Handout 1

Applied Mechanics

MIME 2101N

Introduction + Course Planning

Dr. C. SubramanianLecturerMechanical SectionEngineering DepartmentShinas College of Technology

Dr .C .Subramanian Applied Mechanics I

Mechanics is a physical science which deals with bodies at rest or motion under the action of forces.

Mechanics is all about

Mechanical Aerospace Bio Mechanics Aerodynamics

APPLIED MECHANICS This class is fundamental to your

success in any course which involves solids or fluids. I expect to help you be successful in this class and for you to leave this class well prepared to succeed in dynamics and strength of materials and subsequently machine design, structural analysis.Prerequisites

Vectors, calculus don’t worry it is not so difficult


Conduct of the Course

• Two Theory hours per week and two Practical hours per week including lectures and tutorial sessions.

• Solve lots of problems

• There will be assignments, 4 quizzes (Best Two) one mid term exams and a final examination. There will be a great deal of hands on and observed problem solving in the class. History has shown Regular attendance is necessary to be successful in the class. Quizzes will be short (12 minutes) at the end of class.



Objectives OutcomesThis course should enable the student to

A student who satisfactorily complete the course should be able to:

1.Understand the laws and the principles that govern static.2. Perceive the basic concept in the field of this subject.3. Model and analyze static engineering problems.4. Lay the ground for various courses in engineering.

1. Recognize common equilibrium problems.2. Grasp the condition for transitional and rotational equilibrium and form the proper equation of equilibrium3. Use the pictorial representation of equilibrium situation in terms of free -body diagram.4. Realize the difference between equilibrium force and the resultant force.5. Distinguish between the various forces and stresses arising in a problem such as the internal, external, tensile, compressive, direct , shear and other loading conditions, etc.6. Define centroid, center of gravity and center of mass of a rigid body and appreciate their location and significances.7. Define moment of inertia of mass and area and grasping methods of computing each about any axis.8. Handle various structural problems and utilizing sections and joint methods.9. Distinguish between various types of friction.10. Analyze beams in terms of shearing forces and bending moment under various boundary conditions.11. Carry out laboratory experiment to verify the conditions of equilibrium of forces, analyze beams, determine coefficient of static and kinetic friction and other topics related to the static’s of bodies, frames, etc

Dr.C.Subramanian Applied Mechanics I

Type of Course Course work Midterm Exam

Final Exam

Pure theoretical (3T)

30T marks(30%)

20T marks(20%)

50T marks(50%)

Pure practical (6P)

30P marks(30%)

50P+ 20P* marks(50% )+ (20%)

Mixed (2T+2P) 15T+15P marks(30%)

20T marks(20%)

35T+ 15P# marks(35%) + (15%)

Mixed (1T+4P) 10T+20P(30%)

5T+15P(20%)

15T+35P(50%)

Assessment Plan

MARKS DISTRIBUTION FOR ENGINEERING COURSES

THEORY + PRACTICAL BASED COURSES// ASSESSMENT METHOD


Type of Assessment Marks (%)

Course work

Quizzes 10 marks

Structured assignments 05 marks

15T+15P marks (30%)15 T

Technical assignments

(Technical reports) 15 marks15P

Mid- Term Examination 20 marks 20T marks (20%)

Final Examination

Theory exam 35 marks

35T+ 15P# marks (50%)35 T

#Continuous assessment (Practical)

A minimum of 5 practical assessments

10 marks

Criteria

(Individual performance,

team work, time frame,

attendance) 05 marks

15 P#

Grand Total 100%

What is Mechanics?

Mechanics is a physical science which deals with bodies at rest or motion under the action of forces.

Mechanics is an applied science

Categories of Mechanics:

Rigid bodies- Statics or DynamicsDeformable bodiesFluids

Continuum hypothesis



What is rigid body?

Rigid Bodies: Rigid bodies are those which do not deform under the action of applied forces

Deformable Bodies: Deformable bodies are those which deform under the action of applied forces.

What is continuum hypothesis?

F1

F2

F4

F3

Continuum mechanics – no voids are present “material is continuous”

Basic assumption in solving mechanics problem

Engineering Mechanics

Mechanics of Solids Mechanics of Fluids

Deformable Bodies Rigid Bodies

Statics(Body is at rest)

Dynamics(Body is in motion)

Kinematics Kinetics

Classification of study of Engineering Mechanics

F

Effect of forces on objects either at rest or in motion - Engineering Mechanics F

Force tends to move the body

Force tends to rotate the body

Introduction to Mechanics


Effect of forces


1. Recognize common equilibrium problems.2. Grasp the condition for transitional and rotational equilibrium and form the proper equation of equilibrium3. Use the pictorial representation of equilibrium situation in terms of free -body diagram.4. Realize the difference between equilibrium force and the resultant force.

Out come coverage

1.Understand the laws and the principles that govern static.2. Perceive the basic concept in the field of this subject

Objective coverage


Scalar and Vector Quantities

Scalar Quantities : Physical quantity which defined only by magnitude. Scalar quantities are volume, density, speed, energy, mass, time and work.

For example: Volume of the bottle is 1 litre Mass of the car is 1700kg.

Vector Quantities: These are defined by both magnitude and direction. Vector quantities are displacement, velocity, acceleration, force, moment and momentum.

Vector Approach for solving Problems

A vector may be represented by a straight line, the length of line being directly proportional to the magnitude of the quantity and the direction of the line being in the same direction as the line of action of the quantity. An arrow is used to denote the sense of the vector, that is, for a horizontal vector, say, whether it acts from left to right or vice-versa.

Force: Action of one body on another; characterized by its point of application, magnitude, line of action, and sense


Fundamental Principles of Mechanics

• Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or continue to move in a straight line.Importance of first law

•Deals with equilibrium of particle or body•Involved mostly in solving static problems

Idealization as a particle

Particle. A particle has a mass but the size can be neglected during analysis .

Illustrate with video


Newton’s Second Law: A particle will have an acceleration proportional to a nonzero resultant applied force.

amF

Importance of second lawInvolved mostly in solving dynamic problemsImportance of mass was recognized

Newton’s Third Law: The forces of action and reaction between two particles have the same magnitude and line of action with opposite sense.Importance of Third law•Involved mostly in solving static problems•Basis for drawing the free body diagram indicating the action and reaction


Parallelogram Law of Addition

It states that if two vectors (Say P and Q) acting at a point to be represented in magnitude and direction by the two adjacent sides of parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram at that point.

R = P + Q

Triangle rule for vector addition

R = P + Q

Law of Cosine

Let us consider a triangle ABC with sides a, b, & c, and included angles α, β, & γ where α, β, & γ ≠90°

α

β

γ

a

b

c

a2 = b2 +c2 – 2bc cos α. b2 = c2 +a2 – 2ca cos βc2 = a2 +b2 – 2ab cos γ.

Law of Sine

Sinγc

Sinβb

sinαa


Force system

Coplanar Force System Non-Coplanar Force System

Coplanar Collinear Force System

Coplanar Concurrent Force System

Coplanar Non-Concurrent Force System

Coplanar Parallel Force System

Coplanar Unlike Parallel Force System

Coplanar Like Parallel Force System

Non-Coplanar Concurrent Force System

Non-Coplanar Parallel Force System

Non-Coplanar Non Concurrent Non Parallel Force System

TYPES OF FORCE SYSTEM



The system in which the forces, whose lines of action lie on the same line and in the same plane is called coplanar collinear force system


The system in which the forces meet at one point and lie in the same plane is called coplanar concurrent force system. The concurrent forces may or may not be collinear.

Non-Coplanar concurrent Force System

The system in which the forces meet at one point and lie in a different plane is called coplanar concurrent force system



Non-Coplanar concurrent Force System

The two forces act on a bolt at A. Determine their resultant.

Problem 1

A parallelogram with sides equal to P and Q is drawn to scale. The magnitude and direction of the resultant or of the diagonal to the parallelogram are measured

35N 98 R

A triangle is drawn with P and Q head-to-tail and to scale. The magnitude and direction of the resultant or of the third side of the triangle are measured.

35N 98 R

Graphical Solution


Trigonometric solution - Apply the triangle rule.

155cosN60N402N60N40

cos222

222 BPQQPR

AA

RQBA

BR

AQ

2004.15

N73.97N60155sin

sinsin

sinsin

N73.97R

Law of Sines

04.35

Law of Cosines,



A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is 5000 N directed along the axis of the barge, determine

The tension in each of the ropes for a = 45o

N2600N3700 21 TT

• Graphical solution - Parallelogram Rule with known resultant direction and magnitude, known directions for sides.

Trigonometric solution - Triangle Rule with Law of Sines

105sin

N5000

30sin45sin21 TT

N2590N3660 21 TT

Problem 2


Using the triangle rule and the Law of Cosines,Have:β = 180° − 45°β = 135°Then:R2 = 900 2 + 600 2 − 2 (900) (600) cos 135°or R = 1390.57 N

Using the Law of Sines,600 /sinγ = 1390.57/sin135or γ = 17.7642°and α = 90° − 17.7642°α = 72.236°

Tutorial Problem 1

Three problems will be given for tutorial class separately

Rectangular Components of a Force: Unit Vectors

• Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components.

Fx and Fy are referred to as the scalar components of

jFiFF yx

F

• May resolve a force vector into perpendicular components so that the resulting parallelogram is a rectangle. are referred to as rectangular vector components and

yx FFF

yx FF

and

• Define perpendicular unit vectors which are parallel to the x and y axes.

ji

and


Addition of Forces by Summing Components

SQPR

• Wish to find the resultant of 3 or more concurrent forces,

jSQPiSQP

jSiSjQiQjPiPjRiR

yyyxxx

yxyxyxyx

• Resolve each force into rectangular components

x

xxxxF

SQPR

• The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces.

y

yyyy

F

SQPR

x

yyx R

RRRR 122 tan

• To find the resultant magnitude and direction,


Four forces act on bolt A as shown. Determine the resultant of the force on the bolt.

SOLUTION:

• Resolve each force into rectangular components.

• Calculate the magnitude and direction of the resultant.

• Determine the components of the resultant by adding the corresponding force components.


Problem 3

SOLUTION:• Resolve each force into rectangular

components.

9.256.96100

0.1100110

2.754.2780

0.759.129150

4

3

2

1

F

F

F

F

compycompxmagforce

22 3.141.199 R N6.199R

• Calculate the magnitude and direction.

N1.199

N3.14tan 1.4

• Determine the components of the resultant by adding the corresponding force components.

1.199xR 3.14yR


Follow the same procedure and solve problem 1 in the class

Tutorial Problem 2


Knowing that α = 65°, determine the resultant of thethree forces shown

Tutorial Problem 3

Selecting the x axis along aa , we write′Rx = ΣFx = 300 N + (400 N)cosα + (600 N)sinα (1)Ry = ΣFy = (400 N)sinα − (600 N)cosα (2)

Start with the tail and end with a head

Come on do it use the calculators properly

Substitute α = 65°, Come on do it use the calculators properly

Beer and Johnston Ex. Pbm 41

Equilibrium of a Particle

• When the resultant of all forces acting on a particle is zero, the particle is in equilibrium.

• Particle acted upon by two forces:- equal magnitude- same line of action- opposite sense

• Particle acted upon by three or more forces:- graphical solution yields a closed polygon- algebraic solution

00

0

yx FF

FR

• Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line.



Free-Body Diagrams

Free-Body Diagram: A sketch showing only the forces on the selected particle.

Space Diagram: A sketch showing the physical conditions of the problem.


Conditions of Equilibrium (or) Equations of Equilibrium of particle Equilibrium equations are written as•Σ Fx = 0•Σ Fy = 01.The algebraic sum of the magnitudes of the vertical components of all the forces acting on the body is zero.2.The algebraic sum of the magnitudes of the horizontal components of all the forces acting on the body is zero.


Problem 4

Two cables tied together at C are loaded as shown. Knowing that W = 190 N, determine the tension (a) in cable AC, (b) in cable BC.

(a) TAC = 169.6 N(b) TBC = 265 N

Free-Body Diagram at C

I will do the problem in the class


Tutorial Problem 5Two traffic signals are temporarily suspended from a cable as shown. Knowing that the signal at B weighs 300 N, determine the weight of the signal at C.

Beer and Johnston Ex. Pbm 48

I will do the problem along with you. come on attempt it

Free-Body Diagram at B

Free-Body Diagram at C

TBC = 565.34 NWC = 97.7 N


Lami's Theorem

Three coplanar, concurrent and non-collinear forcesWhen an object is in static equilibrium, According to the theorem

where A, B and C are the magnitudes of three coplanar, concurrent and non-collinear forces, which keep the object in static equilibrium, and α, β and γ are the angles directly opposite to the forces A, B and C respectively.

Only applicable to

Example :

Free-Body Diagram

100sin

N736

140sin120sinAcAB TT

http://en.wikipedia.org/wiki/Coplanar

http://en.wikipedia.org/wiki/Concurrent

http://en.wikipedia.org/wiki/Coplanarity

http://en.wikipedia.org/wiki/Static_equilibrium

Documents

Handout 1