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ENGINEERING MECHANICS
III Semester: ME
Course Code Category Hours / Week Credits Maximum Marks L T P C CIA SEE Total AMEB03 Core 3 0 0 3 30 70 100 Contact Classes: 45 Tutorial Classes: Nil Practical Classes: Nil Total Classes: 45
COURSE OBJECTIVES: The course should enable the students to: I. Students should develop the ability to work comfortably with basic engineering mechanics concepts required for analyzing static structures. II. Identify an appropriate structural system to studying a given problem and isolate it from its environment, model the problem using good free-body diagrams and accurate equilibrium equations. III. Understand the meaning of centre of gravity (mass)/centroid and moment of Inertia using integration methods and method of moments IV. To solve the problem of equilibrium by using the principle of work and energy, impulse momentum and vibrations for preparing the students for higher level courses such as Mechanics of Solids, Mechanics of Fluids, Mechanical Design and Structural Analysis etc...
MODULE-I INTRODUCTION TO ENGINEERING MECHANICS Classes: 10
Force Systems Basic concepts, Particle equilibrium in 2-D & 3-D; Rigid Body equilibrium; System of Forces, Coplanar Concurrent Forces, Components in Space – Resultant- Moment of Forces and its Application; Couples and Resultant of Force System, Equilibrium of System of Forces, Free body diagrams, Equations of Equilibrium of Coplanar Systems and Spatial Systems; Static Indeterminacy
MODULE-II FRICTION AND BASICS STRUCTURAL ANALYSIS Classes: 09
Types of friction, Limiting friction, Laws of Friction, Static and Dynamic Friction; Motion of Bodies, wedge friction, screw jack & differential screw jack; Equilibrium in three dimensions; Method of Sections; Method of Joints; How to determine if a member is in tension or compression; Simple Trusses; Zero force members; Beams &types of beams; Frames &Machines;
MODULE-III
Classes: 10
Centroid of simple figures from first principle, centroid of composite sections; Centre of Gravity and its implications; Area moment of inertia- Definition, Moment of inertia of plane sections from first principles, Theorems of moment of inertia, Moment of inertia of standard sections and composite sections; Mass moment inertia of circular plate, Cylinder, Cone, Sphere, Hook.
Virtual displacements, principle of virtual work for particle and ideal system of rigid bodies, degrees of freedom. Active force diagram, systems with friction, mechanical efficiency. Conservative forces and potential energy (elastic and gravitational), energy equation for equilibrium. Applications of energy method for equilibrium. Stability of equilibrium.
MODULE-IV PARTICLE DYNAMICS AND INTRODUCTION TO KINETICS
Classes: 08
Particle dynamics- Rectilinear motion; Plane curvilinear motion (rectangular, path, and polar coordinates). 3- D curvilinear motion; Relative and constrained motion; Newton’s 2nd law (rectangular, path, and polar coordinates). Work-kinetic energy, power, potential energy. Impulse-momentum (linear, angular); Impact (Direct and oblique). Introduction to Kinetics of Rigid Bodies covering, Basic terms, general principles in dynamics; Types of motion, Instantaneous centre of rotation in plane motion and simple problems.
Mechanics
It is defined as that branch of science, which describes and predicts the conditions of rest or motion of bodies under the action of forces. Engineering mechanics applies the principle of mechanics to design, taking into account the effects of forces.
Statics
Statics deal with the condition of equilibrium of bodies acted upon by forces.
Rigid body
A rigid body is defined as a definite quantity of matter, the parts of which are fixed in position relative to each other. Physical bodies are never absolutely but deform slightly under the action of loads. If the deformation is negligible as compared to its size, the body is termed as rigid.
Force
Force may be defined as any action that tends to change the state of rest or motion of a body to which it is applied.
The three quantities required to completely define force are called its specification or characteristics. So the characteristics of a force are:
Concentrated force/point load
Distributed force
Line of action of force
The direction of a force is the direction, along a straight line through its point of application in which the force tends to move a body when it is applied. This line is called line of action of force.
Representation of force
Graphically a force may be represented by the segment of a straight line.
Composition of two forces
The reduction of a given system of forces to the simplest system that will be its equivalent is called the problem of composition of forces.
Parallelogram law
If two forces represented by vectors AB and AC acting under an angle α are applied to a body at point A. Their action is equivalent to the action of one force, represented by vector AD, obtained as the diagonal of the parallelogram constructed on the vectors AB and AC directed as shown in thefigure.
P (^2) Q (^2) 2 PQ Cos 90 P
Case-III: If α = 90˚
α = tan-1^ (Q/P)
α P
Resolution of a force
The replacement of a single force by a several components which will be equivalent in action to the given force is called resolution of a force.
Action and reaction
Often bodies in equilibrium are constrained to investigate the conditions.
w
Free body diagram
Free body diagram is necessary to investigate the condition of equilibrium of a body or system. While drawing the free body diagram all the supports of the body are removed and replaced with the reaction forces acting on it.
Superposition and transmissibility
Problem 1: A man of weight W = 712 N holds one end of a rope that passes over a pulley vertically above his head and to the other end of which is attached a weight Q = 534 N. Find the force with which the man’s feet press against the floor.
Problem 2: A boat is moved uniformly along a canal by two horses pulling with forces P = 890 N and Q = 1068 N acting under an angle α = 60˚. Determine the magnitude of the resultant pull on the boat and the angles β and ν.
P = 890 N, α = 60˚ Q = 1068 N
β
α
ν
1068 sin 60
33
890 sin60
27
Resolution of a force
Replacement of a single force by several components which will be equivalent in action to the given force is called the problem of resolution of aforce.
By using parallelogram law, a single force R can be resolved into two components P and Q intersecting at a point on its line of action.
Equilibrium of collinear forces:
Equilibrium law: Two forces can be in equilibrium only if they are equal in magnitude, opposite in direction and collinear in action.
Problem 5: Determine the angles α and β shown in the figure.
Problem 6: Find the reactions R 1 and R 2.
Problem 7: Two rollers of weight P and Q are supported by an inclined plane and vertical walls as shown in the figure. Draw the free body diagram of both the rollers separately.
Lami’s theorem
If three concurrent forces are acting on a body kept in an equllibrium, then each force is proportional to the sine of angle between the other two forces and the constant of proportionality issame.
Ra
sin90 sin 180 sin 90
Problem: A ball of weight Q = 53.4N rest in a right angled trough as shown in figure. Determine the forces exerted on the sides of the trough at D and E if all the surfaces are perfectlysmooth.
Problem: An electric light fixture of weight Q = 178 N is supported as shown in figure. Determine the tensile forces S 1 and S 2 in the wires BA and BC, if their angles of inclination aregiven.
sin135 sin150 sin
S = Psecα
S 1 cos 30 20 sin 60 S 2 sin 30 3 S (^20 3) S 2 2 1 S 2
1
S 2 3 S 1 20 (1)
S 1 sin 30 S 2 cos 30 Sd cos 60 20 S 1 S 2 2
2
S 1 3 S 2 60 (2)
Substituting the value of S 2 in Eq.2, we get
l
Problem: A ball of weight W is suspended from a string of length l and is pulled by a horizontal force Q. The weight is displaced by a distance d from the vertical position as shown in Figure. Determine the angle α, forces Q and tension in the string S in the displacedposition.
d l
d
l^2 d^2 l
Applying Lami’s theorem,
S
d^2 l^2
