MENG 372 Chapter 6 Velocity Analysis

Slides:



Advertisements
Similar presentations
Velocity Analysis with Instant Centers for a Four-bar Mechanism
Advertisements

Mechanics of Machines Dr. Mohammad Kilani
Chapter3 Kinematic Analysis of Mechanisms
Instantaneous Center of Zero Velocity (IC) (Ani Dönme Merkezi)
Instant Centers of Velocities
Acceleration analysis (Chapter 4)
MENG 372 Chapter 3 Graphical Linkage Synthesis
Kinematics of Rigid Bodies
Chapter 15 KINEMATICS OF RIGID BODIES
P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar Introduction Velocity Polygon for a Four-bar Mechanism This presentation shows how to construct.
RIGID BODY MOTION: TRANSLATION & ROTATION (Sections )
PLANAR RIGID BODY MOTION: TRANSLATION & ROTATION
Velocity Polygon for a Crank-Slider Mechanism
1 Instant center - point in the plane about which a link can be thought to rotate relative to another link (this link can be the ground) An instant center.
INSTANTANEOUS CENTER (IC) OF ZERO VELOCITY (Section 16.6)
Mechanics of Machines Dr. Mohammad Kilani
INSTANTANEOUS CENTER OF ZERO VELOCITY Today’s Objectives: Students will be able to: 1.Locate the instantaneous center of zero velocity. 2.Use the instantaneous.
1 All figures taken from Design of Machinery, 3 rd ed. Robert Norton 2003 MENG 372 Chapter 5 Analytical Position Synthesis.
Plane Kinematics of Rigid Bodies
ENGR 215 ~ Dynamics Sections Relative Motion Analysis.
P. Nikravesh, AME, U of A Acceleration Polygon for a Crank-Slider Introduction Acceleration Polygon for a Crank-Slider Mechanism This presentation shows.
Position, Velocity and Acceleration Analysis
Chapter 14 Section 14.3 Curves. x y z To get the equation of the line we need to know two things, a direction vector d and a point on the line P. To find.
1 All figures taken from Design of Machinery, 3 rd ed. Robert Norton 2003 MENG 372 Chapter 4 Position Analysis.
INSTANTANEOUS CENTER OF ZERO VELOCITY
PLANAR RIGID BODY MOTION: TRANSLATION & ROTATION
Kinematics of Rigid Bodies
Uniform Circular Motion A special kind of two-dimensional problem involves objects undergoing uniform circular motion. Uniform circular motion is motion.
Adding Vectors, Rules When two vectors are added, the sum is independent of the order of the addition. This is the Commutative Law of Addition.
RELATIVE MOTION ANALYSIS: ACCELERATION
Angular Momentum Linear Momentum. Always work from Principle! Ex: Work of gravity Principle: dW = F * ds Here: dW = - mg * dy mg y.
Introduction Instant Centers for a Crank-Slider
Velocity Polygon for a Crank-Slider Mechanism
ENGR 214 Chapter 15 Kinematics of Rigid Bodies
Rotational Motion 2 Coming around again to a theater near you.
All figures taken from Design of Machinery, 3 rd ed. Robert Norton 2003 MENG 372 Chapter 11 Dynamic Force Analysis.
Chapter 8 Rotational Motion.
MOTION RELATIVE TO ROTATING AXES
1 All figures taken from Design of Machinery, 3 rd ed. Robert Norton 2003 MENG 372 Chapter 7 Acceleration Analysis.
RELATIVE MOTION ANALYSIS: VELOCITY
PLANAR KINEMATICS OF A RIGID BODY
Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Today’s Objectives: Students will be able to:
RELATIVE MOTION ANALYSIS: VELOCITY
Today’s Objectives: Students will be able to: a) Resolve the acceleration of a point on a body into components of translation and rotation. b) Determine.
P. Nikravesh, AME, U of A Instant Centers for a Crank-Slider Introduction Velocity Analysis with Instant Centers for a Crank-Slider Mechanism (Inversion.
Introduction Instant Centers for a Crank-Slider
Introduction Sometimes, a body has simultaneously a motion of rotation and translation, such as a wheel of a car, a sphere rolling (not slipping) on the.
Today’s Objectives: Students will be able to: 1.Describe the velocity of a rigid body in terms of translation and rotation components. 2.Perform a relative-motion.
DYNAMICS VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Tenth Edition Ferdinand P. Beer E. Russell Johnston, Jr. Phillip J. Cornwell Lecture Notes: Brian P.
VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Seventh Edition Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University CHAPTER.
Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Today’s Objectives: Students will be able to:
INSTANTANEOUS CENTER (IC) OF ZERO VELOCITY (Section 16.6)
RELATIVE MOTION ANALYSIS: VELOCITY (Section 16.5)
RELATIVE MOTION ANALYSIS: VELOCITY
INSTANTANEOUS CENTER OF ZERO VELOCITY
ME321 Kinematics and Dynamics of Machines
Analytical Modeling of Kinematic Linkages
Instant Velocity Centers – A ‘Fast Track’ to Linkage Control
ME321 Kinematics and Dynamics of Machines
Analytical Modeling of Kinematic Linkages, Part 2
1 All figures taken from Design of Machinery, 3 rd ed. Robert Norton 2003 MENG 372 Chapter 3 Graphical Linkage Synthesis.
RELATIVE MOTION ANALYSIS: VELOCITY
RELATIVE MOTION ANALYSIS: ACCELERATION
RELATIVE MOTION ANALYSIS: ACCELERATION
RIGID BODY MOTION (Section 16.1)
RELATIVE MOTION ANALYSIS: VELOCITY
Acceleration analysis (Chapter 4)
Acceleration analysis (Chapter 4)
A body 'm' having relative plane motion with a reference body 'n' is shown in figure. A and B are two points on body 'm'. Absolute.
Presentation transcript:

MENG 372 Chapter 6 Velocity Analysis All figures taken from Design of Machinery, 3rd ed. Robert Norton 2003

Velocity Analysis Definitions Linear Velocity Angular Velocity Velocity of a point Link in pure rotation Velocity is perpendicular to radius of rotation & tangent to path of motion Multiplying by i rotates the vector by 90°

Velocity Analysis Imaginary Real Vector r can be written as: Multiplying by i gives: r Multiplying by i rotates a vector 90° q Real

Velocity Analysis If point A is moving Graphical solution:

Graphical Velocity Analysis (w3 & w4) Given linkage configuration & w2. Find w3 and w4 We know VA and direction of VB and VBA (perpendicular to AB) Draw vector triangle. V=wr. VBA Direction VBA VA VB VBA VB VB Direction

Graphical Velocity Analysis (VC) After finding w3 and w4, find VC VC=VA+VCA Recall that w3 was in the opposite direction as w2 VA Double Scale VC VCA VCA VC

Instant Center A point common to two bodies in plane motion, which has the same instantaneous velocity in each body. In ENGR 214 we found the instant center between links 1 and 3 (point on link 3 with no velocity) Now we also have an instant center between links 2 and 4

Instant Centers Kennedy’s rule: any three links will have three instant centers and they will lie on a straight line The pins are instant centers I13 is from links 1,2,3 and 1,3,4 I24 is from links 1,2,4 and 2,3,4 I13 1 2 3 1 3 4 1 2 4 2 3 4 Links I12 I23 I13 I13 I34 I14 I12 I24 I14 I23 I34 I24 IC’s I24

Instant Centers I13 has zero velocity since link 1 is ground w3 is the same all over link 3 Velocity relative to ground=rw, perpendicular to r VA2=aw2=VA3=pw3 From this, w3 must be in the opposite direction as w2, and smaller in magnitude since p>a I13 VA3 p w3 a VA2 A

Instant Centers I24 has the same velocity on link 2 and link 4 VI2=l2w2=VI4=l4w4 From this, w4 is in the same direction as w2 and smaller in magnitude since l4>l2 w4 VI4 l4 I24 VI2 l2

Instant Centers Practice Problems Power=Tinwin=Toutwout A O4 O2 B

Velocity Analysis of a 4-Bar Linkage Given w2. Find w3 and w4

Velocity Analysis of a 4-Bar Linkage Write the vector loop equation After solving the position analysis, take the derivative or where

Velocity Analysis of a 4-Bar Linkage Take knowns to one side: Take conjugate to get 2nd equation: Put in matrix form: Invert matrix:

Inverted Crank Slider Given w2. Find w3 and Link 3 is a slider link: its effective length, b, changes Given w2. Find w3 and

Inverted Crank Slider Given w2. Find w3 and Write the vector loop equation: After solving the position analysis, take the derivative: To get another equation: or so

Inverted Crank Slider Take conjugate to get second equation: Put in matrix form: Invert:

Velocity of any Point on a Linkage Write the vector for RP Take the derivative Similarly RP

Offset Crank Slider Given w2. Find w3 and b c a