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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 . E>@]"O)=M,
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A. radii of the pitch circles B. transmission ratio C. working pressure angle D7 R;IA-w
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(2) The main failure form of the closed gear drives with soft tooth surfaces is the . ,P`:`XQ>_B
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A. pitting of tooth surfaces B. breaking of gear tooth 1 _&W1o
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C. wear of tooth surfaces D. agglutination of tooth surfaces N W
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(3) The tooth form factor in calculation of the bending fatigue strength of tooth root is independent of the . \xOYa
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D. helix angle of helical gear 'D+njxCk.A
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(4) The contact fatigue strength of tooth surfaces can be improved by way of . JWM4S4yZHR
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A. adding module with not changing the diameter of reference circle }uD*\.
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B. increasing the diameter of reference circle c8l\1 ce?7
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C. adding tooth number with not changing the diameter of reference circle !y1qd
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D. decreasing the diameter of reference circle UZJ s!#P
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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.) H8Ra !FW@
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A. equalize strengths of the two gears B. smooth the gear drive Y,KSr|vG
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C. improve the contact strength of the smaller gear ZB$,\|^ 6
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D. compensate possible mounting error and ensure the length of contact line !&G&
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(6) For a pair of involute spur cylindrical gears, if z1 < z2 , b1 > b2 , then . =, U~
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(7) In a worm gear drive, the helix directions of the teeth of worm and worm gear are the same. p ^I#9(PT
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(8) Because of , the general worm gear drives are not suitable for large power transmission. OYIH**?
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A. the larger transmission ratios B. the lower efficiency and the greater friction loss 8Vm)jnM
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C. the lower strength of worm gear D. the slower rotating velocity of worm gear D14i]
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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 . WD' [|s\
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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 . X=X
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C. out of the driving sheave D. out of the driven sheave 6tT*b@/_o
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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°. g O]jeO
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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. &6#Ft]6~
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(13) In order to , the larger sprocket should normally have no more than 120 teeth. ZuVucP>>_d
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A. reduce moving nonuniformity of a chain drive |$.`4h?
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B. ensure the strength of the sprocket teeth C. limit the transmission ratio 5"I8ric
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D. reduce the possibility that the chain falls off from the sprockets due to wear out of the Tn 3<cO7v
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(14) In order to reduce velocity nonuniformity of a chain drive, we should take . 4n, >EA8 5
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2. (6 points) Shown in the figure is the simplified fatigue limit stress diagram of an element. " -a CF
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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. 2 E*h,Mo
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3. (9 points) Shown in the figure is the translating follower velocity curve of a plate cam mechanism. 6
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(1) Draw acceleration curve of the follower schematically.
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(2) Indicate the positions where the impulses exist, and determine the types of the impulses (rigid impulse or soft impulse). ( Y>|P
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(3) For the position F, determine whether the inertia force exists on the follower and whether the impulse exists. .tH[A[/1 a
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4. (8 points) Shown in the figure is a pair of external spur involute gears. 5eJMu=UpR
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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 , . ngI3.v/R
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5. (10 points) For the elastic sliding and the slipping of belt drives, state briefly: {#C)S&o)6
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(1) the causes of producing the elastic sliding and the slipping. iPeW;=-2Wk
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(2) influence of the elastic sliding and the slipping on belt drives. ScCA8JgY
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(3) Can the elastic sliding and the slipping be avoided? Why? "eG@F
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6. (10 points) A transmission system is as shown in the figure. _% KRZx}
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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. -unQ 4G
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(1) Label the rotating direction of the worm 1. ,wZq ~; 2
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(2) Label the helix directions of the teeth of the helical gears 3, 4 and the worm gears 2, 6. 0-: dzf
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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. ^qLes
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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. S^iT &;,
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(1) Label the action lines of the resultant forces of all the pairs for the position shown. NH4T*R)Vz
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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. _6 @GT
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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. 5> !N)pA
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(1) Calculate the DOF of the mechanism and give the detailed calculating process. Rh%@N.Z*
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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 . 9CAu 0N5<
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(3) Replace the higher pair with lower pairs for the position shown. .>S1do+
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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. WWC& -Ni
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9. (15 points) An offset crank-slider mechanism is as shown in the figure. 5 8 7;2
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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. !m%'aQHH(
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(1) Find a, b, e (the offset). iOSt=-p
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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. vLBu E
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(3) Find the minimum transmission angle gmin of the mechanism, and indicate the corresponding position of the crank 1.
