INTRODUCTION TO ENGINEERING MECHANICS - SPP.pptxSamirsinh Parmar
Engineering Mechanics, Mechanics, Scalars, Vectors, Force system, Measurment units, Concept of force, system of forces, Idealized Mechanics, Fundamental Concepts, Scalar and vector Operations, Accuracy of Engineering Calculation, Problem Solving Approach
Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. Visit us: https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
INTRODUCTION TO ENGINEERING MECHANICS - SPP.pptxSamirsinh Parmar
Engineering Mechanics, Mechanics, Scalars, Vectors, Force system, Measurment units, Concept of force, system of forces, Idealized Mechanics, Fundamental Concepts, Scalar and vector Operations, Accuracy of Engineering Calculation, Problem Solving Approach
Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. Visit us: https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
In Engineering Mechanics the static problems are classified as two types: Concurrent and Non-Concurrent force systems. The presentation discloses a methodology to solve the problems of Concurrent and Non-Concurrent force systems.
This document gives the class notes of Unit 2 stresses in composite sections. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Some basic defintions of the topics used in Strength of Materials subject. Pictorial presentation is more than details. Many examples are provided as well.
Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. Visit us: https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
Introduction to kinesiology (Biomechanics- Physiotherapy) vandana7381
Chapter 1: Introduction to Kinesiology ( Biomechanics) for physical therapy students.
Reference: JOINT STRUCTURE AND FUNCTION - by Pamela K. Levangie.
Easy to understand and with lot of examples.
In Engineering Mechanics the static problems are classified as two types: Concurrent and Non-Concurrent force systems. The presentation discloses a methodology to solve the problems of Concurrent and Non-Concurrent force systems.
This document gives the class notes of Unit 2 stresses in composite sections. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Some basic defintions of the topics used in Strength of Materials subject. Pictorial presentation is more than details. Many examples are provided as well.
Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. Visit us: https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
Introduction to kinesiology (Biomechanics- Physiotherapy) vandana7381
Chapter 1: Introduction to Kinesiology ( Biomechanics) for physical therapy students.
Reference: JOINT STRUCTURE AND FUNCTION - by Pamela K. Levangie.
Easy to understand and with lot of examples.
Study Unit
Ill Engineerin M
Part4
an1cs
By
Andrew Pytel, Ph.D.
Associate Professor, Engineering Mechanics
The Pennsylvania State University
When you complete this study unit, you'll be able to
• Calculate the mass moment of inertia
• Calculate the kinetic energy of a body
• Determine the linear impulse and momentum of a body
• Analyze the equations and conditions used to determine the forces involving rectilinear
translation
• Describe centripetal and centrifugal force
• Describe the forces that impact the rotation of a rigid body without translation
• Explain the motion of a wheel, and calculate the magnitude of the linear acceleration and
friction forces
• Analyze the work-energy method as it applies to the motion and action of a body
iii
PRELIMINARY EXPLANATIONS PERTAINING TO KINETICS .
FORCE-MASS-ACCELERATION METHOD .....
Translation of Rigid Body
Rotation of Rigid Body without Translation
General Plane Motion of Rigid Body
23
WORK-ENERGY METHOD . . . . . . . . . . . . . . . . . . . . . . . . . 53
Application of Method for Translation
Other Applications of Work-Energy Method
IMPULSE-MOMENTUM METHOD . . . . . . .
Rectilinear Translation of Single Body
Collision of Two Bodies
PRACTICE PROBLEMS ANSWERS
EXAMINATION . . . . . . . . . .
. ........... 77
93
95
Engineering Nlechanics, Part 4
PRELIMINARY EXPLANATIONS PERTAINING
TO KINETICS
Scope of This Text
1
1 • In the preceding texts on engineering mechanics, we have discussed
separately the relations of forces in a system and the conditions of mo-
tion of bodies. In this text, we shall consider the relation between the
motion of a body and the force or forces acting on the body to produce
the motion. The basis for the relationship between motion and force
is Newton's second law of motion. However, there are three different
methods of applying this law. These are commonly called the force-
mass acceleration method, the work-energy method, and the impulse-
momentum method. Each method is most useful for solving certain
types of problems.
Statement of Newton's Second Law of Motion
2 • In Engineering Mechanics, Part 1, Newton's second law of motion was
stated as follows:
If a resultant force acts upon a particle, the particle will be accelerated
in the direction of the force. Furthermore, the magnitude of the accel-
eration will be directly proportional to the magnitude of the resultant
force and inversely proportional to the mass of the particle.
Newton's second law can be expressed mathematically by the following
equation:
F
a=k-
m
in which a = magnitude of the acceleration of a particle
k = a numerical factor
F = magnitude of the force acting upon the particle
m = mass of the particle
(1)
The mass of a particle is a measure of the exact amount of matter in
the particle. Any body is composed of a number of particles, and the
mass of a body is the sum of the masses of all the particles.
Concept of Particles and Free Body Diagram
Why FBD diagrams are used during the analysis?
It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
It helps to identify the forces and ensures the correct use of equation of equilibrium.
Note:
Reactions on two contacting bodies are equal and opposite on account of Newton's III Law.
The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces.
Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable.
Physical Meaning of Equilibrium and its essence in Structural Application
The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium.
A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium.
The rigid body in equilibrium means the body is stable.
Equilibrium means net force and net moment acting on the body is zero.
Essence in Structural Engineering
To find the unknown parameters such as reaction forces and moments induced by the body.
In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters.
For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.
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Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...Dr.Costas Sachpazis
Terzaghi's soil bearing capacity theory, developed by Karl Terzaghi, is a fundamental principle in geotechnical engineering used to determine the bearing capacity of shallow foundations. This theory provides a method to calculate the ultimate bearing capacity of soil, which is the maximum load per unit area that the soil can support without undergoing shear failure. The Calculation HTML Code included.
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Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
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Technical Specifications
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
Key Features
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface
• Compatible with MAFI CCR system
• Copatiable with IDM8000 CCR
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
Application
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptxR&R Consult
CFD analysis is incredibly effective at solving mysteries and improving the performance of complex systems!
Here's a great example: At a large natural gas-fired power plant, where they use waste heat to generate steam and energy, they were puzzled that their boiler wasn't producing as much steam as expected.
R&R and Tetra Engineering Group Inc. were asked to solve the issue with reduced steam production.
An inspection had shown that a significant amount of hot flue gas was bypassing the boiler tubes, where the heat was supposed to be transferred.
R&R Consult conducted a CFD analysis, which revealed that 6.3% of the flue gas was bypassing the boiler tubes without transferring heat. The analysis also showed that the flue gas was instead being directed along the sides of the boiler and between the modules that were supposed to capture the heat. This was the cause of the reduced performance.
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Work done in cooperation with James Malloy and David Moelling from Tetra Engineering.
More examples of our work https://www.r-r-consult.dk/en/cases-en/
Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
Immunizing Image Classifiers Against Localized Adversary Attacksgerogepatton
This paper addresses the vulnerability of deep learning models, particularly convolutional neural networks
(CNN)s, to adversarial attacks and presents a proactive training technique designed to counter them. We
introduce a novel volumization algorithm, which transforms 2D images into 3D volumetric representations.
When combined with 3D convolution and deep curriculum learning optimization (CLO), itsignificantly improves
the immunity of models against localized universal attacks by up to 40%. We evaluate our proposed approach
using contemporary CNN architectures and the modified Canadian Institute for Advanced Research (CIFAR-10
and CIFAR-100) and ImageNet Large Scale Visual Recognition Challenge (ILSVRC12) datasets, showcasing
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2. EngineeringEngineering
The profession in which knowledge of the mathematical and
natural sciences gained by study, experience, and practice is
applied with judgment to develop ways to use economically the
materials and forces of nature for the benefit of mankind
OR
The discipline dealing with the art or science of
applying scientific knowledge to practical problems.
3. What is MechanicsWhat is Mechanics
Mechanics is a branch of the physical sciences that is concerned
with the state of rest or motion of bodies that are subjected to the
action of forces.
• Knowledge of Engineering Mechanics is very essential for an engineer in
planning, designing and construction of various types of structures &
machines.
4. MechanicsMechanics
Mechanics is divided into two areas;
• Satatics &
• Dynamics
Satatics deals with the equilibrium of
bodies , those that are either at rest or
move with a constant velocity
OR
Statics is branch of mechanics, which deals
with the forces and their effects, while
acting upon the bodies at rest.
5. MechanicsMechanics
Dynamics
• Dynamics is concerned with the accelerated motion of bodies
OR
• Dynamics is branch of mechanics, which deals with the forces and their
effects, while acting upon the bodies in motion.
6. Rigid BodyRigid Body
• A body is said to be rigid if the position of its various particles remain fixed
relative to one another.
• A rigid body can be considered as a combination of a large number of
particles in which all the particles remain at a fixed distance from one another,
both before and after applying a load.
7. MassMass
It is matter contained in a body. Mass is scalar property of matter that does not
change from one location to another.
8. WeightWeight
• It is force by which the body is attracted towards the center of earth.
• The weight of an object is defined as the force of gravity on the object and
may be calculated as the mass times the acceleration of gravity,
w = mg.
• Since the weight is a force, its SI unit is the Newton.
9. Newton’s laws of motionsNewton’s laws of motions
• Newton's First Law of Motion:
It states that Every object in a state of rest or of uniform motion tends to remain
in that state unless an external force is applied to it.
10. Newton’s laws of motionsNewton’s laws of motions
• Newton's Second Law of Motion:
It states that “the rate of change of momentum is directly proportional to the
force and takes place in the same direction in which the force acts”.
The relationship between an object's mass m, its acceleration a, and the applied
force F is F = ma.
• Newton's Third Law of Motion
For every action there is an equal and opposite reaction.
11. Newton’s Law of Gravitational AttractionNewton’s Law of Gravitational Attraction
• After formulating his three laws of motion, Newton postulated a law governing the
gravitational attraction between any two particles, Stated mathematically,
12. ForceForce
• Force is considered as “push” or “pull” exerted by one body on another.
• Its unit is N.
• When force is applied to an object , the velocity of that object changes. This
change in velocity constitutes an acceleration.
• Force is a vector quantity, therefore a force is completely described by;
magnitude, direction and point of application
13. Types of ForcesTypes of Forces
• Concurrent Force System
The force system in which line of action of
forces intersects through a common point.
• Parallel Force System
The force system in which line of action of
forces are parallel to each other.
14. Types of ForcesTypes of Forces
• Collinear Force System
The force system in which line of action of
forces lie on the same path.
--------------------------------------------
--------------------------------------------
• Coplanar Force System
The force system in which line of action of
forces lie on the same plane.
15. Force systemForce system
• Concentrated Force.
A concentrated force represents the effect of a loading which is assumed to act at a
point on a body. We can represent a load by a concentrated force, provided the area
over which the load is applied is very small compared to the overall size of the body. An
example would be the contact force between a wheel and ground.
16. Units of MeasurementUnits of Measurement
Fundamental Units
• Every quantity is measured in terms of some arbitrary, but internationally accept
units, called fundamental units.
• Length
• Mass and
• Time
Derived Units
• Some units are expressed in terms of other units, which are derived from some
fundamentals units are known as derived units.
• E.g. units of force, area, velocity, acceleration, pressure etc.
17. Units of MeasurementUnits of Measurement
• The basic quantities - length, time, mass, and force are not all independent from
one another; in fact, they are related by Newton’s second law of motion.
• The equality F = ma is maintained only if three of the four units, called base units
(length, time, mass) , are defined and the fourth unit (force) is then derived from
the equation.
18. System of unitsSystem of units
International System of units (SI Units)
•The International System of units, abbr. SI is a modern version of the metric system
which has received worldwide recognition. As shown in Table 1–1, the SI system
defines length in meters (m), time in seconds (s), and mass in kilograms (kg). The unit
of force, called a newton (N), is derived from thus, 1 N is equal to a force required to
give 1 kilogram of mass an acceleration of 1 m/s2
(N = kg x m/s2
).
F = maF = ma
19. System of unitsSystem of units
• If the weight of a body located at the “standard location” is to be determined in newtons,
then Eq. W = mg must be applied. Here measurements give g = 9.806 65 m/s2,
however for
calculations, the value g = 9.81 m/s2
is be used.
• Thus, W = mg (g = 9.81 m/s2
), Therefore, a body of mass 1 kg has a weight of 9.81 N, and
2-kg body weighs 19.62 N, and so on, Fig. (a).
21. System of unitsSystem of units
U.S. Customary system of units (FPS Units)
•In the U.S. Customary system of units (FPS) length is measured in feet (ft), time in
seconds (s), and force in pounds (lb),
•Table 1–1.The unit of mass, called a slug, is derived from
•Hence,1 slug is equal to the amount of matter accelerated at 1 ft/s2
when acted upon
by a force of 1 lb (slug = lb.s2
/ft)
F = maF = ma
22. System of unitsSystem of units
• Therefore, if the measurements are made at the “standard location,” where g = 32.2
ft/s2
then from Eq. W = mg OR m = W/g,
(g = 32.2 ft/s2
)
• So a body weighing 32.2 lb has a mass of 1 slug, a 64.4-lb body has a mass of 2 slugs,
and so on.. .Fig-b
23. Conversion of UnitsConversion of Units
Table 1–2 provides a set of direct conversion factors between FPS and SI units for
the basic quantities.
We also know, in the FPS system,
1 ft = 12 in. (inches),
5280 ft = 1 mi (mile),
1000 lb=1 kip (kilo-pound),and 2000 lb=1ton
24. PrefixesPrefixes
• When a numerical quantity is either very large or very small, the units used to define
its size may be modified by using a prefix.
25. General Procedure for AnalysisGeneral Procedure for Analysis
• The most effective way of learning the principles of engineering mechanics is to
solve problems.
• It is important to always present the work in a logical and orderly manner as
suggested by the following sequence of steps:
• Read the problem carefully and try to correlate the actual physical situation
with the theory studied.
26. General Procedure for AnalysisGeneral Procedure for Analysis
1. Tabulate the problem data and draw any necessary diagrams.
2. Apply the relevant principles, generally in mathematical form. Be sure equations
are dimensionally homogeneous.
3. Solve the necessary equations, and report the answer with three significant
figures.
4. Study the answer with technical judgment and common sense to determine
whether or not it seems reasonable.
34. Finding Unknown AnglesFinding Unknown Angles
• For solving problems it is necessary to find the required unknown angle by applying basic
geometric facts.
• As you have studied in lower classes, the basic facts about angles, triangles and
quadrilaterals, some important are given here.
35. Angle FactsAngle Facts
• Vertical angles have equal measure.
• The sum of adjacent angles on a straight line is 1800
.
36. • The sum of adjacent angles around a point is 3600
• Triangle Facts
The angle sum of any triangle is 1800
(The three angles always add to 180°)
37. Triangle FactsTriangle Facts
• When one angle of a triangle is a right
angle, the other two angles add up to
900
• The exterior angle of a triangle is equal
to the sum of the interior
opposite angles.
38. Triangle FactsTriangle Facts
• Base angles of an isosceles
triangle are equal.
(A triangle containing two equal sides and
angels)
• Each interior angle of
equilateral triangle is 600
(3 equal sides, 3 equal angles, always 60°)
Scalene Triangle: No equal sides, No equal
angles
39. Quadrilateral FactsQuadrilateral Facts
• Opposite angles in a parallelogram are equal.
• We already know that the sum of interior angles
of a triangle is 1800,
and sum of Quadrilateral is
360
40. • For a right triangle:
• a2
+ b2
= c2
• sin(θ) = b/c =opposite side/hypotenuse
• cos(θ) = a/c = near side/hypotenuse
• tan(θ) = b/a =opposite side/near side
41. For a general triangle:
α + β + γ = 180ο
•Sine law:
•Cosine law: