ken youssefi mechanical engineering dept, sjsu 1

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Ken Youssefi Mechanical Engineering Dept, SJSU 1

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Page 1: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU 1

Page 2: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Complex Numbers and Polar Notation

x2 + 1 = 0, x2 + x + 1 = 0, x = ? Euler (1777), i = √ -1

i2 = -1

O AA′

OA′ = - OA

OA′ = i2 OA

i2 represents 180o rotation of a vector

i represents 90o rotation of a vector

Real axis

Imaginary axis r

P

x

iy

Argand Diagram

θ

r = x + iy

x = rcos(θ)

y = rsin(θ)

r = rcos(θ) + i rsin(θ)

real part imaginary part

r = r eiθ

eiθ = cos(θ) + i sin(θ),

Euler’s Formula

e-iθ = cos(θ) - i sin(θ)

2

Page 3: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Closed Loop Vector Equation – Complex Polar Notation

r2 + r3 = r1 + r4

r2eiθ

2 + r3eiθ

3 = r1eiθ

1 + r4eiθ

4

Positive sign convention - all angles are measured with respect to the horizontal line in counterclockwise direction.

r2eiθ

2 + r3eiθ

3 + r4eiθ

4 + r1eiθ

1 = 0

2

4

θ2

r4r2

A

O4

B

3

r1

r3

O2

θ4

θ3

θ1 = 0

r4r2

A

O4

B

3

r1

r3

O2

r2 + r3 + r4 + r1 = 0A

O4

2

B3

4r2

r1

r4

r3

O2

θ2

θ4

θ3

θ1 = 1803

Page 4: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Rotational Operator & Stretch Ratio

P1

θ1

x

iy

r1 = r1eiθ

1

Pj

θj

rj = rjeiθ

j

Φj

θj = θ1 + Φj

rj = (rj / r1) r1 eiθ

1 eiΦj

rj = rj (r1 / r1 ) ei(θ

1 + Φj)

rj = r1 ρ eiΦ

j

Stretch ratioRotational operator

4

Page 5: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Analytical Synthesis – Standard Dyad Form

4 Bar mechanism

A

B

O2O4

2

3

4

P

Left side Right side

r2r4

r3

r′3r″3

Design the left side of the 4 bar → r2 & r′3Design the right side of the 4 bar → r4 & r″3

αj

βj

δj

Closed loop vector equation – complex polar notation

r2 + r′3 + δj = r2 eiβj + r′3 e

iαj

Aj

Pj

r2 eiβj

r′3 eiαj

r2 (eiβj – 1) + r′3 (e

iαj – 1) = δj

Standard Dyad form

Parallel

Left side of the mechanism

A1

O2

2

P1

r2

r′3

5

Page 6: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Analytical Synthesis – Standard Dyad Form

Apply the same procedure to obtain the Dyad equation for the right side of the four bar mechanism.

B

O4

4

P

r4

r″3

Rotation of link 4

r2 (eiβj – 1) + r′3 (e

iαj – 1) = δj

Standard Dyad form for the left side of the mechanism

α → rotation of link 3β → rotation of link 2

r4 (eij – 1) + r″3 (e

iαj – 1) = δj

Standard Dyad form for the right side of the mechanism

→ rotation of link 4 α → rotation of link 3

6

Page 7: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Analytical Synthesis Two Position Motion & Path Generation Mechanisms

α2

β2

δ2

Left side of the mechanism

A1

O2

2

P1

r2

r′3A2

P2

r2 eiβ2

r′3 eiα2

Parallel

r2 (eiβ2 – 1) + r′3 (e

iα2 – 1) = δ2

Dyad equation for the left side of the mechanism. One vector equation or two scalar equations

1. Draw the two desired positions accurately.

Motion generation mechanism, the orientation of link 3 is important (angle alpha)

2. Measure the angle α from the drawing, α2

3. Measure the length and angle of vector δ2

There are 5 unknowns; r2, r′3 and angle β2 and only two scalar equations (Dyad).

Select three unknowns and solve the equations for the other two unknowns

Given; α2 and δ2

Select; β2 and r′3Solve for r2

Two position motion gen. Mech.

Three sets of infinite solution

7

Page 8: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Analytical Synthesis Two Position Motion & Path Generation Mechanisms

Apply the same procedure for the right side of the 4-bar mechanism

r4 (eij – 1) + r″3 (e

iαj – 1) = δj

Given; α2 and δ2

Select;, 2 , r″3

Solve for r4Two position motion gen. Mech.

Given; β2 and δ2

Select; α2 and r′3Solve for r2

Two position path gen. Mech.

Three sets of infinite solution

Path Generation Mechanism (left side of the mechanism)

8

Page 9: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Analytical Synthesis Three Position Motion & Path Generation Mechanisms

δ2

A1

O2

2

P1

r2

r′3

α2

β2

Parallel

P3

A3

α3

β3

δ3

A2

P2

r2 eiβ2

r′3 eiα2

r2 eiβ3

r′3 eiα3

9

Page 10: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Analytical Synthesis Three Position Motion & Path Generation Mechanisms

Three position motion gen. mech.

Given; α2, α3, δ2, and δ3

Select; β2 and β3 Solve for r2 and r′3

Three position motion gen. Mech.

Two sets of infinite solution

2 free choices

4 scalar equations

Dyad equationsr2 (e

iβ2 – 1) + r′3 (eiα2 – 1) = δ2

r2 (eiβ3 – 1) + r′3 (e

iα3 – 1) = δ3

6 unknowns; r2 , r′3 , β2 and β3

10

Page 11: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Analytical Synthesis

Four position motion generation mechanism

Given; α2, α3, α4 δ2, δ3 and δ4

Select; one input angle, β2 or β3 or β4 Solve for r2 and r′3

Four position motion gen. Mech.

One set of infinite solution

Dyad equations

r2 (eiβ2 – 1) + r′3 (e

iα2 – 1) = δ2

r2 (eiβ3 – 1) + r′3 (e

iα3 – 1) = δ3

r2 (eiβ4 – 1) + r′3 (e

iα4 – 1) = δ4

Non-linear equations

7 unknowns; r2 , r′3 , β2 , β3 and β4

1 free choices

6 scalar equations

11

Page 12: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Analytical Synthesis

Dyad equations

r2 (eiβ2 – 1) + r′3 (e

iα2 – 1) = δ2

r2 (eiβ3 – 1) + r′3 (e

iα3 – 1) = δ3

r2 (eiβ4 – 1) + r′3 (e

iα4 – 1) = δ4

Non-linear equations

Five position motion generation mechanism

r2 (eiβ5 – 1) + r′3 (e

iα5 – 1) = δ5

8 unknowns; r2 , r′3 , β2 , β3 , β4 and β5

0 free choice

8 scalar equations

Given; α2, α3, α4, α5, δ2, δ3, δ4, and δ5

Select; 0 choice

Four position motion gen. Mech.

Unique solution, not desirable

Solve for r2 and r′3

12

Page 13: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Analytical Synthesis –Function Generation Mechanism

Freudenstein’s method

r4r2

A

O4

B

3

r1

r3

O2

r1 + r2 + r3 = r4

θ2

θ4

θ3

r1eiθ

1 + r2eiθ

2 + r3eiθ

3 = r4eiθ

4

Real part of the equation

eiθ = cos(θ) + i sin(θ)

Euler equation

= r4 cos(θ4)

r1 sin(θ1) + r2 sin(θ2) + r3 sin(θ3) = r4 sin(θ4)

Imaginary part of the equation

r1 cos(θ1) + r2 cos(θ2) + r3 cos(θ3)

13

Page 14: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Analytical Synthesis –Function Generation Mechanism

θ1 = 180

– r4 cos(θ4) = 0– r1 + r2 cos(θ2) + r3 cos(θ3)

r2 sin(θ2) + r3 sin(θ3) – r4 sin(θ4) = 0

+ r4 cos(θ4)]2[r1 – r2 cos(θ2)[r3 cos(θ3)]

2 =

[r3 sin(θ3)]2 = [– r2 sin(θ2) + r4 sin(θ4)]

2

Add the two equations

r32 = [– r2 sin(θ2) + r4 sin(θ4)]

2 + r4 cos(θ4)]2+ [r1 – r2 cos(θ2)

r32 = (r1)

2 + (r2)

2 + (r4)

2 – 2r1 r2 cos(θ2) + 2r1 r4 cos(θ4) – 2r2 r4 cos(θ2– θ4 )

Expand and simplify

14

Page 15: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU

Analytical Synthesis –Function Generation Mechanism

r32 = (r1)

2 + (r2)

2 + (r4)

2 – 2r1 r2 cos(θ2) + 2r1 r4 cos(θ4) – 2r2 r4 cos(θ2 – θ4)

Divide the above equation by 2r2 r4

K2 = – r1

r2

r1

r4

cos(θ2) – r1

r2

cos(θ4) + (r3)

2 – (r1)2 – (r2)

2 – (r4)

2

2r2 r4

= – cos(θ2 – θ4)

(r3)2 – (r1)

2 – (r2)

2 – (r4)

2

2r2 r4

K3 = K1 =r1

r4

Define

K1cos(θ2) + K2 cos(θ4) + K3 = – cos(θ2 – θ4)

Freudenstein’s equation15

Page 16: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU 16

Example – Continuous Function

Synthesize a four bar mechanism to generate a function y = log x in the

interval 1 x 10. The input crank length should be 50 mm.

Use Chebyshev spacing and three precision points

sj = ½ (so + sn+1) - ½ (sn+1 – so) cos[(2j - 1)π/2n]

r4

r2

A1

O4

B1

3

r1

r3

O2

A5

φ5

φ1

ψ1

B5

ψ5

Select

Input link (crank) start angle = 45o

Input link (crank) end angle = 105o

φ1

φ5

output link (link 4) start angle = 135o

output link (link 4) end angle = 225o

ψ1

ψ5

Precision point

so = start point = 1 sn+1 = end point = 10

j = precision point, n = total number of precision points

Determine the precision points

Page 17: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Example – Continuous Function

Ken Youssefi Mechanical Engineering Dept, SJSU 17

x1 = ½ (1 + 10) – ½ (10 – 1) cos(π/6) = 1.6029

x2 = ½ (1 + 10) – ½ (10 – 1) cos(3π/6) = 5.50

x3 = ½ (1+ 10) – ½ (10 – 1) cos(5π/6) = 9.3971

sj = ½ (so + sn+1) - ½ (sn+1 – so) cos[(2j - 1)π/2n]

Corresponding function values are:

y1 = log x1 = log (1.6029) = .2049

y2 = log x2 = log (5.5) = .7404

y3 = log x3 = log (9.3971) = .9730

Page 18: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU 18

Example – Continuous Function

Boundary condition; x = 1, φ = 45 and x = 10, φ = 105o

φ = (20/3)(x) + 115/3

Assume linear relationship between φ (input angle) and x, and between (output angle) and y. Where a, b, c and d are constants.

φ = a(x) + b,

= c(y) + d, Boundary condition; y = log(1) = 0, = 135

and y = log(10) = 1, = 225o

= 90(y) + 135

45 = a(1) + b

105 = a(10) + b

135 = C(0) + d

225 = C(1) + d

Input link (crank) start angle = 45o

Input link (crank) end angle = 105o

φ1

φ5

output link (link 4) start angle = 135o

output link (link 4) end angle = 225o

ψ1

ψ5

Page 19: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU 19

Example – Continuous Function

pts x y 1 1.00 45.00 0.00 135.00

2 1.6029 49.019 .2049 153.44

3 5.50 75.0 .7404 201.64

4 9.3971 100.98 .9730 222.57

5 10.00 135.00 1.00 225.00

φ = 20/3(x) + 115/3

= 90(y) + 135

Precision points

y1 = log x1 = log (1.6029) = .2049

y2 = log x2 = log (5.5) = .7404

y3 = log x3 = log (9.3971) = .9730

Start

End

Page 20: Ken Youssefi Mechanical Engineering Dept, SJSU 1

Ken Youssefi Mechanical Engineering Dept, SJSU 20

Example – Continuous Function

pts x y

1 1.00 45.00 0.00 135.00

2 1.6029 49.019 .2049 153.44

3 5.50 75.0 .7404 201.64

4 9.3971 100.98 .9730 222.57

5 10.00 135.00 1.00 225.00

K1cos (49.019) + K2 cos (153.44) + K3 = – cos (49.019 – 153.44)

K1cos (75) + K2 cos (201.64) + K3 = – cos (75 – 201.64)

K1cos (100.98) + K2 cos (222.57) + K3 = – cos (100.98 – 222.57)

Freudenstein equationK1cos(φ1) + K2 cos(ψ1) + K3 = – cos(φ1 – ψ1)

K1cos(φ2) + K2 cos(ψ2) + K3 = – cos(φ2 – ψ2)

K1cos(φ3) + K2 cos(ψ3) + K3 = – cos(φ3 – ψ3)

Solve for K1, K2, and K3