Mavzu: Metall to‘sinlarni loyhalash va hisoblash

A  ^N R_v  _s
A  axd
d  11,4a
(19.10)

R_b-poydevor betonning hisobiy qarshiligi.

20.9. rasm. Ustun asoslari

19.1. Masala. Ustunni hisoblang. Ta’sir etaѐtgan hisobiy kuch N3516,0 kN. Ustunni hisobiy uzunligi 4m ga teng. Po‘lat markasi St 3 ps5.
Echim: Ustunning egiluvchanligini 70 deb belgilab, shartli

egiluvchanligini topamiz

  70
 2,34 . Demak, υ – koeffitsient

qiymatini (6.4) formuladan foydalanib hisoblanadi;

 ^R_y  _
²³⁵ 

  1  ^0,073  5,53 ^   
E
^{ 1 } ^{0,073  5,53} ^2,34
210000
 1  0,0668 3,58  0,761

_{ }  


va talab qilingan yuzani aniqlaymiz:
_{A }N _3516  10 _{ 207sm2}

T .K .
  R_y
  _c
0,761 235  0,95

Ustunning talab qilingan inersiya radiusi quyidagi ifoda orqali aniqlanadi:

ⁱT .K .
_ ^lef

 ⁴⁰⁰  5,71sm
70

Ustunning kesim yuzasini shaklini tanlab, uni asosiy o‘lchamlarini aniqlaymiz:

^hT .K .


^vT .K .
_ ⁱT .K .
_1
 ⁱT .K .
 ₂
 ^5,71  13,3sm
0,43
 ^5,71  23,8sm
0,24

Ustunning alohida elementlardan tashqil topgan qo‘shtavr shakliga ega bo‘lgan yuzani tanlab olamiz va varaqasimon prokatli po‘latdan 300x10 devori uchun 380x20x2 tokchalari uchun tanlab olamiz. Ustunning umumiy yuzasi 182 sm² teng.

Tanlab olgan yuzamizni geometrik tavsiflarini aniqlaymiz:

h³  t
 ^{h  t}  2
30³ 1
 ^{30  2,0}  2

x f f
I  ^{w w}  2b t ^{ w}
f _{ }
^{ 2  38  2,0 } 
 ^{ 2250  38912  41162sm 4}

I _y 
12

f

f
b³  t
12
 2 


w

w
t ³  h
12
2 _
 38³  2,0
12
12

 2 

1³  30

12
 ² 

 18293sm ⁴

i_x  


i _y  
 15,0sm


 10sm

X va U o‘qlari bo‘yicha egiluvchanligini aniqlaymiz:

х
_{ } ^lefx
i_x

y
_{ } ^lefy
i _y
 ⁴⁰⁰  26,7
15,0
 ⁴⁰⁰  40
10

Katta egiluvchanligiga qarab shartli egiluvchanlikni aniqlaymiz

  40
 1,34  2,5 va υ koeffitsientini hisoblab,

 ^R_y  _


²³⁵ 

  1  ^0,073  5,53 ^  
E
^{ 1 } ^{0,073  5,53 }  ^1,34
210000
 1  0,06681,551  0,896

_{ }  
tanlab olgan yuzani mustahkamligini tekshiramiz:
 ^N ^{3516  10}  227,0MPa
_min  A   _c 0,896 182  0,95

farqi
^{Ry  } 100%  ^{235  227,0} 100%  3,4%  5%

R_y 235
17.2. Masala. Ustunni yuk ko‘tarish qobiliyatini aniqlang. Agar ustunni hisobiy uzunligi 5m ga teng bo‘lib, kesim yuzasi ikkita №2O shvellerdan iborat bo‘lsa. Po‘lat markasi St 3 ps5. [ №20

Echimi. Ustunning egiluvchanligini aniqlaymiz, buning uchun kesim yuzasining geometrik tavsifnomalarini aniqlaymiz;



х
I  1520  2  3040sm ⁴
A  23,4  2  46,8sm ²

i_x  
 8,07sm

у

у
I  I ⁰  A  a ²  2  113  23,4  5,53²  2  1657sm ⁴

a  v  z _u
 7,6  2,07  5,53sm

i _y  
 5,95cm

^efx


^efy
_ ^lefx
i_x
 ^lefy
i _y
 500
8,07
_ 500
5,95

 62

 84

Katta egiluvchanligiga qarab shartli egiluvchanligini aniqlaymiz

  84 
 2,81

va quyidagi formuladan foydalanib υ koeffitsientini

qiymatini topamiz;

^R_y 
^R_y  
^R_y  ²
235

  1,47 13
E
 ^ 0,371  27,3 ^    ^ 0,0275  5,53 ^  
E E
 1,47 13 
210000

   

 ^ 0,371  27,3 ^{235 }  2,81  ^ 0,0275  5,53 

^{235 }  2,81²  1,47  0,01455  0,957 

  
_ 210000 _{ }

210000 _
 0,1680  0,666
va yuzani yuk ko‘tarish qobiliyatini aniqlaymiz;
N  R_y  A   _c  23,5  46,8  0,666  0,95  695,8kH