Mexanika” kafedrasi

1 – Jadval

Variantlar	Nazorat savollari
1	5, 15, 11, 20, 25, 40	16	2, 9, 16, 19, 26,36
2	1, 4, 21, 12, 39, 26	17	3, 6, 17, 12, 36, 28
3	2, 19, 24, 8, 25, 38	18	4, 7, 18, 11, 33, 34
4	3, 13, 19, 7, 37, 26	19	6, 1, 19, 10, 26,38
5	4, 10, 18, 27, 36,30	20	10, 3, 20, 15, 39,37
6	6, 14, 17, 22, 25,38	21	15, 4, 11, 21, 34,30
7	5, 20, 16, 9, 29, 34	22	4, 19, 12, 9, 55, 50
8	3, 11, 15, 21, 33,30	23	3, 8, 13, 18, 49,56
9	6, 13, 22, 1, 31,32	24	2, 7, 14, 21, 47, 58
10	4, 8, 13, 19, 29,25	25	1, 6, 15, 20, 59,46
11	3, 9, 23, 17, 25,37	26	5, 9, 16, 24, 45,60
12	2, 20, 12, 9, 27,35	27	10, 3, 17, 14, 61,44
13	1, 8, 13, 21, 29,34	28	5, 13, 18, 22, 43, 62
14	5, 10, 14, 20, 33,31	29	12, 4, 19, 15, 63,42
15	1, 7, 15, 18, 26,38	30	3, 8, 20, 12, 41, 64

STATIKA
1 mavzu. Tayanchlardagi reaksiya kuchlarini aniqlash
1-Topshiriq
Sterjendagi reakqiya kuchlarini aniqlash.
Keltirilgan jadvalda (1.1 jadval) konstruksiya chizmalari ko‘rsatilgan.


1.1 jadval

1.1 rasm


1.1 jadval

Variant	kattaliklar
Qiymatlar	G, kN	R, kN	M, kN·m	q, kN/m	α, grad
0	10	5	20	I	30
1	12	4	10	2	15
2	8	6	5	4	45
3	14	3	8	3	60
4	16	8	12	2	30
5	6	7	4	3	60
6	10	6	8	0,5	15
7	6	12	15	4	45
8	4	8	9	1.5	30
9	20	10	6	5	60

Topshiriqni bajarish uchun namuna
1-topshiriq


4 chizma 3 variant
t opshiriq sharti
sterjendagi reaksiya kuchlarini aniqlash.
berilgan: G = 14 kN, P = 4 kN, M =7 kN·m, q = 3 kN/m, α = 60°. Topish kerak: balkadagi reaksiya kuchlarini.


ECHISH
balkaga G, og‘irlik kuchi P,burchak ostida yo‘naltirilgankuch α= 60° burchak ostida ta’sir etadi. Q teng taqsimlangan kuch bo‘lib u Q=q⋅CB=3 kN, ga teng bo‘ladi.

Endi konstruksiyaga ta’sir etuvchi reaksiya kuchlarini x o‘qi uchun yozib chiqamiz .
 =0 (1.5)
Moment quyidagiga teng bo‘ladi.




(1.6) moment tenglamasidan R_Ay ni topamiz

(1.7) moment tenglamasidan esa R_B ni topamiz R_B ni (1.5), tenglamaga qo‘yib R_Ax reaksiya kuchini topamiz.

Tenglamalarni to‘g‘ri ishlanganligini aniqlash uchun U o‘qi bo‘yicha reaksiya tenglamasini tuzamiz va hisoblaymiz.

R_A reaksiya kuchi quyidagi formula yordamida topiladi.



J AVOB:

KINEMATIKA
2 mavzu Nuqta kinematikasi
2- Topshiriq
M nuqtaning berilgan xarakat tenglamasida uning traektoriya kurinishi urgatilishi kerak va t=t₁ sekund moment vakt ichida traektoriyadagi iuktaning xolati, uning tezligi. urinma va normal xamda to`la tezlanishlari shuningdek traektoriyaning egrilik radiusi aniqlansin.
Masalani echish uchun berilganlar 2-jadvaldan olinadi.
x = f₁(t), y = f₂(t) topish kerak

1. 
   Traektoriya tenglamasi topiladi va bu tenglama bilan ifodalanuvchi shakl chiziladi.
   

1. 
   Tezlik vektori va uning berilgan vaktdagi kiymati topiladi xamda shaklda ko’rsatiladi
   

3 Nuqtaning urnnma, normal va tula tezlanishlari, ularning berilgan vaktdagi kiymati topiladi va shaklda ko’rsatiladi.
Traektoriya egrilik radiusi topiladi. Nuqtaning berilgan xarakat tenglamalariga kura traektoriya tenglamasini topish uchun xarakat tenglamalaridagi uzgaruvchi t ni xar xil matematik amallarni bajarish yuli bilan yukotish kerak.
Nuqtaning harakati t a va b,larning koeffitsentlari 2.1 jadvaldan olinadi
2.1 jadvaldan

0.	x = a sin(πt/4)				y = b cos(πt /4)					t1 = 3 c
1.	x = a sin(πt /6)				y = 4 + b cos(πt /6)					t1 = 1 c
2.	x = 10at				y = bt^2					t1 = 0,5 c
3.	x = 1 – a cos t				y = b sint					t1 = π/4 c
4.	x = a cos3t – 1				y = 3 + b sin 3t					t1 = π/18 c
5.	x = a – sin t				y = b + 2 cos t					t1 = π/3 c
6.	x = at				y = at – bt^2					t1 = 0,2 c
7.	x = a cos(2 πt /3)				y = 2 + b sin(2 πt /3)					t1 = 2 c
8.	x = a + 2 sin(πt /4)				y = b + 2 sin(πt /4)					t1 = 1 c
9.	x = at^2 + bt				y = 2bt					t1 = 0,3 c
Tablitsa 2.1
Topshiriq		variant
kattaliklar , m		0	1	2	3	4	5	6	7		3	9
a		2	3	1	4	5	6	3	2		1	4
b		6	1	2	5	4	3	2	5		3	2

Topshiriqni bajarish uchun asosiy ko‘rsatmalar
M nuqtaning traektoriyasi x,u,z koordinata o‘qlarida tasvirlaymiz.
x = f₁ (t), y = f₂(t), z = f₃(t). (2.1)
Agar nuqta tekislikda harakatlansa, Oxy, ikkita o‘qqa nisbatan harakat bo‘ladi.
x = f₁ (t), y = f₂(t). (2.2)
agar nuqtaning harakati ma’lum bo‘lsa, yo‘ldan vaqt bo‘yicha olingan birinchi hosila nuqtaning tezligini beradi.
  (2.3)
Tezlik modulini topamiz.
v = = . (2.4)
yo‘ldan vaqt bo‘yicha ikkinchi tartibli hosila olib tezlanishni aniqlaymiz

  (2.5)

To‘la tezlanish modulini topamiz.
a = = . (2.6)
v
aqtga nol berib nuqtaning boshlang‘ich holatini aniqlaymiz va vaqtning berilgan ondagi qiymatini topib grafikda tasvirlaymiz
(rasm. 2.1).
Nuqtaning urinma tezlanishini aniqlaymiz.
 (2.7)

(2.8)


Bu erda ρ – nuqtaning egrilik radiusi.
a=a_τ+a_n. (2.9)
 = (2.10)

Tangensial tezlanish:

 ,  (2.11)

Topshiriqni bajarish uchun keyingi namuna.
Topshiriq-2
Berilgan.
x = 3sin(π/6) m, y = 2 + 4cos(π /6) m, t₁ =1 s.