mechines of mechinism chapter 2

Chapter 2: Kinematic Analysis – Analytical Approach

Yk

2.1. Introduction

Kinematics is the study of motion without consideration of what causes the motion. In other words, the input motion is assumed to be known and the objective is to find the transformation of this motion. Kinematic analysis comprises the following steps, [1]:

Make a skeletal representation of the real mechanism. Choose a coordinate system. Identify all links by numbers. Identify all angles characterizing link positions. Write a loop-closure equation. Identify input and output variables. Solve the loop-closure equation.

What is a loop-closure equation? After choosing the coordinate system, we construct a closed polygon of vectors representing active lengths of all links comprising the mechanism. Let us demonstrate this trough an example of a four-bar linkage as shown in Figure 1.

Figure 1: The position vector polygon (closed loop)

According to the vectors directions, the loop-closure equation is given by

1 2 3 4 0r r r r+ + + = (1)

1r

3r4r

1r

2r

Chapter 2: Complex Number Approach

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The loop-closure Equation (1); can be written in alternative forms according to the chosen orientations of vectors in the loop; as shown in Figure 2.

Figure 2: Alternative loop-closure configurations

(a) 1 2 3 4r r r r= + + (2)

(b) 2 3 1 4r r r r+ = + (3)

a

b4r

4r

1r

1r

3r

3r

2r

2r


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2.2. Position Analysis

For position analysis, we need to solve the loop-closure equation; which is a vector equation. Vectors can be represented in different algebraic notations; e.g. geometric, matrix, or complex number notations. In this course, we will adopt the complex number representation of vectors. The following is a reminder on the basic algebra of complex numbers:

Figure 3: Complex number notations

The position of point P is defined by the position vector Pr , which can be written in complex number notations as

cos sin jP a jb r j r r er θθ θ= + = + = (a)

where a is the component along the real axis (Re), b is the component along the imaginary axis (Im),

and 1j = −

Velocity of point P is given by

( )j

P jP

jd r evdt d

r jr eet

d r θ θθ

θ= = = + (b)

where jr e θ is the translational velocity due to the rate of change of the length of r, and jjr e θθ is the rotational velocity due to the rate of change of orientation θ of the vector r.

Re

Im

P r

θ

b

a


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General rule of differentiation of vectors: The derivative of a vector, in general, produces two terms; one due to rate of change of its length and another due to rate of change of its orientation.

Acceleration of point P is given by

( )2

2( ) ( 2 )

j

j

j

j

PP

j j

j jj

jr e

jr e jr

dadt

r e

re jr edt

r r e j r r e

e r e

d v θ θ

θ θ

θ

θ

θ

θθ

θ

θ

θ

θ

θθ θ

θ

= = +

= +

= − + +

+ −+ (c)

Now, let us go back to solving the loop-closure equation (e.g. Eq. 1) of the four-bar linkage, which can be expressed in complex number notations as

31 2 4

1 2 3 4

1 2 3 4

1 1 1 1 2 2 2 2

3 3 3 3 4 4 4 4

0

0( cos sin ) ( cos sin )( cos sin ) ( cos sin ) 0

jj j j

r r r r

r e r e r e r er jr r jrr jr r jr

θθ θ θ

θ θ θ θθ θ θ θ

+ + + =

+ + + =+ + +

+ + + + =

(4)

Now, split real and imaginary terms to obtain two scalar equations as

Re: 1 1 2 2 3 3 4 4cos cos cos cos 0r r r rθ θ θ θ+ + + = (5)

Im: 1 1 2 2 3 3 4 4sin sin sin sin 0r r r rθ θ θ θ+ + + = (6)

Equations (5) and (6) can be solved for two unknowns depending on the data available for the problem under consideration.

As another demonstration, consider the slider-crank mechanism. Referring to Figure 4, the loop-closure equation is given by 1 2 3 0r r r+ + = (8) Observe how the angles are defined with respect to the positive x-axis and measured in counterclockwise direction.


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Figure 4: Slider-crank mechanism

EXAMPLE 2.1 Perform position analysis for the following four-bar linkage:

4r

1r

3r

2r

4θ

1θ3θ

2θ


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The following data is given: 1 2 3 4120 , 60 , 140 and 80r mm r mm r mm r mm= = = = , and the

angle between the crank (Link 2) and the fixed link is 2 115oθ = . Determine the orientations of links 3 and 4.

SOLUTION

Constructing the loop-closure as shown in the figure; one can write the loop-closure equations as

1 2 3 4 0r r r r+ + + = (a) This can be expanded as

31 2 41 2 3 4

1 1 1 1 2 2 2 2

3 3 3 3 4 4 4 4

0( cos sin ) ( cos sin )( cos sin ) ( cos sin ) 0

jj j jr e r e r e r er jr r jrr jr r jr

θθ θ θ

θ θ θ θθ θ θ θ

+ + + =+ + +

+ + + + =

Now, splitting into two scalar equations as

Re: 1 2 3 3 4 4cos180 cos115 cos cos 0r r r rθ θ+ + + = (b)

Im: 1 2 3 3 4 4sin180 sin115 sin sin 0r r r rθ θ+ + + = (c)

Now, solving Equations (b) and (c) in two unknowns to get

3 43oθ = and 4 309.2 50.8o oθ = −= ⇐ Are these answers correct? (redo calculations)

Also, redo the same problem using the following two alternative loop-closure configurations. Observe the definitions of angles in each case.

(a) 1 2 3 4r r r r= + +

(b) 2 3 1 4r r r r+ = +


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2.3. Velocity Analysis

Since loop-closure equation is basically a position equation; one can just carry out the differentiation with respect to time to obtain the velocity equation; e.g. for a four-bar linkage

( )

( )1 2 4

1 1 2 2 4 4

1 2 3 4

1 2 4

1 1 1 2 2 2 4 4 4

0

......... 0

........... 0

j j j

j j j j j j

d r r r rdtd r e r e r edtr e jr e r e jr e r e jr e

θ θ θ

θ θ θ θ θ θθ θ θ

+ + + =

+ + =

+ + + + + =

Now, split the real and imaginary parts to get two scalar equations and solve.

EXAMPLE 2.2

Perform velocity analysis for the slider-crank mechanism shown in the figure, where 2 1r cm= ,

3 4r cm= , 2 60oθ = and 2 10 / secradω = .

SOLUTION

The loop-closure equation can be written for the loop shown in the second figure as 1 2 3 0r r r+ + = (a) First we need to do position analysis by solving Eq. (a), or we can just solve the triangle using sine and cosine rules. Now, using cosine rule

2 2 2 2 23 1 2 1 2 1 12 cos60 (4) 1 2 cos60o or r r r r r r= + − ⇒ = + −

Solve to get 1 4.405r cm=


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Also, using the cosine rule we get

323

3

347.5sin(360 ) sin 60

orr θθ

= ⇒ =−

Now, all dimensions are known, and we can start velocity analysis

( )

( )3

1

1 2

1

1 2

1

3

1 2 3

1 1

0

0jj j

j j

d r r rdtd r e r e r e

jte er r

dθ

θθ θ

θ θ

+ + =

+ + =

+ 22

jr e θ+ 322 2 3

jjjr e r e θθθ+ + 33 3 0jjr e θθ+ =

Expand to get 1 1 1 2 2 2 2 3 3 3 3(cos sin ) (cos sin ) (cos sin ) 0r j jr j jr jθ θ θ θ θ θ θ θ+ + + + + =

Split into two scalar equations

Re: 1 2 2 3 3cos180 sin 60 sin 347.5 0or r rθ θ− − = (b)

Im: 1 2 2 3 3sin180 cos60 cos347.5 0or r rθ θ+ + = (c)

where 2 2 10 /rad sθ ω= = . Solving equations (b) and (c) in the two unknowns 1r and 3θ to get

1 9.768 / secr cm= −

3 1.280 / sec 1.28 /rad rad s CWθ = − ⇒

EXAMPLE 2.3

In the mechanism shown, 10 /s in s= − and 20 /s in s= for the position corresponding to φ = 60°. Find φ for that position using the loop equation approach.

SOLUTION


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Using the loop-closure equation corresponding to the figure as

3 1 2r r r= + (a) Differentiating Eq. (a) to get

( )

( )3 21

31 2

3 1 2j j j

d r d r rdt dtd dr e r rdt dt

eeθ θ θ

= +

= +( )3 3 1 1

13 3 1 13j j jjr e jr e r jre e θθ θ θθ θ+ = +

3 3 3(cos60 sin 60) (cos 60 sin 60) ( 10) (cos0 sin 0)r j jr j jθ+ + + = − × +

Re: 3 3 30.5 0.866 10r r θ× − × × = − (b)

Im: 3 3 30.866 0.5 0r r θ× + × × = (c)

Noting that 3 10 / sin 60 11.547r in= =

Solving equations (b) and (c) in the two unknowns 3r and 3θ to get

3 5 / secr in= −

3 0.75 / sec 0.75 /rad rad s CCWθ φ= = ⇒


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2.4. Acceleration Analysis

Carry out the differentiation of the velocity equation with respect to time to obtain the acceleration equation; e.g. for a four-bar linkage

( )

( )

( )1 1

1

1 2

2

1

4

4

1

2 4

1

2

1 2 3 42

2

1 2 42

1 1 1

21 1 1

2

1 1 1

2 2 4 4 4

1

0

......... 0

........... 0

.........................2 .

j j

j j j j

j j j

j j j j

d r r r r

r e jr e

r e jr e r e j

dtd r e r e r edtd r e jr

r e

e r e jr edt

θ θ θ

θ θ θθ θ

θ θ θ

θ

θ

θ

θ

θ

θ θ

θ

+ + + =

+

+

+

+ =

+ + + +

− +

=

44 4 0jjr e θθ =

Now, split the real and imaginary parts to get two scalar equations and solve.

EXAMPLE 2.4

Find the angular acceleration φ for the mechanism of problem 2.3.

SOLUTION

Differentiate the velocity equation as

( )3 3 13 3 3 1

j j jd r e jr e r edt

θ θ θθ+ =

3 13 3 323 3 3 3 3 3 132j jj j j r er e jr e r e jr e θθ θ θ θθ θ θ+ − + =

0 (a)

Expanding Eq. (a) into sine & cosine terms, split the real and imaginary parts to get two scalar

equations in two unknowns 3r and 3θ to get

23 0.65 /rad s CCWθ φ= =


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EXAMPLE 2.5 The general action of a person who is doing pushups can be modeled as a four-bar linkage as shown below. The floor is the base link, and link 4 is the back and legs. Link 2 is the forearm, and link 3 is the upper arm. For the purposes of analysis, the motion that is controlled is the motion of link 3 relative to link 2 (elbow joint). Assume that 3 2 0.6 /rad sω ω= + . Compute the angular velocity and angular acceleration of link 4 if link 2 is oriented at 45° to the horizontal.

SOLUTION a) Position analysis

The loop closure equation may be structured as shown in the figure.

1 4 2 3r r r r+ = + (a) Expand Eq. (a) and split into real and imaginary to get two scalar equations


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Now, the above equation can be simplified as

(b)

Solve Eq. (b) to get 3 4149.12 164.24o oandθ θ= = b) Velocity Analysis Now, differentiate Eq. (a); expand and split Real & Imaginary to get:

(c)

Solve Eqs.(c) to get 2 41.55 / 1.2 /rad s CW and rad s CCWθ θ= =

c) Acceleration Analysis

Similarly, differentiate the velocity equation, expand and split Real & Imaginary to get:

(d)

Solve Eqs.(d) to get 24 3.29 /rad s CCWθ =

Administrator

Rejected

Administrator

Accepted

Administrator

Preliminary Results


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One final Remark:

How about complex mechanisms; i.e. multi-loop mechanisms?

Consider the following mechanism: In this case, the mechanism can be represented by two loop-closure equations as shown below.

2r

BAr

4r

5r 6r

1r

CAr

Now, the two loop-closure equations may be written as

2 4BAr r r+ =

2 5 6 1CAr r r r r+ + = + Note that 3AB ACω ω ω= = and 1r is constant at the value of 0.75”, and therefore will vanish upon differentiation.

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mechines of mechinism chapter 2