###
Question Description

The assignment requires you to do Position, Velocity, acceleration and force analysis to a 4 bar linkage with a slider. We are working on a window slider mechanism (similar to crank slider). I have completed the Position, Velocity and acceleration Analysis. But I am struggling with the force analysis. YOU HAVE TO LOOK AT THE ANLYSIS DONE AND DO IT THE SAME METHOD. YOUR FINAL ANSWER SHOULD BE EQUATION OF F(D/A).

ONLY CALCULATIONS NEEDED, NO REPORT OR ANY WRITING.

Note: I look online extensively but couldn’t find any good solution. So please DON’T ACCEPT THE QUESTION UNLESS IF YOU CAN SOLVE IT 100% CORRECTLY AND THE SAME METHOD THE PREVIOUS WORK HAS BEEN DONE (attached).

I am looking for it ti be solved in the same method as the other analysis has been done and the chapter handout

MAAE 3004:

Dynamics of Machinery

Department of Mechanical and Aerospace Engineering

Carleton University

Chapter 5

Force Analysis of Mechanisms

(Static & Dynamic Force Analysis)

Sections in text: 14.1-14.8, 15.1, 15.4-15.7

Some slides are adapted from Professor F.F. Afagh and Professor R.G.

Langlois MAAE 3004 course slides.

Theory of Machines and Mechanisms, 5e

Uicker, Pennock, Shigley

Copyright © 2017 Oxford University Press

Outline

Introduction

Review (Dynamic Analysis of Rigid Bodies)

Forces Acting on Links

Solution Using Graphical Analysis and Superposition

Analytical Force Balances and Matrix Solution

Introduction:

In the subject of mechanism dynamics, we bring together

kinematics the study of motion and geometry, with kinetics,

the study of the relationship between force and motion, to

derive information on the forces and torques active in

moving machinery. This information is, in turn, essential to

the computation of the stresses internal to the members of

a mechanism and the elastic deflections of the members.

Thus, we make use of the full power of methods described

in Chapters 1 through 4 of this course to calculate the

accelerations and angular accelerations to which mechanism

members are subjected. These are related to the loads

acting on those members via Newton’s Second Law.

Introduction (continued):

In the design of mechanisms and machines, the required

motions are often specified first. The dynamics task is

essentially determining those forces that must be applied

and/or those forces that accompany the motions

The required process is to:

1. Determine masses, centers of mass, and mass moments of

inertia, IG , of the links

2. Perform kinematic analysis to determine linear and angular

accelerations of all links (evaluate “⃗# at the center of mass

for all links as well as $)

⃗

3. Perform force analysis either by:

o graphical analysis and superposition, or by

o analytical force balances and matrix solution

Dynamic Analysis:

The relationship between force and motion that is central to the study of

kinetics is Newton’s second law:

ΣF ⃗= m a ⃗

where Σ F ⃗ is the net force acting on a particle, m is its mass, and a ⃗ is its

acceleration.

This relationship refers to only a single particle and the machine members we

want to model are bodies with distributed mass. We need to derive equations

relating force and torque for a rigid body with distributed mass. This results in

the “Newton-Euler” equations describing the motion of a rigid body. The

description of the motion of the center of mass of the body has exactly the

same form as the equation above. That is, the center of mass moves as

though the entire mass of the body were condensed into a particle at that

point:

Σ F ⃗ = m a G⃗

⃗ is the acceleration of the center of mass.

where a G

Dynamic Analysis (continued):

Rotary motion of the body is described by Euler’s equation:

Σ M G⃗ = IG α ⃗

where ΣMG⃗ is the resultant moment of the force system

acting on the body about the center of mass; IG is the inertia

matrix based on fixed coordinate axes with origin at the

center of mass, and is the angular acceleration of the body

relative to the fixed frame.

In the special case in which the body is rotating about a fixed

point, P, moments may be taken about that point and the

inertia matrix expressed in the same fixed coordinate frame

with origin at P. Euler’s equation then becomes

ΣM ⃗ = I α ⃗

P

P

This is the only case in which the inertia matrix may be based

on a point other than the center of mass.

Dynamic Analysis (continued):

For any case, except rotation of a symmetric body about a

fixed axis of symmetry, it is much more convenient to express

the inertia matrix in a coordinate frame fixed in the moving

body. There is always a set of three orthogonal axes, fixed in

the body with origin at the center of mass, for which the

inertia matrix becomes a diagonal matrix. These axes are

called the principal axes of inertia. If the inertia matrix, IG is

expressed in the principal axes, and the angular velocity, ω ,⃗

and angular acceleration, α ⃗ , are expressed in the same

frame, the equation becomes:

Σ M G⃗ = IG α ⃗ + ω ⃗ × IG ω ⃗

This is the form of Euler’s equation that is most useful for

analyzing spatial motion.

Dynamic Equilibrium of Systems of Rigid Bodies

We restrict consideration to planar motion with one of the

principal axes normal to the plane of motion. In this case

ΣM ⃗ × k̂ = I α ⃗

G

G

For planar motion IG is a scalar quantity, being the moment of

inertia of the body about an axis through the center of mass

normal to the plane of motion, k̂ is a unit vector normal to the

plane of motion, and α ⃗ is the angular acceleration, taken to be

positive in the direction. That is,

α ⃗ = αk ̂

If a body is in motion, the sum of the forces acting on it is equal

to its mass multiplied by the acceleration of its center of mass.

That is,

! #⃗ = %&⃗

D’Alembert’s Principle:

If we introduce the inertia force, “⃗# , such that

“⃗# = −& ‘⃗(

applied on a line passing through the center of mass, and a corresponding

couple,

)# = −*( +⃗

called the inertia moment (torque), and treat these in the same way as any other

external force and moment, then the dynamic equilibrium equations assume the

same form as the static equilibrium equations:

, “⃗ = 0

, ). = 0

where O is any point in the plane about which moments are taken.

D’Alembert’s principle states:

o The vector sum of all external forces and inertia forces acting on a system of

rigid bodies is zero.

o The vector sum of all external moments and inertia moments acting on a

system of rigid bodies is also separately zero.

Combining !” and #” into One Offset Force

The combined effect of all forces and moments acting on a rigid body can be

represented by a single equipollent force and a single equipollent moment

acting about the center of mass

The equipollent force and moment can be

replaced by a single offset force %⃗& that has

the same magnitude and direction as ∑ %⃗

but is shifted by h to cause the same effect

as () .

The offset direction is chosen such that the

offset force results in the same moment

direction as () .

Two-Force Members

A two-force member has only 2 significant forces

acting on it.

o m and I often may be ignored due to small

values

o the two forces must be colinear to satisfy ∑ ” =

0 and must act along the line joining their points

of application

o the two forces must have equal magnitudes but

opposite directions to satisfy ∑ &⃗ = 0

o One of the two forces could be the inertia force.

If an element has pins or hinge supports at both ends and

carries no load in-between, it is called a two-force member.

Two force and one moment member: A rigid body acted on by two forces and a

moment is in static equilibrium only when the two forces form a couple whose

moment is equal in magnitude but in opposite sense to the applied moment

Three-Force Members

A three-force member has three significant

forces acting on it:

the lines of action of all three forces must

intersect at a point (necessary to satisfy

∑ ” = 0).

A special case of the three-force member is

when three forces meet at a pin joint that is

connected between three links. When the

system is in static equilibrium, the sum of the

three forces must be equal to zero.

For example, if the axes of two of the forces

are known, the intersection of those two

axes can assist us in determining the axis of

the third force.

Free-Body Diagram (FBD):

A machine system is considered to be a system of an arbitrary group of

bodies (links), which will be considered rigid. We are involved with

different types of forces in such systems.

a) Reaction Forces: are commonly called the joint forces in machine

systems since the action and reaction between the bodies involved will be

through the contacting kinematic elements of the links that form a joint.

The joint forces are along the direction for which the degree-of-freedom is

restricted. e.g. in constrained motion direction

All lower pairs and their constraint forces:

(a) revolute or turning pair with pair

variable q

(b) prismatic or sliding pair with pair

variable z

(c) cylindric pair with pair variables

q and z

(d) screw or helical pair with pair

variables q and z

(e) planar or flat pair with pair

variables x, z, and q .

(f) spheric pair with pair variables

q, f and y

Free-Body Diagram (FBD):

A free-body diagram is a

sketch or drawing of the

body, isolated from the rest

of the machine and its

surroundings, upon which

the forces and moments are

shown in action.

A free body diagram shows

all forces of all types acting

on this body.

Free-Body Diagram (FBD):

Static Analysis:

Example: If the force P = 0.9 kN, determine the torque !”# that must

be applied to crank 2 to maintain the linkage in static equilibrium.

Free-body diagram:

Two-Force and one moment Member

Solution Using Graphical Analysis and Superposition

Principle of Superposition: “For linear systems the individual responses to

several disturbances or driving functions can be superposed on each other to

obtain the total response of the system.”

The general procedure is:

1. Perform a kinematic analysis of the complete mechanism and extract the

linear acceleration of each link center of mass and angular acceleration.

2. Ignoring inertia, perform a static analysis using free-body diagrams (FBDs)

for each link.

3. Find !” and #$ for each link and the corresponding values &⃗’ and offset

distances h.

4. For each offset force, but now ignoring static loads, analyze forces in the

mechanism. This should be repeated for each offset force.

5. Vectorially add the results of each force analysis for each constraint force

location.

Note that constraint forces are the pair of action and reaction forces between

any two contacting points. Applied forces are those that are applied to links but

developed outside the system of links.

Example: Perform a complete dynamic force analysis of the four-bar linkage

shown in the figure below. The known information is provided on the figure. The

input crank is operating at “constant angular velocity”, !” =60 rad/s ccw, and the

applied force is $⃗% = 40 *̂ lb. The weight of link 3 and 4 are 7.13 lb and 3.42 lb,

respectively., and the mass moments of inertia of links are +,- = 0.25 12. 34. 5 ” ,

+,6 = 0.625 12. 34. 5 ” , and +,- = 0.037 12. 34. 5 ” , respectively.

Step 1. Perform the kinematic analysis to determine the velocity and

subsequently acceleration polygons. Here, the numerical results are given. From

the image the acceleration of link 3 and 4, “⃗#$ and “⃗#% , can be obtained.

Angular accelerations are calculated:

t

AB/A

α3 =

= 148 rad /s 2 ccw

RB/A

α4 =

t

AB/O

4

RB/O4

= 604 rad /s 2 cw

These linear and angular accelerations subsequently will be used to

determine inertia forces and moments respectively on links 3 and 4.

⃗ and M ⃗ as well as F ⃗ and M ⃗

Step 2. Calculate F 13

13

14

14

⃗ = − m a ⃗ = (7.13/32.2)(758) = 168 lb (see direction)

F 13

3 G3

⃗ = I α ⃗ = (0.625 in lbs 2)(148) = 92.5 in lbcw

M13

G3

3

h3 =

| M13 |

= 0.550 in

| F13 |

⃗ = − m a ⃗ = 37 lb (see direction)

F 14

4

G4

⃗ = I α ⃗ = 22.3 in lbccw

M 14

G4

4

| M14 |

h4 =

= 0.602 in

| F14 |

Step 3. Force Analysis Strategy

Ø Use superposition principle to simplify the force analysis into a series of

problems comprised of two and three-force members.

1.

Apply inertia force !” $⃗%& only. Link 4 is a three-force member; link

3 is a two-force member

2.

Apply inertia force !’ $⃗%( only. Link 4 is a two-force member; link 3

is a three-force member

3.

Apply external force )⃗* only. Link 4 is a three-force member; link 3 is

a two-force member

Step 4. Evaluate loads associated with the inertia force −”# %⃗&’ on the freebody diagram of link 4.

LINK 3

TWO-FORCE MEMBER

LINK 4

THREE-FORCE MEMEBER

| F′⃗14 | = 44.3 lb

| F′⃗34 | = 24.3 lb

Step 5. Evaluate loads associated with the inertia force −”# %⃗&’ on the freebody diagram of link 3.

| F′′ ⃗23 | = 145 lb

| F′′ ⃗ 43 | = 94.8 lb

Step 6. Evaluate loads associated with applied force “⃗#

Step 7. Determine resultant load “⃗#$ and required torque %&$

⃗ = F′⃗32 +

F 32

F′′ ⃗32 + F′′′⃗32 = − ( F′⃗32 +

F′′ ⃗32 + F′′′⃗32)

Link 2 is pivoted to ground at O2 where the link center of mass G2 is

located; therefore it will not experience any linear inertia force and it can

be considered at the end of the overall analysis.

⃗

M12

⃗ | = h | F ⃗ | = (1.56 in ) (145 lb) = 226 in . lbcw

| M 12

2

32

Scanned by CamScanner

Scanned by CamScanner

Scanned by CamScanner

Scanned by CamScanner

Scanned by CamScanner

Scanned by CamScanner

Scanned by CamScanner

Scanned by CamScanner

Scanned by CamScanner

Scanned by CamScanner

Scanned by CamScanner

Scanned by CamScanner

Perform displacement, velocity, and acceleration analysis of the mechanism. This may be done

using the methods taught in this course for significant configurations of the mechanism

(minimum four poses with the selection of configurations clearly justified) or alternatively using

analytical kinematic analysis that applies throughout the complete range of motion of the

mechanism

Perform force analysis on the mechanism. This may be done using the methods taught in this

course for significant configurations of the mechanism (minimum four poses with the selection of

configurations clearly justified) or alternatively using analytical force analysis that applies

throughout a complete cycle of motion.

**We offer the bestcustom writing paper services. We have done this question before, we can also do it for you.**

#### Why Choose Us

- 100% non-plagiarized Papers
- 24/7 /365 Service Available
- Affordable Prices
- Any Paper, Urgency, and Subject
- Will complete your papers in 6 hours
- On-time Delivery
- Money-back and Privacy guarantees
- Unlimited Amendments upon request
- Satisfaction guarantee

#### How it Works

- Click on the “Place Order” tab at the top menu or “Order Now” icon at the bottom and a new page will appear with an order form to be filled.
- Fill in your paper’s requirements in the "
**PAPER DETAILS**" section. - Fill in your paper’s academic level, deadline, and the required number of pages from the drop-down menus.
- Click “
**CREATE ACCOUNT & SIGN IN**” to enter your registration details and get an account with us for record-keeping and then, click on “PROCEED TO CHECKOUT” at the bottom of the page. - From there, the payment sections will show, follow the guided payment process and your order will be available for our writing team to work on it.