e\u011fi zorlamak isteyip \u00e7atlak bir kal\u0131b\u0131 de\u011fi\u015ftirmek mi?<\/p>\n\n\n\n
Matematik yanl\u0131\u015f de\u011filse ama model yanl\u0131\u015fsa, bunun uzun vadede size maliyeti nedir?<\/p>\n\n\n\n
Da\u011f\u0131n\u0131k bir at\u00f6lyede kusursuz bir form\u00fcle inanman\u0131n gizli maliyeti<\/h3>\n\n\n\n![\"Da\u011f\u0131n\u0131k](\"https:\/\/cn-hawe.com\/wp-content\/uploads\/2026\/03\/The-hidden-cost-of-believing-a-clean-formula-in-a-messy-shop-environment_w1200.jpg\")
<\/figure>\n\n\n\nForm\u00fcl\u00fcn g\u00f6rmedi\u011fi \u015fey \u015fu: rulolar aras\u0131ndaki \u00e7ekme dayan\u0131m\u0131 fark\u0131, hafif bombe yapm\u0131\u015f bir yatak, a\u015f\u0131nm\u0131\u015f z\u0131mba u\u00e7lar\u0131, s\u0131cak bir \u00f6\u011fleden sonra hidrolik kaymas\u0131. Bunlar\u0131n hi\u00e7biri (V \u2212 MT) \/ 2 denkleminde g\u00f6r\u00fcnmez.<\/p>\n\n\n\n
Ama par\u00e7alar\u0131n\u0131zda g\u00f6r\u00fcn\u00fcr.<\/p>\n\n\n\n
Hava b\u00fckme, pres frenin kendisini kritik de\u011fi\u015fken h\u00e2line getirir. Vuru hassasiyeti. Ko\u00e7 paralelli\u011fi. Arka dayama tutarl\u0131l\u0131\u011f\u0131. Alt b\u00fckmede a\u011f\u0131r i\u015fi kal\u0131p yapar. Hava b\u00fckmede makine yapar. Daha temiz bir form\u00fcl, tolerans dahilinde tekrar edemeyen bir makineyi d\u00fczeltmez. Bu y\u00fczden bir\u00e7ok at\u00f6lye, y\u00fcksek hassasiyetli b\u00fck\u00fcm ve otomasyon i\u00e7in tasarlanm\u0131\u015f tamamen CNC kontroll\u00fc sistemlere y\u00f6nelir\u2014\u015funlardaki gibi CN-HAWE pres b\u00fckme serisi<\/a>\u2014burada vuru derinli\u011fi, paralellik ve tekrarlanabilirli\u011fin kontrol\u00fc tahmini de\u011fil, \u00f6l\u00e7\u00fclebilir bir yetenek h\u00e2line gelir.<\/p>\n\n\n\nGizli maliyet sadece hurda de\u011fildir. Bu s\u00fcre\u00e7 i\u00e7in asla tasarlanmam\u0131\u015f say\u0131lar\u0131 kovalamak i\u00e7in harcanan zamand\u0131r. Bu, sahte bir \u00f6zg\u00fcvendir. Ger\u00e7ek sorun dinamik bir sistemde sabit bir denklemi kullanmakken \u201ck\u00f6t\u00fc b\u00fck\u00fcmler\u201d i\u00e7in operat\u00f6rleri su\u00e7lamakt\u0131r.<\/p>\n\n\n\n
200 tonluk bir pres, cebirinizin ne kadar zarif g\u00f6r\u00fcnd\u00fc\u011f\u00fcyle ilgilenmez. Onun tek \u00f6nemsedi\u011fi \u015fey, z\u0131mban\u0131n nerede durdu\u011fudur.<\/p>\n\n\n\n
Yani kal\u0131p, hava b\u00fckmede yar\u0131\u00e7ap\u0131 ger\u00e7ekten tan\u0131mlam\u0131yorsa, o yar\u0131\u00e7ap asl\u0131nda nereden geliyor?<\/p>\n\n\n\n
V-Kal\u0131\u011f\u0131n Fizi\u011fi: Yar\u0131\u00e7ap Asl\u0131nda Nerede Olu\u015fur<\/h2>\n\n\n\n
Ayn\u0131 1.000\u2033 V-kal\u0131\u011f\u0131 ve 10 numara 60 KSI so\u011fuk haddelenmi\u015f \u00e7eli\u011fi al\u0131n. Hava b\u00fckme ile b\u00fck\u00fclm\u00fc\u015f on par\u00e7a boyunca i\u00e7 yar\u0131\u00e7ap\u0131 \u00f6l\u00e7\u00fcn. 0.433\u2033 elde edemezsiniz. Z\u0131mba ucu yar\u0131\u00e7ap\u0131n\u0131z\u0131 da elde edemezsiniz. Yakla\u015f\u0131k 0.160\u2033 ile 0.200\u2033 aras\u0131nda bir de\u011fere ula\u015f\u0131rs\u0131n\u0131z\u2014kaba olarak kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131n\u0131n \u201320\u2019si civar\u0131nda.<\/p>\n\n\n\n
Bu oran o kadar s\u0131k g\u00f6r\u00fcl\u00fcr ki art\u0131k tesad\u00fcf olmaktan \u00e7\u0131kar.<\/p>\n\n\n\n
Kal\u0131p sihirli bir \u015fekilde o say\u0131y\u0131 damgalad\u0131\u011f\u0131 i\u00e7in de\u011fil. Z\u0131mba burnu o de\u011fere uydu\u011fu i\u00e7in de de\u011fil. Levha kal\u0131p omuzlar\u0131 aras\u0131nda \u00e7\u00f6kerken, y\u00fck alt\u0131nda o a\u00e7\u0131kl\u0131\u011f\u0131 ne kadar a\u015fabilece\u011fine ba\u011fl\u0131 do\u011fal bir e\u011frilik buldu\u011fu i\u00e7in. Kal\u0131p geni\u015fli\u011fi sahneyi haz\u0131rlar. Malzeme gerilimi, bu mesafe boyunca ne kadar s\u0131k\u0131 b\u00fck\u00fclebilece\u011fini belirler. Yar\u0131\u00e7ap, omuzlar aras\u0131nda havada olu\u015fur, n\u00fcfuz derinli\u011fi taraf\u0131ndan kontrol edilir.<\/p>\n\n\n\n
Hava b\u00fckmede, temel i\u00e7 yar\u0131\u00e7ap\u0131 belirleyen \u015fey z\u0131mba ucu veya hesap makineniz de\u011fil, kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131d\u0131r. Geri kalan her \u015fey bunun \u00fczerine in\u015fa edilir.<\/p>\n\n\n\n
E\u011fer (V \u2212 MT)\/2\u2019nin pe\u015findeyseniz, yanl\u0131\u015f mekanizmay\u0131 \u00e7\u00f6z\u00fcyorsunuz demektir.<\/p>\n\n\n\n
Ger\u00e7ek temel \u00e7izgiyi hesaplamak: 8x ila 12x kal\u0131p oran\u0131 ger\u00e7eklik kontrol\u00fc<\/h3>\n\n\n\n
Yakla\u015f\u0131k 1\u2033 V kal\u0131b\u0131n alt\u0131nda, 10-gauge bir ba\u011flant\u0131 par\u00e7as\u0131n\u0131n \u00fczerinde durdum, \u00e7\u00fcnk\u00fc 0.135\u2033 kal\u0131nl\u0131\u011f\u0131ndaki 10\u2011gauge \u00e7elik 8 ile \u00e7arp\u0131l\u0131nca yakla\u015f\u0131k 1.08\u2033 eder. Yeterince yak\u0131n. O 8\u00d7 kural\u0131 bir efsane de\u011fildir. Tona makul tutar ve genellikle sizi kararl\u0131 bir b\u00fckme aral\u0131\u011f\u0131na yerle\u015ftirir.<\/p>\n\n\n\n
Matemati\u011fi geriye de\u011fil, ileriye do\u011fru \u00e7al\u0131\u015ft\u0131r\u0131n.<\/p>\n\n\n\n
E\u011fer o kal\u0131nl\u0131k i\u00e7in 1.000\u2033 V yakla\u015f\u0131k de\u011fer aral\u0131\u011f\u0131ndaysa ve hava b\u00fckme, 60\u2011KSI \u00e7elik i\u00e7in V\u2019nin yakla\u015f\u0131k \u201320\u2019si kadar bir i\u00e7 yar\u0131\u00e7ap \u00fcretiyorsa, daha ko\u00e7 bile metale dokunmadan 0.160\u2033\u20130.200\u2033 yar\u0131\u00e7ap tahmin ediyorsunuz demektir. Bu zaten (V \u2212 MT)\/2\u2019ye hi\u00e7 yak\u0131n de\u011fildir.<\/p>\n\n\n\n
\u015eimdi kal\u0131b\u0131 0.800\u2033e s\u0131k\u0131\u015ft\u0131r\u0131n\u2014yakla\u015f\u0131k 6\u00d7 kal\u0131nl\u0131k. Do\u011fal yar\u0131\u00e7ap\u0131n\u0131z yakla\u015f\u0131k 0.128\u2033\u20130.160\u2033\u2019a d\u00fc\u015fer. 1.250\u2033e a\u00e7\u0131n\u2014yakla\u015f\u0131k 10\u00d7\u2014ve yakla\u015f\u0131k 0.200\u2033\u20130.250\u2033 civar\u0131nda dola\u015f\u0131rs\u0131n\u0131z.<\/p>\n\n\n\n
Oran (8\u00d7, 10\u00d7, 12\u00d7) zarafetle ilgili de\u011fildir. Ayn\u0131 anda iki \u015feyi kontrol etmekle ilgilidir: ayak ba\u015f\u0131na tonaj ve bu a\u00e7\u0131kl\u0131ktan \u00e7\u0131kacak y\u00fczdeye dayal\u0131 yar\u0131\u00e7ap. Fazla dar olursa tonaj h\u0131zla artar. Fazla geni\u015f olursa, bask\u0131 ne isterse istesin yar\u0131\u00e7ap\u0131n\u0131z \u015fi\u015fer.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Hi\u00e7 \u201ctonajdan emin olmak i\u00e7in\u201d kal\u0131b\u0131 a\u00e7t\u0131n\u0131z m\u0131 ve sonra flan\u015f\u0131n\u0131z\u0131n neden e\u015f par\u00e7aya uymad\u0131\u011f\u0131n\u0131 merak ettiniz mi? Bu operat\u00f6r hatas\u0131 de\u011fildi. Bu, geometriydi; siz istemeseniz de yar\u0131\u00e7ap\u0131n\u0131z\u0131 \u00f6l\u00e7eklendirdi.<\/p>\n\n\n\n
Yani kal\u0131p geni\u015fli\u011fi temel y\u00fczdeyi belirliyorsa, malzemenin kendisi \u00f6nceki i\u015finizden daha zor\u2014ya da daha kolay\u2014b\u00fck\u00fcl\u00fcyorsa ne olur?<\/p>\n\n\n\n
\u00c7arpan kaymas\u0131: Paslanmaz \u00e7elik ile yumu\u015fak al\u00fcminyum i\u00e7in matemati\u011fi nas\u0131l ayarlamal\u0131<\/h3>\n\n\n\n
O 60\u2011KSI so\u011fuk haddelenmi\u015f \u00e7eli\u011fi ayn\u0131 1.000\u2033 V i\u00e7inde 304 paslanmazla de\u011fi\u015ftirin. Ayn\u0131 kal\u0131nl\u0131k. Ayn\u0131 z\u0131mba. 90\u00b0\u2019yi elde etmek i\u00e7in ayn\u0131 programlanm\u0131\u015f strok derinli\u011fi.<\/p>\n\n\n\n
Yar\u0131\u00e7ap\u0131 \u00f6l\u00e7\u00fcn.<\/p>\n\n\n\n
A\u00e7\u0131l\u0131yor.<\/p>\n\n\n\n
Paslanmaz \u00e7elik daha y\u00fcksek \u00e7ekme dayan\u0131m\u0131 ta\u015f\u0131r ve agresif bi\u00e7imde i\u015f sertle\u015fmesi g\u00f6sterir. Z\u0131mba a\u015fa\u011f\u0131ya do\u011fru bast\u0131\u011f\u0131nda, malzeme e\u011frili\u011fe yumu\u015fak \u00e7eli\u011fe g\u00f6re daha fazla diren\u00e7 g\u00f6sterir. O \u2019l\u0131k alt banda o kadar kolay sar\u0131lmaz. \u2019ye veya biraz \u00fcst\u00fcne kayd\u0131\u011f\u0131n\u0131 g\u00f6rebilirsiniz. Metal, o daha dar yay\u0131 olu\u015fturmadan \u00f6nce y\u00fcksek i\u00e7 gerilme alt\u0131nda uzamaktad\u0131r.<\/p>\n\n\n\n
\u015eimdi yumu\u015fak 5052 al\u00fcminyumu yerle\u015ftirin. Daha d\u00fc\u015f\u00fck \u00e7ekme dayan\u0131m\u0131. Daha d\u00fc\u015f\u00fck akma noktas\u0131. Daha az diren\u00e7. Ayn\u0131 batma seviyesinde daha derine iner ve \u00e7o\u011fu zaman y\u00fczde aral\u0131\u011f\u0131n\u0131n alt ucuna daha yak\u0131n bir yar\u0131\u00e7ap\u2014hatta baz\u0131 durumlarda biraz alt\u0131nda\u2014\u00fcretir.<\/p>\n\n\n\n
Ayn\u0131 kal\u0131p. Farkl\u0131 \u00e7arpan.<\/p>\n\n\n\n
\u00c7o\u011fu hesaplay\u0131c\u0131n\u0131n g\u00f6rmezden geldi\u011fi kayma budur. V\u2019yi tek de\u011fi\u015fken, kal\u0131nl\u0131\u011f\u0131 ise \u00e7\u0131karma terimi olarak ele al\u0131rlar. Ger\u00e7ekte, \u00e7ekme dayan\u0131m\u0131 o \u201320 aral\u0131\u011f\u0131n\u0131n neresinde olaca\u011f\u0131n\u0131z\u0131 de\u011fi\u015ftirir. Daha g\u00fc\u00e7l\u00fc malzeme sizi aral\u0131\u011f\u0131n \u00fcst taraf\u0131na iter. Daha yumu\u015fak malzeme s\u0131k\u0131la\u015fman\u0131za izin verir.<\/p>\n\n\n\n
\u201c6 mm alt\u0131ndaki 1\u00d7 malzeme\u201d gibi kal\u0131nl\u0131k esasl\u0131 kurallar\u0131n ince levhalarda bazen i\u015fe yar\u0131yor gibi g\u00f6r\u00fcnmesinin nedeni budur. Uygun ebatl\u0131 kal\u0131pta ince yumu\u015fak \u00e7elik, genellikle kal\u0131nl\u0131\u011fa yak\u0131n bir yar\u0131\u00e7ap verir. Ancak bu, de\u011fi\u015fkenlerin hizalanmas\u0131d\u0131r; evrensel bir yasan\u0131n kan\u0131t\u0131 de\u011fil. Mukavemeti veya V geni\u015fli\u011fini de\u011fi\u015ftirin ve o d\u00fczg\u00fcn 1\u00d7 ili\u015fkisi ortadan kaybolur.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Paslanmaz\u0131, yumu\u015fak \u00e7elik i\u00e7in m\u00fckemmel \u00e7al\u0131\u015fan bir kal\u0131pta hi\u00e7 \u00e7al\u0131\u015ft\u0131r\u0131p a\u00e7\u0131yla u\u011fra\u015f\u0131rken yar\u0131\u00e7ap\u0131n yine b\u00fcy\u00fcd\u00fc\u011f\u00fcn\u00fc g\u00f6rd\u00fcn\u00fcz m\u00fc?<\/p>\n\n\n\n
Ve e\u011fer z\u0131mba ucu o yay\u0131 oymuyorsa, ger\u00e7ekte hangi rol\u00fc oynar?<\/p>\n\n\n\n
Tak\u0131m geometrisi vs. malzeme davran\u0131\u015f\u0131: Fiziksel z\u0131mba ucu yar\u0131\u00e7ap\u0131n\u0131z ile nihai i\u00e7 yar\u0131\u00e7ap\u0131n\u0131z\u0131n neden nadiren uyu\u015fmas\u0131 gerekti\u011fi<\/h3>\n\n\n\n
Ayn\u0131 1.000\u2033 V \u00fczerinde 0.062\u2033 z\u0131mban\u0131n burnuna bak. 10\u2011gauge\u2019\u0131 havada b\u00fck. Par\u00e7ay\u0131 \u00f6l\u00e7.<\/p>\n\n\n\n
\u0130\u00e7eride 0.062\u2033 bulamayacaks\u0131n.<\/p>\n\n\n\n
Z\u0131mba sac\u0131 kal\u0131p omuzlar\u0131n\u0131n aras\u0131na iter, ancak malzemeyi hi\u00e7bir zaman tamamen V\u2019nin i\u00e7ine zorlamaz. Burundaki temas noktasald\u0131r. N\u00fcfuz artt\u0131k\u00e7a sac z\u0131mban\u0131n etraf\u0131nda k\u0131smen sar\u0131l\u0131r, ard\u0131ndan omuzlar aras\u0131nda serbest bir a\u00e7\u0131kl\u0131\u011fa ge\u00e7i\u015f yapar. Nihai i\u00e7 yar\u0131\u00e7ap\u0131n b\u00fcy\u00fck k\u0131sm\u0131, z\u0131mban\u0131n damga gibi \u015feklini bast\u0131rmas\u0131ndan de\u011fil, bu a\u00e7\u0131kl\u0131kta olu\u015fan b\u00fck\u00fclmeden meydana gelir.<\/p>\n\n\n\n
N\u00fcfuz derinli\u011fi her \u015feyi de\u011fi\u015ftirir. Milin sadece birka\u00e7 binde biri kadar derinle\u015fmesi sar\u0131lmay\u0131 art\u0131r\u0131r, a\u00e7\u0131kl\u0131k uzunlu\u011funu azalt\u0131r ve a\u00e7\u0131y\u0131 s\u0131k\u0131la\u015ft\u0131r\u0131r\u2014ancak yar\u0131\u00e7ap h\u00e2l\u00e2 kal\u0131p geni\u015fli\u011fi ile malzeme direncinin bir fonksiyonu olarak belirlenir. Malzeme tamamen kal\u0131ba oturana kadar altlama veya bask\u0131 (coining) yapmad\u0131k\u00e7a, z\u0131mba ucu bir kal\u0131p de\u011fil, bir y\u00f6nlendiricidir.<\/p>\n\n\n\n
Bu y\u00fczden do\u011fru hava b\u00fck\u00fcm\u00fcnde z\u0131mba yar\u0131\u00e7ap\u0131n, beklenen i\u00e7 yar\u0131\u00e7aptan k\u00fc\u00e7\u00fck olmas\u0131 gerekir. E\u011filmeyi ba\u015flatmas\u0131 ama y\u00f6nlendirmemesi gerekir. Tam olarak e\u015fle\u015firlerse, ya fark\u0131nda olmadan altlama yap\u0131yorsun ya da bask\u0131 tonaj\u0131yla fl\u00f6rt ediyorsun demektir.<\/p>\n\n\n\n
Pres freni 200 tonluk bir ger\u00e7ek makinesidir. Modelinin fizi\u011fe uyup uymad\u0131\u011f\u0131n\u0131 ortaya \u00e7\u0131kar\u0131r. Hava b\u00fck\u00fcm\u00fcnde fizik \u015funu s\u00f6yler: kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131 temel a\u00e7\u0131kl\u0131\u011f\u0131 tan\u0131mlar, malzemenin \u00e7ekme dayan\u0131m\u0131 bu a\u00e7\u0131kl\u0131k i\u00e7indeki oran\u0131 de\u011fi\u015ftirir ve z\u0131mba yaln\u0131zca bu sisteme ne kadar derin dald\u0131\u011f\u0131n\u0131 kontrol eder.<\/p>\n\n\n\n
E\u011fer \u00e7izimin s\u0131k\u0131 bir yar\u0131\u00e7ap istiyorsa ama kal\u0131p oran\u0131n\u0131n do\u011fal e\u011filimi daha b\u00fcy\u00fck bir yar\u0131\u00e7apa y\u00f6neliyorsa, h\u00e2l\u00e2 sabit bir form\u00fcle g\u00fcvenmeye devam m\u0131 edeceksin\u2014yoksa hurda kutusu bir kez daha oy vermeden \u00f6nce kal\u0131b\u0131 m\u0131 de\u011fi\u015ftireceksin?<\/p>\n\n\n\n
Geri esneme telafisi: Sabit hesaplay\u0131c\u0131lar\u0131n g\u00f6z ard\u0131 etti\u011fi de\u011fi\u015fken standart<\/h2>\n\n\n\n
Ram kalkt\u0131\u011f\u0131 anda 90\u00b0 yumu\u015fak \u00e7elik bir braketi 92\u00b0\u2019de frenin \u00fczerinden \u00e7\u0131kt\u0131\u011f\u0131n\u0131 g\u00f6rd\u00fcm. Ayn\u0131 kal\u0131p. Ayn\u0131 program. Ayn\u0131 operat\u00f6r. 200 ton alt\u0131nda tam isabetti. Be\u015f saniye sonra art\u0131k de\u011fildi.<\/p>\n\n\n\n
Do\u011fal kal\u0131ba ba\u011fl\u0131 yar\u0131\u00e7ap yaz\u0131c\u0131dakiyle e\u015fle\u015fmedi\u011finde asl\u0131nda \u015fu soruyu soruyorsun: y\u00fck alt\u0131ndaki durumu mu \u00e7\u00f6z\u00fcyorsun, yoksa y\u00fck kalkt\u0131ktan sonra m\u00fc\u015fterinin \u00f6l\u00e7t\u00fc\u011f\u00fc durumu mu?<\/p>\n\n\n\n
Z\u0131mban\u0131n V\u2019nin i\u00e7ine g\u00f6m\u00fcl\u00fc oldu\u011fu s\u0131rada g\u00f6rd\u00fc\u011f\u00fcn yar\u0131\u00e7ap ve a\u00e7\u0131, sevk etti\u011fin par\u00e7ada g\u00f6rd\u00fc\u011f\u00fcn yar\u0131\u00e7ap ve a\u00e7\u0131 de\u011fildir. Bas\u0131n\u00e7 kalkt\u0131\u011f\u0131 anda elastik gerinim serbest kal\u0131r. Uzanm\u0131\u015f d\u0131\u015f lifler k\u0131salmaya \u00e7al\u0131\u015f\u0131r. S\u0131k\u0131\u015fm\u0131\u015f i\u00e7 lifler toparlanmaya \u00e7al\u0131\u015f\u0131r. Par\u00e7a a\u00e7\u0131l\u0131r. \u0130\u015fte bu geri esneme etkisidir ve o, senin temiz k\u00fc\u00e7\u00fck (V \u2212 MT)\/2 hesab\u0131na ald\u0131rmaz.<\/p>\n\n\n\n
Hurda kutusu ger\u00e7eklik kontrol\u00fc: Derinlik g\u00f6stergesindeki a\u00e7\u0131y\u0131 m\u00fckemmel tutturup, denet\u00e7inin her flan\u015f\u0131n 1,5\u00b0 a\u00e7\u0131k oldu\u011funu s\u00f6yledi\u011fini g\u00f6rd\u00fc\u011f\u00fcn oldu mu? \u00c7elik ekrana itaat etmedi; fizi\u011fe etti.<\/p>\n\n\n\n
Sabit hesaplay\u0131c\u0131lar, bas\u0131n\u00e7 alt\u0131nda olu\u015fturdu\u011fun geometrinin yerinde kald\u0131\u011f\u0131n\u0131 varsayar. Kalm\u0131yor. Ve e\u011fer plana telafi dahil etmezsen, bitmi\u015f bir yar\u0131\u00e7ap\u0131 de\u011fil ge\u00e7ici bir yar\u0131\u00e7ap\u0131 \u00f6ng\u00f6r\u00fcyorsun demektir.<\/p>\n\n\n\n
As\u0131l m\u00fccadele alt strokta elde etti\u011fin yar\u0131\u00e7apta de\u011fil. Bo\u015fald\u0131ktan sonra ayakta kalan yar\u0131\u00e7apta.<\/p>\n\n\n\n
Geri esneme i\u00e7in fazla b\u00fckme: Rahatlam\u0131\u015f yar\u0131\u00e7ap i\u00e7in mi yoksa y\u00fck alt\u0131ndaki yar\u0131\u00e7ap i\u00e7in mi hesap yap\u0131yorsun?<\/h3>\n\n\n\n
Bir 1.000\u2033 V i\u00e7inde 0.125\u2033 so\u011fuk haddelenmi\u015f par\u00e7a hayal et. Z\u0131mban\u0131n a\u015fa\u011f\u0131dayken 90\u00b0\u2019lik bir a\u00e7\u0131 veren derinli\u011fe kadar iniyorsun. Ram\u0131 yukar\u0131 \u00e7ekiyorsun ve 91.5\u00b0\u2019de oturuyor. Bu, y\u00fck alt\u0131ndaki a\u00e7\u0131n\u0131n yakla\u015f\u0131k 88.5\u00b0 oldu\u011fu anlam\u0131na gelir.<\/p>\n\n\n\n
\u015eimdi kendine sor: form\u00fcl\u00fcn hangi a\u00e7\u0131y\u0131 \u00f6ng\u00f6rd\u00fc?<\/p>\n\n\n\n
Ger\u00e7ek bir 90\u00b0 geometrisini varsayarak ve o 1.5\u00b0\u2019lik geri esnemeyi hesaba katmadan b\u00fckme indirimi, b\u00fckme pay\u0131 ve flan\u015f uzunluklar\u0131 hesaplad\u0131ysan, her flan\u015f uzundur. \u00c7ok de\u011fil. Bir montaj\u0131 mahvedecek kadar.<\/p>\n\n\n\n
Fazla b\u00fckme, hepimizin kulland\u0131\u011f\u0131 kaba bir ara\u00e7t\u0131r. 88.5\u00b0\u2019yi programla ki 90\u00b0\u2019ye gev\u015fesin. Ama i\u015fte tuzak: bu telafi t\u00fcm i\u015fler aras\u0131nda sabit de\u011fildir. Kal\u0131b\u0131 1.250\u2033\u2019ye a\u00e7 ve ayn\u0131 malzeme art\u0131k 2\u00b0 veya daha fazla geri esneyebilir; \u00e7\u00fcnk\u00fc daha b\u00fcy\u00fck yar\u0131\u00e7ap plastik \u015fekil de\u011fi\u015ftirmeyi azalt\u0131r ve kesitte daha fazla elastik enerji birikmesine yol a\u00e7ar. Daha ince malzeme? Daha fazla geri esneme. Daha b\u00fcy\u00fck i\u00e7 yar\u0131\u00e7ap? Daha fazla geri esneme. Bu ili\u015fki, geri esnemenin yaln\u0131zca dayan\u0131mla de\u011fil, yar\u0131\u00e7ap\/kal\u0131nl\u0131k oran\u0131yla artt\u0131\u011f\u0131 so\u011fuk haddelenmi\u015f \u00e7elikte g\u00f6sterilmi\u015ftir.<\/p>\n\n\n\n
Yani 90\u00b0 nominal \u00fczerinden b\u00fckme indirimi hesaplay\u0131p makinede elle 1.5\u00b0 fazla b\u00fckme ekliyorsan, hesab\u0131n\u0131 ikiye b\u00f6lm\u00fc\u015fs\u00fcn demektir. D\u00fcz desen bir \u015fey d\u00fc\u015f\u00fcn\u00fcyor. Ram ba\u015fka bir \u015fey yap\u0131yor.<\/p>\n\n\n\n
Geometrinizi y\u00f6nlendiren say\u0131 hangisi\u2014bask\u0131 a\u00e7\u0131s\u0131 m\u0131, yoksa ger\u00e7ekten programlad\u0131\u011f\u0131n\u0131z a\u00e7\u0131 m\u0131?<\/p>\n\n\n\n
\u00c7ekme dayan\u0131m\u0131 ve akma oran\u0131: Hangi \u00f6zellik ger\u00e7ekten fark yarat\u0131yor?<\/h3>\n\n\n\n
O yumu\u015fak \u00e7eli\u011fi 304 paslanmazla de\u011fi\u015ftirin. Ayn\u0131 kal\u0131nl\u0131k. Ayn\u0131 kal\u0131p. Ayn\u0131 hedef a\u00e7\u0131.<\/p>\n\n\n\n
Daha fazla yaylanma g\u00f6receksiniz. Herkes g\u00f6r\u00fcr. \u0130\u00e7g\u00fcd\u00fc, \u00e7ekme dayan\u0131m\u0131n\u0131 su\u00e7lamakt\u0131r \u00e7\u00fcnk\u00fc sertifikada en b\u00fcy\u00fck say\u0131 odur. Daha y\u00fcksek \u00e7ekme, daha fazla diren\u00e7, daha fazla a\u00e7\u0131lma.<\/p>\n\n\n\n
Ama \u201cayn\u0131\u201d 60\u2011KSI \u00e7eli\u011fin iki ergitmesini \u00e7al\u0131\u015ft\u0131rd\u0131\u011f\u0131n\u0131zda neler oldu\u011funu izleyin. Biri kolay b\u00fck\u00fcl\u00fcr. Di\u011feri fazladan bir derece geri yaylan\u0131r. \u00c7ekme dayan\u0131m\u0131 bir gecede 10 KSI s\u0131\u00e7ramad\u0131. De\u011fi\u015fen \u015fey akma\u2011\u00e7ekme oran\u0131d\u0131r\u2014akma oran\u0131.<\/p>\n\n\n\n
Yaylanma, deformasyonun ne kadar\u0131n\u0131n elastik, ne kadar\u0131n\u0131n plastik oldu\u011funa ba\u011fl\u0131d\u0131r. \u00c7ekme dayan\u0131m\u0131na g\u00f6re y\u00fcksek akma dayan\u0131m\u0131na sahip bir malzeme, plastik deformasyona daha ge\u00e7 girer ve derin \u015fekilde akmadan \u00f6nce daha fazla elastik enerji depolar. Bu depolanan enerji, y\u00fck kalkt\u0131\u011f\u0131nda a\u00e7\u0131n\u0131z\u0131 a\u00e7an \u015feydir.<\/p>\n\n\n\n
Geometri bunu b\u00fcy\u00fct\u00fcr. Daha geni\u015f kal\u0131p a\u00e7\u0131kl\u0131klar\u0131 daha b\u00fcy\u00fck i\u00e7 yar\u0131\u00e7aplar olu\u015fturur. Daha b\u00fcy\u00fck yar\u0131\u00e7ap, ayn\u0131 a\u00e7\u0131 i\u00e7in daha d\u00fc\u015f\u00fck plastik gerinim demektir. Daha d\u00fc\u015f\u00fck plastik gerinim, daha y\u00fcksek oranda elastik toparlanma demektir. Bu y\u00fczden geni\u015f V kal\u0131plarla yap\u0131lan yar\u0131\u00e7ap b\u00fck\u00fcmler, s\u0131k\u0131 6\u00d7 kurulumlara g\u00f6re dramatik bi\u00e7imde geri yaylanabilir.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Tonaj tasarrufu i\u00e7in hi\u00e7 geni\u015f bir kal\u0131p kulland\u0131n\u0131z m\u0131, ekrandaki a\u00e7\u0131y\u0131 tutturdunuz, sonra 10 fitlik par\u00e7a boyunca 3\u00b0 yaylanmay\u0131 kovalar halde buldunuz kendinizi? Bu k\u00f6t\u00fc \u015fans de\u011fildi. Bu, d\u00fc\u015f\u00fck plastik gerinimin kontrol\u00fc elastikli\u011fe geri vermesiydi.<\/p>\n\n\n\n
Peki hangisi daha fazla fark yarat\u0131yor\u2014k\u00e2\u011f\u0131ttaki dayan\u0131m say\u0131lar\u0131 m\u0131 yoksa tak\u0131m se\u00e7iminizle belirledi\u011finiz yar\u0131\u00e7ap\u2011kal\u0131nl\u0131k oran\u0131 m\u0131? Uygulamada geometriniz sahneyi kurar. Malzeme \u00f6zellikleri geri tepmenin ne kadar sert olaca\u011f\u0131n\u0131 belirler.<\/p>\n\n\n\n
Ve o geri tepme de\u011fi\u015firse, g\u00fcvendi\u011finiz t\u00fcm d\u00fcz desen say\u0131lar\u0131 ne olur?<\/p>\n\n\n\n
Hesaplanmam\u0131\u015f yaylanman\u0131n b\u00fck\u00fclme indirimini ve K\u2011Fakt\u00f6r\u00fc\u2019n\u00fc sessizce bozmas\u0131<\/h3>\n\n\n\n
D\u00f6rt b\u00fck\u00fcml\u00fc bir kanal al\u0131n. D\u00f6n\u00fc\u015f flan\u015f\u0131 yok. Her b\u00fck\u00fcm 2\u00b0 geri yaylans\u0131n. Bu dramatik de\u011fil. Baz\u0131 paslanmaz i\u015flerde normaldir.<\/p>\n\n\n\n
\u015eimdi \u00fcst \u00fcste koyun.<\/p>\n\n\n\n
D\u00f6rt b\u00fck\u00fcm \u00d7 2\u00b0 demek, her ad\u0131mda do\u011fru \u015fekilde telafi etmediyseniz son flan\u015f\u0131n ilk referansa g\u00f6re 8\u00b0 d\u0131\u015far\u0131da olmas\u0131 demektir. Her b\u00fck\u00fcm \u201cyeterince yak\u0131n\u201d g\u00f6r\u00fcn\u00fcyordu ama birikmi\u015f hata montaj\u0131 pervane gibi burktu\u011funu g\u00f6rd\u00fcm.<\/p>\n\n\n\n
B\u00fck\u00fclme indirimi ve K\u2011fakt\u00f6r\u00fc bilinen bir i\u00e7 yar\u0131\u00e7ap ve bilinen bir nihai a\u00e7\u0131 varsayar. Yaylanma bunlardan herhangi birini de\u011fi\u015ftirirse ve siz say\u0131lar\u0131 g\u00fcncellemezseniz, n\u00f6tr eksenin konumu ger\u00e7ekte de\u011fi\u015fir ama yaz\u0131l\u0131m\u0131n\u0131zda de\u011fi\u015fmez. Kesti\u011finiz d\u00fcz uzunluk, gev\u015feme sonras\u0131 ger\u00e7ekte var olandan daha k\u00fc\u00e7\u00fck bir yay uzunlu\u011funa dayan\u0131r. Bunu \u00e7oklu b\u00fck\u00fcmlere ve toleranslara uygulad\u0131\u011f\u0131n\u0131zda, hatalar h\u0131zla b\u00fcy\u00fcr.<\/p>\n\n\n\n
Bu y\u00fczden tek tip K\u2011fakt\u00f6rler bir hayaldir. Kal\u0131p geni\u015fli\u011fini de\u011fi\u015ftirin, yar\u0131\u00e7ap\u0131 de\u011fi\u015ftirirsiniz. Yar\u0131\u00e7ap\u0131 de\u011fi\u015ftirin, yaylanmay\u0131 de\u011fi\u015ftirirsiniz. Yaylanmay\u0131 de\u011fi\u015ftirin, nihai a\u00e7\u0131y\u0131 ve etkili b\u00fck\u00fclme pay\u0131n\u0131 de\u011fi\u015ftirirsiniz. Sisteminiz bu d\u00f6ng\u00fcy\u00fc kapatm\u0131yorsa\u2014rahatlam\u0131\u015f a\u00e7\u0131y\u0131 ve yar\u0131\u00e7ap\u0131 \u00f6l\u00e7\u00fcp indirime geri beslemiyorsa\u2014siz 200 ton alt\u0131nda var olan bir par\u00e7a i\u00e7in d\u00fcz levha kesiyorsunuz demektir.<\/p>\n\n\n\n
Pres b\u00fck\u00fcc\u00fc 200 tonluk bir ger\u00e7eklik makinesidir. Malzemenin ger\u00e7ekte ne yapaca\u011f\u0131n\u0131 size s\u00f6yler. Hurda kutusu, dinleyip dinlemedi\u011finizin son yarg\u0131c\u0131d\u0131r.<\/p>\n\n\n\n
Y\u00fck kald\u0131r\u0131ld\u0131ktan sonra hayatta kalan yar\u0131\u00e7ap tek \u00f6nemli olan ise, neden h\u00e2l\u00e2 ko\u00e7 kalkt\u0131\u011f\u0131 anda kaybolan yar\u0131\u00e7ap \u00fczerinden d\u00fcz desenler olu\u015fturuyorsunuz?<\/p>\n\n\n\n
Tahminden Kontrole: \u0130\u00e7 B\u00fck\u00fclme Yar\u0131\u00e7ap\u0131n\u0131 Tahmin Etmek \u0130\u00e7in 4 Ad\u0131ml\u0131 Y\u00f6ntem<\/h2>\n\n\n\n
1.000\u2033 V\u2011kal\u0131p alt\u0131nda 10 numara sacdan bir braketi izledim, ilk par\u00e7a bask\u0131 90\u00b0 ba\u011f\u0131r\u0131rken 92\u00b0 \u00e7\u0131kt\u0131. Programc\u0131 d\u00fcz deseni do\u011fru oldu\u011funu s\u00f6yledi. Hesaplay\u0131c\u0131 \u201ckesin\u201d i\u00e7 yar\u0131\u00e7ap dedi. Hurda kutusu umursamad\u0131.<\/p>\n\n\n\n
Ger\u00e7ek geri yaylanmay\u0131 b\u00fckme indirgeme ve K\u2011fakt\u00f6r\u00fcne geri beslemek istiyorsun ki d\u00fcz par\u00e7a gev\u015femi\u015f par\u00e7ayla tam olarak e\u015fle\u015fsin. G\u00fczel. \u00c7\u00fcnk\u00fc o d\u00f6ng\u00fcy\u00fc kapatana kadar hesaplama yapm\u0131yorsun\u2014sacla kumar oynuyorsun.<\/p>\n\n\n\n
G\u00fczel form\u00fcllere sabr\u0131 olmayan bir 200 tonluk ger\u00e7eklik makinesinde kulland\u0131\u011f\u0131m y\u00f6ntem bu.<\/p>\n\n\n\n
Ad\u0131m 1: Bask\u0131daki \u201cistenen\u201d yar\u0131\u00e7aptan de\u011fil, kal\u0131p oran\u0131ndan ba\u015fla<\/h3>\n\n\n\n
0,125\u2033 yumu\u015fak \u00e7eli\u011fi iki kal\u0131p \u00fczerinde ayarla: biri 0,750\u2033 V, di\u011feri 1,000\u2033 V. Ayn\u0131 z\u0131mba. Ayn\u0131 hedef a\u00e7\u0131. Par\u00e7alar ayn\u0131 i\u00e7 yar\u0131\u00e7apla \u00e7\u0131kmaz. \u00c7\u0131kamazlar. Kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131, sac\u0131n nas\u0131l akmas\u0131na izin verilece\u011finin geometrisini belirler.<\/p>\n\n\n\n
Hava b\u00fckmede, kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131 b\u00fcy\u00fck \u00f6l\u00e7\u00fcde olu\u015facak i\u00e7 yar\u0131\u00e7ap\u0131 belirler. Ger\u00e7ek d\u00fcnyada yayg\u0131n bir ba\u015flang\u0131\u00e7 kural\u0131, daha ince yumu\u015fak \u00e7eliklerde 6:1 V\u2011kal\u0131nl\u0131k, kal\u0131nl\u0131k artt\u0131k\u00e7a 8:1 oran\u0131d\u0131r. Yani 0,125\u2033 malzeme mi? Genellikle 0,750\u2033 ila 1,000\u2033 V aras\u0131ndas\u0131nd\u0131r. Bu oran bir \u00f6neri de\u011fildir\u2014bu, gerinim zarf\u0131d\u0131r. \u00c7ok dar olursa \u00e7atlama riski vard\u0131r. \u00c7ok geni\u015f olursa plastik gerinimi azalt\u0131r ve geri yaylanma etkisini davet edersin.<\/p>\n\n\n\n
\u015eimdi birinin bunun yerine teknik resimden ba\u015flad\u0131\u011f\u0131nda ne oldu\u011funa bak. \u201c0,125 yar\u0131\u00e7ap istiyorum.\u201d G\u00fczel. Ancak tonajdan tasarruf etmek i\u00e7in 1,250\u2033 V se\u00e7ersen, do\u011fal hava formu yar\u0131\u00e7ap\u0131n malzemeye ba\u011fl\u0131 olarak 0,200\u2033+ civar\u0131nda olabilir. Hi\u00e7bir hayal g\u00fcc\u00fc bunu de\u011fi\u015ftiremez. Geometri zaten karar\u0131n\u0131 vermi\u015ftir.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Makinede zaten tak\u0131l\u0131 oldu\u011fu i\u00e7in geni\u015f bir kal\u0131b\u0131 zorlayan, ekrandaki a\u00e7\u0131y\u0131 yakalayan, sonra da gev\u015femi\u015f yar\u0131\u00e7ap d\u00fcz modelin varsayd\u0131\u011f\u0131ndan daha b\u00fcy\u00fck oldu\u011fu i\u00e7in flan\u015flar\u0131n uzad\u0131\u011f\u0131n\u0131 izleyen at\u00f6lyeler g\u00f6rd\u00fcm. Form\u00fcl yanl\u0131\u015f de\u011fildi. Ba\u015flang\u0131\u00e7 varsay\u0131m\u0131 \u00f6yleydi.<\/p>\n\n\n\n
Kal\u0131p geni\u015fli\u011fi gerinim ko\u015fulunu belirliyorsa, neden o kal\u0131ptan ba\u011f\u0131ms\u0131z hedef bir yar\u0131\u00e7apla ba\u015flamak isteyesin ki?<\/p>\n\n\n\n
Ad\u0131m 2: V\u2011a\u00e7\u0131kl\u0131\u011f\u0131ndan teorik yar\u0131\u00e7ap\u0131 tahmin et (ve bunun yaln\u0131zca bir ba\u015flang\u0131\u00e7 de\u011feri oldu\u011funu kabul et)<\/h3>\n\n\n\n
Kal\u0131p se\u00e7ildikten sonra, art\u0131k matematik konu\u015fmana izin var.<\/p>\n\n\n\n
Hava b\u00fckmede z\u0131mba burun tamamen V i\u00e7ine girmez. Z\u0131mba, sac\u0131 kal\u0131p omuzlar\u0131 aras\u0131na iter ancak malzemeyi tamamen V i\u00e7ine zorlamaz. Sac gerilir. Bu gerilme, V\u2011a\u00e7\u0131kl\u0131\u011f\u0131 ve y\u00fck alt\u0131ndaki i\u00e7 yar\u0131\u00e7ap aras\u0131nda \u00f6ng\u00f6r\u00fclebilir bir ili\u015fki yarat\u0131r.<\/p>\n\n\n\n
Yumu\u015fak \u00e7elikte, y\u00fck alt\u0131ndaki i\u00e7 yar\u0131\u00e7ap genellikle V\u2011a\u00e7\u0131kl\u0131\u011f\u0131n\u0131n yakla\u015f\u0131k \u201320\u2019si aras\u0131nda olur. Yani 1,000\u2033 V, pres a\u015fa\u011f\u0131dayken yakla\u015f\u0131k 0,160\u2033\u20130,200\u2033 yar\u0131\u00e7ap \u00fcretebilir. Tam de\u011fil. Bir aral\u0131k. \u00c7\u00fcnk\u00fc kal\u0131nl\u0131k, dayan\u0131m ve kal\u0131p omuz yar\u0131\u00e7ap\u0131 hepsi bunu biraz de\u011fi\u015ftirir.<\/p>\n\n\n\n
Ne yapt\u0131\u011f\u0131m\u0131za dikkat et: yar\u0131\u00e7ap\u0131 \u00f6nce kal\u0131p geni\u015fli\u011fine ba\u011flad\u0131k, (V \u2212 MT)\/2\u2019ye ya da z\u0131mba ucuna de\u011fil. Eski (V \u2212 MT)\/2 form\u00fcl\u00fc, kesin g\u00f6r\u00fcnen say\u0131lar verebilir ama V\/T oran\u0131 6:1\u2019den 8:1\u2019e kayarsa, gerinim da\u011f\u0131l\u0131m\u0131n de\u011fi\u015fir ve o d\u00fczg\u00fcn sonu\u00e7 ger\u00e7ekte h\u0131zla sapar.<\/p>\n\n\n\n
Ve bu h\u00e2l\u00e2 y\u00fck alt\u0131ndaki durumdur.<\/p>\n\n\n\n
Ba\u015flang\u0131\u00e7 yar\u0131\u00e7ap\u0131 200 tonluk bir anl\u0131k g\u00f6r\u00fcnt\u00fcd\u00fcr. Sevk etti\u011fin par\u00e7a ise ko\u00e7 yukar\u0131 kalkt\u0131ktan sonraki halidir. Peki depolanm\u0131\u015f elastik enerji geri \u00e7\u0131kt\u0131\u011f\u0131nda ne olur?<\/p>\n\n\n\n
Ad\u0131m 3: Ger\u00e7ek gev\u015femi\u015f yar\u0131\u00e7ap\u0131 bulmak i\u00e7in malzeme temelli geri yaylanma telafisini uygula<\/h3>\n\n\n\n
0,125\u2033 so\u011fuk haddelenmi\u015f \u00e7eli\u011fi o 1,000\u2033 V i\u00e7inde al. Y\u00fck alt\u0131ndayken 90\u00b0 \u00f6l\u00e7\u00fcyorsun. B\u0131rak\u0131nca 91,5\u00b0 oluyorsun. O 1,5\u00b0 geri yaylanma daha derin bir \u015fey anlat\u0131yor: yar\u0131\u00e7ap da b\u00fcy\u00fcd\u00fc.<\/p>\n\n\n\n
A\u00e7\u0131 a\u00e7\u0131ld\u0131k\u00e7a i\u00e7 yar\u0131\u00e7ap artar. N\u00f6tr eksen biraz kayar \u00e7\u00fcnk\u00fc gerinimin elastik k\u0131sm\u0131 geri kazan\u0131l\u0131r. Akma dayan\u0131m\u0131, \u00e7ekme dayan\u0131m\u0131na g\u00f6re daha y\u00fcksek olan malzemeler, tamamen akmadan \u00f6nce daha fazla elastik enerji depolar. Paslanmaz \u00e7elik genellikle ayn\u0131 kal\u0131pta yumu\u015fak \u00e7eli\u011fe g\u00f6re daha fazla geri tepme yapar. Al\u00fcminyum da yar\u0131\u00e7ap\u2011kal\u0131nl\u0131k oran\u0131 y\u00fckseldi\u011finde seni \u015fa\u015f\u0131rtabilir.<\/p>\n\n\n\n
Ama kitaplar\u0131n atlad\u0131\u011f\u0131 k\u0131s\u0131m \u015fu: geri yaylanma yar\u0131\u00e7ap\u2011kal\u0131nl\u0131k oran\u0131yla g\u00fc\u00e7l\u00fc bi\u00e7imde orant\u0131l\u0131d\u0131r. Kal\u0131b\u0131 daha geni\u015f a\u00e7t\u0131\u011f\u0131nda, i\u00e7 yar\u0131\u00e7ap\u0131 art\u0131r\u0131rs\u0131n. Yar\u0131\u00e7ap\u0131 art\u0131rd\u0131\u011f\u0131nda, belirli bir a\u00e7\u0131 i\u00e7in plastik gerinimi azalt\u0131rs\u0131n. Plastik gerinim azal\u0131nca, elastik toparlanma oran\u0131 artar. Daha fazla geri yaylanma. Bu, geometrinin malzeme davran\u0131\u015f\u0131n\u0131 b\u00fcy\u00fctmesidir.<\/p>\n\n\n\n
Yani telafi s\u0131ras\u0131 \u015fu \u015fekildedir:<\/p>\n\n\n\n
\n- Kal\u0131b\u0131 se\u00e7in (V\/T oran\u0131n\u0131 ve \u015fekil de\u011fi\u015ftirme ko\u015fulunu belirler).<\/li>\n\n\n\n
- Y\u00fck alt\u0131ndaki yar\u0131\u00e7ap\u0131 V de\u011ferinden tahmin edin.<\/li>\n\n\n\n
- Malzeme kalitesine ve yar\u0131\u00e7ap\u2011kal\u0131nl\u0131k oran\u0131na g\u00f6re geri esneme a\u00e7\u0131s\u0131n\u0131 tahmin edin.<\/li>\n\n\n\n
- Bu ekstra a\u00e7\u0131y\u0131, ayarlanm\u0131\u015f bir gev\u015femi\u015f yar\u0131\u00e7apa d\u00f6n\u00fc\u015ft\u00fcr\u00fcn.<\/li>\n<\/ol>\n\n\n\n
\u015eimdi b\u00fckme kesintisi ve K\u2011fakt\u00f6r\u00fcn\u00fcz\u00fc y\u00fck alt\u0131 geometrinizle de\u011fil, gev\u015femi\u015f yar\u0131\u00e7ap ve son a\u00e7\u0131yla g\u00fcncelleyin. \u0130\u015fte kapal\u0131 d\u00f6ng\u00fc budur. E\u011fer \u00f6l\u00e7\u00fclen geri esneme 1.5\u00b0 ise, d\u00fcz modeliniz 88.5\u00b0 y\u00fck alt\u0131nda de\u011fil, 90\u00b0 gev\u015femi\u015f durumdaki par\u00e7a yay uzunlu\u011funu yans\u0131tmal\u0131d\u0131r.<\/p>\n\n\n\n
CN-HAWE\u2019nin \u00fcr\u00fcn portf\u00f6y\u00fcn\u00fcn 100% CNC tabanl\u0131 oldu\u011fu ve lazer kesim, b\u00fckme, oluk a\u00e7ma, kesme gibi y\u00fcksek d\u00fczeyli senaryolar\u0131 kapsad\u0131\u011f\u0131 g\u00f6z \u00f6n\u00fcne al\u0131nd\u0131\u011f\u0131nda, ayr\u0131nt\u0131l\u0131 materyal isteyen okuyucular i\u00e7in, Bro\u015f\u00fcrler<\/a> yararl\u0131 bir tamamlay\u0131c\u0131 kaynakt\u0131r.<\/p>\n\n\n\nHurda Kutusu Ger\u00e7eklik Kontrol\u00fc: D\u00f6rt b\u00fck\u00fcml\u00fc kanallar\u0131n, kal\u0131p oran\u0131 de\u011fi\u015fti\u011fi h\u00e2lde programc\u0131n\u0131n bir tablodan sabit K\u2011fakt\u00f6r\u00fc kulland\u0131\u011f\u0131 i\u00e7in toplam geni\u015flikte 0.060\u2033\u2019dan fazla hatayla ka\u00e7t\u0131\u011f\u0131n\u0131 g\u00f6rd\u00fcm. Ayn\u0131 malzeme. Farkl\u0131 V. Farkl\u0131 gev\u015femi\u015f yar\u0131\u00e7ap. Matematik bunu hi\u00e7 bilemedi.<\/p>\n\n\n\n
E\u011fer geri esneme hem malzeme \u00f6zelliklerine hem de kal\u0131pla olu\u015fturdu\u011funuz yar\u0131\u00e7apa ba\u011fl\u0131ysa, neden tek bir sabit K\u2011fakt\u00f6r her i\u015fte i\u015fe yaras\u0131n?<\/p>\n\n\n\n
Ad\u0131m 4: Tam \u00fcretim deneme\u2011yan\u0131lma yerine kontroll\u00fc bir test b\u00fck\u00fcm\u00fcyle do\u011frulay\u0131n<\/h3>\n\n\n\n
2\u2033 geni\u015fli\u011finde bir \u015ferit kesin. Ayn\u0131 kal\u0131nl\u0131k. \u00dcretimle ayn\u0131 hadde y\u00f6n\u00fc. Se\u00e7ilen kal\u0131pta programlad\u0131\u011f\u0131n\u0131z fazla b\u00fck\u00fcmle b\u00fck\u00fcn\u2014\u00f6rne\u011fin 1.5\u00b0 geri esneme bekliyorsan\u0131z 88.5\u00b0. Gev\u015femi\u015f a\u00e7\u0131y\u0131 \u00f6l\u00e7\u00fcn ve do\u011fru mastarlarla i\u00e7 yar\u0131\u00e7ap\u0131 \u00f6l\u00e7\u00fcn.<\/p>\n\n\n\n
Art\u0131k \u00fc\u00e7 ger\u00e7ek say\u0131n\u0131z var: kal\u0131p geni\u015fli\u011fi, gev\u015femi\u015f a\u00e7\u0131, gev\u015femi\u015f yar\u0131\u00e7ap.<\/p>\n\n\n\n
Bunlar\u0131 b\u00fckme pay\u0131 hesaplaman\u0131za geri besleyin. Hesaplanan flan\u015f uzunluklar\u0131 \u00f6l\u00e7\u00fclen test par\u00e7as\u0131yla e\u015fle\u015fene kadar K\u2011fakt\u00f6r\u00fcn\u00fc ayarlay\u0131n. Kalibre edilmi\u015f bu K, o malzeme serisi, o kal\u0131p, o kal\u0131nl\u0131k, o kurulum i\u00e7in ge\u00e7erli h\u00e2le gelir.<\/p>\n\n\n\n
Bu bir tahmin i\u015fi de\u011fil. Kontroll\u00fc yinelemedir. Y\u00fcz par\u00e7ay\u0131 kurtarmak i\u00e7in bir \u015ferit feda edilir.<\/p>\n\n\n\n
Evet, makineye \u00f6zg\u00fc etkenler \u00f6nemlidir. Kal\u0131p kenar a\u015f\u0131nmas\u0131, pres ko\u00e7 defleksiyonu, bombaj ayarlar\u0131\u2014hepsi par\u00e7an\u0131n davran\u0131\u015f\u0131n\u0131 etkiler. \u0130\u015fte bu y\u00fczden test b\u00fck\u00fcm\u00fc ayn\u0131 makinede, ayn\u0131 tak\u0131m y\u0131\u011f\u0131n\u0131 ve ayn\u0131 kurulumda yap\u0131lmal\u0131d\u0131r. Sadece bir form\u00fcl\u00fc de\u011fil, t\u00fcm sistemi kalibre ediyorsunuz.<\/p>\n\n\n\n
Ger\u00e7ek \u015fu: iyi bir tahmin zinciri bile modelemedi\u011finiz de\u011fi\u015fkenler taraf\u0131ndan bozulabilir.<\/p>\n\n\n\n
Bu y\u00fczden kal\u0131p oran\u0131n\u0131, temel yar\u0131\u00e7ap\u0131, geri esneme telafisini ve do\u011frulanm\u0131\u015f K\u2011fakt\u00f6r\u00fcn\u00fc sabitlemi\u015f olsan\u0131z bile, \u201cm\u00fckemmel\u201d d\u00fcz modelinizi otuz mil (0.030\u2033) sapt\u0131rmaya haz\u0131r bekleyen gizli etkenler hangileridir\u2014ve bunlar\u0131 bulmak i\u00e7in ne kadar malzeme harcamaya haz\u0131rs\u0131n\u0131z?<\/p>\n\n\n\n
CN\u2011HAWE y\u0131ll\u0131k sat\u0131\u015f gelirinin %8%\u2019inden fazlas\u0131n\u0131 ara\u015ft\u0131rma ve geli\u015ftirmeye yat\u0131r\u0131rken, ADH pres b\u00fckme makineleri genelinde Ar\u2011Ge yetenekleriyle faaliyet g\u00f6stermektedir; bir sonraki ad\u0131m ekiple do\u011frudan konu\u015fmaksa, Bize ula\u015f\u0131n<\/a> buraya do\u011fal \u015fekilde oturur.<\/p>\n\n\n\nM\u00fckemmel Matemati\u011fi Bozan G\u00f6r\u00fcnmez De\u011fi\u015fkenler<\/h2>\n\n\n\n
Kal\u0131b\u0131 se\u00e7tiniz. Geri esnemeyi \u00f6l\u00e7t\u00fcn\u00fcz. Test \u015feridi \u00e7izime uyana kadar K\u2011fakt\u00f6r\u00fcn\u00fc ayarlad\u0131n\u0131z.<\/p>\n\n\n\n
Ve \u00fcretim serisi h\u00e2l\u00e2 kay\u0131yor.<\/p>\n\n\n\n
\u0130\u015fte 200 tonluk ger\u00e7ek makinesi ad\u0131n\u0131 burada kazan\u0131yor. \u0130\u00e7ine temiz girdiler, kalibre edilmi\u015f yar\u0131\u00e7aplar, \u00f6l\u00e7\u00fclm\u00fc\u015f fazla b\u00fck\u00fcm besleyebilirsiniz\u2014yine de metalin veya onu tutan donan\u0131m\u0131n i\u00e7inde gizlenen zay\u0131f varsay\u0131mlar\u0131 ortaya \u00e7\u0131kar\u0131r. Form\u00fcl y\u00fcksek sesle hata yapmaz. Par\u00e7alar ise ayaklan\u0131r ve tolerans d\u0131\u015f\u0131na \u00e7\u0131kar.<\/p>\n\n\n\n
CN-HAWE\u2019nin m\u00fc\u015fteri taban\u0131; in\u015faat makineleri, otomotiv \u00fcretimi, gemi in\u015fas\u0131, k\u00f6pr\u00fcler, havac\u0131l\u0131k gibi sekt\u00f6rleri kapsad\u0131\u011f\u0131ndan, burada pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Lazer Kesim Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/p>\n\n\n\nM\u00fckemmel matematik sadece m\u00fckemmel ko\u015fullarda hayatta kal\u0131r.<\/p>\n\n\n\n
Sorun \u015fu ki, bir \u00fcretim at\u00f6lyesinde hi\u00e7bir \u015fey uzun s\u00fcre m\u00fckemmel kalmaz.<\/p>\n\n\n\n
Peki \u201cher \u015feyi do\u011fru yapt\u0131\u011f\u0131n\u0131zda\u201d bile yar\u0131\u00e7ap\u0131n\u0131z\u0131 h\u00e2l\u00e2 de\u011fi\u015ftirebilen tek \u015fey ne kald\u0131?<\/p>\n\n\n\n
Tane y\u00f6n\u00fc: Haddelenme y\u00f6n\u00fc gerekli fazla b\u00fckme a\u00e7\u0131s\u0131n\u0131 nas\u0131l de\u011fi\u015ftirir<\/h3>\n\n\n\n
\u00c7elik izotropik de\u011fildir. Bu, basit bir at\u00f6lye ger\u00e7e\u011fi i\u00e7in kitap diliyle \u015fu anlama gelir: bir y\u00f6nde di\u011ferine g\u00f6re daha kolay b\u00fck\u00fcl\u00fcr.<\/p>\n\n\n\n
Levha haddeleme s\u0131ras\u0131nda, taneler haddeleme y\u00f6n\u00fcnde uzar. Tane y\u00f6n\u00fcne dik b\u00fck\u00fcld\u00fc\u011f\u00fcnde, malzeme bu uzam\u0131\u015f lifler boyunca gerilir. Tane y\u00f6n\u00fcne paralel b\u00fck\u00fcld\u00fc\u011f\u00fcnde ise onlar\u0131 bir fermuar gibi a\u00e7maya \u00e7al\u0131\u015f\u0131rs\u0131n\u0131z. Ayn\u0131 kal\u0131nl\u0131k. Ayn\u0131 kal\u0131p. Farkl\u0131 yaylanma miktar\u0131.<\/p>\n\n\n\n
0.125\u2033 paslanmaz \u00e7eli\u011fin tane y\u00f6n\u00fcne dik olarak tam 90\u00b0 b\u00fck\u00fcl\u00fcp \u2014 ayn\u0131 1.000\u2033 V i\u00e7inde tane y\u00f6n\u00fcyle b\u00fck\u00fcld\u00fc\u011f\u00fcnde neredeyse bir derece daha fazla geri s\u0131\u00e7rad\u0131\u011f\u0131n\u0131 izledim. Ba\u015fka hi\u00e7bir \u015fey de\u011fi\u015fmedi. Ayn\u0131 program. Ayn\u0131 operat\u00f6r. Test par\u00e7as\u0131ndan kalibre edilmi\u015f ayn\u0131 K fakt\u00f6r\u00fc.<\/p>\n\n\n\n
Tek fark y\u00f6nelimdi.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Test \u015feridiniz tane y\u00f6n\u00fcne dik kesilmi\u015f, ancak \u00fcretim bo\u015fluklar\u0131 malzemeden tasarruf etmek i\u00e7in tane y\u00f6n\u00fc uzun eksende olacak \u015fekilde yerle\u015ftirilmi\u015fse, ilk darbeden \u00f6nce fazla b\u00fckme telafiniz zaten yanl\u0131\u015ft\u0131r. Presin hangi y\u00f6n\u00fcn daha ucuz yerle\u015ftirildi\u011fi umurunda de\u011fildir.<\/p>\n\n\n\n
A\u00e7\u0131 sens\u00f6rleriyle donat\u0131lm\u0131\u015f modern makineler ger\u00e7ek zamanl\u0131 otomatik d\u00fczeltme yapabilir. G\u00fczel. Bu da noktay\u0131 kan\u0131tlar. Tane \u00f6nemli olmasayd\u0131, her b\u00fckmede \u00f6l\u00e7\u00fcm yapmalar\u0131 gerekmezdi. Ancak modern bir \u00fcretim at\u00f6lyesine girip ger\u00e7ekte ne oldu\u011funa bak\u0131n \u2014 h\u00e2l\u00e2 canl\u0131 a\u00e7\u0131 d\u00fczeltmesi olmadan \u00e7al\u0131\u015fan, d\u00fcnk\u00fc rakamlara g\u00fcvenen bir\u00e7ok presle kar\u015f\u0131la\u015f\u0131rs\u0131n\u0131z.<\/p>\n\n\n\n
Metal y\u00f6n\u00fcne ba\u011fl\u0131 olarak sertli\u011fini de\u011fi\u015ftiriyorsa, sabit bir yar\u0131\u00e7ap form\u00fcl\u00fc levhan\u0131n nas\u0131l do\u011fdu\u011funu hi\u00e7 hat\u0131rlam\u0131yormu\u015f gibi davranabilir mi?<\/p>\n\n\n\n
Tonaj tuza\u011f\u0131: Hatal\u0131 bir hesaplamay\u0131 d\u00fczeltmek i\u00e7in makine bas\u0131nc\u0131n\u0131 art\u0131rmak neden daha k\u00f6t\u00fc sorunlar yarat\u0131r<\/h3>\n\n\n\n
A\u00e7\u0131 beklendi\u011finden daha fazla a\u00e7\u0131ld\u0131\u011f\u0131nda g\u00f6rd\u00fc\u011f\u00fcm hamle \u015fudur: tonaj\u0131 art\u0131rmak.<\/p>\n\n\n\n
Daha fazla bas\u0131n\u00e7 kontrol gibi hissettirir. Ama \u00f6yle de\u011fildir.<\/p>\n\n\n\n
Hava b\u00fckmede, tonaj do\u011frudan yar\u0131\u00e7ap\u0131 belirlemez \u2014 kal\u0131p geni\u015fli\u011fi belirler. Z\u0131mba, levhay\u0131 kal\u0131b\u0131n omuzlar\u0131 aras\u0131na iter, ancak malzemeyi tam olarak V i\u00e7ine zorlamaz. Formu, kuvvetle de\u011fil, geometrik olarak olu\u015fturuyorsunuz. Gerekenden fazla bas\u0131n\u00e7 uygulad\u0131\u011f\u0131n\u0131zda a\u00e7\u0131y\u0131 sihirli bir \u015fekilde \u201ckilitlemezsiniz.\u201d Dibe tam oturma, bask\u0131lama ve b\u00fckme hatt\u0131 boyunca tutars\u0131z gerilme riskini ba\u015flat\u0131rs\u0131n\u0131z.<\/p>\n\n\n\n
Art\u0131k kalibre edilmi\u015f hava b\u00fckme matemati\u011finiz k\u0131smi oturma davran\u0131\u015f\u0131yla kar\u0131\u015f\u0131yor.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Operat\u00f6rlerin, yar\u0131m derece yaylanma fark\u0131n\u0131 kapatmak i\u00e7in tonaj ekledi\u011fini, ancak hafif e\u011frilme ve dengesiz temas nedeniyle u\u00e7larda daha dar, ortada daha gev\u015fek a\u00e7\u0131lar olu\u015fturdu\u011funu g\u00f6rd\u00fcm. Par\u00e7alar presden \u00e7\u0131kt\u0131\u011f\u0131nda iyi g\u00f6r\u00fcn\u00fcyordu. Montajda uymad\u0131lar.<\/p>\n\n\n\n
Daha fazla kuvvet k\u00fc\u00e7\u00fck kurulum hatalar\u0131n\u0131 b\u00fcy\u00fct\u00fcr. Hafif ta\u00e7lanma hatas\u0131. Hafif malzeme kal\u0131nl\u0131k fark\u0131. Hafif ko\u00e7 e\u011filmesi. \u00d6ng\u00f6r\u00fclebilir 1.5\u00b0 yaylanma art\u0131k burada 1.2\u00b0, orada 1.8\u00b0 olur. Form\u00fcl\u00fc d\u00fczeltmediniz \u2014 gerilme desenini bulan\u0131kla\u015ft\u0131rd\u0131n\u0131z.<\/p>\n\n\n\n
Telafi y\u00f6nteminiz yaln\u0131zca belirli bir bas\u0131n\u00e7 aral\u0131\u011f\u0131nda \u00e7al\u0131\u015f\u0131yorsa, ger\u00e7ekten telafi mi \u2014 yoksa dar bir aral\u0131kta \u015fans m\u0131?<\/p>\n\n\n\n
Tak\u0131m a\u015f\u0131nmas\u0131 ve hizas\u0131zl\u0131k: Fiziksel kal\u0131p art\u0131k hesap makinenizin girdisiyle uyu\u015fmad\u0131\u011f\u0131nda ne olur?<\/h3>\n\n\n\n
Hesap makineniz, keskin ve tutarl\u0131 omuzlara sahip 1.000\u2033 V\u2019de b\u00fck\u00fcm yapt\u0131\u011f\u0131n\u0131z\u0131 san\u0131yor.<\/p>\n\n\n\n
\u0130ki y\u0131ll\u0131k \u00fcretimden sonra o kal\u0131b\u0131 gidip \u00f6l\u00e7\u00fcn.<\/p>\n\n\n\n
Kal\u0131p omuzlar\u0131 a\u015f\u0131n\u0131r. Hafif\u00e7e \u015fi\u015fer, p\u00fcr\u00fczs\u00fczle\u015fir, birka\u00e7 binde geni\u015fler. \u201c1.000 in\u00e7\u201d V art\u0131k temas noktalar\u0131nda 1.020\u2033 gibi davranabilir. Bu, etkin V\/T oran\u0131n\u0131 geni\u015fletir. Daha geni\u015f oran, y\u00fck alt\u0131ndaki yar\u0131\u00e7ap\u0131n b\u00fcy\u00fcmesi demektir. Daha b\u00fcy\u00fck yar\u0131\u00e7ap ise daha fazla yaylanma anlam\u0131na gelir.<\/p>\n\n\n\n
Matemati\u011finiz h\u00e2l\u00e2 d\u00fcn\u00fcn geometrisini kullan\u0131yor.<\/p>\n\n\n\n
Sonra toleranslar\u0131 bir araya getirin: b\u00f6l\u00fcml\u00fc tak\u0131mda biraz sa\u011f-sol hizas\u0131zl\u0131\u011f\u0131, bir kal\u0131p par\u00e7as\u0131n\u0131n alt\u0131na konulmu\u015f bir takoz, tam oturmam\u0131\u015f bir kelep\u00e7e. Art\u0131k b\u00fck\u00fcm hatt\u0131 boyunca ko\u015fullar e\u015fit de\u011fil. Tek bir K-fakt\u00f6r\u00fc, hareket eden bir hedefi tan\u0131mlamaya \u00e7al\u0131\u015f\u0131yor.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Flan\u015f uzunluklar\u0131 uzun bir par\u00e7a boyunca yirmi ya da otuz binde kadar sapmaya ba\u015flad\u0131\u011f\u0131nda at\u00f6lyeler su\u00e7luyu programc\u0131da arar. \u00c7o\u011fu zaman yeni bir kal\u0131p par\u00e7as\u0131 sorunu \u00e7\u00f6zer. Elektronik tabloya hi\u00e7 \u015fans tan\u0131nmam\u0131\u015ft\u0131r.<\/p>\n\n\n\n
Evet, daha yeni yukar\u0131dan etkili pres freni sistemleri ve daha ak\u0131ll\u0131 bombe sistemleri tarih\u00ee sapma problemlerini azalt\u0131r. \u0130yi makineler hata aral\u0131\u011f\u0131n\u0131 daralt\u0131r. Fizi\u011fi ortadan kald\u0131rmazlar. Tak\u0131mlar h\u00e2l\u00e2 a\u015f\u0131n\u0131r. Y\u00fczeyler h\u00e2l\u00e2 y\u00fck alt\u0131nda \u015fekil de\u011fi\u015ftirir. \u00c7elik h\u00e2l\u00e2 liflidir.<\/p>\n\n\n\n
Pres freni, varsay\u0131mlar\u0131n\u0131z i\u00e7in bir dayan\u0131kl\u0131l\u0131k testidir.<\/p>\n\n\n\n
Kariyerinizin geri kalan\u0131n\u0131 evrensel bir yar\u0131\u00e7ap form\u00fcl\u00fc kovalamaya harcayabilirsiniz ya da makinenin size s\u00fcrekli s\u00f6yledi\u011fini kabul edebilirsiniz: yar\u0131\u00e7ap bir kez hesaplayaca\u011f\u0131n\u0131z bir say\u0131 de\u011fil \u2014 kontrol edip izlemeniz ve d\u00fczeltmeniz gereken bir ko\u015fuldur.<\/p>\n\n\n\n
E\u011fer lif sertli\u011fi de\u011fi\u015ftiriyorsa, tonaj gerilmeyi etkiliyorsa ve tak\u0131m geometrisi zamanla kay\u0131yorsa, neden h\u00e2l\u00e2 tek bir statik denklemin bizi hurda kutusundan koruyabilece\u011fine inan\u0131yoruz?<\/p>\n\n\n\n
Say\u0131lar\u0131n Pe\u015finden Ko\u015fmay\u0131 B\u0131rak\u0131n: Ger\u00e7ekten \u0130\u015fe Yarayan Bir Yar\u0131\u00e7ap Hesaplama Sistemi Kurmak<\/h2>\n\n\n\n
De\u011fi\u015fen bir yar\u0131\u00e7ap\u0131 daha iyi bir hesap makinesiyle d\u00fczeltemezsiniz.<\/p>\n\n\n\n
Bunu, do\u011frulamad\u0131\u011f\u0131n\u0131z s\u00fcrece pres freninin sizi yan\u0131ltaca\u011f\u0131n\u0131 varsayan bir sistem kurarak d\u00fczeltirsiniz.<\/p>\n\n\n\n
Bir 10 gauge braketi \u00fczerinde, 1.000 par\u00e7al\u0131k bir serinin ortas\u0131nda, ketenin \u00f6n\u00fcnden arkas\u0131na ge\u00e7erken rulo sertli\u011fi de\u011fi\u015ftik\u00e7e a\u00e7\u0131lar\u0131n yar\u0131m derece kayd\u0131\u011f\u0131n\u0131 izledim. Form\u00fcl de\u011fi\u015fmedi. V-kal\u0131p de\u011fi\u015fmedi. Malzeme de\u011fi\u015fti. \u0130\u015fte o anda anl\u0131yorsunuz: \u00e7\u00f6zmeye \u00e7al\u0131\u015ft\u0131\u011f\u0131n\u0131z bir say\u0131 de\u011fil \u2014 hareket eden bir s\u00fcreci kontrol ediyorsunuz.<\/p>\n\n\n\n
Pres freni 200 tonluk bir ger\u00e7eklik makinesidir. Varsay\u0131mlardaki zay\u0131fl\u0131klar\u0131, bir \u00e7ekme testinin zay\u0131f \u00e7eli\u011fi ortaya \u00e7\u0131kard\u0131\u011f\u0131 gibi ortaya \u00e7\u0131kar\u0131r. E\u011fer \u201cyar\u0131\u00e7ap form\u00fcl\u00fcn\u00fcz\u201d sadece t\u00fcm y\u0131ld\u0131zlar hizaland\u0131\u011f\u0131nda \u2014 ayn\u0131 \u0131s\u0131 numaras\u0131, ayn\u0131 lif y\u00f6n\u00fc, ayn\u0131 kal\u0131p a\u015f\u0131nmas\u0131, ayn\u0131 tonaj aral\u0131\u011f\u0131 \u2014 i\u015fe yar\u0131yorsa, bir form\u00fcl\u00fcn\u00fcz yoktur. Tesad\u00fcf\u00fcn\u00fcz vard\u0131r.<\/p>\n\n\n\n
Peki bir sistem teori yerine ger\u00e7eklik i\u00e7in in\u015fa edildi\u011finde nas\u0131l g\u00f6r\u00fcn\u00fcr?<\/p>\n\n\n\n
Neden tek bir evrensel form\u00fcl farkl\u0131 malzeme ve kal\u0131nl\u0131klarda her zaman ba\u015far\u0131s\u0131z olur<\/h3>\n\n\n\n
Kal\u0131nl\u0131k s\u0131\u00e7ramalar\u0131yla ba\u015flay\u0131n.<\/p>\n\n\n\n
Orta aral\u0131kta bir plaka i\u00e7in \u2014 diyelim ki 6 ila 12 mm \u2014 yayg\u0131n kal\u0131p oranlar\u0131yla havada b\u00fck\u00fcm yaparken i\u00e7 yar\u0131\u00e7ap\u0131n genellikle malzeme kal\u0131nl\u0131\u011f\u0131n\u0131n yakla\u015f\u0131k 1,5 kat\u0131 civar\u0131nda olu\u015ftu\u011funu g\u00f6receksiniz. 12 mm\u2019nin \u00fczerine \u00e7\u0131kt\u0131\u011f\u0131n\u0131zda ise bu yar\u0131\u00e7ap, V a\u00e7\u0131kl\u0131\u011f\u0131n\u0131 \u201cdo\u011fru\u201d \u00f6l\u00e7ekleseniz bile kal\u0131nl\u0131\u011f\u0131n yakla\u015f\u0131k 3 kat\u0131na t\u0131rmanabilir. Bu bir yuvarlama hatas\u0131 de\u011fildir. Bu, do\u011frusal olmayan bir davran\u0131\u015ft\u0131r.<\/p>\n\n\n\n
(V \u2212 MT)\/2 tayfas\u0131, malzemenin d\u00fczg\u00fcn \u015fekilde \u00f6l\u00e7eklendi\u011fini varsay\u0131yor. Ger\u00e7ek \u00e7elik \u00f6yle de\u011fildir.<\/p>\n\n\n\n
Kal\u0131nl\u0131k artt\u0131k\u00e7a sadece daha fazla malzemeyi b\u00fckm\u00fcyorsun \u2014 kesit boyunca gerinim da\u011f\u0131l\u0131m\u0131n\u0131 da de\u011fi\u015ftiriyorsun. N\u00f6tr eksen kayar. Gerekli tonaj artar. Yaylanma (springback) artar \u00e7\u00fcnk\u00fc d\u0131\u015f lifler daha fazla elastik enerji depolamaktad\u0131r. K\u00e2\u011f\u0131t \u00fczerinde ayn\u0131 V\/T oran\u0131, y\u00fck alt\u0131nda farkl\u0131 bir i\u00e7 gerilim tablosu.<\/p>\n\n\n\n
\u015eimdi \u00e7ekme dayan\u0131m\u0131n\u0131 devreye sokal\u0131m.<\/p>\n\n\n\n
Ayn\u0131 kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131nda d\u00fc\u015f\u00fck akma s\u0131n\u0131r\u0131na sahip 5052 al\u00fcminyum ile y\u00fcksek dayan\u0131ml\u0131 paslanmaz \u00e7elik ayn\u0131 \u015fekilde yaylanmaz. Daha g\u00fc\u00e7l\u00fc malzeme akmadan \u00f6nce daha fazla elastik gerinim depolar. O enerji z\u0131mba kalkt\u0131\u011f\u0131nda sana geri d\u00f6ner. E\u011fer yar\u0131\u00e7ap hesab\u0131n akma dayan\u0131m\u0131n\u0131 bilmiyorsa, k\u00f6rd\u00fcr.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: \u201cAyn\u0131\u201d A36\u2019n\u0131n iki farkl\u0131 ergitme partisinin d\u00f6rt b\u00fck\u00fcde flan\u015f uzunlu\u011funu otuz binde bir in\u00e7 de\u011fi\u015ftirecek kadar farkl\u0131 davrand\u0131\u011f\u0131n\u0131 g\u00f6rd\u00fcm. Hesap tablosu ayn\u0131 girdiler demi\u015fti. Par\u00e7alar ba\u015fka \u015fey s\u00f6yledi.<\/p>\n\n\n\n
Ve donan\u0131m hatas\u0131na hen\u00fcz de\u011finmedik bile. Kal\u0131pta ondal\u0131k milimetre kadar bir hizas\u0131zl\u0131k a\u00e7\u0131n\u0131 yar\u0131m dereceden fazla sapt\u0131rabilir. Evrensel form\u00fcl\u00fcn kusursuz geometriden yola \u00e7\u0131kar. At\u00f6lyen \u00f6yle de\u011fildir.<\/p>\n\n\n\n
E\u011fer kal\u0131nl\u0131k katmanlar\u0131 do\u011frusal olmayan \u015fekilde davran\u0131yor, \u00e7ekme dayan\u0131m\u0131 yaylanmay\u0131 de\u011fi\u015ftiriyor ve tak\u0131m hizas\u0131 her g\u00fcn kay\u0131yorsa, o tek temiz denklem tam olarak neyi temsil ediyor?<\/p>\n\n\n\n
Ger\u00e7ek test verilerinden malzemeye \u00f6zg\u00fc bir yar\u0131\u00e7ap tablosu olu\u015fturmak<\/h3>\n\n\n\n
Tahmin etmeyi b\u0131rak\u0131r, \u00f6l\u00e7meye ba\u015flars\u0131n.<\/p>\n\n\n\n
Kulland\u0131\u011f\u0131n standart kal\u0131p oranlar\u0131n\u0131 se\u00e7 \u2014 ger\u00e7ekten her g\u00fcn kulland\u0131klar\u0131n\u0131. Her malzeme s\u0131n\u0131f\u0131 ve kal\u0131nl\u0131k aral\u0131\u011f\u0131 i\u00e7in, tane y\u00f6n\u00fc not edilmi\u015f test \u015feritleri kes. Kontroll\u00fc bir a\u00e7\u0131ya hava b\u00fck\u00fcm\u00fc yap. Yaylanmadan sonra i\u00e7 yar\u0131\u00e7ap\u0131 yar\u0131\u00e7ap mastarlar\u0131 veya optik \u00f6l\u00e7\u00fcmle \u00f6l\u00e7. 90\u00b0 serbest duruma ula\u015fmak i\u00e7in gereken ger\u00e7ek a\u015f\u0131r\u0131 b\u00fck\u00fcm miktar\u0131n\u0131 kaydet.<\/p>\n\n\n\n
Bunu her malzeme ailesi ve kal\u0131nl\u0131k katman\u0131 i\u00e7in bir kez yap. Her i\u015f i\u00e7in de\u011fil. Her kontroll\u00fc ko\u015ful i\u00e7in bir kez.<\/p>\n\n\n\n
\u015eimdi \u015f\u00f6yle s\u00f6yleyen bir tablo olu\u015fturursun, \u00f6rne\u011fin (varsay\u0131msal rakamlar):<\/p>\n\n\n\n
\n- 0,125\u2033 5052, 8\u00d7T kal\u0131p \u2192 0,160\u2033 ger\u00e7ek i\u00e7 yar\u0131\u00e7ap, 1,0\u00b0 yaylanma<\/li>\n\n\n\n
- 0,125\u2033 304 paslanmaz, 8\u00d7T kal\u0131p \u2192 0,185\u2033 yar\u0131\u00e7ap, 2,2\u00b0 yaylanma<\/li>\n<\/ul>\n\n\n\n
O tablo teori de\u011fildir. Yaral\u0131 tecr\u00fcbedir.<\/p>\n\n\n\n
Sertifikal\u0131 \u00e7ekme verisi olmayan yeni malzeme geldi\u011finde, bir kupon b\u00fck ve nas\u0131l davrand\u0131\u011f\u0131na bak. Pres frende be\u015f dakika ge\u00e7irmek, par\u00e7alar\u0131 be\u015f saat yeniden i\u015flemeye ye\u011fdir. E\u011fer akma de\u011ferin varsa, g\u00fczel \u2014 bunlar\u0131 tablo e\u011filimlerinle kar\u015f\u0131la\u015ft\u0131r. Zamanla kendi makinen, kendi tak\u0131m sistemin ile dayan\u0131m ve yaylanma aras\u0131ndaki desenleri g\u00f6r\u00fcrs\u00fcn.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Bu ad\u0131m\u0131 atlayan at\u00f6lyeler her ilk numuneyi \u00fcretim par\u00e7alar\u0131 \u00fczerinde \u201cayar \u00e7ekerek\u201d ge\u00e7erler. Bu esneklik de\u011fildir. Bu, sac stokla kumar oynamakt\u0131r.<\/p>\n\n\n\n
Ve \u00e7o\u011funun ka\u00e7\u0131rd\u0131\u011f\u0131 nokta \u015fu: tabloya g\u00fcvenmeden \u00f6nce makineyi kontrol etmelisin. G\u00fcnl\u00fck arka dayama do\u011frulamas\u0131. Tak\u0131mlar\u0131n yerine tam oturdu\u011fundan ve temiz oldu\u011fundan emin ol. Kal\u0131p geni\u015fli\u011fini varsay\u0131mla de\u011fil pimlerle kontrol et. Donan\u0131m kayarsa, verin \u00e7\u00fcr\u00fcr.<\/p>\n\n\n\n
Kontrols\u00fcz kurulumla olu\u015fturulmu\u015f tablo yaln\u0131zca d\u00fczenlenmi\u015f bir kurgudur.<\/p>\n\n\n\n
Yani hesaplay\u0131c\u0131 i\u00e7 yar\u0131\u00e7ap\u0131 sordu\u011funda, bir tahmin mi yaz\u0131yorsun \u2014 yoksa pres freninin zaten kan\u0131tlad\u0131\u011f\u0131 bir veritaban\u0131ndan m\u0131 \u00e7ekiyorsun?<\/p>\n\n\n\n
\u201chesap makinesi girdisine\u201d g\u00fcvenmekten \u201cs\u00fcre\u00e7 sahipli\u011fini\u201d \u00fcstlenmeye ge\u00e7i\u015f\u201d<\/h3>\n\n\n\n
Bu, zihniyet de\u011fi\u015fimidir.<\/p>\n\n\n\n
Hesap makinesi yanl\u0131\u015f de\u011fildir. Eksiktir. Fiziksel d\u00fcnyay\u0131 do\u011fru tan\u0131mlad\u0131\u011f\u0131n\u0131z\u0131 varsayar \u2014 ger\u00e7ek kal\u0131p geni\u015fli\u011fi, ger\u00e7ek malzeme davran\u0131\u015f\u0131, do\u011frulanm\u0131\u015f hizalama. \u00c7o\u011fu at\u00f6lye nominal de\u011ferleri girer ve umut eder.<\/p>\n\n\n\n
S\u00fcre\u00e7 sahipli\u011fi, bu girdileri tan\u0131mlamak ve savunmak anlam\u0131na gelir.<\/p>\n\n\n\n
Kal\u0131plar\u0131 rastgele de\u011fi\u015ftirmek yerine V\/T oranlar\u0131n\u0131 standartla\u015ft\u0131r\u0131rs\u0131n\u0131z. Hangi malzemelerin hangi a\u00e7\u0131kl\u0131klarda kullan\u0131laca\u011f\u0131n\u0131 belirlersiniz. Bask\u0131larda lif y\u00f6n\u00fc gerekliliklerini belgelersiniz. Yeni s\u0131caklar\u0131 b\u00fckme testini ge\u00e7ene kadar karantinaya al\u0131rs\u0131n\u0131z. A\u00e7\u0131 sapmas\u0131n\u0131 bir sinyal olarak ele al\u0131rs\u0131n\u0131z \u2014 daha fazla tonaj ile ezilmesi gereken bir rahats\u0131zl\u0131k olarak de\u011fil.<\/p>\n\n\n\n
Ve evrensel bir i\u00e7 yar\u0131\u00e7ap form\u00fcl\u00fc olmad\u0131\u011f\u0131n\u0131 kabul edersiniz.<\/p>\n\n\n\n
Yaln\u0131zca sizin yar\u0131\u00e7ap\u0131n\u0131z vard\u0131r \u2014 sizin kal\u0131plar\u0131n\u0131zda, sizin pres freninizde, sizin malzemelerinizle \u2014 y\u00fck alt\u0131nda do\u011frulanm\u0131\u015f, el kitab\u0131n\u0131n ne s\u00f6yledi\u011fine ald\u0131rmayan 200 tonluk bir ger\u00e7ek makinesiyle.<\/p>\n\n\n\n
\u0130leri ta\u015f\u0131nmas\u0131 gereken tek \u015fey \u015fu: yar\u0131\u00e7ap hesaplad\u0131\u011f\u0131n\u0131z bir say\u0131 de\u011fildir \u2014 tan\u0131mlad\u0131\u011f\u0131n\u0131z bir davran\u0131\u015ft\u0131r.<\/p>\n\n\n\n
Bunu bu \u015fekilde g\u00f6rd\u00fc\u011f\u00fcn\u00fczde, soru \u201cForm\u00fcl nedir?\u201d olmaktan \u00e7\u0131kar ve \u201cS\u00fcrecim form\u00fcl\u00fcn anlam kazanaca\u011f\u0131 kadar s\u0131k\u0131 m\u0131?\u201d haline gelir.\u201d<\/p>\n\n\n\n
\u0130lgili Kaynaklar ve Sonraki Ad\u0131mlar<\/h2>\n\n\n\n\n- Buradaki pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Giyotin Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/li>\n\n\n\n
- Buradaki pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Panel B\u00fckme Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/li>\n\n\n\n
- Buradaki pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Lazer Kaynak Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/li>\n\n\n\n
- Buradaki pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Sac Yuvarlama Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/li>\n\n\n\n
- Buradaki pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, V Kanal A\u00e7ma Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/li>\n\n\n\n
- Buradaki pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Demir \u0130\u015fleme Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/li>\n<\/ul>","protected":false},"excerpt":{"rendered":"
Bir 10 gauge ba\u011flant\u0131 par\u00e7as\u0131n\u0131n \u00fczerinde durdum, 1.000\u2033 V kal\u0131b\u0131, \u00e7izimde a\u00e7\u0131k ve net rakamlar vard\u0131. (V \u2212 MT) \/ 2 form\u00fcl\u00fc, i\u00e7 yar\u0131\u00e7ap\u0131n 0.433\u2033 olmas\u0131 gerekti\u011fini s\u00f6yl\u00fcyordu. Kumpas ise 0.470\u2033 dedi. Her bir par\u00e7a \u00f6yleydi. Hesab\u0131n\u0131 kontrol ettin. Kal\u0131p damgas\u0131n\u0131 kontrol ettin. Malzeme partisinin su\u00e7lu oldu\u011funu d\u00fc\u015f\u00fcnd\u00fcn. Bu arada hurda kutusu sessizce dolmaya ba\u015flad\u0131 [\u2026]<\/p>","protected":false},"author":3,"featured_media":1427,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_breakdance_hide_in_design_set":false,"_breakdance_tags":"","footnotes":""},"categories":[1],"tags":[],"class_list":["post-1426","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/posts\/1426","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/comments?post=1426"}],"version-history":[{"count":1,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/posts\/1426\/revisions"}],"predecessor-version":[{"id":1431,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/posts\/1426\/revisions\/1431"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/media\/1427"}],"wp:attachment":[{"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/media?parent=1426"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/categories?post=1426"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/tags?post=1426"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
<\/figure>\n\n\n\nOturtmal\u0131 b\u00fck\u00fcm, kaba kuvvettir. Malzemeyi kal\u0131ba sokars\u0131n, ta ki tam uyana kadar. Geri yaylanma, hava b\u00fck\u00fcmden \u00fc\u00e7 ila be\u015f kat fazla tonaj uygulad\u0131\u011f\u0131n i\u00e7in ezilerek yok olur. A\u00e7\u0131y\u0131 kal\u0131p belirler. Yar\u0131\u00e7ap fiilen damgalanm\u0131\u015f olur.<\/p>\n\n\n\n
Form\u00fcl\u00fcn ait oldu\u011fu d\u00fcnya bu.<\/p>\n\n\n\n
Ama bir hava b\u00fckme kurulumunda \u201cmatemati\u011fi do\u011fru yapmaya\u201d \u00e7al\u0131\u015f\u0131rken pres tonaj\u0131n\u0131 art\u0131r\u0131rsan\u0131z, kal\u0131plar\u0131 \u00e7atlatabilecek veya ko\u00e7u zorlayabilecek a\u015f\u0131r\u0131 tonajla fl\u00f6rt ediyorsunuz demektir. Operat\u00f6rlerin, bir bijon somununu s\u0131kar gibi yaylanmay\u0131 dengelemek i\u00e7in bas\u0131n\u00e7 ekledi\u011fini g\u00f6rd\u00fcm. Par\u00e7a do\u011fru a\u00e7\u0131y\u0131 tutturabilir. Ancak tak\u0131m, bunun bedelini daha sonra \u00f6der.<\/p>\n\n\n\n
Hava b\u00fckme, o \u015fiddetten ka\u00e7\u0131nmak i\u00e7in tasarlanm\u0131\u015ft\u0131r. Kaba kesinli\u011fi esneklik ve daha d\u00fc\u015f\u00fck tonajla de\u011fi\u015ftirir. Sadece strok derinli\u011fini de\u011fi\u015ftirerek ayn\u0131 kal\u0131pta birden fazla a\u00e7\u0131 \u00e7al\u0131\u015fabilirsiniz. Bu \u00e7ok y\u00f6nl\u00fcl\u00fck, \u00e7o\u011fu at\u00f6lyenin neredeyse her \u015fey i\u00e7in onu kullanmas\u0131n\u0131n nedenidir.<\/p>\n\n\n\n
Yine de, art\u0131k kullanmad\u0131\u011f\u0131m\u0131z bir i\u015flem i\u00e7in tasarlanm\u0131\u015f bir form\u00fcle h\u00e2l\u00e2 tutunuyoruz.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Vuru derinli\u011fini dinamik olarak ayarlamak m\u0131 daha ucuz, yoksa kitab\u00ee ger\u00e7e\u011fi zorlamak isteyip \u00e7atlak bir kal\u0131b\u0131 de\u011fi\u015ftirmek mi?<\/p>\n\n\n\n
Matematik yanl\u0131\u015f de\u011filse ama model yanl\u0131\u015fsa, bunun uzun vadede size maliyeti nedir?<\/p>\n\n\n\n
Da\u011f\u0131n\u0131k bir at\u00f6lyede kusursuz bir form\u00fcle inanman\u0131n gizli maliyeti<\/h3>\n\n\n\n<\/figure>\n\n\n\nForm\u00fcl\u00fcn g\u00f6rmedi\u011fi \u015fey \u015fu: rulolar aras\u0131ndaki \u00e7ekme dayan\u0131m\u0131 fark\u0131, hafif bombe yapm\u0131\u015f bir yatak, a\u015f\u0131nm\u0131\u015f z\u0131mba u\u00e7lar\u0131, s\u0131cak bir \u00f6\u011fleden sonra hidrolik kaymas\u0131. Bunlar\u0131n hi\u00e7biri (V \u2212 MT) \/ 2 denkleminde g\u00f6r\u00fcnmez.<\/p>\n\n\n\n
Ama par\u00e7alar\u0131n\u0131zda g\u00f6r\u00fcn\u00fcr.<\/p>\n\n\n\n
Hava b\u00fckme, pres frenin kendisini kritik de\u011fi\u015fken h\u00e2line getirir. Vuru hassasiyeti. Ko\u00e7 paralelli\u011fi. Arka dayama tutarl\u0131l\u0131\u011f\u0131. Alt b\u00fckmede a\u011f\u0131r i\u015fi kal\u0131p yapar. Hava b\u00fckmede makine yapar. Daha temiz bir form\u00fcl, tolerans dahilinde tekrar edemeyen bir makineyi d\u00fczeltmez. Bu y\u00fczden bir\u00e7ok at\u00f6lye, y\u00fcksek hassasiyetli b\u00fck\u00fcm ve otomasyon i\u00e7in tasarlanm\u0131\u015f tamamen CNC kontroll\u00fc sistemlere y\u00f6nelir\u2014\u015funlardaki gibi CN-HAWE pres b\u00fckme serisi<\/a>\u2014burada vuru derinli\u011fi, paralellik ve tekrarlanabilirli\u011fin kontrol\u00fc tahmini de\u011fil, \u00f6l\u00e7\u00fclebilir bir yetenek h\u00e2line gelir.<\/p>\n\n\n\nGizli maliyet sadece hurda de\u011fildir. Bu s\u00fcre\u00e7 i\u00e7in asla tasarlanmam\u0131\u015f say\u0131lar\u0131 kovalamak i\u00e7in harcanan zamand\u0131r. Bu, sahte bir \u00f6zg\u00fcvendir. Ger\u00e7ek sorun dinamik bir sistemde sabit bir denklemi kullanmakken \u201ck\u00f6t\u00fc b\u00fck\u00fcmler\u201d i\u00e7in operat\u00f6rleri su\u00e7lamakt\u0131r.<\/p>\n\n\n\n
200 tonluk bir pres, cebirinizin ne kadar zarif g\u00f6r\u00fcnd\u00fc\u011f\u00fcyle ilgilenmez. Onun tek \u00f6nemsedi\u011fi \u015fey, z\u0131mban\u0131n nerede durdu\u011fudur.<\/p>\n\n\n\n
Yani kal\u0131p, hava b\u00fckmede yar\u0131\u00e7ap\u0131 ger\u00e7ekten tan\u0131mlam\u0131yorsa, o yar\u0131\u00e7ap asl\u0131nda nereden geliyor?<\/p>\n\n\n\n
V-Kal\u0131\u011f\u0131n Fizi\u011fi: Yar\u0131\u00e7ap Asl\u0131nda Nerede Olu\u015fur<\/h2>\n\n\n\n
Ayn\u0131 1.000\u2033 V-kal\u0131\u011f\u0131 ve 10 numara 60 KSI so\u011fuk haddelenmi\u015f \u00e7eli\u011fi al\u0131n. Hava b\u00fckme ile b\u00fck\u00fclm\u00fc\u015f on par\u00e7a boyunca i\u00e7 yar\u0131\u00e7ap\u0131 \u00f6l\u00e7\u00fcn. 0.433\u2033 elde edemezsiniz. Z\u0131mba ucu yar\u0131\u00e7ap\u0131n\u0131z\u0131 da elde edemezsiniz. Yakla\u015f\u0131k 0.160\u2033 ile 0.200\u2033 aras\u0131nda bir de\u011fere ula\u015f\u0131rs\u0131n\u0131z\u2014kaba olarak kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131n\u0131n \u201320\u2019si civar\u0131nda.<\/p>\n\n\n\n
Bu oran o kadar s\u0131k g\u00f6r\u00fcl\u00fcr ki art\u0131k tesad\u00fcf olmaktan \u00e7\u0131kar.<\/p>\n\n\n\n
Kal\u0131p sihirli bir \u015fekilde o say\u0131y\u0131 damgalad\u0131\u011f\u0131 i\u00e7in de\u011fil. Z\u0131mba burnu o de\u011fere uydu\u011fu i\u00e7in de de\u011fil. Levha kal\u0131p omuzlar\u0131 aras\u0131nda \u00e7\u00f6kerken, y\u00fck alt\u0131nda o a\u00e7\u0131kl\u0131\u011f\u0131 ne kadar a\u015fabilece\u011fine ba\u011fl\u0131 do\u011fal bir e\u011frilik buldu\u011fu i\u00e7in. Kal\u0131p geni\u015fli\u011fi sahneyi haz\u0131rlar. Malzeme gerilimi, bu mesafe boyunca ne kadar s\u0131k\u0131 b\u00fck\u00fclebilece\u011fini belirler. Yar\u0131\u00e7ap, omuzlar aras\u0131nda havada olu\u015fur, n\u00fcfuz derinli\u011fi taraf\u0131ndan kontrol edilir.<\/p>\n\n\n\n
Hava b\u00fckmede, temel i\u00e7 yar\u0131\u00e7ap\u0131 belirleyen \u015fey z\u0131mba ucu veya hesap makineniz de\u011fil, kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131d\u0131r. Geri kalan her \u015fey bunun \u00fczerine in\u015fa edilir.<\/p>\n\n\n\n
E\u011fer (V \u2212 MT)\/2\u2019nin pe\u015findeyseniz, yanl\u0131\u015f mekanizmay\u0131 \u00e7\u00f6z\u00fcyorsunuz demektir.<\/p>\n\n\n\n
Ger\u00e7ek temel \u00e7izgiyi hesaplamak: 8x ila 12x kal\u0131p oran\u0131 ger\u00e7eklik kontrol\u00fc<\/h3>\n\n\n\n
Yakla\u015f\u0131k 1\u2033 V kal\u0131b\u0131n alt\u0131nda, 10-gauge bir ba\u011flant\u0131 par\u00e7as\u0131n\u0131n \u00fczerinde durdum, \u00e7\u00fcnk\u00fc 0.135\u2033 kal\u0131nl\u0131\u011f\u0131ndaki 10\u2011gauge \u00e7elik 8 ile \u00e7arp\u0131l\u0131nca yakla\u015f\u0131k 1.08\u2033 eder. Yeterince yak\u0131n. O 8\u00d7 kural\u0131 bir efsane de\u011fildir. Tona makul tutar ve genellikle sizi kararl\u0131 bir b\u00fckme aral\u0131\u011f\u0131na yerle\u015ftirir.<\/p>\n\n\n\n
Matemati\u011fi geriye de\u011fil, ileriye do\u011fru \u00e7al\u0131\u015ft\u0131r\u0131n.<\/p>\n\n\n\n
E\u011fer o kal\u0131nl\u0131k i\u00e7in 1.000\u2033 V yakla\u015f\u0131k de\u011fer aral\u0131\u011f\u0131ndaysa ve hava b\u00fckme, 60\u2011KSI \u00e7elik i\u00e7in V\u2019nin yakla\u015f\u0131k \u201320\u2019si kadar bir i\u00e7 yar\u0131\u00e7ap \u00fcretiyorsa, daha ko\u00e7 bile metale dokunmadan 0.160\u2033\u20130.200\u2033 yar\u0131\u00e7ap tahmin ediyorsunuz demektir. Bu zaten (V \u2212 MT)\/2\u2019ye hi\u00e7 yak\u0131n de\u011fildir.<\/p>\n\n\n\n
\u015eimdi kal\u0131b\u0131 0.800\u2033e s\u0131k\u0131\u015ft\u0131r\u0131n\u2014yakla\u015f\u0131k 6\u00d7 kal\u0131nl\u0131k. Do\u011fal yar\u0131\u00e7ap\u0131n\u0131z yakla\u015f\u0131k 0.128\u2033\u20130.160\u2033\u2019a d\u00fc\u015fer. 1.250\u2033e a\u00e7\u0131n\u2014yakla\u015f\u0131k 10\u00d7\u2014ve yakla\u015f\u0131k 0.200\u2033\u20130.250\u2033 civar\u0131nda dola\u015f\u0131rs\u0131n\u0131z.<\/p>\n\n\n\n
Oran (8\u00d7, 10\u00d7, 12\u00d7) zarafetle ilgili de\u011fildir. Ayn\u0131 anda iki \u015feyi kontrol etmekle ilgilidir: ayak ba\u015f\u0131na tonaj ve bu a\u00e7\u0131kl\u0131ktan \u00e7\u0131kacak y\u00fczdeye dayal\u0131 yar\u0131\u00e7ap. Fazla dar olursa tonaj h\u0131zla artar. Fazla geni\u015f olursa, bask\u0131 ne isterse istesin yar\u0131\u00e7ap\u0131n\u0131z \u015fi\u015fer.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Hi\u00e7 \u201ctonajdan emin olmak i\u00e7in\u201d kal\u0131b\u0131 a\u00e7t\u0131n\u0131z m\u0131 ve sonra flan\u015f\u0131n\u0131z\u0131n neden e\u015f par\u00e7aya uymad\u0131\u011f\u0131n\u0131 merak ettiniz mi? Bu operat\u00f6r hatas\u0131 de\u011fildi. Bu, geometriydi; siz istemeseniz de yar\u0131\u00e7ap\u0131n\u0131z\u0131 \u00f6l\u00e7eklendirdi.<\/p>\n\n\n\n
Yani kal\u0131p geni\u015fli\u011fi temel y\u00fczdeyi belirliyorsa, malzemenin kendisi \u00f6nceki i\u015finizden daha zor\u2014ya da daha kolay\u2014b\u00fck\u00fcl\u00fcyorsa ne olur?<\/p>\n\n\n\n
\u00c7arpan kaymas\u0131: Paslanmaz \u00e7elik ile yumu\u015fak al\u00fcminyum i\u00e7in matemati\u011fi nas\u0131l ayarlamal\u0131<\/h3>\n\n\n\n
O 60\u2011KSI so\u011fuk haddelenmi\u015f \u00e7eli\u011fi ayn\u0131 1.000\u2033 V i\u00e7inde 304 paslanmazla de\u011fi\u015ftirin. Ayn\u0131 kal\u0131nl\u0131k. Ayn\u0131 z\u0131mba. 90\u00b0\u2019yi elde etmek i\u00e7in ayn\u0131 programlanm\u0131\u015f strok derinli\u011fi.<\/p>\n\n\n\n
Yar\u0131\u00e7ap\u0131 \u00f6l\u00e7\u00fcn.<\/p>\n\n\n\n
A\u00e7\u0131l\u0131yor.<\/p>\n\n\n\n
Paslanmaz \u00e7elik daha y\u00fcksek \u00e7ekme dayan\u0131m\u0131 ta\u015f\u0131r ve agresif bi\u00e7imde i\u015f sertle\u015fmesi g\u00f6sterir. Z\u0131mba a\u015fa\u011f\u0131ya do\u011fru bast\u0131\u011f\u0131nda, malzeme e\u011frili\u011fe yumu\u015fak \u00e7eli\u011fe g\u00f6re daha fazla diren\u00e7 g\u00f6sterir. O \u2019l\u0131k alt banda o kadar kolay sar\u0131lmaz. \u2019ye veya biraz \u00fcst\u00fcne kayd\u0131\u011f\u0131n\u0131 g\u00f6rebilirsiniz. Metal, o daha dar yay\u0131 olu\u015fturmadan \u00f6nce y\u00fcksek i\u00e7 gerilme alt\u0131nda uzamaktad\u0131r.<\/p>\n\n\n\n
\u015eimdi yumu\u015fak 5052 al\u00fcminyumu yerle\u015ftirin. Daha d\u00fc\u015f\u00fck \u00e7ekme dayan\u0131m\u0131. Daha d\u00fc\u015f\u00fck akma noktas\u0131. Daha az diren\u00e7. Ayn\u0131 batma seviyesinde daha derine iner ve \u00e7o\u011fu zaman y\u00fczde aral\u0131\u011f\u0131n\u0131n alt ucuna daha yak\u0131n bir yar\u0131\u00e7ap\u2014hatta baz\u0131 durumlarda biraz alt\u0131nda\u2014\u00fcretir.<\/p>\n\n\n\n
Ayn\u0131 kal\u0131p. Farkl\u0131 \u00e7arpan.<\/p>\n\n\n\n
\u00c7o\u011fu hesaplay\u0131c\u0131n\u0131n g\u00f6rmezden geldi\u011fi kayma budur. V\u2019yi tek de\u011fi\u015fken, kal\u0131nl\u0131\u011f\u0131 ise \u00e7\u0131karma terimi olarak ele al\u0131rlar. Ger\u00e7ekte, \u00e7ekme dayan\u0131m\u0131 o \u201320 aral\u0131\u011f\u0131n\u0131n neresinde olaca\u011f\u0131n\u0131z\u0131 de\u011fi\u015ftirir. Daha g\u00fc\u00e7l\u00fc malzeme sizi aral\u0131\u011f\u0131n \u00fcst taraf\u0131na iter. Daha yumu\u015fak malzeme s\u0131k\u0131la\u015fman\u0131za izin verir.<\/p>\n\n\n\n
\u201c6 mm alt\u0131ndaki 1\u00d7 malzeme\u201d gibi kal\u0131nl\u0131k esasl\u0131 kurallar\u0131n ince levhalarda bazen i\u015fe yar\u0131yor gibi g\u00f6r\u00fcnmesinin nedeni budur. Uygun ebatl\u0131 kal\u0131pta ince yumu\u015fak \u00e7elik, genellikle kal\u0131nl\u0131\u011fa yak\u0131n bir yar\u0131\u00e7ap verir. Ancak bu, de\u011fi\u015fkenlerin hizalanmas\u0131d\u0131r; evrensel bir yasan\u0131n kan\u0131t\u0131 de\u011fil. Mukavemeti veya V geni\u015fli\u011fini de\u011fi\u015ftirin ve o d\u00fczg\u00fcn 1\u00d7 ili\u015fkisi ortadan kaybolur.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Paslanmaz\u0131, yumu\u015fak \u00e7elik i\u00e7in m\u00fckemmel \u00e7al\u0131\u015fan bir kal\u0131pta hi\u00e7 \u00e7al\u0131\u015ft\u0131r\u0131p a\u00e7\u0131yla u\u011fra\u015f\u0131rken yar\u0131\u00e7ap\u0131n yine b\u00fcy\u00fcd\u00fc\u011f\u00fcn\u00fc g\u00f6rd\u00fcn\u00fcz m\u00fc?<\/p>\n\n\n\n
Ve e\u011fer z\u0131mba ucu o yay\u0131 oymuyorsa, ger\u00e7ekte hangi rol\u00fc oynar?<\/p>\n\n\n\n
Tak\u0131m geometrisi vs. malzeme davran\u0131\u015f\u0131: Fiziksel z\u0131mba ucu yar\u0131\u00e7ap\u0131n\u0131z ile nihai i\u00e7 yar\u0131\u00e7ap\u0131n\u0131z\u0131n neden nadiren uyu\u015fmas\u0131 gerekti\u011fi<\/h3>\n\n\n\n
Ayn\u0131 1.000\u2033 V \u00fczerinde 0.062\u2033 z\u0131mban\u0131n burnuna bak. 10\u2011gauge\u2019\u0131 havada b\u00fck. Par\u00e7ay\u0131 \u00f6l\u00e7.<\/p>\n\n\n\n
\u0130\u00e7eride 0.062\u2033 bulamayacaks\u0131n.<\/p>\n\n\n\n
Z\u0131mba sac\u0131 kal\u0131p omuzlar\u0131n\u0131n aras\u0131na iter, ancak malzemeyi hi\u00e7bir zaman tamamen V\u2019nin i\u00e7ine zorlamaz. Burundaki temas noktasald\u0131r. N\u00fcfuz artt\u0131k\u00e7a sac z\u0131mban\u0131n etraf\u0131nda k\u0131smen sar\u0131l\u0131r, ard\u0131ndan omuzlar aras\u0131nda serbest bir a\u00e7\u0131kl\u0131\u011fa ge\u00e7i\u015f yapar. Nihai i\u00e7 yar\u0131\u00e7ap\u0131n b\u00fcy\u00fck k\u0131sm\u0131, z\u0131mban\u0131n damga gibi \u015feklini bast\u0131rmas\u0131ndan de\u011fil, bu a\u00e7\u0131kl\u0131kta olu\u015fan b\u00fck\u00fclmeden meydana gelir.<\/p>\n\n\n\n
N\u00fcfuz derinli\u011fi her \u015feyi de\u011fi\u015ftirir. Milin sadece birka\u00e7 binde biri kadar derinle\u015fmesi sar\u0131lmay\u0131 art\u0131r\u0131r, a\u00e7\u0131kl\u0131k uzunlu\u011funu azalt\u0131r ve a\u00e7\u0131y\u0131 s\u0131k\u0131la\u015ft\u0131r\u0131r\u2014ancak yar\u0131\u00e7ap h\u00e2l\u00e2 kal\u0131p geni\u015fli\u011fi ile malzeme direncinin bir fonksiyonu olarak belirlenir. Malzeme tamamen kal\u0131ba oturana kadar altlama veya bask\u0131 (coining) yapmad\u0131k\u00e7a, z\u0131mba ucu bir kal\u0131p de\u011fil, bir y\u00f6nlendiricidir.<\/p>\n\n\n\n
Bu y\u00fczden do\u011fru hava b\u00fck\u00fcm\u00fcnde z\u0131mba yar\u0131\u00e7ap\u0131n, beklenen i\u00e7 yar\u0131\u00e7aptan k\u00fc\u00e7\u00fck olmas\u0131 gerekir. E\u011filmeyi ba\u015flatmas\u0131 ama y\u00f6nlendirmemesi gerekir. Tam olarak e\u015fle\u015firlerse, ya fark\u0131nda olmadan altlama yap\u0131yorsun ya da bask\u0131 tonaj\u0131yla fl\u00f6rt ediyorsun demektir.<\/p>\n\n\n\n
Pres freni 200 tonluk bir ger\u00e7ek makinesidir. Modelinin fizi\u011fe uyup uymad\u0131\u011f\u0131n\u0131 ortaya \u00e7\u0131kar\u0131r. Hava b\u00fck\u00fcm\u00fcnde fizik \u015funu s\u00f6yler: kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131 temel a\u00e7\u0131kl\u0131\u011f\u0131 tan\u0131mlar, malzemenin \u00e7ekme dayan\u0131m\u0131 bu a\u00e7\u0131kl\u0131k i\u00e7indeki oran\u0131 de\u011fi\u015ftirir ve z\u0131mba yaln\u0131zca bu sisteme ne kadar derin dald\u0131\u011f\u0131n\u0131 kontrol eder.<\/p>\n\n\n\n
E\u011fer \u00e7izimin s\u0131k\u0131 bir yar\u0131\u00e7ap istiyorsa ama kal\u0131p oran\u0131n\u0131n do\u011fal e\u011filimi daha b\u00fcy\u00fck bir yar\u0131\u00e7apa y\u00f6neliyorsa, h\u00e2l\u00e2 sabit bir form\u00fcle g\u00fcvenmeye devam m\u0131 edeceksin\u2014yoksa hurda kutusu bir kez daha oy vermeden \u00f6nce kal\u0131b\u0131 m\u0131 de\u011fi\u015ftireceksin?<\/p>\n\n\n\n
Geri esneme telafisi: Sabit hesaplay\u0131c\u0131lar\u0131n g\u00f6z ard\u0131 etti\u011fi de\u011fi\u015fken standart<\/h2>\n\n\n\n
Ram kalkt\u0131\u011f\u0131 anda 90\u00b0 yumu\u015fak \u00e7elik bir braketi 92\u00b0\u2019de frenin \u00fczerinden \u00e7\u0131kt\u0131\u011f\u0131n\u0131 g\u00f6rd\u00fcm. Ayn\u0131 kal\u0131p. Ayn\u0131 program. Ayn\u0131 operat\u00f6r. 200 ton alt\u0131nda tam isabetti. Be\u015f saniye sonra art\u0131k de\u011fildi.<\/p>\n\n\n\n
Do\u011fal kal\u0131ba ba\u011fl\u0131 yar\u0131\u00e7ap yaz\u0131c\u0131dakiyle e\u015fle\u015fmedi\u011finde asl\u0131nda \u015fu soruyu soruyorsun: y\u00fck alt\u0131ndaki durumu mu \u00e7\u00f6z\u00fcyorsun, yoksa y\u00fck kalkt\u0131ktan sonra m\u00fc\u015fterinin \u00f6l\u00e7t\u00fc\u011f\u00fc durumu mu?<\/p>\n\n\n\n
Z\u0131mban\u0131n V\u2019nin i\u00e7ine g\u00f6m\u00fcl\u00fc oldu\u011fu s\u0131rada g\u00f6rd\u00fc\u011f\u00fcn yar\u0131\u00e7ap ve a\u00e7\u0131, sevk etti\u011fin par\u00e7ada g\u00f6rd\u00fc\u011f\u00fcn yar\u0131\u00e7ap ve a\u00e7\u0131 de\u011fildir. Bas\u0131n\u00e7 kalkt\u0131\u011f\u0131 anda elastik gerinim serbest kal\u0131r. Uzanm\u0131\u015f d\u0131\u015f lifler k\u0131salmaya \u00e7al\u0131\u015f\u0131r. S\u0131k\u0131\u015fm\u0131\u015f i\u00e7 lifler toparlanmaya \u00e7al\u0131\u015f\u0131r. Par\u00e7a a\u00e7\u0131l\u0131r. \u0130\u015fte bu geri esneme etkisidir ve o, senin temiz k\u00fc\u00e7\u00fck (V \u2212 MT)\/2 hesab\u0131na ald\u0131rmaz.<\/p>\n\n\n\n
Hurda kutusu ger\u00e7eklik kontrol\u00fc: Derinlik g\u00f6stergesindeki a\u00e7\u0131y\u0131 m\u00fckemmel tutturup, denet\u00e7inin her flan\u015f\u0131n 1,5\u00b0 a\u00e7\u0131k oldu\u011funu s\u00f6yledi\u011fini g\u00f6rd\u00fc\u011f\u00fcn oldu mu? \u00c7elik ekrana itaat etmedi; fizi\u011fe etti.<\/p>\n\n\n\n
Sabit hesaplay\u0131c\u0131lar, bas\u0131n\u00e7 alt\u0131nda olu\u015fturdu\u011fun geometrinin yerinde kald\u0131\u011f\u0131n\u0131 varsayar. Kalm\u0131yor. Ve e\u011fer plana telafi dahil etmezsen, bitmi\u015f bir yar\u0131\u00e7ap\u0131 de\u011fil ge\u00e7ici bir yar\u0131\u00e7ap\u0131 \u00f6ng\u00f6r\u00fcyorsun demektir.<\/p>\n\n\n\n
As\u0131l m\u00fccadele alt strokta elde etti\u011fin yar\u0131\u00e7apta de\u011fil. Bo\u015fald\u0131ktan sonra ayakta kalan yar\u0131\u00e7apta.<\/p>\n\n\n\n
Geri esneme i\u00e7in fazla b\u00fckme: Rahatlam\u0131\u015f yar\u0131\u00e7ap i\u00e7in mi yoksa y\u00fck alt\u0131ndaki yar\u0131\u00e7ap i\u00e7in mi hesap yap\u0131yorsun?<\/h3>\n\n\n\n
Bir 1.000\u2033 V i\u00e7inde 0.125\u2033 so\u011fuk haddelenmi\u015f par\u00e7a hayal et. Z\u0131mban\u0131n a\u015fa\u011f\u0131dayken 90\u00b0\u2019lik bir a\u00e7\u0131 veren derinli\u011fe kadar iniyorsun. Ram\u0131 yukar\u0131 \u00e7ekiyorsun ve 91.5\u00b0\u2019de oturuyor. Bu, y\u00fck alt\u0131ndaki a\u00e7\u0131n\u0131n yakla\u015f\u0131k 88.5\u00b0 oldu\u011fu anlam\u0131na gelir.<\/p>\n\n\n\n
\u015eimdi kendine sor: form\u00fcl\u00fcn hangi a\u00e7\u0131y\u0131 \u00f6ng\u00f6rd\u00fc?<\/p>\n\n\n\n
Ger\u00e7ek bir 90\u00b0 geometrisini varsayarak ve o 1.5\u00b0\u2019lik geri esnemeyi hesaba katmadan b\u00fckme indirimi, b\u00fckme pay\u0131 ve flan\u015f uzunluklar\u0131 hesaplad\u0131ysan, her flan\u015f uzundur. \u00c7ok de\u011fil. Bir montaj\u0131 mahvedecek kadar.<\/p>\n\n\n\n
Fazla b\u00fckme, hepimizin kulland\u0131\u011f\u0131 kaba bir ara\u00e7t\u0131r. 88.5\u00b0\u2019yi programla ki 90\u00b0\u2019ye gev\u015fesin. Ama i\u015fte tuzak: bu telafi t\u00fcm i\u015fler aras\u0131nda sabit de\u011fildir. Kal\u0131b\u0131 1.250\u2033\u2019ye a\u00e7 ve ayn\u0131 malzeme art\u0131k 2\u00b0 veya daha fazla geri esneyebilir; \u00e7\u00fcnk\u00fc daha b\u00fcy\u00fck yar\u0131\u00e7ap plastik \u015fekil de\u011fi\u015ftirmeyi azalt\u0131r ve kesitte daha fazla elastik enerji birikmesine yol a\u00e7ar. Daha ince malzeme? Daha fazla geri esneme. Daha b\u00fcy\u00fck i\u00e7 yar\u0131\u00e7ap? Daha fazla geri esneme. Bu ili\u015fki, geri esnemenin yaln\u0131zca dayan\u0131mla de\u011fil, yar\u0131\u00e7ap\/kal\u0131nl\u0131k oran\u0131yla artt\u0131\u011f\u0131 so\u011fuk haddelenmi\u015f \u00e7elikte g\u00f6sterilmi\u015ftir.<\/p>\n\n\n\n
Yani 90\u00b0 nominal \u00fczerinden b\u00fckme indirimi hesaplay\u0131p makinede elle 1.5\u00b0 fazla b\u00fckme ekliyorsan, hesab\u0131n\u0131 ikiye b\u00f6lm\u00fc\u015fs\u00fcn demektir. D\u00fcz desen bir \u015fey d\u00fc\u015f\u00fcn\u00fcyor. Ram ba\u015fka bir \u015fey yap\u0131yor.<\/p>\n\n\n\n
Geometrinizi y\u00f6nlendiren say\u0131 hangisi\u2014bask\u0131 a\u00e7\u0131s\u0131 m\u0131, yoksa ger\u00e7ekten programlad\u0131\u011f\u0131n\u0131z a\u00e7\u0131 m\u0131?<\/p>\n\n\n\n
\u00c7ekme dayan\u0131m\u0131 ve akma oran\u0131: Hangi \u00f6zellik ger\u00e7ekten fark yarat\u0131yor?<\/h3>\n\n\n\n
O yumu\u015fak \u00e7eli\u011fi 304 paslanmazla de\u011fi\u015ftirin. Ayn\u0131 kal\u0131nl\u0131k. Ayn\u0131 kal\u0131p. Ayn\u0131 hedef a\u00e7\u0131.<\/p>\n\n\n\n
Daha fazla yaylanma g\u00f6receksiniz. Herkes g\u00f6r\u00fcr. \u0130\u00e7g\u00fcd\u00fc, \u00e7ekme dayan\u0131m\u0131n\u0131 su\u00e7lamakt\u0131r \u00e7\u00fcnk\u00fc sertifikada en b\u00fcy\u00fck say\u0131 odur. Daha y\u00fcksek \u00e7ekme, daha fazla diren\u00e7, daha fazla a\u00e7\u0131lma.<\/p>\n\n\n\n
Ama \u201cayn\u0131\u201d 60\u2011KSI \u00e7eli\u011fin iki ergitmesini \u00e7al\u0131\u015ft\u0131rd\u0131\u011f\u0131n\u0131zda neler oldu\u011funu izleyin. Biri kolay b\u00fck\u00fcl\u00fcr. Di\u011feri fazladan bir derece geri yaylan\u0131r. \u00c7ekme dayan\u0131m\u0131 bir gecede 10 KSI s\u0131\u00e7ramad\u0131. De\u011fi\u015fen \u015fey akma\u2011\u00e7ekme oran\u0131d\u0131r\u2014akma oran\u0131.<\/p>\n\n\n\n
Yaylanma, deformasyonun ne kadar\u0131n\u0131n elastik, ne kadar\u0131n\u0131n plastik oldu\u011funa ba\u011fl\u0131d\u0131r. \u00c7ekme dayan\u0131m\u0131na g\u00f6re y\u00fcksek akma dayan\u0131m\u0131na sahip bir malzeme, plastik deformasyona daha ge\u00e7 girer ve derin \u015fekilde akmadan \u00f6nce daha fazla elastik enerji depolar. Bu depolanan enerji, y\u00fck kalkt\u0131\u011f\u0131nda a\u00e7\u0131n\u0131z\u0131 a\u00e7an \u015feydir.<\/p>\n\n\n\n
Geometri bunu b\u00fcy\u00fct\u00fcr. Daha geni\u015f kal\u0131p a\u00e7\u0131kl\u0131klar\u0131 daha b\u00fcy\u00fck i\u00e7 yar\u0131\u00e7aplar olu\u015fturur. Daha b\u00fcy\u00fck yar\u0131\u00e7ap, ayn\u0131 a\u00e7\u0131 i\u00e7in daha d\u00fc\u015f\u00fck plastik gerinim demektir. Daha d\u00fc\u015f\u00fck plastik gerinim, daha y\u00fcksek oranda elastik toparlanma demektir. Bu y\u00fczden geni\u015f V kal\u0131plarla yap\u0131lan yar\u0131\u00e7ap b\u00fck\u00fcmler, s\u0131k\u0131 6\u00d7 kurulumlara g\u00f6re dramatik bi\u00e7imde geri yaylanabilir.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Tonaj tasarrufu i\u00e7in hi\u00e7 geni\u015f bir kal\u0131p kulland\u0131n\u0131z m\u0131, ekrandaki a\u00e7\u0131y\u0131 tutturdunuz, sonra 10 fitlik par\u00e7a boyunca 3\u00b0 yaylanmay\u0131 kovalar halde buldunuz kendinizi? Bu k\u00f6t\u00fc \u015fans de\u011fildi. Bu, d\u00fc\u015f\u00fck plastik gerinimin kontrol\u00fc elastikli\u011fe geri vermesiydi.<\/p>\n\n\n\n
Peki hangisi daha fazla fark yarat\u0131yor\u2014k\u00e2\u011f\u0131ttaki dayan\u0131m say\u0131lar\u0131 m\u0131 yoksa tak\u0131m se\u00e7iminizle belirledi\u011finiz yar\u0131\u00e7ap\u2011kal\u0131nl\u0131k oran\u0131 m\u0131? Uygulamada geometriniz sahneyi kurar. Malzeme \u00f6zellikleri geri tepmenin ne kadar sert olaca\u011f\u0131n\u0131 belirler.<\/p>\n\n\n\n
Ve o geri tepme de\u011fi\u015firse, g\u00fcvendi\u011finiz t\u00fcm d\u00fcz desen say\u0131lar\u0131 ne olur?<\/p>\n\n\n\n
Hesaplanmam\u0131\u015f yaylanman\u0131n b\u00fck\u00fclme indirimini ve K\u2011Fakt\u00f6r\u00fc\u2019n\u00fc sessizce bozmas\u0131<\/h3>\n\n\n\n
D\u00f6rt b\u00fck\u00fcml\u00fc bir kanal al\u0131n. D\u00f6n\u00fc\u015f flan\u015f\u0131 yok. Her b\u00fck\u00fcm 2\u00b0 geri yaylans\u0131n. Bu dramatik de\u011fil. Baz\u0131 paslanmaz i\u015flerde normaldir.<\/p>\n\n\n\n
\u015eimdi \u00fcst \u00fcste koyun.<\/p>\n\n\n\n
D\u00f6rt b\u00fck\u00fcm \u00d7 2\u00b0 demek, her ad\u0131mda do\u011fru \u015fekilde telafi etmediyseniz son flan\u015f\u0131n ilk referansa g\u00f6re 8\u00b0 d\u0131\u015far\u0131da olmas\u0131 demektir. Her b\u00fck\u00fcm \u201cyeterince yak\u0131n\u201d g\u00f6r\u00fcn\u00fcyordu ama birikmi\u015f hata montaj\u0131 pervane gibi burktu\u011funu g\u00f6rd\u00fcm.<\/p>\n\n\n\n
B\u00fck\u00fclme indirimi ve K\u2011fakt\u00f6r\u00fc bilinen bir i\u00e7 yar\u0131\u00e7ap ve bilinen bir nihai a\u00e7\u0131 varsayar. Yaylanma bunlardan herhangi birini de\u011fi\u015ftirirse ve siz say\u0131lar\u0131 g\u00fcncellemezseniz, n\u00f6tr eksenin konumu ger\u00e7ekte de\u011fi\u015fir ama yaz\u0131l\u0131m\u0131n\u0131zda de\u011fi\u015fmez. Kesti\u011finiz d\u00fcz uzunluk, gev\u015feme sonras\u0131 ger\u00e7ekte var olandan daha k\u00fc\u00e7\u00fck bir yay uzunlu\u011funa dayan\u0131r. Bunu \u00e7oklu b\u00fck\u00fcmlere ve toleranslara uygulad\u0131\u011f\u0131n\u0131zda, hatalar h\u0131zla b\u00fcy\u00fcr.<\/p>\n\n\n\n
Bu y\u00fczden tek tip K\u2011fakt\u00f6rler bir hayaldir. Kal\u0131p geni\u015fli\u011fini de\u011fi\u015ftirin, yar\u0131\u00e7ap\u0131 de\u011fi\u015ftirirsiniz. Yar\u0131\u00e7ap\u0131 de\u011fi\u015ftirin, yaylanmay\u0131 de\u011fi\u015ftirirsiniz. Yaylanmay\u0131 de\u011fi\u015ftirin, nihai a\u00e7\u0131y\u0131 ve etkili b\u00fck\u00fclme pay\u0131n\u0131 de\u011fi\u015ftirirsiniz. Sisteminiz bu d\u00f6ng\u00fcy\u00fc kapatm\u0131yorsa\u2014rahatlam\u0131\u015f a\u00e7\u0131y\u0131 ve yar\u0131\u00e7ap\u0131 \u00f6l\u00e7\u00fcp indirime geri beslemiyorsa\u2014siz 200 ton alt\u0131nda var olan bir par\u00e7a i\u00e7in d\u00fcz levha kesiyorsunuz demektir.<\/p>\n\n\n\n
Pres b\u00fck\u00fcc\u00fc 200 tonluk bir ger\u00e7eklik makinesidir. Malzemenin ger\u00e7ekte ne yapaca\u011f\u0131n\u0131 size s\u00f6yler. Hurda kutusu, dinleyip dinlemedi\u011finizin son yarg\u0131c\u0131d\u0131r.<\/p>\n\n\n\n
Y\u00fck kald\u0131r\u0131ld\u0131ktan sonra hayatta kalan yar\u0131\u00e7ap tek \u00f6nemli olan ise, neden h\u00e2l\u00e2 ko\u00e7 kalkt\u0131\u011f\u0131 anda kaybolan yar\u0131\u00e7ap \u00fczerinden d\u00fcz desenler olu\u015fturuyorsunuz?<\/p>\n\n\n\n
Tahminden Kontrole: \u0130\u00e7 B\u00fck\u00fclme Yar\u0131\u00e7ap\u0131n\u0131 Tahmin Etmek \u0130\u00e7in 4 Ad\u0131ml\u0131 Y\u00f6ntem<\/h2>\n\n\n\n
1.000\u2033 V\u2011kal\u0131p alt\u0131nda 10 numara sacdan bir braketi izledim, ilk par\u00e7a bask\u0131 90\u00b0 ba\u011f\u0131r\u0131rken 92\u00b0 \u00e7\u0131kt\u0131. Programc\u0131 d\u00fcz deseni do\u011fru oldu\u011funu s\u00f6yledi. Hesaplay\u0131c\u0131 \u201ckesin\u201d i\u00e7 yar\u0131\u00e7ap dedi. Hurda kutusu umursamad\u0131.<\/p>\n\n\n\n
Ger\u00e7ek geri yaylanmay\u0131 b\u00fckme indirgeme ve K\u2011fakt\u00f6r\u00fcne geri beslemek istiyorsun ki d\u00fcz par\u00e7a gev\u015femi\u015f par\u00e7ayla tam olarak e\u015fle\u015fsin. G\u00fczel. \u00c7\u00fcnk\u00fc o d\u00f6ng\u00fcy\u00fc kapatana kadar hesaplama yapm\u0131yorsun\u2014sacla kumar oynuyorsun.<\/p>\n\n\n\n
G\u00fczel form\u00fcllere sabr\u0131 olmayan bir 200 tonluk ger\u00e7eklik makinesinde kulland\u0131\u011f\u0131m y\u00f6ntem bu.<\/p>\n\n\n\n
Ad\u0131m 1: Bask\u0131daki \u201cistenen\u201d yar\u0131\u00e7aptan de\u011fil, kal\u0131p oran\u0131ndan ba\u015fla<\/h3>\n\n\n\n
0,125\u2033 yumu\u015fak \u00e7eli\u011fi iki kal\u0131p \u00fczerinde ayarla: biri 0,750\u2033 V, di\u011feri 1,000\u2033 V. Ayn\u0131 z\u0131mba. Ayn\u0131 hedef a\u00e7\u0131. Par\u00e7alar ayn\u0131 i\u00e7 yar\u0131\u00e7apla \u00e7\u0131kmaz. \u00c7\u0131kamazlar. Kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131, sac\u0131n nas\u0131l akmas\u0131na izin verilece\u011finin geometrisini belirler.<\/p>\n\n\n\n
Hava b\u00fckmede, kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131 b\u00fcy\u00fck \u00f6l\u00e7\u00fcde olu\u015facak i\u00e7 yar\u0131\u00e7ap\u0131 belirler. Ger\u00e7ek d\u00fcnyada yayg\u0131n bir ba\u015flang\u0131\u00e7 kural\u0131, daha ince yumu\u015fak \u00e7eliklerde 6:1 V\u2011kal\u0131nl\u0131k, kal\u0131nl\u0131k artt\u0131k\u00e7a 8:1 oran\u0131d\u0131r. Yani 0,125\u2033 malzeme mi? Genellikle 0,750\u2033 ila 1,000\u2033 V aras\u0131ndas\u0131nd\u0131r. Bu oran bir \u00f6neri de\u011fildir\u2014bu, gerinim zarf\u0131d\u0131r. \u00c7ok dar olursa \u00e7atlama riski vard\u0131r. \u00c7ok geni\u015f olursa plastik gerinimi azalt\u0131r ve geri yaylanma etkisini davet edersin.<\/p>\n\n\n\n
\u015eimdi birinin bunun yerine teknik resimden ba\u015flad\u0131\u011f\u0131nda ne oldu\u011funa bak. \u201c0,125 yar\u0131\u00e7ap istiyorum.\u201d G\u00fczel. Ancak tonajdan tasarruf etmek i\u00e7in 1,250\u2033 V se\u00e7ersen, do\u011fal hava formu yar\u0131\u00e7ap\u0131n malzemeye ba\u011fl\u0131 olarak 0,200\u2033+ civar\u0131nda olabilir. Hi\u00e7bir hayal g\u00fcc\u00fc bunu de\u011fi\u015ftiremez. Geometri zaten karar\u0131n\u0131 vermi\u015ftir.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Makinede zaten tak\u0131l\u0131 oldu\u011fu i\u00e7in geni\u015f bir kal\u0131b\u0131 zorlayan, ekrandaki a\u00e7\u0131y\u0131 yakalayan, sonra da gev\u015femi\u015f yar\u0131\u00e7ap d\u00fcz modelin varsayd\u0131\u011f\u0131ndan daha b\u00fcy\u00fck oldu\u011fu i\u00e7in flan\u015flar\u0131n uzad\u0131\u011f\u0131n\u0131 izleyen at\u00f6lyeler g\u00f6rd\u00fcm. Form\u00fcl yanl\u0131\u015f de\u011fildi. Ba\u015flang\u0131\u00e7 varsay\u0131m\u0131 \u00f6yleydi.<\/p>\n\n\n\n
Kal\u0131p geni\u015fli\u011fi gerinim ko\u015fulunu belirliyorsa, neden o kal\u0131ptan ba\u011f\u0131ms\u0131z hedef bir yar\u0131\u00e7apla ba\u015flamak isteyesin ki?<\/p>\n\n\n\n
Ad\u0131m 2: V\u2011a\u00e7\u0131kl\u0131\u011f\u0131ndan teorik yar\u0131\u00e7ap\u0131 tahmin et (ve bunun yaln\u0131zca bir ba\u015flang\u0131\u00e7 de\u011feri oldu\u011funu kabul et)<\/h3>\n\n\n\n
Kal\u0131p se\u00e7ildikten sonra, art\u0131k matematik konu\u015fmana izin var.<\/p>\n\n\n\n
Hava b\u00fckmede z\u0131mba burun tamamen V i\u00e7ine girmez. Z\u0131mba, sac\u0131 kal\u0131p omuzlar\u0131 aras\u0131na iter ancak malzemeyi tamamen V i\u00e7ine zorlamaz. Sac gerilir. Bu gerilme, V\u2011a\u00e7\u0131kl\u0131\u011f\u0131 ve y\u00fck alt\u0131ndaki i\u00e7 yar\u0131\u00e7ap aras\u0131nda \u00f6ng\u00f6r\u00fclebilir bir ili\u015fki yarat\u0131r.<\/p>\n\n\n\n
Yumu\u015fak \u00e7elikte, y\u00fck alt\u0131ndaki i\u00e7 yar\u0131\u00e7ap genellikle V\u2011a\u00e7\u0131kl\u0131\u011f\u0131n\u0131n yakla\u015f\u0131k \u201320\u2019si aras\u0131nda olur. Yani 1,000\u2033 V, pres a\u015fa\u011f\u0131dayken yakla\u015f\u0131k 0,160\u2033\u20130,200\u2033 yar\u0131\u00e7ap \u00fcretebilir. Tam de\u011fil. Bir aral\u0131k. \u00c7\u00fcnk\u00fc kal\u0131nl\u0131k, dayan\u0131m ve kal\u0131p omuz yar\u0131\u00e7ap\u0131 hepsi bunu biraz de\u011fi\u015ftirir.<\/p>\n\n\n\n
Ne yapt\u0131\u011f\u0131m\u0131za dikkat et: yar\u0131\u00e7ap\u0131 \u00f6nce kal\u0131p geni\u015fli\u011fine ba\u011flad\u0131k, (V \u2212 MT)\/2\u2019ye ya da z\u0131mba ucuna de\u011fil. Eski (V \u2212 MT)\/2 form\u00fcl\u00fc, kesin g\u00f6r\u00fcnen say\u0131lar verebilir ama V\/T oran\u0131 6:1\u2019den 8:1\u2019e kayarsa, gerinim da\u011f\u0131l\u0131m\u0131n de\u011fi\u015fir ve o d\u00fczg\u00fcn sonu\u00e7 ger\u00e7ekte h\u0131zla sapar.<\/p>\n\n\n\n
Ve bu h\u00e2l\u00e2 y\u00fck alt\u0131ndaki durumdur.<\/p>\n\n\n\n
Ba\u015flang\u0131\u00e7 yar\u0131\u00e7ap\u0131 200 tonluk bir anl\u0131k g\u00f6r\u00fcnt\u00fcd\u00fcr. Sevk etti\u011fin par\u00e7a ise ko\u00e7 yukar\u0131 kalkt\u0131ktan sonraki halidir. Peki depolanm\u0131\u015f elastik enerji geri \u00e7\u0131kt\u0131\u011f\u0131nda ne olur?<\/p>\n\n\n\n
Ad\u0131m 3: Ger\u00e7ek gev\u015femi\u015f yar\u0131\u00e7ap\u0131 bulmak i\u00e7in malzeme temelli geri yaylanma telafisini uygula<\/h3>\n\n\n\n
0,125\u2033 so\u011fuk haddelenmi\u015f \u00e7eli\u011fi o 1,000\u2033 V i\u00e7inde al. Y\u00fck alt\u0131ndayken 90\u00b0 \u00f6l\u00e7\u00fcyorsun. B\u0131rak\u0131nca 91,5\u00b0 oluyorsun. O 1,5\u00b0 geri yaylanma daha derin bir \u015fey anlat\u0131yor: yar\u0131\u00e7ap da b\u00fcy\u00fcd\u00fc.<\/p>\n\n\n\n
A\u00e7\u0131 a\u00e7\u0131ld\u0131k\u00e7a i\u00e7 yar\u0131\u00e7ap artar. N\u00f6tr eksen biraz kayar \u00e7\u00fcnk\u00fc gerinimin elastik k\u0131sm\u0131 geri kazan\u0131l\u0131r. Akma dayan\u0131m\u0131, \u00e7ekme dayan\u0131m\u0131na g\u00f6re daha y\u00fcksek olan malzemeler, tamamen akmadan \u00f6nce daha fazla elastik enerji depolar. Paslanmaz \u00e7elik genellikle ayn\u0131 kal\u0131pta yumu\u015fak \u00e7eli\u011fe g\u00f6re daha fazla geri tepme yapar. Al\u00fcminyum da yar\u0131\u00e7ap\u2011kal\u0131nl\u0131k oran\u0131 y\u00fckseldi\u011finde seni \u015fa\u015f\u0131rtabilir.<\/p>\n\n\n\n
Ama kitaplar\u0131n atlad\u0131\u011f\u0131 k\u0131s\u0131m \u015fu: geri yaylanma yar\u0131\u00e7ap\u2011kal\u0131nl\u0131k oran\u0131yla g\u00fc\u00e7l\u00fc bi\u00e7imde orant\u0131l\u0131d\u0131r. Kal\u0131b\u0131 daha geni\u015f a\u00e7t\u0131\u011f\u0131nda, i\u00e7 yar\u0131\u00e7ap\u0131 art\u0131r\u0131rs\u0131n. Yar\u0131\u00e7ap\u0131 art\u0131rd\u0131\u011f\u0131nda, belirli bir a\u00e7\u0131 i\u00e7in plastik gerinimi azalt\u0131rs\u0131n. Plastik gerinim azal\u0131nca, elastik toparlanma oran\u0131 artar. Daha fazla geri yaylanma. Bu, geometrinin malzeme davran\u0131\u015f\u0131n\u0131 b\u00fcy\u00fctmesidir.<\/p>\n\n\n\n
Yani telafi s\u0131ras\u0131 \u015fu \u015fekildedir:<\/p>\n\n\n\n
\n- Kal\u0131b\u0131 se\u00e7in (V\/T oran\u0131n\u0131 ve \u015fekil de\u011fi\u015ftirme ko\u015fulunu belirler).<\/li>\n\n\n\n
- Y\u00fck alt\u0131ndaki yar\u0131\u00e7ap\u0131 V de\u011ferinden tahmin edin.<\/li>\n\n\n\n
- Malzeme kalitesine ve yar\u0131\u00e7ap\u2011kal\u0131nl\u0131k oran\u0131na g\u00f6re geri esneme a\u00e7\u0131s\u0131n\u0131 tahmin edin.<\/li>\n\n\n\n
- Bu ekstra a\u00e7\u0131y\u0131, ayarlanm\u0131\u015f bir gev\u015femi\u015f yar\u0131\u00e7apa d\u00f6n\u00fc\u015ft\u00fcr\u00fcn.<\/li>\n<\/ol>\n\n\n\n
\u015eimdi b\u00fckme kesintisi ve K\u2011fakt\u00f6r\u00fcn\u00fcz\u00fc y\u00fck alt\u0131 geometrinizle de\u011fil, gev\u015femi\u015f yar\u0131\u00e7ap ve son a\u00e7\u0131yla g\u00fcncelleyin. \u0130\u015fte kapal\u0131 d\u00f6ng\u00fc budur. E\u011fer \u00f6l\u00e7\u00fclen geri esneme 1.5\u00b0 ise, d\u00fcz modeliniz 88.5\u00b0 y\u00fck alt\u0131nda de\u011fil, 90\u00b0 gev\u015femi\u015f durumdaki par\u00e7a yay uzunlu\u011funu yans\u0131tmal\u0131d\u0131r.<\/p>\n\n\n\n
CN-HAWE\u2019nin \u00fcr\u00fcn portf\u00f6y\u00fcn\u00fcn 100% CNC tabanl\u0131 oldu\u011fu ve lazer kesim, b\u00fckme, oluk a\u00e7ma, kesme gibi y\u00fcksek d\u00fczeyli senaryolar\u0131 kapsad\u0131\u011f\u0131 g\u00f6z \u00f6n\u00fcne al\u0131nd\u0131\u011f\u0131nda, ayr\u0131nt\u0131l\u0131 materyal isteyen okuyucular i\u00e7in, Bro\u015f\u00fcrler<\/a> yararl\u0131 bir tamamlay\u0131c\u0131 kaynakt\u0131r.<\/p>\n\n\n\nHurda Kutusu Ger\u00e7eklik Kontrol\u00fc: D\u00f6rt b\u00fck\u00fcml\u00fc kanallar\u0131n, kal\u0131p oran\u0131 de\u011fi\u015fti\u011fi h\u00e2lde programc\u0131n\u0131n bir tablodan sabit K\u2011fakt\u00f6r\u00fc kulland\u0131\u011f\u0131 i\u00e7in toplam geni\u015flikte 0.060\u2033\u2019dan fazla hatayla ka\u00e7t\u0131\u011f\u0131n\u0131 g\u00f6rd\u00fcm. Ayn\u0131 malzeme. Farkl\u0131 V. Farkl\u0131 gev\u015femi\u015f yar\u0131\u00e7ap. Matematik bunu hi\u00e7 bilemedi.<\/p>\n\n\n\n
E\u011fer geri esneme hem malzeme \u00f6zelliklerine hem de kal\u0131pla olu\u015fturdu\u011funuz yar\u0131\u00e7apa ba\u011fl\u0131ysa, neden tek bir sabit K\u2011fakt\u00f6r her i\u015fte i\u015fe yaras\u0131n?<\/p>\n\n\n\n
Ad\u0131m 4: Tam \u00fcretim deneme\u2011yan\u0131lma yerine kontroll\u00fc bir test b\u00fck\u00fcm\u00fcyle do\u011frulay\u0131n<\/h3>\n\n\n\n
2\u2033 geni\u015fli\u011finde bir \u015ferit kesin. Ayn\u0131 kal\u0131nl\u0131k. \u00dcretimle ayn\u0131 hadde y\u00f6n\u00fc. Se\u00e7ilen kal\u0131pta programlad\u0131\u011f\u0131n\u0131z fazla b\u00fck\u00fcmle b\u00fck\u00fcn\u2014\u00f6rne\u011fin 1.5\u00b0 geri esneme bekliyorsan\u0131z 88.5\u00b0. Gev\u015femi\u015f a\u00e7\u0131y\u0131 \u00f6l\u00e7\u00fcn ve do\u011fru mastarlarla i\u00e7 yar\u0131\u00e7ap\u0131 \u00f6l\u00e7\u00fcn.<\/p>\n\n\n\n
Art\u0131k \u00fc\u00e7 ger\u00e7ek say\u0131n\u0131z var: kal\u0131p geni\u015fli\u011fi, gev\u015femi\u015f a\u00e7\u0131, gev\u015femi\u015f yar\u0131\u00e7ap.<\/p>\n\n\n\n
Bunlar\u0131 b\u00fckme pay\u0131 hesaplaman\u0131za geri besleyin. Hesaplanan flan\u015f uzunluklar\u0131 \u00f6l\u00e7\u00fclen test par\u00e7as\u0131yla e\u015fle\u015fene kadar K\u2011fakt\u00f6r\u00fcn\u00fc ayarlay\u0131n. Kalibre edilmi\u015f bu K, o malzeme serisi, o kal\u0131p, o kal\u0131nl\u0131k, o kurulum i\u00e7in ge\u00e7erli h\u00e2le gelir.<\/p>\n\n\n\n
Bu bir tahmin i\u015fi de\u011fil. Kontroll\u00fc yinelemedir. Y\u00fcz par\u00e7ay\u0131 kurtarmak i\u00e7in bir \u015ferit feda edilir.<\/p>\n\n\n\n
Evet, makineye \u00f6zg\u00fc etkenler \u00f6nemlidir. Kal\u0131p kenar a\u015f\u0131nmas\u0131, pres ko\u00e7 defleksiyonu, bombaj ayarlar\u0131\u2014hepsi par\u00e7an\u0131n davran\u0131\u015f\u0131n\u0131 etkiler. \u0130\u015fte bu y\u00fczden test b\u00fck\u00fcm\u00fc ayn\u0131 makinede, ayn\u0131 tak\u0131m y\u0131\u011f\u0131n\u0131 ve ayn\u0131 kurulumda yap\u0131lmal\u0131d\u0131r. Sadece bir form\u00fcl\u00fc de\u011fil, t\u00fcm sistemi kalibre ediyorsunuz.<\/p>\n\n\n\n
Ger\u00e7ek \u015fu: iyi bir tahmin zinciri bile modelemedi\u011finiz de\u011fi\u015fkenler taraf\u0131ndan bozulabilir.<\/p>\n\n\n\n
Bu y\u00fczden kal\u0131p oran\u0131n\u0131, temel yar\u0131\u00e7ap\u0131, geri esneme telafisini ve do\u011frulanm\u0131\u015f K\u2011fakt\u00f6r\u00fcn\u00fc sabitlemi\u015f olsan\u0131z bile, \u201cm\u00fckemmel\u201d d\u00fcz modelinizi otuz mil (0.030\u2033) sapt\u0131rmaya haz\u0131r bekleyen gizli etkenler hangileridir\u2014ve bunlar\u0131 bulmak i\u00e7in ne kadar malzeme harcamaya haz\u0131rs\u0131n\u0131z?<\/p>\n\n\n\n
CN\u2011HAWE y\u0131ll\u0131k sat\u0131\u015f gelirinin %8%\u2019inden fazlas\u0131n\u0131 ara\u015ft\u0131rma ve geli\u015ftirmeye yat\u0131r\u0131rken, ADH pres b\u00fckme makineleri genelinde Ar\u2011Ge yetenekleriyle faaliyet g\u00f6stermektedir; bir sonraki ad\u0131m ekiple do\u011frudan konu\u015fmaksa, Bize ula\u015f\u0131n<\/a> buraya do\u011fal \u015fekilde oturur.<\/p>\n\n\n\nM\u00fckemmel Matemati\u011fi Bozan G\u00f6r\u00fcnmez De\u011fi\u015fkenler<\/h2>\n\n\n\n
Kal\u0131b\u0131 se\u00e7tiniz. Geri esnemeyi \u00f6l\u00e7t\u00fcn\u00fcz. Test \u015feridi \u00e7izime uyana kadar K\u2011fakt\u00f6r\u00fcn\u00fc ayarlad\u0131n\u0131z.<\/p>\n\n\n\n
Ve \u00fcretim serisi h\u00e2l\u00e2 kay\u0131yor.<\/p>\n\n\n\n
\u0130\u015fte 200 tonluk ger\u00e7ek makinesi ad\u0131n\u0131 burada kazan\u0131yor. \u0130\u00e7ine temiz girdiler, kalibre edilmi\u015f yar\u0131\u00e7aplar, \u00f6l\u00e7\u00fclm\u00fc\u015f fazla b\u00fck\u00fcm besleyebilirsiniz\u2014yine de metalin veya onu tutan donan\u0131m\u0131n i\u00e7inde gizlenen zay\u0131f varsay\u0131mlar\u0131 ortaya \u00e7\u0131kar\u0131r. Form\u00fcl y\u00fcksek sesle hata yapmaz. Par\u00e7alar ise ayaklan\u0131r ve tolerans d\u0131\u015f\u0131na \u00e7\u0131kar.<\/p>\n\n\n\n
CN-HAWE\u2019nin m\u00fc\u015fteri taban\u0131; in\u015faat makineleri, otomotiv \u00fcretimi, gemi in\u015fas\u0131, k\u00f6pr\u00fcler, havac\u0131l\u0131k gibi sekt\u00f6rleri kapsad\u0131\u011f\u0131ndan, burada pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Lazer Kesim Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/p>\n\n\n\nM\u00fckemmel matematik sadece m\u00fckemmel ko\u015fullarda hayatta kal\u0131r.<\/p>\n\n\n\n
Sorun \u015fu ki, bir \u00fcretim at\u00f6lyesinde hi\u00e7bir \u015fey uzun s\u00fcre m\u00fckemmel kalmaz.<\/p>\n\n\n\n
Peki \u201cher \u015feyi do\u011fru yapt\u0131\u011f\u0131n\u0131zda\u201d bile yar\u0131\u00e7ap\u0131n\u0131z\u0131 h\u00e2l\u00e2 de\u011fi\u015ftirebilen tek \u015fey ne kald\u0131?<\/p>\n\n\n\n
Tane y\u00f6n\u00fc: Haddelenme y\u00f6n\u00fc gerekli fazla b\u00fckme a\u00e7\u0131s\u0131n\u0131 nas\u0131l de\u011fi\u015ftirir<\/h3>\n\n\n\n
\u00c7elik izotropik de\u011fildir. Bu, basit bir at\u00f6lye ger\u00e7e\u011fi i\u00e7in kitap diliyle \u015fu anlama gelir: bir y\u00f6nde di\u011ferine g\u00f6re daha kolay b\u00fck\u00fcl\u00fcr.<\/p>\n\n\n\n
Levha haddeleme s\u0131ras\u0131nda, taneler haddeleme y\u00f6n\u00fcnde uzar. Tane y\u00f6n\u00fcne dik b\u00fck\u00fcld\u00fc\u011f\u00fcnde, malzeme bu uzam\u0131\u015f lifler boyunca gerilir. Tane y\u00f6n\u00fcne paralel b\u00fck\u00fcld\u00fc\u011f\u00fcnde ise onlar\u0131 bir fermuar gibi a\u00e7maya \u00e7al\u0131\u015f\u0131rs\u0131n\u0131z. Ayn\u0131 kal\u0131nl\u0131k. Ayn\u0131 kal\u0131p. Farkl\u0131 yaylanma miktar\u0131.<\/p>\n\n\n\n
0.125\u2033 paslanmaz \u00e7eli\u011fin tane y\u00f6n\u00fcne dik olarak tam 90\u00b0 b\u00fck\u00fcl\u00fcp \u2014 ayn\u0131 1.000\u2033 V i\u00e7inde tane y\u00f6n\u00fcyle b\u00fck\u00fcld\u00fc\u011f\u00fcnde neredeyse bir derece daha fazla geri s\u0131\u00e7rad\u0131\u011f\u0131n\u0131 izledim. Ba\u015fka hi\u00e7bir \u015fey de\u011fi\u015fmedi. Ayn\u0131 program. Ayn\u0131 operat\u00f6r. Test par\u00e7as\u0131ndan kalibre edilmi\u015f ayn\u0131 K fakt\u00f6r\u00fc.<\/p>\n\n\n\n
Tek fark y\u00f6nelimdi.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Test \u015feridiniz tane y\u00f6n\u00fcne dik kesilmi\u015f, ancak \u00fcretim bo\u015fluklar\u0131 malzemeden tasarruf etmek i\u00e7in tane y\u00f6n\u00fc uzun eksende olacak \u015fekilde yerle\u015ftirilmi\u015fse, ilk darbeden \u00f6nce fazla b\u00fckme telafiniz zaten yanl\u0131\u015ft\u0131r. Presin hangi y\u00f6n\u00fcn daha ucuz yerle\u015ftirildi\u011fi umurunda de\u011fildir.<\/p>\n\n\n\n
A\u00e7\u0131 sens\u00f6rleriyle donat\u0131lm\u0131\u015f modern makineler ger\u00e7ek zamanl\u0131 otomatik d\u00fczeltme yapabilir. G\u00fczel. Bu da noktay\u0131 kan\u0131tlar. Tane \u00f6nemli olmasayd\u0131, her b\u00fckmede \u00f6l\u00e7\u00fcm yapmalar\u0131 gerekmezdi. Ancak modern bir \u00fcretim at\u00f6lyesine girip ger\u00e7ekte ne oldu\u011funa bak\u0131n \u2014 h\u00e2l\u00e2 canl\u0131 a\u00e7\u0131 d\u00fczeltmesi olmadan \u00e7al\u0131\u015fan, d\u00fcnk\u00fc rakamlara g\u00fcvenen bir\u00e7ok presle kar\u015f\u0131la\u015f\u0131rs\u0131n\u0131z.<\/p>\n\n\n\n
Metal y\u00f6n\u00fcne ba\u011fl\u0131 olarak sertli\u011fini de\u011fi\u015ftiriyorsa, sabit bir yar\u0131\u00e7ap form\u00fcl\u00fc levhan\u0131n nas\u0131l do\u011fdu\u011funu hi\u00e7 hat\u0131rlam\u0131yormu\u015f gibi davranabilir mi?<\/p>\n\n\n\n
Tonaj tuza\u011f\u0131: Hatal\u0131 bir hesaplamay\u0131 d\u00fczeltmek i\u00e7in makine bas\u0131nc\u0131n\u0131 art\u0131rmak neden daha k\u00f6t\u00fc sorunlar yarat\u0131r<\/h3>\n\n\n\n
A\u00e7\u0131 beklendi\u011finden daha fazla a\u00e7\u0131ld\u0131\u011f\u0131nda g\u00f6rd\u00fc\u011f\u00fcm hamle \u015fudur: tonaj\u0131 art\u0131rmak.<\/p>\n\n\n\n
Daha fazla bas\u0131n\u00e7 kontrol gibi hissettirir. Ama \u00f6yle de\u011fildir.<\/p>\n\n\n\n
Hava b\u00fckmede, tonaj do\u011frudan yar\u0131\u00e7ap\u0131 belirlemez \u2014 kal\u0131p geni\u015fli\u011fi belirler. Z\u0131mba, levhay\u0131 kal\u0131b\u0131n omuzlar\u0131 aras\u0131na iter, ancak malzemeyi tam olarak V i\u00e7ine zorlamaz. Formu, kuvvetle de\u011fil, geometrik olarak olu\u015fturuyorsunuz. Gerekenden fazla bas\u0131n\u00e7 uygulad\u0131\u011f\u0131n\u0131zda a\u00e7\u0131y\u0131 sihirli bir \u015fekilde \u201ckilitlemezsiniz.\u201d Dibe tam oturma, bask\u0131lama ve b\u00fckme hatt\u0131 boyunca tutars\u0131z gerilme riskini ba\u015flat\u0131rs\u0131n\u0131z.<\/p>\n\n\n\n
Art\u0131k kalibre edilmi\u015f hava b\u00fckme matemati\u011finiz k\u0131smi oturma davran\u0131\u015f\u0131yla kar\u0131\u015f\u0131yor.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Operat\u00f6rlerin, yar\u0131m derece yaylanma fark\u0131n\u0131 kapatmak i\u00e7in tonaj ekledi\u011fini, ancak hafif e\u011frilme ve dengesiz temas nedeniyle u\u00e7larda daha dar, ortada daha gev\u015fek a\u00e7\u0131lar olu\u015fturdu\u011funu g\u00f6rd\u00fcm. Par\u00e7alar presden \u00e7\u0131kt\u0131\u011f\u0131nda iyi g\u00f6r\u00fcn\u00fcyordu. Montajda uymad\u0131lar.<\/p>\n\n\n\n
Daha fazla kuvvet k\u00fc\u00e7\u00fck kurulum hatalar\u0131n\u0131 b\u00fcy\u00fct\u00fcr. Hafif ta\u00e7lanma hatas\u0131. Hafif malzeme kal\u0131nl\u0131k fark\u0131. Hafif ko\u00e7 e\u011filmesi. \u00d6ng\u00f6r\u00fclebilir 1.5\u00b0 yaylanma art\u0131k burada 1.2\u00b0, orada 1.8\u00b0 olur. Form\u00fcl\u00fc d\u00fczeltmediniz \u2014 gerilme desenini bulan\u0131kla\u015ft\u0131rd\u0131n\u0131z.<\/p>\n\n\n\n
Telafi y\u00f6nteminiz yaln\u0131zca belirli bir bas\u0131n\u00e7 aral\u0131\u011f\u0131nda \u00e7al\u0131\u015f\u0131yorsa, ger\u00e7ekten telafi mi \u2014 yoksa dar bir aral\u0131kta \u015fans m\u0131?<\/p>\n\n\n\n
Tak\u0131m a\u015f\u0131nmas\u0131 ve hizas\u0131zl\u0131k: Fiziksel kal\u0131p art\u0131k hesap makinenizin girdisiyle uyu\u015fmad\u0131\u011f\u0131nda ne olur?<\/h3>\n\n\n\n
Hesap makineniz, keskin ve tutarl\u0131 omuzlara sahip 1.000\u2033 V\u2019de b\u00fck\u00fcm yapt\u0131\u011f\u0131n\u0131z\u0131 san\u0131yor.<\/p>\n\n\n\n
\u0130ki y\u0131ll\u0131k \u00fcretimden sonra o kal\u0131b\u0131 gidip \u00f6l\u00e7\u00fcn.<\/p>\n\n\n\n
Kal\u0131p omuzlar\u0131 a\u015f\u0131n\u0131r. Hafif\u00e7e \u015fi\u015fer, p\u00fcr\u00fczs\u00fczle\u015fir, birka\u00e7 binde geni\u015fler. \u201c1.000 in\u00e7\u201d V art\u0131k temas noktalar\u0131nda 1.020\u2033 gibi davranabilir. Bu, etkin V\/T oran\u0131n\u0131 geni\u015fletir. Daha geni\u015f oran, y\u00fck alt\u0131ndaki yar\u0131\u00e7ap\u0131n b\u00fcy\u00fcmesi demektir. Daha b\u00fcy\u00fck yar\u0131\u00e7ap ise daha fazla yaylanma anlam\u0131na gelir.<\/p>\n\n\n\n
Matemati\u011finiz h\u00e2l\u00e2 d\u00fcn\u00fcn geometrisini kullan\u0131yor.<\/p>\n\n\n\n
Sonra toleranslar\u0131 bir araya getirin: b\u00f6l\u00fcml\u00fc tak\u0131mda biraz sa\u011f-sol hizas\u0131zl\u0131\u011f\u0131, bir kal\u0131p par\u00e7as\u0131n\u0131n alt\u0131na konulmu\u015f bir takoz, tam oturmam\u0131\u015f bir kelep\u00e7e. Art\u0131k b\u00fck\u00fcm hatt\u0131 boyunca ko\u015fullar e\u015fit de\u011fil. Tek bir K-fakt\u00f6r\u00fc, hareket eden bir hedefi tan\u0131mlamaya \u00e7al\u0131\u015f\u0131yor.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Flan\u015f uzunluklar\u0131 uzun bir par\u00e7a boyunca yirmi ya da otuz binde kadar sapmaya ba\u015flad\u0131\u011f\u0131nda at\u00f6lyeler su\u00e7luyu programc\u0131da arar. \u00c7o\u011fu zaman yeni bir kal\u0131p par\u00e7as\u0131 sorunu \u00e7\u00f6zer. Elektronik tabloya hi\u00e7 \u015fans tan\u0131nmam\u0131\u015ft\u0131r.<\/p>\n\n\n\n
Evet, daha yeni yukar\u0131dan etkili pres freni sistemleri ve daha ak\u0131ll\u0131 bombe sistemleri tarih\u00ee sapma problemlerini azalt\u0131r. \u0130yi makineler hata aral\u0131\u011f\u0131n\u0131 daralt\u0131r. Fizi\u011fi ortadan kald\u0131rmazlar. Tak\u0131mlar h\u00e2l\u00e2 a\u015f\u0131n\u0131r. Y\u00fczeyler h\u00e2l\u00e2 y\u00fck alt\u0131nda \u015fekil de\u011fi\u015ftirir. \u00c7elik h\u00e2l\u00e2 liflidir.<\/p>\n\n\n\n
Pres freni, varsay\u0131mlar\u0131n\u0131z i\u00e7in bir dayan\u0131kl\u0131l\u0131k testidir.<\/p>\n\n\n\n
Kariyerinizin geri kalan\u0131n\u0131 evrensel bir yar\u0131\u00e7ap form\u00fcl\u00fc kovalamaya harcayabilirsiniz ya da makinenin size s\u00fcrekli s\u00f6yledi\u011fini kabul edebilirsiniz: yar\u0131\u00e7ap bir kez hesaplayaca\u011f\u0131n\u0131z bir say\u0131 de\u011fil \u2014 kontrol edip izlemeniz ve d\u00fczeltmeniz gereken bir ko\u015fuldur.<\/p>\n\n\n\n
E\u011fer lif sertli\u011fi de\u011fi\u015ftiriyorsa, tonaj gerilmeyi etkiliyorsa ve tak\u0131m geometrisi zamanla kay\u0131yorsa, neden h\u00e2l\u00e2 tek bir statik denklemin bizi hurda kutusundan koruyabilece\u011fine inan\u0131yoruz?<\/p>\n\n\n\n
Say\u0131lar\u0131n Pe\u015finden Ko\u015fmay\u0131 B\u0131rak\u0131n: Ger\u00e7ekten \u0130\u015fe Yarayan Bir Yar\u0131\u00e7ap Hesaplama Sistemi Kurmak<\/h2>\n\n\n\n
De\u011fi\u015fen bir yar\u0131\u00e7ap\u0131 daha iyi bir hesap makinesiyle d\u00fczeltemezsiniz.<\/p>\n\n\n\n
Bunu, do\u011frulamad\u0131\u011f\u0131n\u0131z s\u00fcrece pres freninin sizi yan\u0131ltaca\u011f\u0131n\u0131 varsayan bir sistem kurarak d\u00fczeltirsiniz.<\/p>\n\n\n\n
Bir 10 gauge braketi \u00fczerinde, 1.000 par\u00e7al\u0131k bir serinin ortas\u0131nda, ketenin \u00f6n\u00fcnden arkas\u0131na ge\u00e7erken rulo sertli\u011fi de\u011fi\u015ftik\u00e7e a\u00e7\u0131lar\u0131n yar\u0131m derece kayd\u0131\u011f\u0131n\u0131 izledim. Form\u00fcl de\u011fi\u015fmedi. V-kal\u0131p de\u011fi\u015fmedi. Malzeme de\u011fi\u015fti. \u0130\u015fte o anda anl\u0131yorsunuz: \u00e7\u00f6zmeye \u00e7al\u0131\u015ft\u0131\u011f\u0131n\u0131z bir say\u0131 de\u011fil \u2014 hareket eden bir s\u00fcreci kontrol ediyorsunuz.<\/p>\n\n\n\n
Pres freni 200 tonluk bir ger\u00e7eklik makinesidir. Varsay\u0131mlardaki zay\u0131fl\u0131klar\u0131, bir \u00e7ekme testinin zay\u0131f \u00e7eli\u011fi ortaya \u00e7\u0131kard\u0131\u011f\u0131 gibi ortaya \u00e7\u0131kar\u0131r. E\u011fer \u201cyar\u0131\u00e7ap form\u00fcl\u00fcn\u00fcz\u201d sadece t\u00fcm y\u0131ld\u0131zlar hizaland\u0131\u011f\u0131nda \u2014 ayn\u0131 \u0131s\u0131 numaras\u0131, ayn\u0131 lif y\u00f6n\u00fc, ayn\u0131 kal\u0131p a\u015f\u0131nmas\u0131, ayn\u0131 tonaj aral\u0131\u011f\u0131 \u2014 i\u015fe yar\u0131yorsa, bir form\u00fcl\u00fcn\u00fcz yoktur. Tesad\u00fcf\u00fcn\u00fcz vard\u0131r.<\/p>\n\n\n\n
Peki bir sistem teori yerine ger\u00e7eklik i\u00e7in in\u015fa edildi\u011finde nas\u0131l g\u00f6r\u00fcn\u00fcr?<\/p>\n\n\n\n
Neden tek bir evrensel form\u00fcl farkl\u0131 malzeme ve kal\u0131nl\u0131klarda her zaman ba\u015far\u0131s\u0131z olur<\/h3>\n\n\n\n
Kal\u0131nl\u0131k s\u0131\u00e7ramalar\u0131yla ba\u015flay\u0131n.<\/p>\n\n\n\n
Orta aral\u0131kta bir plaka i\u00e7in \u2014 diyelim ki 6 ila 12 mm \u2014 yayg\u0131n kal\u0131p oranlar\u0131yla havada b\u00fck\u00fcm yaparken i\u00e7 yar\u0131\u00e7ap\u0131n genellikle malzeme kal\u0131nl\u0131\u011f\u0131n\u0131n yakla\u015f\u0131k 1,5 kat\u0131 civar\u0131nda olu\u015ftu\u011funu g\u00f6receksiniz. 12 mm\u2019nin \u00fczerine \u00e7\u0131kt\u0131\u011f\u0131n\u0131zda ise bu yar\u0131\u00e7ap, V a\u00e7\u0131kl\u0131\u011f\u0131n\u0131 \u201cdo\u011fru\u201d \u00f6l\u00e7ekleseniz bile kal\u0131nl\u0131\u011f\u0131n yakla\u015f\u0131k 3 kat\u0131na t\u0131rmanabilir. Bu bir yuvarlama hatas\u0131 de\u011fildir. Bu, do\u011frusal olmayan bir davran\u0131\u015ft\u0131r.<\/p>\n\n\n\n
(V \u2212 MT)\/2 tayfas\u0131, malzemenin d\u00fczg\u00fcn \u015fekilde \u00f6l\u00e7eklendi\u011fini varsay\u0131yor. Ger\u00e7ek \u00e7elik \u00f6yle de\u011fildir.<\/p>\n\n\n\n
Kal\u0131nl\u0131k artt\u0131k\u00e7a sadece daha fazla malzemeyi b\u00fckm\u00fcyorsun \u2014 kesit boyunca gerinim da\u011f\u0131l\u0131m\u0131n\u0131 da de\u011fi\u015ftiriyorsun. N\u00f6tr eksen kayar. Gerekli tonaj artar. Yaylanma (springback) artar \u00e7\u00fcnk\u00fc d\u0131\u015f lifler daha fazla elastik enerji depolamaktad\u0131r. K\u00e2\u011f\u0131t \u00fczerinde ayn\u0131 V\/T oran\u0131, y\u00fck alt\u0131nda farkl\u0131 bir i\u00e7 gerilim tablosu.<\/p>\n\n\n\n
\u015eimdi \u00e7ekme dayan\u0131m\u0131n\u0131 devreye sokal\u0131m.<\/p>\n\n\n\n
Ayn\u0131 kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131nda d\u00fc\u015f\u00fck akma s\u0131n\u0131r\u0131na sahip 5052 al\u00fcminyum ile y\u00fcksek dayan\u0131ml\u0131 paslanmaz \u00e7elik ayn\u0131 \u015fekilde yaylanmaz. Daha g\u00fc\u00e7l\u00fc malzeme akmadan \u00f6nce daha fazla elastik gerinim depolar. O enerji z\u0131mba kalkt\u0131\u011f\u0131nda sana geri d\u00f6ner. E\u011fer yar\u0131\u00e7ap hesab\u0131n akma dayan\u0131m\u0131n\u0131 bilmiyorsa, k\u00f6rd\u00fcr.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: \u201cAyn\u0131\u201d A36\u2019n\u0131n iki farkl\u0131 ergitme partisinin d\u00f6rt b\u00fck\u00fcde flan\u015f uzunlu\u011funu otuz binde bir in\u00e7 de\u011fi\u015ftirecek kadar farkl\u0131 davrand\u0131\u011f\u0131n\u0131 g\u00f6rd\u00fcm. Hesap tablosu ayn\u0131 girdiler demi\u015fti. Par\u00e7alar ba\u015fka \u015fey s\u00f6yledi.<\/p>\n\n\n\n
Ve donan\u0131m hatas\u0131na hen\u00fcz de\u011finmedik bile. Kal\u0131pta ondal\u0131k milimetre kadar bir hizas\u0131zl\u0131k a\u00e7\u0131n\u0131 yar\u0131m dereceden fazla sapt\u0131rabilir. Evrensel form\u00fcl\u00fcn kusursuz geometriden yola \u00e7\u0131kar. At\u00f6lyen \u00f6yle de\u011fildir.<\/p>\n\n\n\n
E\u011fer kal\u0131nl\u0131k katmanlar\u0131 do\u011frusal olmayan \u015fekilde davran\u0131yor, \u00e7ekme dayan\u0131m\u0131 yaylanmay\u0131 de\u011fi\u015ftiriyor ve tak\u0131m hizas\u0131 her g\u00fcn kay\u0131yorsa, o tek temiz denklem tam olarak neyi temsil ediyor?<\/p>\n\n\n\n
Ger\u00e7ek test verilerinden malzemeye \u00f6zg\u00fc bir yar\u0131\u00e7ap tablosu olu\u015fturmak<\/h3>\n\n\n\n
Tahmin etmeyi b\u0131rak\u0131r, \u00f6l\u00e7meye ba\u015flars\u0131n.<\/p>\n\n\n\n
Kulland\u0131\u011f\u0131n standart kal\u0131p oranlar\u0131n\u0131 se\u00e7 \u2014 ger\u00e7ekten her g\u00fcn kulland\u0131klar\u0131n\u0131. Her malzeme s\u0131n\u0131f\u0131 ve kal\u0131nl\u0131k aral\u0131\u011f\u0131 i\u00e7in, tane y\u00f6n\u00fc not edilmi\u015f test \u015feritleri kes. Kontroll\u00fc bir a\u00e7\u0131ya hava b\u00fck\u00fcm\u00fc yap. Yaylanmadan sonra i\u00e7 yar\u0131\u00e7ap\u0131 yar\u0131\u00e7ap mastarlar\u0131 veya optik \u00f6l\u00e7\u00fcmle \u00f6l\u00e7. 90\u00b0 serbest duruma ula\u015fmak i\u00e7in gereken ger\u00e7ek a\u015f\u0131r\u0131 b\u00fck\u00fcm miktar\u0131n\u0131 kaydet.<\/p>\n\n\n\n
Bunu her malzeme ailesi ve kal\u0131nl\u0131k katman\u0131 i\u00e7in bir kez yap. Her i\u015f i\u00e7in de\u011fil. Her kontroll\u00fc ko\u015ful i\u00e7in bir kez.<\/p>\n\n\n\n
\u015eimdi \u015f\u00f6yle s\u00f6yleyen bir tablo olu\u015fturursun, \u00f6rne\u011fin (varsay\u0131msal rakamlar):<\/p>\n\n\n\n
\n- 0,125\u2033 5052, 8\u00d7T kal\u0131p \u2192 0,160\u2033 ger\u00e7ek i\u00e7 yar\u0131\u00e7ap, 1,0\u00b0 yaylanma<\/li>\n\n\n\n
- 0,125\u2033 304 paslanmaz, 8\u00d7T kal\u0131p \u2192 0,185\u2033 yar\u0131\u00e7ap, 2,2\u00b0 yaylanma<\/li>\n<\/ul>\n\n\n\n
O tablo teori de\u011fildir. Yaral\u0131 tecr\u00fcbedir.<\/p>\n\n\n\n
Sertifikal\u0131 \u00e7ekme verisi olmayan yeni malzeme geldi\u011finde, bir kupon b\u00fck ve nas\u0131l davrand\u0131\u011f\u0131na bak. Pres frende be\u015f dakika ge\u00e7irmek, par\u00e7alar\u0131 be\u015f saat yeniden i\u015flemeye ye\u011fdir. E\u011fer akma de\u011ferin varsa, g\u00fczel \u2014 bunlar\u0131 tablo e\u011filimlerinle kar\u015f\u0131la\u015ft\u0131r. Zamanla kendi makinen, kendi tak\u0131m sistemin ile dayan\u0131m ve yaylanma aras\u0131ndaki desenleri g\u00f6r\u00fcrs\u00fcn.<\/p>\n\n\n\n
Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Bu ad\u0131m\u0131 atlayan at\u00f6lyeler her ilk numuneyi \u00fcretim par\u00e7alar\u0131 \u00fczerinde \u201cayar \u00e7ekerek\u201d ge\u00e7erler. Bu esneklik de\u011fildir. Bu, sac stokla kumar oynamakt\u0131r.<\/p>\n\n\n\n
Ve \u00e7o\u011funun ka\u00e7\u0131rd\u0131\u011f\u0131 nokta \u015fu: tabloya g\u00fcvenmeden \u00f6nce makineyi kontrol etmelisin. G\u00fcnl\u00fck arka dayama do\u011frulamas\u0131. Tak\u0131mlar\u0131n yerine tam oturdu\u011fundan ve temiz oldu\u011fundan emin ol. Kal\u0131p geni\u015fli\u011fini varsay\u0131mla de\u011fil pimlerle kontrol et. Donan\u0131m kayarsa, verin \u00e7\u00fcr\u00fcr.<\/p>\n\n\n\n
Kontrols\u00fcz kurulumla olu\u015fturulmu\u015f tablo yaln\u0131zca d\u00fczenlenmi\u015f bir kurgudur.<\/p>\n\n\n\n
Yani hesaplay\u0131c\u0131 i\u00e7 yar\u0131\u00e7ap\u0131 sordu\u011funda, bir tahmin mi yaz\u0131yorsun \u2014 yoksa pres freninin zaten kan\u0131tlad\u0131\u011f\u0131 bir veritaban\u0131ndan m\u0131 \u00e7ekiyorsun?<\/p>\n\n\n\n
\u201chesap makinesi girdisine\u201d g\u00fcvenmekten \u201cs\u00fcre\u00e7 sahipli\u011fini\u201d \u00fcstlenmeye ge\u00e7i\u015f\u201d<\/h3>\n\n\n\n
Bu, zihniyet de\u011fi\u015fimidir.<\/p>\n\n\n\n
Hesap makinesi yanl\u0131\u015f de\u011fildir. Eksiktir. Fiziksel d\u00fcnyay\u0131 do\u011fru tan\u0131mlad\u0131\u011f\u0131n\u0131z\u0131 varsayar \u2014 ger\u00e7ek kal\u0131p geni\u015fli\u011fi, ger\u00e7ek malzeme davran\u0131\u015f\u0131, do\u011frulanm\u0131\u015f hizalama. \u00c7o\u011fu at\u00f6lye nominal de\u011ferleri girer ve umut eder.<\/p>\n\n\n\n
S\u00fcre\u00e7 sahipli\u011fi, bu girdileri tan\u0131mlamak ve savunmak anlam\u0131na gelir.<\/p>\n\n\n\n
Kal\u0131plar\u0131 rastgele de\u011fi\u015ftirmek yerine V\/T oranlar\u0131n\u0131 standartla\u015ft\u0131r\u0131rs\u0131n\u0131z. Hangi malzemelerin hangi a\u00e7\u0131kl\u0131klarda kullan\u0131laca\u011f\u0131n\u0131 belirlersiniz. Bask\u0131larda lif y\u00f6n\u00fc gerekliliklerini belgelersiniz. Yeni s\u0131caklar\u0131 b\u00fckme testini ge\u00e7ene kadar karantinaya al\u0131rs\u0131n\u0131z. A\u00e7\u0131 sapmas\u0131n\u0131 bir sinyal olarak ele al\u0131rs\u0131n\u0131z \u2014 daha fazla tonaj ile ezilmesi gereken bir rahats\u0131zl\u0131k olarak de\u011fil.<\/p>\n\n\n\n
Ve evrensel bir i\u00e7 yar\u0131\u00e7ap form\u00fcl\u00fc olmad\u0131\u011f\u0131n\u0131 kabul edersiniz.<\/p>\n\n\n\n
Yaln\u0131zca sizin yar\u0131\u00e7ap\u0131n\u0131z vard\u0131r \u2014 sizin kal\u0131plar\u0131n\u0131zda, sizin pres freninizde, sizin malzemelerinizle \u2014 y\u00fck alt\u0131nda do\u011frulanm\u0131\u015f, el kitab\u0131n\u0131n ne s\u00f6yledi\u011fine ald\u0131rmayan 200 tonluk bir ger\u00e7ek makinesiyle.<\/p>\n\n\n\n
\u0130leri ta\u015f\u0131nmas\u0131 gereken tek \u015fey \u015fu: yar\u0131\u00e7ap hesaplad\u0131\u011f\u0131n\u0131z bir say\u0131 de\u011fildir \u2014 tan\u0131mlad\u0131\u011f\u0131n\u0131z bir davran\u0131\u015ft\u0131r.<\/p>\n\n\n\n
Bunu bu \u015fekilde g\u00f6rd\u00fc\u011f\u00fcn\u00fczde, soru \u201cForm\u00fcl nedir?\u201d olmaktan \u00e7\u0131kar ve \u201cS\u00fcrecim form\u00fcl\u00fcn anlam kazanaca\u011f\u0131 kadar s\u0131k\u0131 m\u0131?\u201d haline gelir.\u201d<\/p>\n\n\n\n
\u0130lgili Kaynaklar ve Sonraki Ad\u0131mlar<\/h2>\n\n\n\n\n- Buradaki pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Giyotin Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/li>\n\n\n\n
- Buradaki pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Panel B\u00fckme Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/li>\n\n\n\n
- Buradaki pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Lazer Kaynak Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/li>\n\n\n\n
- Buradaki pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Sac Yuvarlama Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/li>\n\n\n\n
- Buradaki pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, V Kanal A\u00e7ma Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/li>\n\n\n\n
- Buradaki pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Demir \u0130\u015fleme Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/li>\n<\/ul>","protected":false},"excerpt":{"rendered":"
Bir 10 gauge ba\u011flant\u0131 par\u00e7as\u0131n\u0131n \u00fczerinde durdum, 1.000\u2033 V kal\u0131b\u0131, \u00e7izimde a\u00e7\u0131k ve net rakamlar vard\u0131. (V \u2212 MT) \/ 2 form\u00fcl\u00fc, i\u00e7 yar\u0131\u00e7ap\u0131n 0.433\u2033 olmas\u0131 gerekti\u011fini s\u00f6yl\u00fcyordu. Kumpas ise 0.470\u2033 dedi. Her bir par\u00e7a \u00f6yleydi. Hesab\u0131n\u0131 kontrol ettin. Kal\u0131p damgas\u0131n\u0131 kontrol ettin. Malzeme partisinin su\u00e7lu oldu\u011funu d\u00fc\u015f\u00fcnd\u00fcn. Bu arada hurda kutusu sessizce dolmaya ba\u015flad\u0131 [\u2026]<\/p>","protected":false},"author":3,"featured_media":1427,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_breakdance_hide_in_design_set":false,"_breakdance_tags":"","footnotes":""},"categories":[1],"tags":[],"class_list":["post-1426","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/posts\/1426","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/comments?post=1426"}],"version-history":[{"count":1,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/posts\/1426\/revisions"}],"predecessor-version":[{"id":1431,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/posts\/1426\/revisions\/1431"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/media\/1427"}],"wp:attachment":[{"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/media?parent=1426"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/categories?post=1426"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/tags?post=1426"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
<\/figure>\n\n\n\nForm\u00fcl\u00fcn g\u00f6rmedi\u011fi \u015fey \u015fu: rulolar aras\u0131ndaki \u00e7ekme dayan\u0131m\u0131 fark\u0131, hafif bombe yapm\u0131\u015f bir yatak, a\u015f\u0131nm\u0131\u015f z\u0131mba u\u00e7lar\u0131, s\u0131cak bir \u00f6\u011fleden sonra hidrolik kaymas\u0131. Bunlar\u0131n hi\u00e7biri (V \u2212 MT) \/ 2 denkleminde g\u00f6r\u00fcnmez.<\/p>\n\n\n\n
Ama par\u00e7alar\u0131n\u0131zda g\u00f6r\u00fcn\u00fcr.<\/p>\n\n\n\n
Hava b\u00fckme, pres frenin kendisini kritik de\u011fi\u015fken h\u00e2line getirir. Vuru hassasiyeti. Ko\u00e7 paralelli\u011fi. Arka dayama tutarl\u0131l\u0131\u011f\u0131. Alt b\u00fckmede a\u011f\u0131r i\u015fi kal\u0131p yapar. Hava b\u00fckmede makine yapar. Daha temiz bir form\u00fcl, tolerans dahilinde tekrar edemeyen bir makineyi d\u00fczeltmez. Bu y\u00fczden bir\u00e7ok at\u00f6lye, y\u00fcksek hassasiyetli b\u00fck\u00fcm ve otomasyon i\u00e7in tasarlanm\u0131\u015f tamamen CNC kontroll\u00fc sistemlere y\u00f6nelir\u2014\u015funlardaki gibi CN-HAWE pres b\u00fckme serisi<\/a>\u2014burada vuru derinli\u011fi, paralellik ve tekrarlanabilirli\u011fin kontrol\u00fc tahmini de\u011fil, \u00f6l\u00e7\u00fclebilir bir yetenek h\u00e2line gelir.<\/p>\n\n\n\n Gizli maliyet sadece hurda de\u011fildir. Bu s\u00fcre\u00e7 i\u00e7in asla tasarlanmam\u0131\u015f say\u0131lar\u0131 kovalamak i\u00e7in harcanan zamand\u0131r. Bu, sahte bir \u00f6zg\u00fcvendir. Ger\u00e7ek sorun dinamik bir sistemde sabit bir denklemi kullanmakken \u201ck\u00f6t\u00fc b\u00fck\u00fcmler\u201d i\u00e7in operat\u00f6rleri su\u00e7lamakt\u0131r.<\/p>\n\n\n\n 200 tonluk bir pres, cebirinizin ne kadar zarif g\u00f6r\u00fcnd\u00fc\u011f\u00fcyle ilgilenmez. Onun tek \u00f6nemsedi\u011fi \u015fey, z\u0131mban\u0131n nerede durdu\u011fudur.<\/p>\n\n\n\n Yani kal\u0131p, hava b\u00fckmede yar\u0131\u00e7ap\u0131 ger\u00e7ekten tan\u0131mlam\u0131yorsa, o yar\u0131\u00e7ap asl\u0131nda nereden geliyor?<\/p>\n\n\n\n Ayn\u0131 1.000\u2033 V-kal\u0131\u011f\u0131 ve 10 numara 60 KSI so\u011fuk haddelenmi\u015f \u00e7eli\u011fi al\u0131n. Hava b\u00fckme ile b\u00fck\u00fclm\u00fc\u015f on par\u00e7a boyunca i\u00e7 yar\u0131\u00e7ap\u0131 \u00f6l\u00e7\u00fcn. 0.433\u2033 elde edemezsiniz. Z\u0131mba ucu yar\u0131\u00e7ap\u0131n\u0131z\u0131 da elde edemezsiniz. Yakla\u015f\u0131k 0.160\u2033 ile 0.200\u2033 aras\u0131nda bir de\u011fere ula\u015f\u0131rs\u0131n\u0131z\u2014kaba olarak kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131n\u0131n \u201320\u2019si civar\u0131nda.<\/p>\n\n\n\n Bu oran o kadar s\u0131k g\u00f6r\u00fcl\u00fcr ki art\u0131k tesad\u00fcf olmaktan \u00e7\u0131kar.<\/p>\n\n\n\n Kal\u0131p sihirli bir \u015fekilde o say\u0131y\u0131 damgalad\u0131\u011f\u0131 i\u00e7in de\u011fil. Z\u0131mba burnu o de\u011fere uydu\u011fu i\u00e7in de de\u011fil. Levha kal\u0131p omuzlar\u0131 aras\u0131nda \u00e7\u00f6kerken, y\u00fck alt\u0131nda o a\u00e7\u0131kl\u0131\u011f\u0131 ne kadar a\u015fabilece\u011fine ba\u011fl\u0131 do\u011fal bir e\u011frilik buldu\u011fu i\u00e7in. Kal\u0131p geni\u015fli\u011fi sahneyi haz\u0131rlar. Malzeme gerilimi, bu mesafe boyunca ne kadar s\u0131k\u0131 b\u00fck\u00fclebilece\u011fini belirler. Yar\u0131\u00e7ap, omuzlar aras\u0131nda havada olu\u015fur, n\u00fcfuz derinli\u011fi taraf\u0131ndan kontrol edilir.<\/p>\n\n\n\n Hava b\u00fckmede, temel i\u00e7 yar\u0131\u00e7ap\u0131 belirleyen \u015fey z\u0131mba ucu veya hesap makineniz de\u011fil, kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131d\u0131r. Geri kalan her \u015fey bunun \u00fczerine in\u015fa edilir.<\/p>\n\n\n\n E\u011fer (V \u2212 MT)\/2\u2019nin pe\u015findeyseniz, yanl\u0131\u015f mekanizmay\u0131 \u00e7\u00f6z\u00fcyorsunuz demektir.<\/p>\n\n\n\n Yakla\u015f\u0131k 1\u2033 V kal\u0131b\u0131n alt\u0131nda, 10-gauge bir ba\u011flant\u0131 par\u00e7as\u0131n\u0131n \u00fczerinde durdum, \u00e7\u00fcnk\u00fc 0.135\u2033 kal\u0131nl\u0131\u011f\u0131ndaki 10\u2011gauge \u00e7elik 8 ile \u00e7arp\u0131l\u0131nca yakla\u015f\u0131k 1.08\u2033 eder. Yeterince yak\u0131n. O 8\u00d7 kural\u0131 bir efsane de\u011fildir. Tona makul tutar ve genellikle sizi kararl\u0131 bir b\u00fckme aral\u0131\u011f\u0131na yerle\u015ftirir.<\/p>\n\n\n\n Matemati\u011fi geriye de\u011fil, ileriye do\u011fru \u00e7al\u0131\u015ft\u0131r\u0131n.<\/p>\n\n\n\n E\u011fer o kal\u0131nl\u0131k i\u00e7in 1.000\u2033 V yakla\u015f\u0131k de\u011fer aral\u0131\u011f\u0131ndaysa ve hava b\u00fckme, 60\u2011KSI \u00e7elik i\u00e7in V\u2019nin yakla\u015f\u0131k \u201320\u2019si kadar bir i\u00e7 yar\u0131\u00e7ap \u00fcretiyorsa, daha ko\u00e7 bile metale dokunmadan 0.160\u2033\u20130.200\u2033 yar\u0131\u00e7ap tahmin ediyorsunuz demektir. Bu zaten (V \u2212 MT)\/2\u2019ye hi\u00e7 yak\u0131n de\u011fildir.<\/p>\n\n\n\n \u015eimdi kal\u0131b\u0131 0.800\u2033e s\u0131k\u0131\u015ft\u0131r\u0131n\u2014yakla\u015f\u0131k 6\u00d7 kal\u0131nl\u0131k. Do\u011fal yar\u0131\u00e7ap\u0131n\u0131z yakla\u015f\u0131k 0.128\u2033\u20130.160\u2033\u2019a d\u00fc\u015fer. 1.250\u2033e a\u00e7\u0131n\u2014yakla\u015f\u0131k 10\u00d7\u2014ve yakla\u015f\u0131k 0.200\u2033\u20130.250\u2033 civar\u0131nda dola\u015f\u0131rs\u0131n\u0131z.<\/p>\n\n\n\n Oran (8\u00d7, 10\u00d7, 12\u00d7) zarafetle ilgili de\u011fildir. Ayn\u0131 anda iki \u015feyi kontrol etmekle ilgilidir: ayak ba\u015f\u0131na tonaj ve bu a\u00e7\u0131kl\u0131ktan \u00e7\u0131kacak y\u00fczdeye dayal\u0131 yar\u0131\u00e7ap. Fazla dar olursa tonaj h\u0131zla artar. Fazla geni\u015f olursa, bask\u0131 ne isterse istesin yar\u0131\u00e7ap\u0131n\u0131z \u015fi\u015fer.<\/p>\n\n\n\n Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Hi\u00e7 \u201ctonajdan emin olmak i\u00e7in\u201d kal\u0131b\u0131 a\u00e7t\u0131n\u0131z m\u0131 ve sonra flan\u015f\u0131n\u0131z\u0131n neden e\u015f par\u00e7aya uymad\u0131\u011f\u0131n\u0131 merak ettiniz mi? Bu operat\u00f6r hatas\u0131 de\u011fildi. Bu, geometriydi; siz istemeseniz de yar\u0131\u00e7ap\u0131n\u0131z\u0131 \u00f6l\u00e7eklendirdi.<\/p>\n\n\n\n Yani kal\u0131p geni\u015fli\u011fi temel y\u00fczdeyi belirliyorsa, malzemenin kendisi \u00f6nceki i\u015finizden daha zor\u2014ya da daha kolay\u2014b\u00fck\u00fcl\u00fcyorsa ne olur?<\/p>\n\n\n\n O 60\u2011KSI so\u011fuk haddelenmi\u015f \u00e7eli\u011fi ayn\u0131 1.000\u2033 V i\u00e7inde 304 paslanmazla de\u011fi\u015ftirin. Ayn\u0131 kal\u0131nl\u0131k. Ayn\u0131 z\u0131mba. 90\u00b0\u2019yi elde etmek i\u00e7in ayn\u0131 programlanm\u0131\u015f strok derinli\u011fi.<\/p>\n\n\n\n Yar\u0131\u00e7ap\u0131 \u00f6l\u00e7\u00fcn.<\/p>\n\n\n\n A\u00e7\u0131l\u0131yor.<\/p>\n\n\n\n Paslanmaz \u00e7elik daha y\u00fcksek \u00e7ekme dayan\u0131m\u0131 ta\u015f\u0131r ve agresif bi\u00e7imde i\u015f sertle\u015fmesi g\u00f6sterir. Z\u0131mba a\u015fa\u011f\u0131ya do\u011fru bast\u0131\u011f\u0131nda, malzeme e\u011frili\u011fe yumu\u015fak \u00e7eli\u011fe g\u00f6re daha fazla diren\u00e7 g\u00f6sterir. O \u2019l\u0131k alt banda o kadar kolay sar\u0131lmaz. \u2019ye veya biraz \u00fcst\u00fcne kayd\u0131\u011f\u0131n\u0131 g\u00f6rebilirsiniz. Metal, o daha dar yay\u0131 olu\u015fturmadan \u00f6nce y\u00fcksek i\u00e7 gerilme alt\u0131nda uzamaktad\u0131r.<\/p>\n\n\n\n \u015eimdi yumu\u015fak 5052 al\u00fcminyumu yerle\u015ftirin. Daha d\u00fc\u015f\u00fck \u00e7ekme dayan\u0131m\u0131. Daha d\u00fc\u015f\u00fck akma noktas\u0131. Daha az diren\u00e7. Ayn\u0131 batma seviyesinde daha derine iner ve \u00e7o\u011fu zaman y\u00fczde aral\u0131\u011f\u0131n\u0131n alt ucuna daha yak\u0131n bir yar\u0131\u00e7ap\u2014hatta baz\u0131 durumlarda biraz alt\u0131nda\u2014\u00fcretir.<\/p>\n\n\n\n Ayn\u0131 kal\u0131p. Farkl\u0131 \u00e7arpan.<\/p>\n\n\n\n \u00c7o\u011fu hesaplay\u0131c\u0131n\u0131n g\u00f6rmezden geldi\u011fi kayma budur. V\u2019yi tek de\u011fi\u015fken, kal\u0131nl\u0131\u011f\u0131 ise \u00e7\u0131karma terimi olarak ele al\u0131rlar. Ger\u00e7ekte, \u00e7ekme dayan\u0131m\u0131 o \u201320 aral\u0131\u011f\u0131n\u0131n neresinde olaca\u011f\u0131n\u0131z\u0131 de\u011fi\u015ftirir. Daha g\u00fc\u00e7l\u00fc malzeme sizi aral\u0131\u011f\u0131n \u00fcst taraf\u0131na iter. Daha yumu\u015fak malzeme s\u0131k\u0131la\u015fman\u0131za izin verir.<\/p>\n\n\n\n \u201c6 mm alt\u0131ndaki 1\u00d7 malzeme\u201d gibi kal\u0131nl\u0131k esasl\u0131 kurallar\u0131n ince levhalarda bazen i\u015fe yar\u0131yor gibi g\u00f6r\u00fcnmesinin nedeni budur. Uygun ebatl\u0131 kal\u0131pta ince yumu\u015fak \u00e7elik, genellikle kal\u0131nl\u0131\u011fa yak\u0131n bir yar\u0131\u00e7ap verir. Ancak bu, de\u011fi\u015fkenlerin hizalanmas\u0131d\u0131r; evrensel bir yasan\u0131n kan\u0131t\u0131 de\u011fil. Mukavemeti veya V geni\u015fli\u011fini de\u011fi\u015ftirin ve o d\u00fczg\u00fcn 1\u00d7 ili\u015fkisi ortadan kaybolur.<\/p>\n\n\n\n Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Paslanmaz\u0131, yumu\u015fak \u00e7elik i\u00e7in m\u00fckemmel \u00e7al\u0131\u015fan bir kal\u0131pta hi\u00e7 \u00e7al\u0131\u015ft\u0131r\u0131p a\u00e7\u0131yla u\u011fra\u015f\u0131rken yar\u0131\u00e7ap\u0131n yine b\u00fcy\u00fcd\u00fc\u011f\u00fcn\u00fc g\u00f6rd\u00fcn\u00fcz m\u00fc?<\/p>\n\n\n\n Ve e\u011fer z\u0131mba ucu o yay\u0131 oymuyorsa, ger\u00e7ekte hangi rol\u00fc oynar?<\/p>\n\n\n\n Ayn\u0131 1.000\u2033 V \u00fczerinde 0.062\u2033 z\u0131mban\u0131n burnuna bak. 10\u2011gauge\u2019\u0131 havada b\u00fck. Par\u00e7ay\u0131 \u00f6l\u00e7.<\/p>\n\n\n\n \u0130\u00e7eride 0.062\u2033 bulamayacaks\u0131n.<\/p>\n\n\n\n Z\u0131mba sac\u0131 kal\u0131p omuzlar\u0131n\u0131n aras\u0131na iter, ancak malzemeyi hi\u00e7bir zaman tamamen V\u2019nin i\u00e7ine zorlamaz. Burundaki temas noktasald\u0131r. N\u00fcfuz artt\u0131k\u00e7a sac z\u0131mban\u0131n etraf\u0131nda k\u0131smen sar\u0131l\u0131r, ard\u0131ndan omuzlar aras\u0131nda serbest bir a\u00e7\u0131kl\u0131\u011fa ge\u00e7i\u015f yapar. Nihai i\u00e7 yar\u0131\u00e7ap\u0131n b\u00fcy\u00fck k\u0131sm\u0131, z\u0131mban\u0131n damga gibi \u015feklini bast\u0131rmas\u0131ndan de\u011fil, bu a\u00e7\u0131kl\u0131kta olu\u015fan b\u00fck\u00fclmeden meydana gelir.<\/p>\n\n\n\n N\u00fcfuz derinli\u011fi her \u015feyi de\u011fi\u015ftirir. Milin sadece birka\u00e7 binde biri kadar derinle\u015fmesi sar\u0131lmay\u0131 art\u0131r\u0131r, a\u00e7\u0131kl\u0131k uzunlu\u011funu azalt\u0131r ve a\u00e7\u0131y\u0131 s\u0131k\u0131la\u015ft\u0131r\u0131r\u2014ancak yar\u0131\u00e7ap h\u00e2l\u00e2 kal\u0131p geni\u015fli\u011fi ile malzeme direncinin bir fonksiyonu olarak belirlenir. Malzeme tamamen kal\u0131ba oturana kadar altlama veya bask\u0131 (coining) yapmad\u0131k\u00e7a, z\u0131mba ucu bir kal\u0131p de\u011fil, bir y\u00f6nlendiricidir.<\/p>\n\n\n\n Bu y\u00fczden do\u011fru hava b\u00fck\u00fcm\u00fcnde z\u0131mba yar\u0131\u00e7ap\u0131n, beklenen i\u00e7 yar\u0131\u00e7aptan k\u00fc\u00e7\u00fck olmas\u0131 gerekir. E\u011filmeyi ba\u015flatmas\u0131 ama y\u00f6nlendirmemesi gerekir. Tam olarak e\u015fle\u015firlerse, ya fark\u0131nda olmadan altlama yap\u0131yorsun ya da bask\u0131 tonaj\u0131yla fl\u00f6rt ediyorsun demektir.<\/p>\n\n\n\n Pres freni 200 tonluk bir ger\u00e7ek makinesidir. Modelinin fizi\u011fe uyup uymad\u0131\u011f\u0131n\u0131 ortaya \u00e7\u0131kar\u0131r. Hava b\u00fck\u00fcm\u00fcnde fizik \u015funu s\u00f6yler: kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131 temel a\u00e7\u0131kl\u0131\u011f\u0131 tan\u0131mlar, malzemenin \u00e7ekme dayan\u0131m\u0131 bu a\u00e7\u0131kl\u0131k i\u00e7indeki oran\u0131 de\u011fi\u015ftirir ve z\u0131mba yaln\u0131zca bu sisteme ne kadar derin dald\u0131\u011f\u0131n\u0131 kontrol eder.<\/p>\n\n\n\n E\u011fer \u00e7izimin s\u0131k\u0131 bir yar\u0131\u00e7ap istiyorsa ama kal\u0131p oran\u0131n\u0131n do\u011fal e\u011filimi daha b\u00fcy\u00fck bir yar\u0131\u00e7apa y\u00f6neliyorsa, h\u00e2l\u00e2 sabit bir form\u00fcle g\u00fcvenmeye devam m\u0131 edeceksin\u2014yoksa hurda kutusu bir kez daha oy vermeden \u00f6nce kal\u0131b\u0131 m\u0131 de\u011fi\u015ftireceksin?<\/p>\n\n\n\n Ram kalkt\u0131\u011f\u0131 anda 90\u00b0 yumu\u015fak \u00e7elik bir braketi 92\u00b0\u2019de frenin \u00fczerinden \u00e7\u0131kt\u0131\u011f\u0131n\u0131 g\u00f6rd\u00fcm. Ayn\u0131 kal\u0131p. Ayn\u0131 program. Ayn\u0131 operat\u00f6r. 200 ton alt\u0131nda tam isabetti. Be\u015f saniye sonra art\u0131k de\u011fildi.<\/p>\n\n\n\n Do\u011fal kal\u0131ba ba\u011fl\u0131 yar\u0131\u00e7ap yaz\u0131c\u0131dakiyle e\u015fle\u015fmedi\u011finde asl\u0131nda \u015fu soruyu soruyorsun: y\u00fck alt\u0131ndaki durumu mu \u00e7\u00f6z\u00fcyorsun, yoksa y\u00fck kalkt\u0131ktan sonra m\u00fc\u015fterinin \u00f6l\u00e7t\u00fc\u011f\u00fc durumu mu?<\/p>\n\n\n\n Z\u0131mban\u0131n V\u2019nin i\u00e7ine g\u00f6m\u00fcl\u00fc oldu\u011fu s\u0131rada g\u00f6rd\u00fc\u011f\u00fcn yar\u0131\u00e7ap ve a\u00e7\u0131, sevk etti\u011fin par\u00e7ada g\u00f6rd\u00fc\u011f\u00fcn yar\u0131\u00e7ap ve a\u00e7\u0131 de\u011fildir. Bas\u0131n\u00e7 kalkt\u0131\u011f\u0131 anda elastik gerinim serbest kal\u0131r. Uzanm\u0131\u015f d\u0131\u015f lifler k\u0131salmaya \u00e7al\u0131\u015f\u0131r. S\u0131k\u0131\u015fm\u0131\u015f i\u00e7 lifler toparlanmaya \u00e7al\u0131\u015f\u0131r. Par\u00e7a a\u00e7\u0131l\u0131r. \u0130\u015fte bu geri esneme etkisidir ve o, senin temiz k\u00fc\u00e7\u00fck (V \u2212 MT)\/2 hesab\u0131na ald\u0131rmaz.<\/p>\n\n\n\n Hurda kutusu ger\u00e7eklik kontrol\u00fc: Derinlik g\u00f6stergesindeki a\u00e7\u0131y\u0131 m\u00fckemmel tutturup, denet\u00e7inin her flan\u015f\u0131n 1,5\u00b0 a\u00e7\u0131k oldu\u011funu s\u00f6yledi\u011fini g\u00f6rd\u00fc\u011f\u00fcn oldu mu? \u00c7elik ekrana itaat etmedi; fizi\u011fe etti.<\/p>\n\n\n\n Sabit hesaplay\u0131c\u0131lar, bas\u0131n\u00e7 alt\u0131nda olu\u015fturdu\u011fun geometrinin yerinde kald\u0131\u011f\u0131n\u0131 varsayar. Kalm\u0131yor. Ve e\u011fer plana telafi dahil etmezsen, bitmi\u015f bir yar\u0131\u00e7ap\u0131 de\u011fil ge\u00e7ici bir yar\u0131\u00e7ap\u0131 \u00f6ng\u00f6r\u00fcyorsun demektir.<\/p>\n\n\n\n As\u0131l m\u00fccadele alt strokta elde etti\u011fin yar\u0131\u00e7apta de\u011fil. Bo\u015fald\u0131ktan sonra ayakta kalan yar\u0131\u00e7apta.<\/p>\n\n\n\n Bir 1.000\u2033 V i\u00e7inde 0.125\u2033 so\u011fuk haddelenmi\u015f par\u00e7a hayal et. Z\u0131mban\u0131n a\u015fa\u011f\u0131dayken 90\u00b0\u2019lik bir a\u00e7\u0131 veren derinli\u011fe kadar iniyorsun. Ram\u0131 yukar\u0131 \u00e7ekiyorsun ve 91.5\u00b0\u2019de oturuyor. Bu, y\u00fck alt\u0131ndaki a\u00e7\u0131n\u0131n yakla\u015f\u0131k 88.5\u00b0 oldu\u011fu anlam\u0131na gelir.<\/p>\n\n\n\n \u015eimdi kendine sor: form\u00fcl\u00fcn hangi a\u00e7\u0131y\u0131 \u00f6ng\u00f6rd\u00fc?<\/p>\n\n\n\n Ger\u00e7ek bir 90\u00b0 geometrisini varsayarak ve o 1.5\u00b0\u2019lik geri esnemeyi hesaba katmadan b\u00fckme indirimi, b\u00fckme pay\u0131 ve flan\u015f uzunluklar\u0131 hesaplad\u0131ysan, her flan\u015f uzundur. \u00c7ok de\u011fil. Bir montaj\u0131 mahvedecek kadar.<\/p>\n\n\n\n Fazla b\u00fckme, hepimizin kulland\u0131\u011f\u0131 kaba bir ara\u00e7t\u0131r. 88.5\u00b0\u2019yi programla ki 90\u00b0\u2019ye gev\u015fesin. Ama i\u015fte tuzak: bu telafi t\u00fcm i\u015fler aras\u0131nda sabit de\u011fildir. Kal\u0131b\u0131 1.250\u2033\u2019ye a\u00e7 ve ayn\u0131 malzeme art\u0131k 2\u00b0 veya daha fazla geri esneyebilir; \u00e7\u00fcnk\u00fc daha b\u00fcy\u00fck yar\u0131\u00e7ap plastik \u015fekil de\u011fi\u015ftirmeyi azalt\u0131r ve kesitte daha fazla elastik enerji birikmesine yol a\u00e7ar. Daha ince malzeme? Daha fazla geri esneme. Daha b\u00fcy\u00fck i\u00e7 yar\u0131\u00e7ap? Daha fazla geri esneme. Bu ili\u015fki, geri esnemenin yaln\u0131zca dayan\u0131mla de\u011fil, yar\u0131\u00e7ap\/kal\u0131nl\u0131k oran\u0131yla artt\u0131\u011f\u0131 so\u011fuk haddelenmi\u015f \u00e7elikte g\u00f6sterilmi\u015ftir.<\/p>\n\n\n\n Yani 90\u00b0 nominal \u00fczerinden b\u00fckme indirimi hesaplay\u0131p makinede elle 1.5\u00b0 fazla b\u00fckme ekliyorsan, hesab\u0131n\u0131 ikiye b\u00f6lm\u00fc\u015fs\u00fcn demektir. D\u00fcz desen bir \u015fey d\u00fc\u015f\u00fcn\u00fcyor. Ram ba\u015fka bir \u015fey yap\u0131yor.<\/p>\n\n\n\n Geometrinizi y\u00f6nlendiren say\u0131 hangisi\u2014bask\u0131 a\u00e7\u0131s\u0131 m\u0131, yoksa ger\u00e7ekten programlad\u0131\u011f\u0131n\u0131z a\u00e7\u0131 m\u0131?<\/p>\n\n\n\n O yumu\u015fak \u00e7eli\u011fi 304 paslanmazla de\u011fi\u015ftirin. Ayn\u0131 kal\u0131nl\u0131k. Ayn\u0131 kal\u0131p. Ayn\u0131 hedef a\u00e7\u0131.<\/p>\n\n\n\n Daha fazla yaylanma g\u00f6receksiniz. Herkes g\u00f6r\u00fcr. \u0130\u00e7g\u00fcd\u00fc, \u00e7ekme dayan\u0131m\u0131n\u0131 su\u00e7lamakt\u0131r \u00e7\u00fcnk\u00fc sertifikada en b\u00fcy\u00fck say\u0131 odur. Daha y\u00fcksek \u00e7ekme, daha fazla diren\u00e7, daha fazla a\u00e7\u0131lma.<\/p>\n\n\n\n Ama \u201cayn\u0131\u201d 60\u2011KSI \u00e7eli\u011fin iki ergitmesini \u00e7al\u0131\u015ft\u0131rd\u0131\u011f\u0131n\u0131zda neler oldu\u011funu izleyin. Biri kolay b\u00fck\u00fcl\u00fcr. Di\u011feri fazladan bir derece geri yaylan\u0131r. \u00c7ekme dayan\u0131m\u0131 bir gecede 10 KSI s\u0131\u00e7ramad\u0131. De\u011fi\u015fen \u015fey akma\u2011\u00e7ekme oran\u0131d\u0131r\u2014akma oran\u0131.<\/p>\n\n\n\n Yaylanma, deformasyonun ne kadar\u0131n\u0131n elastik, ne kadar\u0131n\u0131n plastik oldu\u011funa ba\u011fl\u0131d\u0131r. \u00c7ekme dayan\u0131m\u0131na g\u00f6re y\u00fcksek akma dayan\u0131m\u0131na sahip bir malzeme, plastik deformasyona daha ge\u00e7 girer ve derin \u015fekilde akmadan \u00f6nce daha fazla elastik enerji depolar. Bu depolanan enerji, y\u00fck kalkt\u0131\u011f\u0131nda a\u00e7\u0131n\u0131z\u0131 a\u00e7an \u015feydir.<\/p>\n\n\n\n Geometri bunu b\u00fcy\u00fct\u00fcr. Daha geni\u015f kal\u0131p a\u00e7\u0131kl\u0131klar\u0131 daha b\u00fcy\u00fck i\u00e7 yar\u0131\u00e7aplar olu\u015fturur. Daha b\u00fcy\u00fck yar\u0131\u00e7ap, ayn\u0131 a\u00e7\u0131 i\u00e7in daha d\u00fc\u015f\u00fck plastik gerinim demektir. Daha d\u00fc\u015f\u00fck plastik gerinim, daha y\u00fcksek oranda elastik toparlanma demektir. Bu y\u00fczden geni\u015f V kal\u0131plarla yap\u0131lan yar\u0131\u00e7ap b\u00fck\u00fcmler, s\u0131k\u0131 6\u00d7 kurulumlara g\u00f6re dramatik bi\u00e7imde geri yaylanabilir.<\/p>\n\n\n\n Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Tonaj tasarrufu i\u00e7in hi\u00e7 geni\u015f bir kal\u0131p kulland\u0131n\u0131z m\u0131, ekrandaki a\u00e7\u0131y\u0131 tutturdunuz, sonra 10 fitlik par\u00e7a boyunca 3\u00b0 yaylanmay\u0131 kovalar halde buldunuz kendinizi? Bu k\u00f6t\u00fc \u015fans de\u011fildi. Bu, d\u00fc\u015f\u00fck plastik gerinimin kontrol\u00fc elastikli\u011fe geri vermesiydi.<\/p>\n\n\n\n Peki hangisi daha fazla fark yarat\u0131yor\u2014k\u00e2\u011f\u0131ttaki dayan\u0131m say\u0131lar\u0131 m\u0131 yoksa tak\u0131m se\u00e7iminizle belirledi\u011finiz yar\u0131\u00e7ap\u2011kal\u0131nl\u0131k oran\u0131 m\u0131? Uygulamada geometriniz sahneyi kurar. Malzeme \u00f6zellikleri geri tepmenin ne kadar sert olaca\u011f\u0131n\u0131 belirler.<\/p>\n\n\n\n Ve o geri tepme de\u011fi\u015firse, g\u00fcvendi\u011finiz t\u00fcm d\u00fcz desen say\u0131lar\u0131 ne olur?<\/p>\n\n\n\n D\u00f6rt b\u00fck\u00fcml\u00fc bir kanal al\u0131n. D\u00f6n\u00fc\u015f flan\u015f\u0131 yok. Her b\u00fck\u00fcm 2\u00b0 geri yaylans\u0131n. Bu dramatik de\u011fil. Baz\u0131 paslanmaz i\u015flerde normaldir.<\/p>\n\n\n\n \u015eimdi \u00fcst \u00fcste koyun.<\/p>\n\n\n\n D\u00f6rt b\u00fck\u00fcm \u00d7 2\u00b0 demek, her ad\u0131mda do\u011fru \u015fekilde telafi etmediyseniz son flan\u015f\u0131n ilk referansa g\u00f6re 8\u00b0 d\u0131\u015far\u0131da olmas\u0131 demektir. Her b\u00fck\u00fcm \u201cyeterince yak\u0131n\u201d g\u00f6r\u00fcn\u00fcyordu ama birikmi\u015f hata montaj\u0131 pervane gibi burktu\u011funu g\u00f6rd\u00fcm.<\/p>\n\n\n\n B\u00fck\u00fclme indirimi ve K\u2011fakt\u00f6r\u00fc bilinen bir i\u00e7 yar\u0131\u00e7ap ve bilinen bir nihai a\u00e7\u0131 varsayar. Yaylanma bunlardan herhangi birini de\u011fi\u015ftirirse ve siz say\u0131lar\u0131 g\u00fcncellemezseniz, n\u00f6tr eksenin konumu ger\u00e7ekte de\u011fi\u015fir ama yaz\u0131l\u0131m\u0131n\u0131zda de\u011fi\u015fmez. Kesti\u011finiz d\u00fcz uzunluk, gev\u015feme sonras\u0131 ger\u00e7ekte var olandan daha k\u00fc\u00e7\u00fck bir yay uzunlu\u011funa dayan\u0131r. Bunu \u00e7oklu b\u00fck\u00fcmlere ve toleranslara uygulad\u0131\u011f\u0131n\u0131zda, hatalar h\u0131zla b\u00fcy\u00fcr.<\/p>\n\n\n\n Bu y\u00fczden tek tip K\u2011fakt\u00f6rler bir hayaldir. Kal\u0131p geni\u015fli\u011fini de\u011fi\u015ftirin, yar\u0131\u00e7ap\u0131 de\u011fi\u015ftirirsiniz. Yar\u0131\u00e7ap\u0131 de\u011fi\u015ftirin, yaylanmay\u0131 de\u011fi\u015ftirirsiniz. Yaylanmay\u0131 de\u011fi\u015ftirin, nihai a\u00e7\u0131y\u0131 ve etkili b\u00fck\u00fclme pay\u0131n\u0131 de\u011fi\u015ftirirsiniz. Sisteminiz bu d\u00f6ng\u00fcy\u00fc kapatm\u0131yorsa\u2014rahatlam\u0131\u015f a\u00e7\u0131y\u0131 ve yar\u0131\u00e7ap\u0131 \u00f6l\u00e7\u00fcp indirime geri beslemiyorsa\u2014siz 200 ton alt\u0131nda var olan bir par\u00e7a i\u00e7in d\u00fcz levha kesiyorsunuz demektir.<\/p>\n\n\n\n Pres b\u00fck\u00fcc\u00fc 200 tonluk bir ger\u00e7eklik makinesidir. Malzemenin ger\u00e7ekte ne yapaca\u011f\u0131n\u0131 size s\u00f6yler. Hurda kutusu, dinleyip dinlemedi\u011finizin son yarg\u0131c\u0131d\u0131r.<\/p>\n\n\n\n Y\u00fck kald\u0131r\u0131ld\u0131ktan sonra hayatta kalan yar\u0131\u00e7ap tek \u00f6nemli olan ise, neden h\u00e2l\u00e2 ko\u00e7 kalkt\u0131\u011f\u0131 anda kaybolan yar\u0131\u00e7ap \u00fczerinden d\u00fcz desenler olu\u015fturuyorsunuz?<\/p>\n\n\n\n 1.000\u2033 V\u2011kal\u0131p alt\u0131nda 10 numara sacdan bir braketi izledim, ilk par\u00e7a bask\u0131 90\u00b0 ba\u011f\u0131r\u0131rken 92\u00b0 \u00e7\u0131kt\u0131. Programc\u0131 d\u00fcz deseni do\u011fru oldu\u011funu s\u00f6yledi. Hesaplay\u0131c\u0131 \u201ckesin\u201d i\u00e7 yar\u0131\u00e7ap dedi. Hurda kutusu umursamad\u0131.<\/p>\n\n\n\n Ger\u00e7ek geri yaylanmay\u0131 b\u00fckme indirgeme ve K\u2011fakt\u00f6r\u00fcne geri beslemek istiyorsun ki d\u00fcz par\u00e7a gev\u015femi\u015f par\u00e7ayla tam olarak e\u015fle\u015fsin. G\u00fczel. \u00c7\u00fcnk\u00fc o d\u00f6ng\u00fcy\u00fc kapatana kadar hesaplama yapm\u0131yorsun\u2014sacla kumar oynuyorsun.<\/p>\n\n\n\n G\u00fczel form\u00fcllere sabr\u0131 olmayan bir 200 tonluk ger\u00e7eklik makinesinde kulland\u0131\u011f\u0131m y\u00f6ntem bu.<\/p>\n\n\n\n 0,125\u2033 yumu\u015fak \u00e7eli\u011fi iki kal\u0131p \u00fczerinde ayarla: biri 0,750\u2033 V, di\u011feri 1,000\u2033 V. Ayn\u0131 z\u0131mba. Ayn\u0131 hedef a\u00e7\u0131. Par\u00e7alar ayn\u0131 i\u00e7 yar\u0131\u00e7apla \u00e7\u0131kmaz. \u00c7\u0131kamazlar. Kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131, sac\u0131n nas\u0131l akmas\u0131na izin verilece\u011finin geometrisini belirler.<\/p>\n\n\n\n Hava b\u00fckmede, kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131 b\u00fcy\u00fck \u00f6l\u00e7\u00fcde olu\u015facak i\u00e7 yar\u0131\u00e7ap\u0131 belirler. Ger\u00e7ek d\u00fcnyada yayg\u0131n bir ba\u015flang\u0131\u00e7 kural\u0131, daha ince yumu\u015fak \u00e7eliklerde 6:1 V\u2011kal\u0131nl\u0131k, kal\u0131nl\u0131k artt\u0131k\u00e7a 8:1 oran\u0131d\u0131r. Yani 0,125\u2033 malzeme mi? Genellikle 0,750\u2033 ila 1,000\u2033 V aras\u0131ndas\u0131nd\u0131r. Bu oran bir \u00f6neri de\u011fildir\u2014bu, gerinim zarf\u0131d\u0131r. \u00c7ok dar olursa \u00e7atlama riski vard\u0131r. \u00c7ok geni\u015f olursa plastik gerinimi azalt\u0131r ve geri yaylanma etkisini davet edersin.<\/p>\n\n\n\n \u015eimdi birinin bunun yerine teknik resimden ba\u015flad\u0131\u011f\u0131nda ne oldu\u011funa bak. \u201c0,125 yar\u0131\u00e7ap istiyorum.\u201d G\u00fczel. Ancak tonajdan tasarruf etmek i\u00e7in 1,250\u2033 V se\u00e7ersen, do\u011fal hava formu yar\u0131\u00e7ap\u0131n malzemeye ba\u011fl\u0131 olarak 0,200\u2033+ civar\u0131nda olabilir. Hi\u00e7bir hayal g\u00fcc\u00fc bunu de\u011fi\u015ftiremez. Geometri zaten karar\u0131n\u0131 vermi\u015ftir.<\/p>\n\n\n\n Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Makinede zaten tak\u0131l\u0131 oldu\u011fu i\u00e7in geni\u015f bir kal\u0131b\u0131 zorlayan, ekrandaki a\u00e7\u0131y\u0131 yakalayan, sonra da gev\u015femi\u015f yar\u0131\u00e7ap d\u00fcz modelin varsayd\u0131\u011f\u0131ndan daha b\u00fcy\u00fck oldu\u011fu i\u00e7in flan\u015flar\u0131n uzad\u0131\u011f\u0131n\u0131 izleyen at\u00f6lyeler g\u00f6rd\u00fcm. Form\u00fcl yanl\u0131\u015f de\u011fildi. Ba\u015flang\u0131\u00e7 varsay\u0131m\u0131 \u00f6yleydi.<\/p>\n\n\n\n Kal\u0131p geni\u015fli\u011fi gerinim ko\u015fulunu belirliyorsa, neden o kal\u0131ptan ba\u011f\u0131ms\u0131z hedef bir yar\u0131\u00e7apla ba\u015flamak isteyesin ki?<\/p>\n\n\n\n Kal\u0131p se\u00e7ildikten sonra, art\u0131k matematik konu\u015fmana izin var.<\/p>\n\n\n\n Hava b\u00fckmede z\u0131mba burun tamamen V i\u00e7ine girmez. Z\u0131mba, sac\u0131 kal\u0131p omuzlar\u0131 aras\u0131na iter ancak malzemeyi tamamen V i\u00e7ine zorlamaz. Sac gerilir. Bu gerilme, V\u2011a\u00e7\u0131kl\u0131\u011f\u0131 ve y\u00fck alt\u0131ndaki i\u00e7 yar\u0131\u00e7ap aras\u0131nda \u00f6ng\u00f6r\u00fclebilir bir ili\u015fki yarat\u0131r.<\/p>\n\n\n\n Yumu\u015fak \u00e7elikte, y\u00fck alt\u0131ndaki i\u00e7 yar\u0131\u00e7ap genellikle V\u2011a\u00e7\u0131kl\u0131\u011f\u0131n\u0131n yakla\u015f\u0131k \u201320\u2019si aras\u0131nda olur. Yani 1,000\u2033 V, pres a\u015fa\u011f\u0131dayken yakla\u015f\u0131k 0,160\u2033\u20130,200\u2033 yar\u0131\u00e7ap \u00fcretebilir. Tam de\u011fil. Bir aral\u0131k. \u00c7\u00fcnk\u00fc kal\u0131nl\u0131k, dayan\u0131m ve kal\u0131p omuz yar\u0131\u00e7ap\u0131 hepsi bunu biraz de\u011fi\u015ftirir.<\/p>\n\n\n\n Ne yapt\u0131\u011f\u0131m\u0131za dikkat et: yar\u0131\u00e7ap\u0131 \u00f6nce kal\u0131p geni\u015fli\u011fine ba\u011flad\u0131k, (V \u2212 MT)\/2\u2019ye ya da z\u0131mba ucuna de\u011fil. Eski (V \u2212 MT)\/2 form\u00fcl\u00fc, kesin g\u00f6r\u00fcnen say\u0131lar verebilir ama V\/T oran\u0131 6:1\u2019den 8:1\u2019e kayarsa, gerinim da\u011f\u0131l\u0131m\u0131n de\u011fi\u015fir ve o d\u00fczg\u00fcn sonu\u00e7 ger\u00e7ekte h\u0131zla sapar.<\/p>\n\n\n\n Ve bu h\u00e2l\u00e2 y\u00fck alt\u0131ndaki durumdur.<\/p>\n\n\n\n Ba\u015flang\u0131\u00e7 yar\u0131\u00e7ap\u0131 200 tonluk bir anl\u0131k g\u00f6r\u00fcnt\u00fcd\u00fcr. Sevk etti\u011fin par\u00e7a ise ko\u00e7 yukar\u0131 kalkt\u0131ktan sonraki halidir. Peki depolanm\u0131\u015f elastik enerji geri \u00e7\u0131kt\u0131\u011f\u0131nda ne olur?<\/p>\n\n\n\n 0,125\u2033 so\u011fuk haddelenmi\u015f \u00e7eli\u011fi o 1,000\u2033 V i\u00e7inde al. Y\u00fck alt\u0131ndayken 90\u00b0 \u00f6l\u00e7\u00fcyorsun. B\u0131rak\u0131nca 91,5\u00b0 oluyorsun. O 1,5\u00b0 geri yaylanma daha derin bir \u015fey anlat\u0131yor: yar\u0131\u00e7ap da b\u00fcy\u00fcd\u00fc.<\/p>\n\n\n\n A\u00e7\u0131 a\u00e7\u0131ld\u0131k\u00e7a i\u00e7 yar\u0131\u00e7ap artar. N\u00f6tr eksen biraz kayar \u00e7\u00fcnk\u00fc gerinimin elastik k\u0131sm\u0131 geri kazan\u0131l\u0131r. Akma dayan\u0131m\u0131, \u00e7ekme dayan\u0131m\u0131na g\u00f6re daha y\u00fcksek olan malzemeler, tamamen akmadan \u00f6nce daha fazla elastik enerji depolar. Paslanmaz \u00e7elik genellikle ayn\u0131 kal\u0131pta yumu\u015fak \u00e7eli\u011fe g\u00f6re daha fazla geri tepme yapar. Al\u00fcminyum da yar\u0131\u00e7ap\u2011kal\u0131nl\u0131k oran\u0131 y\u00fckseldi\u011finde seni \u015fa\u015f\u0131rtabilir.<\/p>\n\n\n\n Ama kitaplar\u0131n atlad\u0131\u011f\u0131 k\u0131s\u0131m \u015fu: geri yaylanma yar\u0131\u00e7ap\u2011kal\u0131nl\u0131k oran\u0131yla g\u00fc\u00e7l\u00fc bi\u00e7imde orant\u0131l\u0131d\u0131r. Kal\u0131b\u0131 daha geni\u015f a\u00e7t\u0131\u011f\u0131nda, i\u00e7 yar\u0131\u00e7ap\u0131 art\u0131r\u0131rs\u0131n. Yar\u0131\u00e7ap\u0131 art\u0131rd\u0131\u011f\u0131nda, belirli bir a\u00e7\u0131 i\u00e7in plastik gerinimi azalt\u0131rs\u0131n. Plastik gerinim azal\u0131nca, elastik toparlanma oran\u0131 artar. Daha fazla geri yaylanma. Bu, geometrinin malzeme davran\u0131\u015f\u0131n\u0131 b\u00fcy\u00fctmesidir.<\/p>\n\n\n\n Yani telafi s\u0131ras\u0131 \u015fu \u015fekildedir:<\/p>\n\n\n\n \u015eimdi b\u00fckme kesintisi ve K\u2011fakt\u00f6r\u00fcn\u00fcz\u00fc y\u00fck alt\u0131 geometrinizle de\u011fil, gev\u015femi\u015f yar\u0131\u00e7ap ve son a\u00e7\u0131yla g\u00fcncelleyin. \u0130\u015fte kapal\u0131 d\u00f6ng\u00fc budur. E\u011fer \u00f6l\u00e7\u00fclen geri esneme 1.5\u00b0 ise, d\u00fcz modeliniz 88.5\u00b0 y\u00fck alt\u0131nda de\u011fil, 90\u00b0 gev\u015femi\u015f durumdaki par\u00e7a yay uzunlu\u011funu yans\u0131tmal\u0131d\u0131r.<\/p>\n\n\n\n CN-HAWE\u2019nin \u00fcr\u00fcn portf\u00f6y\u00fcn\u00fcn 100% CNC tabanl\u0131 oldu\u011fu ve lazer kesim, b\u00fckme, oluk a\u00e7ma, kesme gibi y\u00fcksek d\u00fczeyli senaryolar\u0131 kapsad\u0131\u011f\u0131 g\u00f6z \u00f6n\u00fcne al\u0131nd\u0131\u011f\u0131nda, ayr\u0131nt\u0131l\u0131 materyal isteyen okuyucular i\u00e7in, Bro\u015f\u00fcrler<\/a> yararl\u0131 bir tamamlay\u0131c\u0131 kaynakt\u0131r.<\/p>\n\n\n\n Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: D\u00f6rt b\u00fck\u00fcml\u00fc kanallar\u0131n, kal\u0131p oran\u0131 de\u011fi\u015fti\u011fi h\u00e2lde programc\u0131n\u0131n bir tablodan sabit K\u2011fakt\u00f6r\u00fc kulland\u0131\u011f\u0131 i\u00e7in toplam geni\u015flikte 0.060\u2033\u2019dan fazla hatayla ka\u00e7t\u0131\u011f\u0131n\u0131 g\u00f6rd\u00fcm. Ayn\u0131 malzeme. Farkl\u0131 V. Farkl\u0131 gev\u015femi\u015f yar\u0131\u00e7ap. Matematik bunu hi\u00e7 bilemedi.<\/p>\n\n\n\n E\u011fer geri esneme hem malzeme \u00f6zelliklerine hem de kal\u0131pla olu\u015fturdu\u011funuz yar\u0131\u00e7apa ba\u011fl\u0131ysa, neden tek bir sabit K\u2011fakt\u00f6r her i\u015fte i\u015fe yaras\u0131n?<\/p>\n\n\n\n 2\u2033 geni\u015fli\u011finde bir \u015ferit kesin. Ayn\u0131 kal\u0131nl\u0131k. \u00dcretimle ayn\u0131 hadde y\u00f6n\u00fc. Se\u00e7ilen kal\u0131pta programlad\u0131\u011f\u0131n\u0131z fazla b\u00fck\u00fcmle b\u00fck\u00fcn\u2014\u00f6rne\u011fin 1.5\u00b0 geri esneme bekliyorsan\u0131z 88.5\u00b0. Gev\u015femi\u015f a\u00e7\u0131y\u0131 \u00f6l\u00e7\u00fcn ve do\u011fru mastarlarla i\u00e7 yar\u0131\u00e7ap\u0131 \u00f6l\u00e7\u00fcn.<\/p>\n\n\n\n Art\u0131k \u00fc\u00e7 ger\u00e7ek say\u0131n\u0131z var: kal\u0131p geni\u015fli\u011fi, gev\u015femi\u015f a\u00e7\u0131, gev\u015femi\u015f yar\u0131\u00e7ap.<\/p>\n\n\n\n Bunlar\u0131 b\u00fckme pay\u0131 hesaplaman\u0131za geri besleyin. Hesaplanan flan\u015f uzunluklar\u0131 \u00f6l\u00e7\u00fclen test par\u00e7as\u0131yla e\u015fle\u015fene kadar K\u2011fakt\u00f6r\u00fcn\u00fc ayarlay\u0131n. Kalibre edilmi\u015f bu K, o malzeme serisi, o kal\u0131p, o kal\u0131nl\u0131k, o kurulum i\u00e7in ge\u00e7erli h\u00e2le gelir.<\/p>\n\n\n\n Bu bir tahmin i\u015fi de\u011fil. Kontroll\u00fc yinelemedir. Y\u00fcz par\u00e7ay\u0131 kurtarmak i\u00e7in bir \u015ferit feda edilir.<\/p>\n\n\n\n Evet, makineye \u00f6zg\u00fc etkenler \u00f6nemlidir. Kal\u0131p kenar a\u015f\u0131nmas\u0131, pres ko\u00e7 defleksiyonu, bombaj ayarlar\u0131\u2014hepsi par\u00e7an\u0131n davran\u0131\u015f\u0131n\u0131 etkiler. \u0130\u015fte bu y\u00fczden test b\u00fck\u00fcm\u00fc ayn\u0131 makinede, ayn\u0131 tak\u0131m y\u0131\u011f\u0131n\u0131 ve ayn\u0131 kurulumda yap\u0131lmal\u0131d\u0131r. Sadece bir form\u00fcl\u00fc de\u011fil, t\u00fcm sistemi kalibre ediyorsunuz.<\/p>\n\n\n\n Ger\u00e7ek \u015fu: iyi bir tahmin zinciri bile modelemedi\u011finiz de\u011fi\u015fkenler taraf\u0131ndan bozulabilir.<\/p>\n\n\n\n Bu y\u00fczden kal\u0131p oran\u0131n\u0131, temel yar\u0131\u00e7ap\u0131, geri esneme telafisini ve do\u011frulanm\u0131\u015f K\u2011fakt\u00f6r\u00fcn\u00fc sabitlemi\u015f olsan\u0131z bile, \u201cm\u00fckemmel\u201d d\u00fcz modelinizi otuz mil (0.030\u2033) sapt\u0131rmaya haz\u0131r bekleyen gizli etkenler hangileridir\u2014ve bunlar\u0131 bulmak i\u00e7in ne kadar malzeme harcamaya haz\u0131rs\u0131n\u0131z?<\/p>\n\n\n\n CN\u2011HAWE y\u0131ll\u0131k sat\u0131\u015f gelirinin %8%\u2019inden fazlas\u0131n\u0131 ara\u015ft\u0131rma ve geli\u015ftirmeye yat\u0131r\u0131rken, ADH pres b\u00fckme makineleri genelinde Ar\u2011Ge yetenekleriyle faaliyet g\u00f6stermektedir; bir sonraki ad\u0131m ekiple do\u011frudan konu\u015fmaksa, Bize ula\u015f\u0131n<\/a> buraya do\u011fal \u015fekilde oturur.<\/p>\n\n\n\n Kal\u0131b\u0131 se\u00e7tiniz. Geri esnemeyi \u00f6l\u00e7t\u00fcn\u00fcz. Test \u015feridi \u00e7izime uyana kadar K\u2011fakt\u00f6r\u00fcn\u00fc ayarlad\u0131n\u0131z.<\/p>\n\n\n\n Ve \u00fcretim serisi h\u00e2l\u00e2 kay\u0131yor.<\/p>\n\n\n\n \u0130\u015fte 200 tonluk ger\u00e7ek makinesi ad\u0131n\u0131 burada kazan\u0131yor. \u0130\u00e7ine temiz girdiler, kalibre edilmi\u015f yar\u0131\u00e7aplar, \u00f6l\u00e7\u00fclm\u00fc\u015f fazla b\u00fck\u00fcm besleyebilirsiniz\u2014yine de metalin veya onu tutan donan\u0131m\u0131n i\u00e7inde gizlenen zay\u0131f varsay\u0131mlar\u0131 ortaya \u00e7\u0131kar\u0131r. Form\u00fcl y\u00fcksek sesle hata yapmaz. Par\u00e7alar ise ayaklan\u0131r ve tolerans d\u0131\u015f\u0131na \u00e7\u0131kar.<\/p>\n\n\n\n CN-HAWE\u2019nin m\u00fc\u015fteri taban\u0131; in\u015faat makineleri, otomotiv \u00fcretimi, gemi in\u015fas\u0131, k\u00f6pr\u00fcler, havac\u0131l\u0131k gibi sekt\u00f6rleri kapsad\u0131\u011f\u0131ndan, burada pratik se\u00e7enekleri de\u011ferlendiren ekipler i\u00e7in, Lazer Kesim Makinesi<\/a> ilgili bir sonraki ad\u0131md\u0131r.<\/p>\n\n\n\n M\u00fckemmel matematik sadece m\u00fckemmel ko\u015fullarda hayatta kal\u0131r.<\/p>\n\n\n\n Sorun \u015fu ki, bir \u00fcretim at\u00f6lyesinde hi\u00e7bir \u015fey uzun s\u00fcre m\u00fckemmel kalmaz.<\/p>\n\n\n\n Peki \u201cher \u015feyi do\u011fru yapt\u0131\u011f\u0131n\u0131zda\u201d bile yar\u0131\u00e7ap\u0131n\u0131z\u0131 h\u00e2l\u00e2 de\u011fi\u015ftirebilen tek \u015fey ne kald\u0131?<\/p>\n\n\n\n \u00c7elik izotropik de\u011fildir. Bu, basit bir at\u00f6lye ger\u00e7e\u011fi i\u00e7in kitap diliyle \u015fu anlama gelir: bir y\u00f6nde di\u011ferine g\u00f6re daha kolay b\u00fck\u00fcl\u00fcr.<\/p>\n\n\n\n Levha haddeleme s\u0131ras\u0131nda, taneler haddeleme y\u00f6n\u00fcnde uzar. Tane y\u00f6n\u00fcne dik b\u00fck\u00fcld\u00fc\u011f\u00fcnde, malzeme bu uzam\u0131\u015f lifler boyunca gerilir. Tane y\u00f6n\u00fcne paralel b\u00fck\u00fcld\u00fc\u011f\u00fcnde ise onlar\u0131 bir fermuar gibi a\u00e7maya \u00e7al\u0131\u015f\u0131rs\u0131n\u0131z. Ayn\u0131 kal\u0131nl\u0131k. Ayn\u0131 kal\u0131p. Farkl\u0131 yaylanma miktar\u0131.<\/p>\n\n\n\n 0.125\u2033 paslanmaz \u00e7eli\u011fin tane y\u00f6n\u00fcne dik olarak tam 90\u00b0 b\u00fck\u00fcl\u00fcp \u2014 ayn\u0131 1.000\u2033 V i\u00e7inde tane y\u00f6n\u00fcyle b\u00fck\u00fcld\u00fc\u011f\u00fcnde neredeyse bir derece daha fazla geri s\u0131\u00e7rad\u0131\u011f\u0131n\u0131 izledim. Ba\u015fka hi\u00e7bir \u015fey de\u011fi\u015fmedi. Ayn\u0131 program. Ayn\u0131 operat\u00f6r. Test par\u00e7as\u0131ndan kalibre edilmi\u015f ayn\u0131 K fakt\u00f6r\u00fc.<\/p>\n\n\n\n Tek fark y\u00f6nelimdi.<\/p>\n\n\n\n Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Test \u015feridiniz tane y\u00f6n\u00fcne dik kesilmi\u015f, ancak \u00fcretim bo\u015fluklar\u0131 malzemeden tasarruf etmek i\u00e7in tane y\u00f6n\u00fc uzun eksende olacak \u015fekilde yerle\u015ftirilmi\u015fse, ilk darbeden \u00f6nce fazla b\u00fckme telafiniz zaten yanl\u0131\u015ft\u0131r. Presin hangi y\u00f6n\u00fcn daha ucuz yerle\u015ftirildi\u011fi umurunda de\u011fildir.<\/p>\n\n\n\n A\u00e7\u0131 sens\u00f6rleriyle donat\u0131lm\u0131\u015f modern makineler ger\u00e7ek zamanl\u0131 otomatik d\u00fczeltme yapabilir. G\u00fczel. Bu da noktay\u0131 kan\u0131tlar. Tane \u00f6nemli olmasayd\u0131, her b\u00fckmede \u00f6l\u00e7\u00fcm yapmalar\u0131 gerekmezdi. Ancak modern bir \u00fcretim at\u00f6lyesine girip ger\u00e7ekte ne oldu\u011funa bak\u0131n \u2014 h\u00e2l\u00e2 canl\u0131 a\u00e7\u0131 d\u00fczeltmesi olmadan \u00e7al\u0131\u015fan, d\u00fcnk\u00fc rakamlara g\u00fcvenen bir\u00e7ok presle kar\u015f\u0131la\u015f\u0131rs\u0131n\u0131z.<\/p>\n\n\n\n Metal y\u00f6n\u00fcne ba\u011fl\u0131 olarak sertli\u011fini de\u011fi\u015ftiriyorsa, sabit bir yar\u0131\u00e7ap form\u00fcl\u00fc levhan\u0131n nas\u0131l do\u011fdu\u011funu hi\u00e7 hat\u0131rlam\u0131yormu\u015f gibi davranabilir mi?<\/p>\n\n\n\n A\u00e7\u0131 beklendi\u011finden daha fazla a\u00e7\u0131ld\u0131\u011f\u0131nda g\u00f6rd\u00fc\u011f\u00fcm hamle \u015fudur: tonaj\u0131 art\u0131rmak.<\/p>\n\n\n\n Daha fazla bas\u0131n\u00e7 kontrol gibi hissettirir. Ama \u00f6yle de\u011fildir.<\/p>\n\n\n\n Hava b\u00fckmede, tonaj do\u011frudan yar\u0131\u00e7ap\u0131 belirlemez \u2014 kal\u0131p geni\u015fli\u011fi belirler. Z\u0131mba, levhay\u0131 kal\u0131b\u0131n omuzlar\u0131 aras\u0131na iter, ancak malzemeyi tam olarak V i\u00e7ine zorlamaz. Formu, kuvvetle de\u011fil, geometrik olarak olu\u015fturuyorsunuz. Gerekenden fazla bas\u0131n\u00e7 uygulad\u0131\u011f\u0131n\u0131zda a\u00e7\u0131y\u0131 sihirli bir \u015fekilde \u201ckilitlemezsiniz.\u201d Dibe tam oturma, bask\u0131lama ve b\u00fckme hatt\u0131 boyunca tutars\u0131z gerilme riskini ba\u015flat\u0131rs\u0131n\u0131z.<\/p>\n\n\n\n Art\u0131k kalibre edilmi\u015f hava b\u00fckme matemati\u011finiz k\u0131smi oturma davran\u0131\u015f\u0131yla kar\u0131\u015f\u0131yor.<\/p>\n\n\n\n Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Operat\u00f6rlerin, yar\u0131m derece yaylanma fark\u0131n\u0131 kapatmak i\u00e7in tonaj ekledi\u011fini, ancak hafif e\u011frilme ve dengesiz temas nedeniyle u\u00e7larda daha dar, ortada daha gev\u015fek a\u00e7\u0131lar olu\u015fturdu\u011funu g\u00f6rd\u00fcm. Par\u00e7alar presden \u00e7\u0131kt\u0131\u011f\u0131nda iyi g\u00f6r\u00fcn\u00fcyordu. Montajda uymad\u0131lar.<\/p>\n\n\n\n Daha fazla kuvvet k\u00fc\u00e7\u00fck kurulum hatalar\u0131n\u0131 b\u00fcy\u00fct\u00fcr. Hafif ta\u00e7lanma hatas\u0131. Hafif malzeme kal\u0131nl\u0131k fark\u0131. Hafif ko\u00e7 e\u011filmesi. \u00d6ng\u00f6r\u00fclebilir 1.5\u00b0 yaylanma art\u0131k burada 1.2\u00b0, orada 1.8\u00b0 olur. Form\u00fcl\u00fc d\u00fczeltmediniz \u2014 gerilme desenini bulan\u0131kla\u015ft\u0131rd\u0131n\u0131z.<\/p>\n\n\n\n Telafi y\u00f6nteminiz yaln\u0131zca belirli bir bas\u0131n\u00e7 aral\u0131\u011f\u0131nda \u00e7al\u0131\u015f\u0131yorsa, ger\u00e7ekten telafi mi \u2014 yoksa dar bir aral\u0131kta \u015fans m\u0131?<\/p>\n\n\n\n Hesap makineniz, keskin ve tutarl\u0131 omuzlara sahip 1.000\u2033 V\u2019de b\u00fck\u00fcm yapt\u0131\u011f\u0131n\u0131z\u0131 san\u0131yor.<\/p>\n\n\n\n \u0130ki y\u0131ll\u0131k \u00fcretimden sonra o kal\u0131b\u0131 gidip \u00f6l\u00e7\u00fcn.<\/p>\n\n\n\n Kal\u0131p omuzlar\u0131 a\u015f\u0131n\u0131r. Hafif\u00e7e \u015fi\u015fer, p\u00fcr\u00fczs\u00fczle\u015fir, birka\u00e7 binde geni\u015fler. \u201c1.000 in\u00e7\u201d V art\u0131k temas noktalar\u0131nda 1.020\u2033 gibi davranabilir. Bu, etkin V\/T oran\u0131n\u0131 geni\u015fletir. Daha geni\u015f oran, y\u00fck alt\u0131ndaki yar\u0131\u00e7ap\u0131n b\u00fcy\u00fcmesi demektir. Daha b\u00fcy\u00fck yar\u0131\u00e7ap ise daha fazla yaylanma anlam\u0131na gelir.<\/p>\n\n\n\n Matemati\u011finiz h\u00e2l\u00e2 d\u00fcn\u00fcn geometrisini kullan\u0131yor.<\/p>\n\n\n\n Sonra toleranslar\u0131 bir araya getirin: b\u00f6l\u00fcml\u00fc tak\u0131mda biraz sa\u011f-sol hizas\u0131zl\u0131\u011f\u0131, bir kal\u0131p par\u00e7as\u0131n\u0131n alt\u0131na konulmu\u015f bir takoz, tam oturmam\u0131\u015f bir kelep\u00e7e. Art\u0131k b\u00fck\u00fcm hatt\u0131 boyunca ko\u015fullar e\u015fit de\u011fil. Tek bir K-fakt\u00f6r\u00fc, hareket eden bir hedefi tan\u0131mlamaya \u00e7al\u0131\u015f\u0131yor.<\/p>\n\n\n\n Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Flan\u015f uzunluklar\u0131 uzun bir par\u00e7a boyunca yirmi ya da otuz binde kadar sapmaya ba\u015flad\u0131\u011f\u0131nda at\u00f6lyeler su\u00e7luyu programc\u0131da arar. \u00c7o\u011fu zaman yeni bir kal\u0131p par\u00e7as\u0131 sorunu \u00e7\u00f6zer. Elektronik tabloya hi\u00e7 \u015fans tan\u0131nmam\u0131\u015ft\u0131r.<\/p>\n\n\n\n Evet, daha yeni yukar\u0131dan etkili pres freni sistemleri ve daha ak\u0131ll\u0131 bombe sistemleri tarih\u00ee sapma problemlerini azalt\u0131r. \u0130yi makineler hata aral\u0131\u011f\u0131n\u0131 daralt\u0131r. Fizi\u011fi ortadan kald\u0131rmazlar. Tak\u0131mlar h\u00e2l\u00e2 a\u015f\u0131n\u0131r. Y\u00fczeyler h\u00e2l\u00e2 y\u00fck alt\u0131nda \u015fekil de\u011fi\u015ftirir. \u00c7elik h\u00e2l\u00e2 liflidir.<\/p>\n\n\n\n Pres freni, varsay\u0131mlar\u0131n\u0131z i\u00e7in bir dayan\u0131kl\u0131l\u0131k testidir.<\/p>\n\n\n\n Kariyerinizin geri kalan\u0131n\u0131 evrensel bir yar\u0131\u00e7ap form\u00fcl\u00fc kovalamaya harcayabilirsiniz ya da makinenin size s\u00fcrekli s\u00f6yledi\u011fini kabul edebilirsiniz: yar\u0131\u00e7ap bir kez hesaplayaca\u011f\u0131n\u0131z bir say\u0131 de\u011fil \u2014 kontrol edip izlemeniz ve d\u00fczeltmeniz gereken bir ko\u015fuldur.<\/p>\n\n\n\n E\u011fer lif sertli\u011fi de\u011fi\u015ftiriyorsa, tonaj gerilmeyi etkiliyorsa ve tak\u0131m geometrisi zamanla kay\u0131yorsa, neden h\u00e2l\u00e2 tek bir statik denklemin bizi hurda kutusundan koruyabilece\u011fine inan\u0131yoruz?<\/p>\n\n\n\n De\u011fi\u015fen bir yar\u0131\u00e7ap\u0131 daha iyi bir hesap makinesiyle d\u00fczeltemezsiniz.<\/p>\n\n\n\n Bunu, do\u011frulamad\u0131\u011f\u0131n\u0131z s\u00fcrece pres freninin sizi yan\u0131ltaca\u011f\u0131n\u0131 varsayan bir sistem kurarak d\u00fczeltirsiniz.<\/p>\n\n\n\n Bir 10 gauge braketi \u00fczerinde, 1.000 par\u00e7al\u0131k bir serinin ortas\u0131nda, ketenin \u00f6n\u00fcnden arkas\u0131na ge\u00e7erken rulo sertli\u011fi de\u011fi\u015ftik\u00e7e a\u00e7\u0131lar\u0131n yar\u0131m derece kayd\u0131\u011f\u0131n\u0131 izledim. Form\u00fcl de\u011fi\u015fmedi. V-kal\u0131p de\u011fi\u015fmedi. Malzeme de\u011fi\u015fti. \u0130\u015fte o anda anl\u0131yorsunuz: \u00e7\u00f6zmeye \u00e7al\u0131\u015ft\u0131\u011f\u0131n\u0131z bir say\u0131 de\u011fil \u2014 hareket eden bir s\u00fcreci kontrol ediyorsunuz.<\/p>\n\n\n\n Pres freni 200 tonluk bir ger\u00e7eklik makinesidir. Varsay\u0131mlardaki zay\u0131fl\u0131klar\u0131, bir \u00e7ekme testinin zay\u0131f \u00e7eli\u011fi ortaya \u00e7\u0131kard\u0131\u011f\u0131 gibi ortaya \u00e7\u0131kar\u0131r. E\u011fer \u201cyar\u0131\u00e7ap form\u00fcl\u00fcn\u00fcz\u201d sadece t\u00fcm y\u0131ld\u0131zlar hizaland\u0131\u011f\u0131nda \u2014 ayn\u0131 \u0131s\u0131 numaras\u0131, ayn\u0131 lif y\u00f6n\u00fc, ayn\u0131 kal\u0131p a\u015f\u0131nmas\u0131, ayn\u0131 tonaj aral\u0131\u011f\u0131 \u2014 i\u015fe yar\u0131yorsa, bir form\u00fcl\u00fcn\u00fcz yoktur. Tesad\u00fcf\u00fcn\u00fcz vard\u0131r.<\/p>\n\n\n\n Peki bir sistem teori yerine ger\u00e7eklik i\u00e7in in\u015fa edildi\u011finde nas\u0131l g\u00f6r\u00fcn\u00fcr?<\/p>\n\n\n\n Kal\u0131nl\u0131k s\u0131\u00e7ramalar\u0131yla ba\u015flay\u0131n.<\/p>\n\n\n\n Orta aral\u0131kta bir plaka i\u00e7in \u2014 diyelim ki 6 ila 12 mm \u2014 yayg\u0131n kal\u0131p oranlar\u0131yla havada b\u00fck\u00fcm yaparken i\u00e7 yar\u0131\u00e7ap\u0131n genellikle malzeme kal\u0131nl\u0131\u011f\u0131n\u0131n yakla\u015f\u0131k 1,5 kat\u0131 civar\u0131nda olu\u015ftu\u011funu g\u00f6receksiniz. 12 mm\u2019nin \u00fczerine \u00e7\u0131kt\u0131\u011f\u0131n\u0131zda ise bu yar\u0131\u00e7ap, V a\u00e7\u0131kl\u0131\u011f\u0131n\u0131 \u201cdo\u011fru\u201d \u00f6l\u00e7ekleseniz bile kal\u0131nl\u0131\u011f\u0131n yakla\u015f\u0131k 3 kat\u0131na t\u0131rmanabilir. Bu bir yuvarlama hatas\u0131 de\u011fildir. Bu, do\u011frusal olmayan bir davran\u0131\u015ft\u0131r.<\/p>\n\n\n\n (V \u2212 MT)\/2 tayfas\u0131, malzemenin d\u00fczg\u00fcn \u015fekilde \u00f6l\u00e7eklendi\u011fini varsay\u0131yor. Ger\u00e7ek \u00e7elik \u00f6yle de\u011fildir.<\/p>\n\n\n\n Kal\u0131nl\u0131k artt\u0131k\u00e7a sadece daha fazla malzemeyi b\u00fckm\u00fcyorsun \u2014 kesit boyunca gerinim da\u011f\u0131l\u0131m\u0131n\u0131 da de\u011fi\u015ftiriyorsun. N\u00f6tr eksen kayar. Gerekli tonaj artar. Yaylanma (springback) artar \u00e7\u00fcnk\u00fc d\u0131\u015f lifler daha fazla elastik enerji depolamaktad\u0131r. K\u00e2\u011f\u0131t \u00fczerinde ayn\u0131 V\/T oran\u0131, y\u00fck alt\u0131nda farkl\u0131 bir i\u00e7 gerilim tablosu.<\/p>\n\n\n\n \u015eimdi \u00e7ekme dayan\u0131m\u0131n\u0131 devreye sokal\u0131m.<\/p>\n\n\n\n Ayn\u0131 kal\u0131p a\u00e7\u0131kl\u0131\u011f\u0131nda d\u00fc\u015f\u00fck akma s\u0131n\u0131r\u0131na sahip 5052 al\u00fcminyum ile y\u00fcksek dayan\u0131ml\u0131 paslanmaz \u00e7elik ayn\u0131 \u015fekilde yaylanmaz. Daha g\u00fc\u00e7l\u00fc malzeme akmadan \u00f6nce daha fazla elastik gerinim depolar. O enerji z\u0131mba kalkt\u0131\u011f\u0131nda sana geri d\u00f6ner. E\u011fer yar\u0131\u00e7ap hesab\u0131n akma dayan\u0131m\u0131n\u0131 bilmiyorsa, k\u00f6rd\u00fcr.<\/p>\n\n\n\n Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: \u201cAyn\u0131\u201d A36\u2019n\u0131n iki farkl\u0131 ergitme partisinin d\u00f6rt b\u00fck\u00fcde flan\u015f uzunlu\u011funu otuz binde bir in\u00e7 de\u011fi\u015ftirecek kadar farkl\u0131 davrand\u0131\u011f\u0131n\u0131 g\u00f6rd\u00fcm. Hesap tablosu ayn\u0131 girdiler demi\u015fti. Par\u00e7alar ba\u015fka \u015fey s\u00f6yledi.<\/p>\n\n\n\n Ve donan\u0131m hatas\u0131na hen\u00fcz de\u011finmedik bile. Kal\u0131pta ondal\u0131k milimetre kadar bir hizas\u0131zl\u0131k a\u00e7\u0131n\u0131 yar\u0131m dereceden fazla sapt\u0131rabilir. Evrensel form\u00fcl\u00fcn kusursuz geometriden yola \u00e7\u0131kar. At\u00f6lyen \u00f6yle de\u011fildir.<\/p>\n\n\n\n E\u011fer kal\u0131nl\u0131k katmanlar\u0131 do\u011frusal olmayan \u015fekilde davran\u0131yor, \u00e7ekme dayan\u0131m\u0131 yaylanmay\u0131 de\u011fi\u015ftiriyor ve tak\u0131m hizas\u0131 her g\u00fcn kay\u0131yorsa, o tek temiz denklem tam olarak neyi temsil ediyor?<\/p>\n\n\n\n Tahmin etmeyi b\u0131rak\u0131r, \u00f6l\u00e7meye ba\u015flars\u0131n.<\/p>\n\n\n\n Kulland\u0131\u011f\u0131n standart kal\u0131p oranlar\u0131n\u0131 se\u00e7 \u2014 ger\u00e7ekten her g\u00fcn kulland\u0131klar\u0131n\u0131. Her malzeme s\u0131n\u0131f\u0131 ve kal\u0131nl\u0131k aral\u0131\u011f\u0131 i\u00e7in, tane y\u00f6n\u00fc not edilmi\u015f test \u015feritleri kes. Kontroll\u00fc bir a\u00e7\u0131ya hava b\u00fck\u00fcm\u00fc yap. Yaylanmadan sonra i\u00e7 yar\u0131\u00e7ap\u0131 yar\u0131\u00e7ap mastarlar\u0131 veya optik \u00f6l\u00e7\u00fcmle \u00f6l\u00e7. 90\u00b0 serbest duruma ula\u015fmak i\u00e7in gereken ger\u00e7ek a\u015f\u0131r\u0131 b\u00fck\u00fcm miktar\u0131n\u0131 kaydet.<\/p>\n\n\n\n Bunu her malzeme ailesi ve kal\u0131nl\u0131k katman\u0131 i\u00e7in bir kez yap. Her i\u015f i\u00e7in de\u011fil. Her kontroll\u00fc ko\u015ful i\u00e7in bir kez.<\/p>\n\n\n\n \u015eimdi \u015f\u00f6yle s\u00f6yleyen bir tablo olu\u015fturursun, \u00f6rne\u011fin (varsay\u0131msal rakamlar):<\/p>\n\n\n\n O tablo teori de\u011fildir. Yaral\u0131 tecr\u00fcbedir.<\/p>\n\n\n\n Sertifikal\u0131 \u00e7ekme verisi olmayan yeni malzeme geldi\u011finde, bir kupon b\u00fck ve nas\u0131l davrand\u0131\u011f\u0131na bak. Pres frende be\u015f dakika ge\u00e7irmek, par\u00e7alar\u0131 be\u015f saat yeniden i\u015flemeye ye\u011fdir. E\u011fer akma de\u011ferin varsa, g\u00fczel \u2014 bunlar\u0131 tablo e\u011filimlerinle kar\u015f\u0131la\u015ft\u0131r. Zamanla kendi makinen, kendi tak\u0131m sistemin ile dayan\u0131m ve yaylanma aras\u0131ndaki desenleri g\u00f6r\u00fcrs\u00fcn.<\/p>\n\n\n\n Hurda Kutusu Ger\u00e7eklik Kontrol\u00fc: Bu ad\u0131m\u0131 atlayan at\u00f6lyeler her ilk numuneyi \u00fcretim par\u00e7alar\u0131 \u00fczerinde \u201cayar \u00e7ekerek\u201d ge\u00e7erler. Bu esneklik de\u011fildir. Bu, sac stokla kumar oynamakt\u0131r.<\/p>\n\n\n\n Ve \u00e7o\u011funun ka\u00e7\u0131rd\u0131\u011f\u0131 nokta \u015fu: tabloya g\u00fcvenmeden \u00f6nce makineyi kontrol etmelisin. G\u00fcnl\u00fck arka dayama do\u011frulamas\u0131. Tak\u0131mlar\u0131n yerine tam oturdu\u011fundan ve temiz oldu\u011fundan emin ol. Kal\u0131p geni\u015fli\u011fini varsay\u0131mla de\u011fil pimlerle kontrol et. Donan\u0131m kayarsa, verin \u00e7\u00fcr\u00fcr.<\/p>\n\n\n\n Kontrols\u00fcz kurulumla olu\u015fturulmu\u015f tablo yaln\u0131zca d\u00fczenlenmi\u015f bir kurgudur.<\/p>\n\n\n\n Yani hesaplay\u0131c\u0131 i\u00e7 yar\u0131\u00e7ap\u0131 sordu\u011funda, bir tahmin mi yaz\u0131yorsun \u2014 yoksa pres freninin zaten kan\u0131tlad\u0131\u011f\u0131 bir veritaban\u0131ndan m\u0131 \u00e7ekiyorsun?<\/p>\n\n\n\n Bu, zihniyet de\u011fi\u015fimidir.<\/p>\n\n\n\n Hesap makinesi yanl\u0131\u015f de\u011fildir. Eksiktir. Fiziksel d\u00fcnyay\u0131 do\u011fru tan\u0131mlad\u0131\u011f\u0131n\u0131z\u0131 varsayar \u2014 ger\u00e7ek kal\u0131p geni\u015fli\u011fi, ger\u00e7ek malzeme davran\u0131\u015f\u0131, do\u011frulanm\u0131\u015f hizalama. \u00c7o\u011fu at\u00f6lye nominal de\u011ferleri girer ve umut eder.<\/p>\n\n\n\n S\u00fcre\u00e7 sahipli\u011fi, bu girdileri tan\u0131mlamak ve savunmak anlam\u0131na gelir.<\/p>\n\n\n\n Kal\u0131plar\u0131 rastgele de\u011fi\u015ftirmek yerine V\/T oranlar\u0131n\u0131 standartla\u015ft\u0131r\u0131rs\u0131n\u0131z. Hangi malzemelerin hangi a\u00e7\u0131kl\u0131klarda kullan\u0131laca\u011f\u0131n\u0131 belirlersiniz. Bask\u0131larda lif y\u00f6n\u00fc gerekliliklerini belgelersiniz. Yeni s\u0131caklar\u0131 b\u00fckme testini ge\u00e7ene kadar karantinaya al\u0131rs\u0131n\u0131z. A\u00e7\u0131 sapmas\u0131n\u0131 bir sinyal olarak ele al\u0131rs\u0131n\u0131z \u2014 daha fazla tonaj ile ezilmesi gereken bir rahats\u0131zl\u0131k olarak de\u011fil.<\/p>\n\n\n\n Ve evrensel bir i\u00e7 yar\u0131\u00e7ap form\u00fcl\u00fc olmad\u0131\u011f\u0131n\u0131 kabul edersiniz.<\/p>\n\n\n\n Yaln\u0131zca sizin yar\u0131\u00e7ap\u0131n\u0131z vard\u0131r \u2014 sizin kal\u0131plar\u0131n\u0131zda, sizin pres freninizde, sizin malzemelerinizle \u2014 y\u00fck alt\u0131nda do\u011frulanm\u0131\u015f, el kitab\u0131n\u0131n ne s\u00f6yledi\u011fine ald\u0131rmayan 200 tonluk bir ger\u00e7ek makinesiyle.<\/p>\n\n\n\n \u0130leri ta\u015f\u0131nmas\u0131 gereken tek \u015fey \u015fu: yar\u0131\u00e7ap hesaplad\u0131\u011f\u0131n\u0131z bir say\u0131 de\u011fildir \u2014 tan\u0131mlad\u0131\u011f\u0131n\u0131z bir davran\u0131\u015ft\u0131r.<\/p>\n\n\n\n Bunu bu \u015fekilde g\u00f6rd\u00fc\u011f\u00fcn\u00fczde, soru \u201cForm\u00fcl nedir?\u201d olmaktan \u00e7\u0131kar ve \u201cS\u00fcrecim form\u00fcl\u00fcn anlam kazanaca\u011f\u0131 kadar s\u0131k\u0131 m\u0131?\u201d haline gelir.\u201d<\/p>\n\n\n\n Bir 10 gauge ba\u011flant\u0131 par\u00e7as\u0131n\u0131n \u00fczerinde durdum, 1.000\u2033 V kal\u0131b\u0131, \u00e7izimde a\u00e7\u0131k ve net rakamlar vard\u0131. (V \u2212 MT) \/ 2 form\u00fcl\u00fc, i\u00e7 yar\u0131\u00e7ap\u0131n 0.433\u2033 olmas\u0131 gerekti\u011fini s\u00f6yl\u00fcyordu. Kumpas ise 0.470\u2033 dedi. Her bir par\u00e7a \u00f6yleydi. Hesab\u0131n\u0131 kontrol ettin. Kal\u0131p damgas\u0131n\u0131 kontrol ettin. Malzeme partisinin su\u00e7lu oldu\u011funu d\u00fc\u015f\u00fcnd\u00fcn. Bu arada hurda kutusu sessizce dolmaya ba\u015flad\u0131 [\u2026]<\/p>","protected":false},"author":3,"featured_media":1427,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_breakdance_hide_in_design_set":false,"_breakdance_tags":"","footnotes":""},"categories":[1],"tags":[],"class_list":["post-1426","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/posts\/1426","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/comments?post=1426"}],"version-history":[{"count":1,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/posts\/1426\/revisions"}],"predecessor-version":[{"id":1431,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/posts\/1426\/revisions\/1431"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/media\/1427"}],"wp:attachment":[{"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/media?parent=1426"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/categories?post=1426"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/cn-hawe.com\/tr\/wp-json\/wp\/v2\/tags?post=1426"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}V-Kal\u0131\u011f\u0131n Fizi\u011fi: Yar\u0131\u00e7ap Asl\u0131nda Nerede Olu\u015fur<\/h2>\n\n\n\n
Ger\u00e7ek temel \u00e7izgiyi hesaplamak: 8x ila 12x kal\u0131p oran\u0131 ger\u00e7eklik kontrol\u00fc<\/h3>\n\n\n\n
\u00c7arpan kaymas\u0131: Paslanmaz \u00e7elik ile yumu\u015fak al\u00fcminyum i\u00e7in matemati\u011fi nas\u0131l ayarlamal\u0131<\/h3>\n\n\n\n
Tak\u0131m geometrisi vs. malzeme davran\u0131\u015f\u0131: Fiziksel z\u0131mba ucu yar\u0131\u00e7ap\u0131n\u0131z ile nihai i\u00e7 yar\u0131\u00e7ap\u0131n\u0131z\u0131n neden nadiren uyu\u015fmas\u0131 gerekti\u011fi<\/h3>\n\n\n\n
Geri esneme telafisi: Sabit hesaplay\u0131c\u0131lar\u0131n g\u00f6z ard\u0131 etti\u011fi de\u011fi\u015fken standart<\/h2>\n\n\n\n
Geri esneme i\u00e7in fazla b\u00fckme: Rahatlam\u0131\u015f yar\u0131\u00e7ap i\u00e7in mi yoksa y\u00fck alt\u0131ndaki yar\u0131\u00e7ap i\u00e7in mi hesap yap\u0131yorsun?<\/h3>\n\n\n\n
\u00c7ekme dayan\u0131m\u0131 ve akma oran\u0131: Hangi \u00f6zellik ger\u00e7ekten fark yarat\u0131yor?<\/h3>\n\n\n\n
Hesaplanmam\u0131\u015f yaylanman\u0131n b\u00fck\u00fclme indirimini ve K\u2011Fakt\u00f6r\u00fc\u2019n\u00fc sessizce bozmas\u0131<\/h3>\n\n\n\n
Tahminden Kontrole: \u0130\u00e7 B\u00fck\u00fclme Yar\u0131\u00e7ap\u0131n\u0131 Tahmin Etmek \u0130\u00e7in 4 Ad\u0131ml\u0131 Y\u00f6ntem<\/h2>\n\n\n\n
Ad\u0131m 1: Bask\u0131daki \u201cistenen\u201d yar\u0131\u00e7aptan de\u011fil, kal\u0131p oran\u0131ndan ba\u015fla<\/h3>\n\n\n\n
Ad\u0131m 2: V\u2011a\u00e7\u0131kl\u0131\u011f\u0131ndan teorik yar\u0131\u00e7ap\u0131 tahmin et (ve bunun yaln\u0131zca bir ba\u015flang\u0131\u00e7 de\u011feri oldu\u011funu kabul et)<\/h3>\n\n\n\n
Ad\u0131m 3: Ger\u00e7ek gev\u015femi\u015f yar\u0131\u00e7ap\u0131 bulmak i\u00e7in malzeme temelli geri yaylanma telafisini uygula<\/h3>\n\n\n\n
\n
Ad\u0131m 4: Tam \u00fcretim deneme\u2011yan\u0131lma yerine kontroll\u00fc bir test b\u00fck\u00fcm\u00fcyle do\u011frulay\u0131n<\/h3>\n\n\n\n
M\u00fckemmel Matemati\u011fi Bozan G\u00f6r\u00fcnmez De\u011fi\u015fkenler<\/h2>\n\n\n\n
Tane y\u00f6n\u00fc: Haddelenme y\u00f6n\u00fc gerekli fazla b\u00fckme a\u00e7\u0131s\u0131n\u0131 nas\u0131l de\u011fi\u015ftirir<\/h3>\n\n\n\n
Tonaj tuza\u011f\u0131: Hatal\u0131 bir hesaplamay\u0131 d\u00fczeltmek i\u00e7in makine bas\u0131nc\u0131n\u0131 art\u0131rmak neden daha k\u00f6t\u00fc sorunlar yarat\u0131r<\/h3>\n\n\n\n
Tak\u0131m a\u015f\u0131nmas\u0131 ve hizas\u0131zl\u0131k: Fiziksel kal\u0131p art\u0131k hesap makinenizin girdisiyle uyu\u015fmad\u0131\u011f\u0131nda ne olur?<\/h3>\n\n\n\n
Say\u0131lar\u0131n Pe\u015finden Ko\u015fmay\u0131 B\u0131rak\u0131n: Ger\u00e7ekten \u0130\u015fe Yarayan Bir Yar\u0131\u00e7ap Hesaplama Sistemi Kurmak<\/h2>\n\n\n\n
Neden tek bir evrensel form\u00fcl farkl\u0131 malzeme ve kal\u0131nl\u0131klarda her zaman ba\u015far\u0131s\u0131z olur<\/h3>\n\n\n\n
Ger\u00e7ek test verilerinden malzemeye \u00f6zg\u00fc bir yar\u0131\u00e7ap tablosu olu\u015fturmak<\/h3>\n\n\n\n
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\u201chesap makinesi girdisine\u201d g\u00fcvenmekten \u201cs\u00fcre\u00e7 sahipli\u011fini\u201d \u00fcstlenmeye ge\u00e7i\u015f\u201d<\/h3>\n\n\n\n
\u0130lgili Kaynaklar ve Sonraki Ad\u0131mlar<\/h2>\n\n\n\n
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