99级MMT期末考卷答案.docx

Solution of 《MMT 》for Grade 99. !(!)? Draw the kinematic diagrams of the mechanism using a suitable scale.(4%) Solution: suppose scale 2:1

(2) . Locate all instant centres of the linkage for the position shown. (4%)

(3) Shown in Figure are a pair of involute profiles C2 with the common normal n —n passing through the contact point k. Draw in the Figure and point out the following: (5 %)

1) The pitch circles of both gears with

their radii rf,忖；

2) Find out the point M2 on profile

C2 which engages with profile Ci at

point Ci ?

January, 14, 2001

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2. Calculate the degree of freedom of

the mechanism shown. Indicate

compound hinge, passive DOF ，

redundant constraint before the

calculation of the DOF.(6%) Solution:

The points for attention as shown in

the figure.

(4) . The standard parameters of helical gears are on the normal plane. The standard parameters of spur bevel gears are on the large end plane. The standard parameters of a worm gear drive are on the mid ___________ plane. (3%) (5) . Ratchet mechanism can transform ________ oscillation _

_______

m otion into single intermittently rotation ___________ motion.

Geneva mechanism can transform continuous rotation ______________ motion into single intermittentlv rotation motion*

Which two conditions should be satisfied if a double universal joint is used to connect two shafts in the same plane and the ratio G between the input shaft 1 and the output shaft 3 is identically equal to one?(4%)

The two conditions are as following shown.

⑴卩 12= P23 (Shaft 2 is the intermediate shaft).

(2) The two yokes of the shaft 2 must lie on the same plane.

6. In the dynamically equivalent model of a mechanical system ，if the equivalent link is a rotating link, the equivalent moment of force is calculated by the equality of the sum of the powers and the equivalent moment of inertia is calculated by the equality of the sum of the kinetic energy ? If the mean revolution of the link is n m = 600i7min, the coefficient of speed fluctuation is 8 = 0.05, the maximal and minimal revolutions of the link are n max = 615 r/min, n min = 585 r/min. (6 %) (7)< In the gear-linkage shown below, the crank AB is a driven The output pinion

5 is located on the shaft A. Analyze the motion transmission route and draw its structural block diagram* What is the combination pattern of the combined mechanism?(6%)

F=3n - 2P L-P H=3X6-2X8-1=1

3.The first line of the subroutine LINK, RRR, and RPR is:

SUB

LINK(XA，YA,VAX,VAY,AAX，AAY,Q，

W,E,LAB,XB,YB,VBX,VBY,ABX,ABY),

SUB RRR(XA, YA, VAX, VAY, AAX, AAY, XC, YC, VCX, VCY, ACX, ACY, LAB, LCB, QAB, WAB, EAB, QCB, WCB, ECB), and

SUB RPR(M,XA, YA, VAX, V4Y, AAX, AAY, XC, YC, VCX, VCY, ACX, ACY, Q, W,E)

respectively.

In the six-bar mechanism shown below，X E=0,Y E=0, XB=41mm, Y R=0,X F=0,Y F二34mm，LED=14mm, L D A=39mm, LBA=28mm, ZADC=35°, LDc=15mm，LFG=55mm. The crank ED rotates at a constant speed of 10 rad/sec> A main program is required to analyze the output motions of the point G.The mechanism will be analyzed for the whole cycle when the driver ED rotates from 0° to 360° with a step size of 5°.(12%)

02

03

0)1

//

Solution: The Structural block diagram as shown the fig. Main program

FOR Ql=0 TO 360 STEP 5

CALLLINK(0,0,0,0,0,0,Ql*PI/180,10,0,14,XD,YD,VDX,VDY,ADX,ADY)

CALL RRR(XD,YD,VDX,VDY,ADX,ADY,41,0,0,0,0,039,28,Q2,W2,E2,Q3,W3,E3)

CALLLINK(XD,YD,VDX,VDY,ADX,ADY,Q2+35*PI/180,W2,E2,XC,YC,VCX,VCY,ACX,ACY) CALL RPR(0,0,?34,0,0,(),0,XC,YC,VCX,VCY,ACX,ACY,QFGWFGEFG)

CALLLINK(0,?34,0,0,0,0,QFGWFGEFG55,XGYGVGX,VGY,AGX,AGY)

PRINT Q1 ,XG,YG,VGX,VGY,AGX,AGY

NEXT QI

END

4(1). In the revolute four-bar mechanism, dimensions of all links are as shown. Link AB is a driver.(6%)

(a)Determine the type of the revolute four-bar mechanism?

(b)Find and label the crank acute angle 0 between the two limiting positions.

(c)Find and label the minimum transmission angle y m in?

(d)Find and label the pressure angle a of the mechanism at the position shown?

Solution:

The mechanisms is crank-rocker mechanisms.

The crank acute angle 0, the minimum transmission angle y m i n and the pressure a are shown in the figure.

4(2). In a crank-slider mechanism^ two sets of corresponding positions between the slider and the crank AB are known. At the first position of the mechanism ， the three points A, B h and Ci locate on a straight line. Find the second position B 2 of the revolute B.(6%)

B 2

B 2

Bi ///

cr

Cl

C 2

-

45°

0)

A

M

5.In a plate cam with a translating offset knife-edge follower, the cam contour consists of one straight line AD and three circular arcs: AB，BC，and CD. Their centres are points O, M, and O, respectively- The cam rotates clockwise-(9%)

(1)Draw prime circle and indicate its radius rp.

(2)Indicate the pressure angle a for the position shown.

(3)Indicate the displacement S of the follower during the cam rotation angle of 45°

from the position shown<

6.In the gear train shown in the Figure, the numbers of teeth are given as Zi = 15,

Z2 = Z4 =30, Z3 = 75, Z3J 80, Calculate the gear ratio “?

Solution:

(1)Dividing the gear train

The structure block-scheme can be shown in the following figure?

(2)D eriving the train ratios of subtrains independently In the gear train with fixed axes, we have

Z B=ZL=_￡L =_Z5=_5

n y z] 15

In the differential gear train, we have

(3)F inding the train ratio

i\5= n[/n5=25

7. A pair of external involute spur gears have the parameters as Zi =19, Z 2 = 116, m = 4mm ，a = 20°, dn = 70.00mm, h a * = 1, working center distance a ,=270mm. Questions: (13%) 1) Calculate the modification coefficient Xi for the pinion;

2) Which type of corrected gear pair does it belong to( standard gear pair, with

reference centre distance or with modified centre distance)? Why?

3) Calculate x 2,畑 畑闻，p a 2 for gear 2.

Solution:

(l)For

dn= d{- 2冶=mzi- 2(h ： + ) tn SO

we have

(2) For

and

70.00 = 4x[ 19-2 x( 1+0.25-xi)] x\ =0.5 a = m (zi + Z2 )/2 = 4x (19 + 116)/2 = 270 mm = a' x\ = 0.5/0 so the gear drive may be height correction gear pair.

(3) X2= -x\ = -0.5 尸2=加 ^2/2 = 4 x 116/2 = 232 mm

尸b2 =『2 cos a = 232 x cos 20° = 218.009 mm

心2=八 -rf 7 - c m = 270 一 70.00/2 一 0.25 x 4 = 234.000 mm

a a 2 = arccos (外2仇2) = arcos (218.009/234.000) = 21.305°

Pa2 =(亡2一《2)0 5 = (234.0002 - 218.0092 )05 =85.018 mm

8. Three unbalanced masses m2 and m3 exist on three transverse planes 15 2 and 3, respectively. Their mass-radius products are: miri=2kgmm ? m2r2=6kgmm,

m3r3=3kgmm, respectively. The locations are as shown. Suppose the system is to be balanced fully by two balancing mass-radius products, Pbi and Pb3, on the planes 1 and 3, respectively. Determine the amounts and angular locations of the two

balancing mass-radius products-

W 2i=l2xW 2/( 12+24)=l 2x6/36=2 kgmm

W 23=24x W 2/(l 2+24)=24 x6/36=4 kgmm

(2) Balancing:

Plane 1 W] + W 21 + W bl = 0 illustrating, we have

Wbi = 2^2 kginm

0bi=-135°

Plane 3 + VV 23 + IV b 3 = 0

illustrating, we have

Wb3 = 1 kgmm

0b 3=-90° (?) (T )

Solution: (1) Dividing

Plane 1 Plane 3