不会吧? f~NS{gL*
LDr?'M!D
zMv`
<m%
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the .
CPJ21^
pAOKy
A. radii of the pitch circles B. transmission ratio C. working pressure angle {Dpsr` &
,{PN6B
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . ] 0X|_bU
Pvc)-A
A. pitting of tooth surfaces B. breaking of gear tooth ^
8 }P_
q`1tUd 4G
C. wear of tooth surfaces D. agglutination of tooth surfaces }_nBe
gv
y||
n9
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . 7Mh'x:p
peVY2\1>R
A. tooth number B. modification coefficient C. module n+8YTjd
N|yA]dg[
D. helix angle of helical gear =FzmifTc
9
Zo s;
(4) The contact fatigue strength of tooth surfaces can be improved by way of . ".>#Qp%
6]1RxrAV
A. adding module with not changing the diameter of reference circle xRWfZ3E
#
PB^rniYh
B. increasing the diameter of reference circle c,r6+oX
=78y*`L
C. adding tooth number with not changing the diameter of reference circle k M/:n
w
~+*Vd~U
D. decreasing the diameter of reference circle X?haHM#]
&E]) sJ0
(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.) Vx!ZF+
@wXYza0|d
A. equalize strengths of the two gears B. smooth the gear drive vu.ug$T
[\ku,yd%0
C. improve the contact strength of the smaller gear niS\0ZA
Xn"#Zy_
D. compensate possible mounting error and ensure the length of contact line }!K
#
!S<p"
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . QO$18MBcc
J)g
+I
A. B. C. D. w8+phN(-M
@u+LF]MY
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. [p4([ef
'
bSTori5
A. certainly B. not always C. certainly not {v!w2p@
HA
c"pG
(8) Because of , the general worm gear drives are not suitable for large power transmission. iw.F8[})
=n.&N
A. the larger transmission ratios B. the lower efficiency and the greater friction loss *=md!^x`
Vb\^xdL>
C. the lower strength of worm gear D. the slower rotating velocity of worm gear r6D3u(kMb
uaD+G:{[
(9) In a belt drive, if v1, v2 are the pitch circle velocities of the driving pulley and the driven pulley respectively, v is the belt velocity, then . l/?bXNt
Nl4uQ_"
A. B. C. D. * k\;G?
B|syb!g
(10) In a belt drive, if the smaller sheave is a driver, then the maximum stress of belt is located at the position of going . -Nr*na^H9#
M3eSj`c3
A. into the driving sheave B. into the driven sheave DW,Z})9
VaonG]Ues
C. out of the driving sheave D. out of the driven sheave <'T DOYb
r{*Qsaw
(11) In a V-belt drive, if the wedge angle of V-belt is 40°,then the groove angle of V-belt sheaves should be 40°. fG?a"6~
]'<}kJtN.
A. greater than B. equal to C. less than D. not less than Yl
1l$[A$
U+
-;(Fh~
(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. OZKZv,
1Uah IePf
A. slack side B. tight side &i805,lx
_Kli~$c& M
(13) In order to , the larger sprocket should normally have no more than 120 teeth. )eX{a/Be
A(XX2f!i
A. reduce moving nonuniformity of a chain drive -g"Wi@Qr
oh:.iL}j
B. ensure the strength of the sprocket teeth C. limit the transmission ratio 3#uc+$[
dF&@
q,
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the A/ Sj>Y1j
16ip:/5
chain ErQGVE;zk
aM
K\&yZD
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . %xXb5aY
3z k},8fu
A. the less z1 and the larger p B. the more z1 and the larger p M^n^wz
'O
7>w%#
C. the less z1 and the smaller p D. the more z1 and the smaller p M>*xbBl
I
'ha=PeVn
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch)
msq2/sS~
[.;I}
(15) In design of a chain drive, the pitch number of the chain should be .
(2
P&@!|
"`KT7
A. even number B. odd number C. prime number nwi8>MG
#=ko4?Wr(
D. integral multiple of the tooth number of the smaller sprocket MMQ\V(C
y+',jM
]0")iY_
%YG[?"P'
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. 1,n\Osd
AI$\wp#aw
If the maximum working stress of the element is 180MPa, the minimum working stress is -80MPa. Find the angle q between the abscissa and the line connecting the working stress point to the origin. 0|mCk
&'m&'wDt:
Ibv`/8xh
Jb|dpu/e
dFk$rr>q
C jz(-018
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. ?0d#O_la
3
/9ZU_y4&3f
(1) Draw acceleration curve of the follower schematically. cZ,}1?!
e'3y^Vg
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). &3 x
[0DV
xE6hE'rh.O
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. riSgb=7q9
lG`%4}1
dSE"G>l8
(J.U{N v
S3y('
PeF
*dxE
( dP
0>ce~KU
acow
.|KBQ
MI
a6 "-,Kg
(~?p`g+I.P
KAnq8B!
h
4. (8 points) Shown in the figure is a pair of external spur involute gears. Upz?x{>x
-o@L"C>
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 , . &&
b;Wr
X?'pcYSL
rw[I
oyr-
sL#MYW5E
{n(/ c33
cT
nC
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: BR:Mcc
D=}\]Krmay
(1) the causes of producing the elastic sliding and the slipping. YVPLHwh/5
*$@u`nM
(2) influence of the elastic sliding and the slipping on belt drives. FzG>iC}
rbrh;\<jM
(3) Can the elastic sliding and the slipping be avoided? Why? ~,.;2K73
<,y> W!
XfN(7d0
Hz>Dp
!
"NTiQ}i
k`J|]99Wb
-s]@8VJA"
S1U[{R?,
]88qjKL
J|`.d
46
Q"H/RMo-
J M`[|"R%
iU AY
c:""&>Z
<,D*m+BWn
^!gq_x
1x%B`d
[v`kqL~
?$l|];m)-
x_r*<?OZ
\>X!n2rLZe
w^U{e
xo
Xt
+9z
c3r`T{Kf
s;$f6X
{8h[Bd
?'_6M4UKa
7*"Jx}eM
Keem\/
ESYF4-d+
~]q>}/&YLo
"c|Rpzs[
6. (10 points) A transmission system is as shown in the figure. A\#z<h[>
"%\hDL;
The links 1, 5 are worms. The links 2, 6 are worm gears. The links 3, 4 are helical gears. The links 7, 8 are bevel gears. The worm 1 is a driver. The rotation direction of the bevel gear 8 is as shown in the figure. The directions of the two axial forces acting on each middle axis are opposite. I2Imb9k~B
cnI5G!
(1) Label the rotating direction of the worm 1. ^39lUKL
~?FhQd\Q
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. @&
Bh!_TWc
;j7G$s9
I]E 3&gnC
qR_>41JU"
kwL|gO1L
5doi4b>]!
Tjhy@3
D}LM(s3li7
');vc~C
N9-7YQ`D
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. W\NC3]
bd~m'cob>
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. "~^0
V~ORb1
(1) Label the action lines of the resultant forces of all the pairs for the position shown. yCvtglAJ4
bWAa:
r
(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. }cr'o"4
N)GHQlgH
b&mA1w[W]
dpw-a4o}
]i@73h YT
mXOI"B9Sq
_eOC,J<-~
J\Pb/9M/
Z17b=xJw
Y~85Z0l
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. ~x\uZ^:
9<.O=-1~
(1) Calculate the DOF of the mechanism and give the detailed calculating process. wW1VOj=6V"
uC^
)#Y\"
(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 . uR5+")r@S
C$TU
TS
(3) Replace the higher pair with lower pairs for the position shown. uBx\xeI
N68$b#9Ry
(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. *VsVCUCz5*
MmN{f~Kq9
`}l%61n0
Zek@
xr;]
mG&A_/e!9
eV|N@
4@@Sh`E:
E(4c&
FT-.gi0
|VWT4*K
w5HIR/kP
Wn(!6yid
SWV*w[X<X
iAu/ t
=#V^t$
4@
= UH3.
e.n(NW
1 P
!)4W
~V8z%s@
+)sX8zb*gY
n^4R]9U
\0b}Z#'0
A9MM^jV8
QM0B6F
a_>|Ny6{
9. (15 points) An offset crank-slider mechanism is as shown in the figure. :25LQf^nz
H#SQ>vyAV
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. !{lH*
CQ!pt@|d
(1) Find a, b, e (the offset). y/Xs+ {x
H$h#n~W~
(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. I%3[aBz4
ljKIxSvCFp
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.