不会吧? or`D-x)+@
BC)1FxsGf
uC|bC#;
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . WS.lDMYE7
MmQk@~
A. radii of the pitch circles B. transmission ratio C. working pressure angle gs(ZJO1 /L
GGU>={D)
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . ]Jnrs
9*S9~
A. pitting of tooth surfaces B. breaking of gear tooth .naSK`J,`
k !V@Q!>,
C. wear of tooth surfaces D. agglutination of tooth surfaces r<[G~n
l^2m7 7)
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . 'F^1)Ga$
E/2_@&U
:}
A. tooth number B. modification coefficient C. module y=2nV
9z+ZFIf7d
D. helix angle of helical gear EGzlRSgO
n^Q-K}!T/
(4) The contact fatigue strength of tooth surfaces can be improved by way of . 1cS*T>`
t{e}3}LEd
A. adding module with not changing the diameter of reference circle $lg{J$
h8
?>NX}~2cf
B. increasing the diameter of reference circle U
U3o (Yq
xHe"c<
C. adding tooth number with not changing the diameter of reference circle qG3MyK%O\
C6qGCzlG`
D. decreasing the diameter of reference circle HfEl
TC:3f
a>&dAo}
(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.) q}C;~nMD
NbK67p:
A. equalize strengths of the two gears B. smooth the gear drive ^sF(IV[>
$XcuU
sG
C. improve the contact strength of the smaller gear Qx8O&C?Ti
l)f 2T@bHl
D. compensate possible mounting error and ensure the length of contact line jxZ_-1
+&.39q!
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . G[]h1f!
2BO"mc<#$
A. B. C. D. XdE|7=+s
S<g~VK!Tt
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. SmV}Wf
k/i&e~! \
A. certainly B. not always C. certainly not vBV_aB1{
@DKph!cr
(8) Because of , the general worm gear drives are not suitable for large power transmission. ~BgNMO;|
K/$5SN1
A. the larger transmission ratios B. the lower efficiency and the greater friction loss T3t
w.yh
=WK's8FB;8
C. the lower strength of worm gear D. the slower rotating velocity of worm gear 9u>X,2gUR
!T.yv5ge'
(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 . tcD5"ALJ
i<!1s%i}
A. B. C. D. q*}$1 zb
vb$i00?
(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 . .#LHj}
u
Ci?RuZ"
A. into the driving sheave B. into the driven sheave OKnpG*)u=g
jWK>=|)=c
C. out of the driving sheave D. out of the driven sheave 58]t iP"
Mr@{3do$
(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°. E0eZal],
5KW
n >n
A. greater than B. equal to C. less than D. not less than L"}2Y3
bLO^5` 6
(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. ygxaT"3"=
Zig3WiD&
A. slack side B. tight side fwSI"cfM
[pz1f!Wn
(13) In order to , the larger sprocket should normally have no more than 120 teeth. B
\.05<
B~%SB/eu
A. reduce moving nonuniformity of a chain drive ! ~+mf^D
"Jg*
/F
B. ensure the strength of the sprocket teeth C. limit the transmission ratio _!k\~4U
X`7O%HiX/`
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the R".*dC,0'B
,9W!cD+0
chain =!=DISPo
hmv*IF.
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . B
Wzo|isv
Z
)X(
A. the less z1 and the larger p B. the more z1 and the larger p l'?(4N
)Hqn
C. the less z1 and the smaller p D. the more z1 and the smaller p YGZa##i
+`TwBN,kp-
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) nu6v@<<F>
]F4|@+\9
(15) In design of a chain drive, the pitch number of the chain should be . nW ]T-!
]u%Y
8kBe
A. even number B. odd number C. prime number dEfP272M
h[gKyxZ/t
D. integral multiple of the tooth number of the smaller sprocket 9iGp0_J
B &)wJG
8VMD304
p:Zhg{sF
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. 5,du2
V4|l7
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. #|6M*;l N|
3EyVoS6D
uD4$<rSHb
j8?rMD~
OZno 3H
n
zC!Pb{IaH
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. zmU>
7@ mP;K0
(1) Draw acceleration curve of the follower schematically. ]dnB,
86~q pN
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). 7niI65
"XKd#ncP
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. nPjN\Es6
d0-T\\U
uG4Q\,R
JPkI+0
?^
`EI}g
?yAjxoE~?
Ufe@G\uyI
G":u::hR
<_-8)abK
_E
xd:
?.,2EC=+
Y !AQ7F
4. (8 points) Shown in the figure is a pair of external spur involute gears. .0]Odf:@
jLreN#:9
The driving gear 1 rotates clockwise with angular velocity while the driven gear 2 rotates counterclockwise with angular velocity . , are the radii of the base circles. , are the radii of the addendum circles. , are the radii of the pitch circles. Label the theoretical line of action , the actual line of action , the working pressure angle and the pressure angles on the addendum circles , . 1'9YY")#
cy#N(S[ 1
D,,
x<JG|
9E]7Etfw
"9!CsloWhz
%ysfFE
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: Iih]q
Dhp|%_>
(1) the causes of producing the elastic sliding and the slipping. Q#
*Pjl
xi"Ug41)
(2) influence of the elastic sliding and the slipping on belt drives. "6o5x&H
ah0
(3) Can the elastic sliding and the slipping be avoided? Why? w}``2djR'W
7VA
et
KZ/2#`
>O}J*4A>+#
%Dm:|><V$b
Z1M{5E
pt-
1>Ui
y7Sj^muBY
{-)*.l=
=Cj
N=FM
UvM_~qo
,N`D{H"F
z+6%Ya&ls
Gu@Znh-D
7.tEi}O&_g
k;?E,!{
~TfQuIvQB
Ld3!2g2y7&
HrS
\Yd
0oe82
TkSeD
P
JLH,:2
A J"/T+g_
<<.%Gk
>@c~ M
g
tV*`g
rE&+fSBD
JW-!m8
A%c)=(,
!_SIq`5]@
W^AY:#eX~Q
nH% 1lD?:
6. (10 points) A transmission system is as shown in the figure. K7N.gT*4
"ZLujpZcG
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_Y*Z`Xb
2>!?EIE7
(1) Label the rotating direction of the worm 1. Y94/tjt
nlkQ'XGAI
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. wz(D
}N5
uepL"%.@7|
a4L0Itrp
X ?l F,p
3JR1If
oK3aW6
R1=ir# U|D
J$Qm:DC5
, JUP
(ATCP#lF
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. q4rDAQyPO
r!7e:p JLO
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. [Ifhh2
f;"6I
(1) Label the action lines of the resultant forces of all the pairs for the position shown. -=A W. Zo
a|v}L,
(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. K9J"Q4pEC
k-LT'>CWl
yU@~UCmja
X0.-q%5
Q:x:k+O-
&z\]A,=Tc
-|DSfI#j
PZdYkbj
yO6
_Gq{
Z-{!Z;T)z
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. elgCPX&:W
9o7d3 ir)
(1) Calculate the DOF of the mechanism and give the detailed calculating process. 3PonF4
Dk'EKT-
(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 . %acy%Sy
=oF6|\]{;
(3) Replace the higher pair with lower pairs for the position shown. !_`T8pJ`
!j(KbAhWZ
(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. p0.?R
sK:,c5^
8i;N|:WdH
I_yIVw;
6bg+U`&g
0;)6ZU
K)qbd~<\
Y\T*8\h_[
&F}1\6{fL
=3=KoH/'
F.=uJdl.!
e![Q1!r
AEwb'
8m A6l0
04;y%~,}U/
lkg-l<c\J
1#qCD["8
+29;T0>a
Za!c=(5
BH0rT})
^TWN_(-@
n @L!{zY
8"LaP3U
l``1^&K
Re1@2a>
N0RFPEQ~
9. (15 points) An offset crank-slider mechanism is as shown in the figure. ,.FTw,<
z% /ww
7H
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. lO $M6l
\Di~DN1
(1) Find a, b, e (the offset). @?]-5 ~3;
4_.k Q"'DH
(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. &<Gq-IN
IPxfjBC+J
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.