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     . E&eY79  
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A. radii of the pitch circles     B. transmission ratio     C. working pressure angle ^'a#FbMtt  
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(2)       The main failure form of the closed gear drives with soft tooth surfaces is the       . lo(
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A. pitting of tooth surfaces     B. breaking of gear tooth OF+4Mq  
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C. wear of tooth surfaces     D. agglutination of tooth surfaces m|F1_Gg z  
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(3)       The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the       . kS8?N`2}LV  
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A. tooth number   B. modification coefficient   C. module yCvtglAJ4  
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D. helix angle of helical gear _p*8ke  
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(4)       The contact fatigue strength of tooth surfaces can be improved by way of     . >%om[]0E  
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A. adding module with not changing the diameter of reference circle Z|%h-~  
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B. increasing the diameter of reference circle ;=jF9mV.  
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C. adding tooth number with not changing the diameter of reference circle OyF=G^w  
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D. decreasing the diameter of reference circle >&KH!:OX|  
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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.) .R)P |@z L  
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A. equalize strengths of the two gears         B. smooth the gear drive hm! J@  
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C. improve the contact strength of the smaller gear zR?R,k)m  
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D. compensate possible mounting error and ensure the length of contact line !SW0iq[7j  
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(6)       For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then     . X]
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(7)       In a worm gear drive, the helix directions of the teeth of worm and worm gear are     the same. $qQYxx@  
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(8)       Because of       , the general worm gear drives are not suitable for large power transmission. ZyUcL_  
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A. the larger transmission ratios   B. the lower efficiency and the greater friction loss ,SR7DiYg  
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C. the lower strength of worm gear   D. the slower rotating velocity of worm gear ~~OFymQ%?q  
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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       . x950,`zy  
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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     . u[b
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A. into the driving sheave       B. into the driven sheave qgrJi +WZ  
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C. out of the driving sheave       D. out of the driven sheave ?uq`|1`  
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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°. 5g7@Dj,.  
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A. greater than     B. equal to     C. less than     D. not less than '> "{yi-  
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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. t WI-  
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A. slack side   B. tight side U N9hZ>9  
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(13)   In order to     , the larger sprocket should normally have no more than 120 teeth. Jl/wP  
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A. reduce moving nonuniformity of a chain drive MET9rT  
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B. ensure the strength of the sprocket teeth     C. limit the transmission ratio pE5v~~9Ikv  
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D. reduce the possibility that the chain falls off from the sprockets due to wear out of the S+T/(-W  
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(14)   In order to reduce velocity nonuniformity of a chain drive, we should take       . UkL1h7}a\  
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A. the less z1 and the larger p     B. the more z1 and the larger p Og\k5.! ,  
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C. the less z1 and the smaller p     D. the more z1 and the smaller p QUrPV[JQ  
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(Where z1 is the tooth number of the smaller sprocket, p is the chain pitch) ){;XI2
 
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(15)   In design of a chain drive, the pitch number of the chain should be     . aB,-E>+  
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A. even number   B. odd number   C. prime number (>al-vZ6A  
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D. integral multiple of the tooth number of the smaller sprocket  ;l$$!PJ  
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2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. SVCh!/qe\  
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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. >US*7m }  
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3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. Itl8#LpLM  
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(1)   Draw acceleration curve of the follower schematically. CFtQPTw  
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(2)   Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). yM_/_V|G  
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(3)   For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. pjeNBSu6  
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4. (8 points) Shown in the figure is a pair of external spur involute gears. I Xc
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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 , . EGu%;[  
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5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: )_#V>cvNG  
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(1)   the causes of producing the elastic sliding and the slipping. I%T+H[,  
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(2)   influence of the elastic sliding and the slipping on belt drives. oW^b,{~V  
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(3)   Can the elastic sliding and the slipping be avoided? Why? P{Nvt/%  
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6. (10 points) A transmission system is as shown in the figure. YZ~MByu  
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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. g*$2qKm  
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(1)   Label the rotating direction of the worm 1.  /
 
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(2)   Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. 6]7c
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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. hYoUZ'4  
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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. J&8KIOz14Z  
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(1)   Label the action lines of the resultant forces of all the pairs for the position shown. 868X/lL  
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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. Gxw1P@<F:  
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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. D0Q9A]bD;  
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(1)   Calculate the DOF of the mechanism and give the detailed calculating process. nXT`7  
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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 . ]y@A =nR  
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(3)   Replace the higher pair with lower pairs for the position shown. JX&
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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. `G$>T#Dq  
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9. (15 points) An offset crank-slider mechanism is as shown in the figure. atZe`0  
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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. x+Yo#u22  
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(1)   Find a, b, e (the offset). OkC.e')Vx  
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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. J!om"h  
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(3)   Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.