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1-topshiriq tekis masalani kuchlanishlar funksiyasi (eri funksiyasi) yordamida yechish tekis masala elastiklik nazariyasida ikkita tekis masala farqlanadi (x o y tekisligida qaraymiz) 1. tekis deformatsiyalangan holat, bu holda deformatsiyalar tekislikda bo’lib ya’ni z o’qi yo’nalishdagilari nolga teng bo’ladi: , , . bu masalada berilgan jismni koordinata o’qlaridan z o’qi yo’nalishidagi o’lchami boshqa o’lchamlariga nisbatan juda katta va barcha ta’sir etuvchi kuchlar z o’qiga perpendikulyar bo’lgan tekislikda hamda z ning o’zgarishidan bog’liq emas deb qaraladi. (masalan uzun plastinka, tayanch devor, plotina). 4.1-rasm 2. umumlashgan tekis kuchlanish holati, bu holda tekislikdagi , , kuchlanishlar nolga teng. bunday masalaga yon qirralari bo’ylab kuch ta’siri qo’yilgan yupqa plastinkani hisoblash masalasi kiradi(4.2-rasm). plastinka qalinligini boshq o’lchamlariga nisbatan kichik lekin chekli aniq miqdor uchun quydagilar o’rinli bo’ladi: a) kuchlanishlar bu tekislikda nolga teng; b) , , noma’lum kuchlanishlar plastinka qalinligi bo’ylab tekis taqsimlangan. tekis masalani yechishda (2.1 – 2.8) tenglamalarda dan bog’liq hamma hadlar olib tashlanadi. …

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yenti. (4.3) ifodalarni (4.2) tenglamaga qo’ysak, quyidagiga kelamiz: (4.4) muvozanat differensial tenglamasi ; (4.5) bu yerda va –hajmiy kuchlar. hajmiy kuchlar desak, (4.5) tenglamani qanoatlantaruvchi kuchlanish funksiyasi ni kiritamiz. (4.6) (4.6)ni (4.4) ga qo’ysak, quyidagiga kelamiz (4.7) yoki (4.8) bu yerda -bigarmonik operator: (4.9) chegaraviy shartlar esa quyidagi ko’rinishga ega bo’ladi: (4.10) (4.8) tenglamani yechib, (4.6) formuladan kuchlanishlarni, guk qonuni (4.3) ga ko’ra deformasiyalarni topish mumkin. (4.1) ni integrallab, (4.10) chegaraviy shartlarni hisobga olib, ko’chishlar topiladi. 4.3. mustaqil ish №2. elastiklik nazariyasi tekis masalasini kuchlanishlar funksiyasi yordamida yechish topshiriqni bajarish tartibi masala. to’g’ri to’rtburchak shaklda birlik qalinlikdagi plastinka berilgan (4.3-rasm), kuchlanish funksiyasi 4.1-jadvalda va uning parametrlari qiymatlari 4.2-jadvalda keltirilgan. hajmiy kuchlar hisobga olinmaganda quyidagilarni toping: 1. berilgan funksiyani tekis masala yechimi bo’lishini tekshiring. 2. berilgan funksiya yordamida kuchlanishlar ifodasini toping. 3. kuchlanishlar epyurasini quyidagi hollar uchun quring a) x normalli kesimda kuchlanishlarni; b) y normalli kesimda – kuchlanishlarni (x va …

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li hosilalarni tekis masala uchun o’rinli bigarmonik tenglama (4.7) ga qo’yamiz: . berilgan funksiya tekis masala yechimi bo’ladi. 2. kuchlanishlarni aniqlaymiz. shartga ko’ra: =0; =0. 3. kuchlanishlar epyuralarini quramiz . a) =1 kesim uchun. (to’g’ri chiziq tenglamasi). bo’lganada . , (kvadratik parabola tenglamasi). bo’lganda bo’lganda hol uchun y ni qiymatini topamiz: b) =0,2 kesimda. (to’g’ri chiziq). va (kvadratik parobola). va hol uchun x ni qiymanini topamiz: aniqlangan holatlar asosida kuchlanishlar epyuralarini quramiz: 4. sirt kuchlarini plastinka hamma tomonlari uchun aniqlaymiz va epuyralarini quramiz. chap tomonda. tenglamasi: =0. tashqi normalni koordinata o’qlari bilan hosil qilgan burchaklarini koordinata o’qlari musbat yo’nalishi bilan soat strelkasiga qarama-qarshi yo’nalishda burilgan holda hisoblaymiz. bu tomonda x oqiga parallel kuch yo’q. (parabola). va 4.4-rasm. kuchlanishlar epyurasi. o’ng tomonda: ( to’g’ri chiziq). va yuqori tomonda: ( kvadratik parobola). va hol uchun y ni qiymatini topamiz ( to’g’ri chiziq ). va 4.5-rasm. sirt kuchlari epyurasi pastki tomonda: ( kvadratik …

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1-topshiriq tekis masalani kuchlanishlar funksiyasi (eri funksiyasi) yordamida yechish tekis masala elastiklik nazariyasida ikkita tekis masala farqlanadi (x o y tekisligida qaraymiz) 1. tekis deformatsiyalangan holat, bu holda deformatsiyalar tekislikda bo’lib ya’ni z o’qi yo’nalishdagilari nolga teng bo’ladi: , , . bu masalada berilgan jismni koordinata o’qlaridan z o’qi yo’nalishidagi o’lchami boshqa o’lchamlariga nisbatan juda katta va barcha ta’sir etuvchi kuchlar z o’qiga perpendikulyar bo’lgan tekislikda hamda z ning o’zgarishidan bog’liq emas deb qaraladi. (masalan uzun plastinka, tayanch devor, plotina). 4.1-rasm 2. umumlashgan tekis kuchlanish holati, bu holda tekislikdagi , , kuchlanishlar nolga teng. bunday masalaga yon qirralari bo’ylab kuch ta’siri qo’yilgan yupqa plastinkani...

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