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(1) The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the . AvRZf-Geg
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A. radii of the pitch circles B. transmission ratio C. working pressure angle j#Ly!%d
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(2) The main failure form of the closed gear drives with soft tooth surfaces is the . B~- VGT2o
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A. pitting of tooth surfaces B. breaking of gear tooth wd~e3%JM
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C. wear of tooth surfaces D. agglutination of tooth surfaces ,AGK O,w
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(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the .
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A. tooth number B. modification coefficient C. module Df||#u=n
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D. helix angle of helical gear P9:7_Vc
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(4) The contact fatigue strength of tooth surfaces can be improved by way of . iBwl(,)?m2
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A. adding module with not changing the diameter of reference circle 7.CzS
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B. increasing the diameter of reference circle 60[f- 0X
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C. adding tooth number with not changing the diameter of reference circle vlw2dY@^
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D. decreasing the diameter of reference circle g}{Rk>k
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(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.) bd_U%0)pi1
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A. equalize strengths of the two gears B. smooth the gear drive !lm^(SS
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C. improve the contact strength of the smaller gear Yt++?
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D. compensate possible mounting error and ensure the length of contact line d~g
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(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . bnWIB+%_
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A. B. C. D. r9%4q4D?>9
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(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. zbddn4bW9
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A. certainly B. not always C. certainly not A{: a kK
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(8) Because of , the general worm gear drives are not suitable for large power transmission. 1jh^-d5
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A. the larger transmission ratios B. the lower efficiency and the greater friction loss N1\u~%AT"
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C. the lower strength of worm gear D. the slower rotating velocity of worm gear J*.Nf)i
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(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 .
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(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 . 4^w>An6
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(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°. ,`P,))
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(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.
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A. slack side B. tight side @-qxNw
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(13) In order to , the larger sprocket should normally have no more than 120 teeth. pk}*0
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A. reduce moving nonuniformity of a chain drive
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B. ensure the strength of the sprocket teeth C. limit the transmission ratio wx}\0(]Gl
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D. reduce the possibility that the chain falls off from the sprockets due to wear out of the s]5wzbF O
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(14) In order to reduce velocity nonuniformity of a chain drive, we should take . u<!!%C~+=
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A. the less z1 and the larger p B. the more z1 and the larger p :n0(g B
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(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) u\^<V)
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(15) In design of a chain drive, the pitch number of the chain should be . q,aWF5m@
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2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. }=s@y"["
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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. 4%\L8:
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3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. VG2TiR1
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(1) Draw acceleration curve of the follower schematically. A[htG\A` 0
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(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). ecA0z
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(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. }(op;7
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4. (8 points) Shown in the figure is a pair of external spur involute gears. jY87NHg
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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 , . 6M+~{9(S
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5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: 2ZIf@C{P.
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(1) the causes of producing the elastic sliding and the slipping. qL A
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(2) influence of the elastic sliding and the slipping on belt drives. !Q cgTW)T
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(3) Can the elastic sliding and the slipping be avoided? Why? Ls*.=ARq
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6. (10 points) A transmission system is as shown in the figure. !H?#~{
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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. CgEeO,N]j
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(1) Label the rotating direction of the worm 1. Fp@eb8Pl
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(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. &?zJ|7rh@|
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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. Qj0@^LA
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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 *X{D v
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(1) Label the action lines of the resultant forces of all the pairs for the position shown. /H!I90
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(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. M[TgNWl/[
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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. pw&l.t6
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(1) Calculate the DOF of the mechanism and give the detailed calculating process. P9Rq'u
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(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 . :5h&f
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(3) Replace the higher pair with lower pairs for the position shown. r=qb[4HiV
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(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. =_L"x~0I-
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9. (15 points) An offset crank-slider mechanism is as shown in the figure. IjNE1b$
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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. Hyc19|
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(1) Find a, b, e (the offset). FnL~8otPF'
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(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. UI hB
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(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.