單軌抓斗起重機設(shè)計（3噸）【含CAD圖紙、說明書】

程序本程序的編制是為繪制起升機構(gòu)減速箱第 I 軸、第 II 軸彎矩圖及校核第 I 軸危險截面強度。依據(jù)軸受力分析，每一軸在 XOZ 平面，YOZ 平面均受力作用，故每一軸在這兩平面內(nèi)都可以畫出彎矩圖。同時，第 I 軸有三個危險截面，在程序中對這三個截面一一校核。編制程序時，為了避免過多的重復(fù)性，只較詳細(xì)的寫出第 I 軸在 YOZ 平面中的彎矩圖及第I 軸第 I 截面的校核過程，其余均簡略寫出。程序如下：10 PRINT “ji suan di yi zhou suo shou zai he”20 PRINT30 SM=15.31*10^3:YJ=.9940 DJ=300:DS=15:D0=DS+DJ:M4=SM*d0/(2*YJ)50 I3=4.09:YC=.965:M3=M4/I3/YC60 I2=3.67:M2=M3/I2/YC70 I1=5.67:M1=M2/I1/YC80 MM=2.5:ZZ=12:BB=8.100001/180*3.1416:D=MM*ZZ:D1=D/SQR(1-SIN(BB)^2):P1=2*M1/D190 AA=20/180*3.1416:Q1=P1*TAN(AA)/SQR(1-SIN(BB)^2):P1=2*M1/D1100 N1=P1*TAN(BB)110 PRINT “d0,d,d1,m1,m2,m3,m4,p1,q1,n1,sm=”; D0,D,D1,M1,M2,M3,M4,P1,Q1,N1,SM120 PRINT130 PRINT “hui chu di yi zhou wan ju tu”140 PRINT150 XX=2*(34+418)-310.35:X1=34+418160 AY=418*P1/X1:BY=P1-AY170 FOR I=0 T0 10000180 NEXT I190 SCREEN 2:CLS:KEY ON200 PSET(80,80)210 FOR X=0 T0 X1 STEP 2220 IF X34 THEN 260230 DEF FNA(X)=AY*X240 Y0=FNA(X)/1000:X0=X250 GOTO 280260 DEF FNB(X)=AY*X-P1*(X-34)270 Y0=FNB(X)/1000:X0=X280 LINE-(80+X0,80+Y0)290 NEXT X300 LINE(80,80)-(80+XX,80)310 AX=-(Q1*418-N1*D1/2)/X1:BX=-(Q1+AX)320 FOR X=0 T0 X1 STEP 2330 IF X34 THEN 370340 DEF FNC(X)=AX*X350 Y0=FNC(X)/1000:X0=X360 GOTO 390370 DEF FND(X)=AX*X+Q1*(X-34)-N1*D1/2380 Y0=FND(X)/1000:X0=X390 LINE-(80+X0,80+Y0)400 NEXT X410 PRINT “ay,by,ax,bx=”;AY,BY,AX,BX420 FOR I=1 T0 10000430 NEXT I440 CLS450 PRINT “di yi zhou wei xian jie mian jiao he”460 PRINT470 PRINT “di yi jie mian”480 MN=0:M1X=AY*21/2:M1Y= AX*21/2:M11=SQR(M1X^2+M1Y^2)490 W1=.1*25^3:B1M=M11/W1:N0=2.7500 F1=52.25*9.8:KB=2.22:EB=.8:N01=BF1/(KB/EB*B1M)510 IF N01=N0 THEN 540520 PRINT “gai jie mian bu he ge”530 GOTO 560540 PRINT “gai jie mian he ge”550 PRINT560 PRINT “di er jie mian”570 KB2=1:EB2=.77580 M2X=AX*34+N1*D1:M2Y= AY*34:M22=SQR(M2X^2+M2Y^2)590 W2=.1*25.95 3:B1M=M22/W2:600 BF2=BF1*EB2610 TF2=25.3*9.8:KT=1:ET=.77:FT=.1620 T2M=M22/2/W2；NT2=TF2/（KT/ET+FT）/T2M：NB2=BF2/B2M630 N02=NB2*NT2/SQR（NB2^2+NT2^2）640 IF N02=N0 THEN 680650 PRINT “gai jie mian bu he ge”660 PRINT670 GOTO 700680 PRINT “gai jie mian he ge”690 PRINT700 PRINT “di san jie mian”710 KB3=1:ET3=.8:RT3=.1720 W3=.2*23^3:T3M=M1/W3:730 N03=2*TR1/(KT3/ET3+RT3)/T3M740 IF N03=N0 THEN 780750 PRINT “gai jie mian bu he ge”760 PRINT770 GOTO 810780 PRINT “gai jie mian he ge”790 PRINT800 PRINT810 PRINT “di er zhou wan ju tu ”820 PRINT830 P2=P1:N2=N1:Q2=Q1840 D3=48.48:P3=2*M2/D3850 A0=20/180*3.1416:Q3=2*M2/D3*TAN(A0)/SQR(1-SIN(BB)^2):Q4=Q3860 N3=P3*TAN(BB):N4=N3870 Y0=(17+41/60+1/3600)/180*3.1416:P2X=P2*SQR(1-SIN(Y0)^2):Q2X=Q2*SIN(Y0)880 D2=171.72:D21=D2/2*SIN(Y0):P2Y=P2*SIN(Y0):Q2Y=Q2*P2X/P2890 YI=34.5:Y2=46:Y3=38.5:YY=Y1+Y2+Y3900 CX=(P2X*(Y2+Y3)+N2*D21+Q3*Y3+N3*D3/2-Q2X*(Y2+Y3))/YY910 DX=P2X+Q3-CX-Q2X920 PRINT “CX,DX,YY= ”; CX,DX,YY930 FOR I=1 T0 10000940 NEXT I950 SCREEN 2:CLS960 PSET(80,80)970 FOR Y=0 T0 YY STEP .5980 IF YY1 THEN 1020990 DEF FNP(Y)=-CX*Y1000 Y0=FNP(Y)/2000:X0=2*Y1010 GOTO 10801020 IF YY1+Y2 THEN 10601030 DEF FNQ(Y)=-CX*Y+(P2X-Q2X)*(Y-Y1)+ N2*D21/21040 Y0=FNQ(Y)/2000:X0=Y*21050 GOTO 10801060 DEF FNW(Y)=-CX*Y+(P2X-Q2X)*(Y-Y1)+N3*D3/2+Q3*(Y-Y1-Y2)+N2*D21/21070 Y0=FNW(Y)/2000:X0=Y*21080 LINE-(80+X0,80+Y0)1090 NEXT Y1100 LINE (80,80)-( 80+YY*2,80)1110 PSET(80,80)1120 D22=171.72/2*SQR(1-SIN(Y0)^2)1130 CY=(P3*Y3+(P2Y+Q2Y) *(Y2+Y3)-N2*D22/2)/YY1140 DY=P3+P2Y+Q2Y-CY1150 FOR Y=0 T0 YY STEP .51160 IF YY1 THEN 12001170 DEF FNR(Y)=+CY*Y1180 Y0=FNR(Y)/2000:X0=Y*21190 GOTO 12601200 IF YY1+Y2 THEN 12401210 DEF FNE(Y)=+CY*Y-(P2Y-Q2Y)*(Y-Y1)+ N2*D22/21220 Y0=FNE(Y)/2000:X0=Y*21230 GOTO 12601240 DEF FNR(Y)=+CY*Y-(P2Y-Q2Y)*(Y-Y1)+ N2*D22/2-P3*(Y-Y1-Y2)1250 Y0=FNR(Y)/2000:X0=Y*21260 LINE-(80+X0,80+Y0)1270 NEXT Y1280 END