1、合肥工业大学05齿轮机构及其设计答案齿轮机构及其设计1、设有一渐开线标准齿轮 z=20, m =8mm, a =20o,ha =1,试求:1)其齿廓曲线在 分度圆及齿顶圆上的曲率半径 P、Pa及齿顶圆压力角ota; 2)齿顶圆齿厚Sa及基圆 齿厚Sb; 3)若齿顶变尖(sa=0)时,齿顶圆半径r;又应为多少?解 1)求 P、Pa、aad = mz = 8 20 =160mm . , i _ - .、 da =m(z 2ha) =8 (20 2 1) = 176mmdb =dcosa =160cos20 = 150.36mm:=rbtga =75.175tg20 = 27.36mmaa -cos
2、A(rb/ra) =cos(75.175/88) =31 19.3:a =rbtg:a =75.175tg31 19.3 = 45.75mm2)求 Sa、sbsaSbra 二m 88 .=s -2ra(invaa -inva) 176(inv31 19.3 -inv20 ) = 5.56mmr 2 808 二= cosa(s mz inva) =cos20 ( 8 20 inv20 ) = 14.05mm23)求当s; =0时r:rasa = s - -2ra(invaa -inva ) = 0 rs . inva; = inva =0.093444 2r由渐开线函数表查得:aa =35 28
3、.5r rb/cosa:、75.175/cos35 28.5: -92.32mm a b a2、试问渐开线标准齿轮的齿根圆与基圆重合时,其齿数 z应为多少,又当齿数大于以上求得的齿数时,基圆与齿根圆哪个大?解db = mz cosa* *df = m(z,,2ha 2c )由df至db有* *z)2(ha +c) = 2(1+0.25)=41.451 -cosa 1 -cos20当齿根圆与基圆重合时, z =41.45当z之42时,根圆大于基圆。3、一个标准直齿圆柱齿轮的模数 m=5mm,压力角a =20o,齿数z=18。如图所示,设将直径相同的两圆棒分别放在该轮直径方向相对的齿槽中, 圆棒与
4、两侧齿廓正好切于分度圆上,试求1)圆棒的半径 ; 2)两圆棒外顶点之间的距离(即棒跨距) l。1 二m / 2 二KOPKOP(r )X = 2 mz/2 2z二 180 二52z 二rp =NP -NKp= rb(tan25 -tg20 )=4.33mmr r. 、l =2 b +rD i = 101.98mm 34 , z3 + z4 343 1, 2和3, 4正传动,X3 X4 X1 X21)1, 2标准(等变位) 3, 4正传动 2 3, 4标准(等变位)4 1, 2 和 3, 4 负传动,X1 +X2 A X3 + X45 1, 2负传动,3, 4负传动方案,较佳8、在某牛头刨床中,
5、有一对外啮合渐开线直齿圆柱齿轮传动。 已知:Z1=17, Z2=118,O ,* * , 一 .一, .一、一 m=5mm, a =20 , ha =1, c =0.25, a =337.5mm。现已发现小齿轮严重磨损,拟将其报废,大齿轮磨损较轻(沿齿厚方向两侧总的磨损量为 0.75mm),拟修复使用,并要求新设计小齿轮的齿顶厚尽可能大些,问应如何设计这一对齿轮?解1)确定传动类型m 5 .a= (z+z2)= (17+118) = 337.5mm,因a =a故应采用等移距变位传动2 22)确定变位系数冶 0.75Xi = - X2 = = =0.2062mtg : 2 5tg 20故 x1
6、=0.206, x2 = -0.2063)几何尺寸计算小齿轮大齿轮d1 =mZ) =5x17 = 85mmd2 = mz? = 5黑 118= 590mm. ,.* . 、ha1 =(ha +x1)m=(1 +0.206)X5 = 6.03mm. 八*. 、ha2 = (ha +X2)ml=(1 -0.206) x5 = 3.97mm, 八* . * 、hf1 =(ha +c -X1)m=(1 +0.25-0.206) m 5 =5.22mm. 八* . * 、hf2 =(ha +c -x2)m=(1 + 0.25 + 0.206) m 5 = 7.28mmda1 =d1 +2ha1 =85
7、+2M6.03 = 97.06mmda2 =d2 +2ha2 =590 +2父3.97 =597.94mmdf1 =d1 -2hf1 =85-2x5.22= 74.56mmdf2 =d2 -2hf2 =590 -2x 7.28=575.44 mmdb1 =d1cosa =85cos20 口 = 79.87m mdb2 =d2cosa =590cos20 口 =554.42 mmns1 =m(+2x1tgs)2jr=5(金+2父0,206tg 20) =8.61n& = m(3 + 2x2tg)ji= 5(- -2M0.206tg20口) = 7.1e =m(- -2x1tga)231= 5(g
8、-2M0.206tg20)=7.1jie2 = m( 2x2tga )2ji= 5(3+29206tg 20)=8.61p1=G+ei=nm=5 = 15.71mmp2 = s2 +e2 = nm = 5n = 15.71mm O * *9、设已知一对斜齿轮传动, zi=20, Z2=40, mn =8mm, an =20 , han =1, Cn=0.25, B=30mm,并初取3 =15,试求该传动的中心距 a(a值应圆整为个位数为 。或5, 并相应重算螺旋角3卜几何尺寸、当量齿数和重合度。解1)计算中心距a初取 P =151 则 a = mn 口 (z1 + z2)= 8(20 + 40
9、) = 248,4662cos 2cos15mn (Zi Z2) 8(20 40)取 a = 250mm,贝U P = arccos L = arccos 乙=16 15 372a 2 2502)计算几何尺寸及当量齿数尺寸名称小齿轮人齿车匕分度圆直径d1 =mnz1 /cosP = 166.67mmd2 = 333.33mm齿顶圆直径da1 =d1 + 2ha = 182.67mmda2 = 349.33齿根圆直径df1 =d1 -2hf = 146.67mmdf2 = 313.33基圆直径db1 =d1cosc(t = 155.85mmdb2 = 311.69mm齿顶局、齿根局*ha = h
10、amn =8mmha = (ha + c )mn = 10mm法向及端面齿厚sn = nmn/2 = 12.57mmst =nmn /(2cosP) = 13.09mm法面及端凿齿距pn =nmn = 25.14mmpt = pn/cosP = 26.19mm当量齿数zv1 =z,cos3 P =22.6142 = z2/cos3P =22.613)计算重合度叫二 t =arctg(tg 二 n/cos ) =arctg(tg20 /cos16 15 37) =20 4549ti =arccosdbi/da2)= arccosQ55.84/182.67) =31 2649:at2 =arcco
11、s(db2/da2) = arccos011.69/349.33) =26 50 33,_ Z(tg: 1 tg: t) Z2(tg:-2 tg: t) % = 2 二20(tg31 26 49 -tg20 45 49 ) 40(tg 26 50 33 -tg20 45 49 ) / ”二 -1.592 二;一:=Bsin 7:mn =30sin16 15 37 /8 =0.332,=;/;:_ =1.59 0.332 =1.9210、设计一铳床进给系统中带动工作台转动的阿基米德蜗杆传动。要求 i12=20.5,* *m=5mm, a =20 , ha =1, c =0.2,求蜗轮蜗杆传动的基
12、本参数 亿1、4、q、丫 1、82)、几何尺寸(dd2、da1、da2)和中心距a。解1)确定基本参数选取Z1=2 (因为当i12 =14.530.5时,一般推荐 乙=2。)z2 =i124 =20.5 2 =41查表确定 d1=50mm,计算 q = d1/m = 50/5=101 =arctg(mz/d1) =arctg(5 2/50) =11 18362 - S =11 18 362)计算几何尺寸d1 =50mm,d2 = mz2 = 205mmda1 = d1 2ha =60mmda2 = d2 2ha = 215mmdf1 =d1 -2hf = 38mmdf2 = d2 -2hf = 193mm3)中心距a=a =万(z1 z2) = (10 41) = 127.5mm11、在图示的各蜗轮蜗杆传动中,蜗杆均为主动,试确定图示蜗杆、蜗轮的转向或 螺旋线的旋向。