Slider crank Force analysis (Mechanical engineering 4th year course)

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
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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.

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