mech 335 - lecture pack 2

Position Analysis

• Useful indices for Position Analysis:

a) Types of Mechanism:

• Grashof vs. Non-Grashof:

• Grashof: at least 1 pair of adjacent links is capable of full rotation

• Test for Grashof mech: “Grashof’s Law”

• “The sum of the shortest and longest links’ lengths is less than the sum of the lengths of the remaining two links”

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Position Analysis

• Useful indices for Position Analysis:

a) Types of Mechanism:

• 4-bar mechanisms can be classified based on Grashof criterion

• For those that meet the criterion, the identity of the shortest link defines the mechanism type

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Position Analysis

• Example: Types of 4-bar

– Grashof 4-bar with rshort = r2

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Crank-Rocker

Position Analysis

• Example: Types of 4-bar

– Grashof 4-bar with rshort = r1

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Drag-Link

Position Analysis

• Example: Types of 4-bar

– Grashof 4-bar with rshort = r3 (or 4)

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Rocker-Rocker

Note that the extreme angular positions of the crank and follower do not necessarily coincide

Position Analysis

• Example: Types of 4-bar

– Any non-Grashof 4-bar is a Triple-Rocker

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Again, extreme angular positions of the crank and follower do not necessarily coincide

Also, there are points where the motion of the mechanism diverges, i.e. one or more outputs are possible for a given crank input

Which way will link 3 rotate when the input is rotated CW from this position?

How about link 4?

Position Analysis

• Useful indices for Position Analysis:

b) Inversions of a mechanism

• Inversions change which of the mechanism’s links is fixed

• An n-link mechanism has n inversions

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Position Analysis

• Example: Inversions of the slider-crank

– Standard slider-crank

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Position Analysis

• Example: Inversions of the slider-crank

– Freeing the slide, and fixing link 2

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Position Analysis

• Example: Inversions of the slider-crank

– Or, fixing link 3 (note: 4 is still free to rotate)

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Position Analysis

• Example: Inversions of the slider-crank

– Or, fixing 4 (1 slides, but does not rotate, w.r.t. 4)

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Position Analysis

• Useful indices for Position Analysis:

c) Transmission (γ) and Deviation (δ) angles

• Transmission Angle (γ): The acute angle between the directions of the velocity-difference vector of the driving point , and the absolute velocity vector of the driven point

• Deviation Angle (δ): The angle between the absolute direction of travel of the driven point and the direction of the force exerted by the driving link

MECH 335 Lecture Notes © R.Podhorodeski, 2009

INPUT

VB3

A1,2

B2,3

3

1

2

44

INPUT

A1,2

B2,3

3

1

2

44

γVB

2/A

2

INPUT

A1,2

B2,3

3

1

2

44

Position Analysis

• Example:

– Transmission Angle (γ):

MECH 335 Lecture Notes © R.Podhorodeski, 2009

The acute angle between the directions of:

and

INPUT

VB3

A1,2

B2,3

3

1

2

44

γVB

2/A

2

the velocity-difference vector of the driving point

the absolute velocity vector of the driven point

INPUT

A1,2

B2,3

3

1

2

44

FB2→B3

INPUT

VB3

A1,2

B2,3

3

1

2

44

INPUT

A1,2

B2,3

3

1

2

44

Position Analysis

• Example:

– Deviation Angle (δ):

MECH 335 Lecture Notes © R.Podhorodeski, 2009

The angle between:

and

INPUT

VB3

A1,2

B2,3

3

1

2

44

FB2→B3

δThe absolute direction of travel of the driven point

the direction of the force exerted by the driving link

Displacement Analysis

• Any single-loop, planar mechanism will have two unknowns (dependent variables)

• Need a method to solve for these variables in terms of known mechanism parameters

• 2 main methods presented:

– Graphical

– Analytical

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Displacement Analysis

• Example: 4-bar mech.– Vectors Ri used to

represent links

– All link lengths known

– Input (independent) variable is crank angle θ2

– What are the dependent variables?

MECH 335 Lecture Notes © R.Podhorodeski, 2009

A

B

A0 B0R1

R2

R3

R4

θ2

Displacement Analysis

• Example: 4-bar mech.– Vectors Ri used to

represent links

– All link lengths known

– Input (independent) variable is crank angle θ2

– What are the dependent variables?• Need a method to solve for

these variables

MECH 335 Lecture Notes © R.Podhorodeski, 2009

A

B

A0 B0R1

R2

R3

R4

θ2

θ3

θ4

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

A

A0 B0R1

R2

θ2

• Solution lies at the intersection of the possiblepositions of links 3 & 4

• Graphical (approximate) method

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

A

A0 B0

θ2

• Solution lies at the intersection of the possiblepositions of links 3 & 4

• Note presence of 2 possible solutions!

• Graphical (approximate) method

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

A

B

A0 B0

θ2

θ3

θ4

• Solution lies at the intersection of the possiblepositions of links 3 & 4

• Note presence of 2 possible solutions!

• Unknown angles are measured from drawing

• Graphical (approximate) method

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

• Analytical methods

– Graphical methods inexact, inconvenient for repeated analysis

– 2 Analytical methods presented:

• Geometric analysis

• Loop-closure

4

• Example 1: Geometric analysis

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

A

B

A0 B0

R1

R2

R3

R4

θ2

arg(D)Δ

θ3

φ

γ

Dθ4

• Loop Closure – Based Displacement Analysis

– Represent each link by a vector,

– Sum vectors to close the loop

– Split the real and imaginary components of this equation to yield two equations by applying:

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

4

• Example 2: Loop Closure Analysis

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

A

B

A0 B0

R1

R2

R3

R4

θ2

θ3

θ4

4

• Example 2: Loop Closure Analysis

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

A

B

A0 B0

R1

R2

R3

R4

θ2

θ3

θ4

EQUATION FROM REAL PARTS

EQUATION FROM IMAGINARY PARTS

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• Solution is correct, but “weak”, since multiple angles have the same sine

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Example 3: Loop Closure, offset slider-crank

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• A stronger solution can be found by writing:

4

• Iterative Loop-Closure based displacement solution

– If the dependent variables are inaccurate

– An iterative algorithm for closure can then be based on:

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

and

• Example: Newton-Raphson iterative closure

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

Loop closure error:

Input:Unknowns: ,

i.e.

• Example: Newton-Raphson iterative closure

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

If is correct, Estimate of can be improved by iteration, i.e.

Input:Unknowns: ,

i.e.

2

3

1

0

• Example: Newton-Raphson iterative closure

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

If is correct, Estimate of can be improved by iteration, i.e.

Input:Unknowns: ,

i.e.

2

3

1

0

• Example: Newton-Raphson iterative closure

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

For this example:

Then, from (3):Input:Unknowns: ,

i.e.

2

3

1

0

Input:Unknowns: ,

i.e.

• Example: Newton-Raphson iterative closure

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

Inverting (4), substituting into (5), and then into (2) gives:

Where is given in (1).Equation (6) would be used to iteratively improve the estimates of , until:

2

3

1

0

• Numerical Example:

» Suppose we start by guessing:

» Analysis tolerance:

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0

• Numerical Example:

• Now we iterate from that starting point

Displacement Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

2

3

1

0