传动轴受力简图:
yFy1Fz1z0MxaB3F22aCF·sinαMx2aDFy2EMx1FxaAG2Fz2G1cosα·FDPMx?F2??9549?260.551Nm2nDPMx1?F1?1?95491?150.056Nm2n D2Me2?F??Mx?Mx1?110.495Nm2
3F1
F?1473.275NF解得:1?1000.371NF2?868.504N
Fy1?4?G2?3?Fcos??1??G1?3F1由此可得方程: y方向:
G3?Fcos??3?Fy2?4?(G1?3F1)5?0
Fz1?4?3F2?3?F?sin??0z方向:
3F2?Fsin??3?Fz2?4?0
Fy1??286.321N可解得支座反力为:
Fy2?4343.266N
Fz1??1581.412N Fz2?436.791N 6
扭矩图:
Mx/Nm0AGBHCDIEJFa2a3a4a5ax150.056260.551
弯矩图:
z方向:
z/N1014.10a2a3a4a5ax436.7911591.412
My/Nm262.0730AGBHCDIEJFa2a3a4a5ax954.847
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y方向:
y/N32000aa2a3a4a5ax286.321886.3211142.152Mz/Nm0AGBHCDIEJFa2a3a4a5a171.792x1235.3771920.669设计等直轴轴的直径:
a2aaaABDEFGHIJ
根据第三强度理论:?22r3???4??12WM2x?My?M2z????
分别代入:
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Mxe??150.056Nm,Mye?0,Mze??1920.669NmMxd??260.551Nm,Myd?262.074Nm,Mzd??1235.377Nm Mxb??260.551Nm,Myb??954.847Nm,Mzb??171.792Nm且其中
????80MPa,则可求得:
?1?62.598mm,?2?54.757mm,?3?50.384mm
?1?2?3???1.1 可算得: 则可取?1?64mm,由
?2?3?4?2?58.182mm,?3?52.893mm,?4?48.084mm
取:?1?64mm,?2?58mm,?3?53mm,?4?48mm满足安全条件且能
使阶梯轴过度圆弧r?2。
求齿轮轴的挠度:
xoy平面:
Mz/NmG2·3a/4AGBaHC2aD3aIE4aJF5ax
Mz/NmCosα·F·3a/4AGBaHC2aD3aIE4aJF5ax 9
Mz/NmAGBaHC2aD3aIE4aJF5ax(G1+3F1)·a
Mc/Nm3a/4AGBaHC2aD3aIE4aJF5ax
在y方向应用图乘法: 可知:d??1?64mm,E?210MPa,I??McEI?d464
fcy??1311231513113?[G2a?a??a?G2?a??2a??a???G2?a?2a?a???G2EI426342123423411311117?a??a??a?F?a?cos?(3a??a?a???a)?(G1?3F1)?a?a?a?]22422228173373?[G2a?Facos??(G1?3F1)a3]EI1248??2.822?10?3m??2.822mm
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