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Mechanism Kinematics & Dynamics and Vibrational Modeling Dr. Robert L. Williams II Mechanical Engineering, Ohio University

NotesBook Supplement for ME 3011 Kinematics & Dynamics of Machines © 2014 Dr. Bob Productions [email protected] people.ohio.edu/williar4

k

x(t)

k

x(t)

m x(t)

These notes supplement the ME 3011 NotesBook by Dr. Bob This document presents supplemental notes to accompany the ME 3011 NotesBook. The outline given in the Table of Contents on the next page dovetails with and augments the ME 3011 NotesBook outline and hence is incomplete here.

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ME 3011 Supplement Table of Contents 1. INTRODUCTION................................................................................................................................ 4 1.3 VECTORS. CARTESIAN RE-IM REPRESENTATION (PHASORS) ............................................................. 4 1.6 MATRICES .......................................................................................................................................... 6 2. KINEMATICS ANALYSIS .............................................................................................................. 15 2.1 POSITION KINEMATICS ANALYSIS .................................................................................................... 15 2.1.1 Four-Bar Mechanism Position Analysis .................................................................................. 15 2.1.1.1 Tangent Half-Angle Substitution Derivation and Alternate Solution Method .................. 15 2.1.1.3 Four-Bar Mechanism Solution Irregularities ..................................................................... 20 2.1.1.4 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 KINEMATICS ANALYSIS .................................................................................................. 42 2.2.2 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 KINEMATICS ANALYSIS .......................................................................................... 53 2.3.2 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 KINEMATICS TOPICS ............................................................................................................ 67 2.4.1 Link Extensions Graphics......................................................................................................... 67 2.5 JERK KINEMATICS ANALYSIS ........................................................................................................... 69 2.5.1 Jerk Analysis Introduction ....................................................................................................... 69 2.5.2 Mechanism Jerk Analysis ......................................................................................................... 72 2.6 BRANCH SYMMETRY IN KINEMATICS ANALYSIS.............................................................................. 73 2.6.1 Four-Bar Mechanism ............................................................................................................... 73 2.6.2 Slider-Crank Mechanism.......................................................................................................... 75 2.7 KINEMATICS ANALYSIS EXAMPLES ................................................................................................. 77 2.7.1 Term Example 1: Four-Bar Mechanism .................................................................................. 77 2.7.2 Term Example 2: Slider-Crank Mechanism ............................................................................. 88 3. DYNAMICS ANALYSIS .................................................................................................................. 95 3.1 DYNAMICS INTRODUCTION .............................................................................................................. 95 3.2 MASS, CENTER OF GRAVITY, AND MASS MOMENT OF INERTIA ....................................................... 96 3.4 FOUR-BAR MECHANISM INVERSE DYNAMICS ANALYSIS ............................................................... 108 3.5 SLIDER-CRANK MECHANISM INVERSE DYNAMICS ANALYSIS ....................................................... 111 3.6 INVERTED SLIDER-CRANK MECHANISM INVERSE DYNAMICS ANALYSIS....................................... 113 3.7 MULTI-LOOP MECHANISM INVERSE DYNAMICS ANALYSIS............................................................ 121 3.8 BALANCING OF ROTATING SHAFTS ................................................................................................ 126 3.9 INVERSE DYNAMICS ANALYSIS EXAMPLES.................................................................................... 130 3.9.1 Single Rotating Link Inverse Dynamics Example .................................................................. 130

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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............................................................................................................................................ 145 4.1.1 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 ............................................................................................................................................. 173 4.2.1 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 ............................................................................................................. 191 6.2 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 ................................................................... 205 6.5 ELECTRICAL CIRCUITS MODELING................................................................................................. 237 6.6 MULTI-DOF VIBRATIONAL SYSTEMS MODELS ............................................................................... 246

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1. Introduction 1.3 Vectors. Cartesian Re-Im Representation (Phasors) Here is an alternate vector representation.

P  Pei The phasor Pe i is a polar representation for vectors, where P is the length of vector P , e is the natural logarithm base, i  1 is the imaginary operator, and  is the angle of vector P . ei gives the direction of the length P, according to Euler’s identity.

ei  cos  i sin  ei 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 iˆ ) and the imaginary (Im) axis is along Y (or ˆj ).  P   P cos   i P  P(cos   i sin  )   Re      Pe P P sin    Im    P   P cos   P  P(cos  iˆ  sin  ˆj )   X      PY   P sin  

A strength of Cartesian Re-Im representation using phasors is in taking time derivatives of vectors – the derivative of the exponential is easy ( d ds (e s )  e s ). d 2 P d 2 Pei  dt 2 dt 2 d 2 P d  i  ( Pe  iPei ) dt 2 dt d 2 P  i  Pe  iPei  iPei  iPei  i 2 P 2 ei dt 2 d 2 P  i  Pe  2iPei  iPei  P 2 ei dt 2  cos   2 P sin   Psin  P 2 cos  d 2 P P     dt 2  P sin   2 Pcos  Pcos  P 2sin 

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Where we had to use extensions of Euler’s identity ei  cos   i sin  iei  i cos   i 2 sin    sin   i cos  i 2 ei  i sin   i 2 cos    cos   i sin 

Compare this double-time-derivative with the XY approach. d 2P d 2  dt 2 dt 2

 Pcos     Psin 

d 2 P d  P cos  Psin     dt 2 dt  P sin  Pcos  cos  Psin  Psin  Psin  P 2 cos  d 2P  P     dt 2  P sin  Pcos  Pcos  Pcos  P 2sin   cos   2 P sin   Psin  P 2 cos  d 2 P P     dt 2  P sin   2 Pcos  Pcos  P 2sin 

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

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

d dP (t ) i (t ) dei (t )  dei ( t ) ( P (t )ei (t ) )  e  P (t )  P(t )ei (t )  P (t ) dt dt dt dt

chain rule

de i ( t ) de i ( t ) d (t )   ie i ( t ) (t ) dt d (t ) dt

The result for this example is

d ( P(t )ei (t ) )  P (t )ei (t )  P(t )iei (t )(t ) dt

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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.  a11 a12  a1n  a a22  a2 n  21   A           am1 am 2  amn  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 are not used in position analysis which requires a non-linear solution). Special Matrices square matrix (m = n = 3)

 a11  A   a21  a31

diagonal matrix

 a11  A   0  0

identity matrix

1 0 0   I 3   0 1 0 0 0 1 

transpose matrix

 A

symmetric matrix

 A    A

a12 a22 a32 0 a22 0

 a11   a12  a13

T

T

a21 a22 a23

 a11   a12  a13

column vector (3x1 matrix)

 x1   X    x2  x   3

row vector

X 

(1x3 matrix)

T

  x1

x2

a13  a23  a33  0 0  a33 

a31  a32  a33 

a12 a22 a23

x3 

(switch rows & columns)

a13  a23  a33 

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Matrix Addition

add like terms and keep the results in place

a b   e c d    g    Matrix Multiplication with Scalar

f  a  e b  f   h  c  g d  h 

multiply each term and keep the results in place

a b  ka kb  k    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.

C    A B  (mxn)  (mxp )( pxn)

That is, in a matrix multiplication product, the number of columns p in the left-hand matrix must equal the number of rows p in the right-hand matrix. If this condition is not met, the matrix multiplication is undefined and cannot be done. The size of the resulting matrix [C] is from the number of rows m of the left-hand matrix and the number 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 lefthand matrix and down the columns of the right-hand matrix (use your index fingers from the left and right hands).

Example

g c     ag  bh  ci  h  f     dg  eh  fi   i  (2x1)  (2x3)(3x1) a b C    d e 

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

8

Matrix Multiplication Examples

 1 2 3  A  4 5 6  

7 8   B   9 8   7 6 

C    A B  7 1 2 3    9  4 5 6 7   7  18  21   28  45  42

8 8  6  8  16  18   46 42   32  40  36  115 108 (2x2)  (2x3)(3x2)

 D    B  A 7 8  1 2 3  9 8   4 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 

(3x3)  (3x2)(2x3)

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Matrix Inversion

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

C    A  B  , solve for [B].

Given

C    A  B  

 A C    A  A B    I  B    B 1

1

  B    A

1

C 

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

 A A

1

  A

1

 A   I 

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

 A

1

where A



adjoint( A) A

is the determinant of [A].

adjoint( A)  cofactor( A)

T

cofactor(A)

a ij  (  1) i  j M ij

minor

minor Mij is the determinant of the submatrix with row i and column j removed.

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

C    A  B  

C  B 

1

  A B  B 

1

  A I    A

  A  C  B 

1

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

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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 matrix determinant is det(A). If a nonhomogeneous system of n linear equations in n unknowns is dependent, the coefficient matrix A is singular, and the determinant of matrix A is zero. In this case no unique solution exists to these equations. On the other hand, if the matrix determinant is non-zero, then the matrix is nonsingular, 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 d 

 A   c 

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 the current row and its column and retaining the remaining terms as an n–1 x n–1 square matrix, each of whose determinant must also be evaluated in the process. The pivot term (the entry in the cross-out row and 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 options will yield identical results). e f d f d e A  a b c h k g k g h a b c     A   d e f   g h k    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 1 and 2 outside the matrix, multiplying the downward diagonals with + signs and multiplying the upward diagonals with – signs (clearly the result is the same as in the above formula). a A d

b e

c f

a d

b 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 P 1  P 2 . iˆ P 1  P 2  p1x p2 x

ˆj

kˆ

p1 y p2 y

p1z p2 z

p1 y  iˆ p2 y

p1z p  ˆj 1x p2 z p2 x

p1z p2 z

p1x  kˆ p2 x

 p p  p1z p2 y  p1 y  1 y 2 z    p1x p2 z  p1z p2 x  p2 y    p1x p2 y  p1 y p2 x 

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System of Linear Equations

We can solve n linear equations in n unknowns with the help of a matrix. Below is an example for n = 3. a11 x1  a12 x2  a13 x3  b1 a21 x1  a22 x2  a23 x3  b2 a31 x1  a32 x2  a33 x3  b3

Where aij are the nine known numerical equation coefficients, xi are the three unknowns, and bi are the three known right-hand-side terms. Using matrix multiplication backwards, this is written as  A  x  b .  a11 a  21  a31

a12 a22 a32

a13   x1   b1      a23   x2   b2  a33   x3  b3 

where  a11  A   a21  a31

a12 a22 a32

a13  a23  a33 

is the matrix of known numerical coefficients

 x1   x   x2  x   3

is the vector of unknowns to be solved and

 b1  b  b2  b   3

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

There is a unique solution  x   A b only if [A] has full rank. If not, A  0 (the determinant of coefficient 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] are linearly independent). Gaussian Elimination is more robust and more computationally efficient than matrix inversion to solve the problem  A  x  b for {x}. 1

12

Matrix Example – solve linear equations

Solution of 2x2 coupled linear equations.

x1  2 x2  5

1 2   x1   5       6 4   x2  14 



6 x1  4 x2  14 1 2   6 4

5  14

 x1    x2 

 A  

b  

 x  

 x   A b 1

A  1  4   2  6   8

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

 A

1

check



1 A

 4 2  1/ 2 1/ 4   6 1    3/ 4 1/ 8    

 A A

1

  A

1

1 0  0 1 

 A   I2   

 x1   1/ 2 1/ 4   5  1         x2   3 / 4 1/ 8 14  2 

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 %------------------------------% Matrices.m - matrix examples % Dr. Bob, ME 3011 %------------------------------clear;

clc;

A1 = diag([1 2 3]) A2 = eye(3)

% 3x3 diagonal matrix % 3x3 identity matrix

A3 = [1 2;3 4]; A4 = [5 6;7 8]; Add = A3 + A4

% matrix addition

k = 10; MultSca = k*A3

% matrix-scalar multiplication

Trans = A4'

% matrix transpose (swap rows and columns)

A5 = [1 2 3;4 5 6]; A6 = [7 8;9 8;7 6]; A7 = A5*A6 A8 = A6*A5

% define two matrices

A9 b dA9 invA9 x x1 x2 Check xG

% % % % % %

who whos

= = = = = = = = =

[1 2;6 4]; [5;14]; det(A9) inv(A9) invA9*b x(1); x(2); A9*x A9\b

% matrix-matrix multiplication

matrix for linear equations solution define RHS vector calculate determinant of A calculate the inverse of A solve linear equations extract answers

% check answer – should be b % Gaussian elimination is more efficient % display the user-created variables % user-created variables with dimensions

The first solution of the linear equations above uses the matrix inverse. To solve linear equations, Gaussian Elimination is more efficient (more on this in the dynamics notes later) and more robust numerically; Gaussian elimination implementation is given in the third to the last line of the above m-file (with the back-slash). Since the equations are linear, there is a unique solution (assuming the equations are linearly independent, i.e. the matrix is not near a singularity) and so both solution methods will yield the same answer.

14

Output of Matrices.m A1 = 1 0 0

0 2 0

0 0 3

1 0 0

0 1 0

0 0 1

Add = 6 10

8 12

A2 =

MultSca = 10 20 30 40 Trans = 5 6

7 8

A7 = 46 115

42 108

A8 = 39 41 31

54 58 44

69 75 57

dA9 = -8 invA9 = -0.5000 0.7500 x =

1 2

Check =

xG = 1 2

5 14

0.2500 -0.1250

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 analytical/trigonometric method. Defining parameter t to be

half-angle

substitution

using

an

  t  tan   2 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(a  b)  cos a cos b  sin a sin b sin(a  b)  sin a cos b  cos a sin b To derive the cos term as a function of t, we start with

   cos   cos    2 2 The cosine sum of angles formula yields     cos   cos 2    sin 2   2 2   Multiplying by a ‘1’, i.e. cos 2   over itself yields 2

    cos 2    sin 2   2  2  cos 2     1  tan 2     cos 2    cos          2  2  2 2      cos   2

  The cosine squared term can be divided by another ‘1’, i.e. cos 2    sin 2    1 . 2 2

16

    cos 2         2  cos   1  tan 2      2    cos 2     sin 2           2  2    Dividing top and bottom by cos 2   yields 2

    1      cos   1  tan 2      2   1  tan 2          2   Remembering the earlier definition for t, this result is the first derivation we need, i.e. cos  

1 t2 1 t2

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

   sin   sin    2 2 The sine sum of angles formula yields             sin   sin   cos    cos   sin    2sin   cos   2 2 2 2 2 2 Multiplying top and bottom by cosine yields

  sin    2  cos 2     2 tan    cos2    sin   2         2 2 2 cos   2 From the first derivation we learned 1   cos 2     2  1  tan 2      2

17

Substituting this term yields

    1    sin   2 tan     2  1  tan 2         2  Remembering the earlier definition for t, this result is the second derivation we need, i.e.

sin  

2t 1 t 2

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

E cos  F sin  G  0 arises often in the position solutions for mechanisms and robots. It appeared in the 4 solution for the four-bar mechanism in the ME 3011 NotesBook and was solved using the tangent half-angle substitution. Next we present an alternative and simpler solution to this equation. We make two simple trigonometric substitutions based on the figure below.

Clearly from this figure we have

E

cos 

sin 

E2  F 2

In the original equation we divide by

E E F 2

2

F E2  F 2

E 2  F 2 and rearrange.

cos 

F E F 2

2

sin 

G E2  F 2

The two simple trigonometric substitutions yield

cos cos  sin  sin 

G E2  F 2

Applying the sum-of-angles formula cos(a  b)  cos a cos b  sin a sin b yields

cos(  ) 

G E2  F 2

19

And so the solution for  is



  2 2  E F 

1,2    cos1 

G

where

F 

  tan1   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 is double-valued. Both solutions are correct. We expected these two solutions from the tangent-half-angle substitution 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

1 

G E F 2

2

 1

or

1 

G E F 2

2

 1

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

G E  0 Four-bar mechanism position singularity E  2r4 (r1c1  r2 c2 ) G  r12  r22  r32  r42  2r1r2 cos(1   2 )

For simplicity, let 1 = 0 (just rotate the entire four-bar mechanism model for zero ground link angle). G  E  r12  r22  r32  r42  2 r1 r4  2 r2 ( r4  r1 ) c 2  0

I have encountered two example four-bar mechanisms with this G  E  0 singularity. Case 1

When r1  r4 and r2  r3 , G  E  r12  r22  r22  r12  2 r12  2 r2 ( r1  r1 ) c 2  0 ALWAYS, regardless of 2.

Example Given r1  10, r2  6, r3  6, r4  10 ; this mechanism is ALWAYS singular.

To fix this let

r1  10, r2  5.9999, r3  6.0001, r4  10 and MATLAB will be able to calculate the position analysis reliably at every input angle.

Case 2

When r1  2 r3 and r4  2r2 , and furthermore 3r3  5r2 , G  E  4r32  r22  r32  4r22  8r2 r3  4r2 (r2  r3 ) c 2 100 2 2 25 2 40 8 r2  r2  r2  4r22  r22  r22 c2 9 9 3 3 8   r22 c 2 3 

This G  E  0 occurs only when  2  90 . Case 2 is much less general than case 1. Example Given r1  10, r2  3, r3  5, r4  6 ; this mechanism is singular when  2  90 . To fix this ignore  2  90 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 link rotatability 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 a rotation that is a subset of this full rotation. This section presents more information on Grashof’s Law and 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 link lengths 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 the Grashof 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.

r2  r1  r3  r4 r4  r1  r2  r3

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 O4 A1B1 and O2 A2 B2 , respectively. Note from our standard notation, r1  O2O4 , r2  O2 A , r3  AB , and r4  O4 B .

B

B1

B2 A

A2

A1 O2

O4

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 input link. 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. (r3  r4 ) 2  r12  r22  2r1r2 cos  2 L  r12  r22  (r3  r4 ) 2   2r1r2  

 2 L   cos 1 

 with symmetry about r1.

23

Given r1  10, r2  6, r3  8, r4  7

Joint Limit Example 1

L  S  P  Q ( 10  6  8  7 ) so we predict only double rockers from this Non-Grashof Mechanism. 102  62  (8  7) 2    cos 1  0.742  137.9  2(10)(6)  

 2 L   cos 1 

This method can also be used to find angular limits on link 4 when it is a rocker. In this case links 2 and 3 align.

102  7 2  (6  8) 2    cos 1  0.336  109.6  2(10)(7)  

   cos 1 

 4 L  180    70.4 In this example, the allowable input and output angle ranges are: 137.9   2  137.9

70.4   4  289.6

This example is shown graphically in the ME 3011 NotesBook, in the Grashof’s Law section (2. NonGrashof 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 other and the other link 2 limit occurs when links 3 and 4 stretch out in a straight line. See Example 4 (and Example 3 for a similar situation with the output link 4 limits).

24

Joint Limit Example 2

Given r1  10, r2  4, r3  8, r4  7

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. 102  42  (8  7) 2  1    cos  1.3625 2(10)(4)  

 2 L   cos 1 

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).

102  72  (4  8)2  1    cos  0.036  88.0 2(10)(7)  

   cos 1 

 4min  180    92.0 For 4max, links 2 and 3 are instead folded upon each other (their absolute angles are different by ).

102  7 2  (4  8) 2  1    cos  0.95  18.2 2(10)(7)  

   cos1 

 4min  180    161.8 In this example, the output angle range is 92.0   4  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 r1  11.18, r2  3, r3  8, r4  7 (in) and 1  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 L  120.1 (links 2 and 3 stretched in a line)  4 L  172.5 (links 2 and 3 folded upon each other in a line)

The output angle range is 120.1   4  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 Term Example 1 in the ME 3011 NotesBook.

26

Given r1  10, r2  8, r3  4, r4  7

Joint Limit Example 4

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 limits are 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 are identical). 102  82  (7  4) 2  1    cos  0.969  14.4 2(10)(8)  

 2min   cos 1 

For 2max, links 3 and 4 are instead stretched in a straight line (their absolute angles are different by  as in Example 1). 102  82  (7  4) 2   cos 1  0.269  74.4  2(10)(8)  

8

8

7

7

6

6 Y (m)

Y (m)

 2max   cos 1 

5

5

4

4

3

3

2

2

1

1

0 0

1

2

3

4

5 X (m)

6

7

2min Diagram

8

9

10

0 0

1

2

3

4

5 X (m)

2max Diagram

6

7

8

9

10

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).

102  72  (8  4)2  1    cos  0.036  88.0 2(10)(7)  

   cos 1 

 4min  180    92.0 For 4max, links 2 and 3 are instead folded upon each other (their absolute angles are different by ).

102  7 2  (8  4) 2     cos   cos1  0.95  18.2  2(10)(7)   1

8

8

7

7

6

6 Y (m)

Y (m)

 4min  180    161.8

5

5

4

4

3

3

2

2

1

1

0 0

1

2

3

4

5 X (m)

6

7

4min Diagram

8

9

10

0 0

1

2

3

4

5 X (m)

4max Diagram

6

7

8

9

10

28 200

150

100

50 st

0

3 1

-50

4 1

nd

 (deg)

3 2

st nd

4 2

-100

-150

-200

-250 10

20

30

40

2 (deg)

50

60

70

80

In this plot we can see the minimum and maximum values we just calculated for links 2 and 4. Note at  2 min  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 ( r1  7, r2  10, r3  4, r4  8 , 1d. Grashof double rocker).

Grashof’s Law only predicts the rotatability of the input and output links; it says nothing about the rotatability of the coupler link 3 – in this case, what is the rotatability of the coupler link? (In this case the coupler link S rotates fully, proving that the relative motion is the same amongst all four-bar mechanism 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 Position Analysis Using Polynomial Discriminants”, Journal of Mechanisms, Transmissions, and Automation in Design, Transactions of the ASME, 109(2): 178-182.

29

2.1.2 Slider-Crank Mechanism Position Analysis 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.

r3c3  x  r2c2 r3s3  h  r2 s2 We can square and add to eliminate 3, similar to the four-bar mechanism solution approach. r32 c32  x 2  2 xr2 c2  r22 c22 r32 s32  h 2  2hr2 s2  r22 s22 r32  x 2  h 2  r22  2 xr2 c2  2 hr2 s2

This quadratic equation in x has the following form: a 1

ax  bx  c  0

b  2 r2 c2

2

c  r22  r32  h 2  2hr2 s2

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

x1,2  r2c2  r32  h2  r22s22  2hr2s2 Then 3 is found from a ratio of the Y to X equations.  3  atan2( h  r2 s 2 , x1,2  r2 c 2 ) 1,2

This alternate solution yields identical results as the earlier solution approach in the ME 3011 NotesBook for the right (  3 , x1 ) and left (  3 , x 2 ) branches. 1

2

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 since you 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 piston walls for the slider-crank mechanism graphics, since this was not required for the four-bar mechanism.

Outside the loop: Lp Hp Xp Yp

= = = =

put a number here; put a number here; [-1 -1 1 1]*Lp/2; [-1 1 1 -1]*Hp/2;

% length of piston (slider link) % height of piston

This establishes the rectangular corner coordinates for the slider link, centered at the origin of your coordinate frame. It can be done once, outside the loop. Instead of typing numbers for Lp and Hp, I scale them to a fraction of r2, for generality in different-sized slider-crank mechanisms. Note I only included 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. You can 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 Ypt Xpb Ypb

= = = =

[-1000 [h+wall/2 [-1000 [h-wall/2

1000]; h+wall/2]; 1000]; h-wall/2];

% fixed piston walls

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 r2 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, they may 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); plot(th2/DR,th3/DR); subplot(212); plot(th2/DR,x);

% 2x1 arrangement of plots, first plot % 2x1 arrangement of plots, second plot

Now, you can use the standard axis labels, linetypes, titles, axis limits, grid, etc., for each plot within a subplot (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); plot( . . . );

% m x n arrangement of plots, ith 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 Position Analysis

This slider-crank mechanism inversion 2 is an inversion of the standard zero-offset slider-crank mechanism where the sliding direction is no longer the ground link, but along the rotating link 4. Ground link length r1 and input link length r2 are fixed; r4 is a variable. The slider link 3 is attached to the end of link 2 via an R joint and slides relative to link 4 via a P joint. This mechanism converts rotary input to linear motion and rotary motion output. Practical applications include certain doors/windows opening/damping mechanisms. The inverted slider-crank is also part of quick-return mechanisms.

Step 1. Draw the Kinematic Diagram

3 4

2 1 r1 constant ground link length r2 constant input link length r4 variable output link length

2 variable input angle 4 variable output angle L4 constant total output link length

Link 1 is the fixed ground link. Without loss of generality we may force the ground link to be horizontal. 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

r1, 1 = 0, r2; plus 1-dof position input 2

Find

r4 and 4

33

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

4

2 1

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

r 2  r1  r 4 Step 5. Write the XY Components for the Vector-Loop-Closure Equation. Separate the one vector equation into its two X and Y scalar components.

r2c2  r1  r4c4 r2 s2 

r4 s4

Step 6. Solve for the Unknowns from the XY equations. There are two coupled nonlinear equations in the two unknowns r4, 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 easier than the four-bar mechanism. First rewrite the above XY equations to isolate the unknowns on one side.

r4c4  r2c2  r1 r4 s4  r2 s2 A ratio of the Y to X equations will cancel r4 and solve for 4. r4 s4 rs  2 2 r4c4 r2c2  r1

 4  atan2( r2 s2 , r2 c2  r1 ) Then square and add the XY equations to eliminate 4 and solve for r4.

r4   r12  r22  2rr 1 2c2

34

Note the same r4 formula results from the cosine law. Alternatively, the same r4 can be solved from either the X or Y equations after is 4 known. X)

r4 

r2 c2  r1 c4

Y)

r4 

r2 s2 s4

Both of these r4 alternatives are valid; however, each is subject to a different artificial mathematical singularity (  4  90 and  4  0,180 , respectively), so only the former square-root formula should be used for r4, which has no artificial singularity. The X algorithmic singularity  4  90 never occurs unless r2  r1 , 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 and r4   r1  r2  2rr 1 2c2 ,  4   . 2

2

However, the negative r4 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 be greater than the maximum value of the variable r4.

Slider Limits

The slider reaches its minimum and maximum displacements when 2 = 0 and , respectively. Therefore, the slider limits are r1  r2  r4  r1  r2 . Thus, the fixed length L4 must be greater than r1  r2 . In addition we require r1  r2 for full rotation.

35

Graphical Solution

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

Draw the known ground link (points O2 and O4 separated by r1 at the fixed angle 1 = 0).



Draw the given input link length r2 at the given angle 2 (this defines point A).



Draw a line from O4 to point A.



Measure the unknown values of r4 and 4.

36

Inverted Slider-Crank Mechanism Position Analysis: Term Example 3 Given: r1  0.20

r2  0.10 L4  0.32

m

1  0 Snapshot Analysis (one input angle) Given this mechanism and  2  70 , calculate 4 and r4.

4 (deg)

r4 (m)

150.5

0.191

This Term Example 3 position solution is demonstrated in the figure below. 0.2

0.15

Y (m)

0.1

0.05

0

-0.05

-0.1 -0.1

-0.05

0

0.05 X (m)

0.1

Term Example 3 Position Snapshot

0.15

0.2

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 r4 (m) and 4 (deg), for all 0    2  360  , for Term Example 3. 210 200

4 (deg)

190 180 170 160 150 0

50

100

150

50

100

150

200

250

300

350

200

250

300

350

0.3

r4 (m)

0.25

0.2

0.15

0.1 0

2 (deg)

Term Example 3 4 and r4

 

4 varies symmetrically about 180  , being 180  at  2  0,180, 360 . r4 varies like a negative cosine function with minimum displacement r1  r2  0.1 at  2  0, 360 and maximum displacement r1  r2  0.3 . Since r1 is twice r2 in this example, whenever  2  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, r4  ( 3 2)0.2  0.173 m. There

is

another

right

triangle

that

shows

up

for

 2  90 ;

in

these

cases

r4  0.22  0.12  0.224 m and  4  153.4, 206.6  , respectively. Check all of these special values in the F.R.O.M. plot results.

38

2.1.4 Multi-Loop Mechanism Position Analysis

Thus far we have presented position analysis for the single-loop four-bar, slider-crank, and inverted slider-crank mechanisms. The position analysis for mechanisms of more than one loop is handled using the same general procedures developed for the single loop mechanisms. A good rule of thumb is to look for four-bar (or slider-crank) parts of the multi-loop mechanism as we already know how 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 shown in 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 to our standard four-bar mechanism model. Since we number the links the same as in the four-bar, and if we define the angles identically, the position analysis solution is identical to the four-bar presented earlier. 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 r5 about point C must intersect the circle of radius r6 about point D to form point E (yielding two possible intersections in general). The analytical solution is very similar to the standard four-bar position solution, as we will show. For multi-loop mechanisms, the number of solution branches for position analysis increases compared to the single-loop mechanisms. Most single-loop mechanisms mathematically have two solution branches. For multi-loop mechanisms composed of multiple single-loop mechanisms, the number of solution branches is 2n, where n is the number of mechanism loops. For the two-loop Stephenson 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 mechanism using the formal position analysis steps presented earlier. Assume link 2 is the input link. Step 1. Draw the Kinematic Diagram

r1 r2 r2a r3 r4 r4a r5 r6

this is done in the figure above.

constant ground link length constant input link length to point A constant input link length to point C constant coupler link length, loop I constant output link length, loop I constant input link length to point D constant coupler link length, loop II constant output link length, loop II

1 2 2 3 4 4 5 6

constant ground link angle variable input link angle constant angle on link 2 variable coupler link angle, loop I variable output link angle, loop I constant angle on link 4 variable coupler link angle, loop II 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 shown in the kinematic diagram.

Step 2. State the Problem Given

r1, 1, r2, r3, r4, r2a, r4a, r5, r6, 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 hand from the right horizontal to the link vector (tail-to-head with the right-hand thumb at the vector tail and right-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 by starting and ending at the same points, but choosing a different path.

r 2  r 3  r1  r 4 r 2 a  r 5  r1  r 4 a  r 6 Note an alternative to the second vector loop equation is r 2b  r 5  r 3  r 4b  r 6 . See if you can identify

r 2b and r 4b , plus their angles 2b and 4b. Step 5. Write the XY Components for each Vector-Loop-Closure Equation. Separate the two vector equations into four XY scalar component equations.

r2c2  r3c3  rc 1 1  r4c4 r2 s2  r3s3  r1s1  r4 s4 r2a c2a  r5c5  rc 1 1  r4 a c4 a  r6 c6 r2a s2a  r5 s5  r1s1  r4a s4a  r6 s6 where

2a  2   2 4a  4   4

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 E cos4  F sin 4  G  0 , 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 4a  4   4 . 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 E2 cos6  F2 sin 6  G2  0 , where terms E2, F2, and G2 are known functions of constants and the known

angles 2a  2   2 and 4a  4   4 . 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-loop mechanisms. One can perform a compound Grashof analysis when four-bars are the component mechanisms. For the two-loop Stephenson I six-bar mechanism, the second loop may constrain the first loop (e.g. it may change an expected crank motion of the input link to a rocker). This is an important issue 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 solved graphically, by drawing the mechanism, determining the mechanism closure, and measuring the unknowns. 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 r1 at the fixed angle 1).  Draw the given input link length r2 at the given angle 2 to yield point A.  Draw a circle of radius r3, centered at point A.  Draw a circle of radius r4, 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 r2a at angle 2a  2   2 from point O2.  Draw the link r4a at angle 4a  4   4 from point O4.  Draw a circle of radius r5, centered at point C.  Draw a circle of radius r6, 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 Velocity Kinematics Analysis 2.2.2 Three-Part Velocity Formula Moving Example

Given initial positions

V

0x

V0 y V

P

P0 y

0x

L    0 0 0 0 (m, rad) and constant velocities

  0.2 0.1 0.3 0.4 (m/s, rad/s), simulate this motion and determine V P at each

instant. tf = 5 and t  0.1 sec was used. The initial and final snapshots, with their three-part velocity diagrams, are shown below. 4

1

3.5

0.9 0.8

3

0.7 2.5 Vy (m/s)

Y (m)

0.6 2 1.5

0.5 0.4

1

0.3

0.5

0.2

0

0.1 0

0.5

1

1.5 2 X (m)

2.5

3

3.5

0 -0.5

4

Initial Kinematic Diagram

0 Vx (m/s)

0.5

Initial Three-Part Velocity Diagram

4

1

3.5

0.9 0.8

3

0.7 2.5 Vy (m/s)

Y (m)

0.6 2 1.5

0.5 0.4

1

0.3

0.5

0.2

0

0.1

xL

0

0.5

1

1.5 2 X (m)

2.5

Final Kinematic Diagram

3

3.5

4

0 -0.5

V

VP

V0 0 Vx (m/s)

Final Three-Part Velocity Diagram

0.5

43

Three-Part Velocity Moving Example Plots 2

0.45

1.8

0.4 V

0x

1.6 P P

0y

0x

V

0y



L

Velocity Terms (m/s and rad/s)

Position Terms (m and rad)

1.4

V

0.35

 1.2

1

0.8

0.3

0.25

0.2

0.15

0.6 0.1

0.4

0.05

0.2

0 0

0.5

1

1.5

2

2.5 time (sec)

3

3.5

4

4.5

0 0

5

0.5

1

1.5

Position Terms

2

2.5 time (sec)

3

3.5

4

4.5

5

Velocity Terms

What is the relationship between these plots?

0.5

0.5

0.4

0.4

0.3

V Point P Velocity (m/s)

0.2

V

Sliding and Tangential Velocities (m/s)

0.3 Px Py

0.1 0 -0.1 -0.2

0.1 0 -0.1

V V

-0.2

V V

-0.3

-0.3

sx sy tx ty

-0.4

-0.4

-0.5

-0.5

0

0.2

0.5

1

1.5

2

2.5 time (sec)

3

3.5

4

Point P Translational Velocity

Constant velocity terms V0 x V0 y V

4.5

5

0

0.5

1

1.5

2

2.5 time (sec)

3

3.5

4

4.5

5

Sliding and Tangential Velocity Components

  0.2 0.1 0.3 0.4 lead to non-constant point P velocity

(due to the nonlinear position kinematics).

44

2.2.3 Four-Bar Mechanism Velocity Analysis

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

a  r3 s3 b  r4 s4

3 

ce  bf ae  bd

4 

c  r22 s2

af  cd ae  bd

d  r3c3 e  r4c4 f  r22c2

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

3 

r2 sin(4  2 ) 2 r3 sin(4  3 )

4 

r2 sin(3  2 ) 2 r4 sin(4  3 )

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:

sin(4  3 )  0

 4  3  0 ,180 ,

Physically, this happens when links 3 and 4 are straight out or folded on top of each other, as explained in 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 r4 is a variable so r4  0 in this problem. Step 1. The inverted slider-crank mechanism Position Analysis must first be complete.

Given r1, 1 = 0, r2, and 2, we solved for r4 and 4 Step 2. Draw the inverted slider-crank mechanism Velocity Diagram.

3 4

2 1

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

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

Step 3. State the Problem Given

the mechanism r1, 1 = 0, r2 the position analysis 2, r4, 4 1-dof of velocity input 2

Find

the velocity unknowns r4 and 4

46

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

Here are the inverted slider-crank mechanism position equations from earlier. r 2  r1  r 4

r2 c2  r1  r4 c4 r2 s2 

r4 s4

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

r22 s2  r4c4  r44 s4 r22c2  r4 s4  r44c4

These two linear equations in two unknowns can be written in matrix form.  c4  r4 s4   r4    r22 s2   s      4 r4 c4  4   r22 c2  Step 5. Solve the velocity equations for the unknowns r4 , 4 .  r4  1  r4 c4    4  r4   s4

r  ( s c  c s )  r4 s4   r22 s2   2 2 2 4 2 4      r  (s s  c c )  c4   r22 c2   2 2 2 4 2 4  r4  

 r22 sin( 4   2 )   r4        r22 cos( 4   2 )  4    r4  

where we have used the trigonometric identities: cos  a  b   cos a cos b  sin a sin b sin  a  b   sin a cos b  cos a sin b

sin(a)   sin(a) cos(a)  cos(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 r4 , 4 . c

 A    s4 

4

 r4 s4  r4 c4 

A  r4 c42  (  r4 s42 )  r4 (c42  s42 )  r4  0 Physically, assuming r1  r2 as in the full rotation condition from the inverted slider-crank mechanism position analysis, this is impossible, i.e. r4 never goes to zero. This matrix determinant A  r4 was used in the solution of the previous page.

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

Given

r1  0.20 r2  0.10 L4  0.32

 2  70 (m)

 4  150.5

(deg and m)

r4  0.191

1  0

Snapshot Analysis Given this mechanism position analysis plus 2  25 rad/s, calculate r4 , 4 for this snapshot.  0.870 0.094   r4  2.349   0.493 0.166      0.855    4    r4   2.47     4   2.18 

(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 r4 (bottom, m/s), for all 0    2  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 ). 10 5

 4 (rad/s)

0 -5 -10 -15 -20 -25 0

50

100

150

50

100

150

200

250

300

350

200

250

300

350

2

r4d (m/s)

1 0 -1 -2 0

2 (deg)

Term Example 3 F.R.O.M., 4 and r4

As expected, 4 is zero at the max and min for 4 (at  2  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. r4 is zero at the beginning, middle, and end of motion and is max at  2  60 .

49

2.2.6 Multi-Loop Mechanism Velocity Analysis

Thus far we have presented velocity analysis for the single-loop four-bar, slider-crank, and inverted slider-crank mechanisms. The velocity analysis for mechanisms of more than one loop is handled 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.

Stephenson I 6-Bar Mechanism

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 is essentially another four-bar velocity solution. As in all velocity analysis, the velocity solution for multi-loop mechanisms is a linear analysis yielding a unique solution (assuming the given mechanism position is not singular) for each position solution 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 mechanism using 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 r1, 1, r2, r3, r4, r2a, r4a, r5, r6, 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 previous Stephenson 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 a single angular velocity for the whole link. ri  0 for all links since all links are of fixed length (no sliders).

Step 3. State the Problem Given

the mechanism r1, 1, r2, r3, r4, r2a, r4a, r5, r6, 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 loop closure equations from position analysis, in XY component form.

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

r 2  r 3  r1  r 4 r 2 a  r 5  r1  r 4 a  r 6

XY scalar equations

r2c2  r3c3  rc 1 1  r4c4 r2 s2  r3s3  r1s1  r4 s4 r2a c2a  r5c5  rc 1 1  r4 a c4 a  r6 c6 r2a s2a  r5 s5  r1s1  r4a s4a  r6 s6

where

2 a  2   2 4 a  4   4

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

r22 s2  r33s3  r44 s4

r22c2  r33c3  r44c4 These equations can be written in matrix form.  r3 s3  r4 s4  3   r22 s2    r c     3 3 r4 c4  4   r22 c2  The first time derivative of the Loop II position equations is

r2a2a s2a  r55 s5  r4a4a s4a  r66 s6 r2a2a c2a  r55c5  r4a4a c4a  r66c6

These equations can be written in matrix form.  r5 s5  r6 s6  5  r2 a2 s2 a  r4 a4 s4 a    r c      5 5 r6 c6  6   r2 a2 c2 a  r4 a4 c4 a  where we have used

2a  2 4a  4

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) 3   r3 s3   4    r3c3

1

 r4 s4    r22 s2    r4 c4   r22 c2 

Loop II (similar to the standard four-bar mechanism) 5   r5 s5   6    r5 c5

1

 r6 s6    r2 a2 s2 a  r4 a4 s4 a    r6 c6   r2 a2 c2 a  r4 a 4 c4 a 

Remember, Gaussian elimination is more efficient and robust than the matrix inverse. Also, these equations 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 standard four-bar mechanism, i.e. when links 3 and 4 either line up or fold upon each other, causing a link 2 joint limit. For the second loop, the singularity condition is similar, occurring when links 5 and 6 either line up 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

P

Given initial positions

V

0x

0.2

V0 y 0.05

0x

P0 y

L    0

0

0

0

(m, rad), initial velocities

   0 0 0 0 (m/s, rad/s) and constant accelerations

V

0.15

A

A0 y

0x

A  

0.1 (m/s , rad/s ), simulate this motion and determine AP at each instant. tf = 5 2

2

and t  0.1 sec was used; the initial and final snapshots, with their five-part acceleration diagrams, are shown below. 4

1.2

3.5 1 3 0.8 2

Ay (m/s )

Y (m)

2.5 2 1.5

0.6

0.4

1 0.5

0.2 0 0

0.5

1

1.5 2 X (m)

2.5

3

3.5

0 -0.8

4

Initial Kinematic Diagram

-0.2 2 A (m/s )

0.4

Initial Five-Part Acceleration Diagram

4

1.2

3.5 1 3 0.8 2

Ay (m/s )

Y (m)

2.5 2 1.5

0.6

xL 0.4

1

2 x V

 x ( x L) 0.5 0.2 AP

0 0

0.5

1

1.5 2 X (m)

2.5

3

Final Kinematic Diagram

3.5

4

0 -0.8

A0 -0.2 2 Ax (m/s )

A 0.4

Final Five-Part Acceleration Diagram

54

Five-Part Acceleration Moving Example Plots 2.5

1

0.9 2

0.8 V Velocity Terms (m/s and rad/s)

Position Terms (m and rad)

P

0x

P0y

1.5

L



1

0x

V0y

0.7

V

 0.6

0.5

0.4

0.3 0.5

0.2

0.1 0 0

0.5

1

1.5

2

2.5 time (sec)

3

3.5

4

4.5

0 0

5

0.5

Position Terms

1

1.5

2

2.5 time (sec)

3

3.5

4

4.5

5

3

3.5

4

4.5

5

Velocity Terms

0.25 A A

0x

0.4

0y

A

 0.2

2

Point P Acceleration (m/s )

2

Acceleration Terms (m/s and rad/s )

0.2

2

0.15

0.1

0

APx

-0.2

A

Py

-0.4

0.05

-0.6

0 0

0.5

1

1.5

2

2.5 time (sec)

3

3.5

4

4.5

5

Acceleration Terms

-0.8 0

0.5

1

1.5

2

2.5 time (sec)

Point P Translational Acceleration

What is the relationship amongst the first three plots? Constant acceleration terms  A0 x

A0 y

A    0.2

0.05

0.15

0.1 lead to non-constant point P

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

55

2.3.3 Four-Bar Mechanism Acceleration Analysis

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

a  r3 s3 b  r4 s4

Ce  bF 3  ae  bd

aF  Cd 4  ae  bd

C  r2 2 s2  r222 c2  r332 c3  r442 c4 d  r3c3 e  r4 c4 F  r2 2 c2  r222 s2  r332 s3  r442 s4

Backsubstituting the terms a, b, C, d, e, F yields the following equivalent solutions, which simplify and nicely using the sum-of-angles formulae sin( a  b )  sin a cos b  cos a sin b cos( a  b)  cos a cos b  sin a sin b and better display the structure of the solutions.

3 

r2 2 sin(4  2 )  r222cos(4  2 )  r332cos(4  3 )  r442 r3 sin(4  3 )

r2 2 sin(3  2 )  r222cos(3  2 )  r332  r442cos(4  3 ) 4  r4 sin(4  3 )

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:

sin(4  3 )  0

 4  3  0 ,180 ,

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

56

2.3.4 Slider-Crank Mechanism Acceleration Analysis Derivative/Integral Relationships

x (m)

When one variable is the derivative of another, recall the relationships from calculus. For example: dx(t ) x(t )  x(t )  x0   x (t )dt dt dx (t )  x(t )  x (t )  x0    x(t )dt dt

0.2

0.1 0

50

100

150

200

250

300

350

50

100

150

200

250

300

350

50

100

150

200

250

300

350

3

xd (m/s)

2 1 0 -1 -2 0

2

xdd (m/s )

50

0

-50 0

2 (deg)

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 area values accurately.

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 r1, 1 = 0, r2, and 2 we solved for r4 and 4; then given 2 we solved for r4 and 4.

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

3 4

2 1

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

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

Step 3. State the Problem Given

the mechanism r1, 1 = 0, r2 the position analysis 2, r4, 4 the velocity analysis 2, r4 , 4 1-dof acceleration input 2

Find

the acceleration unknowns r4 and 4

58

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

Here are the inverted slider-crank mechanism velocity equations.

r4c4  r44 s4  r22 s2 r4 s4  r44c4  r22c2 The first time derivative of the velocity equations is given below.  r4 c4  2r44 s4  r4 4 s4  r442 c4   r2 2 s2  r222 c2  r4 s4  2r44 c4  r4 4 c4  r442 s4  r2 2 c2  r222 s2

These equations can be written in matrix form. r4  r2 2 s2  r222 c2  2r44 s4  r442 c4  c4  r4 s4     s    2 2  4 r4 c4   4   r2 2 c2  r22 s2  2r44 c4  r44 s4 

r4 ,  4 . Step 5. Solve the acceleration equations for the unknowns  r4  1  r4c4       4  r4   s4

r4 s4  r2 2 s2  r222 c2  2r44 s4  r442 c4    c4   r2 2 c2  r222 s2  2r44c4  r442 s4 

   2 2  r4  r2 2 sin( 4   2 )  r22 cos( 4   2 )  r44              r  cos(   )  r  2sin(   )  2r    4  2 2 4 2 2 2 4 2 4 4 r4   where we have again used the trigonometric identities: cos  a  b   cos a cos b  sin a sin b

sin(a)   sin(a)

sin  a  b   sin a cos b  cos a sin b

cos(a)  cos(a)

A major amount of algebra and trigonometry is required to get the final analytical solution for  r4 ,  4 above. Interestingly, the link 4 Coriolis term 2r44 cancelled in r4 while the link 4 centripetal term r4 42 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 the singularity condition is identical (see the singularity discussion in the inverted slider-crank mechanism velocity section – the only singularity is when r4 goes to zero; this will never to occur if r1  r2 ).

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

Given

r1  0.20 r2  0.10 L4  0.32

1  0

 2  70 (m)

2  25

 4  150.5 (deg and m) 

r4  0.191

r4  2.47 (rad/s and m/s)

4  2.18

Snapshot Analysis

r4 ,  4 for this Given this mechanism position and velocity analyses, plus  2  0 rad/s2, calculate  snapshot. r4   16.873  0.870 0.094     0.493 0.166     48.957    4   r4   9.46        4  267.14

(m/s2 and rad/s2)

r4 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

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

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 r4 (bottom,

m/s2), for all 0    2  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 ). 1000

4 (rad/s2)

500

0

-500

-1000 0

50

100

150

50

100

150

200

250

300

350

200

250

300

350

2

r4dd (m/s )

100

50

0

-50 0

2 (deg)

Term Example 3 F.R.O.M., 4 and r4

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. r4 is maximum (and minimum) at the beginning, middle, and end since the r4 curve is steepest at those points; r4 is zero when the r4 curve is flat, i.e.  2  60 .

61

Derivative/Integral Relationships

When one variable is the derivative of another, recall the relationships from calculus (the derivative 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:

4 (t ) 

d4 (t ) dt

 4 (t )   40   4 (t )dt

 4 (t ) 

d4 (t ) dt

4 (t )  40    4 (t )dt

 4 (d e g )

200 180 160 0

50

100

150

200

250

300

350

50

100

150

200

250

300

350

50

100

150

200

250

300

350

 4 (ra d /s )

10 0 -10 -20 0

 4 (ra d /s 2 )

1000 500 0 -500 -1000 0

2 (deg)

62

r4 (t ) 

dr4 (t ) dt

r4 (t )  r40   r4 (t )dt

 r4 (t ) 

dr4 (t ) dt

r4 (t )  r40    r4 (t )dt

r 4 (m )

0.25 0.2 0.15 0.1 0

50

100

150

200

250

300

350

50

100

150

200

250

300

350

50

100

150

200

250

300

350

r 4 d (m /s)

2 1 0 -1 -2 0

r 4 d d (m /s 2 )

100 50 0 -50 0

These plots are all from Term Example 3.

2 (deg)

63

2.3.6 Multi-Loop Mechanism Acceleration Analysis

Thus far we have presented acceleration analysis for the single-loop four-bar, slider-crank, and inverted slider-crank mechanisms. The acceleration analysis for mechanisms of more than one loop is handled using the same general procedures developed for the single loop mechanisms. This section presents acceleration analysis for the two-loop Stephenson I six-bar mechanism shown below as an example multi-loop mechanism. It follows the position and velocity analyses for the same mechanism presented earlier.

Stephenson I 6-Bar Mechanism

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

64

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

Step 1. The Stephenson I six-bar mechanism Position and Velocity Analyses must first be complete. Given r1, 1, r2, r3, r4, r2a, r4a, r5, r6, 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 ri  0 for all links since all links are of fixed length (no a single angular acceleration for the whole link.  sliders).

Step 3. State the Problem Given

the mechanism r1, 1, r2, r3, r4, r2a, r4a, r5, r6, 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 the four scalar XY velocity equations.

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

XY scalar velocity equations

r22 s2  r33s3  r44 s4 r22c2  r33c3  r44c4

r2a2a s2a  r55 s5  r4a4a s4a  r66 s6 r2a2a c2a  r55c5  r4a4a c4a  r66c6

2 a  2   2 4 a  4   4

where

The first time derivative of the Loop I velocity equations is identical to that for the standard fourbar mechanism.  r2 2 s2  r222 c2  r3 3 s3  r332 c3   r4 4 s4  r442 c4 r2 2 c2  r222 s2  r3 3c3  r332 s3  r4 4 c4  r442 s4

These equations can be written in matrix form.  r3 s3  r c  3 3

 r4 s4   3   r2 2 s2  r222 c2  r332 c3  r442 c4     r4 c4   4   r2 2 c2  r222 s2  r332 s3  r442 s4 

The first time derivatives of the Loop II velocity equations are:  r2 a 2 a s2 a  r2 a22a c2 a  r5 5 s5  r552 c5   r4 a 4 a s4 a  r4 a42a c4 a  r6 6 s6  r662 c6 r2 a 2 a c2 a  r2 a22a s2 a  r5 5 c5  r552 s5  r4 a 4 a c4 a  r4 a42a s4 a  r6 6 c6  r662 s6

These equations can be written in matrix form.  r5 s5  r c  5 5

 r6 s6   5    r2 a 2 s2 a  r2 a22 c2 a  r552 c5  r4 a 4 s4 a  r4 a42c4 a  r662 c6     r6 c6   6   r2 a 2 c2 a  r2 a22 s2 a  r552 s5  r4 a 4 c4 a  r4 a42 s4 a  r662 s6 

where we have used

2a  2 2a  2 and since 2 and 4 are constant angles. 4a  4 4a  4

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)  3   r3 s3    4    r3c3

1

 r4 s4    r2 2 s2  r222 c2  r332 c3  r442 c4    r4 c4   r2 2 c2  r222 s2  r332 s3  r442 s4 

Loop II (similar to the standard four-bar mechanism)  5   r5 s5    6    r5 c5

1

 r6 s6    r2 a 2 s2 a  r2 a22 c2 a  r552 c5  r4 a 4 s4 a  r4 a42 c4 a  r662 c6    r6 c6   r2 a 2 c2 a  r2 a22 s2 a  r552 s5  r4 a 4 c4 a  r4 a42 s4 a  r662 s6 

Remember, Gaussian elimination is more efficient and robust than the matrix inverse. Also, these equations 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 to zero. 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 velocity solutions. 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 standard four-bar mechanism, i.e. when links 3 and 4 either line up or fold upon each other, causing a link 2 joint limit. For the second loop, the singularity condition is similar, occurring when 5 and 6 either line up or fold upon each other. These conditions also cause problems for the velocity and position analyses, so the 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 the entire range of motion. However, these methods have focused on the basic models shown below. What if your term project requires animation of modifications of these basic mechanisms with extensions from the existing rigid links?

Four-Bar Mechanism

Offset Slider-Crank Mechanism

Four-bar mechanism link 3 extensions

D B A C

Here are the kinematics equations (we already presented the point C kinematics equations in the ME 3011 NotesBook).  d x  bx  rDB cos(3   3 D )   cx  ax  rCA cos( 3   3C )  C  D    d y   by  rDB sin( 3   3 D )  c y   a y  rCA sin(3   3C )  where

 ax  r cos  2  A  2  a y   r2 sin  2 

bx  r cos 1  r4 cos  4  B 1  by   r1 sin 1  r4 sin  4 

In this simple straight-line case, use  3C  180  and  3 D  0 . 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');

68

Four-bar mechanism link 4 extensions

C

B A

O4

D The kinematics equations are given below.  cx  bx  rCB cos( 4   4C )  C   c y   by  rCB sin( 4   4C ) 

 d x  O4 x  rDO4 cos( 4   4 D )  D    d y   O4 y  rDO4 sin( 4   4 D ) 

In this simple straight-line case, use  4C  0 and  4 D  180  . bx and by were given above and O 4 is: r cos 1  O4   1   r1 sin 1 

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.

69

2.5 Jerk Kinematics Analysis Jerk is the time rate of change of the acceleration (and hence the second and third time rates of change of the velocity and position, respectively). Again, this jerk time rate of change may describe a change 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 this section. Jerk analysis is the fourth step in kinematics analysis. It is not required for standard NewtonEuler 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, and acceleration analyses must be completed first. Jerk is the first time derivative of the acceleration, the second 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. d A(t ) d 2 V (t ) d 3 P (t ) J (t )    dt dt 2 dt 3

SI units:

m s3

d  (t ) d 2  (t ) d 3 (t )   dt dt 2 dt 3

SI units:

rad s3

 (t ) 

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

L, L

Y L

O



,L

P



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

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 d AP d  AOX  A cos   2V  sin   L sin   L cos     JP   dt dt  AOY  A sin   2V  cos   L cos   L 2 sin  

Recall the two-part position, three-part velocity, and five-part acceleration formula results below. P P  PO  L  P  L cos     OX   POY  L sin   V P  V O V    L V  V cos   L sin     OX  VOY  V sin   L cos   AP  AO  A  2  V    L      L   AOX  A cos   2V  sin   L sin   L 2 cos     2  AOY  A sin   2V  cos   L cos   L sin  

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



d  d 2  d 3  2  3 dt dt dt

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

d A d 2V d 3 L  2  3 dt dt dt

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

 

d  d 2  d 3  2  3 dt dt dt

d L d 2 L d 3 L  L  2  3 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

d d (2V  sin  )  2 A sin   2V  sin   2V  (sin  ) dt dt d (sin  ) d  2 A sin   2V  sin   2V  d dt 2  2 A sin   2V  sin   2V  cos 

d d d ( L 2 cos  )  V  2 cos   L ( 2 ) cos   L 2 (cos  ) dt dt dt 2 d ( ) d d (cos  ) d  V  2 cos   L cos   L 2 d  dt d dt 2 3  V  cos   2 L cos   L sin 

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

72

2.5.2 Mechanism Jerk Analysis Generic Mechanism Jerk Analysis Problem Statement

Given the mechanism, complete position, velocity, and acceleration analyses, and one-dof of jerk input, calculate the jerk unknowns. For a given branch of a known mechanism, this will yield linear equations so a matrix-vector approach may be used to obtain the unique solution (assuming no singularity). Since jerk kinematics is not required for dynamics analysis or machine design (generally), this solution is beyond the scope of the class. For our jerk needs, we may use the time differentiation approach of the previous subsection.

Snap, what a happy sound Snap is the happiest sound I found You may clap, rap, tap, slap, But Snap makes the world go round Snap, crackle, pop – Rice Krispies!

I say it’s Crackle, the crispy sound You gotta have Crackle or the clock’s not wound Geese cackle, feathers tickle, belts buckle, beets pickle, But Crackle makes the world go round Snap, crackle, pop – Rice Krispies! I insist that Pop’s the sound The best is missed unless Pop’s around You can’t stop hoppin’ when the cereal’s poppin’ Pop makes the world go round Snap, crackle, pop – Rice Krispies! -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 the right-branch only (slider-crank mechanism). What do the kinematics results look like for the crossed and left branches? Are there any relationships amongst the various analyses for the two branches? The reader is left to draw their own conclusions. 2.6.1 Four-Bar Mechanism

Given:

r1  10

1  0

r2  4

rCA  4

r3  8

3  0 2  10 (constant)

r4  7 Open Branch

Crossed Branch

74

Open Branch

Crossed Branch

75

2.6.2 Slider-Crank Mechanism r2  2, r3  6, h  0, rCA  3,  3  0, 2  10 (constant) Given: 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 to demonstrate the rotational and translational position, velocity, and acceleration kinematics analyses in the ME 3011 NotesBook. Term Example 1 is for the four-bar mechanism and Term Example 2 is for the slider-crank mechanism. 2.7.1 Term Example 1: Four-Bar Mechanism Four-Bar Mechanism Position Analysis – Term Example 1

r1  11.18 r2  3

Given:

r1  0.284 r2  0.076

in

r3  8

r3  0.203

m

r4  0.178

r4  7

and 1  10.3 (Ground link is 11" over and 2" up). Also given rCA  5 in (0.127 m) and  3  36.9 for the coupler link point of interest. Snapshot Analysis (for one given input angle 2)

Given this mechanism and  2  30 , calculate 3 ,4 ,  , and PC for both branches. Results: E  0.076 F  0.005 G  0.036

branch

t

 3

 4

 

PC

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

0.25 0.2 0.15

Y (m)

0.1 0.05 0 -0.05 -0.1 -0.1

0

0.1 X (m)

0.2

0.3

0.2

0.3

Open Branch (t(2))

0.25 0.2 0.15

Y (m)

0.1 0.05 0 -0.05 -0.1 -0.1

0

0.1 X (m)

Crossed Branch (t(1)) Term Example 1 Snapshot

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 analysis unknowns for the entire range of mechanism motion. The plot below gives 3 (red), 4 (green), and  (blue), all deg, for all 0   2  360 , for Term Example 1, the open branch only. 180

160

140

i (deg)

120

3 100

4 

80

60

40

20 0

50

100

150

2 (deg)

200

3, 4, and 

250

300

350

80

The plot below gives the initial (and final) animation position, for  2  0,360 . It also gives the coupler curve for the open branch, plotting PCY vs. PCX in green circles.

0.25

0.2

0.15

Y (m)

0.1

0.05

0

-0.05

-0.1

-0.1

-0.05

0

0.05

0.1 X (m)

0.15

P C Coupler Curve Term Example 1 F.R.O.M. Position Results

0.2

0.25

0.3

81

Four-bar mechanism velocity example – Term Example 1 continued

Given r1 = 0.284, r2 = 0.076, r3 = 0.203, r4 = 0.178, rCA = 0.127 m, and 1  10.3 , 2  30 ,

3  53.8 ,  4  121.7 ,  3  36.9 . 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 V C for this instant in motion (snapshot).  0.164 0.151 3  0.760   0.120 0.094      1.316    4  

3   8.073    4  3.729 

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

0.265 VC    (m/s) 1.330

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 analysis unknowns for the entire range of mechanism motion. The plot below gives 3 (red) and 4 (green) (rad/s) for all 0   2  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 ). 10

8

6

4

 i (rad/s)

2

0

-2

3

-4

4

-6

-8

-10 0

50

100

150

2 (deg)

3 and 4

200

250

300

350

83

The plot below gives the absolute translational coupler point C velocity for all 0   2  360 , for Term Example 1, for the open branch only. 2

1.5

1 VCX VCY VC (m/s)

0.5

0

-0.5

-1

-1.5 0

50

100

150

2 (deg)

200

250

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

300

350

84

Four-bar mechanism acceleration example – Term Example 1 continued

Given r1 = 0.284, r2 = 0.076, r3 = 0.203, r4 = 0.178, rCA = 0.127 m, and 1  10.3 , 2  30 ,

3  53.8 ,  4  121.7 ,  3  36.9 ; 2  20 (constant), 3  8.073 , 4  3.729 rad/s. This is the open branch of the position and velocity example (Term Example 1).

Snapshot Analysis

Given this mechanism position and velocity analysis, plus  2  0 rad/s2, calculate  3 ,  4 for this instant in motion (snapshot).

 0.164 0.151 3  35.433 0.120 0.093    23.785   4   3   7.994     4  243.018

Both are positive, so they are ccw in direction. These results are the absolute angular accelerations of links 3 and 4 with respect to the ground link. The coupler point translational acceleration vector is:

27.230 2 AC    m/s 23.490

85

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

A more meaningful result from acceleration analysis is to solve and plot the acceleration analysis unknowns for the entire range of mechanism motion. The plot below gives 3 (red) and 4 (green), (rad/s2), for all 0   2  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 ). 250

200

3

150

4

100

2 i (rad/s )

50

0

-50

-100

-150

-200

-250 0

50

100

150

2 (deg)

3 and 4

200

250

300

350

86

The plot below gives the translational coupler point acceleration for all 0   2  360 , for Term Example 1, the open branch only. 30

20

2

AC (m/s )

10

0

-10 ACX ACY -20

-30

0

50

100

150

2 (deg)

200

250

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

300

350

87

Derivative/Integral Relationships

When one variable is the derivative of another, recall the relationships from calculus. e.g.: d (t ) 4 (t )  4  4 (t )   40   4 (t )dt dt

 4 (t ) 

d 4 (t ) dt

4 (t )  40    4 (t )dt

4 (rad)

3

2.5

2 0

50

100

150

200

250

300

350

50

100

150

200

250

300

350

50

100

150

200

250

300

350

 4 (rad/s)

10 5 0 -5 -10 0 300 2 4 (rad/s )

200 100 0 -100 -200 0

2 (deg)

The value of 4 at any point is the slope of the 4 curve at that point. The value of 4 at any point is 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 area values accurately.

88

2.7.2 Term Example 2: Slider-Crank Mechanism Slider-Crank Mechanism Position Analysis – Term Example 2

Given:

r2  4

r2  0.102

r3  8 in

r3  0.203 m

h3

h  0.076

Snapshot Analysis (one input angle)

Given this mechanism and 2  30 , calculate x and 3 for both branches. Results:

3 (deg)

branch

x (m)

right

0.290

7.1

left

–0.113

172.9

0.2

0.2

0.1

0.1 Y (m)

Y (m)

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

0

0

-0.1

-0.1

-0.2

-0.2

-0.1

0

0.1 X (m)

0.2

0.3

-0.1

Right Branch

0

0.1 X (m)

Left Branch Term Example 2 Snapshot

0.2

0.3

89

Full-Range-Of-Motion (F.R.O.M.) Analysis – Term Example 2

A more meaningful result from position analysis is to solve and plot the position analysis unknowns for the entire range of mechanism motion. The top plot gives 3 (deg) and the bottom plot gives x (m), for all 0   2  360 , for Term Example 2, the right branch only.

70 60

3 (deg)

50 40 30 20 10 0 -10 0

50

100

150

50

100

150

200

250

300

350

200

250

300

350

0.3

x (m)

0.25 0.2 0.15 0.1 0.05 0

2 (deg)

3 and x

90

The plot below gives the initial (and final) animation position, for  2  0,360 . It also gives the coupler curve to scale for the right branch, plotting PCY vs. PCX in green. In this case the coupler point is taken to be the midpoint of coupler link 3. 0.2

0.15

0.1

Y (m)

0.05

0

-0.05

-0.1

-0.15

-0.2 -0.05

0

0.05

0.1

0.15 X (m)

0.2

0.25

0.3

0.35

P C Coupler Curve Term Example 2 F.R.O.M. Position Results

For Term Example 2, the slider translation limits are 0.067  x  0.295 , as seen in the x plot above, calculated from the equations in the 3011 NotesBook.

91

Slider-crank mechanism velocity example – Term Example 2 continued

Given r2 = 0.102, r3 = 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. Snapshot Analysis (one input angle)

Given this mechanism position analysis plus 2  15 rad/s (+ so ccw), calculate 3 , x for this instant in time (snapshot).  0.025 1  3   0.762   0.202 0   x    1.320       3   6.577      x   0.601

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

92

Full-Range-Of-Motion (F.R.O.M.) Analysis – Term Example 2 continued

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 0   2  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 ). 10

 3 (rad/s)

5

0

-5

-10 0

50

100

150

50

100

150

200

250

300

350

200

250

300

350

3

xd (m/s)

2 1 0 -1 -2 0

2 (deg)

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

93

Slider-crank mechanism acceleration example – Term Example 2 continued

Given r2 = 0.102, r3 = 0.203, h = 0.076 m, and 2  30 ,  3  7.2 , x = 0.290 m; and 2  15 , 3  6.55 rad/s, x   0.60 m/s. This is the right branch of the position and velocity example for the slider-crank mechanism of Term Example 2.

Snapshot Analysis (one input angle)

x for this Given this mechanism position and velocity analysis plus  2  0 rad/s2, calculate  3 ,  snapshot in time.  0.025 1   3   28.590     0.202 0       x  12.557   3   62.329     x  30.148  

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

94

Full-Range-Of-Motion (F.R.O.M.) Analysis – Term Example 2 continued

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 0   2  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 to time t via 2  2t ). 150 100

2 3 (rad/s )

50 0 -50 -100 -150 -200 -250 0

50

100

150

50

100

150

200

250

300

350

200

250

300

350

2

xdd (m/s )

50

0

-50 0

2 (deg)

Term Example 2 F.R.O.M. Acceleration Results, 3 and x

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 inertial force F 0  mAG and a vector inertial moment M 0   I G  . Centrifugal force mr 2 , directed away from the center of rotation, is an example of an inertial force vector. It’s not really a force but a felt effect of an inertia in acceleration. Using D’Alembert’s principle, the right-hand side of the translational and rotational dynamics equations is subtracted to the other side of the equation. Then the forces and moments 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 to consider statics problems as a subset of dynamics problems, with zero accelerations.

R  mAG  0 R  FO  0 T  r  R  IG   0 T rRMO  0

Dynamics Humor from xkcd

xkcd.com

96

3.2 Mass, Center of Gravity, and Mass Moment of Inertia This 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.  M G  I GZ   F  m AG 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 IGZ. translational rotational

m mass

IGZ

PCG center of gravity center of gravity

Mass

In Newton’s Second Law

 F  mA

G

mass moment of inertia

, 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. 1 2 Mass is also a measure of the storage of translational kinetic energy KET  mv and the units are kg. 2 Examples for m, CG, IG

System of particles

General rigid body

Rectangular rigid body Mass calculation System of particles

General rigid body

N

m   mi

m

i 1



dm

body

Rectangular rigid body

using  

m  

b/2

m , V



h/2

b / 2  h / 2

dm   dV ,

so

m



body

tdxdy   tbh

dm  



dV

body

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 a vector quantity and the units are length units, m. CG calculation System of particles P CG 

m r m i

m x m m y Y  m

X CG  X  i

i i i

Cartesian components

i

YCG

i

i

i

General rigid body

PCG 



X CG  X 

rdm

body

Cartesian components

 dm

YCG  Y 

 xdm x

 dm

 ydm y

 dm

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

X  

 xdm

Y 

x

 dm



xdV m



x

   



m

b/2

b / 2

 th m



b/2

b / 2



xdx

b/2

 th x 2



m 2

 th  b



xthdx

b / 2 2

 2m  4



b  0 4  2



 ydm y

 dm 

m y

ydV



h/2

m

h / 2

 tb m



ytbdy

h/2

h / 2

h/2

 tb y 2 m 2

 tb  h

ydy

h / 2 2

 2m  4



h2  0 4 

 X  0  P CG        Y  0 

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

98

Mass Moment of Inertia IG This is not the same as area moment of inertia (IA) for beam bending, which is recalled below.

I Ax   y 2 dA

I Ay   x 2 dA

y

units: I A  m 4

x

In Euler’s rotational dynamics equation

M

G

 I GZ  , the mass moment of inertia IGZ is the

proportionality constant. IGZ is also a measure of rotational inertia, i.e. the resistance to change in rotational motion according to Newton’s First Law. Also, it is a measure of how hard it is to accelerate about certain axes in rotation. IGZ is also a measure of the storage of rotational kinetic energy, 1 KER  IGZ  2 , and its units are kg-m2. 2 What is the mass moment of inertia, a scalar, vector, matrix, or something else? Answer – it is a tensor.

Mass Moment of Inertia IG calculation System of particles

I axis   mi ri 2

where ri 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, the second moment (IGZ) terms do not cancel since the squared terms are all positive. General rigid body inertia tensor (symmetric)

I axis

I XX 



( y 2  z 2 )dm

body

 I XX   I XY  I XZ

IYY 

 (x

I XY IYY IYZ 2

I XZ  IYZ  I ZZ 

 z 2 )dm

I ZZ 

body

body

What is the only term that matters for XY planar motion? Answer – IZZ. In the yardstick example:

I GZ  I GY  I GX



also

I OZ  I GZ

( x 2  y 2 )dm

99

Rectangular rigid body

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

I GZ 



body

( x 2  y 2 )dm  

b/ 2



h /2

 b /2  h /2

 y3  t   x2 y   b /2  3  b /2

( x 2  y 2 )  tdxdy

  dx   h /2  h /2

 2  h  h  1  h3  h3    t   x     dx     b /2 8    2 2  3 8  hx 3 h3 x b /2  b /2  h3  2    t   hx   dx   t    b /2  3 12  12  b / 2     b /2

 h  b3 b3  h3  b b    t         8  12  2 2   3 8  b3 h bh3   tbh 2  t   (b  h 2 )  12 12 12  

IGZ 

m 2 2 (b  h ) 12

(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 IGZ automatically for each 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 parallel axes through different points as follows. I ZZO  I ZZG  md 2

where d is the scalar distance separating the axis of interest O from the axis through the CG. Notice IZZG is as small as possible. Any IZZO must be greater, due to the term md2 which is always positive. Parallel axis theorem example Rectangular rigid body (where axis O is the corner)

I ZZO  I ZZG  md 2  b2 h2  m 2 (b  h 2 )  m    12  4 4   b2 b2 h2 h2   m      12 4 12 4  

 b2 h2   m    3 3  m  (b 2  h 2 ) 3 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. n

mT  m1  m2    mn   mi i 1

P CGT

 n  mi xi     m1 x1  m2 x2    mn xn  i 1   m  mT  X T      T    n   YT   m1 y1  m2 y2    mn yn   m y   i i mT    i 1    mT  n

I GZT  I GZ 1  m1d12  I GZ 2  m2 d 22    I GZn  mn d n2   I GZi  mi di2 i 1

The subscript ‘T’ indicates total (or combined) and di is the distance between the combined CG ( P CG ) T

and the CG of body i ( PCG i   xi

yi  ). These equations are obtained from the mass, CG, and IGZ equations for particles, where now each particle is instead a rigid body. T

101

Example 1. Two rectangles joined as shown below

Given

b1 = 2.2 x h1 = 0.1 x t1 = 0.005 (m) b2 = 1.0 x h2 = 0.8 x t2 = 0.005 (m) the material is steel with a mass density of  = 7850 kg/m3

The equations are:

mT  m1  m2

P CGT

 m1 x1  m2 x2   mT  X     T I GZT  I GZ 1  m1d12  I GZ 2  m2 d 22  m y m y  2 2  YT   1 1  mT  where di is the distance from CGi to the combined CG P CGT .

The results are: m1 = 8.64

0.05 PCG1    1.10  d1 = 0.70 m IGZ1  1 b12  h12  3.49 12





m2 = 31.40

mT = 40.04 (kg)

0.60  P CG 2    1.80  d2 = 0.19 (m) m IGZ 2  2 b22  h22  4.29 12

0.48 PCGT    (m) 1.65 





IGZT = 13.15 (kgm2)

2

Y (m)

1.5

1

0.5

0 0

0.5

1

1.5

2

X (m)

Example 1. Two Unequal Steel Rectangles

(here b1 is vertical and b2 is horizontal)

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:

mT  2m

P CGT

 b  3h    X    T 4   YT   3b  h   4 

IGZT 

b2  h2 is the distance from CGi to P CGT . 8

where d1  d 2  The results are: m1 = m2 = 15.70 0.10  1.20 PCG1   PCG 2     1.90 1.00  d1 = d2 = 0.71 (m) m IGZ1  IGZ 2  (b2  h2 )  5.29 12

5 m(b2  h2 ) 12

mT = 31.40 (kg) 0.65 PCGT    (m) 1.45  IGZT = 26.43 (kgm2)

2

Y (m)

1.5

1

0.5

0 0

0.5

1

1.5

2

X (m)

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

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 (kgm2) 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 (m4). 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 below for general rigid bodies.

m



dm



rdm

body

P CG 

I ZZ 

body

 dm 

( x 2  y 2 )dm

body

All of these terms require double integrals over the rigid body in the XY plane. Both center of gravity and mass moment of inertia depend on the origin of the chosen XYZ Cartesian coordinate system. For general 3D rigid bodies, mass moment of inertia is a 3x3 tensor. For planar mechanism dynamics we only need one scalar term out of these 9 terms, IZZ. 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 inertia about axis O is related to the mass moment of inertia about axis G is via the parallel axis theorem, where d is the scalar distance (vector length) between axes G and O in the XY plane: I OZ  I GZ  md 2

104

Mass Properties for Planar Links

Name

Model

Mass (kg)

Center of Gravity (m)

Mass Moment of Inertia (kg-m2)

m

0   0

0

m

L   0

Y point mass

m O

X

Y point mass on massless rod

L

m

O

I GZ  0 I OZ  mL2

X

Y slender rod m

L O

X

Y L

rectangular parallelopiped

 Lht

h O

Y square

L   2  0 

L  2    h  2 

I OZ

m ( L2  h 2 ) 12 m ( L2  h 2 )  3

I GZ  I OZ

X

s s

O

mL2 12 mL2  3

I GZ 

X

 s 2t

s  2    s  2 

ms 2 6 2ms 2  3

I GZ  I OZ

105

Mass Properties for Planar Links (continued)

Name

Model

Mass (kg)

Center of Gravity (m)

Mass Moment of Inertia (kg-m2)

Y

I GZ

R cylinder

 R2t

O Y

R   R

I OZ

mR 2  2 5mR 2  2

X Ro

hollow cylinder

Ri

O

  Ro2  Ri2  t

 R0     R0 

m

R   R

m ( Ro2  Ri2 ) 2 m (5 Ro2  Ri2 )  2

I GZ  I OZ

X

Y R thin ring

O

Y triangle

I OZ  3mR 2

X

a bht

h

2

O

I GZ  mR 2

b

a  b   3     h   3 

m ( a 2  ab  b 2  h 2 ) 18 2 m ( a  ab  b 2  h 2 )  6

I GZ  I OZ

X

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-level dynamics book, or in any other textbook, nor in any Internet search.

106

Mass Properties for Cylindrical Links

Now, the previous nine shapes for which the mass properties were summarized are all planar shapes, with a constant thickness t in the Z direction (except for the point mass, point mass on massless rod, and slender rod, whose Z dimensions are unimportant). Here we give one more shape, a cylinder that is 3D but useful for planar slider-crank mechanisms and other planar mechanisms with a prismatic joint and sliding cylindrical piston. The cylinder given on the previous page was arranged with the circle in the XY plane. Now we need to rotate this so the rectangular projection of the cylinder is the XY plane. The mass moment of inertia is quite different from that of the rectangular parallelepiped, due to the effect of the radius R in this case, as opposed to the constant thickness t in the rectangular parallelepiped case. Not all pistons are solid, so we also include a similar model for the hollow piston cylinder, with outer radius Ro and inner radius Ri.

Name

Model

Mass (kg)

Center of Gravity (m)

Y L R

piston cylinder

 R 2 L O

X

L   2  R 

Y

hollow piston cylinder

L Ri O

Ro

  Ro2  Ri2  L X

L   2  R 

Mass Moment of Inertia (kg-m2) m ( L2  3 R 2 ) 12 m (4 L2  15 R 2 )  12

I GZ  I OZ

m ( L2  3 Ro2  3 Ri2 ) 12 2 m (4 L  15 Ro2  3 Ri2 )  12

I GZ  I OZ

107

English Units for Mass

ME 3011 uses SI units exclusively. However, many of you perform your capstone project work using English units, which is fine, since we live in the U.S. In the 1970s the U.S. government mandated a 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. Using standard 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. The same unit, pound (lb), applies both to force (lbf) and mass (lbm), depending on the context. Please always 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 or inches 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 are using feet or blobs if you are using inches. If you have estimated your masses in lbm, simply divide by 32.2 to get slugs, or divide by 386.1 to get blobs. Finally, from the above we have the following units equivalents:

1 lb f

 1

slug  ft s2

so

1 slug

 1

lb f  s 2 ft

1 lb f

 1

blob  in s2

so

1 blob

 1

lb f  s 2 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

 r    R cos( 2   2 )  r 12   12 X    G 2   r12Y    RG 2 sin( 2   2 ) 

 r    R cos( 4   4 )  r 14   14 X    G 4   r14Y    RG 4 sin( 4   4 ) 

 r  r cos  2  r 32  r12  r 2   12 X 2   r12Y  r2 sin  2 

 r  r cos  4  r 34  r14  r 4   14 X 4   r14Y  r4 sin  4 

AG 2

 AG 2 X    RG 2 2 sin( 2   2 )  RG 222 cos( 2   2 )     2  AG 2Y   RG 2 2 cos( 2   2 )  RG 22 sin( 2   2 )  AG 4

 AG 4 X    RG 4 4 sin( 4   4 )  RG 442 cos( 4   4 )     2  AG 4Y   RG 4 4 cos( 4   4 )  RG 44 sin( 4   4 )   F   F cos E 4  F E4   E4X    E4   FE 4Y   FE 4 sin E 4   r   rE 4 cos rE 4  r E4   E4X      rE 4Y   rE 4 sin rE 4 

M E 4  M E 4kˆ

109

Four-bar mechanism inverse dynamics matrix equation  1   0  r12Y   0  0   0  0   0  0 

0

1

0

0

0

0

0

1  r12 X 0 0 0

0 r32Y 1 0 r23Y

1 r32 X 0 1 r23 X

0 0 1 0  r43Y

0 0 0 1 r43 X

0 0 0 0 0

0 0 0 0 0

0 0 0

0 0 0

0 0 0

1 0 r34Y

0 1 r34 X

1 0 r14Y

0 1 r14 X

0   F21 X   m2 AG 2 X       0   F21Y   m2 ( AG 2Y  g )   1   F32 X   IG 2 Z 2     0   F32Y   m3 AG 3 X  FE 3 X      0   F43 X    m3 ( AG 3Y  g )  FE 3Y   0   F43Y   I G 3 Z  3  rE 3 X FE 3Y  rE 3Y FE 3 X  M E 3      0   F14 X   m4 AG 4 X  FE 4 X      0   F14Y   m4 ( AG 4Y  g )  FE 4Y  0    2   I G 4 Z  4  rE 4 X FE 4Y  rE 4Y FE 4 X  M E 4 

 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 only links 3 and 4 first, this is 6 equations in 6 unknowns – this is verified by looking at the original 9x9 matrix and noting three 6x1 columns of zeros (1, 2, 9) in rows 4 through 9. Here is a more efficient solution. The reduced 6x6 set of equations for links 3 and 4 are given below.

 1  0   r23Y   0  0   0

0 1 r23 X 0 0 0

1 0 r43Y 1 0 r34Y

0 1 r43 X 0 1 r34 X

0 0 0 1 0 r14Y

m3 AG 3 X  FE 3 x    F32 X     F   m3 ( AG 3Y  g )  FE 3 y    32Y     F43 X   I G 3Z  3  rE 3 x FE 3 y  rE 3 y FE 3 x  M E 3     m4 AG 4 X  FE 4 x F 43 Y       F14 X    m4 ( AG 4Y  g )  FE 4 y     r14 X   F14Y   I G 4 Z  4  rE 4 x FE 4 y  rE 4 y FE 4 x  M E 4  0 0 0 0 1

 A34 v34   b34  Solve for six unknowns v34    A34  b34  and then use F32X and F32Y in the following 3x3 set of linear equations, from the link 2 FBD, very similar to the single rotating link. 1

 1  0   r12Y

1

0 1  r12 X

0   F21 X   m2 AG 2 X  F32 X       0   F21Y    m2 ( AG 2Y  g )  F32Y      1    2   I G 2 Z  2  r32Y F32 X  r32 X F32Y 

R.L. Williams II, 2009, “Partial Decoupling of the Matrix Method for Planar Mechanisms Inverse Dynamics”, CD Proceedings 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:

F21 X  m2 AG 2 X  F32 X F21Y  m2 ( AG 2Y  g )  F32Y

 2  I G 2 Z  2  r32Y F32 X  r32 X F32Y  r12Y F21X  r12 X F21Y Matrix inversion requires approximately

3n3 and Gaussian elimination requires approximately log n

(n2  1)n 2  n multiplications/divisions2. 3 Number of Multiplications/Divisions for Four-bar Inverse Dynamics Solution Method 9x9 6x6 plus decoupled link 2 Reduction in cost

Inversion 2292 840 63%

Gaussian 321 113 65%

Reduction 86% 87%

There is a substantial 65% reduction in computational cost for Gaussian elimination with the 6x6 plus decoupled link 2 method. Also, the numerical accuracy may also improve with this method since 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  1  0   r12Y   0  0   0  0   0

0

1

0

0

0

1

0

1

0

0

 r12 X

 r32Y

r32 X

0

0

0

1

0

1

0

0

0

1

0

1

0

r23Y

r23 X

 r43Y

r43 X

0

0

0

1

0

0

0

0

0

1

m2 AG 2 X 0   F21 X         m2 ( AG 2Y  g ) 0 0   F21Y     IG 2 Z 2 0 1   F32 X       m3 AG 3 X  FE 3 X 0 0   F32Y        m3 ( AG 3Y  g )  FE 3Y 0 0  F43 X     0 0   F43Y   I G 3 Z  3  rE 3 X FE 3Y  rE 3Y FE 3 X  M E 3      m4 AG 4 X  FE 4 X   0   F14Y         m4 g  FE 4Y 1 0    2    0

 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. If we consider the FBDs of only links 3 and 4 first, this is 5 equations in 5 unknowns – this is verified by looking 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.  1  0   r23Y   0  0

0

1

0

1

0

1

 r23 X 0

 r43Y 1

r43 X 0

0

0

1

m3 AG 3 X  FE 3 x 0   F32 X         m3 ( AG 3Y  g )  FE 3 y 0   F32Y       0   F43 X    I G 3 Z  3  rE 3 x FE 3 y  rE 3 y FE 3 x  M E 3    m4 AG 4 X  FE 4 x     F43Y       m4 g  FE 4 y 1   F14Y   

 A34 v34   b34  Solve for five unknowns v34    A34  linear equations, from the link 2 FBD.

1

 1  0   r12Y

0 1  r12 X

b34 

and then use F32X and F32Y in the following 3x3 set of

0   F21 X   m2 AG 2 X  F32 X       0   F21Y    m2 ( AG 2Y  g )  F32Y     1    2   I G 2 Z  2  r32Y F32 X  r32 X F32Y 

112

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

F21 X  m2 AG 2 X  F32 X F21Y  m2 ( AG 2Y  g )  F32Y

 2  I G 2 Z  2  r32Y F32 X  r32 X F32Y  r12Y F21X  r12 X F21Y 3n3 Matrix inversion requires approximately and Gaussian elimination requires approximately log n

(n2  1)n 2  n multiplications/divisions (Nering, 1974). 3 Number of Multiplications/Divisions for Slider-Crank Inverse Dynamics Solution Method 8x8 5x5 plus decoupled link 2 Reduction in cost

Inversion 1701 544 68%

Gaussian 232 72 69%

Reduction 86% 87%

There is a substantial 69% reduction in computational cost for Gaussian elimination with the 5x5 plus decoupled link 2 method. Also, the numerical accuracy may also improve with this method since 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 try a simpler approach – assume the link 3 mass is small and use only FBDs for links 2 and 4. We further assume 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

F ij

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

r ij

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

114

Step 3. State the Problem Given r1, 1 = 0, r2 2, r4, 4 2, r4 , 4 2, r4 , 4 assume zero external forces and moments

F21 , F42 , F14 and 2

Find

Step 4. Derive the Newton-Euler Dynamics Equations. Newton's Second Law Link 2

Link 4

F

2

 F 42  F 21  W 2  m 2 A G 2

F

4

 F 14  F 42  W 4  m 4 A G 4

Euler's Rotational Dynamics Equation Link 2

Link 4

M

G2z

  2  r 42  F 42  r 12  F 21  I G 2 Z  2

M

G4z

 r 14  F 14  r 24  F 42  I G 4 Z  4

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, F42 is always perpendicular to link 4 and there is only one unknown from this vector, the magnitude F42.  F   F cos( 4   2)  F 42   42 X    42   F42Y   F42 sin( 4   2) 

115

Step 5. Derive the XYZ scalar dynamics equations from the vector dynamics equations. Link 2

F42 X  F21 X  m2 AG 2 X F42Y  F21Y  m2 ( AG 2Y  g )

 2  (r42 X F42Y  r42Y F42 X )  (r12 X F21Y  r12Y F21 X )  I G 2 Z  2 Link 4

F14 X  F42 X  m4 AG 4 X F14Y  F42Y  m4  AG 4Y  g 

 r14 X F14Y  r14Y F14 X    r24 X F42Y  r24Y F42 X   IG 4 Z 4 Express these scalar equations in matrix/vector form. The simplified inverted slider-crank mechanism inverse dynamics matrix equation is given below.

 1  0   r12Y   0  0   0

0 1 r12 X 0 0 0

c 0 s 0 r42 X s  r42Y c 0 c 1 s 0 r24 X s  r24Y c r14Y

0 0 0 0 1 r14 X

c  cos( 4   2)  where:       s   sin( 4   2) 

 A v  b

0  F21 X   m2 AG 2 X  0  F21Y  m2 ( AG 2Y  g )  1   F42   I G 2 Z 2     0  F14 X   m4 AG 4 X  0  F14Y  m4 ( AG 4Y  g )      0   2   I G 4 Z 4 

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.

v   A b 1

Matrix/vector solution 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 to this 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 through 6. Here is a more efficient solution. The reduced 3x3 set of equations for link 4 is given below. c 1   0 s   r24 X s  r24Y c  r14Y

0   F42   m4 AG 4 X      1   F14 X   m4 ( AG 4Y  g )  r14 X   F14Y   I G 4 Z  4 

 A4 v4   b4  Solve for three unknowns v4    A4 

1

b4  and then use F42 in the following 3x3 set of linear equations,

from the link 2 FBD.  1  0   r12Y

0

1  r12 X

0   F21 X   m2 AG 2 X  cF42       0   F21Y    m2 ( AG 2Y  g )  sF42  1    2   I G 2 Z  2  (r42 X s  r42Y c) F42 

We do not need a matrix solution here since the X and Y force equations are decoupled. The solution is identical to that of the four-bar mechanism. F21 X   m2 AG 2 X  cF42 F21Y   m2 ( AG 2Y  g )  sF42

 2  I G 2 Z  2  r12Y F21 X  r12 X F21Y  (r42 X s  r42Y c) F42

117

Matrix inversion requires approximately

3n3 and Gaussian elimination requires approximately log n

(n2  1)n 2  n multiplications/divisions (Nering, 1974). 3 Method

6x6 3x3 plus decoupled link 2 Reduction in cost

Number of Multiplications/Divisions Inversion Gaussian 833 106 177 24 79% 77%

Reduction 87% 86%

There is a substantial 77% reduction in computational cost for Gaussian elimination with the 3x3 plus decoupled link 2 method. Also, the numerical accuracy may also improve with this method since we needn’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 force and 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 force and moment for the inverted slider-crank mechanism is identical to the four-bar in notation and terms. Ground link force/moment diagram

Shaking force  F  F14 X  FS  F21  F41  F21  F14   21 X   F21Y  F14Y 

Shaking moment

M S   2  r21  F21  r41  F14   2  r21 X F21Y  r21Y F21 X  r41 X F14Y  r41Y F14 X

118

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

Given

r1  0.20 r2  0.10 L4  0.32

 2  70  4  150.5 (deg and m)

(m)

r4  0.19

1  0 2  25

2  0

r4  2.47 (rad/s and m/s)

 r4  9.46 (rad/s2 and m/s2)

4  2.18

 4  267.14

In this problem the external forces and moments are zero for both links 2 and 4. In inverse dynamics 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 fixed dynamics parameters. m2 = 0.47 and m4 = 1.51 (kg) IGZ2 = 0.0004 and IGZ4 = 0.013 (kg-m2) 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. 0 0.49 0 0  1  0 1 0.87 0 0   0.05 0.02 0.01 0 0  0 0.49 1 0  0  0 0 0.87 0 1  0.03 0.08 0.14 0  0

0   F21 X   5.03  0   F21Y   9.21  1   F42   0     0   F14 X  30.77  0   F14Y  41.82      0    2   3.47 

The answer is:  F21 X   35.35   F   62.69   21Y     F42  61.47      F14 X   0.46   F14Y   11.65        2   1.10 

 F  35.81 The associated vector shaking force and moment are FS   SX     (N) M S   8.53kˆ (Nm) F 51.03   SY  

119

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

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 all 0    2  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 ).

30

2 AVG

20

RMS

2 (Nm)

10

0

-10

-20

0

50

100

150

2 (deg)

200

Input Torque 2

250

300

350

120

150 FSX

100

FSY

FS (N)

50

0

-50

-100

-150

-200

0

50

100

150

2 (deg)

200

250

300

350

Shaking Force

20

15

10

MS (Nm)

5

0

-5

-10

-15

-20 0

50

100

150

2 (deg)

200

Shaking Moment

250

300

350

121

3.7 Multi-loop Mechanism Inverse Dynamics Analysis The Matrix Method can be applied to any planar mechanism inverse dynamics problem. Here are the five two-loop six-bar mechanisms from Dr. Bob’s on-line Mechanism Atlas.

Stephenson I 6-Bar Mechanism

Stephenson II 6-Bar Mechanism

Watt I 6-Bar Mechanism

Stephenson III 6-Bar Mechanism

Watt II 6-Bar Mechanism

122

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

123

The Watt II six-bar mechanism inverse dynamics 15x15 matrix-vector equation is given below.  1  0   r12Y   0  0   0  0   0  0   0   0  0   0  0   0

0

1

0

0

0

0

0

0

0

0

0

0

0

1  r12 X 0

0  r32Y 1

1 r32 X 0

0 0 1

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0

0

0  r43Y

1

r23Y

1  r23 X

r43 X

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0 0

0 0 0

0 0 0

1 0 r34Y

0 1  r34 X

1 0 r14Y

0 1

0 1

r14 X

1 0  r54Y

r54 X

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

1 0 r45Y

0 1  r45 X

1 0 r65Y

0 1 r65 X

0 0 0

0 0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

1 0

0 1

1 0

0 1

0

0

0

0

0

0

0

0

0

r56Y

 r56 X

 r16Y

r16 X

 A v  b

0   F21 X   m2 AG 2 X  0   F21Y  m2 ( AG 2Y  g )  1   F32 X   I G 2 Z  2      0   F32Y   m3 AG 3 X  0   F43 X   m3 ( AG 3Y  g )      0   F43Y   I G 3 Z  3  0   F14 X   m4 AG 4 X       0   F14Y   m4 ( AG 4Y  g )  0   F54 X   I G 4 Z  4      0   F54Y   m5 AG 5 X   0   F65 X   m5 ( AG 5Y  g )  0   F65Y   I G 5 Z  5      0   F16 X   m6 AG 6 X  0   F16Y   m6 ( AG 6Y  g )      0    2   I G 6 Z  6 

124

Step 6. Solve for the unknowns (continued) Like the four-bar mechanism, it is possible to partially decouple the solution to this problem. If we consider the FBDs of only links 5 and 6 first, this is 6 equations in 6 unknowns – this is verified by looking at the original 15x15 matrix and noting nine 6x1 columns of zeros (1, 2, 3, 4, 5, 6, 7, 8, 15) in the six rows 10 through 15. Here is a more efficient solution. The reduced 6x6 set of equations for links 5 and 6 is given below. 0 1 0 0 0   F54 X   m5 AG 5 X   1  0 1 0 1 0 0   F54Y   m5 ( AG 5Y  g )    r45Y r45 X r65Y r65 X 0 0   F65 X   I G 5 Z  5       0 0 1 0 1 0 F 65 Y     m6 AG 6 X   0 1 0 0 0 1   F16 X  m6 ( AG 6Y  g )       0 r56Y r56 X r16Y r16 X   F16Y   I G 6 Z  6   0

 A56 v56   b56  Solve for the six unknowns v56    A56 

1

b56  .

Second, consider the FBDs of only links 3 and 4: this

is 6 equations in 6 unknowns – this is verified by looking at the original 15x15 matrix and noting seven 6x1 columns of zeros (1, 2, 11, 12, 13, 14, 15) in the six rows 4 through 9. Recognizing that now F54X and F54Y are now known from the above 6x6 partial solution, here is a more efficient solution. The reduced 6x6 set of equations for links 3 and 4 is given below. 0 1 0 0 0   F32 X   m3 AG 3 X  1   0      1 0 1 0 0   F32Y   m3 ( AG 3Y  g )    r23Y r23 X r43Y r43 X  IG 3Z3 0 0   F43 X       1 m4 AG 4 X  F54 X 0 0 1 0   F43Y    0   0      1 m4 ( AG 4Y  g )  F54Y 0 0 0 1 F14 X      r34Y r34 X r14Y r14 X   F14Y   I G 4 Z  4  r54Y F54 X  r54 X F54Y  0  0

 A34 v34   b34  Solve for six unknowns v34    A34 

b34 

equations, from the link 2 FBD. 0  1  0 1   r12Y  r12 X

0   F21 X   m2 AG 2 X  F32 X       0   F21Y    m2 ( AG 2Y  g )  F32Y      1    2   I G 2 Z  2  r32Y F32 X  r32 X F32Y 

1

and then use F32X and F32Y in the following 3x3 set of linear

We do not need a matrix solution here since the X and Y force equations are decoupled. The solution is identical to that of the four-bar mechanism. F21 X  m2 AG 2 X  F32 X

F21Y  m2 ( AG 2Y  g )  F32Y

 2  I G 2 Z  2  r32Y F32 X  r32 X F32Y  r12Y F21X  r12 X F21Y

125

Matrix inversion requires approximately

3n3 and Gaussian elimination requires approximately log n

(n2  1)n 2  n multiplications/divisions (Nering, 1974). 3 Number of Multiplications/Divisions Method

15x15 6x6 twice, plus decoupled link 2 Reduction in cost

Inversion 8609 1672 81%

Gaussian 1345 183 86%

Reduction 84% 89%

There is an astonishing 86% reduction in computational cost for Gaussian elimination with the methods using 6x6 inversion twice, plus the decoupled link 2 solution. Also, the numerical accuracy may also improve with this method since we needn’t do unnecessary calculations with the sixteen 6x1 columns of zeros.

Any mechanism with a dyad of binary links may be decoupled in this manner. Thus, the method is similar and the computational complexity identical for the Stephenson I, Stephenson III, Watt I, and Watt 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 unknown vector force from 5 and 6, then link 2 independently). But links 3, 4, 5, and 6 can be solved first independently of link 2: a 12x12 solution followed by the standard link 2 solution. The computational savings is not as impressive as in the former six-bar cases.

Number of Multiplications/Divisions, Stephenson II Six-Bar Method

15x15 12x12 plus decoupled link 2 Reduction in cost

Inversion 8609 4811 44%

Gaussian 1345 723 46%

Reduction 84% 85%

There is a 46% reduction in computational cost for Gaussian elimination with the 12x12 plus decoupled link 2 method. Also, the numerical accuracy may also improve with this method since we needn’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. Moments about the rotating shaft must be balanced statically.

1) Static Balance

M 2) Dynamic Balance

Z

mgr cos   mB grB cos   0

0

mr  mB rB

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 centripetal acceleration directed inward. F I   m AC   m (  (  r ))

the vector magnitude is FI  mr 2

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

F I  F IB  0 The original and balance inertial forces must be equal in magnitude and opposite in direction. This vector balance condition is equivalent to the two scalar equations below.

F F

X

0

mr 2 cos   mB rB 2 cos   0

Y

0

mr 2 sin   mB rB 2 sin   0

127

Both equations yield the same condition as the static balance case, namely: mr  mB rB 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.  X  0  P CG        Y  0 

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

m x m

YCG  Y 

i i i

m y m i

i

i

Let us consider a system of four point masses. There are two equations to satisfy: 4

4

 mi ri cos i  0

 m r sin 

i 1

where xi  ri cos i , yi  ri sin i , and the

i 1

i i

i

0

4

m i 1

i

term in the denominator cancels out.

If we fix each miri, 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.

m1rc 1 1  m2 r2 c2  m3r3c3  m4 r4 c4  0

r2c2  r3c3  rc 1 1  r4c4

m1r1s1  m2 r2 s2  m3r3 s3  m4 r4 s4  0

r2 s2  r3 s3  r1s1  r4 s4

128

Let Ri = miri; 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 associated figure.

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.  FX  0 MX  0

F F

Y

0

Z

0

M M

Y

0

Z

0

Let us consider each in turn. 4

4

i 1

i 1

 FX   mi ri 2 cosi  mi ri cos i  0 4

 F   m r Y

F

Z

F The  F The

X

Z

and

F

Y

i 1

i i

4

2

sin i  mi ri sin i  0 i 1

00

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

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

129

4

 M X   mi ri 2 sin i Li 0 i 1

M M The

M

X

and

M

Y

4

Y

Z

  mi ri 2 cos i Li 0 i 1

00

equations must be solved to satisfy dynamic balancing. The

M

Z

equation

yields nothing because all inertial forces pass through the axis of rotation. We fixed each miri, we previously determined i, and the 2 term divides out. Therefore, we have two equations in the four unknowns Li; arbitrarily fix L1, L2 and solve for L3, L4. The result is two linear equations in the two unknowns.

m1r1 sin 1L1  m2 r2 sin 2 L2  m3r3 sin 3 L3  m4 r4 sin 4 L4  0 m1r1 cos1L1  m2 r2 cos2 L2  m3r3 cos3 L3  m4 r4 cos4 L4  0  m3 r3 sin  3  m r cos  3  33

m4 r4 sin  4   L3    m1r1 sin 1 L1  m2 r2 sin  2 L2     m4 r4 cos  4   L4    m1r1 cos 1 L1  m2 r2 cos  2 L2 

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 to demonstrate the inverse dynamics analyses in the ME 3011 NotesBook. The first example is for the single rotating link inverse dynamics (whose kinematics were not yet presented). Term Example 1 is for the four-bar mechanism and Term Example 2 is for the slider-crank mechanism. The latter two examples are continuations for the Term Example 1 and 2 kinematics examples presented earlier in this ME 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, FE = 150 N, E = 0, ME = 0 Nm.

r12  rE  0.5 m

Calculated terms

IGZ = 0.17 kgm2

Snapshot inverse dynamics analysis

At   150 , given this link, motion, and external force, calculate F12 X , F12Y , and F S , M S . AGx  4330 m AGy  2500

s2

0 0  F12 X   8510   1      0 1 0  F12Y   4980  0.250 0.433 1     37.5 

 8510  F S  F 21  F 12    N  4980

 F12 X   8510      F12Y    4980  N, Nm         66.5 

M S    66.5kˆ Nm

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

A more meaningful result from inverse dynamics analysis is to solve and plot the dynamics unknowns 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 the Y green. Is the static link weight (mg) significant in this problem?

131

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

2

AGX and AGY (m/s )

5000

0

-5000 0

50

150

200 (deg) 

250

300

350

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

4

1

100

x 10

FSX and FSY (N)

0.5

0

-0.5

-1 0

50

100

150

200 (deg) 

250

300

350

The plot above gives the Shaking Force F S for all 0     360  . The X component is red and the Y green. The Shaking Moment M S is simply the negative of the driving torque  plot shown next.

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 be supplied 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.

Tau (red) with average (green) and rms (blue) 150 100

 (Nm)

50 0 -50 -100 -150 0

50

100

150

200  (deg)

250

300

350

Single Rotating Link Torque Results

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

 AVG 

 0  1   2     k k 1

 RMS 

 02  12   22     k2 k 1

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');

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 been using all along in Term Example 1 (previously called point C) is the CG of the rectangular link shown below for inverse dynamics.

0.3 0.25 0.2

Y (m)

0.15 0.1 0.05 0 -0.05 -0.1 -0.1

0

0.1 X (m)

0.2

0.3

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 r1 = 0.284, r2 = 0.076, r3 = 0.203, r4 = 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 ri 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 m2  0.015 , m3  0.327 , m4  0.036 (kg) and I G 2 Z  7.9 106 , I G 3 Z  1.8  103 , I G 4 Z  9.5 105 (kgm2). All external forces and moments are zero but gravity, g = 9.81 m/s2, is included. Snapshot Analysis (one input angle) 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 F S , M S for this snapshot.

0 1 0 0 0 0 0  1  0 0 1 0 0 0 0 1   0.019 0.033 0.019 0.033 0 0 0 0  0 0 1 0 0 0 1  0  0 0 0 0 1 0 0 1  0 0 0 0.127 0.002 0.037 0.122  0  0 0 0 0 0 1 0 1  0 0 0 0 0 1 1  0  0 0 0 0 0.076 0.047 0.076 0.047 

0   F21x   0.202  0   F21 y   0.034  1   F32 x   0      0   F32 y   8.955      0   F43 x   4.497   0   F43 y   0.015      0   F14 x   0.638      0   F14 y   0.095  0    2   0.023 

The solution is accomplished by Gaussian elimination, or v   A b , or by the reduced 6x6 plus decoupled link 2 method (see the on-line ME 3011 NotesBook Supplement). All methods yield the same results. Snapshot answers: 1

 F21x   6.20   F  10.08   21 y     F32 x   5.99       F32 y   10.11  v   F43 x   2.96 (N, Nm)  F   5.61   43 y     F14 x  3.60  F     14 y   5.52    2   0.43

9.80 FS    (N) 4.56

M S  1.68kˆ (Nm)

135

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

A more meaningful result from inverse dynamics analysis is to solve and plot the dynamics unknowns 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 0   2  360 . The X component is red and the Y component is green. 30

20

2

AG3 (m/s )

10

0

-10 AG3X AG3Y -20

-30

0

50

100

150

2 (deg)

200

Term Example 1, AG 3

250

300

350

136 0.8

0.6

2 (Nm)

0.4

0.2

0

-0.2

-0.4

0

50

100

150

2 (deg)

200

250

300

350

Term Example 1 Inverse Dynamics, 2

The plot above gives the required driving torque  2 (Nm) for all 0   2  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 AVG  0.003 and the root-mean-square torque value (blue)

 2 RMS  0.354 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).  2 RMS 

 22   22   22     22 0

1

2

k 1

k

137

The plot below gives the shaking force F S (N) results for all 0   2  360 . The X component is red and the Y component is green. 10

5 FSX FSY

FS (N)

0

-5

-10

0

50

100

150

2 (deg)

200

250

Term Example 1 Inverse Dynamics, F S

300

350

138

The plot below gives the shaking moment M S (Nm) results for all 0   2  360 . There is only the Z component since a planar moment is a kˆ vector. 2

1.5

1

MS (Nm)

0.5

0

-0.5

-1

-1.5

-2 0

50

100

150

2 (deg)

200

250

Term Example 1 Inverse Dynamics, M S

300

350

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.

0.2 0.15 0.1 Y (m)

0.05 0 -0.05 -0.1 -0.15 -0.2 -0.25

-0.1

0

0.1 X (m)

0.2

0.3

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 r2 = 0.102, r3 = 0.203, h = 0.076 m, 2  30 , 3  7.1 , x = 0.29 m, 2  15 2 rad/s (constant), 3  6.58 rad/s, x   0.60 m/s,  2  0 ,  3  62.33 rad/s2, and  x  30.15 m/s . All moving links are wood, with mass density   830.4 kg/m3. Links 2 and 3 have rectangular dimensions ri x 0.019 x 0.013 thick (m; i=2,3); link 4 has rectangular dimensions 0.076 x 0.019 x 0.013 thick (m), as shown on the previous page. The calculated inertia parameters are m2 = 0.020, m3 = 0.041, m4 = 0.015 (kg) and IG2Z = 1.819e-005, IG3Z = 1.418e-004 (kgm2). The CGs all lie at their respective link centers. There is a constant external force of 1 N acting at the center of the piston end, directed horizontally 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.

Snapshot Analysis (one input angle)

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 F S , M S for this instant (snapshot). 0 1 0 0 0 0 0   F21 X  0.202   1  0 1 0 1 0 0 0 0   F21Y   0.084    -0.025 0.044 -0.025 0.044 0 0 0 1   F32 X   0       1 0 0 1 0 0 0   F32Y   1.018   0     0 0 0 1 0 1 0 0   F43 X   0.168    0 -0.013 0.101 -0.013 0.101 0 0   F43Y   0.009   0      0 1 0 0 0 0 0.2 0   F14Y   0.540    0 0 0 0 1 1 0    2   0.150   0

The solution is accomplished by Gaussian elimination, or v   A

1

b , or by the reduced 5x5 plus

decoupled link 2 method (see the on-line ME 3011 NotesBook Supplement). All methods yield the same results. Snapshot answers:  F21 X   0.736   F   0.121  21Y     F32 X   0.534       F32Y  0.037  v       (N, Nm) F  0.484 43 X      F43Y   0.131       F14Y   0.281      0.039    2  

 0.680  FS    (N) 0.401

M S  0.116kˆ (Nm)

141

Full-Range-Of-Motion (F.R.O.M.) Analysis – Term Example 2 continued

A more meaningful result from inverse dynamics analysis is to solve and plot the dynamics unknowns for the entire range of mechanism motion. Prior to solving the inverse dynamics problem, the plot below gives the CG translational acceleration results for link 3 for all 0   2  360 . Here CG3 is taken as the midpoint of link 3. The X component is red and the Y component is green. 25 20 15 10

2

AG3 (m/s )

5 0 -5 AG3X

-10

AG3Y

-15 -20 -25 -30 0

50

100

150

200

2 (deg)

Term Example 2, AG3

250

300

350

142

0.25

0.2

0.15

2 (Nm)

0.1

0.05

0

-0.05

-0.1

0

50

100

150

200

250

2 (deg)

300

350

Term Example 2 Inverse Dynamics, 2

The plot above gives the required driving torque  2 (Nm) for all 0   2  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 AVG  0.004 and the root-mean-square torque value (blue)  2 RMS  0.099 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).

 2 RMS 

 22   22   22     22 0

1

2

k 1

k

143

The plot below gives the shaking force F S (N) results for all 0   2  360 . The X component is red and the Y component is green. 1.5

1 FSX

0.5

FSY

0

FS (N)

-0.5

-1

-1.5

-2

-2.5

-3 0

50

100

150

2 (deg)

200

250

Term Example 2 Inverse Dynamics, F S

300

350

144

The plot below gives the shaking moment M S (Nm) result for all 0   2  360 . There is only the Z component since a planar moment is a kˆ vector. 0.15

0.1

MS (Nm)

0.05

0

-0.05

-0.1

0

50

100

150

2 (deg)

200

250

Term Example 2 Inverse Dynamics, M S

300

350

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 and chain drives are related to gear mechanisms. Robot joint example

Gear Classification

Externally-meshing Spur Gears

Internally-meshing Spur Gears

146

Helical (Parallel Shaft)

Rack & Pinion

Straight Bevel Gears

Helical (Crossed Shaft)

Worm and Gear

Spiral Bevel Gears

147

V-Belt Drive

Chain Drive

Toothed Belt Drive

Bicycle Sprockets

Herringbone Gears

Automotive Hypoid Gears

148

Automotive Gear Train

Planetary Gear Train

Non-Circular Gears

Automotive Differential

Planetary Gear Train – Aircraft

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 control system. 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 preloads the 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 output shaft, 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. n

IN WG N FS   OUT FS N FS  N CS

where WG stands for wave generator, FS stands for flexible spline, and CS stands for circular spline. For example, if NFS = 200 and NCS = 202, the gear ratio is n

N FS IN 200    100 OUT N FS  N CS 200  202

which means that the output shaft rotates 100 times slower than the input shaft, but the output shaft carries 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 the angular velocity and torque of the output shaft are in the opposite direction of the angular velocity and torque of the input shaft.

151

4.1.2 Gear Ratio

Classification of gear ratios

If n  1

2  1  2  1 The output has reduced speed and increased torque.

This is the electric motor / robot joint case, where n >> 1.

If n  1

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 as high as 1.5.

If n  1

2  1  2  1

This case is called an idler, where the output speed and torque are unchanged, but the direction reverses (for external spur gears)

152

Gear ratio examples – Bicycle Transmissions

Gear Ratio:

n

N OUT N R F  R    N IN N F R  F

Cannondale M400 mountain bike

rear teeth

Ni 11 12 14 16 18 21 24 28 32

44 0.25 0.27 0.32 0.36 0.41 0.48 0.55 0.64 0.73

front teeth 32 0.34 0.38 0.44 0.50 0.56 0.66 0.75 0.88 1.00

22 0.50 0.55 0.64 0.73 0.82 0.95 1.09 1.27 1.45

The Cannondale mountain bike has a traditional front/rear derailleur transmission with three gears in the front and nine gears in the rear, for a total combination of 27 gears. The standard BikeE recumbent bike has a traditional derailleur transmission with seven gears only in the rear. Instead of a traditional derailleur transmission in the front with three gears, the BikeE has an internal hub planetary gear arrangement with three selectable ratios of 1.2913 : 1 (high), 1 : 1 (medium), and 0.7 : 1 (low). The BikeE then has a single chain ring in front to drive the rear chain rings. The standard BikeE original front single chain ring had 46 teeth – I changed this to a smaller front chain ring of 34 teeth for more granny gear in order to climb Mulligan Hill; the cost of this is loss of high gear for the bike path. I designed the lowest gear to mimic the lowest Cannondale gear ratio since I knew that granny gear climbed well. The Cannondale mountain bike has a standard wheel size of 26” diameter and the BikeE has a rear wheel size of 20” diameter. We must consider this difference in wheel sizes to compute the overall effective BikeE gear ratios. The table below reflects this calculation.

153

BikeE standard recumbent bike

rear teeth

Ni 11 13 15 18 21 24 28

34 1.2913 : 1 0.33 0.38 0.44 0.53 0.62 0.71 0.83

front teeth 34 1:1 0.42 0.50 0.57 0.69 0.80 0.92 1.07

34 0.7 : 1 0.60 0.71 0.82 0.98 1.15 1.31 1.53

I was able to obtain a used deluxe BikeE recumbent bike from noted luthier Dan Erlewine. I decided 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 my new 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 into account in the table below. BikeE deluxe recumbent bike

rear teeth

Ni 11 13 15 18 21 24 28

46 1.2913 0.24 0.28 0.33 0.39 0.46 0.53 0.61

front teeth 46 1 0.31 0.37 0.42 0.51 0.59 0.68 0.79

46 0.7 0.44 0.52 0.61 0.73 0.85 0.97 1.13

We see that the standard BikeE that was designed to equal the granny gear of the Cannondale (it was 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 out in 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, one can mate any number of spur gears in a gearbox, or gear train. The leftmost gear is the driving gear and the rightmost is the output gear. All intermediate gears are first the driven gear and then the driving gear as we proceed from left to right. Let us calculate the overall gear ratio nGT. nGT 

 IN OUT

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 change direction. We should have included the +/– signs, by inspection. For gear trains composed of externally-meshing spur gear pairs: odd number of gears the output is in the same direction as the input even 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 the driving and driven gears are distinct, because each pair is rigidly attached to the same shaft. Again, let us calculate the overall gear ratio.

nGT 

 IN OUT

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 input even 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 each other (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 linkage vary wildly over the range of motion. Velocity Ratio

VR 

OUT r 1   IN  IN rOUT n

Torque Ratio

TR 

IN r   OUT  n OUT rIN

The velocity ratio is the inverse of the gear ratio n and the torque ratio is the same as the gear ratio 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 engineer to allow for these effects; this is called clearance), the angular velocity ratio will still be constant to satisfy 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 socalled base circle, always keeping it tangent to the circle. The figure below has th = [0:5:60]*DR and rb = 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.

1

Y (m)

0.5

0

-0.5

-1

-1

-0.5

0

0.5

1

1.5

X (m)

Involute Function Construction

In polar coordinates (, r), the parametric equations for the involute of a circle are given below.   tan t  t r  rb sec t where t is the independent parameter and rb is the base circle radius. In Cartesian coordinates (xI, yI): xI  r cos   rb sec t cos(tan t  t )

yI  r sin   rb sec t sin(tan t  t )

158

Involute Function Example

The plot below shows the involute function for a circle of rb = 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.

4

3

2

Y (m)

1

0

-1

-2

-3

-4 -6

-5

-4

-3

-2 X (m)

-1

Involute Function of a Circle

0

1

2

159 3

The next three figures from Norton (2008) show important gear geometry for planar spur gear design and standardization. In these figures, the pink pinion is the driving or input gear, and the gray gear is the driven or output gear.

Norton (2008) Base Circle

The involute function starts from this circle.

Pitch Circle

A fictitious circle (you cannot see it on a spur gear) with theoretical pure rolling in contact between two cylinders of the pinion and gear.

Pitch Point

The contact point between the two pitch circles.

Pressure Angle

The angle between the common normal (also called axis of transmission) of the two meshing teeth and the velocity of the pitch point (the tangent to both pitch circles). The point of contact slides along this line. A similar angle is defined for cams and followers.

The relationship among the base circle radius rb, pitch circle radius rp, pressure angle  is rb  rp cos 

3

R. Norton, 2008, Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, McGraw-Hill.

160

Norton (2008) The length of contact is measured along the axis of transmission. The beginning of contact is when the tip of the driven gear tooth intersects the axis of transmission. The end of contact is when the tip of driving gear tooth intersects the axis of transmission, as shown in the figure above. Only one or two 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 gear ratio 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 constant rb). 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. rb  rp cos 

Again, rb is fixed, and rp and  both increase – the cosine function maintains the constant rb.

162

Spur Gear Standardization

Gear standardization is used to allow interchangeability in manufacturing and to allow meshing of different size gears (different pitch radii and number of teeth) to achieve desired gear ratios. For two spur 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

pd 

N d

Where N is the number of teeth and d is the pitch diameter, both for each gear.

module

m

d 1  N pd

Module is the SI version of diametral pitch (it is the inverse). SI gears are not interchangeable with English system gears because of different tooth proportion standards.

circular pitch

pc 

d N

Circular pitch is the circumferential distance (arc length) between teeth along the pitch circle 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

pc  2t  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 backlash in 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, and then an impact occurs, which is bad for gear dynamics. This is a non-linear effect in robotics. On earth gravity 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 direct meshing 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 another revolute joint to the planet gear(s) on the other side. Planetary gear trains have two-dof, so two inputs must be given to control the mechanism. For instance, one can drive the sun gear and the arm link with independent 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-planetary gear trains.

Conventional Gear Pair

Planetary Gear Train

M  3(3  1)  2(2)  1(1)  1

M  3(4  1)  2(3)  1(1)  2

n

IN rOUT NOUT   OUT rIN N IN

n

IN ? OUT

We present the tabular method below to determine the gear ratio for various planetary gear trains.

166

Planetary Gear Train Applications

Airplane Propeller Transmission

Planetary gear train applications include airplane propellers, some automotive transmissions and differentials, 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 S  0 . 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 S P

G

=

A

+

G/A

=

A A A

+

G/A

+

G/A –A

First, fill in the given information. G S P

G 0

Each row must add up according to the relative velocity equation G S P

G

=

0

A A A

Now we can fill in down the relative column, using a simple gear ratio (relative to the arm). P / A N N N  S  P / A   S S / A  S  A S / A NP NP NP G S P

G 0

=

A A A

+

G/A –A NS A NP

169

The last row must add up according to the relative equation, to finish the table. G S P

G

A A A

=

0  NS  1  A  NP 

+

G/A –A NS A NP

 N  Therefore, the absolute angular velocity of the planetary gear is P  1  S   A . Since the sign is  NP  positive, it has the same direction as A (CCW).

Calculate Effective Gear Ratio n

 IN  NP 1  A    OUT  P 1  N S NP  NS NP

Example

NS  10

S  0

N P  40  

P  1  n

Check n 

 A  100 rpm, CCW

10  100  125 rpm, CCW 40 

NP 40   0.8 N P  N S 40  10

A  P

P 

A n



100  125 0.8

So we see that with S  0 , the gear ratio is not higher than the conventional gear train. n Let us include S  0 next.

N P 40  . N S 10

170

Now we present a more general system with S  0 .

Table Method Analysis

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 N N N N  S P / A   S S / A   S (S   A )  S ( A  S ) NP NP NP NP S / A 4) The P equation must add up; therefore:  N  N N P   A  S ( A  S )  1  S   A  S S . NP NP  NP  G S

G * S

P

 NS  NS S 1  A  NP  NP 

NS  10

+

* A

S  100 rpm, CW

N P  40

 

P  1 

2)

A * A

NS (A  S ) NP

NS  40  

 A  100 rpm, CCW

10  10 100   100   125  25  150 rpm, CCW 40  40

S  100 rpm, CW

NP  10 40 

NS  40

40



S  125 rpm, CCW

NP  10  

40 

Still not a big ratio.

 A  100 rpm, CCW

P  1  100  (100)  500  400  900 rpm, CCW 10 10

3)

G/A S   A

S  0

Examples

1)

=

That’s a big ratio.

 A  100 rpm, CCW

40

P  1  100  (125)  500  500  0 10 10 

In Example 3, the Sun and Arm rotational velocities cancel so the absolute angular velocity of the Planet is zero.

171

Table Method Analysis

Now let us consider an even more general system with S  0 , also adding an internal-teeth ring gear.

1) The given information is starred (*). 2) The first two rows are identical to the case above. 3) Fill down the right column using a simple gear train, relative to arm A. R / A N N N N N  P R / A   P P / A   P S ( A  S )  S ( A  S ) NR P / A NR NR NP NR 4) The R equation must add up; therefore:  N  N N R   A  S ( A  S )  1  S   A  S S . NR NR  NR 

G * S

G S P R

 1    1  

NS  A  NP  NS  A  NR 

=

NS S NP NS S NR

A * A

+

G/A S   A

* A

NS (A  S ) NP

* A

NS (A  S ) NR

172

Examples

1) NS  10

S  0 plus ring gear N P  40

NR  100  

10  

S  100 rpm, CW  A  100 rpm, CCW 10

(100)  110  10  120 rpm, CCW R   1  100  100 100

2) NS  40

N P  20

NR  80

S  100 rpm, CW  A  200 rpm, CW 

P   1  

 

40 

NS  NS S  A  NP  NP

40

P  1   (200)  (100)  600  200  400 rpm, CW 20 20 



R   1  

 

R  1 

NS NR

 NS S  A  NR 

40  40  (200)  (100)  300  50  250 rpm, CW 80  80

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 and difficult to produce. Also, the cam contact and wear properties are worse than for linkages. Cam Classification

Mabie & Reinholtz (19874) a) disk cam with knife-edge follower d) disk cam with oscillating roller follower

4

b) disk cam with radial roller follower e) disk cam with radial flat-faced follower

H.H. Mabie and C.F. Reinholtz, 1987, Mechanisms and Dynamics of Machinery, Wiley.

c) disk cam with offset radial roller follower f) disk cam with oscillating flatfaced follower

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 J2, 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 a mechanism. 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  f  2  , 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 ( )

176

4.2.2 Cam Motion Profiles

Up to this point, we have been mostly concerned with mechanism analysis: given a mechanism design and its input parameters, determine the position, velocity, acceleration, and dynamics behavior. With cams we must consider mechanism synthesis for the first time: given the motion requirements (follower motion and timing with the input cam angle), design the cam. The first step is to determine a smooth cam follower motion profile. In general a cam follower has 4 motion zones (rise, dwell, fall, dwell), as shown below.

When the motion transitions between different motion functions, we must ensure smooth motion.

Fundamental Law of Cam Design

The cam function must be continuous through the first and second derivatives of displacement across the entire motion interval. Which means: Position, velocity, and acceleration must be continuous for the entire 360  of cam rotation. The jerk function (the derivative of the acceleration) must be finite, but need not be continuous.

If the Fundamental law of Cam Design is satisfied, the resulting dynamic performance will be acceptable for high-speed cam/follower operation. If not, there will be performance degradation due to noise, vibrations, high wear, etc. There is a cyclical impulse hammering at each point in the cam cycle when acceleration is not continuous (even worse if position or velocity is not continuous).

177

S V A J Diagrams

In cam synthesis (design), we are only given the total motion range and perhaps some timing requirements. It is the engineer’s job to determine the position curves and to match the velocity and acceleration across junctions. Position is automatically matched by shifting the dependent function axes. 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 angular velocity,    t ). The slope of a function is the value of its derivative at a point in time (or ). Therefore, for continuous velocity and acceleration curves, the slopes of the position and velocity curves must match across 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 input links that start and stop at zero velocity and acceleration. In fact, I adapted the input motion curves from cam 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. vi ( i   i )  vi 1 ( i 1  0) apply to all functions / junctions.

Match A

slope of V must match across junctions. ai ( i   i )  ai 1 ( i 1  0) 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:

1 f1 (1 )  A012 2

f 2 ( 2 )  V0 2

f 3 ( 3 )  0

V:

v1 (1 )  A01

v2 ( 2 )  V0

v3 ( 3 )  0

A:

a1 (1 )  A0

a2 ( 2 )  0

a3 ( 3 )  0

J:

j1 (1 )  0

j2 ( 2 )  0

j3 ( 3 )  0

Match S at junction B

just shift the vertical axis up.

Match V at junction B v1 (1  1 )  v2 ( 2  0)

A0 1  V0

Try to match A at junction B: a1 (1  1 )  a2 ( 2  0)

A0  0

so V0  A0 1

A0  0 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 (constant velocity) rise. half-cycloidal function constant velocity function  1   S f1 (1 )  L1  1  sin 1  f 2 ( 2 )  V0 2 1   1    L  V v1 (1 )  1 1  cos 1  v2 ( 2 )  V0 1  1   L    A a1 (1 )  21  sin 1  a2 ( 2 )  0 1  1   2L    J j1 (1 )  3 1  cos 1  j2 ( 2 )  0 1  1 

Match S at junction B Match V at junction B

just shift the vertical axis up

v1 (1  1 )  v2 ( 2  0)

1  L1  1  cos  V 1  1  0

so V0 

2L1

1

Match A at junction B

 L1  1  0=0  sin 0 12  1  In this case the acceleration function is continuous because the half-cycloidal function ensures that 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. a1 (1  1 )  a2 ( 2  0)

Half Cycloid 1

0 0 0.05

3

J (m/deg ) A (m/deg2)

V (m/deg)

S (m)

2

0 -4 0x 10

20

40

60

80

20

40

60

80

20

40

60

80

20

40

60

80

5

0 -5 0x 10 2 0 -2 0

 (deg) Half-Cycloidal Rise (to connect with constant velocity)

180

Cam Follower Motion Profile Examples Example 3 We now specify a full-cycloidal function (sinusoidal in cam angle). This will rise all the way to meet a dwell smoothly; it satisfies the Fundamental Law of Cam Design. This is the same function used in term project input link motion specification earlier, when starting at stopping at rest.  21  1 S f1 (1 )  L1  1  sin f 2 ( 2 )  0  1   1 2 L1  21  v2 ( 2 )  0 1  cos  1  1  2 L1  21  A a1 (1 )  sin a2 ( 2 )  0  2  1  1  4 2 L1  21  J j1 (1 )  j2 ( 2 )  0  cos  3 1  1  The full-cycloidal function plots are shown below, rising through the displacement to connect with a dwell. v1 (1 ) 

V

Full Cycloid 1

0 0 0.05

3

2

J (m/deg ) A (m/deg )

V (m/deg)

S (m)

2

0 -3 0x 10 2

20

40

60

80

20

40

60

80

20

40

60

80

20

40

60

80

0 -2 -4 0x 10 2 0 -2 0

 (deg)

Full-Cycloidal Rise (to connect with a dwell)

Note that the derivatives above are with respect to  (deg). To find the time derivatives, use the chain rule (e.g., for velocity, multiplying by d (t ) dt  (t )   , a constant). However, for a constant , it is customary to use the derivatives with respect to  for cam design. The full-cycloidal function matches the ensuing dwell: the displacement functions are made to match, and the velocity and acceleration are zero at the end of the full-cycloidal function and the start of the ensuing dwell. The jerk does not match, but the discontinuity in jerk is finite, which satisfies the Fundamental Law of Cam Design.

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 the engineer, now we must determine the cam contour. Is it as simple as polar-plotting S = f() vs. cam angle  ? No – that approach would not account for the face width of the cam follower, i.e. the contact points are not along radial lines in general. We will 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 cam angle through geometry.

182

R  x cos   y sin  L   x sin   y cos  Notice that

dR   x sin   y cos  L d L 

d df (C  f ( ))  d d

To calculate the follower flat-face length, double the maximum of L from above. It is doubled because by symmetry the contact point will change to the other side at   180 . To summarize thus far:

R  C  f ( )

L

df d

This is sufficient to manufacture the cam since it is machined with , R, L coordinates. If we want to know the cam contour in Cartesian coordinates, we must solve the relationships for x, y. In matrix form:

 cos sin    x  R  sin  cos   y    L       This special coefficient matrix [A] is orthonormal, which means both columns and rows are perpendicular to each other and both columns and rows are unit vectors. One unique property of 1 T orthonormal matrices is  A   A . The Cartesian cam contour solution is thus:

 x  cos  sin   R  R cos  L sin          y   sin  cos   L   R sin   L cos 

x  (C  f  ) cos  

df sin  d

y  (C  f  ) sin  

df cos  d

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.

dx dy  0 d d

will cause a cusp.

dx df df d2 f  (C  f  ) sin   cos   cos   2 sin  d d d d dy df df d2 f  (C  f  ) cos   sin   sin   2 cos  d d d d  dx d2 f    C  f    2 d d 

  sin  

 dy d2 f   C  f    2 d d 

  cos  

dx dy   0 occurs simultaneously only when d d C  f ( ) 

d2 f 0 d 2

Therefore, to avoid cusps on the entire cam contour, we must ensure that d2 f C  f ( )  2  0 d Note that C is always positive and f() starts and ends at zero and never goes negative. So the sum of these positive terms and the sometimes-negative second derivative of the cam motion profile must always 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 symmetric full-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, and acceleration curves are continuous. The jerk is not continuous, but it remains finite over all cam angles. 50 40 S

30 20 10 0 0

50

100

150

200

250

300

350

50

100

150

200

250

300

350

50

100

150

200

250

300

350

50

100

150

200

250

300

350

V

50

0

-50 0

100

A

50 0 -50 -100 0

J

500

0

-500 0



185

Choosing a minimum cam radius of C = 100 mm, the resulting cam contour is shown below.

150

Y (mm)

100

50

0

-50

-100 -150

-100

-50 0 X (mm) Cam Cartesian Contour

50

100

186

Let us check the cusp avoidance plot. To avoid cusps in this cam, we require that

C  S ( )  A( )  C  f ( ) 

d2 f 0 d 2

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. 250

200

C+S+A

150

100

50

0 0

50

100

150



200

Cam Cusp Avoidance Plot

250

300

350

187

200

150

100

Y (m)

50

0

-50

-100

-150

-200 -200

-150

-100

-50

0 X (m)

50

100

Cam/Follower Animation Snapshot for  2  60 (CW)

150

200

188

5. Mechanical Vibrations Introduction 5.2 Mechanical Vibrations Definitions This section presents some important definitions for mechanical vibrations concepts, used throughout the ME 3011 and ME 3012 NotesBooks. You may already be familiar with some these terms – if some are unfamiliar, don’t freak out, we will discuss them later. mechanical vibrations

Continual, periodic motions about an equilibrium point by a machine or machine component, in some cases desired but in others unwanted.

degrees of freedom

dof – the number of independent parameters required to fully specify the location of a device. The number of different ways something can move.

statics

The study of forces/torques without regard to motion. force/moment equilibrium in structures.

kinematics

The study of motion without regard to forces/torques.

dynamics

The study of motion with regard to forces/torques.

free-body diagram

FBD, a diagram drawn out of context for each separate mass or inertia with all external and internal forces and moments shown to give context.

cycle

A repeating unit of the vibration.

period

T, time for one cycle, sec.

cyclical frequency

f, number of cycles per second, the inverse of time period, 1 / T, Hz.

circular frequency

  2 f , scalar measure of rotational rate, i.e. the magnitude of the

The study of

angular velocity vector, rad/sec. natural frequency

The rate at which an unforced, undamped vibratory system tends to vibrate, either cyclical frequency fn, or circular frequency n.

spring

Idealized massless mechanical element that provides the oscillatory motion in a vibrational system.

dashpot

Idealized massless mechanical element that provides the energy dissipation in a vibrational system.

mass

Idealized mechanical element that provides the inertia in a vibrational system.

mass moment of inertia

Idealized mechanical element that provides the rotational inertia in a vibrational system.

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 a virtual 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 or lag 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. There must 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 forcing function, 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 is also 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 fd, or circular d.

logarithmic decrement

The natural logarithm of the ratio of two consecutive points, separated by the damped time period Td, on a unforced underdamped vibratory wave.

beat frequency

Vibrational phenomenon wherein two frequencies in a dynamic system are close 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 resulting amplitude of the steady-state vibration becomes very large.

190

frequency response

The plots of system amplitude and phase angle vs. independent variable input frequency.

critical shaft speed

The angular shaft velocity that excites the natural frequency of a rotating dynamic system. When driven at the critical speed, resonance occurs, a condition generally to be avoided at all costs.

transmissibility

The ratio of the maximum force transmitted to the base over the maximum input force magnitude in a vibratory dynamic system.

vibration mode

Characteristic shape at which a dynamic system will vibrate. Multi-dof systems 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

2)

u1 (t )  y1 (t ) and u2 (t )  y2 (t )

then

u1 (t )  u2 (t )  y1 (t )  y2 (t )

u (t )  y (t )

then

 u (t )   y (t )

homogeneity

if

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 translational springs 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.

f(t)

f(t)

D

d

Gd 4 k 8nD3

where G d n D

material shear modulus wire diameter number of spring coils diameter of spring coiling

N/m2 m unitless m

This type of coiled physical spring is common; however, there are many mechanical systems without a clear spring such as above. Instead, their elastic properties are provided by distributed material in the system. Here we present several such distributed springs that are modeled as linear springs 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 force is applied and the displacement measured at the end of the cantilevered beam and in the center of the other beams.

192

Cylindrical Rod Spring. An elastic rod has an equivalent spring constant along the longitudinal axis of the rod. From Strength of Materials, the linear spring constant k for small displacements is given below. r

f(t)

f(t)

f(t)

f(t)

r1

r2

L

L

k

EA L

where E A r L

N/m2

material Young’s modulus rod cross-sectional area,  r 2 radius of rod length of rod

m2 m m

So the cylindrical rod spring constant for a constant circular cross section becomes E r2 k L If the elastic rod is tapered with different radii r1 and r2 on opposite ends of the rod, the equivalent spring constant is E rr 1 2 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 small displacements is:

m

h L

x(t) k

3EI L3

where E I

material Young’s modulus beam area moment of inertia, bh 3 12

b h L

beam width beam height length of beam

So the cantilevered beam spring constant becomes k

Ebh3 4L3

N/m2 m4 m m m

193

Doubly-Simply-Supported Beam Spring. A rectangular beam simply-supported at both ends has an equivalent spring constant for small vertical displacements. From Strength of Materials, the linear spring constant k for small displacements is given below. m

h

L x(t)

k

48EI L3

where E b h L

material Young’s modulus beam width beam height length of beam

N/m2 m m m

So the doubly-simply-supported beam spring constant becomes 4Ebh3 k 3 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 equivalent spring constant for small vertical displacements. From Strength of Materials, the linear spring constant k for small displacements is given below.

m

h

L x(t)

k

192EI L3

where E b h L

material Young’s modulus beam width beam height length of beam

N/m2 m m m

So the doubly-fixed beam spring constant becomes 16Ebh3 k L3 This doubly-fixed beam is 64 times stiffer in vertical displacement than the cantilever beam, and 4 times stiffer than the simply-supported beam presented earlier.

194

Air Spring. An enclosed volume of air has an equivalent spring constant for small displacements. From Fluid Mechanics for compressible air, the linear spring constant k for small displacements is given below. f(t)

k where

 P A V

 PA2 V

ratio of specific heats enclosed air pressure enclosed air cross-sectional area enclosed air volume

unitless N/m2 m2 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:

E  200 GPa = 200 109 N / m2

G  79 GPa = 79 109

N / m2

Let us now check the units for the previous six spring constants.

Gd 4 Nm4 N   8nD3 m2m3 m

EA Nm2 N   L m2m m

3EI Nm4 N   L3 m2m3 m

4Ebh3 N(m)m3 N  2 3  L3 mm m

16Ebh3 N(m)m3 N  2 3  L3 mm m

 PA2

Nm4 N  2 3 V mm m

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 be constructed of a coiled elastic material, spiraling in one plane. An elastic cylindrical rod also provides a torsional spring effect when torque about the longitudinal axis of the rod. Additionally, a fixed-simplysupported beam also provides a torsional spring effect.

195

Torsional Rod Spring. An elastic rod has an equivalent spring constant about the longitudinal axis of the rod. From Strength of Materials, the torsional spring constant kR for small angular displacements is given below. r

 (t)

 (t)

L

G r4 kR  2L where G r L

N/m2 m m

material shear modulus radius of rod length of rod

If the torsional shaft is hollow, the torsional spring stiffness constant is G (ro4  ri4 ) kR  2L

where ro and ri are the outer and inner shaft radii, respectively. Fixed-Simply-Supported Torsional Beam Spring. A rectangular beam rigidly fixed at one end and simply supported on the other has an equivalent torsional spring constant for small angular displacements. From Strength of Materials, the torsional spring constant kR for small displacements is given below. (t)

m

h

L

kR 

4EI L

where E b h L

material Young’s modulus beam width beam height or thickness length of beam

N/m2 m m m

So the fixed-simply-supported torsional beam spring constant becomes Ebh3 kR  3L

196

Let us now check the units for these torsional spring constants.

G r4 Nm4 Nm Nm    2L m2m 1 rad

Ebh3 N m4 Nm Nm  = = 3L m2 m 1 rad

Recall that rad doesn’t count as a unit so we are free to include it above, yielding the correct SI units Nm/rad for torsional spring constant.

Non-massless Spring. For translational mechanical systems in which the distributed mass of the spring is significant and cannot be ignored, we use the following formula for the lumped system mass.

mE 

mS m 3

kg

Where m is the modeled point mass, mS is the spring mass, and mE is the calculated equivalent point mass to include in the model with a massless spring. The spring constant k does not change and is calculated as shown earlier in this subsection. For the cantilever and simply-supported beam spring models, the equivalent mass is given below (left and right below, respectively, and mB is the distributed beam mass, all kg).

mE  0.23mB  m

mE 

mB m 2

197

Calculation of Damping Constants

In general the damping coefficient for a physical system is harder to identify and calculate than the 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 included to account for friction losses in the system, using a simple linear model (the law for linear dashpots where the friction force is proportional to the velocity). Also see the later section on logarithmic decrement, where the damping coefficient c can be estimated from experimental observations. Dashpot. The piston-based dampers indicated so far represent physical dashpots with a piston moving (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 linear viscous damping coefficient c for small velocities is given below.

f(t)

r w

h

c where

 r h w

fluid viscosity piston radius piston height wall thickness ro – r

6 r 3h  w  1   w3  r  N-s/m2 m m m

198

Coulomb damping. Coulomb damping arises from the so-called dry friction of sliding. This is a nonlinear effect that can be approximated by the following equivalent linear viscous damping coefficient ceq.

v(t) m f

f

ceq  where ff

 N B



friction force N coefficient of friction normal force amplitude of vibration frequency of vibration

4ff B



4 N B

N unitless N m rad/s

Flat plates. When two flat plates move relative to each other, separated by a layer of viscous fluid, damping arises. From Fluid Mechanics, the linear viscous damping coefficient c for small velocities is given below. f(t) w

ceq  where

 A w

A w

fluid viscosity N-s/m2 area of smaller flat plate m2 viscous fluid layer thickness m

These three damping examples are all for translational dampers, so their viscous damping coefficient c units should all be N-s/m. The SI units for fluid viscosity are N-s/m2 and the Coulomb friction coefficient  is unitless. Let us now check the units for the previous three damping coefficients.

6r 3h  w  Ns m3m  m  Ns 1      w3  r  m2 m3  m  m

4 N N Ns   B m rad m s

All units are correct for a damping coefficient, i.e. N-s/m.

 A Ns m2 w



2

m m



Ns m

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 the reciprocating motion is rotary about the center axis, instead of translational. From Fluid Mechanics, the linear rotational viscous damping coefficient cR for small angular velocities is given below.

 (t)

r w

h

h0

h r  cR  2 r 3     w 8h0 

where

 r h w h0

fluid viscosity piston radius piston height wall thickness ro – r height from base to rotating piston

N-s/m2 m m m m

200

Journal bearing. A journal bearing also has a fluid providing rotational damping. From Fluid Mechanics, the linear rotational viscous damping coefficient cR for small angular velocities is given below.

 r

w

cR 

where

 w r



2 wr 3

fluid viscosity base thickness bearing radius journal bearing fluid clearance

 N-s/m2 m m m

Let us now check the units for these torsional damping coefficients. h r  Ns 3  m  Nms 2 r     2 m    Nms  rad m  w 8h0  m 3

2wr3



Ns m4 Nms  2  Nms  m m rad

Recall that rad doesn’t count as a unit so we are free to include it above, yielding the correct SI units Nm-s/rad for torsional damping coefficient.

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.

x(t) k1

k2

m

1-dof Translational Mechanical System with Parallel Springs, Alternate

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 spring derivation is now presented.

f  f1  f 2 f  k1 x  k2 x  (k1  k2 ) x Therefore, the equivalent spring constant for parallel springs case is again:

kpar 

f  k1  k2 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:

kser2 

k1k2 k1  k2

kser3 

k1k2 k3 k1k2  k1k3  k2 k3

kser4 

k1k2k3k4 k1k2k3  k1k2 k4  k1k3k4  k2 k3k4

Actually, the original general formula below yields the same answers, with less computation:

kser 

1 n

1

i 1

i

k

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 ME 3011 NotesBook, for the two pendulum models shown below. The figure on the left shows a pendulum consisting 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 point mass of m2.

O

O g

g

m

m1

L

L

 (t)

 (t)

m2 Two Additional Pendulum System Diagrams

Only the mass moment of inertia IOZ changes; for the cases on the left and right above, respectively, the formulae are given below. m L2 mL2 I OZ  1  m2 L2 I OZ  3 3 The associated models are: 3g (t)   (t)  0 2L

(t ) 

3(m1  2m2 ) g  (t )  0 2(m1  3m2 )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 rotational viscous damping coefficient at the rotational bearing.

mL2(t )  cR(t )  mgL (t )  0

(t ) 

cR  g  (t )   (t )  0 2 mL L

And while not common, it is also possible to include an input forcing function, input torque (t) at the bearing.

203

(t) 

cR  1 g  (t)   (t)  2  (t) 2 mL L mL

Finally, if we model a horizontal pendulum, the restoring torque of gravity disappears and we have the dynamic model of a gravity-neutral single rotating link.

(t) 

cR  1  (t)  2  (t) 2 mL mL

204

6.2.4 Uniform Circular Motion Uniform 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.

Y

P r (t)

 O

X

Uniform Circular Motion Diagram

In this UCM case no differential equations or solutions are required. Simultaneously simple harmonic motion (SHM) is generated in the X with a pure cosine wave (with no phase angle) and in the Y 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. x(t )  r cos  (t )  r cos t y(t )  r sin  (t )  r sin t

MATLAB UCM Animation

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 their respective parallel and series cases. Considering the parallel-case figure shown below,

x(t) k2

k1 m

c1

f(t) c2

1-dof Translational Mechanical System with Parallel Springs and Dashpots

the system model is:

mx(t )  (c1  c2 ) x (t )  (k1  k2 ) x(t )  f (t )

Considering the series-case figure shown below,

x(t) k1

k2

c1

c2

m

f(t)

1-dof Translational Mechanical System with Series Springs and Dashpots

the system model is:  cc   kk  mx(t )   1 2  x (t )   1 2  x(t )  f (t )  c1  c2   k1  k2 

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 translational systems presented on the last page, with regard to parallel and series arrangements of rotational springs and dashpots. Considering the parallel case, the system model is given below. J (t )  (c R1  c R 2 ) (t )  ( k R1  k R 2 ) (t )   (t )

Considering the series case, the system model is given below.  c c   k k  J(t )   R1 R 2  (t )   R1 R 2   (t )   (t )  cR1  cR 2   k R1  k R 2 

207

Compound m-c-k translational systems with parallel and series springs and dampers

Certain real-world systems may require more complicated models with combinations of springs and 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 dashpots are not shown but they have the same arrangement.

x(t) k1

k2

k3

k4

m

f(t)

The second figure shows a simplified diagram for this case.

x(t) ka

kb

m

f(t)

Using the parallel- and then serial-springs equivalent spring formulas derived in the ME 3011 NotesBook:

k a  k1  k3 kb  k 2  k 4

keq 

k a kb k k k k k k k k  1 2 2 3 1 4 3 4 k a  kb k1  k2  k3  k4

The system model is now given. mx(t )  ceq x (t )  k eq x (t )  f (t )

where ceq 

c1c2  c2 c3  c1c4  c3c4 c1  c2  c3  c4

This example is for a translational system, but equivalent results exist for an analogous rotational system.

208

4. Series springs and dampers in parallel

The figure below shows two sets of series springs arranged in parallel. For clarity, the dashpots are not shown but they have the same arrangement.

x(t) k1

k2

k3

k4

m

f(t)

The second figure shows a simplified diagram for this case.

x(t) ka kb

m

f(t)

Using the series- and then parallel-springs equivalent spring formulas derived in the ME 3011 NotesBook:

ka  kb 

k1k 2 k1  k 2 k3 k 4 k3  k 4

keq  ka  kb 

kk k k k  k k k  k k k  k 2 k3 k 4 k1k2  3 4  1 2 3 1 2 4 1 3 4 k1  k2 k3  k4 k1k3  k2 k4  k1k 4  k 2 k3

The system model is given below. mx(t )  ceq x (t )  k eq x (t )  f (t )

Where ceq 

c1c2 c3  c1c2 c4  c1c3c4  c2 c3c4 c1c3  c2 c4  c1c4  c2 c3

This example is for a translational system, but equivalent results exist for an analogous rotational system.

209

5. Electric motor with vibration isolation

FBD

a (t)

k

J

k

Often a vibration isolation system will be designed to minimize the transference of shaking forces and shaking moments from rotating machinery to the ground. The rotating electric motor shown above, whose moment of inertia about the motor shaft axis is J, is supported by four identical springs (2 shown, 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 when the 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 right translational spring are in tension in applying the restoring torques.

 M  J (t ) 4 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.

J(t )  4ka 2 (t )  0

210

6. Sprung mass/cylinder

r

FBDs

(t)

m k m x(t)

In this dynamic system, m1 is the point mass suspended by cable about a cylindrical pulley of mass m2 and radius r. The other end of the cable is attached to a linear translational spring with constant 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 zero when x(t) is zero. First, draw the FBDs. Using Newton’s Second Law for the point mass:  FX  m1g  T  m1x(t ) Using Euler’s Rotational Dynamics Law for the cylinder:  M Z  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 J  m2 r 2 2 . Now substitute x(t )  r (t ) and  x(t )  r(t ) into the above equations. m g  T  m r(t ) 1

Tr  kr (  r (t )) 

1

m2 r 2   (t ) 2

We can combine the above two equations to eliminate the cable tension T. T  m1 g  m1r(t ) m1 gr  m1r 2(t )  kr (  r (t )) 

m2 r 2   (t ) 2

At static equilibrium we have m1 gr  kr , so the final 1-dof model for this system is  m1r 2(t )  kr 2 (t ) 

m2 r 2   (t ) 2

m2     m1    (t )  k (t )  0 2  

211

7. Doubly-sprung torsional rod

FBD

k (t)

kR

y(t)

L k

In this dynamic system, m is the distributed mass of the slender rigid rod of length L. The rod’s left end is connected to ground by a torsional spring kR and the rod’s right end is supported by two identical linear translational springs k as shown. We measure angular displacement (t) in a righthanded sense from the horizontal rod position. To linearize we will make a small angle assumption; we further 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:

M

Z

 kR (t )  2kLy(t )  J(t )

The mass moment of inertia of the slender rod about its end is J  mL2 3 . Also, y(t )  L sin  (t ) , leading to:

mL2   (t )  kR (t )  2kL2 sin  (t )  0 3 Applying the small angle assumption, we have sin  (t )   (t ) and the final 1-dof model for this system is

mL2   (t )   kR  2kL2  (t )  0 3

212

8. Sprung rolling cylinder

FBD

x(t) r

k

(t)

m

In this dynamic system, a cylinder of mass m and radius r rolls without slipping on a horizontal surface. 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 (Ff is the force due to friction):

F

X

 Ff  kx(t )  mx(t )

Using Euler’s Rotational Dynamics Law for the cylinder:

M

Z

 Ff r  J(t )

We can combine the above two equations to eliminate the friction force Ff: F f  mx(t )  kx (t ) ( mx(t )  kx(t )) r  J(t )

The mass moment of inertia of the cylindrical pulley is J  mr 2 2 . (t )    x(t ) r into the above equation to express the motion in terms of x(t).

(mx(t )  kx(t ))r 

mr 2   x(t )    2  r 

The final 1-dof model for this system is

3 mx(t )  kx(t )  0 2

Now substitute

213

9. Doubly-sprung rotating cylinder

r

FBD

(t)

m k

k

In this dynamic system, a rotating cylindrical pulley of mass m and radius r is suspended from the ground link as shown. A massless inextensible cable is passed over the pulley and connected to two grounded 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-handrule (positive CCW). First, draw the FBD. Using Euler’s Rotational Dynamics Law for the cylinder:

M

Z

 T1 r  T2 r  J (t )

Where the left cable tension is T1  k (  r (t )) and the right cable tension is T2  k (  r (t )) . Also, the mass moment of inertia of the cylindrical pulley is J  mr 2 2 .

kr (  r (t ))  kr (  r (t )) 

The final 1-dof model for this system is

m(t )  4k (t )  0

mr 2   (t ) 2

214

10. Sprung pendulum

FBD

L0 k L (t) m 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 pendulum rod. We measure angular displacement (t) with the right-hand-rule (positive CCW) such that it is zero when the pendulum is vertical (down). First, draw the FBD. Using Euler’s Rotational Dynamics Law for the pendulum:

M

Z

 mgL sin  (t )  a cos  (t )ka sin  (t )  J(t )

Where kasin (t) is the amount the spring stretches. The mass moment of inertia of the pointmass pendulum about the point of rotation is J  mL2 . mL2(t )  (mgL  ka 2 cos  (t ))sin  (t )  0 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(t) 1

We are implicitly ignoring the vertical motion of the spring. The final 1-dof model for this system is

mL2(t )  (mgL  ka 2 ) (t )  0

215

11. Doubly-sprung rolling cylinder

FBD

x(t) k

k r

a

(t)

m

In this dynamic system, a cylinder of mass m and radius r rolls without slipping on a horizontal surface. 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 (Ff is the force due to friction):

F

X

  Ff  k (  (r  a) (t ))  k (  (r  a) (t ))  mx(t )  mr(t )

Using Euler’s Rotational Dynamics Law for the cylinder:

M

Z

 F 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 J  mr 2 2 . We can combine the above two equations to eliminate the friction force Ff:

Ff  mr(t )  2k (r  a) (t ) 2

mr  (mr(t )  2k (r  a) (t ))r  2ka(r  a) (t )   (t ) 2

The final 1-dof model for this system is

3 2  mr  (t )  2k (r  a)2 (t )  0 2

216

12. Hung-sprung mass/cylinder

FBDs

k

r

(t)

m

x  (t) m x(t)

In this dynamic system, m1 is the point mass suspended by an inextensible cable about a cylindrical 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 measure displacements 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. Using Newton’s Second Law for the point mass:

F

 m1 g  T1  m1 x1 (t )

X

Using Newton’s Second Law for the cylinder:

F

X

 T1  T2  m2 g  k (  x2 (t ))  m2  x2 (t )

Where  is the static deflection of the spring due to gravity. Using Euler’s Rotational Dynamics Law for the cylinder:

M

Z

 T1 r  T2 r  J (t )

The mass moment of inertia of the cylindrical pulley is J  m2 r 2 2 . Substituting the mass moment of inertia and eliminating the cable tensions T1 and T2 amongst these three equations yields

217

T1  m1 g  m1 x1 (t ) T2  m2  x2 (t )  m2 g  k (  x2 (t ))  T1 2(m1 g  m1 x1 (t ))  m2  x2 (t )  m2 g  k (  x2 (t )) 

m2 r   (t ) 2

At static equilibrium we have

T1  T2  m2 g  k (2m1  m2 ) g  k So the model simplifies to

m2r   (t )  2m1x1 (t )  m2 x2 (t )  kx2 (t )  0 2 Note that from kinematics we know x1 (t )  2 x2 (t ) and x2 (t )  r (t ) . equations yields the final 1-dof model for this system.

Substituting the kinematic

3   x1 (t )  kx1 (t )  0  4m1  m2   2   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

k a

r

(t)

m m x(t)

In this dynamic system, m1 is the point mass suspended by cable about a cylindrical pulley of mass m2 and radius r. The pulley is grounded via a linear translational spring with constant k, attached distance 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-handrule (positive CCW) such that it is zero when x(t) is zero. First, draw the FBDs. Using Newton’s Second Law for the point mass:  FX  m1g  T  m1x(t ) Using Euler’s Rotational Dynamics Law for the cylinder:  M Z  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 J  m2 r 2 2 . We can combine the above two equations to eliminate the cable tension T. T  m1 g  m1  x(t ) x(t ))r  ka (  a (t ))  (m1 g  m1 

m2 r 2   (t ) 2

At static equilibrium we have m1 gr  k a , so the model becomes

m1rx(t )  ka 2 (t ) 

m2 r 2   (t ) 2

x(t )  r(t ) in the above equation, the final 1Now using the kinematic relationships x(t )  r (t ) and  dof model for this system is

m2  2   2  m1   r  (t )  ka  (t )  0 2 

219

14. Cylinder rolling in a big cylinder

R

FBD

2(t) r 1(t)

m

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 the angular displacement of the cylinder CG from the vertical. Both angles are measured with the righthand-rule (positive CCW) such that they are zero when the cylinder is at static equilibrium at the bottom of the well. We define the moving X axis such that it is perpendicular to the radial direction, pointing to the left. First, draw the FBD. Using Newton’s Second Law for the cylinder:  FX  mg sin 2 (t )  Ff  mx(t ) Using Euler’s Rotational Dynamics Law for the cylinder:  M Z  F f r  J (1 (t )  2 (t )) Note we must use the absolute angular acceleration 1 (t )  2 (t ) in the above equation. The mass moment of inertia of the cylindrical pulley is J  mr 2 2 . Substituting the mass moment of inertia and eliminating the friction force Ff between these two equations yields Ff  mg sin  2 (t )  mx(t ) mg sin  2 (t )  mx(t ) 

mr  (1 (t )  2 (t )) 2

Note that from kinematics we know  x (t )   ( R  r )2 (t ) and 1 (t )   R 2 (t ) r . kinematic equations yields 3( R  r )  2 (t )  g sin 2 (t )  0 2

Substituting the

We apply a small angle assumption such that sin  2 (t )   2 (t ) . The final linearized 1-dof model for this system is 3( R  r )  2 (t )  g2 (t )  0 2

220

15. Doubly-sprung pulleys/mass

FBDs

k

x(t)

k

x(t)

m x(t)

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 translational springs with constants k1 and k2, respectively. We measure displacements x(t) (positive downward), x1(t), and x2(t) as shown. The kinematics relationship is:

x(t )  2 x1 (t )  2 x2 (t ) From a FBD for pulley 1 and pulley 2: 2T (t )  k1 x1 (t )

2T (t )  k2 x2 (t ) Substituting the above relationships into the kinematics relationship yields 4T (t ) 4T (t ) x(t )   k1 k2 Looking at the point mass we can write k EQ x (t )  T (t )

Where kEQ is the equivalent spring constant.

kEQ  The final 1-dof model for this system is

T (t ) k1k2  x(t ) 4  k1  k2  mx(t )  k EQ x(t )  0

mx(t ) 

k1k2 x(t )  0 4  k1  k2 

221

16. Rolling cylinder/pendulum

FBD

r m2

L (t) m1 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. M1 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) with the 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 of zero velocity. Therefore let us take point C as the moment center for deriving the dynamic model  M CZ  JC(t ) . The mass moment of inertia of the compound rigid link, using the parallel axis theorem, is: m2 r 2 3 2 J C  m1R   m2 r 2  m1R2  m2 r 2 2 2 Where, by use of a vector loop-closure equation, R 2  L2  2 Lr cos  (t )  r 2 . Using Euler’s Rotational Dynamics Law for the cylinder:  M CZ  m1gL sin  (t )  JC(t )

3  2 2    m1R  m2 r   (t )  m1 gL sin  (t )  0 2   3  2 2 2    m1 ( L  2 Lr cos  (t )  r )  m2 r   (t )  m1 gL sin  (t )  0 2   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(t) 1 The final 1-dof model for this system is

3  2 2    m1 (L  r)  m2r  (t )  m1gL (t )  0 2  

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. g (t )   (t )  0 L

We also derived the dynamic model for a slender-rod pendulum with distributed mass and also for a slender-rod pendulum with distributed mass plus point mass. Here we present the dynamic models for 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 rG 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 JG

FBD

(t) 

J O  J G  mr

2 G

mgrG  (t)  0 JO

Note that if we define a virtual point mass pendulum length of LEQ  J O mrG , 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.

g

(t)

r

Semi-circular 4r rG  3 mr 2 JO  2

(t ) 

(t)

(t)

m

8g  (t )  0 3 r

g

g

g

m

m

r

m

s

L

(t) s

h

Circular mr 2 JG  2 3mr 2 JO  2

(t ) 

2g  (t )  0 3r

Rectangular m( L2  h 2 ) JG  12 m(4 L2  h 2 ) JO  12

(t ) 

6 gL  (t )  0 4L2  h2

Square ms 2 JG  6 5ms 2 JO  12

(t ) 

6g  (t )  0 5s

223

18. Beams with a point mass

Earlier we derived the equivalent spring constants kEQ for three massless beams with point masses attached as shown, the cantilevered beam, the simply-supported beam, and the rigidly-clamped beam. For all three cases the general model is mx(t )  k EQ x (t )  0 . m

h L

h x(t)

Cantilevered beam

mx(t ) 

Ebh3 x(t )  0 4 L3

L x(t) Simply-supported beam

mx(t ) 

4Ebh3 x(t )  0 L3

m

h

L x(t) Doubly-fixed beam 16Ebh3  mx(t )  x(t )  0 L3 Where, for all three cases: E material Young’s modulus b beam width h beam height L length of beam

m

224

19. Rotating imbalance

Machines with rotating parts are common in industry. The figure below represents a simplified idealized machine with an offset mass m1 that is unbalanced in rotation. The total mass of the machine is 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 ki and ci coefficients add up to yield the overall k and c). The machine is constrained to move in the vertical direction only. We measure displacement x(t) in the vertical (up) direction, from the gravityloaded spring location.

m1

x(t)

e t

m2 k

c

The model is the same we have derived earlier for the standard m-c-k translational mechanical system.

m2  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 e2 (directed inward) and thus the inertial force of the rotating mass is m1e 2 (directed outward). The angle of rotation is  t and so the vertical component of the rotating imbalance is:

f (t )  m1e 2 sin t And so the final 1-dof model for this system is

m2  x(t )  cx (t )  kx(t )  m1e 2 sin t

225

20. m-c-k system with displacement input

u(t) k c

FBD

x(t)

m

Some machines can be modeled as the standard linearized, lumped-parameter, constant coefficient, second-order m-c-k system, but with a base displacement input in place of the standard force f(t) input (see the figure above). In this case, the base is not fixed, but harmonically-moved via base displacement u(t) in units of meters. We measure displacement x(t) in the same manner (positive to the right, from the neutral position of the spring, this time with u(t) = 0). There is no input forcing function f(t). The moving base portion is massless. The model is similar to what we have previously derived. Assume that displacements x(t) and u(t) are small and positive at this instant in time, with x(t)  u(t) . We know that the model we derive based on these assumptions will hold good for all small displacement motions, as both x(t) and u(t) vibrate at different phases in different directions. With this assumption, the spring is in tension, and the damper behaves similarly. First, draw the FBD. Using Newton’s Second Law for mass m:  FX  k ( x(t )  u(t ))  c( x(t )  u(t ))  mx(t )

mx(t )  cx (t )  kx(t )  cu (t )  ku (t ) We can genericize this model similar to previous derivations by using the dimensionless k c as follows. and natural frequency n  damping ratio   m 2 km  x(t )  2n x (t )  n2 x(t )  2nu (t )  n2u (t ) It may be convenient in some cases to use a new variable in the above model, the relative difference in displacements, z(t)  x(t)  u(t) . Then we have mz(t)  cz(t)  kz(t)  mu(t) Where now mu(t) plays the role of input forcing function.

If we specify a sinusoidal input

displacement u(t )  Asin t then the model becomes mz(t )  cz (t )  kz (t )  mA 2 sin t Note that this version of the model is identical in form to the forced, damped m-c-k system where the force amplitude F has been replaced by mA2 .

226

21. Horizontal sprung pendulum

FBD

(t) m

x(t)

b

a

k

In the figure above, a horizontal simple pendulum has a point mass on the end of a massless rigid rod 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 displacement variable x(t) as the extension/compression of the spring as shown. First, draw the FBD. Using Euler’s Rotational Dynamics Law for the pendulum:

M

Z

 mga cos  (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 is J  ma2 .

ma 2(t )  bk (  x(t ))  mga cos  (t )  0 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  (t )  1 .

ma 2(t )  bk (  x(t ))  mga  0 At static equilibrium we have mga  kb , so the simplified model for this system is

ma 2(t )  bkx (t )  0 For a small displacement x(t) the approximation x(t )  b sin  (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

ma 2(t )  kb 2 (t )  0

227

22. Sprung cylinder

FBD

k (t) x(t)

r

y(t)

In this dynamic system, a cylinder of mass m and radius r is suspended via a cable on one side and connected to ground via a linear translational spring with constant k on the other side. We measure displacement 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 Second Law for the cylinder:  FX  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 for the cylinder:  M Z  T (t ) r  k (  x (t )) r  J (t ) The mass moment of inertia of the cylindrical pulley is J  mr 2 2 . Substituting the mass moment of inertia and eliminating the cable tension T(t) between these two equations yields

T (t )  mg  k (  x(t ))  my(t ) mr   (t ) mg  2k (  x(t ))  my(t )  2 At static equilibrium we have

k  T 

mg 2

So the model simplifies to

mr   (t )  my(t )  2kx(t )  0 2 Note that from kinematics we know x(t)  2y(t) and y(t)  r(t) . Substituting the kinematic equations yields the final 1-dof model for this system. 3  m (t )  4k (t )  0 2

228

229

23. Vertical sprung pendulum with damping k

FBD

xb(t) b a

c

(t) L

m

In the figure above, a simple pendulum rotating about point O has a point mass on the end of a massless rigid rod of length L. The rod extends by length b to the other side of point O to connect to the ground 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 xb (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 x a (t )  a(t ) . First, draw the FBD. Using Euler’s Rotational Dynamics Law for the pendulum:

M

OZ

 mgL sin  (t )  kb (t )b cos  (t )  ca(t )a cos  (t )  J(t )

The mass moment of inertia of the pendulum about the axis of rotation is J  mL2 .

mL2(t )  mgL sin  (t )  kb 2 (t ) cos  (t )  ca 2(t ) cos  (t )  0 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(t) 1

Then the final 1-dof model for this system is

mL2(t )  ca 2(t )  (mgL  kb 2 ) (t )  0

230

24. Mass suspended via tensed string

L1

L2

m L

T

T x(t)

L1

L2

In this system a point mass m is suspended between two strings of tension T. Assume that for small motions the tension can be considered to be constant. Also assume there is only vertical displacement, i.e. ignore horizontal motion of the mass. We measure displacement x(t) positive vertically (up).

Using Newton’s Second Law for the point mass:

F

  f  mx(t )

X

Where f is the restoring force due to both strings:

f  T1 sin 1 (t )  T2 sin  2 (t )  T (sin 1 (t )  sin  2 (t )) Assuming small displacement we also have small angular motions. For small angular motions sin (t)  tan (t) and then we have

f  T(

x(t ) x(t ) x(t ) x(t )   )  T( ) L1 L2 L1 L  L1

The final 1-dof model for the tensed-string system is

mx(t )  T (

1 1 ) x(t )  0  L1 L  L1

mx(t )  T (

L ) x(t )  0 L1 ( L  L1 )

231

25. Manometer

x(t)

x(t)

A manometer is used for measuring pressure. The manometer shown has a uniform crosssectional 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:

F

X

 w(t )  mx(t )

Where w(t) is the weight acting due to the difference in height in the manometer. Therefore:

w(t)  gV (t)  2gAx(t) Since the total liquid mass m in the manometer is  AL , the final 1-dof manometer dynamics model is

mx(t )  2 gAx(t )  0

 ALx(t )  2 gAx(t )  0 Lx(t )  2 gx(t )  0

232

26. Cylinder in water bath

r h

x(t)

A solid buoyant cylinder is immersed in a water bath. Derive the vibratory dynamic equation of motion assuming purely vertical motion with no rotations. The cylinder has a height h and radius r and thus a mass of m  V   r 2 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 fR due to the water is the weight of water displaced. f R (t )  W VW g  W g r 2 x (t )

The vibratory equation of motion is

mx(t )  kx(t )  0

 r 2 hx(t )  W g r 2 x(t )  0  hx(t )  W gx(t )  0

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, to show one type of engineering mechanics modeling for muscles. In this class we will generally treat the effect of muscles more simply, i.e. as simple cables in tension with no mass, spring constants, or dynamics of their own. The model in this section was suggested by Dr. Scott Hooper (OU Biological Sciences) and the equations were derived and the simulations performed by Elvedin Kljuno (OU Fulbright Scholar from Bosnia). Dr. Hooper uses this model to study neural muscular control in the stomachs of lobsters, but he says 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 represents the load the muscle is lifting vs. gravity g. The muscle itself is represented by linear elastic spring stiffness k1, in parallel with linear elastic spring stiffness k2 that is in series with linear dissipative dashpot b. The absolute displacement of the dashpot end and the muscle end are measured by coordinates 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 in displacements relative to the gravity-stretched position, but biologists need to include the absolute spring lengths as well. The actuator FA represents the contractile element of the muscle and Fm is the force generated by the muscle. Note: this derivation assumes zero pennation angle, i.e. zero angle  between the muscle fibers and the tendons.

Dynamic Skeletal Muscle Diagram

234

The parallel elastic component k1, provided by the muscle membranes, gives spring resistance when passively stretched. The series elastic component k2 represents the tendons, storing elastic energy when a muscle is stretched. The tension-generating contractile effect of the sarcomeres, modeled by actuator force FA, is in parallel with the membranes and in series with the tendons. From the left figure below, the equation for the left spring is:

k1 ( y  L1R )  F1

(1)

From the right figure above, the equations for the right spring and dashpot are:

k2 ( y  x  L2R )  F2 bx  F2  FA

(2, 3)

From the FBD shown below, the dynamics equation for the mass is obtained using Newton’s Second law:

F

vert

 ma vert

 F1  F2  Fm  mg  my

(4)

235

Substituting (1) and (2) into (4), and also substituting (2) into (3) yields the following coupled second/first order linear vibrational dynamics model for skeletal muscle.

my  k1 ( y  L1R )  k2 ( y  x  L2R )  mg  Fm bx  k2 ( y  x  L2R )  FA

(5, 6)

Equations (5, 6) can be written in the following form:

 y

1 k 1 F (k1  k2 ) y  2 x  (k1 L1R  k2 L2R )  g  m m m m m k k F k x  2 x  2 y  A  2 L2R b b b b

(7, 8)

As mentioned above, (7, 8) can be simplified by choosing more adequate coordinates that will measure 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 of time given time functions for FA and Fm. In the case of a simple form of FA and Fm the solution could be found using reduction and decoupling by a formation of a single third-order linear ordinary differential equation.

 y (t ) 

k2 kk (k  k )  y(t )  1 2 y (t )  1 2 y(t )  b m mb

Fm (t ) k2 ( FA (t )  Fm (t )  mg  k1L1R )  m mb

Equation (9) can be solved for y(t) which is then substituted into (8) to directly solve for x(t).

(9)

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 will never 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 the spring 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 your resulting model.

237

6.5 Electrical Circuits Modeling This section presents modeling of electrical circuits. Like mechanical translational and rotational systems, electrical circuits have dynamics described by IVP ODEs, including input and output variables that 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 these mechanical and electrical ODEs (in Chapters 3 and 4). Many electrical circuits can exhibit electrical vibrations, i.e. oscillation of the electrical output variable 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 Laws to derive the models for electrical circuits. Kirchoff’s Current Law (KCL) states that the sum of currents flowing into any circuit node is

zero. Kirchoff’s Voltage Law (KVL) states that the sum of voltages around any circuit loop is zero. m

n

i (t)  0

v (t)  0

KCL Diagram

KVL Diagram

j 1

j

k 1

j

Electrical Circuit Elements In order to use Kirchoff’s Laws to derive models for electrical circuits, we need the following relationships. They relate the current i(t) flowing through and the voltage v(t) across the three standard electrical elements. A capacitor C stores electrical energy, a resistor R dissipates electrical energy, and an inductor L oscillates electrical energy. Note the equations in the last two columns of each row are equivalent, solved for the current i(t) or the voltage v(t). element

notation

units

i(t)

capacitor

C

Farads

iC (t )  C

resistor

R

Ohms

iR (t ) 

inductor

L

Henrys

iL (t ) 

dvC (t ) dt vR (t ) R

1 vL (t )dt L

v(t)

vC (t ) 

1 iC (t )dt C

vR (t )  RiR (t )

vL (t )  L

diL (t ) 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 this day, 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. The input 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).

+ i(t)

iR(t)

R

iL(t)

L

iC (t)

C

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:

i(t )  iR (t )  iL (t )  iC (t )  0 iC (t )  iR (t )  iL (t )  i (t ) Substituting the current through each element as a function of voltage v(t) from the previous table, we obtain the model for the current-driven parallel R-L-C circuit.

C

dv(t ) 1 1  v(t )   v(t )dt  i(t ) dt R L

This model is a linear, lumped-parameter, constant-coefficient, integro-differential equation. It is second-order, since, although the highest derivative order is first-order, there are two time differentiation steps between the integral and the first derivative. This equation is written in standard form, where the output variable v(t) appears with its derivative and integral on the left-hand-side and the input forcing function i(t) appears on the right-hand-side of the equation. The external current input is i(t) (amps) and the output is voltage v(t) (volts).

240

Force-Current Analogy We can make an analogy between the mechanical and electrical modeling worlds by comparing the 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 of velocity 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 )

J(t )  cR(t )  k R (t )   (t )

mv(t )  cv(t )  k  v(t )dt  f (t )

J  (t )  cR (t )  k R   (t )dt   (t )

C

dv(t ) 1 1  v(t )   v(t )dt  i(t ) dt R L

Compare the mechanical systems ODEs to the parallel R-L-C circuit model (repeated above). Translational Mechanical dv(t ) f S (t )  m dt

Rotational Mechanical d (t )  S (t )  J dt

dissipate energy

f D (t )  cv(t )

 D (t )  cR (t )

oscillate energy

fO (t )  k  v(t )dt

 O (t )  kR  (t )dt

Energy Role

store energy

Electrical

iS (t )  C iD (t )  iO (t ) 

dv(t ) dt

1 v(t ) R

1 v(t )dt L

Force-Current Analogy Translational 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

cR

1/R

stiffness

k

kR

1/L

Variable Type

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. The input is voltage v(t), and the output is current i(t). The current i(t) flowing through all elements is the same. The input voltage source v(t) increases as shown negative to positive; the voltage drops across each of the circuit elements, positive to negative in each case.

R v(t)

+ -

i(t)

L

C Series R-L-C Circuit Diagram

Using Kirchoff’s Voltage Law (the sum of voltages around the circuit loop is zero), we have:

v(t )  vR (t )  vL (t )  vC (t )  0 vL (t )  vR (t )  vC (t )  v(t ) Substituting the voltage drop across each element as a function of current i(t) from the first table in this section, we obtain the model for the voltage-driven series R-L-C circuit.

L

di(t ) 1  Ri(t )   i(t )dt  v(t ) dt C

This model is a linear, lumped-parameter, constant-coefficient, integro-differential equation. Again, it is second-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 and integral on the left-hand-side and the input forcing function v(t) appears on the right-hand-side of the equation. The external input is voltage v(t) (volts) and the output is current i(t) (amps).

243

Force-Voltage Analogy

We can make a different analogy between the mechanical and electrical modeling worlds by comparing 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 )

J(t )  cR(t )  k R (t )   (t )

mv(t )  cv(t )  k  v(t )dt  f (t )

J  (t )  cR (t )  k R   (t )dt   (t )

L

di(t ) 1  Ri(t )   i(t )dt  v(t ) dt C Force-Voltage Analogy

Translational 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

cR

R

stiffness

k

kR

1/C

Variable Type

244

Subset Electrical Circuit 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

C

dv(t ) 1 1  v(t )   v(t )dt  i(t ) dt R L

Series R-L-C Circuit

L

di(t ) 1  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

L

Parallel R-C Circuit

Series R-C Circuit

C

dv(t ) 1  v(t )  i(t ) dt R

Parallel L-C Circuit

C

dv(t ) 1   v(t )dt  i(t ) dt L

di (t )  Ri (t )  v(t ) dt

Ri(t ) 

1 i(t )dt  v(t ) C

Series L-C Circuit

L

di(t ) 1  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) and electrical 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 substitution relationships.

 (t )   v(t )dt (t )  v(t ) (t ) 

dv(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

Lq(t )  Rq (t ) 

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

Rq (t ) 

Parallel L-C Circuit

Series L-C Circuit

1 C(t )   (t )  i(t ) L

Lq(t ) 

1 q(t )  v(t ) C

1 q(t )  v(t ) C

1 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 degreeof-freedom (dof). For multiple independent masses, draw a free-body diagram (FBD) for each mass and derive 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 be coupled, where n is the number of degrees-of-freedom. 1. Two-Mass Vibrational System

y1(t) u1(t)

k1 m1

c1

y2(t) u2(t)

k2

c2

m2

u1(t) k1 y1(t)

u2(t) k 2 (y2 (t) - y1(t))

m1

m2 c 2 (y2 (t) - y1(t))

c1 y1(t)

Two-mass system free-body diagrams m1  y1 (t )   c1  c2  y1 (t )   k1  k2  y1 (t )  c2 y 2 (t )  k2 y2 (t )  u1 (t ) m2  y2 (t )  c2 y 2 (t )  k2 y2 (t )  c2 y1 (t )  k2 y1 (t )  u2 (t )

 m1 0 

y1  t    c1  c2 0       y2  t    c2 m2   

c2   y1  t    k1  k 2   c2   y 2  t     k 2

 k 2   y1  t    u1  t         k 2   y2  t   u2  t  

247

2. Automotive Shock Model

This is a ¼ car suspension model. M2 is a point mass representing one-fourth of the car mass (excluding wheels/tires). Spring constant k2 and viscous dashpot coefficient c2 represent the shock spring and damping fluid friction dissipation elements. Point mass m1 represents one wheel/tire mass and spring constant k1 and viscous dashpot coefficient c1 represent the spring effect and energy dissipation quality of the tire, respectively. The input is the road displacement z(t), and there are two displacement 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.

y2(t)

m2 k2

y1(t)

c2 m1

k1

z(t)

c1

m1  y1 (t )   c1  c2  y1 (t )   k1  k2  y1 (t )  c2 y 2 (t )  k2 y2 (t )  c1 z (t )  k1 z (t ) m2  y2 (t )  c2 y 2 (t )  k2 y2 (t )  c2 y1 (t )  k2 y1 (t )  0

 m1  0

y1  t    c1  c2 0       y2  t    c2 m2   

c2   y1  t    k1  k 2   c2   y 2  t     k 2

 k 2   y1  t   c1 z (t )  k1 z (t )        k 2   y2  t   0  

248

3. Another Two-Mass Vibrational System

y1(t)

y2(t)

f(t) k1 m1

c

k2 m2

m1  y1 (t )  k1 y1 (t )  k2[ y2 (t )  y1 (t )]  f (t ) m2  y2 (t )  cy2 (t )  k2[ y2 (t )  y1 (t )]  0

 m1  0

y1  t    0 0   y1  t    k1  k 2 0        y2  t    0 c   y 2  t     k 2 m2   

 k 2   y1  t    f  t        k 2   y2  t    0 

249

4. Inverted Pendulum

m2

Y

 (t)

g L

X f(t)

m1

w(t)

Coupled nonlinear differential equations (t )  m2 L cos  (t )(t )  m2 L sin  (t )(t ) 2  f (t ) ( m1  m2 ) w (t )  m2 gL sin  (t )  0 m2 L2(t )  m2 L cos  (t ) w

Coupled linearized differential equations (t )  m2 L(t )  f (t ) (m1  m2 ) w (t )  m2 L2(t )  m2 gL (t )  0 m2 Lw

 m1  m2  m L 2 

(t )  0 m2 L   w 0   w(t )   f (t )        2   m2 L   (t )  0 m2 gL   (t )   0 

250

5. Ball and Beam

) p(t

 (t)

 (t)

Coupled nonlinear differential equations  Jb  p (t )  m g sin  (t )  m p (t ) (t ) 2  0  r 2  m     2  m p (t )  J  J b  (t )  2 m p (t ) p (t ) (t )  m g p (t ) cos  (t )   (t )

Coupled linearized differential equations  Jb  p (t )  m g (t )  r 2  m      m p (t ) 2  J  J b  (t )  m g p (t )

 Jb r 2  m    p (t )   0 0        2 m p (t )  J  J b   (t )   mg 0 

 0   (t )

mg   p (t )   0     0   (t )   (t ) 

251

6. Proof-Mass Actuator q(t)

M k

n(t)

f(t) e

J

 (t) m

Coupled nonlinear differential equations ( M  m)q(t )  kq (t )  me((t ) cos  (t )   2 (t ) sin  (t ))  0 ( J  me 2 )(t )  meq(t ) cos  (t )  n(t )

Coupled linearized differential equations ( M  m)q(t )  kq (t )  me(t )  0 ( J  me 2 )(t )  meq(t )  n(t )

me  q(t )  k 0 q(t )   0  M  m    me   (t )  n(t ) 2    ( ) 0 0  J me t         

252

7. Two Masses suspended via Tensed Strings

m1

m2

L T

L T

T

x1(t) L

L

T x2(t)

L

L 2T T x1 (t )  x2 (t )  0 L L 2T T m2  x2 (t )  x2 (t )  x1 (t )  0 L L m1 x1 (t ) 

 m1 0 

x1  t    2T L  T L   x1  t   0    0            x2  t     T L 2T L   x2  t   m2    0 

253

8. Double Pendulum

L1 g

 1 (t) m1

L2  2 (t) m2

Coupled nonlinear differential equations (m1  m2 ) L121 (t )  m2 L1 L22 (t )  ( m1  m2 ) gL1 sin 1 (t )  0 m2 L1 L21 (t )  m2 L222 (t )  m2 gL2 sin  2 (t )  0

Coupled linearized differential equations (m1  m2 ) L121 (t )  m2 L1 L22 (t )  (m1  m2 ) gL11 (t )  0 m2 L1 L21 (t )  m2 L222 (t )  m2 gL2 2 (t )  0 (m1  m2 ) L12   m2 L1 L2

m2 L1 L2  1 (t )   (m1  m2 ) gL1    0 m2 L22  2 (t )  

0  1 (t )  0     m2 gL2   2 (t )  0 

If m1 = m2 and L1 = L2  2 1 1 (t )   2 g L  1 1      0    2 (t )  

0  1 (t )  0     g L   2 (t )  0 

254

9. Sprung Double Pendulum

L0

L0 k L

L

1 (t)

2 (t) m

m

Coupled nonlinear differential equations m1 L121 (t )  m1 gL1 sin 1 (t )  kL20 (1 (t )   2 (t ))  0 m2 L222 (t )  m2 gL2 sin  2 (t )  kL20 ( 2 (t )  1 (t ))  0

Coupled linearized differential equations m1 L121 (t )  m1 gL11 (t )  kL20 (1 (t )   2 (t ))  0 m2 L222 (t )  m2 gL2 2 (t )  kL20 ( 2 (t )  1 (t ))  0

 m1 L12   0

0  1 (t )   m1 gL1  kL20    m2 L22  2 (t )   kL20

 1 (t )  0  kL20   2 m2 gL2  kL0   2 (t )  0 

If m1 = m2 and L1 = L2  g k  L0 2     1 0  1 (t )   L m  L   0 1       2    2 (t )   k  L0    m  L 

2 k  L0       m  L   1 (t )  0    2 g k  L0    2 (t )  0   L m  L  

255

10. Doubly-sprung mass/cylinder

r

(t)

m k k m x(t)

m1  x (t )  k1 x (t )  k1r (t )  0 m2 r 2   (t )  ( k1  k 2 ) r 2 (t )  k1rx (t )  0 2

 m1 0 

0    k1r   x(t )  0  x(t )   k1          2 m2 r 2   (t )    k1r (k1  k2 )r 2   (t )  0 

256

11. Sprung Dual Cylinders k r

m

(t) k r

(t) k

m

m1r12  1 (t )  (k1  k2 )r121 (t )  k2 r1r2 2 (t )  0 2 m2 r22   2 (t )  (k2  k3 )r22 2 (t )  k2 r1r21 (t )  0 2  m1r12   2   0

 0   1 (t )   ( k1  k2 ) r12     m2 r22  2 (t )    k2 r1r2 2 

 k2 r1r2  1 (t )  0     ( k2  k3 ) r22   2 (t )  0 

If m1 = m2, and r1 = r2, and k1 = k2 = k3  4k   m   1 0   1 (t )     0 1      2 (t )    2k  m

2k  m  1 (t )  0     4k   2 (t )  0  m 



257

12. Sprung Cart Pendulum

x(t) k

k m1

(t) m2

(m1  m2 )  x(t )  m2 L(t )  2kx(t )  0 m2 L2(t )  m2 Lx(t )  m2 Lg (t )  0  m1  m2  mL  2

m2 L    x(t )   2k  2    m2 L   (t )   0

0   x(t )  0     m2 Lg   (t )  0 

258

13. Translating rotating rigid body

This is a 2-dof system, still constrained to vibrate in the vertical direction, with small-angle rotational vibrations as well.

m, J

(t)

x(t) L1

L2

k1

k2

mx(t )  (k1  k2 ) x(t )  ( k1 L1  k2 L2 ) (t )  0 J(t )  (k1 L1  k2 L2 ) x(t )  (k1 L12  k2 L22 ) (t )  0 x(t )   k1  k2  m 0     0 J  (t )     k L  k L     1 1 2 2

k1 L1  k2 L2   x(t )  0     k1 L12  k2 L22   (t )  0 

259

14. Doubly sprung cylinder mass

x2(t) x1(t) r

k2

(t)

k1 m2

m1

(m1 

m2 m x1 (t )  2  x2 (t )  (k1  k2 ) x1 (t )  k2 x2 (t )  0 )  2 2 m 3m  2  x1 (t )  2  x2 (t )  k2 x1 (t )  k2 x2 (t )  0 2 2

m2   m1  2    m2 2 

m2  x1 (t )   k1  k2 2      x2 (t )   k2 3m2    2 



k2   x1 (t )  0    k2   x2 (t )  0

260

15. Cylinder rolling in a sprung big cylinder

(t)

r1

r2 m2 x(t)

m1

k

Coupled nonlinear differential equations

(m1  m2 )  x(t )  m2 (r1  r2 )(t )sin  (t )  kx(t )  0 3  x(t )sin  (t )  (r1  r2 )(t )  g sin  (t )  0 2 The linearized differential equations become decoupled. ( m1  m2 )  x (t )  kx (t )  0 3 (r1  r2 )(t )  g (t )  0 2

261

16. Dual torsional disks coupled by a spring

k a m2

r2

a

 2(t)

k R2

m1

r1

 1(t)

k R1

m1r12  1 (t )  (k R1  ka 2 )1 (t )  ka 2 2 (t )  0 2 m2 r22   2 (t )  (k R 2  ka 2 ) 2 (t )  ka 21 (t )  0 2  m1r12   2   0

 0   1 (t )   k R1  ka 2    2 m2 r22  2 (t )    ka 2 

 ka 2  1 (t )  0     k R 2  ka 2   2 (t )  0 

262

17. Dual point masses

f(t) x2(t) L k2

m2

L c

x1(t) m1

k1

m1 x1 (t )  (k1  k2 ) x1 (t )  2k2 x2 (t )  f (t ) m2  x2 (t )  cx2 (t )  4k2 x2 (t )  2k2 x1 (t )  0 x1 (t )  0 0   x1 (t )   k1  k2  m1 0         0 m     0 c   x2 (t )   2k2 2   x2 (t )  

2k2   x1 (t )   f (t )     4k2   x2 (t )   0 

263

18. Rotating sprung point masses

k m

 (t)

L

L

m

 (t)

k m x(t) k

mx(t )  2kx(t )  kL (t )  0 2mL2(t )  2kL2 (t )  kLx(t )   (t ) 0    x(t )   2k kL   x(t )   0  m  0 2mL2  (t )  kL 2kL2   (t )   (t )        

264

19. Three-Mass Vibrational System y1(t) u1(t)

k1

c1

m1

y3(t)

y2(t) u2(t)

k2 m2

c2

u3(t)

k3

c3

m3

k4

c4

m1  y1 (t )  (c1  c2 ) y1 (t )  (k1  k2 ) y1 (t )  c2 y 2 (t )  k2 y2 (t )  u1 (t ) m2  y2 (t )  (c2  c3 ) y 2 (t )  (k2  k3 ) y2 (t )  c2 y1 (t )  k2 y1 (t )  c3 y3 (t )  k3 y3 (t )  u2 (t ) m3  y3 (t )  (c3  c4 ) y3 (t )  (k3  k4 ) y3 (t )  c3 y 2 (t )  k3 y2 (t )  u3 (t )  m1  0  0

0

m2 0

0    y1 (t )   c1  c2    y2 (t )     c2 0    m3    y3 (t )   0

 c2 c2  c3  c3

  y1 (t )   k1  k 2    c3   y 2 (t )     k 2 c3  c4   y 3 (t )   0 0

 k2 k 2  k3  k3

  y1 (t )   u1 (t )       k3   y2 (t )   u2 (t )  k3  k 4   y3 (t )   u3 (t )  0

265

20. Three Masses suspended via Tensed Strings

m1

m2

L

T

m3

L

L

T

T

T

L T T

x2(t)

x1(t) L

L

x3(t) L

m1 x1 (t ) 

m2  x2 (t ) 

0 m2 0

2T T x1 (t )  x2 (t )  0 L L

2T T T x2 (t )  x1 (t )  x3 (t )  0 L L L

m3  x3 (t ) 

 m1 0   0

L

2T T x3 (t )  x2 (t )  0 L L

0    x1  t    2T L  T L 0   x1  t   0          0    x2  t      T L 2T L  T L   x2  t    0  m3    x3  t    0  T L 2T L   x3  t   0 

266

21. Triple Pendulum

L1 g

 1 (t) m1

L2  2 (t) m2

L3  3 (t)

m3

Let m1 = m2 = m3 and L1 = L2 = L3 3L1 (t )  2 L2 (t )  L3 (t )  3 g1 (t )  0 2 L1 (t )  2 L2 (t )  L3 (t )  2 g 2 (t )  0 L (t )  L (t )  L (t )  g (t )  0 1

2

 3g   3 2 1 1 (t )   L  2 2 1  (t )    0   2   1 1 1 3 (t )   0 

3

3

0 2g L 0

 0  1 (t )  0     0   2 (t )   0   (t )  0 g   3    L 

267

22. Sprung Triple Pendulum

L0

L0

L0

k

k L

L 1 (t)

L

2 (t)

3 (t)

m

m

m

where L0 = L / 2. kL2 kL2 )1 (t )   2 (t )  0 4 4 kL2 kL2 kL2 ) 2 (t )  1 (t )  3 (t )  0 mL22 (t )  ( mgL  2 4 4 kL2 kL2 2  ) 3 (t )   2 (t )  0 mL  3 (t )  (mgL  4 4 mL21 (t )  (mgL 

g k   1 0 0  1 (t )   L 4m 0 1 0   (t )     k    2   4m 0 0 1  3 (t )    0 

k 4m g k  L 2m k  4m 

   1 (t )  0  k       2 (t )   0  4m     (t )  0  g k  3     L 4m  0

268

23. Sprung Cylinder with Point Masses

k

m

r (t)

r/3

x(t) k

m m

x(t)

x(t) 3m1 x1 (t )  (k1  9k2 ) x1 (t )  2m1 x2 (t )  6k2 x2 (t )  3k2 x3 (t )  0 (2m1  m2 )  x2 (t )  4k2 x2 (t )  2m1 x1 (t )  6k2 x1 (t )  2k2 x3 (t )  0 m3  x3 (t )  k2 x3 (t )  3k2 x1 (t )  2k2 x2 (t )  0

 3m1  2 m 1   0

2m1 2m1  m2 0

0    x1  t    k1  9k 2    0    x2  t     6k 2 m3    x3  t    3k 2

6 k 2 4k2 2k2

3k 2   x1  t   0      2k 2   x2  t    0  k 2   x3  t   0 

269

Common Model Features

Considering all multi-dof vibrational models we have derived, some general comments can be made about the models.

1. When gathering the unknown i and its time derivatives on one side of the ODE i, there will never be a change in sign amongst these terms; generally they will all be positive. Generally the unknown j appears on the ODE i with a negative sign.

2. The effect of gravity always cancels out. That is, any mg static weight terms will load the springs with static deflections 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 your resulting model.

4. Considering the matrix-vector forms of the coupled ODEs a. The matrix multiplying the vector of second-order derivatives is called the mass matrix. It is square, symmetric, and positive-definite. b. The matrix multiplying the vector of first-order derivatives is called the damping matrix. It is square and symmetric. c. The matrix multiplying the vector of zeroth-order unknowns is called the stiffness matrix. It is square and symmetric.