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Mechanism Kinematics & Dynamics and

Vibrational Modeling

Dr. Robert L. Williams IIMechanical Engineering, Ohio University

NotesBook Supplement forME 3011 Kinematics & Dynamics of Machines

© 2014 Dr. Bob Productions

[email protected] people.ohio.edu/williar4

These notes supplement the ME 3011 NotesBook by Dr. Bob

This document presents supplemental notes to accompany the ME 3011 NotesBook. The outlinegiven in the Table of Contents on the next page dovetails with and augments the ME 3011 NotesBook

outline and hence is incomplete here.

mk

x(t)

k

x (t) x (t)



2

ME 3011 Supplement Table of Contents

1. INTRODUCTION................................................................................................................................ 4

1.3 VECTORS. CARTESIAN R E-IM R EPRESENTATION (PHASORS) ............................................................. 41.6 MATRICES .......................................................................................................................................... 6

2. KINEMATICS ANALYSIS .............................................................................................................. 15

2.1 POSITION K INEMATICS A NALYSIS .................................................................................................... 152.1.1 Four-Bar Mechanism Position Analysis .................................................................................. 15

2.1.1.1 Tangent Half-Angle Substitution Derivation and Alternate Solution Method .................. 152.1.1.3 Four-Bar Mechanism Solution Irregularities ..................................................................... 202.1.1.4 Grashof’s Law and Four-Bar Mechanism Joint Limits ..................................................... 21

2.1.2 Slider-Crank Mechanism Position Analysis ............................................................................. 29 2.1.3 Inverted Slider-Crank Mechanism Position Analysis .............................................................. 32 2.1.4 Multi-Loop Mechanism Position Analysis ............................................................................... 38

2.2 VELOCITY K INEMATICS A NALYSIS .................................................................................................. 422.2.2 Three-Part Velocity Formula Moving Example ....................................................................... 42 2.2.3 Four-Bar Mechanism Velocity Analysis .................................................................................. 44 2.2.5 Inverted Slider-Crank Mechanism Velocity Analysis............................................................... 45 2.2.6 Multi-Loop Mechanism Velocity Analysis ................................................................................ 49

2.3 ACCELERATION K INEMATICS A NALYSIS .......................................................................................... 532.3.2 Five-Part Acceleration Formula Moving Example .................................................................. 53 2.3.3 Four-Bar Mechanism Acceleration Analysis ........................................................................... 55 2.3.4 Slider-Crank Mechanism Acceleration Analysis ...................................................................... 56 2.3.5 Inverted Slider-Crank Mechanism Acceleration Analysis ....................................................... 57 2.3.6 Multi-Loop Mechanism Acceleration Analysis ........................................................................ 63

2.4 OTHER K INEMATICS TOPICS ............................................................................................................ 672.4.1 Link Extensions Graphics ......................................................................................................... 67

2.5 JERK K INEMATICS A NALYSIS ........................................................................................................... 692.5.1 Jerk Analysis Introduction ....................................................................................................... 69 2.5.2 Mechanism Jerk Analysis ......................................................................................................... 72

2.6 BRANCH SYMMETRY IN K INEMATICS A NALYSIS .............................................................................. 732.6.1 Four-Bar Mechanism ............................................................................................................... 73 2.6.2 Slider-Crank Mechanism.......................................................................................................... 75

2.7 K INEMATICS A NALYSIS EXAMPLES ................................................................................................. 772.7.1 Term Example 1: Four-Bar Mechanism .................................................................................. 77 2.7.2 Term Example 2: Slider-Crank Mechanism ............................................................................. 88

3. DYNAMICS ANALYSIS .................................................................................................................. 95

3.1 DYNAMICS I NTRODUCTION .............................................................................................................. 953.2 MASS, CENTER OF GRAVITY, AND MASS MOMENT OF I NERTIA ....................................................... 963.4 FOUR -BAR MECHANISM I NVERSE DYNAMICS A NALYSIS ............................................................... 1083.5 SLIDER -CRANK MECHANISM I NVERSE DYNAMICS A NALYSIS ....................................................... 1113.6 I NVERTED SLIDER -CRANK MECHANISM I NVERSE DYNAMICS A NALYSIS....................................... 1133.7 MULTI-LOOP MECHANISM I NVERSE DYNAMICS A NALYSIS ............................................................ 1213.8 BALANCING OF R OTATING SHAFTS ................................................................................................ 1263.9 I NVERSE DYNAMICS A NALYSIS EXAMPLES .................................................................................... 130

3.9.1 Single Rotating Link Inverse Dynamics Example .................................................................. 130



3

3.9.2 Term Example 1: Four-Bar Mechanism ................................................................................ 133 3.9.3 Term Example 2: Slider-Crank Mechanism ........................................................................... 139

4. GEARS AND CAMS ....................................................................................................................... 145

4.1 GEARS ............................................................................................................................................ 1454.1.1 Gear Introduction ................................................................................................................... 145 4.1.2 Gear Ratio .............................................................................................................................. 151

4.1.3 Gear Trains ............................................................................................................................ 154 4.1.4 Involute Spur Gear Standardization ...................................................................................... 156 4.1.5 Planetary Gear Trains ........................................................................................................... 165

4.2 CAMS ............................................................................................................................................. 1734.2.1 Cam Introduction ................................................................................................................... 173 4.2.2 Cam Motion Profiles .............................................................................................................. 176 4.2.3 Analytical Cam Synthesis ....................................................................................................... 181

5. MECHANICAL VIBRATIONS INTRODUCTION .................................................................... 188

5.2 MECHANICAL VIBRATIONS DEFINITIONS ....................................................................................... 188

6. VIBRATIONAL SYSTEMS MODELING.................................................................................... 191

6.1 ZEROTH-ORDER SYSTEMS ............................................................................................................. 1916.2 SECOND-ORDER SYSTEMS ............................................................................................................. 201

6.2.1 Translational m-c-k System Dynamics Model ........................................................................ 201 6.2.3 Pendulum System Dynamics Model ....................................................................................... 202 6.2.4 Uniform Circular Motion ....................................................................................................... 204

6.4 ADDITIONAL 1-DOF VIBRATIONAL SYSTEMS MODELS ................................................................... 2056.5 ELECTRICAL CIRCUITS MODELING ................................................................................................. 2376.6 MULTI-DOF VIBRATIONAL SYSTEMS MODELS ............................................................................... 246



4

1. Introduction

1.3 Vectors . Cartesian Re-Im Representation (Phasors)

Here is an alternate vector representation.

iP Pe

The phasor iPe

is a polar representation for vectors, where P is the length of vector P , e is the

natural logarithm base, 1i is the imaginary operator, and is the angle of vector P . ie

gives the

direction of the length P, according to Euler’s identity.

cos sinie i

ie

is a unit vector in the direction of vector P .

Phasor Re- Im representation of a vector is equivalent to Cartesian XY representation, where the real ( Re)

axis is along X (or i ) and the imaginary ( Im) axis is along Y (or j ).

cos(cos sin )

sin

cosˆ ˆ(cos sin )

sin

Re i

Im

X

Y

P PP P i Pe

P P

P PP P i j

P P

A strength of Cartesian Re- Im representation using phasors is in taking time derivatives of vectors – the

derivative of the exponential is easy ( ( )s sd ds e e ).2 2

2 2

2

2

22 2

2

22

2

22

2

( )

2

cos 2 sin sin cos

sin 2 cos cos

i

i i

i i i i i

i i i i

d P d Pe

dt dt

d P d Pe iP e

dt dt

d PPe iP e iP e iP e i P edt

d PPe iP e iP e P e

dt

P P P Pd P

dt P P P

2sinP



5

Where we had to use extensions of Euler’s identity

2

2 2

cos sin

cos sin sin cos

sin cos cos sin

i

i

i

e i

ie i i i

i e i i i

Compare this double-time-derivative with the XY approach.

2 2

2 2

2

2

22

2 2

2

2

cos

sin

cos sin

sin cos

cos sin sin sin cos

sin cos cos cos sin

cos 2 sin

Pd P d

Pdt dt

P Pd P d

dt dt P P

P P P P Pd P

dt P P P P P

P P Pd P

dt

2

2

sin cos

sin 2 cos cos sin

P

P P P P

We obtain the same result, but the Re- Im phasor time differentiation is made in compact vector notationalong the way.

Above we used the product and chain rules of time differentiation.

product rule ( ) ( )

( ) ( ) ( )( )( ( ) ) ( ) ( ) ( )

i t i t i t i t i t d dP t de de

P t e e P t P t e P t dt dt dt dt

chain rule ( ) ( )

( )( )( )

( )

i t i t i t de de d t

ie t dt d t dt

The result for this example is

( ) ( ) ( )( ( ) ) ( ) ( ) ( )i t i t i t d P t e P t e P t ie t

dt



6

1.6 Matrices

Matrix an m x n array of numbers, where m is the number of rows and n is the number of columns.

11 12 1

21 22 2

1 2

n

n

m m mn

a a a

a a a A

a a a

Matrices may be used to simplify and standardize the solution of n linear equations in n unknowns(where m = n). Matrices are used in velocity, acceleration, and dynamics linear equations (matrices arenot used in position analysis which requires a non-linear solution).

Special Matrices

square matrix (m = n = 3) 11 12 13

21 22 23

31 32 33

a a a

A a a a

a a a

diagonal matrix 11

22

33

0 0

0 0

0 0

a

A a

a

identity matrix 3

1 0 0

0 1 0

0 0 1

I

transpose matrix 11 21 31

12 22 32

13 23 33

T

a a a

A a a a

a a a

(switch rows & columns)

symmetric matrix 11 12 13

12 22 23

13 23 33

T

a a a

A A a a a

a a a

column vector (3x1 matrix) 1

2

3

x

X x

x

row vector (1x3 matrix) 1 2 3

T X x x x



7

Matrix Addition add like terms and keep the results in place

a b e f a e b f

c d g h c g d h

Matrix Multiplication with Scalar multiply each term and keep the results in place

a b ka kbk

c d kc kd

Matrix Multiplication

C A B

In general, A B B A

The row and column indices must line up as follows.

( x ) ( x )( x )

C A B

m n m p p n

That is, in a matrix multiplication product, the number of columns p in the left-hand matrix must equalthe number of rows p in the right-hand matrix. If this condition is not met, the matrix multiplication isundefined and cannot be done.

The size of the resulting matrix [C ] is from the number of rows m of the left-hand matrix and thenumber of columns n of the right-hand matrix, m x n.

Multiplication proceeds by multiplying like terms and adding them, along the rows of the left-hand matrix and down the columns of the right-hand matrix (use your index fingers from the left andright hands).

Example

(2x1) (2x3)(3x1)

ga b c ag bh ci

C hd e f dg eh fi

i

Note the inner indices ( p = 3) must match, as stated above, and the dimension of the result is dictated bythe outer indices, i.e. m x n = 2x1.



8

Matrix Multiplication Examples

1 2 3

4 5 6 A

7 8

9 8

7 6

B

7 8

1 2 39 8

4 5 67 6

7 18 21 8 16 18 46 42

28 45 42 32 40 36 115 108

C A B

(2x2) (2x3)(3x2)

7 81 2 3

9 84 5 6

7 6

7 32 14 40 21 48 39 54 69

9 32 18 40 27 48 41 58 75

7 24 14 30 21 36 31 44 57

D B A

(3x3) (3x2)(2x3)



9

Matrix Inversion

Since we cannot divide by a matrix, we multiply by the matrix inverse instead. Given

C A B , solve for [ B].

C A B

1 1 A C A A B

I B

B

1

B A C

Matrix [ A] must be square (m = n) to invert.

1 1

A A A A I

where [ I ] is the identity matrix, the matrix 1 (ones on the diagonal and zeros everywhere else). Tocalculate the matrix inverse use the following expression.

1 adjoint( ) A

A A

where A is the determinant of [ A].

adjoint( ) cofactor( ) T

A A

cofactor( A) ( 1)i j

ij ija M

minor minor M ij is the determinant of the submatrix with row i andcolumn j removed.

For another example, given C A B , solve for [ A]

C A B

1 1C B A B B

A I

A

1

A C B

In general the order of matrix multiplication and inversion is crucial and cannot be changed.



10

Matrix Determinant The determinant of a square n x n matrix is a scalar. The matrix determinant is undefined for a

non-square matrix. The determinant of a square matrix A is denoted det( A) or A . The determinant

notation should not be confused with the absolute-value symbol. The MATLAB function for matrixdeterminant is det(A).

If a nonhomogeneous system of n linear equations in n unknowns is dependent, the coefficientmatrix A is singular, and the determinant of matrix A is zero. In this case no unique solution exists tothese equations. On the other hand, if the matrix determinant is non-zero, then the matrix is non-singular, the system of equations is independent, and a unique solution exists.

The formula to calculate a 2 x 2 matrix determinant is straight-forward.

a b

Ac d

A ad bc

To calculate the determinant of 3 x 3 and larger square matrices, we can expand about any one

row or column, utilizing sub-matrix determinants. Each sub-determinant is formed by crossing out thecurrent row and its column and retaining the remaining terms as an n–1 x n–1 square matrix, each ofwhose determinant must also be evaluated in the process. The pivot term (the entry in the cross-out rowand column) multiplies the sub-matrix determinants, and there is an alternating + / – / + / – etc. sign pattern. Here is an explicit example for a 3 x 3 matrix, expanding about the first row (all other optionswill yield identical results).

a b c

A d e f

g h k

( ) ( ) ( )

e f d f d e A a b c

h k g k g h

a ek hf b dk gf c dh ge

For a 3 x 3 matrix only, the determinant can alternatively be calculated as shown, by copying columns 1and 2 outside the matrix, multiplying the downward diagonals with + signs and multiplying the upwarddiagonals with – signs (clearly the result is the same as in the above formula).

( ) ( ) ( )

a b c a b

A d e f d e

g h k g h

aek bfg cdh gec hfa kdb a ek hf b kd fg c dh ge

A common usage of the 3 x 3 matrix determinant is to calculate the cross product 1 2P P .

1 2 1 2

1 1 1 11 1

1 2 1 1 1 1 2 1 2

2 2 2 22 2

2 2 2 1 2 1 2

ˆˆ ˆ

ˆˆ ˆ y z z y

y z x y x z

x y z x z z x

y z x y x z

x y z x y y x

i j k p p p p p p p p p p

P P p p p i j k p p p p p p p p p p

p p p p p p p



11

System of Linear Equations

We can solve n linear equations in n unknowns with the help of a matrix. Below is an examplefor n = 3.

11 1 12 2 13 3 1

21 1 22 2 23 3 2

31 1 32 2 33 3 3

a x a x a x b

a x a x a x ba x a x a x b

Where aij are the nine known numerical equation coefficients, xi are the three unknowns, and bi are thethree known right-hand-side terms. Using matrix multiplication backwards, this is written as

A x b .

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

a a a x b

a a a x b

a a a x b

where

11 12 13

21 22 23

31 32 33

a a a

A a a a

a a a

is the matrix of known numerical coefficients

1

2

3

x x x

x

is the vector of unknowns to be solved and

1

2

3

b

b b

b

is the vector of known numerical right-hand-side terms.

There is a unique solution 1

x A b

only if [ A] has full rank. If not, 0 A (the determinant ofcoefficient matrix [ A] is zero) and the inverse of matrix [ A] is undefined (since it would require dividing by zero; in this case the rank is not full, it is less than 3, which means not all rows/columns of [ A] arelinearly independent). Gaussian Elimination is more robust and more computationally efficient thanmatrix inversion to solve the problem A x b for { x}.



12

Matrix Example – solve linear equations

Solution of 2x2 coupled linear equations.

1 2

1 2

2 5

6 4 14

x x

x x

1

2

1 2 5

6 4 14

x

x

1 2

6 4 A

1

2

x x

x

5

14b

1

x A b

1 4 2 6 8 A

The determinant of [ A] is non-zero so there is a unique solution.

1 4 2 1/ 2 1/ 41

6 1 3/ 4 1/ 8 A

A

check 1 1

2

1 0

0 1 A A A A I

1

2

1/ 2 1/ 4 5 1

3/ 4 1/ 8 14 2

x

x

answer.

Check this solution by substituting the answer { x} into the original equations A x b and ensuring

the required original {b} results.

1 2 1 1(1) 2(2) 5

6 4 2 6(1) 4(2) 14



13

Same Matrix Examples in MATLAB

%- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -% Mat r i ces. m - mat r i x exampl es% Dr . Bob, ME 3011 %- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

cl ear ; cl c;

A1 = di ag( [ 1 2 3] ) % 3x3 di agonal mat r i x A2 = eye( 3) % 3x3 i dent i t y mat r i x

A3 = [ 1 2; 3 4] ; % mat r i x addi t i on A4 = [ 5 6; 7 8] ;Add = A3 + A4

k = 10; % mat ri x- scal ar mul t i pl i cat i on Mul t Sca = k*A3

Tr ans = A4' % mat r i x t r anspose ( swap rows and col umns)

A5 = [ 1 2 3; 4 5 6] ; % def i ne t wo mat r i ces A6 = [ 7 8; 9 8; 7 6] ;A7 = A5*A6 % mat r i x- mat r i x mul t i pl i cat i on A8 = A6*A5

A9 = [ 1 2; 6 4] ; % mat r i x f or l i near equat i ons sol ut i on b = [ 5; 14] ; % def i ne RHS vector dA9 = det ( A9) % cal cul at e det er mi nant of A i nvA9 = i nv( A9) % cal cul at e t he i nver se of A x = i nvA9*b % sol ve l i near equat i ons x1 = x( 1) ; % ext r act answer s x2 = x( 2) ;Check = A9*x % check answer – shoul d be b xG = A9\ b % Gaussi an el i mi nat i on i s mor e ef f i ci ent

who % di spl ay t he user - creat ed var i abl eswhos % user - cr eat ed var i abl es wi t h di mensi ons

The first solution of the linear equations above uses the matrix inverse. To solve linearequations, Gaussian Elimination is more efficient (more on this in the dynamics notes later) and morerobust numerically; Gaussian elimination implementation is given in the third to the last line of theabove m-file (with the back-slash).

Since the equations are linear, there is a unique solution (assuming the equations are linearlyindependent, i.e. the matrix is not near a singularity) and so both solution methods will yield the sameanswer.



14

Output of Matrices.m

A1 =1 0 00 2 00 0 3

A2 =

1 0 00 1 00 0 1

Add =6 8

10 12

Mul t Sca =10 2030 40

Tr ans =

5 76 8

A7 =46 42

115 108

A8 =39 54 6941 58 7531 44 57

dA9 = - 8

i nvA9 =- 0. 5000 0. 25000. 7500 - 0. 1250

x = 12

Check = 514

xG = 12



15

2. Kinematics Analysis

2.1 Position Kinematics Analysis

2.1.1 Four-Bar Mechanism Position Analysis

2.1.1.1 Tangent Half-Angle Substitution Derivation and Alternate Solution Method

Tangent half-angle substitution derivation

In this subsection we first derive the tangent half-angle substitution using ananalytical/trigonometric method. Defining parameter t to be

tan2

t

i.e. the tangent of half of the unknown angle , we need to derive cos and sin as functions of parameter t . This derivation requires the trigonometric sum of angles formulae.

cos( ) cos cos sin sin

sin( ) sin cos cos sin

a b a b a b

a b a b a b

To derive the cos term as a function of t , we start with

cos cos2 2

The cosine sum of angles formula yields

2 2cos cos sin2 2

Multiplying by a ‘1’, i.e. 2cos2

over itself yields

2 2

2 2 2

2

cos sin

2 2cos cos 1 tan cos2 2 2

cos2

The cosine squared term can be divided by another ‘1’, i.e. 2 2cos sin 12 2

.



16

2

2

2 2

cos2

cos 1 tan2

cos sin2 2

Dividing top and bottom by2

cos 2

yields

2

2

1cos 1 tan

21 tan

2

Remembering the earlier definition for t , this result is the first derivation we need, i.e.

2

21cos1

t t

To derive the sin term as a function of t , we start with

sin sin2 2

The sine sum of angles formula yields

sin sin cos cos sin 2sin cos2 2 2 2 2 2

Multiplying top and bottom by cosine yields

2 2

sin2

sin 2 cos 2 tan cos2 2 2

cos

2

From the first derivation we learned

2

2

1cos

21 tan

2



17

Substituting this term yields

2

1sin 2 tan

21 tan

2

Remembering the earlier definition for t , this result is the second derivation we need, i.e.

2

2sin

1

t

t

The tangent half-angle substitution can also be derived using a graphical method as in the figure below.



18

Alternate solution method

The equation form

cos sin 0 E F G

arises often in the position solutions for mechanisms and robots. It appeared in the 4 solution for thefour-bar mechanism in the ME 3011 NotesBook and was solved using the tangent half-anglesubstitution.

Next we present an alternative and simpler solution to this equation. We make two simpletrigonometric substitutions based on the figure below.

Clearly from this figure we have

2 2cos E

E F

2 2sin F

E F

In the original equation we divide by 2 2 E F and rearrange.

2 2 2 2 2 2cos sin

E F G

E F E F E F

The two simple trigonometric substitutions yield

2 2cos cos sin sin

G

E F

Applying the sum-of-angles formula cos( ) cos cos sin sina b a b a b yields

2 2cos( )

G

E F



19

And so the solution for is

1

1,22 2

cos G

E F

where

1tan F

E

and the quadrant-specific inverse tangent function atan2 must be used in the above expression for .

There are two solutions for , indicated by the subscripts 1,2, since the inverse cosine function isdouble-valued. Both solutions are correct. We expected these two solutions from the tangent-half-anglesubstitution approach. They correspond to the open- and crossed-branch solutions (the engineer must

determine which is which) to the four-bar mechanism position analysis problem.

For real solutions for to exist, we must have

2 21 1

G

E F

or

2 21 1

G

E F

If this condition is violated for the four-bar mechanism, this means that the given input angle 2 is beyond its reachable limits (see Grashof’s Law).



20

2.1.1.3 Four-Bar Mechanism Solution Irregularities

Four-bar mechanism position singularity 0G E

4 1 1 2 2

2 2 2 2

1 2 3 4 1 2 1 2

2 ( )

2 cos( )

E r r c r c

G r r r r r r

For simplicity, let 1 = 0 (just rotate the entire four-bar mechanism model for zero ground link angle).

2 2 2 2

1 2 3 4 1 4 2 4 1 22 2 ( ) 0G E r r r r r r r r r c

I have encountered two example four-bar mechanisms with this 0G E singularity.

Case 1

When 1 4r r and 2 3

r r , 2 2 2 2 2

1 2 2 1 1 2 1 1 22 2 ( ) 0G E r r r r r r r r c ALWAYS, regardless

of 2.

ExampleGiven 1 2 3 410, 6, 6, 10r r r r ; this mechanism is ALWAYS singular. To fix this let

1 2 3 410, 5.9999, 6.0001, 10r r r r and MATLAB will be able to calculate the position

analysis reliably at every input angle.

Case 2

When 1 32r r and 4 22r r , and furthermore 3 23 5r r ,

2 2 2 2

3 2 3 2 2 3 2 2 3 2

2 2 2 2 2 2

2 2 2 2 2 2 2

2

2 2

4 4 8 4 ( )c

100 25 40 84 c

9 9 3 3

8c

3

G E r r r r r r r r r

r r r r r r

r

This 0G E occurs only when2 90 . Case 2 is much less general than case 1.

Example

Given 1 2 3 410, 3, 5, 6r r r r ; this mechanism is singular when2 90 . To fix this ignore

290 or set your

2 array to avoid these values.



21

2.1.1.4 Grashof’s Law and Four-Bar Mechanism Joint Limits

Grashof’s Law

Grashof’s Law was presented in the ME 3011 NotesBook to determine the input and output linkrotatability in a four-bar mechanism. Applying Grashof’s Law we determine if the input and output

links are a crank (C) or a rocker (R). A crank enjoys full 360 degree rotation while a rocker has arotation that is a subset of this full rotation. This section presents more information on Grashof’s Lawand then the next subsection presents four-bar mechanism joint limits.

Grashof's condition states "For a four-bar mechanism, the sum of the shortest and longest linklengths should not be greater than the sum of two remaining link lengths". With a given four-bar

mechanism, the Grashof Condition is satisfied if L S P Q where S and L are the lengths of the

shortest and longest links, and P and Q are the lengths of the other two intermediate-sized links. If theGrashof condition is satisfied, at least one link will be fully rotatable, i.e. can rotate 360 degrees.

For a four-bar mechanism, the following inequalities must be satisfied to avoid locking of the

mechanism for all motion.

2 1 3 4

4 1 2 3

r r r r

r r r r

With reference to the figure below, these inequalities are derived from the fact that the sum of

two sides of a triangle must be greater than the third side, for triangles 4 1 1O A B and 2 2 2

O A B , respectively.

Note from our standard notation, 1 2 4r O O , 2 2r O A , 3r AB , and 4 4r O B .

A

B

A

O

2

2 O4

A1

B1

B2



22

Four-Bar Mechanism Joint Limits

If Grashof's Law predicts that the input link is a rocker, there will be rotation limits on the inputlink. These joint limits occur when links 3 and 4 are aligned. As shown in the figure below, there will be two joint limits, symmetric about the ground link.

To calculate the joint limits, we use the law of cosines.

2 2 2

3 4 1 2 1 2 2

2 2 21 1 2 3 4

2

1 2

( ) 2 cos

( )cos

2

L

L

r r r r r r

r r r r

r r

with symmetry about r 1.



23

Joint Limit Example 1 Given 1 2 3 410, 6, 8, 7r r r r

L S P Q (10 6 8 7 )

so we predict only double rockers from this Non-Grashof Mechanism.

2 2 2

1 1

2

10 6 (8 7)cos cos 0.742 137.92(10)(6)

L

This method can also be used to find angular limits on link 4 when it is a rocker. In this case links 2 and3 align.

2 2 2

1 1

4

10 7 (6 8)cos cos 0.336 109.6

2(10)(7)

180 70.4 L

In this example, the allowable input and output angle ranges are:

2137.9 137.9 470.4 289.6

This example is shown graphically in the ME 3011 NotesBook, in the Grashof’s Law section (2. Non-Grashof double rocker, first inversion).

Caution The figure on the previous page does not apply in all joint limit cases. For Grashof

Mechanisms with a rocker input link, one link 2 limit occurs when links 3 and 4 fold upon each otherand the other link 2 limit occurs when links 3 and 4 stretch out in a straight line. See Example 4 (andExample 3 for a similar situation with the output link 4 limits).



24


L S P Q (10 4 8 7 )

Since the S link is adjacent to the fixed link, we predict this Grashof Mechanism is a crank-rocker.

Therefore, there are no 2 joint limits.

2 2 2

1 1

2

10 4 (8 7)cos cos 1.3625

2(10)(4) L

which is undefined, thus confirming there are no 2 joint limits.

There are limits on link 4 since it is a rocker. For 4min, links 2 and 3 are stretched in a straight line(their absolute angles are identical).

2 2 21 1

4min

10 7 (4 8)cos cos 0.036 88.02(10)(7)

180 92.0

For 4max, links 2 and 3 are instead folded upon each other (their absolute angles are different by ).

2 2 2

1 1

4min

10 7 ( 4 8)cos cos 0.95 18.2

2(10)(7)

180 161.8

In this example, the output angle range is

492.0 161.8

and 2 is not limited. This example is shown graphically in the ME 3011 NotesBook, in the Grashof’s

Law section (1a. Grashof crank-rocker).



25

Joint Limit Example 3 Given 1 2 3 411.18, 3, 8, 7r r r r (in) and1 10.3

L S P Q (11.18 3 8 7 )

This is the four-bar mechanism from Term Example 1 and it is a four-bar crank-rocker Grashof

Mechanism. There are no limits on 2 since link 2 is a crank.

The 4 limits are

4 120.1 L (links 2 and 3 stretched in a line)

4 172.5 L

(links 2 and 3 folded upon each other in a line)

The output angle range is

4120.1 172.5

and 2 is not limited. This example is NOT shown graphically in the ME 3011 NotesBook Grashof’s

Law section. However, these 4 limits are clearly seen in the F.R.O.M. plot for angle 4 in TermExample 1 in the ME 3011 NotesBook.



26


L S P Q (10 4 8 7 )

so we predict this Grashof Mechanism is a double-rocker (S opposite fixed link). The 2 joint limitsare no longer symmetric about the ground link, as was the case in the Non-Grashof Mechanism double

rocker (Example 1). For 2min, links 3 and 4 are folded upon each other (their absolute angles areidentical).

2 2 2

1 1

2min

10 8 (7 4)cos cos 0.969 14.4

2(10)(8)

For 2max, links 3 and 4 are instead stretched in a straight line (their absolute angles are different by asin Example 1).

2 2 21 1

2max

10 8 (7 4)

cos cos 0.269 74.42(10)(8)

2min Diagram2max Diagram

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

X (m)

Y ( m )

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

X (m)

Y ( m )



27

This behavior reverses for the 4 joint limits. For 4min, links 2 and 3 are stretched in a straight line(their absolute angles are identical).

2 2 2

1 1

4min

10 7 (8 4)cos cos 0.036 88.0

2(10)(7)

180 92.0

For 4max, links 2 and 3 are instead folded upon each other (their absolute angles are different by ).

2 2 2

1 1

4min

10 7 (8 4)cos cos 0.95 18.2

2(10)(7)

180 161.8

4min Diagram4max Diagram

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

X (m)

Y ( m )

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

X (m)

Y ( m )



28

In this plot we can see the minimum and maximum values we just calculated for links 2 and 4.

Note at 2min 14.4

, 3 138.6

and 3 221.4

are the same angle.

Again, this example is NOT shown graphically in the ME 3011 NotesBook Grashof’s Law section.However, a similar case with the same dimensions, in different order, is shown in the ME 3011

NotesBook ( 1 2 3 47, 10, 4, 8r r r r , 1d. Grashof double rocker).

Grashof’s Law only predicts the rotatability of the input and output links; it says nothing aboutthe rotatability of the coupler link 3 – in this case, what is the rotatability of the coupler link? (In thiscase the coupler link S rotates fully, proving that the relative motion is the same amongst all four-barmechanism inversions, though the absolute motion with respect to the possible 4 ground links is very

different.)

For more information, see:

R.L. Williams II and C.F. Reinholtz, 1987, “Mechanism Link Rotatability and Limit PositionAnalysis Using Polynomial Discriminants”, Journal of Mechanisms, Transmissions, and Automation

in Design, Transactions of the ASME, 109(2): 178-182.

10 20 30 40 50 60 70 80-250

-200

-150

-100

-50

0

50

100

150

200

2 (deg)

( d e g )

3 1

st

3 2

nd

4 1

st

4 2

nd



29

2.1.2 Slider-Crank Mechanism Posi tion Analys is

Step 6. Solve for the unknowns – alternate solution

Here are the same slider-crank mechanism position analysis XY component equations, rearranged

to isolate the 3 terms.

3 3 2 2

3 3 2 2

r c x r c

r s h r s

We can square and add to eliminate 3, similar to the four-bar mechanism solution approach.

2 2 2 2 2

3 3 2 2 2 2

2 2 2 2 2

3 3 2 2 2 2

2

2

r c x xr c r c

r s h hr s r s

2 2 2 2

3 2 2 2 2 22 2r x h r xr c hr s

This quadratic equation in x has the following form:

2 0ax bx c 2 2

2 2 2

2 3 2 2

1

2

2

a

b r c

c r r h hr s

There are two solutions for x, corresponding to the right and left branches.

2 2 2 2

1,2 2 2 3 2 2 2 22 x r c r h r s hr s

Then 3 is found from a ratio of the Y to X equations.

1,23 2 2 1,2 2 2atan2( , )h r s x r c

This alternate solution yields identical results as the earlier solution approach in the ME 3011 NotesBook for the right (

13 1, x ) and left (23 2, x ) branches.



30

Slider-Crank Mechanism Snapshot and F.R.O.M. MATLAB m-files

No sample m-files are given in the ME 3011 NotesBook for the slider-crank mechanism sinceyou can readily adapt the snapshot and F.R.O.M. m-files given for the four-bar mechanism previously.

However, below we include a partial m-file to show how to draw the slider and fixed pistonwalls for the slider-crank mechanism graphics, since this was not required for the four-bar mechanism.

Outside the loop:

Lp = put a number here; % length of piston (slider link) Hp = put a number here; % height of piston Xp = [-1 -1 1 1]*Lp/2;Yp = [-1 1 1 -1]*Hp/2;

This establishes the rectangular corner coordinates for the slider link, centered at the origin of yourcoordinate frame. It can be done once, outside the loop. Instead of typing numbers for Lp and Hp, I

scale them to a fraction of r 2, for generality in different-sized slider-crank mechanisms. Note I onlyincluded the four corner points – MATLAB patch (below) closes the rectangular figure, i.e. back to

the starting point.

Inside the loop (right after the plot command where links 2 and 3 are drawn to the screen)

patch(Xp+x(i),Yp+h,'g'); % draw piston to screen

where x(i) is the variable horizontal slider displacement and h is the constant vertical offset. These

position parameters shift the piston coordinates from the origin to the correct location in each loop. Youcan use any piston color you like (I show green here, 'g').

Further, to draw the horizontal lines representing the piston walls:

Outside the loop

Xpt = [-1000 1000]; % fixed piston walls Ypt = [h+wall/2 h+wall/2];Xpb = [-1000 1000];Ypb = [h-wall/2 h-wall/2];

Inside the loop (right after the plot command where links 2 and 3 are drawn to the screen)

line(Xpt,Ypt,'LineWidth',2); line(Xpb,Ypb,'LineWidth',2);

Set the piston wall width wall to allow a small clearance between the piston and the walls. Again, it

can be scaled to a small fraction of r 2 for generality. The -1000 and 1000 coordinates used above are

to extend the piston wall lines off the screen to the left and to the right.



31

MATLAB subplot feature

In a slider-crank mechanism full-range-of-motion (F.R.O.M.) simulation you will need to plot

both 3 and x vs. the independent variable 2. Since the units of 3 (deg) and x (m) are dissimilar, theymay not fit clearly on the same plot. In this situation you should use a sub-plot arrangement.

Outside the F.R.O.M. loop you can do the subplot in this way:

subplot(211); % 2x1 arrangement of plots, first plot plot(th2/DR,th3/DR);subplot(212); % 2x1 arrangement of plots, second plot plot(th2/DR,x);

Now, you can use the standard axis labels, linetypes, titles, axis limits, grid, etc., for each plot within asubplot (repeat these formatting commands after each plot statement above to use similar formatting

for each). These options are not shown, for clarity.

The generalized usage of subplot is shown below.

subplot(mni); % m x n arrangement of plots, ith plot plot( . . . );

As seen in the example syntax above, the integers need not be separated by spaces or commas.However, I believe they may be so separated if you desire.



32

2.1.3 Inverted Slider-Crank Mechanism Posi tion Analysis

This slider-crank mechanism inversion 2 is an inversion of the standard zero-offset slider-crankmechanism where the sliding direction is no longer the ground link, but along the rotating link 4.Ground link length r 1 and input link length r 2 are fixed; r 4 is a variable. The slider link 3 is attached tothe end of link 2 via an R joint and slides relative to link 4 via a P joint. This mechanism converts rotaryinput to linear motion and rotary motion output. Practical applications include certain doors/windowsopening/damping mechanisms. The inverted slider-crank is also part of quick-return mechanisms.

Step 1. Draw the Kinematic Diagram

r 1 constant ground link length 2 variable input angle

r 2 constant input link length 4 variable output angler 4 variable output link length L4 constant total output link length

Link 1 is the fixed ground link. Without loss of generality we may force the ground link to behorizontal. If it is not so in the real world, merely rotate the entire inverted slider-crank mechanism so it

is horizontal. Both angles 2 and 4 are measured in a right-hand sense from the horizontal to the link.

Step 2. State the Problem

Given r 1, 1 = 0 , r 2; plus 1-dof position input 2

Find r 4 and 4

2

1

4

3



33

Step 3. Draw the Vector Diagram. Define all angles in a positive sense, measured with the right handfrom the right horizontal to the link vector (tail-to-head; your right-hand thumb is located at the vectortail).

Step 4. Derive the Vector-Loop-Closure Equation. Starting at one point, add vectors tail-to-head untilyou reach a second point. Write the VLCE by starting and ending at the same points, but choosing adifferent path.

2 1 4r r r

Step 5. Write the XY Components for the Vector-Loop-Closure Equation. Separate the one vectorequation into its two X and Y scalar components.

2 2 1 4 4

2 2 4 4

r c r r c

r s r s

Step 6. Solve for the Unknowns from the XY equations. There are two coupled nonlinear equations in

the two unknowns r 4,

4. Unlike the standard slider-crank mechanism, there is no decoupling of X and

Y . However, unlike the four-bar mechanism, there is only one unknown angle so the solution is easierthan the four-bar mechanism. First rewrite the above XY equations to isolate the unknowns on one side.

4 4 2 2 1

4 4 2 2

r c r c r

r s r s

A ratio of the Y to X equations will cancel r 4 and solve for 4.

4 4 2 2

4 4 2 2 1

r s r s

r c r c r

4 2 2 2 2 1atan2( , )r s r c r

Then square and add the XY equations to eliminate 4 and solve for r 4.

2 2

4 1 2 1 2 22r r r rr c

2

1

4



34

Note the same r 4 formula results from the cosine law. Alternatively, the same r 4 can be solved from

either the X or Y equations after is 4 known.

X ) 2 2 14

4

r c r r

c

Y ) 2 2

4

4

r sr

s

Both of these r 4 alternatives are valid; however, each is subject to a different artificial mathematicalsingularity (

4 90 and4 0,180 , respectively), so only the former square-root formula should be

used for r 4, which has no artificial singularity. The X algorithmic singularity4 90 never occurs

unless 2 1r r , which is to be avoided (see below), but the Y algorithmic singularity occurs twice per full

range of motion.

Technically there are two solution sets – the one above and2 2

4 1 2 1 2 22r r r rr c , 4 .

However, the negative r 4 is not practical and so only the one solution set (branch) exists, unlike most planar mechanisms with two or more branches.

Full-rotation condition

For the inverted slider-crank mechanism to rotate fully, the fixed length of link 4, L4, must begreater than the maximum value of the variable r 4.

Slider Limits

The slider reaches its minimum and maximum displacements when 2 = 0 and , respectively.

Therefore, the slider limits are 1 2 4 1 2r r r r r . Thus, the fixed length L4 must be greater than 1 2r r .In addition we require 1 2r r for full rotation.



35

Graphical Solution

The Inverted Slider-Crank mechanism position analysis may be solved graphically, by drawingthe mechanism, determining the mechanism closure, and measuring the unknowns. This is an excellentmethod to validate your computer results at a given snapshot.

Draw the known ground link (points O2 and O4 separated by r 1 at the fixed angle 1 = 0).

Draw the given input link length r 2 at the given angle 2 (this defines point A).

Draw a line from O4 to point A.

Measure the unknown values of r 4 and 4.



37

Full-Range-Of-Motion (F.R.O.M.) Analysis: Term Example 3

A more meaningful result from position analysis is to report the position analysis unknowns for

the entire range of mechanism motion. The subplots below gives r 4 (m) and 4 (deg), for all

20 360 , for Term Example 3.

Term Example 3 4 and r4

4 varies symmetrically about 180 , being 180

at2 0,180,360 . r 4 varies like a negative cosine

function with minimum displacement 1 2 0.1r r at2 0,360 and maximum displacement 1 2 0.3r r

. Since r 1 is twice r 2 in this example, whenever2 60,300 , a perfect 30 60 90

triangle is

formed; the relative angle between links 2 and 4 is 90 which corresponds to the max and min of

4 150,210 , respectively. At these special points, 4 ( 3 2)0.2 0.173r m.

There is another right triangle that shows up for2 90 ; in these cases

2 2

4 0.2 0.1 0.224r m and4 153.4,206.6 , respectively. Check all of these special values in the

F.R.O.M. plot results.

0 50 100 150 200 250 300 350150

160

170

180

190

200

210

4 ( d e g )

0 50 100 150 200 250 300 3500.1

0.15

0.2

0.25

0.3

2 (deg)

r 4

m



38

2.1.4 Multi-Loop Mechanism Position Analysis

Thus far we have presented position analysis for the single-loop four-bar, slider-crank, andinverted slider-crank mechanisms. The position analysis for mechanisms of more than one loop ishandled using the same general procedures developed for the single loop mechanisms. A good rule ofthumb is to look for four-bar (or slider-crank) parts of the multi-loop mechanism as we already knowhow to solve the complete position analyses for these.

This section presents position analysis for the two-loop Stephenson I six-bar mechanism shown below as an example multi-loop mechanism. This is one of the five possible six-bar mechanisms shownin the on-line Mechanisms Atlas.

Stephenson I 6-Bar Mechanism

We immediately see that the bottom loop of the Stephenson I six-bar mechanism is identical toour standard four-bar mechanism model. Since we number the links the same as in the four-bar, and ifwe define the angles identically, the position analysis solution is identical to the four-bar presentedearlier. With the complete position analysis of the bottom loop thus solved, we see that points C and D can be easily calculated. Then the solution for the top loop is essentially another four-bar solution:graphically, the circle of radius r 5 about point C must intersect the circle of radius r 6 about point D toform point E (yielding two possible intersections in general). The analytical solution is very similar tothe standard four-bar position solution, as we will show.

For multi-loop mechanisms, the number of solution branches for position analysis increasescompared to the single-loop mechanisms. Most single-loop mechanisms mathematically have twosolution branches. For multi-loop mechanisms composed of multiple single-loop mechanisms, thenumber of solution branches is 2n, where n is the number of mechanism loops. For the two-loopStephenson I six-bar mechanism, the number of solution branches for the position analysis problem is 4,two from the standard four-bar part, and two for each of these branches from the upper loop.



39

Now let us solve the position analysis problem for the two-loop Stephenson I six-bar mechanismusing the formal position analysis steps presented earlier. Assume link 2 is the input link.

Step 1. Draw the Kinematic Diagram this is done in the figure above.

r 1 constant ground link length 1 constant ground link angle

r 2

constant input link length to point A 2 variable input link angle

r 2a constant input link length to point C 2 constant angle on link 2

r 3 constant coupler link length, loop I 3 variable coupler link angle, loop I

r 4 constant output link length, loop I 4 variable output link angle, loop I

r 4a constant input link length to point D 4 constant angle on link 4

r 5 constant coupler link length, loop II 5 variable coupler link angle, loop II

r 6 constant output link length, loop II 6 variable output link angle, loop II

As usual, all angles are measured in a right-hand sense from the absolute horizontal to the link, as shownin the kinematic diagram.


Given r 1, 1 , r 2 , r 3 , r 4 , r 2a , r 4a , r 5 , r 6, 2, 4; plus 1-dof position input 2

Find 3, 4, 5, 6

Step 3. Draw the Vector Diagram. Define all angles in a positive sense, measured with the right handfrom the right horizontal to the link vector (tail-to-head with the right-hand thumb at the vector tail andright-hand fingers towards the arrow in the vector diagram below).



40

Step 4. Derive the Vector-Loop-Closure Equations. One VLCE is required for each mechanism loop.Start at one point, add vectors tail-to-head until you reach a second point. Write each vector equation bystarting and ending at the same points, but choosing a different path.

2 3 1 4

2 5 1 4 6a a

r r r r

r r r r r

Note an alternative to the second vector loop equation is 2 5 3 4 6b br r r r r . See if you can identify

2br and 4br , plus their angles 2b and 4b.

Step 5. Write the XY Components for each Vector-Loop-Closure Equation. Separate the two vectorequations into four XY scalar component equations.

2 2 3 3 1 1 4 4

2 2 3 3 1 1 4 4

r c r c rc r cr s r s r s r s

2 2 5 5 1 1 4 4 6 6

2 2 5 5 1 1 4 4 6 6

a a a a

a a a a

r c r c rc r c r c

r s r s r s r s r s

where2 2 2

4 4 4

a

a

Step 6. Solve for the Unknowns from the four XY Equations. The four coupled nonlinear equations in

the four unknowns 3, 4, 5, 6 can be solved in two stages, one for each mechanism loop.

Loop I. This solution is identical to the standard four-bar mechanism solution for 3, 4,

summarized here from earlier. From the first two XY scalar equations above, isolate 3 terms, square

and add both equations to obtain one equation in one unknown 4. This equation has the form

4 4cos sin 0 E F G , where terms E , F , and G are known functions of constants and the input angle

2. Solve this equation for two possible values of 4 using the tangent half-angle substitution. The two

4 values correspond to the open and crossed branches. Then return to the original XY scalar equations

with 3 terms isolated, divide the Y by the X equations, and solve for 3 using the atan2 function,substituting the solved values for 4. One unique 3 will result for each of the two possible 4 values.

Loop II. The method is analogous to the Loop I solution above. Since 4 is now known, we also

know 4 4 4a . From the second two XY scalar equations above, isolate 5 terms, square and add

both equations to obtain one equation in one unknown 6. This equation is of the form

2 6 2 6 2cos sin 0 E F G , where terms E 2, F 2, and G2 are known functions of constants and the known

angles 2 2 2a and 4 4 4a . Solve this equation for two possible values of 6 using the tangent



41

half-angle substitution. The two 6 values correspond to open and crossed branches, for each of the

Loop I open and crossed branches. Then return to the original XY scalar equations with 5 terms

isolated, divide the Y by the X equations, and solve for 5 using the atan2 function, substituting the

solved values for 6. One unique 5 will result for each of the two possible 6 values from each 4.

Branches. There are two 5, 6 branches for each of the two 3, 4 branches, so there are four

overall mechanism branches for the two-loop Stephenson I six-bar mechanism.

Full-rotation condition

The range of motion of a multi-loop mechanism may be more limited than that of single-loopmechanisms. One can perform a compound Grashof analysis when four-bars are the componentmechanisms. For the two-loop Stephenson I six-bar mechanism, the second loop may constrain the firstloop (e.g. it may change an expected crank motion of the input link to a rocker). This is an importantissue in design of multi-loop mechanisms if the input link must still rotate fully.

Graphical Solution

The two-loop Stephenson I six-bar mechanism position analysis may readily be solvedgraphically, by drawing the mechanism, determining the mechanism closure, and measuring theunknowns. This is an excellent method to validate your computer results at a given snapshot.

Loop I (this part is identical to the standard four-bar graphical solution)

Draw the known ground link (points O2 and O4 separated by r 1 at the fixed angle 1).

Draw the given input link length r 2 at the given angle 2 to yield point A.

Draw a circle of radius r 3, centered at point A.

Draw a circle of radius r 4, centered at point O4. These circles intersect in general in two places to yield two possible points B.

Connect the two branches and measure the unknown angles 3 and 4.

Loop II (this part is a modification of the standard four-bar graphical solution). Start with the end of the procedure above, on the same drawing. For each solution branch above, perform the following steps.

Draw the link r 2a at angle 2 2 2a from point O2.

Draw the link r 4a at angle 4 4 4a from point O4.

Draw a circle of radius r 5, centered at point C .

Draw a circle of radius r 6, centered at point D.

These circles intersect in general in two places to yield two possible points E . Connect the two branches and measure the unknown values 5 and 6.

In general, there are four overall position solution branches.



42

2.2 Veloci ty Kinematics Analysis

2.2.2 Three-Part Velocity Formula Moving Example

Given initial positions 0 00 0 0 0

x yP P L (m, rad ) and constant velocities

0 0 0.2 0.1 0.3 0.4

x yV V V (m/s, rad /s), simulate this motion and determine PV at each

instant. t f = 5 and 0.1t sec was used. The initial and final snapshots, with their three-part velocity

diagrams, are shown below.

Initial Kinematic Diagram Initial Three-Part Velocity Diagram

Final Kinematic Diagram Final Three-Part Velocity Diagram

0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.5

1

1.5

2

2.5

3

3.5

4

X (m)

Y ( m )

-0.5 0 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Vx (m/s)

V y

( m / s )

0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.5

1

1.5

2

2.5

3

3.5

4

X (m)

Y

( m )

-0.5 0 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Vx (m/s)

V y

( m / s )

V0

V x L

VP



43

Three-Part Velocity Moving Example Plots

Position Terms Velocity Terms

What is the relationship between these plots?

Point P Translational Velocity Sliding and Tangential Veloci ty Components

Constant velocity terms 0 0 0.2 0.1 0.3 0.4 x yV V V lead to non-constant point P velocity

(due to the nonlinear position kinematics).

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

time (sec)

P o s i t i o n T e r m s

( m a

n d r a d )

P0x

P0y

L

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

time (sec)

V e l o c i t y T e r m s ( m / s a n d r a d / s )

V0x

V0y

V

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

time (sec)

P o i n t P V e l o c i t y ( m / s )

VPx

VPy

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

time (sec)

S l i d i n g a n d T a n g e n t i a l V e l o c i t i e s

( m / s )

Vsx

Vsy

Vtx

Vty



44

2.2.3 Four-Bar Mechanism Veloci ty Analysis

The four-bar mechanism velocity solution presented in the 3011 NotesBook used the simplifyingterms a – f . Here is the solution presented in the 3011 NotesBook:

3

ce bf

ae bd

4

af cd

ae bd

3 3

4 4

2 2 2

3 3

4 4

2 2 2

a r s

b r sc r s

d r c

e r c

f r c

Backsubstituting the terms a – f yields the following equivalent solutions, which simplify nicely usingthe sum-of-angles formula sin( ) sin cos cos sina b a b a b and better display the structure of the

solutions.

2 4 23 2

3 4 3

sin( )

sin( )

r

r

2 3 24 2

4 4 3

sin( )

sin( )

r

r

These solution forms clearly show the singularity condition derived for this problem in the 3011 NotesBook, i.e. the mechanism is singular and these solutions are both infinite when:

4 3sin( ) 0 4 3 0 ,180 ,

Physically, this happens when links 3 and 4 are straight out or folded on top of each other, as explainedin the 3011 NotesBook (corresponding to joint limits on the input link 2).



45

2.2.5 Inverted Slider-Crank Mechanism Velocity Analysis

Again, link 2 (the crank) is the input and link 4 is the output. Remember r 4 is a variable so

4 0r in this problem.

Step 1. The inverted slider-crank mechanism Position Analysis must first be complete.

Given r 1, 1 = 0 , r 2, and 2, we solved for r 4 and 4

Step 2. Draw the inverted slider-crank mechanism Velocity Diagram.

where i (i = 2,4) is the absolute angular velocity of link i. 4r is the slider velocity along link 4.

3 4 since the slider cannot rotate relative to link 4.


Given the mechanism r 1, 1 = 0 , r 2

the position analysis 2, r 4, 4

1-dof of velocity input 2

Find the velocity unknowns 4r and 4

2

1

4

3



46

Step 4. Derive the velocity equations. Take the first time derivative of the vector loop closureequations from position analysis, in XY component form.

Here are the inverted slider-crank mechanism position equations from earlier.

2 1 4

2 2 1 4 4

2 2 4 4

r r r

r c r r c

r s r s

The first time derivative of the position equations is given below.

2 2 2 4 4 4 4 4

2 2 2 4 4 4 4 4

r s r c r s

r c r s r c

These two linear equations in two unknowns can be written in matrix form.

4 4 4 4 2 2 2

4 4 4 4 2 2 2

c r s r r s

s r c r c

Step 5. Solve the velocity equations for the unknowns 4 4,r .

2 2 2 4 2 4

4 4 4 4 4 2 2 2

2 2 2 4 2 4

4 4 4 2 2 24

4

2 2 4 24

2 2 4 2

4

4

( )1

( )

sin( )cos( )

r s c c sr r c r s r s

r s s c cs c r cr

r

r r r

r

where we have used the trigonometric identities:

cos cos cos sin sin

sin sin cos cos sin

a b a b a b

a b a b a b

sin( ) sin( )

cos( ) cos( )

a a

a a

The units are all correct in the solution above, m/s and rad /s, respectively.



47

Inverted slider-crank mechanism singularity condition When does the solution fail? This is an inverted slider-crank mechanism singularity, when the

determinant of the coefficient matrix goes to zero. The result is dividing by zero, resulting in infinite

4 4,r .

4 4 4

4 4 4

c r s A

s r c

2 2 2 2

4 4 4 4 4 4 4 4( ) ( ) 0 A r c r s r c s r

Physically, assuming1 2r r as in the full rotation condition from the inverted slider-crank mechanism

position analysis, this is impossible, i.e. r 4 never goes to zero.

This matrix determinant 4 A r was used in the solution of the previous page.

Inverted slider-crank mechanism velocity example – Term Example 3 continued

Given

1

2

4

1

0.20

0.10

0.32

0

r

r

L

(m)

2

4

4

70

150.5

0.191r

(deg and m)

Snapshot AnalysisGiven this mechanism position analysis plus 2 25 rad /s, calculate 4 4,r for this snapshot.

4

4

0.870 0.094 2.349

0.493 0.166 0.855

r

4

4

2.47

2.18

r

(m/s and rad /s)

Both are positive, so the slider link 3 is currently traveling up link 4 and link 4 is currently rotating in the

ccw direction, which makes sense from the physical problem.



48

Full-Range-Of-Motion (F.R.O.M.) Analysis – Term Example 3 continued

A more meaningful result from velocity analysis is to report the velocity analysis results for the

entire range of mechanism motion. The subplot below gives 4 (top, rad /s) and 4r (bottom, m/s), for all

20 360 , for Term Example 3. Since 2 is constant, we can plot the velocity results vs. 2 (since it

is related to time t via 2 2t ).

Term Example 3 F.R.O.M., 4 and 4r

As expected, 4 is zero at the max and min for 4 (at2 60 ); also, 4 has a large range of nearly

constant positive velocity near the middle of motion – this can be seen in a MATLAB animation. 4r

iszero at the beginning, middle, and end of motion and is max at

2 60 .

0 50 100 150 200 250 300 350-25

-20

-15

-10

-5

0

5

10

4

( r a d / s )

0 50 100 150 200 250 300 350

-2

-1

0

1

2

2 (deg)

r 4 d

( m / s )



49

2.2.6 Multi-Loop Mechanism Velocity Analysis

Thus far we have presented velocity analysis for the single-loop four-bar, slider-crank, andinverted slider-crank mechanisms. The velocity analysis for mechanisms of more than one loop ishandled using the same general procedures developed for the single loop mechanisms.

This section presents velocity analysis for the two-loop Stephenson I six-bar mechanism shown below as an example multi-loop mechanism. It follows the position analysis for the same mechanism presented earlier.


The bottom loop of the Stephenson I six-bar mechanism is identical to the standard four-bar

mechanism model and so the velocity analysis solution is identical to the four-bar presented earlier.With the complete velocity analysis of the bottom loop thus solved, the solution for the top loop isessentially another four-bar velocity solution.

As in all velocity analysis, the velocity solution for multi-loop mechanisms is a linear analysisyielding a unique solution (assuming the given mechanism position is not singular) for each positionsolution branch considered. The position analysis must be complete prior to the velocity solution.

Now let us solve the velocity analysis problem for the two-loop Stephenson I six-bar mechanismusing the formal velocity analysis steps presented earlier. Again, assume link 2 is the input link.



50

Step 1. The Stephenson I six-bar mechanism Position Analysis must first be complete.

Given r 1, 1 , r 2 , r 3 , r 4 , r 2a , r 4a , r 5 , r 6, 2, 4, and 2 we solved for 3, 4, 5, 6.

Step 2. Draw the Stephenson I six-bar mechanism Velocity Diagram.This should include all the information from the position diagram, plus the new velocity

information. For clarity, we show only the new velocity information here. Refer to the previousStephenson I position kinematics diagram for complete information.

where i (i = 2,3,4,5,6) is the absolute angular velocity of link i. Triangular links 2 and 4 each have asingle angular velocity for the whole link. 0i

r for all links since all links are of fixed length (no

sliders).


Given the mechanism r 1, 1 , r 2 , r 3 , r 4 , r 2a , r 4a , r 5 , r 6, 2, 4,

the position analysis 2, 3, 4, 5, 6,

1-dof velocity input 2

Find the velocity unknowns 3, 4, 5, 6



51

Step 4. Derive the velocity equations. Take the first time derivative of each of the two vector loopclosure equations from position analysis, in XY component form.

Here are the Stephenson I six-bar mechanism position equations.

Vector equations

2 3 1 4

2 5 1 4 6a a

r r r r

r r r r r

XY scalar equations

2 2 3 3 1 1 4 4

2 2 3 3 1 1 4 4

r c r c rc r c

r s r s r s r s

2 2 5 5 1 1 4 4 6 6

2 2 5 5 1 1 4 4 6 6

a a a a

a a a a

r c r c rc r c r c

r s r s r s r s r s

where2 2 2

4 4 4

a

a

The first time derivatives of the Loop I position equations are identical to those for the standardfour-bar mechanism.

2 2 2 3 3 3 4 4 4

2 2 2 3 3 3 4 4 4

r s r s r s

r c r c r c

These equations can be written in matrix form.

3 3 4 4 3 2 2 2

3 3 4 4 4 2 2 2

r s r s r s

r c r c r c

The first time derivative of the Loop II position equations is

2 2 2 5 5 5 4 4 4 6 6 6

2 2 2 5 5 5 4 4 4 6 6 6

a a a a a a

a a a a a a

r s r s r s r s

r c r c r c r c


5 5 6 6 5 2 2 2 4 4 4

5 5 6 6 6 2 2 2 4 4 4

a a a a

a a a a

r s r s r s r s

r c r c r c r c

where we have used2 2

4 4

a

a

since 2 and 4 are constant angles.



52

Step 5. Solve the velocity equations for the unknowns 3, 4, 5, 6.

The two mechanism loops decouple so we find 3 and 4 from Loop I first and then use 4 to

find 5 and 6 from Loop II. The solutions are given below.

Loop I (identical to the standard four-bar mechanism)

1

3 3 4 43 2 2 2

3 3 4 44 2 2 2

r s r s r s

r c r c r c

Loop II (similar to the standard four-bar mechanism)

1

5 5 5 6 6 2 2 2 4 4 4

6 5 5 6 6 2 2 2 4 4 4

a a a a

a a a a

r s r s r s r s

r c r c r c r c

Remember, Gaussian elimination is more efficient and robust than the matrix inverse. Also, theseequations may be solved algebraically instead of using matrix methods to yield the same answers.

Stephenson I six-bar mechanism singularity condition

The velocity solution fails when the determinant of either coefficient matrix above goes to zero.The result is dividing by zero, resulting in infinite angular velocities for the associated loop.

For the first loop, the singularity condition is identical to the singularity condition of the standardfour-bar mechanism, i.e. when links 3 and 4 either line up or fold upon each other, causing a link 2 jointlimit. For the second loop, the singularity condition is similar, occurring when links 5 and 6 either lineup or fold upon each other. These conditions also cause angle limit problems for the position analysis,so the velocity singularities are known problems.



53

2.3 Acceleration Kinematics Analysis

2.3.2 Five-Part Acceleration Formula Moving Example

Given initial positions 0 0 0 0 0 0 x yP P L (m, rad ), initial velocities

0 0 0 0 0 0 x y

V V V (m/s, rad /s) and constant accelerations 0 0 x y A A A

0.2 0.05 0.15 0.1 (m/s2, rad /s2), simulate this motion and determine P A at each instant. t f = 5

and 0.1t sec was used; the initial and final snapshots, with their five-part acceleration diagrams, are

shown below.

Initial Kinematic Diagram Initial Five-Part Acceleration Diagram

Final Kinematic Diagram Final Five-Part Acceleration Diagram

0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.5

1

1.5

2

2.5

3

3.5

4

X (m)

Y ( m )

-0.8 -0.2 0.40

0.2

0.4

0.6

0.8

1

1.2

A m/s2

A y

( m / s

2 )

0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.5

1

1.5

2

2.5

3

3.5

4

X (m)

Y ( m )

-0.8 -0.2 0.40

0.2

0.4

0.6

0.8

1

1.2

Ax (m/s

2)

A y

( m / s

2 )

A0

A

2 x V

x L

x ( x L)

AP



54

Five-Part Acceleration Moving Example Plots

Position Terms Velocity Terms

Acceleration Terms Point P Translational Acceleration

What is the relationship amongst the first three plots?

Constant acceleration terms 0 0 0.2 0.05 0.15 0.1 x y

A A A lead to non-constant point P

acceleration (due to nonlinear position kinematics and centripetal acceleration).

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

2

2.5

time (sec)

P o s i t i o n T e r m s ( m a n

d r a d )

P0x

P0y

L

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

time (sec)

V e l o c i t y T e r m s ( m / s a n d r a d / s )

V0x

V0y

V

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.05

0.1

0.15

0.2

0.25

time (sec)

A c c e l e r a t i o n T e r m s ( m / s

2 a

n d r a d / s

2 )

A0x

A0y

A

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

time (sec)

P o i n t P A c c e l e r a t i o n ( m / s

2 )

APx

APy



55

2.3.3 Four-Bar Mechanism Acceleration Analysis

The four-bar mechanism acceleration solution presented in the 3011 NotesBook used thesimplifying terms a, b, C , d , e, F . Here is the solution presented in the 3011 NotesBook:

3

Ce bF

ae bd

4

aF Cd

ae bd

3 3

4 4

2 2 2

2 2 2 2 2 2 3 3 3 4 4 4

3 3

4 4

2 2 2

2 2 2 2 2 2 3 3 3 4 4 4

a r s

b r sC r s r c r c r c

d r c

e r c

F r c r s r s r s

Backsubstituting the terms a, b, C , d , e, F yields the following equivalent solutions, which simplifynicely using the sum-of-angles formulae sin( ) sin cos cos sina b a b a b and

cos( ) cos cos sin sina b a b a b and better display the structure of the solutions.

2 2 2

2 2 4 2 2 2 4 2 3 3 4 3 4 43

3 4 3

2 2 2

2 2 3 2 2 2 3 2 3 3 4 4 4 34

4 4 3

sin( ) cos( ) cos( )

sin( )

sin( ) cos( ) cos( )

sin( )

r r r r

r

r r r r

r

These solution forms clearly show the singularity condition derived for this problem in the 3011 NotesBook, i.e. the mechanism is singular and these solutions are both infinite when:

4 3sin( ) 0 4 30 ,180 ,

Physically, this happens when links 3 and 4 are straight out or folded on top of each other, as explainedin the 3011 NotesBook (corresponding to joint limits on the input link 2).



56

2.3.4 Slider-Crank Mechanism Acceleration Analysis

Derivative/Integral Relationships

When one variable is the derivative of another, recall the relationships from calculus. For example:

( )( )

dx t x t

dt

0( ) ( ) x t x x t dt

( )( )

dx t x t

dt

0

( ) ( ) x t x x t dt

Term Example 2 F.R.O.M. Slider Results, x, x , x

The value of x at any point is the slope of the x curve at that point. The value of x at any point is

the integral of the x curve up to that point (the value of x at any point is the area under the x curve up

to that point plus the initial value x0). A similar relationship exists for x and x .

These graphs are plotted vs. 2, but the same type of relationships hold when plotted vs. time t

since 2 is constant. This is the Term Example 2 F.R.O.M. result.

Note these curves should be plotted vs. time t instead of 2 in order to see the true slope and areavalues accurately.

0 50 100 150 200 250 300 350

0.1

0.2

x ( m )

0 50 100 150 200 250 300 350-2

-1

0

1

2

3

x d ( m / s )

0 50 100 150 200 250 300 350-50

0

50

2 (deg)

x d d ( m / s 2 )



57

2.3.5 Inverted Slider-Crank Mechanism Acceleration Analysis

Again, link 2 (the crank) is the input and link 4 is the output.

Step 1. The inverted slider-crank mechanism Position and Velocity Analyses must first be complete.

Given r 1, 1 = 0 , r 2, and 2 we solved for r 4 and 4; then given 2 we solved for4

r and 4.

Step 2. Draw the inverted slider-crank mechanism Acceleration Diagram.

where i (i = 2,4) is the absolute angular acceleration of link i. 4r is the slider acceleration along link 4.

3 4 since the slider cannot rotate relative to link 4.


Given the mechanism r 1, 1 = 0 , r 2

the position analysis 2, r 4, 4

the velocity analysis 2, 4r , 4

1-dof acceleration input 2

Find the acceleration unknowns 4r and 4

2

1

4

3



58

Step 4. Derive the acceleration equations. Take the first time derivative of the inverted slider-crankmechanism velocity equations from velocity analysis, in XY component form.

Here are the inverted slider-crank mechanism velocity equations.

4 4 4 4 4 2 2 2

4 4 4 4 4 2 2 2

r c r s r s

r s r c r c

The first time derivative of the velocity equations is given below.2 2

4 4 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2

2 2

4 4 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2

2

2

r c r s r s r c r s r c

r s r c r c r s r c r s

These equations can be written in matrix form.2 2

4 4 4 4 2 2 2 2 2 2 4 4 4 4 4 4

2 24 4 4 4 2 2 2 2 2 2 4 4 4 4 4 4

2

2

c r s r r s r c r s r c

s r c r c r s r c r s

Step 5. Solve the acceleration equations for the unknowns 4 4,r .

2 24 4 4 4 4 2 2 2 2 2 2 4 4 4 4 4 4

2 24 4 44 2 2 2 2 2 2 4 4 4 4 4 4

2 2

4 2 2 4 2 2 2 4 2 4 4

24 2 2 4 2 2 2 4 2 4 4

4

21

2

sin( ) cos( )

cos( ) sin( ) 2

r r c r s r s r c r s r c

s cr r c r s r c r s

r r r r

r r r

r

where we have again used the trigonometric identities:

cos cos cos sin sin

sin sin cos cos sin

a b a b a b

a b a b a b

sin( ) sin( )

cos( ) cos( )

a a

a a

A major amount of algebra and trigonometry is required to get the final analytical solution for

4 4,r above. Interestingly, the link 4 Coriolis term4 42r cancelled in 4

r while the link 4 centripetal

term 2

4 4r cancelled in 4. The units are all correct, m/s2 and rad /s2, respectively.



59

Inverted slider-crank mechanism singularity condition

The acceleration problem has the same coefficient matrix [ A] as the velocity problem, so thesingularity condition is identical (see the singularity discussion in the inverted slider-crank mechanism

velocity section – the only singularity is when r 4 goes to zero; this will never to occur if 1 2r r ).

Inverted slider-crank mechanism acceleration example – Term Example 3 continued

Given

1

2

4

1

0.20

0.10

0.32

0

r

r

L

(m)

2

4

4

70

150.5

0.191r

(deg and m)

2

4

4

25

2.47

2.18

r

(rad /s and m/s)

Snapshot Analysis

Given this mechanism position and velocity analyses, plus 2 0 rad /s2, calculate 4 4,r for this

snapshot.

4

4

0.870 0.094 16.873

0.493 0.166 48.957

r

4

4

9.46

267.14

r

(m/s2 and rad /s2)

4r is negative, so the slider link 3 is currently slowing down its positive velocity up link 4 and 4 is

positive, so the link 4 angular velocity is currently increasing in the ccw direction.



60


A more meaningful result from acceleration analysis is to report the acceleration analysis results

for the entire range of mechanism motion. The subplot below gives 4 (top, rad /s2) and 4

r (bottom,

m/s2), for all20 360 , for Term Example 3. Again, since 2 is constant, we can plot the

acceleration results vs. 2 (since it is related to time t via 2 2t ).

Term Example 3 F.R.O.M., 4 and 4r

As expected, 4 is zero at the beginning, middle, and end since the 4 curve flattens out at those

points. The maximum (and minimum) 4 values correspond to the greatest slopes for 4. 4r is

maximum (and minimum) at the beginning, middle, and end since the 4r curve is steepest at those points; 4r is zero when the 4r curve is flat, i.e.

2 60 .

0 50 100 150 200 250 300 350-1000

-500

0

500

1000

4

( r a d / s

2 )

0 50 100 150 200 250 300 350-50

0

50

100

2 (deg)

r 4 d d

( m / s

2 )



61


When one variable is the derivative of another, recall the relationships from calculus (thederivative is the slope of the above curve at each point; the integral is the area under the curve up to that point, taking into account the initial value). For example:

44

( )( )

d t t dt

4 40 4( ) ( )t t dt

44

( )( )

d t t

dt

4 40 4( ) ( )t t dt

0 50 100 150 200 250 300 350

160

180

200

4

( d e g )

0 50 100 150 200 250 300 350

-20

-10

0

10

4

( r a d / s )

0 50 100 150 200 250 300 350-1000

-500

0

500

1000

2 (deg)

4

( r a d / s 2 )



62

44

( )( )

dr t r t

dt 4 40 4( ) ( )r t r r t dt

44

( )( )

dr t r t

dt

4 40 4( ) ( )r t r r t dt

These plots are all from Term Example 3.

0 50 100 150 200 250 300 3500.1

0.15

0.2

0.25

r 4

( m

)

0 50 100 150 200 250 300 350

-2

-1

0

1

2

r 4

d

( m

/ s )

0 50 100 150 200 250 300 350-50

0

50

100

2 (deg)

r 4 d

d

( m

/ s 2 )



63

2.3.6 Multi-Loop Mechanism Acceleration Analysis

Thus far we have presented acceleration analysis for the single-loop four-bar, slider-crank, andinverted slider-crank mechanisms. The acceleration analysis for mechanisms of more than one loop ishandled using the same general procedures developed for the single loop mechanisms.

This section presents acceleration analysis for the two-loop Stephenson I six-bar mechanismshown below as an example multi-loop mechanism. It follows the position and velocity analyses for thesame mechanism presented earlier.


The bottom loop of the Stephenson I six-bar mechanism is identical to the standard four-bar

mechanism model and so the acceleration analysis solution is identical to the four-bar presented earlier.With the complete acceleration analysis of the bottom loop thus solved, the solution for the top loop isessentially another four-bar acceleration solution.

As in all acceleration analysis, the acceleration solution for multi-loop mechanisms is a linearanalysis yielding a unique solution (assuming the given mechanism position is not singular) for eachsolution branch considered. The position and velocity analyses must be complete prior to theacceleration solution.



64

Now let us solve the acceleration analysis problem for the two-loop Stephenson I six-barmechanism using the formal acceleration analysis steps presented earlier. Again, assume link 2 is theinput link.

Step 1. The Stephenson I six-bar mechanism Position and Velocity Analyses must first be complete.

Given r 1, 1 , r 2 , r 3 , r 4 , r 2a , r 4a , r 5 , r 6, 2, 4, 2, and 2, we solved for 3, 4, 5, 6, 3, 4, 5, 6.

Step 2. Draw the Stephenson I six-bar mechanism Acceleration Diagram.This should include all the information from the position and velocity diagrams, plus the new

acceleration information. For clarity, we show only the new acceleration information here. Refer to the previous Stephenson I position and velocity kinematics diagrams for complete information.

where i (i = 2,3,4,5,6) is the absolute angular acceleration of link i. Triangular links 2 and 4 each have

a single angular acceleration for the whole link. 0ir for all links since all links are of fixed length (no

sliders).


Given the mechanism r 1, 1 , r 2 , r 3 , r 4 , r 2a , r 4a , r 5 , r 6, 2, 4,the position analysis 2, 3, 4, 5, 6,

the velocity analysis 2, 3, 4, 5, 6,

1-dof acceleration input 2

Find the acceleration unknowns 3, 4, 5, 6



65

Step 4. Derive the acceleration equations. Take the first time derivative of both sides of each of thefour scalar XY velocity equations.

The Stephenson I six-bar mechanism velocity equations are given below.

XY scalar velocity equations

2 2 2 3 3 3 4 4 4

2 2 2 3 3 3 4 4 4

r s r s r s

r c r c r c

2 2 2 5 5 5 4 4 4 6 6 6

2 2 2 5 5 5 4 4 4 6 6 6

a a a a a a

a a a a a a

r s r s r s r s

r c r c r c r c

where2 2 2

4 4 4

a

a

The first time derivative of the Loop I velocity equations is identical to that for the standard four- bar mechanism.

2 2 2

2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4

2 2 2

2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4

r s r c r s r c r s r c

r c r s r c r s r c r s


2 2 23 3 4 4 3 2 2 2 2 2 2 3 3 3 4 4 4

2 2 2

3 3 4 4 4 2 2 2 2 2 2 3 3 3 4 4 4

r s r s r s r c r c r c

r c r c r c r s r s r s

The first time derivatives of the Loop II velocity equations are:

2 2 2 2

2 2 2 2 2 2 5 5 5 5 5 5 4 4 4 4 4 4 6 6 6 6 6 6

2 2 2 2

2 2 2 2 2 2 5 5 5 5 5 5 4 4 4 4 4 4 6 6 6 6 6 6

a a a a a a a a a a a a

a a a a a a a a a a a a

r s r c r s r c r s r c r s r c

r c r s r c r s r c r s r c r s


2 2 2 25 5 6 6 5 2 2 2 2 2 2 5 5 5 4 4 4 4 4 4 6 6 6

2 2 2 25 5 6 6 6 2 2 2 2 2 2 5 5 5 4 4 4 4 4 4 6 6 6

a a a a a a a a

a a a a a a a a

r s r s r s r c r c r s r c r c

r c r c r c r s r s r c r s r s

where we have used2 2

4 4

a

a

and

2 2

4 4

a

a

since 2 and 4 are constant angles.



66

Step 5. Solve the acceleration equations for the unknowns 3, 4, 5, 6.

The two loops decouple so we find 3 and 4 from Loop I first and then use 4 to find 5 and 6 from Loop II. The solutions are given below.

Loop I (identical to the standard four-bar mechanism)

1 2 2 23 3 4 43 2 2 2 2 2 2 3 3 3 4 4 4

2 2 23 3 4 44 2 2 2 2 2 2 3 3 3 4 4 4

r s r s r s r c r c r c

r c r c r c r s r s r s

Loop II (similar to the standard four-bar mechanism)

1 2 2 2 25 5 5 6 6 2 2 2 2 2 2 5 5 5 4 4 4 4 4 4 6 6 6

2 2 2 26 5 5 6 6 2 2 2 2 2 2 5 5 5 4 4 4 4 4 4 6 6 6

a a a a a a a a

a a a a a a a a

r s r s r s r c r c r s r c r c

r c r c r c r s r s r c r s r s

Remember, Gaussian elimination is more efficient and robust than the matrix inverse. Also, theseequations may easily be solved algebraically instead of using matrix methods.

Stephenson I six-bar mechanism singularity condition

The acceleration solution fails when the determinant of either coefficient matrix above goes tozero. The result is dividing by zero, resulting in infinite angular accelerations for the associated loop. Note the two coefficients matrices in the acceleration solutions are identical to those for the velocitysolutions. Therefore, the acceleration singularity conditions are identical to the velocity singularity

conditions.

For the first loop, the singularity condition is identical to the singularity condition of the standardfour-bar mechanism, i.e. when links 3 and 4 either line up or fold upon each other, causing a link 2 jointlimit. For the second loop, the singularity condition is similar, occurring when 5 and 6 either line up orfold upon each other. These conditions also cause problems for the velocity and position analyses, sothe acceleration singularities are known problems.



67

2.4 Other Kinematics Topics

2.4.1 Link Extensions Graphics

Using the methods presented thus far, we can use MATLAB to animate mechanisms for theentire range of motion. However, these methods have focused on the basic models shown below. Whatif your term project requires animation of modifications of these basic mechanisms with extensions fromthe existing rigid links?

Four-Bar Mechanism Offset Slider-Crank Mechanism

Four-bar mechanism link 3 extensions

Here are the kinematics equations (we already presented the point C kinematics equations in the ME3011 NotesBook).

3 3

3 3

cos( )

sin( )

x x CA C

y y CA C

c a r

c a r

C 3 3

3 3

cos( )

sin( )

x x DB D

y y DB D

d b r

d b r

D

where2 2

2 2

cos

sin

x

y

a r

a r

A

1 1 4 4

1 1 4 4

cos cos

sin sin

x

y

b r r

b r r

B

In this simple straight-line case, use3 180C

and3 0 D

. Here is partial MATLAB code for link 3

extensions animation.x2 = [0 ax(i)]; % coordinates of link 2 y2 = [0 ay(i)];x3 = [cx(i) dx(i)]; % coordinates of link 3 y3 = [cy(i) dy(i)];x4 = [r1x bx(i)]; % coordinates of link 4 y4 = [r1y by(i)];figure; plot(x2,y2,'r',x3,y3,'g',x4,y4,'b');

C

D

A

B



68

Four-bar mechanism link 4 extensions

The kinematics equations are given below.

4 4

4 4

cos( )

sin( )

x x CB C

y y CB C

c b r

c b r

C 4

4

4 4 4

4 4 4

cos( )

sin( )

x DO D x

y y DO D

O r d

d O r

D

In this simple straight-line case, use4 0C

and4 180 D

. b x and b y were given above and4

O is:

1 1

1 1

cos

sin

r

r

4O

Here is partial MATLAB code for link 4 extensions animation.x2 = [0 ax(i)]; % coordinates of link 2 y2 = [0 ay(i)];x3 = [ax(i) bx(i)]; % coordinates of link 3 y3 = [ay(i) by(i)];x4 = [cx(i) dx(i)]; % coordinates of link 4

y4 = [cy(i) dy(i)];figure; plot(x2,y2,'r',x3,y3,'g',x4,y4,'b');

Of course, one can use combinations of these graphics approaches as necessary. Also, use patch for drawing solid polygonal links rather than straight lines. You can also use patch for

drawing solid circles. This method is also extendable to non-straight-line links by using the appropriate

angles.

C

D

A

B

O4



69

2.5 Jerk Kinematics Analysis

Jerk is the time rate of change of the acceleration (and hence the second and third time rates ofchange of the velocity and position, respectively). Again, this jerk time rate of change may describe achange in magnitude of acceleration, a change in direction of acceleration, or both. What names have been given to the next three position derivatives after jerk? The answer is given somewhere in thissection. Jerk analysis is the fourth step in kinematics analysis. It is not required for standard Newton-Euler dynamics analysis. However, it is useful for the following items.

Input link motion specification

Cam motion profiles and cam design

Smooth motion control as in elevators

Jerk can be important for kinematic motion analysis in general. Position, velocity, andacceleration analyses must be completed first. Jerk is the first time derivative of the acceleration, thesecond time derivative of the velocity, and the third time derivative of the position. Like all the terms preceding it, jerk is also a vector quantity. There are both translational and rotational jerk terms.

2 3

2 3

( ) ( ) ( )( )

d A t d V t d P t J t

dt dt dt SI units: 3

m

s

2 3

2 3

( ) ( ) ( )( )

d t d t d t t

dt dt dt

SI units: 3

rad

s

2.5.1 Jerk Analysis Introduction

In this section we will derive the n-part jerk formula, showing the most general jerk terms possible for planar devices.

n-part Jerk Derivation Figure

This is a four-dof system consisting of a translating/rotating rigid rod with a slider. The samesystem was used for the two-part position, three-part velocity, and five-part accelerations formuladerivations in the ME 3011 NotesBook. Find the total jerk of the point P, which is on the slider.

Y

L

X

L, L, L

P

O

PO



70

Start with the five-part acceleration formula from before and take another time derivate of the XY components. What is n? (Hint – clearly n must be greater than 5.)

2

2

cos 2 sin sin cos

sin 2 cos cos sin

OX PP

OY

A A V L Ld A d J

dt dt A A V L L

Recall the two-part position, three-part velocity, and five-part acceleration formula results below.

2

2

cos

sin

cos sin

sin cos

2

cos 2 sin sin cos

sin 2 cos cos sin

P O

OX

OY

P O

OX

OY

P O

OX

OY

P P L

P L

P L

V V V L

V V L

V V L

A A A V L L

A A V L L

A A V L L

The angle , angular velocity , angular acceleration , and angular jerk are all changing with respectto time (only the planar case is this simple; the spatial rotation case is more complicated).

2 3

2 3

d d d

dt dt dt

The rod length L, sliding velocity V , sliding acceleration A, and sliding jerk J are all changing withrespect to time.

2 3

2 3

d A d V d L J

dt dt dt

Here are the same relationships, using the dot notation to indicate time differentiation.

2 3

2 3

d d d

dt dt dt

2 3

2 3

d L d L d L L

dt dt dt



71

Product and Chain Rules of Differentiation

Again, we’ll need to use the product and chain rules repeatedly in jerk analysis derivations.

Product rule

( ) ( )( ( ) ( )) ( ) ( )

d dx t dy t x t y t y t x t dt dt dt

x, y are both functions of time

Chain rule

( ( )) ( )( ( ( )))

d df x t dx t f x t

dt dx dt

f is a function of x, which is an implicit function of t

Examples

2

( 2 sin ) 2 sin 2 sin 2 (sin )

(sin )2 sin 2 sin 2

2 sin 2 sin 2 cos

d d V A V V

dt dt

d d A V V

d dt

A V V

2 2 2 2

22 2

2 3

( cos ) cos ( )cos (cos )

( ) (cos )cos cos

cos 2 cos sin

d d d L V L L

dt dt dt

d d d d V L L

d dt d dt

V L L

Many terms will combine (like in the Coriolis acceleration case). Check the resulting units of allcomponents to check your results. The full n-part jerk derivation is left to the interested student.



72

2.5.2 Mechanism Jerk Analys is

Generic Mechanism Jerk Analysis Problem Statement

Given the mechanism, complete position, velocity, and acceleration analyses, and one-dof of jerkinput, calculate the jerk unknowns.

For a given branch of a known mechanism, this will yield linear equations so a matrix-vectorapproach may be used to obtain the unique solution (assuming no singularity).

Since jerk kinematics is not required for dynamics analysis or machine design (generally), thissolution is beyond the scope of the class.

For our jerk needs, we may use the time differentiation approach of the previous subsection.

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-Old Kellogg’s Advertisement



73

2.6 Branch Symmetry in Kinematics Analysis

We have been doing F.R.O.M. analysis for the open-branch only (four-bar mechanism) and theright-branch only (slider-crank mechanism). What do the kinematics results look like for the crossedand left branches? Are there any relationships amongst the various analyses for the two branches? Thereader is left to draw their own conclusions.

2.6.1 Four-Bar Mechanism

Given:

1

2

3

4

10

4

8

7

r

r

r

r

1

3

2

0

4

0

10 (constant)

CAr

Open Branch Crossed Branch



74

Open Branch Crossed Branch



75

2.6.2 Slider-Crank Mechanism

Given: 2 3 3 22, 6, 0, 3, 0, 10 (constant)CA

r r h r

Right Branch Left Branch



76

We see a great deal of symmetry for both four-bar and slider crank mechanism kinematic

analyses. The coupler point curves and all plots have either horizontal midpoint 2 180 or vertical

zero point flip-symmetry (some have both).

However, for the x and x slider-crank mechanism results, this symmetry is not immediately

evident. This special symmetry is revealed when first cutting the plots at 2 180 , and then performing

the horizontal and/or vertical flipping.



77

2.7 Kinematics Analysis Examples

This section presents complete snapshot and full-range-of-motion (F.R.O.M.) examples todemonstrate the rotational and translational position, velocity, and acceleration kinematics analyses inthe ME 3011 NotesBook. Term Example 1 is for the four-bar mechanism and Term Example 2 is forthe slider-crank mechanism.

2.7.1 Term Example 1: Four-Bar Mechanism

Four-Bar Mechanism Position Analysis – Term Example 1

Given:

1

2

3

4

11.18

3

8

7

r

r

r

r

in

1

2

3

4

0.284

0.076

0.203

0.178

r

r

r

r

m

and 1 10.3

(Ground link is 11" over and 2" up). Also given 5CAr in (0.127 m) and 3 36.9

forthe coupler link point of interest.

Snapshot Analysis (for one given input angle2)

Given this mechanism and 230 , calculate 3 4, , , and

C P for both branches. Results:

0.076

0.005

0.036

E

F

G

branch t3 4 C P

open 1.799 53.8 121.7 67.9 (0.064, 0.165)

crossed –1.570 313.0 245.0 67.9 (0.191, 0.016)

These two branch solutions are demonstrated in the figures below. The length units are m. Note

is identical for both branches due to the conventions presented earlier.



78

Open Branch (t(2))

Crossed Branch (t(1))

Term Example 1 Snapshot

-0.1 0 0.1 0.2 0.3

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

X (m)

Y ( m )

-0.1 0 0.1 0.2 0.3

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

X (m)

Y ( m )



79

Full-Range-Of-Motion (F.R.O.M.) Analysis – Term Example 1

A more meaningful result from position analysis is to solve and plot the position analysisunknowns for the entire range of mechanism motion.

The plot below gives 3 (red ), 4 (green), and ( blue), all deg, for all 20 360 , for Term

Example 1, the open branch only.

3, 4, and

0 50 100 150 200 250 300 35020

40

60

80

100

120

140

160

180

2 (deg)

i

( d e g )

3

4



80

The plot below gives the initial (and final) animation position, for 20,360 . It also gives the

coupler curve for the open branch, plotting PCY vs. PCX in green circles.

C P

Coupler Curve

Term Example 1 F.R.O.M. Posi tion Results

-0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

X (m)

Y ( m )



81

Four-bar mechanism velocity example – Term Example 1 continued

Given r 1 = 0.284, r 2 = 0.076, r 3 = 0.203, r 4 = 0.178, r C A = 0.127 m, and 110.3 , 2

30 ,

353.8 , 4

121.7 , 336.9 . This is the open branch of the Term Example 1 four-bar mechanism.

Snapshot Analysis

Given this mechanism position analysis plus 2 20 rad /s (positive, which indicates ccw),

calculate 3 4, , and C V for this instant in motion (snapshot).

3

4

0.164 0.151 0.760

0.120 0.094 1.316

3

4

8.073

3.729

Both 3 and 4 are negative, so they are in the cw direction for this snapshot. These results are theabsolute angular velocities of links 3 and 4 with respect to the ground link. The associated coupler pointtranslational velocity vector for this snapshot is:

0.265

1.330C V

(m/s)



82

Full-Range-Of-Motion (F.R.O.M.) Analysis: Term Example 1 continued

A more meaningful result from velocity analysis is to solve and plot the velocity analysisunknowns for the entire range of mechanism motion.

The plot below gives 3 (red ) and 4 (green) (rad /s) for all 20 360 , for Term Example 1,

for the open branch only. For all of Term Example 1, assume the 2 given above is constant. Since 2 is constant, we can plot the velocity results vs. 2 (since it is related to time t via 2 2t ).

3 and 4

0 50 100 150 200 250 300 350-10

-8

-6

-4

-2

0

2

4

6

8

10

2 (deg)

i

( r a d / s )

3

4



83

The plot below gives the absolute translational coupler point C velocity for all 20 360 , for

Term Example 1, for the open branch only.

C V

Term Example 1 F.R.O.M. Velocity Results

0 50 100 150 200 250 300 350-1.5

-1

-0.5

0

0.5

1

1.5

2

2 (deg)

V C

( m / s )

VCX

VCY



85


A more meaningful result from acceleration analysis is to solve and plot the acceleration analysisunknowns for the entire range of mechanism motion.

The plot below gives 3 (red ) and 4 (green), (rad /s2), for all 20 360 , for Term Example

1, open branch. In the Term Example 1 velocity section it was assumed that the given 2 is constant,which means that the given 2 is always zero. Since 2 is constant, we can plot the acceleration results

vs. 2 (since it is related to time t via 2 2t ).

3 and 4

0 50 100 150 200 250 300 350-250

-200

-150

-100

-50

0

50

100

150

200

250

2 (deg)

i

( r a d / s

2 )

3

4



86

The plot below gives the translational coupler point acceleration for all 20 360 , for Term

Example 1, the open branch only.

C A

Term Example 1 F.R.O.M. Acceleration Results

0 50 100 150 200 250 300 350

-30

-20

-10

0

10

20

30

2 (deg)

A C

( m / s 2 )

ACX

ACY



87


When one variable is the derivative of another, recall the relationships from calculus. e.g.:

44

44

( )( )

( )( )

d t t

dt

d t t dt

4 40 4

4 40 4

( ) ( )

( ) ( )

t t dt

t t dt

The value of 4 at any point is the slope of the 4 curve at that point. The value of 4 at any pointis the integral of the 4 curve up to that point (the value of 4 at any point is the area under the 4 curve

up to that point plus the initial value 40). A similar relationship exists for 4 and 4.

These graphs are plotted vs. 2, but the same type of relationships hold when plotted vs. time t

since 2 is constant. This is the Term Example 1 F.R.O.M. result, but 4 was changed from deg to rad for better comparison.

These curves should be plotted vs. time t instead of 2 in order to see the true slope and areavalues accurately.

0 50 100 150 200 250 300 3502

2.5

3

4

( r a d )

0 50 100 150 200 250 300 350-10

-5

0

5

10

4

( r a d / s )

0 50 100 150 200 250 300 350-200

-100

0

100

200

300

4

( r a d / s 2 )

2 (deg)



88

2.7.2 Term Example 2: Slider-Crank Mechanism

Slider-Crank Mechanism Position Analysis – Term Example 2

Given:

2

3

4

8

3

r

r

h

in

2

3

0.102

0.203

0.076

r

r

h

m

Snapshot Analysis (one input angle)

Given this mechanism and 230 , calculate x and 3 for both branches. Results:

branch x ( m) 3 ( deg)

right 0.290 7.1

left –0.113 172.9

These two branch solutions are demonstrated in the figures below (length units are m).

Right Branch Left Branch

Term Example 2 Snapshot

-0.1 0 0.1 0.2 0.3

-0.2

-0.1

0

0.1

0.2

X (m)

Y

( m )

-0.1 0 0.1 0.2 0.3

-0.2

-0.1

0

0.1

0.2

X (m)

Y ( m )



90

The plot below gives the initial (and final) animation position, for 20,360 . It also gives the

coupler curve to scale for the right branch, plotting PCY vs. PCX in green. In this case the coupler point istaken to be the midpoint of coupler link 3.

C P Coupler Curve

Term Example 2 F.R.O.M. Posi tion Results

For Term Example 2, the slider translation limits are 0.067 0.295 x , as seen in the x plot

above, calculated from the equations in the 3011 NotesBook.

-0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

X (m)

Y ( m )



91

Slider-crank mechanism velocity example – Term Example 2 continued

Given r 2 = 0.102, r 3 = 0.203, h = 0.076 m, and 2 30 , 3 7.2 , x = 0.290 m. This is the right

branch of the slider-crank position example of Term Example 2.


Given this mechanism position analysis plus 2 15 rad /s (+ so ccw), calculate3 , x for this

instant in time (snapshot).

3

3

0.025 1 0.762

0.202 0 1.320

6.577

0.601

x

x

These results are the absolute rotational and translational velocities of links 3 and 4 with respect to thefixed ground link. Both are negative, so the coupler link 3 is currently rotating in the clockwise directionand the slider 4 is currently translating to the left .



92


A more meaningful result from velocity analysis is to solve and plot the velocity analysis

unknowns for the entire range of mechanism motion. The subplot arrangement below gives 3 (top,

rad /s) and x (bottom, m/s), for all 20 360 , for Term Example 2, the right branch only. For all of

Term Example 2, assume the 2 given above is constant. Since 2 is constant, we can plot the velocity

results vs. 2 (since 2 changes linearly, as it is related to time t via 2 2t ).

Term Example 2 F.R.O.M. Velocity Results, 3 and x

0 50 100 150 200 250 300 350-10

-5

0

5

3

( r a d / s )

0 50 100 150 200 250 300 350-2

-1

0

1

2

3

2 (deg)

x d ( m / s )



93

Slider-crank mechanism acceleration example – Term Example 2 continued

Given r 2 = 0.102, r 3 = 0.203, h = 0.076 m, and 2 30 , 3 7.2 , x = 0.290 m; and 2 15 ,

3 6.55 rad /s, 0.60 x m/s. This is the right branch of the position and velocity example for the

slider-crank mechanism of Term Example 2.


Given this mechanism position and velocity analysis plus 2 0 rad /s2, calculate 3 , x for this

snapshot in time.

3

3

0.025 1 28.590

0.202 0 12.557

62.329

30.148

x

x

These results are the absolute angular and linear accelerations of links 3 and 4 with respect to the fixedground link.



94


A more meaningful result from acceleration analysis is to solve and plot the acceleration analysis

unknowns for the entire range of slider-crank mechanism motion. The top plot gives 3 (rad /s2) and the

bottom plot gives x (m/s2), for all 20 360 , for Term Example 2, for the right branch only. In the

Term Example 2 velocity section it was assumed that the given 2 is constant, which means that the

given 2 is always zero. Since 2 is constant, we can plot the velocity results vs. 2 (since it is related totime t via 2 2t ).

Term Example 2 F.R.O.M. Acceleration Resul ts, 3 and x

0 50 100 150 200 250 300 350-250

-200

-150

-100

-50

0

50

100

150

3

( r a d / s

2 )

0 50 100 150 200 250 300 350-50

0

50

2 (deg)

x d d ( m / s

2 )



95

3. Dynamics Analysis

3.1 Dynamics Introduction

D’Alembert’s Principle

We can convert dynamics problems into statics problems by the inclusion of a vector inertialforce 0 GF mA and a vector inertial moment 0 G M I . Centrifugal force 2

mr , directed away

from the center of rotation, is an example of an inertial force vector. It’s not really a force but a felteffect of an inertia in acceleration. Using D’Alembert’s principle, the right-hand side of the translationaland rotational dynamics equations is subtracted to the other side of the equation. Then the forces andmoments balance to zero as in statics, when the inertial forces are included in the FBD.

We won’t use this method, but it is mentioned for completeness. We would instead to prefer toconsider statics problems as a subset of dynamics problems, with zero accelerations.

0

0

G

O

R mA

R F

0

0

G

O

T r R I

T r R M

Dynamics Humor from xkcd

xkcd.com



96

3.2 Mass, Center of Gravity, and Mass Moment of InertiaThis section presents a thorough review of mass, center of gravity, and mass moment of inertia,

for use in the translational and rotational dynamics equations for each FBD.

GF m A GZ G M I

Newton’s Second Law Euler’s Rotational Dynamics Equation

Newton’s Second Law requires m and CG, Euler’s Rotational Dynamics Equation requires CG and I GZ .

m PCG I GZ

translational mass center of gravityrotational center of gravity mass moment of inertia

Mass

In Newton’s Second LawGF m A , the mass m is the proportionality constant. Mass is

measure of translational inertia – a resistance to change in motion according to Newton’s First Law.

Mass is also a measure of the storage of translational kinetic energy21

2

T KE mv and the units are kg.

Examples for m, CG, I G

System of particles General rigid body

Rectangular rigid body

Mass calculation

System of particles General rigid body

1

N

i

i

m m

body

m dm


usingm

V , dm dV , so

body body

m dm dV

/ 2 / 2

/ 2 / 2

b h

b hm tdxdy tbh

m V (an obvious result)



97

Center of Gravity (CG, G) The CG is the point at which a body is balanced with respect to gravity. It is also the point at

which the body weight acts. The CG is also called the mass center, center of mass, and centroid. It is avector quantity and the units are length units, m.

CG calculation

System of particles

ii

CG

i

m r P

m

Cartesian components

i i

CG

i

i i

CG

i

m x X X

m

m yY Y

m

General rigid body

body

CG

rdm

Pdm

Cartesian components

xCG

y

CG

xdm

X X dm

ydm

Y Y dm

Rectangular rigid body Using an XY coordinate frame at the geometric center, the CG is calculated below.

/ 2

/ 2

/ 2

/ 2

/ 22

/ 2

2 2

2

02 4 4

x

x

b

b

b

b

b

b

xdm

X dm

xdV m

xthdxm

th xdx

m

th x

m

th b bm

/ 2

/ 2

/ 2

/ 2

/ 22

/ 2

2 2

2

02 4 4

y

y

h

h

h

h

h

h

ydm

Y dm

ydV m

ytbdym

tb ydy

m

tb y

m

tb h hm

0

0CG

X P

Y

For a homogeneous, regular geometric body of uniform thickness, the CG is the geometric center.



98

Mass Moment of Inertia I G

This is not the same as area moment of inertia ( I A) for beam bending, which is recalled below.

2

Ax

y

I y dA 2

Ay

x

I x dA units: 4

A I m

In Euler’s rotational dynamics equation GZ G M I , the mass moment of inertia I GZ is the proportionality constant. I GZ is also a measure of rotational inertia, i.e. the resistance to change inrotational motion according to Newton’s First Law. Also, it is a measure of how hard it is to accelerateabout certain axes in rotation. I GZ is also a measure of the storage of rotational kinetic energy,

21

2 R GZ

KE I , and its units are kg-m2.

What is the mass moment of inertia, a scalar, vector, matrix, or something else? Answer – it is a tensor.

Mass Moment of Inertia I G calculation

System of particles2

axis i i I m r

where r i is the scalar perpendicular distance from the axis to the ith particle. With squaring, all terms

will be positive, so there can be no canceling like for the CG. If the first moment (CG) is balanced, thesecond moment ( I GZ ) terms do not cancel since the squared terms are all positive.

General rigid body inertia tensor (symmetric)

XX XY XZ

axis XY YY YZ

XZ YZ ZZ

I I I I I I I

I I I

2 2( ) XX

body

I y z dm 2 2( )

YY

body

I x z dm 2 2( )

ZZ

body

I x y dm

What is the only term that matters for XY planar motion? Answer – I ZZ .

In the yardstick example:GZ GY GX

I I I also OZ GZ I I



99


Using an XY coordinate frame at the CG, I GZ is calculated below.

/2 /22 2 2 2

/2 /2

/23/22

/2/2

3 3/2

2

/2

/23 3 3

/22

/2/2

3 3

( ) ( )

3

1

2 2 3 8 8

12 3 12

3 8

b h

GZ b h

body

hb

bh

b

b

bb

bb

I x y dm x y tdxdy

yt x y dx

h h h ht x dx

h hx h xt hx dx t

h b bt

3

3 32 2

8 12 2 2

( )12 12 12

h b b

b h bh tbht b h

2 2( )12

GZ

m I b h (using m V tbh )

units: mass times distance squared, kg-m2

This formula agrees with the result given in tables.

How do we find mass, center of gravity, and mass moment of inertia in the real world?

From tables – for example, see the three tables at the end of this section.

CAD packages (such as SolidEdge and AutoCAD) calculate m, CG, and I GZ automatically foreach link drawn, once material is associated with the 3D model.



100

Parallel Axis Theorem

The mass moment of inertia through the CG is related to mass moments of inertia of parallelaxes through different points as follows.

2

ZZO ZZG I I md

where d is the scalar distance separating the axis of interest O from the axis through the CG. Notice I ZZG

is as small as possible. Any I ZZO must be greater, due to the term md

2

which is always positive.

Parallel axis theorem example

Rectangular rigid body (where axis O is the corner)

2

2 22 2

2 2 2 2

2 2

2 2

( )12 4 4

12 4 12 4

3 3

( )3

ZZO ZZG I I md

m b hb h m

b b h h

m

b hm

mb h

Combining multiple rigid bodies into a rigid link

To combine any number of bodies n of known material, shape, size, and location into one rigid

body, use the following equations for mass, center of gravity, and mass moment of inertia.

1 2

1

n

T n i

i

m m m m m

1 1 2 2 1

1 1 2 2

1

n

i i

n n i

T T T

CGT nn nT

i i

iT

T

m xm x m x m x

m m X P

m y m y m yY m y

m

m

2 2 2 21 1 1 2 2 2

1

n

GZT GZ GZ GZn n n GZi i i

i

I I m d I m d I m d I m d

The subscript ‘T ’ indicates total (or combined) and d i is the distance between the combined CG (T CGP )

and the CG of body i ( i

T

CG i iP x y ). These equations are obtained from the mass, CG, and I GZ

equations for particles, where now each particle is instead a rigid body.



101

Example 1. Two rectangles joined as shown below

Given b1 = 2.2 x h1 = 0.1 x t 1 = 0.005 (m)b2 = 1.0 x h2 = 0.8 x t 2 = 0.005 (m)

the material is steel with a mass density of = 7850 kg/m3

The equations are:

1 2T m m m

1 1 2 2

1 1 2 2

T T

CGT

T

T

m x m x

m X P

m y m yY

m

2 2

1 1 1 2 2 2GZT GZ GZ I I m d I m d

where d i is the distance from CGi to the combined CG CGT P .

The results are:m1 = 8.64 m2 = 31.40 mT = 40.04 (kg)

1

0.05

1.10CGP

2

0.60

1.80CGP

0.48

1.65CGT P

(m)

d 1 = 0.70 d 2 = 0.19 (m)

2 211 1 1 3.49

12GZ

m I b h

2 222 2 2 4.29

12GZ

m I b h I GZT = 13.15 (kgm

2)

Example 1. Two Unequal Steel Rectangles (here b1 is vertical and b2 is horizontal)

0 0.5 1 1.5 20

0.5

1

1.5

2

X (m)

Y ( m )



102

Example 2. Two equal rectangles joined in the same manner

Given b = 2.0 x h = 0.2 x t = 0.005 (m) (twice)

= 7850 kg/m3

The analytical equations for this special case are:

2T

m m 3

4

3

4

T

CGT

T

b h X P

b hY

2 25

( )12

GZT I m b h

where2 2

1 28

b hd d

is the distance from CGi to CGT P .

The results are:m1 = m2 = 15.70 mT = 31.40 (kg)

1

0.10

1.00CGP

2

1.20

1.90CGP

0.65

1.45CGT P

(m)

d 1 = d 2 = 0.71 (m)

2 2

1 2 ( ) 5.2912

GZ GZ

m I I b h I GZT = 26.43 (kgm

2)

Example 2. Two Equal Steel Rectangles(again, b1 is vertical and b2 is horizontal, with b1 = b2 = b here)

0 0.5 1 1.5 20

0.5

1

1.5

2

X (m)

Y ( m )



103

Tables of Mass, Center of Gravity, and Mass Moment of Inertia

The tables below present the mass (kg), center of gravity (m), and mass moment of inertia (kg-m

2) for some common planar link shapes.

Note that mass moment of inertia (kg-m2) is not the same as area moment of inertia for beam bending (m

4). The former represents rotational inertia while the latter is a measure of resistance to beam

bending.

We assume that all link materials are homogeneous and uniformly distributed, with mass density

(kg/m3), all links have regular geometry, and all links have a constant thickness t in the Z direction.

The general equations for mass, center of gravity, and mass moment of inertia are given belowfor general rigid bodies.

2 2( )

body

body

CG

ZZ

body

m dm

rdm

Pdm

I x y dm

All of these terms require double integrals over the rigid body in the XY plane. Both center ofgravity and mass moment of inertia depend on the origin of the chosen XYZ Cartesian coordinatesystem.

For general 3D rigid bodies, mass moment of inertia is a 3x3 tensor. For planar mechanismdynamics we only need one scalar term out of these 9 terms, I ZZ .

In the drawings below, the planar reference Cartesian coordinate system is shown, with origin O,and the standard symbol is used for center of gravity, denoted as point G. The mass moment of inertiaabout axis O is related to the mass moment of inertia about axis G is via the parallel axis theorem, whered is the scalar distance (vector length) between axes G and O in the XY plane:

2

OZ GZ

I I md



104

Mass Properties for Planar Links

Name Model Mass ( kg)Center of

Gravity ( m)

Mass Moment of

Inertia ( kg- m2)

point massm 0

0

0

point mass onmassless rod m

0

L

2

0GZ

OZ

I

I mL

slender rodm

2

0

L

2

2

12

3

GZ

OZ

mL

I

mL I

rectangular parallelopiped Lht

2

2

L

h

2 2

2 2

( )

12

( )

3

GZ

OZ

m L h I

m L h I

square2s t

2

2

s

s

2

2

6

2

3

GZ

OZ

ms I

ms I

X

Y

O

m

X

Y

m L

O

X

Y

L

O

X

Y L

O

h

X

Y

O

s

s



105

Mass Properties for Planar Links (continued)

Name Model Mass ( kg)

Center of

Gravity

( m)

Mass Moment of Inertia

( kg- m2)

cylinder2 R t

R

R

2

22

5

2

GZ

OZ

mR I

mR I

hollowcylinder 2 2

o i R R t

0

0

R

R

2 2

2 2

( )

2

(5 )

2

o iGZ

o iOZ

m R R I

m R R I

thin ringm

R

R

2

23

GZ

OZ

I mR

I mR

triangle

2

bht

3

3

a b

h

2 2 2

2 2 2

( )

18

( )

6

GZ

OZ

m a ab b h I

m a ab b h I

The formula for the mass moment inertia of a triangle was derived via double integral over the body by Ohio University Ph.D. student Elvedin Kljuno – it could not be found in any sophomore-leveldynamics book, or in any other textbook, nor in any Internet search.

X

Y

O

R

X

Y

O

Ri

Ro

X

Y

O

R

X

Y

O

h

a

b



107

English Units for Mass

ME 3011 uses SI units exclusively. However, many of you perform your capstone project workusing English units, which is fine, since we live in the U.S. In the 1970s the U.S. government mandateda change to the SI system – this failed spectacularly (why?).

One big benefit of the SI system is seen in the units for Newton’s Second Law, F = ma. Usingstandard SI units, this equation uses all ones (1s):

1 Newton accelerates 1 kg 1 m/s2

Sadly the English units DO NOT behave with ones in Newton’s Second Law:

1 lbf DOES NOT accelerate 1 lbm 1 ft/s2

1 lbf DOES NOT accelerate 1 lbm 1 in/s2

Further, the English system has another confusion which does not exist for the SI system. Thesame unit, pound (lb), applies both to force (lbf ) and mass (lbm), depending on the context. Pleasealways use the correct subscript for clarity. Happily, a mass of 1 lbm does weigh 1 lbf at standard gravity(g = 32.2 ft/s2 or 386.1 in/s2).

Now we present the standard English mass units; there are two, depending on if you use feet orinches for the length unit.

1 lbf accelerates 1 slug 1 ft/s2

1 lbf accelerates 1 blob 1 in/s2

WTF?!? slug? blob? I promise you I am not making this up. A slug is a rather large mass,equivalent to 32.2 lbm (14.6 kg). A blob is even larger, equivalent to 12 slugs, 386.1 lbm (175.1 kg).

In conclusion, do not use lbm in dynamics equations for your project. Instead use slugs if you areusing feet or blobs if you are using inches. If you have estimated your masses in lbm, simply divide by32.2 to get slugs, or divide by 386.1 to get blobs.

Finally, from the above we have the following units equivalents:

f 2

slug ft1 lb 1s

so2

f lb s1 slug 1

ft

f 2

blob in1 lb 1

s

so

2

f lb s1 blob 1

in



108

3.4 Four-Bar Mechanism Inverse Dynamics Analysis

Here are the link 2 and 4 details for the four-bar inverse dynamics matrix from the NotesBook.

Link 2 details Link 4 details

2 2 212

12

2 2 212

12 2 2

32 12 2

12 2 2

cos( )

sin( )

cos

sin

G X

GY

X

Y

Rr r

Rr

r r r r r

r r

4 4 414

14

4 4 414

14 4 4

34 14 4

14 4 4

cos( )

sin( )

cos

sin

G X

GY

X

Y

Rr r

Rr

r r r r r

r r

22 2 2 2 2 2 2 2 2

2 22 2 2 2 2 2 2 2 2

sin( ) cos( )

cos( ) sin( )

G X G G

G

G Y G G

A R R A

A R R

24 4 4 4 4 4 4 4 4

4 24 4 4 4 4 4 4 4 4

sin( ) cos( )

cos( ) sin( )

G X G G

G

G Y G G

A R R A

A R R

4

4

4 4 4

44 4 4

44

4

4 4

cos

sin

cos

sin

E

E

E X E E

E E Y E E

E r E X

E

E Y E r

F F F

F F

r r r

r r

4 4ˆ

E E M M k



109

Four-bar mechanism inverse dynamics matrix equation

21

21

12 12 32 32 32

32

43

23 23 43 43 43

14

14

34 34 14 14 2

1 0 1 0 0 0 0 0 0

0 1 0 1 0 0 0 0 0

0 0 0 0 1

0 0 1 0 1 0 0 0 0

0 0 0 1 0 1 0 0 0

0 0 0 0 0

0 0 0 0 1 0 1 0 0

0 0 0 0 0 1 0 1 0

0 0 0 0 0

X

Y

Y X Y X X

Y

X

Y X Y X Y

X

Y

Y X Y X

F

F

r r r r F

F

F

r r r r F

F

F

r r r r

2 2

2 2

2 2

3 3 3

3 3 3

3 3 3 3 3 3 3

4 4 4

4 4 4

4 4 4 4 4 4 4

( )

( )

( )

G X

G Y

G Z

G X E X

G Y E Y

G Z E X E Y E Y E X E

G X E X

G Y E Y


m A

m A g

I

m A F

m A g F

I r F r F M

m A F

m A g F

I r F r F M

A v b

Step 6. Solve for the unknowns (alternate solution)

It is possible to partially decouple the solution to this problem1. If we consider the FBDs of onlylinks 3 and 4 first, this is 6 equations in 6 unknowns – this is verified by looking at the original 9x9matrix and noting three 6x1 columns of zeros (1, 2, 9) in rows 4 through 9. Here is a more efficientsolution. The reduced 6x6 set of equations for links 3 and 4 are given below.

3 3 332

3 3 332

3 3 3 3 3 3 323 23 43 43 43

4 4 443

14

34 34 14 14 14

1 0 1 0 0 0

( )0 1 0 1 0 0

0 0

0 0 1 0 1 0

0 0 0 1 0 10 0

G X E x X

G Y E yY

G Z E x E y E y E x E Y X Y X X

G X E xY

X

Y X Y X Y

m A F F

m A g F F

I r F r F M r r r r F

m A F F

F r r r r F

4 4 4

4 4 4 4 4 4 4

( )G Y E y

G Z E x E y E y E x E

m A g F I r F r F M

34 34 34 A v b

Solve for six unknowns 1

34 34 34v A b

and then use F 32 X and F 32Y in the following 3x3 set of linear

equations, from the link 2 FBD, very similar to the single rotating link.

21 2 2 32

21 2 2 32

12 12 2 2 2 32 32 32 32

1 0 0

0 1 0 ( )1

X G X X

Y G Y Y

Y X G Z Y X X Y

F m A F

F m A g F r r I r F r F

1 R.L. Williams II, 2009, “Partial Decoupling of the Matrix Method for Planar Mechanisms Inverse Dynamics”, CDProceedings of the ASME International Design Technical Conferences, 33rd Mechanisms and Robotics Conference, Paper #DETC2009-87054, San Diego CA, August 30-September 2.



110

We do not need a matrix solution here since the X and Y force equations are decoupled. The solution is:

21 2 2 32

21 2 2 32

2 2 2 32 32 32 32 12 21 12 21

( )

X G X X

Y G Y Y

G Z Y X X Y Y X X Y

F m A F

F m A g F

I r F r F r F r F

Matrix inversion requires approximately33

log

n

n and Gaussian elimination requires approximately

22( 1)

3

n nn

multiplications/divisions2.

Number of Multiplications/Divisions for Four-bar Inverse Dynamics Solution

Method Inversion Gaussian Reduction

9x9 2292 321 86%

6x6 plus decoupled link 2 840 113 87%

Reduction in cost 63% 65%

There is a substantial 65% reduction in computational cost for Gaussian elimination with the6x6 plus decoupled link 2 method. Also, the numerical accuracy may also improve with this methodsince we needn’t do unnecessary calculations with the three 6x1 columns of zeros.

2 E.D. Nering, 1974, Elementary Linear Algebra, W.B. Saunders Company, Philadelphia: 38-39.



111

3.5 Slider-Crank Mechanism Inverse Dynamics Analysis

Slider-crank mechanism inverse dynamics matrix equation

21 2 2

21 2 2

3212 12 32 32

32

43

4323 23 43 43

14

2

1 0 1 0 0 0 0 0

( )0 1 0 1 0 0 0 0

0 0 0 1

0 0 1 0 1 0 0 0

0 0 0 1 0 1 0 0

0 0 0 0

0 0 0 0 1 0 0

0 0 0 0 0 1 1 0

X G X

Y G Y

X GY X Y X

Y

X

Y Y X Y X

Y

F m A

F m A g

F I r r r r

F

F

F r r r r

F

2 2

3 3 3

3 3 3

3 3 3 3 3 3 3

4 4 4

4 4

( )

Z

G X E X

G Y E Y


G X E X

E Y

m A F

m A g F

I r F r F M

m A F

m g F

A v b

Step 6. Solve for the unknowns (alternate solution)

Like the four-bar mechanism, it is possible to partially decouple the solution to this problem1. Ifwe consider the FBDs of only links 3 and 4 first, this is 5 equations in 5 unknowns – this is verified bylooking at the original 8x8 matrix and noting three 5x1 columns of zeros (1, 2, 8) in rows 4 through 8.Here is a more efficient solution. The reduced 5x5 set of equations for links 3 and 4 are given below.

3 3 332

3 3 332

3 3 3 3 3 3 34323 23 43 43

4 4 443

4 414

1 0 1 0 0

( )0 1 0 1 0

0

0 0 1 0

0 0 0 1 1

G X E x X

G Y E yY

G Z E x E y E y E x E X Y X Y X

G X E xY

E yY

m A F F

m A g F F

I r F r F M F r r r r

m A F F

m g F F

34 34 34 A v b

Solve for five unknowns 1

34 34 34v A b

and then use F 32 X and F 32Y in the following 3x3 set of

linear equations, from the link 2 FBD.

21 2 2 32

21 2 2 32

12 12 2 2 2 32 32 32 32

1 0 0

0 1 0 ( )

1

X G X X

Y G Y Y

Y X G Z Y X X Y

F m A F

F m A g F

r r I r F r F



112

We do not need a matrix solution here since the X and Y force equations are decoupled. The solution isidentical to that of the four-bar mechanism.

21 2 2 32

21 2 2 32

2 2 2 32 32 32 32 12 21 12 21

( )

X G X X

Y G Y Y

G Z Y X X Y Y X X Y

F m A F

F m A g F

I r F r F r F r F


log

n


22( 1)

3

n nn

multiplications/divisions (Nering, 1974).

Number of Multiplications/Divisions for Slider-Crank Inverse Dynamics Solution


8x8 1701 232 86%



There is a substantial 69% reduction in computational cost for Gaussian elimination with the5x5 plus decoupled link 2 method. Also, the numerical accuracy may also improve with this methodsince we needn’t do unnecessary calculations with the three 5x1 columns of zeros.



113

3.6 Inverted Slider-Crank Mechanism Inverse Dynamics Analysis

This problem can be solved with a 9x9 matrix, after eliminating a redundant equation. Let’s trya simpler approach – assume the link 3 mass is small and use only FBDs for links 2 and 4. We furtherassume zero external forces and moments.

Step 1. The inverted slider-crank Position, Velocity, and Acceleration Analyses must be complete.

Step 2. Draw the Inverted Slider-Crank Mechanism Free-Body Diagrams

ijF unknown vector internal joint force of link i acting on link j.

ijr known vector moment arm pointing to the joint connection with link i from the CG of link j.



114


Given r 1, 1 = 0 , r 2

2, r 4, 4

2, 4r , 4

2, 4r , 4

assume zero external forces and moments

Find 21F , 42F , 14F and 2

Step 4. Derive the Newton-Euler Dynamics Equations.

Newton's Second Law

Link 2

2 42 21 22 2 GF F F W m A

Link 4

4 14 42 44 4 GF F F W m A

Euler's Rotational Dynamics Equation

Link 2

2 42 42 12 212 22G z G Z M r F r F I

Link 4

4 14 14 24 42 44G z G Z M r F r F I

Count the number of unknowns and the number of equations: 6 scalar equations and 7 scalar unknowns.

We need an additional equation; let us assume zero friction between links 2 and 4. Therefore, 42F is

always perpendicular to link 4 and there is only one unknown from this vector, the magnitude F 42.

42 42 4

42

42 42 4

cos( 2)sin( 2)

X

Y

F F F F F



115

Step 5. Derive the XYZ scalar dynamics equations from the vector dynamics equations.

Link 2

42 21 2 2

42 21 2 2

2 42 42 42 42 12 21 12 21 2 2

( )

( ) ( )

X X G X

Y Y G Y

X Y Y X X Y Y X G Z

F F m A

F F m A g

r F r F r F r F I

Link 4

14 42 4 4

14 42 4 4

14 14 14 14 24 42 24 42 4 4

X X G X

Y Y G Y

X Y Y X X Y Y X G Z

F F m A

F F m A g

r F r F r F r F I

Express these scalar equations in matrix/vector form. The simplified inverted slider-crank mechanisminverse dynamics matrix equation is given below.

2 221

2 221

2 212 12 42 42 42

4 414

4 414

4 424 24 14 14 2

1 0 0 0 0

( )0 1 0 0 0

0 0 1

0 0 1 0 0

( )0 0 0 1 0

0 0 0

G X X

G Y Y

G Z Y X X Y

G X X

G Y Y

G Z X Y Y X

m Ac F

m A gs F

I r r r s r c F

m Ac F

m A gs F

I r s r c r r

where:4

4

cos( 2)

sin( 2)

c

s

A v b



116

Step 6. Solve for the unknowns The coefficient matrix [ A] is dependent on the mechanism geometry (i.e. the angles from the

position kinematics solution). The right-hand-side vector {b} is dependent on inertial terms and gravity.

Matrix/vector solution 1

v A b

MATLAB v = inv(A)*b; % Solution via matrix inverse

Using Gaussian elimination is more efficient and robust.

MATLAB v = A\b; % Solution via Gaussian elimination

The solution to the internal forces and input torque are contained in the components of v. To save these

values for later plotting, use the following MATLAB code, inside the i loop.

f21x(i) = v(1);f21y(i) = v(2);

tau2(i) = v(6);

Like the four-bar and slider-crank mechanisms, it is possible to partially decouple the solution tothis problem. If we consider the FBDs of only 4 first, this is 3 equations in 3 unknowns – this is verified by looking at the original 6x6 matrix and noting three 3x1 columns of zeros (1, 2, 6) in rows 4 through6. Here is a more efficient solution. The reduced 3x3 set of equations for link 4 is given below.

42 4 4

14 4 4

24 24 14 14 14 4 4

1 0

0 1 ( )

G X

X G Y

X Y Y X Y G Z

c F m A

s F m A g

r s r c r r F I

4 4 4 A v b

Solve for three unknowns 1

4 4 4v A b

and then use F 42 in the following 3x3 set of linear equations,

from the link 2 FBD.

21 2 2 42

21 2 2 42

12 12 2 2 2 42 42 42

1 0 0

0 1 0 ( )

1 ( )

X G X

Y G Y

Y X G Z X Y

F m A cF

F m A g sF

r r I r s r c F


21 2 2 42

21 2 2 42

2 2 2 12 21 12 21 42 42 42

( )

( )

X G X

Y G Y

G Z Y X X Y X Y

F m A cF

F m A g sF

I r F r F r s r c F



117


log

n


22( 1)

3

n nn


Number of Multiplications/Divisions


6x6 833 106 87%



There is a substantial 77% reduction in computational cost for Gaussian elimination with the 3x3 plusdecoupled link 2 method. Also, the numerical accuracy may also improve with this method since weneedn’t do unnecessary calculations with the three 3x1 columns of zeros.

Step 7. Calculate Shaking Force and Moment

After the basic inverse dynamics problem is solved, we can calculate the vector shaking forceand vector shaking moment, which is the force/moment reaction on the ground link due to the motion,inertia, weight, and external loads (which we assumed to be zero in this problem). The shaking forceand moment for the inverted slider-crank mechanism is identical to the four-bar in notation and terms.

Ground link force/moment diagram

Shaking force

21 14

21 41 21 14

21 14

X X

S

Y Y

F F F F F F F

F F

Shaking moment

2 21 21 41 14

2 21 21 21 21 41 14 41 14

S

X Y Y X X Y Y X

M r F r F

r F r F r F r F



118

Inverted slider-crank mechanism inverse dynamics example – Term Example 3 continued

Given

1

2

4

1

0.20

0.10

0.32

0

r

r

L

(m)

2

4

4

70

150.5

0.19r

(deg and m)

2

4

4

25

2.47

2.18

r

(rad /s and m/s)

2

4

4

0

9.46

267.14

r

(rad /s2 and m/s2)

In this problem the external forces and moments are zero for both links 2 and 4. In inversedynamics we ignore the slider link mass and inertia. The mechanism links 2 and 4 are uniform,

homogeneous rectangular solids made of steel ( = 7850 kg/m3) with a constant thickness of 2 cm and

link widths of 3 cm. The CGs are in the geometric center of each link. This yields the following fixeddynamics parameters.m2 = 0.47 and m4 = 1.51 (kg) I GZ2 = 0.0004 and I GZ4 = 0.013 (kg-m

2)

Snapshot Analysis

Given the previous mechanism position, velocity, and acceleration analyses, solve the inverse

dynamics problem for this snapshot (2 70 ). The matrix-vector equation to solve is given below.

21

21

42

14

14

2

1 0 0.49 0 0 0 5.03

0 1 0.87 0 0 0 9.21

0.05 0.02 0.01 0 0 1 00 0 0.49 1 0 0 30.77

0 0 0.87 0 1 0 41.82

0 0 0.03 0.08 0.14 0 3.47

X

Y

X

Y

F

F

F F

F

The answer is:

21

21

42

14

14

2

35.35

62.69

61.47

0.4611.65

1.10

X

Y

X

Y

F

F

F

F F

The associated vector shaking force and moment are35.81

51.03

SX

S

SY

F F

F

( N ) ˆ8.53S M k ( Nm)



119


A more meaningful result is to solve and plot the inverse dynamics analysis results for the entire

range of mechanism motion. The plots below give the inverse dynamics results for all20 360 ,

for Term Example 3. Since 2 is constant, we can plot the velocity results vs. 2 (since it is related to

time t via 2 2t ).

Input Torque 2

0 50 100 150 200 250 300 350

-20

-10

0

10

20

30

2 (deg)

2

( N m )

2

AVG

RMS



120

Shaking Force

Shaking Moment

0 50 100 150 200 250 300 350

-200

-150

-100

-50

0

50

100

150

2 (deg)

F S

( N )

FSX

FSY

0 50 100 150 200 250 300 350

-20

-15

-10

-5

0

5

10

15

20

2 (deg)

M S

( N m )



121

3.7 Multi-loop Mechanism Inverse Dynamics Analysis

The Matrix Method can be applied to any planar mechanism inverse dynamics problem. Hereare the five two-loop six-bar mechanisms from Dr. Bob’s on-line Mechanism Atlas.


Stephenson II 6-Bar Mechanism Stephenson III 6-Bar Mechanism

Watt I 6-Bar Mechanism Watt II 6-Bar Mechanism



For example, here are the Watt II six-bar mechanism FBDs, ignoring external forces and moments.



124

Step 6. Solve for the unknowns (continued)

Like the four-bar mechanism, it is possible to partially decouple the solution to this problem. Ifwe consider the FBDs of only links 5 and 6 first, this is 6 equations in 6 unknowns – this is verified bylooking at the original 15x15 matrix and noting nine 6x1 columns of zeros (1, 2, 3, 4, 5, 6, 7, 8, 15) inthe six rows 10 through 15. Here is a more efficient solution. The reduced 6x6 set of equations for links5 and 6 is given below.

54 5 5

54 5 5

45 45 65 65 65 5 5

65 6 6

16 6 6

56 56 16 16 16 6 6

1 0 1 0 0 0

0 1 0 1 0 0 ( )

0 0

0 0 1 0 1 0

0 0 0 1 0 1 ( )

0 0

X G X

Y G Y

Y X Y X X G Z

Y G X

X G Y

Y X Y X Y G Z

F m A

F m A g

r r r r F I

F m A

F m A g

r r r r F I

56 56 56 A v b

Solve for the six unknowns 1

56 56 56v A b

. Second, consider the FBDs of only links 3 and 4: thisis 6 equations in 6 unknowns – this is verified by looking at the original 15x15 matrix and noting seven6x1 columns of zeros (1, 2, 11, 12, 13, 14, 15) in the six rows 4 through 9. Recognizing that now F 54 X and F 54Y are now known from the above 6x6 partial solution, here is a more efficient solution. Thereduced 6x6 set of equations for links 3 and 4 is given below.

3 332

3 332

23 23 43 43 3 343

4 4 5443

4 4 5414

34 34 14 14 4 4 54 5414

1 0 1 0 0 0

0 1 0 1 0 0 ( )

0 0

0 0 1 0 1 0

0 0 0 1 0 1 ( )0 0

G X X

G Y Y

Y X Y X G Z X

G X X Y

G Y Y X

Y X Y X G Z Y X Y

m AF

m A gF

r r r r I F

m A F F

m A g F F r r r r I r F r F

54 54 X Y F

34 34 34 A v b

Solve for six unknowns 1

34 34 34v A b

and then use F 32 X and F 32Y in the following 3x3 set of linear

equations, from the link 2 FBD.

21 2 2 32

21 2 2 32

12 12 2 2 2 32 32 32 32

1 0 0

0 1 0 ( )

1

X G X X

Y G Y Y

Y X G Z Y X X Y

F m A F

F m A g F

r r I r F r F


21 2 2 32

21 2 2 32

2 2 2 32 32 32 32 12 21 12 21

( )

X G X X

Y G Y Y

G Z Y X X Y Y X X Y

F m A F

F m A g F

I r F r F r F r F



125


log

n


22( 1)

3

n nn


Number of Multiplications/Divisions


15x15 8609 1345 84%

6x6 twice, plus decoupled link 2 1672 183 89%


There is an astonishing 86% reduction in computational cost for Gaussian elimination with themethods using 6x6 inversion twice, plus the decoupled link 2 solution. Also, the numerical accuracymay also improve with this method since we needn’t do unnecessary calculations with the sixteen 6x1columns of zeros.

Any mechanism with a dyad of binary links may be decoupled in this manner. Thus, the methodis similar and the computational complexity identical for the Stephenson I, Stephenson III, Watt I, andWatt II six-bar mechanisms.

The Stephenson II six-bar mechanism does not include a dyad of binary links and so it cannot be solved like the other 4 six-bar mechanisms (first links 5 and 6, then links 3 and 4 with one unknownvector force from 5 and 6, then link 2 independently). But links 3, 4, 5, and 6 can be solved firstindependently of link 2: a 12x12 solution followed by the standard link 2 solution. The computationalsavings is not as impressive as in the former six-bar cases.

Number of Multiplications/Divisions, Stephenson II Six-Bar


15x15 8609 1345 84%



There is a 46% reduction in computational cost for Gaussian elimination with the 12x12 plusdecoupled link 2 method. Also, the numerical accuracy may also improve with this method since weneedn’t do unnecessary calculations with the three 12x1 columns of zeros.



126

3.8 Balancing of Rotating Shafts

If high-speed shafts are unbalanced, this can lead to the following problems.

unwanted vibrations

shaking forces

wear noise

safety concerns

comfort of users/riders

less efficient shorter service life

Let us start with the balancing of a single idealized point mass.

1) Static Balance Moments about the rotating shaft must be balanced statically.

0 Z M cos cos 0 B Bmgr m gr B Bmr m r

2) Dynamic Balance Inertial forces due to motion must also be balanced.

Inertial forces are not actual forces but are effects of acceleration, e.g. centripetal force. Assuming

constant input angular velocity , the inertial force is directed outward, opposite to the centripetalacceleration directed inward.

( ( )) I C F m A m r the vector magnitude is 2

I F mr

For dynamic balance, we again add a balance mass. The dynamic balance condition is:

0 I IBF F

The original and balance inertial forces must be equal in magnitude and opposite in direction. Thisvector balance condition is equivalent to the two scalar equations below.

0

0

X

Y

F

F

2 2

2 2

cos cos 0

sin sin 0

B B

B B

mr m r

mr m r



127

Both equations yield the same condition as the static balance case, namely:

B Bmr m r

So, if a single mass is balanced statically, it is also balanced dynamically.

Now let us include a system of idealized point masses attached to the same rotating shaft.

1) Static Balance The rotating shaft will be balanced statically if the system CG lies on the axis of rotation.

0

0CG

X P

Y

Recall the scalar CG equations for a system of point masses:

i i

CG

i

m x X X

m

i i

CG

i

m yY Y

m

Let us consider a system of four point masses. There are two equations to satisfy:

4

1

cos 0i i i

i

m r

4

1

sin 0i i i

i

m r

where cosi i i x r , sini i i

y r , and the4

1

i

i

m

term in the denominator cancels out.

If we fix each mir i, these are 2 equations in the four unknowns i. Therefore, arbitrarily fix 1, 2

and solve for 3, 4. This is equivalent to solving the four-bar linkage position problem.

1 1 1 2 2 2 3 3 3 4 4 4

1 1 1 2 2 2 3 3 3 4 4 4

0

0

m rc m r c m r c m r c

m r s m r s m r s m r s

2 2 3 3 1 1 4 4

2 2 3 3 1 1 4 4

r c r c rc r c

r s r s r s r s



128

Let Ri = mir i; the vectors 1 and 4 are reversed compared to the four-bar mechanism. So we know how to

solve these equations. The shaft will be statically balanced for any shaft angle . Here is the associatedfigure.

2) Dynamic Balance For dynamic balance, we must consider the front view in addition to the previously-shown side

view.

If the following conditions are satisfied assuming constant , we will have dynamic balance.

00

0

X

Y

Z

F

F

F

00

0

X

Y

Z

M

M

M

Let us consider each in turn.

4 42

1 1

4 42

1 1

cos cos 0

sin sin 0

0 0

X i i i i i i

i i

Y i i i i i i

i i

Z

F m r m r

F m r m r

F

The X

F andY

F equations are already satisfied by the static balancing, because 2 divides out.

The Z

F equation yields nothing because all inertial forces are in the XY plane.



129

42

1

42

1

sin 0

cos 0

0 0

X i i i ii

Y i i i i

i

Z

M m r L

M m r L

M

The X

M andY

M equations must be solved to satisfy dynamic balancing. The Z

M equation

yields nothing because all inertial forces pass through the axis of rotation.

We fixed each mir i, we previously determined i, and the

2

term divides out. Therefore, wehave two equations in the four unknowns Li; arbitrarily fix L1, L2 and solve for L3, L4. The result is twolinear equations in the two unknowns.

1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4

1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4

sin sin sin sin 0

cos cos cos cos 0

m r L m r L m r L m r L

m r L m r L m r L m r L

3 3 3 4 4 4 3 1 1 1 1 2 2 2 2

3 3 3 4 4 4 4 1 1 1 1 2 2 2 2

sin sin sin sin

cos cos cos cos

m r m r L m r L m r L

m r m r L m r L m r L

Solve for L3, L4 and the system will be balanced statically and dynamically.



130

3.9 Inverse Dynamics Analysis Examples

This section presents complete snapshot and full-range-of-motion (F.R.O.M.) examples todemonstrate the inverse dynamics analyses in the ME 3011 NotesBook. The first example is for thesingle rotating link inverse dynamics (whose kinematics were not yet presented). Term Example 1 isfor the four-bar mechanism and Term Example 2 is for the slider-crank mechanism. The latter twoexamples are continuations for the Term Example 1 and 2 kinematics examples presented earlier in thisME 3011 NotesBook Supplement.

3.9.1 Single Rotating Link Inverse Dynamics Example

Given: L = 1, h = 0.1 m, m = 2 kg, = 100 rad/s, = 0, F E = 150 N, E = 0, M E = 0 Nm.

Calculated terms 12 0.5 E r r m I GZ = 0.17 kgm

2

Snapshot inverse dynamics analysis

At 150 , given this link, motion, and external force, calculate 12 12, , X Y F F and ,S S F M .

4330

2500

Gx

Gy

A

A

2

m

s

12

12

1 0 0 8510

0 1 0 4980

0.250 0.433 1 37.5

X

Y

F

F

12

12

8510

4980

66.5

X

Y

F

F

N, Nm

21 12

8510

4980S F F F

N ˆ66.5S M k Nm

Full-Range-Of-Motion (F.R.O.M.) Analysis

A more meaningful result from inverse dynamics analysis is to solve and plot the dynamicsunknowns for the entire range of mechanism motion. For the same example as the snapshot we specify

that the given is constant. Prior to solving the inverse dynamics problem, the CG translational

acceleration results for all 0 360 are given in the plot below. The X components are red and theY green. Is the static link weight (mg) significant in this problem?



131

The plot above gives the Shaking Force S F for all 0 360 . The X component is red and the Y

green. The Shaking Moment S M is simply the negative of the driving torque plot shown next.

0 50 100 150 200 250 300 350-5000

0

5000

(deg)

A G X

a n d

A G Y

( m / s

2 )

CG Acceleration, X (red) and Y (green)

0 50 100 150 200 250 300 350-1

-0.5

0

0.5

1x 10

4

(deg)

F S X

a n d

F S Y

( N

)

Shaking Force, X (red) and Y (green)



132

The plot below gives the required driving torque (Nm, red ) for all 0 360 , assuming

the given is constant, for the same example as the snapshot. This shows the torque that must besupplied by an external DC servomotor to cause the specified motion. Also plotted is the average torque

(green) AVG = 0 and the root-mean-square (RMS) torque value ( blue) RMS = 106.1 Nm.

Single Rotating Link Torque Results

Here are the calculations for the average and root-mean-square torques.

0 1 2

1

k AVG

k

2 2 2 2

0 1 2

1

k

RMS k

where k +1 is the total number of elements in the array (since the counting index k starts at zero).

MATLAB can be used to calculate and plot the average and root-mean-square torques on the plot of for easy comparison.

tauAVG = mean(tau); % after the for loop tauRMS = norm(tau)/sqrt(k+1);wuns = ones(1,length(th)); % to plot a constant line plot(th/DR,tau,'r',th/DR,tauAVG*wuns,'g',th/DR,tauRMS*wuns,'b');

0 50 100 150 200 250 300 350-150

-100

-50

0

50

100

150

(deg)

( N

m )

Tau (red) with average (green) and rms (blue)



133

3.9.2 Term Example 1: Four-Bar Mechanism

Figure for Term Example 1 Inverse Dynamics

The coupler link 3 is a rectangle of dimensions 0.203 x 0.152 (m). The triangle tip we have beenusing all along in Term Example 1 (previously called point C ) is the CG of the rectangular link shown below for inverse dynamics.

-0.1 0 0.1 0.2 0.3-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

X (m)

Y ( m )



134

Four-bar mechanism inverse dynamics example – Term Example 1 continued This is the mechanism from Term Example 1 (open branch), with kinematics solutions as

presented before. Given r 1 = 0.284, r 2 = 0.076, r 3 = 0.203, r 4 = 0.178, 1 10.3 , 2 30 , 3 53.8 ,

4 121.7 , RG2 = 0.038, RG3 = 0.127, RG4 = 0.089 (m), 2 0 , 3 36.9 , 4 0 , 2 20 , 3 8.09 ,

4 3.73 (rad /s), 2 0 , 3 8.65 , and 4 244.4 (rad /s2).

All moving links are wood, with mass density 830.4 kg/m3. Links 2 and 4 have rectangular

dimensions r i x 0.019 x 0.013 thick (m; i=2,4); link 3 has rectangular dimensions 0.203 x 0.152 x 0.013

thick (m), as shown on the previous page. The calculated mass and inertia parameters are 2 0.015m ,

3 0.327m , 4 0.036m (kg) and 6

2 7.9 10G Z I , 3

3 1.8 10G Z I , 5

4 9.5 10G Z I (kgm

2). All

external forces and moments are zero but gravity, g = 9.81 m/s2, is included.


At 230 , given this mechanism and motion, calculate the four vector internal joint forces, the

driving torque 2 , and the shaking force and moment ,S S F M for this snapshot.

21

21

32

32

43

43

14

14

1 0 1 0 0 0 0 0 00 1 0 1 0 0 0 0 0

0.019 0.033 0.019 0.033 0 0 0 0 1

0 0 1 0 1 0 0 0 0

0 0 0 1 0 1 0 0 0

0 0 0.127 0.002 0.037 0.122 0 0 0

0 0 0 0 1 0 1 0 0

0 0 0 0 0 1 0 1 0

0 0 0 0 0.076 0.047 0.076 0.047 0

x

y

x

y

x

y

x

y

F

F

F

F

F

F

F

F

2

0.2020.034

0

8.955

4.497

0.015

0.638

0.095

0.023

The solution is accomplished by Gaussian elimination, or 1

v A b

, or by the reduced 6x6

plus decoupled link 2 method (see the on-line ME 3011 NotesBook Supplement). All methods yield thesame results. Snapshot answers:

21

21

32

32

43

43

14

14

2

6.20

10.08

5.99

10.112.96

5.61

3.60

5.52

0.43

x

y

x

y

x

y

x

y

F

F

F

F

v F

F

F

F

( N , Nm)9.80

4.56S F

( N ) ˆ1.68S M k ( Nm)



135


A more meaningful result from inverse dynamics analysis is to solve and plot the dynamicsunknowns for the entire range of four-bar mechanism motion. Prior to solving the inverse dynamics

problem, the plot below shows the CG translational acceleration results for link 3 for all 20 360 .

The X component is red and the Y component is green.

Term Example 1, 3G A

0 50 100 150 200 250 300 350

-30

-20

-10

0

10

20

30

2 (deg)

A G 3

( m / s

2 )

AG3X

AG3Y



136

Term Example 1 Inverse Dynamics, 2

The plot above gives the required driving torque 2 ( Nm) for all 20 360 , for the Term

Example 1 mechanism, assuming the given 2 20 rad /s is constant. This plot shows the torque (red )

that must be supplied in all configurations by an external DC servomotor to cause the specified motion.

Also plotted is the average torque (green) 2 0.003 AVG and the root-mean-square torque value ( blue)

2 0.354 RMS Nm. The root-mean-square (RMS) torque is more meaningful than the average torque

since its terms do not cancel each other (k +1 is the number of elements in the 2 array).

0 1 2

2 2 2 2

2 2 2 2

21

k

RMS k

0 50 100 150 200 250 300 350

-0.4

-0.2

0

0.2

0.4

0.6

0.8

2 (deg)

2

( N m )



138

The plot below gives the shaking moment S M ( Nm) results for all 20 360 . There is only

the Z component since a planar moment is a k vector.

Term Example 1 Inverse Dynamics, S M

0 50 100 150 200 250 300 350

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2 (deg)

M

S

( N m )



139

3.9.3 Term Example 2: Slider-Crank Mechanism

Figure for Term Example 2 Inverse Dynamics

The Term Example 2 slider-crank mechanism is shown below at the starting (or ending) position,

with zero (or 360 ) input angle 2.

-0.1 0 0.1 0.2 0.3

-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

X (m)

Y ( m )



140

Slider-crank mechanism inverse dynamics example – Term Example 2 continued

This is the mechanism from Term Example 2 (right branch), with kinematics solutions as

presented before. Given r 2 = 0.102, r 3 = 0.203, h = 0.076 m, 230 , 3

7.1 , x = 0.29 m, 2 15

rad /s (constant), 3 6.58 rad /s, 0.60 x m/s, 2 0 , 3 62.33 rad /s2, and 30.15 x m/s2.

All moving links are wood, with mass density 830.4 kg/m3. Links 2 and 3 have rectangular

dimensions r i x 0.019 x 0.013 thick (m; i=2,3); link 4 has rectangular dimensions 0.076 x 0.019 x 0.013thick (m), as shown on the previous page. The calculated inertia parameters are m2 = 0.020, m3 = 0.041,m4 = 0.015 (kg) and I G2 Z = 1.819e-005, I G3 Z = 1.418e-004 (kgm

2). The CGs all lie at their respective linkcenters. There is a constant external force of 1 N acting at the center of the piston end, directedhorizontally to the left; gravity is included but all other external forces and moments are zero. We

assume the coefficient of kinetic friction between the piston and the fixed wall is = 0.2.


At 2 30 , given this mechanism and motion, calculate the four vector internal joint forces, the

driving torque 2 , and the shaking force and moment ,S S F M for this instant (snapshot).

21

21

32

32

43

43

14

2

1 0 1 0 0 0 0 0 0.202

0 1 0 1 0 0 0 0 0.0

-0.025 0.044 -0.025 0.044 0 0 0 1

0 0 1 0 1 0 0 0

0 0 0 1 0 1 0 0

0 0 -0.013 0.101 -0.013 0.101 0 0

0 0 0 0 1 0 0.2 00 0 0 0 0 1 1 0

X

Y

X

Y

X

Y

Y

F

F

F

F

F

F

F

84

0

1.018

0.168

0.009

0.5400.150

The solution is accomplished by Gaussian elimination, or 1

v A b

, or by the reduced 5x5 plus

decoupled link 2 method (see the on-line ME 3011 NotesBook Supplement). All methods yield thesame results. Snapshot answers:

21

21

32

32

43

43

14

2

0.736

0.121

0.5340.037

0.484

0.131

0.281

0.039

X

Y

X

Y

X

Y

Y

F

F

F F

vF

F

F

( N , Nm)0.680

0.401S F

( N ) ˆ0.116S M k ( Nm)



142

Term Example 2 Inverse Dynamics, 2

The plot above gives the required driving torque 2 ( Nm) for all20 360 , for the Term

Example 2 slider-crank mechanism, right branch only, assuming the given 2 15 rad /s is constant.

This plot shows the torque (red ) that must be supplied in all configurations by an external DC

servomotor to cause the specified motion. Also plotted is the average torque (green) 2 0.004 AVG

and the root-mean-square torque value ( blue) 2 0.099 RMS Nm. The root-mean-square (RMS) torque is

more meaningful than the average torque since its terms do not cancel each other (k +1 is the number of

elements in the 2 array).

0 1 2

2 2 2 2

2 2 2 2

21

k

RMS k

0 50 100 150 200 250 300 350

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

2 (deg)

2

( N m )



143

The plot below gives the shaking force S F ( N ) results for all 20 360 . The X component

is red and the Y component is green.

Term Example 2 Inverse Dynamics, S F

0 50 100 150 200 250 300 350-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

.5

2 (deg)

F S

( N )

FSX

FSY



144

The plot below gives the shaking moment S M ( Nm) result for all 20 360 . There is only

the Z component since a planar moment is a k vector.

Term Example 2 Inverse Dynamics, S M

0 50 100 150 200 250 300 350

-0.1

-0.05

0

0.05

0.1

0.15

2 (deg)

M

S

( N m )



145

4. Gears and Cams

4.1 Gears

4.1.1 Gear Introduction

Gears are used to transfer motion between rotating shafts in machinery, mechanisms, robots,vehicles, toys, and other electromechanical systems. Gears cause changes in angular velocity, torque,and direction. Gears are used in various applications, from can openers to aircraft carriers. Belt andchain drives are related to gear mechanisms.

Robot joint example

Gear Classification

Externally-meshing Spur Gears Internally-meshing Spur Gears



146

Helical (Parallel Shaft) Helical (Crossed Shaft)

Rack & Pinion Worm and Gear

Straight Bevel Gears Spiral Bevel Gears



147

V-Belt Drive Chain Drive

Toothed Belt Drive Bicycle Sprockets

Herringbone Gears Automotive Hypoid Gears



148

Automotive Gear Train Automotive Differential

Planetary Gear Train Planetary Gear Train – Aircraft

Non-Circular Gears Harmonic Gearing



149

Harmonic Gearing

“The harmonic gear allows high reduction ratios with concentric shafts and with very low

backlash and vibration. It is based on a very simple construction utilizing metal’s elasto-mechanical property.”

“Harmonic drive transmissions are noted for their ability to reduce backlash in a motion controlsystem. How they work is through the use of a thin-walled flexible cup with external splines on its lip,

placed inside a circular thick-walled rigid ring machined with internal splines. The external flexible

spline has two fewer teeth than the internal circular spline. An elliptical cam enclosed in an antifriction ball bearing assembly is mounted inside the flexible cup and forces the flexible cup splines to push

deeply into the rigid ring at two opposite points while rotating. The two contact points rotate at a speed

governed by the difference in the number of teeth on the two splines. This method basically preloadsthe teeth, which reduces backlash.”

roymech.co.uk

Harmonic Gear Sketch

roymech.co.uk



150

The wave generator is attached to the input shaft, the flexible spline is attached to the outputshaft, and the circular spline is fixed.

Harmonic Gear Diagram

roymech.co.uk

For harmonic gearing, the gear ratio is also calculated from the numbers of teeth in each gear.

IN WG FS

OUT FS FS CS

N n

N N

where WG stands for wave generator, FS stands for flexible spline, and CS stands for circular spline.For example, if N FS = 200 and N CS = 202, the gear ratio is

IN FS

OUT FS CS

200100

200 202

N n

N N

which means that the output shaft rotates 100 times slower than the input shaft, but the output shaftcarries 100 times more torque than the input shaft. Therefore, this example would be good for the robot

joint case, i.e. reducing speed and increasing torque, with n >> 1. The negative sign indicates theangular velocity and torque of the output shaft are in the opposite direction of the angular velocity andtorque of the input shaft.



151

4.1.2 Gear Ratio

Classification of gear ratios

If 1n 2 1

2 1

The output has reduced speed and increased torque.

This is the electric motor / robot joint case, where n >> 1.

If 1n 2 1

2 1

The output has increased speed and reduced torque.

This is the bicycle transmission case, except for some granny gears where n can be ashigh as 1.5.

If 1n 2 1

2 1

This case is called an idler, where the output speed and torque are unchanged, but thedirection reverses (for external spur gears)



153

BikeE standard recumbent bike

front teeth

34 34 34

N i 1.2913 : 1 1 : 1 0.7 : 1

11 0.33 0.42 0.60

13 0.38 0.50 0.71

15 0.44 0.57 0.82

rear teeth 18 0.53 0.69 0.98

21 0.62 0.80 1.15

24 0.71 0.92 1.31

28 0.83 1.07 1.53

I was able to obtain a used deluxe BikeE recumbent bike from noted luthier Dan Erlewine. Idecided to keep the front chain ring of 46 teeth (numbers of teeth in the rear chain ring and the planetary

gear ratios in the rear hub are identical between the standard and deluxe BikeE models). This means mynew deluxe BikeE doesn’t climb as well as my modified standard BikeE, but it flies much faster on the bike path in high gear than the standard BikeE! Again, the difference in wheel diameter is taken intoaccount in the table below.

BikeE deluxe recumbent bike

front teeth

46 46 46

N i 1.2913 1 0.7

11 0.24 0.31 0.44

13 0.28 0.37 0.52

15 0.33 0.42 0.61

rear teeth 18 0.39 0.51 0.73

21 0.46 0.59 0.85

24 0.53 0.68 0.97

28 0.61 0.79 1.13

We see that the standard BikeE that was designed to equal the granny gear of the Cannondale (itwas exceeded, 1.53 vs. 1.45). However, the mountain bike still climbs better in granny gear, since your

legs are positioned above the pedals in the mountain bike case, and your legs are positioned straight outin front of you in the recumbent bike case.

Unlike the robot joint example, bicycle gearing generally has n < 1 and so the transmission

increases angular velocity

decreases torque by the gear ratio n. The exception is the granny gears with n > 1.



154

4.1.3 Gear Trains

To obtain a higher gear ratio than practical with a single pair of standard involute spur gears, onecan mate any number of spur gears in a gearbox, or gear train. The leftmost gear is the driving gear andthe rightmost is the output gear. All intermediate gears are first the driven gear and then the driving gearas we proceed from left to right. Let us calculate the overall gear ratio nGT .

IN GT

OUT n

Example

We can find the overall gear ratio by canceling neighboring intermediate angular velocities.

Each term in the above product may be replaced by its known number of teeth ratio.

All intermediate ratios cancel, so

We could have done the same with pitch radii instead of number of teeth because they are in direct proportion.

So, the intermediate gears are idlers. Their number of teeth effect cancels out, but they do changedirection. We should have included the +/– signs, by inspection. For gear trains composed ofexternally-meshing spur gear pairs:

odd number of gears the output is in the same direction as the inputeven number of gears the output is in the opposite direction as the input



155

Improved gear train

That gear train concept did not work. Now let us mate any number of spur gears, where thedriving and driven gears are distinct, because each pair is rigidly attached to the same shaft. Again, letus calculate the overall gear ratio.

IN GT

OUT

n

Example

Again, we use the equation

But now the gears rigidly attached to the same shaft have the same angular velocity ratio, so

The general formula for this case is

Again, we must consider direction

For gear trains composed of externally-meshing spur gear pairs:odd number of pairs the output is in the opposite direction as the inputeven number of pairs the output is in the same direction as the input



156

4.1.4 Involute Spur Gear Standardization

Rolling Cylinders

Mating spur gears are based on two pitch circles rolling without slip. These are fictitious circles,i.e. you cannot look on a gear to see them. The actual gear teeth both roll and slide with respect to eachother (via the two-dof gear joint).

Fundamental Law of Gearing

The angular velocity between the gears of a gearset must remain constant throughout the mesh .

From our study of linkage velocity, we know this is no easy feat. Velocity ratios in a linkagevary wildly over the range of motion.

Velocity Ratio1

VR OUT IN

IN OUT

r

r n

Torque Ratio TR IN OUT

OUT IN

r n

r

The velocity ratio is the inverse of the gear ratio n and the torque ratio is the same as the gearratio n defined previously. The torque ratio is also called Mechanical Advantage (MA).

Involute Function

Standard spur gears have an involute tooth shape. If the gears’ center distance is not perfect(tolerances, thermal expansion, wear – in design the center distance is increased slightly by the engineerto allow for these effects; this is called clearance), the angular velocity ratio will still be constant tosatisfy the Fundamental Law of Gearing.



157

The involute of a circle is a curve generated by unwrapping a taut sting from the circumference of a so-called base circle, always keeping it tangent to the circle. The figure below has th = [0:5:60]*DR

and r b = 1 m. The red circle is the gear base circle, the blue lines are the taut tangent construction lines,and the green curve is the involute function.

Involute Function Construction

In polar coordinates ( , r ), the parametric equations for the involute of a circle are given below.tan

secb

t t

r r t

where t is the independent parameter and r b is the base circle radius. In Cartesian coordinates ( x I , y I ):

cos sec cos(tan )

sin sec sin(tan )

I b

I b

x r r t t t

y r r t t t

-1 -0.5 0 0.5 1 1.5

-1

-0.5

0

0.5

1

X (m)

Y ( m )



158

Involute Function Example

The plot below shows the involute function for a circle of r b = 1 m. The parameter range is t =

[0:1:80]*DR; near the circle the involute points are very close to each other and farther away the

step size increases dramatically.The involute function is symmetric (try t = [0:5:360]*DR ) but watch out for those

intermediate steps). Gear teeth only require the involute near the base circle, with two symmetric sides.

Involute Function of a Circle

-6 -5 -4 -3 -2 -1 0 1 2-4

-3

-2

-1

0

1

2

3

4

X (m)

Y ( m )



160

Norton (2008)

The length of contact is measured along the axis of transmission. The beginning of contact iswhen the tip of the driven gear tooth intersects the axis of transmission. The end of contact is when thetip of driving gear tooth intersects the axis of transmission, as shown in the figure above. Only one ortwo teeth are in contact at any one time for standard spur gears.

For harmonic gearing, many teeth are in contact at any one time, which provides a higher gearratio in a smaller package.



161

Norton (2008)

Increasing the center distance increases the pressure angle and increases the pitch circle

radii, but doesn’t change the base circles (obviously – the gears are made based on their own constantr b). Thanks to the involute tooth shape, increasing the center distance does not affect the angular

velocity ratio. This is why the involute function is so widely used in spur gears.

The relationship from before still applies with an increase in center distance.

cosb pr r

Again, r b is fixed, and r p and both increase – the cosine function maintains the constant r b.



162

Spur Gear Standardization

Gear standardization is used to allow interchangeability in manufacturing and to allow meshingof different size gears (different pitch radii and number of teeth) to achieve desired gear ratios. For twospur gears to mesh, they must have

1) the same pressure angle (see previous figures and definition)

2) the same diametral pitch (see the equation below)

3) standard tooth proportions (see the figure below)

diametral pitch d

N p

d

Where N is the number of teeth and d is the pitch diameter, both for each gear.

module1

d

d m

N p

Module is the SI version of diametral pitch (it is the inverse). SI gears are notinterchangeable with English system gears because of different tooth proportionstandards.

circular pitch c

d

p N

Circular pitch is the circumferential distance (arc length) between teeth along the pitchcircle of a spur gear.



163

from Norton (2008)

Standard involute tooth proportions

Addendum is radial distance from pitch circle to top land of tooth.

Dedendum is radial distance from pitch circle to bottom land of tooth (not to the base circle).

Clearance is radial distance from bottom land to mating gear top land (radial backlash).

Face width is thickness of tooth and gear (mating widths needn’t be the same).

Tooth thickness t is the circumferential arc length of each tooth. It is related to the circular

pitch pc and backlash (next page) b by

2c p t b



164

Backlash

Backlash B is the distance between mating teeth measured along the pitch circle circumference.Backlash can be thought of as circumferential clearance. All real-world gears must have some backlashin order to still function despite real-world problems of manufacturing tolerances, thermal expansion,wear, etc. However, one must minimize backlash for smooth operation. For example, robot joints must be driven both directions. Upon changing direction, nothing happens until the backlash is passed, andthen an impact occurs, which is bad for gear dynamics. This is a non-linear effect in robotics. On earthgravity tends to load the backlash for predictable effects. In space however, the backlash is less predictable.



165

4.1.5 Planetary Gear Trains

Planetary gear trains are also called epicyclic gear trains. The sun gear rotates about a fixed axis.Each planet gear rotates about its own axis and also orbits the sun gear. This happens with directmeshing of teeth, unlike celestial planetary motion. The arm link (which is a rigid body with no teeth)carries the planet(s) around the sun. The arm has a revolute joint to the sun gear on one side and anotherrevolute joint to the planet gear(s) on the other side. Planetary gear trains have two-dof, so two inputsmust be given to control the mechanism. For instance, one can drive the sun gear and the arm link withindependent external motors. Alternatively, the sun gear may be fixed and the arm link driven. Using planetary gear trains, one can obtain a higher gear ratio in a smaller package, compared to non-planetarygear trains.

Conventional Gear Pair Planetary Gear Train

3(3 1) 2(2) 1(1) 1 M 3(4 1) 2(3) 1(1) 2 M

IN OUT OUT

OUT IN IN

r N n

r N

? IN

OUT

n

We present the tabular method below to determine the gear ratio for various planetary gear trains.



166

Planetary Gear Train Applications

Airp lane Propel ler Transmission

Planetary gear train applications include airplane propellers, some automotive transmissions anddifferentials, machine tools, hoists, and a hub-enclosed multi-speed bicycle transmission.

Old-Fashioned Pencil Sharpener

appauto.files.wordpress.com



167

Planetary Gear Box in Stages

This is a commercial planetary gear set with 8 possible ratios (4:1, 5:1, 16:1, 20:1, 25:1, 80:1,100:1, 400:1).

servocity.com



168

Table Method Analysis

First, consider a simplified planetary gear system with 0S . With the sun gear as the fixed

link, we have a 1-dof system. Given the input A, calculate the absolute value of P. The Table

Method is based on the following relative velocity equation.

/G A G A

This is a vector equation, but since all rotations are about the Z axis, we just use the magnitudes and + for CCW and – for CW , according to the right-hand-rule. This equation is written for each gear in the

system (replace G with the appropriate index). A stands for arm. ,G A are the absolute angular

velocities of a gear and the arm link. /G A is the relative velocity of a gear with respect to the moving

arm link. Construct a table as below; each row is the relative equation written for a different gear.

G G = A + G/A

S

P

First, fill in the given information.

G G = A + G/A

S 0 A

P A

Each row must add up according to the relative velocity equation

G G = A + G/A

S 0 A – A

P A

Now we can fill in down the relative column, using a simple gear ratio (relative to the arm).

/

/

P A S

S A P

N

N

/ /

S S P A S A A

P P

N N

N N

G G = A + G/A

S 0 A – A

P A S A

P

N

N



169

The last row must add up according to the relative equation, to finish the table.

G G = A + G/A

S 0 A – A

P1 S

A

P

N

N

A S A

P

N

N

Therefore, the absolute angular velocity of the planetary gear is 1 S P A

P

N

N

. Since the sign is

positive, it has the same direction as A (CCW ).

Calculate Effective Gear Ratio

1

1

IN A P

S OUT P P S

P

N n

N N N

N

Example

10S N 40P

N 0S 100 A

rpm, CCW

101 100 125

40P

rpm, CCW

400.8

40 10P

P S

N n

N N

Check A

P

n

100125

0.8 A

Pn

So we see that with 0S , the gear ratio is not higher than the conventional gear train.

4010

P

S

N n N

.

Let us include 0S next.



170

Table Method Analysis Now we present a more general system with 0S .

1) The given information is starred (*).

2) The S equation must add up; therefore /S A S A .

3) Fill down the right column using a simple gear train, relative to arm A./

/

P A S

S A P

N

N

/ / ( ) ( )S S S

P A S A S A A S

P P P

N N N

N N N

4) The P equation must add up; therefore:

( ) 1S S S P A A S A S

P P P

N N N

N N N

.

G G = A + G/A

S * S * A S A

P1 S S

A S

P P

N N

N N

* A ( )S

A S

P

N

N

Examples 0S

1) 10S N 40P N 100S rpm, CW 100 A rpm, CCW

10 10

1 100 100 125 25 15040 40P

rpm, CCW Still not a big ratio.

2) 40S N 10P

N 100S rpm, CW 100 A rpm, CCW

40 401 100 ( 100) 500 400 900

10 10P

rpm, CCW That’s a big ratio.

3) 40S N 10P N 125S rpm, CCW 100 A rpm, CCW

40 401 100 (125) 500 500 0

10 10P

In Example 3, the Sun and Arm rotational velocities cancel so the absolute angular velocity of the Planet is zero.



172

Examples 0S plus ring gear

1) 10S N 40P N 100 R N 100S rpm, CW 100 A rpm, CCW

10 101 100 ( 100) 110 10 120

100 100 R

rpm, CCW

2) 40S N 20P

N 80 R N 100S

rpm, CW 200 A rpm, CW

1 S S P A S

P P

N N

N N

40 40

1 ( 200) ( 100) 600 200 40020 20P

rpm, CW

1 S S R A S

R R

N N

N N

40 401 ( 200) ( 100) 300 50 250

80 80 R

rpm, CW



173

4.2 Cams

4.2.1 Cam Introduction

Applications

Compared to linkages, it is easier to design desired motion with cams, but it is more expensive anddifficult to produce. Also, the cam contact and wear properties are worse than for linkages.

Cam Classification

Mabie & Reinholtz (19874

)

a) disk cam with knife-edgefollower

b) disk cam with radial rollerfollower

c) disk cam with offset radialroller follower

d) disk cam with oscillatingroller follower

e) disk cam with radial flat-facedfollower

f) disk cam with oscillating flat-faced follower

4 H.H. Mabie and C.F. Reinholtz, 1987, Mechanisms and Dynamics of Machinery, Wiley.



174

Disk cams with followers

Norton (2008)

Norton (2008)



175

Norton (2008)

Two cam/follower systems are shown above with equivalent four-bar and slider-crank

mechanism models.

Caution These are instantaneous equivalents only, i.e. the virtual link lengths for both the four-bar and

slider-crank models change with cam mechanism configuration.

Degrees of Freedom (mobility)

A cam joint is a J 2, i.e. it has two-dof since it allows both rolling and sliding, like a gear joint.

Function Generation

In function generation, the output parameter is a continuous function of the input parameter in amechanism. With linkages, we can only satisfy a function exactly at a finite number of points: 3, 4, or 5,usually. For example, for a four-bar linkage with 4 2 f , this function is only exact at a few points.

With a cam and follower mechanism, however, we can satisfy function generation at infinite

points. is the cam input angle and the output is S for a reciprocating (translating) follower and the

output is for an oscillating (rotating) follower.( )S f ( ) f



177

S V A J Diagrams

In cam synthesis (design), we are only given the total motion range and perhaps some timingrequirements. It is the engineer’s job to determine the position curves and to match the velocity andacceleration across junctions. Position is automatically matched by shifting the dependent functionaxes. Draw S V A J diagrams vs. time to graphically see if the Fundamental Law of Cam Design is

satisfied for candidate curves. We can plot vs. time or vs. input cam angle (assuming constant angularvelocity, t ).

The slope of a function is the value of its derivative at a point in time (or ). Therefore, forcontinuous velocity and acceleration curves, the slopes of the position and velocity curves must matchacross all junctions. The slope of the acceleration can be discontinuous (leading to finite jumps in jerk), but the acceleration itself must be continuous.

Cam motion curves are very much like the input link motion curves discussed earlier, for inputlinks that start and stop at zero velocity and acceleration. In fact, I adapted the input motion curves fromcam motion curve design.

Generic Cam-Follower Motion Profile Figure

Define each separate function so the value is zero at the initial angle, which is zero. Then to put the

whole cam motion profile together, just shift the and S axes.

Match S easy, just shift the S axis.

Match V slope of S must match across junctions.

1 1( ) ( 0)i i i i i

v v

apply to all functions / junctions.

Match A slope of V must match across junctions.

1 1( ) ( 0)i i i i i

a a

apply to all functions / junctions.



178

Cam Follower Motion Profile Examples Example 1

rise – dwell portion. Specify parabolic (constant acceleration) to straight line (constant velocity) rise,followed by a dwell.

parabolic function constant velocity function dwell

S: 2

1 1 0 1

1( )

2 f A

2 2 0 2( ) f V 3 3( ) 0 f

V: 1 1 0 1( )v A

2 2 0( )v V 3 3( ) 0v

A: 1 1 0( )a A

2 2( ) 0a 3 3( ) 0a

J: 1 1( ) 0 j

2 2( ) 0 j 3 3( ) 0 j

Match S at junction B just shift the vertical axis up.

Match V at junction B

1 1 1 2 2( ) ( 0)v v 0 1 0 A V so 0 0 1V A

Try to match A at junction B:

1 1 1 2 2( ) ( 0)a a 0 0 A

0 0 A is impossible, or else the parabola is degenerate, which we cannot allow. This case violates the

Fundamental Law of Cam Design since the acceleration function cannot be made to be continuous at junction B. Therefore, this cam motion profile example cannot be used for cam design.

We have a bigger problem at junction C , between functions 2 and 3: the velocity function cannot be made to be continuous at junction C . Discontinuous velocity is one level worse than discontinuous

acceleration; either renders the resulting cam motion profile unacceptable.

Example 1 Plots



179

Cam Follower Motion Profile Examples Example 2 Let us fix the rise portion only, at junction B. Then the problem at junction C can be fixed using

symmetry. We specify a half-cycloidal function (sinusoidal in cam angle) to a straight line (constantvelocity) rise.

half-cycloidal function constant velocity function

S 1 11 1 1

1 1

1( ) sin f L

2 2 0 2( ) f V

V 1 11 1

1 1

( ) 1 cos L

v

2 2 0( )v V

A 1 11 1 2

1 1

( ) sin L

a

2 2( ) 0a

J 2

1 11 1 3

1 1

( ) cos L

j

2 2( ) 0 j

Match S at junction B just shift the vertical axis up

Match V at junction B

1 1 1 2 2( ) ( 0)v v 1 10

1 1

1 cos L

V

so 1

0

1

2 LV

Match A at junction B

1 1 1 2 2( ) ( 0)a a 1 1

2

1 1

sin 0 L

0 = 0

In this case the acceleration function is continuous because the half-cycloidal function ensuresthat the acceleration is zero at the end of the function range. This case obeys the Fundamental Law of

Cam Design and so this cam motion profile example portion can be used for cam design.

Half-Cycloidal Rise (to connect with constant velocity)

0 20 40 60 800

1

2

S ( m )

Half Cycloid

0 20 40 60 800

0.05

V ( m / d e g )

0 20 40 60 800

5

x 10-4

A ( m

/ d e g

2 )

0 20 40 60 80

-2

0

2

x 10-5

(deg)

J ( m / d e g

3 )



181

4.2.3 Analytical Cam Synthesis

Disk Cam with Radial Flat-Faced Follower

Assume a valid cam motion profile has been designed according to the Fundamental Law of

Cam Design; i.e. we now have continuous S, V , A curves. Given the motion profile found by theengineer, now we must determine the cam contour.

Is it as simple as polar-plotting S = f ( ) vs. cam angle ? No – that approach would not accountfor the face width of the cam follower, i.e. the contact points are not along radial lines in general. Wewill use kinematic inversion to simplify the synthesis process.

DCRFFF Figure

As seen in the figure, the radius R out to the flat-faced follower (not to the point of contact x, y) is:( ) R C f

where C is the minimum cam radius, a design variable, and S = f ( ) is the given cam motion profile.The radius R and the flat-faced follower length L can be related to the contact point x, y and the camangle through geometry.



183

Minimum Cam radius to Avoid Cusps

A cusp is when the cam becomes pointed or undercut. Clearly, this must be avoided for good

cam motion. The cusp condition is that for a finite , there is no change in x, y.

0dx dy

d d

will cause a cusp.

2

2

2

2

( )sin cos cos sin

( )cos sin sin cos

dx df df d f C f

d d d d

dy df df d f C f

d d d d

2

2

2

2

sin

cos

dx d f C f d d

dy d f C f

d d

0dx dy

d d occurs simultaneously only when

2

2( ) 0d f C f

d

Therefore, to avoid cusps on the entire cam contour, we must ensure that2

2( ) 0

d f C f

d

Note that C is always positive and f ( ) starts and ends at zero and never goes negative. So the sum ofthese positive terms and the sometimes-negative second derivative of the cam motion profile mustalways be greater than zero to avoid cusps or undercutting in the practical cam you are designing.



184

Disk Cam with Radial Flat-Faced Follower Design Example

Specify a full-cycloidal rise with a total lift of 50 mm, followed by a high dwell, a symmetricfull-cycloidal return with a total fall of 50 mm, and then a low dwell. Each of these four motion steps

occurs for 90 of cam shaft rotation.

The full-cycloidal rise and fall cam motion profile associated with this specification is shown below. Clearly, this satisfies the Fundamental Law of Cam Design because the position, velocity, andacceleration curves are continuous. The jerk is not continuous, but it remains finite over all cam angles.

0 50 100 150 200 250 300 3500

10

20

30

40

50

S

0 50 100 150 200 250 300 350

-50

0

50

V

0 50 100 150 200 250 300 350

-100

-50

0

50

100

A

0 50 100 150 200 250 300 350-500

0

500

J



185

Choosing a minimum cam radius of C = 100 mm, the resulting cam contour is shown below.

Cam Cartesian Contour

-150 -100 -50 0 50 100-100

-50

0

50

100

150

X (mm)

Y ( m

m )



186

Let us check the cusp avoidance plot. To avoid cusps in this cam, we require that

2

2( ) ( ) ( ) 0

d f C S A C f

d

As seen in the plot below, this inequality is satisfied for the entire range of motion, so this cam design is

acceptable with respect to avoiding cusps and undercutting.

Cam Cusp Avoidance Plot

0 50 100 150 200 250 300 3500

50

100

150

200

250

C + S + A



187

Cam/Follower Animation Snapshot for2 60 (CW)

-200 -150 -100 -50 0 50 100 150 200-200

-150

-100

-50

0

50

100

150

200

X (m)

Y ( m )



189

damping A linear model for energy loss due to friction and other energy dissipaters,for either translational or rotational motion. Damping is provided by avirtual viscous dashpot (an automotive shock system is a real example).

amplitude The maximum magnitude of a sinusoidal oscillation.

phase angle Angle in a single sinusoidal function that indicates the amount of lead orlag of a cosine or sine wave relative to the standard cosine or sine wave.

differential equation An equation in which the unknown appears with its time derivatives.

initial conditions Given values at time t = 0 for the output and its time derivatives. Theremust be as many initial conditions as the order of the differential equation.

input The external forcing function that drives a dynamic system.

output The variable of interest in motion of a dynamic system.

free vibrations Vibrations in the absence of a forcing function.

homogeneous solution Transient solution due to initial conditions and the external input forcingfunction, yielding zero in the ODE.

transient vibrations Vibrations that disappear given enough time.

forced vibrations Vibrations due to an external input forcing function.

particular solution Steady-state solution due to the external input forcing function. There isalso a transient solution due to the external input forcing function.

steady-state vibrations Vibrations that do not disappear as time increases.

forcing function The input actuation provided by an external source.

driving frequency The rate at which an external input forces a vibratory system, either

cyclical f , or circular .

damped frequency The rate at which an unforced, damped vibratory system tends to vibrate,

either cyclical f d , or circular d .

logarithmic decrement The natural logarithm of the ratio of two consecutive points, separated bythe damped time period T d , on a unforced underdamped vibratory wave.

beat frequency Vibrational phenomenon wherein two frequencies in a dynamic system areclose in value, causing a vibration within a vibrational envelope.

resonance This condition occurs when the driving frequency is the same as, or

close to, the system’s natural frequency n. In this case the resultingamplitude of the steady-state vibration becomes very large.



190

frequency response The plots of system amplitude and phase angle vs. independent variableinput frequency.

critical shaft speed The angular shaft velocity that excites the natural frequency of a rotatingdynamic system. When driven at the critical speed, resonance occurs, acondition generally to be avoided at all costs.

transmissibility The ratio of the maximum force transmitted to the base over the maximuminput force magnitude in a vibratory dynamic system.

vibration mode Characteristic shape at which a dynamic system will vibrate. Multi-dofsystems have multiple modes that sum to yield the total time response.

A linear system is one in which all governing equations (differential, algebraic) are linear.Linear systems satisfy the principles of linear superposition and homogeneity. Let u(t ) be the input

and y(t ) be the output. Further, ( ) ( )u t y t indicates a system yielding output y(t ) given input u(t ).

1) linear superposition

if 1 1( ) ( )u t y t and 2 2( ) ( )u t y t then 1 2 1 2( ) ( ) ( ) ( )u t u t y t y t

2) homogeneity

if ( ) ( )u t y t then ( ) ( )u t y t

where is any constant.



191

6. Vibrational Systems Modeling

6.1 Zeroth-Order Systems

Calculation of Spring Constants

Helical Coiled Spring. The jagged springs indicated so far represent physical translationalsprings coiled from an elastic wire into a helical shape about a central axis. For this physical spring,from Strength of Materials, the linear spring constant k for small displacements is given below.

4

38

Gd k

nD

whereG material shear modulus N/m2

d wire diameter mn number of spring coils unitless D diameter of spring coiling m

This type of coiled physical spring is common; however, there are many mechanical systemswithout a clear spring such as above. Instead, their elastic properties are provided by distributedmaterial in the system. Here we present several such distributed springs that are modeled as linearsprings for small displacements, as follows.

For the following 4 translational springs, the following method is used. For each given physical

situation, we find the displacement under an applied force F , from Strength of Materials. Then the

spring stiffness constant k is found from Hooke’s Law F = k : k = F / . Below we assume that the forceis applied and the displacement measured at the end of the cantilevered beam and in the center of theother beams.

f(t)

d

D f(t)



192

Cylindrical Rod Spring. An elastic rod has an equivalent spring constant along the longitudinalaxis of the rod. From Strength of Materials, the linear spring constant k for small displacements is given below.

EAk

L

where E material Young’s modulus N/m

2

A rod cross-sectional area, 2r m2

r radius of rod m L length of rod m

So the cylindrical rod spring constant for a constant circular cross section becomes2

E r k

L

If the elastic rod is tapered with different radii r 1 and r 2 on opposite ends of the rod, the equivalentspring constant is

1 2 E rr k

L

Cantilever Beam Spring. A rectangular cantilevered beam has an equivalent spring constant

for small vertical displacements. From Strength of Materials, the linear spring constant k for smalldisplacements is:

3

3 EI k

L

where E material Young’s modulus N/m2

I beam area moment of inertia,3

12bh m4

b beam width mh beam height m L length of beam m

So the cantilevered beam spring constant becomes3

34

Ebhk

L

f(t)r

L

f(t) f(t)

r

L

f(t)

1

r 2

m

L x(t)

h



193

Doubly-Simply-Supported Beam Spring. A rectangular beam simply-supported at both endshas an equivalent spring constant for small vertical displacements. From Strength of Materials, thelinear spring constant k for small displacements is given below.

3

48 EI k

L

where E material Young’s modulus N/m2


So the doubly-simply-supported beam spring constant becomes

3

34 Ebhk

L

This doubly-simply-supported beam is 16 times stiffer in vertical displacement than the cantilever beam.

Doubly-Fixed Beam Spring. A rectangular beam rigidly fixed at both ends has an equivalentspring constant for small vertical displacements. From Strength of Materials, the linear spring constantk for small displacements is given below.

3

192 EI k

L

where E material Young’s modulus N/m

2


So the doubly-fixed beam spring constant becomes3

3

16 Ebhk

L

This doubly-fixed beam is 64 times stiffer in vertical displacement than the cantilever beam, and 4 timesstiffer than the simply-supported beam presented earlier.

m

L x(t)

h

m

L x(t)

h



194

Air Spring. An enclosed volume of air has an equivalent spring constant for smalldisplacements. From Fluid Mechanics for compressible air, the linear spring constant k for smalldisplacements is given below.

2PA

k V

where

ratio of specific heats unitlessP enclosed air pressure N/m2

A enclosed air cross-sectional area m2

V enclosed air volume m3

These six spring constant examples are all for translational springs, so their spring constant k units should all be N/m. The material Young’s Modulus E and Shear Modulus G for steel are:

9 2200 GPa = 200 10 N / m E 9 279 GPa = 79 10 N / mG

Let us now check the units for the previous six spring constants.

4 4

3 2 3

Nm N

8 m m m

Gd

nD

2

2

Nm N

m m m

EA

L

4

3 2 3

3 Nm N

m m m

EI

L

3 3

3 2 3

4 N(m)m N

m m m

Ebh

L

3 3

3 2 3

16 N(m)m N

m m m

Ebh

L

2 4

2 3

Nm N

m m m

PA

V

All units are correct for a translational spring, i.e. N/m.

We have also used torsional springs in our mechanical systems models. Torsional springs can beconstructed of a coiled elastic material, spiraling in one plane. An elastic cylindrical rod also provides atorsional spring effect when torque about the longitudinal axis of the rod. Additionally, a fixed-simply-supported beam also provides a torsional spring effect.

f(t)



195

Torsional Rod Spring. An elastic rod has an equivalent spring constant about the longitudinalaxis of the rod. From Strength of Materials, the torsional spring constant k R for small angulardisplacements is given below.

4

2 R

G r k

L

whereG material shear modulus N/m

2

r radius of rod m L length of rod m

If the torsional shaft is hollow, the torsional spring stiffness constant is

4 4( )

2

o i R

G r r k

L

where r o and r i are the outer and inner shaft radii, respectively.

Fixed-Simply-Supported Torsional Beam Spring. A rectangular beam rigidly fixed at one endand simply supported on the other has an equivalent torsional spring constant for small angulardisplacements. From Strength of Materials, the torsional spring constant k R for small displacements is

given below.

4 R

EI k

L

where

E material Young’s modulus N/m2

b beam width mh beam height or thickness m L length of beam m

So the fixed-simply-supported torsional beam spring constant becomes3

3 R

Ebhk

L

r

L

(t) (t)

m

L

(t)h



197

Calculation of Damping Constants

In general the damping coefficient for a physical system is harder to identify and calculate thanthe mass and spring constants. This subsection presents 4 equivalent damping coefficient cases. Many physical systems have no identifiable damping dashpot, but a viscous damping coefficient c is includedto account for friction losses in the system, using a simple linear model (the law for linear dashpotswhere the friction force is proportional to the velocity).

Also see the later section on logarithmic decrement, where the damping coefficient c can beestimated from experimental observations.

Dashpot. The piston-based dampers indicated so far represent physical dashpots with a pistonmoving (translating in a reciprocating fashion) in a viscous fluid to remove energy from the system,such as in an automotive shock absorber. For this physical dashpot, from Fluid Mechanics, the linearviscous damping coefficient c for small velocities is given below.

3

3

61

r h wc

w r

where

fluid viscosity N-s/m2

r piston radius mh piston height mw wall thickness r o – r m

f(t)

h

r

w



199

Rotational dashpot. We have also used torsional dashpots in our mechanical systems models.Torsional dashpots can be constructed in a similar manner to a translational dashpot, but thereciprocating motion is rotary about the center axis, instead of translational. From Fluid Mechanics, thelinear rotational viscous damping coefficient c R for small angular velocities is given below.

3

0

28

R

h r c r

w h

where

fluid viscosity N-s/m2 r piston radius mh piston height mw wall thickness r o – r m

h0 height from base to rotating piston m

(t)

h

r

w

h 0



200

Journal bearing. A journal bearing also has a fluid providing rotational damping. From FluidMechanics, the linear rotational viscous damping coefficient c R for small angular velocities is given below.

32 R

wr c

where

fluid viscosity N-s/m2 w base thickness mr bearing radius m

journal bearing fluid clearance m

Let us now check the units for these torsional damping coefficients.

3 3

2

0

Ns m Nms2 m Nms

8 m m rad

h r r

w h

3 4

2

2 Ns m Nms Nms

m m rad

wr

Recall that rad doesn’t count as a unit so we are free to include it above, yielding the correct SI units N-m-s/rad for torsional damping coefficient.

r

w



201

6.2 Second-Order Systems

6.2.1 Translational m-c-k System Dynamics Model

Equivalent Springs

There is an alternate parallel springs case, shown in the diagram below.

1-dof Translational Mechanical System wi th Parallel Springs, Al ternate

From the FBD (not shown), again there is a common displacement x(t ). Assuming a small positive displacement x(t ), both spring restoring forces are negative (to the left). The equivalent springderivation is now presented.

1 2

1 2 1 2( )

f f f

f k x k x k k x

Therefore, the equivalent spring constant for parallel springs case is again:

par 1 2

f

k k k x

This result is identical to the original parallel springs case presented in the ME 3011 NotesBook.

The equivalent-spring formulas for 2, 3, and 4 springs in series are:

1 2ser2

1 2

k k k

k k

1 2 3ser3

1 2 1 3 2 3

k k k k

k k k k k k

1 2 3 4ser4

1 2 3 1 2 4 1 3 4 2 3 4

k k k k k

k k k k k k k k k k k k

Actually, the original general formula below yields the same answers, with less computation:

ser

1

1

1n

i i

k

k

x(t)

mk k 1 2



202

6.2.3 Pendulum System Dynamics Model

Other Pendulum Models

Linearized ODE models can be derived that are very similar to the simple pendulum in the ME3011 NotesBook, for the two pendulum models shown below. The figure on the left shows a pendulumconsisting of a slender rod of length L, with distributed mass m. The figure on the right shows a pendulum consisting of a slender rod of length L, with distributed mass m1, plus an end-mounted pointmass of m2.

Two Addit ional Pendulum System Diagrams

Only the mass moment of inertia I OZ changes; for the cases on the left and right above, respectively, theformulae are given below.

2

3OZ

mL I

221

23

OZ

m L I m L

The associated models are:3

( ) ( ) 02

gt t

L

1 2

1 2

3( 2 )( ) ( ) 0

2( 3 )

m m gt t

m m L

Where, for both slender rod cases, in the sum of moments we used the fact that the slender rod weight(mg and m1g, respectively), acts at the slender rod CG, at a radius of L/2 from point O.

It is possible to account for friction in the simple pendulum system by including a rotationalviscous damping coefficient at the rotational bearing.

2

2

( ) ( ) ( ) 0

( ) ( ) ( ) 0

R

R

mL t c t mgL t

c gt t t

mL L

And while not common, it is also possible to include an input forcing function, input torque (t )at the bearing.

g m L

O

(t)2

1

gm

m(t)

O

L



203

2 2

1( ) ( ) ( ) ( ) R

c gt t t t

mL L mL

Finally, if we model a horizontal pendulum, the restoring torque of gravity disappears and wehave the dynamic model of a gravity-neutral single rotating link.

2 21( ) ( ) ( ) Rct t t

mL mL



204

6.2.4 Uniform Circular MotionUniform circular motion (UCM), wherein a point P at the end of a fixed-length link of length r

rotating in the XY plane about a fixed point O with a constant angular velocity (see figure below)generates Simple Harmonic Motion, like an unforced, undamped m-k translational mechanical system.

Uniform Circular Motion Diagram

In this UCM case no differential equations or solutions are required. Simultaneously simpleharmonic motion (SHM) is generated in the X with a pure cosine wave (with no phase angle) and in theY with a pure sine wave (with no phase angle); see the figure below, generated by UCM.m . Since

angular velocity is constant, the angle linearly increases with time, ( )t t . With 0 = 0 at t = 0, this

system will generate SHM in the X and Y directions indefinitely, repeating every 360 deg or 2 rad.

( ) cos ( ) cos

( ) sin ( ) sin

x t r t r t

y t r t r t

MATLAB UCM Animation

r

(t)

P

O X

Y



205

6.4 Additional 1-dof Vibrational Systems Models

1-dof Vibrational Systems

Derive the vibrational equation of motion for each given mechanical system below.

1. m-c-k translational systems with parallel and series springs and dampers

The ME 3011 NotesBook presented the derivation of equivalent translational springs for the parallel springs and series springs cases. Translational dashpots also follow the same formulas for theirrespective parallel and series cases.

Considering the parallel-case figure shown below,

1-dof Translational Mechanical System with Parallel Springs and Dashpots

the system model is:

1 2 1 2( ) ( ) ( ) ( ) ( ) ( )mx t c c x t k k x t f t

Considering the series-case figure shown below,

1-dof Translational Mechanical System with Series Springs and Dashpots

the system model is:

1 2 1 2

1 2 1 2

( ) ( ) ( ) ( )c c k k

mx t x t x t f t c c k k

x(t)

m

k k 1 2

c

c1

2

f(t)

x(t)

m

k k 1 2

c1 f(t)c2



206

2. J-cR-kR rotational systems with parallel and series springs and dampers

The same type of second-order models result for the rotational systems as in the translationalsystems presented on the last page, with regard to parallel and series arrangements of rotational springsand dashpots.

Considering the parallel case, the system model is given below.

1 2 1 2( ) ( ) ( ) ( ) ( ) ( ) R R R R J t c c t k k t t

Considering the series case, the system model is given below.

1 2 1 2

1 2 1 2

( ) ( ) ( ) ( ) R R R R

R R R R

c c k k J t t t t

c c k k



207

Compound m-c-k translational systems wi th parallel and series springs and dampers

Certain real-world systems may require more complicated models with combinations of springsand dampers in parallel and in series. There are infinite possible combinations of such models, dictated by the real-world component arrangement. The next two sections present two such possibilities.

3. Parallel springs and dampers in series

The figure below shows two sets of parallel springs arranged in series. For clarity, the dashpotsare not shown but they have the same arrangement.

The second figure shows a simplified diagram for this case.

Using the parallel- and then serial-springs equivalent spring formulas derived in the ME 3011 NotesBook:

1 3

2 4

a

b

k k k

k k k

1 2 2 3 1 4 3 4

1 2 3 4

a beq

a b

k k k k k k k k k k k

k k k k k k

The system model is now given.

( ) ( ) ( ) ( )eq eq

mx t c x t k x t f t

where 1 2 2 3 1 4 3 4

1 2 3 4

eq

c c c c c c c cc

c c c c

This example is for a translational system, but equivalent results exist for an analogous rotationalsystem.

1

3

k x(t)

mk

2

4

k

k f(t)

k

x(t)

mk

f(t)a b



208

4. Series springs and dampers in parallel

The figure below shows two sets of series springs arranged in parallel. For clarity, the dashpotsare not shown but they have the same arrangement.

The second figure shows a simplified diagram for this case.

Using the series- and then parallel-springs equivalent spring formulas derived in the ME 3011 NotesBook:

1 2

1 2

3 4

3 4

a

b

k k k

k k

k k k

k k

3 4 1 2 3 1 2 4 1 3 4 2 3 41 2

1 2 3 4 1 3 2 4 1 4 2 3

eq a b

k k k k k k k k k k k k k k k k k k k

k k k k k k k k k k k k

The system model is given below.

( ) ( ) ( ) ( )eq eqmx t c x t k x t f t

Where 1 2 3 1 2 4 1 3 4 2 3 4

1 3 2 4 1 4 2 3

eq

c c c c c c c c c c c cc

c c c c c c c c

This example is for a translational system, but equivalent results exist for an analogous rotationalsystem.

1

3

k

x(t)

mk

2

4

k

k f(t)

k x(t)

mk

f(t)

a

b



209

5. Electric motor with vibration isolation FBD

Often a vibration isolation system will be designed to minimize the transference of shakingforces and shaking moments from rotating machinery to the ground. The rotating electric motor shownabove, whose moment of inertia about the motor shaft axis is J , is supported by four identical springs (2shown, 2 symmetric into the page on the back side of the motor) of spring constant k . We measure

angular displacement (t ) according to the right-hand-rule (positive CCW) such that (t ) is zero whenthe system is horizontal. First, draw the FBD.

Using Euler’s Rotational Dynamics Law for the motor shaft, in words we can say that the

excitation torque caused by the inertia J and angular acceleration ( )t of the motor shaft must be

balanced by the restoring torque due to the four translational springs through the distance a. Assuming positive angular acceleration, the two left translational springs are in compression and the two righttranslational spring are in tension in applying the restoring torques.

( )

4 ( ) ( )

M J t

kax t J t

Where x(t ) is the total displacement of each spring. Now substitute ( ) ( ) x t a t (approximation

for small displacement) into the above equation to yield the final 1-dof model for this system.

2( ) 4 ( ) 0 J t ka t

J k

a

(t)

k



210

6. Sprung mass/cylinder FBDs

In this dynamic system, m1 is the point mass suspended by cable about a cylindrical pulley ofmass m2 and radius r . The other end of the cable is attached to a linear translational spring withconstant k . We measure displacement x(t ) (positive downward) from the neutral position of the spring.

We also measure angular displacement (t ) with the right-hand-rule (positive CCW) such that it is zerowhen x(t ) is zero. First, draw the FBDs.

Using Newton’s Second Law for the point mass:

1 1 ( ) X

F m g T m x t

Using Euler’s Rotational Dynamics Law for the cylinder:

( ( )) ( ) Z

M Tr kr x t J t

Where ( ) x t is the total displacement of the spring ( is the static deflection of the spring due

to gravity). Also, the mass moment of inertia of the cylindrical pulley is

2

2 2 J m r . Now substitute( ) ( ) x t r t and ( ) ( ) x t r t into the above equations.

1 1

2

2

( )

( ( )) ( )2

m g T m r t

m r Tr kr r t t

We can combine the above two equations to eliminate the cable tension T .

1 1

22 2

1 1

( )

( ) ( ( )) ( )2

T m g m r t

m r m gr m r t kr r t t

At static equilibrium we have 1m gr kr , so the final 1-dof model for this system is

22 2 2

1

21

( ) ( ) ( )2

( ) ( ) 02

m r m r t kr t t

mm t k t

m

m

k

x(t)

r (t)



211

7. Doubly-sprung torsional rod FBD

In this dynamic system, m is the distributed mass of the slender rigid rod of length L. The rod’sleft end is connected to ground by a torsional spring k R and the rod’s right end is supported by two

identical linear translational springs k as shown. We measure angular displacement (t ) in a right-handed sense from the horizontal rod position. To linearize we will make a small angle assumption; wefurther ignore the small horizontal motion and gravity loading of the springs. First, draw the FBD.

Using Euler’s Rotational Dynamics Law for the slender rod:

( ) 2 ( ) ( ) Z R M k t kLy t J t

The mass moment of inertia of the slender rod about its end is 2 3 J mL . Also, ( ) sin ( ) y t L t ,

leading to:

22( ) ( ) 2 sin ( ) 0

3 R

mLt k t kL t

Applying the small angle assumption, we have sin ( ) ( )t t and the final 1-dof model for this system

is

2

2( ) 2 ( ) 03

R

mLt k kL t

L

k

y(t)

(t)

k

k R



212

8. Sprung roll ing cylinder FBD

In this dynamic system, a cylinder of mass m and radius r rolls without slipping on a horizontalsurface. The cylinder center is attached to the ground via a linear translational spring with constant k .We measure displacement x(t ) (positive to the right) from the neutral position of the spring. We also

measure angular displacement (t ) with the right-hand-rule (positive CCW) such that it is zero when x(t )is zero. First, draw the FBD.

Using Newton’s Second Law for the cylinder (F f is the force due to friction):

( ) ( ) X f F F kx t mx t


( ) Z f M F r J t

We can combine the above two equations to eliminate the friction force F f :

( ) ( )

( ( ) ( )) ( )

f F mx t kx t

mx t kx t r J t

The mass moment of inertia of the cylindrical pulley is 2 2 J mr . Now substitute

( ) ( )t x t r into the above equation to express the motion in terms of x(t ).

2

( )( ( ) ( ))2

mr x t mx t kx t r r

The final 1-dof model for this system is

3( ) ( ) 0

2mx t kx t

m

k

x(t)

r (t)



213

9. Doubly-sprung rotating cylinder FBD

In this dynamic system, a rotating cylindrical pulley of mass m and radius r is suspended fromthe ground link as shown. A massless inextensible cable is passed over the pulley and connected to twogrounded linear translational springs of equal spring constant k . At static equilibrium both springs are

pre-tensioned by a vertical displacement . We measure angular displacement (t ) with the right-hand-rule (positive CCW). First, draw the FBD.


1 2 ( ) Z

M T r T r J t

Where the left cable tension is 1 ( ( ))T k r t and the right cable tension is 2 ( ( ))T k r t .

Also, the mass moment of inertia of the cylindrical pulley is 2 2 J mr .

2

( ( )) ( ( )) ( )2

mr kr r t kr r t t


( ) 4 ( ) 0m t k t

m

k

r (t)

k



214

10. Sprung pendulum FBD

In this dynamic system, a standard point-mass m pendulum is tied to the ground through a linear

translational spring with constant k , connected a distance L0 along the length L of the massless pendulumrod. We measure angular displacement (t ) with the right-hand-rule (positive CCW) such that it is zerowhen the pendulum is vertical (down). First, draw the FBD.

Using Euler’s Rotational Dynamics Law for the pendulum:

sin ( ) cos ( ) sin ( ) ( ) Z M mgL t a t ka t J t

Where sin ( )ka t is the amount the spring stretches. The mass moment of inertia of the point-

mass pendulum about the point of rotation is 2 J mL .

2 2

( ) ( cos ( ))sin ( ) 0mL t mgL ka t t

This is a nonlinear model; in addition to the sine of the regular pendulum, now we also have a cosine

term of the unknown angular displacement (t ). To linearize let us assume small angle motion.

sin ( ) ( )t t

cos ( ) 1t

We are implicitly ignoring the vertical motion of the spring. The final 1-dof model for this system is

2 2( ) ( ) ( ) 0mL t mgL ka t

m

k

L

(t)

L0



215

11. Doubly-sprung rolling cylinder FBD

In this dynamic system, a cylinder of mass m and radius r rolls without slipping on a horizontalsurface. The cylinder is attached off-center (distance a from the center to the springs connection point)to the ground via two identical linear translational springs with constant k . We measure displacement x(t ) (positive to the left) from the equilibrium position of the springs. At static equilibrium both springs

are pre-tensioned by an equal horizontal displacement . We also measure angular displacement (t )with the right-hand-rule (positive CCW) such that it is zero when the springs are in their equilibrium position. First, draw the FBD.

Using Newton’s Second Law for the cylinder (F f is the force due to friction):

( ( ) ( )) ( ( ) ( )) ( ) ( ) X f F F k r a t k r a t mx t mr t


( ( ) ( )) ( ( ) ( )) ( ) Z f

M F r ka r a t ka r a t J t

Where the left spring force tension is ( ( ) ( ))k r a t and the right cable tension is

( ( ) ( ))k r a t . Also, the mass moment of inertia of the cylinder is 2 2 J mr . We can combine theabove two equations to eliminate the friction force F f :

2

( ) 2 ( ) ( )

( ( ) 2 ( ) ( )) 2 ( ) ( ) ( )2

f F mr t k r a t

mr mr t k r a t r ka r a t t


2 23( ) 2 ( ) ( ) 0

2mr t k r a t

m

k x(t)

r (t)

k

a



216

12. Hung-sprung mass/cylinder FBDs

In this dynamic system, m1 is the point mass suspended by an inextensible cable about acylindrical pulley of mass m2 and radius r . The other end of the cable is rigidly attached to the ground.The pulley is attached to ground via a linear translational spring with constant k . We measuredisplacements x1(t ) and x2(t ) (both positive downward) from the neutral position of the spring. We also

measure angular displacement (t ) with the right-hand-rule (positive CCW) such that it is zero when x(t )is zero. First, draw the FBDs.


1 1 1 1( ) X F m g T m x t

Using Newton’s Second Law for the cylinder:

1 2 2 2 2 2( ( )) ( ) X F T T m g k x t m x t

Where is the static deflection of the spring due to gravity. Using Euler’s Rotational Dynamics Law forthe cylinder:

1 2 ( ) Z

M T r T r J t

The mass moment of inertia of the cylindrical pulley is 22 2 J m r . Substituting the mass moment of

inertia and eliminating the cable tensions T 1 and T 2 amongst these three equations yields

m

m

k

x (t)

r (t)

x (t)



217

1 1 1 1

2 2 2 2 2 1

21 1 1 2 2 2 2

( )

( ) ( ( ))

2( ( )) ( ) ( ( )) ( )2

T m g m x t

T m x t m g k x t T

m r m g m x t m x t m g k x t t

At static equilibrium we have

1 2 2

1 2(2 )

T T m g k

m m g k

So the model simplifies to

21 1 2 2 2( ) 2 ( ) ( ) ( ) 0

2

m r t m x t m x t kx t

Note that from kinematics we know 1 2( ) 2 ( ) x t x t and 2( ) ( ) x t r t . Substituting the kinematic

equations yields the final 1-dof model for this system.

1 2 1 1

34 ( ) ( ) 0

2m m x t kx t

Interestingly, the same exact equation form results, regardless of whether the model variable is x1(t ), as

above, or x2(t ), or (t ).



218

13. Horizontally-sprung mass/cylinder FBDs

In this dynamic system, m1 is the point mass suspended by cable about a cylindrical pulley ofmass m2 and radius r . The pulley is grounded via a linear translational spring with constant k , attacheddistance a from the rotational center of the pulley. We measure displacement x(t ) (positive downward)

from the neutral position of the spring. We also measure angular displacement (t ) with the right-hand-

rule (positive CCW) such that it is zero when x(t ) is zero. First, draw the FBDs.


1 1 ( ) X F m g T m x t


( ( )) ( ) Z M Tr ka a t J t

Where ( )a t is the total displacement of the spring ( is the static deflection of the spring due

to gravity). Also, the mass moment of inertia of the cylindrical pulley is 2

2 2 J m r . We can combine

the above two equations to eliminate the cable tension T .

1 1

2

21 1

( )

( ( )) ( ( )) ( )2

T m g m x t

m r m g m x t r ka a t t

At static equilibrium we have 1m gr k a , so the model becomes

22 2

1 ( ) ( ) ( )2

m r m rx t ka t t

Now using the kinematic relationships ( ) ( ) x t r t and ( ) ( ) x t r t in the above equation, the final 1-

dof model for this system is

2 221 ( ) ( ) 0

2

mm r t ka t

m

m

k

x(t)

r (t)a



219

14. Cylinder rolling in a big cylinder FBD

In this dynamic system, a cylinder of mass m and radius r rolls without slipping in a concave

cylindrical surface of radius R as shown. 1(t ) is the angular displacement of the cylinder and 2(t ) is theangular displacement of the cylinder CG from the vertical. Both angles are measured with the right-

hand-rule (positive CCW) such that they are zero when the cylinder is at static equilibrium at the bottomof the well. We define the moving X axis such that it is perpendicular to the radial direction, pointing tothe left. First, draw the FBD.

Using Newton’s Second Law for the cylinder:

2sin ( ) ( ) X f F mg t F mx t


1 2( ( ) ( )) Z f

M F r J t t

Note we must use the absolute angular acceleration1 2( ) ( )t t in the above equation. The mass

moment of inertia of the cylindrical pulley is 2 2 J mr . Substituting the mass moment of inertia and

eliminating the friction force F f between these two equations yields

2

2 1 2

sin ( ) ( )

sin ( ) ( ) ( ( ) ( ))2

f F mg t mx t

mr mg t mx t t t

Note that from kinematics we know2( ) ( ) ( ) x t R r t and 1 2( ) ( )t R t r . Substituting the

kinematic equations yields

2 23( ) ( ) sin ( ) 0

2 R r t g t

We apply a small angle assumption such that 2 2sin ( ) ( )t t . The final linearized 1-dof model for this

system is

2 2

3( )( ) ( ) 0

2

R r t g t

m

r

(t)

R

1

(t)2



220

15. Doubly-sprung pulleys/mass FBDs

In this dynamic system, m is the point mass suspended by an inextensible cable about two pulleys of negligible mass as shown. Each pulley is connected to the ground via linear translationalsprings with constants k 1 and k 2, respectively. We measure displacements x(t ) (positive downward), x1(t ), and x2(t ) as shown.

The kinematics relationship is:

1 2( ) 2 ( ) 2 ( ) x t x t x t

From a FBD for pulley 1 and pulley 2:

1 1

2 2

2 ( ) ( )

2 ( ) ( )

T t k x t

T t k x t

Substituting the above relationships into the kinematics relationship yields

1 2

4 ( ) 4 ( )( )

T t T t x t

k k

Looking at the point mass we can write

( ) ( ) EQk x t T t

Where k EQ is the equivalent spring constant.

1 2

1 2

( )

( ) 4 EQ

k k T t k x t k k


1 2

1 2

( ) ( ) 0

( ) ( ) 04

EQmx t k x t

k k mx t x t

k k

mk

x(t)

k

x (t) x (t)



221

16. Rolling cylinder/pendulum FBD

In this dynamic system, a standard point-mass pendulum is rigidly connected to a cylinder that is

free to rotate without slipping on a horizontal surface. M 1 is the point mass and the cylinder has mass m2 and radius r . The length of the massless pendulum rod is L. We measure angular displacement (t ) withthe right-hand-rule (positive CCW) such that it is zero when the pendulum is vertical (down). First,draw the FBD.

At any instant of the motion, the rotation is about (moving) point C , the instantaneous center ofzero velocity. Therefore let us take point C as the moment center for deriving the dynamic model

( )CZ C M J t . The mass moment of inertia of the compound rigid link, using the parallel axis

theorem, is:2

2 2 2 221 2 1 2

3

2 2C

m r J m R m r m R m r

Where, by use of a vector loop-closure equation, 2 2 22 cos ( ) R L Lr t r . Using Euler’s

Rotational Dynamics Law for the cylinder:

1

2 2

1 2 1

2 2 2

1 2 1

sin ( ) ( )

3( ) sin ( ) 0

2

3( 2 cos ( ) ) ( ) sin ( ) 0

2

CZ C M m gL t J t

m R m r t m gL t

m L Lr t r m r t m gL t

This is a nonlinear model; in addition to the sine of the regular pendulum, now we also have a cosineterm of the unknown angular displacement (t ). To linearize let us assume small angle motion.

sin ( ) ( )t t

cos ( ) 1t


2 2

1 2 1

3( ) ( ) ( ) 0

2m L r m r t m gL t

m

L

(t)

1

m2

r



222

17. Rigid-body pendula Earlier we derived the linearized dynamic model for a simple pendulum consisting of a massless

rigid rod of length L and a point mass m.

( ) ( ) 0g

t t L

We also derived the dynamic model for a slender-rod pendulum with distributed mass and alsofor a slender-rod pendulum with distributed mass plus point mass. Here we present the dynamic modelsfor pendula of various other rigid-body shapes, including for the general case. The pendulum mass is m,O is the fixed point of rotation, G is the pendulum center of mass, and r G is the distance between these

two points. As before, we measure angular displacement (t ) with the right-hand-rule (positive CCW)such that it is zero when the pendulum is vertical (down). Draw the FBD for the general rigid-body pendulum below.

General Rigid-Body Pendulum FBD

2

G

O G G

J

J J mr ( ) ( ) 0G

O

mgr t t

J

Note that if we define a virtual point mass pendulum length of EQ O G

L J mr , we could use the simple

point-mass pendulum model for any rigid body pendulum. Now we can apply these results to specific

rigid-body pendula below.

Semi-circular Circular Rectangular Square

2

43

2

G

O

r r

mr J

2

2

2

3

2

G

O

mr J

mr J

2 2

2 2

( )12

(4 )

12

G

O

m L h J

m L h J

2

2

6

5

12

G

O

ms J

ms J

8( ) ( ) 0

3

gt t

r

2( ) ( ) 0

3

gt t

r

2 2

6( ) ( ) 0

4

gLt t

L h

6( ) ( ) 0

5

gt t

s

r

g

(t)

m

r

g

(t)

m

h

g

(t)

m

L

g

(t)

m

s

s



223

18. Beams with a point mass

Earlier we derived the equivalent spring constants k EQ for three massless beams with pointmasses attached as shown, the cantilevered beam, the simply-supported beam, and the rigidly-clamped beam. For all three cases the general model is ( ) ( ) 0 EQ

mx t k x t .

Cantilevered beam Simply-supported beam

3

3( ) ( ) 0

4

Ebhmx t x t

L

3

3

4( ) ( ) 0

Ebhmx t x t

L

Doubly-fixed beam3

3

16( ) ( ) 0

Ebhmx t x t

L

Where, for all three cases: E material Young’s modulusb beam widthh beam height L length of beam

m

L x(t)

hm

L x(t)

h

m

L x(t)

h



224

19. Rotating imbalance

Machines with rotating parts are common in industry. The figure below represents a simplifiedidealized machine with an offset mass m1 that is unbalanced in rotation. The total mass of the machineis m2, supported by equivalent spring and damping coefficients k and c, respectively, to the ground(there may be four machine mounts, one on each corner of the machine, in parallel, so their individual k i and ci coefficients add up to yield the overall k and c). The machine is constrained to move in thevertical direction only. We measure displacement x(t ) in the vertical (up) direction, from the gravity-loaded spring location.

The model is the same we have derived earlier for the standard m-c-k translational mechanicalsystem.

2 ( ) ( ) ( ) ( )m x t cx t kx t f t

Where now the forcing function f (t ) is supplied by the rotating imbalance, mass m1 at an offset of e, as

shown. Assume the rotation is at a constant angular velocity of . Then the centripetal acceleration of

the rotating mass is 2e (directed inward) and thus the inertial force of the rotating mass is 2

1m e

(directed outward). The angle of rotation is t and so the vertical component of the rotating imbalanceis:

2

1( ) sin f t me t

And so the final 1-dof model for this system is

2

2 1( ) ( ) ( ) sinm x t cx t kx t m e t

m1

m2

k c

x(t)

e

t



226

21. Horizontal sprung pendulum FBD

In the figure above, a horizontal simple pendulum has a point mass on the end of a massless rigidrod of length a. The rod extends by length b to the other side to connect to the ground through a linear

translational spring with constant k . We measure angular displacement (t ) with the right-hand-rule(positive CCW) such that it is zero when the pendulum is horizontal. We further measure displacementvariable x(t ) as the extension/compression of the spring as shown. First, draw the FBD.


cos ( ) ( ( )) ( ) Z M mga t bk x t J t

Where ( ) x t is the total displacement of the spring ( is the static deflection of the spring in

tension due to gravity). Also, the mass moment of inertia of the pendulum about the axis of rotation is2

J ma .

2 ( ) ( ( )) cos ( ) 0ma t bk x t mga t

This is a nonlinear model, due to the cosine term of the unknown angular displacement (t ). To

linearize let us assume small angle motion cos ( ) 1t .

2 ( ) ( ( )) 0ma t bk x t mga

At static equilibrium we have mga kb , so the simplified model for this system is

2 ( ) ( ) 0ma t bkx t

For a small displacement x(t ) the approximation ( ) sin ( ) x t b t holds good. Once again assuming

small angle motion, to linearize we use sin ( ) ( )t t , so the final 1-dof model for this system is

2 2( ) ( ) 0ma t kb t

m

k

a

(t)

x(t)b



227

22. Sprung cylinder FBD

In this dynamic system, a cylinder of mass m and radius r is suspended via a cable on one sideand connected to ground via a linear translational spring with constant k on the other side. We measuredisplacement x(t ) (positive downward) from the neutral position of the spring and y(t ) is the

displacement of the cylinder center. We also measure angular displacement (t ) with the right-hand-rule(positive CCW) such that it is zero when x(t ) is zero. First, draw the FBD. Using Newton’s SecondLaw for the cylinder:

( ( )) ( ) ( ) X F mg k x t T t my t

Where is the static deflection of the spring due to gravity. Using Euler’s Rotational Dynamics Law forthe cylinder:

( ) ( ( )) ( ) Z

M T t r k x t r J t

The mass moment of inertia of the cylindrical pulley is 2 2 J mr . Substituting the mass moment of

inertia and eliminating the cable tension T (t ) between these two equations yields

( ) ( ( )) ( )

2 ( ( )) ( ) ( )2

T t mg k x t my t

mr mg k x t my t t

At static equilibrium we have

2

mgk T

So the model simplifies to

( ) ( ) 2 ( ) 02

mr t my t kx t

Note that from kinematics we know ( ) 2 ( ) x t y t and ( ) ( ) y t r t . Substituting the kinematic equations

yields the final 1-dof model for this system.

3( ) 4 ( ) 0

2m t k t

k

x(t)r

(t)

y(t)



228



229

23. Vertical sprung pendulum with damping FBD

In the figure above, a simple pendulum rotating about point O has a point mass on the end of amassless rigid rod of length L. The rod extends by length b to the other side of point O to connect to theground through a horizontal linear translational spring with constant k . We measure angular

displacement (t ) with the right-hand-rule (positive CCW) such that it is zero when the pendulum is

vertical. We further measure displacement variable ( ) ( )b x t b t as the compression/extension of the

spring as shown. A viscous dashpot is connected by length a down along the rod to provide damping,

with velocity variable ( ) ( )a x t a t . First, draw the FBD.


sin ( ) ( ) cos ( ) ( ) cos ( ) ( )OZ M mgL t kb t b t ca t a t J t

The mass moment of inertia of the pendulum about the axis of rotation is 2 J mL .

2 2 2( ) sin ( ) ( )cos ( ) ( )cos ( ) 0mL t mgL t kb t t ca t t

This is a nonlinear model; in addition to the sine of the regular pendulum, now we also have a cosine

term of the unknown angular displacement (t ). To linearize let us assume small angle motion.

sin ( ) ( )t t

cos ( ) 1t

Then the final 1-dof model for this system is

2 2 2( ) ( ) ( ) ( ) 0mL t ca t mgL kb t

m

k

L

(t)

ca

b

x (t)b



230

24. Mass suspended via tensed string

In this system a point mass m is suspended between two strings of tension T . Assume that forsmall motions the tension can be considered to be constant. Also assume there is only verticaldisplacement, i.e. ignore horizontal motion of the mass. We measure displacement x(t ) positivevertically (up).


( ) X F f mx t

Where f is the restoring force due to both strings:

1 1 2 2 1 2sin ( ) sin ( ) (sin ( ) sin ( )) f T t T t T t t

Assuming small displacement we also have small angular motions. For small angular motionssin ( ) tan ( )t t and then we have

1 2 1 1

( ) ( ) ( ) ( )( ) ( ) x t x t x t x t

f T T L L L L L

The final 1-dof model for the tensed-string system is

1 1

1 1

1 1( ) ( ) ( ) 0

( ) ( ) ( ) 0( )

mx t T x t L L L

Lmx t T x t

L L L

m

L

L 1 L 2

L 1 L 2

x(t)

T T



231

25. Manometer

A manometer is used for measuring pressure. The manometer shown has a uniform cross-

sectional area A, a liquid column of length L, and a liquid density of . We measure displacement x(t ) positive vertically as shown, relative to atmospheric pressure, where the two columns have equal height.

Using Newton’s Second Law for the fluid mass:

( ) ( ) X F w t mx t

Where w(t ) is the weight acting due to the difference in height in the manometer. Therefore:

( ) ( ) 2 ( )w t gV t gAx t

Since the total liquid mass m in the manometer is AL , the final 1-dof manometer dynamics

model is

( ) 2 ( ) 0

( ) 2 ( ) 0

( ) 2 ( ) 0

mx t gAx t

ALx t gAx t

Lx t gx t

x(t)

x(t)



232

26. Cylinder in water bath

A solid buoyant cylinder is immersed in a water bath. Derive the vibratory dynamic equation ofmotion assuming purely vertical motion with no rotations. The cylinder has a height h and radius r and

thus a mass of 2m V r h , where is the cylinder material density. Further let W be the water

density. Displacement x(t ) is the amount of the cylinder height under water at any instant, as shown.From Archimedes’ principle, the time-varying restoring force f R due to the water is the weight of waterdisplaced.

2( ) ( ) R W W W f t V g g r x t

The vibratory equation of motion is

2 2

( ) ( ) 0

( ) ( ) 0

( ) ( ) 0

W

W

mx t kx t

r hx t g r x t

hx t gx t

x(t)

h

r



233

27. Human skeletal muscle model

This section presents a common single skeletal muscle dynamics model, including the figure,equations, state-space form, and simulations. This is a vibrational model, included for completeness, toshow one type of engineering mechanics modeling for muscles. In this class we will generally treat theeffect of muscles more simply, i.e. as simple cables in tension with no mass, spring constants, ordynamics of their own.

The model in this section was suggested by Dr. Scott Hooper (OU Biological Sciences) and theequations were derived and the simulations performed by Elvedin Kljuno (OU Fulbright Scholar fromBosnia). Dr. Hooper uses this model to study neural muscular control in the stomachs of lobsters, but hesays it can be scaled to adequately model skeletal muscle in many animals, including humans.

The skeletal muscle vibrational dynamics model is shown below. The lumped mass m representsthe load the muscle is lifting vs. gravity g. The muscle itself is represented by linear elastic springstiffness k 1, in parallel with linear elastic spring stiffness k 2 that is in series with linear dissipativedashpot b. The absolute displacement of the dashpot end and the muscle end are measured bycoordinates x and y, respectively. Lengths L1R and L2R (not shown) are the resting lengths (not stretched

by gravity) of springs 1 and 2, respectively. In vibrations it is more common to measure change indisplacements relative to the gravity-stretched position, but biologists need to include the absolute springlengths as well. The actuator F A represents the contractile element of the muscle and F m is the force

generated by the muscle. Note: this derivation assumes zero pennation angle, i.e. zero angle betweenthe muscle fibers and the tendons.

Dynamic Skeletal Muscle Diagram



234

The parallel elastic component k 1, provided by the muscle membranes, gives spring resistancewhen passively stretched. The series elastic component k 2 represents the tendons, storing elasticenergy when a muscle is stretched. The tension-generating contractile effect of the sarcomeres,modeled by actuator force F A, is in parallel with the membranes and in series with the tendons.

From the left figure below, the equation for the left spring is:

1 1R 1

( )k y L F (1)

From the right figure above, the equations for the right spring and dashpot are:

2 2R 2

2 A

( )k y x L F

bx F F

(2, 3)

From the FBD shown below, the dynamics equation for the mass is obtained using Newton’s Secondlaw:

1 2 m

vert vert F ma

F F F mg my

(4)



235

Substituting (1) and (2) into (4), and also substituting (2) into (3) yields the following coupledsecond/first order linear vibrational dynamics model for skeletal muscle.

1 1R 2 2R m

2 2R A

( ) ( )

( )

my k y L k y x L mg F

bx k y x L F

(5, 6)

Equations (5, 6) can be written in the following form:

2 m1 2 1 1R 2 2R

2 2 A 22R

1 1( ) ( )

k F y k k y x k L k L g

m m m m

k k F k x x y L

b b b b

(7, 8)

As mentioned above, (7, 8) can be simplified by choosing more adequate coordinates that willmeasure the deviation from the static position only. The form shown above is formed intentionally for

biological systems, using absolute coordinates and initial spring lengths.

The system of differential equations (7, 8) can be solved for coordinates x and y as functions oftime given time functions for F A and F m. In the case of a simple form of F A and F m the solution could be found using reduction and decoupling by a formation of a single third-order linear ordinarydifferential equation.

2 1 2 1 2 21 1

( )( )( ) ( ) ( ) ( ) ( ( ) ( ) )m

A m R

F t k k k k k k y t y t y t y t F t F t mg k L

b m mb m mb

(9)

Equation (9) can be solved for y(t ) which is then substituted into (8) to directly solve for x(t ).



236

Common Model Features

Considering all vibrational models we have derived, including the many cases in this section,some general comments can be made about all models.

1.

When gathering the unknown and its time derivatives on one side of the ODE, there willnever be a change in sign amongst these terms; generally they will all be positive.

2. The effect of gravity always cancels out. That is, any mg static weight terms will load thespring with a static deflection and these terms always subtract out of the final ODE.

3. As always, the resulting units must be consistent. This is a good check to perform on yourresulting model.



237

6.5 Electrical Circuits ModelingThis section presents modeling of electrical circuits. Like mechanical translational and rotational

systems, electrical circuits have dynamics described by IVP ODEs, including input and output variablesthat are functions of time. We will make analogies between the mechanical and electrical systems, i.e.the models they yield are very similar. Hence, we can use the same techniques for solving thesemechanical and electrical ODEs (in Chapters 3 and 4).

Many electrical circuits can exhibit electrical vibrations, i.e. oscillation of the electrical outputvariable of interest, which is why their modeling is included here.

Kirchoff’s Laws

Instead of Newton’s and Euler’s dynamics laws for mechanical systems, we use Kirchoff’s Lawsto derive the models for electrical circuits.

Kirchoff’s Current Law (KCL) states that the sum of currents flowing into any circuit node iszero.

Kirchoff’s Voltage Law (KVL) states that the sum of voltages around any circuit loop is zero.

1

( ) 0n

j

j

i t

1

( ) 0m

j

k

v t

KCL Diagram KVL Diagram

Electrical Circuit Elements

In order to use Kirchoff’s Laws to derive models for electrical circuits, we need the followingrelationships. They relate the current i(t ) flowing through and the voltage v(t ) across the three standardelectrical elements. A capacitor C stores electrical energy, a resistor R dissipates electrical energy, andan inductor L oscillates electrical energy. Note the equations in the last two columns of each row areequivalent, solved for the current i(t ) or the voltage v(t ).

element notation units i( t) v( t)

capacitor C Farads

( )( ) C

C

dv t i t C

dt

1

( ) ( )C C v t i t dt

C

resistor R Ohms

( )( ) R

R

v t i t

R ( ) ( ) R Rv t Ri t

inductor L Henrys

1( ) ( )

L Li t v t dt L

( )

( ) L L

di t v t L

dt



238

Additional Electrical Circuit Variables

In some electrical circuit modeling you may encounter two more variable types, electrical charge

and electrical flux. As shown in Dr. Bob’s on-line Atlas of Models and Transfer Functions,

www.ohio.edu/people/williar4/html/PDF/ModelTFAtlas.pdf

electrical charge q(t ) is the time integral of current i(t ) and electrical flux (t ) is the time integral of

voltage v(t ).

( ) ( )q t i t dt

( ) ( )t v t dt

Convention for Electron Flow

When early electricity experimenters and early electrical engineer pioneers established the

convention for the flow of electrons composing current, they got it backwards. This error persists to thisday, because it really doesn’t matter in the equations.

xkcd.com



239

Current-Driven Parallel R- L-C Electrical Circuit Model

The system diagram for the current-driven parallel R- L-C electrical circuit is shown below. Theinput is current i(t ), and the output is voltage v(t ). The voltage across the current generator, the resistor R, the inductor L, and the capacitor C are all identical, v(t ).

Parallel R-L-C Circuit Diagram

Using Kirchoff’s Current Law (the sum of currents into any circuit node is zero), we have:

( ) ( ) ( ) ( ) 0

( ) ( ) ( ) ( )

R L C

C R L

i t i t i t i t

i t i t i t i t

Substituting the current through each element as a function of voltage v(t ) from the previous table, weobtain the model for the current-driven parallel R- L-C circuit.

( ) 1 1( ) ( ) ( )

dv t C v t v t dt i t

dt R L

This model is a linear, lumped-parameter, constant-coefficient, integro-differential equation. It issecond-order, since, although the highest derivative order is first-order, there are two time differentiationsteps between the integral and the first derivative. This equation is written in standard form, where theoutput variable v(t ) appears with its derivative and integral on the left-hand-side and the input forcingfunction i(t ) appears on the right-hand-side of the equation. The external current input is i(t ) (amps) andthe output is voltage v(t ) (volts).

i(t) R L C v(t)

+

-

i (t) L

i (t) R

i (t)C



240

Force-Current Analogy

We can make an analogy between the mechanical and electrical modeling worlds by comparingthe translational (and rotational) mechanical system with the current-driven parallel R- L-C circuit model.This is called the force-current analogy.

We first rewrite the left-hand-side of the translational mechanical system model in terms ofvelocity v(t ) instead of displacement x(t ). We can also compare to the rotational mechanical system,

rewritten in terms of angular velocity (t ) instead of angular displacement (t ).

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

mx t cx t kx t f t

mv t cv t k v t dt f t

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

R R

R R

J t c t k t t

J t c t k t dt t

( ) 1 1( ) ( ) ( )


dt R L

Compare the mechanical systems ODEs to the parallel R- L-C circuit model (repeated above).

Energy Role TranslationalMechanical

RotationalMechanical

Electrical

store energy( )

( )S

dv t f t m

dt

( )( )

S

d t t J

dt

( )( )

S

dv t i t C

dt

dissipate energy ( ) ( ) D f t cv t ( ) ( ) D R

t c t 1

( ) ( ) Di t v t

R

oscillate energy ( ) ( )O f t k v t dt ( ) ( )

O Rt k t dt 1

( ) ( )Oi t v t dt

L

Force-Current Analogy

Variable TypeTranslational

Mechanical System

Rotational

Mechanical System

R- L-C Parallel

Circuit

input (through) f (t ) (t ) i(t )

output (across) v(t ) (velocity) (t ) v(t ) (voltage)

inertia m J C

damping c c R 1 / R

stiffness k k R 1 / L

Caution The Force-Voltage Analogy, presented next, based on the voltage-driven series R- L-C circuit

model, is different.



241



242

Voltage-Driven Series R- L-C Electrical Circuit Model

The system diagram for the voltage-driven series R- L-C electrical circuit is shown below. Theinput is voltage v(t ), and the output is current i(t ). The current i(t ) flowing through all elements is thesame. The input voltage source v(t ) increases as shown negative to positive; the voltage drops acrosseach of the circuit elements, positive to negative in each case.

Series R-L-C Circuit Diagram

Using Kirchoff’s Voltage Law (the sum of voltages around the circuit loop is zero), we have:

( ) ( ) ( ) ( ) 0

( ) ( ) ( ) ( )

R L C

L R C

v t v t v t v t

v t v t v t v t

Substituting the voltage drop across each element as a function of current i(t) from the first table in thissection, we obtain the model for the voltage-driven series R- L-C circuit.

( ) 1( ) ( ) ( )

di t L Ri t i t dt v t dt C

This model is a linear, lumped-parameter, constant-coefficient, integro-differential equation. Again, it issecond-order since there are two time differentiation steps between the integral and the first derivative.This equation is written in standard form, where the output variable i(t ) appears with its derivative andintegral on the left-hand-side and the input forcing function v(t ) appears on the right-hand-side of theequation. The external input is voltage v(t ) (volts) and the output is current i(t ) (amps).

v(t) L

R

+

-i(t)

C



243

Force-Voltage Analogy

We can make a different analogy between the mechanical and electrical modeling worlds bycomparing the translational (and rotational) mechanical system with the voltage-driven series R- L-C circuit model. This is called the force-voltage analogy.

We again use the translational and rotational mechanical system models, rewritten in terms of

velocity v(t ) instead of displacement x(t ) and rewritten in terms of angular velocity (t ) instead of

angular displacement (t ).

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

mx t cx t kx t f t

mv t cv t k v t dt f t

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

R R

R R

J t c t k t t

J t c t k t dt t

( ) 1( ) ( ) ( )

di t L Ri t i t dt v t

dt C

Force-Voltage Analogy

Variable TypeTranslational

Mechanical System

Rotational

Mechanical System

R- L-C Series

Circuit

input (through) f (t ) (t ) v(t ) (voltage)

output (across) v(t ) (velocity) (t ) i(t )

inertia m J L

damping c c R R

stiffness k k R 1/C



244

Subset Electrical Circui t Models

This subsection presents subset models derived from the standard current-driven parallel R- L-C electrical circuit and the voltage-driven series R- L-C electrical circuit models.

Parallel R-L-C Circuit Series R-L-C Circuit

( ) 1 1( ) ( ) ( )


dt R L

( ) 1( ) ( ) ( )

di t L Ri t i t dt v t

dt C

Parallel R-L Circuit Series R-L Circuit

1 1( ) ( ) ( )v t v t dt i t

R L

( )( ) ( )

di t L Ri t v t

dt

Parallel R-C Circuit Series R-C Circuit

( ) 1( ) ( )

dv t C v t i t

dt R

1( ) ( ) ( ) Ri t i t dt v t

C

Parallel L-C Circuit Series L-C Circuit

( ) 1( ) ( )

dv t C v t dt i t

dt L

( ) 1( ) ( )

di t L i t dt v t

dt C



245

Parallel R-L-C models based on flux (t) and series R-L-C models based on charge q(t)

Recall that, in an electrical circuit, electrical flux (t ) is the time integral of voltage v(t ) andelectrical charge q(t ) is the time integral of current i(t ). The subset models for the current-driven parallel R- L-C electrical circuit and voltage-driven series R- L-C electrical circuit can thus be rewritten as

follows, using flux (t ) as the output variable in place of voltage v(t ) and using charge q(t ) as the output

variable in place of current i(t ), respectively. Here are the flux/voltage and charge/current substitutionrelationships.

( ) ( )

( ) ( )

( )( )

t v t dt

t v t

dv t t

dt

( ) ( )

( ) ( )

( )( )

q t i t dt

q t i t

di t q t

dt

Parallel R-L-C Circuit Series R-L-C Circuit

1 1( ) ( ) ( ) ( )C t t t i t

R L

1( ) ( ) ( ) ( ) Lq t Rq t q t v t

C

Parallel R-L Circuit Series R-L Circuit

1 1( ) ( ) ( )t t i t

R L

( ) ( ) ( ) Lq t Rq t v t

Parallel R-C Circuit Series R-C Circuit

1( ) ( ) ( )C t t i t

R

1( ) ( ) ( ) Rq t q t v t

C

Parallel L-C Circuit Series L-C Circuit

1( ) ( ) ( )C t t i t

L

1( ) ( ) ( ) Lq t q t v t

C

In the Force-Current Analogy the new output variable would be flux (t ) instead of voltage v(t )and in the Force-Voltage Analogy the new output variable would be charge q(t ) instead of current i(t ).The rest of these analogies are unchanged.



246

6.6 Multi-dof Vibrational Systems Models

This chapter presents some mechanical system models for systems with greater than one degree-of-freedom (dof). For multiple independent masses, draw a free-body diagram (FBD) for each mass andderive an ODE using Newton’s Second Law or Euler’s Rotational Dynamics Law. Generally there will be one second-order ODE for each independent mass. Normally the resulting set of n ODEs will becoupled, where n is the number of degrees-of-freedom.

1. Two-Mass Vibrational System

Two-mass system free-body diagrams

1 1 1 2 1 1 2 1 2 2 2 2 1

2 2 2 2 2 2 2 1 2 1 2

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

m y t c c y t k k y t c y t k y t u t

m y t c y t k y t c y t k y t u t

1 1 1 11 1 2 2 1 2 2

2 2 2 22 2 2 2 2

0

0

y t y t y t u t m c c c k k k

y t y t y t u t m c c k k

m

y (t)1 y (t)

2

u (t)1 u (t)

2

2m1

c2c1

k 2k 1

m1 m2

1 2 12

u (t)1 u (t)2

k y (t)1

1c y (t)

1

k (y (t) - y (t))

2 12c (y (t) - y (t))



247

2. Automotive Shock Model

This is a ¼ car suspension model. M 2 is a point mass representing one-fourth of the car mass(excluding wheels/tires). Spring constant k 2 and viscous dashpot coefficient c2 represent the shockspring and damping fluid friction dissipation elements. Point mass m1 represents one wheel/tire massand spring constant k 1 and viscous dashpot coefficient c1 represent the spring effect and energydissipation quality of the tire, respectively. The input is the road displacement z(t ), and there are twodisplacement outputs y1(t ) and y2(t ). The output y2(t ) relates to passenger comfort. Since this is a passive system, there are no force inputs.

1 1 1 2 1 1 2 1 2 2 2 2 1 1

2 2 2 2 2 2 2 1 2 1

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) 0

m y t c c y t k k y t c y t k y t c z t k z t

m y t c y t k y t c y t k y t

1 1 11 1 2 2 1 2 2 1 1

2 2 22 2 2 2 2

0 ( ) ( )

0 0

y t y t y t m c c c k k k c z t k z t

y t y t y t m c c k k

y (t)

m1

1

1c

m2

1k

y (t)2

z(t)

2c

2k



248

3. Another Two-Mass Vibrational System

1 1 1 1 2 2 1

2 2 2 2 2 1

( ) ( ) [ ( ) ( )] ( )

( ) ( ) [ ( ) ( )] 0

m y t k y t k y t y t f t

m y t cy t k y t y t

1 1 11 1 2 2

2 2 22 2 2

0 0 0

0 0 0

y t y t y t m k k k f t

y t y t y t m k k c

y (t)

f(t)

1 y (t)

2

m1 m

2

k 1

k 2

c



249

4. Inverted Pendulum

Coupled nonlinear differential equations

2

1 2 2 2

2

2 2 2

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

( ) cos ( ) ( ) sin ( ) 0

m m w t m L t t m L t t f t

m L t m L t w t m gL t

Coupled linearized differential equations

1 2 2

2

2 2 2

( ) ( ) ( ) ( )

( ) ( ) ( ) 0

m m w t m L t f t

m Lw t m L t m gL t

1 2 2

2

2 2 2

0 0( ) ( ) ( )

0( ) ( ) 0

m m m L w t w t f t

m L m L m gLt t

Y

f(t) X

g

L

m

w(t)

1

m2

(t)



250

5. Ball and Beam


2

2

2

( ) sin ( ) ( ) ( ) 0

( ) ( ) 2 ( ) ( ) ( ) ( ) cos ( ) ( )

b

b

J m p t m g t m p t t

r

m p t J J t m p t p t t m g p t t t


2

2

( ) ( ) 0

( ) ( ) ( ) ( )

b

b

J m p t m g t

r

m p t J J t m g p t t

2

2

( ) 0 ( ) 00

( ) 0 ( ) ( )0 ( )

b

b

p t mg p t J r m

t mg t t m p t J J

p ( t )

(t)

(t)



251

6. Proof-Mass Actuator


2

2

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

( ) ( ) ( ) cos ( ) ( )

M m q t kq t me t t t t

J me t meq t t n t


2

( ) ( ) ( ) ( ) 0

( ) ( ) ( ) ( )

M m q t kq t me t

J me t meq t n t

2

( ) 0 ( ) 0

( ) 0 0 ( ) ( )

M m me q t k q t

me J me t t n t

m

J

k

M

e

f(t)

q(t)

n(t)

(t)



252

7. Two Masses suspended via Tensed Strings

1 1 1 2

2 2 2 1

2( ) ( ) ( ) 0

2( ) ( ) ( ) 0

T T m x t x t x t

L L

T T m x t x t x t

L L

1 11

2 22

0 2 0

0 2 0

x t x t m T L T L

x t x t m T L T L

m L

x (t)

T T

1 m L

2

L

1 x (t)2

L L L

T T



253

8. Double Pendulum


2

1 2 1 1 2 1 2 2 1 2 1 1

2

2 1 2 1 2 2 2 2 2 2

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

( ) ( ) sin ( ) 0

m m L t m L L t m m gL t

m L L t m L t m gL t


2

1 2 1 1 2 1 2 2 1 2 1 1

2

2 1 2 1 2 2 2 2 2 2

( ) ( ) ( ) ( ) ( ) 0

( ) ( ) ( ) 0

m m L t m L L t m m gL t

m L L t m L t m gL t

21 2 1 11 2 1 2 1 2 1

22 2 22 1 2 2 2 2

( ) 0 ( ) 0( ) ( )

0 ( ) 0( )

m m gL t m m L m L L t

m gL t m L L m L t

If m1 = m2 and L1 = L2

11

22

( )2 1 2 0 0( )

( )1 1 0 0( )

t g Lt

t g Lt

L

1

1(t)

m1

L

2

2(t)

m2

g



254

9. Sprung Double Pendulum


2 2

1 1 1 1 1 1 0 1 2

2 2

2 2 2 2 2 2 0 2 1

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

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

m L t m gL t kL t t

m L t m gL t kL t t


2 2

1 1 1 1 1 1 0 1 2

2 2

2 2 2 2 2 2 0 2 1

( ) ( ) ( ( ) ( )) 0

( ) ( ) ( ( ) ( )) 0

m L t m gL t kL t t

m L t m gL t kL t t

2 2211 1 0 01 1 1

2 2220 2 2 02 2 2

( ) 00 ( )

( ) 00 ( )

t m gL kL kLm L t

t kL m gL kLm L t

If m1 = m2 and L1 = L2

2 2

0 0

11

2 222 0 0

( )1 0 0( )

( )0 1 0( )

L Lg k k

t L m L m Lt

t t L Lk g k

m L L m L

m

k

L

L0

m

L

(t)

L0

1 (t)2



256

11. Sprung Dual Cylinders

22

1 1 1 1 2 1 1 2 1 2 2

222 2

2 2 3 2 2 2 1 2 1

( ) ( ) ( ) ( ) 02

( ) ( ) ( ) ( ) 02

m r t k k r t k r r t

m r t k k r t k r r t

2

1 12

11 2 1 2 1 21

2222 1 2 2 3 222 2

0( ) 0( )( )2

( ) 0( )( )0

2

m r

t k k r k r r t

t k r r k k r t m r

If m1 = m2, and r 1 = r 2, and k 1 = k 2 = k 3

11

22

4 2

( )1 0 0( )

( )0 1 2 4 0( )

k k

t t m m

t k k t

m m

m k

r (t)

k m

r (t)

k



257

12. Sprung Cart Pendulum

1 2 22

2 2 2

( ) ( ) ( ) 2 ( ) 0( ) ( ) ( ) 0

m m x t m L t kx t m L t m Lx t m Lg t

1 2 2

2

2 2 2

2 0( ) ( ) 0

0( ) ( ) 0

m m m L k x t x t

m L m L m Lgt t

m

k x(t)

(t)

k

1

m2



258

13. Translating rotating rigid body

This is a 2-dof system, still constrained to vibrate in the vertical direction, with small-anglerotational vibrations as well.

1 2 1 1 2 2

2 2

1 1 2 2 1 1 2 2

( ) ( ) ( ) ( ) ( ) 0

( ) ( ) ( ) ( ) ( ) 0

mx t k k x t k L k L t

J t k L k L x t k L k L t

1 2 1 1 2 2

2 2

1 1 2 2 1 1 2 2

0 ( ) ( ) 0

0 ( ) ( ) 0

k k k L k Lm x t x t

k L k L k L k L J t t

m, J

k

(t)

k 1 2

x(t)

L L1 2



259

14. Doubly sprung cylinder mass

2 21 1 2 1 2 1 2 2

2 21 2 2 1 2 2

( ) ( ) ( ) ( ) ( ) ( ) 02 2

3( ) ( ) ( ) ( ) 0

2 2

m mm x t x t k k x t k x t

m m x t x t k x t k x t

2 21

1 1 2 2 1

2 2 2 22 2

( ) ( ) 02 2

( ) ( )3 0

2 2

m mm

x t k k k x t

x t k k x t m m

m

k

x (t)

r (t)2

2

k 1

m1

2

x (t)1



260

15. Cylinder rolling in a sprung big cylinder


1 2 2 1 2

1 2

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

3( )sin ( ) ( ) ( ) sin ( ) 0

2

m m x t m r r t t kx t

x t t r r t g t

The linearized differential equations become decoupled.

1 2

1 2

( ) ( ) ( ) 0

3( ) ( ) ( ) 02

m m x t kx t

r r t g t

m

r

(t)

2

2

r 1

k

m1 x(t)



261

16. Dual torsional disks coupled by a spring

22 21 1

1 1 1 2

22 22 2

2 2 2 1

( ) ( ) ( ) ( ) 02

( ) ( ) ( ) ( ) 02

R

R

m r t k ka t ka t

m r t k ka t ka t

2

1 12 2

111

2 222222 2

0( ) 0( )2

( ) 0( )0

2

R

R

m r

t k ka kat

t ka k kat m r

m

k

r (t)a1

1

k R11m

r (t)a2

2

k R22



262

17. Dual point masses

1 1 1 2 1 2 2

2 2 2 2 2 2 1

( ) ( ) ( ) 2 ( ) ( )

( ) ( ) 4 ( ) 2 ( ) 0

m x t k k x t k x t f t

m x t cx t k x t k x t

1 1 1 1 2 2 1

2 2 2 2 2 2

0 ( ) ( ) 2 ( )0 0 ( )

0 ( ) ( ) 2 4 ( )0 0

m x t x t k k k x t f t

m x t x t k k x t c

k c

m

x (t)

1k 2

1m2

L L

1 x (t)

2

f(t)



263

18. Rotating sprung point masses

2 2

( ) 2 ( ) ( ) 0

2 ( ) 2 ( ) ( ) ( )

mx t kx t kL t

mL t kL t kLx t t

2 2

0 ( ) 2 ( ) 0

0 2 ( ) 2 ( ) ( )

m x t k kL x t

mL t kL kL t t

k

m

x(t)

L L

k

mm

k

(t)

(t)



264

19. Three-Mass Vibrational System

1 1 1 2 1 1 2 1 2 2 2 2 1

2 2 2 3 2 2 3 2 2 1 2 1 3 3 3 3 2

3 3 3 4 3 3 4 3 3 2 3 2 3

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )


m y t c c y t k k y t c y t k y t c y t k y t u t


1 1 1 2 2 1 1 2 2 1 1

2 2 2 2 3 3 2 2 2 3 3 2 2

3 3 3 3 4 3 3 3 4 3 3

0 0 ( ) 0 ( ) 0 ( ) ( )

0 0 ( ) ( ) ( ) ( )

0 0 ( ) 0 ( ) 0 ( ) (

m y t c c c y t k k k y t u t

m y t c c c c y t k k k k y t u t

m y t c c c y t k k k y t u

)t

y (t)

m1 m2 m3

k 1 k 2 k 3

c1 c2 c3

k 4

c4

1 y (t)

2 y (t)

3

u (t)1

u (t)2

u (t)3



265

20. Three Masses suspended via Tensed Strings

1 1 1 2

2 2 2 1 3

3 3 3 2

2( ) ( ) ( ) 0

2( ) ( ) ( ) ( ) 0

2( ) ( ) ( ) 0

T T m x t x t x t

L L

T T T m x t x t x t x t

L L L

T T m x t x t x t

L L

1 1 1

2 2 2

3 3 3

0 0 2 0 0

0 0 2 0

0 0 0 2 0

m x t T L T L x t

m x t T L T L T L x t

m x t T L T L x t

m L

x (t)

T T

1 m L

2

L

2 x (t)3

L L L

m3

L

T T

T T

L

x (t)1



266

21. Triple Pendulum

Let m1 = m2 = m3 and L1 = L2 = L3

1 2 3 1

1 2 3 2

1 2 3 3

3 ( ) 2 ( ) ( ) 3 ( ) 0

2 ( ) 2 ( ) ( ) 2 ( ) 0

( ) ( ) ( ) ( ) 0

L t L t L t g t

L t L t L t g t

L t L t L t g t

1 1

2 2

3 3

30 0

3 2 1 ( ) ( ) 02

2 2 1 ( ) 0 0 ( ) 0

1 1 1 ( ) ( ) 0

0 0

g

Lt t g

t t

Lt t g

L

L

1

1(t)

m1

L

2

2(t)

m2

L

3

3(t)

m3

g



267

22. Sprung Trip le Pendulum

where L0 = L / 2.

2 22

1 1 2

2 2 22

2 2 1 3

2 22

3 3 2

( ) ( ) ( ) ( ) 04 4

( ) ( ) ( ) ( ) ( ) 02 4 4

( ) ( ) ( ) ( ) 04 4

kL kLmL t mgL t t

kL kL kLmL t mgL t t t

kL kLmL t mgL t t

1 1

2 2

3 3

04 41 0 0 ( ) ( ) 0

0 1 0 ( ) ( ) 04 2 4

0 0 1 ( ) ( ) 0

04 4

g k k

L m mt t k g k k

t t m L m m

t t k g k

m L m

m

k

L

L0

m

L

(t)

L0

m

k

L

L0

1 (t)2 (t)3



268

23. Sprung Cylinder with Point Masses

1 1 1 2 1 1 2 2 2 2 3

1 2 2 2 2 1 1 2 1 2 3

3 3 2 3 2 1 2 2

3 ( ) ( 9 ) ( ) 2 ( ) 6 ( ) 3 ( ) 0

(2 ) ( ) 4 ( ) 2 ( ) 6 ( ) 2 ( ) 0

( ) ( ) 3 ( ) 2 ( ) 0

m x t k k x t m x t k x t k x t

m m x t k x t m x t k x t k x t

m x t k x t k x t k x t

1 1 1 1 2 2 2 1

1 1 2 2 2 2 2 2

3 3 2 2 2 3

3 2 0 9 6 3 0

2 2 0 6 4 2 0

0 0 3 2 0

m m x t k k k k x t

m m m x t k k k x t

m x t k k k x t

m

m

k

x (t)

r

(t)

k

r/3

m

x (t)

x (t)



269

Common Model Features

Considering all multi-dof vibrational models we have derived, some general comments can bemade about the models.

mechanism kinematics & dynamics.pdf

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