不会吧? <}1GYeP
/4,U@s)"/
0<]]q[
pr
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . N1X;&qZDd
AI3\eH+
A. radii of the pitch circles B. transmission ratio C. working pressure angle 1Ewg_/R
sW!MV v
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . s7:w>,v/
xO;Qr.3PX
A. pitting of tooth surfaces B. breaking of gear tooth Fzn#>`qG
;;n=(cM|z
C. wear of tooth surfaces D. agglutination of tooth surfaces M[KYt"v
HG[gJ7
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . l6HT}x7OiH
~)]n67Or~
A. tooth number B. modification coefficient C. module ;Yn_*M/*
8YkCTJfBGu
D. helix angle of helical gear l
?gh7m_ej
KPI[{T\`ZM
(4) The contact fatigue strength of tooth surfaces can be improved by way of . G>W:3y
Bt"*a=t;
A. adding module with not changing the diameter of reference circle "
{
Nw K
X$ 76#x
B. increasing the diameter of reference circle _<l 9j;6
~Ec@hz]js
C. adding tooth number with not changing the diameter of reference circle Aq{7WA
jp+#N
pH
D. decreasing the diameter of reference circle W^dRA xVX
5fud:k
(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.) $
iU~p
&3'zG)
A. equalize strengths of the two gears B. smooth the gear drive <N\#6m
\|Ya*8V
C. improve the contact strength of the smaller gear ~
aRcA|`
(1?k_!)T
D. compensate possible mounting error and ensure the length of contact line wvg>SfV,e
"=XRonQZ
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . "Yf?33UNZ
h*?/[XY
A. B. C. D. Y%]&h#F
Nyo,6 AA
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. t&Z:G<;
;/<J.
A. certainly B. not always C. certainly not _K|513I
H1g"09?h6o
(8) Because of , the general worm gear drives are not suitable for large power transmission. + Ek('KOF
b-,]21
A. the larger transmission ratios B. the lower efficiency and the greater friction loss bLi>jE.%.
(_O_zu8_
C. the lower strength of worm gear D. the slower rotating velocity of worm gear mbRN W
ce P1mO
(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 . Gr?[s'Ze
0q
^dpM
A. B. C. D. SH5G
d>qxaX;
(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 . x9!vtrM\Zr
d^.fB+)A3
A. into the driving sheave B. into the driven sheave sS|N.2*
BnLWC
C. out of the driving sheave D. out of the driven sheave .5x+FHu7
v/yt C/WH"
(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°. i@M^9|Gh
zbDM+;
A. greater than B. equal to C. less than D. not less than U3yIONlt
:9`T.V<?
(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. ,@Izx
V5{^R+_)Ya
A. slack side B. tight side &LS&O
B)M& FO
(13) In order to , the larger sprocket should normally have no more than 120 teeth. e/;chMCq
^^I3%6UY
A. reduce moving nonuniformity of a chain drive MMfcY
3#%
>"
Z^8J
B. ensure the strength of the sprocket teeth C. limit the transmission ratio Ppzd.=E
XKU+'Tz
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the |QvG;{!
YEWHr>&Z
chain :_q
pCQB<6&1N
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . ,xew3c'(W
o AkF
A. the less z1 and the larger p B. the more z1 and the larger p
bg'B^E3
M[ (mH(j
C. the less z1 and the smaller p D. the more z1 and the smaller p IwIk;pB O
DG x9 \8^
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) Eg"DiI)7
ai(<"|(
(15) In design of a chain drive, the pitch number of the chain should be . ^IxT.g
oVsj
Q
A. even number B. odd number C. prime number L)'JkX J
U
n#7@8,
D. integral multiple of the tooth number of the smaller sprocket 8z^?PZ/
!<!sB)
:
5G3uN+\
8{HeHU
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. fv!l {
:z^ps0
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~g_rcG
C+Wb_
GdavCwJ
CJ_X:Frj)
GV1\8OG7
$V~r*#$.
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. cI4%zeR
WIXzxI<)
(1) Draw acceleration curve of the follower schematically. !Y-MUZ$f
-vI?b#
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). 5j`"@C5;O
p2PD';"
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. {a9Z<P
|v8 >22y
TH/!z,(>
={b/s31H:
]Ole#L
z}Q
P"_x/C(]@J
Z;XR%n8
JGSeu =)
-:m;ePK
JiP]FJ;
_< 69d
fu9
y3`
4. (8 points) Shown in the figure is a pair of external spur involute gears. ;=hl!CB
[+gX6
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 , . e}f!zA
RX>kOp29
a<sEd p
p{!aRB%
0gHJ%m9s
$2Ox;+
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: (SoV2[|
>
zh%CF$
(1) the causes of producing the elastic sliding and the slipping. H]"Z_n_
;%]Q%7
(2) influence of the elastic sliding and the slipping on belt drives. [f.[C5f%"'
Hm`9M.5b
(3) Can the elastic sliding and the slipping be avoided? Why? "f~S3 ?^!2
HV&N(;@
e<l Wel
N+b"LZc
0fXdE ;M3
qU
x!-DMY
~!,'z
Z5[ t/
)1lR;fD
h;^h[q1'
Bb]pUb
KZ`d3ad
EHcqj;@m
w%VHq z$
aoco'BR F
hBLJKSv
IJs*zzR
$9+}$lpPd
OlRBvfoh8
_wCp.[3?t
|Ag~k? QC
rV"<1y:g
G! zV=p
e8a_)TU?
R6h(mPYA
Ol~sCr
+Ys<V
]V<[W,*(5
;B^G<
82YTd(yB
>+7+ gSD#:
\Bw9%P~ G
6. (10 points) A transmission system is as shown in the figure. $iEM$
@7"xDgA
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. B$ +YK%I
`5r*4N<
(1) Label the rotating direction of the worm 1. YKJk)%;+w
x4CrWm
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. hvtg_w6K
J|W~\(W6i
czi$&(N0w$
Gx|$A+U
NN\% X3ri"
kUUN2
<T)9mJYr
V()s!w
izebQVQO*
>)R7*^m{'
7. (12 points) A planar cam-linkage mechanism is as shown in the figure with the working resistant force Q acting on the slider 4. /+1+6MqRn*
&$lz@Z
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. 7Y&W^]UZ0t
89v9BWF
(1) Label the action lines of the resultant forces of all the pairs for the position shown. WogJ~N,d53
u>
XCE|D*
(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. E
y1mlW
-8-
JfxD-9U^>u
T6=, A }t-
s$4!?b$tw
kl4FVZof
!f2f
gX
0k>bsn/j
\L>3E#R-Q
5<wIJ5t
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. ;{[&&qMwU
?){V7<'?y
(1) Calculate the DOF of the mechanism and give the detailed calculating process. FloCR=^H
}enm#0Ha
(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 . m X{_B!j^
87l(a,#J
(3) Replace the higher pair with lower pairs for the position shown. u&z5)iU
b-ZC~#?|b
(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. ~,3+]ts='\
\CNv,HUm3
A:Pp;9wl
_;/onM
>Vc;s!R
m<qPj"g~L
Mhp6,JL
~iI4v#0
YaU)66=u
DdDw
Mq
fGS5{dti
D32~>J.F
:'p+Ql~c
;|7]%Z}%
|rW,:&;
R@zl?>+
f/+UD-@%m
BN\Y
N
vU|=" #
kD1[6cJ!=.
/WqiGkHV*
wP:ab
(b!`klQ
cvSr><(
q8 jI
y@
=XAFW
9. (15 points) An offset crank-slider mechanism is as shown in the figure. W<']Q_su
N-C=O
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. ?
}t[
lJ y\Ky(*
(1) Find a, b, e (the offset). *Y"j 0Yob
67,@*cK3?J
(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. WBOebv
{[W [S@+
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.