I. Aşağıdaki ekran alıntıları adams ın klasik arayüzü kullanılarak oluşturulduğu için şekil 1 dikkate alınarak arayüz klasiğe dönüştürülmelidir.

(1)

Adams’ta yeni bir sayfa açarken Gravity kısmını No gravity yaparak yerçekimi ivmesini ihmal edebiliriz. Yapacağımız uygulamada da uzuvlara ait ağırlık kuvvetleri dikkate alınmayacağı için bu kısım önemlidir. Ayrıca working directory kısmından da çıkaracağımız veya kaydedeceğimiz dosyaların kaydedileceği yeri seçebiliriz.

I. Aşağıdaki ekran alıntıları adams’ın klasik arayüzü kullanılarak oluşturulduğu için şekil 1 dikkate alınarak arayüz klasiğe dönüştürülmelidir.

Şekil 1

(2)

Numaranıza denk gelen mekanizmayı çizdikten sonra 200 d/dk’lık giriş hızı (1200 deg/sn) şekil 2’de göserildiği gibi tahrik uzvuna girilmelidir.

Şekil 2

Mekanizma oluşturulup bir çevrim yapacak şekilde çalıştırıldıktan sonra grafik (adams post processor) menüsüne girilip, istenen grafikler görülebilir. Mesela şekil 3’te Part_2 olarak isimlendirilen uzvun kütle merkezi ivmesinin x bileşenine ait değişim θ

₂₁

açısına göre görülmektedir.

Şekil 3

(3)

Şekil 4’te ise giriş uzvuna etki eden momentin (torkun) değişimi N.mm biriminde görülmektedir. (MOTION_1-Element_Torque-Z)

Şekil 4

Grafikleri oluşturan sayısal değerleri elde etmek için şekil 5’te görülen menülere girerek Table seçeneği tıklanır.

Şekil 5

Table seçeneği tıklandıktan sonra açılan pencere şekil 6’da görülmektedir. Type

kısmı, Table; File Name, çıkarılan dosyanın ismi (yaptığım uygulamada

part_2’nin kütle merkezinin ivmesinin x bileşenini aldığım için ismi

part2_ivme_xbileseni olarak girdim), Plot da değerlerini elde edeceğimiz grafiği

(4)

tanımlamaktadır (Plot kısmına sağ tıklayıp guesses kısmında gördüğümüz son plot, ekranda görünen grafiktir). Son olarak Format, spreadsheet yapılarak OK’e basılır. Dosya önceden tanımladığımız çalışma ortamına kaydedilmiştir. Çalışma ortamını ayarlamak için File-Select Directory komutlarını takip edebilirsiniz.

Şekil 6

Dosya açıldığında grafiğe ait tüm sayısal değerler görülebilir durumdadır.

Buradan, θ

21

giriş açı değerinize karşılık ivmenin x bileşeni değerini okuyabilirsiniz. Örneğin şekil 7’ye göre tahrik uzvunun 18 derecelik konumu için part_2 uzvuna ait ivmenin x bileşeni -3,128851x10

⁴

mm/sn

²

olarak elde edilir. Eğer grafikteki yatay bileşeni tahrik uzvunun açı değişimi cinsinden tanımlamayıp simülasyon zamanı olarak aldıysanız aşağıdaki dönüşümü kullanarak açınıza denk gelen zamanı (t) tespit edebilirsiniz.

θ

₂₁

/1200=t θ

₂₁

: Numaranıza göre hesaplanan açı t:zaman

(5)

Şekil 7

I. Malzemesini değiştirmek istediğimiz uzva çift tıklayarak veya sağ tıklayıp modify seçeneğini tıklayarak açılan diyalog penceresinden Material Type kısmına sağ tıklayıp şekil 8’de görülen işlemleri yaparak uzvun malzemesini değiştirebiliriz. Daha sonra Show calculated inertia butonunu tıklayarak uzva ait kütle ve kütlesel atalet momentleri değerlerini elde edebiliriz. Kütlesel atalet momentleri değerlerinden bizim dikkate almamız gereken değer I

zz

değeridir (şekil 9).

Şekil 8

(6)

Şekil 9

Bütün bu işlemler, (uzuv malzemesini değiştirmek hariç) bütün uzuvlar için yapılarak, her uzvun; kütle merkezinin ivmesinin x bileşeni, y bileşeni, açısal ivme ve tork değerleri, numaranıza göre belirlenen θ

₂₁

açısı anı için bulunacak ve çözümlerde kullanılacaktır. Ayrıca gerekli olan kütle ve kütlesel atalet momenti (I

_zz

) değerleri de yukarıda gösterildiği şekilde bulunup analitik çözümde kullanılacaktır.

Elde edilen bu değerler, çözüm için bilinenlerimiz olmaktadır.

Yararlanılabilecek örnek uygulamalar ileri sayfalarda bulunmaktadır.

(7)

AME 352 FORCE ANALYSIS

P.E. Nikravesh 7-16

Example 7

The mass and moment of inertia for the links of this slider-crank are given. A known force F acts on the slider block, and an unknown torque T acts on the crank. In the depicted configuration, the angular velocity and acceleration of the crank are given. The objective is to find the magnitude and the direction of the unknown torque.

Based on the given angular velocity and acceleration of the crank, polygons are constructed and the angular velocity and acceleration of all the links are found (magnitudes and directions—ω₂, α₂ and α₃ are CCW, and ω₃ is CW). Then, in a process similar to that of the static equilibrium (refer to Example 1), the FBD for each link is constructed.

A

B O2

T

F

G₂ G₃

G₄

V_G

4

O _V

V_G

2

V_G

3

A_G

4 O _A

A_G₂ A

B A_G

3

Link 2: Link 3: Link 4:

F₁ F₂

F₃ F₄ T

a b G₂ A_G

2 b

a

α₂

F₅ F₆ F₃

F₄

c d G₃ A_G

3

d

c _α

3

F₇ F₅ F₆ G₄ F

T₈ A_G

4

F₁+F₃=m₂A_G

2( x )

F₂+F₄ =m₂A_G

2( y )

aF₁− bF₂− aF₃+bF₄+T=I_G

2α₂

−F₃+F₅=m₃A_G

3( x )

−F₄+F₆=m₃A_G

3( y )

cF₃+dF₄+cF₅+dF₆=I_G

3α₃

−F₅+F=m₄A_G

4

−F₆+F₇=0 T₈=0 Note that

A_G

2( x ), A_G

2( y ), A_G

3( x ), A_G

3( y ) and A_G

4 are negative, where α₂ and α₃ are positive.

These equations are put into matrix form. This set of 9 equations in 9 unknowns can be solved by any preferred numerical method.

1 0 1 0 0 0 0 0 0

0 1 0 1 0 0 0 0 0

a −b −a b 0 0 0 0 1

0 0 −1 0 1 0 0 0 0

0 0 0 −1 0 1 0 0 0

0 0 c d c d 0 0 0

0 0 0 0 −1 0 0 0 0

0 0 0 0 0 −1 1 0 0

0 0 0 0 0 0 0 1 0

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

⎥⎥ F₁ F₂ F₃ F₄ F₅ F₆ F₇ T₈ T

⎧

⎨

⎪⎪

⎩

⎪⎪

⎫

⎬

⎪⎪

⎭

⎪⎪

=

m₂A_G

2( x )

m₂A_G

2( y )

I_G

2α₂ m₃A_G

3( x )

m₃A_G

3( y )

I_G

3α₃ m₄A_G

4 − F

0 0

⎧

⎨

⎪⎪

⎪

⎩

⎪⎪

⎪

⎫

⎬

⎪⎪

⎪

⎭

⎪⎪

⎪ Coulomb Friction

Coulomb friction can be included between two contacting surfaces in a dynamic force analysis. Given the dynamic (kinetic) coefficient of friction,

μ

( k ), the friction force can be

described as the product of the coefficient of friction and the reaction force normal to the contacting surfaces. The friction force must act in the opposite direction of the motion. The process is illustrated through a simple example.

(8)

53

Chapter 3

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. Position, Velocity, and Acceleration Analyses must first be complete.

Step 2. Draw the Free-Body Diagrams:

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

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

(9)

54 Step 3. State the Problem:

Given: r1, θ1 = 0, r2

θ2, r₄, θ4

ω2, r , ₄ ω4

α2, r , ₄ α4

assume zero external forces and moments Find: F₂₁, F₄₂, F₁₄ and τ2

Step 4. Derive the Newton-Euler Dynamics Equations.

Newton's 2^nd Law:

Link 2:

2 42 21 2 2 G2

F =F −F +W =m A

∑

Link 4:

4 14 42 4 4 G4

F =F −F +W =m A

∑

Euler's Equation:

Link 2:

2 2 42 42 12 21 2 2

G z G Z

M = +τ r ×F −r ×F =I α

∑

Link 4:

4 14 14 24 42 4 4

G z G Z

M =r ×F −r ×F =I α

∑

Count # of unknowns and # of equations: six scalar equations and seven 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 we have only one unknown from it, F₄₂:

( )

42 4

42 42

42 4

42

cos 2

sin 2

X Y

F F F

F F

θ π θ π +

⎧ ⎫

⎧ ⎫ ⎪ ⎪

=⎨⎩ ⎬ ⎨⎭ ⎩=⎪ + ⎬⎪⎭

(10)

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

Link 2:

( )

( ) ( )

42 21 2 2

2 42 42 42 42 12 21 12 21 2 2

X X G X

Y Y G Y

X Y Y X X Y Y X G Z

F F m A

F F m A g

r F r F r F r F I

τ α

− =

− = +

+ − − − =

Link 4:

( )

( ) ( )

14 42 4 4

14 14 14 14 24 42 24 42 4 4

X X G X

Y Y G Y

X Y Y X X Y Y X G Z

F F m A

F F m A g

r F r F r F r F I α

− =

− = +

− − − =

Express these scalar equations into matrix/vector form.

Inverted slider-crank mechanism inverse dynamics matrix equation:

( )

2 2

21

2 2

21

2 2

12 12 42 42 42

4 4

14

4 4

14

4 4

24 24 14 14 2

1 0 0 0 0

0 1 0 0 0

0 0 1

0 0 1 0 0

0 0 0 1 0

0 0 0

G X X

G Y Y

G Z

Y X X Y

G X X

G Y Y

G Z

X Y Y X

m A

c F

m A g

s F

I

r r r s r c F

m A

c F

m A g

s F

I r s r c r r

α

α τ

− ⎧ ⎫

⎡ ⎤ ⎧ ⎫

⎪ ⎪

⎢ − ⎥ ⎪ ⎪ ⎪ + ⎪

⎢ ⎥ ⎪ ⎪

⎪ ⎪

⎢ − − ⎥ ⎪⎪ ⎪⎪ ⎪= ⎪

⎢ − ⎥ ⎨ ⎬ ⎨ ⎬

⎢ ⎥ ⎪ ⎪ ⎪ ⎪

⎢ − ⎥ ⎪ ⎪ ⎪ + ⎪

⎢ ⎥ ⎪ ⎪ ⎪ ⎪

− + −

⎢ ⎥ ⎪ ⎪ ⎪

⎣ ⎦ ⎩ ⎭ ⎩ ⎭⎪

where:

( )

(

⁴⁴

)

cos 2

sin 2

c s

θ π θ π +

⎧ ⎫

⎧ ⎫ ⎪= ⎪

⎨ ⎬ ⎨⎪ + ⎬⎪

⎩ ⎭ ⎩ ⎭

[ ]

^{A v}

{ } { }

⁼ ^b

(11)

56 Step 6. Solve for the unknowns

Coefficient matrix [A] dependent on geometry (kinematics solutions – angles). RHS {b}

dependent on inertial terms and gravity.

Simultaneous matrix solution:

{ }

^v ⁼

[ ]

^A ⁻¹

{ }

^b

Matlab: v = inv(A)*b; % solution via inverse Actually, using Gaussian elimination is more efficient and robust.

Matlab: v = A\b; % Gaussian elimination

Solution to internal forces and input torque are contained in the components of v. Matlab, 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:

( )

42 4 4

14 4 4

24 24 14 14 14 4 4

1 0

0 1

G X

X G Y

X Y Y X Y G Z

c F m A

s F m A g

r s r c r r F I α

−

⎡ ⎤ ⎧ ⎫ ⎧ ⎫

⎪ ⎪ ⎪ ⎪

⎢ − ⎥⎨ ⎬ ⎨= + ⎬

⎢ ⎥⎪ ⎪ ⎪ ⎪

⎢− + − ⎥

⎣ ⎦ ⎩ ⎭ ⎩ ⎭

[ ]

^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:

( )

21 2 2 42

12 12 2 2 2 42 42 42

1 0 0

0 1 0

1

X G X

Y G Y

Y X G Z X Y

F m A cF

F m A g sF

r r τ I α r s r c F

⎧ ⎫

− −

⎡ ⎤ ⎧ ⎫

⎪ ⎪ ⎪ ⎪

⎢ − ⎥⎨ ⎬ ⎨= + − ⎬

⎢ ⎥

⎪ ⎪ ⎪ ⎪

⎢ − ⎥ − −

⎣ ⎦ ⎩ ⎭ ⎩ ⎭

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):

( )

21 2 2 42

2 2 2 12 21 12 21 42 42 42

X G X

Y G Y

G Z Y X X Y X Y

F m A cF

F m A g sF

I r F r F r s r c F

τ α

= − +

= − + +

= − + − −

(12)

57

Matrix inversion requires approximately 3 3

log n

n and Gaussian elimination requires approximately

(

² ¹

)

²

3

n n

− n

+ multiplications/divisions¹.

Number of Multiplications/Divisions

Method Inversion Gaussian Reduction

6x6 833 106 87%

3x3 plus decoupled link 2 177 24 86%

Reduction in cost 79% 77%

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 4-bar in notation and terms.

Ground link force/moment diagram:

Shaking force:

21 14

21 41 21 14

21 14

X X

S

Y Y

F F

F F F F F

F F

−

⎧ ⎫

= + = − = ⎨⎩ − ⎬⎭

Shaking moment:

2 21 21 41 14

2 21 21 21 21 41 14 41 14

S

X Y Y X X Y Y X

M r F r F

r F r F r F r F

τ τ

= − + × − ×

= − + − − +

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

(13)

58 Inverted slider-crank mechanism inverse dynamics example: Term Example 3 cont.

Given:

1

2

4

1

0.20 0.10 0.32 0 r r L θ

=

(m)

2

4

70 150.5

0.19 r

θ θ

=

(deg and m)

2

4

25 2.47 2.18 r

ω ω

=

(rad/s and m/s)

2

4

0 9.46 267.14 r

α α

=

= −

=

(rad/s² and m/s²)

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 (ρ = 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 (kgm²) Snapshot Analysis

Given this mechanism position, velocity, and acceleration analyses, solve the inverse dynamics problem for this snapshot (θ₂ =70 ). The matrix-vector equation to solve is:

21

42

14

2

1 0 0.49 0 0 0 5.03

0 1 0.87 0 0 0 9.21

0.05 0.02 0.01 0 0 1 0

0 0 0.49 1 0 0 30.77

0 0 0.87 0 1 0 41.82

0 0 0.03 0.08 0.14 0 3.47

X Y

F F F F F τ

− − ⎧ ⎫ −

⎡ ⎤ ⎧ ⎫

⎪ ⎪

⎢ − − ⎥⎪ ⎪ ⎪ − ⎪

⎢ ⎥ ⎪ ⎪

⎪ ⎪

⎢− ⎥⎪ ⎪ =⎪ ⎪

⎢ ⎥ ⎨ ⎬ ⎨− ⎬

⎢ ⎥ ⎪ ⎪ ⎪ ⎪

⎢ ⎥ ⎪ ⎪ ⎪− ⎪

⎢ ⎥ ⎪ ⎪ ⎪ ⎪

− ⎪ ⎪

⎣ ⎦⎩ ⎭ ⎩ ⎭

The answer is:

21

42

14

2

35.35 62.69 61.47

0.46 11.65 1.10

X Y

F F F F

F τ

⎧ ⎫ ⎧ ⎫

⎪ ⎪ ⎪ ⎪

⎪ ⎪ ⎪− ⎪

⎪ ⎪ =

⎨ ⎬ ⎨− ⎬

⎪ ⎪ ⎪ ⎪

⎪ ⎪ ⎩ ⎭

⎩ ⎭

The associated vector shaking force and moment are: ^35.81 51.03

SX S

SY

F F F

⎧ ⎫ ⎧ ⎫

=⎨ ⎬ ⎨= ⎬

⎩ ⎭

⎩ ⎭ (N) M_S = −8.53kˆ (Nm)

(14)

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

A more meaningful result is to report the inverse dynamics analysis results for the entire range of mechanism motion. The plots below give the inverse dynamics results for all 0 ≤θ₂ ≤360 , for Term Example 3. Again, since ω2 is constant, we can plot the velocity results vs. θ2 (since it is related to time t via θ₂ =ω₂t).

0 50 100 150 200 250 300 350

-20 -10 0 10 20 30

θ₂ (deg) τ 2 (Nm)

τ₂

τ_AVG

τ_RMS

Input Torque τ2

(15)

60

0 50 100 150 200 250 300 350

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

θ₂^(deg) F S (N)

F_SX F_SY

Shaking Force

0 50 100 150 200 250 300 350

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

θ₂ (deg) MS (Nm)

Shaking Moment

(16)

61 3.7 Multi-loop Mechanism Inverse Dynamics Analysis

The matrix method can be applied to any mechanism inverse dynamics problem. Here are the five two-loop six-bar mechanisms from the Mechanism Atlas in the Introduction:

2 3 4

5 6

1

2 3

4 5

6

1

Stephenson I Stephenson II

1 2

3

4

5

6

Stephenson III

1 2

3

4

5 6

2 1

3 4

5

6

Watt I Watt II

(17)

62

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

(18)

63 Watt II six-bar mechanism inverse dynamics 15x15 matrix equation:

12 12 32 32

23 23 43 43

34 34 14 14 54 54

1 0 1 0 0 0 0 0 0 0 0 0 0 0 0

0 1 0 1 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 0 1 0 0 0 0 0 0 0 0 0 0

0 0 0 1 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 1 0 1 0 1 0 0 0 0 0 0

0 0 0 0 0 1 0 1 0 1 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 1

Y X Y X

Y X Y X Y X

r r r r

r r r r r r

−

− −

−

− −

−

− − −

−

21 21 32 32 43 43 14 14 54 54

45 45 65 65

56 56 16 16

0 1 0 0 0 0

0 0 0 0 0 0 0 0 0 1 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 1 0 1 0 0

0 0 0 0 0 0 0 0 0 0 0 1 0 1 0

0 0 0 0 0 0 0 0 0 0 0

X Y X Y X Y X Y X Y

Y X Y X

F F F F F F F F F F

r r r r

⎡ ⎤

⎢ ⎥

⎢ − ⎥

⎢ ⎥

⎢ − − ⎥

⎢ ⎥

⎢ − ⎥

⎢ ⎥

− −

⎢ ⎥

⎣ ⎦

( )

2 2

3 3

4 4

5 5

65

5 5

65

6 6

16

6 6

16

6 6

2

G X G Y G Z

G X X G Y

Y G Z X G X Y G Y

G Z

m A

m A g

I m A

m A g

I m A

m A g

I m A

m A g

F

I F

m A F

m A g

F

I α

α

α τ

⎧ ⎫

⎪ ⎪

⎪ ⎪ ⎪ + ⎪

⎪ ⎪

⎪

⎪ ⎪

⎪

⎪ ⎪ +

⎪

⎪ ⎪

⎪

⎪ ⎪

⎪

⎪ ⎪

⎪

⎪ ⎪

⎪ ⎪ ⎪= +

⎨ ⎬ ⎨ ⎬

⎪ ⎪ ⎪

⎪ ⎪ ⎪ +

⎪ ⎪ ⎪

⎪ ⎪ ⎪ +

⎪ ⎪ ⎪

⎩ ⎭ ⎩

⎪⎪

⎪⎪⎪

⎪⎪

⎪⎭

[ ]

^{A v}

{ } { }

⁼ ^b

(19)

64 Step 6. Solve for the unknowns (cont.)

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:

( )

54 5 5

45 45 65 65 65 5 5

65 6 6

16 6 6

56 56 16 16 16 6 6

1 0 1 0 0 0

0 1 0 1 0 0

0 0

0 0 1 0 1 0

0 0 0 1 0 1

0 0

X G X

Y G Y

Y X Y X X G Z

Y G X

X G Y

Y X Y X Y G Z

F m A

F m A g

r r r r F I

F m A

F m A g

r r r r F I

α α

−

⎡ ⎤ ⎧ ⎫ ⎧ ⎫

⎢ − ⎥ ⎪ ⎪ ⎪ + ⎪

⎢ ⎥ ⎪ ⎪ ⎪ ⎪

⎢ − − ⎥ ⎪⎪ ⎪ ⎪⎪ ⎪= ⎪⎪

⎢ − ⎥ ⎨ ⎬ ⎨ ⎬

⎢ ⎥ ⎪ ⎪ ⎪ ⎪

⎢ − ⎥ ⎪ ⎪ ⎪ + ⎪

⎢ ⎥ ⎪ ⎪ ⎪ ⎪

− −

⎢ ⎥ ⎪ ⎪ ⎪ ⎪

⎣ ⎦ ⎩ ⎭ ⎩ ⎭

[ ]

^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 F_54X and F_54Y are now known from the above 6x6 partial solution, here is a more efficient solution:

( )

3 3

32

3 3

32

23 23 43 43 43 3 3

4 4 54

43

4 4 54

14

34 34 14 14 14 4 4 54 54 54

1 0 1 0 0 0

0 1 0 1 0 0

0 0

0 0 1 0 1 0

0 0 0 1 0 1

0 0

X G X Y G Y

Y X Y X X G Z

G X X

Y

G Y Y

X

Y X Y X Y G Z Y X X

m A F

m A g

F

r r r r F I

m A F

F

m A g F

F

r r r r F I r F r F

α

−

⎡ ⎤ ⎧ ⎫

⎢ − ⎥ ⎪ ⎪ +

⎢ ⎥ ⎪ ⎪

⎢ − − ⎥ ⎪⎪ ⎪⎪ =

⎢ − ⎥ ⎨ ⎬ −

⎢ ⎥ ⎪ ⎪

⎢ − ⎥ ⎪ ⎪ + −

⎢ ⎥ ⎪ ⎪

− − + −

⎢ ⎥ ⎪⎩ ⎪⎭

⎣ ⎦ ^54Y

⎧ ⎫

⎪ ⎪

⎨ ⎬

⎪ ⎪

⎩ ⎭

[ ]

^A³⁴

{ } { }

^v³⁴ ⁼ ^b³⁴

Solve for six unknowns

{ }

^v³⁴ ⁼

[ ]

^A³⁴ ⁻¹

{ }

^b³⁴ and then use F32X and F32Y in the following 3x3 set of linear equations, from the link 2 FBD:

( )

21 2 2 32

12 12 2 2 2 32 32 32 32

1 0 0

0 1 0

1

X G X X

Y G Y Y

Y X G Z Y X X Y

F m A F

F m A g F

r r τ I α r F r F

− −

⎡ ⎤ ⎧ ⎫ ⎧ ⎫

⎪ ⎪ ⎪ ⎪

⎢ − ⎥⎨ ⎬ ⎨= + − ⎬

⎢ ⎥⎪ ⎪ ⎪ ⎪

⎢ − ⎥ + −

⎣ ⎦ ⎩ ⎭ ⎩ ⎭

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):

( )

21 2 2 32

2 2 2 32 32 32 32 12 21 12 21

X G X X

Y G Y Y

G Z Y X X Y Y X X Y

F m A F

F m A g F

I r F r F r F r F

τ α

= − +

= − + +

= + − − +

(20)

65 Matrix inversion requires approximately

3 3

log n

n and Gaussian elimination requires approximately

(

² ¹

)

²

3

n n

− n

+ multiplications/divisions².

Number of Multiplications/Divisions

15x15 8609 1345 84%

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

There is an astonishing 86% reduction in computational cost for Gaussian elimination with the 6x6 twice, plus decoupled link 2 method. 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 & 6, then links 3 & 4 with one unknown vector force from 5 & 6, then link 2 independently). But links 3, 4, 5, and 6 can be solved first independently of link 2: 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

15x15 8609 1345 84%

12x12 plus decoupled link 2 4811 723 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.

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