航空发动机强度振动上机作业题3

航空发动机强度振动上机作业题题目三班级：140411姓名：苏雨学号：14041032一：题目要求3-1某转子叶片根部固定，其材料密度ρ=2850kg/m3,弹性模量E=71.54GPa，叶片长0.1m,各截面位置、面积、惯性矩列于下表，试求其前3阶固有静频。二：分析公式如题目所示，已知材料密度，弹性模量，各截面位置、面积、惯性矩,需要求解3阶静频率。求解静频公式如下：1.弹性线归一化2.振型逼近法3.二阶振型和固有频率求解4.三阶振型和固有频率求解三：编程计算程序使用c语言编写，源代码如下：#include<stdio.h>#include<math.h>#include<stdlib.h>intmain(void){floatrou=2850;floatE=71540000000;floatX[11]={0.0,0.01,0.02,0.03,0.04,0.05,0.06,0.07,0.08,0.09,0.10};floatA[11]={0.00017,0.000146,0.000126,0.000109,0.000096,0.000086,0.000077,0.000073,0.00007,0.000068,0.000068};floatI[11]={0.000000000279,0.000000000212,0.000000000157,0.000000000108,0.000000000084,0.000000000061,0.000000000045,0.000000000037,0.000000000032,0.000000000030,0.000000000030};floatAb[10];floatIb[10];inti=0;while(i<=9){ Ab[i]=((A[i+1]+A[i])/2); i=i+1;}i=0;while(i<=9){ Ib[i]=((I[i+1]+I[i])/2); i=i+1;}floatY0js[10]={0,0.1,0.2,0.4,0.5,0.6,0.7,0.8,0.9,1};floatY0sj[10]={0,0,0,0,0,0,0,0,0,0};floatY0xxx[10]={0,0,0,0,0,0,0,0,0,0};floatone[10]={0,0,0,0,0,0,0,0,0,0};floattwo[10]={0,0,0,0,0,0,0,0,0,0};floatthree[10]={0,0,0,0,0,0,0,0,0,0};floatfour[10]={0,0,0,0,0,0,0,0,0,0};floatwucha=10;floatxz=2.58;inta=0;intb=0;intc=0;intd=0;inte=0;intf=0;intg=0;inth=0;intj=0;intk=0;intp=0;intq=0;while(wucha>=0.000001)//给定拟合精确度{ q=0; while(9>=q){ one[q]=0; two[q]=0; three[q]=0; four[q]=0; q=q+1;}//这步是给数组清零，千万不能忘！ a=0;b=9;while(9>=a){ b=a; while(9>=b+1){ one[a]=one[a]+(Ab[b+1]*Y0js[b+1]*0.01); b=b+1; }a=a+1; }//第一重循环c=0;d=0;while(9>=c){ d=c; while(9>=d+1){ two[c]=two[c]+(one[d+1]*0.01); d=d+1;}c=c+1;}//第二重循环e=0;f=0;while(9>=e){ f=0; while(e>=f){ three[e]=three[e]+two[f]*0.01*(1/Ib[f]); f=f+1;}e=e+1;}//第三重循环g=0;h=0;while(9>=g){ h=0; while(g>=h){ four[g]=four[g]+three[h]*0.01; h=h+1;}g=g+1;}//第四重循环k=0;while(9>=k){ Y0sj[k]=four[k]/four[9]; Y0xxx[k]=four[k];k=k+1;}//求出实际y0，以便和假设yo对比迭代wucha=0;j=0;while(j<=9){ wucha=wucha+fabs(Y0js[j]-Y0sj[j]); j=j+1;}//假设的y0与求出y0之间的误差p=0;while(9>=p){ Y0js[p]=Y0sj[p];p=p+1;}//令实际值等于假设值，再次迭代运算}floatomega=0;omega=5000*sqrt((1/four[9]));printf("一阶固有静频为：%.5fHZ\n",omega/(2*3.1415926));floatY2js[10]={0,-0.1,-0.3,-0.5,-0.4,-0.2,-0.1,0.5,0.8,1};floatY2[10]={0,0,0,0,0,0,0,0,0,0};floatxiuzhen=4;floatY2sj[10]={0,0,0,0,0,0,0,0,0,0};floatb11=0;floatC21=0;floata21=0;a21=10;while(fabs(a21)>=0.00000001)//给定拟合精确度{ b11=0; C21=0; a21=0; i=0;while(i<=9){ b11=b11+2850*Ab[i]*Y0xxx[i]*Y0xxx[i]*0.01; i++;}i=0;while(i<=9){ C21=C21+2850*Ab[i]*Y0xxx[i]*Y2js[i]*0.01; i++;}a21=C21/b11;i=0;while(i<=9){ Y2[i]=Y2js[i]-a21*Y0xxx[i]; i++;}p=0;while(9>=p){ Y2js[p]=Y2[p];p=p+1;}//令实际值等于假设值，再次迭代运算}i=0;while(i<=9){Y2sj[i]=Y2[i]/Y2[9];i++; }//归一化q=0; while(9>=q){ one[q]=0; two[q]=0; three[q]=0; four[q]=0; q=q+1;}//这步是给数组清零，千万不能忘！a=0;b=0;while(9>=a){ b=a; while(9>=b+2){ one[a]=one[a]+(Ab[b+2]*Y2sj[b+2]*0.01); b=b+1;}a=a+1;}//第一重循环c=0;d=0;while(9>=c){ d=c; while(9>=d+1){ two[c]=two[c]+(one[d+1]*0.01); d=d+1;}c=c+1;}//第二重循环e=0;f=0;while(9>=e){ f=0; while(e>=f){ three[e]=three[e]+two[f]*0.01*(1/Ib[f]); f=f+1;}e=e+1;}//第三重循环g=0;h=0;while(9>=g){ h=0; while(g>=h){ four[g]=four[g]+three[h]*0.01; h=h+1;}g=g+1;}//第四重循环omega=sqrt((E/rou)*(1/four[9]));omega=omega*xz;printf("二阶固有静频为：%.5fHZ\n",omega/(2*3.1415926));floatY3js[10]={0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1};floatY3[10]={0,0,0,0,0,0,0,0,0,0};floatY3sj[10]={0,0,0,0,0,0,0,0,0,0};b11=0;floatb12=0;floatb21=0;floatb22=0;floatC31=0;floatC32=0;floata31=0;floata32=0;a32=10;floatshoulian=10;while(shoulian>=0.0001)//给定拟合精确度{ while(i<=9){ Y3js[i]=Y3[i]; i++;} b12=0;b21=0;b22=0;C31=0;C32=0;a31=0;a32=0;i=0;while(i<=9){ b11=b11+2850*Ab[i]*Y0xxx[i]*Y0xxx[i]*0.01; i++;}i=0;while(i<=9){ b12=b12+2850*Ab[i]*Y0xxx[i]*Y2[i]*0.01; i++;}i=0;while(i<=9){ b21=b21+2850*Ab[i]*Y0xxx[i]*Y2[i]*0.01; i++;}i=0;while(i<=9){ b22=b22+2850*Ab[i]*Y2[b]*Y2[i]*0.01; i++;}i=0;while(i<=9){ C31=C31+2850*Ab[i]*Y0xxx[i]*Y3js[i]*0.01; i++;}i=0;while(i<=9){ C32=C32+2850*Ab[i]*Y2[i]*Y3js[i]*0.01; i++;}a31=C31/b11;a32=C32/b22;i=0;while(i<=9){ Y3[i]=Y3js[i]-a31*Y0xxx[i]-a32*Y2[i]; i++;}shoulian=fabs(Y3js[0]-Y3[0]);}q=0; while(9>=q){ one[q]=0; two[q]=0; three[q]=0; four[q]=0; q=q+1;}//这步是给数组清零，千万不能忘！a=0;b=0;while(9>=a){ b=a; while(9>=b+2){ one[a]=one[a]+(Ab[b+2]*Y3[b+2]*0.01); b=b+1;}a=a+1;}//第一重循环c=0;d=0;while(9>=c){ d=c; while(9>=d+1){ two[c]=two[c]+(one[d+1]*0.01); d=d+1;}c=c+1;}//第二重循环e=0;f=0;while(9>=e){ f=0; while(e>=f){ three[e]=three[e]+two[f]*0.01*(1/Ib[f]); f=f+1;}e=e+1;}//第三重循环g=0;h=0;while(9>=g)