转子动力学作业(传递矩阵法)(共8页).doc

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1、精选优质文档-倾情为你奉上%求解转子系统前三个临界转速和主振型的传递矩阵法clcclear%等截面轴参数l1=0.12;d=0.04;A=pi*d*d/4;%轮盘参数D=0.5;h=0.025;%盘轴材料参数（忽略轴的质量）a=1;u=0.3;rou=7800;E=2.0e11;G=E/(2*(1+u);I=pi*(d4)/64;K1=2.0e7;v1=6*E*I/(a*G*A*l1*l1);mi=rou*pi*D2/4*h;%轮盘的集质量Jp=mi*D2/8; Jd=Jp/2;Ji=Jp-Jd;%参数的数组形式L=l1 l1 l1 l1 l1 l1 l1 l1 l1 l1 l1 l1 l1

2、0 0;M=0 mi mi mi mi mi mi 0 0 0 0 0 mi mi 0;K=K1 0 0 0 0 0 0 K1 0 0 0 K1 0 0 0;v=v1 v1 v1 v1 v1 v1 v1 v1 v1 v1 v1 v1 v1 0 0;J=0 Ji Ji Ji Ji Ji Ji 0 0 0 0 0 Ji Ji 0; k=0;Tit=第一阶频率的振型和弯矩图;第二阶频率的振型和弯矩图;第三阶频率的振型和弯矩图;for w=0:0.01:4000;for i=1:15;T(:,:,i)=1+(L(i)3)*(1-v(i)*(M(i)*w2-K(i)/(6*E*I) L(i)+L(i)2

3、*J(i)*w2/(2*E*I) L(i)2/(2*E*I) L(i)3*(1-v(i)/(6*E*I); (L(i)2)*(M(i)*w2-K(i)/(2*E*I) 1+L(i)*J(i)*w2/(E*I) L(i)/(E*I) L(i)2/(2*E*I); L(i)*(M(i)*w2-K(i) J(i)*w2 1 L(i); M(i)*w2-K(i) 0 0 1;endH=T(:,:,1);for i2=2:15; H=T(:,:,i2)*H;end F=H(3,1)*H(4,2)-H(3,2)*H(4,1); if F*(-1)k 0 %求解临界转速 k=k+1; wi(k)=w; w=

4、wi(k) ni(k)=wi(k)*30/pi;endendfor i1=1:3; w=wi(i1); for j=1:14;T(:,:,j)=1+(L(j)3)*(1-v(j)*(M(j)*w2-K(j)/(6*E*I) L(j)+L(j)2*J(j)*w2/(2*E*I) L(j)2/(2*E*I) L(j)3*(1-v(j)/(6*E*I); (L(j)2)*(M(j)*w2-K(j)/(2*E*I) 1+L(j)*J(j)*w2/(E*I) L(j)/(E*I) L(j)2/(2*E*I); L(j)*(M(j)*w2-K(j) J(j)*w2 1 L(j); M(j)*w2-K(j)

5、 0 0 1;endH=T(:,:,1);for j=2:15; H=T(:,:,j)*H; end b=-H(4,1)/H(4,2); X(:,1)=(1 b 0 0); for n=2:16; X(:,n)=T(:,:,n-1)*X(:,n-1); %相邻两质点右边的传递关系 end for j1=1:15; y(j1)=X(1,j1); z(j1)=X(3,j1); x(j1)=(j1-1)*l1; end y(16)=X(1,16);x(16)=1.56;z(16)=X(3,16); y=y/max(abs(y);%归一化z=z/max(abs(z);subplot(3,1,i1)pl

6、ot(x,y,b-,x,z,r:)title(Tit(i1,:) xlabel(轴长),ylabel(不平衡值)axis(0,1.56,-1.2,1.2)grid onz;endlegend(振型,弯矩) niwini = 1.0e+004 * 0.1468 0.2065 0.5254 1.3837 2.3759 2.3832 3.1036 3.5473wi = 1.0e+003 *0.1537 0.2162 0.5502 1.4490 2.4881 2.4956 3.2501 3.7147%转子系统的不平衡响应clcclearww=153.68 216.23 550.22 1449 2488

7、.1 2495.6 3250.1 3714.7 %前8阶固有频率n=ww*30/pi %前8阶转频wi=0.9*ww(1) (ww(1)+ww(2)/2 %0.9w(1)和（w(1)+w(2)/2）%等截面轴参数l1=0.12;d=0.04;A=pi*d*d/4;%轮盘参数D=0.5;h=0.025;%盘轴材料参数（忽略轴的质量）rou=7800;E=2.0e11;I=pi*(d4)/64;K1=2.0e7;m=rou*pi*D2/4*h;%轮盘的集质量Jp=m*D2/8; Jd=Jp/2;J1=Jp-Jd;u1=0.8e-4;%参数的数组形式L=l1 l1 l1 l1 l1 l1 l1 l1

8、 l1 l1 l1 l1 l1 l1;M=0 m m m m m m 0 0 0 0 0 m m;K=K1 0 0 0 0 0 0 K1 0 0 0 K1 0 0;J=0 J1 J1 J1 J1 J1 J1 0 0 0 0 0 J1 J1;Tit=wi(1)时的振动响应图;wi(2)时的振动响应图;U=0 0 0 0 0 0 0 0 0 0 0 0 0 u1; for i=1:2 w=wi(i); n(i)=w*30/pi for j=1:14;T(:,:,j)=1+(L(j)3)*(M(j)*w2-K(j)/(6*E*I) L(j)+L(j)2*J(j)*w2/(2*E*I) L(j)2/(

9、2*E*I) L(j)3/(6*E*I) L(j)3/(6*E*I)*U(j)*w2; (L(j)2)*(M(j)*w2-K(j)/(2*E*I) 1+L(j)*J(j)*w2/(E*I) L(j)/(E*I) L(j)2/(2*E*I) L(j)2/(2*E*I)*U(j)*w2; L(j)*(M(j)*w2-K(j) J(j)*w2 1 L(j) L(j)*U(j)*w2; M(j)*w2-K(j) 0 0 1 U(j)*w2;0 0 0 0 1; endG=T(:,:,1);for j1=2:14; H=T(:,:,j1)*G; G=H;endD1=H(3 4,1 2);B=H(3 4,

10、5 2);B(:,1)=-B(:,1);C=H(3 4,1 5);C(:,2)=-C(:,2);b=det(B)/det(D1); c=det(C)/det(D1); X(:,1)=(b c 0 0 1); for n=2:14; X(:,n)=T(:,:,n-1)*X(:,n-1); %相邻两质点右边的传递关系 end y(1)=X(1,1); x(1)=0; for j2=2:13; y(j2)=X(1,j2); x(j2)=x(j2-1)+L(j2-1); end y(14)=X(1,14);x(14)=1.56; xi=0:0.05:1.56; yi=interp1(x,y,xi,sp

11、line); subplot(2,1,i) plot(xi,yi, b-o,LineWidth,1.5)title(Tit(i,:) xlabel(轴长),ylabel(不平衡值) grid on endww = 1.0e+003 * 0.1537 0.2162 0.5502 1.4490 2.4881 2.4956 3.2501 3.7147n = 1.0e+004 * 0.1468 0.2065 0.5254 1.3837 2.3760 2.3831 3.1036 3.5473wi = 138.3120 184.9550n = 1.0e+003 * 0.0150 1.7662专心-专注-专业