Prizmatik sterjenlarning ko‘ndalang tebranishlari
![Mav zu: PRIZMA TIK STERJ EN LA RN IN G KO’N DA LA N G
TEBRA N ISHLA RI
Uzluk siz k o’ndalang y uk lanish
(10) yechimga ega bo’lgan holda uzluksiz taqsimlangan qiymati
davriy o’zgaruvchi yuklanish ta’sirida balkaning tebranishlarini
o’rganish mumkin. Faraz qilaylik Psin ω tdc - dc uzunlikdagi elementga
ta’sir etuvchi yuklanish bo’lsin (
P - c dan bog’liq biror funksiya). (10)
yechimda
P0 ning o’rniga pdc ni qo’yib va c bo’yicha 0 dan l gacha
integrallab izlanayotgan yechimni topamiz.
P = const bo’lgan holda
balkaga tekis taqsimlangan vaqt davomida o’zgaruvchi yuklanish
ta’sir etadi. Yuqoriy aytilgan integrallashni bajaramiz:
y= 4gpl 4
qπ ∑k=1
∞ sin (2k+1)πx
l ⋅sin ω t
(2k+1)[(2k+1)4b2π4−ω2l2]
− 4gpl 6ω
qb π3 ∑k=1
∞ sin (2k+1)πx
l ⋅sin b(2k+1)2π2t
l2
(2k+1)2[(2k+1)4b2π4−ω2l4]
.
(14)
Yuklanish intensivligi sekin o’zgarganda
ω kichik va u uchun
quyidagi ifodani olamiz:
y= 4pl 4
EJ π6∑k=1
∞ sin (2k+1)πx
l
(2k+1)5 .
Bu balkaning statik egilishiga mos keladi.
Agar tebranish chastotasi xususiy tebranish chastotasidan
kichik bo’lsa, eng katta dinamik egilishni (13) formula yordamida
topish mumkin.
2-masala. Balk a bo’y lab harak at lanuv chi o’zgarmas k uch t a’sirida
t ebranish](https://docx.uz/documents/9adfe53d-621a-4a5d-a824-0c6bfba6370f/page_1.png?v=1)
![(15) yechimda x= l/2 va i= 1 ga mos hadlarni ifodalab, (16)
formala yordamida balka o’rtasining egilishini topamiz.
(y)x=l/2=− Pg
qπa tcos πat
l + Pgl
qπ 2a2sin πat
l +
+2Plg
qπ 2a2∑k=1
∞
(−1)ksin (2k+1)πt
l
(2k+1)2[(2k+1)2−1]
−
− 2Pl g
qπ 2a2∑k=1
∞
(−1)ksin (2k+1)2πt
l
(2k+1)2[(2k+1)2−1]
.
(18)
Bu ifodadagi yig’indilarning eng katta egilish qiymatiga ta’siri
juda kam. Shuning uchun (18) da dastlabki ikkita had bilan
chegaralanib,
P kuch o’ng tayanchga yetgandagi t= l/a momentga
mos keluvchi eng katta egilishni topamiz. Eng katta egilishning
qiymati
ymax = Pl 3/π3EJ . Buni eng katta statik egilish fst= Pl 3/48 EJ
bilan taqqoslab quyidagiga ega bo’lamiz:
(fd)max
fst
= 48
π3≈ 1,55
.
Amaliyotda odatda asosiy tip tebranish davri
T vaqtning 2l/a
oralig’idan ko’p marta kichik bo’ladi va demak
al miqdor bπ ga
nisbatan kichik.
al :bπ = α belgilash kiritib (15) yechimni quyidagiga
ifodalash mumkin.
y= 2Pl 3
EJ π4∑i=0
∞ sin iπx
l sin iπ at
l
i2(i2− α2)
− 2Pl 3
EJ π4α∑i=1
∞ sin iπx
l sin bi π2t
l2
i3(i2−α2)
.(19)
P
kuch chap uchidan с masofada ta’sir etayotgan momentda
balka o’rtasining egilishini topish uchun (19) yechimda
x= l/2 ,
at = c
deb olish kerak. 3](https://docx.uz/documents/9adfe53d-621a-4a5d-a824-0c6bfba6370f/page_4.png?v=1)
![U holda (y)x=l/2= fd= 2Pl 3
Es π4∑k=0
∞ (−1)ksin (2k+1)πc
l
(2k+1)2[(2k+1)2−α2]
− 2Pl 3
EJ π4α∑k=0
∞ (−1)ksin (2k+1)2bπ 2t
l
(2k+1)2[(2k+1)2− α2]
.
Agar
P kuchchapuchdan c masofagaqo’yilgandao’rtadagistatikegil
ishni
fst kabi, P kucho’rtagaqo’yilgandagistatikegilishniesa Fst
kabibelgilasakdinamikegilishyetarlichaaniqlikdaquyidgianiqlanishimu
mkin:
fd= fst
1−α2−αF stsin bπ 2t
l2 .
(20)
Engnoqulayholda
(fd)max = Fst
1−α2+αF st.
(21)
4](https://docx.uz/documents/9adfe53d-621a-4a5d-a824-0c6bfba6370f/page_5.png?v=1)
Prizmatik sterjenlarning ko‘ndalang tebranishlari
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