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  • 大小: 4KB
    文件類型: .m
    金幣: 1
    下載: 0 次
    發布日期: 2021-06-18
  • 語言: Matlab
  • 標簽: m.m??

資源簡介

四階變步長龍格庫塔matlab程序(齒輪系統動力學),適用于機械動力學

資源截圖

代碼片段和文件信息

clear;
z1=10;
z2=10;
m=5;
d1=m*z1;
d2=m*z2;
R1=d1/2;
R2=d2/2;
S=pi*m/2;
ha=m;
hf=1.25*m
h=2.25*m;
B=0.4*d1;
Alpha=20*pi/180;
Rb1=R1*cos(Alpha);
Rb2=R2*cos(Alpha);
Ra1=R1+ha;
Ra2=R2+ha;
Rf1=R1-hf;
Rf2=R2-hf;
Pb=pi*m*cos(Alpha);
thi0=tan(20*pi/180)-Alpha;
n=100
for?i=1:n
????
????
r1k(i)=Rb1+(i-1)*(Ra1-Rb1)/(n-1);
Ak1(i)=acos(Rb1/r1k(i));
thi_k1(i)=tan(Ak1(i))-Ak1(i);
phi1(i)=S/(2*R1)-(thi_k1(i)-thi0);
x1(i)=r1k(i)*cos(phi1(i));
y1(i)=r1k(i)*sin(phi1(i));
S1(i)=r1k(i)*phi1(i)*2;
r2k(i)=Rb2+(i-1)*(Ra2-Rb2)/(n-1);
Ak2(i)=acos(Rb2/r2k(i));
thi_k2(i)=tan(Ak2(i))-Ak2(i);
phi2(i)=S/(2*R2)-(thi_k2(i)-thi0);
x2(i)=r2k(i)*cos(phi2(i));
y2(i)=r2k(i)*sin(phi2(i));
S2(i)=r2k(i)*phi2(i)*2;
end
figure(‘color‘[1?1?1]);
figure(1)
plot(r1kS1‘r‘)
xlabel(‘輪齒上任意圓周/mm‘)
ylabel(‘齒厚/mm‘)
figure(‘color‘[1?1?1]);
figure(2)
plot(x2y2);hold?on
plot(r2kS2‘r‘)
A_a1=acos(Rb1/Ra1);
A_a2=acos(Rb2/Ra2);
epsilon=(z1*(tan(A_a1)-tan(Alpha))+z2*(tan(A_a2)-tan(Alpha)))/(2*pi);
n1=500;
Tz=60/(z1*n1);
t1=Tz*(epsilon-1);
t2=Tz*(2-epsilon);
P=5500;
T=P/(2*pi*n1/60);
Fn=2*T/(d1*cos(Alpha));
Ft=Fn*cos(Alpha);
m1=2;
ratio=z2/z1;
B1B2=pi*m*epsilon*cos(Alpha);
PN2=Rb2*tan(Alpha);
PN1=Rb1*tan(Alpha);
N2B2=Rb2*tan(A_a2);
N1B1=Rb1*tan(A_a1);
PB2=N2B2-PN2;
PB1=N1B1-PN1;
N1B2=PN1-PB2;
N2B1=PN2-PB1;
CB2=B1B2-Pb;
Rs1=sqrt(Rb1^2+N1B2^2);
Rs2=sqrt(Rb2^2+N2B1^2);
for?i=1:n
????x(i)=(i-1)*B1B2/(n-1);
????if?x(i)????????SN(i)=PB2-x(i);
????else
????????SN(i)=x(i)-PB2;
????end
????um(i)=pi*n1/(30*ratio)*(R2*sin(Alpha)+SN(i)/2*(1-ratio));
????Ru1(i)=N1B2+x(i);
????Ru2(i)=N2B2-x(i);
????Ru(i)=Ru1(i)*?Ru2(i)/(Ru1(i)+?Ru2(i));
end
n2=100;
E=200e9;
I=B*S^3/12;
for?i=1:n
????%??x(i)=(i-1)*B1B2/(n2-1);
?????Rk1(i)=sqrt(Rb1^2+(N1B2+x(i))^2);
?????Rk2(i)=sqrt(Rb2^2+(N2B2+x(i))^2);
?????h1(i)=Rk1(i)-Rf1+0.25*m;
?????h2(i)=Rk2(i)-Rf2+0.25*m;
?????kk10(i)=3*E*I/h1(i)^3;
?????kk20(i)=3*E*I/h2(i)^3;
?????K_t1(i)=kk10(i)*kk20(i)/(kk10(i)+kk20(i));
?????K_t(i)=?K_t1(i);
?????if?x(i)?????????kk=i
?????end
?????K_t2(i-kk+1)=K_t1(i);
?????F_L(i)=Fn;
?????if?x(i)?????????kk1=i;
?????end
end
%?for?i=1:kk
%?????

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