不会吧? (`Q_^Bfyl
(
j~trpe,
atWAhN
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . h3
:dO|Z
[KjL`
A. radii of the pitch circles B. transmission ratio C. working pressure angle a0x/ ?)DO
xo*[
g`N
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . o
Gi{d5
gL;tyf1P
A. pitting of tooth surfaces B. breaking of gear tooth YNi3oG]h
Cdd
+I5~
C. wear of tooth surfaces D. agglutination of tooth surfaces T5di#%
: s
~2*8pb 4
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . ocT.2/~d
A PR%ZpG
A. tooth number B. modification coefficient C. module gRdg3qvU
FucLcq2Z
D. helix angle of helical gear Wc)f:]7
r-a/vx#
(4) The contact fatigue strength of tooth surfaces can be improved by way of . f
i3 <
Kr]`.@/.S
A. adding module with not changing the diameter of reference circle Q!VPk~~(
L?n*b
B. increasing the diameter of reference circle }Ik{tUS$
(9J,Qs[;
C. adding tooth number with not changing the diameter of reference circle 5? s$(Lt~
nEM>*;iE
D. decreasing the diameter of reference circle <j.bG 7
:d<;h:^_
(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.) \5_7!.
2Ek6YNx
A. equalize strengths of the two gears B. smooth the gear drive F}A@H<?
M ,8r{[2
C. improve the contact strength of the smaller gear x(z[S$6Y\
'@u/] ra:
D. compensate possible mounting error and ensure the length of contact line /)xG%J7H
4XDR?KUM
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . OedL?4
9/MU
zt
A. B. C. D. \c+)Y}:D
Te~"\`omJ3
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. m u(HNj
!Ljs9 =UF
A. certainly B. not always C. certainly not dfe 9)m>
glh2CRUj
(8) Because of , the general worm gear drives are not suitable for large power transmission. 8h~v%aZ1
:*e0Z2=
A. the larger transmission ratios B. the lower efficiency and the greater friction loss k'wF+>
G.O0*E2V
C. the lower strength of worm gear D. the slower rotating velocity of worm gear l'VgS:NT
y>u+.z a|
(9) In a belt drive, if v1, v2 are the pitch circle velocities of the driving pulley and the driven pulley respectively, v is the belt velocity, then . <oPo?r|oM|
r^&{0c&o
A. B. C. D. t(Cd
oE,6
4-=> >#
P
(10) In a belt drive, if the smaller sheave is a driver, then the maximum stress of belt is located at the position of going . iqghcY)
i)\`"&.j>N
A. into the driving sheave B. into the driven sheave "tUXYY
^DVr>u
C. out of the driving sheave D. out of the driven sheave 9'Y~! vY
Q$W0>bUP
(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°. JTpKF_Za<
0[xu
m
A. greater than B. equal to C. less than D. not less than n
<kcK
Wk'KN o
(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. ${KDGJ,^
2;5E
H0
A. slack side B. tight side T^aEx.`O}`
7<AHQ<#@
(13) In order to , the larger sprocket should normally have no more than 120 teeth. P`5@$1CJ
z*l3O~mZ
A. reduce moving nonuniformity of a chain drive A"\kdxC
j.sxyW?3
B. ensure the strength of the sprocket teeth C. limit the transmission ratio 9cWl/7;zXO
cG,B;kMjo
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the 5Qo\0YH
$f*N
chain ,
4jkTQ*@2
@xmO\
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . m-dne/%_
J8J~$DU\Gv
A. the less z1 and the larger p B. the more z1 and the larger p $s4 rG=q
KFdV_e5lU
C. the less z1 and the smaller p D. the more z1 and the smaller p _jR%o1Y}
CK 3]]{
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) rF2`4j&!
1<fS&)^W
(15) In design of a chain drive, the pitch number of the chain should be . `Ff3H$_*
m(s(2
wq"f
A. even number B. odd number C. prime number VljAAt
4v33{sp
D. integral multiple of the tooth number of the smaller sprocket "2i{ L '
Y
_m4:9p
U,LW(wueT
II3)Cz}xRG
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. 8s/gjEwA
}Xr-xh\v
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. \6-x~%xK
-GD_xk
TUIj-HSe
WM
.JoQ
--d<s
2qo=ud
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. ah1d0eP
}=z_3JfO
(1) Draw acceleration curve of the follower schematically. swYlp
SG_^Rd9
D
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). ;&q}G1
Y ( x_bJ
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. X^ckTIdR
MP(R2y
7Ab&C&3
k__i Jsk
oGM Ls
^hzlR[
AD#]PSB
nDuf<mw
u~[HC)4(0
usH9dys,
gyj.M`+y
&E0L7?l
4. (8 points) Shown in the figure is a pair of external spur involute gears. s:3 altv
5Ds/^fA
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 , . 5b45u 6
|x@)%QeC
NAjY,)>'K
JBfDz0P
;iR(
Ir
I$/*Pt];
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: 9pUvw_9MY
:f%FM&b
(1) the causes of producing the elastic sliding and the slipping. CF"$&+ s9
7K.&zn
(2) influence of the elastic sliding and the slipping on belt drives. Z?X0:WK
_J(n~"eR
(3) Can the elastic sliding and the slipping be avoided? Why? 9<u^.w
x{{QS$6v
S$J}>a#Ry
Atl`J.;G
2gz}]_
D^Ahw"X)
:K.%^ag=j
w}+#w8hu
NbOeF7cq+
!OWPwBm;
r>73IpJI
(Mi]vK.4
hY
2nT
Gr9/@U+
F*V<L
X-1<YG
C
yg e
u8ofg
cFYE
Rs
+rlJq
[h>|6%sW
kN3 T/96
)|;*[S4
s_]p6M
5
o:VixZf
u] C/RDTH
xlPUum-o
TU ]Ed*'&
~wnTl[:
)"?6Es SF
y*2R#jTA
@$FE}j_
_^4\z*x
6. (10 points) A transmission system is as shown in the figure. dz/'
m7
{MmHR
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. `|K,E
WqN=D5
(1) Label the rotating direction of the worm 1. 8@%Xd^
}Ictnb
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. [vki^M5i|Z
KDwz!:ye
$ q*kD#;mh
PoMkFG6
SsA;T5:6
\~xI#S@
Ke[doQ#c
g-'y_'%0G
h}xUZ:
4X7y}F.J
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. K%LDOVE8e
~`2w
u
l
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. 8D]:>[|E
Mq) n=M
(1) Label the action lines of the resultant forces of all the pairs for the position shown. /=Ug}%.
,2S
<#p!
(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. FzmCS@yA
O$x-&pW`g
SCeZt [
q!W~>c!
[]Cvma1\
>7!aZO
Bp\io$(%
;a!o$y
4zghM<
La%\-o
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. .G+}Kn9!
$n `Zvl2
(1) Calculate the DOF of the mechanism and give the detailed calculating process. uZZ[`PA(
gT @YG;
(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 . zcJ]US
..FUg"sSO
(3) Replace the higher pair with lower pairs for the position shown. B0=:A
p@i U}SUaE
(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. f<SSg*A;
*B
}vYX
?3SlvKI}H`
z+" :,#
v ;MI*!E
Oi
BK
yE9JMi0
|+6Z+-.Hg
n!Y.?mU6
5YS`v#+
mMN oR]
@<
vDR">
}"m@~kg=
L*FmJ{Yf
bG^eP:r
$L</{bXW
"vLqYc4$
tHoFnPd\|
yWzvE:!)
! FHNKh
e}[$ =
dUO~dV1
ULU
]k#
GQF7]j/
EVq<gGy
2#+@bk>^{
9. (15 points) An offset crank-slider mechanism is as shown in the figure. R63"j\0
s ^)W?3t]
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. B/5C jHz
a*
SJHBB
(1) Find a, b, e (the offset). EiJSLL
1M3U)U
(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. ~]"}s(J;
TP^.]IO-
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.