Engineering Mechanics for First Year B.E. Degree Students

1. Engineering Mechanics for First Year B.E. Degree Students

2. COURSECONTENT IN BRIEF • Introduction. • Resultant of concurrent and non-concurrent coplanar forces. • Equilibrium of concurrent and non-concurrent coplanar forces. • Analysis of plane trusses. • Friction. • Centroid and Moment of Inertia. • Resultant and Equilibrium of concurrent non-coplanar forces. • Rectilinear and Projectile motion. • Newton’s second law, D’Alembert’s principle, banking and super elevation. • Work, Energy, and Power. • Impulse- Momentum principle.

3. Books for Reference 1.Engineering Mechanics, by Meriam & Craige, John Wiley & Sons. 2.Engineering Mechanics, by Irwing Shames, Prentice Hall of India. 3.Mechanics for Engineers, by Beer and Johnston, McGraw Hills Edition 4.Engineering Mechanics, by K.L. Kumar, Tata McGraw Hills Co.

4. INTRODUCTION CHAPTER – I 1 Definition of Mechanics : In its broadest sense the term ‘Mechanics’ may be defined as the ‘Science which describes and predicts the conditions of rest or motion of bodies under the action of forces’. This Course on Engineering Mechanics comprises of Mechanics of Rigid bodies and the sub-divisions that come under it.

5. 2 Branches of Mechanics Engineering Mechanics Mechanics of Solids Mechanics of Fluids Rigid Bodies Deformable Bodies Ideal Fluids Compressible Fluids Viscous Fluids Dynamics Statics Theory of Elasticity Theory of Plasticity Strength of Materials Kinematics Kinetics

6. Fundamental Concepts and Axioms 3 Rigid body : It is defined as a definite amount of matter the parts of which are fixed in position relative to one another. Actually solid bodies are never rigid; they deform under the action of applied forces. In those cases where this deformation is negligible compared to the size of the body, the body may be considered to be rigid.

7. 4 Particle A body whose dimensions are negligible when compared to the distances involved in the discussion of its motion is called a ‘Particle’. For example, while studying the motion of sun and earth, they are considered as particles since their dimensions are small when compared with the distance between them.

8. 5 Space The concept of space is associated with the notion of the position of a point, defined using a frame of reference, with respect to which the position of the point is fixed through three measures specific to the frame of reference. These three measures are known as the co-ordinates of the point, in that particular frame of reference. z y x

9. 6 Mass: It is a measure of the quantity of matter contained in a body. It may also be treated as a measure of inertia, or resistance to change the state of rest, or of uniform motion along a straight line, of a body. Two bodies of the same mass will be attracted by the earth in the same manner. Continuum : A particle can be divided into molecules, atoms, etc. It is not feasible to solve any engineering problem by treating a body as a conglomeration of such discrete particles. A body is assumed to be made up of a continuous distribution of matter. This concept is called ‘Continuum’.

10. 7 Force • It is that agent which causes or tends to cause, changes or tends to change the state of rest or of motion of a mass. A force is fully defined only when the following four characteristics are known: • Magnitude • Direction • Point of application and • Line of action.

11. 8 Scalars and Vectors A quantity is said to be a ‘scalar’ if it is completely defined by its magnitude alone. Example : Length, Area, and Time. Whereas a quantity is said to be a ‘vector’ if it is completely defined only when its magnitude and direction are specified. Example : Force, Velocity, and Acceleration.

12. Classification of force system 9 Force system Coplanar Forces Non-Coplanar Forces Concurrent Non-concurrent Concurrent Non-concurrent Likeparallel Unlike parallel Likeparallel Unlike parallel A force that can replace a set of forces, in a force system, and cause the same ‘external effect’ is called the Resultant. ( More detailed discussion on Resultant will follow in Chapter 2 )

13. B C P2 R   O A P1 10 Axioms of Mechanics • Parallelogram law of forces : It is stated as follows : ‘If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, then the resultant of these two forces is represented in magnitude and direction by the diagonal of the parallelogram passing through the same point.’ Contd..

14. B C P2 R   O A P1 11 Contd.. In the above figure, P1 and P2, represented by the sides OA and OB have R as their resultant represented by the diagonal OC of the parallelogram OACB. It can be shown that the magnitude of the resultant is given by: R = P12 + P22 + 2P1P2Cos α Inclination of the resultant w.r.t. the force P1 is given by:  = tan-1 [( P2 Sin ) / ( P1 + P2 Cos  )]

15. 12 (2)   Principle of Transmissibility :  It is stated as follows : ‘The external effect of a force on a rigid body is the same for all points of application along its line of action’. P P A B P P O For example, consider the above figure. The motion of the block will be the same if a force of magnitude P is applied as a push at A or as a pull at B. The same is true when the force is applied at a point O.

16. 13 (3) Newton’s Laws of motion: (i)First Law : If the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest) or will move with constant speed in a straight line (if originally in uniform motion).  (ii) Second Law : If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of the resultant and in the direction of this resultant i.e., F α a ,or F = m.a , where F, m, and a, respectively represent the resultant force, mass, and acceleration of the particle.  (iii) Third law:   The forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.

17. 14 Note :   1. ‘Axioms’ are nothing but principles or postulates that are self – evident facts which cannot be proved mathematically but can only be verified experimentally and/or demonstrated to be true. 2. The three basic quantities of mechanics are length, time, and force. Throughout this Course we adopt SI units and therefore they are expressed in meters, seconds, and Newtons, written as m, s, and N respectively.    3. The ‘external effect’ of a force on a body is manifest in a change in the state of inertia of the body. While the ‘internal effect’ of a force on a body is in the form of deformation.

18. Y-Direction R F2 Fy  F1 Fx X-Direction F3 15 CHAPTER – 2 RESULTANT OF CONCURRENT COPLANAR FORCES Composition of forces and Resolution of force Resultant, R :It is defined as that single force which can replace a set of forces, in a force system, and cause the same external effect. In the above diagram F1, F2, F3 form a system of concurrent coplanar forces. If R is the resultant of the force system, then its magnitude and direction are given by: Contd..

19. 16 Contd.. (i)   Magnitude, R =  (Fx)2 + (Fy)2    (ii) Direction, θ = tan –1(Fy / Fx) , where:  ΣFx = Algebraic summation of x-components of all individual forces.   ΣFy = Algebraic summation of y-components of all individual forces.  θ = Angle measured to the resultant w.r.t. x-direction.  The process of obtaining the resultant of a given force system is called ‘Composition of forces’. Note: The orientation of x-y frame of reference is arbitrary. It may be chosen to suit a particular problem.

20. 17 Component of a force : Fx  F F Fy Fy Fy F   Fx Fx Fig. 3 Fig. 1 Fig. 2 Component of a force, in simple terms, is the effect of a force in a certain direction. A force can be split into infinite number of components along infinite directions. Usually, a force is split into two mutually perpendicular components, one along the x-direction and the other along y-direction (generally horizontal and vertical, respectively). Such components that are mutually perpendicular are called ‘Rectangular Components’.    The process of obtaining the components of a force is called ‘Resolution of a force’.

21. 18 Sign Convention for force components: The adjacent diagram gives the sign convention for force components, i.e., force components that are directed along positive x-direction are taken +ve for summation along the x-direction. y y x +ve x +ve Also force components that are directed along +ve y-direction are taken +ve for summation along the y-direction. Oblique Components of a force: When the components of a force are not mutually perpendicular they are called ‘Oblique Components’. Consider the following case. Contd..

22. 19 Let F1 and F2 be the oblique components of a force F. The components F1 and F2 can be found using the ‘triangle law of forces’, which states as follows: ‘If two forces acting at a point can be represented both in magnitude and direction, by the two sides of a triangle Contd.. F2 F F1   taken in tip to tail order, the third side of the triangle represents both in magnitude and direction the resultant force F, the sense of the same is defined by its tail at the tail of the first force and its tip at the tip of the second force’. F1 / Sin  = F2 / Sin  = F / Sin(180 -  - ) F  F2  F1

23. 20 NumericalProblems & Solutions (1A) Obtain the resultant of the concurrent coplanar forces acting as shown in Fig. 1A. Solution: ∑ Fx = + 15 Cos 15 – 75 – 45 Sin 35 + 60 Cos 40 ∑ Fy = + 15 Sin 15 + 105 – 45 Cos 35 – 60 Sin 40 = + 33.453 kN 105kN +ve 15kN 150 = -40.359 kN 75 kN = 40.359 kN 400 350 60 kN 45kN Fig.1A +ve Contd..

24. 105 kN (1A) 15kN 150 75 kN 400 350 60 kN 45 kN Fig.1A 21 Contd.. R = ( ∑Fx )2 + (∑Fy)2 = (- 40.359)2 + (33.453) 2 Θ = tan-1(∑Fy/ ∑Fx) R Answer: ∑Fy Magnitude,R = 52.42 kN Inclination,Θ = 39.69 o Θ (w.r.t. X – direction) ∑Fx

25. 50kN 100kN º 120 2 3 º 1 30 2 75kN 25kN Fig. 1B 22 Obtain the resultant of the concurrent coplanar forces acting as shown in Fig. 1B. (1B) 50 kN 100 kN 2 26.31o 3 α Solution: 30o 1 β º α = tan-1(2/3)=33.69 75 kN 2 β= tan-1(2/1) = 63.43º 25kN ∑Fx = -50 Cos 26.31- 100 Cos33.69 – 25 Cos 63.43 + 75 Cos 30 Contd.. -74.26kN = 74.26kN + ve

26. 23 Contd.. 50 kN 100 kN (1B) 2 26.31o 3 α β 30o 1 75 kN 2 25kN ∑FY = 50sin26.31- 100sin 33.69 – 75sin30 – 25sin63.43 + ve = -93.17kN = 93.17kN Contd..

27. ∑Fx Θ ∑Fy R 24 Contd.. 50kN (1B) 100kN 26.31o 33.69º 30o 63.43º 75 kN 25kN Answers: R = (∑Fx) 2 + (∑Fy) 2 = 119.14 kN Θ = tan-1(∑Fy / ∑Fx)= 51.44o

28. 200N 150N F5 110 º 50° 20º α R =250 N 45° 50N 120N 25 (2) 150N 200N A system of concurrent coplanar forces has five forces of which only four are shown in Fig.2. If the resultant is a force of magnitude R = 250 N acting rightwards along the horizontal, find the unknown fifth force. 110º 50° 45º Fig. 2 50N 120N Solution: - Assume the fifth force F5 in the first quadrant, at an angle α, as shown. - The 150 N force makes an angle of 20o w.r.t. horizontal Contd..

29. 200N 150N F5 110 º 50° 20º α R =250 N 45° 50N 120N 26 Contd.. ∑FX = R +ve 200 cos 50 – 150 cos 20 – 50cos 45 +F5 cos α = 250.  F5 cos α = +297.75 N Contd..

30. 200N 150N F5 110 º 50° 20º α R =250 N 45° 50N 120N 27 Contd.. ∑FY = 0. F5 sin α + 200sin 50 + 150 sin 20 – 120 + 50 sin 45 = 0 F5 sin α= -119.87N = 119.87N +ve tan α = F5sinα/F5cos α =0.402 α = 21.90º F5= 320.97N F5cosα= 297.75N Answers α = 21.90º F5 = 320.97N F5sinα= 119.87N

31. 28 (3) A system of concurrent coplanar forces has four forces of which only three are shown in Fig.3. If the resultant is a force R = 100N acting as indicated, obtain the unknown fourth force. 75N 25N 60° 70° 45° 40° Fig. 3 50N R=100N Contd..

32. 29 Contd.. - Assume the fourth force F4 in the 1st quadrant, making an angle α as shown 75N 25N F4 60° 70° α 45° 40° R=100N 50N Fx = -Rcos40 +ve F4cosα + 75cos70 – 50cos45 – 25sin60 = -100cos40 Or, F4cosα = - 45.25N ; or, F4cos α = 45.25N

33. 30 Contd.. 75N 25N F4 60° α 70° 45° 40° R=100N 50N Fy = -Rsin40 F4sinα + 75sin70+25cos60+50sin45 = - 100sin40  F4sinα = -182.61N ; or, F4sin α = 182.61N +ve Contd..

34. F4cosα = 45.25N F4sinα = α= 76.08º 182.61N F4=188.13N 31 Contd.. Answers:  = tan-1(F4sin /F4cos) = 76.08º & F4 =188.13N

35. 32 .(4) The resultant of a system of concurrent coplanar forces is a force acting vertically upwards. Find the magnitude of the resultant, and the force F4 acting as shown in Fig. 4. 10 kN F4 70° 60° 30° 45° 5 kN 15 kN Fig. 4 Contd..

36. 33 Contd.. R Solution: F4 10 kN 70° 60° 30° 45° 5 kN 15 kN F4 sin70 – 10cos 60 – 15cos 45 – 5cos 30 = 0; or, F4sin70 = 19.94 ∑Fx = 0 Fig. 4 +ve F4 = 21.22kN Contd..

37. 34 Contd.. 10 kN R F4 Solution: 70° 60° 30° 45° 5 kN 15 kN ∑Fy = +R +ve Fig. 4 Answers: F4cos70 + 10sin60 – 15sin45 + 5sin30 = +R  +R - 0.342F4 = 0.554 Substituting for F4 , R= +7.81kN F4 = 21.22 kN R= +7.81kN

38. 100N Q 40° 70° 60° 45° 50N P Fig. 5 35 (5) Obtain the magnitudes of the forces P and Q if the resultant of the system shown in Fig. 5 is zero . Contd..

39. 36 Contd.. 100N Q 40° 70° 60° 50N 45° P For R to be = zero, ∑Fx = 0 and ∑ Fy = 0 ∑Fx = 0 : -Psin45 – Qcos40 + 100cos70 + 50cos60 = 0 Or, 0.707P + 0.766Q = 59.2 Fig. 5 +ve Contd..

40. 100N Q 40° 70° 60° 45° 50N P Fig. 5 37 Contd.. Solving (a) & (b) Answers: ∑Fy = 0 -Pcos45 + Qsin40 + 100sin70 – 50sin60 = 0 or, -0.707P + 0.642Q = -50.67 P = 77.17 N & Q = 6.058N +ve (b)

41. 38 (6) Forces of magnitude 50N and 100N are the oblique components of a force F. Obtain the magnitude and direction of the force F. Refer Fig.6. 100N 50N 30° Fig. 6 Contd..

42. 39 Contd.. (6) 100N 100N Y-AXIS X -AXIS 50N 30° 30° 50N Fig. 6 Rotating the axes to have X parallel to 50N, ∑Fx = +50 + 100cos30 = +136.6N ∑ Fy = +100sin30 = +50N +ve Contd.. +ve

43. 40 Contd.. 100N (6) 100N Y-AXIS X -AXIS 50N 30° 30° 50N Fig. 6 F = 145.46N θ = 20.1º w r t X direction (50N force) F= (∑Fx)2+(∑Fy)2 Y-AXIS θ = tan-1[(∑Fx)2+(∑Fy)2] F X -AXIS θ 50N

44. 41 (7)Resolve the 3kN force along the directions P and Q. Refer Fig. 7. Q 3kN 45° 60° 30° P Contd.. Fig. 7

45. Q P 3kN 45º Q 45º 45º 60° X – Axis 30° P 42 Contd.. Fig. 7 3kN Move the force P parallel to itself to complete a triangle. Using sine rule, P/sin45 = Q/sin90 = 3/sin45 Answer : P = 3kN, and Q = 4.243kN

46. 1. A body of negligible weight, subjected to two forces F1= 1200N, and F2=400N acting along the vertical, and the horizontal respectively, is shown in Fig.1. Find the component of each force parallel, and perpendicular to the plane. F1 = 1200N Y X F2 = 400 N 3 4 FIG. 1 43 EXERCISEPROBLEMS Ans : F1X = -720 N, F1Y = -960N, F2X = 320N, F2Y = -240N

47. F2 = 390 N Y 12 X 5 30º 40º F1 = 300 N F3 = 400 N FIG. 2 44 2. Determine the X and Y components of each of the forces shown in FIG.2. (Ans : F1X = 259.81 N, F1Y= -150 N, F2X= -150N, F2Y= 360 N, F3X = -306.42 N, F3Y= -257.12N )

48. 600N 800N 40º 20º 30º 200N FIG. 3 45 3. Obtain the resultant of the concurrent coplanar forces shown in FIG.3 (Ans: R = 522.67 N, θ = 68.43º)

49. α R = 300 N X - direction 20º FIG. 4 T1 46 • 4. A disabled ship is pulled by means of two tug boats as shown in FIG. 4. If the resultant of the two forces T1 and T2 exerted by the ropes is a 300 N force acting parallel to the X – direction, find : • Force exerted by each of the tug boats knowing α = 30º. • The value of α such that the force of tugboat 2 is minimum, • while that of 1 acts in the same direction. • Find the corresponding force to be exerted by tugboat 2. T2 ( Ans: a. T1= 195.81 N, T2 = 133.94 N b. α= 70º, T1 = 281.91 N, T2(min) = 102.61 N )

50. P 20º R 40º Q = 5 kN FIG. 5 47 5. An automobile which is disabled is pulled by two ropes as shown in FIG. 5. Find the force P and resultant R, such that R is directed as shown in the figure. (Ans: P = 9.4 kN , R = 12.66 kN)