string2.m

加强环的应力和位移计算程序,可以求出每隔5度位置处圆环的应力弯矩和位移.

t1=0:5*pi/180:2*pi/9;
f=26238;
r1=5.598;
r2=5.570;
r3=5.764;      %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%    R0=r3
th=0.025;
detaR=0.388;
ex=2.1e11;
u=0.3;
G=ex/2/(1+u);
wz=0.0006273;
I=0.0001217;
A=th*detaR;
%r=A/th/(log(r1/r2));
%S=A*(r3-r)*r3
X1=0.0224*f/3/pi;
Q1=-X1*sin(t1);
N1=-X1*cos(t1);
SHEAR1=Q1/A;
NORMAL11=N1/A;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
ff3=inline('(1-cos(x)).^2','x');
U3=quad(ff3,0,pi,1e-3)/pi;
ff4=inline('(sin(x)-sin(2*pi/9)).*(1-cos(x))','x');
U4=quad(ff4,2*pi/9,pi,1e-3);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
moment1=-0.0224*f*r3*(1-cos(t1))/(3*pi)
stress11=moment1/wz;
t2=2*pi/9:5*pi/180:pi;
moment2=-f*r3*(sin(t2)-sin(2*pi/9))-0.0224*f*r3*(1-cos(t2))/(3*pi)
stress22=moment2/wz;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
N2=f*sin(t2)-X1*cos(t2);
Q2=-f*cos(t2)-X1*sin(t2);
SHEAR2=Q2/A;
NORMAL22=N2/A;
NORMAL1=stress11+NORMAL11
NORMAL2=stress22+NORMAL22
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%ff11=inline('2*26238*(0.0224/3/pi)^2*5.764^3*(1-cos(x)).^2','x');
%U11=quad(ff11,0,2*pi/9,1e-3);
%ff22=inline('(0.0224/3/pi*(1-cos(x))+sin(x)-sin(2*pi/9)).^2*5.764^3*2*26238','x');
%U22=quad(ff22,2*pi/9,pi,1e-3);
%disp=(U11+U22)/2/ex/I
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
ff11=inline('2*26238*(0.0224/3/pi)^2*5.764^3*(1-cos(x)).^2','x');
for i=1:9
U11(i)=quad(ff11,t1(i),2*pi/9,1e-3);
ff22=inline('(0.0224/3/pi*(1-cos(x))+sin(x)-sin(2*pi/9)).^2*5.764^3*2*26238','x');
U22=quad(ff22,2*pi/9,pi,1e-3);
DEFLE1(i)=(U11(i)+U22)/2/ex/I
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for i=1:29
U33(i)=quad(ff22,t2(i),pi,1e-3);
DEFLE2(i)=U33(i)/2/ex/I
end