Course Notebook - The University of Tulsa

Transcription

ME 3212: Mechanisms
Course Notebook
Instructor:
Jeremy S. Daily, Ph.D., P.E.
Fall 2012
Contents
1. Syllabus
1.1. Course Bulletin Description . . . . . . . . .
1.2. Objectives . . . . . . . . . . . . . . . . . . .
1.3. Course Outline . . . . . . . . . . . . . . . .
1.4. Course Policies . . . . . . . . . . . . . . . .
1.4.1. Text Books . . . . . . . . . . . . . .
1.4.2. Grading Procedures . . . . . . . . . .
1.4.3. Exam Policy . . . . . . . . . . . . .
1.4.4. Computer Usage . . . . . . . . . . .
1.4.5. Late Submission and Absences . . . .
1.4.6. Class Conduct . . . . . . . . . . . .
1.4.7. Academic Misconduct . . . . . . . .
1.4.8. Center for Student Academic Support
2. Introduction To Mechanisms
2.1. Mechanisms Vocabulary . . . . . . . .
2.2. Common Mechanisms . . . . . . . . .
2.3. Kinematic Pairs (a.k.a. Joints) . . . . .
2.3.1. Low Order Pairs . . . . . . . .
2.3.2. High Order Pairs . . . . . . . .
2.4. Degrees of Freedom . . . . . . . . . . .
2.4.1. Mobility . . . . . . . . . . . .
2.4.2. Kutzbach Criteria . . . . . . . .
2.5. Grashof’s Law for Four-bar Mechanisms
2.5.1. Special Cases . . . . . . . . . .
2.5.2. Inversions . . . . . . . . . . . .
2.6. Homework Problem Set 1 . . . . . . . .
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3. Position Analysis
3.1. Loop Closure Equations . . . . . . . . .
3.1.1. Derivation of the Law of Cosines
3.1.2. Derivation of the Law of Sines . .
3.1.3. Inverted Slider Crank . . . . . . .
3.1.4. Offset Slider Crank . . . . . . . .
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Contents
3.2.
3.3.
3.4.
3.5.
3.6.
3.7.
3.8.
3.9.
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3.1.5. Four Bar Mechanism . . . . . . . . . .
3.1.5.1. Open Closure . . . . . . . .
3.1.5.2. Cross Closure . . . . . . . .
Coupler Curves . . . . . . . . . . . . . . . . .
3.2.1. Matlab Implementation . . . . . . . . .
3.2.2. Excel Implementation . . . . . . . . .
3.2.2.1. Standard Algebraic Solution .
3.2.2.2. Use Excel Solver . . . . . .
3.2.2.3. SolidWorks Implementation .
Homework Problem Set 2 . . . . . . . . . . . .
Newton-Raphson Method . . . . . . . . . . . .
3.4.1. Rocking Slider Crank . . . . . . . . . .
Multi Loop Mechanisms . . . . . . . . . . . .
Toggle and Limit Positions . . . . . . . . . . .
Transmission Angle . . . . . . . . . . . . . . .
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4. Mechanism Synthesis
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4.1. Geometric Constraint Programming . . . . . . . . . . . . . . . . . . . . . . . . 59
4.2. Homework Problem Set 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
5. Velocity Analysis
5.1. Vector Operations . . . . . . . . . . . . .
5.1.1. Dot Product . . . . . . . . . . . .
5.1.2. Cross Product . . . . . . . . . . .
5.1.3. Derivatives of Vector Products . .
5.2. Velocity with a Rotating Reference Frame
5.3. Graphical Analysis . . . . . . . . . . . .
5.3.2. Four-Bar Mechanism . . . . . . .
5.4. Analytical Analysis . . . . . . . . . . . .
5.4.2. Four Bar Mechanism . . . . . . .
5.5. Homework Problem Set 6 . . . . . . . . .
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6. Acceleration Analysis
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6.1. Accelerations in Four-bar Mechanisms . . . . . . . . . . . . . . . . . . . . . . . 93
6.2. Inverted Slider Crank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
6.3. Homework Problem Set 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
Contents
7. Cams
7.1. Types of Cam Followers . . . . . . . . . .
7.1.1. Flat Faced Radial . . . . . . . . . .
7.1.2. Offset Roller Follower . . . . . . .
7.1.3. Barrel Cam with Roller Follower . .
7.1.4. Heavy Truck Brake Cams (S-Cams)
7.2. Cam Follower Motion . . . . . . . . . . . .
7.2.1. Displacement . . . . . . . . . . . .
7.2.2. Velocity . . . . . . . . . . . . . . .
7.2.3. Acceleration . . . . . . . . . . . .
7.2.4. Jerk . . . . . . . . . . . . . . . . .
7.3. Cam Follower Profiles . . . . . . . . . . .
7.3.1. Constant Acceleration (Parabolic) .
7.3.2. Harmonic Motion . . . . . . . . . .
7.3.3. Cycloidal Motion . . . . . . . . . .
7.4. Cam Design . . . . . . . . . . . . . . . . .
7.5. Homework Problem Set 8 . . . . . . . . . .
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8. Gears
8.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . .
8.1.1. Cog and Lantern Gears . . . . . . . . . . . . . . .
8.1.2. Common Types of Gears . . . . . . . . . . . . . .
8.2. Fundamental Law of Gearing . . . . . . . . . . . . . . . .
8.3. Conjugate Profiles and Involutometry . . . . . . . . . . .
8.3.1. The Involute Curve . . . . . . . . . . . . . . . . .
8.3.2. Cycloidal Profiles . . . . . . . . . . . . . . . . . .
8.3.3. Gear Sizing and Terminology . . . . . . . . . . .
8.4. Homework Problem Set 9 . . . . . . . . . . . . . . . . . .
8.5. Gear Train Analysis . . . . . . . . . . . . . . . . . . . . .
8.5.1. Taxonomy of gear trains: . . . . . . . . . . . . . .
8.5.2. Gear Train Ratio . . . . . . . . . . . . . . . . . .
8.5.3. Idler Gears . . . . . . . . . . . . . . . . . . . . .
8.5.4. Compound Gear Trains . . . . . . . . . . . . . . .
8.6. Homework Problem Set 10 . . . . . . . . . . . . . . . . .
8.7. Reverted Gear Train Design . . . . . . . . . . . . . . . .
8.7.1. Design Considerations . . . . . . . . . . . . . . .
8.8. Planetary Gear Trains . . . . . . . . . . . . . . . . . . . .
8.8.1. Advantages . . . . . . . . . . . . . . . . . . . . .
8.8.2. Vector Approach to Planetary Gear Train Analysis
8.8.3. Tabular Planetary Gear Train Analysis . . . . . . .
8.8.4. Compound Planetary Gear Trains . . . . . . . . .
8.8.5. Plotting Planetary Gears . . . . . . . . . . . . . .
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Contents
5
8.9. Differential Gear Trains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
8.9.1. Ackerman Steering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
A. Mechanisms Design Project
A.1. Project Proposal . . . . . . . . . . .
A.2. Hand Sketches . . . . . . . . . . .
A.3. Geometric Constraint Programming
A.4. SolidWorks Parts and Assembly . .
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145
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1. Syllabus
Instructor: Dr. Jeremy S. Daily
• E-mail: jeremy-daily@utulsa.edu
Phone: 918-631-3056
Office: 2080 Stephenson
Office Hours: Tuesday and Thursday 9:30 - 12:00. Otherwise, drop in or schedule an appointment.
Classroom: U4
Days: Tuesday and Thursday
Time: 12:30 - 1:20 PM
This course notebook is required for the course and costs $20. It includes a 3-ring binder. While
the pages in here are designed to help you take notes, additional writing space will be required.
Therefore, loose-leaf paper is recommended to augment the notebook.
This notebook can also be accessed (but not printed) in electronic format at
http://personal.utulsa.edu/~jeremy-daily/ME3212/MechanismsCourseNotebook.
pdf
1.1. Course Bulletin Description
Displacement, velocity and acceleration analysis of linkages, cams, and gear trains. An introduction to synthesis. Computer simulation and design of planar mechanisms using modern
engineering software culminating in a design project. Prerequisites: ES 2023 (Dynamics).
This required two-credit hour course is offered once a year, typically at the beginning (fall
semester) of the junior year.
1.2. Objectives
This section of the syllabus is designed to give the student a larger picture of the purpose of
this class. This course is designed to provide the students the analytical skill necessary for
1.3. Course Outline
analyzing mechanism motion analysis. In addition to analysis, student will learn mechanism
synthesis using geometric constraint programming techniques. It is important for the student to
learn to use modern computer tools to aid in mechanism design and analysis. The familiarity
with rigid body analysis in modern computer analysis programs will give students a competitive
advantage in the current marketplace. The course covers position, velocity, and acceleration
analysis of planar linkages and slider crank mechanisms. Grashof’s laws, Kutzbach criterion,
and the Newton-Raphson method. Cam profiles and cam follower motion is studied along with
gears and gearing including epicyclic gear trains.
1.3. Course Outline
The following is a dynamically updated schedule for the class. It is a shared Google calendar
http://www.google.com/calendar (Search for ME 3212)
and can be integrated into your own personal calendaring system. While the calendar can be
updated and changed as the course goes on, the schedule will remain fairly rigid during the
semester. All changes and details concerning specific events and items on the schedule will be
updated through the Google calendar for this course.
The calendar ID is ge5k33qntvmv453ct92ti5c0c4@group.calendar.google.com
1.4. Course Policies
1.4.1. Text Books
In addition to this course notebook the following books are useful.
Required: Mechanics of Machines by W. L. Cleghorn, Oxford University Press, 2005, ISBN
0195154525
Reference: Theory of Machines and Mechanisms, 3rd Edition by J.J. Uicker, G.R. Pennock,
and J.E. Shigley, Oxford University Press, ISBN 019515598
Other Resources: http://kmoddl.library.cornell.edu/index.php
1.4.2. Grading Procedures
The following table gives the weights of the different aspects of the graded material for this class:
Homework: 10%
Design Project: 30%
7
Exam 1: 20%
Exam 2: 20%
Final Exam: 30%
90-100 = A, 80-89 = B, 70-79 = C, 60-69 = D, < 60 = F
The instructor reserves the right to lower the minimum requirements for each letter grade.
Grades will be kept on WebCT and will be updated on a regular basis.
1.4.3. Exam Policy
Exams are open book and open notes; closed computer.
1.4.4. Computer Usage
Matlab and other specialty software will be used for labs and homework for data analysis and
plotting. These programs are available in the undergraduate computer lab. SolidWorks and Ansys
should be available in the computer labs as well.
SolidWorks is available to all ME students for installation on there personal computer. The media
is available under the TU shared space at S:\ENS\Mechanical Engineering\SolidWorks.
1.4.5. Late Submission and Absences
Late submission of homework will receive no score. Late computer projects will receive no score.
Exams have mandatory attendance. Make-up exams will be offered only under very exceptional
circumstances provided prior permission from the instructor is obtained. Neatness and clarity of
presentation will be given due consideration while grading homework, computer projects, and
exams.
1.4.6. Class Conduct
Please do whatever necessary to maintain a friendly, pleasant and business-like environment so
that it will be a positive learning experience for everyone. Please turn off all cell phone ringers
or any other device that could spontaneously make noise.
8
1.4.7. Academic Misconduct
All students are expected to practice and display a high level of personal and professional integrity. During examinations each student should conduct himself in a way that avoids even the
appearance of cheating. Any homework or computer problem must be entirely the student’s own
work. Consultation with other students is acceptable; however, copying homework from one
another will be considered academic misconduct. Any academic misconduct will be dealt with
under the policies of the College of Engineering and Natural Sciences. This could mean a failing
grade and/or dismissal. The policy of the University regarding withdrawals and incompletes will
be strictly adhered to.
1.4.8. Center for Student Academic Support
Students with disabilities should contact the Center for Student Academic Support to self-identify
their needs in order to facilitate their rights under the Americans with Disabilities Act. The center
for Student Academic Support is located in Lorton Hall, Room 210. All students are encouraged
to familiarize themselves with and take advantage of services provided by the Center for Student
Academic Support such as tutoring, academic counseling, and developing study skills. The
Center for Student Academic Support provides confidential consultations to any student with
academic concerns as well as to students with disabilities.
9
M E 3 2 1 2 : M e c h a n i s m s, US Holidays
Tue Aug 21, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
C a l e n d a r : ME 3212: Mechanisms
C r e a t e d b y : jeremydaily@gmail.com
Thu Aug 23, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Tue Aug 28, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Thu Aug 30, 2012
All day
HW 1 Due
Thu Aug 30, 2012 - Fri Aug 31, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Mon Sep 3, 2012
All day
Labor Day
Mon Sep 3, 2012 - Tue Sep 4, 2012
C a l e n d a r : US Holidays
Tue Sep 4, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Thu Sep 6, 2012
All day
HW 2 Due
Thu Sep 6, 2012 - Fri Sep 7, 2012
12:30pm - 1:20pm
Mechanisms Computer Lecture
W h e r e : KEP L1
Tue Sep 11, 2012
All day
Patriot Day
Tue Sep 11, 2012 - Wed Sep 12, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Thu Sep 13, 2012
All day
HW 3 Due
Thu Sep 13, 2012 - Fri Sep 14, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Tue Sep 18, 2012
All day
Project Proposal Due
Tue Sep 18, 2012 - Wed Sep 19, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Thu Sep 20, 2012
All day
HW 4 Due
Thu Sep 20, 2012 - Fri Sep 21, 2012
12:30pm - 1:20pm
W h e r e : KEP L1
Tue Sep 25, 2012
12:30pm - 1:20pm
W h e r e : KEP L1
Thu Sep 27, 2012
12:30pm - 1:20pm
Mechanis m s E x a m 1
W h e r e : KEP U4
Tue Oct 2, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Thu Oct 4, 2012
All day
HW 5 Due
Thu Oct 4, 2012 - Fri Oct 5, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Mon Oct 8, 2012
All day
Columbus Day
Mon Oct 8, 2012 - Tue Oct 9, 2012
Tue Oct 9, 2012
All day
Project Hand Sketches Due
Tue Oct 9, 2012 - Wed Oct 10, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Thu Oct 11, 2012
All day
HW 6 due
Thu Oct 11, 2012 - Fri Oct 12, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Tue Oct 16, 2012
All day
No Class
Tue Oct 16, 2012 - Wed Oct 17, 2012
12:30pm - 1:20pm
Mechanisms Project Time
W h e r e : KEP L1
Thu Oct 18, 2012
All day
No Class
Thu Oct 18, 2012 - Fri Oct 19, 2012
12:30pm - 1:20pm
Mechanisms Project Time
W h e r e : KEP L1
Tue Oct 23, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Thu Oct 25, 2012
All day
HW 7 Due
Thu Oct 25, 2012 - Fri Oct 26, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Tue Oct 30, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Wed Oct 31, 2012
All day
Halloween
Wed Oct 31, 2012 - Thu Nov 1, 2012
Thu Nov 1, 2012
12:30pm - 1:20pm
Mechanis m s E x a m 2
W h e r e : KEP U4
Sun Nov 4, 2012
All day
Daylight Saving Time Ends
Sun Nov 4, 2012 - Mon Nov 5, 2012
Tue Nov 6, 2012
All day
Election Day
Tue Nov 6, 2012 - Wed Nov 7, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Thu Nov 8, 2012
All day
HW 8 Due
Thu Nov 8, 2012 - Fri Nov 9, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Sun Nov 11, 2012
All day
Veterans Day
Sun Nov 11, 2012 - Mon Nov 12, 2012
Tue Nov 13, 2012
All day
Geometric Constraint Program Due
Tue Nov 13, 2012 - Wed Nov 14, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Thu Nov 15, 2012
All day
HW 9 Due
Thu Nov 15, 2012 - Fri Nov 16, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Thu Nov 22, 2012
All day
Thanksgiving
Thu Nov 22, 2012 - Fri Nov 23, 2012
Tue Nov 27, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Thu Nov 29, 2012
12:30pm - 1:20pm
Mechanisms Lecture
W h e r e : KEP U4
Fri Nov 30, 2012
All day
HW 10 Due
Fri Nov 30, 2012 - Sat Dec 1, 2012
Mon Dec 3, 2012
All day
Mechanisms Final Design Project Due
Mon Dec 3, 2012 - Tue Dec 4, 2012
Mon Dec 10, 2012
1pm - 3:25pm
Mechanisms Final
Mon Dec 24, 2012
All day
Christmas Eve
Mon Dec 24, 2012 - Tue Dec 25, 2012
2. Introduction To Mechanisms
Mechanisms is a the study of rigid body motion. The concepts in this course are restricted to
analyzing the motion and does not consider the cause of motion. In the field of dynamics, there
are a few broad categories as shown in Fig. 2.1.
Figure 2.1.: Hierarchy of Mechanics
2.1. Mechanisms Vocabulary
Fill in the appropriate definitions for the following terms.
Kinematics:
2.2. Common Mechanisms
Kinetics:
Mechanism:
Machine:
Linkage:
Link:
Joint:
Skeleton Diagram:
2.2. Common Mechanisms
• Slider Crank
– Example: Air compressor
• Four-Bar
11
2.3. Kinematic Pairs (a.k.a. Joints)
12
– Example: Washing Machine Rocker
• Belts and Gears
– Example: Transmission
• Cams
– Example: Internal combustion engine valve train
2.3.1. Low Order Pairs
• Spherical (G)
–
–
–
–
–
• Revolute (R)
–
–
–
–
–
• Cylindrical (C)
–
–
• Prismatic (P)
–
–
–
See
Cleghorn’s
Section
1.4
• Helical or Screw (S)
–
–
–
• Planar or Flat (F)
–
–
–
• Revolute 2
–
–
2.3.2. High Order Pairs
• Rolling-no slip
–
–
–
• Rolling/rotating with slip
– Cams, Link against plane
∗
∗
– Pin-In-Slot
∗
∗
• Rotating Pairs
– Gears, Friction Drives
–
–
13
2.4. Degrees of Freedom
14
• Wrapping Pairs
– Example: Belt on Pulley (sheave)
–
–
• The number of inputs needed to get an output.
•
• For planar links there are:
• For spatial links there are:
2.4.1. Mobility
Calculating the number of degrees of freedom for a mechanism is determining its mobility.
2.4.2. Kutzbach Criteria
The formula to calculate mobility is
m = 3(n − 1) − 2 j1 − j2
where
n. . .
j1 . . .
j2 . . .
if m ≥ 1, then
if m = 0, then
if m ≤ −1, then
Remember, rolling pairs count as a 2 d.o.f. joint.
Example: Consider the planar slider crank mechanism shown in Fig. 2.2a. Determine the mobility using the Kutzbach Criteria. Determine the number of links, the number of single degree of
freedom joints, and the number of 2 d.o.f. joints.
See
Cleghorn’s
Section
1.5
y
15
A
θ2
θ3
O2
%
B
(a) Planar Slider Crank Mechanism
%
(b) Four-bar slider
%
%
(c) Four-bar linkage
Figure 2.2.: Determine the mobility of the mechanisms shown above.
x
16
Grashof’s law is a method to categorize four-bar mechanisms based on the ability for a link to
make a complete rotation compared to the other links. In a four-bar mechanism there are four
possible link lengths:
• s. . .
• l. . .
• p. . .
• q. . .
s+l < p+q
(2.1)
If Eq. 2.1 is satisfied, then
If s is the input link, then
If s is the frame (base) link, then
If s is the coupling link, then
Example: Can the shortest link make a full revolution in a four-bar mechanism where s = 4,
l = 9, p = 6, and q = 6?
2.5.1. Special Cases
Change point mechanism
Parallelogram four-bar mechanisms
17
2.5.2. Inversions
• Every mechanism has a “ground” or “base” or “frame” link that is fixed.
•
•
Let’s do an example to find the inversions for a four-bar mechanism where s = 1, l = 8, p = 6,
and q = 6 using the grids shown in Fig. 2.3.
Check Grashof’s Criteria:
6
6
5
5
4
4
3
3
2
2
1
1
0
0
(a) Crank Rocker
(b) Crank Rocker
6
6
5
5
4
4
3
3
2
2
1
1
0
0
(c) Double Crank
(d) Double Rocker
Figure 2.3.: Inversions of a four-bar mechanism.
See
Cleghorn’s
Section
1.7
2.6. Homework Problem Set 1
1. The link lengths of a planar four-bar linkage are 1, 3, 5, and 5 inches.
a) Assemble the links in all possible combinations and sketch (with a ruler and compass
on engineering graph paper) the four inversions of each.
b) Describe each inversion by name (e.g., crank-rocker, drag-link, rocker-rocker, changepoint)
c) Do these linkages follow Grashof’s Law?
2. Book Problem P1.1.
3. Book Problem P1.5.
4. Complete the “Introduction to SolidWorks” Tutorial. The tutorial can be found by accessing the SolidWorks Help Menu. Under the Getting Started Tutorial Category, perform the
Introduction to SolidWorks Tutorial. Turn in a print of the drawing you create.
18
Due on: _____________________________
19
3. Position Analysis
The goal of a position analysis is to describe any position of any point in any mechanism configuration. The mechanical engineering skill from learning how to do a position analysis is to learn
the following concepts:
• Vector analysis
• Computer programming and mathematical modeling
• Numerical methods for root finding
• Develop the skill of abstracting physical objects using mathematics
A locus is defined as
3.1. Loop Closure Equations
The loop closure equations are fundamental to modeling mechanisms. The vectors that describe
the components must add to zero when the links form a loop:
n
∑ ~Ri = 0
i=1
where n. . .
Consider a loop of three fixed lengths: A, B, and C with angles α , β , and γ .
β
A
C
γ
α
B
See
Cleghorn
Section
4.2
21
Let’s define the x-axis to be along length B and the vectors defining the loop to go in a clockwise
direction.
β
A
C
y
γ
x
α
B
Breaking these vectors into components gives the following six expressions. Keep in mind all
angles are defined from the positive x-axis and are positive when measure in a counterclockwise
direction.
Ax :
Ay :
Bx :
By :
Cx :
Cy :
Now all the components in the x direction can be summed to zero and all the y components can
be summed to zero:
x:
y:
Ax + Bx +Cx = 0
Ay + By +Cy = 0
or
x:
y:
3.1.1. Derivation of the Law of Cosines
In Section 3.1 the concept of the loop closure equations was presented. If two sides, say C and
B, are known and the angle between them α is known, then the loop closure equations can be
solved for the length of A. The following procedures demonstrate solution techniques for the
loop closure equations.
22
1. Rewrite each component equation to move the unknown and unneeded angle as a single
term on one side:
x:
y:
2. Square each component equation and add them together. Invoking the relationship
sin2 θ + cos2 θ = 1
x:
y:
Sum:
3. Now the unknown angle γ has been eliminated and only the remaining length A is unknown. Expand the binomial terms:
4. Simplify and solve for A
A=
p
C2 + B2 − 2BC cos α
C=
q
B=
q
Similarly
and
A2 + B2 − 2AB cos γ
A2 +C2 − 2AC cos β
3.1.2. Derivation of the Law of Sines
If only one length is known and two angles are known, then the solution technique of the loop
closure equations requires a slightly different approach. This time, an unknown length needs to
be eliminated from the loop closure equations. The following procedure will derive the Law of
Sines from the loop closure equations. Let’s say we know α , γ and B. If two angles are known
in a triangle, then the other is also known because the sum of all the angles must be π radians
(180 degrees).
1. Start by arranging the loop closure equations to eliminate C by moving all terms with C on
the left hand side.
x:
y:
23
2. Divide the x equation into the y equation:
y
:
x
3. Eliminate the unknown length and cross multiply:
4. Arrange so B sin α is the only term on the left hand side:
5. Recall that β = π − α − γ . Take the sine of both angles and recall a trigonometric identity:
sin β
sin β
sin β
sin β
=
=
=
=
sin(π − α − γ )
sin(π ) cos(α + γ ) − cos(π ) sin(α + γ )
0 − (−1) sin(α + γ )
sin(α ) cos(γ ) + cos(α ) sin(γ )
6. Notice the right hand side of the equation matches the term from the loop closure equations.
Make the appropriate substitution:
B sin α = A sin β
This is the law of sines. A similar formulation will give the familiar ratios:
sin β
sin γ
sin α
=
=
A
B
C
(3.1)
3.1.3. Inverted Slider Crank
An Inverted Slider-Crank is a mechanism that most used as an actuator. For example, a hydraulic
or pneumatic cylinder can be modeled as an Inverted Slider Crank. Fig. shows a photo of an
electric linear actuator and how it is modeled as an inverted Slider-Crank.
This section describes a simple mechanism to understand the concepts of position analysis.
Many texts will label the axis as real and imaginary instead of x and y. Either is valid.
See
Cleghorn’s
Section
4.3.2
24
(a) Retracted
(b) Extended
Figure 3.1.: A pneumatic cylinder is an example of an inverted slider-crank where the stroke and
angle can change.
D
A
y
r
θ3
θ2
%C
d
%B
x
Figure 3.2.: Inverted slider crank. Let a be the fixed length from A to C (the crank length).
What is the mobility of the inverted slider crank in Fig. 3.2?
Another way to set up the loop closure equations is to find two different paths to the same point
in the mechanism. This formulation may be a little easier depending on how the angles are
defined. Consider two different loop closure formulations shown in Fig. 3.3 on the next page.
Both formulations describe the same physical system, so they should ultimately produce the
same solution.
25
26
y
~a
~r
θ3
θ2
x
d~
(a) ~a +~r + d~ = 0
y
~a
~r
θ3
θ2
x
d~
(b) ~a = d~ +~r
Figure 3.3.: Loop closure formulations for the inverted slider crank shown in Fig. 3.2.
For the inverted slider crank shown in Fig. 3.2, the following variables are known: a = 0.15 m,
d = 0.20 m and θ2 = 35◦ . The goal of the position analysis is to find r and θ3 .
1. Break the vectors into components
a) For the loop closure equation from Fig. 3.3a on the preceding page
x:
y:
b) For the loop closure equation from Fig. 3.3b on the previous page
x:
y:
2. Rearrange the loop closure equations to have all terms of θ3 on one side. Notice that either
formulation will give the same results.
x:
y:
3. Square each equation and add them together to eliminate θ3 :
4. Expand the squared terms:
5. Solve for r:
6. Divide the equations in Step 2 to eliminate r:
7. Compute the arctangent and adjust for the correct quadrant. Use the atan2 function.
The above example is also implemented in Mathematica and the Mathematica notebook can be
downloaded from
http://personal.utulsa.edu/~jeremy-daily/ME3212/InvertedSliderCrank.
nb
27
3.1.4. Offset Slider Crank
Given:
r2 =
r3 =
e=
θ2 =
Find:
θ3 and xB
1. Write the Loop equations:
2. Break into components
a) x :
b) y :
3. Rearrange y equation to solve for θ3
4. Substitute into x expression and solve for xB
Note: For a standard slider crank. . .
28
29
3.1.5. Four Bar Mechanism
The four bar mechanism is a very versatile device and is the building block for many other
mechanisms. A four bar linkage in conjunction with a slider crank is frequently employed in
lifting mechanisms and construction equipment.
Consider the crank rocker in Fig. 3.4 where r1 = 10 cm, r2 = 4 cm, r3 = 12 cm, r4 = 8 cm, and
θ2 = 120◦ . The goal is to find θ3 , θ4 , and the transmission angle.
B
r3
θ3
r4
A
y
r2
θ2
%O
2
r1
%O
θ4
x
4
Figure 3.4.: Crank rocker four-bar mechanism.
1. Consider the triangles formed by drawing a line from A to O4 and label it s.
2. Use the Law of Cosines (LOC) to determine s
3. Determine β :
4. Solve for ψ using LOC
5. Solve for λ using LOC
See
Cleghorn’s
Section
4.3.3
6. Solve for θ3
7. Solve for θ4
8. Determine the transmission angle, γ using LOC.
3.1.5.1. Open Closure
30
3.1.5.2. Cross Closure
See course website for a Mathematica notebook with the solutions to this example.
http://personal.utulsa.edu/~jeremy-daily/ME3212/4barPositions.nb
31
3.2. Coupler Curves
32
3.2. Coupler Curves
The goal of a coupler curve analysis is to determine the locus of some arbitrary point P on a
mechanism.
Physically a mechanism can have many different shapes for a single skeleton diagram.
Let’s add onto the example of Section 3.1.4:
Point P is. . .
Analysis Steps:
1. Write down loop closure equations and solve for θ3 .
2. Write the equations for the vector describing point P.
See
Cleghorn’s
Section
4.2
3.2. Coupler Curves
3.2.1. Matlab Implementation
A computer is useful to plot the curves with θ2 as the independent variable. Make sure both axis
have the same scale to get the true shape of the coupler curve. Below is some example Matlab
code to generate the coupler curve.
1 %ME 3 2 1 2 : Mechanisms
%In −c l a s s Example
3 %O f f s e t s l i d e r cr a n k −d i r e c t s o l u t i o n
5 c l c %c l e a r s c r e e n
c l e a r a l l %c l e a r memory
7 c l o s e a l l %c l o s e a l l f i g u r e windows
9 %i n p u t known v a l u e s
r2 =0.5
11 r 3 = 1 . 2
r4 =1.3
13 e = 0 . 3
15 %d e f i n e t h e t a 2 f r o m 0 t o 360 i n 5 d e g r e e i n c r e m e n t s u s i n g r a d i a n s
t h e t a 2 =[0: pi / 3 6 : 2 ∗ pi ] ;
17
%s o l v e f o r t h e t a 3
19 t h e t a 3 = a s i n ( ( e+ r 2 ∗ s i n ( t h e t a 2 ) ) / r 3 ) ;
21 %s o l v e f o r xB
xB= r 2 ∗ c o s ( t h e t a 2 )+ r 3 ∗ c o s ( t h e t a 3 ) ;
23 yB= z e r o s ( s i z e ( xB ) ) ; %T h i s i s n e e d e d f o r t h e c o u p l e r c u r v e
25 %D e t e r m i n e C o u p l e r c u r v e
xp =
r2 ∗ cos ( t h e t a 2 ) − r4 ∗ cos ( t h e t a 3 ) ;
27 yp = e + r 2 ∗ s i n ( t h e t a 2 ) + r 4 ∗ s i n ( t h e t a 3 ) ;
29 f o r n = 1 : l e n g t h ( t h e t a 2 ) %I t e r a t e t h r o u g h t h e v a l u e s o f t h e t a 2
31
33
%d e f i n e k e y p o i n t s on mechanism i n a
%f a s h i o n t o p l o t t h e c o n f i g u r a t i o n
p t s x = [ 0 , r 2 ∗ c o s ( t h e t a 2 ( n ) ) , xB ( n ) , xp ( n ) ] ;
p t s y =[ e , e+ r 2 ∗ s i n ( t h e t a 2 ( n ) ) , 0 , yp ( n ) ] ;
35
%p l o t t h e c o u p l e r c u r v e p o i n t s and one mechanism o r i e n t i a t i o n
33
3.2. Coupler Curves
37
F= f i g u r e ( 1 ) ; %open new f i g u r e window 1
39
p l o t ( xp , yp , ’ . ’ , p t s x , p t s y , ’−o ’ , xB , yB , ’ y ’ )
41
%m o d i f y t h e p l o t
axis equal
x l a b e l ( ’ x−p o s i t i o n (m) ’ )
y l a b e l ( ’ y−p o s i t i o n (m) ’ )
t i t l e ( [ ’ C o u p l e r Curve : S l i d e r c r a n k shown w i t h ’ . . .
num2str ( t h e t a 2 ( n ) ∗ 1 8 0 / p i ) ’ d e g r e e i n p u t ’ ] )
i f t h e t a 2 ( n )== p i / 3 %f i n d t h e p o s i t i o n when t h e t a 2 i s 60 d e g r e e s
s a v e a s ( F , [ ’ C o u p l e r C u r v e _ ’ num2str ( t h e t a 2 ( n ) ∗ 1 8 0 / p i ) ’ . p d f ’ ] )
end
43
45
47
49
51
53
%C a p t u r e t h e c u r r e n t p l o t t o make an a n i m a t i o n
s e t ( gca , ’ P l o t B o x A s p e c t R a t i o ’ , [ 5 1 2 384 7 2 0 ] )
M( n )= g e t f r a m e ( F , [ 0 0 512 3 8 4 ] ) ;
end
55 %C r e a t e a m o vi e f i l e :
m o v i e 2 a v i (M, ’ O f f s e t S l i d e r C r a n k . a v i ’ )
57
F= f i g u r e %open new f i g u r e window
59 p l o t ( t h e t a 2 ∗ 1 8 0 / pi , xB , ’− ’ )
x l a b e l ( ’ Crank Angle , \ t h e t a _ 2 ( deg ) ’ )
61 y l a b e l ( ’ S l i d e D i s t a n c e , x_B (m) ’ )
t i t l e ( ’ S l i d e p o s i t i o n as a f u n c t i o n of i n p u t angle ’ )
63 g r i d on
axis tight
65 s a v e a s ( F , ’ S l i d e P o s i t i o n . p d f ’ )
The output generated by the above code is shown in Fig. 3.5 on the following page.
34
3.2. Coupler Curves
35
Slide position as a function of input angle
Coupler Curve: Slider crank shown with 60 degree input
1.6
1.5
1.5
1.4
1.3
B
Slide Distance, x (m)
y−position (m)
1
0.5
0
1.2
1.1
1
0.9
0.8
−0.5
0.7
−1.5
−1
−0.5
0
x−position (m)
0.5
1
1.5
0
50
100
150
200
Crank Angle, θ (deg)
250
300
350
2
(a)
(b)
Figure 3.5.: Output graph from Matlab code for the coupler curves of an Offset slider crank
mechanism.
3.2.2. Excel Implementation
Consider the following four-bar mechanism where
r1 =
,r2 =
, r3 =
Coupler Curve Equations:
xp =
yp =
Trigonometric Identities:
, r4 =
, r5 =
, r31 =
, θ2 =
.
3.2. Coupler Curves
3.2.2.1. Standard Algebraic Solution
36
3.2. Coupler Curves
37
3.2.2.2. Use Excel Solver
Open Closure
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
B
C
ME!3212:!Mechanisms
Four Bar!Linkage!Analysis
Parameter Value
Units
r1
10 in
r2
3 in
12 in
r3
r4
8 in
theta2
theta3
theta4
D
E
F
Plot!Coordinates
Upper!Leg
Vertex
X
1
0
2
1.5000
3 4.98362767
Radians
Degrees
2.094
120.000
1.000
57.296
2.000
114.592
Vertex
3*
4
Difference
residual
Lower!Leg
X
6.6708
10.0000
G
Y
0
2.5981
12.6957
Y
7.2744
0.0000
1.6872E+00 5.4213E+00
5.6778E+00
14
12
10
8
Upper!Leg
6
Lower!Leg
4
2
0
4
2
0
2
4
6
8
10
12
3.2. Coupler Curves
38
3.2. Coupler Curves
39
A
B
C
ME!3212:!Mechanisms
Parameter Value
Units
10 in
r1
r2
3 in
r3
12 in
r4
8 in
1
2
3
4
5
6
7
8
9
10 theta2
11 theta3
12 theta4
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
4
33
34
D
E
F
Plot!Coordinates
Upper!Leg
Vertex
X
1
0
2
1.5000
3 9.233836209
Radians
Degrees
2.094
120.000
0.464
26.557
1.667
95.496
Vertex
3*
4
Difference
residual
Lower!Leg
X
9.2338
10.0000
3.7786E 06
8.7571E 06
G
Y
0
2.5981
7.9632
Y
7.9632
0.0000
7.8999E 06
9
8
7
6
5
Upper!Leg
4
Lower!Leg
3
2
1
0
2
0
2
4
6
8
10
12
3.2. Coupler Curves
40
A
B
C
ME!3212:!Mechanisms
Parameter Value
Units
r1
10
in
r2
3
in
12
in
r3
r4
8
in
1
2
3
4
5
6
7
8
9
10 theta2
11 theta3
12 theta4
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
4
33
34
Cross Closure
D
E
F
Plot!Coordinates
Upper!Leg
X
Vertex
Radians
Degrees
=2*PI()/3
=DEGREES(B10)
0.4635153062=DEGREES(B11)
1.6667142836=DEGREES(B12)
G
1
2
3
0
=B5*COS(B10)
=F5+B6*COS(B11)
Y
0
=B5*SIN(B10)
=G5+B6*SIN(B11)
3*
4
Lower!Leg
X
=F11+B7*COS(B12)
=B4
Y
=B7*SIN(B12)
0
Vertex
Difference =F6 F10
=G6 G10
residual
=SQRT(F13^2+G13^2)
9
8
7
6
5
Upper!Leg
4
Lower!Leg
3
2
1
0
2
0
2
4
6
8
10
12
3.2. Coupler Curves
41
A
B
C
ME!3212:!Mechanisms
Parameter Value
Units
10 in
r1
r2
3 in
r3
12 in
r4
8 in
1
2
3
4
5
6
7
8
9
10 theta2
11 theta3
12 theta4
13
14
15
16
17
18
19
20
21
22
5
23
24
25
26
27
28
29
30
31
32
33
34
D
Radians
Degrees
2.094
120.000
0.500
28.648
1.000
57.296
E
F
Plot!Coordinates
Upper!Leg
Vertex
X
1
0
2
1.5000
3 9.030990743
Vertex
3*
4
Difference
residual
Lower!Leg
X
14.3224
10.0000
G
Y
0
2.5981
3.1550
Y
6.7318
0.0000
5.2914E+00 3.5767E+00
6.3869E+00
4
2
0
0
5
10
15
20
Upper!Leg
2
Lower!Leg
4
6
8
3.2. Coupler Curves
42
A
B
C
ME!3212:!Mechanisms
Parameter Value
Units
10 in
r1
r2
3 in
r3
12 in
r4
8 in
1
2
3
4
5
6
7
8
9
10 theta2
11 theta3
12 theta4
13
14
15
16
17
18
19
20
21
22
4
23
24
25
26
27
28
29
30
31
32
33
34
D
E
F
Plot!Coordinates
Upper!Leg
Vertex
X
1
0
2
1.5000
3 5.884873205
Radians
Degrees
2.094
120.000
0.908
52.019
2.111
120.957
Vertex
3*
4
Difference
residual
Lower!Leg
X
5.8849
10.0000
1.8606E 06
1.9300E 06
G
Y
0
2.5981
6.8604
Y
6.8604
0.0000
5.1288E 07
4
2
0
2
0
2
4
6
8
10
12
Upper!Leg
2
Lower!Leg
4
6
8
3.2. Coupler Curves
3.2.2.3. SolidWorks Implementation
43
1. An offset slider crank mechanism is driven by rotating the crank. The axis of the slider is
1 inch below the x-axis, the crank is 2.5 inches, and the connecting rod is 7 inches. Solve
for the position of the slider as a function of the crank angle θ2 . Write a Matlab program
that plots the position of the slider for a complete revolution of the crank. Turn in your
hand analysis, Matlab program (the .m file) and a properly labeled plot.
http://personal.utulsa.edu/~jeremy-daily/ME3212/HW2Prob1.avi
2. Using the figure of book problem P4.10, plot the locus of Point C for a complete revolution
of link 2. Also, plot on the same graph the configuration of the mechanism when θ2 = 10◦ .
For now, ignore the questions regarding velocity and acceleration. Be sure to use the command axis equal to plot the configuration of the mechanisms to scale.
3. In book problem P4.18a, plot the locus of point C for a complete revolution of link 2
for the open closure configuration. Also, plot on the same graph the configuration of the
mechanism when θ2 = 30◦ . Draw the coupler as a triangle (△BCD). For now, ignore
the questions regarding velocity and acceleration. A movie showing the positions of the
mechanism is available at:
4. Perform the “Assembly Mates” SolidWorks Tutorial under the Building Models Tutorial.
Turn in a print of your completed part.
44
Due on: __________________________
45
3.4. Newton-Raphson Method
What if you can’t solve for the positions by trigonometry and algebra?
3.4.1. Rocking Slider Crank
Consider the following example:
Given: r2 , xo , R, θ2
Find: r and θ3
Write down the Loop closure equations:
x:
y:
Rearrange to eliminate r:
This equation has just 1 unknown θ3 but no analytical solution. Therefore, we’ll solve using the
Newton-Raphson method.
46
47
The Newton-Raphson method
•
•
•
If we can rewrite an expression such that f (x) = 0, then the NR method can be used to find x.
It is an iterative scheme such that
(n+1)
x
(n)
=x
f (x(n) )
− ′ (n)
f (x )
where
n is
For this example,
f=
f′ =
The algorithm is as follows:
1. Initialize the criteria: let ∆θ3 = 1
(1)
2. Guess a value for θ3
3. Do while |∆θ3 | > 0.000175
(n)
(n)
4. Calculate f (θ3 ) and f ′ (θ3 )
5. ∆θ3 = − f / f ′
(n+1)
6. θ3
(n)
= θ 3 + ∆θ 3
7. Increment n by 1 and loop back to step 3.
8. Continue to complete the solution by solving for
a) r =
or
r=
b) x =
9. Increment θ2 and repeat.
In Matlab, the algorithm for the example looks like the following:
%Newton−Rahpson Method f o r s o l v i n g l o o p c l o s u r e e q u a t i o n s
3 %Dr . Jer em y D a i l y
%
5 clc ; clear all ; close al l
7 %knowns :
r 2 = 1 . 2 %m e t e r s
9 x0 = 1 . 8
R= . 4
11
t h e t a 2 = [ 0 : p i / 1 0 1 : 4 . ∗ p i ] ; %Two c y c l e s o f t h e c r a n k
13
t h e t a 3 = 0 ; %g u e s s a s o l u t i o n f o r f i r s t i t e r a t i o n
15 f o r n = 1 : l e n g t h ( t h e t a 2 )
d e l t a t h e t a 3 = 1 ; %s e t t h e l o o p c o n d i t i o n a l
17
%g u e s s t h e same s o l u t i o n a s t h e t i m e b e f o r e
19
w h i l e norm ( d e l t a t h e t a 3 ) > 0 . 0 0 0 1 7 5 %0 . 0 1 d e g r e e s
f = ta n ( t h e t a 3 ( n ) ) . ∗ ( x0+R∗ t h e t a 3 ( n)− r 2 ∗ c o s ( t h e t a 2 ( n ) ) ) . . .
−r 2 ∗ s i n ( t h e t a 2 ( n ) ) ;
d f d t h e t a 3 = ( 1 . / ( c o s ( t h e t a 3 ( n ) ) . ^ 2 ) ) . ∗ ( x0+R∗ t h e t a 3 ( n ) . . .
−r 2 ∗ c o s ( t h e t a 2 ( n ) ) ) + R∗ c o s ( t h e t a 3 ( n ) ) ;
d e l t a t h e t a 3 =−f / d f d t h e t a 3 ;
%u p d a t e t h e s o l u t i o n u n t i l i t i s w i t h i n t o l e r a n c e
t h e t a 3 ( n )= t h e t a 3 ( n )+ d e l t a t h e t a 3 ;
end
t h e t a 3 ( n +1)= t h e t a 3 ( n ) ;
21
23
25
27
29
end
31
%S i n c e t h e l a s t command i n t h e l o o p c r e a t e d an e x t r a t h e t a 3 e n t r y
33 %we m u s t remove i t by a s s i g n i n g i t t o t h e em p t y s e t
theta3 (n +1)=[];
35
%S o l v e f o r t h e r e m a i n i n g unkowns
37 r _ s i n e s = r 2 ∗ s i n ( t h e t a 2 ) . / s i n ( t h e t a 3 ) ; %law o f s i n e s
r _ c o s i n e s = s q r t ( r 2 . ^ 2 + ( x0+R∗ t h e t a 3 ) . ^ 2 . . .
39
− 2∗ r 2 . ∗ ( x0+R . ∗ t h e t a 3 ) . ∗ c o s ( t h e t a 2 ) ) ; %law o f c o s i n e s
x=R∗ t h e t a 3 ;
48
49
41 p l o t ( t h e t a 2 , t h e t a 3 , ’ : ’ , t h e t a 2 , r _ s i n e s , ’− ’ , t h e t a 2 , r _ c o s i n e s , . . .
’−− ’ , t h e t a 2 , x , ’ −. ’ )
43 x l a b e l ( ’ \ t h e t a _ 2 ( r a d i a n s ) ’ )
ylabel ( ’ Position variables ’ )
45 l e g e n d ( ’ \ t h e t a _ 3 ( r a d ) ’ , ’ r : s i n e s (m) ’ , ’ r : c o s i n e s (m) ’ , ’ x (m) ’ )
The graph generated by the code is shown in Fig. 3.6.
3.5
θ3 (rad)
r:sines (m)
r:cosines (m)
x (m)
3
2.5
Position variables
2
1.5
1
0.5
0
−0.5
−1
−1.5
0
2
4
6
θ2 (radians)
8
10
Figure 3.6.: Output of Newton-Raphson example. What are the spikes from?
12
14
How does Newton-Raphson work?
The N-R method requires a good first guess or:
•
•
How to make a good initial guess:
•
•
50
1. Use the Newton-Raphson method to solve for θ3 by hand for the four-bar linkage shown
in Fig. 3.4 on page 29. Compare the result from the N-R method to the analytical solution
for θ2 = 2π /3 radians (120 degrees). Then, create a Matlab program that implements the
N-R method and plots θ3 against θ2 for every 5 degrees of a complete revolution of the
crank. Plot the solution from the N-R method with a dot and the solution from the analytical method with a square. Generate two plots: one for the open closure configuration and
one for the cross closure configuration. The graphs you generate should look like the ones
in Fig. 3.7 on page 54. The Matlab code used to plot the results is as follows:
1 F= f i g u r e ( 1 ) ;
p l o t ( t h e t a 2 ∗ 1 8 0 / pi , t h e t a 3 ∗ 1 8 0 / pi , ’ . ’ , . . .
3
t h e t a 2 ∗ 1 8 0 / pi , t h e t a 3 _ o p e n ∗ 1 8 0 / pi , ’ s ’ )
%No t e : t h e t a 3 _ o p e n m u s t be d e f i n e d i n o r d e r f o r t h i s t o work
5 l e g e n d ( ’ Newton−Raphson ’ , ’ A n a l y t i c a l ’ , ’ L o c a t i o n ’ , ’ N o rt h Wes t ’ )
x l a b e l ( ’ I n p u t Crank Angle , \ t h e t a _ 2 [ deg ] ’ )
7 y l a b e l ( ’ C o u p l e r Angle , \ t h e t a _ 3 [ deg ] ’ )
axis tight
9 s e t ( gca , ’ XTick ’ , [ 0 : 3 0 : 3 6 0 ] )
g r i d on
11 t i t l e ( ’ C o u p l e r a n g l e i n t h e open c l o s u r e c o n f i g u r a t i o n ’ )
51
52
2. Eccentric Cam Analysis (See figure below):
a) Plot the locus of point P for a complete turn of the cam when r1 = 100 mm, r2 = 150
mm. e = 20 mm, and R = 40 mm. Hint: use N-R to determine θ2 .
b) Plot the y position of point P as a function of θ3 .
c) Determine the maximum and minimum of θ2 by hand.
An animation of the motion of this mechanism is available at
Hints:
r and R form a right angle.
R and e do not form a right angle (the angle changes).
Due on:___________________
P
r2
R
r
y
e
%
O2
O3
θ2
%
θ3
x
r1
3. Perform the “Advanced Design” SolidWorks Tutorial. Turn in a print of your completed part.
53
54
Coupler angle in the open closure configuration
65
Newton−Raphson
Analytical
60
55
3
Coupler Angle, θ [deg]
50
45
40
35
30
25
20
0
30
60
90
120
150
180
210
240
Input Crank Angle, θ2 [deg]
270
300
330
360
(a) Open Closure Configuration
Coupler angle in the cross closure configuration
Newton−Raphson
Analytical
−20
−25
Coupler Angle, θ3 [deg]
−30
−35
−40
−45
−50
−55
−60
−65
0
30
60
90
120
150
180
210
240
Input Crank Angle, θ [deg]
270
300
330
360
2
(b) Cross Closure Configuration
Figure 3.7.: Graphs for the Newton-Raphson and analytical solution for θ3
3.6. Multi Loop Mechanisms
55
3.6. Multi Loop Mechanisms
Multi loop mechanisms are analyzed by constructing the loop closure equations for all the elementary loops. Open chains can also be considered.
Consider the following example:
Calculate the mobility using the Kutzbach Criteria:
Determine the known and unknown variables:
There are four equations after breaking up the loop closure equations:
x1 :
y1 :
x2 :
y2 :
See
Cleghorn’s
Section
4.2
3.7. Toggle and Limit Positions
56
3.7. Toggle and Limit Positions
Mechanical Advantage
Consider an offset slider crank in the “dead” center positions.
Top center:
Bottom center:
The criteria to find a toggle position are:
Loop Closure Equations:
x:
y:
Substitute θ2 = θ3 into the y equation:
See
Cleghorn’s
Section
2.6
3.8. Transmission Angle
57
Substitute θ2 = θ3 + π into the y equation:
3.8. Transmission Angle
For a slider crank, the transmission angle is γ .
γ=
Goal: find the maximum and minimum γ and corresponding θ2 .
Rewrite Loop equations:
x:
y:
but,
and,
so
x:
y:
To find extrema, find values of θ2 when
x′ :
y′ :
Consider only the numerator:
θ2 =
So the transmission angles are:
dγ
d θ2
= 0. Use implicit differentiation:
See
Cleghorn’s
Section
2.7
1. Using the offset slider-crank mechanism shown in Fig. 4.4 of Cleghorn’s text, determine
the extreme values of the transmission angle, γ . Hint: Write an expression for γ in terms
of θ2 , differentiate with respect to θ2 , and let d γ /d θ2 = 0 for extremes.
2. For a four bar mechanism where r1 = 400 mm, r2 = 200 mm, r3 = 500 mm, and r4 = 400
mm,
a) Determine θ2 , θ3 , θ4 , and γ for both limit positions.
b) Draw the mechanisms in each limit position to scale.
c) Determine the total rocking angle (∆θ4 ).
Ans: 78◦
3. Perform the “Animation” SolidWorks Tutorial. Turn in a print of your completed part.
Due on: ________________
58
4. Mechanism Synthesis
The design or creation of a mechanism to achieve the desired motion.
Type Synthesis:
Number Synthesis:
Dimensional Synthesis:
Classical Analysis:
4.1. Geometric Constraint Programming
Everyone should download their own copy of:
Kinzel et al., “Kinematic synthesis for finitely separated positions using geometric constraint
programming.” Journal of Mechanical Design (2006)
Meet in the Computer Lab (L1) to see the following demonstration:
Goal: pick up an object with a scoop and dump it at a point 3” higher and 4” over.
Approach: design a four bar mechanism that has a coupler that will follow 5 precision points and
directions.
To synthesize a mechanism that satisfies these constraints, follow these instructions:
1. Create a new Part in SolidWorks.
2. Change the document properties to the IPS system. This is found under the Tools, Options,
Document Properties menu.
3. Under the System Options tab (Tools ⊲ Options), click on the Relations/Snaps branch and
make sure the Automatic Relations box is unchecked.
60
4. Create a new sketch on any plane.
61
5. Create 5 3-point arcs. They do not have to be the same size yet and placement is arbitrary
at this time. These arcs will represent the scoop.
62
6. Make all the arcs equal by selecting them while holding the Shift key. Then click on the
Equal button under the Add Relations Pane.
63
7. Create centerlines an connect each end of the arc.
8. Make the center of the arc coincident with the line.
64
9. Position and dimension the arcs in appropriate locations. The position will describe locations along the path of travel. Therefore, the arc must be facing up at the bottom location
and tilted at the drop off location. The other points are used to guide the path of travel.
Start by making the right edge of one of the arcs coincident with the origin and fixing its
location.
65
10. Draw congruent triangles from each of the arc endpoints. To ensure congruency, make all
corresponding legs of the triangle equal. The tip of the triangle will represent the connection to one of the links.
66
11. Draw a perimeter circle through three of the five tips of the triangles.
12. Make the points coincident with the circle.
67
13. Drag the other two points close to the circle and make them coincident. All five points
should be coincident with the circle.
68
14. Create another set of congruent triangles to represent the other connecting point for the
coupler.
69
15. Draw the other circle and maneuver the locations of the arc and the triangles to get a compact package. This may be difficult and you may have to delete some fixed constraints.
70
16. Once the position is where you like, Select All and make a Block (Tools ⊲ Make Block).
71
17. Draw the frame (link 1) by connecting the two circle centers. Fix the ends. The draw
moving links 2 and 4. Trace the coupler. Add dimensions to keep the lengths fixed.
72
18. Drag the mechanism to check its motion.
All dimensions should be determined and the mechanism synthesis is completed. The mechanism can be actuated by a motor for continuous rotation or by a cylinder or cam to rock back and
forth.
73
Construct the headlight cover in Example 11.3 of Cleghorn’s book using SolidWorks. Demonstrate that it works by turning in three prints from SolidWorks: the open position, an intermediate
position, and the closed position. Clearly identify all relevant details of this design. These details
include: link lengths, rotation positions, and coupler angles.
74
5. Velocity Analysis
5.1. Vector Operations
5.1.1. Dot Product
5.1.2. Cross Product
5.1.3. Derivatives of Vector Products
Draw two vectors. The first vector, ~r1 , describes the position of a point at time t1 . The second
vector,~r2 , describes the position at time t2 . Draw tangent and normal unit vectors at each position.
~
Also, draw ∆r.
To determine velocity, take the limit:
Relate inertial unit vectors to rotating unit vectors:
Determine the velocity based on the product rule:
But~r can also be written in terms of a tangent-normal system:
76
Take the derivative with respect to time to get the velocity:
77
5.3. Graphical Analysis
78
Start with a scale drawing of the mechanism.
8
See
Cleghorn’s
Example
3.2
Example of an Inverted Slider Crank.
7
6
5
4
3
2
1
0
Relative velocity equivalence equations:
~vP =~vP/A +~vA =~vP/B +~vB
~vP/B =
with numbers:
~vP/A =
with numbers:
Break into components:
~vP =
Draw the velocity polygon:
1. Draw Construction Lines
2. Scale Known Velocity
3. Draw Perpendiculars
See
Cleghorn’s
Section
2.2
79
4. Measure Magnitudes
x
0v
reference
Goal: Determine the angular velocity of link 4.
5.3.2. Four-Bar Mechanism
Example of a four-bar mechanism: Determine the angular velocities of link 3, link 4, and the
velocity of point C.
where r1 = 600 mm, r2 = O2 A = 140 mm, r3 = 690 mm, r4 = 400 mm, r5 = 200 mm, r6 = 200
mm, θ2 = 240◦ , θ3 = 44◦ , θ4 = 116◦ , θ̇2 = ω2 = 50 rad/sec (constant).
See
Cleghorn’s
Section
3.3
80
B
C
y
r5
r6
r4
r3
θ4
θ2
O2%
r1
θ̇2
O4%
x
θ3
A
Figure 5.1.: Four-Bar Mechanism Example
Use the space allocated in Fig. 5.2 to construct the velocity polygon for this example. Recall
velocity equivalence:
~vB =~vA +~vB/A =~vO4 +~vB/O4
1. Determine the angles for the lines of action.
2. Draw v~A
3. Draw a construction line at the angle of vB through Ov .
4. Construct a line through vA perpendicular to r3 to show the line of action of vB/A
5. Find the intersection of the vB construction line and the vB/A construction line. Draw the
vector representing ~vB/A and measure its length.
6. Determine angular velocities
7. Velocity of C:
8. Draw construction line for ~vC/B perpendicular to BC and going through vB
9. Draw construction line for ~vC/A
10. Check magnitudes
11. Measure to determine ~vC
81
5.4. Analytical Analysis
82
x
0v
reference
Figure 5.2.: Workspace for graphical velocity determination for
the four-bar mechanism in Fig. 5.1.
Notes on velocity images:
• The velocity image has a triangle...
• The ratio...
• If no angular velocity, then...
• The point 0v ...
• Determining absolute velocity ...
Once the position vectors are known, then they can be differentiated with time to get the velocity
equations. A couple examples show this technique.
83
Determine θ˙4 and ṙ in the following inverted slider crank mechanism:
C
A
r
y
θ̇2
%O
θ4
4
O2
%
r1
r1 =
r2 =
r=
θ2 =
θ4 =
θ̇2 =
where
1. Loop Equations:
a) x :
b) y :
2. Differentiate with respect to time:
a)
b)
r2
θ2
x
3. Arrange in matrix form:
4. Solve the linear equation.
a) On TI-89:
b) In Matlab:
c) By Hand:
Compare graphical solution to analytical solution:
Variable
Graphical
Analytical
We can write a computer program to solve the above system for various angles of θ2 and a
constant angular velocity. An example in Matlab is as follows:
%I n v e r t e d S l i d e r Crank
3 %F i n d i n g t h e v e l o c i t i e s o f l i n k s
%Dr . Jer em y D a i l y
5
%c l o s e an i n i t i a l i z e s y s t e m
7 clc
84
clear all
9 close all
11 %I n v e r t e d s l i d e r c r a n k mechanism
%knowns :
13 r 1 =45
r 2 =25
15 omega2 =20
17 t h e t a 2 = l i n s p a c e ( 0 , 2 ∗ pi , 1 5 0 ) ;
19 %p o s i t i o n s o l u t i o n :
r = s q r t ( r 1 ^2+ r 2 ^2−2∗ r 1 ∗ r 2 ∗ c o s ( pi −t h e t a 2 ) ) ;
21 t h e t a 4 = a c o s ( ( r . ^ 2 + r 1 . ^ 2 − r 2 . ^ 2 ) . / ( 2 ∗ r ∗ r 1 ) ) ;
t h e t a 4 = a ta n2 ( r 2 ∗ s i n ( t h e t a 2 ) , r 1 + r 2 ∗ c o s ( t h e t a 2 ) ) ;
23
%v e l o c i t y a n a l y s i s
25 f o r i = 1 : l e n g t h ( t h e t a 2 )
A=[ c o s ( t h e t a 4 ( i ) ) −r ∗ s i n ( t h e t a 4 ( i ) ) ;
27
sin ( theta4 ( i ))
r ∗ cos ( t h e t a 4 ( i ) ) ] ;
C=[ −r 2 ∗ omega2 ∗ s i n ( t h e t a 2 ( i ) )
r 2 ∗ omega2 ∗ c o s ( t h e t a 2 ( i ) ) ] ’ ;
29
v=A \ C ;
r d o t ( i )= v ( 1 ) ;
31
omega4 ( i )= v ( 2 ) ;
end
33
%P l o t t h e r e s u l t s
35 F= f i g u r e ( 1 )
subplot (2 ,2 ,1)
37 p l o t ( t h e t a 2 ∗ 1 8 0 / pi , t h e t a 4 )
x l a b e l ( ’ \ t h e t a _ 2 ( deg ) ’ )
39 y l a b e l ( ’ \ t h e t a _ 4 ( r a d ) ’ )
g r i d on
41 a x i s t i g h t
%s e t s t h e c u r r e n t a x i s t o h a ve t i c k s e v e r y 45 deg
43 s e t ( gca , ’ X t i c k ’ , [ 0 : 4 5 : 3 6 0 ] )
45 s u b p l o t ( 2 , 2 , 2 )
p l o t ( t h e t a 2 ∗ 1 8 0 / pi , r )
47 x l a b e l ( ’ \ t h e t a _ 2 ( deg ) ’ )
y l a b e l ( ’ s l i d e r p o s i t i o n (mm) ’ )
49 g r i d on
85
axis tight
51 s e t s t h e c u r r e n t a x i s t o h av e t i c k s e v e r y 45 deg
s e t ( gca , ’ X t i c k ’ , [ 0 : 4 5 : 3 6 0 ] )
53
subplot (2 ,2 ,3)
55 p l o t ( t h e t a 2 ∗ 1 8 0 / pi , omega4 )
x l a b e l ( ’ \ t h e t a _ 2 ( deg ) ’ )
57 y l a b e l ( ’ \ omega_4 ( r a d / s ) ’ )
g r i d on
s e t ( gca , ’ X t i c k ’ , [ 0 : 4 5 : 3 6 0 ] )
61
subplot (2 ,2 ,4)
63 p l o t ( t h e t a 2 ∗ 1 8 0 / pi , r d o t )
x l a b e l ( ’ \ t h e t a _ 2 ( deg ) ’ )
65 y l a b e l ( ’ s l i d e r v e l o c i t y (mm/ s ) ’ )
g r i d on
s e t ( gca , ’ X t i c k ’ , [ 0 : 4 5 : 3 6 0 ] )
69
saveas (F , ’ In v ert ed Sl i d erCran k V el o c i t y . pdf ’ )
86
87
70
slider position (mm)
θ (rad)
0.5
4
0
60
50
40
30
−0.5
0
45
90
135 180 225 270 315 360
θ (deg)
0
45
90
2
2
6
slider velocity (mm/s)
400
4
2
0
4
ω (rad/s)
135 180 225 270 315 360
θ (deg)
−2
−4
200
0
−200
−400
−6
0
45
90
135 180 225 270 315 360
θ2 (deg)
0
45
90
135 180 225 270 315 360
θ2 (deg)
Figure 5.3.: Output of the velocity analysis program of an inverted slider crank.
5.4.2. Four Bar Mechanism
Recall the four-bar mechanism from Section 5.3.2 on page 79.
1. To find θ˙3 and θ̇4 by starting with the loop closure equations:
a) x :
b) y :
2. Differentiate each equation with respect to time:
a) ẋ :
b) ẏ
3. Arrange in matrix form:
4. Solve the linear equation.
a) On TI-89:
b) In Matlab:
c) By Hand:
For the velocity of point C:
5. Write a vector equation from O2 to C:
6. Break in to x and y components.
a) x :
b) y :
7. Differentiate with respect to time.
vc,x =
vc,y =
8. Substitute the appropriate values.
88
89
90
This is a long homework problem set and should be started early.
1. For the slider-crank drawn below, solve for the angular velocity of the slider and the velocity of B when r1 = 125 mm, r2 = 75 mm, r3 = AB = 400 mm, θ2 = 30◦ , θ̇2 = −60 rad/sec.
Solve analytically and graphically by hand.
y
A
r
r2
x
θ̇2
θ2
r3
θ3
O2
O4
r1
Ans: ṙ = 1452.4 mm/s and θ˙3 = −22 rad/s
2. For the four-bar linkage below, solve for the angular velocities of links 3 and 4. Also
determine the magnitude and direction of the velocity of point C. Solve this problem analytically and check your answer using a CAD program. Given: r1 = 100 mm, r2 = 150
mm, r3 = AB = 250 mm, r4 = 250 mm, r5 = BC = 150 mm, r6 = AC = 300, θ2 = 105◦ ,
and θ˙2 = 56 rad/s (ccw). Note: you will have to do a position analysis first.
B
91
B
r3
r5
θ3 φ
A
r6
r2
C
r4
y
θ̇2
θ2
O4
O2%
%
θ4
x
r1
Ans: θ̇4 = 45.51 rad/s, |vc | = 9.025 m/s and ∠vc = 137.9◦
3. Determine the velocities of points B, C, and D along with the angular velocity of the connecting rod (△ABC) in a double slider crank where r2 = 2 in., r3 = 10 in, CA = 4 in,
CD = 8 in., θ2 = −120◦ , θ˙2 = ω2 = 42 rad/s cw, and O2 is at (−3, 0).
10
9
8
7
6
5
4
3
2
1
0
Ans: |vc | = 67.9 in/s and ∠vc = 154◦
92
6. Acceleration Analysis
Graphical techniques are discussed in Cleghorn’s Chapter 3. This section is focused on analytical
techniques. Simply put, the acceleration is the time derivative of the velocity equations.
~ = m~a
Accelerations are needed to determine forces: F
6.1. Accelerations in Four-bar Mechanisms
Start with velocity equations from the example in Section 5.4.2 on page 87.
x:
y:
Differentiate with respect to time to get accelerations:
x:
y:
Arrange in matrix form:
θ̈2 =
Substitute numbers:
See
Cleghorn’s
Section
4.4.2
6.2. Inverted Slider Crank
Solve for θ̈3 and θ̈4 on a TI-85:
For the acceleration of point C:
Write down the velocity equations.
vc,x =
vc,y =
Differentiate for acceleration:
ac,x =
ac,y =
Substitute in the numbers:
Continue with the example from Section 5.4.1 on page 83.
The velocity loop equations:
vx :
vy :
Differentiate with respect to time to determine acceleration:
94
Arrange in matrix form:
Substitute the numbers with θ̈2 =
Coriolis acceleration:
95
96
1. For the inverted slider crank shown below,
a) Differentiate the velocity equations to get the acceleration equations and put into
matrix form.
b) Solve for r̈ and θ̈3 .
c) Setup and differentiate the position equations to get the velocity of point B.
d) Finally, solve for the acceleration magnitude and direction of point B.
y
A
r
θ2
a2
θ̇2
a3
O4
O2
x
θ3
d
where a2 = 75 mm, a3 = 400 mm, d = 125 mm, r = 193.6 mm, θ2 = 150◦ , θ3 = 11.17◦ ,
and θ̇2 = 60rad/sec (constant). The position equations are as follows:
a2 cos θ2 + r cos θ3 =d
a2 sin θ2 − r sin θ3 =0
and the velocity equations are:
−a2 θ̇2 sin θ2 + ṙ cos θ3 − rθ̇3 sin θ3 =0
a2 θ̇2 cos θ2 − ṙ sin θ3 − rθ̇3 cos θ3 =0
Ans: θ̈3 = −120 rad/s/s and r̈ = −161.87 m/s/s
B
97
2. For the 4-bar linkage shown below,
a) Differentiate the velocity equations to get the acceleration equations and put into
matrix form.
b) Solve for θ̈3 and θ̈4 .
c) Setup and differentiate the position equations to get the velocity of point C.
d) Finally, setup the equations for the velocity of point C and solve for the acceleration
of point C.
B
a3
a5
θ3
A
φ
a6
a2
C
a4
y
θ̇2
θ2
O4
O2%
%
θ4
x
d
where a2 = 150 mm, a3 = 250 mm, a4 = 250 mm, d = 100 mm, a5 = 150 mm, a6 = 300
mm, θ2 = 105◦ , θ3 = 20.13◦, θ4 = 67.47◦, φ = 29.93◦, θ̇2 = ω2 = 56 rad/sec (constant),
θ̇3 = 27.84 rad/s, and θ˙4 = 45.51 rad/s. The position equations are as follows:
a2 cos θ2 + a3 cos θ3 − a4 cos θ4 =d
a2 sin θ2 + a3 sin θ3 − a4 sin θ4 =0
and the velocity equations are:
−a2 θ̇2 sin θ2 − a3 θ̇3 sin θ3 + a4 θ̇4 sin θ4 =0
a2 θ̇2 cos θ2 + a3 θ̇3 cos θ3 − a4 θ̇4 cos θ4 =0
Ans: θ̈3 = −73.02 rad/s/s and θ̈4 = −625.8 rad/s/s
3. Perform the “SolidWorks Motion” tutorial. Turn in a print of the time history of the contact
force.
98
7. Cams
A cam transmits rotary motion to another link by direct contact.
Vocabulary:
• Cam:
• Follower:
• Cam Profile:
• Trace Point:
• Pitch Curve:
• Base Circle:
• Prime Circle:
• Pressure Angle:
• Rise:
• Dwell:
• Fall:
Cam systems are:
•
•
•
See
Cleghorn’s
Figure 7.1
for cam
types.
7.1. Types of Cam Followers
•
•
•
7.1.1. Flat Faced Radial
8
7
6
5
4
3
2
1
0
100
7.1.2. Offset Roller Follower
8
7
6
5
4
3
2
1
0
7.1.3. Barrel Cam with Roller Follower
8
7
6
5
4
3
2
1
0
101
7.2. Cam Follower Motion
7.1.4. Heavy Truck Brake Cams (S-Cams)
8
7
6
5
4
3
2
1
0
7.2. Cam Follower Motion
7.2.1. Displacement
y = f (θ )
7.2.2. Velocity
v = ẏ =
7.2.3. Acceleration
a = ÿ =
7.2.4. Jerk
...
jerk = y =
102
7.3. Cam Follower Profiles
103
Determine the geometric properties of the cam follower motion.
Kinematic properties require knowing ______________________.
7.3.1. Constant Acceleration (Parabolic)
A follower is to have the following motion:
Interval
Initial Angle
Fundamental Equations:
Final Angle
Type of Motion
Initial Displacement
Final Displacement
Parabolic
0
A
Const. Velocity
A
B
Parabolic
B
C
Dwell
C
C
Parabolic
C
D
Parabolic
D
0
See
Cleghorn’s
Section
7.5.2 and
7.5.3
104
f (θ )
60
40
20
θ
30
60
90
120
150
180
210
240
270
300
330
360
f ′ (θ )
θ
30
60
90
120
150
180
210
240
270
300
330
360
f ′′ (θ )
θ
30
60
90
120
150
180
210
240
270
300
330
360
Parabolic Equation Derivations:
105
106
7.3.2. Harmonic Motion
Interval
Initial Angle
Final Angle
Type of Motion
Final Displacement
Harmonic
0
A
Const. Velocity
A
B
Harmonic
B
C
Dwell
C
C
Harmonic
C
0
See
Cleghorn’s
Section
7.5.4
107
f (θ )
60
40
20
θ
30
60
90
120
150
180
210
240
270
300
330
360
f ′ (θ )
θ
30
60
90
120
150
180
210
240
270
300
330
360
f ′′ (θ )
θ
30
60
90
120
150
180
210
240
270
300
330
360
Harmonic Equation Derivations:
108
109
7.3.3. Cycloidal Motion
Interval
Initial Angle
Final Angle
Type of Motion
Final Displacement
Cycloidal
0
A
Dwell
A
A
Cycloidal
A
B
Dwell
B
B
Cycloidal
B
0
See
Cleghorn’s
Section
7.5.5
110
f (θ )
60
40
20
θ
30
60
90
120
150
180
210
240
270
300
330
360
f ′ (θ )
θ
30
60
90
120
150
180
210
240
270
300
330
360
f ′′ (θ )
θ
30
60
90
120
150
180
210
240
270
300
330
360
Cycloidal Equation Derivations:
111
7.4. Cam Design
112
7.4. Cam Design
Given the following table for the cam follower function specification:
Interval
Initial Angle
Final Angle
Type of Motion
Initial Output
Final Output
1
2
3
4
0
90
210
300
90
210
300
360
Harmonic
Dwell
Cycloidal
Parabolic
0
0.4 rad
0.4 rad
0.1 rad
0.4 rad
0.4 rad
0.1 rad
0
The mechanism is a disk cam with a flat face pivoting follower.
From Cleghorn’s Figure 7-19 (top of pg 279):
6
X = 130 cm, Y = 100 cm, e = 50 cm, rb = 70 cm
See
Cleghorn’s
Example
7.1
5
4
3
2
1
0
f (θ ), rad
The follower motion profile is as follows:
0.30
0.20
0.10
θ
30
60
90
120
150
180
210
240
270
300
330
360
7.4. Cam Design
Derivation of polar profile of the disk cam.
1. Determine αo
6
5
4
3
2
1
0
2. Write the Loop closure equations:
a) xc :
b) yc :
3. Take the derivative w.r.t. θ
a) x′c :
b) y′c :
4. Substitute and solve for δc :
113
7.4. Cam Design
5. Determine the point of contact
a) xc =
b) yc =
6. Determine the radial length and orientation
a) rc =
b) γc =
7. Determine the angle with respect to the disk cam:
114
7.4. Cam Design
115
8. Plot the cam profile in polar coordinates:
Cam profile in polar form
90
150
120
60
100
150
30
50
180
0
210
330
240
300
270
Figure 7.1.: Cam profile in polar coordinates. The inner circle shows the base radius and the
outer circle shows the maximum dimensions.
7.4. Cam Design
9. The following program implements this example:
%Mechanisms : ME3212
2 %Dr . Jer em y D a i l y
%D i s k Cam P r o f i l e D e s i g n
4
%T h i s Program t a k e s t h e d e s i r e d cam f o l l o w e r m o t i o n and d e s i g n s
6 %a d i s k cam p r o f i l e f o r a r o t a t i n g f l a t f o l l o w e r . T h i s i s
%s i m i l a r t o e x a m p l e 7 . 1 i n Cl eg h o r n ’ s t e x t . I t i s f o r
8 %a r o t a t i n g f l a t f o l l o w e r cam ( t o p o f pg 2 7 9 ) .
10 c l c %c l e a r s t h e command window
c l e a r a l l %c l e a r s memory
12 c l o s e a l l %c l o s e s a l l f i g u r e windows
14 %C o n s t r u c t cam f o l l o w e r p r o f i l e
16 %Harmonic r i s e f r o m 0 t o 100 d e g r e e s ( e v e r y d e g r e e )
L=.4
18 b e t a =100∗ p i / 1 8 0
t h e t a = [ 0 : p i / 1 8 0 : b e t a ] ; %m u s t be i n r a d i a n s
20 f =L∗(1 − c o s ( p i ∗ t h e t a / b e t a ) ) / 2 ; %f o l l o w e r d i s p l a c e m e n t
f p r i m e = p i ∗L∗ s i n ( p i ∗ t h e t a / b e t a ) / ( 2 ∗ b e t a ) ; %f o l l o w e r m o t i o n s l o p e
22 f d o u b l e p r i m e = p i ^2∗L∗ c o s ( p i ∗ t h e t a / b e t a ) / 2 / b e t a ^ 2 ; %f o l l o w e r a c c e l
24 %d w e l l f r o m 100 t o 160 deg
t h e t a s t a r = [ 0 : 1 : 6 0 ] ∗ pi / 1 8 0 ;
26 s s t a r =L∗ o n e s ( s i z e ( t h e t a s t a r ) ) ;
f p r i m e s t a r =zeros ( si z e ( t h e t a s t a r ) ) ;
28 t h e t a =[ t h e t a t h e t a s t a r + t h e t a ( end ) ] ;
30 %b u i l d t h e f o l l o w e r m o t i o n p r o f i l e s by c o n c a t e n a t i n g f u n c t i o n s
f =[ f s s t a r ] ;
32 f p r i m e =[ f p r i m e f p r i m e s t a r ] ;
f d o u b l e p r i m e =[ f d o u b l e p r i m e z e r o s ( s i z e ( s s t a r ) ) ] ;
34
%C y c l o i d a l F a l l f r o m 160 t o 300 d e g r e e s
36 L2 = . 1
L=L−L2
38 b e t a =140∗ p i / 1 8 0
t h e t a s t a r = [ 0 : p i / 1 8 0 : b e t a ] ; %m u s t be i n r a d i a n s
40 s s t a r =L2+L∗(1 − t h e t a s t a r / b e t a + s i n ( 2 ∗ p i ∗ t h e t a s t a r / b e t a ) / ( 2 ∗ p i ) ) ;
116
7.4. Cam Design
f p r i m e s t a r = −L / b e t a ∗(1 − c o s ( 2 ∗ p i ∗ t h e t a s t a r / b e t a ) ) ;
42 f d o u b l e p r i m e s t a r =−2∗ p i ∗L∗ s i n ( 2 ∗ p i ∗ t h e t a s t a r / b e t a ) / b e t a ^ 2 ;
t h e t a =[ t h e t a t h e t a ( end )+ t h e t a s t a r ] ;
44
%b u i l d t h e f o l l o w e r m o t i o n p r o f i l e s by c o n c a t e n a t i n g f u n c t i o n s
46 f =[ f s s t a r ] ;
f p r i m e =[ f p r i m e f p r i m e s t a r ] ;
48 f d o u b l e p r i m e =[ f d o u b l e p r i m e f d o u b l e p r i m e s t a r ] ;
50 %P a r o b o l i c f a l l f r o m 300 t o 360 d e g r e e s
L=L2
52 b e t a =60∗ p i / 1 8 0
54 %p a r t 1
t h e t a s t a r = [ 0 : p i / 1 8 0 : b e t a / 2 ] ; %m u s t be i n r a d i a n s
56 s s t a r = L∗(1 −2∗( t h e t a s t a r / b e t a ) . ^ 2 ) ;
f p r i m e s t a r =−4∗L∗ t h e t a s t a r / b e t a ^ 2 ;
58 f d o u b l e p r i m e s t a r =−4∗L / b e t a ^2∗ o n e s ( s i z e ( t h e t a s t a r ) ) ;
f =[ f s s t a r ] ;
f d o u b l e p r i m e =[ f d o u b l e p r i m e f d o u b l e p r i m e s t a r ] ;
64
%p a r t 2
66 t h e t a s t a r =[ b e t a / 2 : p i / 1 8 0 : b e t a ] ; %m u s t be i n r a d i a n s
s s t a r = L∗(2 −4∗( t h e t a s t a r / b e t a ) + 2 ∗ ( t h e t a s t a r / b e t a ) . ^ 2 ) ;
68 f p r i m e s t a r =−4∗L / b e t a + 4∗L∗ t h e t a s t a r / b e t a ^ 2 ;
f d o u b l e p r i m e s t a r =4∗L / b e t a ^2∗ o n e s ( s i z e ( t h e t a s t a r ) ) ;
70
%remove r e p e a t e d e l e m e n t t o e n s u r e t h e a r r a y s a r e t h e same s i z e
72 s s t a r ( 1 ) = [ ] ;
fprimestar (1)=[];
74 f d o u b l e p r i m e s t a r ( 1 ) = [ ] ;
f =[ f s s t a r ] ;
f d o u b l e p r i m e =[ f d o u b l e p r i m e f d o u b l e p r i m e s t a r ] ;
80
t h e t a s t a r = [ 0 : pi / 1 8 0 : beta ] ;
82 t h e t a =[ t h e t a t h e t a s t a r + t h e t a ( end ) ] ;
117
7.4. Cam Design
84 %P l o t t h e cam f o l l o w e r p r o f i l e
F= f i g u r e ( 1 )
86 s u b p l o t ( 3 , 1 , 1 )
p l o t ( t h e t a ∗ 1 8 0 / pi , f )
g r i d on
90 x l a b e l ( ’Cam a n g l e \ t h e t a , deg ’ )
ylabel ( ’ f ( t h e t a ) , rad ’ )
92 t i t l e ( ’ K i n e m a t i c c o e f f i c i e n t s f o r Cam F o l l o w e r R o t a t i o n ’ )
subplot (3 ,1 ,2)
94 p l o t ( t h e t a ∗ 1 8 0 / pi , f p r i m e )
axis tight
96 g r i d on
x l a b e l ( ’Cam a n g l e \ t h e t a , deg ’ )
98 y l a b e l ( ’ f ^ \ p r i m e ( \ t h e t a ) ’ )
subplot (3 ,1 ,3)
100 p l o t ( t h e t a ∗ 1 8 0 / pi , f d o u b l e p r i m e )
axis tight
102 g r i d on
x l a b e l ( ’Cam a n g l e \ t h e t a , deg ’ )
104 y l a b e l ( ’ f ^ { \ p r i m e \ p r i m e } ( \ t h e t a ) ’ )
s a v e a s ( F , ’ Cam Fo l l o w erM o t i o n . e p s ’ )
106
%Now t h a t t h e cam f o l l o w e r k i n e m a t i c p r o f i l e s a r e known ,
108 %we can d e s i g n t h e d i s k p r o f i l e t h a t g e n e r a t e s t h a t m o t i o n .
110 %G i ven cam s y s t e m v a l u e s :
r b =70 %cm
112 X=130 %cm
Y=100 %cm
114 d =200 %cm ( t h i s i s t h e l e n g t h o f t h e f o l l o w e r
e =50 %cm
116
%Compute k e y p a r a m e t e r s f o r cam s y s t e m
118 a l p h a 0 = p i /2 − a ta n2 (Y , X)− a c o s ( ( r b +e ) / s q r t (X^2+Y^ 2 ) )
a l p h a = a l p h a 0 + f ; %c o n v e r t f t o r a d i a n s
120 d e l t a c =(X∗ c o s ( a l p h a )−Y∗ s i n ( a l p h a ) ) . / ( 1 + f p r i m e ) ;
122 %d e t e r m i n e t h e p o i n t o f c o n t a c t b e t w e e n t h e cam and t h e f o l l o w e r
xc=X−d e l t a c . ∗ c o s ( a l p h a )− e ∗ s i n ( a l p h a ) ;
124 yc=Y+ d e l t a c . ∗ s i n ( a l p h a )− e ∗ c o s ( a l p h a ) ;
118
7.4. Cam Design
126 r c = s q r t ( xc . ^ 2 + yc . ^ 2 ) ;
gammac= a ta n2 ( yc , xc ) ;
128
%d e t e r m i n e t h e l o c a t i o n o f t h e r a d i a l i n r e l a t i o n t o a
130 %r e f e r e n c e on t h e r o t a t i n g cam :
b e t a =gammac−t h e t a ;
132
134 b i g R a d i u s =max ( r c )
136 F= f i g u r e ( 2 )
p o l a r ( beta , r c )
138 ho l d on
p o l a r ( beta , o n e s ( s i z e ( b e t a ) ) ∗ rb , ’ g ’ )
140 p o l a r ( beta , o n e s ( s i z e ( b e t a ) ) ∗ b i g R a d i u s , ’ r ’ )
ho l d o f f
142 t i t l e ( ’Cam p r o f i l e i n p o l a r form ’ )
saveas (F , ’ CamProfile_polar . eps ’ )
144
F= f i g u r e ( 3 )
146 p l o t ( [ 0 r c . ∗ c o s ( b e t a ) ] , . . .
[ 0 r c . ∗ s i n ( b e t a ) ] , r b ∗ c o s ( b e t a ) , r b ∗ s i n ( b e t a ) , ’−− ’ , . . .
148
bigRadius ∗ cos ( beta ) , bigRadius ∗ s i n ( beta ) , ’ : ’ )
ho l d on
150 i =1
p l o t ( [ X X−e ∗ s i n ( a l p h a ( i ) ) xc ( i ) . . .
152
X−e ∗ s i n ( a l p h a ( i )) − d∗ c o s ( a l p h a ( i ) ) ] ,
[Y Y−e ∗ c o s ( a l p h a ( i ) ) yc ( i ) . . .
154
Y−e ∗ c o s ( a l p h a ( i ) ) + d∗ s i n ( a l p h a ( i ) ) ] , ’−ok ’ )
axis equal
156 ho l d o f f
x l a b e l ( ’ x ( cm ) ’ )
158 y l a b e l ( ’ y ( cm ) ’ )
saveas (F , ’ CamSystemConfiguration . eps ’ )
160
F= f i g u r e ( 4 )
162 a x i s ([ −100 160 −120 1 8 0 ] )
axis equal
164 v= a x i s
%P l o t t h e m o t i o n o f t h e cam
166 f o r i = 1 : l e n g t h ( t h e t a )
119
7.4. Cam Design
168
170
172
174
176
p l o t ( [ 0 r c . ∗ cos ( beta + t h e t a ( i ) ) ] , . . .
[0 r c . ∗ s i n ( beta + t h e t a ( i ) ) ] , . . .
[X X−e ∗ s i n ( a l p h a ( i ) ) xc ( i ) . . .
X−e ∗ s i n ( a l p h a ( i )) − d ∗ c o s ( a l p h a ( i ) ) ] , . . .
[Y Y−e ∗ c o s ( a l p h a ( i ) ) yc ( i ) . . .
Y−e ∗ c o s ( a l p h a ( i ) ) + d ∗ s i n ( a l p h a ( i ) ) ] , ’−ok ’ )
axis (v)
t i t l e ( ’ S i m u l a t e d Cam Motion ’ )
x l a b e l ( ’ x ( cm ) ’ )
y l a b e l ( ’ y ( cm ) ’ )
M( i )= g e t f r a m e ( ) ;
178 end
%make a m o vi e
180 m o v i e 2 a v i (M, ’ CamMotion . a v i ’ )
The graphs generated with this program are shown in Figs. 7.2-7.3
A movie of this cam system can be seen at:
http://personal.utulsa.edu/~jeremy-daily/ME3212/CamMotion.avi
120
7.4. Cam Design
121
Kinematic coefficients for Cam Follower Rotation
f(theta), rad
0.4
0.3
0.2
0.1
0
0
50
100
150
200
Cam angle θ, deg
250
300
350
0
50
100
150
200
Cam angle θ, deg
250
300
350
0
50
100
150
200
Cam angle θ, deg
250
300
350
f′(θ)
0.2
0
−0.2
f′′(θ)
0.5
0
−0.5
Figure 7.2.: Cam Follower Motion Profiles for the example.
7.4. Cam Design
122
100
80
60
40
y (cm)
20
0
−20
−40
−60
−80
−100
−100
−50
0
50
100
x (cm)
Figure 7.3.: Cam Follower Motion Profiles for the example. The dashed line represents the base
circle. The cam rotates counterclockwise.
123
1. Plot the acceleration profile from the displacement diagram in figure 7.21 of Cleghorn’s
text. Also, indicate the values of the jerk function at key points on the graph.
2. Given a base circle diameter of 4 cm, and angular velocity of 20 rad/s, an offset of 1.5 cm,
and the follower motion described in the table below:
Interval Initial Angle Final Angle Type of Motion Rise or Return
1
2
3
4
0
120
240
330
120
240
330
360
Cycloidal
Dwell
Harmonic
Dwell
1.2 cm
0
-1.2 cm
0
a) Plot the displacement, velocity, acceleration, and jerk profiles for a complete turn of
the cam.
b) Determine the disk cam profile. Plot this profile in rectangular (x, y) coordinates
using Matlab. Show the center of rotation.
c) Determine the minimum width of the cam follower face.
d) Determine the profile if the base diameter was 0.5 inches. Are there any issues with
this cam?
8. Gears
8.1. Introduction
8.1.1. Cog and Lantern Gears
8.1. Introduction
8.1.2. Common Types of Gears
• Spur Gears
• Helical Gears
• Miter Gears
• Herringbone Gears
• Plain Bevel Gears
• Spiral Bevel Gears
• Hypoid Gears
• Worm and Wheel Gears
125
8.2. Fundamental Law of Gearing
8.2. Fundamental Law of Gearing
8.3. Conjugate Profiles and Involutometry
126
127
8.3.1. The Involute Curve
b
For two gears in a mesh:
See
Cleghorn’s
Section
5.5
b
b
8.3.2. Cycloidal Profiles
8.3.3. Gear Sizing and Terminology
Know what the following terms mean and understand how to calculate them
• Center to Center Distance
• Addendum
• Dedendum
• Clearance
• Circular Pitch
• Diametrical Pitch
• Module
• Pressure angle
• Contact Ratio
• Backlash
• Interference
• Undercutting
128
1. (Book Problem 5.5) Two spur gears have a diametrical pitch of 5 /in, a magnitude of the
speed ratio of 0.20, and a center-to-center distance of 12 in. How many teeth do the gears
have?
2. (Book Problem 5.8) A gear with a diametrical pitch of 3 /in and 20◦ full depth involute
form has 35 teeth and meshes with a 70-tooth gear. The larger gear rotates at 300 rpm CW.
Both gears have external teeth. Determine
a) the pitch circle diameters
b) the base circle radius of the smaller gear
c) the circular pitch
d) the center-to-center distance
e) the rotational speed of the smaller gear
f) the contact ratio
129
8.5. Gear Train Analysis
8.5.1. Taxonomy of gear trains:
1. Ordinary
a) Simple
b) Compound
2. Planetary (Epicyclic)
a) Single Stage
b) Multiple Stage
130
131
8.5.2. Gear Train Ratio
8.5.3. Idler Gears
8.5.4. Compound Gear Trains
Example: Transmission
See
Cleghorn’s
Example
6.1
1. Book Problem P6.1
132
8.7. Reverted Gear Train Design
8.7. Reverted Gear Train Design
A reverted gear train is an in-line gear train.
8.7.1. Design Considerations
• Don’t go too small
• Keep a “hunting” condition
133
8.8. Planetary Gear Trains
134
Example: Rescue Winch Gears, Automatic Transmissions
8.8.1. Advantages
1.
2.
3.
4.
5.
8.8.2. Vector Approach to Planetary Gear Train Analysis
b
b
b
b
Define some Vectors:
~rA :
~rP :
~rA/P :
~rB :
~rB/P :
135
136
8.8.3. Tabular Planetary Gear Train Analysis
b
b
b
b
1.
2.
3.
See
Cleghorn’s
Section
4.4.2
137
Example:
Ns = 34, NR = 78, N p = 22, the sun gear is the input, the ring gear is fixed, and the planet carrier
is the output. Build and analysis table:
Component
Gear 1 (Sun) ω1
Crank (Carrier) ωarm
Gear 3 (Planet) ω3
Gear 4 (ring) ω4
All Turn at x
rpm
Crank is
Fixed, Sun
turns at y
Absolute
Speeds
Example:
Ns = 34, NR = 78, N p = 22, the sun gear is fixed, the ring gear is the output, and the planet carrier
is the input. Use the same analysis table as before.
138
8.8.4. Compound Planetary Gear Trains
6
5
4
3
2
1
0
-1
-2
-3
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
Example: an electric winch gear train where NS1 = 34, NP1 = 22, NS2 = 26, NP2 = 30, and
NR = 78.
See
Cleghorn’s
Section
6.5
Table 1: First Stage
Component
Gear 1 (Sun) ω1
All Turn at x
rpm
Crank is
Fixed, Sun
turns at y
Absolute
Speeds
Table2: Second Stage
Gear 3 (Planet) ω3
Gear 4 (ring) ω4
Component
All Turn at x
rpm
Crank is
Fixed, Sun
turns at y
Absolute
Speeds
Gear 1 (Sun) ω1
139
Gear 3 (Planet) ω3
Gear 4 (ring) ω4
140
8.8.5. Plotting Planetary Gears
The following program will draw the layout of a three-gear planetary gear train:
2
4
6
8
10
12
14
16
18
r 1 =3
r 2 =2
r 3 = r 1 +2∗ r 2
X1 = 3 ; Y1 = 7 ;
X2 = 9 ; Y2 = 7 ;
X3=X1 ; Y3=Y1 ;
t h e t a 1 =.5
t h e t a 2 =.6
t h e t a 3 =.7
t h e t a = l i ns pa ce (0 ,2∗ pi ) ;
x1 =[X1 X1+ r 1 ∗ c o s ( t h e t a 1 + t h e t a
y1 =[Y1 Y1+ r 1 ∗ s i n ( t h e t a 1 + t h e t a
p l o t ( x1 , y1 , x2 , y2 , x3 , y3 )
axis equal
)];
)];
)];
)];
)];
)];
141
14
12
10
8
6
4
2
0
-4
-2
0
2
4
6
8
10
Figure 8.1.: Output from the code listing to plot the configuration of a planetary gear train.
8.9. Differential Gear Trains
142
8.9.1. Ackerman Steering
b
See
Cleghorn’s
Section
6.5
143
Example:When a vehicle is in a turn, determine a) the speed of each wheel and b) the angular
velocity of the planet carrier, c) the driveshaft speed.
b
144
A. Mechanisms Design Project
A computer mechanism design project will be given in class. You will have the choice of desiging
one of the following:
• Moving batters tee
• Runner friendly stroller
• Air-bag module near-deployment mechanism
• Remote motorcycle braking system
• Scissor lift safety mechanism
A.1. Project Proposal
A 1 page design proposal in a formal memo format is required from each student that clearly
identifies the following:
• Definition of the project
• Customer requirements
• Design constraints
• Project schedule
• Linkage concepts (i.e. 4-bar, slider-crank, or some combination of linkages)
A.2. Hand Sketches
The hand sketch will have the skeleton diagram all parts of the mechanism sketched out with
appropriate labels. The lengths of the links are not known yet and should be represented as
variables. The assembly will need to be sketched with key locations. Finally, the loop closure
vectors and position vectors will need to be defined.
All sketches must be made with quad ruled graph paper, a straightedge and a compass. The
sketches must be neat and well labeled. Only one or two parts should be sketched on an 8.5x11
inch page.
Neatness will be 25% of the grade for these sketches.
Hand Sketches are due on: ______________
Use geometric constraint programming techniques to determine critical dimensions of the mechanism. Use a single rigid triangle to represent the couplers.
Use the Tools -> Options dialog to change units and turn off automatic constraints.
The print out of the geometric constraint programming analysis and a list of all critical dimensions corresponding to the variables in the hand sketch are to be turned in. Also, the hand
sketches are to be corrected (if necessary) and resubmitted. Be sure to include the location of the
link attachments to the frame.
The geometric constraint programming analysis is due on:________________________
A.4. SolidWorks Parts and Assembly
Create SolidWorks parts for all components of the mechanism to include bearings, pins, bolts
and the frame. The product will work in only 2 dimensions but will be constructed in three
dimensions. All joints must be in double shear to maintain strength and reduce out of plane
twisting. An example of building a joint would be to use flange bearings and a shoulder bolts as
seen in Figs. A.1 on the next page and A.2 on page 148Another option is to use regular hex bolts
and rod ends as seen in Fig. .The following components can be obtained from an industrial supply
company, the ones shown herein are available at http://www.mcmaster.com. Also, CAD
drawings for some parts are available from some vendors.
All parts must be compatible so you should pick thicknesses to match the bearings and shoulder
bolts.
A hydraulic cylinder is available for your design model in two parts: the cylinder and the ram.
Both parts must be assembled to make the system work. A clevis pin should be used to attach
the cylinder base to the frame and a female thread rod end should be used to attach the ram to the
lifting plates. An example of a rod end is shown in Fig. A.3 on page 149. The technical diagram
of the rod end is shown in Fig. A.4 on page 150.
Connect the rod end to the lifting link using a shaft held in place with retaining rings. The
retaining rings are shown in Fig. A.5 on page 151.
An example of the technical drawing for the shoulder screw is available at www.mcmaster.com
and is seen in Fig. A.6 on page 152.
146
147
Figure A.1.: Flanged Sleeve Bearing description from http://www.mcmaster.com. This is an oil
impregnated bronze bearing that does not need lubrication.
Produce a Bill of Materials that includes the following entries: Quantity, Description, Source,
Part number, Price. If something is not available from a vendor, then mark it as “in-house.” The
part numbers should correspond to the labels in your original hand drawings. For example
Qty
Description
Source
Part Number
Price
Total
8
2
..
.
1”x1-1/2” Shoulder Bolts
Rear Guide Links
..
.
McMaster-Carr
In-house
..
.
91259A132
r2
..
.
$16.82
TBD
..
.
$134.56
..
.
Grand Total
Produce full page engineering drawings to manufacture each part. Include all pertinent dimensions. Drawings are not required for items available from a vendor, only items in the bill of
materials whose source is in-house need to be drawn.
Create a SolidWorks assembly to demonstrate the functionality of the mechanism. No parts
should bind or interfere. Apply a linear motor to the hydraulic cylinder and produce an animation
of the assembly.
Technical drawings of each part, an image of the assembly in the low position, an image of the
assembly in the up position, and a bill of materials are due on the last day of classes.
Figure A.2.: Shoulder Screw description from http://www.mcmaster.com. There are also metric
equivalent sizes that can be specified.
148
Figure A.3.: Rod end description from http://www.mcmaster.com. These can be found in both
right hand thread and left hand thread configurations. A jam nut should always be
used to secure the rod end.
149
Figure A.4.: Rod End drawing to interface with the hydraulic cylinder.
150
Figure A.5.: External retaining ring description and drawing from www.mcmaster.com. These
are also known as snap rings.
151
Figure A.6.: Technical Drawing of a Shoulder Screw from www.mcmaster.com.
152

Course Notebook - The University of Tulsa

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