不会吧？ H}@|ucM"\
t b
i;X=5
P o_9M4k
U
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . C VU DN2
-Rvxjy)[N
A. radii of the pitch circles B. transmission ratio C. working pressure angle 3}+/\:q*
@N(jd($E
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . /}VQzF
<hea%6
A. pitting of tooth surfaces B. breaking of gear tooth AVJk
}/L YI
C. wear of tooth surfaces D. agglutination of tooth surfaces }n.h)Oz
Za} |Ee
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . f 4 _\F/
5A)w.i&V
A. tooth number B. modification coefficient C. module BOWTH{KR<<
8iCI s=06
D. helix angle of helical gear ELPJ}moWZ
{,
|"Rpd
(4) The contact fatigue strength of tooth surfaces can be improved by way of . X=hgLK^3<,
*"5a5.`%,
A. adding module with not changing the diameter of reference circle y"hM6JI
d{he
B. increasing the diameter of reference circle q^cF D
&SM$ oy#?
C. adding tooth number with not changing the diameter of reference circle P=@lkF!\#
Q. Xs Y.{
D. decreasing the diameter of reference circle pxbNeqK@p
s,k1KTXg<B
(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.) =,
0a3D6b
F:g{rm[
A. equalize strengths of the two gears B. smooth the gear drive z]YhQIU4n8
'k67$H
C. improve the contact strength of the smaller gear EVL;"
##_Za6/n
D. compensate possible mounting error and ensure the length of contact line O gS8.wX
"6d bRo5%
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . @e,Zmx
{|h"/
A. B. C. D. w3N%J>4_E
iq:[+
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. ,m?D\Pru
F"H!CJJu&
A. certainly B. not always C. certainly not ] )L'Rk#4
5ecz'eA%
(8) Because of , the general worm gear drives are not suitable for large power transmission. N)*e^ Nfb
D:E_ h
A. the larger transmission ratios B. the lower efficiency and the greater friction loss "V0:Lq
FBAC9}V"
C. the lower strength of worm gear D. the slower rotating velocity of worm gear kV3j}C"
x2sOEkcQ
(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 . !O6Is' %B
1!wEXH(
A. B. C. D. yn[ZN-H~
9sj W
(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 . SWsv,
mMAr8~ A=
A. into the driving sheave B. into the driven sheave t/WnDR/fM
7B\(r~f`t
C. out of the driving sheave D. out of the driven sheave r_,m\'~s !
YIQ]]q8R!L
(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°. /_OZ1jX
<`_OpNxqW
A. greater than B. equal to C. less than D. not less than 7 eQoc2X2
^s ~n[
(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. %)Uvf`Xhh4
E
D^rWE_
A. slack side B. tight side ?U/Wio$@
Y+u-J4bj
(13) In order to , the larger sprocket should normally have no more than 120 teeth. {dA
~#fW<
ho]!G4
98
A. reduce moving nonuniformity of a chain drive Y `7#[g
# G +
B. ensure the strength of the sprocket teeth C. limit the transmission ratio TflS@Z7C
W+f&%En
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the B+e_Y\B u
,J (5@8(>a
chain wz* A <iU
r d 6F"W
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . 8yWu{'G
8f,'p}@!d
A. the less z1 and the larger p B. the more z1 and the larger p HA9Nr.NqC@
|aS~"lImh
C. the less z1 and the smaller p D. the more z1 and the smaller p <duBwkiG
s"=F^#
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) |)?aH2IL
1t[;` iZ
(15) In design of a chain drive, the pitch number of the chain should be . H6ky)kF&
{yHB2=nI
A. even number B. odd number C. prime number RO%tuU,-
[-E{}FL|
D. integral multiple of the tooth number of the smaller sprocket -eR!qy:.]5
5X `w&(]m
jom} _
4+5OR&kxZ
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. GX7 eRqz >
sb}K%
-
If the maximum working stress of the element is 180MPa, the minimum working stress is -80MPa. Find the angle q between the abscissa and the line connecting the working stress point to the origin. @ .Z[M
'9-axIj70
Mhv1K|4s
>@)*S n9"
@ ],6SKbG6
O t *K+^I
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. 3$?9uMl#
NK7H,V}T
(1) Draw acceleration curve of the follower schematically. 5) d,G
9
4 _Qa =T8
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). }B I~am_
t7u m
[
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. a=J@y K
UF&0
& `@
r6d0x
H GO#e
/d0Q>v.g
,f8}q]FTA
)XLj[6j0
N]1V1c$G*
$}qDV>
qo
|d,F-9
iw
.S~@BI(|<
LP.HS'M~u
4. (8 points) Shown in the figure is a pair of external spur involute gears. 2y3?!^$
UWQtvQ
f
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 , . +7|
[b
@GN(]t&3
b;S6'7Jf9
fNm E, ~
X AQGG >
_Zh2eXWdjM
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: $Mdbt o~ <
R<"2%oY
(1) the causes of producing the elastic sliding and the slipping. Z;ZuS[Z
A
wk" l[cH>
(2) influence of the elastic sliding and the slipping on belt drives. L
;
6b+I
dZPW2 yf
(3) Can the elastic sliding and the slipping be avoided? Why? EJ1Bq>u 7
10mK}HT>4B
z 8IPhE@
3 pB}2 ]
GxxDY]!
ODm&&W#*
>^dyQyK
v2d<o[[C
BWW q4mdb{
_}D%iJg#
.Y;b)]@f
:pRF*^eU
m+?N7
[!EXMpq'
yf!,4SUkU
nTPB,QE<
Pg`JQC|
;+sl7qlA4
t-WjL@$F/
o]<jZ_|gB
PY?8 [A+
6JhMkB^h
L7] ]ZAH!1
Ea-bC:>
b^x07lO
`z9J`r= I
:$?Q D
`?"6l5d.]
B8 H75sz
>.e+S?o
8u>gbdU
NgVR,G|1
6. (10 points) A transmission system is as shown in the figure. v8m`jxII64
=\ Iu$2r`
The links 1, 5 are worms. The links 2, 6 are worm gears. The links 3, 4 are helical gears. The links 7, 8 are bevel gears. The worm 1 is a driver. The rotation direction of the bevel gear 8 is as shown in the figure. The directions of the two axial forces acting on each middle axis are opposite. a[s KE?
yzR=A%V8A
(1) Label the rotating direction of the worm 1. *)'V vu<
=xL )$DTg)
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. /aY pIMi9}
L&kr {7q
<^8*
<;PaG
XN#&NT{t}
$ J1f.YE
C5*xQlCq}
ut_pHj@
/7/0x ./{
1$pb (OK
gl8Ib<{
7. (12 points) A planar cam-linkage mechanism is as shown in the figure with the working resistant force Q acting on the slider 4. S_ b/DO
^A9 M;q
The magnitude of friction angle j (corresponding to the sliding pair and the higher pair) and the dashed friction circles (corresponding to all the revolute pairs) are as shown in the figure. The eccentric cam 1 is a driver and rotates clockwise. The masses of all the links are neglected. ;b0NGa(k
D#Kuo$
(1) Label the action lines of the resultant forces of all the pairs for the position shown. `VT>M@i/
_mJG
5(|
(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. W\ cjd d
taWqS q!
+v7) 1y
!wE}(0B Tx
l dqLM
!%)]56(
vScjq5 "p
s wcd&~9r
f&KdlpxKv
-i jC_ `>
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. & ,: h)
9902+pW
(1) Calculate the DOF of the mechanism and give the detailed calculating process. G`0V)S
g>JL DQdc
(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 . ";SiL {Z
(s<s@`
(3) Replace the higher pair with lower pairs for the position shown. i^msjA
L%"LlS g
(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. 6gL-OJNo
LeOP;#
}H> ^o9
4T6 {Y
64>krmVIe
@(g_<@Jz
uVE.,)xz
PNs ~[
Pe7 3g %
$;N* c H~
*cuuz
i&
\y`+B*\i
kR CQv-*
b;sVls
\NZIEu)5?
w' MGA
*iBTI+"]
OJ |r6
,l YE
^& @w$
EP'h@zdz
Q/ uwQ o/
[qUN 4x5b
jp8@vdRg
,mhQ"\ +C
)a 9 ]US^
9. (15 points) An offset crank-slider mechanism is as shown in the figure. y&} E~5O
x3jb%`o#!
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-OA([;/
!tfb*@{;'
(1) Find a, b, e (the offset). U*Ge<(v$
L^2 FQti>
(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. D]w!2k%V
5^']+5_vb
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
