(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . e G =Hyc
5Ta<$ t
A. radii of the pitch circles B. transmission ratio C. working pressure angle Fzt{^%\`
'M VE5
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . 3E)
X(WJY
.}=gr+<bf
A. pitting of tooth surfaces B. breaking of gear tooth 6Q"fRXM
l`>|XUf6
C. wear of tooth surfaces D. agglutination of tooth surfaces U["'>&B
.hPk}B/KV
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . r01Z
0>
# 3UrGom
A. tooth number B. modification coefficient C. module "MM)AY*b
.FYxVF.
D. helix angle of helical gear %k3A`ClW
C%&A9(j G
(4) The contact fatigue strength of tooth surfaces can be improved by way of . \USl 9*E
2 z E gAc
A. adding module with not changing the diameter of reference circle (`'(`x#
[A\D uJx
B. increasing the diameter of reference circle S,0h
&A9
l/xpAx
C. adding tooth number with not changing the diameter of reference circle \0 j-p
- 8zdkm8k
D. decreasing the diameter of reference circle | -l9 Z
3R>"X c
(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.) 3kz O
VZ
M!tR>NMH
A. equalize strengths of the two gears B. smooth the gear drive L$t.$[~L
vC^Ul
C. improve the contact strength of the smaller gear do[w&`jw 8
w<LV5w+
D. compensate possible mounting error and ensure the length of contact line {[ pzqzL6
A$3ll| %j
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . Cu! S|Xj.
Z'~/=a)7
A. B. C. D. (M# m BS
.4J7 ^l
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. SFjR SMi
8b"vXNB.f
A. certainly B. not always C. certainly not 5SEGV|
%
w eMufT
(8) Because of , the general worm gear drives are not suitable for large power transmission. mtf><YU
QL\3|' a
A. the larger transmission ratios B. the lower efficiency and the greater friction loss h AJ^(|
*JOp)e0 b
C. the lower strength of worm gear D. the slower rotating velocity of worm gear 6?'; ip
ZEB,Q ~
(9) In a belt drive, if v1, v2 are the pitch circle velocities of the driving pulley and the driven pulley respectively, v is the belt velocity, then . ]P2Wa
=)f.Yf|A*
A. B. C. D. fskc'%x
2]5ux!Lqln
(10) In a belt drive, if the smaller sheave is a driver, then the maximum stress of belt is located at the position of going . :cA8[!
L2%n pps
A. into the driving sheave B. into the driven sheave q4niA
z
J V>;
C. out of the driving sheave D. out of the driven sheave \98|.EG
gb26Y!7%
(11) In a V-belt drive, if the wedge angle of V-belt is 40°，then the groove angle of V-belt sheaves should be 40°. % O*)'ni
Bam 4%G5
A. greater than B. equal to C. less than D. not less than }
O:l]O`
# "TL *p
(12) When the centerline of the two sheaves for a belt drive is horizontal, in order to increase the loading capacity, the preferred arrangement is with the on top. zLjQ,Lp.I
(X|lK.W y
A. slack side B. tight side m$bNQ7
cK t8e^P
(13) In order to , the larger sprocket should normally have no more than 120 teeth. 41o ~5 :&
ZRYHsl{F+
A. reduce moving nonuniformity of a chain drive I z)~h>-F
5dhT?/qvc
B. ensure the strength of the sprocket teeth C. limit the transmission ratio Cc` )P>L
Mh5 =]O+
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the #}zL? s^G
v,mn=Q&9
chain si|b>R&Z
}'Yk
#Q
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . Z#|IMmT;*=
n*i1QC
A. the less z1 and the larger p B. the more z1 and the larger p ;Z}V}B
+vCW${U
C. the less z1 and the smaller p D. the more z1 and the smaller p 1R"ymWg"
ue#Y h
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) j%OnLTZ
7 = x]p
(15) In design of a chain drive, the pitch number of the chain should be . &d^u$Y5
X?Mc"M
A. even number B. odd number C. prime number j{u! /FD
J|2OmbJ e
D. integral multiple of the tooth number of the smaller sprocket b
ZEyP
W
/|2#s%|-=
C) "|sG
pC'GKk 8
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. ;n ,@[v
30v 3C7o=
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. 1 k}U+
Q@-7{3
-S"$S16D
oG )JH)!
t3)nG8>
)
@]H:=Q'gj
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. `_6!nk q8
Jy x6{O j
(1) Draw acceleration curve of the follower schematically. G2%%$7Jj
F- k3F80=
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). J ['paHSF
~ (xIG
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. |-hzvuSX
-O2ZrJ!q
xFpMn}CD
N4%q-fi
Y" s1z<?
9&=~_,wJd
8 VJUaL@
Y?2I
/
)DsC:cP
1pC!F ;9Oo
e g #.f`
: luVsQ
4. (8 points) Shown in the figure is a pair of external spur involute gears. bE2^sx`(
0A@ 'w*=
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 , . ewqfs/
_PrK6M@"L
C`3}7qi|C
S*6P=O*
Cs7YD
~,
, gk49z9
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: |k:MXI
j0!Z 20
(1) the causes of producing the elastic sliding and the slipping. "W;Gv I
0.C y4sH'
(2) influence of the elastic sliding and the slipping on belt drives. d6-a\]gF
kt.z,<w5O
(3) Can the elastic sliding and the slipping be avoided? Why? { wRs V=*
M ()&GlNs
^4_)a0Kcm,
[s$vY~_
x):cirwkl
4bqi &h3
* F %Wf
*Hy-D</w%
Bs(\e^ }
$^?"/;8P5
R9nW5f
Nf
H=[
eO
;<d("Yz:@Z
4Tgy2[D?q
)
WavG1
4 Ei*\:
YKKZRl Q o
;X,|I)
U8 qb2'a8
WG>Nm89
=V~p QbZ
N|-M|1w96
=]jc{Y%o
(~ro_WC/I
a )w
*
5r) ndW,aN
Pj(Dl C7G,
8AL`<8$
sI\v} $(~
8h"Val|qP
QS(aA*D
W;%$7&+0
6. (10 points) A transmission system is as shown in the figure. Ai>= n;
I)7STzlMj.
The links 1, 5 are worms. The links 2, 6 are worm gears. The links 3, 4 are helical gears. The links 7, 8 are bevel gears. The worm 1 is a driver. The rotation direction of the bevel gear 8 is as shown in the figure. The directions of the two axial forces acting on each middle axis are opposite. ^#-d^ )f;
n!Hj4~T0
(1) Label the rotating direction of the worm 1. yC #%fgQ r
3 s%K
w,z
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. aE aU_f /
1
,Ams
1f }YKT
f6/<lS oW
#BPJ
RNXd
1`6kc9f.
h D5NX
y~ 2C2'7
"rl(
%~Op
pVm]<jO
7. (12 points) A planar cam-linkage mechanism is as shown in the figure with the working resistant force Q acting on the slider 4. 4*0C_F@RX
(r6'q 0[
The magnitude of friction angle j (corresponding to the sliding pair and the higher pair) and the dashed friction circles (corresponding to all the revolute pairs) are as shown in the figure. The eccentric cam 1 is a driver and rotates clockwise. The masses of all the links are neglected. pH9 xyN[:a
/]1$Soo
(1) Label the action lines of the resultant forces of all the pairs for the position shown. 6y+_ x'
#zf ,%IYF
(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. `{ KdmWhW
1G'D
'
W(jOD,QMB
=?3b3PZn
@?2ES@G+Ji
@fQvAok
c4f3Dr'xw
9oU1IT9
b!<\#[
A4
>~;=
j~
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. gs"w
0[$
rT|wZz9$@
(1) Calculate the DOF of the mechanism and give the detailed calculating process. 8B`w!@hf
%Ze]6TP/><
(2) List the calculating expressions for finding the angular velocity ratios and for the position shown, using the method of instant centers. Determine the rotating directions of and . N mA6L+
5S*aZ1t18
(3) Replace the higher pair with lower pairs for the position shown. `F$lO2 #k
Aw] kQ\P&
(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. @R_ON"h
u[HamGxx$u
$-p9cyk
@\WeI"^F8
R2A#2{+H
DB+oCE< .#
Kk).KgR
'5(T0Ws/w
-zzoz x]S=
$)eS Gslz
2w
Ja:= $
fDRG+/q(+
cua ( w
\
B'AXv 6
?G<ISiABQC
h\2iArw8
{.?pl]Zl6
s\< @v7A
&;sW4j
nt
Dr76+9'i
!b=W> 5h
8rjD1<
+* j8[ sz
|MOz> 1<a
0":k[y
Uu52uR
9. (15 points) An offset crank-slider mechanism is as shown in the figure. 4*D"*kR;
EqGpo_
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. )\j
dF-s
E [6:}z<
(1) Find a, b, e (the offset). a4gX@&it_k
s2GF*{
(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. `uUzBV.FR
u1Slu%^e
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
