不会吧？ TH"<6*f2L
DzbcLg%:W
1! j ^
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . ,Yn$X
n
5 <B*
A. radii of the pitch circles B. transmission ratio C. working pressure angle gj7'4 3
?W
QFB2,k6jN
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . SmS6B5j\R
l1.eAs5U
A. pitting of tooth surfaces B. breaking of gear tooth r]'[qaP
Y>{%,d#s_
C. wear of tooth surfaces D. agglutination of tooth surfaces _&N:%;9uD
u/CR7Y
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . e< G[!m
/uPMzl
A. tooth number B. modification coefficient C. module ? <w[ZWytm
y#P _ }Kfo
D. helix angle of helical gear &Z}}9d d
-dw/wH f"
(4) The contact fatigue strength of tooth surfaces can be improved by way of . J! "m{ 8-
MM~4D
A. adding module with not changing the diameter of reference circle bWZ
oGFT
k#2b3}(,
B. increasing the diameter of reference circle rc}=`D`
ZC &~InN
C. adding tooth number with not changing the diameter of reference circle yn2k!2]&T<
^J@
Xsl
D. decreasing the diameter of reference circle RRtOBrIedI
Zf?jnD A
(5) In design of cylindrical gear drives, b1 = b2 +(5~10)mm is recommended on purpose to . (Where b1, b2 are the face widths of tooth of the smaller gear and the large gear respectively.) R;w1& Z
oTeQY[%$
A. equalize strengths of the two gears B. smooth the gear drive +H _ /
<*opVy^
C. improve the contact strength of the smaller gear VjS A& R
\U?{m)N
D. compensate possible mounting error and ensure the length of contact line z6OJT6<'
l 4cTN
@E
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . HF%)ip+
z h4m`}p
A. B. C. D. iu'r c/=V
SA6.g2pFz
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. YfH+kDT
Il#9t?/
A. certainly B. not always C. certainly not K/LoHWy+n*
`L "{sW6S
(8) Because of , the general worm gear drives are not suitable for large power transmission. t-w4rXvF
xw rleB
A. the larger transmission ratios B. the lower efficiency and the greater friction loss <k2]GI-}h
T8J[B( )L
C. the lower strength of worm gear D. the slower rotating velocity of worm gear z,=k F I
2;=xH t
(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 . [y$sJF7;I
]+i~Cbj
A. B. C. D. fwi( qx1=}
Zae$M0)
(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 . E:ocx2dp
8_F 5c@7
A. into the driving sheave B. into the driven sheave u@-x3%W
S-k:+ 4
C. out of the driving sheave D. out of the driven sheave "i ;c )ZP
V0x;*)\PYm
(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°. $Y5m"wySZ
[q{[Avq f
A. greater than B. equal to C. less than D. not less than mxJ& IV
yk| < P\
(12) When the centerline of the two sheaves for a belt drive is horizontal, in order to increase the loading capacity, the preferred arrangement is with the on top. [~<X|_L G
9I; d>%
A. slack side B. tight side e$ 3 {URg
@Js^=G2
(13) In order to , the larger sprocket should normally have no more than 120 teeth. lU[" ZFP
\' gb{ JO
A. reduce moving nonuniformity of a chain drive j%m9y_rg}
N1 }#6YNw
B. ensure the strength of the sprocket teeth C. limit the transmission ratio aI l}|n"
R"9^FQ 13
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the .MzVc42<
? %9-5"U[
chain .BR2pf|R
;|.^ _Xs
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . {WE1^&Vk-}
LK}Ih@ f
A. the less z1 and the larger p B. the more z1 and the larger p 0GtL6M@pP
zF9SZ#{a
C. the less z1 and the smaller p D. the more z1 and the smaller p L|wD2iw
) ]/gu\90
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) 3 %ppvvQ
1Vl RdDg
(15) In design of a chain drive, the pitch number of the chain should be . b ?p <y`
! z^%$;p
A. even number B. odd number C. prime number EqN <" " 2
3C,G~)=
x
D. integral multiple of the tooth number of the smaller sprocket raWs6b4 Q
,cgFdOM.
:!SV pCt3
LH=gNFgzt
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. rd*`8B
\3M
<_73
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. XX85]49`%
uwQ~4
9g7d:zG
+-DF3(
Q]JWWKt6rV
#\|Ac*>
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. jL
}bG D
;Y6XX_
(1) Draw acceleration curve of the follower schematically. F9ys.Bc
z\d{A7
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). _d|C O
Wq5 }SM
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. %^ bHQB%
9HsiAi*
&kR*J<)V
S55h}5Y
M_I.Y1|
e^LjB/<Th
W )_|jpd[
UmR\2
cs
V< Ib#rd'
xU1_L*tu '
_VUG!?_D$5
{\WRW}iO
4. (8 points) Shown in the figure is a pair of external spur involute gears. yTxrbE
0.aIcc
The driving gear 1 rotates clockwise with angular velocity while the driven gear 2 rotates counterclockwise with angular velocity . , are the radii of the base circles. , are the radii of the addendum circles. , are the radii of the pitch circles. Label the theoretical line of action , the actual line of action , the working pressure angle and the pressure angles on the addendum circles , . S-f
.NC}:i
7es<%H
hCOCX
_
y+aKk6(_W
? K ,d
,M.phRJ-`
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: l(y,lK=YP1
]#\De73K
(1) the causes of producing the elastic sliding and the slipping. q'Nafa&a)
TE5J
@I
(2) influence of the elastic sliding and the slipping on belt drives. (e_z*o)\T
N/!( `Z,
(3) Can the elastic sliding and the slipping be avoided? Why? #`ZBA>FLaQ
2lCgUe)
N
zIA)se
Js
9SBTeJ$RZ
%a?\y_a=b
= fy\W=c
/^L <q
j s9^~:Tw
BM{*5Lf
~>H ,~</`
["#H/L]3
RrT`]1".
kL F~^/
&lg
zNC9g%
{;|pcx\L6~
N 4z[=b>
O')=]6CQ*
&Dt=[yqeG
tZv^u uEp3
!C|Z+w9Y
,6X__Z#rGT
w?zy/+N~
5
1N /XE k
XdmpfUR,13
J[0 5T1
}30Sb &"
)o~/yB7
# S/n3
U?d
I
pKpB
B=;kC#Emtf
Rnoz[1y?0
6. (10 points) A transmission system is as shown in the figure. :4-, Ru1C"
?hGE[.(eh]
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. |O 6/p7+.
7G^Q2w
(1) Label the rotating direction of the worm 1. M]p-<R\
-b{<V rZ
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. u8.
Tu7~
XVqkw@Ia4!
W6}>iB
:N _]*>
~-6_-Y|
f,E7eL@
14 ,t
/hu>MZ(\
+B1&bOb
lya},_WCq
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. $l"%o9ICG
&.?E[db"h
The magnitude of friction angle j (corresponding to the sliding pair and the higher pair) and the dashed friction circles (corresponding to all the revolute pairs) are as shown in the figure. The eccentric cam 1 is a driver and rotates clockwise. The masses of all the links are neglected. w EM=Tr/h
"Kp#Lx
(1) Label the action lines of the resultant forces of all the pairs for the position shown. ' nGUm[vh
-$o0P'Vx
(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. / @"{u0
~?Ky{jah:^
P9 '`
2c
|&JeJ0 k>~
jhf3(hx&F
"}qs +
$Q8
&TM}E
S Dil\x
.>F4s_6l
_ C?Wk:Y@
8. (15 points) In the gear-linkage mechanism shown in the figure, the link 1 is a driver and rotates clockwise; the gear 4 is an output link. : gU5C Um
3r kcIVO
(1) Calculate the DOF of the mechanism and give the detailed calculating process. jW`JTh oq
8i`>],,ch
(2) List the calculating expressions for finding the angular velocity ratios and for the position shown, using the method of instant centers. Determine the rotating directions of and . :}v-+eIQ
K#wA ;
(3) Replace the higher pair with lower pairs for the position shown. s/IsrcfM
G$i)ELs
(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. WsV"`ij#
!B%em%Tv
pVy =rS
-
0Xh_.PF
\s5Uvws
a P
MMFw T(l<1
w_56y8Pd4
mgjJNzclL
`3rwqcxA
0!|d .jZI
&ry*~"xoh
/YvXyi>^"%
]5Uuz? :e
^}=)jLS
'K,\
u< BU4c/p
o{
s2T)2
Lj1l ]OD
Kq")\Ha,f
m@2E ~m
lYx _8x2
") |/\ w,
9jkz83/+<
;t&q |}x"
*'/,
9. (15 points) An offset crank-slider mechanism is as shown in the figure. 3Zb%-_%j
Z,).)y#B
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. +K 4XMf
2}vi bDq p
(1) Find a, b, e (the offset). v ccH(T
r6DLShP-Ur
(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.
s
fM"!{7
ZoG @"vr2
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
