## What is Engineering Mechanics

Table of Contents

**What is Engineering Mechanics**

### What is Engineering Mechanics

Mechanics is the study of forces that act on bodies and the resultant motion that those bodies experience. With roots in physics and mathematics, Engineering Mechanics is the basis of all the mechanical sciences: civil engineering, materials science and engineering, mechanical engineering and aeronautical and aerospace engineering.

Engineering Mechanics provides the “building blocks” of statics, dynamics, strength of materials, and fluid dynamics. Engineering mechanics is the the discipline devoted to the solution of mechanics problems through the integrated application of mathematical, scientific, and engineering principles. Special emphasis is placed on the physical principles underlying modern engineering design.

Engineering Mechanics students are also encouraged to engage in undergraduate research with a faculty member. As a result, Engineering Mechanics students are prepared for careers at the forefront of a wide variety of fields, including the aerospace, electronics, automotive, manufacturing, software, and computer industries. The curriculum also provides excellent preparation for graduate school in many different engineering disciplines.

**BASICS AND STATICS OF PARTICLES**

** Introduction to Mechanics**

Continuum mechanics is concerned with motion and deformation of material objects, called bodies, under the action of forces. If these objects are solid bodies, the respective subject area is termed solid mechanics, if they are fluids, it is fluid mechanics or fluid dynamics. The mathematical equations describing the fundamental physical laws for both solids and fluids are alike, so the different characteristics of solids and fluids have to be expressed by constitutive equations.

Obviously, the number of different constitutive equations is huge considering the large number of materials. All of this can be written using a unified mathematical framework and common tools. In the following we concentrate on solids. Continuum mechanics is a phenomenological field theory based on a fundamental hypothesis called continuum hypothesis. The governing equations comprise material independent principles namely,

**Mechanics**

- v Body of Knowledge which Deals with the Study and Prediction of the State of Rest or Motion of articles and Bodies under the action of Forces
- v
**Kinematics**, being a purely geometrical description of motion and deformation of material bodies; - v
**Kinetics**, addressing forces as external actions and stresses as internal reactions; - v
**Balance equations**for conservation of mass, momentum and energy; and material dependent laws, the - v
**Constitutive equations**.

Altogether, these equations form an initial boundary value problem.

#### What is Engineering Mechanics

**Laws of Mechanics**

**1 Newton’s law**

**Law I**

Each body remains in its state of rest or motion uniform in direction until it is made to change this state by imposed forces.

**Law II**

The change of motion is proportional to the imposed driving force and occurs along a straight line in which the force acts.

**Law III**

To every action there is always an equal reaction: or the mutual interactions of two bodies are always equal but directed contrary.

**2 Lami’s theorem**

It states that, “If there forces acting at a point are in equilibrium, each force will be proportional to the sine of the angle between the other two forces.”

Suppose the three forces P, Q and Rare acting at a point 0 and they are in equilibrium as shown in Fig.

Let a = Angle between force P and Q.

β =Angle between force Q and R.

y = Angle between force R and P.

Then according to Lame’s theorem, P is proportional sine of angle between Q and R a sin β. P / sin β = constant.

Similarly Q / sin γ = constant,

R / sin a = constant P / sin β = Q / sin γ = R / sin a

**Example-1: In a jib crane, the jib and the tie rod are 5 m and 4 m long respectively. the height of crane post in 3 m and the ties red remains horizontal. Determine the forces produced in jib and tie rod when a load of 2 kn in suspended at the crane head.**

**Solution:** From figure

sin q = 3/5 = 0.6

q = 36.87^{o}

Let P_{1} and P_{2} be the forces developed in jib and tie rod respectively. the three forces P_{1}, P_{2} and W are shown in figure with the angle between the forces calculated from the given directions. The line of action of forces P_{1}, P_{2} and weigh W meet at the point C, and therefore Lami’s theorem is

applicable. That gives:

_{∴}P_{1}/sin 270^{o} = P_{2}/sin 53.13^{o} = 2/si n36.87^{o}

P_{1} = 2 × sin 270 ^{o} / sin 36.87^{o} = 2 × 1/0. = –3.33 kN

P_{2} = 2 × sin 53.13 ^{o} / sin 36.87^{o} = 2 × 0.8 / 0.6 = 2.667 kN

The –ve sign indicates that the direction of force P_{1} is opposite to that shown in figure obviously the tie rod will be under tension and jib will in compression.

**Examples-2: A string ABCD w hose extremity A is fixed has weights W**_{1} and W_{2} attached to it at B and C, and passes round a smooth peg at D carrying a weight of 800 N at the free end E shown in Figure. If in a state of equilibrium, BC is horizontal and AB and CD make angles of 150^{o} and 120^{o} respectively wit h BC, make calculation for (a) The tension i n portion AB, BC, CD and DE of the string. (b) t he value of weights W_{1} and W_{2} (c) The load o n the peg at D

_{1}and W

_{2}attached to it at B and C, and passes round a smooth peg at D carrying a weight of 800 N at the free end E shown in Figure. If in a state of equilibrium, BC is horizontal and AB and CD make angles of 150

^{o}and 120

^{o}respectively wit h BC, make calculation for (a) The tension i n portion AB, BC, CD and DE of the string. (b) t he value of weights W

_{1}and W

_{2}(c) The load o n the peg at D

**Solution: **Let T_{1}, T_{2}, T_{3}, T_{4}be the tension in segments AB, BC, CD and DE of the string.

Under equilibrium condition, T_{5} = T_{4} = 800 N

Applying Lami’s theorem at point B,

T_{1}/sin90^{o} = T_{2}/sin120^{o} = W_{1}/sin150^{o}

T_{1} = T_{2} sin 90^{o} / sin 120^{o} = 400 × 1/0.866 = 461.89 N

W_{1} = T_{2} sin 150^{o} / sin 120^{o} = 400 × 0.5 / 0.866 = 230.95 N

(c) Load on peg at D = T_{3} sin 60^{o} + W

= 800 sin 60^{o} + 800 = 692.82 + 800

= 1492.82 N

**3 Parallelogram Law of forces**

The law of parallelogram of forces is used to determine the resultant* of two forces acting at a point in a plane. It states, “If two forces, acting at a point be represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point.”

Let two forces P and Q act at a point 0 as shown in Fig. 1.3. The force P is represented in magnitude

and direction by OA whereas the force Q is presented in magnitude and direction by OB. Let the angle between the two forces be ‘a’. The resultant of these two forces will be obtained in magnitude

and direction by the diagonal (passing through O) of the parallelogram of which OA and OB are two adjacent sides. Hence draw the parallelogram with OA and OB as adjacent sides as shown in Fig. The resultant R is represented by OC in magnitude and direction.

**Magnitude of Resultant (R)**

From C draw CD perpendicular to OA produced.

Let a = Angle between two forces P and Q = LAOB

Now < DAC = < LAOB (Corresponding angles)

In parallelogram OACB, AC is parallel and equal to OB .

AC=Q.

In triangle ACD,

AD = AC cos a = Q cos a and CD =AC sin a= Q sin a.

In triangle OCD,

OC^{2} = OD^{2} + DC^{2}.

But OC = R, OD = OA +AD = P + Q cos a and DC =Q sin a

R ^{2} = (P + Q cos a)^{2} + (Q sin a)^{2} = p^{2} + Q^{2} cos^{2} a+ 2PQ cos < X+ Q^{2} sin^{2} a

= p^{2} + Q^{2} (cos^{2} a+ sin^{2} a)+ 2PQ cos a

= P^{2} + Q^{2} + 2PQ cos a

R = √ p^{2} + Q^{2} + 2PQ cos a

**Direction of Resultant**

Let θ = Angle made by resultant with OA.

Then from triangle OCD,

tan θ = CD / OD = Q sin a / P + Q cos a θ = tan ^{-1} ( Q sin a / P + Q cos a)

The direction of resultant can a lso be obtained by using sine rule [In triangle O AC, OA = P, AC = Q, OC = R, angle O AC = (180 – a), angle ACO = 180- [θ + 180 – a] = (a- θ)]

sin θ / AC = sin (180 –a) OC = sin (a – θ) / OA sin θ / Q = sin (180 – a) / R = sin (a – θ) / P

Two cases are important.

**1st Case**. If the two forces P and Q act at right angles, then a = 90° we get the magnitude of resultant as

the direction of resultant is obtained as θ = tan ^{-1} ( Q sin a / P + Q sin a)

= tan ^{-1} ( Q sin 90^{0} / P + Q cos 90 ^{0} ) = tan ^{-1} Q /P

**2nd Case. **The two forces P and Q are equal and are acting at an angle a between them. Then the magnitude and direction of resultant is given as

**4 Law of Triangle of Forces**

It states that, “if three forces acting at a point be represented in magnitude and direction by the three sides of a triangle, take n in order, they will be in equilibrium.”

**5 Vectorial representation of forces**

A force can be represented as a vector. Forces and vectors share three major characteristics:

- v Magnitude
- v Direction
- v Location

The simple support structure in Figure can be used to illustrate the three characteristics that make a force equivalent to a vector.

**6 How Forces Are Represented**

There are two ways in which forces can be represented in written form:

1. Scalar Notation

2. Vector Notation

3. The method used depends on the type of problem being solved and the easiest approach to finding a solution.

Scalar notation is useful when describing a force as a set of orthogonal force components. For example: Fx = 15N, Fy = 20N, Fz = 10N.

Vector notation is useful when vector mathematics are to be applied to a problem, such as addition or multiplication. Vector notation is somewhat simple in form:

F = 15i + 20j + 10k N.

The N term represents the unit of force, Newtons in this instance.

**7 Addition of Forces**

Multiple forces can be applied at a point. These forces are known as concurrent forces and can be added together to form a resultant force. If the component forces are orthogonal, then the magnitude of the resultant force can be determined by taking the Square Root of the Sum of the Squares (SRSS). The SRSS method is an extension of the Pythagorean Theorem to three dimensions.

**8 Dot Product**

Sometimes it is useful t o know how much of a force is acting in a dire ction other than the X,Y,or Z direction. Such a case might involve a bolted connection with a slotted hole, as shown in Figure.

Suppose that the force i n Figure 1.5.1 was F=40i + 10j + 30kNewtons, and that the X’ axis was oriented along the unit vect or e=0.6i + 0.5j – 0.624k. The projection of F onto X’ would be the force component acting in the direction of the slot. Now consider the possibility that the connection would slip if Fx’ was greater th an 35 N. Will the connection slip? The dot p roduct will help us answer this question.

**9 Resultant of Three Force Vectors**

Three force vectors (F1, F2, F3) are simmultaneously applied at point A. The resultant of these three forces is F. Determin e F3 such that: F = 3.2*i* + 5.5*j* + 2.9*k*. Write F3 in vector notation.

**10 Equation of equilibrium**

A particle is in equilibrium if it is stationary or it moves uniformly relative to an inertial frame of reference. A body is in equilibrium if all the particles that may be considered to comprise the body are in equilibrium. One can study the equilibrium of a part of the body by isolating the part for analysis. Such a body is called a free body.

We make a free body diagram and show all the forces from the surrounding that act on the body. Such a diagram is called a free-body diagram. For example, consider a ladder resting against a smooth wall and floor. The free body diagram of ladder is shown in the right. When a body is in equilibrium, the resultant of all forces acting on it is zero. Thus that resultant force ** R** and the resultant couple

**are both zero, and we have the equilibrium equations**

*M R*These requirements are necessary and sufficient conditions.

Let us understand equation for different type of force systems.

**Types of system of forces 1. Collinear forces :**

In this system, line of action of forces act along the same line is called collinear forces. For example consider a rope is being pulled by two players as shown in figure F1 F2

**2. Coplanar forces**

When all forces acting on the body are in the same plane the forces are coplanar

**3. Coplanar Concurrent force system**

A concurrent force system contains forces whose lines-of action meet at same one point. Forces may be **tensile (pulling)** or Forces may be **compressive (pushing)**

**4. Non Concurrent Co-Planar Forces**

A system of forces acting on the same plane but whose line of action does not pass through the same point is known as non concurrent coplanar forces or system for example a ladder resting against a wall and a man is standing on the rung but not on the center of gravity.

**5. Coplanar parallel forces**

When the forces acting on the body are in the same plane but their line of actions are parallel to each other known as coplanar parallel forces for example forces acting on the beams and two boys are sitting on the sea saw.

**6. Non coplanar parallel forces**

In this case all the forces are parallel to each other but not in the same plane,for example the force acting on the table when a book is kept on it.

**11 Forces in Space**

The resultant**, R** of two o r more forces in space is obtained by summing their rectangular components i.e.

**12 Equilibrium of a particle in space**

A particle is in equilibrium if the resultant of ALL forces acting on the particle is equal to zero

(Newton’s first law is that a bod y at rest is not subjected to any unbalanced forces).

**Equilibrium equations in component form: **In a rectangular coordinate system the equilibrium equations can be represented by three scalar equations:

To apply equilibrium equations we must account for all known and unknown forces acting on the particle. The best way to do this is to draw a free-body diagram of the particle.

**13 The principle of transmissibility**

If a force, acting at a point on a rigid body, is shifted to any other point which is on the line of action of the force, the external effect of the force on the body remains unchanged”.

**14 Single Equivalent Force**

When a number of forces are acting on a rigid body, then these forces can be replaced by a single force which has. the same effect on the rigid body as that of all the forces acting together, then this single, force is known as ‘Single Equivalent Force’. This single equivalent force is also known as resultant of several forces. Hence a single force which can replace a number of forces acting on a rigid body, without causing any change in external effects on the body, is known as single equivalent force (or resultant force )

What is Engineering Mechanics