机械原理课后答案第6章作业.ppt

6-1 图示为一钢制圆盘，盘厚b=50mm，位置Ⅰ处有一直径Φ=50mm的通孔，位置Ⅱ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γ=7.8g/cm3）;2）解析法：m1r1+ m2r2+mbrb=0 ΣFx=0：m1r1 cos135°+ m2r2 cos(210°-180°)+ (mbrb)x =0 ΣFy=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(kg·mm) ∴ mb = mbrb/rb =109.09/200= 0.545(kg) θb =arctan[(mbrb)y / (mbrb)x]= arctan[(-104.16) /(-32.44)]=252.7° mb =πd2/4 · b ·γ ∴ 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Ⅰ及Ⅱ中的平衡质量mbⅠ及mbⅡ的回转半径均为50cm，试求此mbⅠ及mbⅡ的大小和方位。;解：1）图解法：(m1r1)Ⅰ+(m2r2)Ⅰ+(m3r3)Ⅰ+(m4r4)Ⅰ+(mbrb)Ⅰ=0 (m1r1)Ⅰ=m1r1=400kg·cm (m2r2)Ⅰ=m2r2(l23+l34)/l=300kg·cm (m3r3)Ⅰ=m3r3 l34/l=133.3kg·cm (m4r4)Ⅰ=0 取μW=10kg·cm/mm作矢量多边形，图b； 则：(mbrb)Ⅰ=WbⅠ · μW=28×10=280(kg·cm) ∴ mbⅠ =5.6kg 量取θbⅠ = 6°;(m1r1)Ⅱ+(m2r2)Ⅱ+(m3r3)Ⅱ+(m4r4)Ⅱ+(mbrb)Ⅱ=0 (m1r1)Ⅱ=0 (m2r2)Ⅱ=m2r2l12/l=150kg·cm (m3r3)Ⅱ=m3r3 (l12+l23) /l=266.7kg·cm (m4r4)Ⅱ= m4r4 =300kg·cm 取μW=10kg·cm/mm作矢量多边形，图c； 则：(mbrb)Ⅱ=WbⅡ · μW=37×10=370(kg·cm) ∴ mbⅡ=7.4kg 量取θbⅡ = 145°;2）解析法：(m1r1)Ⅰ+(m2r2)Ⅰ+(m3r3)Ⅰ+(m4r4)Ⅰ+(mbrb)Ⅰ=0 (m1r1)Ⅰ=m1r1=400kg·cm (m2r2)Ⅰ=m2r2(l23+l34)/l=kg·cm (m3r3)Ⅰ=m3r3 l34/l=133.3kg·cm (m4r4)Ⅰ=0 ΣFx=0：400cos120°+300cos240°+133.3cos300°+(mbrb)Ⅰx =0 ΣFy=0：400sin120°+ 300sin240°+133.3sin300°+ (mbrb)Ⅰy =0 (mbrb)Ⅰx = 283.35 (mbrb)Ⅰy = 28.84 ∴ (mbrb)Ⅰ = [(mbrb)Ⅰx2+ (mbrb)Ⅰy2 ]1/2 =284.81(kg·mm) ∴ mbⅠ = (mbrb)Ⅰ/rb=284.81/50= 5.70(kg) θbⅠ =arctan[(mbrb)Ⅰy / (mbrb)Ⅰx]= arctan(28.84 / 283.35)]=5.81°;(m1r1)Ⅱ+(m2r2)Ⅱ+(m3r3)Ⅱ+(m4r4)Ⅱ+(mbrb)Ⅱ=0 (m1r1)Ⅱ=0 (m2r2)Ⅱ=m2r2l12/l=150kg·cm (m3r3)Ⅱ=m3r3 (l12+l23) /l=266.7kg·cm (m4r4)Ⅱ= m4r4 =300kg·cm ΣFx=0：150cos240°+266.7cos300°+300cos30°+(mbrb)Ⅱx =0 ΣFy=0：150sin240°+ 266.7sin300°+300sin30°+ (mbrb)Ⅱy =0 (mbrb) Ⅱx