不会吧? C_dsYuQ5R
L^yQb4$&M
+2O('}t
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . 4%LG9hS
g2L
A. radii of the pitch circles B. transmission ratio C. working pressure angle 5)'
_3r
D{Y~kV|
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . J5)e 7
iN8?~T}w
A. pitting of tooth surfaces B. breaking of gear tooth &{9'ylv-B)
7Gnslp?[U
C. wear of tooth surfaces D. agglutination of tooth surfaces 1>jG*tr
Jy@cMq2
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . FXV=D_G}
pRx^O
F(3
A. tooth number B. modification coefficient C. module V2g,JFp&
/%po@Pm#I
D. helix angle of helical gear 6[$kEKOY=
)MZ]c)JD^
(4) The contact fatigue strength of tooth surfaces can be improved by way of . X_C9Z
UBOCd[
A. adding module with not changing the diameter of reference circle t;Jt+k~
d8<Lk9H9R
B. increasing the diameter of reference circle QtJe){(z+
Ky|Hi3?
C. adding tooth number with not changing the diameter of reference circle WLiY:X(+|
cE 8vSQ%
D. decreasing the diameter of reference circle twz
8xj4N%PA
(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.) {*~aVw {k
Fly@"W4a
A. equalize strengths of the two gears B. smooth the gear drive EAF<PMb
-kZz,pNQ,
C. improve the contact strength of the smaller gear MVjc.^
#I%< 1c%XA
D. compensate possible mounting error and ensure the length of contact line j:"+/5rV8
{-^>)
iJqt
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . H|0B*i@81
&_,^OE}K_:
A. B. C. D. 4S 2I]d
R"5/
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. \>cZ=
Mki(,Y|1~
A. certainly B. not always C. certainly not fB1JU1
AbZ:AJ(
(8) Because of , the general worm gear drives are not suitable for large power transmission. dLD"Cx
aP8Im1<A
A. the larger transmission ratios B. the lower efficiency and the greater friction loss o|]xj'
H4k`wWOk
C. the lower strength of worm gear D. the slower rotating velocity of worm gear (S5'iksx
~#I
1!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 . ~S|Vd
)K@ 20Q+0K
A. B. C. D. RK'3b/T
q5lRc=.b[
(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 . }-zx4<4BH
3(^9K2.s}
A. into the driving sheave B. into the driven sheave &HFMF
)NA
T]Tz<w W(
C. out of the driving sheave D. out of the driven sheave :U?P~HI
MuZ\<;W$
(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°. SjlkKulMF
v
bb mmv
A. greater than B. equal to C. less than D. not less than +R2
(d#?\
(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. r}f-.Fo
g'pE z
A. slack side B. tight side r2""p
Q=e?G300#L
(13) In order to , the larger sprocket should normally have no more than 120 teeth. 'QCvN
b6
Ao{wd1
A. reduce moving nonuniformity of a chain drive 3XL0Pm
CXUNdB
B. ensure the strength of the sprocket teeth C. limit the transmission ratio ZR..>=
c?q#?K
aF
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the z W+wtYV4
/&kZ)XOi
chain 4`Fbl]Q
_;'}P2&Q
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . rzqCQZHL5
v+Mt/8
A. the less z1 and the larger p B. the more z1 and the larger p B"+Ygvxb
$#KSvo{otI
C. the less z1 and the smaller p D. the more z1 and the smaller p <~:
g
@
Chj0wWZ>
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) QF%@MK0zC
[fO]oTh
(15) In design of a chain drive, the pitch number of the chain should be . P_b00",S
^]aD
LjD
A. even number B. odd number C. prime number ?5/Sa
N18Zsdrp
D. integral multiple of the tooth number of the smaller sprocket 61gZZM
`^_.E:f
@<alWBS
E"u>&uPH
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. zH|!O!3"4
rRsLl/d
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. DM>j@(uWF
)ZiJl5l@
R
jh/M`|
-.z~u/uL
|@d7o]eM|
?]s
j!7
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. >L=l
{F6
p
d]a*)m&
(1) Draw acceleration curve of the follower schematically. tSQ>P -O
jUjQ{eT
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). nwV\[E
[qid4S~r,&
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. uW
MAXGL
KeXQ'.x5O
TQnMPELh"
^*R
r x
mtJI#P
lm@<i4%$F
E#_}y}7JY
qY0GeE>N
(5'qEi
ea
Yo:>m*31
-*tP_=- Dg
*S{%+1F
4. (8 points) Shown in the figure is a pair of external spur involute gears. WFLT[j!1
n; !t?jnf.
The driving gear 1 rotates clockwise with angular velocity while the driven gear 2 rotates counterclockwise with angular velocity . , are the radii of the base circles. , are the radii of the addendum circles. , are the radii of the pitch circles. Label the theoretical line of action , the actual line of action , the working pressure angle and the pressure angles on the addendum circles , . VN;Sz,1Z
5H?`a7q N
eK3J9;X
~M43#E[oOF
2<Ub[R
Q{mls
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: f@>27&'WV
GD.mB[f*
(1) the causes of producing the elastic sliding and the slipping. RI cA)I.
8ae]tX5$
(2) influence of the elastic sliding and the slipping on belt drives. blxH`O!
nQg6
j Zf
(3) Can the elastic sliding and the slipping be avoided? Why? jskATA
/
E
jBEZL|_
&|Z:8]'P
USe"1(|E
F]SexP4:A
Wc03Sv&FZ
fm:/}7s
)"
|g&=
K'n^,
t
]?mWnEi!z
Cl4y9|
F!j@b!J8
=4+UX*&i?.
4$d|}aj
H
rWi9'
6
=@V4V} ?
)d0&iE`@
/]<0`nI.
xWn.vSos
+z;xl-*[
[$9 sr=3:
;^O^&<
{Kd9}CDAZ
=5I1[p;
E)H8jBm6w
G-bG}9vc]
6bo,x
nLQ
3s3@1>
f1hi\p0q
>sPu*8D40a
uSM4:!8
(3c,;koRR
6. (10 points) A transmission system is as shown in the figure. -H'_%~OV(
REnRpp$
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*@^c$&P
#
e+%;5\
(1) Label the rotating direction of the worm 1. \QHe 0?6
4x e:+sA.N
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. vlp]!7v
V |#B=W
7$,["cJX
VGWqy4m
S;i^ucAF
3f
eI
p^5B_r:
Xb#x^?|
tvf"w`H
'[`.&-;
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.
2Z7smDJ
+g7Iu! cA
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. .<rL2`C[c
hKt
AvTg
(1) Label the action lines of the resultant forces of all the pairs for the position shown. HR.S.(t[_
q,Q
MvUK:
(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. qaG%PH}a
I;uZ/cZ|/
Xb)XV$0
K&D
-1u
s/`4]B;2U
}1N)3~
0Q]x[;!k
7:mM`0g!
O0WzDD
2EQ:mjxk
8. (15 points) In the gear-linkage mechanism shown in the figure, the link 1 is a driver and rotates clockwise; the gear 4 is an output link. DK*2d_
vRn^n
(1) Calculate the DOF of the mechanism and give the detailed calculating process. +%U
XI$v
_BHb0zeot
(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 .
OHp 121
sBZKf8 @/
(3) Replace the higher pair with lower pairs for the position shown. wEKm3mY;
z=n"cE[KtB
(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. x8b w#
hb_YdnG
kpxd+w
p`33`25
b==<7[8
HQQc<7c",
}"Hf/{E$_"
/G G QO$'
+ :iNoDz
2+sNt6B2
Z3~$"V*ZB{
B07(15y]
xh_6@}D2J
S=eY`,'#R
?P0$n 7,
pW
w aN4
:&&Ps4\Sq
C;G~_if4PR
ZK{VQ~
)O#>ONm^
d]0fgwwGC
#wk'&XsC#z
A^bg*t,
]wER&/v"
u6awcn
R1 hb-
9. (15 points) An offset crank-slider mechanism is as shown in the figure. ncu
&<j }U
7'J}|m{7
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. L
XJ"
ct
CsQ}eW8uEf
(1) Find a, b, e (the offset). ~+q$TV
]uWx<aDB
(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. (`<l" @:_*
[JO'ta
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.