(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . 3pv1L~ ZI
2T%f~yQ^
A. radii of the pitch circles B. transmission ratio C. working pressure angle n9/0W%X>
J$/ BH\
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . ; ,-Vap z
X|1YG ZJ
A. pitting of tooth surfaces B. breaking of gear tooth $(L 7/M
N&h!14]{ Z
C. wear of tooth surfaces D. agglutination of tooth surfaces 8 (h
N_Ld,J%g
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . j"u)/A8*
6%INNIyAWa
A. tooth number B. modification coefficient C. module .)
;:K
Tn-]0hWkP
D. helix angle of helical gear 'R'a/ZR`B7
TC{Qu;`H+U
(4) The contact fatigue strength of tooth surfaces can be improved by way of . E(LE*J
SY)$2RC+}
A. adding module with not changing the diameter of reference circle iQ"F`C
}V;]c~Q/H
B. increasing the diameter of reference circle Y(Ezw ! a
<1.mm_pw
C. adding tooth number with not changing the diameter of reference circle swL|Ff`$
l*H"]6cXRL
D. decreasing the diameter of reference circle aBi:S3 qk
8~#Q *
(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.) XqH<)B
]
" 1Aus
A. equalize strengths of the two gears B. smooth the gear drive Rc P5].^T
hP<qK Vy
C. improve the contact strength of the smaller gear JH H&@C n
>v,j ;[(
D. compensate possible mounting error and ensure the length of contact line } _;!E@
zRFM/IYC
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . ?}
tQaj
*S_Iza #&x
A. B. C. D. KZfRiCZ
{U-EBXV
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. (SA*9%
TJ:Lz]l >
A. certainly B. not always C. certainly not oF
xVK
TL gVu Y
(8) Because of , the general worm gear drives are not suitable for large power transmission. ".D +#
2Kl
F| Q#KwN
A. the larger transmission ratios B. the lower efficiency and the greater friction loss Dx'e+Bm
-;"l 5oX
C. the lower strength of worm gear D. the slower rotating velocity of worm gear 1_lL?S3,a@
&_j4q
(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 . C ,hsr
{hz :[
A. B. C. D. ;xwQzu%M>5
#mYxO
(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 . Shz;)0To
F9W5x= EK\
A. into the driving sheave B. into the driven sheave j 83
V$
Le
e IA=?k.y
C. out of the driving sheave D. out of the driven sheave :TI1tJS~*
+)S X
(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°. 2wQ
CQ"
SZvsJ)
A. greater than B. equal to C. less than D. not less than A4' aB0^
aDuanGC/V
(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. f6 nltZ
PAkW[;GSDh
A. slack side B. tight side +F|[9o
z
Je,8{J |e
(13) In order to , the larger sprocket should normally have no more than 120 teeth. F1>,^qyG6
c&&UT-Z
A. reduce moving nonuniformity of a chain drive uE(w$2Wi
whCv9)x
B. ensure the strength of the sprocket teeth C. limit the transmission ratio :%{7Q $Xv<
4<|u~n*JF
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the 7[L C*nrr
&k vVMn ok
chain \m&:J >^
k"J?-1L
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . -|mABHjx*
Q"O _h
A. the less z1 and the larger p B. the more z1 and the larger p y+a&swd2(U
RMXP)[
C. the less z1 and the smaller p D. the more z1 and the smaller p J$0*K+m
GfDA5v[
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) C$8=HM3
#
V +e
(15) In design of a chain drive, the pitch number of the chain should be . Ax
^9J)C
5zU$_ M
A. even number B. odd number C. prime number /-T%y uU
Oaui@q
D. integral multiple of the tooth number of the smaller sprocket T~la,>p|}
Ap /WgVw ;
gIeo7>u
~ 1p
f ?
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. "'% x|nB
b~r ?#2K
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. n}2} 4^
C4y<+G.`
I8m(p+Z=
1*Z}M%
:H~r
_>E
gCmGFQE-f
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. \-^3Pe,
;kY=
}=9
(1) Draw acceleration curve of the follower schematically. `| fF)kI
xaPTTa
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). @^%# ]x,:
b cQ$S;U)
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. MD%86m{Sg=
1_A< nt?'R
uJi|@{V
_]B' C
Q WOd&=:
ST O6cNi
@ hk~8y]rz
#U^@)g 6
1g
*4 e
bEM-^SR
a4c~ThbI
?>2k>~xlQ
4. (8 points) Shown in the figure is a pair of external spur involute gears. gVO[R6C 5C
+g(QF
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 , . *f1MgP*GKF
n<?:!f`
>=wlS\:"
60Y&)UR
~h -0rE
IG
6yt
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: 3#W>
< yBZsSj
(1) the causes of producing the elastic sliding and the slipping. q'[5h>Pa
5Osx__6 $t
(2) influence of the elastic sliding and the slipping on belt drives. F8f@^LVM/
+g%kr~w=
(3) Can the elastic sliding and the slipping be avoided? Why? Hr:WE + '
y^?7de}
\IIR2Xf,K
) m [0,
eVS 6#R]'m
V0XQG}
xM
'bb5
K} LmU{/t/
g ?qm >X
HTA Jn _
S{&,I2
aO
nDhr;/"i
qS:h v&~
Y xp.`
A!\ g!*
x\T 9V~8a
OrNi<TY>
7 .
`Fe g.
TCWy^8LA
`fYICp
&NM.}f
L|A}A[ P
V( u#8M
+uZ,}J
}86&?
0j.
g<(3wL,"
B >2" O
CobMagPhr
uy
Z
; a Xu
<?yf<G'$
J8IdQ:4^l
6. (10 points) A transmission system is as shown in the figure. ts|dk%
,t?c=u\5
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. /q[5-
96c
z 3((L
(1) Label the rotating direction of the worm 1. P5;n(E(19
*WTmS2?'h
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. f 0 "N
b~$B 0o)
- QPM$
bzj9U>eY
XL =Y~7b
<=O/_Iu
(
'?$N.lj$d
B8=r^!jEL
,EsPm'`?A/
o4Ny9s
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. 7c.96FA
.C HET]
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. 7 tZvz `\
+,8j]<wpo
(1) Label the action lines of the resultant forces of all the pairs for the position shown. yLFZo"r
3$ 'eDa[
(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. Wnf3[fV6P
l( Y
U9dp
<q}w, XU
# saK8; tp
--D&a;CO}
V?[dg^*0
YZ0y_it)
LvM;ZfA Ev
th
:I31
'nx";[6(
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. 2bnF#-(
(9 !/bX<
(1) Calculate the DOF of the mechanism and give the detailed calculating process. <i1.W !%
\c1NIuJR
(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 . :6T 8\W
HxE`"/~.7k
(3) Replace the higher pair with lower pairs for the position shown. M~ !DQ1u
n1E^8[~ '
(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. H?H(=
RJ63"F $
q@+#CUa&n
n[f<]4<
f7du
1k3
Q >QES-.l
Vi:<W0:
6 @X j
.ou#BWav/
USrBi[_ci\
i0jR~vF
{B
>cdxe3I\
y5!fbmf
B098/`r
cy%S 5Rz
wMx# dP4W8
I `I+7~t
s.{n
xk.
H6Dw5vG
"l
&}+^*X
E4; @P']`
!J/fJW>m6
xA7Aw 0
h!4jl0 oX]
WKDa]({k%
'G|M
_ e
9. (15 points) An offset crank-slider mechanism is as shown in the figure. y$h.k"x`
jSKhWxL;'
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. li j >u
#l<un<
(1) Find a, b, e (the offset). NJqjW
Bnw Yyh
(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. +yO^,{8SE
B J,D1E
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
