機械原理紅唐大放著礦業(yè)課后答案

習(xí)題 在題6－1圖所示的凸輪機構(gòu)中，μL(實際尺寸/圖上尺寸)＝10，凸輪1為主動件，推桿畫出凸輪的理論基圓并量取半徑r0（參考答案：r0＝139.5mm）；α＝21°標(biāo)出并量取從動件的位移s（參考答案：s＝93mm）；習(xí)題 在題6－1圖所示的凸輪機構(gòu)中，μL(實際尺寸/圖上尺寸)＝10，凸輪1為主動件，推桿畫出凸輪的理論基圓并量取半徑r0（參考答案：r0＝139.5mm）；α＝21°標(biāo)出并量取從動件的位移s（參考答案：s＝93mm）；偏心距e的引入對受力是否有利（參考答案：有利。3yC431eδ232B1BrB02xφψωAω11DASP1146－36－16－2量出基圓半徑r0＝139.5mm；該位置時從動件的位移s＝93mm；Fr3C4en2BαFr＝Mdω1/V3＝Md/(ω11PAn6－2在6－2圖所示的凸輪機構(gòu)中，μL(實際尺寸/圖上尺寸)＝101為主動件，推桿2為從動件，已知凸輪1的基圓半徑r0，從動件在推程[0，80oS＝b·sin[9δ/(4π)](mm)，b為常數(shù)6－1試推導(dǎo)凸輪在推程階段的輪廓方程（S＝b·sin[9δ/(4π)](mmdS/dδ9b/(4π)cos[9δ/(4π)]，x(r0S)sin(dS/d)cos，y(r0S)cos(dS/d)sin該機構(gòu)的壓力角α（參考答案：α＝0；1SshS)sin(dS/d)cosxy(rS)sin(dS/d)cosxy(rS)cos(dS/d)sin0(4)求常數(shù)b80o＝80o×π/180o＝108.144y3δ2C21B3BxψφDAS4S106－26－3 α＝31°畫出擺桿2的擺角ψ（參考答案：ψ＝29°；2的運動規(guī)律（kA2bL3sinkB2L3dbcos L2d2b22bdcos，2arctan2[(kk)kk2k2kk 3A [L3cos(dbcos)]2(L3sinbsin)2(R1R22L2(dbcos)22Lcos(dbcos)2Lbsinsinb2sin2(R333 2L2d2b22bdcos2dLcos2bLcoscos2bLsinsin(R3333 (2bLsin)sin2L(dbcos)cos[(R L2d2b22bdcos])33 3 2bLsin，k2LL2d2b22bdcos2dLcos2bLcoscos2bLsinsin(R3333 (2bLsin)sin2L(dbcos)cos[(R L2d2b22bdcos])33 3 2bLsin，k2L(dbcos)，k(R L2d2b22bdcos)A3 3kAsinkBcoskC2arctan2[(k k2k2)/(kkkA 3C4Sh[ sin(2π1e T200S0.04[3 1BmT S e20.062r0.058 T Md1ST)sinecosxTy(SS)cosesin6－1 xxrgcosyyrsing Srb3S（S0rbbrb，b1，S3C2Dθ2Aβ030LKδ3δbωδ1ψ1Cδrbbω11b46－66－53SaSBA brbcos(b0)[rb(0)rg]cos(b00.5π)L3 brbcos(b0)[rb(0)rg]cos(b00.5π)L36－7在6－7圖ab、cd、efR1，三段小圓弧fa、bc、de的半徑都為R2、O2＝O2O3＝O3O1＝R1－R2，從動件2與凸輪1接觸的寬度BC＝R1＋2處于回程，如圖(a)所示；當(dāng)1ab、de2接觸，abyyAAC22ψnne3cωb1φω2bθDxaxBedd1fφω1c1nnEyyAAnnψψcddeebθθcxxOO11afbnnyyAALLbnnθθxxcbedcdnn6－74Lψ32處于推程，如圖(b)所示；當(dāng)1cd2接觸，cd段圓弧位于2處于推程，如圖(b)所示；當(dāng)1cd2接觸，cd段圓弧位于1)確定凸輪推程起點（下停歇終點、推程終點（上停歇起點、回程起點（上停歇終點）與回 (0.5R10.5R2)/(0.5R0.5R)]2 由圖由圖))eiRei(0.5π)S)ei(1.5π)(R0.5(R 22 (R1R2)cosR2cos(0.5π)S2cos0.5(R1R2)cos(1.5π)Lcos(R1R2)sinR2sin(0.5π)S2sin0.5(R1R2)sin(1.5π)Lsin(R1R2)cosR2sinS2cos0.5(R1R2)sinLcos(R1R2)sinR2cosS2sin0.5(R1R2)cosLsin(R1R2)cossinR2sin2S2cossin0.5(R1R2)sin2Lcossin(R1R2)sincosR2cos2S2sincos0.5(R1R2)cos2Lsincos(R1R2)cossinR2sin20.5(R1R2)sin2(R1R2)sincosR2cos20.5(R1R2)cos2LcossinLsin(R1R2)cossinR20.5(R1R2)(R1R2)sincosLcossinLsincos(R1R2)cossin(R1R2)sincos0.5(R1R2)LcossinLsincos[(R1R2)cosLcos]sin[(R1R2)sinLsin]cos0.5(R1R2)kA(R1R2)cosLcoskAsinkBcoskCkB(R1R2)sinLsin，kC0.5(R1R2sin2x/(1x2)，cos(1x2)/(1x2kA2x/(1x2)kB(1x2)/(1x2)kC2kAxkBx2kkkx2(kBkC)x22kAxkBkC 2arctan2[(kk2k2k2)/(kk 5 BB1212ψφCψφAOC1A336－8 BB1212ψφCψφAOC1A336－8α＝0°畫出擺桿2的擺角ψb（參考答案：ψb＝25°；2的運動規(guī)律（kA(bcosL3kBbsinkCR12arctan2[(k k2k2kkkA 作圖得擺桿2的擺角ψb＝25°；beiRei(0.5π)Lei0S132bcosR1cos(0.5π)L3cos0S2bsinR1sin(0.5π)S2bcosR1sinL3S2bsinR1cosS2bcossinR1sin2L3sinS2cosbsincosR1cos2S2sinbsincosR1cos2(bcossinR1sin2L3sin)bsincosR1