(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . YuoIhT
RE08\gNIt
A. radii of the pitch circles B. transmission ratio C. working pressure angle #E_<}o
}gd'pgN"t
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . .k|\xR
OLqynY
A. pitting of tooth surfaces B. breaking of gear tooth 9H+Q/Q*-a
4Wa*Pcj
C. wear of tooth surfaces D. agglutination of tooth surfaces 6Q]JY,+
v8C4BuwA
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . ff00s+
`
d
RqheX
A. tooth number B. modification coefficient C. module kX+98?h-C
F#.ph?W
D. helix angle of helical gear f|~X}R
Af`Tr6)
(4) The contact fatigue strength of tooth surfaces can be improved by way of . 1had8K-
-TS,~`O
A. adding module with not changing the diameter of reference circle >oC{YYcK
/NxuNi;5
B. increasing the diameter of reference circle _7dp(R
dAy\IfZX=
C. adding tooth number with not changing the diameter of reference circle "+Sq}WR
@3bVjQ`4f
D. decreasing the diameter of reference circle l+X\>,
DhD##5a
(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.) ):y^
g:
@#;*e] 1a
A. equalize strengths of the two gears B. smooth the gear drive 8~&=vc
,Z}
ST|$u
C. improve the contact strength of the smaller gear y9w,Su2
NQqNBI?cr
D. compensate possible mounting error and ensure the length of contact line M }q;\}
($W%&(:/
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . }}k*i0
yB 'C9wEH
A. B. C. D. .tQ(q=#
s!>9od6^
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. r3KNRr@
YHMJ5IM@.
A. certainly B. not always C. certainly not $: |`DCC
Wu(^k25
(8) Because of , the general worm gear drives are not suitable for large power transmission. 1}+b4"7]
23 #JmR
A. the larger transmission ratios B. the lower efficiency and the greater friction loss VT=gb/W6)a
B<n[yiJ}
C. the lower strength of worm gear D. the slower rotating velocity of worm gear }X/YMgJ
c]bG5
(9) In a belt drive, if v1, v2 are the pitch circle velocities of the driving pulley and the driven pulley respectively, v is the belt velocity, then . rS8\Vf]F
gGx(mX._L?
A. B. C. D. t1yOAbI
kGYsjhL\d
(10) In a belt drive, if the smaller sheave is a driver, then the maximum stress of belt is located at the position of going . T<NOLfk66
G6G-qqXy6
A. into the driving sheave B. into the driven sheave V>GJO (9
}maD8,:t
C. out of the driving sheave D. out of the driven sheave {9.UeVz
{2q0Ko<
(11) In a V-belt drive, if the wedge angle of V-belt is 40°,then the groove angle of V-belt sheaves should be 40°. } # L_R
a|.IAxJ
A. greater than B. equal to C. less than D. not less than X%kJ3
{
Fq5u%S
(12) When the centerline of the two sheaves for a belt drive is horizontal, in order to increase the loading capacity, the preferred arrangement is with the on top. cMnN} '
DS>&|zF5l
A. slack side B. tight side ;Bj&9DZd
"@/pQoLy
(13) In order to , the larger sprocket should normally have no more than 120 teeth. :@ VC Kq!
Vsw:&$
A. reduce moving nonuniformity of a chain drive :N"&o(^
/!A"[Tyt
B. ensure the strength of the sprocket teeth C. limit the transmission ratio JYA>Q&
yNn=r;FZQ
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the &zJ*afi)
F]r'j
ZL
chain jYFmL_{
s/s&d pT*
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . n:) [%on
A9u>bWIE7
A. the less z1 and the larger p B. the more z1 and the larger p @G=7A;-pv0
fjl9*
C. the less z1 and the smaller p D. the more z1 and the smaller p vNuws_
I"x|U[*B
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) g1"ZpD
un -h%-e|
(15) In design of a chain drive, the pitch number of the chain should be . u a_w5o7
b7I0R;Zj
A. even number B. odd number C. prime number |9IOZ>H9
/2c(6h
D. integral multiple of the tooth number of the smaller sprocket <OKc?[
qGEp 6b H
ty
rP[y
Zl3e=sg=
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. xP&7i'ag
<!UnH6J.b
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. ]H}2|~c
{"{]S12N
L3xN#W;m7
F`x_W;\
"0#(<zb|
tp7$t#
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. nQ+5jGP1
aEIz,^3
(1) Draw acceleration curve of the follower schematically. DgW@v[#BK=
.IdbaH
_a
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). w(S~}'Sg*P
?#(LH\$l_
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. 4=|Q2qgFV
Xnjl {`
~//E'V-
%,02i@Fc
Aixe?A_x
(Q%
@]
s_fe4K
0LL0\ly]
3Q=\W<Wu
2v4K3O60G
FP=-
jf/
!0Nf`iCQ(
4. (8 points) Shown in the figure is a pair of external spur involute gears. (P-^ PNz&
!JrKTB%
The driving gear 1 rotates clockwise with angular velocity while the driven gear 2 rotates counterclockwise with angular velocity . , are the radii of the base circles. , are the radii of the addendum circles. , are the radii of the pitch circles. Label the theoretical line of action , the actual line of action , the working pressure angle and the pressure angles on the addendum circles , . M3@qhEf?vk
w[gt9]}N
/Z^a,%1
"ckK{kS4~
}l$zZ>.\H
1aRTvaGo
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: Tt~[hC
h
sa o &
(1) the causes of producing the elastic sliding and the slipping. 2 >O [Y1
Z@Q*An
(2) influence of the elastic sliding and the slipping on belt drives. k"DZ"JC
Lv%t*s2$/
(3) Can the elastic sliding and the slipping be avoided? Why? Evm3Sm!S
~210O5^
MW rhVn{R
/Ls|'2J<$
TU~y;:O
J
.p e3L7g
jH<Sf: Y(
#`|N
m3b
dNt
^lx
A)n_ST0
61kSCu
C;JW\J~W
aanS^t0
O8\f]!O(
PsNrCe%e
Y3[KS;_fr9
-d4v:Jab
aIa<,
=bD.5,F)
KM[0aXOtv
|A*4Fuc&
[#@l
sI
#b428-
`O2P&!9&
N(4y}-w$
LBD],Ba!
nm<VcCc
o[Qb/ 7
Lwi"K8.u
[sH3REE1h
7$%G3Q|)L
N}n3 +F
6. (10 points) A transmission system is as shown in the figure. H0"'j
d
O(%6/r`L,k
The links 1, 5 are worms. The links 2, 6 are worm gears. The links 3, 4 are helical gears. The links 7, 8 are bevel gears. The worm 1 is a driver. The rotation direction of the bevel gear 8 is as shown in the figure. The directions of the two axial forces acting on each middle axis are opposite. r{r~!=u
|z3!3?%R
(1) Label the rotating direction of the worm 1. T8g\_m
#
lqH/>`>
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. :atd_6
:t`W&z41
sdO8;v>
xva
e^gr
bl>W i@GL
pm=O.)g4`
IxUj(l1Fm
v4Ag~Ev
cx
3,v/zcV
-E6av|c,F
7. (12 points) A planar cam-linkage mechanism is as shown in the figure with the working resistant force Q acting on the slider 4. ?,riwDI 2
nHbi{,3
The magnitude of friction angle j (corresponding to the sliding pair and the higher pair) and the dashed friction circles (corresponding to all the revolute pairs) are as shown in the figure. The eccentric cam 1 is a driver and rotates clockwise. The masses of all the links are neglected. zP0<4E$M`
XQ[\K6X5
(1) Label the action lines of the resultant forces of all the pairs for the position shown. 3 $%#n*
ZMMx)}hS
(2) Label the rotation angle d of the cam 1 during which the point C moves from its highest position to the position shown in the figure. Give the graphing steps and all the graphical lines. gh[q*%#
Xp<O
8wd["hga<%
|7,$.MK-
@
g&p(XuN
sdCG}..`
:,p3&2I
7^rT-f07
@6DKw;Q
2l V`UIa
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. 1S(oi
|j'@no_rv
(1) Calculate the DOF of the mechanism and give the detailed calculating process. -6_<]
+foz
E?
(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 . +K'YVB
U}
K($+ILZ
(3) Replace the higher pair with lower pairs for the position shown. R$}Hv
?\4kV*/Cqz
(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. 7mi=Xa:U
4?(=?0/[
A6_ER&9$>N
g1F9IB42@<
{7IZN< e
!T)_(}|6}
Wn;%B].I
mBrZ{hqS
/;q3Q#
7Y.yl F:
&?@U_e
mLi
${.
:(z
It 3@
Cd>
~Odclrs
H_Hr=_8}-
qa:muW
FDBNKQV
hQzT
=0
J~@W":v
I L&PN`#
%AT/g&M&1#
@1k-h;`,
8*VQw?{Uee
KxJDAP
L)H/t6}i
*gwo.s
9. (15 points) An offset crank-slider mechanism is as shown in the figure. s/;S2l$`
~-yq,x
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. "[_gRe*2
7
0R_O&f-k
(1) Find a, b, e (the offset). gZz5P>^
gazX2P[D
(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. {*F8'6YQ$
|mw3v>
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.