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MỤC LỤC MATLAB
4. Nụng Văn Vỡn (2008), Động lực học chuyển động mỏy kộo – ụ tụ.
MỤC LỤC MATLAB
MỤC LỤC MATLAB
%DAC TINH DONG CO Dieden %============================= clc
clear all
%=== DONG CO MK xich (De tai thay Que) ======================= NeH= 30; % ml - Cong suat danh nghia cua dong co
neH= 2000 ; % v/ph - So vong quay danh nghia
MeH= NeH*0.735*1000/(neH*pi/30)/9.81; % kGm - Mo men quay danh ngh?a nemax= 1.1*neH; %v/ph - So vong quay cuc dai
Memax= 1.25*MeH ; %kGm - Mo men cuc dai
neM= 800 ; % v/ph - So vong quay khi mo men dat cuc dai Memax nemin=400;
%---%== NHANH TU DIEU CHINH ===== %== NHANH TU DIEU CHINH =====
% Nhanh tu dieu chinh: Me1=(nemax-ne)*tgB; neH < ne <= nemax
% Nhanh qua tai : Me2=MeH*(0.875 + 0.97*(ne/neH)-0.85*(ne/neH).^2); % ( neM< ne<=neH)
tgB= MeH/(nemax-neH); % Goc doc cua nhanh tu dieu chinh a1= -tgB; b1= nemax*tgB; % Nhanh tu dieu chinh
%Me1= a1*ne +b1; ne1= neH:nemax; Me1= a1*ne1 + b1;
Ne1= Me1*9.81.*ne1*pi/30/1000*1.36 ; %ma luc %=== NHANH QUA TAI ===========
% Me2= MeH*(a + b*ne/neH + c*(ne/neH)^2) xmin=neM/neH;
A=[1 1 1
0 1 2*xmin
1 xmin xmin^2 ];B=[1 ; 0 ; Memax/MeH]; B=[1 ; 0 ; Memax/MeH];
abc= A\B ; % Cac he so ham Momen M[kGm]; ne[v/ph]
a2=abc(1); b2=abc(2); c2=abc(3); % kQ: a= 0.9279; b= 0.7927; c= -0.7206 %---
ne2= nemin:neH;
Me2=MeH*(a2 + b2*ne2/neH + c2*(ne2/neH).^2); Ne2= Me2*9.81.*ne2*pi/30/1000*1.36 ; %ma luc Mmax=max(Me2);
Nmax=max(Ne2);
%================== DUONG DAC TINH TOC DO ======================figure(1), hold on figure(1), hold on
plot(ne1,Me1,'r','linewidth',1.5) plot(ne2,Me2,'r','linewidth',1.5) plot(ne1,Ne1,'linewidth',1.5) plot(ne2,Ne2,'linewidth',1.5) grid on, box on
axis([1200 2400 0 35])
title('DUONG DAC TINH TOC DO DONG CO Diezel lap tren MK ??') xlabel('toc do quay ne[v/ph]')
ylabel('Me[kGm], Ne[maluc]') text(1220,14,'Me')
text(1220,24,'Ne')
%================== DUONG DAC TINH TAI TRONG ======================m=length(Me1); m=length(Me1); for i=1:m M1(i)=Me1(1+m-i); n1(i)=ne1(1+m-i); end i=find(Me2==max(Mmax)); n2=length(Me2);
Me22=Me2(i:n2)'; ne22=ne2(i:n2); Ne22=Ne2(i:n2); n=length(Me22); (adsbygoogle = window.adsbygoogle || []).push({});
for i=1:n
M2(i+1)=Me22(1+n-i); n2(i+1)=ne22(1+n-i);
%N2= M2*9.81.*n2*pi/30/1000*1.36 ; %ma luc end
M2(1)=M1(end); n2(1)=n1(end);
A=polyfit(M2,n2,2); a2=A(1); a1=A(2);a0=A(3); %--- M1=0:M2(1); b1=-(nemax-n2(1))/M2(1); b0=nemax; b1=1.3*b1; %--- A=[a2 (a1-b1) (a0-b0)]; x=roots(A);
MeH= 10.825; M1=0:0.01:MeH; n1=b1*M1+b0;
N1=M1*9.81.*n1*pi/30/1000*1.36 ; %ma luc Memax=max(M2);
M2=M1(end):0.01: Memax; n2=a2*M2.^2+a1*M2+a0;
N2=M2*9.81.*n2*pi/30/1000*1.36 ; %ma luc
%---figure(2), hold on figure(2), hold on
dne=500; dNe=10; k=dne/dNe;
plot(M1,n1,'r',M2,n2,'r','linewidth',1.5) plot(M1,N1*k,'b',M2,N2*k,'b','linewidth',1.5) grid on, box on
axis([ 0 1.2*Memax 0 2400 ]) for Ne=0:dNe:NeH
ch= num2str(Ne); text(16.5,Ne*k,ch); end
text(16.2,35*k,'Ne[ml]')
title('DUONG DAC TINH TAI TRONG DONG CO Diezel lap tren MK ??') xlabel('Momen Me [kGm]')
ylabel('ne[v/ph]') text(5,2200,'ne') text(5,18*k,'Ne')
%======================= IN CAC HE SO HOI QUI ==============%clc %clc
fprintf('CAC HE SO HOI QUI cho DUONG DAC TINH TAI TRONG \n') fprintf('Momen quay danh nghia MeH= %6.3f kGm\n',MeH ) fprintf('Momen quay cuc dai Memax= %6.3f kGm\n',Memax ) fprintf('-Nhanh tu dieu chinh: ne = a1*Me + b1 khi Me<= MeH\n') fprintf(' a1= %9.4f ;b1= %8.4f \n',b1,b0)
%fprintf(' a1= %6.4f \n',b1) %fprintf(' b1= %6.4f \n',b0)
fprintf('-Nhanh qua tai: ne = a2*Me^2+ b2*Me + c2 khi Me>MeH\n')
fprintf(' a2= %9.4f ;b2= %8.4f ;c2= %6.4f \n',a2,a1,a0) %fprintf(' a2= %6.4f \n',a2)
%fprintf(' b2= %6.4f \n',a1) %fprintf(' c2= %6.4f \n',a0)
% =================================================% XAY DUNG DUONG DAC TINH KEO LY THUYET CUA MAY KEO XICH % XAY DUNG DUONG DAC TINH KEO LY THUYET CUA MAY KEO XICH %================================================== % Sinh vien thuc hien:
%---clc clc
clear all
% Loai may keo: MAY KEO XICH CAO SU (De tai Thay Que) %======== NHAP CAC SO LIEU BAN DAU =========================== G= 820; % kG - trong luong mk
r=0.22; % m - ban kinh banh chu dong f=0.08; % he so can lan
Pf=f*G; % kG - luc can lan fi=1; % He so bam;
Pfi=fi*G; % kG - Luc bam (adsbygoogle = window.adsbygoogle || []).push({});
% Cac he so thuc nghiem tinh do truot D=Aln(B/(C-Pm))[%] A=8.22; C=Pfi;B=C;
%--- DONG CO ---
MeHne= 10.825; % kGm - Mo men quay danh nghia cho ne Memax= 13.415; % kGm - Mo men quay cuc dai
%CAC HE SO HOI QUI CUA HAM Mome
%Ham momen Nhanh tu dieu chinh: Me1= a1*ne1 +b1
%Ham momen Nhanh qua tai: Me2= a2*ne2^2 +b2*ne2 + c2 a1= -24.2266; b1= 2200.0000 ;
a2= -136.3629; b2= 2925.3759 ; c2= -13763.1558 ; %--- Ti so truyen ---