精品课件!《机械原理》_机械原理(双语)试卷B卷答案

四川理工学院试卷 B（至学年第学期）

课程名称：机械原理（双语）

命题教师：胡莲君

适用班级：

注意事项：

1、满分100分。要求卷面整洁、字迹工整、无错别字。

2、考生必须将姓名、班级、学号完整、准确、清楚地填写在试卷规定的地方，否则视为废

卷。

3、考生必须在签到单上签到，若出现遗漏，后果自负。

4、如有答题纸，答案请全部写在答题纸上，否则不给分；考完请将试卷和答题卷分别一同

交回，否则不给分。

答案

一、Checkpoint (mark your answer on the answer sheet by True or False)

(20 points 2 points/each )

1. The grade of a mechanism is defined as the highest grade of Assur group in the mechanism. ( T )

2. Any mechanism which has determined motions can be assembled from

a basic mechanism by connecting Assur group to the determined links using outer pairs, group by group. This is the composition principle of mechanism ( T )

3. Inversion of a mechanism can change the relative motion between its links ( F )

4. If a mechanism has quick return characteristics, then we should set the faster stroke as working stroke. ( F )

5. In a cam mechanism, the sine acceleration motion curve would produce soft impulse ( F )

6. A rotor is dynamically balanced if it has been statically balanced. ( F )

7.If crank- rocker mechanism(rocker is driver) reach it’s dead point, then transmission angle γis equal to zero . ( T )

8. If the efficiency of mechanismη≤0, then the mechanism is self-lock mechanism. ( T )

9. The pressure angle remains the same at all points on the tooth profiles of the rack (a k=a), which is also called the normal pressure angle. ( T )

10. The proper meshing conditions for external helical gear is m n1=m n2= m n , a n1=a n2=a n and β1=－β2 ( T ) 二．Calculate the degree of freedom of the mechanism as shown in Fig (10 points ) Solution:

A ：passive DOF

n=8, pl=11, ph=1

F=3n-2Pl-Ph (3 points ) =3x8-2x11-1

=1 (7 points )

三.In the cam mechanism with inline translating Flat-faced follower shown in Fig, the cam is a circular disk (The radius of the circular disk r=50mm ). l A0=202mm Supposing that the angular velocity

ω1 of the cam is known, ω1=10rad/s, the velocity V 2 of the follower 2 is to be found when θ=45°by instant center method. （15 points ）

Solution:

Because point c is instant centre between Cam 1 and follower 2, (6 points ) then:

V C2= V C1=ω1.L AC =10x202x 2/2=200mm/S

Because follower 2 is a translating follower Then:

V 2= V C2=200mm/S (3 points )

四. An offset slider-crank mechanism is as shown in Fig. Supposing that the length a=120mm ，b=600mm ，offset e=120mm . find out the coefficient K of travel speed variation and The working stroke H by graphic method (15 points)

Solution:

Select u l =1/10, draw out two limiting positions (AC1, AC2)of slider-crank mechanism

AC1=AB+BC

AC2=BC-BC (6 points )

THEN θ=∠C1AC2=4.90 (3 points ) K=（180+θ）/（180-θ）=1.06 (3 points ) H=C1C2/u l =245 mm (3 points )

五. How many teeth would an external

standard spur involute gear have when its dedendum circle and its base

circle coincide? Which one is bigger as the number of teeth increases? (10 points)

Solution:

d b =dcos α=mz cos α (2 points )

d f =d-2h a =m （z-2h a *-2c *） (2 points ) when d b = d f

mz cos α=m （z-2h a *-2c *

）

external standard spur involute gear α=200

, h a *=1, c *=0.25 (2 points )

then z=41.46 (2 points ) since z is a integer ,so:

when z ≥42时 d b

六. The numbers of teeth of the gears in the gear train shown in Fig are given as Z 1=20, Z 2=40, Z 2’=20 ,Z 3=30and Z 4=80. Calculate the train ratio i lH (15 points)

solution:

1．Dividing the gear train gear train with fixed axes : gear1_gear2 planetary gear train:

gear 2’_gear3_carrier H_ gear4 (5 points) 2. Calculate the train ratio i lH gear train with fixed axes : i 12=n 1/n 2=-z 2/z 1=-2

n 2=- n 1/2 (1) (2 points) planetary gear train:

A C1

C2

e

4/10'24'2'24'24'2-=-=-

=--=

--=

Z Z n n n n n n n n n i H

H

H H

H H

(4 points)

2

'2n n = H n n 52= (2) (2 points)

(2 points)

七．The equivalent resistant moment of force of a machine is as shown in Fig with the period φT =2π. The equivalent driving moment of force M d of a machine is constant, The average speed of the equivalent link is ωm =20 rad/s . The equivalent moment of inertia of all the links except the flywheel is negligible. The allowable coefficient of speed fluctuation is [δ]=0.01. Find the minimum moment of inertia J F of the flywheel on the equivalent link

Note: J F ≥?Wmax / (ωm 2[δ]) (15 points)

solution:

1).find M d

2πM d =400×(π/2+π/4) M d =200×(1/2+1/4) =150 N.m (4 points) W1=(-400+150) π/2=-125π W2=(150-0) π/2=75π W3=(-400+150) π/4=-62.5π W4=(150-0) 3π/4=112.5π ΔWmax=125π

J F ≥?Wmax / (ωm 2[δ]) =125π/(202×0.01)=125π/4=31.25π=98.125kgm 2

M r φ

10

2/15112-==

-==H

H H n n i n n n