常见的带传动的张紧装置[1]有定期张紧装置;自动张紧装置;采用张紧轮[2]的张紧装置等几种。解题要点:-|||-(1)三分力方向如图b所示。-|||-(2)求出齿轮主要尺寸及所传递的转矩T1:-|||-_(1)=m(z)_(1)=4times 28mm=112mm-|||-_(2)=m(z)_(2)=4times 48mm=192mm-|||-_(1)=9.55times (10)^6dfrac (P)(n)=9.55times (10)^6dfrac (3)(960)=29844 N'cdot mm-|||-=dfrac ({z)_(1)}({z)_(2)}=dfrac (48)(28)=1.7143-|||-cos (theta )_(1)=dfrac (u)(sqrt {1+{u)^2}}=dfrac (1.7143)(sqrt {1+{(1.7143))^2}}=0.8638-|||-_(1)=(30)^circ 15'14''-|||-sin (theta )_(1)=sin (30)^circ 15'14''=0.5038-|||-(3)求F1、、22、Fa2的大小:-|||-_(12)=(F)_(11)=dfrac (2{T)_(1)}((1-0.54{a)_(1))(d)_(1)}=dfrac (2times 2984)((1-0.5times 0.3)times 112)=627N-|||-_({a)_(2)}=(F)_(n)=(F)_(n)tan acos (a)_(1)=627times 0.364times 0.863=197N-|||-_(n2)=(F)_(u1)=(F)_(u)tan alpha sin (8)_(1)=627times 0.364times 0.5038=115N-|||-图a所示为一对标准斜齿圆柱齿轮-蜗杆组成的二级传动,小齿轮为主动轮。-|||-已知:①蜗轮的节圆直径d4,螺旋角β,端面压力角a;②蜗轮传递的转矩T4-|||-(N·m)。-|||-1-|||-4 4-|||-= n-|||-一Ⅲ Fu F-|||-F e= Ⅲ-|||-F-|||-3 Fr3 3-|||-么 ll ll-|||-少 O2-|||-1 1-|||-i 2-|||-ì 2-|||-a) b)-|||-试求:试求:(1)蜗杆、齿轮1、2的螺旋线方向以及轴Ⅰ、Ⅱ的旋转方向(用箭头在图中标出)。(2)Ft3、Fr3、Fa3及Ft4、Fr4、Fa4的方向已标于图b中。解题要点:-|||-(1)三分力方向如图b所示。-|||-(2)求出齿轮主要尺寸及所传递的转矩T1:-|||-_(1)=m(z)_(1)=4times 28mm=112mm-|||-_(2)=m(z)_(2)=4times 48mm=192mm-|||-_(1)=9.55times (10)^6dfrac (P)(n)=9.55times (10)^6dfrac (3)(960)=29844 N'cdot mm-|||-=dfrac ({z)_(1)}({z)_(2)}=dfrac (48)(28)=1.7143-|||-cos (theta )_(1)=dfrac (u)(sqrt {1+{u)^2}}=dfrac (1.7143)(sqrt {1+{(1.7143))^2}}=0.8638-|||-_(1)=(30)^circ 15'14''-|||-sin (theta )_(1)=sin (30)^circ 15'14''=0.5038-|||-(3)求F1、、22、Fa2的大小:-|||-_(12)=(F)_(11)=dfrac (2{T)_(1)}((1-0.54{a)_(1))(d)_(1)}=dfrac (2times 2984)((1-0.5times 0.3)times 112)=627N-|||-_({a)_(2)}=(F)_(n)=(F)_(n)tan acos (a)_(1)=627times 0.364times 0.863=197N-|||-_(n2)=(F)_(u1)=(F)_(u)tan alpha sin (8)_(1)=627times 0.364times 0.5038=115N-|||-图a所示为一对标准斜齿圆柱齿轮-蜗杆组成的二级传动,小齿轮为主动轮。-|||-已知:①蜗轮的节圆直径d4,螺旋角β,端面压力角a;②蜗轮传递的转矩T4-|||-(N·m)。-|||-1-|||-4 4-|||-= n-|||-一Ⅲ Fu F-|||-F e= Ⅲ-|||-F-|||-3 Fr3 3-|||-么 ll ll-|||-少 O2-|||-1 1-|||-i 2-|||-ì 2-|||-a) b)-|||-试求:4.图所示为二级斜齿圆柱齿轮减速器。已知:齿轮1的螺旋线方向和轴Ⅲ的转向,齿轮2的参数mn=3mm,z2=57,β2 =14°;齿轮3的参数mn=5mm,z3=21。试求:(1)为使轴Ⅱ所受的轴向力最小,齿轮3应选取的螺旋线方向,并在图b上标出齿轮2和齿轮3的螺旋线方向;(2)在图b上标出齿轮2、3所受各分力的方向;解题要点:-|||-(1)三分力方向如图b所示。-|||-(2)求出齿轮主要尺寸及所传递的转矩T1:-|||-_(1)=m(z)_(1)=4times 28mm=112mm-|||-_(2)=m(z)_(2)=4times 48mm=192mm-|||-_(1)=9.55times (10)^6dfrac (P)(n)=9.55times (10)^6dfrac (3)(960)=29844 N'cdot mm-|||-=dfrac ({z)_(1)}({z)_(2)}=dfrac (48)(28)=1.7143-|||-cos (theta )_(1)=dfrac (u)(sqrt {1+{u)^2}}=dfrac (1.7143)(sqrt {1+{(1.7143))^2}}=0.8638-|||-_(1)=(30)^circ 15'14''-|||-sin (theta )_(1)=sin (30)^circ 15'14''=0.5038-|||-(3)求F1、、22、Fa2的大小:-|||-_(12)=(F)_(11)=dfrac (2{T)_(1)}((1-0.54{a)_(1))(d)_(1)}=dfrac (2times 2984)((1-0.5times 0.3)times 112)=627N-|||-_({a)_(2)}=(F)_(n)=(F)_(n)tan acos (a)_(1)=627times 0.364times 0.863=197N-|||-_(n2)=(F)_(u1)=(F)_(u)tan alpha sin (8)_(1)=627times 0.364times 0.5038=115N-|||-图a所示为一对标准斜齿圆柱齿轮-蜗杆组成的二级传动,小齿轮为主动轮。-|||-已知:①蜗轮的节圆直径d4,螺旋角β,端面压力角a;②蜗轮传递的转矩T4-|||-(N·m)。-|||-1-|||-4 4-|||-= n-|||-一Ⅲ Fu F-|||-F e= Ⅲ-|||-F-|||-3 Fr3 3-|||-么 ll ll-|||-少 O2-|||-1 1-|||-i 2-|||-ì 2-|||-a) b)-|||-试求:提示:轴Ⅱ用深沟球轴承。解题要点:(1)根据轴Ⅲ转向nⅢ,可在图解a上标出nⅡ和nⅢ的转向。而齿轮2应为左旋,已标于图解b上。(2)根据主动轮左、右手定则判断出Fa1、Fa3;根据齿轮2是从动轮,齿轮3是主动轮判断Ft2、Ft3;根据径向力指向各自轴心的原则,判断径向力Fr2、Fr3;的方向。Fa1、Fa3、Ft2、Ft3、Fr2、Fr3已在啮合点画出(图解b)。5.图所示为直齿锥齿轮–斜齿圆柱齿轮减速器,齿轮1主动,转向如图示。锥齿轮的参数为mn=2mm,z1=20,z2=40,解题要点:-|||-(1)三分力方向如图b所示。-|||-(2)求出齿轮主要尺寸及所传递的转矩T1:-|||-_(1)=m(z)_(1)=4times 28mm=112mm-|||-_(2)=m(z)_(2)=4times 48mm=192mm-|||-_(1)=9.55times (10)^6dfrac (P)(n)=9.55times (10)^6dfrac (3)(960)=29844 N'cdot mm-|||-=dfrac ({z)_(1)}({z)_(2)}=dfrac (48)(28)=1.7143-|||-cos (theta )_(1)=dfrac (u)(sqrt {1+{u)^2}}=dfrac (1.7143)(sqrt {1+{(1.7143))^2}}=0.8638-|||-_(1)=(30)^circ 15'14''-|||-sin (theta )_(1)=sin (30)^circ 15'14''=0.5038-|||-(3)求F1、、22、Fa2的大小:-|||-_(12)=(F)_(11)=dfrac (2{T)_(1)}((1-0.54{a)_(1))(d)_(1)}=dfrac (2times 2984)((1-0.5times 0.3)times 112)=627N-|||-_({a)_(2)}=(F)_(n)=(F)_(n)tan acos (a)_(1)=627times 0.364times 0.863=197N-|||-_(n2)=(F)_(u1)=(F)_(u)tan alpha sin (8)_(1)=627times 0.364times 0.5038=115N-|||-图a所示为一对标准斜齿圆柱齿轮-蜗杆组成的二级传动,小齿轮为主动轮。-|||-已知:①蜗轮的节圆直径d4,螺旋角β,端面压力角a;②蜗轮传递的转矩T4-|||-(N·m)。-|||-1-|||-4 4-|||-= n-|||-一Ⅲ Fu F-|||-F e= Ⅲ-|||-F-|||-3 Fr3 3-|||-么 ll ll-|||-少 O2-|||-1 1-|||-i 2-|||-ì 2-|||-a) b)-|||-试求:;斜齿圆柱齿轮的参数为mn=3mm,z3=20,z4=60。试求:(1)画出各轴的转向;(2)为使轴Ⅱ所受轴向力最小,标出齿轮3、4的螺旋线方向;(3)画出轴Ⅱ上齿轮2、3所受各力的方向;解题要点:-|||-(1)三分力方向如图b所示。-|||-(2)求出齿轮主要尺寸及所传递的转矩T1:-|||-_(1)=m(z)_(1)=4times 28mm=112mm-|||-_(2)=m(z)_(2)=4times 48mm=192mm-|||-_(1)=9.55times (10)^6dfrac (P)(n)=9.55times (10)^6dfrac (3)(960)=29844 N'cdot mm-|||-=dfrac ({z)_(1)}({z)_(2)}=dfrac (48)(28)=1.7143-|||-cos (theta )_(1)=dfrac (u)(sqrt {1+{u)^2}}=dfrac (1.7143)(sqrt {1+{(1.7143))^2}}=0.8638-|||-_(1)=(30)^circ 15'14''-|||-sin (theta )_(1)=sin (30)^circ 15'14''=0.5038-|||-(3)求F1、、22、Fa2的大小:-|||-_(12)=(F)_(11)=dfrac (2{T)_(1)}((1-0.54{a)_(1))(d)_(1)}=dfrac (2times 2984)((1-0.5times 0.3)times 112)=627N-|||-_({a)_(2)}=(F)_(n)=(F)_(n)tan acos (a)_(1)=627times 0.364times 0.863=197N-|||-_(n2)=(F)_(u1)=(F)_(u)tan alpha sin (8)_(1)=627times 0.364times 0.5038=115N-|||-图a所示为一对标准斜齿圆柱齿轮-蜗杆组成的二级传动,小齿轮为主动轮。-|||-已知:①蜗轮的节圆直径d4,螺旋角β,端面压力角a;②蜗轮传递的转矩T4-|||-(N·m)。-|||-1-|||-4 4-|||-= n-|||-一Ⅲ Fu F-|||-F e= Ⅲ-|||-F-|||-3 Fr3 3-|||-么 ll ll-|||-少 O2-|||-1 1-|||-i 2-|||-ì 2-|||-a) b)-|||-试求:解题要点:(1)轴Ⅱ、Ⅲ的转向已示于图中;(2)齿轮3、4的螺旋线方向已示于题图解中,即z3为右螺旋,z4为左螺旋;(3)齿轮2、3所受各力 Ft2、Fr2、Fa2及Ft3、Fr3、Fa3已示于图解图中;为什么连续传动的闭式蜗杆传动必须进行热平衡计算? 答:由于蜗杆传动效率低,工作时发热量大。在闭式传动中,如果产生的热量不能及时散逸,将因油温不断升高使润滑油稀释,从而增大摩擦损失,甚至发生胶合。所以进行热平衡计算。6 在图中,标出未注明的蜗杆(或蜗轮)的螺旋线旋向及蜗杆或蜗轮的转向,并绘出蜗杆或蜗轮啮合点作用力的方向(用三个分力表示)。解题要点:-|||-(1)三分力方向如图b所示。-|||-(2)求出齿轮主要尺寸及所传递的转矩T1:-|||-_(1)=m(z)_(1)=4times 28mm=112mm-|||-_(2)=m(z)_(2)=4times 48mm=192mm-|||-_(1)=9.55times (10)^6dfrac (P)(n)=9.55times (10)^6dfrac (3)(960)=29844 N'cdot mm-|||-=dfrac ({z)_(1)}({z)_(2)}=dfrac (48)(28)=1.7143-|||-cos (theta )_(1)=dfrac (u)(sqrt {1+{u)^2}}=dfrac (1.7143)(sqrt {1+{(1.7143))^2}}=0.8638-|||-_(1)=(30)^circ 15'14''-|||-sin (theta )_(1)=sin (30)^circ 15'14''=0.5038-|||-(3)求F1、、22、Fa2的大小:-|||-_(12)=(F)_(11)=dfrac (2{T)_(1)}((1-0.54{a)_(1))(d)_(1)}=dfrac (2times 2984)((1-0.5times 0.3)times 112)=627N-|||-_({a)_(2)}=(F)_(n)=(F)_(n)tan acos (a)_(1)=627times 0.364times 0.863=197N-|||-_(n2)=(F)_(u1)=(F)_(u)tan alpha sin (8)_(1)=627times 0.364times 0.5038=115N-|||-图a所示为一对标准斜齿圆柱齿轮-蜗杆组成的二级传动,小齿轮为主动轮。-|||-已知:①蜗轮的节圆直径d4,螺旋角β,端面压力角a;②蜗轮传递的转矩T4-|||-(N·m)。-|||-1-|||-4 4-|||-= n-|||-一Ⅲ Fu F-|||-F e= Ⅲ-|||-F-|||-3 Fr3 3-|||-么 ll ll-|||-少 O2-|||-1 1-|||-i 2-|||-ì 2-|||-a) b)-|||-试求:7图示蜗杆传动,蜗杆1主动,其转向如图示,螺旋线方向为右旋。试决定: 1)蜗轮2的螺旋线方向及转向n2。2)蜗杆、蜗轮受到的各力(Ft、Fr、Fα)解题要点:-|||-(1)三分力方向如图b所示。-|||-(2)求出齿轮主要尺寸及所传递的转矩T1:-|||-_(1)=m(z)_(1)=4times 28mm=112mm-|||-_(2)=m(z)_(2)=4times 48mm=192mm-|||-_(1)=9.55times (10)^6dfrac (P)(n)=9.55times (10)^6dfrac (3)(960)=29844 N'cdot mm-|||-=dfrac ({z)_(1)}({z)_(2)}=dfrac (48)(28)=1.7143-|||-cos (theta )_(1)=dfrac (u)(sqrt {1+{u)^2}}=dfrac (1.7143)(sqrt {1+{(1.7143))^2}}=0.8638-|||-_(1)=(30)^circ 15'14''-|||-sin (theta )_(1)=sin (30)^circ 15'14''=0.5038-|||-(3)求F1、、22、Fa2的大小:-|||-_(12)=(F)_(11)=dfrac (2{T)_(1)}((1-0.54{a)_(1))(d)_(1)}=dfrac (2times 2984)((1-0.5times 0.3)times 112)=627N-|||-_({a)_(2)}=(F)_(n)=(F)_(n)tan acos (a)_(1)=627times 0.364times 0.863=197N-|||-_(n2)=(F)_(u1)=(F)_(u)tan alpha sin (8)_(1)=627times 0.364times 0.5038=115N-|||-图a所示为一对标准斜齿圆柱齿轮-蜗杆组成的二级传动,小齿轮为主动轮。-|||-已知:①蜗轮的节圆直径d4,螺旋角β,端面压力角a;②蜗轮传递的转矩T4-|||-(N·m)。-|||-1-|||-4 4-|||-= n-|||-一Ⅲ Fu F-|||-F e= Ⅲ-|||-F-|||-3 Fr3 3-|||-么 ll ll-|||-少 O2-|||-1 1-|||-i 2-|||-ì 2-|||-a) b)-|||-试求:题解分析:1)蜗轮的螺旋线方向—右旋,蜗轮的转向n2—见题解图所示。2)蜗杆受力Ft1、Fr1、Fα1及蜗累受力Ft2、Fr2、Fα2——见题解图所示。解题要点:-|||-(1)三分力方向如图b所示。-|||-(2)求出齿轮主要尺寸及所传递的转矩T1:-|||-_(1)=m(z)_(1)=4times 28mm=112mm-|||-_(2)=m(z)_(2)=4times 48mm=192mm-|||-_(1)=9.55times (10)^6dfrac (P)(n)=9.55times (10)^6dfrac (3)(960)=29844 N'cdot mm-|||-=dfrac ({z)_(1)}({z)_(2)}=dfrac (48)(28)=1.7143-|||-cos (theta )_(1)=dfrac (u)(sqrt {1+{u)^2}}=dfrac (1.7143)(sqrt {1+{(1.7143))^2}}=0.8638-|||-_(1)=(30)^circ 15'14''-|||-sin (theta )_(1)=sin (30)^circ 15'14''=0.5038-|||-(3)求F1、、22、Fa2的大小:-|||-_(12)=(F)_(11)=dfrac (2{T)_(1)}((1-0.54{a)_(1))(d)_(1)}=dfrac (2times 2984)((1-0.5times 0.3)times 112)=627N-|||-_({a)_(2)}=(F)_(n)=(F)_(n)tan acos (a)_(1)=627times 0.364times 0.863=197N-|||-_(n2)=(F)_(u1)=(F)_(u)tan alpha sin (8)_(1)=627times 0.364times 0.5038=115N-|||-图a所示为一对标准斜齿圆柱齿轮-蜗杆组成的二级传动,小齿轮为主动轮。-|||-已知:①蜗轮的节圆直径d4,螺旋角β,端面压力角a;②蜗轮传递的转矩T4-|||-(N·m)。-|||-1-|||-4 4-|||-= n-|||-一Ⅲ Fu F-|||-F e= Ⅲ-|||-F-|||-3 Fr3 3-|||-么 ll ll-|||-少 O2-|||-1 1-|||-i 2-|||-ì 2-|||-a) b)-|||-试求:8.图示为一标准蜗杆传动,蜗杆主动,螺旋线方向为左旋,转矩T1=2500N·mm,模数m=4mm,压力角解题要点:-|||-(1)三分力方向如图b所示。-|||-(2)求出齿轮主要尺寸及所传递的转矩T1:-|||-_(1)=m(z)_(1)=4times 28mm=112mm-|||-_(2)=m(z)_(2)=4times 48mm=192mm-|||-_(1)=9.55times (10)^6dfrac (P)(n)=9.55times (10)^6dfrac (3)(960)=29844 N'cdot mm-|||-=dfrac ({z)_(1)}({z)_(2)}=dfrac (48)(28)=1.7143-|||-cos (theta )_(1)=dfrac (u)(sqrt {1+{u)^2}}=dfrac (1.7143)(sqrt {1+{(1.7143))^2}}=0.8638-|||-_(1)=(30)^circ 15'14''-|||-sin (theta )_(1)=sin (30)^circ 15'14''=0.5038-|||-(3)求F1、、22、Fa2的大小:-|||-_(12)=(F)_(11)=dfrac (2{T)_(1)}((1-0.54{a)_(1))(d)_(1)}=dfrac (2times 2984)((1-0.5times 0.3)times 112)=627N-|||-_({a)_(2)}=(F)_(n)=(F)_(n)tan acos (a)_(1)=627times 0.364times 0.863=197N-|||-_(n2)=(F)_(u1)=(F)_(u)tan alpha sin (8)_(1)=627times 0.364times 0.5038=115N-|||-图a所示为一对标准斜齿圆柱齿轮-蜗杆组成的二级传动,小齿轮为主动轮。-|||-已知:①蜗轮的节圆直径d4,螺旋角β,端面压力角a;②蜗轮传递的转矩T4-|||-(N·m)。-|||-1-|||-4 4-|||-= n-|||-一Ⅲ Fu F-|||-F e= Ⅲ-|||-F-|||-3 Fr3 3-|||-么 ll ll-|||-少 O2-|||-1 1-|||-i 2-|||-ì 2-|||-a) b)-|||-试求:,蜗杆头数Z1=2,蜗杆直径系数q=10,蜗轮齿数Z2=54,传动效率解题要点:-|||-(1)三分力方向如图b所示。-|||-(2)求出齿轮主要尺寸及所传递的转矩T1:-|||-_(1)=m(z)_(1)=4times 28mm=112mm-|||-_(2)=m(z)_(2)=4times 48mm=192mm-|||-_(1)=9.55times (10)^6dfrac (P)(n)=9.55times (10)^6dfrac (3)(960)=29844 N'cdot mm-|||-=dfrac ({z)_(1)}({z)_(2)}=dfrac (48)(28)=1.7143-|||-cos (theta )_(1)=dfrac (u)(sqrt {1+{u)^2}}=dfrac (1.7143)(sqrt {1+{(1.7143))^2}}=0.8638-|||-_(1)=(30)^circ 15'14''-|||-sin (theta )_(1)=sin (30)^circ 15'14''=0.5038-|||-(3)求F1、、22、Fa2的大小:-|||-_(12)=(F)_(11)=dfrac (2{T)_(1)}((1-0.54{a)_(1))(d)_(1)}=dfrac (2times 2984)((1-0.5times 0.3)times 112)=627N-|||-_({a)_(2)}=(F)_(n)=(F)_(n)tan acos (a)_(1)=627times 0.364times 0.863=197N-|||-_(n2)=(F)_(u1)=(F)_(u)tan alpha sin (8)_(1)=627times 0.364times 0.5038=115N-|||-图a所示为一对标准斜齿圆柱齿轮-蜗杆组成的二级传动,小齿轮为主动轮。-|||-已知:①蜗轮的节圆直径d4,螺旋角β,端面压力角a;②蜗轮传递的转矩T4-|||-(N·m)。-|||-1-|||-4 4-|||-= n-|||-一Ⅲ Fu F-|||-F e= Ⅲ-|||-F-|||-3 Fr3 3-|||-么 ll ll-|||-少 O2-|||-1 1-|||-i 2-|||-ì 2-|||-a) b)-|||-试求:=0.75。试确定:1)蜗轮2的转向及螺旋线方向;2)作用在蜗杆、蜗轮上的各力的方向(在图中标出)。