Avtoyo’l temirbeton ko’prikning qirqilgan qovurg’ali diagmasiz

Ta‟sir chiziqlarining ordinatalari

I-Varant A-11 yukning ikki palasasi xafsizlik to’sig’iga maksimal darajada yaqinlashtiilgan

O-to’sin uchun

KQK_A=0.5(0.5842+0.2294+0.6(0.0929-000.018882))=0.429

KQK_AT=0.5(0.5842+0.2294+0.0929-0.0182)=0.444

1-to’sin uchun

KQK_A=0.5(0.3323+0.3426+0.6(0.277+0.1436))=0.464

KQK_PT=0.5(0.3323+0.3426+0.277+0.1436)=0.548

2-to’sin

KQK_A=0.5(0.6(0.1232+0.2889)+0.3545+0.2998)=0.451

KQK_AT=0.5(0.1232+0.2889+0.3545+0.2998)=0.533

II-Varant A-11 yukning ikki palasasi xafsizlik to’sig’iga maksimal darajada yaqinlashtiilgan va ' ushbu yuk tratvardagi piyodalar bilan birgalikda ko’riladi.

O-to’sin uchun

KQK_A=0.5(0.2791+0.0432)=0.161

KQK_AT=0.5(0.2791+0.0432)=0.161

KQK_T=0.89

1-to’sin uchun

KQK_A=0.5(0.3665+0.2532+0.6(0.1728+0.0603))=0.38

KQK_AT=0.5(0.3665+0.2532+0.1728+0.0603)=0.426

KQK_T=0.27

2-to’sin uchun

KQK_P=0.5(0.2651+0.3783+0.6(0.3207+0.1961))=0.477

KQK_PT=0.5(0.2651+0.3783+0.3207+0.1961)=0.58

KQK_T=0

Nk 80 yuki qatnovi qismining chetigajoylashtiriladi

O-to’sin uchun KQK_k=0.5(0.2977+0.0166)=0.157

1-to’sin uchun KQK_k=0.5(0.3754+0.2057)=0.291

2-to’sin uchun KQK_k=0.5(0.2562+0.3443)=0.3

Y=1/n+a²/(2Эa²+nl²/3*G_bl_k/E_bI_b) yoki G_b=0.42E_b yoki G_b=0.42E_b ekanligini hisobga olganda

Y=1/n+a²/(2Эa²+14nl²/3*G_bl_k/I_b)

Alohida olingan to’sining enertsiya momentlari I_b va I_k ilgari hisoblab chiqilgan

I_b=118.12*10^-3m⁴

I_k=7.5*10^-3m⁴

Ko’prik oraliq qurilmasining ko’ndalang hisobidan 6 dona to’sin joylashgan

Ea²=2.3²+6.3²+10.5²=154.35m²

Y=1/6+10.5²/(2*154.35+0.14*6*23.4²*(7.5*10^-3/118.2*10^-3))

Y₁=1/6+0.326=0.433

Y₁=1/6-0.3263= -0.1596

26-rasm. To‘sin oralig‘i o‘rtasidagi maksimal eguvchi moment Mtaxni aniqlash uchun
sxemalar (o‘lchamlar metrlarda).

Y₁=L/4=5.85

|Y₂=y+(0.5L-1.5)/0.5L (0.5*23.4-1.5)/0.5*23.4=6.1

Nk 80 g’ildirakdagi ostidagi ordinatalar

Nk-80 y1=5.85

Y₃=Y₄=Y₁*(0.5l-1.2)/0.5l=5.85*(0.5*23.4-1.2)/0.5*23.4=5.25 m

Y₅=Y₁*(0.5l-1.2-1.2)/0.5l=5.85*(0.5*23.4-1.2-1.2)/0.5*23.4=4.65 m

Dinamik koefitsientlar

A-11 yuk uchun :

(1+M)A=1+(45-л)/135=1+(45-23.4)/135=1.16

A14 yuki va tratuardagi piyodalar to’dasidan (tratuardan kengligi b_t=1metr ) 1-to’sin oralig’i o’rtasidagi eguvchi momentni aniqlaymiz

M₁=gwm+(1+M)*(Yfa*q_pol*KQKAWM+Y_at*P_at*KQ_kpt(Y₁+Y₂))+Y₁+P_tKQKtWM=28.3*68445+1.16[1.2*11*0.38*68.445+1.266*110*0.426(5.86+5.1)]+1.2*3.532*1*0.27*68.445=1932.88+1151.794+78.326=3163 kNm

M_1n=24*68.445+11*0.38*68.445+110*0.426*10.95+3.532*0.27*68.445=1636.52+286.1+513.117+65*272=2501.01 kNm

M₂=1932.88+1.16(1.2*1.1*0.477*68.445+1.266*110*0.58*10.35)=1932.88+ +01525.86=3458.74 kNm

M_2n=1636.52+11*0.477*68.445+110*0.58*10.95=1636.52+359.131+6919.41=2694.26 kNm

Birinchi uchastka yuzasi

W=(1+0.833)/2*3.9=3.574 m²

va uning polosali yuk uchun ko’ndalang qo’yish koefitsienti

KQK_fi=(0.691+0.477)/2=0.5784

To’sin oralig’ining o’rtasidagi keltirilgan kesimning geometrik xarakteriskalari

A_red=A_b+n₁A_p=6591.2+6*42.39=6845.54 m²

A_p=(5*8+3*18+1*28)/9=13.55 sm

Keltirilgan kesimning ostki qirraga nisbatan statik momenti

S_red=S_b+n₁A_pa_p=518613.28+6*42*3.9*13.55=522059 sm³

To’g’ri chiziqli armatura tratuarlari uchun

W=170*W_6p/R_v=170*9.37/28=56.89 MPa

To’sinni eguvchi moment bo’yicha mustahkamlikka hisoblash : To’sinlar uchun V40 (M500) sinfli beton qabul qilingan . Ushbu betonning siqilishga bo’lgan hisobiy qarshiligi R_b=20 MPa cho’zilishga bo’lgan hisobiy qarshiligi R_bt=1.27 MPa R_bn=29MPa R_bmcl=21.5 MPa R_bmsr=17.5 MPa R_b+ser=21 MPa R_bsh=3.6 MPa . Doimiy va vaqtinchalik yuklardan hosil bo’lgan eng katta eguvchi moment M=3458.74 Kn*m to’sin oralig’ining o’rtasida bo’ladi

A_p^t=1.1*m/R_p(h_d-0.5h_f=1.1*(3458.74*10⁵)/1100*10²(0.87*120-0.5*18)=36.26 sm²

Har biri 24 ta diametrli 5mm li simdan iborat bo’lgan 9dona tutam qabul qilamiz . Bir tutamdagi armaturaning yuzasi

П*0.5²/4*24=4.71sm² har bittasi uchun

9dona tutamdagi armaturaning yuzasi A_r=9*4.71=42.39sm²>A_r^t=36.26sm²
Tayanchlar yonida ko’ndalang kuchni kamaytirish va tayanch uchastkalarining darzbardoshligini oshirish maqsadida tayanchdan 7.5 metr masofaga 3 dona tutamni ustki zonaga qarab biriktiramiz . Bunda birinchi ikkinchi va uchinchi qatordan bittadan tutam yuqoriga qarab bukiladi.

29-rasm. To‘sinda oldindan zo‘riqtirilgan armaturani joylashtirish sxemasi

Tutamlarning to‗sin o‗qiga nisbatan bukilish burchaklarini aniqlaymiz:

uchinchi qator


tg

Ikkinchi qator tg

Poliganel armature tutamlari uchun

Б₂=170*8.13/28=49.36 MPa

Ekspltatsiya bosqichida rolig’ida to’sin o’rtasidagi kesimdagi oldindan zo’riqtirish kuchlanishni topamiz

To’g’iri chiziqli armatura tutamlaridan

Birinchi qator tg

tutamlar burilishinging o’rtacha burchagi

a_cp=(5.86+5.1+4.34)/3=5.1⁰=0.089 rad

To‘sin oralig‘ining o‘rtasidagi keltirilgan kesimning geometrik

xarakteristikalari

A_rad=A_b+nA_p=6591.2+6*42.39=6845.54 m²

A_p=(5*8+3*18+2*18)/9

S_rad=S_b+nP_pa_p=51813.28+6*42.39*13.55=522059 m²

Ostki qismiga nisbatan holat

Y_red^Hg=S_red/A_red=522059.59/6845.54=76.26 m

Ustki qismiga nisbatan holat

Y_red^Bg=h- Y_red^Hg=120-76.26=43.74m

I_red=I_b+P_b(Y_red^Hg- Y_red^Bg)²=+n P_b(Y_red^Hg-a_p)²=118.12*10⁵+6591.2(76.26-78.7)²+

+6*42.39(76.26-13.55)²=128.51*10⁵m⁴

B₀=b_red-b_p=1100-290.32=809.68 MPa

Poliyanol armature tutamlarida

B₀=1100-306.99=793.01 MPa

N₀=906.57*10^-10(4*4.71+3*4.17*0.996)=2983.84 kN

e_o=33.48 sm

b_bnm=8.89 MPa

b₂=170*8.89/28=5396 MPa

Xulosa

Ushbu kurs loyha ishini qilib chiqish mobaynida ko’p narsalarni o’rganib chiqdim loyhalanayotgan joyni muhiti iqlimi o’rganib chiqish kerak ekan bu juda katta ahamiyatga ega ekan men shu kurs ishini qilish davomida o’zimga kerakli ko’p narsalarni o’rganib oldimm va ushbu kurs loyha ishi qilib chiqish menga juda yoqdi o’zim hohlab bajardim menga buni bajarish yoqdi va o’zim juda qiziqib qoldim xulosa qilib shuni qaytarishim mumkinki “ Qurilish mexanikasi va konsturuksiyalari “ fanidan menga berilgan variyant asosida Avtomobil yo’l toifam III toifali edi 21-metrli va o’qlar asasida masofam 115 sm vaqtinchalik yuklar esa A-11 ga xisoblab toptim ko’ndalang kesimlarini va armaturalar chizib juda ko’p bilim va ko’nikmalar ega bo’ldim bu kurs loyha ishini konsturiksiyaning og’irligi va unga tushadigan doimiy va vaqtinchalik yuklarni aniqladim va bilim va ko’nikmalarga ega bo’ldim.