(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . 'c<@SVF{Zz
dM wVgc:
A. radii of the pitch circles B. transmission ratio C. working pressure angle I fZaK([
FZ^byIS[
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . I$`Vw >
ZtEHP`Iin
A. pitting of tooth surfaces B. breaking of gear tooth [21tT/
K (plzQ3
C. wear of tooth surfaces D. agglutination of tooth surfaces *0R=(Gy
=fHt |}.K
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . Z<r&- !z
kx1-.~)p(z
A. tooth number B. modification coefficient C. module r\[HR ^`
't{=n
[
D. helix angle of helical gear y Ht63z8'
,h]N*Z-I"
(4) The contact fatigue strength of tooth surfaces can be improved by way of . -lMC{~h\(S
SQ5*?u\
A. adding module with not changing the diameter of reference circle r+0)l:{.
aXC`yQ?
B. increasing the diameter of reference circle ]jHB' Y
@-b}iP<T
C. adding tooth number with not changing the diameter of reference circle v& XG4 &
G)4 3Y!
D. decreasing the diameter of reference circle ]NAPvw#p
Y\ p
yl
(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.) TTE#7\K~B
Op%^dwVG(v
A. equalize strengths of the two gears B. smooth the gear drive 1C$^S]v%a
~+l C %R
C. improve the contact strength of the smaller gear { J0^S
2C xdNj
D. compensate possible mounting error and ensure the length of contact line i ^IvT
6:EO
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . %?wuKZLnc
`z=MI66Nl
A. B. C. D. e ;6Sj
huTJ
a2
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. Q49 |,ou[H
v l*RR
oJ
A. certainly B. not always C. certainly not 8Peqm?{5Y5
*Cb(4h -
(8) Because of , the general worm gear drives are not suitable for large power transmission. J5o"JRJ"
tRv#%>fj
A. the larger transmission ratios B. the lower efficiency and the greater friction loss 1Xs! ew)>
NfO0^^"
C. the lower strength of worm gear D. the slower rotating velocity of worm gear 7]v- 2
*
h NoN=J
(9) In a belt drive, if v1, v2 are the pitch circle velocities of the driving pulley and the driven pulley respectively, v is the belt velocity, then . 'Aqmf+Mm
X!
]~]%K$y
A. B. C. D. ry<
P LRN
I &xRK'
(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 . >'T%=50YH
7 uDUZdJy
A. into the driving sheave B. into the driven sheave DuE>KX{<!R
zm}1~A
C. out of the driving sheave D. out of the driven sheave d G rA 18
xl9l>k6,
(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°. dg!sRm1iZ:
-?(RoWv@X&
A. greater than B. equal to C. less than D. not less than _NkN3f5 1L
q38; w~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. 3edAI&a5
_SaK]7}m!
A. slack side B. tight side <#Lw.;(U;k
?gtkf[0B|
(13) In order to , the larger sprocket should normally have no more than 120 teeth. + ND9###
3pp
w_?k
A. reduce moving nonuniformity of a chain drive [Ok 8l='
P2p^jm
B. ensure the strength of the sprocket teeth C. limit the transmission ratio c`.:"i" k3
E)P1`X
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the yU.0'r5uR
,54<U~Lg:
chain US|vYd}u+
FCWphpz
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . _: L *{=N
6bGD8 ;
A. the less z1 and the larger p B. the more z1 and the larger p ,7(/Il
9
$-RhCnE
C. the less z1 and the smaller p D. the more z1 and the smaller p IMZKlU3
^ztf:'l@C
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) DX"; v
J
;$QJnQ"R
(15) In design of a chain drive, the pitch number of the chain should be . v#|c.<].
]S L&x:/-
A. even number B. odd number C. prime number ECScx02
EV FfXv^
D. integral multiple of the tooth number of the smaller sprocket 2YKM 9Ks
fhmr*E'J
.;b >
T
C]fTV{
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. p|d9 g
^
@<G/H|f
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. _6!/}Fm
(NdgF+'=
<6C9R>
-ZP&zOsDr
_o
TT3[7 P
C|\^uR0
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. zD^f%p ["#
//lZmyP?
(1) Draw acceleration curve of the follower schematically.
~8t}*oV
@$d\5Q( G
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). q7&yb.<KD.
.)> /!|i
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. ]HV~xD7\
I`W-RWZ
z< 6P3x|
R+K&<Rz
; axa ZV
p@8krOo`
~,!hE&LE~
OF<n T
qH%L"J
.8qzU47E
92aDHECo
K-5"#
4. (8 points) Shown in the figure is a pair of external spur involute gears. Ag#5.,B
-
wGzXp5
dl
The driving gear 1 rotates clockwise with angular velocity while the driven gear 2 rotates counterclockwise with angular velocity . , are the radii of the base circles. , are the radii of the addendum circles. , are the radii of the pitch circles. Label the theoretical line of action , the actual line of action , the working pressure angle and the pressure angles on the addendum circles , . |g\. 5IM#W
.l( r8qY#
TXD^Do5^
0Dm`Ek3A7x
nhP~ jJn
/i
k)4]>
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: fteyG$-s
_WDBG
(1) the causes of producing the elastic sliding and the slipping. GATP
b*FU*)<4.
(2) influence of the elastic sliding and the slipping on belt drives. XuFm4DEJ
"` jZ(+
(3) Can the elastic sliding and the slipping be avoided? Why? @8M'<tr<z
n$hqNsM
U/3e,`c
#>E3' 5b
nXM9Px!
] m/@wW 9
KF:]4`$
7dufY
} }
mq{$9@ 3
6"7:44O;G
zfP[1
2[8fFo>
A{
i][1N
=0 C l
kH>vD = q>
8"?Vcw &
7 E!";HT
k lP{yxU'n
6: |!1Pg5
}{#;;5 KrB
|z!q
r }i
Q8;x9o@
p
A=v lC?&Z
YZ"+c&V"
DU/9/ I?~
fB;&n
+X* F<6mZ
l#2r.q^$|
o ;[C(OS
8Z9MD<RLw
m6bWmGn GC
dAY I D E
6. (10 points) A transmission system is as shown in the figure. b[J-ja.
%K&+~CJE
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. AI^AK0.L
L<f-Ed9|
(1) Label the rotating direction of the worm 1. jsk:fh0~M
^G|w8t+^
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. U7
ns MD
1zb$5 {,|
jGk7=}nw
3n.+_ jQ>s
k1^ V ?O
rtB |N-
/
p"U
4 " pS
~hS .\h
zrLhQ3V#>
7. (12 points) A planar cam-linkage mechanism is as shown in the figure with the working resistant force Q acting on the slider 4. &GXtdO>;Zv
,3w I~ j=
The magnitude of friction angle j (corresponding to the sliding pair and the higher pair) and the dashed friction circles (corresponding to all the revolute pairs) are as shown in the figure. The eccentric cam 1 is a driver and rotates clockwise. The masses of all the links are neglected. [TRGIGtq
EsT
0"{
(1) Label the action lines of the resultant forces of all the pairs for the position shown. { M`
) v${& H
(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. +<)tql*
@;||p eU
MB}
:GY?
am]3
"V>
II.<S C
o:Ln._bj
L=d$"Q
5? ^]1P_
I|$'Q$m~
0v;ve
8. (15 points) In the gear-linkage mechanism shown in the figure, the link 1 is a driver and rotates clockwise; the gear 4 is an output link. BfX%|CWh
.?A'6
(1) Calculate the DOF of the mechanism and give the detailed calculating process. D*|(
p6v1&
Jm?l59bv
v
(2) List the calculating expressions for finding the angular velocity ratios and for the position shown, using the method of instant centers. Determine the rotating directions of and . J09*v )L
wU bmzP.
(3) Replace the higher pair with lower pairs for the position shown. ERk kS Tp
o z Vpfs
(4) Disconnect the Assur groups from the mechanism and draw up their outlines. Determine the grade of each Assur group and the grade of the mechanism. DKR<W.!*t
B<A:_'g
3QH(4N
8a7YHUL<3i
L+_8QK <
TV?
^c?{5
#U\$@4D
Pv# KmSA9
AD(xaQ&T
M\I_{Q?_
E`~i-k f
O!o <P5X^
0(\p<q q
D^N[=q99&e
{K+.A 9!
S3; lKr
@ ozm;
: sFo
fD0{ 5
(:JjQ`i
ej<`CQ
}N@n{bu+
%U-Qsy8|D)
6I"C~&dt
D6Dn&/>Zp
nVNs ][
9. (15 points) An offset crank-slider mechanism is as shown in the figure. *aI~W^N3
&u("|O)w$
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. V>AS%lXj
NfOp=X?Y
(1) Find a, b, e (the offset). L(;.n>
/
zqY)dk
(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. 2!BsEvB(
A4}#U=3tI
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
