不会吧？ Dw
y|mxlFn
5JhpBx/>o=
uA cvUN-@
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . :^FH.6}x
Ds{bYK_y
A. radii of the pitch circles B. transmission ratio C. working pressure angle p4kK"
\l n
u #Bj#y!
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . B~,?Gbl+g
+?{LLD*2e
A. pitting of tooth surfaces B. breaking of gear tooth i~*6JB|
=5:kV /p
C. wear of tooth surfaces D. agglutination of tooth surfaces b{X.lz0
jR*iA3LDo
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . :`0,f ?cE
8: uh0
A. tooth number B. modification coefficient C. module gV*4{ d`
B}aW y &D
D. helix angle of helical gear T~4HeEG>uH
U@(8)[?nxn
(4) The contact fatigue strength of tooth surfaces can be improved by way of . >,rzPc)
jP<6Q|5F
A. adding module with not changing the diameter of reference circle u{dkUG1ia
dN){w _
B. increasing the diameter of reference circle WMFn #.aY5
h645;sb0
C. adding tooth number with not changing the diameter of reference circle d[E= HN
`[ZA#8Ma
D. decreasing the diameter of reference circle V=+p8nE0
"`'+@KlE
(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.) UY:Be8C A
dLf
;g}W
A. equalize strengths of the two gears B. smooth the gear drive s,D GFK
") fgQ3XZ
C. improve the contact strength of the smaller gear vJW`aN1<I3
m
7+=w>o
D. compensate possible mounting error and ensure the length of contact line -'i[/{
Aiqb*v$
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . =E~_F >SD
h5^We"}+
A. B. C. D. jank<Q &w
^Je *k)COn
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. TdoH(( nY
i-x /h -
A. certainly B. not always C. certainly not Zh?n;n}
@Z96902<t
(8) Because of , the general worm gear drives are not suitable for large power transmission. t V7{j'If
"7?x aGh8
A. the larger transmission ratios B. the lower efficiency and the greater friction loss ^SwU]e
J/[PA[Rf
C. the lower strength of worm gear D. the slower rotating velocity of worm gear V. o*`V
>Q)S-4i R
(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&~*;an7
gER(&L 4[
A. B. C. D. rE\&FVx
U.|0y =
(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 . C Y)[{r
L_IvR 4:j~
A. into the driving sheave B. into the driven sheave 3LVL5y7|
m QtOx
C. out of the driving sheave D. out of the driven sheave Btc[
_ ^^5
(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°. }oii|
=,#^
*(VwD)*
A. greater than B. equal to C. less than D. not less than xf{=~j/L
LM".]f!,
(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. hrbeTtqi
) KY U[
A. slack side B. tight side qp
]s VY
Uz7V2r%]
(13) In order to , the larger sprocket should normally have no more than 120 teeth. x}N
1Wl=8g
S,C/l1s
A. reduce moving nonuniformity of a chain drive kgRgHkAH~
cLMFC1=b
B. ensure the strength of the sprocket teeth C. limit the transmission ratio .~4DlT
[xPO'@Y
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the 0a)LZp|
rf$ eg
chain Qexv_:C
}4xz, oN
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . 4&E &{<;
e&=T`
A. the less z1 and the larger p B. the more z1 and the larger p Il<ezD{
=]sM,E,n
C. the less z1 and the smaller p D. the more z1 and the smaller p g U v`G
.BZ
3>]F3<
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) qg8T}y>
`~ R%}ID
(15) In design of a chain drive, the pitch number of the chain should be . &0euNHH;sL
B,ZLX/ c9
A. even number B. odd number C. prime number EeS VY
<ApzcyC
D. integral multiple of the tooth number of the smaller sprocket n6
AP6PK7
q,_ 1?A)
ITEd[
@^d
>*[Bq;
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. 4[m4u6z=
P 4jg]g
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. wLH
[rwPr
,W~a%8*
G+f@m,
Y,C3E>}Dq
-<WQ>mrB&
lzStJ,NPqn
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. 0 6S-3bis
B *
j
AD2
(1) Draw acceleration curve of the follower schematically. QBfsdu<@^
! &OybjQ
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). Y ;WHjW(K
^3
'7
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. ?Wm.'S'to
GT} =(sD L
OTe0[p6v
<UcbBcW,
nWAx!0G
8P'zQ:#RV
i''[ u
a,sU-w!X'
Dpa PRA)x
qm1; ^j& y
Rv T>{G~
(PmaVwF
4. (8 points) Shown in the figure is a pair of external spur involute gears. ,#P eK(
Z
^tF
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 , . YI*Av+Z)
sKT GZA
'8 ~E
<ZV
!fn
;-1yG@KG
420cJ{;A
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: v~N8H+! d
MDCK@?\
(1) the causes of producing the elastic sliding and the slipping. E}V8+f54S
SqhG\qE{Qj
(2) influence of the elastic sliding and the slipping on belt drives. [D=3:B&f
:N<ZO`l?
(3) Can the elastic sliding and the slipping be avoided? Why? ?>V4pgGCE
~97T0{E3
8Ilg[Drj*
sILkTzs w
u[_~ !y
V3'QA1$
=Zcbfo_&
b ojx:g
i6p0(OS&D
FL+^r6DQ
_
FcfNF
DbdxHuKa>
);*A$C9RA
5 fK#*(x
C ebl"3Q
"Ldi<xq%xl
s mLX NO
_h4{Sx
:?VM1!~ga
Z _oBZ
s
N b2]}; O
<!#6c :(Q
ho#]i$b}f2
;Jex#+H(:D
J; pn5k~3
d7+YCi?
gw0b>E8gZ&
] 8sVXZ
F2yc&mXyk
sGjYL>*
Hn%n>Bnl
,}i`1E 1=
6. (10 points) A transmission system is as shown in the figure. 85U ')LY
&h .E
B
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.
"wi=aV9j
FI1R7A
(1) Label the rotating direction of the worm 1. hX\z93an
Fv \yhR
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. IKhpe5}
`a}!t=~#w
0*S2_&Q)
|A".Mo_5
Wpo:'?!(M^
ZOGH.`
8<}f:9/
PYDf|S7
KD9Y
&KC^ Vn3Nj
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.
Xk[;MZ[
n5i#GvO^
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. D&q-L[tA@
H~1&hF"d
(1) Label the action lines of the resultant forces of all the pairs for the position shown. p!<Y 'G
1A;>@4iC0
(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. vy?Zz<c;
FVOR~z
}ie\-V
ml $" C
{S4^;Va1
'&{`^l/ MH
6XPf0G l
1Wzm51RU
ZW\}4q;[A
W$= MuF7R
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. Tl9KL%9
{?c `0C
(1) Calculate the DOF of the mechanism and give the detailed calculating process. _]W
{)=ap
HY[eo/nM1d
(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 . 1DPgiIG~
NLZTIZCK
(3) Replace the higher pair with lower pairs for the position shown. kWs "v6B
J0@
^h
(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. wmh[yYWc
w@U`@})r.
FFGG6r
NsYeg&>`
iT#)i3
)r _zM~jI
"
]|I;I"b
Gr
WzgO
^|KX)g
/8 / 2#`3R
1RC(T{\x
Z3Le?cMt^
_B4H"2}[Y
ZsjDe {TH
8jyg1NN D
~ x`BV+R
e-vwve
9L+dN%C
\
UCOe
Y> }[c
zX+NhT
TB
8RC7 Ei
d\Q~L 3x
3=( Gb
hVPSW# .d
tY=sl_
9. (15 points) An offset crank-slider mechanism is as shown in the figure. jr$]kLY
0'ha!4h3Z
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. Hk)IV"[R
|y+<|fb,a
(1) Find a, b, e (the offset). -X#Zn>#
t
e
e
(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. z:C
VzK,
FM\yf ]'
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
