西安建筑科技大学：《材料力学》课程习题与答案_第3章 扭转

第4次作业: 3-1 M1=9.55× 20050.6kNm,M2=287kNm,M3=0.57KNm M4=1.05kNm,M5=0.38Nm 0.86kN. m (-) 0. 38kN. m 1. 43kN.m 2.01k 3.14 M=zmW=60×2×50=1471875Nmm=147kNm Mn1.47×120 =1847kW 955955 3-9 T=0.2×400=80Nm d≥3 16T,16×80×10 vr[Tl 314x40≈21.7mm 齿轮啮合力F1 2×0.2×400 0.8kN 200 起吊重量F1×350=P×250,P=08x350 =1.12k 第5次作业: 3-12 16T 16T 直径比 d md mD[-a]D 0.839

第 4 次作业： 3-1 M k N m M k N m M k N m M k N m M k N m 1.05 . , 0.38 . 0.86 . , 2.87 . , 0.57 . , 200 18 9.55 4 5 1 2 3 = = =  = = = 3-3 k W M n P M WP N mm k N m 18.47 9.55 1.47 120 9.55 . 50 1471875 . 1.47 . 16 3.14 60 3 max =  = = = =   = = 3-9 (1) mm T d T N m 21.7 3.14 40 16 80 10 [ ] 16 0.2 400 80 . 3 3 3 =     = =  =   (2) F P P k N F k N 1.12 250 0.8 350 350 250, 0.8 200 2 0.2 400 1 1 =   =  = =   = 起吊重量 齿轮啮合力 第 5 次作业： 3-12 直径比：   , 1 0.839 1 16 16 3 4 3 3 4 = − = − =     D d D T d T 0.86kN.m (+) (-) 0.38kN.m 1.43kN.m 2.01kN.m

A,D2( 总量比 0.51 A 0.8392 刚度比 1D(1-a2) =1.19 0.8394 T=18kN.m 16×18×10° 45.2MPa<[r] 3.14×1403×1 140 180 046°/m<[r] 丌 BC =14kN 16×14×10° m=in=314×100=713Mmn(1 a=22×180=102°/m< 3-23 m=1.5,a=0.294,B=0.346,v=0.858 =40.1MPa W0.346×603 r'=vr=0.858×40.1=344MPa T180 3×10° 180 =0.000565°/mm=0.565°/m GL.x80×103×0.294×6043.14

1.19 0.839 (1 ) 1 0.51 0.839 (1 ) 1 A A 4 4 4 4 4 2 1 2 2 2 2 2 2 1 = − = − = = = − = − = =     d D I I d D 刚度比 总量比 3-16 AE: 046 / [ ] 180 45.2 [ ] 140 100 3.14 140 1 16 18 10 18 . 1 1 1 4 3 6 1 1 max 1      =  =  =                  −   = = = m G I T MPa W T T k N m o P P BC: 1.02 / [ ] 180 71.3 [ ] 3.14 100 16 14 10 14 . 2 2 2 3 6 2 2 max 2      =  =  =     = = = m GI T MPa W T T k N m o P P 3-23 mm m G I T MPa MPa W T m o o t t 0.000565 / 0.565 / 3.14 180 80 10 0.294 60 180 3 10 ' ' 0.858 40.1 34.4 40.1 0.346 60 3 10 1.5, 0.294, 0.346, 0.858 3 4 6 max 3 6 max  = =     =  = = =  = =   = = = = = =         