《机械设计-吴昌林》课后习题答案(部分)
2知某钢制零件 受弯曲变应力的作用,其中,最大工作应力 00小工作应力 50险截面上的应力集中系数 k=尺寸系数 =表明状态系数 =1。材料的 s=7500=58050试求: ( 1)绘制材料的简化极限应力图,并在图上标出工作应力点的位置; ( 2) 求材料在该应力状态下的疲劳极限应力 r; ( 3)按疲劳极限应力和安全系数分别校核 此零件是否安全(取 An s=7500=5805000 50k=1, (1) (2) r. (3) if is 1) M P 25)50(200(2121 m a x M P 5)50200(2121 m a x a xm r 2) 2)1)( )()1)( 10 0100 Ds 502580)750 )580750(350750580) is m ,( sG),0( 1)2,2( 00 8035022 0 01 )( 1 M P )75125(350)( 1 4) . P aS r 200 M B. a x SS r 350)( 1 3 3、 3 6、 3 8 3 3 A of is as of on I, a3 to be of of I 15. to of of on (1) of 550000*P2/3=9550000*P3/n3 n2=on 2= So 3 (2) of to 333222 , 333222 2,2 of 3333222 c o s,c o s (be by ) 3332222 s )60*3/(15s i n*30*4s i ns i n 0222 333 zm zm 3= =05624 3 6 An of 0430r/u=16 is 08 55 1) of ( 1) 408 555( 2) 16(c), 200 17(c), 60 . N、 11 63005(143016060 8812 18, 121 Z . 19, 121 Y . 4, 2.1 m S 。 M P a 1200m 1 H Z12 M P a 2360 m 1 F 12 2) is 3) ( 1) s 1161 ( 2) ( 3) ( 4) =12°, 0, 30× 29, 1=2 0, 6, d 5) To to be 1, K= o o s 3311 zz o o s 3322 zz v 14, 15, 1 Y 2 22 Y c o 121 12c o 2 = 7, 2mm 2930(12c o )(c o 1 n a = 165 5529151652 )12930(2a r c c o (a r c c o s 21 a n 52915c o s 302c o s 11 n 52915c o s 1292c o s 22 n ( × o×× ×× ; ×× .×× ( ) m / s 1 to b = d b1= 5 10) =( 6) ) 11, Z ; 2, ; Z ; 552915c o sc o s Z =695 =12005) 3 8 1 a of 18, 36, m=2b=1330r/40005. 30 250 90 210. 1) of ( 1) 5. 30 25040. 90 21000. ( 2) 16(b), 60050 17(b), 2010 . N、 11 4000(93016060 8812 18, 121 Z . 19, 121 Y . 4, 1.1 m S 。 M P a 600m i Z 502 M P a 2220 m i *2*2102 2) is or (If is its to 3) (s 161161 4) o 、 K= 11, Z ; 2, ; : 118/36= , =2*18=36 R= (2)/ =18/ R =b/R=13/ M P a 321 1 21 P ( 250 N 1、 4 3 4 1 s 、 3 2、 4 3 in a 0470 r/20 r/2 0 1 05 55 5 m/s, (88). 2 P 6, M P . i=n1/470/120=2 10. 3 010 8778 7 3 z z . 41 z ; 4 1) on T 41z , . 12 2) K 。 3) E s. M 160 4) m , 1d d 32221 6 6 60(7162501 )z Z(9 1, 312 , m , d , 41z ,q . 5) m / 22 6) a . 1 7) "42917)80 r c t a n ()a r c t a n ( 011 d 1) vS /2917c o 47080c o 11 2) v 9, 2010v 3) 201"42917( "40 0 tg tg v 4) 00 00(33.0 5) 20 . t 15 W/( 011 01000)1(1000 6 t is be 5 4 5a in a 132P 1440 r/i 2. is 00 1) c 6 c = .8)) c=440r/. 3) 00 =96002)1(12 dd 2004) v m / 440100100060 11 d 25m/s. 5) a, d 210 600 400 300,500 02122100 4 )()(22 a 4004 )100200()200100( =800+471+d = 1400 (1250)he of a 4612)002)( 00 385) 6) s 1 21 a dd (165>120)( 7) 00 00 200)1(1 2 d dd v=s, 1=167°, d=1400( 000 z = 6. 8) 0 220 00)00 KF c =9) on Q 0Q 6 6 is V =3000N T=510 N m 1=128, 20 he is 5. 300C. (1) (2) of of (3) 1)轴承型号 7300C 指内径未定; 2)按比例画出轴系装配图,不画轴承座 及轴承盖; 3)注意轴的转向及各分力的方向。 1 45, M P 360, 650 s 650 , b =602 ) T=510 N m, 1=128 N 1000022 t 012c o s o F N F 2) 3) n , 3241 N c, 1000/100 )( M = 100/1000N=m 000/)( (=3000× (250+160)/1000250/1000=b, 603000)( In , c, 1000/100)( M = 100/1000N=m 3. to of ) 2 c, 222)(2)()( c, 222)(2)()( is 2) 22 )( c, 2222)()( = m b, 80)( 2222)()( = m 3) he c d W , b ( 3 1 (c, 31) V 0 4 he on V d=50 (is 4036 mm 07 is 2-4 mm 70mm 312C (械设计课程设计, 030 1 5(is 2016 5 is as M P 74800003 0m M P a 725100002 3 M M P 360, 650 s -5 7 60 60 70 75 65 136 15 5 31 31 116 10 15 50 5 100 250 160 50 ) 650( a a 0 . 7 8= So is 7 3( 7a in a =60d=160n=960r/80o, is be 1. 80° 2. l/d=l/d o l=(l/d)×d= 160=. p, v, P 0000 100060 m / a 4. of on p,v,pv 5 5. L o 4 =20s r=20×10 6×900 )10)1 v 2 4 2 50 6. -5 P 2 l/d=1, so (2m m R of is 7. f=Q m 001 So is is -2(8 of to =15o. d = 35n= 1800 r/ 3390 N, 1040 N. A = 870 N. d = 35So 307C 26800N, 34200N = . 2 = : S 1= 3390N =1695 N S 2= 1040N =520 N 3 a . to =520 + 870 = 1390 . 8得 06010 3616 If is to 307