(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . (J\Qo9Il
_iW-i
A. radii of the pitch circles B. transmission ratio C. working pressure angle +L<x0-&
l|~SVk|
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . k[;(@e@c
X5Ff2@."y|
A. pitting of tooth surfaces B. breaking of gear tooth Ke,$3Yx
9N{"ob
Z
C. wear of tooth surfaces D. agglutination of tooth surfaces hS:j$je
#
>L^W7^
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . >&2n\HR\
ao.v]6a
A. tooth number B. modification coefficient C. module d"JI4)%
IBe0?F #
D. helix angle of helical gear !JVpR]lWS
iWu^m+"k
(4) The contact fatigue strength of tooth surfaces can be improved by way of . %y}l^P5z
:6X?EbXhK
A. adding module with not changing the diameter of reference circle f1UGDC<p9
!aub@wH3
B. increasing the diameter of reference circle l:rT{l=8*
h2]gA_T`
C. adding tooth number with not changing the diameter of reference circle - Ado-'aaS
KF4}cM=.5
D. decreasing the diameter of reference circle BSib/)p
.kkrU
(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.) {
]CO;5:
"G3zl{
?GP
A. equalize strengths of the two gears B. smooth the gear drive 8h| 9;%
my^2}>wi
C. improve the contact strength of the smaller gear <X,0\U!lL
J|A:C[7 2
D. compensate possible mounting error and ensure the length of contact line hB)TH'R{:
2XN];,{
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . wkwsBi
"dO>P*k,
A. B. C. D.
4Cdl^4(LT
?g21U97Q
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. 5,<:|/r
$J]b+Bp
A. certainly B. not always C. certainly not T+y3Ph--^
.% rB-vO:g
(8) Because of , the general worm gear drives are not suitable for large power transmission. ,`ZIW
XvIrO]F-
A. the larger transmission ratios B. the lower efficiency and the greater friction loss M!
e$h?vB
2
}^fhMS
C. the lower strength of worm gear D. the slower rotating velocity of worm gear )j[rm
_
IqUp Y
(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 . xxlYn9ke
DgQw9`WA
A. B. C. D. P;#}@ /E
>,Zf3M
(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 . uNV\_'9>Y
>%~%O`+
A. into the driving sheave B. into the driven sheave ;be2sTo
`6.rTs$<
C. out of the driving sheave D. out of the driven sheave 7;UUS1
MU ;
L
7^
(11) In a V-belt drive, if the wedge angle of V-belt is 40°,then the groove angle of V-belt sheaves should be 40°. Fj`6v"h
=psX2?%L
A. greater than B. equal to C. less than D. not less than h7( R/R f
Kug_0+gI
(12) When the centerline of the two sheaves for a belt drive is horizontal, in order to increase the loading capacity, the preferred arrangement is with the on top. 7>a-`"`O
eN/o}<(e
A. slack side B. tight side yJ(p-3O5
Hvn{aLa.
(13) In order to , the larger sprocket should normally have no more than 120 teeth. R(?g+:eCpM
Hq{i-z+
A. reduce moving nonuniformity of a chain drive ZE ())W"
F9sVMV
B. ensure the strength of the sprocket teeth C. limit the transmission ratio +w
;2k w
@w)Vt$+b]
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the 0i[,`>-Av
p*T`fOL
chain 6n{`t/
Ut;,Z
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . Q0)6 2[cMm
E1U~ew
A. the less z1 and the larger p B. the more z1 and the larger p #Z'r;YOzs
73kF=*m
C. the less z1 and the smaller p D. the more z1 and the smaller p $"d< F3k
K5O#BBX=
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) ##gq{hgjb$
9V]\,mD=
(15) In design of a chain drive, the pitch number of the chain should be . zj ;'0Zu
|U*wMYC
A. even number B. odd number C. prime number n o*p`a
*
8Bvc#+B
D. integral multiple of the tooth number of the smaller sprocket mHqw,28}
j Y>BU&
WD=#. $z$
>4gGb)
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. ;(cqaB
/8\gT(@
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. tta\.ic
[b-wak})aD
FSaCbs(
EgAM
,\
Ax*=kZmH|
[whX),3>
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. VKcO]_W1
q*,g
(1) Draw acceleration curve of the follower schematically. R;2tb7 o
0{0|M8
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). w
)A@
7B@[`>5?%L
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. Z5 lE*z
!;";L5()
BU-m\Kf)
zJE$sB.f
%UEV['=
dd+[FU
?}y{tav=
g`6_Ao8
qJY'"_Q{
CRK%%;=>
$i s|B9B
Rj&V~or
4. (8 points) Shown in the figure is a pair of external spur involute gears. Y7Q
IFY's~
dgb#PxOMH
The driving gear 1 rotates clockwise with angular velocity while the driven gear 2 rotates counterclockwise with angular velocity . , are the radii of the base circles. , are the radii of the addendum circles. , are the radii of the pitch circles. Label the theoretical line of action , the actual line of action , the working pressure angle and the pressure angles on the addendum circles , . y6.Q\=
Ep/kb-~-
eK%~`Y
4a'GWzUtS
&X0qH8W
o.qeF4\d6
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: +bi%4DA
^j"*-)R
(1) the causes of producing the elastic sliding and the slipping. gwvy$H
m"MTw@}SJ;
(2) influence of the elastic sliding and the slipping on belt drives. -"u9s[L{
?n$;l-m[
(3) Can the elastic sliding and the slipping be avoided? Why? (A.%q1h
Y b57Xu
&"v h=Z-
#!,tId
l1X&Nw1W
bL`O k
B dHLow
i#$N
,kt
\ "193CW!
Do3;-yp>`
Y|bCbaF
B2UQO4[w
nK*$P +[R
;M '?k8L
D'y/pv}!
2"
(vjnfH
LD]>_P83
$E;`Y|r%WK
V!=1 !"}OG
rSU%!E+|<
`$|!h-"
!QoOL<(){
31b9pi}nf
RFyeA.
N
;5 W|#{I
7[)4k7
"y@B|
"qEHK;
o ~"?K2@T
JCNZtWF
CA[-\>J7y
*RKYdwnb
6. (10 points) A transmission system is as shown in the figure. 2[i:bksjW
XBv:$F.>$
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$I% 1
e=KA|"vxh
(1) Label the rotating direction of the worm 1. xk=5q|u_-
F Vkb9(WW
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. Y*iYr
2?;
$fSV8 n;Y
&d]%b`EXq
Sw~jyUEr
[>xwwm
]dB6--
7r#U^d(
j!9p#JK#u
1t{h)fwi
E! ;giPq*n
7. (12 points) A planar cam-linkage mechanism is as shown in the figure with the working resistant force Q acting on the slider 4. #|76dU
f Co- ony
The magnitude of friction angle j (corresponding to the sliding pair and the higher pair) and the dashed friction circles (corresponding to all the revolute pairs) are as shown in the figure. The eccentric cam 1 is a driver and rotates clockwise. The masses of all the links are neglected. w=>~pYASH
v5U\E`)s
(1) Label the action lines of the resultant forces of all the pairs for the position shown. ! 9B| `
CYA#:
(2) Label the rotation angle d of the cam 1 during which the point C moves from its highest position to the position shown in the figure. Give the graphing steps and all the graphical lines. (h> Jz
p.,o@GcL~
txM R[o_
a-\\A[E
Srw ciF
5`CPaJT$
Kr $R "
J9J[.6k8
1(6B|w5+
MkJ}dncg*
8. (15 points) In the gear-linkage mechanism shown in the figure, the link 1 is a driver and rotates clockwise; the gear 4 is an output link. Am=D kkP%
[9S\3&yoh
(1) Calculate the DOF of the mechanism and give the detailed calculating process. m[S6pqz
-kY7~yS7
(2) List the calculating expressions for finding the angular velocity ratios and for the position shown, using the method of instant centers. Determine the rotating directions of and . %Hdg,NH
&Npv~Iy
(3) Replace the higher pair with lower pairs for the position shown. 4?YhqJ
bZ c&uq_
(4) Disconnect the Assur groups from the mechanism and draw up their outlines. Determine the grade of each Assur group and the grade of the mechanism. `c@KlL*!Q
Qd"{2>
[OJ@{{U%
.m_yx{FZ=
l/[0N@r~
}5dYmny
_urv
We
a x1
V~-<VM6
I=)Hb?qT~
D&_Ir>"\
VQE8hQ37
dA-ik
J \U}U'qP
U085qKyCw
m#Rgelhk.
]$/TsN
~6!{\un
)f]E<*k'E
Z#.J>_u
)
u
)+;(Vd
[f["9(:
[ 9)9>-
/NFv?~</k
"l~wzPY)
g2L^cP>2
9. (15 points) An offset crank-slider mechanism is as shown in the figure. 1zNH[
7eb^^a?
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. TS+itU62
Z4NNrA#
(1) Find a, b, e (the offset). n%PHHu
'
6i"pJ0%
(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. ],ioY
*4G
)70-q yA
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.