看不到什么东西啊 LpmspIPvf
fwq|8^S@
}MIH{CMH
(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . Hzj*X}X#K
&\"fH+S
A. radii of the pitch circles B. transmission ratio C. working pressure angle 25ayYO%PTc
! 4oIx`
(2) The main failure form of the closed gear drives with soft tooth surfaces is the . j}VOr >xz
vlPE 8U
=
A. pitting of tooth surfaces B. breaking of gear tooth gx6&'${=#
;9I#> u
C. wear of tooth surfaces D. agglutination of tooth surfaces K[kmfXKu
4O`h%`M
(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . }w-`J5E
q#
Tb)x8-0
A. tooth number B. modification coefficient C. module YSwD#jO0
k9n
D. helix angle of helical gear E{6}'FG+A
s Y6'y'a95
(4) The contact fatigue strength of tooth surfaces can be improved by way of . 0* AlLwO
+>3c+h,%.
A. adding module with not changing the diameter of reference circle ;VuB8cnL`
@i68%6 H`?
B. increasing the diameter of reference circle IW1\vfe
`
o3d@Vc
C. adding tooth number with not changing the diameter of reference circle M
/DTD98'N
*]eZ Y
D. decreasing the diameter of reference circle "pPNlV]UA^
6,)y{/ENC
(5) In design of cylindrical gear drives, b1 = b2 +(5~10)mm is recommended on purpose to . (Where b1, b2 are the face widths of tooth of the smaller gear and the large gear respectively.) e||_j
<^\r9Qxl
A. equalize strengths of the two gears B. smooth the gear drive #TWc` 8
&wuV}S 7
C. improve the contact strength of the smaller gear ~"mZ0 E
YXtGuO\q
D. compensate possible mounting error and ensure the length of contact line Ueeay^zN
|gVO Iq
(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . U %ESuq#
k:TfE6JZ
A. B. C. D. j VZi_de
:#/bA&
(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. i njmP9ed
zK 2wLX
A. certainly B. not always C. certainly not ?lK!OyCkc
wu zz%9;@B
(8) Because of , the general worm gear drives are not suitable for large power transmission. nnG2z
@$-
'oz =
{ ;
A. the larger transmission ratios B. the lower efficiency and the greater friction loss #CPP dU$
vb- .^l
C. the lower strength of worm gear D. the slower rotating velocity of worm gear Hl;p>>n
6p " c ^
(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 . qN[7zsaj
n vPE
N
A. B. C. D. Q: O>k CDV
|y% ].y)
(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 . ql~{`qoD~
X u):.0I
A. into the driving sheave B. into the driven sheave pq3 A%|
wYA/<0'yH
C. out of the driving sheave D. out of the driven sheave F2$bUY
&f W=5'
(11) In a V-belt drive, if the wedge angle of V-belt is 40°，then the groove angle of V-belt sheaves should be 40°. c@-
K
6I.m c
A. greater than B. equal to C. less than D. not less than " ^baiN@ac
9r nk\`E
(12) When the centerline of the two sheaves for a belt drive is horizontal, in order to increase the loading capacity, the preferred arrangement is with the on top. 0; v~5|r
lK,=`xe
A. slack side B. tight side `Kw"XGT
*ZR@ z80i
(13) In order to , the larger sprocket should normally have no more than 120 teeth. `4.sy +2
;3|Lw<D5;
A. reduce moving nonuniformity of a chain drive fG2&/42
J
=)- Q?1q
B. ensure the strength of the sprocket teeth C. limit the transmission ratio BZ+ mO
Ws*
P MK.0
D. reduce the possibility that the chain falls off from the sprockets due to wear out of the -;)SER3Wq4
@k +G
Cf
chain o9 LD6$
%DRy&k/T
(14) In order to reduce velocity nonuniformity of a chain drive, we should take . aE\BAbD7
s`iNbW ="
A. the less z1 and the larger p B. the more z1 and the larger p [du>ff
boF4d'g"
C. the less z1 and the smaller p D. the more z1 and the smaller p re M
U a j`
(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) ]X:
rby$
!+A"Lej
(15) In design of a chain drive, the pitch number of the chain should be . an=+6lIl
nG ^M 2)(8
A. even number B. odd number C. prime number J7?)$,ij%
sM fFm@\ N
D. integral multiple of the tooth number of the smaller sprocket ^[seK)
S=
P\X=*
Wt+y-ES
UJz4>J F
2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. ^?6
W<
G{f`K^
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. qsft*&
HhWwc#B
}+dM1 O
/ h7>Z9T
FU5LY XCs
r%T Lv
3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. V<HOSB7
J\xz^%p
(1) Draw acceleration curve of the follower schematically. Zr.\`mG4f
=wIdC3Ph
(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). 'z + $3\5L
B_."?*|w
(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. uG=t?C 6
;/?Z<[ B
]^*_F
Jb3>vCIn
3R<ME c
<OR.q
o%V
@D'w
P\q <d
%w '/n>]j
X+dR<GN+YX
lL:KaQ 0E
Qb! !J4| !
4. (8 points) Shown in the figure is a pair of external spur involute gears. .)i O Du
:"M9*XeHO
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 , . Iz6 ss(UJ
gt
=j5
x[vp oB+c
Ci]'G>F@"
L !bfh`
v2/ yw,
5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: 42b. 7 E
rQ`\JE&`
(1) the causes of producing the elastic sliding and the slipping. |. J,8~ x
9U#\nXM
(2) influence of the elastic sliding and the slipping on belt drives. x`]Of r'
'8s>rH5[V
(3) Can the elastic sliding and the slipping be avoided? Why? +<S9E'gT3V
,9=gVW{
J?._/RL8-
O!:QJ
^8 d
*-?Wc z
tRLE,(S,-
8{R&EijC
*,"jF!C&[
ki\B!<uv
Ba/RO36&c
SXQ@;= ]xV
r VUUH!
py]KT Rzy
qv
*3A?uzr
-7k[Vg?
^uia`sOP4
3I|O^
#S%4?
<O.Kqk*
nq
>L8?=>>?\
L~IE,4
(4~X}:
ba8 6 N
bMsECA&
y=w`w
>%
}g >kpa0c
YJd8 l>mz
^ 00{Hd6
fuMN"T 6%+
A#(`9
H8&p<=
fYB*6Xb,w
6. (10 points) A transmission system is as shown in the figure. }S
*/b1
&iNS?1a%f=
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. /Pk:4,
+i. u< T
(1) Label the rotating direction of the worm 1. !{( ls<
+ % XnMl
(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. ePEe?o4;
vM(Xi
p7
T`\]!>eb
Iyyo3a wc
w>[T&0-N
s]Gd-j
#Ao !>qCE
{E51Kv&_
=u=Kw R
Uyk,. *8"
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. XQ?fJWLU
b+{r! D}~
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. Xwq]f :@V
q@l(Qol
(1) Label the action lines of the resultant forces of all the pairs for the position shown. 6+3 $:?
AbfZ++aJ
(2) Label the rotation angle d of the cam 1 during which the point C moves from its highest position to the position shown in the figure. Give the graphing steps and all the graphical lines. 0Z
HDBh
xS8 ,W
((<` zx
! E0! -UpY
3-|3`(
OJ1tV% E
:( , mL2[
0V`[Zgf
1e}8LH7
"0Q1qZ
8. (15 points) In the gear-linkage mechanism shown in the figure, the link 1 is a driver and rotates clockwise; the gear 4 is an output link. BBnq_w"a
9
v?l
(1) Calculate the DOF of the mechanism and give the detailed calculating process. }k7'"`#?"
=&vFVIhWcf
(2) List the calculating expressions for finding the angular velocity ratios and for the position shown, using the method of instant centers. Determine the rotating directions of and . !SxG(*u
O0hu qF$K
(3) Replace the higher pair with lower pairs for the position shown. JOD/Raq.1k
dRaO Gm)
(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. =\3Tv
R_N:#K.M
vmX"+sHz$]
XP3x Jm3
fex,z%}p
)uheV,ZnY
FN^FvQ
9Q.#\
09 h.1/
0
n
vSvk
dl~|Izm
j
!H^-d}q
'LS z f/w
\6!W05[ Q
`T& jPA9eY
@)6jE !LC
5h {`<W
vbG]mMJ
c&nh>oN
,,4
GNbBC
t18$x "\4k
W yhhCR=;
s^8u&y)3
\cJa;WM>
+0l`
5."d
6<76H
9. (15 points) An offset crank-slider mechanism is as shown in the figure. # ]Y>KX2HG
c a<"
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. x.mrCJn)
""iaGH+Cxw
(1) Find a, b, e (the offset). x2Lq=zwJ
V@-) \RZm
(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. }f} } A =
nw: -J1kWR
(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
