THEORY OF MACHINES (Relative velocity & acceleration ......THEORY OF MACHINES (Relative velocity & acceleration) -٤ -Example 1.3: In Figure 1.7, the velocity of A is known, and it

THEORY OF MACHINES (Relative velocity & acceleration)

- ١ -

CHAPTER 0ne RELATIVE VELOCITY & ACCELERATION

1. VELOCITY AND ACCELERATION DIAGRAMS: 1.1 RELATIVE VELOCITY METHOD When accelerations of mechanisms are studied, however, it is necessary to determine the relative velocities between points on a link. For this reason, the relative velocity method of obtaining velocities is the most useful method. This section involves the construction of velocity diagrams, which needs to be done accurately and to a suitable scale. Students should use a compass, protractor and triangles and possess the necessary drawing skills.

To explain the method, considering the two moving points A and B in Figure 1.1, the absolute velocity VB of point B is equal to the vector sum of the absolute velocity VA of point A and the velocity VB/A of B relative to A. VB = VA ↦ VB/A

Therefore, it is necessary to know both the velocity VA of A and the velocity VB/A of B relative to A to find the velocity VB of B.

In Figure 1.2a, the velocity of A is known and the direction only of B is known. Avelocity diagram is constructed as follows:

1. Arbitrarily locate the origin o (lower-case letter), which represents a fixed point. All vectors originating at o represent absolute velocities.

2. The known absolute velocity VA is laid off from o.

3. The direction of the absolute velocity VB is drawn through o.

4. The direction of the relative velocity VB/A is known to be ⊥ to the line connecting A and B, and it must connect the termini of the two absolute velocities VA and VB. Therefore, a line is drawn through the known terminus of VA and in a direction ⊥ to line AB.

5. This establishes the magnitudes of VB and VB/A.

Note that VB/A points toward VB. A vector of opposite sense would represent VA/B. Note also that the absolute velocities emanate from the origin o and that terminus of each is labeled with a lower-case letter corresponding to the point involved.

From the velocity diagram in Figure 1.2b it is evident that VB = VA ↦ VB/A

VA = VB ↦ VA/B

Figure 1.1

Figure 1.2


- ٢ -

Example 1.1: In the mechanism shown in Figure 1.3, the velocity of A is known and it is required to find the velocities of B, C, D, E, and F. TO FIND VB:

1. The given velocity VA is laid off from origin, as shown in Figure 1.3a'. 2. The direction of VB (⊥ to link 4) is drawn to indefinite length through the origin. 3. The direction of VB/A (⊥ to line AB) is drawn through the terminus of VA. 4. The intersection of the VB and VB/A direction lines determines the magnitudes of both.

TO FIND VC (Figure 1.3b):

1. Draw a line through the terminus of VA in the direction of VC/A (⊥ to CA). VC = VA ↦ VC/A

2. Draw a line through the terminus of VB in the direction of VC/B (⊥ to CB). VC = VB ↦ VC/B

3. The terminus of VC is at the intersection of these two lines, as shown in Figure 1.3b. TO FIND VD: Since A, D, and B lie along a straight line on link 3, the relative velocities VD/A and VD/B cannot be used together to get VD, because they are parallel (coincide) and will yield no intersection. Therefore, VD/C must be used in combination with VD/A (or VD/B). This is done as follows:

1. Draw a line through the terminus of VC in the direction of VD/C (⊥ to DC). VD = VC ↦ VD/C

2. Draw a line through the terminus of VA in the direction of VD/A (⊥ to DA). This coincides with the VB/A line already drawn, since A, D, and B lie on a straight line.

VD = VA ↦ VD/A 3. The terminus of VD is established by the intersection of these

two lines, as shown in Figure 1.3c. TO FIND VD BY PROPORTION: An easier way to locate VD is to take the advantage of the fact that the velocity diagram is an image of the actual mechanism, as shown in Figure 1.4. Distances along the images are proportional to corresponding distances along the links. Therefore,

Figure 1.3 (a')

Fig1.4


- ٣ -

(AD/AB) = (ad/ab) Similarly, the velocities of points E on link 2 and F on link 4 can be determined: (O2E/O2A) = (oe/oa) and (O4F/O4B) = (of/ob) Example 1.2: In the mechanism shown in Figure 1.5 (Slider-crank mechanism), the velocity of A is known, and it is required to find the velocities of B and C. TO FIND VB (direction known):

1. Draw vector VA in velocity diagram. 2. Lay off direction of VB. 3. From terminus of VA, lay off direction of VB/A (⊥ to line BA). 4. The intersection of VB/A and VB lines determines the magnitudes of both (VB = VA ↦ VB/A).

TO FIND VC:

1. Draw a line through the terminus of VA in direction of VC/A (⊥ to line CA). VC = VA ↦ VC/A

2. Draw a line through the terminus of VB in the direction of VC/B (⊥ to line CB). VC = VB ↦ VC/B

3. The intersection of VC/A with VB/A determines VC. In the above example, if it were required to find the velocity of a point such as D on link 3, it would be simple to extend the image in the velocity diagram by proportion, (AB/AD) = (ab/ad), as shown in the figure. 1.1.1 VELOCITIES OF POINTS ON A ROLLING BODY: Let disk 2 in Figure 4.6 is rolling on body 1. Then body 2 is rotating about point P in link1 at the instant. The center of the disk will have a velocity VC = ω * R; where R is the radius and ω is the angular velocity of the disk. Any other point on the disk, such as Q will have a velocity relative to C which is VQ/C = CQ *ω. Vector VQ/C must be ⊥ to the radius of rotation CQ. The absolute velocity of Q is then VQ = VC ↦ VQ/C.

Figure 1.5

Fig.4.6


- ٤ -

Example 1.3: In Figure 1.7, the velocity of A is known, and it is required to find the velocity of C. It is assumed that no slipping occurs, Therefore, VA = VB, and the mechanism could be represented by the equivalent four-bar mechanism shown in dashed lines. 4.1.2 VELOCITIES OF SLIDING LINKS: The solutions of sliding link problems are based on the fact that the only possible relative velocity that can exist between two sliding surfaces is along their common tangent line. This establishes the direction of relative motion, which provides sufficient information to complete the velocity diagram. Figure 1.8 shows two sliding problems solved.

2

3

4

1

1

O2

O4

Figure 1.7

Figure 1.8

O2 O3

O2 O4


- ٥ -

Figure 1.10

O4 O6

VELOCITIES OF COMPLEX MECHANISM: Example 1.4: In Figure 1.9a, the velocity of point A is given and the velocities of points B and C are required.

Solution Ex 1.4:

1. Draw vector oa in the velocity diagram (Figure 1.9b). 2. Lay out the direction of ob (⊥ to link 6). 3. Through a, lay out the direction of ab (⊥ to link 3). 4. The intersection of the ob and ab direction lines locates b. 5. Lay out the direction of oc (⊥ to link 4). 6. Through b, lay out the direction of bc (⊥ to link 5). 7. The intersection of the oc and bc direction lines locates c.

Example 1.5: In Figure 4.10a, the velocity of point A is given and the velocities of points B, C, and D are required. Solution Ex 1.5: In this example, neither the direction nor the magnitude of VC is known, nor is the magnitude of VC/A known. The trial-and-error approach may be made to determine the velocities of points on link 5will be used here.

1. Draw oa in the velocity diagram (Figure 1.10b).

2. Lay out the direction for ob and oc. 3. Choose a trial position for b

labeling it b*, i.e. any length ob* as shown in Figure 1.10b can be assumed for VB.

4. Then VD will be od* which is ⊥ to link 4 and b*d* is ⊥ to BD.

5. Lay out c* in the same relative position along b*d* that C has relative to B and D, that is,

(BC/BD) = (b*c*/b*d*) 6. Now a line may be drawn from o to c*, thereby establishing the direction of VC.

Figure 1.9

O4 O6

THEORY OF MACHINES 7. Through a, draw the direction of line

line drawn in step 6 locates c. 8. Through c, draw a line parallel to b

Note: An alternative approach to this problem would be to determine the dire Example 1.6: Figure 1.11, shows a variable stroke engine mechanism. The lengths of the cranks OA and QB are 90 mm and 45respectively. The diameters of wheels with centers O and Q are 250 mm and 120 mm respectively. Other lengths are shown in the diagram in mm. There is a rolling contact between the two wheels. If OA rotates at 100 rpm, determine the

(i) Velocity of the slider D (ii) Angular velocity of links BC and CD.

(Relative velocity & acceleration) the direction of line ac (⊥ to link 3). The intersection of this line and the

b*d*, which will locate the correct positions for

An alternative approach to this problem would be to determine the direction VC by locating centre

, shows a variable ngths of the

45 mm respectively. The diameters of wheels with

mm Other lengths are shown in the

contact between the two wheels. If OA

Angular velocity of links BC and CD.

Figure 1.11

). The intersection of this line and the

, which will locate the correct positions for b and d.

by locating centre 15.


- ٧ -

Example 1.7: In the mechanism shown below, the slider D is constrained to move on a horizontal path. The crank OA is rotated in a ccw direction at a speed of 150 RPM. The dimensions of the links are as follows: OA = 180 mm, CB = 240 mm, and BD = 540 mm. For the given configuration find the velocity of slider D and the angular velocity of BD. Solution Ex 1.7: ωOA = π*150/30 = 15.707 rad/s VA = 15.707 * 180 = 2827.43 mm/s = 282.743 cm/s

360mm

105mm

A

O

B

D

C

From the velocity diagram the following velocities and angular velocities can be determined:

VD=14.8924/0.2=74.46 cm/secVB/A= 147.932 cm/secϖBA= 147.932/36.0= 4.109 rad/sec ccwVD/B= 201.161 cm/secϖDB= 201.161/54.0= 3.725 rad/sec cwVB= 206.5835 cm/secϖCB= 8.606 rad/sec ccwVelocity diagram

Scale = 0.2mm/cm/sec

b

d

a

o,c


- ٨ -

Example 1.8: A cam with oscillating follower is shown below, where the angular velocity of the cam is indicated. Find the angular velocity of link 4. Solution Ex 1.8: By equivalent linkage method: VA = 5*68.2 = 341 mm/s SV = 0.146627 mm/mm/s From velocity diagram: VP4 = 214.56 mm/s ω4 = 214.56/191 = 1.123 rad/s cw

2 =5 rad/s

22 = 2.5 rad/s

2

O2

Path P4 describes on link 2

P2 ,P 4

3

O4

1

4

O2

Path P4 describes on link 2 4

3P2 ,P4

O4

A

2

Velocity diagramSV=0.146627 mm/mm/so

p4 a

VP4


- ٩ -

Figure 1.12

240 mm

160 mm

Example 1.9: Figure 1.12 shows the mechanism of a molding press in which O2A= 80 mm, AB = 320 mm, O4B = 120 mm, CD = 320 mm, BC = 80 mm. For the configuration shown, construct the velocity diagram and determine VD/C, VC, and the angular velocity of link O2A. Q2) Find the velocity ratio of the two co-axial shafts in the gear shown in Figure 2: when (a) the wheel A1 is fixed and S1 is the driver, (b) the wheel S2 is fixed. The tooth numbers of the gears are S1 = 40, A1 = 120, S2 = 30, A2 = 100. Solution Ex 1.9: VD = 0.2 m/s (given) With SV = 100 mm /m/s the velocity diagram will be drawing as shown in the figure below. From the velocity diagram: VD = 0.2 m/s

VA = 20.0075/100 = 0.2 m/sVD/C = 6.963/100 = 0.0693 m/sVC = 19.83/100 = 0.198 m/sϖO2A = VA / O2A = 0.2/80 = 2.5 rad/s ccw

A

B

C

O4

O2

D

c*b*

a*

Velocity polygonSV = 100mm/ m/s

acb

o

d

VD VAVB

VC


- ١٠ -

Example 1.10: The absolute velocity of J in in the mechanism shown is 450 cm /sec while the disk is slowing down with an acceleration of 10 rad/sec2.

(a) Construct the velocity diagram and determine ω3, VD, VB and VC.

(b) Determine the angular acceleration

of link 4.

Solution Ex 1.10: VJ = 450 cm/s = 4.5 m/s ∴ω2 = VJ/OJ = 4.5/.45 = 10 rad/sec cw ∴ VB = ω2 * OB = 10 * 0.3 = 3 m/s

Now draw the velocity diagram with suitable scale. Let SV = 20 mm/m/s; From the velocity diagram as shown below, the following can be obtained:

VD = 62.78 / 20 = 3.139 m/s VC = 34.51 / 20 = 1.77255 m/s VC/B = 58.23 / 20 = 2.9115 m/s VD/B = 70.67 / 20 = 3.5335 m/s VD/C = 28.27 / 20 = 1.4135 m/s

From the mechanism diagram, we can determine BC = 1.236 m. The angular velocity of link 3 can be determined as follows:

ω3 = VC/B / BC = 2.9115 / 1.236 = 2.3556 rad/s ccw Or ω3 = VD/B / DB = 3.5335 / 1.5 = 2.3557 rad/s ccw Or ω3 = VD/C / DC = 1.4135 / 0.6 = 2.3558 rad/s ccw After that all the normal component of acceleration will be calculated as follows:

onlydirectionknowna

s/m33.36.0

4135.1a

onlydirectionknowna

s/m858.6236.1

9115.2a

onlydirectionknowna

s/m3237.85.1

5335.3a

onlydirectionknowna

s/m422.166.0

129.3a

s/m33.0*10OB*a

s/m303.0

9OBVa

tD/C

22

nD/C

tB/C

22

nB/C

tB/D

22

nB/D

tD

22

nD

2tB

22Bn

B

=

==

=

==

=

==

=

==

===

===

α


- ١١ -

Now draw the acceleration diagram with Sa = 3 mm/m/s2. From the acceleration diagram, the angular acceleration of link 4 can be obtained:

ccws/rad6333.156.038.9

s/m38.9314.28a

cws/rad327.366.0/7967.21

s/m7967.21339.65a

23

2tD/C

24

2tD

==∴

==

==∴

==

α

α

t

t

t

t

t

n

n

n

n

aC/D

aD/B

aD/B

aC/B

aC/B aB

aB

aDaD

Mechanism diagramSm = 75 mm/m

Velocity diagramSv = 20 mm/m/s

2Acceleration diagramSa = 3 mm/m/s

aC/B

aD/B

aC/D aD

aB

c

b

o

d

4

1

1

2

3

D

C

B

A

JO

VC

VD/C

VD/BVC/B

VB

cd

b

o


- ١٢ -

Problems (Relative velocity): Q1) (a) In Figure P1.1, link 4 rolls on link 1. Construct the velocity polygon, ω2 =144 rad/s. (b) Determine ω3, ω4 in radians per second. Q2) (a) Construct the velocity polygon for Figure P1.2. ω2 = 150 rad/s. Determine the velocity of

slider 6 in meters per second. (b) Determine ω3, ω4 and ω5 in radians per second. Q3) The velocity of point B of the linkage shown in Figure P1.3 is 40 m/s. Find the velocity of point A and

the angular velocity of link 3.

Q4) Crank 2 of the mechanism shown in Figure P1.4 is driven at ω2 = 60 rad/s cw. Find the velocities of points B and C and the angular velocity of link 3 and 4. O2A=150mm, AB=300mm, O4O2=75mm, O4B=300mm, AD=150mm, DC=100mm. Q5) The inversion of the slider-crank mechanism shown in Figure P1.5 is driven by link 2 at ω2 = 60 rad/s ccw. Find the velocity of

point B and the angular velocities of links 3 and 4. O2A=75mm, AB=400mm, O2O4=125mm. Q6) Find the velocity of point C and the angular

velocity of link 3 of the mechanism shown in Figure P1.6. Link 2 is the driver and rotates at 8 rad/s ccw. O2A=150mm, AB=O4B=250mm, O2O4=75mm, AC=300mm, CB=100mm.

Figure P1.1

Figure P1.2

AB=400 mm

Figure P1.3

Figure P1.4

Figure P1.5

Figure P1.6


- ١٣ -

Q7) Find the velocities of points B, C, and D of double slider mechanism shown in Figure P1.7 if the crank 2 rotates at 42 rad/s cw. O2A=2in, AB=10in, AC=4in, BC=7in, CD=8in, (Take 1 in= 25.4mm). Q8) A Scotch-yoke mechanism shown in Figure P1.8. It is driven by crank 2 at ω2 = 36 rad/s ccw. Find the velocity of the crosshead, link 4. O2A=250mm. Q9) In the steam engine mechanism as shown in Figure P1.9, the crank AB rotates at 200 rpm in cw direction. Find the velocity of C, D, E, F, and P. The dimensions of the various links are AB = 12 cm, BC = 48 cm, CD = 18 cm, DE = 36 cm, EF = 12 cm and FP = 36 cm. Q10) If V6 = 1.081 m/s down ward, Figure P1.2. Construct the velocity diagram and determine the velocity of point VC. Q11) Two rolling wheels are connected by link 3 as shown in Figure P1.10. Wheel 2 rolls with uniform angular velocity and M moves to the left by 6.25 m/sec. (a) Construct the velocity diagram and find VR, VQ, and ω3. (b) Find aR, and α3. Q12) For the position of the mechanism shown in Figure P1.11, find the velocity of the slider B and the angular velocity of link AB if the velocity of the slider A is 3 m/s.

Figure P1.7

Figure P1.8

P

F

E

D

C

B

A

Figure P1.9

45°

60°

R

1 2

5½

1

1

φ 2½

φ 3

Figure P1.10: (All dimensions in meter)

Figure P1.11


- ١٤ -

RELATIVE ACCELERATION METHOD The relative acceleration method of analyzing accelerations of parts in a mechanism is based on the following principles:

1. That all motions are considered instantaneous 2. That the instantaneous motion of a point may considered pure rotation 3. That the acceleration of a point much more easily analyzed if it is resolved into two

rectangular components, one normal and one tangent to its path 4. That the relative velocities as well as the absolute velocities of the various points in the

mechanism are available.

ACCELERATION DIRECTIONS The directions of accelerations of points moving with curvilinear motion are not always known. It is necessary to regard accelerations as made up of two rectangular components: the normal component an and the tangential component at. As shown in Figure 4.13, the normal component is directed toward the center of rotation, and the tangential component is perpendicular to the component, or tangent to the path of the point. It is evident that the acceleration is vector sum of these two components: that is, a = an ↦ at Figure 4.14 shows a link with two points A and B, neither of which is fixed. If the link has angular acceleration α, the normal and tangential acceleration components for the acceleration aB/A of a point B relative to A are shown in Figure 4.14a. The normal component anB/A points toward A, and the tangential component atB/A is ⊥ to the line AB. In Figure 4.14b, the components for the acceleration of point A relative to B are shown. The magnitudes of these acceleration components may be expressed as follows:

αα ∗=∗=

==

ABaBAa

ABV

aAB

Va

tBA

tAB

BAnBA

ABnAB

//

2/

/

2/

/)()(

Example 4.11: In Figure 4.15a, crank 2 is rotating ccw at 75 rpm and is slowing down at the rate of 15 rad/s2. It is required to find the accelerations of point A and B and the angular velocities and accelerations of links 3 and 4. O2O4=370mm, O2A=365mm, AB=1067mm, O4B=762mm.

Figure 4.13

Figure 4.14


- ١٥ -

Solution Ex.4.10:

1. Draw the velocity diagram as shown in Figure 4.15b.

sm865252smm252865857365AOV

rad85760

7528660

N2

22A

2

/./..**

sec/.*.

====

===

ω

πω

a. Lay out VA from the original o (⊥ to link 2). b. Lay out the direction of VB from the original o (⊥ to link 4). c. Lay out the direction of VB/A from the terminus of VA (⊥ to link 3). d. The intersection of the VB and VB/A direction lines determines the magnitudes of

both. e. After that, all the linear and angular velocities can be determined.

VB = 46.9945/10= 4.69945 m/s ω4= 4.69945/0.762= 6.16 rad/sec ccw VB/A= 45.3530/10= 4.53530 m/s ω3= 4.5353/1.0670= 4.25 rad/sec ccw

2. Write the acceleration equation for aB:

tAB

nAB

tA

nA

tB

nB

ABAB

aaaaaaaaa

//

/

aaaa

a

=

=

Figure 4.15


- ١٦ -

3. Determine the magnitudes and directions of the various terms:

)3(*

)3(/27268.19067.1*25.4*

)2(/475.5365.0*15*

)2(/4922.22365.0*

)4(*

)4(/9145.28762.0*16.6762.0*

3/

2223/

222

223

2

244

2224

4

2

linktoonlyknowndirectionABa

linktoparallelsmABa

linktosmAOa

linktoparallelsmAO

Va

linktoonlyknowndirectionBOa

linktoparallelsmBO

Va

tAB

nAB

tA

AnA

tB

BnB

⊥=

===

⊥===

===

⊥=

====

α

ω

α

ω

α

ω

4. Draw the acceleration diagram to obtain aB (Figure 4.15c).

a. Lay out nBa from the original o (║to link 4). b. Through the terminus of nBa draw a ⊥ line of indefinite length representing the

direction of tBa , whose magnitude is unknown. c. Again starting from the origin o, lay out nAa (║to link 2). d. From the terminus of nAa and ⊥ to it, lay out tAa . This establishes the acceleration aA

of A, which represents the known acceleration, aA= 23.148 m/sec2. e. From the terminus of aA, lay out n ABa / (║to link 3). f. Through the terminus of n ABa / and ⊥ to it, draw a line of indefinite length

representing the direction of t ABa / , whose magnitude is unknown. g. The intersection of two indefinite-length lines draw in steps (b) and (f),

representing tBa and t ABa / , respectively, determines aB, which equals about 38.4 m/s2.

5. Compute the angular accelerations:

cwsrad46337620

525BO

a

cwsrad021606710924517

ABa

2

4

tB

4

2t

AB3

/..

.

/.../

===

===

α

α


- ١٧ -

Example 4.12: In Figure 4.16a, crank 2 rotates ccw with a uniform angular velocity of 30 rad/sec. It is required to find the linear accelerations of points A, B, C, and D and the angular velocity and acceleration of link 3. O2A=75 mm, AD=25 mm, AB= 229 mm, AC= 1/3 AB.

Figure 4.16


- ١٨ -

Solution Ex.4.12: 1. Draw the velocity diagram as shown in Figure 4.16b.

smsmmAOV A

/25.2/225030*75* 22

==== ω

VB = direction known only (D to path of link 4) VB/A= direction known only (⊥ to AB)

2. Write the acceleration equation for aB:

tAB

nAB

tA

nAB

ABAB

aaaaa

aaa

//

/

aaa

a

=

=


)(*

)(/..

.

),(

)(/...

)(

/

//

3linktoonlyknowndirectionABa

3linktoparallelsm674112290

635051AB

Va

0uniformis0a

2linktoparallelsm5670750252

AOVa

sliderofpathtoParallelonlyknowndirectiona

3t

AB

222

ABnAB

22tA

22

2

2An

A

B

⊥=

===

==

===

=

α

αω

4. Draw the acceleration diagram to obtain aB (Figure 4.16c). a. Lay out the direction aB (

tBB aa = ) through the original o (║to path of slider).

b. Again starting from the origin o, lay out Aa ( nAA aa = ) (║to link 2). c. From the terminus of Aa , lay out n ABa / . d. Through the terminus of n ABa / and ⊥ to it, draw a line of indefinite length

representing the direction of t ABa / , whose magnitude is unknown. The intersection of this line with the aB direction line drawn in step a determine aB

5. Obtain aC and aD (Figure 4.16c). Points c and d in the acceleration diagram are located by proportion. 6. Determine the angular velocity and acceleration of link 3:

ccwsrad752012290202346

ABa

cwrad1472290

635051AB

V

2t

AB3

AB3

/...

sec/..

.

/

/

===

===

α

ω

Example 4.13: In Figure 4.17a, crank 2 rotates ccw with a uniform angular velocity of 240 rpm. It is required to find the linear accelerations of points B, C, and D and the angular velocity and acceleration of link 3 and link 5.


- ١٩ -

Ab=100mm, BD=500mm, BC=400mm

Solution Ex.4.13:

Figure 4.17


- ٢٠ -

1. Draw the velocity diagram as shown in Figure 4.17b.

smsmmABV

rad

B /513.2/274.2513132.25*100*

sec/1327.2560

240*2

2

2

====

==

ω

πω

VC = direction known only (║to path of slider 4) VC/B= direction known only (⊥ to CB) VD = direction known only (║to path of slider 6) VD/B= direction known only (⊥ to DB)

2. Write the acceleration equation for aC and aD:

tBD

nBD

tB

nBD

BDBD

tBC

nBC

tB

nBC

BCBC

aaaaaaaa

aaaaaaaa

//

/

//

/

aaa

a

aaa

a

=

==

=


)(*

)(/..

.

)()(*

)(/..

.

),(

)(/..

.

)(

/

//

/

//

3linktoonlyknowndirectionDBa

Btoward5linktoparallelsm517250

1218751DB

Va

6sliderofpathtoParallelonlyknowndirectiona3linktoonlyknowndirectionCBa

Btoward3linktoparallelsm1234840

80261CB

Va

0uniformis0a

Atoward2linktoparallelsm151696310

5132BAVa

4sliderofpathtoParallelonlyknowndirectiona

5t

BD

222

BDnBD

D

3t

BC

222

BCnBC

22tB

222

BnB

C

⊥=

===

=

⊥=

===

==

===

=

α

α

αω

4. Draw the acceleration diagram to obtain aC and aD (Figure 4.17c).

a. Lay out the direction aC ( tCC aa = ) through the original o (║to path of slider 4).

b. Again starting from the origin o, lay out Ba ( nBB aa = ) (║to link 2 toward A).

c. From the terminus of Ba , lay out n BCa / .

d. Through the terminus of n BCa / and ⊥ to it, draw a line of indefinite length

representing the direction of t BCa / , whose magnitude is unknown. The intersection

of this line with the aC direction line drawn in step a determine aC.

e. Lay out the direction aD through the original o (║to path of slider 6).

f. From the terminus of Ba , lay out n BDa / .


- ٢١ -

g. Through the terminus of n BDa / and ⊥ to it, draw a line of indefinite length

representing the direction of t BDa / , whose magnitude is unknown. The intersection

of this line with the aD direction line drawn in step e determine aD.

5. Obtain aB, aC and aD (Figure 4.17c), which are 63.15 m/s2, 44.89 m/s2 and 8.53 m/s2

respectively.

6. Determine the angular velocity and acceleration of link 3 and link 5:

cwsrad062512050

0312560DB

a

cwrad24375250

1218751DB

V

cwsrad77610940

9105543CBa

cwrad5065440

80261CB

V

2t

BD5

BD5

2t

BC3

BC3

/..

.

sec/..

.

/..

.

sec/..

.

/

/

/

/

===

===

===

===

α

ω

α

ω

Example 4.14: In the link ABC in Figure 4.18a, AB = 600 mm, BC = 225 mm. A and B are

attached by pin joints to the sliding blocks. If, for the position where BD = 375 mm, A is

sliding towards D with a velocity of 6 m/s and a retardation of 150 m/s2, find the acceleration of

C and angular acceleration of the link.

Solution Ex 4.14:


- ٢٢ -

2

n

t aB/A

aB/A

t

(b)

(a)

2

2aC = 129.7702/0.5 = 259.54 m/sαAB = aB/A/AB = 175.38/0.6 = 292.3 rad/s

Acceleration diagramSA=0.5 mm/m/s

aB d

c

a

b

aA

aC

Answer Q2:

VCc

VC = 6.95518 m/sVC/A = 8.035 m/sVB/A =5.8438 m/s VB =5.6739 m/s

VB/A

VA

VB

Velocity diagramSV=10 mm/m/s

b

a

d

Mechanism diagramSM=0.2 mm/mm

C

B

A

D

Figure 4.18


- ٢٣ -

4.2.2 CORIOLIS ACCELERATION:

Whenever a point in one body moves along a path on a second body, and if the second body is

rotating, then the acceleration of the point in the first body relative to a coincident point in the

second body will have a Coriolis component.

In Figure 4.19 let P3 be a point on slider 3 which is moving along the path OF in body 2. Let P2

be a fixed point on the path and let P3 and P2 be coincident at the instant. The angular velocity

for body 2 is ω2 and hence also for the path. The path is again shown in Figure 4.19, where

23 / PPV is the velocity of P3 relative to P2. In a time interval dt, line OF will rotate through an

angle dθ to position OF'. During this time P2 moves to P2' and point P3 moves to P3' as shown

in Figure 4. 20.

θdBPBPArc )( '2'

3 =

But dtdanddtVBP PP 22/'

2 3ωθ ==

Thus 2

2/'

3 )(23 dtVBP PP ω= (1)

For a displacement with constant acceleration

2'3 )(21 dtABP = (2)

From Equations 1 and 2

22

2/ )(21)(

23dtadtV PP =ω

Or (3)

which is called the Coriolis component of acceleration for point P3.

The relationship between23 / PP

V , ω2, and 2/ 232 ωPPV for the case of Figure 4.19 is shown in Figure

4.19a. If 23 / PP

V is toward the center O, the relationship will be that of Figure 4.19b. The rule is

as follows: the Coriolis acceleration is the direction of23 / PP

V , after the latter has been rotated

90° in the direction of the angular velocity of the path.

From Equation 3 we note that if either 23 / PP

V or ω2, or both, are zero, then there will no Coriolis

component of acceleration.

2/ 232 ωPPc Va =


- ٢٤ -

Figure 4.19

Figure 4.20


- ٢٥ -

Figure 4.22

The general case of relative motion of two bodies in a plane is illustrated in Figure 4.21. The

absolute acceleration of point P3 is

2323 / PPPP

aaa →=

Or 2/// 2323232233 2 ωPPt

PPn

PPtP

nP

tP

nP Vaaaaaa aaaaa = (4)

Figure 4.21

where the Coriolis component 2VP3/P2ω2 is part of the acceleration of P3 relative to P2.

Example 4.15: A quick-return mechanism is shown in

Figure 4.22. Link 2 is the driver, and the angular

velocity and acceleration of link 4 are to be found.

Solution Ex 4.15:

Let P2 and P4 be fixed points on links 2 and 4 which are

coincident at the instant, then

ω2 = 2π*(9.5/60) = 0.995 rad/s

( ) smPOVP /151.0995.0*152.0* 2222 === ω

a

a

a

a


- ٢٦ -

Scaling values from the velocity polygon in Figure 4.23, we find smVP /0742.04 = and

smV PP /131.024 / = . Then

ccwsradPOVP /144.0

514.00742.0

444

4 ===ω

Before we can find the angular acceleration of link 4, it is necessary to find the acceleration of

P4. Similar to Equation 4 we may write

2/// 242424224 2 ωPPt

PPn

PPtP

nPP Vaaaaa aaaa=

In order to solve this equation it is necessary to know the radius of curvature of the path which

P4 describes on body 2. This path is not known. However, the path which P2 describes on 4 is

straight line along the link. We can use this path if we write the equation for2P

A . Then

××−×−×××××

= 4//0

/44

0

424242222 ωPP

tPP

nPP

tntP

nP Vaaaaaa aaaaa (5)

where 222

2

/15.0152.0151.02

2sm

POV

a PnP ===

00 22 == αbecauseatP

244

2

/0107.0514.0

0742.044

smPO

Va PnP ===

( ) unknownisbutPOa tP 44444 αα=

( ) 0131.0

22/

/42

42=

∝==

RV

a PPn PP

Figure 4.23


- ٢٧ -

Figure 4.24

24/

/

/0377.0144.0*131.0*22

,

42

42

smV

andunknownisa

PP

tPP

==ω

Thus the magnitudes of the tangential components are the only unknowns Equation 5. Their

values may be found by drawing the acceleration polygon which is shown in Figure 4.24 or

4.25. Then 2/0921.04

sma tp = , and

2

444 /179.0514.0

0921.04 sradPO

a tp ===α

Figure 4.25

Example 4.16: A cam with oscillating follower is shown in Figure 4.26, where the angular

velocity and acceleration of the cam are indicated. The angular acceleration of link 4 is wanted.

n

2p a =

t

42p /p a

n

t

4p a

p 4

a

2p

4p /p 2

42 4p /p

a

a

2V ω

o

4p

2p


- ٢٨ -

The radius of curvature of the path is 138 mm and equals the radius of the cam outline plus the

roller radius.

Solution Ex 4.16:

( ) smPOVP /417.05*0833.0* 2222 === ω

and the velocity polygon is shown in Figure 4.26. smVP /213.04 = and smV PP /533.024 / = . Then

cwsradPOVP /12.1

191.0213.0

444

4 ===ω

Next, in order to find the angular acceleration of link 4, we must find the acceleration of P4.

Thus

×××−×−××××

= 2//0

/22

0

242424442 ωPP

tPP

nPP

tntP

nP Vaaaaaa aaaaa (6)

where

244

2

/238.0191.0213.04

4sm

POV

a PnP ===

( ) 22222 /08.25*0833.022 smPOanP === ω

( ) 2222 /208.05.2*0833.0*2 smPOatP === α

( ) 22

2

2/

/ /06.2138.0533.024

24sm

CPV

a PPn PP ===

22/ /33.55*533.0.0*22 24 smV PP ==ω

Thus the magnitudes of the tangential components are the only unknowns Equation 6.

Figure 4.28 shows the acceleration polygon. 2/97.14

sma tp = and then

cwsradPOa tp 2

444 /31.10191.0

97.14 ===α

THEORY OF MACHINES

Figure 4.2

Figure 4.2

Figure 4.28: Acceleration diagram

(Relative velocity & acceleration)

4.26: Mechanism diagram

4.27: Velocity diagram


- ٣٠ -

Problems (Relative acceleration): Q132) For the data given in the Figure P4.12, Find the velocity and acceleration of points B and C. AB=16 cm, AC=10 cm, BC=8 cm Q14) For the straight-line mechanism shown in Figure P4.13, ω2=20 rad/sec cw and α2=140 rad/sec2. Determine the velocity and acceleration of point B and the angular acceleration of link 3. O2A=AC=AB=100mm Q15) In the Figure P4.13, the slider 4 is moving to the left with a constant velocity of 20 m/s2. Find the velocity and acceleration of link 2. Q16) Solve (Q3) for the acceleration of point A and angular acceleration of link 3. Q17) In the mechanism shown in Figure P4.14 the crank BC, 100mm long, turns at 300 rpm about center C offset 75mm from the horizontal line of stroke of A. The connecting rod BA=300 mm. Find the velocity and acceleration of point A.

Q18) In the mechanism shown in Figure P4.15, the guide is part of the fixed link and its centerline is a circular arc of radius R. Determine the magnitude of the angular velocity of the slider when ω2= 1 rad/sec, also determine aB and α4. Q19) In the mechanism shown in Figure P4.16, Determine the relative velocity and acceleration of the two sliders when ω2 is uniform and equal to 32 rad/sec cw and θ = 50°. Q20) (a) Construct the velocity diagram for the mechanism of Figure P4.17 and determine the velocity of point D. (b) Construct the acceleration diagram using a unit value of the angular velocity of the driving link (α2= 0). Calculate the acceleration of point D. Q21) Crank 2 in Figure P4.18 rotates ccw with an angular velocity of 3 rad/sec and speeding up with an acceleration of 50 rad/sec2. (a) Find VA and VB. (b) Find ω3 and ω4. (c) Find aA and aB. (d) Find α3 and α4. Q22) Rework Example 4.7 when the crank OA is rotated in a ccw direction at a speed of 150 RPM and at an angular acceleration of 40 rad/s2 also in ccw direction. Find the acceleration of slider D and the angular acceleration of BD.

VA=1.6 m/s

aA= 4 m/s2

A

C

B

2

ω2=24 rad/sec

α2=160 rad/sec2

15°

Figure P4.12

Figure P4.14

Figure P4.13

2


- ٣١ -

Figure P4.15

Figure P4.16

Figure P4.17

Q23) Crank 2 in Figure P4.19 rotates ccw with an angular velocity of 24 rad/sec and speeding up with an acceleration of 300 rad/sec2. (a) Find the velocities of points A, B, C, and D. (b) Find ω3 and ω4. (c) Find aA and aB, aC, and aD. (d) Find α3 and α4. Q24) Crank 2 in Figure P4.20 rotates cw with an angular velocity of 6 rad/sec and speeding down with an acceleration of 50 rad/sec2. (a) Find the velocities of points A, B, and C. (b) Find ω3 and ω4. (c) Find aA and aB, and aC. (d) Find α3 and α4. Q25) (a) Construct the velocity and acceleration polygons for the mechanism shown in Figure P4.2, when ω2 = 150 rad/s and α2 = 300 rad/s2 cw. (b) Determine the velocity and acceleration of slider 6.

Figure P4.18

Figure P4.19

Figure P4.20

THEORY OF MACHINES Q26) (a) Construct the velocity and acceleration diagrams for Figure P (b) Determine ω3, ω6, α3, and α6.

Q27) In Figure P4.22, disk 4 is driven by link through block 3, which is pivoted on 2instantaneous velocity of slide of block 3

(a) Construct the velocity and acceleration polygons for points (b) Determine α4.

Q28) In the mechanism shown in Figure Pmeans of the sliding block at B. AB = 120When the crank is horizontal, as shown, and is rotating at velocity of slider F, (b) the angular velocity of the link CDE, and (c) the acceleration of slider F.

(Relative velocity & acceleration) Construct the velocity and acceleration diagrams for Figure P4.21.

Figure P4.21

is driven by link 2 sliding in guides on 1 as shown. The drive is 2 at point P3. The velocity of link 2 is constant. The 3 on link 4 is 38.1 m/s toward the center of

Construct the velocity and acceleration polygons for points O4, P3, and P4.

Figure P2.22

) In the mechanism shown in Figure P2.23, the crank AB drives the bent link CDE by 120 mm, CD = 90 mm, DE = 450 mm, EF =

When the crank is horizontal, as shown, and is rotating at 60 rpm anticlockwise, fvelocity of slider F, (b) the angular velocity of the link CDE, and (c) the acceleration of slider

as shown. The drive is is constant. The

m/s toward the center of 4.

, the crank AB drives the bent link CDE by mm, EF = 450 mm.

rpm anticlockwise, find (a) the velocity of slider F, (b) the angular velocity of the link CDE, and (c) the acceleration of slider


- ٣٣ -

Figure P2.24

Figure P2.25

Q29) Crank 2 in Figure P2.24 rotates ccw at 11rpm. (a) Find the velocities of points A, B, and C on link 4. (b) Find the velocity of link 4. (c) Find the acceleration of points A, B, and C on link 4. (d) Find the angular acceleration of link 4.

Q30) The dimensions of the various links in the mechanism shown in Figure P2.25, are as follows: OA = 175 mm; AB = 180 mm; AD = 500 mm; and BC = 325 mm. Find: (a) the velocity ratio between C and D when OB is vertical and crank OA rotates at a uniform speed of 120 rpm, (b) acceleration of D and C.

Figure P2.23

Documents

THEORY OF MACHINES (Relative velocity & acceleration ......THEORY OF MACHINES (Relative velocity & acceleration) -٤ -Example 1.3: In Figure 1.7, the velocity of A is known, and it