不会吧？ dE[X6$H[
Lnn^j#n
"k [$euV
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . Fgwe `[
= r`E%P:
A. radii of the pitch circles B. transmission ratio C. working pressure angle ^
5 "yY2}-
'5T:*Yh
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . Avljrds+7
5>1c4u`x
A. pitting of tooth surfaces B. breaking of gear tooth }Sh@.3*
{
}"
<
C. wear of tooth surfaces D. agglutination of tooth surfaces cqZ lpm$c
DBvozTsF~
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . #>5T,[{?j
|P9)*~\5
A. tooth number B. modification coefficient C. module 0H}O6kU
M M @&Q aK
D. helix angle of helical gear a{By U%
T`I4_x
(4) The contact fatigue strength of tooth surfaces can be improved by way of . nM8'="$
$Zrc- tkV
A. adding module with not changing the diameter of reference circle * a(GG
TPN1Rnt0`
B. increasing the diameter of reference circle \Q?#^< O
U Rr{J}5
C. adding tooth number with not changing the diameter of reference circle v&66F`
mI^S% HT
D. decreasing the diameter of reference circle RM25] hx
="g9
>
(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.) $ O;a~/T
= |
r%
lx
A. equalize strengths of the two gears B. smooth the gear drive ?*QL;[n1
@_#]
7
C. improve the contact strength of the smaller gear [/J(E\9
/)We g1b
D. compensate possible mounting error and ensure the length of contact line ZZeF1y[q
+5Y;JL<%/
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . %W2
o`W$
ae+*gkPv8
A. B. C. D. 5*y6{7FLp
;l@Ge` &u
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. NQd0$q
R8j\CiV17
A. certainly B. not always C. certainly not T'14OU2N{Y
o<Rrr,
(8) Because of , the general worm gear drives are not suitable for large power transmission. +'9xTd
3pkx3tp{
A. the larger transmission ratios B. the lower efficiency and the greater friction loss P[P]oT.N
|f[:mO
C. the lower strength of worm gear D. the slower rotating velocity of worm gear 7I:<i$)V
M9(Kxux#
(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 . T FK#ign
W=!D[G R
A. B. C. D. kE{-h'xADD
3TT?GgQ
(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 . ].x`Fq3
{MmK:C
A. into the driving sheave B. into the driven sheave xcXnd"YYE
$ &0\BvS
C. out of the driving sheave D. out of the driven sheave :h<QM$P<
"l-b(8n
(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°. m^1'aO_;q
.:TSdusr~
A. greater than B. equal to C. less than D. not less than 2NWQiSz
(fC U+
(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(5&%}QFQ
7V;wCm#b
A. slack side B. tight side 9*xv
,Yz8
0/b3]{skK
(13) In order to , the larger sprocket should normally have no more than 120 teeth. ]iH~ 1 [
2~W8tv0^b2
A. reduce moving nonuniformity of a chain drive `&o>7a;
sj% \ lq
B. ensure the strength of the sprocket teeth C. limit the transmission ratio aumM\rY
[O^mG
9
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the 6vebGf
e4Jx%v?_P
chain 4[f>kY%[
q
`^ 5<
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . oa8xuFu(n
vI+X9C?
A. the less z1 and the larger p B. the more z1 and the larger p Q`[J3-Q*{
iig@ $
i#
C. the less z1 and the smaller p D. the more z1 and the smaller p 3^p;'7x
{t QZqqdn@
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) E|om C_h
OE_A$8L
(15) In design of a chain drive, the pitch number of the chain should be . ?<eH!MHF
$ 2PpG|q
A. even number B. odd number C. prime number l
*>, :y
>r5s>A[YC
D. integral multiple of the tooth number of the smaller sprocket nM6/c
eU m,=s
G}p* oz~
?cD2EX%(
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. E iWsVic[
g~Agy
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. w2e 9Ue~WH
:x /L.Bz
k&S
I -jxj
Ob>M]udn
M|uWSG
U
fAN)SE"
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. ju {\7X5
$ g+q;Y~i0
(1) Draw acceleration curve of the follower schematically. bf1EMai"
m03]SF(#3
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). } qn@8}
|>L|7>J{<d
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. ~k[q:$ T
etkKVr;Kv
pX v@ QD#!
>@wyiBU
|#_`aT"
X9
oxni#
5<?s86GHh'
#Kd^t =k
KoL3CA"N
s GdlS&08(
g<&n V>wF
xesZ 7{ o
4. (8 points) Shown in the figure is a pair of external spur involute gears. 4 O OH
3O
="*:H)
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 , . FJl_2
vOKWi:-U
_K5R?"H0
S"wn0B$"
i^zncDMA
[^4)3cj7}
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: h
&R1"
byW9]('e
(1) the causes of producing the elastic sliding and the slipping. 9< $n'g
w xx3']:
(2) influence of the elastic sliding and the slipping on belt drives. QSxR@ hC
NEb M >1>^
(3) Can the elastic sliding and the slipping be avoided? Why? XzBnj7E
I}S~,4
h /a|-V}m&
WE|-zo
SZhW)0
x]=s/+Y
UHh7x%$n
,sqx xq
OdrnPo {
k nsTy0]
_3'FX# xc
@*VfG CQ(
$xK2M
$gv3Up"U
^z%ShmM&LZ
Yi5^# G
>$R Q
iW
# |N^
y _7lSo8<
?m(]@6qa
1 [g -f ,
WE*L=_zDS
xRI7_8Jpyn
,DZo E~
e` QniTkT
5vmc'Om
r+D ?_Lk
z[0+9
=<Y
sYG:\> }ie
wi gs1
sMn)[k
vX
j TyR+#Wn
6. (10 points) A transmission system is as shown in the figure. Z"pC
DW)
l+vD`aJ 3
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. J\@W+/# dF
%6%~`((4
(1) Label the rotating direction of the worm 1. )-s9CWJv
w wuM!Z+
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. Lhz*o6)
@fc-[pv
1o8C4?T&
Hh1]\4D, 4
mp:xR ^5c
0R-J
\
2bnYYQ14
:
gr^T L1(
bl3?C
k6$Ft.0d1Z
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. ZamOYkRX
\:m1{+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. ;TC "n!ew
Xmm) z
(1) Label the action lines of the resultant forces of all the pairs for the position shown. 0 L $[w
i,B<k 0W9
(2) Label the rotation angle d of the cam 1 during which the point C moves from its highest position to the position shown in the figure. Give the graphing steps and all the graphical lines. ,x=S)
t
vk4Q2P
srQGqE~
eev-";c
AnK~<9WQj
dSI <s^n
>/4N :=.h
}0Ie Kpu5
mKBO<l{S
ni8 5Ne$
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. c`@";+|r
a D+4uGN
(1) Calculate the DOF of the mechanism and give the detailed calculating process. 1Ror1%Q"?
4a=QTq0 p
(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 . h343$,))u
klR\7+lK
(3) Replace the higher pair with lower pairs for the position shown. W\2 ']7}e
3r
Y\y+m
(4) Disconnect the Assur groups from the mechanism and draw up their outlines. Determine the grade of each Assur group and the grade of the mechanism. 61w
({F
` {qt4zd0
B@U'7` v
nMK$&h,{
}xry
U&ytZ7iB
THlQifA!
Z9.0#Jnu
Eq%f`Qg+1E
I~;w Q
Re2&qxE
tpb lm|sW
mPhu#oK'f
:..WL;gC
ap8q`a{j^
I tb_ H
@mJ~?d95v
?;_*8Doq-a
;?k<L\zaw
OIJT~Z}
sPG500=)
|Iw glb!k
O ,9^R
5|=J\Lp2I
4y#XX[2Wj
ALy7D*Z]w
9. (15 points) An offset crank-slider mechanism is as shown in the figure. .$H"j>
641P )
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. BM/o7%]n
w|6/ i/X
(1) Find a, b, e (the offset). P1Z"}Qw
k
#zDY*kj
(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. aop
Z-^
DnFzCJ
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
