这个应该算是课堂作业了，利用当量正态法来求解以下问题，求管道的可靠性指标，管材承载能力R服从对数正态分布，环向应力服从正态分布，地面载荷服从极值I型分布。

Contents

§ JC METHOD

§ [original]

§ [the itertation]

§ [R tranfer] transfer R togauss distribution

§ [N2 transfer] transfer N2to gauss distribution

§ [getcos]

§ [loop] get loop variables

JC METHOD

AUTHOR:DUBAOLEI@GMAIL.COMDATE:2014-04-02

functionexer()

%EXER Summary of this function goes here

%   Detailedexplanation goes here

[original]

%Done

clc;

Rmu=309.2;CVr=0.17;Rdelta=52.6;

N1mu=53;CVn1=0.07;N1delta=3.7;

N2mu=70;CVn2=0.29;N2delta=20.3;

%the iteration number

i=1

%first step

i =

    1

[the itertation]

while(i<6)

   ifi==1

       rstar=Rmu;

       rmu2=Rmu;

      n1star=N1mu;

       n1mu2=N1mu;

      n1delta2=N1delta;

      n2star=N2mu;

       n2mu2=N2mu;

      n2delta2=N2delta;

       Cvr=0.17

   end

   %R loop

  rmu2=getRmu2(rstar,rmu2,Cvr)

  rdelta2=getRdelta2(rstar,Cvr)

  Cvr=rdelta2/rmu2

   %N2 loop

  [Fn2,fn2,n2mu2,n2delta2]=getN2(n2mu2,n2delta2,n2star)

   %get the direction cos

   [cosr cosn1cosn2]=gettheta(rdelta2,n1delta2,n2delta2)

   %get the loop results

  beta=getloopbeta(rmu2,n1mu2,n2mu2,rdelta2,n1delta2,n2delta2)

  rstar=getloopRstar(rmu2,rdelta2,beta,cosr)

  n1star=getloopN1star(n1mu2,n1delta2,beta,cosn1)

   n2star=getloopN2star(n2mu2,n2delta2,beta,cosn2)

   %show the result

   disp('__________________________________________________________n')

   %increse the loop variables

   i=i+1

end

Cvr =

   0.1700

end

[R tranfer] transfer R to gauss distribution

%DONE

functionRm=getRmu2(Rst,Rm,Cvr)

%H1 this function get the average number

Rm=Rst*(1+log(Rm/(sqrt(1+Cvr^2)))-log(Rst));

end

functionRdel=getRdelta2(Rst,Cvr)

%H1 this function get the delta

Rdel=Rst*(sqrt(log(1+Cvr^2)));

end

rmu2 =

 304.7954

rdelta2 =

  52.1901

Cvr =

   0.1712

rmu2 =

 249.0042

rdelta2 =

  24.2745

Cvr =

   0.0975

rmu2 =

 239.8020

rdelta2 =

  18.2344

Cvr =

   0.0760

rmu2 =

 238.4973

rdelta2 =

  16.8698

Cvr =

   0.0707

rmu2 =

 237.8386

rdelta2 =

  16.4174

Cvr =

   0.0690

[N2 transfer] transfer N2 to gauss distribution

%DONE

function[Fn2,fn2,N2m,N2del]=getN2(N2m,N2del,N2st)

delta=0.78*N2del;

mu=N2m-0.577*delta;

Fn2=exp(-exp(-(N2st-mu)/delta));

fn2=1/delta*exp(-(N2st-mu)/delta)*Fn2;

z=norminv(Fn2,0,1);

z1=normpdf(z,0,1);

N2del=z1/fn2;

N2m=N2st-z*N2del;

end

Fn2 =

   0.5703

fn2 =

   0.0202

n2mu2 =

  66.5603

n2delta2 =

  19.4162

Fn2 =

   0.8801

fn2 =

   0.0074

n2mu2 =

  57.3282

n2delta2 =

  26.9336

Fn2 =

   0.9848

fn2 =

  7.1587e-04

n2mu2 =

  17.4598

n2delta2 =

  53.3693

Fn2 =

   0.9852

fn2 =

  3.5342e-04

n2mu2 =

 -62.2161

n2delta2 =

 106.0635

Fn2 =

   0.9701

fn2 =

  3.5625e-04

n2mu2 =

-179.5856

n2delta2 =

 190.5970

[getcos]

function[cosRtheta cosN1theta cosN2theta]=gettheta(Rdel,N1del,N2del)

%H1 this function get the direction cos

%first normalized

% syms R N1 N2;

% f=sym('R-N1-N2');

%subs(f,{'R','N1','N2'},{'(R-Rmu)/Rdelta','(N1-N1mu)/Rdelta','(N2-N2mu)/N2delta'});

pd1=1;

pd2=-1;

pd3=-1;

sum1=((pd1*Rdel)^2+(pd2*N1del)^2+(pd3*N2del)^2)^0.5;

%then get the cos

cosRtheta=(-pd1*Rdel)/sum1;

cosN1theta=(-pd2*N1del)/sum1;

cosN2theta=(-pd3*N2del)/sum1;

end

cosr =

  -0.9352

cosn1 =

   0.0663

cosn2 =

   0.3479

cosr =

  -0.6660

cosn1 =

   0.1015

cosn2 =

   0.7390

cosr =

  -0.3226

cosn1 =

   0.0655

cosn2 =

   0.9443

cosr =

  -0.1570

cosn1 =

   0.0344

cosn2 =

   0.9870

cosr =

  -0.0858

cosn1 =

   0.0193

cosn2 =

   0.9961

[loop] get loop variables

%DONE

functionbeta=getloopbeta(Rm,N1m,N2m,Rdel,N1del,N2del)

beta=(Rm-N1m-N2m)/(sqrt(Rdel^2+N1del^2+N2del^2));

end

function Rstar=getloopRstar(Rm,Rdel,Beta,costhetaR)

Rstar=Rm+Rdel*Beta*costhetaR;

end

functionN1star=getloopN1star(N1m,N1del,Beta,costhetaN1)

N1star=N1m+N1del*Beta*costhetaN1;

end

functionN2star=getloopN2star(N2m,N2del,Beta,costhetaN2)

N2star=N2m+N2del*Beta*costhetaN2;

end

beta =

   3.3192

rstar =

 142.7962

n1star =

  53.8142

n2star =

  88.9819

__________________________________________________________n

i =

    2

beta =

   3.8049

rstar =

 187.4883

n1star =

  54.4292

n2star =

 133.0591

__________________________________________________________n

i =

    3

beta =

   2.9962

rstar =

 222.1762

n1star =

  53.7257

n2star =

 168.4505

__________________________________________________________n

i =

    4

beta =

   2.3052

rstar =

 232.3925

n1star =

  53.2937

n2star =

 179.0989

__________________________________________________________n

i =

    5

beta =

   1.9046

rstar =

 235.1556

n1star =

  53.1363

n2star =

 182.0194

__________________________________________________________n

i =

    6

Published with MATLAB® R2013b