不会吧? Cqd\n#d/~
Y ^^4n$
sn"((BsO<
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . %QP[/5vQ
B{Q}^Mcxy
A. radii of the pitch circles B. transmission ratio C. working pressure angle U>tR :)
20Zxv!
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . ZP";B^J
~J
\qkQ
A. pitting of tooth surfaces B. breaking of gear tooth bBIh}aDN
}`=7%b`-?
C. wear of tooth surfaces D. agglutination of tooth surfaces {4_s:+v0
B=A!hXNa
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . _k:8ib2TQ
Snr(<u
A. tooth number B. modification coefficient C. module 8eqTA8$?
4DL;Y
D. helix angle of helical gear BX_yC=S
s;A7:_z#7
(4) The contact fatigue strength of tooth surfaces can be improved by way of . @$
7 GrT
ky2]%cw
A. adding module with not changing the diameter of reference circle ^ZMbJe%L
FJn-cR.n
B. increasing the diameter of reference circle =$OGHc
Nu7>G
C. adding tooth number with not changing the diameter of reference circle h$5[04.Q
XFWpHe_ L
D. decreasing the diameter of reference circle O)jD2X?
*}2L4]
(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.)
;JxL>K(
fkI 5~Y|
A. equalize strengths of the two gears B. smooth the gear drive m%0-3
c(
,HP }}K+S
C. improve the contact strength of the smaller gear DS?.'"n[u
Sp>g77@
D. compensate possible mounting error and ensure the length of contact line [9
MH"\
FKO2UY#&7
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . j(%N.f6
prZ
,4\
A. B. C. D. tWn
dAM(U7
QIiy\E%
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. ??F* Z" x
y#'hOSR2
A. certainly B. not always C. certainly not =}Tm8b0
8 m"k3:e^
(8) Because of , the general worm gear drives are not suitable for large power transmission. "{~5QO
T-N>w;P
A. the larger transmission ratios B. the lower efficiency and the greater friction loss #gbB// <
F*P0=DD
C. the lower strength of worm gear D. the slower rotating velocity of worm gear 8m")
)i-
6+$2rS$1V
(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 . coAXYn
-leX|U}k
A. B. C. D. LO=U?`)q
zamMlmls^
(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 . ~
*&\5rPb
%mC@}
A. into the driving sheave B. into the driven sheave :7K
a4
+Kk1[fh-
C. out of the driving sheave D. out of the driven sheave poBeEpbs
Z8E<^<|
(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°. *X.1b!
,)`_?^\$f
A. greater than B. equal to C. less than D. not less than E`TZ:W]r,
Um+_S@h
(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. LSta]81B4L
?I?G+(bq
A. slack side B. tight side h,{Q%sqO
DvH-M3
(13) In order to , the larger sprocket should normally have no more than 120 teeth. g@#he95 }
8~(xi<"e
A. reduce moving nonuniformity of a chain drive jf3Zy:*K
7
.29'
B. ensure the strength of the sprocket teeth C. limit the transmission ratio +JEr
c)%
SrKitSG
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the TKnWhB/J
m%
L!eR
chain m]?Z_*1
btg= # u
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . r6FTpOF
U =J5lo
A. the less z1 and the larger p B. the more z1 and the larger p d.+*o
\d"\7SA
C. the less z1 and the smaller p D. the more z1 and the smaller p "=a3"/u
!` 1h *}
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) 4r
$#-
#czInXTTx
(15) In design of a chain drive, the pitch number of the chain should be . 6SCjlaGW5
W5u5!L/
A. even number B. odd number C. prime number \~#\ [r_
!`_f
D. integral multiple of the tooth number of the smaller sprocket N[
=I
k^oSG1F
jnqp"
Ult>
Ij(dgY
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. uL)MbM]
"c[> >t
If the maximum working stress of the element is 180MPa, the minimum working stress is -80MPa. Find the angle q between the abscissa and the line connecting the working stress point to the origin. 6Y2,fW8i,
c^=
q(V
SGl|{+(A
b]]k\b
a >fA-@
?`TJ0("z"
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. L@\t]
~
W^elzN(
(1) Draw acceleration curve of the follower schematically. b#I,Z+0ry
4Ah
FE@
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). `G ":y[Q
:(enaHn#~
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. *?Eu{J){7%
Y#lAG@$
pW ~;B*hF
%;4#?.W8
wSjy31
f@Mku0VT
'0 Ys`Qo
o,k#ft<
p3fVw]N
"Io-%Su+
2;]tIt d1
_`Kh8G
{e
4. (8 points) Shown in the figure is a pair of external spur involute gears. IF>v
-Z
7#
'j>]
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 , . M/V"Ke"N
Q9y|1Wg1W
Nt+UL/
1]
N
cHCcc
.WOF:Nu4
q&ed4{H<
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: 7({.kD6
]H:K$nmX
(1) the causes of producing the elastic sliding and the slipping. 7;}TNK\+v
!oV'
(2) influence of the elastic sliding and the slipping on belt drives. Fo"'[`
fv3)#>Dgp>
(3) Can the elastic sliding and the slipping be avoided? Why? jM@@N.
}=3W(1cu-
>|T?87
Qzk/oHs
]
)x z
I(WND/&
zGFW?|o<
ldoN!J
,v1-y
?kB
]*;+ U6/?
4sK|l|W
1[`l`Truz
M/sqOhg
Mz{ Rh+gS
skP_us~
TAJ 9Y<
Js#c9l{{
<F.Ol/'h
\
ZDT=?
.y[K =p3
t
@19a6:Co
=v5(*$"pd"
0_y%Qj^e
;09J;sf
o-'i)pp
!V
-SV`+X
qK1V!a2
/7"1\s0 U
+m7x>ie)
=<_5gR
&{=~)>h
xNqQbkF
6. (10 points) A transmission system is as shown in the figure.
]D-48o0
yEqmB4^-
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. V=*J9~K
I~qiF%?d
(1) Label the rotating direction of the worm 1. ry]7$MQyV
EH "g`r
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. K]Cvk%
"AK3t'
jF*
CzEn_ZMb
hw'2q9J|
_Lb& 2PAG
r=Q5=(hn
|\
(/dXXP
.>'Z9.Xnk
Qd4T?5 vG
Aii[=x8
7. (12 points) A planar cam-linkage mechanism is as shown in the figure with the working resistant force Q acting on the slider 4. WO9/rF_
/nP=E
The magnitude of friction angle j (corresponding to the sliding pair and the higher pair) and the dashed friction circles (corresponding to all the revolute pairs) are as shown in the figure. The eccentric cam 1 is a driver and rotates clockwise. The masses of all the links are neglected. qi2dT
B
)\;r
V';
(1) Label the action lines of the resultant forces of all the pairs for the position shown. UIj/Id
\IfgL$+
(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. //@=Q!MW
P[K=']c
]c2| m}I{:
@ 1A_eF
WCu%@hh=h
z0[ZO1Fo(
wS0bk<(
U9 *2< c
)x &@j4,
q)b?X
^
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. q*nz4QTOE
yo,!u\^x
(1) Calculate the DOF of the mechanism and give the detailed calculating process. Q$L(fHkw
5z ^UQq
(2) List the calculating expressions for finding the angular velocity ratios and for the position shown, using the method of instant centers. Determine the rotating directions of and . ~{G:,|`
X2YOD2<v
(3) Replace the higher pair with lower pairs for the position shown. MwqT`;lb
4"x;XVNM[
(4) Disconnect the Assur groups from the mechanism and draw up their outlines. Determine the grade of each Assur group and the grade of the mechanism. QS*cd|7J;
C}'="g^=sl
y?#J`o-
O
'2$!thm
yGG
B
*;E+9^:V
:yjK*"T|OD
?lN8~Ze
)2Ru!l#
6oh@$.ThG
1 3`0d
{V%O4/
IKzRM|/
bTQa'y`3
zZ"U9!T
s[2ZxCrCw
]'!xc9KGR
2, "q_d'V
uV}WSoq[
F:o#
EH*ym#Y
{_XrZ(y/
B5,QJ W*
v<3o[m q
Q|=
Q]$d
yaRcBT?
9. (15 points) An offset crank-slider mechanism is as shown in the figure. %:w% o$
?(/j<,m^
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. u2-@?yt
H8-D'q>R
(1) Find a, b, e (the offset). Go\} A:|s
of'ZNQ/
(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. -<51CD w,
q]N?@l]
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.