Torsion na sanduna da sassa na bakin ciki-banga

** Abubuwan da ke cikin ƙwararrun ma’ajin ajiya:** rubutu da ƙididdiga suna isar da gabatarwar ka’idar tarihi kuma ba aiki ba ne, ƙididdigewa a tsaye, duban ƙarfin ɗaukar nauyi, kimanta abin da ke akwai, ko umarnin gini ko gyarawa. Dole ne a duba zayyanawa, zato da ƙimayar ƙima da ƙayyadaddun tsarin, abu, lissafi, yanayin iyaka da ƙa’idodi masu dacewa. Ƙididdigar ɓarna, kwanciyar hankali, gajiya, haɗin gwiwa da damuwa saboda hana lankwasawa ya kamata injiniya mai lasisi yayi, tare da kulawa mai zaman kanta lokacin da ake buƙata.

A yau, Savo Kusić yana mai da hankali kan gilashin katako, itace-aluminum Taga, Taga na al’ada, kofa da buƙatun magana. Wannan labarin ya kasance a matsayin tarihin ƙwararrun ma’ajin tarihi kuma baya wakiltar sabis ɗin ƙira.

Saint-Venant ka’idar torsion (torsion kyauta)

Ma’auni daban-daban

Sl.1

Bari a gyara sanda madaidaiciya a daya daga cikin iyakarta x=0kuma bari a ɗora shi a ɗayan ƙarshen tare da lokacin jujjuyawa M (fig.1). Sannan kowane sashinsa yana jujjuya ta wani kusurwa α(x) a kusa da wani layi daya daidai da axis na sandar, wanda zamu dauka a matsayin x-axis. Idan duk sassan sun yi daidai da juna, kowane madaidaiciyar layi mai layi daya da axis x zai shiga cikin coil, kuma kusurwar α zata kasance daidai da nisa daga ƙayyadaddun ƙarshen: α = ϑx, daga abin da keɓancewar ɓangaren a cikin jirgin na sashin ya tashi (Fig.2):

v = - r zunubi__ψ = - ϑ x z,

_w = r a cos__ψ = ϑ x y.  (1a)

Adadin ϑ ana kiransa kusurwar torsion (girma: tsayin-1, juyawa kowane tsayin raka’a). Matsakaicin kusurwa yayi daidai da lokacin torsion M:

ϑ = M/GJT.   (2)

Sl.2

Matsakaicin ma’auni na GJT, taurin kai, daidai yake da samfurin G shear modules da factor JT, wanda ke wakiltar tasirin sifa da girman sashin.

Wannan jujjuyawar sashin ba shine kawai nakasar sandar ba. Bugu da kari, akwai kuma lankwasawa na giciye-section, i.e. zuwa gudun hijira u a cikin shugabanci na x\ -axis, wanda yana da guda girma ga madaidaicin maki na duk giciye-sections kuma saboda haka ne kawai wani aiki na daidaitawa a cikin jirgin na sashe, kuma shi ne in ba haka ba daidai da lodi M kwana: σsion.

u = u(y,z) = _ϑ ∙ φ(y,z), dφ/dx=0.  (1b)

Idan an gabatar da maganganun da ke sama a cikin daidaitattun daidaiton ƙaura, ana samun cewa babban damuwa na yau da kullun daidai yake da sifili, yana barin abubuwa biyu ne kawai ke damuwa τxy da τxz:

Daya.3

Ana samun bambancin daidaito na matsalar torsion bisa ga ma’auni (3) sanya:

Daya.4kuma i4b

da warware waɗannan bambance-bambancen ma’auni don abubuwan da ba a sani ba τxy da τxz ta hanyar gabatar da aikin damuwa F(y,z) cewa:

τxy = - dF/dz,   τxz = + dF/dy.  (5)

Tare da wannan, equation (4a) gamsuwa iri ɗaya, kuma daga lissafin (4b) ana samun ma’aunin bambancin da ake buƙata

d2F/da2+ d2F/dz2=2G__ϑ (6)

Yanayin iyaka da ake buƙata don ƙayyadaddun bayani na wannan ma’auni yana samuwa ne daga abin da ake bukata cewa babu damuwa a kan ambulaf na sanda, daga abin da, saboda yanayin haɗin gwiwa na ƙarfin juzu’i, ya biyo baya cewa damuwa a cikin sashin jirgin ba dole ba ne ya kasance yana da wani ɓangaren da ya dace da shi tare da kwane-kwane. Daga nan, fig.2:

_dy/dz = _τxy/_τxz = - (dF/dz) / (dF/dy)

saboda haka

dF/dy dy + dF/dz dz =0,

watau jimlar bambancin aikin F dole ne ya zama sifili, wato, dole ne ya zama F=const. tare da kwane-kwane.

Idan sashin yana da kwane-kwane ɗaya kawai, ana iya sanya shi kawai F=0, saboda tare da aikin ƙarfin lantarki, wanda a zahiri muna sha’awar abubuwan da aka samo asali ne kawai, ƙari na sabani ba ya taka rawa. Ga sassan da ke da kwane-kwane fiye da ɗaya, wannan batu ya fi rikitarwa.

Kamar yadda aka gani daga lissafin (5) sakamakon matsananciyar shear τxy da τxz a madaidaicin mahadar sabani daidai yake da girman girman aikin danniya F, amma yana tsaye da shi, watau yana da madaidaicin shugabanci zuwa layin F=const.

Ta hanyar warware daidaiton bambancin (6)) ana samun ƙarfin lantarki τ azaman ayyukan y, z da ϑ. Ayyukan ƙididdiga shine, ba shakka, ƙaddamar da kusurwar torsion ϑ da damuwa, musamman ma matsakaicin matsakaici a matsayin aiki na lokacin torsional, sabili da haka, ƙaddamar da ƙaddamar da sashin JT daga lissafin (2) da juriya juriya WT = Mmaxτ. Don samun waɗannan, kawai dole ne a bayyana M azaman aikin F:

Duka taƙaice guda biyu na wannan haɗin kai, wanda ya kamata a ɗauka a kan dukan sashe, za a iya canza shi ta hanyar haɗin kai. Idan, alal misali, na farkonsu ya fara haɗawa tare da layin z=const. daga wurin da ke kan kwane-kwane y=y1 har zuwa y=y2(fig.3) kuma kuyi la’akari da cewa a cikin waɗannan abubuwan biyu F=0, ana samu

Ana samun darajar iri ɗaya daga haɗin na biyu, don haka

Daya.7

Sl.3

Sauran lokuta na cikakkun sassan

Bambancin equation (6) shine ma’auni mara daidaituwa na yuwuwar ka’idar. Sabili da haka, don magance matsalar torsion a cikin yanayin ƙetare na sabani, akwai hanyoyin da za a iya amfani da ka’idar, wanda ke ba da damar samun shi koyaushe. Waɗannan hanyoyin, duk da haka, gabaɗaya suna da sarƙaƙƙiya da yawa waɗanda ba za a iya amfani da su ba a kan shari’o’in ɗaiɗaikun waɗanda suka taso a aikace. Saboda haka, tebur mai zuwa yana nuna sakamakon irin waɗannan ƙididdiga don sassa da yawa.

Don sassan giciye kusa da da’ira, JT=F ana iya sanyawa4/40_Ip_, inda F shine yankin kuma Ip shine lokacin rashin aiki na sashin.

** Bayanan kula akan ma’auni da tebur: ** ana adana maganganu da ƙima a cikin hotunan da aka bincika azaman tushen tarihi. Rubutun OCR, alamomin fihirisa da dubawa ba su da abin dogaro ga lissafi. Kafin amfani, kowace dabara, naúrar, ƙididdiga da yanki na inganci yakamata a duba su a cikin tushen ƙwararru mai iko da ƙa’idar aiki.

Analogy tare da membrane

Kyakkyawan taimako don kimanta matsalolin ƙwanƙwasa, kuma a lokaci guda ana ba da hanya don tabbatar da daidaitattun ƙayyadaddun su ta hanyar kwatankwacin aiki tsakanin aikin damuwa F na matsalar torsional da juzu’i mai jujjuyawa na membrane. Bari a sami matsakaicin buɗewa akan bangon lebur na jirgin ruwa wanda ya yi daidai da ɓangaren giciye na sanda da aka ɗora a cikin torsion. Lokacin da membrane ya shimfiɗa a kan wannan buɗewar kuma an haifar da ƙananan matsa lamba p a cikin jirgin ruwa, membrane yana kumbura waje, ta yadda ma’anarsa ξ(y,z) gamsar da daidaitattun daidaito.

inda N shine tashin hankali na sama a cikin membrane wanda sojojin capillary suka haifar (girma: karfi da tsawon raka’a). Wannan ma’auni ya zo daidai da lissafin (6), da yanayin iyaka ξ=0daidai yake da na aikin wutar lantarki. Ta wannan hanyar, ana iya samun aikin damuwa F(y,z) tare da taimakon wannan madubi tare da membrane kuma ta haka ne za a warware matsalar torsion. Danniya τ yayi daidai da gangaren membrane dangane da y-z\ -jirgin sama, don haka wuri mafi tsayi shine inda sandar da aka fallasa ga torsion ke da mafi girman damuwa.

Sl.4

Sl.5

Wannan hanya ta membrane ta dace musamman don samun ƙimayar ƙima don sassan bakin bango. Lokacin da, alal misali, an raba sashin I-kashi bisa ga fig.4a kan rectangles uku kuma ga kowane ɗayansu ana yin kwatancen tare da membrane daban, jimlar juzu’i na fitowar da aka samu zai zama ɗan ƙarami kaɗan fiye da ƙarar da ke daidai da aikin ƙarfin lantarki. Don haka, lokacin da kauri ke b1 da b2 karami, kusan

JT \2JT1+ JT2,

ku JT1 da JT2 abubuwan taurin kai na mutum rectangles waɗanda za a iya ƙididdige su bisa tebur1. Ana samun ingantacciyar hanya idan aka sanya JT don hakarkarinsa2 = a2b32/3, don haka, lokacin da ake ɗaukar haƙarƙarin a matsayin wani ɓangare na rectangle mai tsayi sosai. Ana samun madaidaicin ƙimar bisa ga Trayer da March \ daga

JT =2JT1+1/3a2b32+2 da D4,_

inda D shine diamita mafi girman da’irar da aka rubuta kamar yadda aka nuna a fig.4, kuma α ana iya ɗauka daga fig.5. Don sashin T yana samuwa kamar haka

_JT = JT1+JT2+ A D4,

ku JT2 rabin darajar da aka lissafta bisa ga tebur1za a=2_a2_, b= b2. Don sassan L (fig.6) yana kama da haka

JT = JT1+ JT2+ _ β D4

ku JT1 ga hannun mai kauri kai tsaye daga tebur1, yayin da _JT2_domin an ƙididdige ƙafar bakin ciki kamar sashin T, kuma an ɗauko β daga fig.7.

Geometry na L-section tare da girman ƙafafu, kauri da radius na zagaye

Sl.6

Sl.7

Lokacin ƙididdige damuwa, lokacin torsional ya kasu kashi huɗu na rectangles: ɓangaren da ya faɗi akan flanges M1= M JT2/JT; Sannan ana ƙididdige damuwa mai ƙarfi a tsakiyar tsakiya na mafi tsayin gefen kowane rectangle bisa tebur1. Mafi girma daga cikin waɗannan matsalolin bazai buƙatar zama mafi girman damuwa da ke bayyana a sashin ba. Wato, idan ba a zagaye kusurwar abin da ke faruwa ba, ana samun τ=∞ a wannan wuri, ta yadda don ƙananan radius na zagaye, ana iya sa ran damuwa mafi girma a nan.

** Bayanin Injiniyan:** Mathematics singularity na ingantacciyar kusurwa ba lasisi ba ne don ƙididdige ainihin dalla-dalla ta wannan madaidaicin tsari kaɗai. Haƙiƙanin juzu’i na jujjuyawa, juriya, filastik, walda, damuwa saura, kwanciyar hankali na gida da gajiya suna buƙatar ƙirar ƙirar da ta dace da tabbatarwar ƙwararru.

Don kusurwa mai gefe ɗaya wanda kaurin gefensa shine t bisa ga Trefftz\ yana kan zagayen diamita r:

ku da τ0 Alamar madaidaicin ƙarfin lantarki da aka ƙididdige kowace ƙafa ɗaya bisa tsarin da aka kwatanta a nan. Don manyan radiyon zagaye, τ= ana iya ɗauka2__τ0.

Bututun Karo Na Siriri (Tsarin Bradt)

A cikin bututu masu bakin ciki, ƙumburi mai ƙyalli yana da siffar da aka nuna a cikin fig.8, wanda ya ƙunshi jirgin sama sama da sararin samaniya a cikin bututu da kuma wani wuri mai zurfi wanda ke kewaye da shi kuma ya kwanta a saman sashin tasiri na bangon bututu; dole ne wannan fili mai karkata ya kasance mai zurfi idan ganuwar bututun ya fi bakin ciki a wurin da ya dace. A cikin wannan yanayin, ana rarraba matsalolin daɗaɗɗa daidai daidai a kan kaurin ganuwar kuma suna da bambanci da kauri na ganuwar, watau shear force T = τt ba ya canzawa tare da kewayen sashin.

Sl.8

Abubuwan da ke sama t_∙ds_ suna karɓar wani ɓangare na lokacin torsional

dM = t_∙ds∙τ∙h,_

don haka jimlar lokacin shine

Daya.8

inda filin FR ke rufe ta tsakiyar layi mai alamar layi da dige a cikin fig.8. Ana samun shi daga nan

Daya.9- Tsarin Bredt na farko

Ana samun taurin kai cikin sauƙi lokacin da aikin nakasawa ya bayyana a gefe guda ta hanyar juzu’i da kusurwar torsion kuma a gefe guda ta hanyar damuwa mai ƙarfi. Ta wannan hanyar, ana samun shi don ɓangaren sanda mai tsayi l

don haka, gabatar da ƙimar τ bisa ga ƙididdiga (8),

wato idan aka kwatanta da ma’auni (2):

Daya.10

Idan kauri daga cikin ganuwar ya kasance iri ɗaya a kusa da shi, to yana da sauƙi (2. Tsarin Bredt):

Daya.11

Matsaloli na al’ada a cikin ƙwanƙwasa saboda lankwasawa sashe

Kalmomin asali da ma’auni daban-daban

Maɓallin ɓangaren u, equation (1b), yana ƙayyade karkatar sassan daga jiragen su. Lokacin da lokacin torsion ya kasance akai-akai, wannan lanƙwasawa iri ɗaya ce a duk sassan kuma damuwa na yau da kullun σx baya bayyana.

A lokuta da yawa, curvature na giciye-section, wanda ya kamata tashi a kan tushen _St. Ka’idar Venant ba zai yiwu ba. Idan lokacin jujjuyawar ya canza ba zato ba tsammani a cikin sashe guda ɗaya, lanƙwasawa mai girma daban-daban zai faru a ɓangarorin biyu na ɓangaren, kuma ci gaba da nakasar za ta damu. Ƙarshen sandar ɗaya na iya zama manne don haka ba a cire lanƙwasa sashin gaba ɗaya a wannan lokacin. A irin waɗannan lokuta, ya zama dole don ƙarin _St. Ka’idodin Venant. Ƙarin damuwa σx a cikin hanyar axis na sanda da ƙarin ƙarfi mai ƙarfi T (ƙuƙuwar yankewa T/t) suna bayyana a cikin sandar, waɗanda sune mafi girma a cikin sashin da ci gaba na nakasawa ya karye kuma wanda ke raguwa da nisa daga wannan sashe. Game da sanduna masu ƙarfi da bututu, wannan ka’idar tana da rikitarwa, kuma ƙarin ƙarfin lantarki yana raguwa da sauri ta yadda ƙudurinsu ba ya da fa’ida a mafi yawan lokuta. Don sanduna masu buɗe ido na bakin ciki, wannan gyaran yana da mahimmanci kuma ana iya ƙididdige shi cikin sauƙi.

Sl.9

Sl.9yana nuna layin da tsakiyar bangon bangon da sashin jirgin sama suka haɗu (layin tsakiyar sashin). O shine tsakiyar jujjuyawar dangi na sassa biyu maƙwabta. Idan nisan su dx ne, kusurwar torsion shine ϑ dx, don haka wurin sabani na tsakiyar layi yana fuskantar ƙaura u ϑ dx. Wannan ƙaura yana da ɓangaren rt ϑ dx a cikin hanyar tangent zuwa sashin, kuma shine zamewar sigar tsakiyar farfajiyar bangon (la’akari da cewa _u=_φϑ):

γ=__ϑ (rt + d__φ/ds).

Tunda damuwa mai ƙarfi a tsakiyar bangon bango ba komai bane, dole ne kuma ya zama γ=0, don haka ne

Wannan tsarin yana ƙunshe da sigogi na sabani guda uku: asalin da aka ƙididdige tsayin arc s da kuma haɗin gwiwar tsakiyar juyawar dangi O, wanda rt ya dogara da shi. Za a iya zaɓar wurin farawa wanda aka ƙididdige tsayin baka s ta yadda ma’anar curvature ta zama sifili:

kuma za’a iya tantance ma’auni na cibiyar O ta yadda lokutan da aka lankwashe sashin ba su zama sifili ba:

Ana iya nuna cewa center of rotation don haka ƙayyadaddun ya zo daidai da cibiyar shear wacce ke a nan da aka bayyana da kuma ayyana ta wata hanya.

Idan an hana lanƙwasawa na ɓangaren da waɗannan alamu suka bayyana gaba ɗaya ko kuma an hana su, damuwa na al’ada σx a cikin madaidaicin sandar dole ne ya bayyana. Bincike na kusa ya nuna cewa waɗannan matsalolin za a iya ɗauka su yi daidai da aikin curvature φ: σx=k__φ, inda φ ya dogara ne akan s da k kawai akan x. Dangane da dokar Hooke:

σx = E_∂u/∂x = Eφ dϑ/dx_, (12)

dole ne ya zama k=E d__ϑ/dx. Matsaloli na al’ada σx ana kiran sashe na lanƙwasawa. Idan aka yi la’akari da ma’auni na sama, sun samar da tsarin ma’auni na ma’auni a kowane sashe, tsarin karfi saboda lankwasa sashe.

Jimlar karfin juyi ya ƙunshi sassa biyu: ɗaya daga cikinsu M1=GJT__ϑ yana tasowa saboda matsananciyar juzu’i da St. Ka’idar Venant, da sauran M2 yana faruwa ne saboda lankwasa sashin:

Daya.13

Tun lokacin da aka bayar da M azaman aikin x, wannan magana tana wakiltar ma’auni na banbanta don kusurwar torsion ϑ. Da zarar an warware wannan ma’auni, ana iya ƙididdige M1 da M2, don haka duk damuwa da damuwa.

Magani da aikace-aikace

Lokacin da lokacin torsion M ya kasance akai-akai, cikakken bayani na lissafin (13) shine:

Matsakaicin haɗin kai _C1, C2_Dole ne a ƙayyade daga sharuɗɗan iyaka biyu.

a) Ƙarshen manne da ƙarfi x=0(fig.1)

A wannan karshen shine u≡0, don haka bisa ga lissafin (1b) ϑ=0. Idan sanda ya yi tsayi sosai (αl>>1), sannan C1=0kuma C2= -M/GJT. A cikin yanayin gajeren sanduna, bayyanar danniya saboda lankwasawa na giciye ya shafi yanayin da ke gefen dama na sandar. Idan ba a hana lanƙwasawa na sashin ba a wurin, za a sami σx≡0da dϑ/dx=0. Ana samun shi daga nan

b) Matsakaicin nauyin torsion bisa ga fig.10

A gefen hagu na sandar, lokacin jujjuyawar shine -M/2, kuma a hannun dama +M/2. A cikin yanayin ɓarke ​​​​kyauta, sassan hagu da dama na tsakiyar sanda za a karkatar da su a wata hanya ta gaba, saboda haka ba za a kiyaye ci gaba da nakasar ba. Saboda haka, a wurin x=0Dole ne a canja wurin duk lokacin torsional ta hanyar matsalolin da ke faruwa saboda lanƙwasa ɓangaren giciye, don haka a cikin wannan ɓangaren M1=0don haka ϑ=0. A wannan lokaci, tsarin dakarun yana aiki don tabbatar da cewa sashin ya kasance a tsaye, ko da yake sassan da ke kwance daga tsakiyar igiya. Idan, ban da haka, ana buƙatar cewa a ƙarshen sandar matsalolin sune σx≡0, Ana samun ƙarin yanayin iyaka ga kowane rabin sandar, don haka yana da rabin dama

Sl.10

c) Lokacin waje M a wani wuri na sabani (fig.11)

A wannan yanayin, maganin ϑ na sassan biyu na sanda dole ne a rubuta shi daban, don haka jimlar ƙididdiga huɗu C ya bayyana.1, C2, C3, C4. Bugu da kari, lokacin Ma, Mb ba a san su ba. Akwai ma’auni guda shida: yanayin ci gaba don ϑ (wanda aka bayyana ta u) da na dϑ/dx (wanda aka bayyana ta σx) a wurin lodi, yanayi ɗaya a kowane ƙarshen sanda (misali matsi mai ƙarfi ϑ=0, ko lankwasawa mara hankali, dϑ/dx=0), yanayin ma’auni ta lokuta -Ma+Mb=M kuma a ƙarshe yanayin cewa sassan ƙarshen ba sa juyawa game da juna: ∫ ϑdx =0.

Sl.11

** Bayanin ƙarshe:** zaɓin ƙirar torsion ya dogara da sashin buɗewa ko rufe, siririn, hanyar tallafi, wurin shigar da lokacin da yuwuwar lanƙwasawa kyauta. Don ainihin kashi, ƙayyadaddun ƙayyadaddun kayan aiki da iyakokin sabis, cikakkun bayanai na gida, haɗin gwiwa, gajiya da kwanciyar hankali dole ne a duba su, ba kawai wata keɓewar dabara daga labarin ba.