Control Charts for Attributes
Chapter 7
Week 6 and 7
IE 2076 Besterfield-Sacre
1
Objectives
Understand the statistical basis for attribute CC
Know how to set up p and np CC
Know how to set up c and u CC
Know what to do with varying sample sizes
Understand
Introduction to TQM
IE2076 Besterfield-Sacre
1
TQM Framework
Company
Organization
Firm
Philosophy
and
Principles
Design
of Product,
Service,
and/or Process
Manufacturing
of Product or
Delivery of
Service
Inspection of
Product or
Service
Customer
Supplier
S T R A T E G I C
P L A N N I N G
From Design to Action:
Developing a Corporate Strategy
Quality management tools and a simple system of reference
can help build sound approaches based on real priorities
days when the extension of trend curves would
produ
Appendix VI Factors for Constructing Variables Control Charts
Chart for Averages Chart for Standard Deviations Chart for Ranges :3:_
Obscrvations Factors for Factors for Factors for g.
in Com Limits M W M Factors for Control Limits '.:
Smplc
TQM
7 New Tools
Week 3/Day 2
7 Management Planning Tools
(the New Tools)
01/16/17
Not really new
Most have their roots in post WWII
Operations Research work and Japanese TQC
1972-1979 Society for QC Technique
Development refined and tested and created
an
TQM
Quality Control Methods
Week 4
Days 1 and 2
1
Process Variation
Natural Random Variation
inherent in all processes, can only reduce by
improving the process capability but cannot eliminate
this is also known as chance causes
Assignable
can be a
Six Sigma
Week 8 & 9
ASQ Definition
A method that provides organizations tools to improve the
capability of their business processes.
This increase in performance and decrease in process variation
lead to:
defect reduction and
improvement in profits,
Process Capability
Week 6!
http:/www.supplychaindigital.com/procurement/2068/Motorolas-Six-Si
gma-Journey:-In-pursuit-of-perfection
LOWER SPEC
LIMIT (LSL)
PROCESS
Control chart
UPPER SPEC
LIMIT (USL)
Meeting the Specs
Meeting the Specs
PROCESS
UPPER SPEC
TQM
7 Magnificent Tools
Week 3/Day 1
The Old Seven
The First Seven
The Basic Seven
Your Toolbox
Method
Plan-Do-Check-Act
PDCA or PDSA
Tools
Old tools
New tools
P-D-C-A/P-D-S-A
Plan
Act
Do
Study
https:/www.youtube.com/watch?v=8T1sYPrQqvY
http:/med
Total Quality Management
Mary Besterfield-Sacre
Week 1/Day 1
1
TQM Framework
Company
Organization
Firm
Design
of Product,
Service,
and/or Process
Philosophy
and
Principles
Manufacturing
of Product or
Delivery of
Service
Inspection of
Product or
Service
Cu
INTERNATIONAL ISLAMIC UNIVERSITY ISLAMABAD
ENRONS FALL
CASE STUDY
IRAM SARFRAZ(3508), SAHAR BASHIR (3543), KOMAL SHAFIQUE (3542)
ENRONS FALL
Q1: What are the systematic, corporate and individual issues raised by this case?
Answer:
Systematic Issues:
Vital
161
F
+
Solving for v(t1 ), we obtain the velocity of the center of mass of the system
the instant after impact as
F
+
v(t1 ) =
m
2ghEx
m+M
(4.113)
Next, we know that the impact is perfectly inelastic. Consequently, the coecient of restitution is zero,
160
Chapter 4. Kinetics of a System of Particles
Equating the energy of the block at times t0 and t1 we have that
1
2
mv1 (t1 ) mgh
2
0=
(4.103)
Solving Eq. (4.103) for v1 (t1 ), we obtain
v1 (t1 ) = 2gh
(4.104)
From Eq. (4.104), the velocity of the block
152
Chapter 4. Kinetics of a System of Particles
Therefore, the resultant force acting on the block-particle system is given as
FT = F + N + (M + m)g = F Ex + NEy + (M + m)gEy
(4.39)
Eq. (4.39) simplies to
FT = F Ex + N + (M + m)g Ey
(4.40)
a
Setting FT e
159
The velocity of the center of mass of the block-plate system in reference frame
F is then given as
F
v=
mv1 + Mv2
m F v1 + M F v2
=
Ex
m+M
m+M
(4.93)
Kinetics
Phase 1: During Descent of Block
The only force acting on the block during its descent is th
158
Chapter 4. Kinetics of a System of Particles
Question 43
A block of mass m is dropped from a height h above a plate of mass M as shown
in Fig. P4-3. The plate is supported by three linear springs, each with spring
constant K, and is initially in stati
155
Rearranging, we have
(mv0 cos + Fx )Ex + (Fy mv0 sin )Ey = mvmx Ex + mvmy Ey
(4.66)
We then obtain the following two scalar equations:
mv0 cos + Fx
Fy mv0 sin
= mvmx
(4.67)
= mvmy
(4.68)
Application of Linear Impulse and Linear Momentum to the Entire
157
Simplifying this last expression, we have that
Fx = mv0 cos
M
m+M
(4.85)
Also, from Eq. (4.68) we have that
Fy mv0 sin = mvmy = mvy = 0
(4.86)
Solving this last equation for Fy , we obtain
Fy = mv0 sin
(4.87)
Consequently, the impulse exerted by the
156
Chapter 4. Kinetics of a System of Particles
Rearranging Eq. (4.74), we obtain
mv0 cos Ex + (P mv0 sin )Ey = (mvmx + MvMx )Ex + (mvmy + MvMy )Ey
(4.75)
We then obtain the following two scalar equations:
mv0 cos
= mvmx + MvMx
(4.76)
P mv0 sin
= mvmy
154
Chapter 4. Kinetics of a System of Particles
Solution to Question 42
First, let F be a xed reference frame. Then, choose the following coordinate
system xed in reference frame F :
Origin at Location of Block
Ex To The Left
Ez Into Page
Ey = E z E x
(4
153
Eq. (4.49) simplies to
ml 2 cos + ml sin mg R cos = 0
(4.50)
Then, multiplying Eq. (4.50) by sin , we obtain
ml 2 cos sin + ml sin2 mg sin R sin = 0
(4.51)
Next, multiplying Eq. (4.34) by cos , we obtain
F cos + R sin cos = M x cos
(4.52)
Rearranging
151
Now we have that
F = F Ex
(4.28)
R = Rer
(4.29)
N = NEy
(4.30)
Mg = MgEy
(4.31)
Then the resultant force acting on the block is given as
FO = F + R + N + Mg = F Ex + NEy + Rer + MgEy
(4.32)
Using the expression for er from Eq. (4.3), we have that
FO =
150
Chapter 4. Kinetics of a System of Particles
Now we have that
A
d F
vP /O
dt
F A
F vP /O
= le
(4.22)
= ez le = l 2 er
(4.23)
Adding Eq. (4.22) and Eq. (4.23) gives
F
aP /O = l 2 er + le
(4.24)
Then, adding Eq. (4.7) and Eq. (4.24), we obtain the acce
Chapter 4
Kinetics of a System of Particles
Question 41
A particle of mass m is connected to a block of mass M via a rigid massless
rod of length l as shown in Fig. P4-1. The rod is free to pivot about a hinge
attached to the block at point O. Furthermore
148
Chapter 4. Kinetics of a System of Particles
Solution to Question 41
Kinematics
Let F be a reference frame xed to the ground. Then, choose the following
coordinate system xed in reference frame F :
Ex
Ez
Ey
Origin at O at t = 0
=
To the Right
=
Into P
149
The velocity of point P in reference frame F is then given as
F
vP =
F
F
d
d
rP /O = F vO + F vP /O
(rO ) +
dt
dt
(4.10)
Now we already have F vO from Eq. (4.6). Next, since rP /O is expressed in the basis cfw_er , e , ez and cfw_er , e , ez rotates
144
Chapter 3. Kinetics of Particles
Finally, the force of gravity is given as
mg = mgey
(3.584)
Then, adding Eq. (3.576), Eq. (3.583), and Eq. (3.584), the resultant force acting
on the particle is then obtained as
(2x/a)ex + ey
+ Nb ez K(x x0 )ex mgey
(
143
Therefore,
en = eb et = ez
ex + (2x/a)ey
2x
1+
a
2
=
(2x/a)ex + ey
2x
1+
a
2
(3.573)
Suppose now that we dene
=
Then we can write
en =
2x
a
1+
2
(2x/a)ex + ey
(3.574)
(3.575)
The reaction force exerted by the parabola on the particle is then given as