Siniq chiziqli murakkab balkalar.
Ichki kuchlar epyurasini siniq
chiziqli sterjenlar uchun qurishda, qo‘zg‘almas sanoq sistemasi yanada
noqulay bo‘lganligi sababli, u deyarli qo‘llanilmaydi. Tekis masala
holatida ham bu tipdagi sterjenlarning ko‘ndalang kesimlarida eguvchi
moment
M
va ko‘ndalang kuch
Q
dan tashqari bo‘ylama kuch
N
ham
paydo bo‘lishi mumkin. Bu holda 3 ta ichki kuch
M, Q, N
larning
epyurasini qurish kerak bo‘ladi.
1-misol.
Bir nechta sterjendan tashkil topgan (2.20-rasm) sterjenli
sistema uchun ichki kuchlarning epyurasi qurilsin. P = 4tk, q = 2tk /m,
a
= 6 m, b = 3 m, c = 2 m, d = 4m .
Qo‘zg‘almas Y
0
, Z
0
sanoq sistemasidan foydalanib tayanch
reaksiyalarini aniqlaymiz:
∑
Z
0
= – q·d = 0, = qd = 2·4 = 8 tk;
∑
Y
0
= – P = 0, = P = 4 tk;
20
2
4
6
4
2
3
4
2
M
,
0
2
0
0
0
=
⎟
⎠
⎞
⎜
⎝
⎛ −
⋅
⋅
+
⋅
−
=
⎟
⎠
⎞
⎜
⎝
⎛ −
+
−
=
=
⎟
⎠
⎞
⎜
⎝
⎛ −
+
−
−
=
ΣΜ
d
a
qd
Р
b
d
a
qd
Р
b
M
tk·m
.
56
2.20-rasm. Turli xil kuchlar bilan yuklangan siniq chiziqli sterjenlar
uchun ichki kuch epyuralarini qurish.
Bu sistema 4 ta uchastkadan iborat.
Sistemaning har bir uchastkasida hosil bo‘ladigan ichki kuchlarni
aniqlash uchun yuqorida (2.8-rasmda) keltirilgan qoidadan
foydalanamiz.
Birinchi uchastkada – I:
(
)
.
0
,
0
2
1
1
1
1
2
1
1
d
z
N
qz
Q
qz
M
≤
≤
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
=
=
−
=
Bundan, z
1
= 0 da, M
1
= 0, Q
1
= 0, N
1
=0;
z
1
= 2m da, M
1
= -4 tk·m, Q
1
= 4 tk, N
1
=0;
z
1
= 4m da, M
1
= -16 tk·m, Q
1
= 8 tk, N
1
=0.
III
z
1
II
I
z
2
d
IV
g
1
R
v
1
R
z
4
M
o
P
z
3
d
q
q
q
z
3
z
2
z
1
z
4
+
B
-
-
N epyurasi
+
-
B
A
8
8
Q epyurasi
4
A
4
M
epyurasi
28
28
16
A
B
16
26
y
o
P
q
b
c
M
o
g
1
R
0
z
o
v
R
1
0
а
d
57
Ikkinchi uchastkada – II:
(
)
с
z
qd
N
Q
d
d
q
М
≤
≤
⎪
⎪
⎭
⎪
⎪
⎬
⎫
−
=
=
⋅
⋅
−
=
2
2
2
2
0
,
0
2
Bundan, z
2
= 0da, M
2
= -16tk·m, Q
2
= 0, N
2
=-8tk;
z
2
=2m, M
2
=-16tk·m, Q
2
=0, N
2
=-8tk.
Uchinchi uchastkada – III:
(
)
b
z
qd
N
P
Q
Pz
d
qd
M
≤
≤
⎪
⎪
⎭
⎪
⎪
⎬
⎫
−
=
=
−
−
=
3
3
3
3
3
0
,
2
Bundan, z
3
= 0da, M
3
= -16tk·m, Q
3
= 4tk, N
3
= -8tk;
z
3
=3m da, M
3
=-28tk·m, Q
3
=4tk, N
3
= -8tk.
To‘rtinchi uchastkada – IV:
(
)
a
z
R
N
R
Q
z
R
M
M
o
≤
≤
⎪
⎭
⎪
⎬
⎫
−
=
−
=
−
=
4
v
1
4
g
1
4
4
g
1
4
0
,
Bundan, z
4
= 0da, M
4
= 20tk·m, Q
4
= -8tk, N
4
= -4tk;
z
2
=6m da, M
3
= -28tk·m, Q
4
= -8tk, N
4
= -4tk.
Olingan natijalar bo‘yicha
M, Q,
Ν
ichki
kuchlarning epyurasini
quramiz (2.20-rasm).
Siniq chiziqli sterjenlar uchun qurilgan ichki kuchlar epyurasini
qurishda, sterjen tugunlarining muvozanatini ham tekshirib ko‘rish kerak
bo‘ladi.
Buning uchun tugun yaqinidan o‘tkazilgan 2 ta kesim orqali
sterjendan bu tugun ajratib olinib, unga kesimlardagi ichki kuchlar va
tugunga ta’sir etuvchi tashqi kuchlar qo‘yiladi (2.21a,b-rasm).
Bu holda ichki kuch
Q, N, M
larning ishoralari quyidagi tartibda
olinadi. Agarda
Q
kesib olingan tugunni soat strelkasi bo‘yicha
aylantirsa musbat (+)
N
kuchi esa sterjenni cho‘zsa musbat (+) deb
olinadi. Agarda
M
epyurasi sterjenning cho‘ziluvchi tolalar tomoniga
qurilgan bo‘lsa, u ham musbat (+) ishorali bo‘ladi. Bu qoidalarni
tugunlarga qo‘llab, ularning muvozanatini tekshiramiz.
58
2.21a-rasm. A tugunga ta’sir qiluvchi ichki kuchlar.
2.21b-rasm. B tugunga ta’sir qiluvchi ichki kuchlar.
A
-tugun (o‘ng tomondagi tugun):
Bu tugunning vertikal sterjeniga
M
1
shunday qo‘yilganki, u tashqi
tomondagi tolalarni cho‘zib, soat strelkasi bo‘yicha aylanadi. Gorizontal
sterjenda esa
M
2
yuqoridagi tolalarni cho‘zib, soat strelkasiga qarshi
aylanadi.
Q
1
kuchi vertikal sterjenga qo‘yilgan bo‘lib, u tugunni soat
strelkasi bo‘ylab aylantiradi, ya’ni musbat ishoralidir.
N
2
kuchi epyura
bo‘yicha manfiy ishorali bo‘lib, gorizontal sterjenni siqadi (2.21a-rasm).
Shunga asosan
∑Μ
0
=
Μ
1
– M
2
=16 – 16 = 0;
∑
Z = N
2
-Q
1
= 8 – 8 = 0 bo‘ladi.
B
Q
3
=4 tk
N
3
=8 tk
M
3
=28 tk·m
M
4
=28 tk·m
N
4
=4 tk
Q
4
=8 tk
0
0
A
Q
1
=8 tk
N
2
=8 tk
M
2
=16 tk·m
M
1
=16 tk·m
59
B
-tugun (chap tomondagi tugun): xuddi shunday (2.21b-rasm).
∑Μ
0
=M
3
– M
4
= 28 – 28 = 0;
∑
Z =Q
4
–N
3
= 8 – 8 = 0;
∑
Y = N
4
– N
3
= 4 – 4 = 0.
Demak, tugunlar muvozanatda turibdi, qurilgan epyuralar to‘g‘ri
ekan.
2-misol.
Quyidagi (2.22-rasm) sterjenli sistema uchun ichki
kuchlarning epyurasi qurilsin:
q=2 tk/m, P =5 tk, a=10 m,
с
=1 m, d=1 m , b= 5m.
Tayanch reaksiyalarini aniqlaymiz:
.
t
13
5
1
5
10
)
1
1
(
2
10
10
2
))
1
1
(
10
(
15
)
(
2
))
(
(
,
0
)
(
2
))
(
(
M
;
t
13
5
2
10
2
10
)
1
10
(
5
2
)
(
,
0
2
)
(
M
;
t
15
10
2
5
,
0
g
1
v
1
g
1
v
1
2
2
2
2
2
1
g
1
g
1
0
k
b
d
P
M
d
c
a
a
q
d
c
a
R
R
d
P
M
d
c
a
a
q
d
ñ
a
R
b
R
k
b
a
q
M
d
a
P
R
a
a
q
M
d
a
P
b
R
k
à
q
Ð
R
a
q
P
R
Z
=
−
⋅
+
−
⎟
⎠
⎞
⎜
⎝
⎛
+
−
⋅
+
+
−
−
=
=
−
⋅
+
−
⎟
⎠
⎞
⎜
⎝
⎛
+
−
⋅
+
+
−
⋅
−
=
=
⋅
+
−
⎟
⎠
⎞
⎜
⎝
⎛
+
−
⋅
+
+
−
−
⋅
=
=
+
+
−
−
=
+
+
−
−
=
=
⋅
−
−
−
+
⋅
=
=
⋅
+
−
=
⋅
+
−
=
=
⋅
+
−
−
=
∑
∑
∑
60
a)
b)
d)
2.22-rasm. Tugunga ta’sir qiluvchi ichki kuchlar:
a) sterjen uchastkalari va ichki kuch epyuralari; b) A tugunga ta’sir etuvchi ichki
kuchlar; d) B tugunga ta’sir etuvchi ichki kuchlar.
Tayanch reaksiyalarini to‘g‘ri topilganligini tekshiramiz:
∑
=
+
−
=
+
−
=
0
13
13
2
v
1
R
R
Y
o
Bu sterjenli sistemani 4 uchastkadan iborat deb qarash mumkin.
Birinchi uchastkada
(
)
.
0
,
t
13
0
0
1
2
1
1
1
d
z
k
R
N
Q
M
≤
≤
⎪
⎭
⎪
⎬
⎫
−
=
−
=
=
=
Q
2
=5 tk
N
2
=13 tk
M
2
=5 tk·m
M=10 tk·m
B
N
3
=5 tk
M
3
=50 tk·m
Q
3
=13 tk
N
4
=13 tk
Q
4
=5 tk
M
4
=50 tk·m
A
0
M
3
=15 tk·m
N
2
=5 tk
Q
3
=13 tk
0
R
2
q
z
4
в
1
R
z
3
c
d
R
2
P
M
z
2
d
R
2
z
1
P
I
II
III
IV
г
1
R
z
1
z
2
z
3
z
4
+
-
-
5
13
61
Ikkinchi
uchastkada
(
)
.
0
,
2
2
2
2
2
2
c
z
R
N
P
Q
Р
z
M
≤
≤
⎪
⎭
⎪
⎬
⎫
−
=
=
−
=
Bundan,
z
2
= 0 da, M
2
= 0, Q
2
= 5 tk,
Ν
2
= -13 tk;
z
2
= 1m da, M
2
= -5 tk·m, Q
2
= 5 tk,
Ν
2
= -13 tk ga teng
bo‘ladi.
Uchinchi uchastkada
(
)
.
0
,
3
3
2
3
3
2
3
b
z
P
N
R
Q
M
Р
c
z
R
M
≤
≤
⎪
⎭
⎪
⎬
⎫
−
=
−
=
−
−
=
Bundan,
z
3
= 0 da, M
3
= -15 tk·m, Q
3
= -13 tk,
Ν
3
= -5 tk;
z
3
= 5m da, M
3
= 50 tk·m,
Q
3
= -13 tk,
Ν
3
= -5 tk ga teng
bo‘ladi.
To‘rtinchi uchastkada
(
)
.
0
,
2
4
v
1
4
4
g
1
4
4
4
4
g
1
4
a
z
R
N
z
q
R
Q
z
z
q
z
R
Ì
≤
≤
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
=
⋅
−
=
⋅
−
=
Bundan,
z
4
= 0da, M
4
= 0, Q
4
= 15 tk,
Ν
4
= 13 tk;
z
4
= 10 m da, M
4
= 50 tk·m, Q
4
= -5 tk,
Ν
4
= 13 tk.
M
ning eng katta qiymati Q = 0 bo‘lgan kesimda paydo bo‘lib, u
quyidagicha topilishi mumkin:
a
z
bundan
z
z
q
R
Q
g
d
m
5
,
7
2
15
,
0
2
15
4
4
4
1
4
=
=
=
−
=
⋅
−
=
M
max
qiymatga erishadi, ya’ni:
.
5
2
5
,
7
5
,
7
2
5
,
7
15
2
Ì
4
4
4
1
max
m
tk
z
z
q
z
R
g
⋅
=
⋅
−
⋅
=
⋅
−
⋅
=
Topilgan
M, Q, N
ning qiymatlari bo‘yicha epyura qursak, u 2.22a-
rasmdagi ko‘rinishga ega bo‘ladi. Epyuralarni to‘g‘ri qurilganligini
tekshirish uchun tugunlar muvozanatini tekshiramiz:
Chap tomondagi A tugun (2.22b-rasmdan):
62
∑
Z
o
= Q
4
– N
3
= 5 – 5 = 0;
∑
Y
o
= Q
3
– N
4
= +13 – 13 = 0,
∑Μ
0
= M
4
– M
3
= 50 – 50 = 0;
O‘ng tomondagi B tugun (2.22d-rasmdan):
∑
Z
0
= N
3
– Q
2
= 5 – 5
= 0,
∑
Y
o
= N
2
– Q
3
= 13 – 13 = 0,
∑Μ
0
= -M
3
+ M + M
2
= -15 + 10 + 5 = 0.
Demak, rama tugunlarining muvozanatda ekanligi epyuraning
to‘g‘ri qurilganligini isbotlaydi.
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