%CHUONG TRINH TINH LUC TAC DUNG CUA XE Ô TÔ 3 TRUC % CHAY TREN DUONG CO DANG SONG HINH SIN
clear all
global M m v K C l y0 y1 y2 y3 y1c y2c y3c w ct1 ct2 cl1 cl2 kl1 kl2 kt1
kt2 tc2 tc3
% Cac thong so cua oto va duong
l=1.1; % Khoang cach tu banh truoc den trong tam than xe
l1=6; % Khoang cach tu banh truoc den trong tam than xe
l2=4; % Khoang cach tu trong tam khung gia sau den trong tam xe
m1=600; % Khoi luong truc va banh truoc
m2=30000; % Khoi luong truc va banh truoc
j2=40000; % Khoi luong than xe
m3=1800; % Momen quan tinh cua than xe
j3=350; % Khoi luong khung gia sau
kl1=5600000;
kl2=kl1; % Do cung cua lop sau va lop truoc cua truc truoc
kt1=3000000; % Do cung lo xo tren truc sau
kt2=4000000; % Do cung lo xo tren truc sau
cl1=20000;
cl2=cl1; % He so can nhot cua lop sau va lop truoc cua truc truoc
ct1=10000; % He so can nhot cua giam chan cua truc truoc than xe
ct2=30000; % He so can nhot cua giam chan cua truc sau than xe
Ls=6; % Buoc song mat duong
y0=0.01; %bien do song cua mat duong % Cac ma tran
M=[m2 0 0 0 0; 0 j2 0 0 0; 0 0 m1 0 0; 0 0 0 m3 0;
0 0 0 0 j3]; % Ma tran khoi luong % ma tran do cung K=[kt1+kt2 kt1*l1-kt2*l2 -kt1 -kt2 0; kt1*l1-kt2*l2 kt1*l1^2+kt2*l2^2 -kt1*l1 kt2*l2 0; -kt1 -kt1*l1 kt1+kl1 0 0; -kt2 kt2*l2 0 kt2+2*kl2 0; 0 0 0 0 2*kl2*l^2]; % ma tran do nhot C=[ct1+ct2 ct1*l1-ct2*l2 -ct1 -ct2 0; ct1*l1-ct2*l2 ct1*l1^2+ct2*l2^2 -ct1*l1 ct2*l2 0; -ct1 -ct1*l1 ct1+cl1 0 0; -ct2 ct2*l2 0 ct2+2*cl2 0; 0 0 0 0 2*cl2*l^2];
v=60; % van toc xe chay
w=2*pi*v/3.6/Ls; %tan so goc kich thich tu duong %thoi gian tre cua cac banh xe phia sau
tc2=l1+l2-l/(v*1000/3600); %truc banh xe thu 2
x0=[.0 .0 0 0 0 0 0 .0 0 0]'; %dieu kien ban dau
tf=8; %thoi gian cho xe dao dong
[t,x]=ode45('txk0',tf,x0); %Ham ode45 de giai phuong trinh vi phan mot bac
Xt=x(:,1)+l1*x(:,2);
% chuyen vi va van toc cua cac bo phan
Xs=x(:,1)-l2*x(:,2); Xr=x(:,1); Xv=x(:,4)+l*x(:,5); Xn=x(:,4)-l*x(:,5); Xtc=x(:,6)+l1*x(:,7); Xsc=x(:,6)-l2*x(:,7); Xrc= x(:,6); Xvc=x(:,9)+l*x(:,10); Xnc=x(:,9)-l*x(:,10);
% chuyen dong tuong doi
Zt=Xt-x(:,3); Zs=Xs-x(:,4); Zr=Xr-y1; Zv=Xv-y2; Zn=Xn-y3;
% van toc tuong doi
Ztc=Xtc-x(:,8); Zsc=Xsc-x(:,9); Zrc=Xrc-y1; Zvc=Xvc-y2c; Znc=Xnc-y3c;
% luc dong tac dung tu mat duong nen cac banh va cac banh len dau truoc va dau sau Ft=kt1*Zt+ct1*Ztc; Fs=kt2*Zs+ct2*Zsc; Fr=kl1*Zr+cl1*Zrc; Fv=kl2*Zv+cl2*Zvc; Fn=kl2*Zn+cl2*Znc;
%bieu dien cac dao dong cua cac banh xe va cac bo phan cua doan xe
subplot(2,1,1) plot(t,Ft/1000)
title('LUC TAC DUNG TU BANH TRUOC LEN THAN XE-kN') xlabel('Thoi gian-s')
ylabel('Luc-kN') grid
pause
subplot(2,1,1) plot(t,Fs/1000)
title('LUC TAC DUNG TU GIA SAU LEN THAN XE-kN') xlabel('Thoi gian-s')
ylabel('Luc-kN') grid
pause
subplot(2,1,1) plot(t,Fr/1000)
title('LUC TAC DUNG TU MAT DUONG LEN BANH TRUOC-kN') xlabel('Thoi gian-s')
ylabel('Luc-kN') grid
pause
subplot(2,1,1) plot(t,Fv/1000)
title('LUC TAC DUNG TU MAT DUONG LEN BANH 2-kN') xlabel('Thoi gian-s')
ylabel('Luc-kN') grid
pause
subplot(2,1,1) plot(t,Fn/1000)
title('LUC TAC DUNG TU MAT DUONG LEN BANH 3-kN') xlabel('Thoi gian-s')
ylabel('Luc-kN') grid
pause
subplot(2,1,1) plot(t,Fr/1000)
title('LUC TAC DUNG TU MAT DUONG LeN BANH 3-kN') xlabel('Thoi gian-s')
ylabel('Luc-kN') grid
2.2.Tính lực lớn nhất khi vận tốc thay đổi.
%CHUONG TRINH TINH DAO DONG CUA XE Ô TÔ 3 TR?C % CHAY TREN DUONG CO DANG SONG HINH SIN
clear all
global M m n v K C l y0 y1 y2 y3 y1c y2c y3c w ct1 ct2 cl1 cl2 kl1 kl2 kt1
kt2 tc2 tc3
% Cac thong so cua oto va duong
l=1.1; l1=6; l2=4; m1=600; m2=30000; j2=40000; m3=1800; j3=350; kl1=5600000; kl2=kl1; kt1=3000000; kt2=4000000; cl1=20000; cl2=cl1; ct1=10000; ct2=30000; Ls=6;
y0=0.1;%bien do song cua mat duong % Cac ma tran
M=[m2 0 0 0 0; 0 j2 0 0 0; 0 0 m1 0 0; 0 0 0 m3 0;
0 0 0 0 j3]; % Ma tran khoi luong % ma tran do cung K=[kt1+kt2 kt1*l1-kt2*l2 -kt1 -kt2 0; kt1*l1-kt2*l2 kt1*l1^2+kt2*l2^2 -kt1*l1 kt2*l2 0; -kt1 -kt1*l1 kt1+kl1 0 0; -kt2 kt2*l2 0 kt2+2*kl2 0; 0 0 0 0 2*kl2*l^2]; % ma tran do nhot C=[ct1+ct2 ct1*l1-ct2*l2 -ct1 -ct2 0; ct1*l1-ct2*l2 ct1*l1^2+ct2*l2^2 -ct1*l1 ct2*l2 0; -ct1 -ct1*l1 ct1+cl1 0 0; -ct2 ct2*l2 0 ct2+2*cl2 0; 0 0 0 0 2*cl2*l^2];
v=input('van toc toa xe: v= ');
w=2*pi*v/3.6/Ls; %tan so goc kich thich tu duong %thoi gian tre cua cac banh xe phia sau
tc2=l1+l2-l/(v*1000/3600); %truc banh xe thu 2
tc3=l1+l2+l/(v*1000/3600); %truc banh xe thu 3
x0=[.0 .0 0 0 0 0 0 .0 0 0]'; %dieu kien ban dau
tf=8; %thoi gian cho xe dao dong
[t,x]=ode45('txk0',tf,x0); %Ham ode45 de giai phuong trinh vi phan mot bac
Xt=x(:,1)+l1*x(:,2); Xs=x(:,1)-l2*x(:,2);
Xr=x(:,1); Xv=x(:,4)+l*x(:,5); Xn=x(:,4)-l*x(:,5); Xtc=x(:,6)+l1*x(:,7); Xsc=x(:,6)-l2*x(:,7); Xrc= x(:,6); Xvc=x(:,9)+l*x(:,10); Xnc=x(:,9)-l*x(:,10);
% chuyen dong tuong doi
Zt=Xt-x(:,3); Zs=Xs-x(:,4); Zr=Xr-y1; Zv=Xv-y2; Zn=Xn-y3;
% van toc tuong doi
Ztc=Xtc-x(:,8); Zsc=Xsc-x(:,9); Zrc=Xrc-y1; Zvc=Xvc-y2c; Znc=Xnc-y3c;
% luc dong tac dung tu mat duong nen cac banh va cac banh len dau truoc va dau sau Ft=kt1*Zt+ct1*Ztc; Fs=kt2*Zs+ct2*Zsc; Fr=kl1*Zr+cl1*Zrc; Fv=kl2*Zv+cl2*Zvc; Fsc=kl2*Zn+cl2*Znc;
%tinh luc cuc dai
nt=size(Ft);%kich thuoc cua vecto luc tac dung tu banh xe truoc len than
Ftmax=max (Ft(.8*nt:nt,1));%luc cuc dai tac dung tu banh xe truoc len than
ns=size(Fs);%kich thuoc cua vecto luc tac dung tu gia sau len than
Fsmax=max (Fs(.8*ns:ns,1));%luc cuc dai tac dung tu gia sau len than
nr=size(Fr);%kich thuoc cua vecto luc tac dung tu mat duong len banh truoc
Frmax=max (Fr(.8*nr:nr,1));%luc cuc dai tac dung tu mat duong len banh truoc
nv=size(Fv);%kich thuoc cua vecto luc tac dung tu mat duong len banh 2
Fvmax=max (Fv(.8*nv:nv,1));%luc cuc dai tac dung tu mat duong len banh 2
nsc=size(Fsc);%kich thuoc cua vecto luc tac dung tu mat duong len banh 3
Fscmax=max (Fsc(.8*nsc:nsc,1));%luc cuc dai tac dung tu mat duong len banh 3
Sau đây là lệnh thao tác trên cửa sổ Commod Window trên Malab.
%tong hop lai coppy tat 4 ket qua sau vào:
Fmax40 = 1.0e+006 *[0.1594 0.2189 1.0348 0.4906 0.1380] %day la cac luc cuc dai theo thu tu Ftmax,Fsmax,Frmax,Fvmax,Fsc(sau cung tuc Fn cua c day) tai v=40km/h.
Fmax50 = 1.0e+006 *[0.4697 0.7215 1.1174 0.4635 0.2758] %day la cac luc cuc dai theo thu tu Ftmax,Fsmax,Frmax,Fvmax,Fsc(sau cung tuc Fn cua c day) tai v=50km/h
Fmax60 = 1.0e+006 *[0.2156 0.3383 -0.2076 1.0839 -0.1420] %day la cac luc cuc dai theo thu tu Ftmax,Fsmax,Frmax,Fvmax,Fsc(sau cung tuc Fn cua c day) tai v=60km/h
Fmax70 = 1.0e+006 *[0.0586 0.1429 0.3458 1.0852 0.1891] %day la cac luc cuc dai theo thu tu Ftmax,Fsmax,Frmax,Fvmax,Fsc(sau cung tuc Fn cua c day) tai v=70km/h
%luc cuc dai thay doi theo van toc:
Ftmax=[Fmax40(1,1) Fmax50(1,1) Fmax60(1,1) Fmax70(1,1)];%day la ma tran Ftmax tai tat ca van toc40,50,60km/h.
Fsmax=[Fmax40(1,2) Fmax50(1,2) Fmax60(1,2) Fmax70(1,2)];%day la ma tran Fsmax tai tat ca van toc40,50,60km/h
Frmax=[Fmax40(1,3) Fmax50(1,3) Fmax60(1,3) Fmax70(1,3)];%day la ma tran Frmax tai tat ca van toc40,50,60km/h
Fvmax=[Fmax40(1,4) Fmax50(1,4) Fmax60(1,4) Fmax70(1,4)];%day la ma tran Fvmax tai tat ca van toc40,50,60km/h
Fscmax=[Fmax40(1,5) Fmax50(1,5) Fmax60(1,5) Fmax70(1,5)];%day la ma tran Fscmax (Fsau cung hay Fn cua chu day) tai tat ca van toc40,50,60km/h
v=[40 50 60 70];%day la khai bao cac van toc de ve do thi. %sau do nhap (lenh ve do thi (v,Fmax)):
plot(v,Ftmax,'r',v,Fsmax,'g',v,Frmax,'y',v,Fvmax,'c',v,Fscmax,'b')
3.Chương trình trị riêng.
clear all
global M m v K C l y0 y1 y2 y3 y1c y2c y3c w ct1 ct2 cl1 cl2 kl1 kl2 kt1
kt2 tc2 tc3
% Cac thong so cua oto va duong
l=1.1; l1=6; l2=4; m1=600; m2=30000; j2=40000; m3=1800; j3=350; kl1=5600000; kl2=kl1; kt1=3000000; kt2=4000000; cl1=20000; cl2=cl1; ct1=10000; ct2=30000; Ls=6; y0=0.01; % Cac ma tran M=[m2 0 0 0 0; 0 j2 0 0 0; 0 0 m1 0 0; 0 0 0 m3 0;
0 0 0 0 j3]; % Ma tran khoi luong % ma tran do cung K=[kt1+kt2 kt1*l1-kt2*l2 -kt1 -kt2 0; kt1*l1-kt2*l2 kt1*l1^2+kt2*l2^2 -kt1*l1 kt2*l2 0; -kt1 -kt1*l1 kt1+kl1 0 0; -kt2 kt2*l2 0 kt2+2*kl2 0; 0 0 0 0 2*kl2*l^2]; % ma tran do nhot C=[ct1+ct2 ct1*l1-ct2*l2 -ct1 -ct2 0; ct1*l1-ct2*l2 ct1*l1^2+ct2*l2^2 -ct1*l1 ct2*l2 0; -ct1 -ct1*l1 ct1+cl1 0 0; -ct2 ct2*l2 0 ct2+2*cl2 0; 0 0 0 0 2*cl2*l^2];
[X,U]=eig(K,M); % cac tri rieng cua he
w=sqrt(U); %cac tan so dao dong rieng cua he(rad)
f=diag(w)/(2*pi)%cac tan so dao dong rieng cua he(Hz
vch=3.6*f*Ls %cac van toc cong huong cua he
disp('cac vecto rieng') disp(X)
subplot(3,2,1) plot(X(:,1)) title('mode1') subplot(3,2,2)
plot(X(:,2)) title('mode2') subplot(3,2,3) plot(X(:,3)) title('mode3') subplot(3,2,4) plot(X(:,4)) title('mode4') subplot(3,2,5) plot(X(:,5)) title('mode5')
Biểu đồ trị riêng vec to riêng
1 2 3 4 5 0 2 4 6x 10 -3 mode1 1 2 3 4 5 -0.02 -0.01 0 0.01 mode2 1 2 3 4 5 -0.05 0 0.05 mode3 1 2 3 4 5 -0.05 0 0.05 mode4 1 2 3 4 5 0 0.05 0.1 mode5
TÀI LIỆU THAM KHẢO
1-GIÁO TRÌNH : DAO ĐỘNG KĨ THUẬT – GV.Nguyễn Bá Nghị (2011)-Trường Đại Học Giao Thông Vận Tải Hà Nội.
2-BÀI GIẢNG : CẤU TẠO Ô TÔ – GV.Trương Mạnh Hùng (2006)- Trường Đại Học Giao Thông Vận Tải Hà Nội.
3-BÀI GIẢNG : CƠ SỞ VỀ MATLAB – Biên Soạn.Nguyễn Thị Hồng Thúy. 4-CÁC TRANG WED THAM KHẢO - https://vi.wikipedia.org/