不会吧？ }
|
jw
P5pu
J X{_,2*$
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . YLuf2ja}X
}b/P\1#z
A. radii of the pitch circles B. transmission ratio C. working pressure angle KtGbpcS$f
l2I%$|)d
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . K~B
3 FV -&Y
A. pitting of tooth surfaces B. breaking of gear tooth h
J0U-m
m=SI *V
C. wear of tooth surfaces D. agglutination of tooth surfaces 7VL|\^Y `q
m.< u !MI
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . ?'_6M4UKa
B9Hib1<8
A. tooth number B. modification coefficient C. module MdmN7 >
V1~@
D. helix angle of helical gear </u=<^i re
: q?#$?
(4) The contact fatigue strength of tooth surfaces can be improved by way of . %D5F7wB
P-3f51 Q
A. adding module with not changing the diameter of reference circle kDI?v6y5
_^NyLI%
B. increasing the diameter of reference circle T6QRr}8`/J
M X8|; t
C. adding tooth number with not changing the diameter of reference circle !& 9(D ^
:mW<
E
D. decreasing the diameter of reference circle /;Hr{f jl{
mEsb_3?#+
(5) In design of cylindrical gear drives, b1 = b2 +(5~10)mm is recommended on purpose to . (Where b1, b2 are the face widths of tooth of the smaller gear and the large gear respectively.) zRou~Kxi
.Z
67
A. equalize strengths of the two gears B. smooth the gear drive ],k~t5+
X-(4/T+v
C. improve the contact strength of the smaller gear _OyQ:>M6P
(:HT|gK oE
D. compensate possible mounting error and ensure the length of contact line `$ S&:Q,
{: |3V 7X
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . /s}
"0/Y\
1 l"2 ~k
A. B. C. D. ,(pp+hNq
6#!CBY^ {
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. %l!?d`?
ao"2kqa)r
A. certainly B. not always C. certainly not PXkpttIE]M
Cj<8r S4+
(8) Because of , the general worm gear drives are not suitable for large power transmission. S`U8\KTi
(8)9S6
A. the larger transmission ratios B. the lower efficiency and the greater friction loss ,1#? 0q
o_BTo5]
C. the lower strength of worm gear D. the slower rotating velocity of worm gear vxj:Y '}
}o*A>le
(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 . rB$~,q&.V
'<BLkr# @
A. B. C. D. $R/@8qnP
W
suwR`2
(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 . $t.oGd@N
Hf?@<4
A. into the driving sheave B. into the driven sheave EXCE^
Vw
Rz g;GH
C. out of the driving sheave D. out of the driven sheave B_Q{B|eEt&
aC\f;&P >
(11) In a V-belt drive, if the wedge angle of V-belt is 40°，then the groove angle of V-belt sheaves should be 40°. Lz. khE<
x2"1,1%H7
A. greater than B. equal to C. less than D. not less than .i RKuBM/
Lc{arhN
(12) When the centerline of the two sheaves for a belt drive is horizontal, in order to increase the loading capacity, the preferred arrangement is with the on top. S > ~
f.
.Rk8qR
B
A. slack side B. tight side I.jqC2G
ZyUcL_
(13) In order to , the larger sprocket should normally have no more than 120 teeth. g2|Myz)
E{u6< B*
A. reduce moving nonuniformity of a chain drive hLLSmW (
5;/n`Bd
B. ensure the strength of the sprocket teeth C. limit the transmission ratio #y:D{%Wp
x950,`zy
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the 8~C_ng-wn
UPuoIfuqI
chain u;_h %z5K
u[b |QR=5
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . \sAaVdZJH(
lI=<lmM0|/
A. the less z1 and the larger p B. the more z1 and the larger p 5[
zN M
{JgY-#R?{(
C. the less z1 and the smaller p D. the more z1 and the smaller p
]$@a.#}
^HC 6v;K
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) W "'6 M=*
AEyvljv
(15) In design of a chain drive, the pitch number of the chain should be . b5NVQ8Mq
]
4dl6T
A. even number B. odd number C. prime number ' k,2 *.A
YEx gUE|
D. integral multiple of the tooth number of the smaller sprocket U N9hZ> 9
tqGrhOt
h@Q^&%w
cV+?j}"*+
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. g g
/`{
w 9dkJ o
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. :o
'x?]
4Z)s8sD KW
BX >L7 n
\ vA*dQ-
G !q[NRu
4ijtx)SA
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. w'
7sh5
AFrJ zh:V[
(1) Draw acceleration curve of the follower schematically. bKQho31a'
I!sT=w8V
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). n>Ff tVZNJ
JVc{vSa!rm
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. y\?ey'o
n4K! Wv&u
#<*Vc6pC
[9; @1I<x
4>KF`?%4
7~7L5PRW
MEDh
9d5$cV
QNNURf\[(
V b0T)C
( S[z
oZ'a}kF
4. (8 points) Shown in the figure is a pair of external spur involute gears. p+d-7'?I
DR# " 3
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 , . 28MMH
Q
?.~ 1%l !
98[uRywI
M(-)\~9T
%8'8XDq^8
|L
2 >|4
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: 384n1?
{2\Y%Y'}*
(1) the causes of producing the elastic sliding and the slipping. ] .d2C J'
78& |^sq
(2) influence of the elastic sliding and the slipping on belt drives. ?
wS}'
r4yz{^G
(3) Can the elastic sliding and the slipping be avoided? Why? u ?^V4 +V
iN;Pg _Kq
RT [p!xL
p Cz6 [*kC
=KE7NXu]-
bZu'5 +(@
mqSVd^
jv)+qmqo!
"HOZ2_(o
s QDgNJbU
)%(ZFn}
)G P;KUVae
%cF`x_h[j
]H1mj#EWU
i<uU_g'M
v+ $3
is.t,&H4P]
Cfi{%,em
o o'9ZE/%
C ,| &
>" W^|2R
LU3p CM{
&PMQ ]B
:jiEn
y
Ycr3$n]e
HE}0_x.
~Ztn(1N
KzHN|8 $o
(5T>`7g8
xm m,- u
~j!n`#.\
AR | 4^
6. (10 points) A transmission system is as shown in the figure. YidcV lOsO
{gD ED
The links 1, 5 are worms. The links 2, 6 are worm gears. The links 3, 4 are helical gears. The links 7, 8 are bevel gears. The worm 1 is a driver. The rotation direction of the bevel gear 8 is as shown in the figure. The directions of the two axial forces acting on each middle axis are opposite. <\229
4$-R|@,|_
(1) Label the rotating direction of the worm 1. lm`* x=x
"P !
.5B
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. jCqz^5=$
_zj^k$ j
b(K"CL\p
'F W?
XIcUoKg^
8xy8/UBIk0
TXImmkC
\M`qaFan5^
zTng]Mvx
[ycX)iM
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. 7Kw'Y8
!2 t7s96
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. -y. cy'$f
KjGu !B
(1) Label the action lines of the resultant forces of all the pairs for the position shown. eRUdPPq_d
fD!c t; UK
(2) Label the rotation angle d of the cam 1 during which the point C moves from its highest position to the position shown in the figure. Give the graphing steps and all the graphical lines. FC0fe_U(F
12`u[O}\}-
ZTR9e\F
'!m6^*m|c
1T|f<ChIF<
Nh \vWAz9
:FAPH8]
R+LKa Z
tF;0P\i
^s)`UZ<C=
8. (15 points) In the gear-linkage mechanism shown in the figure, the link 1 is a driver and rotates clockwise; the gear 4 is an output link. O3 sV)
g W tc3
(1) Calculate the DOF of the mechanism and give the detailed calculating process. 0B.Gt&O al
tSHW" R
(2) List the calculating expressions for finding the angular velocity ratios and for the position shown, using the method of instant centers. Determine the rotating directions of and . `n&:\I b
R4p Pt
(3) Replace the higher pair with lower pairs for the position shown. Tpl]\L1v-
PK]3uh
(4) Disconnect the Assur groups from the mechanism and draw up their outlines. Determine the grade of each Assur group and the grade of the mechanism. & ijz'Sg 3
8c5YX
zYzV!s2^
3U<m\A1
3;>ls~4
Fi``l )Tt
& M13F>!
GLsa]}m,9
(=j;rfvP
/ l$enexSt
/3,/j)`a
<- \|>r Q
G80N8L m
w;lpJ B\
?+] =|hN
G4(R/<J,BQ
? c#s}IH
e a=E/HR-
hU$a Z
,iV%{*p]
4p"' ox #
H$%MIBz >$
0GUJc}fgvN
x?L[*N_ml
kyHli~Nr"
M`Q$-#E:
9. (15 points) An offset crank-slider mechanism is as shown in the figure. D
/QLp3+o
gWr7^u&q@|
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. NsB]f{7>8+
!+SL=x y!{
(1) Find a, b, e (the offset). 72, m c
]?}>D ?5
(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. SW9
C
8 Q
YdhV
a!Y
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
