Ana kiran_shells manyan masu ɗaukar kaya na saman, waɗanda tsakiyar samansu ba lebur ba ne, sai dai a bayyane yake ɗan kumbura. Halayensu na ƙarfi na tsaya-tsaye sun bambanta ƙwarai da na faranti. Idan suna da siffa mai dacewa, damuwarsu ta fi ƙanƙanta sosai, ƙarfinsu ya fi yawa daidai gwargwado, kuma ƙididdigarsu a mafi yawan lokuta ta fi sauƙi.
Damuwa a kowace harsashi za a iya raba ta zuwa sassa biyu: damuwar membrane da lankwasawa. Damuwa ta membrane, kamar yadda ake yi da damuwa a farantin da aka ɗora a cikin gadonta, za a iya bayyana ta hanyar ƙarfafan sashen a cikin tangential plane – misali, a sashin x=const. ƙarfin al’ada _Nx=_σx*t da ƙarfin yankewa _Nxy=_τxy*t, inda t kaurin harsashi ne. Damuwa saboda lankwasawa iri ɗaya ce da ta faranti kuma ana iya wakilta ta da momentin lankwasawa Mx, momentin torsion Mxy da ƙarfafan yankewa Qx.
Muhimman siffofin harsashi suna dogara ne da gaskiyar cewa, a ƙa’ida, damuwa saboda lankwasawa ba su da girma ko muhimmanci kamar damuwar membrane, don haka sau da yawa za a iya yin watsi da su gaba ɗaya. A wannan hali, duk damuwa ana rarraba su daidai a cikin kaurin harsashi kuma suna aiki tangentially a saman tsakiya. Lissafin da suka dogara da waɗannan zato suna shiga cikin tsarin ka’idar membrane ta harsashi.
Ka’idar membrane ta harsashi masu daidaiton juyawa ƙarƙashin kaya mai daidaiton juyawa
Tsarin gabaɗaya
A matsayin layukan daidaitawa ana ɗaukar meridiyoyi da da’irori masu layi a saman tsakiya na harsashi, don haka ƙa’idodin suna: azimut ϑ (“tsawon ƙasa”) da karkata φ na tangenti zuwa meridiyan dangane da jirgin da ya yi daidai da axis ɗin daidaitawa (sl. 1a). Saboda haka ƙarfin yanka su ne: ƙarfin al’ada Nφ a cikin hanyar meridiyan da ƙarfin al’ada Nϑ a cikin zobe (sl. 1b). Ƙarfin yankewa Nφϑ ba sa bayyana a cikin damuwar daidaitacciyar juyawa. Nauyin a kowace raka’ar fili ana nuna shi kamar haka (sl. 1b): Y tangensiyal zuwa meridiyan, mai kyau a cikin hanyar da φ ke ƙaruwa, Z a tsaye ga harsashi, mai kyau zuwa ga axis ɗin daidaitawa.
Hoto. 1
Fig. 2
Bari a ɗauki radius ɗin lanƙwasa na meridiyan r1. Shi aikin kusurwa φ ne don haka da daidaitawar _r1=r1(_φ) ake ƙayyade siffar meridiyan. Ana iya ɗaukar wannan a matsayin daidaitawar meridiyan kuma tana cewa, misali.
domin da’ira r1=a=const.
ga parabola r1=a/cos3__φ
ga ellips r1=a2b2(a2sin2__φ+b2cos2__φ)-3/2
Domin lissafa ƙarfin al’ada N__φ a cikin alkiblar meridiyan, ya kamata a kafa sharadin daidaito na ɓangaren harsashi da ke sama da da’irar layi ɗaya φ (siffa 2). Sakamakon ƙarfi R na lodin Y da Z da ke aiki a kan wannan ɓangare yana tsaye a tsaye, saboda daidaituwa. Ana samun girmansa ta hanyar haɗa kan saman harsashi daga φ’=0 zuwa φ’=φ. Yankin ɗan ɓangaren harsashi shi ne dF=r1d__φ’*rd__ϑ, don haka ɓangaren tsaye na lodi shine (Ysin__φ’+Zcos__φ’)dF. Ga dukan ɓangaren zobe na harsashi da ke tsakanin da’irori φ’ da φ’+dφ’, sakamakon lodi na waje shi ne:
dR=2__πr (Ysin__φ’+Zcos__φ’) r1d__φ’,
daga nan ta hanyar haɗawa dangane da φ’ ga ɓangaren harsashi da ke sama da da’irar da ta yi daidai da φ’=φ ana samun:
Eq. 1
Wannan nauyi ne da karɓuwa ta ƙarfin al’ada a alkiblar meridiyan, waɗanda ke aiki a kewayen da’irar da aka lura; sashen tsaye na sakamakon waɗannan ƙarfi shi ne 2__πr * N__φ sin__φ, don haka ta daidaitawa daga nan ake samun ƙarfin al’ada da ake nema
Madaidaici. 2
Idan an yanke ɓangaren sama na harsashi a kan da’irar φ=φ0 (misali, buɗe wuta a cikin kubba), iyakar ƙasa ta haɗin kai ita ce, a fili, φ0, ba sifili ba. Idan a gefen harsashin da aka samu ta wannan hanya kuma akwai nauyin R0, to ya kamata a ƙara shi ma ga jimillar sakamakon R (sl. 3).
Hoto. 3
Domin tantance ƙarfin al’ada N__ϑ dole ne a kafa sharaɗin daidaito na ɓangaren kwallo a tsaye da farfajiyar tangential (sl. 4). Jimillar nauyin waje a wannan shugabanci ita ce Z r1 d__φ r d__ϑ. Ƙarfin N__φ*r d__ϑ da ke kwance a cikin farfajiyar meridial suna samar da ƙaramar kusurwa d__φ tsakaninsu, don haka jimillarsu a shugabancin madaidaicin kwallo ita ce N__φ r d__ϑ*dφ. Haka kuma, ƙarfafa Nϑ r1 dφ da ke kwance a cikin farfajiyar da’irar kwance kuma suna samar da kusurwa dϑ tsakaninsu suna da jimillar Nϑ r1 dφ*dϑ wacce ita ma ke kwance a farfajiyar da’irar kwance, sabili da haka tare da madaidaicin kwallo tana ƙulla kusurwa 0,5_π-φ_, don haka a cikin sharaɗin daidaito sai kawai ɓangaren su Nϑ r1 dφ dϑ*sinφ ke shiga. Saboda haka sharaɗin daidaito yana cewa
ko, idan aka raba shi da r r1 dϑ dφ:
Daya.3
Anan, an nuna r2=r/sinφ a matsayin na biyu daga manyan radiyoyin lankwasawa na harsashi.
Hoto. 4
Ɗaukakar zagaye
Idan siffar meridiyan ta samu ta hanyar lissafi da ma’aunin r1=r1(φ) kuma idan an ba da bayyanannun kalmomi na lissafi ga lodin Y(φ) da Z(φ), za a iya tantance ƙarfin yanke a cikin harsashi koyaushe ta hanyar haɗawa a siffar rufe ko kuma ta lambobi, ta amfani da ƙa’idar Simpson-. Ga kupola mai zagaye mai diamita a, a gaba ana ba da ƙarfin yanke ga wasu halayen lodawa:
Siffar ta ƙarshe, wadda ga φ=0 ke ba da manyan ƙarfafan yanke marasa iyaka, tana aiki ne kawai a wani tazara daga wurin da ƙarfin ke kai hari. A kusancin nan da nan na wurin da ƙarfin ke aiki, ko da lokacin da aka rarraba shi a kan saman da’irar mai diamita ƙanana amma tabbatacce, nauyin yawanci ana ɗaukar sa ne ta lanƙwasawa.
Ana iya samo siffofin don ƙofofin da aka buɗe a ɓangaren sama daga waɗanda suka gabata, idan aka ƙara nauyi na ƙirƙira P a ƙolin, wanda girmansa ake ƙayyade ta yadda jimillar ƙarfin al’ada a hanyar meridian da ke aiki a gefen sama na ƙoƙon φ=φ0 ya zama Nφ=0, ko kuma ya yi daidai da wani ƙima da ƙarfafan waje suka ƙayyade da ƙoƙon ke karɓa a nan.
Harsashi mai conical
Hoto 5
A wajen ƙashin conical, gangaren meridiyan φ ba za a iya amfani da shi a matsayin coordinat ba, domin yana da darajar iri ɗaya a duk wurare. Maimakon haka, ana shigar da nisan s daga saman dome, an auna shi tare da layin samarwa (hoto 5). Ƙarfin al’ada a wajen meridiyan bisa ga haka ana nuna shi da Ns. Ana iya samun tsarin da ake buƙata daga waɗanda aka ambata a ƙarƙashin 1 idan aka yi iyakar sauyi. Daga lissafin (2) ana samun haka
a daga ƙa’idar (3)
Nϑ = -r2 Z = -Z s ctga.
Ga muhimman yanayi na nauyi, ana amfani da waɗannan ƙa’idodi:
Hanyar zane don kowane irin siffar meridijan
Hoto. 6
Idan lankwasar meridiana ba a ba ta da bayanin nazari ba, amma, misali, ta hanyar zane, tantance radius na lankwasawa yana da wahala ƙwarai kuma ba daidai ba ƙwarai. A irin wannan yanayi ya fi dacewa a yi amfani da hanyar zane (sl. 6).
Ana rarraba harsashin zuwa manyan matakai da dama na φ=const., ana ƙididdige nauyin Δ_R_ na yankunan zobe da ke tsakanin matakan, kuma ya fi dacewa a raba su nan take da 2_π_. Ana zuba waɗannan ƙarfi ΔR/2π a cikin tsarin ƙarfi (fig. 6b). Idan daga baya, misali, ta wurin rabo 7 aka ja madaidaiciya mai layi ɗaya da tangenti zuwa meridiyan a wurin meridiyan 7, to a kwance ta cikin saman tsarin ƙarfi za ta yanke a kansa tsawon R1/2π sinφ7 wanda daga gare shi, bisa ga ma’auni (2), ta hanyar raba da madaidaicin r ana samun madaidaiciyar ƙarfi a kan hanyar meridiyan.
Don tantance ƙarfi a cikin zobba, ana amfani da sharadin daidaito na ƙarfin kwance a kan wani ɓangaren harsashi. Idan kawai lodin tsaye suke aiki, wannan sharadi yana kasancewa haka (sl. 4):
Idan a nan dφ an maye gurbinsa da bambancin ƙarshe Δφ, wanda ya dace da rarrabuwar harsashin da aka karɓa, kuma idan maimakon r1 dφ aka saka tsayin Δs na element ɗin meridiyan, daga nan ana samun
A cikin baka a gefen dama akwai sashen yanke a kan layi na kwance a cikin shirin ƙarfin. Saboda haka, dukan gefen dama yana wakiltar sassan da ke tsakanin maki na rabo a kan wannan layin. Domin a iya karanta su daidai gwargwado, dole ne a zana shirin ƙarfin a hankali kuma cikin girma mai isasshe.
Zoben ja da zoben matsawa
Kowace harsashin da ke da daidaiton juyawa tana iyakance da da’ira ɗaya ko biyu a kwance (fig. 7). A waɗannan gefuna, a bisa ƙa’ida, tana iya karɓar irin waɗannan lodi da martanin goyon baya ne kawai, waɗanda suke da alkiblar tangenti ga meridian. Idan ƙarin ƙarfi suna da wata alkibla dabam, kamar lodin P daga sashen sama na gini ko martanin S na ƙasan tankin da aka nuna a fig. 97, dole ne a raba waɗannan ƙarfi, kamar yadda aka nuna a hoton, zuwa abubuwa a alkiblar da’irar a kwance da tangenti ga meridian. Bangaren P/sinφ0 ko S/sinφu ya yi daidai da ƙarfin al’ada a alkiblar meridian Nφ da ke aiki a wurin, yayin da don karɓar ƙarfin kwance Pctgφ0 ko Sctgφu ake buƙatar zobe mai karɓar ja, wato matsa lamba, kuma a cikinsa wannan ɗaukar nauyi na radial yana haifar da ƙarfin al’ada + P r0 ctgφ0, wato -S ru ctgφu. Irin waɗannan zobba, ban da a gefunan harsashi, dole ne a sanya su a duk wuraren da layin meridian ke da lanƙwasa. Suna koyaushe haifar da rikicewar yanayin damuwa na membrane.
Fig. 7
Matsin lamba sakamakon lankwasawa a cikin harsasai masu daidaiton juyawa
Tsarin ka’idojin membrane ba su ƙunshe da isasshen adadin dindindin na haɗawa domin a iya cika dukkan sharuɗɗan iyaka da suka dace da aikin da aka kafa da gaske. Musamman ga ƙasusuwa masu daidaiton juyawa, ba zai yiwu ba ta hanyar zaɓin dindindin na haɗawa mai dacewa a sa elongation εϑ = Nϑ/Et a wajen zobe a gefen ƙashin ta zama daidai da elongation ɗin wani ɓangaren gini da aka haɗa da shi da ƙarfi a wannan wuri. Haka kuma wannan ya shafi ƙasusuwa inda a kan wani zobe na kwance curvature na meridiyan, kaurin ƙashi ko nauyi a kowace yanki ke sauyawa kwatsam. Ko da ƙashin ba a ƙarfafa shi da zobe a wani gefe ba, amma a wannan wurin an ɗora masa ƙarfi da ke da sashi a cikin hanyar al’ada ga ƙashin (misali, nauyin P a hoto 7 idan aka yi tunanin an cire zoben), ba zai iya ɗaukar irin waɗannan lodin ta hanyar ƙarfin membrane kaɗai ba. A irin waɗannan lokuta, ƙarfin yanke da moments na lankwasawa (hoto 8) suna taka muhimmiyar rawa wajen rabon ƙarfi a ƙashin, abin da dole ne a yi la’akari da shi a lissafin tsari.
Hoto 8
Daidaitaccen ƙididdigar harsashi wajen lanƙwasawa, sai dai a mafi sauƙin yanayi na musamman na harsashin silinda, aiki ne mai matuƙar wahala. Ga harsashi masu bangon siriri, ana iya gina sauƙaƙƙiyar kuma matuƙar amfani da ka’idar kusanci wacce ta dogara da gaskiya da aka sani daga ƙaƙƙarfan ka’ida, cewa damuwa saboda lanƙwasawa tana takaita a wani yanki mai ɗan faɗi kusa da gefen kuma tana raguwa da sauri yayin da nisan daga gefen ke ƙaruwa. Wannan yana ba da damar a yi watsi da yawan manyan sassa a cikin ainihin daidaitaccen lissafin bambance-bambance, don haka sai ya ɗauki salo mai sauƙi.
Idan a cikin yankin iyaka wanda kawai ke da muhimmanci κ aka maye gurbinsa da ƙimar matsakaici, kuma wannan kusan koyaushe yana yiwuwa, to mafitar ma’aunin ita ce
Jedn. 4 da 5
inda a1, b1, a2, b2 su ne ƙa’idodin haɗawa. Tunda mafita tana da ban sha’awa ne kawai kusa da iyakar φ=φ0, ya dace a ɗauki tazarar kusurwa ω=φ0-φ ko ω=φ-φ0 daga wannan iyaka a matsayin ƙoordinate, sannan ta hanyar canza mafitar da aka ambata a sama a samu kalma ta biyu da aka nuna. A nan kawai ƙa’idodi biyu na haɗawa A, B ko C, ψ suke bayyana, domin ɓangaren mafitar da ke ɗauke da ma’aunin eκω, kuma saboda haka ba ya raguwa tare da ƙaruwa ω, za a iya cire shi daga la’akari.
Bisa ga ma’aunin 5 ana samun waɗannan ƙimomi na ƙarfafan yanke da kuma jujjuyawar κ ta tangantiyar meridian:
Damuwa sakamakon lankwasawa a tankunan siffar silinda mai zagaye
Idan a ɗauka cewa wani tanki mai siffar silinda da tsawo h an cika shi da ruwa (sl. 9), to a tsayin x sama da tankuna biyu matsin ruwan p=γ(h-x). Idan a wannan tsayin, ta hanyar yanke sama biyu a nesa dx, a yanke zobe daga tankin kuma a raba shi a kan diamita guda biyu zuwa kashi biyu, sharadin daidaito na ɗaya daga cikinsu yana ba da ƙarfin membrane a cikin zoben Nφ=pa. Idan kaurin tankin a wannan wuri t ne, tsawaitawar zoben ita ce εφ=Nφ/Et don haka ƙarin radius ɗin tanki (lanƙwasawa) shi ne
A kan ƙasan gefen x=0 haka nan
Hoto 9
Babu ɗaya daga cikin biyun da, a ka’ida, yake daidaita da sharuddan iyaka, domin nakasar bango an hana ta saboda haɗinsa da ƙasan tanki. A nan, a wurin haɗin tsakanin ƙasa da bangon silinda, ana samun ƙarfin transversali Qx0 da moments Mx0 (sl. 11), kuma ya kamata a ƙayyade girmansu ta yadda, tare da ƙarfin membran da aka ƙididdige a cikin harsashi, za su haifar da wannan nakasar a gefen harsashin da ƙasan ke amincewa da ita.
Sl. 10, 11, 12, bi da bi
A hoto 11 an nuna wani ɓangare na harsashi dx*a dφ tare da ƙarfafan da suke aiki a kansa. Sharadin daidaiton ƙarfi a kan alkiblar da ke daidai da saman harsashin yana ba da
dQx a dφ + Nφ dx dφ = p a dφ dx,
daga nan ta hanyar raba da dx dφ:
a Q’x + Nφ = p a.
Daga sharadin cewa moment a kusa da tangent na kwance ga silinda ya zama sifili, ana samun:
dM/dx = M’x = Qx
Ta hanyar kawar da ƙarfin transverse Qx daga waɗannan ƙa’idodi biyu ana samun
a M’’x + Nφ = p a.
Liss. 6
A cikin wannan ma’auni, waɗannan ƙananan ƙarfi na yanke da ba a sani ba za a iya bayyana su ta hanyar karkacewar w. Ga Nφ an riga an samu Nφ=Etw/a, kuma ƙarfin yana daidai da lankwason meridian w’’, wato Mx=Kw’’, inda, kamar yadda yake a faranti, K=Et3/12(1-μ2) ita ce ƙarfafuwar harsashi, wacce za ta iya canzawa da x. Idan an saka waɗannan bayanan ƙarfi na matsakaici a cikin ma’auni (6), ana samun ma’aunin bambanci na ka’idar tanki:
Ga tanki wanda kaurin bango bai canza ba, mafita ita ce:
Daya.7
inda λ4=3(1-μ2)/a2t2. Idan λh ya yi girma (misali, fiye da 3), ya fi dacewa a maimakon ayyukan hyperbolic a gabatar da na exponential, don haka:
Daya.8
Sa’an nan ana iya ƙayyade ƙimomin dindindin A1, B1 ba tare da dogaro da juna ba daga sharuɗɗan iyaka a ƙasan, sannan ƙimomin dindindin A2, B2 daga sharuɗɗan iyaka a saman gefe, inda a ka’ida ake samun A2=B2=0. Bisa ga ƙa’idar (8) da (7), ana samun ƙarfin ƙetare, idan lissafin ya gudana a akasin hanya kuma a ko’ina aka saka ƙa’idar (8) da (7); don haka misali ga A2=B2=0:
Idan bangon tanki an kulle su da ƙarfi a cikin farantin, a kasan tankin wanda kaurinsa ya isa a ɗauke shi mai ƙarfi, dole ne a ƙayyade ƙa’idodin A1, B1 daga yanayin cewa x=0, w=0 da w’=0 sannan ana samun:
Idan ƙasan tanki farantin roba ne ko harsashi, dole a ƙayyade ƙarfin kwance Qx da moments Mx a kan kewayen gefen da ke tsakanin ƙasan da bangon silinda ta hanyoyin ka’idar tsarin da ba a san su kai tsaye ba, bisa ga buƙatar cewa Mx da w’ ga ɓangarorin biyu su kasance daidai.