MECH 335 - Lecture Pack 2
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Transcript of 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