Xodjayorova dilxayot abdiaxatovna deformatsiyalanuvchi muhitdagi silindrik qobiqning buralma tebranishlarini sonli tadqiq etish

aylanish inersiyasini hisobga olib yechish”
Program Xodjayarova_aylanish_inersiyasi;
Uses crt;
const Nz=99; Nt=200; jt=200; kt=10;
type
ff = text;
vec1 = array [0..Nz+1] of real;
var i,j,k,n,it : integer;
u1,u2,u3,x,y,ck,dk,fk,mk,nk,szt,s : vec1;
l,d,ta,d2,ta2,t,t0,t1,
al1,al0,bet0,bet1,Ab,Bb,r0,ak,gam,eta,gam1,eta1,Dat,
ksi1,ksi2,ksi3,a1,a2,a3 ,AL : real;
w1,sigma,tn: array[1..100] of real;
fl,fli : text;
procedure prog3(var y : vec1);
begin
ck[0]:=al1/(al0*d-al1);
dk[0]:=d*Ab/al1;
for i:=1 to Nz do
begin
ck[i]:=1/(mk[i]-nk[i]*ck[i-1]);
dk[i]:=fk[i]-nk[i]*ck[i-1]*dk[i-1];
end;
y[Nz+1]:=(Bb*d+bet1*ck[Nz]*dk[Nz])/(bet0*d+bet1*(ck[Nz]+1));
i:=Nz+1;
while i>=1 do begin i:=i-1; y[i]:=ck[i]*(dk[i]-y[i+1]); end;
end;
function f(t,t1:real): real;
var pi: real;
begin
pi:=3.14159;
if t<=t1 then f:=sin(pi*t/t1) else f:=0;
end;
begin
clrscr;
assign(fl,'xodja.dat'); rewrite(fl);
assign(fli,'xodja.dat'); rewrite(fli);
{Boshlang'ich ma'lumotlar}
l:=10; r0:=1; gam:=0.498; eta:=1-gam; Dat:=0;
AL:=0; a1:=0.6; a2:=0.3; a3:=0.1; ksi1:=0.1; ksi2:=0.3; ksi3:=0.6;
{Boshlang'ich hisoblar}
d:=l/(Nz+1); ta:=0.5*d; d2:=d*d; ta2:=ta*ta; t1:=Nt*ta/4;
{Boshlang'ich shartlar}
for i:=0 to Nz+1 do
begin x[i]:=i*d;
u1[i]:=0; u2[i]:=0; u3[i]:=0; s[i]:=0 end;

t:=0; j:=0; k:=0;n:=0;

for it:=1 to Nt do
begin
t:=t+ta;
{Tenglamani yechish}
for i:=0 to Nz+1 do
begin
{Hamma hadlarda ikkiga ajratish}
s[i]:=s[i]+AL*ta*(a1/ksi1*exp(-ta/ksi1)+a2/ksi2*exp(-ta/ksi2)+
a3/ksi3*exp(-ta/ksi3))*(u2[i+1]-2*u2[i]+u2[i-1])*ta2/d2;

ak:=-gam/d2-Dat*(1-2*gam)/(6*d2*ta2);

mk[i]:=(1/ta2+2*gam/d2+Dat*(1-2*gam)/(3*d2*ta2))/ak;
nk[i]:=1;

fk[i]:=(-s[i]+1/ta2*(2*u2[i]-u1[i])+eta/d2*(u1[i+1]-2*u1[i]+u1[i-1])+

Dat*(1-2*eta)/(6*d2*ta2)*(u1[i+1]-2*u1[i]+u1[i-1]))/ak;
end;
al0:=1; al1:=1; bet0:=1; bet1:=1;
Ab:=f(t,t1); Bb:=0;
prog3(u3);
{Chegaraviy shartlar}
u3[0]:=f(t,t1); u3[Nz+1]:=0;
for i:=1 to Nz do
begin szt[i]:=(u3[i+1]-u3[i-1])/(2*d);

end;
for i:=0 to Nz+1 do begin u1[i]:=u2[i]; u2[i]:=u3[i]; end;

{Pechat}
j:=j+1; k:=k+1;
if k=kt then begin k:=0; n:=n+1;tn[n]:=t; w1[n]:=u3[10]; end;
if j=jt then
begin
j:=0;
for i:=0 to Nz+1 do
begin
writeln(x[i]:10:5,' ',u3[i]:10:5);
writeln(fl,x[i]:10:5,' ',u3[i]:10:5);
end;
end;
end;
close(fl);
for i:=1 to n do writeln(fli, tn[i]:10:4,' ',w1[i]:10:5);
close(fli);
end.


3.1-rasm. Sterjen uchidan z bo’yicha i=10 qadamga mos keluvchi kesim nuqtalarining buralma ko’chishlari

3.2-rasm. Sterjen uchidan z bo’yicha i=50 qadamga mos keluvchi kesim nuqtalarining buralma ko’chishlari.

Olingan natijalarni grafik ko’rinishida tasvirlaymiz.

i=25 Klassik
i=50 Klassik

t

U

3.3-rasm

t

U

i=25 Aylanish inersiyasi
i=50 Aylanish inersiyasi

3.4-rasm
Olingan natijalar 3.3, 3.4-chizmalarda buralma ko’chish va vaqtdan bog’liq grafiklar ko’rinishida keltirilgan. Bunda 3.3-cizmada i=25 va i=50 (z bo’yicha) qadamga mos keluvchi kesim nuqtalarining klassik holi uchun ko’chishlari grafiklari tasvirlangan. 3.4-chizmada shu grafiklar i=25, i=50 qadam uchun, va vaqtdan bog’liq aniqlashtirilgan tebranish tenglamalari uchun grafiklar tasvirlangan.
Bu natijalar asosida qurilgan grafiklar asosida quyidagi xulosalarni chiqarish mumkin.

1. 
   Klassik hol uchun olingan natijalar xususiy holda sterjen uchun olingan natijalarga to’liq mos keladi.
   
2. 
   Ko’ndalang kesimlarda aylanish inersiyasining ta’siri tebranish amplitudalariga sezilarli darajada (20% gacha) kamayishiga olib keladi;
   
3. 
   Jismning kesimlaridagi ko’chishlari amplitudalarining kamayishiga olib keladi. (3.4-rasm). Aylanish inersiyasi hisobga olinganda bu ko’rsatkich klassik holdagiga nisbatan 2-3 barobar katta bo’ladi.
   
4. 
   Aylanish inersiyasini ta’siri elastik jism uchun ham qo’zg’alishlarning koordinata bo’yicha ham asta sekin so’nib borishiga sabab bo’ladi.