- 1. Prepared By Prof. V. V. Nalawade ENGINEERING MECHANICS Prof. V.V.Nalawade 1
- 2. Module 1: Force System 8hrs Module 2 : Equilibrium 7hrs Module 3: Center of Gravity and Moment of Inertia 7 hrs. Module 4: Friction 6hrs Module 5 : Kinematics 6hrs Module 6 : Kinetics 6hrs C u r r i c u l u m 2
- 3. Prof. V.V.Nalawade 3 Engineering Mechanics • Dr. R.K. Bansal • N.H. Dubey • Beer & Johnston • K. L. Kumar • R. V. Kulkarni Applied Mechanics • R. S. Khurmi • Sunil Deo • R. K. Singer Reference Books
- 5. Prerequisite Basic Concepts Resolution & composition of forces Contents of the presentation Prof. V.V.Nalawade 5
- 7. Coordinate Geometry Prof. V.V.Nalawade 7 II Quadrant (-,+) I Quadrant (+,+) III Quadrant (-,-) IV Quadrant (+,-)
- 8. Slope of line Prof. V.V.Nalawade 8 θ (X2-X1) (Y2-Y1) P Q (X2,Y2) (X1,Y1) X Axis Y Axis m = tan θ =
- 9. Few more to Recall…… Trigonometry Pythagoras Theorem Cosine & Sine rule Sign conversion Unit conversion Prof. V.V.Nalawade 9
- 11. Prof. V.V.Nalawade 11 Sr. No. Topic Learning Objective (TLO) CO BL CA Code 1 To recall the basic principles of mechanics CO1 L1 1.2, 1.3 2 To describe the concepts on mechanics and its practical implementation CO1 L2 1.1, 1.2 3 To identify the force system and calculate the resultant of it CO1 L3 1.2, 2.1, 12.2 4 To analyze the numerical of different cases CO1 L3 2.1, 12.2 Learning Outcome: At the end of the topic the student should be able to:
- 12. Branches of Mechanics Engineering Mechanics Statics (Rest) Dynamics (Motion) Kinetics Kinematics EM is the branch of physics which deals with the study of forces and their effect on body when body is at rest or in motion. With Reference to the Cause of motion Without Reference to the Cause of motion Prof. V.V.Nalawade 1 2
- 15. Prof. V.V.Nalawade 15 We learn ENGINEERING MECHANICS
- 17. automobiles, aircrafts, electric motors, robots, television, mobile , satellite, projectile of missiles, launching of rockets, radar communication, trusses, lifting machines like crane, hoist, screw jack, elevator, conveyor belt, cargo ship, submarine, etc. Prof. V.V.Nalawade 17
- 18. Laws of Mechanics The following are the fundamental laws of mechanics: (i) Newton’s first law (ii) Newton’s second law (iii) Newton’s third law (iv) Newton’s gravitational law (v) Law of transmissibility of forces (vi) Parallelogram law of forces Prof. V.V.Nalawade 1 8
- 19. Prof. V.V.Nalawade 1 9 Image source: https://issuu.com/daennagonzalez/docs/newton___s_laws_of_motion
- 24. Prof. V.V.Nalawade 24 Because the amount of acceleration of a body is proportional to the acting force and inversely proportional to the mass of the body
- 26. Prof. V.V.Nalawade 26 Balloon goes up Air goes down
- 29. Prof. V.V.Nalawade 29 1.What happens according to Newton if you let an untied balloon go????
- 30. Prof. V.V.Nalawade 30 1.What happens according to Newton if you let an untied balloon go???? Balloon goes up Air goes down 3 rd Law Air will rush out of the balloon forcing the balloon to move through the air in the opposite direction, but equal in force.
- 31. Prof. V.V.Nalawade 31 2. Describe what happens if you are riding a skateboard and hit something (like a curb) with the front wheels???
- 32. Prof. V.V.Nalawade 32 2. Describe what happens if you are riding a skateboard and hit something (like a curb) with the front wheels??? 1 st Law Your body will keep moving forward and fly off your skateboard since the curb only stops the board, not yourself.
- 33. Prof. V.V.Nalawade 33 3. Describe why you hold your gun next to your shoulder while deer hunting????
- 34. Prof. V.V.Nalawade 34 3. Describe why you hold your gun next to your shoulder while deer hunting???? 3 rd Law When you pull the gun’s trigger, it forces the bullet out of the gun, but at the same time, the gun is forced in the opposite direction of the bullet (towards you). Your shoulder is a new force that is introduced in order to keep your gun from flying away from you.
- 35. Prof. V.V.Nalawade 35 4. Why should we wear seatbelts – use one of Newton’s Laws in your answer?
- 36. Prof. V.V.Nalawade 36 4. Why should we wear seatbelts – use one of Newton’s Laws in your answer? We should wear seatbelts so if we are in an accident our body doesn’t keep moving at the same speed and in the same direction that the car was going. A new force would be introduced to our bodies (the seatbelt) in order to keep our bodies in place.
- 38. Prof. V.V.Nalawade 38 Newton’s third law would tell us that when the rocket pushes out fire with a specific amount of force, the rocket will move in the opposite direction, but with the same amount of force. This is what causes the rocket to shoot up into the air.
- 39. Prof. V.V.Nalawade 39 6. Explain how each of Newton’s laws affects a game of Tug of War.
- 40. Prof. V.V.Nalawade 40 6. Explain how each of Newton’s laws affects a game of Tug of War. •First Law: The rope will stay in the same place until the tugging starts (a new force is introduced) •Second Law: We could measure a team’s force that they can pull the rope with based on their body masses and the acceleration that they are causing the rope to move at. •Third Law: 1 team pulls the rope towards themselves with a certain amount of force and the opposing team is also putting force on the rope. The same amount of force is applied from the ground to the people as they are putting on the ground.
- 41. 41 Concept of force and its measurements Prof. V.V.Nalawade
- 43. 43 1. Magnitude: 2. Direction : Prof. V.V.Nalawade
- 44. 44 3. Point of application : 4. Sense or Nature : Prof. V.V.Nalawade
- 45. Prof. V.V.Nalawade 45 Questions: 1. Define Mechanics. What are the different branches of mechanics? 2. What are the characteristics of force?
- 46. Prof. V.V.Nalawade 46 System of forces
- 47. System of Forces Several forces acting simultaneously upon a body Force System Coplanar Concurrent Collinear Parallel Like Unlike Non Concurrent & Non Parallel (General) Non-coplanar Concurrent Collinear Parallel Like Unlike Non Concurrent & Non Parallel (General) Prof. V.V.Nalawade 4 7
- 48. Coplanar System of Forces 2D 4 8 Prof. V.V.Nalawade
- 49. Non-Coplanar System of Forces 4 9 Prof. V.V.Nalawade
- 50. Composition of forces Forces added to obtain a single force which produces the same effect as the original system of forces. This single force is known as Resultant force. The process of finding the resultant force is called composition of forces. Prof. V.V.Nalawade 50
- 51. Composition of forces There are two methods of finding resultant 1. Analytical method 2. Graphical method Analytical methods are Parallelogram law & Method of Resolution Prof. V.V.Nalawade 51
- 53. Type I: Problems on Composition of Forces by Parallelogram and Triangle Law Sin Sin Sin R P Q Where, R = Resultant of force P & Q θ = Angle Between P & R β = Angle Between P & Q α = Angle Between Q & R 1.1 Law Of Parallelogram:- 1.2 Triangle Law :- Prof. V.V.Nalawade 5 3
- 54. Ex.1. Find the resultant of the following forces • Solution : Method i) By Parallelogram Law Prof. V.V.Nalawade 54 3 N 4N 3 N 4N R α R = 5 N
- 55. Continue…….. Prof. V.V.Nalawade 55 • Solution : Mathod ii) By Triangle Law 3 N 4N R α By Cosine rule R = 5 N By Sine rule
- 56. Prof. V.V.Nalawade 56 70 N 50 N 60ᵒ 70 N 50 N α R α 70 N 50 N 120ᵒ 60ᵒ R
- 58. Prof. V.V.Nalawade 58 500 N 50 300 N 500 N 300 N 50
- 59. Prof. V.V.Nalawade 59 Resolution of forces Definition Problems
- 60. Resolution of forces • The way of representing a single force into number of forces without changing the effect of the force on the body is called as resolution of forces. Prof. V.V.Nalawade 60 R Fx Fy
- 61. Prof. V.V.Nalawade 61 Fx Fy R = 10 = 90
- 62. Ex. 1. Two Forces act at an angle of 120°. The bigger force is of 40N and the resultant is perpendicular to the smaller one. Find the smaller force. Prof. V.V.Nalawade 6 2 F2 = 20 N
- 63. Ex. 2. Resolve the 100 N force acting a 30° to horizontal into two component one along horizontal and other along 120° to horizontal. Prof. V.V.Nalawade 6 3
- 64. Resolution of a force into two mutually perpendicular components (Rectangular Components) • Let a force F be inclined at an angle as shown in fig. We have to resolve it into two mutually perpendicular components Fx along X- Axis and Fy along Y- Axis. Prof. V.V.Nalawade 64 A B O • From point A on the line of action of a force, draw perpendicular AB on X-Axis. • Now we have to calculate the lengths OB & AB. • Length OB represents the magnitude of X component i.e. (Fx) • & Similarly AB represents (Fy)
- 65. Prof. V.V.Nalawade 65 • In ∆ AOB, OB = OA cos θ But OA = F Therefore, OB = F cos θ Lets say OB = Fx, as it is the magnitude of x-component Hence, Fx = F cos θ A B O Fy Fx
- 66. Prof. V.V.Nalawade 66 • In ∆ OBA, AB = OA sin θ But OA = F Therefore, AB = F sin θ Lets say AB = Fy, as it is the magnitude of y- component Hence, Fy = F sin θ A B O Fy Fx
- 67. Resolution of a force into two non perpendicular components (Oblique Components) • A force can also be resolved along the two directions which are not at right angles to each other. • In ∆OAC, Applying sine rule, we get Prof. V.V.Nalawade 67 α β (α+β) F1 F2 F1 F2 O B C A F F2 F1 α β
- 68. Resolution of Force By Perpendicular component cos sin x y F F F F 1st Quad = Fx (+ve) & Fy (-ve) 2nd Quad = Fx (-ve) & Fy (+ve) 3rd Quad = Fx (-ve) & Fy (-ve) 4th Quad = Fx (+ve) & Fy (-ve) Where, Fx = Horizontal component of Force Fy = Vertical Component of force Prof. V.V.Nalawade 6 8
- 69. Q. 2 Find the Component of force 100 N passing through the points (0,2) & (-1,2) Prof. V.V.Nalawade 69 -1 1 -2 2 (-1,2) (0,2) F = 100 N -1 1 -2 2 (-1,2) (0,2) F = 100 N Θ = 0 cos sin x y F F F F Fx = 100N Fy = 0 Θ = 180 Fx = -100N Fy = 0
- 70. Resolution of Force By Non-Perpendicular component 1st Quad = Fx (+ve) & Fy (-ve) 2nd Quad = Fx (-ve) & Fy (+ve) 3rd Quad = Fx (-ve) & Fy (-ve) 4th Quad = Fx (+ve) & Fy (-ve) Prof. V.V.Nalawade 7 0 F F2 F1 α β
- 71. Prof. V.V.Nalawade 71 60° 105° 330° X- axis 2000 N F1 F2
- 72. Prof. V.V.Nalawade 72 X- axis F1 F2 F 30° 60°
- 73. Method of resolution STEPWISE PROCEDURE OF METHOD OF RESOLUTION: i. Resolve all forces horizontally and find the algebraic sum of all the horizontal components (i.e., ΣFx) ii. Resolve all forces vertically and find the algebraic sum of all the vertical components (i.e., ΣFy). iii. The resultant R of the given forces will be given by the equation: iv. The resultant force will be inclined at an angle θ, with the horizontal, such that v. Position of the resultant Prof. V.V.Nalawade 73
- 74. 1.4 Design steps of resolution of concurrent force system :- Case I :- When magnitude and direction of all forces in the force system is given & resultant is to be determined Step 1:- Find ΣFx (Horizontal Component) Step 2:- Find ΣFy (Vertical Component) Step 3:- Find Magnitude of Resultant Step 4:- Find Direction of Resultant Step 5:- Find Position of Resultant 2 2 x y R F F 1 tan y x F F Case II :- When resultant is horizontal Σ Fx = R and Σ Fy = 0 Case III :- When resultant is Vertical Σ Fx = 0 and Σ Fy = R Case IV :- When resultant is Zero Σ Fx = 0 and Σ Fy = 0 Case V :- When magnitude & Direction of resultant is given & magnitude & direction of any one force among the force system is to be determined Σ Fx = R cos θ and Σ Fy = R sin θ θ is measured w.r.t X-axis Prof. V.V.Nalawade 7 4
- 76. Prof. V.V.Nalawade 76 X-Axis Y-Axis 155.8 N 76.6ᵒ
- 77. Prof. V.V.Nalawade 77 132.3ᵒ 45.6 N X-Axis Y-Axis 47.7ᵒ
- 78. Prof. V.V.Nalawade 78 X-Axis Y-Axis 29.09 N 45.89ᵒ
- 79. Prof. V.V.Nalawade 79 132.3ᵒ 45.6 N X-Axis Y-Axis 47.7ᵒ 132.3ᵒ 45.6 N X-Axis Y-Axis 47.7ᵒ X-Axis Y-Axis 29.09 N 45.89ᵒ
- 80. Prof. V.V.Nalawade 80 X-Axis Y-Axis 29.09 N 45.89ᵒ
- 81. Prof. V.V.Nalawade 81 Moment of force Law of moments Varignon’s Theorem Couples Problems
- 82. Moment of forces • The rotational effect produced by force is known as moment of force. • It is equal to the magnitude of force multiplied by the perpendicular distance of the point from the line of action of the force. • M = F x d • Unit N.m , KN.m , N.mm etc., Prof. V.V.Nalawade 82
- 83. Sign Convention Prof. V.V.Nalawade 83 Clockwise (+VE) Anti-Clockwise (-VE) F d O F d O F d O F d O M@O = F x d
- 84. Law of Moments • It states that, “ In equilibrium when no of coplanar forces act on a body, the sum of the clockwise moments@ any point in their plane is equal to the sum of the anticlockwise moments @ the same point. • Algebraic sum Clockwise moments = Algebraic sum Anti-Clockwise moments @ same point Prof. V.V.Nalawade 84
- 85. • Moment about pivot • 500*2 = 1000*1 • 1000 = 1000 • Algebraic sum Clockwise moments = Algebraic sum Anti-Clockwise moments @ same point Prof. V.V.Nalawade 85
- 86. To find beam reaction To find forces in frames Prof. V.V.Nalawade 86 Use of Law of Moments
- 87. Varignon’s theorem of moments • It status that, “ The algebraic sum of moments of all forces about any point is equal to the moments of their resultant about the same point.” • Let ƩMFA = Algebraic sum of moments of all forces about any point A • ƩMRA = Moment of resultant force about same point A • Then ƩMFA = ƩMRA • i.e. F1.x1 + F2.x2 + F3.x3+……..+Fn.Xn = R.x Prof. V.V.Nalawade 87
- 88. Use of Varignon’s theorem of moments Prof. V.V.Nalawade 88 • This theorem is very useful in locating the position of the resultant of non- concurrent forces.
- 89. Examples of Moment of forces • Rotation of door • Tightening of nut by spanner • Compass etc., Prof. V.V.Nalawade 89
- 90. Couple • Two non-colinear, equal, unlike, parallel forces forms a couple. • As the forces are equal & opposite their resultant is ZERO. • Hence couple produces only rotary motion without producing linear motion. Prof. V.V.Nalawade 90
- 91. Prof. V.V.Nalawade 91 Examples of Couple Rotation of Steering Wheel, key, tap etc.,
- 92. Lever Arm OR Arm of the Couple • The distance between two forces of a couple is known as lever arm. • SI unit of couple is same as moment i.e. N.m, N.mm, KN.m, KN.mm etc., Prof. V.V.Nalawade 92 a P P
- 93. Sign Convention Prof. V.V.Nalawade 93 a P P a P P Clockwise (+VE) Anti-Clockwise (-VE)
- 94. Properties of Couple • The resultant of the force of a couple is always ZERO. i.e. R = P – P = 0 • The moment of couple is equal to the product of one of the force and lever arm. i.e. M = P x a Prof. V.V.Nalawade 94 a P P
- 95. Properties of Couple • The moment of couple about any point is constant. • Moment of couple = P x a • Moment of couple @ C = P*AC – P*BC = P*a • Moment of couple @ D = - P*AD + P*BD = P*a Prof. V.V.Nalawade 95 a P P D C B A
- 96. Properties of Couple • A couple can be balanced only by another couple of equal and opposite moment. • Two or more couples are said to be equal when they have same sense or moment • Moment = 100 *1=100 N. m= 10*10=100N.m = 50*2=100 N.m Prof. V.V.Nalawade 96 2 m 50 N 50 N 10 m 10 N 10 N 1 m 100 N 100 N
- 97. Properties of Couple • Couple can only rotate the body but cannot translate the body. • A couple does not have moment centre, like moment of force. Prof. V.V.Nalawade 97
- 100. 1.5 Moment of force:- M= F x d Unit = N.m or KN.m Clockwise = +ve Anticlockwise = -ve Couple:- M= F x d Unit = N.m or KN.m Clockwise = +ve Anticlockwise = -ve Varignon’s Theorem:- Moment of resultant about any point A = Σ of moments of all the forces about same point A. R x d = Σ M Prof. V.V.Nalawade 1 0
- 101. 1.6 Resultant of parallel force system:- Step 1:- Find R = ΣF Step 2:- Find ΣM @ O Step 3:- Apply Varignon’s theorem ΣM @ O = R. d Step 4:- Find position of resultant w.r.t. O i) ΣF (upward) = Σmo (+ve) ii) ΣF (downward) =Σmo (-ve) 1.7 Resultant of general force system:- Step 1:- Find ΣFx (Horizontal Component) Step 2:- Find ΣFy (Vertical Component) Step 3:- Find Magnitude of Resultant Step 4:- Find Direction of Resultant Step 5:- Find ΣM @O Step 6:-:- Apply Varignon’s theorem ΣM @ O = R. d Step 4:- Find position of resultant w.r.t. O 2 2 x y R F F 1 tan y x F F Prof. V.V.Nalawade 1 0
- 102. Method of approach to solve Coplanar (2D) problems Problem 2-D Concurrent Not Equilibrium (Resultant) 1. Parallelogram Law 2. Triangle law 3. Polygon law 4. Method of projections Equilibrium (Unknowns) F =0 x Fy=0 Non-concurrent Not Equilibrium (Resultant) 1.Choose a reference Point 2.Shift all the forces to a point 3.Find the resultant force and couple at that point 4.Reduce the force-couple system to a single force Equilibrium (Unknowns) ΣFx=0 ΣFy=0 ΣMz=0 Prof. V.V.Nalawade 1 0
- 104. Problem – 2D- Concurrent - Resultant Ex.1. Determine the resultant of the following figure Prof. V.V.Nalawade 1 0
- 106. Ex.2. The resultant of the four concurrent forces as shown in Fig acts along Y-axis and is equal to 300N.Determine the forces P and Q. Problem – 2D- Concurrent - Resultant F x 0 9 F y R 300N Prof. V.V.Nalawade 1 0
- 107. F x 800380QSin45 Psin50 0 F y QCos45 PCos50 R 300 Solve quadratic equation and find P = 511 N Q = - 40.3N Problem solution F x 0 Fy R 300N Prof. V.V.Nalawade 1 0
- 109. Determine the resultant of the following figure Problem – 2D- Non Concurrent - Resultant Problem Prof. V.V.Nalawade 1 0
- 110. Resultant of General forces in a plane – Coplanar non-concurrent Step 2: Shift all the forces to a point Step 3: Find the resultant force Step 4: Reduce resultant force Step 1: Choose a reference point Prof. V.V.Nalawade 1 1
- 111. Resultant – Non-concurrent general forces in a plane Step:1: Choose A as reference Point Step:2: Shift all forces to point A Step:4: Reduce it to a single force Step 3: Find resultant force and couple x = 1880/600 x = 3.13m Problem solution: Prof. V.V.Nalawade 1 1
- 112. Determine the resultant of the following figure Problem – 2D- Non Concurrent - Resultant Problem Prof. V.V.Nalawade 1 1
- 113. Example: Resultant – Non-concurrent general forces in a plane Determine the resultant force of the non-concurrent forces as shown in plate and distance of the resultant force from point O͛. Step:2 Step:4 Step:3 Step:1 Problem solution Prof. V.V.Nalawade 1 1
- 118. Exercise 2 Prof. V.V.Nalawade 1 1
- 119. Prof. V.V.Nalawade 11 9 References • Engg. Mechanics ,Timoshenko & Young. • 2.Engg. Mechanics, R.K. Bansal , Laxmi publications • 3.Engineering Mechanics,Fedinand.L.Singer , Harper – Collins. • 4. Engineering Mechanics statics and dynamics, A Nelson, Mc Gra Hill publications • 5. Engg. Mechanics Umesh Regl, Tayal. • 6. Engineering Mechanics by N H Dubey • 7. Engineering Mechanics , statics – J.L.Meriam, 6th Edn – Wiley India Pvt Ltd. • 8. Engineering Mechanics , dynamics – J.L.Meriam, 6th Edn – Wiley India Pvt Ltd. • 9. Mechanics For Engineers , statics - F.P.Beer & E.R.Johnston 5th Edn Mc Graw Hill Publ. • 10.Mechanics For Engineers, dynamics - F.P.Beer & E.R.Johnston – 5th Edn Mc Graw Hill Publ. • 11. www.google.com • 12. http://nptel.iitm.ac.in/
- 120. Prof. V.V.Nalawade 12 0 Thank YOU