东大机械洋卷2

(1)       The center distance separability of a pair of involute spur cylindrical gears implies that a change in center distance does not affect the     . &G55<tRE  
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A. radii of the pitch circles     B. transmission ratio     C. working pressure angle {Z;W|w1t  
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(2)       The main failure form of the closed gear drives with soft tooth surfaces is the       . I_xJ[ALdm  
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A. pitting of tooth surfaces     B. breaking of gear tooth Yl1@gw7  
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C. wear of tooth surfaces     D. agglutination of tooth surfaces > C_! }~  
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(3)       The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the       . )2F:l0
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A. tooth number   B. modification coefficient   C. module F>A-+]X3o  
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D. helix angle of helical gear X ?rJO~5  
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(4)       The contact fatigue strength of tooth surfaces can be improved by way of     . ~`tc
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A. adding module with not changing the diameter of reference circle oF V9t{~j  
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B. increasing the diameter of reference circle #FaR?L![Y  
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C. adding tooth number with not changing the diameter of reference circle zCvR/  
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D. decreasing the diameter of reference circle >wcsJ{I  
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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.) 2b|vb}|t{  
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A. equalize strengths of the two gears         B. smooth the gear drive j?z(fs-  
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C. improve the contact strength of the smaller gear $8it&/JP,  
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D. compensate possible mounting error and ensure the length of contact line Uk*s`Y  
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(6)       For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then     . 7^Onq0ym T  
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(7)       In a worm gear drive, the helix directions of the teeth of worm and worm gear are     the same. U}Hmzb  
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A. certainly       B. not always       C. certainly not .-Ao%A W  
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(8)       Because of       , the general worm gear drives are not suitable for large power transmission. NWNgh/9?  
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A. the larger transmission ratios   B. the lower efficiency and the greater friction loss T\cdtjk  
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C. the lower strength of worm gear   D. the slower rotating velocity of worm gear )O7Mfr  
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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       . ;wv
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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     . v.Q)Obyn  
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A. into the driving sheave       B. into the driven sheave ke/o11LP  
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C. out of the driving sheave       D. out of the driven sheave Crey}A/N  
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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°. ^Nc\D7( l  
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A. greater than     B. equal to     C. less than     D. not less than Mv;7kC7]  
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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. g4~{#P^i  
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A. slack side   B. tight side 5$9j&&R  
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(13)   In order to     , the larger sprocket should normally have no more than 120 teeth. ^srx/6X  
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A. reduce moving nonuniformity of a chain drive = [:ruE  
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B. ensure the strength of the sprocket teeth     C. limit the transmission ratio !$&K~>`  
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D. reduce the possibility that the chain falls off from the sprockets due to wear out of the /J5wwQ (:  
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(14)   In order to reduce velocity nonuniformity of a chain drive, we should take       . S{&;  
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C. the less z1 and the smaller p     D. the more z1 and the smaller p gh3_})8c  
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(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) *<]ulR2  
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(15)   In design of a chain drive, the pitch number of the chain should be     . E ekX|*  
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A. even number   B. odd number   C. prime number CKy' 8I9  
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D. integral multiple of the tooth number of the smaller sprocket !w{4FE74  
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2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. $mgW|TBXCQ  
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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. l1l=
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3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. ^m~&2l\N=  
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(1)   Draw acceleration curve of the follower schematically. =h0vdi%{  
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(2)   Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). $>Qq 7  
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(3)   For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. BX\/Am11
 
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4. (8 points) Shown in the figure is a pair of external spur involute gears. jr,N+K(@T  
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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 , . _%q~K (::  
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5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: S=(<m%f  
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(1)   the causes of producing the elastic sliding and the slipping. rUpe  ;c  
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(2)   influence of the elastic sliding and the slipping on belt drives. f. "\~  
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(3)   Can the elastic sliding and the slipping be avoided? Why? } vmRm*8z  
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6. (10 points) A transmission system is as shown in the figure.  bGRt  
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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. _(W@FS  
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(1)   Label the rotating direction of the worm 1. U1RU2M]v  
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(2)   Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. / 3A6xPOg  
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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. WI8}_){ d  
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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. MP]<m7669*  
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(1)   Label the action lines of the resultant forces of all the pairs for the position shown. t],a1I.gk  
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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. [W[{ 4 Xu  
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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. Vaxg  
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(1)   Calculate the DOF of the mechanism and give the detailed calculating process. Y,]Lk<Hm3  
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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 . #qRoTtMq7  
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(3)   Replace the higher pair with lower pairs for the position shown. j cd<'\;  
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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. .
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9. (15 points) An offset crank-slider mechanism is as shown in the figure. cB&_':F  
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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. *VsGa<V  
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(1)   Find a, b, e (the offset). s AFn.W  
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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. Hn >VPz+I  
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(3)   Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.