To‘g‘ri to‘rtburchak shakldagi plastinka egilishlari va sirti elliptik paraboloid bo‘lgan qobiqlar hisobi masalalari - Педагогика - Дипломные работы

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To‘g‘ri to‘rtburchak shakldagi plastinka egilishlari va sirti elliptik paraboloid bo‘lgan qobiqlar hisobi masalalari

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O’ZBEKISTON RESPUBLIKASI
OLIY VA O’RTA MAXSUS TA’LIM VAZIRLIGI

SHAROF RASHIDOV NOMIDAGI SAMARQAND
DAVLAT UNIVERSITETI

MATEMATIKA FAKULTETI

NAZARIY VA AMALIY MEXANIKA KAFEDRASI
Plastinka va qobiqlar nazariyasi fanidan
MUSTAQIL ISH
Mavzu: Tog’ri to’rtburchak sha kldagi plastinka
egilishlari va sirti elliptik paraboloid bo’lgan qobiqlar
hisobi masalalari

Bajardi:____ -guruh talabasi _________________

_____________________

Qabul qildi: _______________________
Mustaqil ish topshirilgan vaqt _________
Mustaqil ish qabul qilingan vaqt_________

Samarqand 2022

Mustaqil ishni bajarish tartibi: Xar bir talaba avval namunaviy yechib
ko’rsatilgan masalalar bilan tanishib chiqadi so’ngra o’zining jurnaldagi nomeri
bo’yicha o’ziga tegishli bo’lgan masalasining qiymatlarini oladi. Masalan jurnal
bo’yicha 1 -nomerdagi talaba 1 -jadvaldan 1 -satrni va 2 -jadvaldan ham 1 -satrni,
jurnal bo’yicha 10 -nomerda turgan talaba 1 -jadvaldan 1 -satrni 2 -jadvaldan 0 -satrni
oladi. Mustaqil ishni bajarayotgan vaqti namunaviy masalalarni yozish shart emas.
Xar bir talaba chizmalarni chizishda alohida e’tibor berishi, mustaqi l ish yuz
namunadagidek bo’lishi va mustaqil ish oxirida foydalanilgan adabiyotlar ro’yxati
keltirilishi talab etiladi.

Tog ’ri to ’rtburchak shakldagi plastinka egilishlarini hisoblash
Asosiy tushunchalar
A s o y i y t e n g l a m a l a r . C h e g a r a v i y s h a r t l a r .
Plastinka egilishlari uchun muvozanat diierensial tenglama (Sofi Jerman
tenglamasi):
??????4&#3627408484;
??????&#3627408485;4 + 2
??????4&#3627408484;
??????&#3627408485;2??????&#3627408486;2+
??????4&#3627408484;
??????&#3627408486;4 =
&#3627408478;(&#3627408485;,&#3627408486;)
&#3627408439; (1)
bu yerda &#3627408439; = ??????ℎ3
12 (1−??????2) plastinka ko’ndalang kesimi silindrik bikrligi.
Ichki zo’riqish kuchlari quyidagi formulalar orqali aniqlanadi:
а) &#3627408485; va &#3627408486; o’qli normalli ko’dalang kesimdagi eguvchi momentlar mos ravishda:
{

&#3627408448; &#3627408485; = − &#3627408439; (
??????2&#3627408484;
??????&#3627408485;2 + &#3627409160;
??????2&#3627408484;
??????&#3627408486;2)
&#3627408448; &#3627408486; = − &#3627408439; (
??????2&#3627408484;
??????&#3627408486;2 + &#3627409160;
??????2&#3627408484;
??????&#3627408485;2)
(2)
b) k o’ndalang kuch intensivligi:
{

&#3627408452;&#3627408485; = − &#3627408439; (
??????3&#3627408484;
??????&#3627408485;3 + &#3627409160;
??????3&#3627408484;
??????&#3627408485;?????? &#3627408486;2)
&#3627408452;&#3627408486; = − &#3627408439; (
??????3&#3627408484;
??????&#3627408486;3 + &#3627409160;
??????3&#3627408484;
??????&#3627408486;?????? &#3627408485;2)
(3)
с) b urovchi moment intensivligi:
?????? = − &#3627408439; (1 − &#3627409160;)
??????2&#3627408484;
??????&#3627408485;??????&#3627408486; (4)
(1) tenglamani yechishdan hosil bo’ladigan inte grallash o’zgarmaslari
chegravi y shartlardan topiladi. Chegaraviy shartlar chegarada kinematik va statik
parametrlarni berilishiga qarab gruhlarga ajratiladi. Bular:
1. Kinematik – ko’chish va burilish burcha gi berilgan:
&#3627408462;) &#3627408485; &#3627408475;&#3627408476;&#3627408479;&#3627408474;&#3627408462;&#3627408473;&#3627408473;?????? &#3627408472;&#3627408466;&#3627408480;??????&#3627408474;&#3627408465;&#3627408462; &#3627408484; ,??????&#3627408485; =
??????&#3627408484;
??????&#3627408485;
&#3627408463;) &#3627408486; &#3627408475;&#3627408476;&#3627408479;&#3627408474;&#3627408462;&#3627408473;&#3627408473;?????? &#3627408472;&#3627408466;&#3627408480;??????&#3627408474;&#3627408465;&#3627408462; &#3627408484; ,??????&#3627408486; =
??????&#3627408484;
??????&#3627408486;
2. Statik – ichki zo’riqish kuchlari berilgan:

&#3627408462;) &#3627408485; &#3627408475;&#3627408476;&#3627408479;&#3627408474;&#3627408462;&#3627408473;&#3627408473;?????? &#3627408472;&#3627408466;&#3627408480;??????&#3627408474;&#3627408465;&#3627408462; &#3627408448; &#3627408485;,&#3627408452;&#3627408485;,??????
&#3627408463;) &#3627408486; &#3627408475;&#3627408476;&#3627408479;&#3627408474;&#3627408462;&#3627408473;&#3627408473;?????? &#3627408472;&#3627408466;&#3627408480;??????&#3627408474;&#3627408465;&#3627408462; &#3627408448; &#3627408486;,&#3627408452;&#3627408486;,??????
3. Aralash – bir qism ko’chishlar va ichki kuchlar berilgan.
Rasmda plastinka o’rta sirti mahkamlanish i ko’rsatilgan.

Plastinka barcha tomonlari uchun chegaraviy shartlarni keltiramiz:
1. &#3627408485; = 0 bo ’lgan tomon uchun
?????? = 0; &#3627408448; &#3627408485; = − &#3627408439; (
??????2&#3627408484;
??????&#3627408485;2 + &#3627409160;
??????2&#3627408484;
??????&#3627408486;2) = 0, (5)
(aralash guruh).
2. &#3627408485; = &#3627408462; bo’lgan tomon uchun
?????? = 0; &#3627408448; &#3627408485; = − &#3627408439; (
??????2&#3627408484;
??????&#3627408485;2 + &#3627409160;
??????2&#3627408484;
??????&#3627408486;2) = &#3627408448;̅, (6)
(aralash guruh).

3. &#3627408486; = 0 bo’lgan tomon uchun
?????? = 0; ??????&#3627408486; =
??????&#3627408484;
??????&#3627408486; , (7)
(kinematik guruh).
4. &#3627408486; = &#3627408463; bo’lgan tomon uchun
&#3627408448; &#3627408486; = 0; &#3627408452;̅&#3627408486; = − &#3627408439; (
??????3&#3627408484;
??????&#3627408485;3 + (2 − &#3627409160;)
??????3&#3627408484;
??????&#3627408486;3) = 0, (8)
(statik guruh), bu yerda &#3627408452;̅&#3627408486; − keltirilgan ko’ndalang kuch intensivligi.
1-M i s o l . To’gri to’rtburchak shakldagi egiluvchan plastinka berilgan (1 -a,b
rasm ). Egilish funksiyasi va yuklanis h 1 -jadvalda, son qiymatlari esa 2 -jadvalda
berilgan.
Quyidagilarni aniqlang:
1. Plastinka tomonlarini mahkamlanish shartlarini toping.
2. &#3627408438; o’zgarmasni aniqlang
3. Ichki kuchlar ifodasini toping.
4. Ichki zo’riqish kuchlari epyurasini kesimlarning birida quring:
&#3627408485; − kesimdagi &#3627408448; &#3627408485;,&#3627408452;&#3627408485;; &#3627408486; − kesimdagi &#3627408448; &#3627408486;,&#3627408452;&#3627408486;.

1-rasm
1-jadval
Plastinka ko ’rinishi 1- rasmda K &#3627408478;(&#3627408485;,&#3627408486;) yuklanish va ?????? (&#3627408485;,&#3627408486;)egilish
funksiyalari . &#3627408439; ,&#3627408438; va &#3627408478;0− o’zgarmas kattaliklar.
0 1.&#3627408462;) &#3627408479;&#3627408462;&#3627408480;&#3627408474; .&#3627408478; = &#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481; , &#3627408484; = &#3627408438;(&#3627408485; − &#3627408462;)2(&#3627408486; − &#3627408463;)2
1 1.&#3627408462;) &#3627408479;&#3627408462;&#3627408480;&#3627408474; .&#3627408478; = &#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481; , &#3627408484; = &#3627408438;&#3627408485;&#3627408486; (&#3627408485; − &#3627408462;)(&#3627408486; − &#3627408463;)
2 1.&#3627408462;) &#3627408479;&#3627408462;&#3627408480;&#3627408474; .&#3627408478; = &#3627408478;0&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
&#3627408463; ; &#3627408484; = &#3627408438;&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
&#3627408463;
3 1.&#3627408462;) &#3627408479;&#3627408462;&#3627408480;&#3627408474; .&#3627408478; = &#3627408478;0&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
2??????&#3627408486;
&#3627408463; ; &#3627408484; = &#3627408438;&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
2??????&#3627408486;
&#3627408463;
4 1.&#3627408462;) &#3627408479;&#3627408462;&#3627408480;&#3627408474; .&#3627408478; = &#3627408478;0&#3627408480;??????&#3627408475;
2??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
2&#3627408463; ; &#3627408484; = &#3627408438;&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
2&#3627408463;
5 1.&#3627408463;) &#3627408479;&#3627408462;&#3627408480;&#3627408474; .&#3627408478; = &#3627408478;0&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
2&#3627408463; ; &#3627408484; = &#3627408438;&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
2&#3627408463;
6 1.&#3627408463;) &#3627408479;&#3627408462;&#3627408480;&#3627408474; .&#3627408478; = &#3627408478;0&#3627408464;&#3627408476;&#3627408480;
??????&#3627408485;
2&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
&#3627408463; ; &#3627408484; = &#3627408438;&#3627408464;&#3627408476;&#3627408480;
??????&#3627408485;
2&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
&#3627408463;
7 1.&#3627408463;) &#3627408479;&#3627408462;&#3627408480;&#3627408474; .&#3627408478; = &#3627408478;0&#3627408480;??????&#3627408475;
2??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
2&#3627408463; ; &#3627408484; = &#3627408438;&#3627408480;??????&#3627408475;
2??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
2&#3627408463;
8 1.&#3627408463;) &#3627408479;&#3627408462;&#3627408480;&#3627408474; .&#3627408478; = &#3627408478;0&#3627408464;&#3627408476;&#3627408480;
??????&#3627408485;
2&#3627408462; &#3627408480;??????&#3627408475;
2??????&#3627408486;
&#3627408463; ; &#3627408484; = &#3627408438;&#3627408464;&#3627408476;&#3627408480;
??????&#3627408485;
2&#3627408462; &#3627408480;??????&#3627408475;
2??????&#3627408486;
&#3627408463;
9 1.&#3627408463;) &#3627408479;&#3627408462;&#3627408480;&#3627408474; .&#3627408478; = &#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481; ; &#3627408484; = &#3627408438;(&#3627408485;2− &#3627408462;2)(&#3627408486; − &#3627408463;)2
2-jadval
Variant oxirgi raqami &#3627408448;
&#3627409160; &#3627408462; &#3627408463; ℎ &#3627408485; &#3627408486;
0 4 3 0.1 3 2 0.35
1 3 3 0.1 2 2 0.30
2 3 3 0.1 1 1 0.25
3 4 4 0.2 2 3 0.25
4 5 3 0.1 3 1 0.30
5 3 5 0.1 1 3 0.30
6 3 3 0.2 2 2 0.35
7 5 5 0.2 2 2 0.35
8 5 4 0.2 3 1 0.30
9 4 5 0.2 2 3 0.30

N a m u n a v i y m a s a l a :
Berilgan:
&#3627408478;(&#3627408485;,&#3627408486;)= &#3627408478;0&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ; &#3627408484; (&#3627408485;,&#3627408486;)= &#3627408438;&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ; &#3627408462; = 4;
&#3627408463; = 3; ℎ = 0.2; &#3627408485; = 1; &#3627408486; = 1; &#3627409160; = 0.2;
1.&#3627408463;) &#3627408479;&#3627408462;&#3627408480;&#3627408474;
1. Plastinka tomonlari mahkamlanishini topamiz.
a) &#3627408484; (&#3627408485;,&#3627408486;)= 0 chegaraviy shartni tekshiramiz.
&#3627408485; = ± &#3627408462; bo 'lganda &#3627408484; (± &#3627408462;,&#3627408486;)= &#3627408438;&#3627408480;??????&#3627408475; ??????(±&#3627408462;)
&#3627408462; &#3627408464;&#3627408476;&#3627408480; ??????&#3627408486;
&#3627408463; = 0 ve rtikal tayanch bor .
&#3627408486; = ± &#3627408463; bo'lganda &#3627408484; (&#3627408485;,± &#3627408463;)= &#3627408438;&#3627408480;??????&#3627408475; ??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480; ??????(±&#3627408463;)
&#3627408463; = − &#3627408438;&#3627408480;??????&#3627408475; ??????&#3627408485;
&#3627408462; ≠ 0, vertikal
tayanch yo ’q.
b) burilish burchagini tekshiramiz .
??????&#3627408485; =
??????&#3627408484;
??????&#3627408485; = &#3627408438;
??????
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463;
&#3627408485; = ± &#3627408462; bo 'lganda
??????&#3627408485; = &#3627408438;
??????
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????(± &#3627408462;)
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; = − &#3627408438;
??????
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ≠ 0
Bu tomon sharnirli mahkamlangan.
??????&#3627408486; =
??????&#3627408484;
??????&#3627408486; = − &#3627408438;
??????
&#3627408463; &#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
&#3627408463;
&#3627408486; = ± &#3627408463; bo 'lganda
??????&#3627408486; = − &#3627408438;
??????
&#3627408463; &#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????(± &#3627408463;)
&#3627408463; ≠ 0
Bu tomon harakatlanuvchi ( o’qi bo ’ylab ) mahkamlangan .
2. С o’zgarmasni topamiz.
Buning uchun plastinka egilish tenglamasiga &#3627408478;(&#3627408485;,&#3627408486;) yuklanish va &#3627408484; (&#3627408485;,&#3627408486;)
egilish funksiyalari ni qo ’yamiz :
Hosilalarni hisoblaymiz:
??????4&#3627408484;
??????&#3627408485;4 = &#3627408438; (
??????
&#3627408462;)
4
&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ;
??????4&#3627408484;
??????&#3627408485;2??????&#3627408486;2 = &#3627408438; (
??????
&#3627408462;)
2
(
??????
&#3627408463;)
2
&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ;
??????4&#3627408484;
??????&#3627408486;4 = &#3627408438; (
??????
&#3627408463;)
4
&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ;
topilgan ushbu ifodalarni plastinka egilish teglama si
??????4&#3627408484;
??????&#3627408485;4 + 2
??????4&#3627408484;
??????&#3627408485;2??????&#3627408486;2+
??????4&#3627408484;
??????&#3627408486;4 =
&#3627408478;
&#3627408439;
ga qo’yamiz z

&#3627408438;??????4(
1
&#3627408462;4+ 2
1
&#3627408462;2&#3627408463;2+
1
&#3627408463;4)&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; =
&#3627408478;0
&#3627408439; &#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463;
Bundan quyidagini olamiz :
&#3627408438; =
&#3627408478;0
&#3627408439; ??????4( 1
&#3627408462;2+ 1
&#3627408463;2)
2
ekanligi kelib chiqadi .
3. Ichki kuchlar ifodasini topamiz .
Buning uchun zarur bo’ ladigan hosilalarni hisoblaymiz:
??????2&#3627408484;
??????&#3627408485;2 = − &#3627408438;
??????2
&#3627408462;2&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ;
??????2&#3627408484;
??????&#3627408486;2 = − &#3627408438;
??????2
&#3627408463;2&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ;
??????2&#3627408484;
??????&#3627408485;??????&#3627408486; = − &#3627408438;
??????2
&#3627408462;&#3627408463; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
&#3627408463; ;
??????3&#3627408484;
??????&#3627408485;3 = − &#3627408438;
??????3
&#3627408462;3&#3627408464;&#3627408476;&#3627408480;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
&#3627408463; ;
??????3&#3627408484;
??????&#3627408486;3 = &#3627408438;
??????3
&#3627408463;3&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
&#3627408463; ;
??????3&#3627408484;
??????&#3627408485;?????? &#3627408486;2 = − &#3627408438;
??????3
&#3627408462;&#3627408463;2&#3627408464;&#3627408476;&#3627408480;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ;
??????3&#3627408484;
??????&#3627408485;2??????&#3627408486; = &#3627408438;
??????3
&#3627408462;2&#3627408463; &#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
&#3627408463; ;
Bu ifodalar yordamida ichki kuchlarni aniqlaymiz :
&#3627408448; &#3627408485; = &#3627408439;&#3627408438; ??????2(
1
&#3627408462;2+
&#3627409160;
&#3627408463;2)&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; = &#3627408472;1&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ;
&#3627408448; &#3627408486; = &#3627408439;&#3627408438; ??????2(
1
&#3627408463;2+
&#3627409160;
&#3627408462;2)&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; = &#3627408472;2&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ;
&#3627408452;&#3627408485; = &#3627408439;&#3627408438; ??????3(
1
&#3627408462;3+
&#3627409160;
&#3627408462;&#3627408463; 2)&#3627408464;&#3627408476;&#3627408480;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; = &#3627408472;3&#3627408464;&#3627408476;&#3627408480;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ;
&#3627408452;&#3627408486; = − &#3627408439;&#3627408438; ??????2(
1
&#3627408463;3+
&#3627409160;
&#3627408463;&#3627408462; 2)&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
&#3627408463; = − &#3627408472;4&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
&#3627408463; ;
?????? = &#3627408439;&#3627408438; ??????2 1
&#3627408462;&#3627408463; (1 − &#3627409160;)&#3627408464;&#3627408476;&#3627408480;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????&#3627408486;
&#3627408463; = &#3627408472;5&#3627408464;&#3627408476;&#3627408480;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463;
4. Ichki kuchlar epyuralarini quramiz (2 -rasm).
a) &#3627408485; = 1 &#3627408474; bo ’lgan kesimda
&#3627408448; &#3627408485; = &#3627408472;1&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; = 0.707 &#3627408472;1&#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; .
bu yerda:

0.707 &#3627408472;1 = 0.707
&#3627408478;0( 1
&#3627408462;2+ &#3627409160;
&#3627408463;2)
??????2( 1
&#3627408462;2+ 1
&#3627408463;2)
2 = 0.707 ∙
&#3627408478;0(1
42+ 0.2
32)
3.14 2(1
42+ 1
32)
2 = 0.202 &#3627408478;0
&#3627408448; &#3627408485; = 0.202 &#3627408478;0&#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ;
&#3627408452;&#3627408485; = &#3627408472;3&#3627408464;&#3627408476;&#3627408480;
??????
4 &#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; = 0.707 ∙&#3627408472;3&#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ;
0.707 ∙&#3627408472;3 = 0.707 ∙
&#3627408478;0( 1
&#3627408462;3+ 1
&#3627408462;&#3627408463; 2)
?????? ( 1
&#3627408462;2+ 1
&#3627408463;2)
2 = 0.707 ∙
&#3627408478;0(1
42+ 1
4 ∙32)
3.14 ∙(1
42+ 1
32)
2 = 0.325 &#3627408478;0;
&#3627408452;&#3627408485; = 0.325 &#3627408478;0&#3627408464;&#3627408476;&#3627408480;
??????&#3627408486;
&#3627408463; ;
b) &#3627408486; = 1&#3627408474; bo ’lgan kesimda
&#3627408448; &#3627408486; = &#3627408472;2&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
??????
3 = 0.5 ∙&#3627408472;2&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462;
0.5 ∙&#3627408472;2 = 0.5 ∙
&#3627408478;0(1
32+ 0.2
42)
3.14 2∙(1
42+ 1
32)
2 = 0.208 &#3627408478;0;
&#3627408448; &#3627408486; = 0.208 &#3627408478;0&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; ;
&#3627408452;&#3627408486; = − &#3627408472;4&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
??????
3 = − 0.866 &#3627408472;4&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462; = − 0.53 &#3627408478;0&#3627408480;??????&#3627408475;
??????&#3627408485;
&#3627408462;
Ichki kuc hlarni &#3627408485; va &#3627408486; ning turli qiymatlarida hisoblaymiz 3 -jadval
3-jadval
&#3627408485; = 1 &#3627408474; &#3627408472;&#3627408466;&#3627408480;??????&#3627408474;&#3627408465;&#3627408462; &#3627408486; = 1 &#3627408474; &#3627408472;&#3627408466;&#3627408480;??????&#3627408474;&#3627408465;&#3627408462;
&#3627408486; &#3627408448; &#3627408485;∙&#3627408478;0 &#3627408452;&#3627408485;∙&#3627408478;0 &#3627408485; &#3627408448; &#3627408486; ∙&#3627408478;0 &#3627408452;&#3627408486; ∙&#3627408478;0
0 0.202 0.325 0 0 0
± 1 0.101 0.162 ± 2 ± 0.208 ± 0.53
± 1.5 0 0 ± 4 0 0
± 2 − 0.145 − 0.281 − − −
± 3 − 0.202 − 0.325 − − −

2-rasm

Sirti elliptik paraboloid bo’lgan qobiqlar hisobi masalasi .
Asosiy tushunchalar
Yotiq qobiq deb qobiq egriligi uning o’lchamlariga nisbatan yetarlicha kichik
qobiqlarga aytiladi. Masalan &#3627408467;/&#3627408462; ≤ 1/5, bu yerda &#3627408467; − qobiq ko’tarilish balandligi,
&#3627408462; uning tomo ni.
2-Misol. To’g’ri burchakli to’rtburchak panelli o’rta sirti ushbu
&#3627408487; = 2&#3627408467; (
&#3627408485;2
&#3627408462;2+
&#3627408486;2
&#3627408463;2)
funksiya bilan ifodalanuvchi yotiq elliptik paraboloid qobiq egilishini aniqlang .
Yechish. Qobiq muvozanat tenglamalar sistemasini quyidagi ko’rinishda
olamiz.
{
∇??????2Φ (&#3627408485;,&#3627408486;)− &#3627408439; ∇4&#3627408484; (&#3627408485;,&#3627408486;)+ &#3627408461;(&#3627408485;,&#3627408486;)= 0
1
&#3627408440;&#3627409151; ∇4Φ (&#3627408485;,&#3627408486;)+ ∇??????2&#3627408484; (&#3627408485;,&#3627408486;)= 0
(9)
bu yerda
∇??????2= 4&#3627408467; (
1
&#3627408462;2
??????2
??????&#3627408486;2+
1
&#3627408463;2
??????2
??????&#3627408485;2) ; ∇4= (
??????2
??????&#3627408485;2+
??????2
??????&#3627408486;2)
2
; &#3627408439; =
&#3627408440;ℎ3
12 (1 − &#3627409160;2).
Koordinata o’qlari qobiq markaziga joylashtirilgan bo’ lib q obiq tomonlari
&#3627408485; = ± &#3627408462;/2 va &#3627408486; = ± &#3627408463;/2 ga teng bo’lsin:
 bu &#3627408485; = ± &#3627408462;/2 chegarada eguvchi moment &#3627408448; &#3627408485; = − &#3627408439; (??????2&#3627408484;
??????&#3627408485;2+ &#3627409160;??????2&#3627408484;
??????&#3627408486;2) va o’rta
sirtning egriligi ??????&#3627408486; = ??????2&#3627408484;
??????&#3627408486;2 nolga teng;
 &#3627408485; = ± &#3627408462;/2 chegarada diagrammaning cho’zilmasligidan &#3627409152;&#3627408486; uzayish yo’q
va uning yupqaligidan &#3627408449;&#3627408485; normal zo’riqish yo’q. Guk qonuniga ko’ra &#3627408449;&#3627408486; =
??????2Φ
??????&#3627408485;2 = 0 ekanligi kelib chiqadi;
 &#3627408486; = ± &#3627408463;/2 tomon uchun ham yuqoridagi shartlar o’rinli. Agar qobiq
tomonlari tayanish chegrasida gorizontal tashqi kuchlar bo’lmasa, u holda
kuchlanish funksiyasi Φ bu chegarada nolga teng bo’ladi.
Shunday qilib, qobiq tomonlari uchun quyidagi chegaraviy shartlar o’rinli
bo’ladi:
&#3627408485; = ±
&#3627408462;
2 &#3627408481;&#3627408476;&#3627408474;&#3627408476;&#3627408475; &#3627408482;&#3627408464; ℎ&#3627408482;&#3627408475; &#3627408484; = 0,
??????2&#3627408484;
??????&#3627408485;2 = 0,&#3627408449;&#3627408486; =
??????2Φ
??????&#3627408485;2 = 0
&#3627408486; = ±
&#3627408463;
2 &#3627408481;&#3627408476;&#3627408474;&#3627408476;&#3627408475; &#3627408482;&#3627408464; ℎ&#3627408482;&#3627408475; &#3627408484; = 0,
??????2&#3627408484;
??????&#3627408486;2 = 0,&#3627408449;&#3627408485; =
??????2Φ
??????&#3627408486;2 = 0
}

(10 )
Yechimni (9 ) muvoza nat tenglamalari sistemasi va (10 ) chegaraviy shartlarni
qanoatlantiruvchi quyidagi ikki karali trigonometrik qator shaklida izlaymiz.

&#3627408484; (&#3627408485;,&#3627408486;)= ∑ ∑ &#3627408484;&#3627408475;,&#3627408474; &#3627408464;&#3627408476;&#3627408480;
&#3627408474;??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
&#3627408475;??????&#3627408486;
&#3627408463;
∞
&#3627408474;=1
∞
&#3627408475;=1
Φ (&#3627408485;,&#3627408486;)= ∑ ∑ Φ &#3627408475;,&#3627408474; &#3627408464;&#3627408476;&#3627408480;
&#3627408474;??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
&#3627408475;??????&#3627408486;
&#3627408463;
∞
&#3627408474;=1
∞
&#3627408475;=1 }

(11 )
bu yerda &#3627408484;&#3627408475;,&#3627408474; va Φ &#3627408475;,&#3627408474; sonli koeffi tsiy entlar. Tashqi yuklanish &#3627408461;(&#3627408485;,&#3627408486;) ko’rinishini
quyidagi ikki karali trigonometrik qator ko’rinishda tasvirlaymiz:
&#3627408461;(&#3627408485;,&#3627408486;)= ∑ ∑ Z&#3627408475;,&#3627408474; &#3627408464;&#3627408476;&#3627408480;
&#3627408474;??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
&#3627408475;??????&#3627408486;
&#3627408463;
∞
&#3627408474;=1
∞
&#3627408475;=1
(12 )
Bu qatorning koeffi tsiye ntlari bi zga ma’lum bo’lgan Fur’ e qatorlari
nazariyasidan kelib chiqib quyidagi formuladan topiladi.
Z&#3627408475;,&#3627408474; =
4
&#3627408462;&#3627408463; ∫ ∫ &#3627408461;(&#3627408485;,&#3627408486;)&#3627408464;&#3627408476;&#3627408480;
&#3627408474;??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
&#3627408475;??????&#3627408486;
&#3627408463; &#3627408465;&#3627408485;&#3627408465;&#3627408486;
&#3627408463;
0
&#3627408462;
0
(13 )
Bu olingan (11), (12) ifodalarni (9 ) muvozanat tenglamasiga qo’ysak va uni
&#3627408464;&#3627408476;&#3627408480; &#3627408474;??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480; &#3627408475;??????&#3627408486;
&#3627408463; ga qisqartirsak, u holda
??????4&#3627408439; (
&#3627408474; 4
&#3627408462;4 + 2
&#3627408474; 2
&#3627408462;2
&#3627408475;2
&#3627408463;2+
&#3627408475;4
&#3627408463;4) &#3627408484;&#3627408475;,&#3627408474; + 4&#3627408467;??????2(
&#3627408475;2
&#3627408462;2&#3627408463;2+
&#3627408474; 2
&#3627408462;2&#3627408463;2) Φ &#3627408475;,&#3627408474; = Z&#3627408475;,&#3627408474; ;
− 4&#3627408467;??????2(
&#3627408475;2
&#3627408462;2&#3627408463;2+
&#3627408474; 2
&#3627408462;2&#3627408463;2) &#3627408484;&#3627408475;,&#3627408474; +
??????4
&#3627408440;&#3627409151; (
&#3627408474; 4
&#3627408462;4 + 2
&#3627408474; 2
&#3627408462;2
&#3627408475;2
&#3627408463;2+
&#3627408475;4
&#3627408463;4) Φ &#3627408475;,&#3627408474; = 0.
bundan
&#3627408484;&#3627408475;,&#3627408474; =
1
&#3627408440;&#3627409151; ∙
(&#3627408474; 2
&#3627408462;2 + &#3627408475;2
&#3627408463;2)Z&#3627408475;,&#3627408474;
[&#3627408439; ??????4
&#3627408440;&#3627409151; (&#3627408474; 2
&#3627408462;2 + &#3627408475;2
&#3627408463;2)
4
+ 16 &#3627408467;2
&#3627408462;4&#3627408463;4(&#3627408474; 2+ &#3627408475;2)2]
;
Φ &#3627408475;,&#3627408474; =
4
&#3627408462;2&#3627408463;2∙
(&#3627408474; 2+ &#3627408475;2)Z&#3627408475;,&#3627408474;
[&#3627408439; ??????4
&#3627408440;&#3627409151; (&#3627408474; 2
&#3627408462;2 + &#3627408475;2
&#3627408463;2)
4
+ 16 &#3627408467;2
&#3627408462;4&#3627408463;4(&#3627408474; 2+ &#3627408475;2)2]
.
Bu formulalarni qisqaroq ko’rinishda yozamiz.
&#3627408484;&#3627408475;,&#3627408474; =
Z&#3627408475;,&#3627408474; &#3627408462;4
&#3627408440;&#3627408467;2&#3627409151; ∙
&#3627409161;4(&#3627408474; 2&#3627409161;2+ &#3627408475;2)2
??????(&#3627408474; 2&#3627409161;2+ &#3627408475;2)4+ 16 &#3627409161;4(&#3627408474; 2+ &#3627408475;2)2;
Φ &#3627408475;,&#3627408474; =
4Z&#3627408475;,&#3627408474; &#3627408462;4
&#3627408467; ∙
&#3627409161;6(&#3627408474; 2+ &#3627408475;2)
??????(&#3627408474; 2&#3627409161;2+ &#3627408475;2)4+ 16 &#3627409161;4(&#3627408474; 2+ &#3627408475;2).
}

(14 )
bu yerda
&#3627409161; =
&#3627408463;
&#3627408462; ; ?????? =
&#3627408439;
&#3627408467;2&#3627408440;&#3627409151; =
&#3627409151;2
12 (1 − &#3627409160;2)&#3627408467;2.
Agar qobiq kvadrat shaklda bo’lsa, u holda &#3627409161; = 1, s hunga ko’ra,

&#3627408484;&#3627408475;,&#3627408474; =
Z&#3627408475;,&#3627408474; &#3627408462;4
&#3627408440;&#3627408467;2&#3627409151; ∙
1
??????(&#3627408474; 2+ &#3627408475;2)2+ 16 ;
Φ &#3627408475;,&#3627408474; =
Z&#3627408475;,&#3627408474; &#3627408462;4
&#3627408467; ∙
4
??????(&#3627408474; 2+ &#3627408475;2)3+ 16 (&#3627408474; 2+ &#3627408475;2).
}

(15 )
Qobiqdagi ichki zo’riqishlarni quyidagicha aniqlaymiz:
&#3627408449;&#3627408485; =
??????2Φ
??????&#3627408486;2 = − ∑ ∑ Φ &#3627408475;,&#3627408474;
&#3627408475;2??????2
&#3627408463;2 &#3627408464;&#3627408476;&#3627408480;
&#3627408474;??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
&#3627408475;??????&#3627408486;
&#3627408463;
∞
&#3627408474;=1
∞
&#3627408475;=1
;
&#3627408449;&#3627408486; =
??????2Φ
??????&#3627408485;2 = − ∑ ∑ Φ &#3627408475;,&#3627408474;
&#3627408474; 2??????2
&#3627408462;2 &#3627408464;&#3627408476;&#3627408480;
&#3627408474;??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
&#3627408475;??????&#3627408486;
&#3627408463;
∞
&#3627408474;=1
∞
&#3627408475;=1
; (16 )
?????? =
??????2Φ
??????&#3627408485;??????&#3627408486; = − ∑ ∑ Φ &#3627408475;,&#3627408474;
&#3627408474;&#3627408475; ??????2
&#3627408462;&#3627408463; &#3627408480;??????&#3627408475;
&#3627408474;??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
&#3627408475;??????&#3627408486;
&#3627408463;
∞
&#3627408474;=1
∞
&#3627408475;=1
.
&#3627408448; &#3627408485;,&#3627408448; &#3627408486; eguvchi va &#3627408448; &#3627408485;&#3627408486; burovchi momentlar esa quyidagi ko’rinishda bo’ladi:
&#3627408448; &#3627408485; = − &#3627408439; (
??????2&#3627408484;
??????&#3627408485;2 + &#3627409160;
??????2&#3627408484;
??????&#3627408486;2) = &#3627408439; ∑ ∑ ??????2(
&#3627408474; 2
&#3627408462;2 + &#3627409160;
&#3627408475;2
&#3627408463;2) &#3627408484;&#3627408475;,&#3627408474; &#3627408464;&#3627408476;&#3627408480;
&#3627408474;??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
&#3627408475;??????&#3627408486;
&#3627408463;
∞
&#3627408474;=1
∞
&#3627408475;=1
;
&#3627408448; &#3627408486; = − &#3627408439; (
??????2&#3627408484;
??????&#3627408486;2 + &#3627409160;
??????2&#3627408484;
??????&#3627408485;2) = &#3627408439; ∑ ∑ ??????2(
&#3627408475;2
&#3627408463;2+ &#3627409160;
&#3627408474; 2
&#3627408462;2) &#3627408484;&#3627408475;,&#3627408474; &#3627408464;&#3627408476;&#3627408480;
&#3627408474;??????&#3627408485;
&#3627408462; &#3627408464;&#3627408476;&#3627408480;
&#3627408475;??????&#3627408486;
&#3627408463;
∞
&#3627408474;=1
∞
&#3627408475;=1
;
&#3627408448; &#3627408485;&#3627408486; = ?????? = − (1 − &#3627409160;)
??????2&#3627408484;
??????&#3627408485;??????&#3627408486; = − &#3627408439; (1 − &#3627409160;)∑ ∑
&#3627408474;&#3627408475;?????? 2
&#3627408462;&#3627408463; &#3627408484;&#3627408475;,&#3627408474; &#3627408480;??????&#3627408475;
&#3627408474;??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
&#3627408475;??????&#3627408486;
&#3627408463;
∞
&#3627408474;=1
∞
&#3627408475;=1
;
Ko’ndalang kuchlarni hisoblashda quyidagi ifodalarga kelamiz:
&#3627408452;&#3627408485; =
??????&#3627408448; &#3627408485;
??????&#3627408485; +
??????&#3627408448; &#3627408485;&#3627408486;
??????&#3627408486; = − &#3627408439; ∑ ∑ (
&#3627408474; 3??????3
&#3627408462;3 +
&#3627408474; 3&#3627408475;3??????3
&#3627408462;&#3627408463; 2 ) &#3627408484;&#3627408475;,&#3627408474; &#3627408464;&#3627408476;&#3627408480;
&#3627408474;??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
&#3627408475;??????&#3627408486;
&#3627408463;
∞
&#3627408474;=1
∞
&#3627408475;=1

&#3627408452;&#3627408486; =
??????&#3627408448; &#3627408486;
??????&#3627408485; +
??????&#3627408448; &#3627408485;&#3627408486;
??????&#3627408486; = − &#3627408439; ∑ ∑ (
&#3627408475;&#3627408474; 2??????3
&#3627408462;2&#3627408463; +
&#3627408474; 3??????3
&#3627408462;3 ) &#3627408484;&#3627408475;,&#3627408474; &#3627408464;&#3627408476;&#3627408480;
&#3627408474;??????&#3627408485;
&#3627408462; &#3627408480;??????&#3627408475;
&#3627408475;??????&#3627408486;
&#3627408463; .
∞
&#3627408474;=1
∞
&#3627408475;=1

Mustaqil ishlash uchun topshiriqlar
Jadvalda berilgan variantlarga mos qobiqning deformasiyalanganlik -
kuchlanganlik holatini aniqlang

№ &#3627408478;(&#3627408485;,&#3627408486;) &#3627408467; &#3627408462; &#3627408463; ℎ &#3627408474; &#3627408475; Material
nomi
Mexanik
xarakteristikala
ri
&#3627408440;,
10 11 &#3627408451;&#3627408462;
&#3627409160;
1 &#3627408485;&#3627408486; 0.05 1 1 0.01 1 1 Beton 0.2 0.17
2 (1− &#3627408485;)(1− &#3627408486;) 0.02 1/2 1/2 0.01 1 1 Miss 1 0.3
3 2&#3627408485;+ 3 0.08 1 2 0.01 1 1 Cho’yan 1.5 0.25
4 2&#3627408485;+ 1 0.01 2 2 0.01 1 2 Po’lat 2 0.34
5 3&#3627408486;− 2 0.06 2 1 0.02 1 1 Alyumin 0.8 0.35
6 (1− &#3627408486;)&#3627408485; 0.05 1 1 0.03 1 1 Beton 0.2 0.17
7 (1− &#3627408485;)&#3627408486; 0.04 3/2 3/2 0.02 1 1 Alyumin 0.8 0.35
8 (&#3627408485;+ 2)(&#3627408486;+ 1) 0.03 1/2 1/2 0.01 1 1 Alyumin 0.8 0.35
9 &#3627408485;&#3627408486; + 1 0.06 1 1 0.06 1 1 Po’lat 2 0.34
10 3+ &#3627408485;&#3627408486; 0.07 2 1 0.05 2 1 Cho’yan 1.5 0.25
11 5− 2&#3627408485;&#3627408486; 0.08 2 2 0.05 1 1 Miss 1 0.3
12 &#3627408485;+ 5&#3627408486; 0.1 2 2 0.05 1 1 Alyumin 0.8 0.35
13 2− &#3627408486; 0.02 1/2 1/2 0.01 2 2 Cho’yan 1.5 0.25
14 4+ 3&#3627408485; 0.08 1 2 0.02 1 1 Po’lat 2 0.34
15 6− 7&#3627408486; 0.05 2 1 0.02 1 1 Miss 1 0.3
16 &#3627408485;+ 3&#3627408486; 0.1 2 2 0.06 1 1 Beton 0.2 0.17
17 5(&#3627408485;+ &#3627408486;) 0.1 2 2 0.1 1 1 Alyumin 0.8 0.35
18 (3− &#3627408486;)(&#3627408485;+ 2) 0.05 1 1 0.01 1 1 Po’lat 2 0.34
19 (&#3627408486;− 5)(&#3627408485;+ 2) 0.1 2 2 0.07 1 1 Alyumin 0.8 0.35
20 6&#3627408485;&#3627408486; + 4 0.05 2 1 0.02 1 1 Po’lat 2 0.34
21 3&#3627408486;+ 2&#3627408485;&#3627408486; 0.04 3/2 3/2 0.01 1 1 Cho’yan 1.5 0.25
22 &#3627408485;&#3627408486; + 4&#3627408486; 0.1 2 2 0.05 1 1 Miss 1 0.3
23 8&#3627408485;+ 5 0.03 4/2 1/2 0.01 1 2 Alyumin 0.8 0.35
24 7− 2&#3627408486; 0.05 1 2 0.01 2 1 Cho’yan 1.5 0.25
25 4+ &#3627408485;&#3627408486; 0.05 2 2 0.02 1 2 Po’lat 2 0.34
26 (3+ &#3627408486;)(&#3627408485;ғ2 ) 0.03 3/2 3/2 0.01 1 1 Cho’yan 1.5 0.25
27 (&#3627408485;+ 2)(&#3627408486;+ 1) 0.02 2 2 0.05 1 1 Miss 1 0.3
28 2&#3627408485;&#3627408486; + 6 0.05 4/2 1/2 0.01 1 2 Alyumin 0.8 0.35
29 8+ 2&#3627408485;&#3627408486; 0.06 1 2 0.01 2 1 Cho’yan 1.5 0.25
30 7+ 2&#3627408485;&#3627408486; 0.08 2 2 0.02 1 2 Po’lat 2 0.34

A dabiyotlar ro’yxati

1. Александров А. В., Потапов В. Д. Основы теории упругости и
ieZklbqghklb

To‘g‘ri to‘rtburchak shakldagi plastinka egilishlari va sirti elliptik paraboloid bo‘lgan qobiqlar hisobi masalalari

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