CHO'ZILISH - SIQILISHDA STATIK ANIQ MASALALAR USTIDA AMALLAR BAJARISH

Аннотация

Mazkur maqolada cho'zilish va siqilishda statik aniq masalalarni hal qilishning asosiy tamoyillari va metodlari ko'rib chiqiladi. Mualliflar statik kuchlanish taqsimotini aniqlash va kuchlanish o'zgarishlarining hisoblash amallarini bajarish bo'yicha nazariy hamda amaliy yondashuvlarni taqdim etishadi. Ishda cho'zilish va siqilish jarayonlarini modellashtirish, matematik hisoblashlarning aniqligini oshirish usullari va ular texnik amaliyotda qo'llanilishi haqida batafsil ma'lumot berilgan. Kalit so'zlar: cho'zilish, siqilish, statik masalalar, mexanik kuchlanish, matematik, modellashtirish, texnik hisoblash. 338 “JOURNAL OF SCIENCE-INNOVATIVE RESEARCH IN UZBEKISTAN” JURNALI VOLUME 2, ISSUE 12, 2024. DECEMBER ResearchBib Impact Factor: 9.654/2024 ISSN 2992-8869 𝑃 = 55𝑘𝑁 ℓ = 0.3𝑚 N E = 2۰105 mm2 N [σ] = 72 mm2 A = A = A = A3 1 2 2 Topish kerak ? 𝐴 =? ∆ℓ =? Π =? Topilishi kerak 1. Kesish usulidan foydalanib 𝑁− bo‘ylama kuchlarni hisoblash va epyurasini qurish, 2. Normal kuchlanishlarni hisoblash va epyurasini qurish, 3. Kesimda xosil bo’ladigan ko‘chishlarni hisoblash va epyurasini qurish, 4. Cho‘zilish – siqilish deformatsiyasining mustahkamlik shartidan foydalanib xavfli kesim ko’ndalang kesim yuzasini hisoblash, 5. Sterjenning bo‘sh uchidagi ko’chishning son qiymatini hisoblash, 6. Sterjenning bo‘sh uchidagi potensial energiyani xisoblanadi. Yechish: 1. Sterjenni ixtiyoriy joyidan I-I vertikal tekislik bilan x masofada fikran (xayolan) kesib olamiz 1 va chap tomonini muvozanatda qaraymiz. (1.a – rasm) 339 “JOURNAL OF SCIENCE-INNOVATIVE RESEARCH IN UZBEKISTAN” JURNALI VOLUME 2, ISSUE 12, 2024. DECEMBER ResearchBib Impact Factor: 9.654/2024 ISSN 2992-8869 0 ≤ x ≤ ℓ, 1 Muvozanat tenglamasini tuzamiz. ∑F = 0 → −4𝑃+N = 0 xi 1 N = 4𝑃 (cho‘zilish) 1 Ikkinchi uchastkani ham ІІ-ІІ vertikal tekislik bilan x - masofada fikran kesib olamiz va chap 2 tomonini muvozanatda qaraymiz. (1.b – rasm) 0 ≤ x ≤ ℓ, 2 ∑F = 0 → xi −4𝑃+3𝑃+N = 0 2 N = 𝑃 (cho‘zilish) 2 Uchinchi uchastkani xam III-III vertikal tekislik bilan ixtiyoriy x - masofada fikran kesib olamiz 3 va chap tomonini muvozanatda qaraymiz.(1.d – rasm) 340 “JOURNAL OF SCIENCE-INNOVATIVE RESEARCH IN UZBEKISTAN” JURNALI VOLUME 2, ISSUE 12, 2024. DECEMBER ResearchBib Impact Factor: 9.654/2024 ISSN 2992-8869 0 ≤ x ≤ 1.2ℓ 3 ∑F = 0→ −4𝑃+3𝑃+2𝑃+N = 0 xi 3 N = −𝑃 ( siqilish). 3 Epyurani qurish uchun masshtab tanlaymiz, nollik chiziq o‘tkazamiz va ishorani qabul qilamiz. Epyura 𝑁− qurildi. 2. Normal kuchlanishlarni hisoblaymiz va epyurasini quramiz. 𝑁 4P 𝑃 1 𝜎 = = = 4 1 𝐴 A 𝐴 1 𝑁 𝑃 2 𝜎 = = 2 𝐴 𝐴 2 𝑁 P 𝑃 3 𝜎 = = − = −0,5 3 𝐴 2A 𝐴 3 Epyurani qurish uchun nollik chiziq o‘tkazamiz, masshtab tanlab olamiz va ishorani qabul qilamiz. 3. Ko‘chishlarni hisoblaganda K nuqtada ko‘chish nolga teng ekanligini inobatga olamiz, chunki K - qo‘zg‘almas nuqta, qolgan kesimlardagi ko‘chishlarni shu nuqtaga nisbatan hisoblaymiz, demak ko‘chishni o‘ng tomondan boshlab hisoblaymiz. Guk qonuniga binoan. N 𝑥 3 3 ∆ℓ (𝑥 ) = 3 3 𝐸𝐴 3 (0 ≤ 𝑥 ≤ 1.2ℓ) 3 P∙0 𝑥 = 0 → ∆ℓ (0) = = 0 = ∆ℓ 3 3 𝐸 ∙2𝐴 K −P1.2ℓ Pℓ 𝑥 = 1.2ℓ → ∆ℓ (1.2ℓ) = = −0.6 = ∆ℓ 3 3 𝐸 ∙2𝐴 𝐸𝐴 D N 𝑥 2 2 ∆ℓ (𝑥 ) = ∆ℓ + 2 2 D 𝐸 ∙𝐴 2 341 “JOURNAL OF SCIENCE-INNOVATIVE RESEARCH IN UZBEKISTAN” JURNALI VOLUME 2, ISSUE 12, 2024. DECEMBER ResearchBib Impact Factor: 9.654/2024 ISSN 2992-8869 (0 ≤ 𝑥 ≤ ℓ) 2 Pℓ P∙0 Pℓ 𝑥 = 0 → ∆ℓ (0) = −0,6 + = −0,6 = ∆ℓ 2 2 𝐸𝐴 𝐸 ∙𝐴 𝐸𝐴 D Pℓ P∙ℓ Pℓ 𝑥 = ℓ → ∆ℓ (ℓ) = −0,6 + = 0.4 = ∆ℓ 2 2 𝐸𝐴 𝐸 ∙𝐴 𝐸𝐴 𝑩 N 𝑥 1 1 ∆ℓ (𝑥 ) = ∆ℓ + 1 1 D 𝐸𝐴 1 (0 ≤ 𝑥 ≤ ℓ) 1 4P∙0 Pℓ 𝑥 = 0 → ∆ℓ (0) = ∆ℓ + = ∆ℓ = 0.4 1 1 B 𝐸 ∙𝐴 B 𝐸𝐴 4P∙ℓ Pℓ 4P∙ℓ Pℓ 𝑥 = ℓ → ∆ℓ (ℓ) = ∆ℓ + = 0.4 + = 4.4 = ∆ℓ 1 1 B 𝐸 ∙𝐴 𝐸𝐴 𝐸 ∙𝐴 𝐸𝐴 𝑠 Ko‘chishning epyurasini qurish uchun masshtab tanlab olinadi, nollik (neytral) chiziq o‘tkaziladi va ishorasi qabul qilinadi. 4. Xavfli kesim 1- uchastka kesimida eng katta normal kuchlanish bo'lganligi uchun, kesim yuzasi doira bo'lgan sterjen ko'ndalang kesimining yuzasi geometrik harakteristikasi o'rnatiladi. P 𝜎 = 𝜎 = 4 ≤ [𝜎] max 1 𝐴 Tengsizlikning o‘ng tomonidan. 4P 4∙55∙103 𝑁 220∙103 𝐴 ≥ ≥ ≥ ≥ 3050 𝑚𝑚2 [𝜎] 𝑁 72 72 𝑚𝑚2 Ko‘ndalang kesim yuzasi doiradan iborat deb qarab. 𝜋𝑑2 𝐴 = ≥ 3050 𝑚𝑚2 4 4∙3050 𝑑 ≥ √ ≥ √3885 mm ≥ 62.3𝑚𝑚 𝜋 342 “JOURNAL OF SCIENCE-INNOVATIVE RESEARCH IN UZBEKISTAN” JURNALI VOLUME 2, ISSUE 12, 2024. DECEMBER ResearchBib Impact Factor: 9.654/2024 ISSN 2992-8869 Sterjenning diametrini tanlashda jadvaldan GOST bo'yicha 5 ga karrali son olamiz, amalda keyingi kursda, "Mashina detallari" fanidan kurs loyihasini amalga oshirganda, valning aniq hisobida diametri 5 ga karrali son olinadi, chunki val o'rnatiladigan podshipnikning markasida qo'yilgan yukning qiymatiga qarab, zo'ldorli yoki rolikli bo'lishiga qaramasdan ichki diametri 5 ga karrali son bo'ladi. 343 “JOURNAL OF SCIENCE-INNOVATIVE RESEARCH IN UZBEKISTAN” JURNALI VOLUME 2, ISSUE 12, 2024. DECEMBER ResearchBib Impact Factor: 9.654/2024 ISSN 2992-8869 Demak, 𝑑 = 65𝑚𝑚 Sterjenning haqiqiy ko‘ndalang kesim yuzasi 𝜋𝑑2 3.14∙652 𝐴 = = = 3316 𝑚𝑚2 4 4 Keyingi hisoblashlarda ushbu yuzani valn