机械原理课后答案第六章作业.ppt

6-1 图示为一钢制圆盘，盘厚b=50mm，位置处有一直径=50mm的通孔，位置处是一质量m2=0.5kg的重块。为了使圆盘平衡，拟在圆盘上r=200mm处制一通孔。试求此通孔的直径与位置。（钢的密度=7.8g/cm3）,解：m1r1+ m2r2+mbrb=0 m1 =2/4 b =0.766kg m1r1 =76.6kgmm，m2r2 =100kgmm； 1）图解法：取W=2kgmm/mm作矢量多边形； 则：mbrb =Wb W=552=110(kgmm) mb = 0.55kg mb =d2/4 b d=42.2mm 量取b = 252.7,2）解析法：m1r1+ m2r2+mbrb=0 Fx=0：m1r1 cos135+ m2r2 cos(210-180)+ (mbrb)x =0 Fy=0：m1r1 sin135+ m2r2 sin(210-180)+ (mbrb)y =0 (mbrb)x = -(76.6 cos135+ 100cos30) = -32.44 (mbrb)y = -(76.6 sin135+ 100sin30) = -104.16 mbrb= (mbrb)x2+ (mbrb)y2 1/2 =109.09(kgmm) mb = mbrb/rb =109.09/200= 0.545(kg) b =arctan(mbrb)y / (mbrb)x= arctan(-104.16) /(-32.44)=252.7 mb =d2/4 b d=42.2mm,6-2 在图示的转子中，已知各偏心质量m1=10kg， m2=15kg， m3= 20kg， m4=10kg，它们的回转半径分别为r1=40cm， r2= r4= 30 cm， r3=20cm，又已知各偏心质量所在的回转平面间的距离l12 = l23 = l34 = 30cm，各偏心质量的方位角如图。若置于平衡基面及中的平衡质量mb及mb的回转半径均为50cm，试求此mb及mb的大小和方位。,解：1）图解法：(m1r1)+(m2r2)+(m3r3)+(m4r4)+(mbrb)=0 (m1r1)=m1r1=400kgcm (m2r2)=m2r2(l23+l34)/l=300kgcm (m3r3)=m3r3 l34/l=133.3kgcm (m4r4)=0 取W=10kgcm/mm作矢量多边形，图b； 则：(mbrb)=Wb W=2810=280(kgcm) mb =5.6kg 量取b = 6,图b,(m1r1)+(m2r2)+(m3r3)+(m4r4)+(mbrb)=0 (m1r1)=0 (m2r2)=m2r2l12/l=150kgcm (m3r3)=m3r3 (l12+l23) /l=266.7kgcm (m4r4)= m4r4 =300kgcm 取W=10kgcm/mm作矢量多边形，图c； 则：(mbrb)=Wb W=3710=370(kgcm) mb=7.4kg 量取b = 145,图c,2）解析法：(m1r1)+(m2r2)+(m3r3)+(m4r4)+(mbrb)=0 (m1r1)=m1r1=400kgcm (m2r2)=m2r2(l23+l34)/l=kgcm (m3r3)=m3r3 l34/l=133.3kgcm (m4r4)=0 Fx=0：400cos120+300cos240+133.3cos300+(mbrb)x =0 Fy=0：400sin120+ 300sin240+133.3sin300+ (mbrb)y =0 (mbrb)x = 283.35 (mbrb)y = 28.84 (mbrb) = (mbrb)x2+ (mbrb)y2 1/2 =284.81(kgmm) mb = (mbrb)/rb=284.81/50= 5.70(kg) b =arctan(mbrb)y / (mbrb)x= arctan(28.84 / 283.35)=5.81,(m1r1)+(m2r2)+(m3r3)+(m4r4)+(mbrb)=0 (m1r1)=0 (m2r2)=m2r2l12/l=150kgcm (m3r3)=m3r3 (l12+l23) /l=266.7kgcm (m4r4)= m4r4 =300kgcm Fx=0：150cos240+266.7cos300+300cos30+(mbrb)x =0 Fy=0：150sin240+ 266.7sin300+300sin30+ (mbrb)y =0 (mbrb) x = -318.16 (mbrb) y = 210.87 (mbrb)= (mbrb)x2+ (mbrb)y2 1/2 =381.70(kgmm) mb =