看不到什么东西啊 k\HRG@
/G
) ae/+Q8
mgH4)!Z*56
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . CO4*"~']t
k"GW3E;
A. radii of the pitch circles B. transmission ratio C. working pressure angle (MJu3t
@
L7a+ #mGE
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . 4i/ TEHQ
ffuV$#
A. pitting of tooth surfaces B. breaking of gear tooth t=A|
K
bL aD1rnGi
C. wear of tooth surfaces D. agglutination of tooth surfaces HYW+,ts'
#l8CUg~Uj
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . >UR-37g{p
$~ zqt%}
A. tooth number B. modification coefficient C. module
fwXk{P/
> {npg2
D. helix angle of helical gear 7m@pdq5Ub
%Cb8vYz~
(4) The contact fatigue strength of tooth surfaces can be improved by way of . UI>-5,X
*A
c~
A. adding module with not changing the diameter of reference circle 9OXrz}8C
1cO p"!
B. increasing the diameter of reference circle %o+bO}
/9
49Ht I9@
C. adding tooth number with not changing the diameter of reference circle LNk
3=v2M
;?h#',(p
D. decreasing the diameter of reference circle fgTv
wO Sk
fIWQ+E
(5) In design of cylindrical gear drives, b1 = b2 +(5~10)mm is recommended on purpose to . (Where b1, b2 are the face widths of tooth of the smaller gear and the large gear respectively.) q`9~F4\
.yg"!X
A. equalize strengths of the two gears B. smooth the gear drive )k29mqa`
HRx#}hN?+
C. improve the contact strength of the smaller gear v+G}n\F
rYN`u
D. compensate possible mounting error and ensure the length of contact line rw_&t>Ri;
YVu8/D@ o
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . ) : Q5u6
#8nF8J< 4
A. B. C. D. EB@!?=0x
>`,v?<>+
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. !v`q%JW(
y~ ^>my7G
A. certainly B. not always C. certainly not >2$Ehw:K^
|a\s}M1
(8) Because of , the general worm gear drives are not suitable for large power transmission. 7^|3T TK
dVe3h.,[v
A. the larger transmission ratios B. the lower efficiency and the greater friction loss =E e f
b2p<!?
C. the lower strength of worm gear D. the slower rotating velocity of worm gear /xsa-F
]*zF#Vo c
(9) In a belt drive, if v1, v2 are the pitch circle velocities of the driving pulley and the driven pulley respectively, v is the belt velocity, then . `4;<\VYCr
jN5Sc0|b
A. B. C. D. WYa DN:kZf
O8dDoP\F2
(10) In a belt drive, if the smaller sheave is a driver, then the maximum stress of belt is located at the position of going . `I
m;@_J
wQ [2yq
A. into the driving sheave B. into the driven sheave *+@/:$|U
kO2im+y
C. out of the driving sheave D. out of the driven sheave zh50]tX
cteHuRd
(11) In a V-belt drive, if the wedge angle of V-belt is 40°，then the groove angle of V-belt sheaves should be 40°. dHOz;4_
8,kbGlSD
A. greater than B. equal to C. less than D. not less than KJ pM?:
!' sDqBZ&7
(12) When the centerline of the two sheaves for a belt drive is horizontal, in order to increase the loading capacity, the preferred arrangement is with the on top. )k7`!@ID
|AWu0h\keO
A. slack side B. tight side 0!(Ii@m=N
i
[j`'.fj
(13) In order to , the larger sprocket should normally have no more than 120 teeth. <F=j6U7
.3HC*E.e
A. reduce moving nonuniformity of a chain drive twO)b"0
`_v-Y`Z
B. ensure the strength of the sprocket teeth C. limit the transmission ratio ~$Mp >ZB2W
w=^*)jZ8
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the YDQV,`S7
(U 7%Z<
chain C ]r$
>ueJ+sgH
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . qKSM*k ~
@u$oqjK
A. the less z1 and the larger p B. the more z1 and the larger p 2?}(
n1cAI|ZE
C. the less z1 and the smaller p D. the more z1 and the smaller p m ' aakq
, g6.d#c
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) 2ag]p
riBT5
(15) In design of a chain drive, the pitch number of the chain should be . >c1mwZS ;
<~Y4JMr"
A. even number B. odd number C. prime number I|`K;a
U2bb|6j
D. integral multiple of the tooth number of the smaller sprocket A+
f{j
Mv9s
I{RktO;1
kQH!`-n:T
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. -({\eL$n
f?{Y<M~]
If the maximum working stress of the element is 180MPa, the minimum working stress is -80MPa. Find the angle q between the abscissa and the line connecting the working stress point to the origin. DL~LSh
&FK=w] P
d[S!e`,iD
^Krkf4fO
.&[nS<~`
JTb<u C
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. =9)ypI-2
O nc! 5L
(1) Draw acceleration curve of the follower schematically. (Q 'XjN\#
c\x?k<=
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). '<Vvv^Er
,{oANqP
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. $rlIJwqn
ML'R[~|
2
Sr'B;`p
g = %W"v
I%(+tJ
y= &)sq
0P)c)x5
(&x[>):6?
;WzT"yW) T
NH?s
z* `8 1
}8FP5Z'Cf%
4. (8 points) Shown in the figure is a pair of external spur involute gears. :wF(([&4p!
bVVa5? HP
The driving gear 1 rotates clockwise with angular velocity while the driven gear 2 rotates counterclockwise with angular velocity . , are the radii of the base circles. , are the radii of the addendum circles. , are the radii of the pitch circles. Label the theoretical line of action , the actual line of action , the working pressure angle and the pressure angles on the addendum circles , . 1'd "O
@
ZNx{7]=a
9~4Kbmr>q
2AlLcfAW
\q,w)BE
bc 0 |tJc
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: JD'/m
h N0
3F r}8 Dy
(1) the causes of producing the elastic sliding and the slipping. - -]\z* x
hQk mB|];5
(2) influence of the elastic sliding and the slipping on belt drives. B kC(9[Ei
Rc6Rk!^
(3) Can the elastic sliding and the slipping be avoided? Why? A#p@`|H#B
;J`X0Vl$
,%:`Ll
t]$
N[mOJa:
hgh1G7A &
z,NHH):~
+6>Pp[%
9R E;5 0h
|UBJu `%
zb3ir|
im?nR+t+X
D2hvf ^g'*
\2LCpN
s}A]lY
Xrv rN^'
4C2 D wj
WRnUF [y+)
Z BYm
A D
{wm
`
_,'UP>Si
1H ZexV
NfSe(rd
^hgpeu
b! N
`@m=
;S%wPXj&
:4A^~+J
zL{KK 9Or
$L&*0$[]Q
ePF9Vzq
O
8fh'6
!nd*U}q
~vt8|OOo0
6. (10 points) A transmission system is as shown in the figure. ':fVb3A[*d
e5AiIVlv
The links 1, 5 are worms. The links 2, 6 are worm gears. The links 3, 4 are helical gears. The links 7, 8 are bevel gears. The worm 1 is a driver. The rotation direction of the bevel gear 8 is as shown in the figure. The directions of the two axial forces acting on each middle axis are opposite. N@) D,~
;TR.UUT
(1) Label the rotating direction of the worm 1. yc H=L8
4[(P>`Unx
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. &^Xm4r%u_
+XQS
-=
S2Wxf>b t2
Urt N3icph
u=4tW:W,
9\ dC8
|
0&~fY
GB{%4)%6
R[9[lQ'vR
~UL; O\-b0
7. (12 points) A planar cam-linkage mechanism is as shown in the figure with the working resistant force Q acting on the slider 4. Ps TwJLY
^c/.D*J [I
The magnitude of friction angle j (corresponding to the sliding pair and the higher pair) and the dashed friction circles (corresponding to all the revolute pairs) are as shown in the figure. The eccentric cam 1 is a driver and rotates clockwise. The masses of all the links are neglected. 'Yj /M
p^^<Bj kQ
(1) Label the action lines of the resultant forces of all the pairs for the position shown. "EA%!P:d,
g4U%(3,>D
(2) Label the rotation angle d of the cam 1 during which the point C moves from its highest position to the position shown in the figure. Give the graphing steps and all the graphical lines. g<tr |
n
8?ZK^+]y
2r\ f!m'
2'N%KKmJ
L
sN;xHT Y
rBT#Cyl
Z}WMpp^r
=%p 0r z|b
qWy(f|:hYi
Fq9A O~z
8. (15 points) In the gear-linkage mechanism shown in the figure, the link 1 is a driver and rotates clockwise; the gear 4 is an output link. fgq#Oi}
vDqmD{%4N
(1) Calculate the DOF of the mechanism and give the detailed calculating process. 3N[t2Y 1r
G
D$o |l]\
(2) List the calculating expressions for finding the angular velocity ratios and for the position shown, using the method of instant centers. Determine the rotating directions of and . DFcgUEq
I^emH+!MW
(3) Replace the higher pair with lower pairs for the position shown. &iORB
}CIH1
q3P
(4) Disconnect the Assur groups from the mechanism and draw up their outlines. Determine the grade of each Assur group and the grade of the mechanism. 0NL~2Qf_4
*2 P%731n5
Zi/ tax9C
[3"k :
%-1-y]R|
r~nD%H:}P
L 7VD ZCV
>"5 f B
K2|2Ks_CS
:{B D/6
W T~UEK'
4} . PQ {
P|j|0o,8p
IyK^` y
aG_@- -=
r ]s7a?O
` `mnk>/
7M8oI.?C|
H1Jk_@b
)+7|_7
!x
WoV"&9y
{=I:K|&
e 'I/}J
0Wkk$0h9
|qs8(
5z0
atiyQuT6Wh
9. (15 points) An offset crank-slider mechanism is as shown in the figure. 2x5^kN7
OI^??joQ
If the stroke of the slider 3 is H =500mm, the coefficient of travel speed variation is K =1.4, the ratio of the length of the crank AB to the length of the coupler BC is l = a/b =1/3. X Q3"+M_KG
#7Fdm nu`
(1) Find a, b, e (the offset). fV#,<JG
fZ aTckbE
(2) If the working stroke of the mechanism is the slower stroke during which the slider 3 moves from its left limiting position to its right limiting position, determine the rotation direction of the crank 1. J5TT+FQ
dy>5LzqK3
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
