機(jī)械原理紅唐大放著礦業(yè)課后答案

習(xí)題 4－1圖yxDSJ＝0.600kgm22m423B12習(xí)題 4－1圖yxDSJ＝0.600kgm22m423B12機(jī)構(gòu)的受力狀態(tài)為，當(dāng)滑塊5向上運(yùn)動(dòng)時(shí)，工作阻力AC64r分析（參考答案：F12＝F61＝2398N，F(xiàn)2C＝2535＝F63＝13637N，F(xiàn)C4＝12797N，F(xiàn)45＝118655090N，M＝308Nm14－1F54tLCE＝G4F45tcos5.1+F45rF45tsin5.1－F45r＝[(104+185×9.81)sin5.1+35]/cos5.1＝1090F45r＝F65sin5.1+(Fr+G5)＝1090×sin5.1+(104+185×9.81)cos5.1＝11865yxD3B21FφAFCω164BC2F 4CFG4BFFF EA55FC4x＝F54tcos5.1+F54rsin5.1＝35cos5.1+11865sin5.1＝10901EEEFF ＝12797 C4φC4＝180°－arctan(FC4y/FC4x)＝180°－arctan(12747/1090)＝94.887°F12t＝80×9.81(0.800－0.450)cos9.062°/0.800＝339NF12t+F3Csin75.438°＝FC4sin85.8258°+G2cos9.062°FFF ＝12797 C4φC4＝180°－arctan(FC4y/FC4x)＝180°－arctan(12747/1090)＝94.887°F12t＝80×9.81(0.800－0.450)cos9.062°/0.800＝339NF12t+F3Csin75.438°＝FC4sin85.8258°+G2cos9.062°F12r+G2sin9.062°+FC4cos85.8258°＝F3Ccos75.438°F3C＝(FC4sin85.8258°+G2cos9.062°－F12t)/sin75.438°＝13637cos75.438°－80×9.81sin9.062°－12797cos85.8258°＝2374FF2＝2398 FF2＝11865 G2cos9.062)2(F12rG2sin9.062)(339809.81cos9.062)2(2374809.81sin9.062)2＝2535M1F21rLABsin(9.062)F21tLABcos(M123740.150sin(609.062)3390.150cos(609.062)＝308 =0.731QQFV1FQ2F++ αφFF12+F32+3F310=17232FQ0=14067FF31=13553F21=10301Q=10225η＝Q/Q0=14(1)φ＝arctanf＝arctanF31sin()F21sinFF31cos()F21cos0F31sin(2)FcosF31sin()cosF21sincosFF31cos()sinF21cossinF31Fcos/sin(F31Fcos/sin(2)10000cos5.711/sin(35.525.711)F21F31cos()/cos13622cos(35.55.711)/cos5.7112QF21cos10299cos5.711F32F21sin10299sin5.711(2)F310F/sin10000/sin35.5F310cos17221cos35.5Q0(3)QF21cos10299cos5.711F32F21sin10299sin5.711(2)F310F/sin10000/sin35.5F310cos17221cos35.5Q0(3)Q/Q010248/14019 Nm。CCh1＝21.839h3＝35.023223bc3BBMrω131adDDAA444－3解 F31O12左側(cè)時(shí)，21NON3134－4 4－5圖h＝400mm，b＝803(2)計(jì)入慣性力時(shí)，試求圖示位置驅(qū)動(dòng)力矩Md的大?。▍⒖即鸢福篗d＝417NmBF21Q3A4F23＝868hC4－5(2)計(jì)入慣性力時(shí)，試求圖示位置驅(qū)動(dòng)力矩Md的大?。▍⒖即鸢福篗d＝417NmBF21Q3A4F23＝868hC4－5F23＝903S3htan1，V3h1/(cos1)2，V3(cos1)2h1，a3cos212V31cos1sin1a32V31sin1/cos12V31tanV3h1/(cos1)20.410/cos2305.333a32V31tan125.33310tan3061.584 不計(jì)慣性力時(shí)，試用解析法求機(jī)構(gòu)在φ1＝30°、φ1＝60°、φ1＝120°和φ1＝220°計(jì)入慣性力時(shí)，試用解析法求機(jī)構(gòu)在φ1＝30°、φ1＝60°、φ1＝120°和φ1＝220°33B121φ1AA4－6S3L1sin1，x3L1V3L11cos1，Vx3L11sinLsin， L2a2 1 x114SbBm244Mb(m2m3)gL1cos1389.810.110cos3041.0058cos3035.512Mb(m2m3)gL1cos1389.810.110cos6041.0058Mb(m2m3)gL1cos1389.810.110cos3041.0058cos3035.512Mb(m2m3)gL1cos1389.810.110cos6041.0058cos6020.503Mb(m2m3)gL1cos1389.810.110cos12041.0058cos12020.503Mb(m2m3)gL1cos1389.810.110cos22041.0058cos22031.412Lsin0.110100sin11sin11sin305.52a 1 11Lsin0.110100sin11sin11sin609.526a2 1 11Lsin0.110100sin11sin11sin1209.526a2 1 11Lsin0.110100sin11sin11sin2207.0702a 1 11Mb(m2m3)(ga3)L1cos138(9.815.5)0.110cos3055.422NmMb(m2m3)(ga3)L1cos138(9.819.526)0.110cos6040.412Mb(m2m3)(ga3)L1cos138(9.819.526)0.110cos12040.412Mb(m2m3)(ga3)L1cos138(9.817.070)0.110cos2208.774 F43＝646N該機(jī)構(gòu)在ρ＝6mm，φ＝8°時(shí)的自鎖條件（參考答案：α＝4.3°FF解F23＝1300F41＝1143F43＝6464433QVVFFF4－74－84－8圖R＝100mm，A處為固定支承，OA與垂線之間的夾α＝30°，轉(zhuǎn)動(dòng)副A處的摩擦圓半ρ＝8mm，構(gòu)12、23之間的摩擦系數(shù)f＝0.1。試分答案：OAmax＝4.8918×μL＝35mmR＝100RTu＝14OATu＝8.5796ρTu＝2.24φ＝arctanf＝arctanOATumax＝4.8918×μL＝3534－852AR OBQB2 A1QB2 A1 Md1的大?。▍⒖即鸢福篴rctan2[asin/(acosd，S2(asin)acosd)22Md1Mr3acos(S2 Md1的大?。▍⒖即鸢福篴rctan2[asin/(acosd，S2(asin)acosd)22Md1Mr3acos(S2arctan2[asin/(acosdS2asin)acosd223a1cos()/S2，Mr3a1cos()/S2Md11，Md1Mr3acos()/SB1aφ2δ3dAD4－94－104－10圖為一工作臺(tái)升降機(jī)構(gòu)，當(dāng)油缸下腔通入高壓油時(shí)，工作臺(tái)作上升運(yùn)動(dòng)。已知F12＝28175。F65Gwbab10000(33.421.5)33.43563F45Gwa/b1000021.5/33.46437F56c/(cd)353632.5/(32.516.5)4972.5F12gF63fF12(F63fF43e)/g(4972.516643732.9)/10.34NabaGWb5454EDFc63Bd6FC27f CA38g1B4－106eFG4－114－11圖為一種型式的基于平行四邊形機(jī)構(gòu)的裝載機(jī)機(jī)構(gòu)。在舉升階段，BCDE為平行四邊形，從而使鏟斗作平動(dòng)，B、CO1μL＝實(shí)際尺F45的