机航空发动机强度振动上机作业题1

航空发动机强度振动上机作业题题目一班级：140411姓名：苏雨学号：14041032一：题目要求1-1某级涡轮转子的转速为4700r/min,共有68片转子叶片，叶片材料GH33的密度ρ为8.2×3103/kgm，气流参数沿叶高均布，平均半径处叶栅进、出口的气流参数，叶高各截面的重心位置（X，Y,Z),截面面积A，主惯性矩,II以及轴与x轴的夹角α，弯曲应力最大的A，B，C三点的坐标,,,,,AABBCC列于下表，试求叶片各截面上的离心拉伸应力、气动力弯矩、离心力弯矩、合成弯矩及A，B，C三点的弯曲应力和总应力。二：分析公式如题目所示，已知叶片各截面参数，需要求叶片各截面上的离心拉伸应力、气动力弯矩、离心力弯矩、合成弯矩及A，B，C三点的弯曲应力和总应力。先列出所要求量的相关公式。1.离心拉伸应力2.气动力弯矩3.离心力弯矩4.合成弯矩5.弯曲应力6.总应力三：编程计算程序使用c语言编写，源代码如下：#includestdio.h#includemath.h#includestdlib.hintmain(void){floatX[6]={0.53,0.41,0.41,0.40,0.24,0.12};floatY[6]={-0.41,-0.38,-0.30,-0.19,-0.11,-0.02};floatZ[6]={62.8,59.1,56.0,53.0,49.4,45.8};floatA[6]={1.80,2.32,3.12,4.10,5.48,7.05};floatIk[6]={0,0.242,0.304,0.484,0.939,1.802};floatIt[6]={0,6.694,9.332,12.52,17.57,23.74};floatkA[6]={0,-2.685,-2.847,-2.938,-2.899,-2.894};floattA[6]={0,0.797,0.951,1.094,1.232,1.319};floatkB[6]={0,-0.084,-0.205,-0.303,-0.219,-0.302};floattB[6]={0,-0.481,-0.521,-0.655,-0.749,-1.015};floatkC[6]={0,3.728,3.909,4.060,4.366,4.597};floattC[6]={0,0.773,0.824,0.840,1.130,1.305};floatalfa[6]={0.55,0.49,0.44,0.39,0.30,0.22};floatLXLS[6];inti=0,j=0;floathe=0;while(i=5){j=i;he=0;while(j=4){he=he+(A[j+1]*Z[j+1]*Z[j+1]/2-A[j]*Z[j]*Z[j]/2);j=j+1;}LXLS[5-i]=(8.2*1000*492.18*492.18*he)/(10000*A[i]);i=i+1;}i=0;while(i=5){printf(截面%i的离心拉伸应力为：%fMPa\n,i,LXLS[i]/1000000);i++;}floatPx=2864.16,Py=-6831.4;floatQDWJx[6],QDWJy[6];i=0;while(i=5){QDWJx[i]=Py*(0.628-Z[i]/100)*(0.628-Z[i]/100)/2;i++;}i=0;while(i=5){QDWJy[i]=Px*(0.628-Z[i]/100)*(0.628-Z[i]/100)/2;i++;}i=0;printf(\n);while(i=5){printf(截面%i的x方向气动力弯矩为：%fN*m,y方向气动力弯矩为：%fN*m\n,i,QDWJx[i],QDWJy[i]);i++;}floatLXWJx[6],LXWJy[6];i=0;while(i=5){LXWJx[i]=LXLS[i]*Y[i]/100*(A[i]/10000);i++;}i=0;while(i=5){LXWJy[i]=LXLS[i]*X[i]/100*(A[i]/10000);i++;}i=0;printf(\n);while(i=5){printf(截面%i的x方向离心力弯矩为：%fN*m,y方向离心力弯矩为：%fN*m\n,i,LXWJx[i],LXWJy[i]);i++;}floatMx[6],My[6];i=0;while(i=5){Mx[i]=QDWJx[i]+LXWJx[i];i++;}i=0;while(i=5){My[i]=QDWJy[i]+LXWJy[i];i++;}i=0;printf(\n);while(i=5){printf(截面%i的x方向合成弯矩为：%fN*m,y方向合成弯矩为：%fN*m\n,i,Mx[i],My[i]);i++;}floatMk[6],Mt[6];i=0;while(i=5){Mk[i]=Mx[i]*cos(alfa[i])+My[i]*sin(alfa[i]);Mt[i]=-Mx[i]*sin(alfa[i])+My[i]*cos(alfa[i]);i++;}floatWQYLa[6],WQYLb[6],WQYLc[6];i=1;while(i=5){WQYLa[i]=(Mk[i]/Ik[i])*tA[i]-(Mt[i]/It[i])*kA[i];WQYLb[i]=(Mk[i]/Ik[i])*tB[i]-(Mt[i]/It[i])*kB[i];WQYLc[i]=(Mk[i]/Ik[i])*tC[i]-(Mt[i]/It[i])*kC[i];i++;}i=1;printf(\n);while(i=5){printf(截面%i的弯曲应力为:\nA点：%fMPa\nB点：%fMPa\nC点：%fMPa\n,i,WQYLa[i],WQYLb[i],WQYLc[i]);printf(\n);i++;}floatCETAa[6],CETAb[6],CETAc[6];i=1;while(i=5){CETAa[i]=WQYLa[i]+LXLS[i]/1000000;CETAb[i]=WQYLb[i]+LXLS[i]/1000000;CETAc[i]=WQYLc[i]+LXLS[i]/1000000;i++;}i=1;printf(\n);while(i=5){printf(截面%i的总应力为:\nA点：%fMPa\nB点：%fMPa\nC点：%fMPa\n,i,CETAa[i],CETAb[i],CETAc[i]);printf(\n);i++;}system(pause);}程序运行结果如下：四：总结与感悟这次编程作业，可以说是四个大作业题目中最简单，最基础的一个。我利用之前学习过的c语言中数组，循环，判断知识成功解决了这个问题。通过这次编程作业，我学会了变截面叶片各截面上的离心拉伸应力、气动力弯矩、离心力弯矩、合成弯矩及特定点的弯曲应力和总应力的求解方法，在编程过程中对c语言的使用也更加熟练，可以说获益匪浅。