Materiallar

ВВ'  l_2
  l₁   l_{3 .}
2
A

Bu absolyut cho’zilishlarni Guk qonuni yordamida tеgishli noma'lum zo’riqish kuchlari orqali ifodalaymiz:

 l₁

_ N₁ l _,
EF

 l₂

_ N ₂ l
EF


ва  l_3
 N ₃l _.
EF

Bu formulalarni yozishda uchala stеrjеnning matеrialini, ko’ndalang kеsimining yuzalari va uzunliklari bir hil dеb qabul qildik. Bularni (A) tеnglamaga qo’yib, quyidagi qo’shimcha tеnglamani hosil qilamiz:

N  ¹ N  N .
(4.3)

2 ₂ 1 3

Ana endi uchala tеnglamani birgalikda еchib,

7
N₁  P, 12
4
N ₂  P, 12
1
N ₃  P.
12
larni hosil qilamiz.

1-Masala
Chozilish va siqilishda statik aniq masala Reja

1. 
   Bir uchi bilan maxkamlangan sterjin uchaska kesimlari aloxida aloxida qirqilib muvozanat tenglamasi topilsin.
   
2. 
   Kesim uchun boylama kuch N aniqlansin.
   

3. 
   Kesimlar uchun kuchlanish G mutloq defarmasiya (belgilansin) topilsin.
   

l bilan

4. 
   Topilgan qiymatlar asosida boylama kuch, kuchlanish mutloq defarmasiya epyuralari qurilsin.
   

Berilgan


F=1.4t=14k A₁=1.2sm²=12 10^-4
A₂=1.1sm²=11 10^-4


a=1.3m b=1.1m c=1.7m
E=2 10⁸мп

Topish kerak: N-?

G-?

 l -?
Yechish:

1. 
   N bo’ylama kuchni aniqlaymiz:
   

I.I N₁-F₂=0
N= F₂=28kn


II.II N₂-F₂=0 N₂=28kn


III.III N₃+F₁- F₂=0
N₃=-F₁+F₂=14kn

2. 
    kuchlanishni aniqlaymiz:
   

G  ^N
A

1
G  ^N1 A₁

2
G  ^N2
A₂

3
G  ^N3
A_1
 28kn 1.210^4

_ 28kn 1.110^4

_ 14kn 1.210^4

 23.310^4kn/ m²

25.4510^4kn/m²

11.610^4 kn/ m²

3. 
   l mutlaq deformatsiyani aniqlaymiz:
   

 l 
N  l E  A

l  ^{N1 C} 
E  A₁
281.7
210⁸ 1.210^4
19.810^4

l  ^{N2  B}  ^281.1  14.10^4

E  A₂
l  ^{N3 C} 
E A₁
2 10⁸ 1.110^4
141.3



210⁸ 1.210^4


7.5810^4

2-masala
Chozilish va siqilishda statik noaniq masala Reja

1. 
   Ikki tomondan qistirilib maxkamlangan polat sterjin uchun berilgan sistema asosiy sistemaga aylantirilsin hamda muozanat tenglamasi tuzilsin.
   
2. 
   Statik noaniq masala static aniq masalaga aylantirilsin.
   

3. 
   Berilgan polat sterjin uchun N,G,
   
4. 
   Topilgan qiymatlar asosida N,G,
   

^{ l} aniqlansin.
^l epyuryasi topilsin.

5. 
   Mutloq defarmasiya 0ga tengligi aniqlansin
   

l ₌₀

A				A1
B		F			A2
C				A1

Berilgan


F=450 A₁=11*10^-4 A₂= 22*10^-4

a=2.2 m b=2.4 m c=1.8m
E=2 10⁸мп

Topish kerak:

R_A -? R_B -?

N-? G-?
 l _-?
Yechish:

Muvozanat tenglamasini tuzamiz:

R_A-F+R_B=0

 l 
N  l
l ₌₀

E  A
 l ₌ l_F+ l_RB=0


F  C
l ₌
^F E  A

R_B
l 
R_B  a _
R_B  b
 ^{RB  c}  0

E  A₁
E  A₂
E  A₁

 l 
R_B  a _
R_B  b
_ ^R_B ^{ c} _ Fl _{ 0}

E  A₁
E  A₂
E  A₁ E  A₁

^ c ^
_ 1.8 _

F  ^{ }
A
^{450  } 11 10 ^{4 }

_{R }  1  _
^{ }  156


^B a b c _{ 2.2}
2.4
1.8 _

 
A₁ A₂ A₁
^ 11 10 ^4

22 10 ^4
 _
11 10 ^4 _

R_A=F-R_B=450-156=294

1. 
   N bo’lama kuchni aniqlaymiz:
   

I.I N₁+ R_B =0
N= - R_B =-156кн


II.II N₂- R_B =0 N₂=-156кн


III.III N₃-F₁+ R_B =0 N₃=F₁-R_B =294кн

2. 
    kuchlanishni aniqlaymiz:
   

G  ^N A

N ₁
G ₁ 
A_1
  156 _
11  10 ^{ 4}
 14 .19  10 ^{ 4} kí
/ m ²

2
G  ^{N 2}
_  156 _
 7 .1  10 ^{ 4} kí
/ m ²

 4
A ₂ 22  10

N ₃
G ₃ 
A₁

_ 294

11 10 ^4
 26 .73 10 ^4 kn / m ²

3. 
   l mutlaq deformatsiyani aniqlaymiz:
   

 l 
N  l E  A

 l 
N ₁  a _
 156  2 .2
  15 .6  10 ^{ 4}

E  A₁
2  10 ⁸  11
 10 ^{ 4}

 l 
N ₂  b
_  156  2 . 4 _
 8 . 51
.  10 ^{ 4}

E  A ₂
2  10 ⁸  22  10 ^{ 4}

 l 
N ₃  c _

294  1 .8

 24
.05

 10 ^{ 4}

E  A₁

2  10 ⁸  11

 10 ^{ 4}

l ₌ l ₁₊ l ₂₊  l ₃
l  15.610^4  8.5110^4 24.0510^4  0

3-masala

Yassi kesimlarning geometrik harakteristikalari

Berilgan nosimmetrik kesim uchun bosh inersiya momentlari aniqlansin va inersiya o’qlari o’tkazilsin, buning uchun:

1. 
   Og’irlik markazining koordinatalari aniqlansin
   
2. 
   Og’irlik markazi orqali o’tadigan x_c , y_c o’qlarga nisbatan markazdan qochirma inersiya momentlari topilsin
   
3. 
   Bosh markaziy o’qlarning yo’nalishlari aniqlansin
   
4. 
   Bosh markaziy o’qlarga nisbatan inersiya momentlarining qiymatlari topilsin
   
5. 
   Inersiya radiuslari topilsin
   
6. 
   Tekshirish bajarilsin
   

Berilgan:

1. 
   B teng yonli burchak 100*100*10
   

h 10sm 1
b  10sm 1

I  284sm⁴
x
A  19.2sm²

1
0₁
I  74.1sm⁴
I _x  I

y
1 ₁
 179sm⁴

0

0
^y Z  2.83sm
₁ 1
d₁=1.sm

2. 
   E vertical to’g’ri tortburchak
   

b₂  1sm h₂  20sm

2
A  1.0  20  20sm²

I _x2
b h ³
_ 2 2 _
12
10  20³
12

 666.67sm⁴

I _y2
b ³h
_ 2 2 _
12
1.0³  20


12
 1.67sm ⁴

3-shakl

3. 
   Б koshtavir
   

h₃  9sm

3
A  23.4sm ²
b₃  18sm
I _x  1290sm⁴ 3
d₃  5.1sm

I _y  82.6sm⁴ 3

Yechish:

1. 
   Og’irlik markazi orqali o’tadigan x_c , y_c o’qlarga nisbatan markazdan qochirma inersiya momentlarini aniqlaymiz:
   

1
c  b
3

• 
  z₀
  

1
 9  2.83  6.17sm

c₂  b₃

• 
  b₂
  

2
 9  ^1.0  9.5sm
2

c  ^b3
3 ₂
 4.5sm

k₁  z₀  2.83sm

k  ^{h2
2 2} 3
 ²⁰  10sm 2

k₃  h₂

• 
  b₃
  

2
 20  ⁹  15.5sm
2

c
_{x } ^c₁ ^A₁ ^{ c}₂ ^A₂ ^{ c}₃ ^A_{3 } 6.17 19.2  9.5  20  4.5  23.4 _ 118.5  190  105.3 _{ 6.61sm}

A₁  A₂

• 
  A₃
  

19.2  20  23.4
62.6

c
_{y } ^k₁ ^A₁ ^{ k}₂ ^A₂ ^{ k}₃ ^A_{3 } 2.83 19.2  10  20  15.5  23.4 _ 54.3  200  362.7 _{ 9.86sm}

A₁  A₂

• 
  A₃
  

19.2  20  23.4
62.6

2. 
   Bosh markaziy o’qlarga nisbatan inersiya momentlarining qiymatlarini aniqlaymiz:
   

a₁  c₁  x_c  6.17  6.61  0.44sm a₂  c₂  x_c  9.5  6.61  2.89sm a₃  c₃  x_c  4.5  6.61  2.11sm
m₁  k₁  y_c  2.83  9.86  7.03sm m₂  k₂  y_c  10  9.86  0.14sm m₃  k₃  y_c  15.5  9.86  5.64sm

I _xc  I

• 
  a ² A  I
  
• 
  a ² A  I  a ² A 
  

x 1 1 x
1 2
2 2 x₃ 3 3

 179   0.44² 19.2  666.67   2.89²  20  0,288  1290   2.11²  23.4  2410.6sm⁴

I  I  m² A  J

• 
  m² A  I  m² A 
  

c
y y 1 1 y
1 2
2 2 y₃ 3 3

 179   7.03² 19.2  1.67  0.14²  20  82.6  5.64²  23.4  1956.9sm⁴

I _x  I _y

284  74.1

^I x y
1 1
_ 0 0 _
2
 105sm ⁴
2

I _xc y_c  I _{x y}  a m A

• 
  I _{x y}  a m A
  
• 
  I _{x y}  a m A 
  

_{1 1} 1 1 1
_{2 2} 2 2 2
_{3 3} 3 3 3

 105  0.44   7.0319.2  2.89  0.14  20  2.11 5.64  23.4  106

4. 
   Markaziy bosh inersiya o’qlari vaziyatini aniqlaymiz:
   

tg 2 
2  I _xc y_c I _yc  I _xc
2  106

1956.9  2410.6
_{ } 212
453.7

 0.48

4. 
   Markaziy bosh inersiya momentlari qiymatlarini aniqlaymiz:
   

^Iu ,v
_ I _yc  I _{xc }
2

I _y  I _x ²

_{I } c c _
 1956.9  2410.6 _
 106²  2434.2sm

u max _{2 2}
J _y  J _x

^Jv min
_ c c _
2
 2183.8  250.4  1933.4sm⁴

4. 
   Tekshirish:
   

J _yc  J _xc
^{ J} max ^{ J} min

2410.6  1956.9  2434.2  1933.4
4367.5  4367.6