(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . I>PZYh'.T
Nln`fE/Ht
A. radii of the pitch circles B. transmission ratio C. working pressure angle |0`hE;Kt7
7R}9oK_I
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . XZ`:wmc|
:.k ZR;
A. pitting of tooth surfaces B. breaking of gear tooth `v d= ec
"#yJHsu]
C. wear of tooth surfaces D. agglutination of tooth surfaces ,~COZi;R.D
Mi+H#xx16
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . )VSwT x&
V
A<5uk04K
A. tooth number B. modification coefficient C. module IOL L1ar
F.tfgW(A@
D. helix angle of helical gear w~z[wm Okp
O. ,3|
(4) The contact fatigue strength of tooth surfaces can be improved by way of . nd ink$
~e `Bq>
A. adding module with not changing the diameter of reference circle t 42u b
K{@ 3\5<
B. increasing the diameter of reference circle 7Y m(n8
aZ X mlq
C. adding tooth number with not changing the diameter of reference circle }lP;U $
xe4Oxo
D. decreasing the diameter of reference circle z0%\OhuCcf
i
Cj2"T4TN
(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.) AH#4wPx F
ujFzJdp3k
A. equalize strengths of the two gears B. smooth the gear drive p =(@3%k
)@gZ; `n
C. improve the contact strength of the smaller gear ehNzDr\s
t2 3'x0 l
D. compensate possible mounting error and ensure the length of contact line UlcH%pxTt1
T!RT<&
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . ke_ [
j} 1zdA
A. B. C. D. m8R9{LC
Ea6
&~"
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. 0'5/K ,
C x<0 H
A. certainly B. not always C. certainly not KLoE&ds
j;b42G~p
(8) Because of , the general worm gear drives are not suitable for large power transmission. L< ^j"!0
Q=mI 9
A. the larger transmission ratios B. the lower efficiency and the greater friction loss P{OAV+cG
=Prb'8 W
C. the lower strength of worm gear D. the slower rotating velocity of worm gear 1MlU G5
hc#!Lv
(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 . qfY5Ww
$ 8
ncu`vYI.
A. B. C. D. +4;uF]T
(*}
yjUYLZ
(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 . ) J]9 lW&y
l^GP3S
A. into the driving sheave B. into the driven sheave PZxAH9 S?
]i(-I <`
C. out of the driving sheave D. out of the driven sheave 1"Z @Q`}
CTX9zrY*T
(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°. F+R?a+ e
NBl+_/2'w
A. greater than B. equal to C. less than D. not less than v+p {|X-
b@RHc!,>jV
(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. O ~j> ?
'z76 Sa
A. slack side B. tight side KT]Pw\y5
b
vfk
(13) In order to , the larger sprocket should normally have no more than 120 teeth. P96pm6H_;
EY*(Bw
A. reduce moving nonuniformity of a chain drive HhO".GA
mF` B#
B. ensure the strength of the sprocket teeth C. limit the transmission ratio
+)JpUqHa
dJ$"l|$$
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the /M b"V5S(W
G\|,5HE D
chain DNGvpKY@
V_plq6z
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . o7IxJCL=Q
]HpA5q1ck
A. the less z1 and the larger p B. the more z1 and the larger p 'SQG>F Uy
SG6 sw]x
C. the less z1 and the smaller p D. the more z1 and the smaller p L^Jk=8
~]ZpA-*@Ut
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) M Zmb`%BZ
a pE
(15) In design of a chain drive, the pitch number of the chain should be . 3fN. bU9_
bWOS `5
A. even number B. odd number C. prime number @&[T _l
/x3/Ubmz~x
D. integral multiple of the tooth number of the smaller sprocket )BRK ZQN
|_u aS
;WldHaZ9r
=4 h+
M$2
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. 'M3">$N
~o n(3|$
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. >I=2!C
1w
dx ;k`r$w
`9 {mr<
: }`-B 0
x \QY@9
[~&XL0
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. 29reG,>
%Ts PyiYl
(1) Draw acceleration curve of the follower schematically. `_&Vt=7lG
><=gV~7lx
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). `SVmQSwO[
DLCkM*'
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. &7_Qd4=08w
J/3_C6UZ
b MU(?hb
@JRNb=?a
gkHNRAL
fK(}Ce
5^j45'%I
z,|{fKtY}
l9M 0cZ,
Q~Z=(rP20
Zy09L}5 9P
oP4GEr
4. (8 points) Shown in the figure is a pair of external spur involute gears. t3+Py7qv
P bQk<"J1
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 , . >fIk;6<{
m RZ :ie
_.-;5M-
&N i`e<mP
[= |jZVhT
6eokCc"o
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: $t/x;< .H
{f{ZHi|
(1) the causes of producing the elastic sliding and the slipping. i9=& ;_z
./$cMaDJ
(2) influence of the elastic sliding and the slipping on belt drives. |$7vI&m
u~)`&1{%
(3) Can the elastic sliding and the slipping be avoided? Why? +NFzSa l
MJ =)v]a
CwJDmz\tk
{ m| pl
(@#M!'
Oq3A#6~
boo
}u
# M!!CX*k
%e@HZ"V
!@'%G6:.
BW, mw q
@X3{
x\i'I
rcGb[=B f
ujJI
1I
q{ /3V
4H+Ked
&Oq
9!>< <7TS
# 6!
5 2
F _bF
]Qu M< ms
TO-$B8*nq
,j 5&6X=1M
w}{5#
XW?b\!@ $
"6a8s;
vLxQ *50v$
Ys! >+nL|
kPjd_8z2n
zX#%{#9
_sy'.Fo
* H[Iq !@
:pRpv hm
6. (10 points) A transmission system is as shown in the figure. zS< j d~
ql<rU@
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. ?6!]Nl1gr
m4:^} O-#
(1) Label the rotating direction of the worm 1. ESoAz o ,u
7E!7"2e
a
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. *MI*Rz?4
3 DaQo
0N
4'p=p#o
Rte+(- iL
IcI OC8WC
W9oAjO NE
~ 7)A"t
pG"5!42M!
_ IeU+tS
=:+0)t=ao
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. //Ai.Q.J[
,YhwpkL
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. #%@ MGrsK
AOrHU M[I
(1) Label the action lines of the resultant forces of all the pairs for the position shown. ^Y{D^\} ,
(qwdQMj`
(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. 0|D&"/.R#!
>"3>fche
m =qOg>k
&W }<:WH~
9N[EZhW
8&V_$+ U
TFDm5XJ
DG;y6#|p
Ml>( tec
@;wzsh >o
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. nws"RcP+Z
#t!}K_
(1) Calculate the DOF of the mechanism and give the detailed calculating process. zGc(Ef5`M6
ZkB3[$4C=5
(2) List the calculating expressions for finding the angular velocity ratios and for the position shown, using the method of instant centers. Determine the rotating directions of and .
+f4W"t
O uIv e>8
(3) Replace the higher pair with lower pairs for the position shown. 6Gn4asoA
ZEbLL4n
(4) Disconnect the Assur groups from the mechanism and draw up their outlines. Determine the grade of each Assur group and the grade of the mechanism. P"7ow-
b4,jN~ci
6 -,m}Ce\
&d`^ E6#
"r.2] R3
Uz8hANN0_
G@d`F
lj &>cScC
,=[*Lo >O
<rK=9"$y(t
qg,N b
B!aK
Fm j=
F!KV\?eM$
I6OSC&A`
JE%A|R<Jl
c 8'Cq7
WO%h"' iJ
{sS_|sX
-a) T6:e
>BqCkyM9Kf
)}X5u%woV
9d[qh kPu)
,@Z_{,b
kN'Thq/ZE
t;Wotfc[#0
9. (15 points) An offset crank-slider mechanism is as shown in the figure. s'2Rs^,hN
?ei%RWo
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. }]h \/ ,
y U`: IMz
(1) Find a, b, e (the offset). E#h~V5Tf
|>s v8/!
(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. Y-
&|VE2
r>3^kL5UI
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
