Engineering Mechanics :
STATICS
BDA1023
Lecture #02
Group of FKMP Lecturers
University Tun Hussein Onn Malaysia (UTHM),
Faculty of Mechanical and Manufacturing,
Department of Mechanical Engineering
What is Mechanics?
The branch of applied mathematics dea
BDA
10203
Engineering Mechanics: Statics
1
GOALS
This course is designed to
introduce the basic principles
and concepts of statics and its
application in the field of
engineering.
2
LEARNING OUTCOMES
After studying this topic, students will be
able to:
Ap
Microprocessor Systems
[1](8 marks) What happens during a read bus cycle in each of the following states?
a) T1: Addressisplacedonbus;ALE,DT/RandIO/Mactivated2
b) T2: RDsignalandDENactivated;READYsampledattheend
c) T3: IfREADYwaslow,thisisawaitstateandREA
Microprocessors And Assembly Language
Programming
Object1
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Chapter 4 : Microprocessor Development System
POINTS TO REMEMBER
Microprocessor development systems
consist of peripherals, large memory,
mass storag
REVISION QUESTIONS
a) Discuss three 8086 interrupt classes
b) Explain the different types of port addressing modes and give an example for each case.
c)
Give the sum and the flag settings for SF, ZF, CF, OF, and PF after subtracting 3AB1 from each of the
DEDAN KIMATHI UNIVERSITY OF TECHNOLOGY
DEPARTMENT OF ELECTRICAL AND ELECTRICS ENGINEERING
FOUTH YEAR BACHELOR OF SCIENCE
IN MECHANICAL ENGINEERING
Assignment 1
Figure 2 shows a mechatronic system that uses the 8085 microprocessor and Pulse
Width Modulatio
EMG 2312 Metrology
Course Outline
1. Introduction to Metrology: standards of measurements; wavelength standards, line
and end-standards. System of international standards.
2. Mathematical concepts in metrology; errors, precision and accuracy.
3. Linear me
4.0
Part Durability and Life Limitation
Many mechanical components have life limits in their service due to the deterioration of their
durability over time. These limitations are either explicitly defined by mandating the
replacement of the part within a
Intellectual Property (IP) Rights
Intellectual properties are protected through patent, copyright, trade secret, trademark, service
mark, and mask work. Patents protect new, useful, and nonobvious inventions. Copyright
provides protection for an expressio
Product analysis
It is essential to meet the form, fit, and function requirements, and other design details of a
component. In a reverse engineering process, the parts physical features are determined by
measuring its geometric dimensions, and the toleran
Intellectual Property (IP) Rights
Intellectual properties are protected through patent, copyright, trade secret, trademark, service
mark, and mask work. Patents protect new, useful, and nonobvious inventions. Copyright
provides protection for an expressio
Product analysis
It is essential to meet the form, fit, and function requirements, and other design details of a
component. In a reverse engineering process, the parts physical features are determined by
measuring its geometric dimensions, and the toleran
DEDAN KIMATHI UNIVERSITY OF TECHNOLOGY
UNIVERSITY EXAMINATIONS 2015/2016
FIFTH YEAR FIRST SEMESTER EXAMAMINATION FOR THE DEGREE
OF BACHELOR OF SCIENCE IN MECHANICAL ENGINEERING
EMG 2506 REVERSE ENGINEERING
DATE: 10TH SEPTEMBER 2015
TIME: 2.00PM 4.00PM
INS
Dedan Kimathi University
5th Year 1st Semester
EMG 2506
Reverse Engineering
Course description
Importance and purpose of reverse engineering, Intellectual property (IP) rights, Product
analysis, Stages of reverse engineering.
Sketching and detailing of a
QUALITY CONTROL
Introduction
Quality is an important factor when it comes to any product or service. With the high market
competition, quality has become the market differentiator for almost all products and services.
Therefore, all manufacturers and serv
DEDAN KIMATHI UNIVERSITY OF TECHNOLOGY
BSC MECHANICAL ENGINEERING
EMG 2506: REVERSE ENGINEERING
Preared By: Dr. J. N. KERAITA
APRIL 2016
SYLLABUS
Purpose
The purpose of this course is to enable the student to;
1. Appreciate the real world of design and pr
DEDAN KIMATHI UNIVERSITY OF TECHNOLOGY
BSC MECHANICAL ENGINEERING
EMG 2506: REVERSE ENGINEERING
Preared By: Dr. J. N. KERAITA
APRIL 2016
SYLLABUS
Purpose
The purpose of this course is to enable the student to;
1. Appreciate the real world of design and pr
Product analysis
It is essential to meet the form, fit, and function requirements, and other design details of a
component. In a reverse engineering process, the parts physical features are determined by
measuring its geometric dimensions, and the toleran
http:/bama.ua.edu/~mweaver/courses/MechBeh/Day13.pdf
https:/en.wikipedia.org/wiki/Crystal_structure
Theoretical Cleavage stress
E 0
c
d0
Here,
is energy
0
is modulus
E
is inter planar spacing
d0
Inter planar spacing
1
1
1
d0 2 a 2 c 2
1
d0 2
1
0.7752
1
5
Force equation,
s
Fx k s d d 2 2 k mg k k s 2 d 2 d
s d
k kd 2
kd s
ks
k mg k kd
2
2
2
2
s d
s d
41.6 1.5 s
41.6s
0.18 2.5 9.81
2
2
s 1.5
2
0.18 41.6 1.5
0.18
41.6
1.5
2
2
s 1.5
62.4s
16.848
41.6 s
4.4145 11.232
2
2
2
2
s 1
the force elongation curve of a metal shown measured on a cylindrical rod
Force versus total length:
From the diagram,
At force
F 150 kN
, total length is
Stress,
F
A
150 103
0.025 2
4
3.05577 108 N/m 2
305.577 MPa
Strain,
l l0
l0
15.27 15
15
0.018
So
Equation (1)
(7)
Az C y
Equation (6)
Az Bz 692.82 0
Substitute equation (1) in equation (6)
C y Bz 692.82 0
8
Bz C y 692.82
Equation (2),
Bz 1 C x 2 0
Substitute
Bz C y 692.82
in equation (2)
Bz 1 C x 2 0
C y 692.82 1 Cx 2 0
2C x C y 692.82
C x 0.5C
(A)
Joint B
Horizontal equilibrium
TAB cos50 TBC cos 60
TAB cos50 20000cos 60
TAB 15557.238 lb
Vertical equilibrium
WD TAB sin 50 TBC sin 60
15557.238sin 50 20000sin 60
29238 lb
(D)
Equivalent spring constant for spring 1 and 2
Both springs 1 and 2 are
Variables,
Shear stress,
Viscosity,
Velocity,
Height,
N / m 2 ML1T 2
Pa.sec ML1T 1
du m / sec LT 1
dy meter L
Number of
n 43
terms
1
Find the
term
B
C
A du dy
LT
1 1 A
ML T
Compare
A 1 0
1 B
L C ML1T 2
M coefficient
A 1
Compare
T coefficient
(a)
Ratio,
di
t
di d 0 2 t
t
t
d0 2 t
t
3
5 2
32
3 / 32
51.33
Ratio,
di
20
t
so given cylinder thin cylinder
(b)
Hoops stress and longitudinal stress
pd
hoop 0
2t
380 5
3
2
32
10133.33 psi
pd
long 0
4t
380 5
3
4
32
5066.66 psi
Axi
The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the tension in the cable is 1600 Ib, determine
a) The moment about the diagonal AD of the force exerted on the
In truss, each node has two degree of freedom along x and
y directions. Each element has two nodes, so stiffness
matrix has order of
.
4 4
Hence, the stiffness matrix has does not depends on number
of elements of the nodes.
Stiffness matrix,
l2
lm l 2 lm
The conveyor belt is moving to the right at v = 8 ft>s, and at
the same instant the cylinder is rolling counterclockwise at
v = 2 rad>s without slipping. Determine the velocities of
the cylinders center C and point B at this instant
Belt velocity
v 13 i f