航空发动机强度振动上机作业题3

1、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=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

2、<stdlib.h>int main(void)float rou=2850;float E=71540000000;float X11=0.0,0.01,0.02,0.03,0.04,0.05,0.06,0.07,0.08,0.09,0.10;float A11=0.00017,0.000146,0.000126,0.000109,0.000096,0.000086,0.000077,0.000073,0.00007,0.000068,0.000068;float I11=0.000000000279,0.000000000212,0.000000000157,0.0000000

3、00108,0.000000000084,0.000000000061,0.000000000045,0.000000000037,0.000000000032,0.000000000030,0.000000000030;float Ab10;float Ib10;int i=0;while(i<=9) Abi=(Ai+1+Ai)/2); i=i+1; i=0;while(i<=9) Ibi=(Ii+1+Ii)/2); i=i+1; float Y0js10=0,0.1,0.2,0.4,0.5,0.6,0.7,0.8,0.9,1;float Y0sj10=0,0,0,0,0,0,0

4、,0,0,0;float Y0xxx10=0,0,0,0,0,0,0,0,0,0;float one10=0,0,0,0,0,0,0,0,0,0;float two10=0,0,0,0,0,0,0,0,0,0;float three10=0,0,0,0,0,0,0,0,0,0;float four10=0,0,0,0,0,0,0,0,0,0;float wucha=10;float xz=2.58; int a=0;int b=0;int c=0;int d=0;int e=0;int f=0;int g=0;int h=0;int j=0;int k=0;int p=0;int q=0;wh

5、ile(wucha>=0.000001) /给定拟合精确度 q=0; while(9>=q) oneq=0; twoq=0; threeq=0; fourq=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) onea=onea+(Abb+1*Y0jsb+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) twoc=twoc+(oned+1*0.01); d=d+1; c=c+1; /第

6、二重循环 e=0; f=0; while(9>=e) f=0; while(e>=f) threee=threee+twof*0.01*(1/Ibf); f=f+1; e=e+1; /第三重循环 g=0; h=0; while(9>=g) h=0; while(g>=h) fourg=fourg+threeh*0.01; h=h+1; g=g+1; /第四重循环 k=0; while(9>=k) Y0sjk=fourk/four9; Y0xxxk=fourk; k=k+1; /求出实际y0，以便和假设yo对比迭代 wucha=0; j=0; while(j<

7、=9) wucha=wucha+fabs(Y0jsj-Y0sjj); j=j+1; /假设的y0与求出y0之间的误差 p=0; while(9>=p) Y0jsp=Y0sjp; p=p+1; /令实际值等于假设值，再次迭代运算 float omega=0;omega=5000*sqrt(1/four9);printf("一阶固有静频为：%.5fHZn",omega/(2*3.1415926);float Y2js10=0,-0.1,-0.3,-0.5,-0.4,-0.2,-0.1,0.5,0.8,1;float Y210=0,0,0,0,0,0,0,0,0,0;flo

8、at xiuzhen=4;float Y2sj10=0,0,0,0,0,0,0,0,0,0;float b11=0; float C21=0;float a21=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*Abi*Y0xxxi*Y0xxxi*0.01;i+; i=0;while(i<=9) C21=C21+2850*Abi*Y0xxxi*Y2jsi*0.01;i+; a21=C21/b11;i=0;while(i<=9)

9、 Y2i=Y2jsi-a21*Y0xxxi;i+; p=0; while(9>=p) Y2jsp=Y2p; p=p+1; /令实际值等于假设值，再次迭代运算 i=0; while (i<=9) Y2sji=Y2i/Y29; i+; /归一化 q=0; while(9>=q) oneq=0; twoq=0; threeq=0; fourq=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) onea=onea+(Abb+2*Y2sjb+2*0.01); b=b+1; a=a+1; /

10、第一重循环 c=0; d=0; while(9>=c) d=c; while(9>=d+1) twoc=twoc+(oned+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) threee=threee+twof*0.01*(1/Ibf); f=f+1; e=e+1; /第三重循环 g=0; h=0; while(9>=g) h=0; while(g>=h) fourg=fourg+threeh*0.01; h=h+1; g=g+1; /第四重循环 omega=sqrt(

11、E/rou)*(1/four9);omega=omega*xz;printf("二阶固有静频为：%.5fHZn",omega/(2*3.1415926);float Y3js10=0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1;float Y310=0,0,0,0,0,0,0,0,0,0;float Y3sj10=0,0,0,0,0,0,0,0,0,0;b11=0;float b12=0; float b21=0;float b22=0;float C31=0;float C32=0;float a31=0;float a32=0;a32=10;

12、float shoulian=10;while(shoulian>=0.0001) /给定拟合精确度 while(i<=9) Y3jsi=Y3i;i+; b12=0; b21=0; b22=0; C31=0; C32=0; a31=0; a32=0; i=0;while(i<=9) b11=b11+2850*Abi*Y0xxxi*Y0xxxi*0.01;i+; i=0;while(i<=9) b12=b12+2850*Abi*Y0xxxi*Y2i*0.01;i+; i=0;while(i<=9) b21=b21+2850*Abi*Y0xxxi*Y2i*0.01;i

13、+; i=0;while(i<=9) b22=b22+2850*Abi*Y2b*Y2i*0.01;i+; i=0;while(i<=9) C31=C31+2850*Abi*Y0xxxi*Y3jsi*0.01;i+; i=0;while(i<=9) C32=C32+2850*Abi*Y2i*Y3jsi*0.01;i+; a31=C31/b11; a32=C32/b22; i=0;while(i<=9) Y3i=Y3jsi-a31*Y0xxxi-a32*Y2i;i+; shoulian=fabs(Y3js0-Y30); q=0; while(9>=q) oneq=0;

14、 twoq=0; threeq=0; fourq=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) onea=onea+(Abb+2*Y3b+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) twoc=twoc+(oned+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) threee=three

15、e+twof*0.01*(1/Ibf); f=f+1; e=e+1; /第三重循环 g=0; h=0; while(9>=g) h=0; while(g>=h) fourg=fourg+threeh*0.01; h=h+1; g=g+1; /第四重循环 omega=sqrt(E/rou)*(1/four9);omega=omega*xz; omega=sqrt(E/rou)*(1/four9);omega=omega*xiuzhen;printf("三阶固有静频为：%.5fHZn",(2*omega)/(2*3.1415926);system("pause");程序运行结果如下：四：总结与感悟这个程序，我个人认为是四个程序里面难度最大的一个。从开始构思程序，到全部编写完毕，前前后后用了一个多月的时间。这一个月，我几乎将所有的课余精力都投入到了这个程序中。经历了无数次的错误，返工，最终得出了合理的解答。在此我要特别感谢蒋向华老师。蒋老师不厌其烦的给我讲解，帮助我分