A yau, Savo Kusić yana mai da hankali kan gilashin katako, itace-aluminum Taga, Taga na al’ada, kofa da buƙatun magana. Wannan rubutu ya kasance a matsayin tarihin ka’idar tsari kuma ba tayin ƙira ba ne, ƙididdigar kwanciyar hankali ko gyara tsarin.
Bambance-bambancen lissafi na lankwasawa
An ɗauka cewa an gamsu da duk abubuwan da suka dace, musamman
- cewa sandar daidai take
- cewa nauyin P yana aiki a cikin shaft na sanda
- cewa ƙarfin lantarki ya kasance ƙasa da iyakar daidaituwa
- cewa nakasawa suna da ƙananan isa don ba da damar sauƙaƙan ƙididdiga na yau da kullun.
Idan sanda ya lanƙwasa ƙarƙashin aikin ƙarfin P kuma watakila ma saboda nauyin q yana aiki a cikin madaidaiciyar hanya, to dole ne a gamsu da yanayin da aka sani na ma’auni tsakanin nauyin waje da dakarun da ke cikin sashin. Bayan haka, abu ne na dabi’a cewa ya kamata a sami sanannun alaƙa tsakanin lalacewa da ƙarfi a cikin sashin.
Sl.1- Balance
A cikin fig.1an gabatar da ɓangaren sanda mai lanƙwasa tsawon ds. gangaren v’ tangent ƙarami ne ta yadda har zuwa tsari na biyu, ana iya saita tsawon simin ɗin ya zama daidai da tsinkaya ds=dz. Bugu da kari, a cikin gwaje-gwajen da aka ambata a ƙasa, lokuta waɗanda P koyaushe ke canzawa tare da axis na sanda ba za a la’akari da su ba, i.e. ƙarfi na al’ada N≈P=const. A cikin ma’aunan ma’auni guda uku da ake da su don kashi, za a iya yin watsi da takamaiman adadin mambobi. Don alamomi da alamu daga fig.10zauna to a farkon kusantar lissafin
N=P, M’=Q, Q’=-q+Pv’’.
Za a iya kawar da ƙarfin jujjuyawar, sannan M’‘=Pv’’-q yana riƙe. Dokokin Bernoulli M=-EI/R tana ba da ma’aunin da ya ɓace. Lanƙwasa axis na sanda, kamar yadda aka saba, daga ka’idar lanƙwasa katako, an sauƙaƙe kamar haka.
Don haka, yana yiwuwa a kawar da lokacin lanƙwasawa; kuma tare da M = -EI \ * v’’ ana samun ma’auni na banbance na sandar lanƙwasa.
(EI*v’‘)’’ + Pv’’ = q. (1)
A cikin yanayi na musamman na taurin kai tsaye EI=const. an sauƙaƙa lissafin (1) haka yake
EIv’‘’’ + Pv’’ = q. (1a)
Daya. (1) da (1a) suna layi a cikin v bisa ga sauƙaƙan da aka ambata.
A zahiri, shari’ar mafi sauƙi na taurin kai akai-akai EI=const., wanda ke zama tushen misalan da ke ƙasa, yana da mahimmanci musamman.
Shari’ar Euler
Zato na al’adar buckling na gargajiya shine, baya ga abubuwan da aka ambata: p=0, don haka N=const.=P, haka kuma da taurin kai ba tare da juyi ba, wanda ke kwance daidai da gadar sandar. Ma’auni banbanci na yau da kullun (1a) yana cikin tsari na huɗu, madaidaiciya kuma mai kama da juna saboda q=0. Maganin gamayya shine:
v(z) = A_∙_sin__αz + B∙cosαz + C∙z + D, (2)
inda A, B, C, D sune madaidaitan haɗin kai guda huɗu. Siga yana da girman iyaka saboda zato EI≠0. Bugu da ƙari, α≠ yana aiki0don P>0.
Sl.2
Tunda ma’aunin bambancin layi (1a) yayi kama da juna, wannan matsalar kuma ana magance ta ta n_∙__v(z)_, inda n ke zama sabani. Tuni daga rarrabuwar lissafi na ma’aunin bambancin (1a) ana iya ganin girman lankwasawa v ba za a iya tantancewa ba. Don ƙayyade ƙayyadaddun haɗin kai A, B, C, D a cikin Euler\ case (fig. 2) ana samun yanayin kwane-kwane masu zuwa: Saboda goyan bayan da aka bayyana, lokacin lanƙwasawa ba shi da komai a ƙarshen sandar, watau.
v=0 da M=0 don z=0
v=0 da M=0 don z=s.
Saboda M=-EI_∙__v’‘_, ya biyo bayan cewa M=0 shine darajar v’’=0, don haka sharudan kwankwaso hudu sun karanta:
v=0 da v’’=0 don z=0
v=0 da v’’=0 don z=s.
Maganin (2) don haka yana haifar da ma’auni guda huɗu:
Daya.3
Waɗannan ma’auni guda huɗu masu kama da juna suna da abin da ake kira farkon abin da ake kira da rashin fahimta A=B=C=D=0, wanda yayi daidai da v(z)=0. Sanda mara lankwasa v_(z_) = 0 shine matsayi na yau da kullun don sabani, musamman ƙananan ƙimar kaya P. Anan, mun fi sha’awar yanayin da sandar ke lanƙwasa, watau v(z)≠0. Ma’auni (3) ba su da irin wannan ƙaramin bayani tare da A, B, C, D ≠ 0 kawai idan ma’aunin ƙididdiga ya yi daidai da sifili.
Daya.3a
Daga yanayin buckling Δ = α4∙s∙sinαs =0 mog don lissafta mahimman ƙimar PK na kaya. Wannan ma’aunin ya cika ne kawai idan sin αs =0, wato na αKms = mπ tare da m=1,2,3,… Madaidaicin ƙimar ƙarfin Euler shine PKm=EI_∙_α2km = EI(mπ/s)2. Ga kowane ƙima na m, ana samun wani ƙaƙƙarfan ƙarfi. A mafi yawan lokuta, kawai mafi ƙanƙanta daga cikin waɗannan dakarun shine sha’awa, kuma hakan
PK1=EI(π/s)2.
A cikin sigar mai zuwa, za a sami fihirisar m, ko 1 galibi wanda aka tsallake, yana nufin mafi ƙanƙanta na manyan lodi. Abin sha’awa shine layin roba mai dacewa. Daga daya (2) akai-akai B=C=D=0 ana samunsu. Sauran yanayin shine A∙sin__αs=0. Layin buckling na roba na shari’ar da ke sama m=1 shine, saboda haka, v(z) = A_∙sinπz/s_, inda v(z) yake don αK resp. PK aikin eigen. Ba a tantance girman A ba.
Bari mu sake kallon gaba dayan wurin kaya. Don ƙimar sabani na P, musamman ga sojojin P_<_PK1 shine v(z)=0, sandar ta kasance madaidaiciya.
A cikin ka’idar kwanciyar hankali na gargajiya, a ƙarƙashin “ƙara a hankali” lodi P yana nufin jerin ƙimar ƙimar P, inda za a gwada kwanciyar hankali na yanayin ɗaukar nauyi da sake gwadawa ga kowane matakin nauyi P. Yayin gwajin kwanciyar hankali, girman P ya kasance baya canzawa. Idan wata jiha P ta tabbata ta tabbata, to ana ƙara nauyin da wani abu, don haka ana maimaita gwajin kwanciyar hankali. Wannan ƙayyadaddun ƙayyadaddun ƙayyadaddun ƙayyadaddun ƙayyadaddun ya zama dole don kada a sami rudani tare da nau’in kaya, wanda shine tushen tasirin Shanley, wanda a cikinsa yana karuwa a cikin nauyin yayin da ƙananan rikice-rikice ke aiki a lokaci guda. Lokacin nazarin kwanciyar hankali bisa ga Shanley, a cikin yanki na buckling filastik, ƙaramin ƙarfin bifurcation yana faruwa fiye da hanyar la’akari na gargajiya.
Waɗannan la’akari suna ɗaukar hanyar farko ta haɓakar kaya: A ƙarar kaya a hankali P a wannan ma’ana, mutum ya isa ga PK1=EI_∙(π/s)_2 na farko a daya bifurcation batu na bututun ƙarfe a matsayin ma’auni mai sauƙi, tare da madaidaicin ma’auni mai sauƙi, tare da madaidaicin sikela mai sauƙi. layi
v1(z)=f_∙_sin__π__z/s.
f=A yana nuna girman layin da ba a tantance ba. Domin lodi _P*__>_PK1 ya kasance bisa ga ka’idar layin layi Δ≠0, watau ma’auni ba zai yiwu a can ba. Sai kawai don mafi girman rundunonin PK2 da sauransu zai sake zama Δ=0 kuma haka v(z)≠0. Dalilin wannan ƙarshe na ka’idar, wanda bai dace da ainihin ɗabi’a ba, ya ta’allaka ne a cikin daidaita ma’aunin bambancin.
Tushen sandunan da aka matse da sassa biyu
Muna takaita kanmu da sandunan da aka danne kashi biyu, kuma ga matsewar sanduna mai kunshe da sassa uku ko fiye, ana amfani da irin wannan rabo.
Sl.3
Dangane da nau’in haɗin kai, sandunan lattice sun bambanta (fig.3a) da kuma sanduna (fig.3e). Idan ba tare da haɗin giciye ba, kowane ɓangaren sanda zai lanƙwasa da kansa (fig.3b). Idan cikon yana da rauni sosai, to, sandar bel guda ɗaya na iya tanƙwara ba tare da lanƙwasa sandar gaba ɗaya ba. Ga sandar lattice fig.3c za a iya kwatanta: hagu don tsarin hinged, dama don ci gaba da bel; kuma ga firam sanda fig.3f a hagu don zanen haɗi mai laushi, a dama don takaddun haɗin haɗin kai. Daga can, ana iya ganin cewa slimness λ_ yana da babban tasiri1__=s1/i1_. Ga kowane ɓangaren sandan, an sanya I1=F1_∙_i12, s1 shine tazarar filin (fig.3), i1 yana nufin axis1-1(duba Fig.4).
Sl.4
Ƙarfin ƙwanƙwasa na sandunan fili wanda ya dace da layukan buckling na roba na fig. 3d, ko 3g ya kasance batun cikakken bincike na ka’idar. Idan sandan ya ƙunshi sassa na musamman (misali tare da haƙarƙari) ko kuma idan haɗin haɗin gwiwar ya kasance akai-akai har sandar da aka lanƙwasa daga sassa biyu ba ta bambanta da tasirinta da sanda ɗaya ba, to zai zama σ__K_=_π__2__E/λ__y__2. A can yana nuna λ__y=sK/iy (fig. 4). Duk da haka, wannan matsananciyar damuwa ba za a kai ga cikawa a aikace ba saboda babu isassun hanyoyin haɗin kai don hakan, kuma suna da ƙarfi. Ana samun ƙima mai gamsarwa a mafi yawan lokuta idan muka sanya bisa ga tsari F. Engesser\a cewa damuwan lanƙwasawa na sandar da aka matse wanda ya ƙunshi sassa biyu daidai yake
σ__K_=_π__2__E/λ__yi__2.
λyi shine abin da ake kira siriri mai kyau.
Dole ne haɗin haɗin giciye ya isa a adadi kuma yana da ƙarfi sosai don tilastawa sassan sandan biyu suyi aiki tare. Ban da damuwa na biyu, hanyoyin haɗin kai ba su da damuwa muddin sandar ta kasance madaidaiciya. Sai kawai tare da lankwasa cewa dole ne su karbi runduna, kuma dole ne su watsa karfin da ke karuwa tare da lankwasawa. Idan layin na roba ne
karfin juyewa a I=const. daidai yake
Q = M’ = - EI∙v’’.
Tare da fig. Ana iya karanta 5 m nan take Q(z)=PK∙v’(z). Mafi girman darajar a ƙarshen sanda shine
Sl.5
Dole ne hanyoyin haɗin keɓaɓɓu su sami damar ɗaukar wannan ƙarfin juzu’i. Tunda nauyin da ya dace shine ƙarfin PK, ya isa idan iyakar ɗaukar nauyi ya kai ga wannan kuma a cikin haɗin kai - watau a lokaci guda tare da duk sanda. Babu shakka, ba zai zama da amfani ba idan hanyoyin haɗin giciye sun yi girma da ƙarfi fiye da yadda ake buƙata don karɓar Q0. A gefe guda kuma, hanyoyin haɗin yanar gizo waɗanda ba su da rauni sosai za su sa sandar ta gaza da wuri.
Ba za a iya amfani da ma’aunin da ke sama da farko ba saboda ba a san girman kibiya mai ɗaurewa ba. Sanda yana cikin ma’auni na rashin sha’awa a ƙarƙashin PK kuma yana iya, aƙalla a matakin farko na kusantar, ɗaukar ƙimar sabani. Don haka yana da amfani, bisa ga shawarar F. Krohn-a, don sanin girman wannan buckling maxv wanda sandar da aka lanƙwasa ta rasa ƙarfin ɗaukarsa saboda gazawa saboda lankwasawa. Wannan tabbas zai faru idan an kai wurin samar da amfanin gona a gefen juzu’i na lanƙwasa (Fig.6). Lanƙwasawa don wannan yana ƙara girma kuma ba za a iya amfani da lokacin lanƙwasawa daidai ba.
Sl.6
Jimlar damuwa a gefen maƙarƙashiya na lanƙwasa shine bisa ga fig.6don manufa filastik abu
Daga can ya zo mafi girman swagger wanda har yanzu ana iya jurewa
maxQ0 shine madaidaicin ƙarfin juzu’i a ƙarshen sanda.
Idan ana amfani da fom ɗin σK Tetmeier kuma a saka σF=3,1t/cm2da sK=e/2∙ λy wanda yayi daidai da iy≈e/2, sannan Krohn’s kusan ƙimar maxQ ya sami0__F/28(Q cikin t, F a cikin cm2).
Don layukan sabani ana samun damuwa mai ɗaurewa
maxQ0= μK∙SK.
Coefficient μK yana canzawa tare da siririn sanda, kuma ya dogara da kayan. Lokacin da aka raba ma’aunin da ke sama ta hanyar ƙididdigewa na aminci game da buckling ana iya rubuta shi
dozv__Q=μ∙dozvS.
Buckling na tsarin layi
A cikin tsarin tsarin layi, yana da mahimmanci don ƙayyade tsayin lanƙwasa sK \ = β∙h na sanda wanda aka jingina a ƙarshen duka kuma wanda ƙarfinsa shine PK \ = π don kwatantawa.2EI/s2K kamar babba. Za a ba da kwatankwacin “tsawon buckling” don misalan da ke ƙasa.
a) Sanda a manne a ƙarshen biyu (EI=const.)
Sl.7
Ga al’amarin fig.7na buckling shine PK=4π2EI/h2=4,0∙PE, kwatancen “tsawon buckling” sK=h/2da layin buckling na roba v(z)=D(1- ku2πz/h).
b) Sanda kyauta a ƙarshen ɗaya (fig.8)
Sl.8
Ana tsammanin cewa alkiblar P baya canzawa yayin da sanda yake lanƙwasa. Sannan (EI=const.)
PK=π2EI/4h2=PE/4, sK =2h
v(z)=D(1-kosπz/2h).
Sharuɗɗan madanni na ƙarshen kyauta z=h suna nan
M=0ko _v’’0
Q=M’-Pv’=0ko v’‘+a2∙v’=0.
c) Sanda a manne a gefe ɗaya (fig.9)
Sl.9
Tare da yanayin kwane-kwane
v=0kuma v’=0don z=0
v=0kuma v’’=0na z= h
an samo shi don EI=const. yanayin buckling
sin_αh - αh∙cos_αh=0_,_
ko makamancin haka
tgαh=ah. (4)
Tushen wannan ma’auni mai zurfi ana samun su a cikin tebur K. Hayashi’s. Mafi ƙanƙancin ƙarfin ƙugiya shine bisa ga tebur PK1=4,49342/ h2∙ EI =2,046π2EI/h2.
Tsawon kwatancen kwatancen shine sK = h / √2,046=0,699h.
d) Sanda da aka manne a gefe ɗaya tare da cuta (fig.10)
Sl.10
Za mu yi la’akari da yanayin ƙaura na tip na girman kai tsaye. Sharuɗɗan kwane-kwane na hakan
_v= ku0kuma v’=0_na z=0,
v= ku1kuma v’=0 na z=h.
Matsala ce ta ma’auni tare da layin roba
Ƙididdigar kalmar kafin {} - brackets za su zama sifili kuma ƙugiya tana ƙaruwa sama da duk iyakoki idan P →PK (duba yanayin buckling Eq. ()4)). Yana da halayyar halayyar sanda cewa lanƙwasa a tsakiya da ƙananan ƙarshen sandar sun fara ƙarami tare da ƙara nauyin P kuma suna canza alamar su kawai don manyan kaya. Sl.10yana nuna lanƙwasawa a rabin tsayi.
e) Ci gaba da sanda a fadin filayen biyu
Sl.11
Domin bangarorin mafita guda biyu (fig.11) za mu yi amfani da saitin
Kuma_. v(z1) = A1∙ synaz1+ B1∙cosAZ1+C1∙z1+ D1,_
II. v(z2) = A2∙ synaz2+ B2∙cosAZ2+C2∙z2+ D2.
Don sauƙi bari ya kasance (EI)1=(EI)2=EI. Don nemo madaukai takwas A1,…, A2, … yi aiki azaman yanayin kwane-kwane
_v= ku0kuma v’=0_za z1=0,
_v= ku0kuma v’=0_za z2=h,
da kuma yanayin tsaka-tsaki tsakanin yankunan I da II
_v= ku0_za z1=h i na z2=0,
v’ (z1= h) = v’(z2=0),
v’ (z1= h) = v’’(z2=0).
Ana ba da ma’auni na sharadi a cikin tebur da ke ƙasa
Ta hanyar saita ƙayyadaddun ƙididdiga daidai da sifili, ana samun yanayin buckling
Tsawon lankwasawa sK=0,878_h_, yayin da fig.35ko da β=0,699. Ƙarfin buckling ya ragu idan aka kwatanta da2,046:1,297saboda karin fili na biyu da aka rataye. Idan ɗayan filin ya fi tsayi, to, ƙarfin daɗaɗɗa zai sake karuwa zuwa kololuwa.2,04_PE_.
Wannan gabatarwar tarihi na ƙididdiga ba aiki ba ne ko hujja na kwanciyar hankali na ginin. Zato na al’ada na madaidaicin madaidaicin memba, nauyin ɗabi’a da yanki na roba ba dole ba ne ya haɗa da lahani na farko, saura damuwa, eccentricity, taurin haɗin kai, rashin layi na kayan abu, kwanciyar hankali ko, lokacin da ya dace, wuta, aikin girgizar ƙasa da matakan taro. Dole ne injiniyan farar hula mai izini ya yi lissafin ainihin ginin bisa ga ingantattun ƙa’idodi da ƙa’idodi.