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MECH1230 SOLID MECHANICS Dynamics Unit 2 RIGID BODY KINEMATICS AND MECHANISMS

2.1

Contents 2.1

Rigid Body Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.3

2.1.1

Plane Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.5

2.1.2

Relative Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6

2.1.3

Relative Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9

2.2

Instantaneous Centre of Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . 2.11

2.3

Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12 2.3.1

Diagrammatic representation of mechanisms . . . . . . . . . . . 2.12

2.3.2

Alternative mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.15

2.3.3

Analysis of Links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.17 2.3.3.1 A single link . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.17 2.3.3.2 Two connected links. . . . . . . . . . . . . . . . . . . . . . . . . 2.18

2.3.4 2.4

Case studies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.22

Analysis of Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.24 2.4.1

Slider Crank Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.24

2.4.2

The Four Bar Chain Mechanism . . . . . . . . . . . . . . . . . . . . . 2.41

2.4.3

Quick Return Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . 2.43

2.2

School of Mechanical Engineering

MECH 1230

Solid Mechanics

2.1 Rigid Body Kinematics In order to analyse the kinematics of a particle, all that is required is a complete description of the location of the particle at all instants of time. For a rigid body a corresponding description of the motion requires the definition of the location and orientation of the body. Thus a kinematic analysis of a rigid body involves both linear and angular quantities. There are five types of rigid body motion: 1. Translation, which may be one of either: 

rectilinear (a straight line in 3-D space);



curvilinear (along a curve in space but with the body having the same orientation);



co-planar (within a 2-D plane).

2. Rotation about a fixed axis. 3. General plane motion. 4. Rotation about a fixed point. 5. General motion. In this part of the course, general plane motion is studied since there exists a vast range of engineering problems which are amenable to solution through application of the techniques described in the following sections. When considering plane motion, every particle in the body remains in a single plane. As all points along all lines drawn perpendicular to the plane have the same motion, only the motion in a single plane need be considered. This may be, but is not limited to, the plane on which the mass centre lies. As particles cannot move out of the plane of motion, the position of the rigid body in plane motion is completely determined by providing the location of one point and the orientation of one line in the plane of motion. The orientation of the line may be given by: 1. the angle the line makes with a fixed direction, see Figure 2.1, or by; 2. specifying the location of any two points on the line, see Figure 2.2.

2.3

A fixed line across the body

y A 

rA

x

O

Figure 2.1: Position of a point on a rigid body

A fixed line across the body

y B A rB

rA

x

O

Figure 2.2: Position of two points on a rigid body The angular motion of lines in the plane of motion is the same for every straight line in a rigid body. Consider the situation depicted in Figure 2.3 which shows a rigid body on which two lines AB and CD are drawn and which are separated by the angle . The line AB forms an angle AB to the reference direction and the line CD forms an angle CD to the reference direction. Clearly,

CD=AB+.

(2.1)

As the rigid body rotates the angles AB and CD change but as the body is rigid, the angle  remains fixed. Differentiation of equation (2.1) with respect to time gives:

ωCD  θCD  θAB  ωAB ,

(2.2)

where  is the angular velocity or rate of change of angular position. Thus every line in the body has the same angular velocity, . 2.4

D

y



C

CD

B A

AB

 x

O

Figure 2.3: Angular motion of a rigid body

Similarly, the angular acceleration of every line within the body is the same:

 CD  θCD  θAB  ω  AB  αAB , αCD  ω

(2.3)

where  is the angular acceleration or the rate of change of angular velocity.

2.1.1 Plane Motion Consider any two points A and B on a general plane rigid body. Since the body is rigid it is important to appreciate that the distance AB will remain constant regardless of the motion. We view the body at two instants of time, t and t + dt, as shown.

B

y

C

dl

B

C

d

 

dl

B1

 A1

C1

A Time t

Time t + dt

O

x Figure 2.4: Plane motion of a rigid body

2.5

Figure 2.4 shows that the motion over the time interval dt may be thought of as:  

a translation from A to A1 of dl and from B to B of dl; plus, a rotation d about A1 to bring B into position B1.

A point C has been included in the diagram representing another point on the body such that AC is at an angle of  with respect to AB. It can be seen that the angle  remains unchanged throughout the processes of translation and rotation. It follows therefore that in moving from position AB to position A1B1 the body has a velocity of translation given by:

v

dl   l, dt

(2.4)

and a rotational velocity or angular velocity, , given by:

ω

dθ   θ. dt

(2.5)

Hence every point on the body is treated as having a velocity of translation, v, and every line joining any two points the angular velocity, .

2.1.2 Relative Velocity Once again consider a plane rigid body and any two points A and B within it having velocity vA and vb respectively - see Figure 2.5. For a rigid body vA does not have to be equal to vB as there can be rigid body rotation about an axis located at some arbitrary position in the plane.

y

B

vB

B

A A

O

vA

x

Figure 2.5: Relative velocity of two points on a rigid body

2.6

Using the idea of the triangle law for vector addition write the velocity at A and B in component form - one parallel to the direction of AB the other perpendicular to AB - see Figure 2.6, where: v A _ Par  v A cosθ A , vB _ Par  v B cosθ B v A _ Perp  v A sin θ A , vB _ Perp  v B sin θ B .

vB_Perp

y

vB_Par

B

B vA_Perp

vA_Par A

vB

A vA

O

x Figure 2.6: Velocity components

Since the body is rigid the distance between A and B must remain fixed. Consequently vA_Par = vB_Par and the magnitude of the velocity of B relative to A, vBA, is given by: vBA = vB_Perp  vA_Perp,

(2.6)

whose direction is perpendicular to AB - see Figure 2.7.

2.7



y vBA

B d

A

O

x Figure 2.7: Velocity of B relative to A

Thus the motion of B as seen by A is one of rotation about A - it follows therefore that the angular velocity, , of the body is given by:

ω

vBA . d

(2.7)

The absolute velocity of the point B is then given by: vB = vA + vBA.

(2.8)

Clearly, these must be added vectorially - see Figure 2.8. Equation (2.8) states that the absolute velocity of B is the resultant of the absolute velocity of A and the velocity of B relative to A.

Parallel to BA

vBA

vB vA

2.8

Figure 2.8: Vector addition for relative velocities

2.1.3 Relative Acceleration Since the velocities at the points A and B will, in general, vary with time they will accelerate. As in the case of relative velocity, the absolute accelerations aA and aB of the points A and B can be reduced to component form - see Figure 2.9.

y



aB_Par aB_Perp

B d

aA_Perp

aB

aA_Par A

aA

O

x Figure 2.9: Acceleration components

So, the acceleration of B relative to A has two components: ( aB_Par  aA_Par ) ( aB_Perp  aA_Perp )

along AB; normal to AB.

If the angular acceleration of the body, θ , is denoted by , then for a point moving on a circular path it has accelerations R 2 and R, parallel and perpendicular to the radius R from the centre of rotation, respectively. (Recall that an angular velocity gives a tangential, or perpendicular velocity component with respect to AB, and this angular velocity generates an acceleration that is radial or parallel to AB).

2.9

Now considering the motion of B relative to A, which are separated by a distance d , the components of acceleration are given by: ( aB_Par  aA_Par ) =  2 d along AB; ( aB_Perp  aA_Perp ) = d normal to AB. These accelerations are those that an observer at A would see.  

The angular velocity,  is found from the velocities, vA_Perp and vB_Perp and the geometry. The angular acceleration,  is found from the accelerations, aA_Perp and aB_Perp and the geometry.

Now, aB = aA + aBA.

(2.9)

but, aBA has two components (unlike relative velocity which has only one). Also, aA = aB + aAB. So, for the point B, aBA has components d ( normal to AB) and 2 d (parallel to AB) as shown in Figure 2.10. Parallel to BA

aBA_Perp =  d

aA

aBA aB

aBA_Par = 2d

Figure 2.10: Acceleration components of B relative to A

2.2 Instantaneous Centre of Rotation

2.10

Consider the rigid body shown in Figure 2.11. The points A and B have absolute velocity vA and vB, respectively.

If vA  vB the rigid body will have an angular velocity  at a particular instant. So A and B can be considered to be rotating about some point along a perpendicular to the velocity vectors vA and vB at these points. These perpendiculars lie along A A and B B as shown in Figure 2.11. We conclude therefore, that at this particular instant the body must be rotating about the point of intersection of A A and B B - that is, about the point I, the instantaneous centre of rotation. Note that in general vA and vB are not constant, so I is not a fixed point - but a point regarded as being instantaneously at rest. If  is the angular velocity of the body at that instant, then (from v = r:

ω

v A vB  . IA IB

(2.10)

The concept of an instantaneous centre of rotation is a useful and quick way of finding the velocity of the moving parts of a mechanism. Mechanisms are examined more fully subsequently but in the meantime consider the schematic of an up-and-over garage door shown in Figure 2.12 (not to scale). As the door QS opens the slider S moves to the left, as it closes the slider moves to the right. If PR is the pivot arm for the garage door where is the instantaneous centre of rotation of the garage door in the position shown?

2.11

Q

2.3 Mechanisms There exists a vast range of mechanisms, which find application throughout mechanical engineering. Mechanisms are made up of combinations of links, slides, gears and cams which may operate at low or high speeds and may either transmit power or modify motion. The word mechanism is derived from the Greek word meaning machine and in fact dictionary definitions of the word mechanism include "the working parts of a machine" and ä system whose parts work together as in a machine". Mechanisms vary greatly in their size and complexity ranging from simple engine mechanisms to highspeed textile equipment.

2.3.1 Diagrammatic representation of mechanisms A convenient way to represent the kinematic behaviour of a mechanism is to use a socalled stick diagram. A stick diagram accurately conveys the ways in which the various components can move relative to one another. The basic elements of a stick diagram are shown in Figure 2.13 together with some examples of commonly encountered mechanisms in stick form in Figure 2.14 in which the arrows indicate a typical input to the mechanism.

2.12

2.13

2.14

2.3.2 Alternative mechanisms Many mechanisms involve motion in three-dimensional space, for example an excavator shovel, vehicle suspension system (because the spring/damper axis is normally angled and the suspension linkages pivot generating complex 3D motion) or robotic device. Fortunately there are a significant number whose motion can be described in a 2D plane. Some examples found in everyday life are shown in Figure 2.15 and a number of engineering applications are provided in Figure 2.16.

2.15

2.16

2.3.3 Analysis of Links 2.3.3.1 A single link The application of velocity and acceleration components in polar coordinates (see unit 1, page 1.26) may be illustrated through consideration of the crank element of a slider crank mechanism. The crank can be represented as a link, OA, of fixed length, which is pivoted at one end. Figure 2.17 shows how the generalised velocity and acceleration terms simplify for this case.

Figure 2.17

(Note in the diagrams above,    and      ) The restriction of fixed length means that all terms involving r and r must be zero. The velocity and acceleration components shown are very important as they form the basis for the graphical method of mechanism analysis which follows.

2.17

2.3.3.2 Two connected links Now consider a situation involving a second fixed length link, AB, attached to the free end of the pivoted link OA. This might, for example, represent the connecting rod of a slider crank mechanism. The motion of the link AB can involve translation in both X and Y directions and rotation. This is depicted in Figure 2.18.

Figure 2.18

In comparison to AB, the link OA may only rotate about O and is said to be kinematically constrained. This means that the motion of OA is simply a function of the angle 1 and the link geometry. Velocity The direction and magnitude of the velocity of the free end of the link AB can be determined by application of the idea of relative velocity: The velocity of point A relative to ground at O is the absolute velocity of A, which has a magnitude vA given by: v A  OA θ1  OA 1

(2.11)

which is perpendicular to the line OA as shown in Figure 2.18. Calculation of the velocity of point B relative to point A is complicated by the fact that point A is itself moving. Relative to an observation position at A, point B can only have a rω velocity term since its length is fixed. Thus, the velocity of point B relative to point A has a magnitude equal to:

2.18

vBA  ABω2

(2.12)

where 2 is the angular velocity of AB and vBA has a direction which is perpendicular to the link AB, see Figure 2.18. From this follows the important result that allows an expression for the absolute velocity of point B (velocity of B relative to ground at O) to be derived as: vB = vA + vBA.

(2.13)

A pictorial representation of this result is shown in the velocity vector diagram of Figure 2.19, which provides a means of calculating the magnitude and direction of the resulting velocity vB.

vB = vA + vBA

Figure 2.19

Figure 2.18

EXAMPLE TO TRY YOURSELF Using the data provided below in conjunction with the figure, calculate the resultant magnitude and direction of the velocity of point B at this instant.

2.19

Data: OA=0.4 m, AB=0.5 m, 1=20, 2=75, 1  2 rads 1, 2   4.5 rads 1 Answers: vA = 0.8 ms-1, vBA = -2.25 ms-1, vB = 1.91 ms-1, at an angle of 5.1 to the positive x-axis. Acceleration The same principles of relative motion can be applied to the calculation of the accelerations of the interconnected links. The point A on the link OA can only have two components, namely r1 and  rω12 . However if the link OA is rotating with constant angular velocity, 1 then the r1 term is zero, leaving a single component  rω12 directed towards the point O. The acceleration of B relative to A has two components:  

 ABω22 along AB ; ABα2 perpendicular to AB ;

which are shown in Figure 2.20.

2.20

Figure 2.20

The absolute acceleration of B relative to A is given by the vector expression: aB=aA+aBA

(2.14)

from which a diagram showing the acceleration vector, Figure 2.21 (which assumes OA is rotating at constant angular speed), can be drawn. In order to calculate aB, a value for 2 is required which is the instantaneous rotational acceleration of the link AB. Note that calculations of this type refer to one position of the links. This can be thought of as equivalent to analysing a still frame taken from a motion picture.

2.21

Figure 2.21

When considering the dynamic behaviour of a body, or an assembly of bodies, in a machine the first step is to reduce the actual system to an idealised model by making appropriate assumptions. One should remember that it is very important to bear in mind at all times the assumptions made in setting up the model, particularly when interpreting the results of any analyses. If the assumptions are invalid, the results will also be invalid. Once the model has been defined, a free body diagram must be drawn before proceeding further, for each body in the system, showing the forces which act on them. The appropriate equations of motion can then be applied. Next, any kinematic constraints must be identified and the resulting equations then solved for the required variables.

2.22

2.3.4 Case studies In order to demonstrate the techniques used to analyse the kinematic behaviour of plane mechanisms it is appropriate to concentrate on a small number of mechanisms from the vast range available. What follows will concentrate on the analysis of two mechanisms, namely the slider crank and the four bar chain. Together these make an excellent vehicle through which to introduce the basic concepts and techniques associated with the analysis of linkages and mechanisms. The four bar chain typifies a particularly large class of mechanisms. Slider cranks are commonly found in internal combustion engines and a typical layout is shown in Figure 2.22 along with the equivalent stick diagram in Figure 2.23. Note the differences and the fact that the stick diagram captures all the essential kinematic information.

2.23

Rocker arm

Valve

Figure 2.22

The main elements of the slider crank are the piston, crank and connecting rod. These perform the following types of motion: 

Piston - Rectilinear (translational) motion;



Crank - Rotational motion;



Connecting rod - Combination of translational and rotational motion.

2.24

Figure 2.23

2.4 Analysis of Mechanisms In this section we focus our attention on the analysis of the slider crank and four bar chain. The former is a convenient vehicle by which to introduce the basic concepts involved; the latter is fundamental in the design of mechanisms - it is simple, fairly inexpensive, widely used and takes many different shapes and sizes.

2.4.1 Slider Crank Mechanism. This mechanism is very common and therefore of great interest. Pivot of connecting rod about the crank

Connecting Rod

Cylinder

Crankshaft Valves Crank axis Piston

2.25

Method of Analysis First reduce the mechanism to a `Stick' Position Diagram (SPD) as shown below:

There are three approaches that can be adopted to analyse this and other mechanisms: 1. Trigonometric. - Write down analytic expression(s) for position and then differentiate twice to obtain explicit expressions for velocity and acceleration. 

This is a lengthy process even for simple mechanisms.



Almost impossible for complex ones!

2. Graphical - scale drawings. - Gives velocity and acceleration for one position only and hence this process must be repeated many times to produce an overall picture of the motion for one complete cycle. 

Its use as an aid to physical understanding is questionable.



Long winded and tedious.

3. Computational. - This is vector based and is usually implemented on a computer. It is straightforward to write down the displacement vectors and then to perform subsequent velocity and acceleration analysis via numerical methods. This approach is: 

quick (if the programme already exists);



convenient (use the computer to plot out the results);



and flexible (easy to solve for any input, different link lengths, etc).

Clearly the latter approach is the preferred one, particularly if the necessary software is readily available.

2.26

Trigonometric Analysis – Slider Crank For simple mechanisms, algebraic expression(s) may be formulated to describe the position of an output of interest relative to some known input. This expression can then be differentiated once with respect to time to obtain an explicit expression for the velocity and again to derive an explicit expression for the acceleration. Note that simplifying assumptions may be required to effect the differentiation. Consider the slider crank mechanism shown in the figure below, which is assumed to have its crank driven at a constant speed of  rads1. If the output of interest is the piston then an expression for piston displacement x as a function of the crank angle  is required.

The main idea is to find x as a function of , x = f(), and then differentiate this function w.r.t. time, t, to obtain v  x and a  x . Now:

x  lcos   rcosθ 

(2.15)

and clearly (from the sine rule), l r  sin  sin 

Now from equation (2.16) sin 



rsin(θ) = lsin()

(2.16)

r sinθ  and: l



1

cos  1  sin 2  2 since sin 2   cos 2   1.

So

2.27

1

 2 r2 cos   1  2 sin 2 θ  . l  

(2.17)

Hence substituting for cos() from equation (2.17) into (2.15) gives: 1 2

  r x  rcosθ   l  1  2 sin 2 θ  . l   2

(2.18)

The velocity and acceleration of the piston are found by the successive differentiation of equation (2.18) with respect to time. In order to simplify this process it is appropriate to expand the term within the square root sign using a Binomial expansion of the form

l  k m

 mm  1  2  k  ....,  1  mk    1 2 

which leads to:   r2 x  rcosθ   l  1  2 sin 2 θ ; 2l   2 r  rcosθ   l  sin 2 θ ; 2l

(2.19)

where higher order terms are ignored based on the assumption that they will contain terms of diminishing magnitude. Thus since  varies with time, the velocity can be determined using the chain rule as:

x 

dx dx dθ dx     θ, dt dθ dt dθ

(2.20)

such that: dx r2   rsin θ   2sinθ cosθ ; dθ 2l r2   rsin θ   sin2θ . 2l

(2.21)

If the input speed θ is a constant, say , then:

2.28

  r2 x    r sin θ   sin 2θ  ω; 2l   r     ωr sin θ   sin 2θ . 2l  

(2.22)

Similarly, for acceleration:

x 

dx dθ dx  ω , dθ dt dθ

(2.23)

where: dx r     r  cosθ   2 cos2θ , dθ 2l   r     r  cosθ   cos2θ . l  

(2.24)

Combination of equations (2.23) and (2.24) and noting once again that θ  ω leads to the desired result: r   x   ωr cosθ   cos2θ θ; l   r     ω 2 r  cosθ   cos2θ . l  

(2.25)

Equations (2.18), (2.22) and (2.25) can be expressed as a function of time given that  is constant and the crank angle  can be written as:

 = t.

(2.26)

Substitution of (2.26) into equations (2.18), (2.22) and (2.25) leads to the following expressions for the position, velocity and acceleration of the piston under the assumption that  is constant: 1

Postion Velocity Acceleration

 2 r2 x  rcosωt   l  1  2 sin 2 ωt  ; l   r   x   ωr  sinωt   sin2ωt ; 2l   r   x   ω 2 r  cosωt   cos2t . l  

(2.27) (2.28) (2.29)

2.29

Graphical Analysis – Slider Crank This is the traditional approach to solving mechanism problems. There are two types of graphical analysis. The first uses the concept of an instantaneous centre of rotation. Method of Instantaneous Centre The following diagram shows a scaled drawing of the displacements for a slider crank at some particular angle of input, where P is the piston, O the axis of rotation of the crank and A is the pivot between the connecting rod and the crank.

The following steps are involved in performing a graphical analysis: (i) Draw the mechanism to scale (on graph paper) at the desired position. (ii) Locate the instantaneous centre, I, in this particular position. This can be determined as the direction of motion of each end of the connecting rod is known. The end at the piston must move with the piston which is modeled as a slider and hence moves horizontally (in the diagram as drawn). The other end of the connecting rod is at the end of the crank which is rotating with uniform circular

2.30

motion about O, hence the velocity at A must be a tangent to OA. (iii) Calculate vA ( the tangential velocity of the point A = r, where  here refers to the angular speed of the crankshaft). (iv) The angular velocity of the connecting rod (PA) is then given by:

ωrod 

v A ωr  , IA IA

where IA is scaled from the position diagram. (v) The velocity of the piston, vP is given by (again IP is scaled from the position diagram):

vP  rod  IP  v A 

IP . IA

WORKED EXAMPLE Consider the garage door problem introduced earlier, with the dimensions shown. Given that PR rotates about P at the constant angular speed indicated, find the velocity of the door and the slider (at this instant) together with the length d. Note that by defining SF in the position diagram defines the position of the whole mechanism.

2.31

Solution To determine the velocities and distances a Scaled Position Diagram (SPD) must be drawn: (i)

The distance d is easily measured off the above scaled diagram and is found to equal 0.9m.

(ii)

The position of the instantaneous centre, I, is as shown in the above scaled position diagram and lies at the intersection of perpendiculars to the velocity vectors vS and vR from the points S and R respectively.

(iii)

(iv)

20  2π rads 1  0.349rads1  ω. The velocity of 360 R, vR  0.349  0.9 PQω  0.314ms 1 and hence the angular velocity of the v 0.314  0.218rads1 . (Where IR is measured from the SPD). door is = R  IR 1.44 20degs1 

The velocity of the slider, S is (where IS is measured from the SPD): vS v IS 0.8  R  vS  v R  0.314   0.174ms1 . IS IR IR 1.44

2.32

The second is based on drawing scale velocity and acceleration diagrams derived from a scale position diagram. Note that the velocity and acceleration relate to one instantaneous position and so many scale drawings are required to obtain a picture of the kinematic behaviour of a mechanism over a complete cycle. With the advent of computer based methods the graphical approach is declining in its use. However, from an educational standpoint, the graphical method does serve to reinforce the concepts behind vector mechanics. Below, the slider crank on the previous page and a four-bar chain mechanism is analysed in this way for purely illustrative purposes. Graphical Analysis for Kinematics using velocity and acceleration diagrams Another way of finding the velocity and acceleration (at any point) for one position only is as follows:

2.33

(i) Draw a position (stick) diagram of the mechanism to scale (repeated below).

(ii) Construct the velocity vector diagrams (to scale):   

Draw vA = r = ×OA and therefore the magnitude and the direction of this vector are known. vP must be horizontal and therefore only its direction is known. The velocity of P relative to A must be perpendicular to the line PA and thus again only its direction is known.

2.34

vP 90- vPA

vA





the intersection of vP and vPA determines their magnitudes (their directions are already known) and therefore it is possible to read off the value of vP and vPA from the scale of the velocity diagram.

(iii) To determine the accelerations when the mechanism is in this position draw an acceleration diagram to scale: 

 

Consider the case when  = constant  θ = 0, i.e., A has no tangential acceleration but only radial (r θ 2), that is OA2 and hence the magnitude and direction of this vector is known. The acceleration of the point P, aP must be horizontal and so its direction only is known. The acceleration of the point P relative to A has two components. As we are considering the rotation of P relative to A, the component of acceleration parallel to PA is in the direction from P to A and has magnitude PA  2 denoted by a The other component is perpendicular to PA (PA  ) denoted by a. In this case the magnitude and direction of a is known (  = vPA/PA and vPA is known from the velocity diagram), but only the direction of a is known.

2.35

Parallel to PA

 a|| aPA

aA

a

 aP 

The intersection of the vectors a and aP defines the magnitudes of these vectors, and for aPA and from the scaled diagram their values can be obtained (note that their direction is already known).

WORKED EXAMPLE A four bar linkage is shown below in which the crank AB rotates clockwise with a constant angular velocity of 10 rads1. Using velocity and acceleration diagrams calculate the angular velocities and accelerations of the links BC and CD. (Note the diagram is not to scale and fixed pivots A and D should be on the same horizontal line).

2.36

Solution 1. Draw a position diagram (SPD) to scale: From this diagram, measure the angles of the links, angle of BC to the horizontal is 32, and of CD to the horizontal is 24.

B

60

D

A

32

24 C

2. Construct the corresponding velocity diagram as follows:

(i) The velocity of B relative to A is AB  = 30 ×10 Therefore 300 mms1 perpendicular to AB (at 30 to horizontal).

2.37

(ii) The velocity of C relative to D is of an unknown magnitude but its direction must be perpendicular to CD, i.e. at 24 to the vertical. (iii) The velocity of C relative to B is also of unknown magnitude but its direction must be perpendicular to BC, i.e. at 32 to the vertical. (iv) The intersection of the two vectors with unknown magnitude define the velocity of C, 172 mm s-1, and the relative velocity of C with respect to B, 362 mm s-1. The angular velocities of the links BC and DC can now be calculated, i.e., 362 Angular velocity of BC = = 4.83 rad s1 anticlockwise. 75 172 Angular velocity of DC = = 4.91 rad s1 anticlockwise. 35 3. The acceleration diagram can be constructed as follows:

(i) The acceleration of B relative to A is AB 2 = 30 ×100. Therefore 3000 mm s2 along BA. Note that there is no AB ω component

2.38

since  is considered to be constant. (Note however that the angular rotation rates of CD and BC cannot be considered to be constant). (ii) The acceleration of the point C relative to D has two components, of 2 magnitude CD CD along CD = 35 ×(4.91)2 = 843.8 mm s2 and CD  CD perpendicular to CD but its magnitude is as yet unknown. (iii) The acceleration of the point C relative to B also has two components, of magnitude BC  2BC along the link CB = 75 ×(4.83)2 = 1749.7mm s2 and BC  BC is perpendicular to BC but its magnitude is unknown.

(iv) The intersection of the above components now defines the acceleration of C and the unknown components for both CD and BC, i.e., perpendicular component for CD = 560 mm s2 and hence 560  CD  = 16 rad s2 in an anticlockwise direction 35 and the component for BC = 1780 mm s2 and hence 1780  CB  = 23.7 rad s2 anticlockwise. 75

2.39

Computational Analysis This is a modern, vector based approach which is quick, accurate and general, i.e., it gives a full description of position, velocity and acceleration and is therefore now the method of choice. The computational approach relies on the fact that the position vectors associated with any mechanism are straightforward to define. The subsequent velocity and acceleration analysis is performed through numerical methods. The method involves an interesting blend of numerical analysis and computing, and hinges on the use of vector displacement/loop equations. The emphasis here, therefore, is the derivation of such equations for a variety of mechanisms. Once you have these, it is simply a matter of using readily available computer software to solve them. Slider-Crank Arrows indicate positive directions

y



p1 B a2

x a1

q

C

A p2

Variables: Known: Input:

p1, p2 which can be either a length or an angle, and in the case of the slider crank these are enough to completely define the mechanism. a1 and a2, the fixed links the angle of input is q(t).

Displacement Loop Equations Form the displacement loops by projecting the displacement onto the x and y axes to give two equations: x-direction

a1cos(q) + a2cos(p1) + p2

= 0;

y-direction

a1sin(q) + a2sin(p1)

= 0;

(2.30)

2.40

i.e., two equations for two unknowns which are easily solved. Note that you need to be careful of the signs of the various terms. By defining the angles with respect to the positive x-axis and working around the ‘loop’ in an ordered fashion most of the signs will take care of themselves. For example, in the x direction equation above, starting from A the distance is a1 to get to B at an angle of q with respect to the positive x-axis. As q (as drawn) is between 90 and 180, a1 cos(q) will be negative. The next term is from B to C and following the same methodology will be a2 cos(p1) which again will be negative as p1 is between 180 and 270. The final term is from C back to A, and is effectively p2 cos(0), hence +p2. What about p 1 , p 2 , p1 and p2 ? These can be found by differentiating the displacement equations (2.30) with respect to time. Velocity Loop Equations The velocity loop equations are therefore: x-direction

 a1 sin q q  a2 sin  p1  p1  p 2

= 0;

y-direction

a1cosq q  a2 cos p1  p 1

= 0.

(2.31)

To find the two acceleration loop equations for the two unknown accelerations, differentiate equation (2.31) w.r.t. time. This is left as an exercise. In the computational analysis, the loop equations for the displacement, velocity and acceleration for the slider-crank are solved using commercially available kinematics software. The same is done in the case of loop equations derived for the other mechanisms shown earlier. Let us now consider a different mechanism in order to illustrate the graphical and computational methods described above and the mathematics behind them.

2.41

2.4.2 The Four Bar Chain Mechanism

p2 a2

Arrows indicate positive directions

y

B

C

p1



a3

x

a1

q

O

D a4

Variables:

p1, p2

Known:

a1, a2, a3 and a4, the fixed links

Input:

the angle of input is q.

The variables that need to be determined to analyse the mechanism completely are: (a)

p1 and p2 for any given input, q and therefore the displacements of the mechanism are known completely ;

(b)

p 1 and p 2 ;

(c)

p1 and p2 .

Displacement Loop Equations The displacement loop equations for this mechanisms are (working around OBCDO): x-direction a 1cosq   a 2 cosp1   a 3cosp 2   a 4 cos(180  ) = 0; a1cosq   a2 cos p1   a3cos p2   a4 = 0;

y-direction

2.42



a 1sin q   a 2 sin p1   a 3sin p 2   a 4 sin 180 

a1 sinq   a2 sin p1   a3 sin p2 



=0

= 0.

(2.32)

Unlike the displacement loop equations for the slider crank mechanism, these equations are nonlinear (i.e., one cannot write explicit expressions for p1 and p2 as the variables are not separable, in fact these two variables are present as the argument of trigonometric functions which are intrinsically non-linear). The two equations have to be solved numerically which requires initial estimates for the two unknowns p1 and p2 at a known value of input crank angle q. An algorithm must be employed to increase the accuracy of the estimates for p1 and p2 until the desired precision is obtained. This must be done for every possible value of the input q. The displacement information of the mechanism is thus completely defined. Velocity Loop Equations The next stage involves calculating the velocity and acceleration information so that a complete kinematic description of the mechanism is available for the purpose of later analysis. Differentiation of the displacement loop equations with respect to time leads to the velocity loop equations in p 1 and p 2 . x-direction

- a1qsinq   a2 p 1 sin p1   a3 p 2 sin p2  = 0;

y-direction

a1qcosq   a2 p 1cos p1   a3 p 2 cos p2  = 0.

(2.33)

Assuming the angles q, p1 and p2 are known and q has been defined, then solution of the velocity equations for the angular velocities p 1 and p 2 is simple, both expressions are linear simultaneous equations (Cramer’s rule can be applied – for two equations with two unknowns). Acceleration Loop Equations Differentiation of the velocity loop equations with respect to time yields two further equations for the angular accelerations p1 and p2 : x - direction

- a1qsinq   a1q 2 cosq   a2 p1 sin p1 

 a2 p 12 cos p1   a3 p2 sin p2   a3 p 22 cos p2   0;

y - direction

- a1qcosq   a1q 2 sinq   a2 p1cos p1 

 a2 p 12 sin p1   a3 p2 cos p2   a3 p 22 sin p2   0.

(2.34)

2.43

Again, the acceleration loop equations are linear and their solution for p1 and p2 is straightforward particularly as the angles p1 and p2 and velocities p 1 and p 2 are at this stage known. Thus all the kinematic information is contained in the two coordinate variables p1 and p2 and in their derivatives. As demonstrated by the four bar chain, the displacement loop equations are non-linear, simultaneous equations which it is impossible to solve through the derivation of explicit expressions. This situation is in fact commonly encountered and so a numerical, iterative technique is used to solve the displacement loop equations and is built into available kinematics analysis software. There are several numerical methods available to the engineer, however they all work by using a basic trial and error process, which can be explained as follows: 1. Begin the process by guessing a solution for p. 2. Using this estimate, calculate an incremental change, p, to attempt to get closer to the solution. 3. Calculate a new estimate p=p+p 4. Go back to 2 and keep going round the stages (2 to 4) until some convergence condition is met (e.g. |p| < an accuracy requirement) and the estimated solution is close to the actual one Possible refinements to the technique involve: 1. making the iterative process as fast and as efficient as possible so that the number of iterations is minimised, 2. ensuring the algorithm is as robust as possible so that it will deal with a wide range of equation sets (which translates into a wide range of mechanisms), and 3. making it stable so that the algorithm always converges to the correct solution rather that diverging away from it. Certain simple mechanisms such the slider crank have displacement loop equations which are explicit and so can be solved directly without recourse to any iterative solvers.

2.4.3 Quick Return Mechanism WORKED EXAMPLE Write appropriate loop equations in order to analyse the quick return mechanism shown below.

2.44

Solution. The displacement loop equations for the above mechanism are: Considering Loop ACD:

a4cos(p3) + p4 cos(180˚) + (a1 +p1 ) cos(270˚) = 0 a4cos(p3) – p4 = 0; a4sin(p3)+ p4sin(180˚)+ (a1 +p1 ) sin(270˚) = 0. a4sin(p3) – a1 – p1

= 0.

Considering loop ABO: p2cos(p3) + a2cos(180˚+q)+ p1cos(270˚) p2cos(p3) – a2cos(q)

= 0;

= 0;

2.45

p2sin(p3) + a2sin(180˚+q)+ p1sin(270˚)

= 0;

p2sin(p3) – a2sin(q) – p1 = 0. The above loop equations are not the only ones that could have been chosen. For example loop ACD could be replaced by loop OBCD: a2cos(q) + (a4 – p2)cos(p3) – p4 = 0; a2sin(q) + (a4 – p2)sin(p3) – a1 = 0. After all the object is to produce the same number of independent equations as there are unknowns in the problem. A point of special interest is the CofM of the link AC, denoted by G on the above diagram. The coordinates of this point relative to the fixed axis, O are: xG = a2cos(q) – (p2 – a3)cos(p3); yG = a2sin(q) – (p2 – a3)sin(p3). Deriving the loop equations for velocity and acceleration are left as an exercise.

ADDITIONAL EXAMPLES TO TRY 1) The four-bar linkage shown below has the following lengths: A0A =125mm, AB = 275mm, B0B = 225mm and A0B0 = 200mm. For the position when the angle A A0 B0 = 120o calculate the angular velocity of the output and the coupler if the input angular velocity 1 = 25 rad s1 clockwise.

2.46

2.47

2) The figure shown below is that of a variable stroke mechanism which allows the stroke of a reciprocating piston to be varied by altering the position of the fixed point A. Write down a set of displacement loop equations for the system.

2.48