不会吧？ 8})|^%@n
9 t A E#A
z L8 J`W
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . mF~ys{"t
fFHK:n`
A. radii of the pitch circles B. transmission ratio C. working pressure angle {83C,C-
K4Y' B
o4
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . T +(M8 qb
*?p
^6vO
A. pitting of tooth surfaces B. breaking of gear tooth $$\V 2%v
?61L|vr
C. wear of tooth surfaces D. agglutination of tooth surfaces }1,'rm T
\#2
s4RCji
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . ",QYDFFeF
V07VwVD
A. tooth number B. modification coefficient C. module `R
eGnT[
},v&r kwR
D. helix angle of helical gear [.ey_} X8
u?H@C)P
(4) The contact fatigue strength of tooth surfaces can be improved by way of . l\_x( BH
: q(D (mK
A. adding module with not changing the diameter of reference circle )O9f hj )
\FVNXU MU
B. increasing the diameter of reference circle "~_$T@^k>
~Po\ En
C. adding tooth number with not changing the diameter of reference circle +;YE)~R?
4Q@\h=r
D. decreasing the diameter of reference circle 0u>yT?jP
g
ji*Wq
(5) In design of cylindrical gear drives, b1 = b2 +(5~10)mm is recommended on purpose to . (Where b1, b2 are the face widths of tooth of the smaller gear and the large gear respectively.) R08&cd#$
'^e0Ud,
A. equalize strengths of the two gears B. smooth the gear drive xna7kA
re]%f"v:5
C. improve the contact strength of the smaller gear YXgWH'i~
3x P~~j;7
D. compensate possible mounting error and ensure the length of contact line =[ s 8q2V
XMomFW_
@
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . Sw yaYK
z'?SRK5+
A. B. C. D. =sxkr ih
BO[+E ' 2
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. d:A\<F
_MGNKA6JI
A. certainly B. not always C. certainly not D% oueW
3gC\{y!8
(8) Because of , the general worm gear drives are not suitable for large power transmission. $[)6H7!U)
-)p@BtMS
A. the larger transmission ratios B. the lower efficiency and the greater friction loss ^cB49s+{e
yZ?xt't n
C. the lower strength of worm gear D. the slower rotating velocity of worm gear Dq{: R
>/( i3)
(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 .
-PcS(
qsD?dH i7
A. B. C. D. S Cgyp(
ll^Th >
(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 . &B?@@ 6
~C%2t {"
A. into the driving sheave B. into the driven sheave ;7;zhJs1t
t8L< x
C. out of the driving sheave D. out of the driven sheave x eJ9H~^
^RY n8I
(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°. /p[|DJo M
8dV.nO
A. greater than B. equal to C. less than D. not less than p w0Px
Q2JdO 6[96
(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. kqj;l\N
|^YzFrc
A. slack side B. tight side RY >)eGJ
D+bB G
(13) In order to , the larger sprocket should normally have no more than 120 teeth. _<u8%\
uPa/,"p
A. reduce moving nonuniformity of a chain drive !7 e
i1
+ AE&GU
B. ensure the strength of the sprocket teeth C. limit the transmission ratio X=rc3~}f
TYs+XJ'Xj
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the <dju6k7uz
F{#N6,T
chain sN} s61
Tns?m Q
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . >yV)d/
\_m\U.*
A. the less z1 and the larger p B. the more z1 and the larger p b*I& k ":
[}W^4,
C. the less z1 and the smaller p D. the more z1 and the smaller p P<j4\zJ
9@KUqo X
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) K7 t&fDI
Wrf^O2
(15) In design of a chain drive, the pitch number of the chain should be . 7Y-FUZ.`>
|<|28~#
A. even number B. odd number C. prime number
{AU` }*5
]jVIpG M
D. integral multiple of the tooth number of the smaller sprocket p56KS5duI.
!$:0E
y(S
KKeb ioW
4_6W s$x
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. wQo6!H "K
B)j`}7O 06
If the maximum working stress of the element is 180MPa, the minimum working stress is -80MPa. Find the angle q between the abscissa and the line connecting the working stress point to the origin. mhHA!:
Y
P. [6s$J
aH5t.x79b
ifJv~asp
X_-Hrp!h
1=9qAp;?o
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. 'Yaf\Hp
kbIY%\QSO
(1) Draw acceleration curve of the follower schematically. M-&^
Gx(K N57D
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). ++J Bbuzj!
n&2=6$ *,k
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. 3X-{2R/ 3
sx)$= ~o
2-"Lxe6 5f
N[ +o[%A
?{@!!te@3v
gqO%^b)6
t?%}hs\!
DU*g~{8T$
{|Mxvp*Hg
4t +/
9:CVN@E
YWcui+4p}
4. (8 points) Shown in the figure is a pair of external spur involute gears. wa(Wit"-
t(z (-G|&
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 , . Sq?6R}q%
:uT
fhr
(_%J F[W
lk+=2 6>
k<j)?_=`
N)
V7yo?
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: 6gg# Z
<@x+N%C
(1) the causes of producing the elastic sliding and the slipping. 654jS
!
{*ko=77$*
(2) influence of the elastic sliding and the slipping on belt drives. 1 ^ WA
m ISu o
(3) Can the elastic sliding and the slipping be avoided? Why? g k_X u
?^G$;X7 B
AP?{N: +
z hR_qW+
V)?x*R*T)
-fSKJo#}|
8tFyNl`c
%. 6?\w1e
("P mB?20
On9 6N|
8w@W8(3B
z7l;|T
2h? r![
[%)B%h`XGf
DRmh (T
L&gEQDPgq|
Y;/=3T7An
5Ret,~Vs9|
\%4+mgiD
u2.r,<rC*Q
*
SON>BSF
;|
$]Qq
d[ Fr
uF.\dY
\xv
j
BQqpFH9
_ ~#C $-T
hjp
?/i%TQ
3<' Q`H >
N n-6/]d#
"?0 G^zu
uu/7Ie
E~
O>m8hF
6. (10 points) A transmission system is as shown in the figure. #1!BD!u
I;m@cSJ|j
The links 1, 5 are worms. The links 2, 6 are worm gears. The links 3, 4 are helical gears. The links 7, 8 are bevel gears. The worm 1 is a driver. The rotation direction of the bevel gear 8 is as shown in the figure. The directions of the two axial forces acting on each middle axis are opposite. Gt%? [
om39;nk!}
(1) Label the rotating direction of the worm 1. Q c< O; #
7Hlh
(k
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. J1ro\"
~4X!8b_
p`.fYW:p
[Ik
B/Xbw|
@ DKl<F
8;Bwz RtgT
T`<k4ur
ZTz(NS
EK
f2e$BA
B-p5;h>
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. 90696v.
Jj!tRZT
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. xfQ;5
n
;j\$[4W.i
(1) Label the action lines of the resultant forces of all the pairs for the position shown. |N.q[>^R
4~MUc!
(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. 1ucUnNkcV
n/Z =q?_
uXa}<=O
7#*CWh1BNO
(: k`wh&
} j@@
yAkN2
4KB?g7_*
bv" ({:x
e' ;c8WF3E
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. 9cPucKuj
D)$k{v#~
(1) Calculate the DOF of the mechanism and give the detailed calculating process. O@&I.d$
\:28z
(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 . mw.aavB
g{{SY5qDj
(3) Replace the higher pair with lower pairs for the position shown. +T;qv
x6
Ghe=hhZ
(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. ~qxc!k!w4
n-Wv Iy
%uMsXa
-p~B
-,
2AE|N_v8W
S jU6+|l
L1+c v;t
L2tmo-]nw
=]Bm>67"
jlZNANR3
fTcRqov
]sE^=;Pv?
C
iF bk&-g
o
! d0
%J'_c|EQM
$
S/ 8T
uEY 5&wX`
P
V Q%y
YN3uhd[2
m-]"I8 [
`^x^=
og'
&v 88x s
Am
FHn
8 iB}a\]B
R7>@-EG
9`{cX
9. (15 points) An offset crank-slider mechanism is as shown in the figure. yDrJn*
r^
qK
pU.rP
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. w.jATMJ)F
WJWrLu92\U
(1) Find a, b, e (the offset). w6tY6bf}
#;)7 ~ 69
(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. muLTY gaM
70Am]L&M
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
