Solution to Tut2

8/3/2019 Solution to Tut2

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Theory of Machines (ME 220)Solution to Tutorial sheet 2

Solution :1a)

Figure 1

The angular velocity of the link O 2A is given2 100 = rad/sec

The velocity of link O 2A

2100 0.075 7.5 / aov m s= =

The configuration diagram has shown in figure 3 to a convenient scale.Writing the vector equation,

Or2 2bo ba ao

v v v= +

Or4 2 babo ao

v v v= +

Take the vector2ao

v to a convenient scale in the proper direction and sense

bav is perpendicular to BA, draw a line perpendicular to BA through a;

4bov is perpendicular to BO 4, draw a line perpendicular to BO 4 through O 4;

The intersection of the two lines locates the point b .

2 4,o o

a

p


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1) The angular velocity of the coupler AB is

4.3457.86

0.075ba

ba

v

BA = = = rad/sec

2) apv is perpendicular to AP, draw a line perpendicular to AP through a ;

bpv is perpendicular to BP, draw a line perpendicular to BP through b ;

The intersection locates the point p pv = O 4p = 9.18 m/s

Solution: 1b)

Figure 2

2 4,o o

a

b

p


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The velocity of link O 2A

2100 0.10 10 / aov m s= =

The configuration diagram has shown in figure 2 to a convenient scale.Writing the vector equation,

Or2 2bo ba ao

v v v= +

Or4 2 babo ao

v v v= +

Take the vector2ao

v to a convenient scale in the proper direction and sense

bav is perpendicular to BA, draw a line perpendicular to BA through a;

4bov is perpendicular to BO 4, draw a line perpendicular to BO 4 through O 4;

The intersection of the two lines locates the point b.

1) The angular velocity of the coupler AB is6.45

1290.05

ba

ba

v

BA = = = rad/sec

2) apv is perpendicular to AP, draw a line perpendicular to AP through a ;

bpv is perpendicular to BP, draw a line perpendicular to BP through b;

The intersection locates the point p pv = O 4p = 7.69 m/s


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Solution 2

Figure 3


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Firstly find out all instantaneous centre of rotation using Aronhold-Kennedy theorem.These are shown in Figure 3 along with the construction lines (shown in green).

The velocity of crank OA is

6 0.15 0.9 / av OA m s = = =

Referring to Figure 4 below, the instant centre P 12, P 14, and P 24 must be in a straight lineaccording to the Kennedy-Aronhold theorem. The absolute velocity of link 2 and 4 iscommon about P 24.

First consider instant center P 24 as a point of link 2. The velocity Va can be found fromw2 using velocity difference equation about P 12, and the velocity of P 24 can be found fromwhile using graphical construction shown in figure .

Now consider P 24 as a point of link 4 rotating about P 14, knowing V P24 , we can find the

velocity of any other point of link 4. Similarly we can find out the velocity of slider B.

1) The velocity of slider B is 0.79 m/s

2) The angular velocity of link BD is

5 2

5

5

25 1225 15

10.96

82.250.795 /

w w

w

w rad s

= = =

Please note that if we were only interested in finding the velocity of the slider, themethod of instantaneous centres allows us to determine that immediately by locating I26and using that to relate the velocity of body 2 with that of body 5 as shown in Figure 5.


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Figure 4


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Solution II using ICs

Figure 6


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Solution: 3D

G

E

240

210

A

O

9020 o

20 oB

Q

45

185

185

35

180

a

e

f

bc

C

o V

0.34m/s

0.98m/s0.79m/s

d

(b)

Figure 7

The velocity of point A

20.09 0.943 /

60 A N

V m s = =

0.1250.943 1.309 /

0.09 E AOE

V V m sOA

= = =


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0.0451.309 0.982 /

0.06 B F QB

V V m sQF

= = =

Obtain the BV graphically as follows:

Take vector A

V to a convenient scale (fig 7) produce oa to e such that oe/oa = OE/OA .Rotate oe to of so that of is perpendicular to QF . Mark point b on qf such that qb/qf =QB/QF .

Now,V CA is perpendicular to AC , draw a line perpendicular to AC through a ;V CB is perpendicular to BC , draw a line perpendicular to BC through b;These intersect at the point c and V C can be computed.

FurtherV DC is perpendicular to DC , draw a line perpendicular to DC through c;

V D is horizontal, draw a horizontal line OV ;The point of intersection of these locates d . The magnitude of V D can be estimatedgraphically to be 0.34 m/s,

0.120.649

0.185bc

bc

v BC

= = = rad/sec (clockwise)

1.04.17

0.24dc

cd

v DC

= = = rad/s (counter clockwise)


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Solution-4


The velocity of link O 2AV A2 = V A3 = 20 x 0.04 = 0.8 m/s

The configuration diagram has shown in Figure 9 to a convenient scale.Writing the vector equation,

VB4 = V D + V B4D and V B3 = V A3 + V B3A3also V B3 = V B4 + V B3/4

Note that both V B3A3 and V B3/4 are vertical direction (also perpendicular to A 3B3),

therefore the point b 4 is the intersection of horizontal line from O 2 and vertical line froma2. Having determined point , one can draw the velocity image of the wheel. Link 3rotates at the same rate as link 4 (due to the slot constraint) has the same angular velocityand therefore its velocity image is scaled with the same factor as the wheel. We cansketch the image of link 3 as we know the absolute velocity of point A 3. This completesthe velocity diagram.

O 2

25 y

x

37.5 o

A B

C

D

8

40

25

25

0v,d

b , c3 4

b 4

a 2 ,a 3

V =0.8 m/s A

V =0.5 m/sB4

V =0.5 m /sB3/4

V =0.14 m/sB3/A3

Figure 8

The absolute velocity of point B on link AB is 0.707 m/s and its velocity with respect torolling wheel is 0.5 m/s.


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The motion of link 3 can be better visualized by locating the instantaneous centre of rotation of link3 (with respect to ground), i.e., I 13. The instantaneous centre can be foundout using Aronhold-Kennedy theorem as shown below in Figure 9.

Figure 9

(End)

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Solution to Tut2