Applied Mechanics - Lecture notes 5 PDF

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APPLIED MECHANICS – GEC 214 Ajayi, O.O. MNSE, B. Sc (Ife), M.Eng (Nsukka), PhD (in View). Mechanical Engineering Department Covenant University, Ota, Ogun State, Nigeria Office No. P 51 Ext. 3379

I. PRINCIPLES OF MECHANICS

I.1 DEFINITION AND CONCEPT Mechanics can be defined as the branch of physical science which studies the behaviour of a body at rest or in motion under the action of forces.

It is basic to Engineering Sciences because of its use in the analysis and design of engineering components and systems. A thorough knowledge of mechanics is a foundation tool for the understanding of the physical phenomenon of engineering.

I.2 BRANCHES OF MECHANICS Mechanics is a branch of physical sciences which is divided into three parts, namely: (i) Fluid Mechanics; (ii) Mechanics of Deformable Bodies and (iii) Solid Mechanics. Fluid Mechanics: This is the part of mechanics which deals with the behaviour of liquids and gases under the action of balanced and unbalanced forces. It could be statics or dynamics. A sub division of fluid mechanics include: Compressible fluid mechanics and incompressible fluid mechanics. Mechanics of Deformable bodies: This is the part of mechanics commonly referred to as ‘STRENGTH OF MATERIALS’. It studies the reaction of bodies to external forces i.e. it looks into the resulting stress and/or deformation that arise as a result of application of external loads (or forces). It is intermediate between fluid mechanics and solid mechanics. Solid Mechanics: In this part of mechanics, the bodies are assumed to be rigid solids. Although, in practical phenomenon, they are not actually absolute solids; they deform under the

compressible or tensile loads (or forces) to which they are subjected. These deformations are however very small and do not cause problems to the conditions of equilibrium of such bodies. A solid (or rigid) body can therefore be defined as that in which the distance between any two points in the body remains ‘appreciably’ unchanged. Sub division of solid mechanics This include: (i) Statics and (ii) Dynamics. Statics is the study of the behaviour of bodies at rest under the action of balanced forces; while dynamics is the study of the bahaviours of bodies in motion under the action of unbalanced forces. The emphasis in this course shall be on solid mechanics.

BASIC CONCEPTS The basic concepts of mechanics are the concepts of time space mass and force. Concept of time: This answers the question ‘when’. It gives the precise definition of when an event takes place or will take place. Concept of space: This answers the question ‘where. It gives the definition of the position. This definition arises from the frame of reference that all point are defined from an imaginary origin measured out in three direction called xyz axes. Concept of mass: This answers the question ‘how much’ It gives the measure of quantity of a body involved.e.g. 2kg of a body is placed on a tower. Concept of force: This answers the question ‘what is the cause’ It defines what causes an action to take place.

2.0 FORCE Force is generally defined as ‘any effect or action’ that may cause a change in the state of rest or motion of a body. It could be direct or indirect (i.e. at a distance). 2.1 KINDS OF FORCES 1. Contact Forces: These are forces which are in direct contact with a body on which they act e.g. force of push or pull and friction.

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2. Force Fields: these are forces that act on a body from a distance. It can be gravitational, electrical or magnetic in nature. These forces have a space of action around them where any body brought into such space feels the effect of the force creating the field.

2.2 PROPERTIES OF FORCES For adequate description of a force, it must have three basic properties: (i) Magnitude: This is the size of the force represented by a number with a unit e.g. 10N, 50N. (ii) Direction: This is the line of action to which the force is acting (or the effect of the force can be felt) e.g. 10N North 45o South. Diagrammatically it is shown by a line with an arrow head pointing in the line of action. (iii) Point of Action: This is the point on a body on which the force is acting.

10N

A force of 10N, acting on a body at point X in the direction of the east.

2.3 TYPES OF FORCES (1) Distributed Force: This is a force whose effect on a body is spread over many areas of the body e.g. the weight of a beam. (2) Concentrated Force: This is a force whose effect on a body is felt on very small area compared to the entire surface area of the body. Other types of forces include external force and internal force.

2.4 FORCE CONFIGURATION Forces whose lines of action lie on the same plane are called coplanar forces. Spatial or space forces have their lines of action lying in three dimensional space. Concurrent forces have their lines of action passing through a common point, while for non concurrent forces, the lines of action pass through different points. The following configurations are obtainable: (i) concurrent coplanar (ii) Non concurrent coplanar and (iii) Spatial; concurrent and non concurrent.

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2.5 LAWS OF SOLID MECHANICS Owing to the contribution of Sir Isaac Newton to science, Newtonian mechanics still remains the basis of engineering sciences. This gives rise to what is called NEWTON’S LAWS. These laws govern the behaviours of bodies under the action of forces. (1) ‘Newton’s’ First Law: States that a particle will remain at rest (if originally at rest) or continues to move (if originally in motion) in a straight line with uniform velocity if no force is acting on it.. N.B: The concept of ‘NO’ force can actually mean the resultant force is zero.

(2) ‘Newton’s’ Second Law: States that if the resultant force acting on a particle is not zero; the

rate of change of velocity with time is directly proportionally to the resultant force,

and takes place in the direction of the force. Mathematically;

Fα

v t

But

v = a where a = acceleration. t

Thus;

where F = force, V = velocity and t = time

Fα a

Applying the principles of proportionality and dimensionality; F = ma where m = constant of proportionality called mass.

Hence, m = mass of the particle.

(3) ‘Newton’s’ Third Law: States that the force acting on a body produces a reaction which is equal in magnitude and opposite in direction.

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(4) ‘Newton’s Law of Gravitation: States that two bodies of masses M and m attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. Mathematically;

Fα

Mm r2

where F = force of attraction btw bodies, M and m are masses

of the bodies and r2 is the square of their distance apart.

F=

Mm r2

G = constant of proportionality called Universal constant or gravitation

constant.

(5) The parallelogram law for addition of forces: states that if two forces acting on a body can be represented by the two sides of a parallelogram, then the resultant can be represented by the diagonal of the parallelogram in magnitude and direction. Diagrammatically;

(6) Principle of Transmissibility: States that if the point of action of a force is shifted from one point to the other on the same line of action, it will produce no different effect provided the magnitude and direction are retained.

3.0 RESULTANT OF COPLANAR FORCE SYSTEM The Resultant of forces simply implies a single force that gives the same effect in magnitude and direction, as the effect of two or more forces acting together simultaneously on an element. Another name for resultant of a force system is ‘ EQUIVALENT FORCE’: the equivalent force is the vector sum of the individual forces.

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3.1 RESULTANT OF CONCURRENT FORCES. PRINCIPLES The resultant of concurrent coplanar forces can carried out using any of (i) Parallelogram Law of Forces described earlier, (ii) Triangle Law of Forces and (iii) Polygon Rule. Triangle Law of Forces This law sates that ‘if two forces acting on a body can be represented by the sides of a triangle; then the third side drawn into size can represent the resultant in magnitude and direction.’ For example , consider the figure below acted upon by two forces X and Y. X Y

If forces X and Y can be represented by the sides of a rectangle in magnitude and direction, the third side will represent the resultant (sum of X and Y) in magnitude and direction. Hence,

Triangle law of forces like the parallelogram law of forces is used to find the resultant of two coplanar forces

Y

X+Y

R X

Polygon Law of forces This law states that ‘if three or more forces can be used to represent (in magnitude and direction) the sides of a regular polygon; then the line which is drawn to complete the polygon represents the resultant in magnitude and direction.’ For example: consider the body below, acted upon by the system of coplanar forces

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X X

Y

M

Z

If the forces X, Y, Z and M can be used to represent the sides of a polygon, then the line which is drawn to close up the polygon will represent the resultant (sum of X, Y, Z, and M). From the solution of the figure above, it is clear that, in application, polygon of forces is obtained by drawing the forces (tip to base) starting from a particular force and maintaining the respective orientation of the forces. The last side that would close the polygon is the resultant R.

METHODS OF SOLUTION 1. Graphical method of solution: This method makes use of the three laws explained above. It is carried out by graphically drawing the forces in magnitude and direction using a properly chosen line scale. The direction of each force is measured using a protractor. The closing side of the figure represents the resultant. 2. The trigonometric method: This method makes use of the sine and cosine together with the three laws explained above. This method gives a more accurate solution. 3. Rectangular component method: This is carried out by resolving the forces into their rectangular components. i.e into their x and y components. The resultant is the summation of the x and y components. For example, determine the resultant of the two force system below:

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F1

30o

10o F2

The resultant R = ΣRx + Σ Ry And ΣR y = ( f y ) + ( f y ) 1

Where ΣR x = ( f x )1 + ( f x ) 2

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Then, the magnitude of R is found by :

R = Rx2 + Ry2 − ⎛ Ry ⎞ And the direction is found by: tan 1 ⎜ ⎟ ⎝ Rx ⎠

Note: all angles are measured counterclockwise from the positive x-axis.

4.0 MOMENT AND COUPLE Moments of a force: The moment of a force about a point is defined as the product of the force and the perpendicular distance measured from the point along the line of action to the force. Three things are considered in taking moment about a point. Consider the diagram below: O is the fixed point of reference. d is the perpendicular distance from O to F which is the force acting.

The moment of F about O is Mo = F x d

Conventional sign of moment: Clockwise moment is taken to be negative and anticlockwise moment is taken to be positive.

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For systems in equilibrium; Sum of clockwise moment must be equal to sum of anticlockwise moment about a particular point of reference.

Method of determining the moment of a force Depending on the nature of application of a force; there are three methods that can be used to determine the moments of a force (1) by direct method: Using the definition that moment is force multiplied by perpendicular distance. (2)By Varignon’s Theorem: This theorem states that the moment of a force about any point is equal to the sum of the moments produced by the rectangular components of the force. For this theorem to be applied adequately; the forces concerned will first have to be resolved to their rectangular components after which each moment due to individual component is determined and then summed in accordance with the sign convention of moment. (3) By the Principle of Transmissibility: This principle is applied by moving a force from its point of application to any convenient point ( for ease of calculation) along its line of action.

WORKED EXAMPLE Determine the moment of force F about O

F = 24N 20o 0.7m

y O

x 2m 6m

Solution (a) By direct method: Locate a perpendicular distance from F to O. 9

Project F backwards to P where a line from O meets F at 90o F = 24N 20o 0.7m

y Q

O 20

x 2m

o

d P Then,

(since sin 20o =

d = 4sin20o

OP ) OQ

d = 1.4m Therefore, moment Mo = F x d = 24 x 1.4 = 33Nm (b) By Varigno’s Theorem: Resolve F into rectangular components. Fx = Fcos20o = 24cos20o = 22.6N Fy = Fsin20o = 24sin20o

= 8.2N

Thus, Moment Mo = - ( Fx x 0.7 ) + ( Fy x 6 )

NB: clockwise Moment is –ve.

Mo = -22.6 x 0.7 + 8.2 x 6 = 33Nm Hence, Moment is 33Nm anticlockwise.

( c) By Principle of Transmissibility: Take the force to Q, a convenient point along the line of action of b F Moment Mo = F x d = Fy x 4 = 33Nm

y O

Fy Q

0.7m Fx x 2m

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Couple: This is defined as two equal and opposite parallel forces acting on a body. Consider the diagram below. F1 must be equal to F2. F1

x

a

a d

y

F2

Moment of a couple: This is defined as the product of one of the force and the perpendicular distance between the forces. i.e. Moment of couple = F1 x (a + b)

Addition of couples: Two or more couples acting in a plane or parallel planes can be added algebraically by taking into cognizance the sign convention of moment. Equivalent couples: Two couples acting on the plane or parallel planes are said to be equivalent if they have the same moment acting in the same direction.

4.1 Replacing a force with a force – couple system. x

a

a d

y

F Fig. 2 Consider the diagram above. If it is required to move the force F away from X to a point Y at a perpendicular distance d away; without altering its mechanical effect. Then, two equal and opposite forces ( both equal to F) will be placed at Y. Making the point Y have a force and a couple formed by the opposing force F and the original force F at X. See diagram below (fig. 3& 4).

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F

x

a

a

y

d

F

F

The two Fs at Y counsels each other out and thus, maintains the mechanical effect of F at x. F at X forms a couple with F at Y. The moment of the couple is F x d. x

a

a

y

d My F

By this F has been transferred from x to y without any alteration of its mechanical effect. WORKED EXAMPLE Consider the couple system below. (1) Determine the resultant moment of the two couples and (2) Replace the couples by 2 forces applied at point x and y. determine the magnitude of the forces.

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5KN 10KN 30o

4m 30o 10KN 5KN Solution To determine the resultant; resolve the 5KN to its component forces Fy = 5sin30o = 2.5kN Fx = 5cos30o = 4.3kN The couple system becomes 10KN

2.5KN 4.3KN

4m 4.3KN 10KN

2.5KN Resultant Moment = summation of individual moments of forces: ∑M = + (10 x 4) + (-4.3 x 4) + (-2.5 x 4) = 40 – 10 – 17.2 = - 12.8kNm. Moment is 12.8kNm clockwise.

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(2)

x

A

d

F

y

F

4m

c

4m

To cal. line /xy/.

[ xy ] = [ xc ]2 + [ yc ]2

= 42 + 42 = 5.7m

But moment = 12.8kNm F.d = 12.8 F = 12.8 / 5.7 = 2.2kN

5.0 RESULTANT OF NON- CONCURRENT COPLANAR FORCE SYSTEM

As was defined earlier for non-concurrent coplanar forces; there’s not the same point of action for non- concurrent coplanar forces. Therefore, the location of the line of action of the resultant force is not immediately known. Estimating the resultant and its line of action

Resultant: To find the resultant, its magnitude and direction, can be calculated using the rectangular component method introduced earlier. This gives rise to the Rx and Ry components of the resultant and then, its magnitude and resultant and direction can be estimated from: R = Rx2 + R2y and the direction α = tan-1( Ry / Rx)

( in its proper sense).

Line of Action of the Resultant: To find the line of action which can also be called the location of the resultant; the principle of moment is employed.

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This is done by equating the sum of individual moments of the forces to the moment of the ∑M = R. x

resultant i.e.

where ∑ M = sum of moments and x =

distance of resultant From point of reference.

WORKED EXAMPLE

Consider the system of non- concurrent coplanar forces below. F1

F3 F4 ∅3

∅4

F5 F2 To Estimate The Resultant Resolve the forces to their rectangular components. F1 = 0 + ( - F1y)

; F2 = 0 + F2y

;

F3 = (F3cos∅3)x + (F3sin∅3)y

;

F = ( F4cos∅4)x + ( - F4 sin∅4)y ; F5 = 0 - F5y The Resultant R is given by:

R = Rx

+

Ry .

Rx = F1x + F2x + F3x + F4x + F5x Ö Rx = 0 + 0 + F3Cos ∅3 + F4Cos∅4 + 0

And: Ry = F1y + F2y + F3y + F4y + F5y Ö Ry = -F1 + F2 + F3Sin∅3 + F4Sin∅4 - F5

Hence, knowing the values of F1,F2,F3,F4 and F5 can lead to the final determination of Rx and Ry and thus, R in magnitude and direction. To estimate the location or line of Action of Resultant. -

Find the individual moment of the forces about a reference point

-

Sum up all the values of moment found above

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-

Equate this sum to the moment |R|.x

-

Compute x from the values of the moment and magnitude |R|. Thus;

From diagram above; taking moment about point A. M1 = F1X1; M2 = F2X2; M3 = F3X3; M4 = F4X4; M5 = F5X5 ∑MA = M1 + M2 + M3 + M4 + M5 Then; Use the positive value of ∑MA

|Ry|.x = |∑MA| Hence, x = |∑MA| |Ry|

And |Rx|. y = |∑MA|

.’. y = |∑MA| |Rx|

WORKED EXAMPLES

1. Replace the forces acting on the beam in diagram below with a single resultant force and determine its point of application along the beam. 20N

7N 3m

80N

1m

100N 2m

2m B

A Solution

The question is all about finding the resultant and location of resultant (a) To find the resultant, resolve the forces to its rectangular components Rx = 0 i.e. no force in the horizontal direction Ry = -7 - 20 - 80 + 100 = -7N R = p(02 +72) = 7N

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R = 7N in the downward vertical (b) To find the location of Resultant: Take moment about A ∑MA = 0 + (-20) x 3 + (-30 x 4) + 100 x 6 = 0 - 60 - 320 + 600 = 220 Nm .’. ∑MA = 220Nm counterclockwise -To find location for Rx – direction: |Rx|.y = 220

No location since Rx is zero.

-To find location for Ry |Ry|.x = 220 x = 220 = 31.3m 7 Hence, R (in downward vertical direction) is 31.3m ...