PEMP-AML2506. Day 01B. Session Speaker Dr. M. D. Deshpande. M.S. Ramaiah School of Advanced Studies - Bangalore 1 01B

Transcription

1 AML2506 Biomechanics and Flow Simulation Day Review of fengineering i Mechanics Session Speaker Dr. M. D. Deshpande 1

2 Session Objectives At the end of this session the delegate would have understood Basic concepts of Statics Basic concepts of Dynamics 2

3 Session Topics 1. Mechanics and its branches 2. Newton s laws of motion 3. Newton s law of gravitation 4. Force and Moment 5. Centre of Gravity and Moment of Inertia 6. Free body diagram 7. Velocity and Acceleration 8. Free fall of bodies 9. Work, Energy and Power 10. Law of Conservation of fenergy 3

4 Introduction WHAT IS MECHANICS? Mechanics - the physical science which describes or predicts the conditions of rest or motion of bodies under the action of forces. Mechanics is an applied science which explains and predicts the physical phenomenon and thus to lay foundations for engineering applications. It is divided into two parts: Statics & Dynamics. Further Dynamics is divided into two parts: Kinematics & kinetics. 4

5 Parallelogram Law FUNDAMENTAL CONCEPTS Two forces on a body can be replaced by a single force called the resultant by drawing the diagonal of the parallelogram with sides equivalent to the two forces 5

6 FUNDAMENTAL CONCEPTS Principle of Transmissibility The conditions of equilibrium or motion of a body remain unchanged if a force on the body is replaced by aforce of the same magnitude and direction along the line of action of the original force. 6

7 First Law FUNDAMENTAL CONCEPTS (Newton s Three Laws of Motion) A particle originally at rest, or moving with constant velocity, will remain in this state provided the particle is not subjected to an unbalanced force Second Law F=ma Third Law For every action there is an equal and opposite reaction 7

8 FUNDAMENTAL CONCEPTS (Newton s Law of Gravitational Attraction) G = 6.67 x N m 2 /kg 2 8

9 FUNDAMENTAL CONCEPTS (Weight) G = 6.67 x N m 2 /kg 2 9

10 UNITS OF MEASUREMENT (1) SI Units It is the international system of units Modern version of the metric system F = ma F = force in newton (N) m = mass in kg a = acceleration in m/s 2 N = kg. m/s 2 W = mg (g = 9.81 m/s 2 ) 10

11 UNITS OF MEASUREMENT (2) U.S. Customary (FPS) F = ma F = force in pound (lb) m = mass in slug a = acceleration in ft/s 2 slug = lb. s 2 /ft W = mg (g = 32.2 ft/s 2 ) 11

12 UNITS OF MEASUREMENT (3) Name Length Time Mass Force SI Metre m Second s Kilogram kg Newton N=(kg.m/s 2 ) FPS Feet ft Seconds Slug=(lb.s 2 /ft) Pound lb 12

13 Statics 13

14 Statics of particles Force on a Particle: A force represents the action of one body (Note) on another and is generally characterized by its Point of application Magnitude Direction Resultant of Forces: Two forces acting on a particle may be replaced by a single force called resultant force which has the same effect on the particle. 14

15 Moment of a force The moment M A of the force F about an axis through A, orfor short, the moment of F about A, is defined as the product of the magnitude F of the force and of the perpendicular distance d from A to the line of action of F: M A =Fxd A 15

16 Varignon s Theorem The moment of a force about any axis is equal to the sum of the moments of its components about that axis. Fd =Pd 1 +Qd 2 16

17 Moment of a Couple Two forces having the same magnitude and opposite sense, (parallel lines of action), are said to form a couple. M=Fd 1 -Fd 2 (F=-F ) =F(d 1 -d 2 ) = F d 17

18 The sum M is called the moment of the couple. It may be noted that M does not depend upon the choice of A (the point about which moments are taken); M will have the same magnitude and same sense regardless of the location of A. Thus, The moment M of a couple is constant. Its magnitude is equal to the product Fd of the common magnitude F of the two forces and of the distance d between their lines of action. The sense of M is obtained by direct observation. Two couples may be replaced by a single couple of moment equal to the algebraic sum of the moments of the given couples. 18

19 Contd.. Resolution of the given Force into a Force acting at a given Point and a Couple: Any force F acting (at point B in the figure) on a rigid body may be moved to any given point A, provided that a couple M is added; the moment M of the couple must equal the moment of F about A. 19

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21 Q1: Why the truss is not accelerating due to this force? Q2: Are the stresses in the truss same when we replace the original i lforces by the equivalent system? 21

22 Equilibrium of Rigid Bodies Equilibrium of Rigid Bodies: A rigid body is said to be in equilibrium when external forces and moments acting on the body form a system of forces and moments equivalent to zero. Σ F x = 0, Σ F y = 0, Σ M A = 0 Equilibrium of Rigid Bodies in space (3-D): Σ F x = 0, Σ F y = 0, Σ F z = 0 Σ M x = 0, Σ M y = 0, Σ M z = 0 NOTE: In 3-D we have an additional force F z, and two additional moments M y = 0 & M z. 22

23 Two men exert forces of F=80 lb and P=50 lb on the ropes. Determine the moment of each force about A. Which way will the pole rotate, clockwise or counterclockwise? 23

24 MR A F d i i M M R R A A counterclo ckwise clockwise 24

25 Centre of Gravity The centre of gravity ofacollectionofmassesisthepoint where all the weight of the object (elements) can be considered to be concentrated. ΔW 1, ΔW 2, aretheweightsof iht each of theelements t Then ΣF z : W=ΔW 1 + ΔW 2 + ΔW 3 + ΣM y : xw= x 1 ΔW 1 +x 2 ΔW 2 +x 3 ΔW 3 + ΣM x : yw= y y 1 ΔW 1 + y 2 ΔW 2 + y 3 ΔW 3 + W= dw xw= xdw yw= ydw 25

26 Contd.. 26

27 Centroids and Inertia Thecentroid is the centre of mass of a physical object. We can obtain this by taking the average location of the object, along three dimensions (conventionally, length, height, and width). If the object varies in density, we would need to take this variation into account. An object, when thrown with a spin, can be considered to move with a liner velocity (of the centroid) and rotation around its centroid. 27

28 The inertia of an object is the tendency of an object at rest to stay at rest. Hence the mass of a body is a measure of its inertia to stay in a place; larger the mass, smaller is the acceleration for a given force. The inertia of an object suspended from its centroid, or balanced on a fulcrum, is directly related to how widely dispersed the mass is away from its centroid. 28

29 Moment of Inertia The Moment of Inertia (I) isatermusedtodescribethe capacity of a cross-section to resist bending. It is always considered with respect to a reference axis such as X-X or Y-Y Y. It is a mathematical property of a section concerned with a surface area and how that area is distributed about the reference axis. The reference axis is usually a centroidal axis. 29

30 The moment of inertia is also known as the Second Moment of the Area and is expressed mathematically as: I xx = Sum (da)(y 2 ) = Σ (da)(y 2 ) In which: I xx = the moment of inertia around the x axis da = Elemental areas considered on the plane y = the distance between the centroid of da from the x axis 30

31 Mass moment of inertia Mass moment of inertia for a particle: The mass moment of inertia is one measure of the distributionib i of the mass of an object relative to a given axis. O r m The mass moment of inertia is denoted by I and is given for a O single particle of mass m as I r m 2 O 31

32 Radius of gyration: Sometime in place of the mass moment of inertia the radius of gyration k is provided. The mass moment of inertia of the body can be calculated from k using the relation I = M k 2 where M is the total mass of the body. One can interpret t the radius of gyration k as the distance from the axis that one could put a single particle of mass M equal to the mass of the rigid body and have this particle have the same mass moment of inertia as the original body. 32

33 Free-body diagram A sketch showing the physical conditions of the problem is known as Space Diagram A sketch showing the particle or part of the body and all the forces acting on it is called a Free-body diagram 33

34 Free-body diagram 34

35 Free-Body Diagrams 1. Determine the extent of the body to be included. 2. Completely isolate the body from supports and other attached bodies. 3. If internal resultants are desired, pass a sectioning plane through the member at the appropriate location. 4. Sketch the outline of theresulting Free Body. 5. Indicate on the sketch all external applied loads. 6. Clearly indicate the location, magnitude and direction of each load. 35

36 Free-Body Diagrams 7. At supports, connections and section cuts, show unknown forces and couples. 8. Assign a symbol to each unknown. 9. Use sign convention to assign positive sense to unknowns or assign it arbitrarily. 10. Label significant points and dimensions. 11. Show reference axes. 36

37 Free-Body Diagrams Idealized Models Assume material is rigid A pin can be considered for support (A) A roller can be considered for support (B) Weight of the beam is generally neglected when it is small compared to the load the beam supports 37

38 PROBLEM-1 (1/3) Two smooth pipes each having a mass of 300 kg, are supported by the forks of the tractor, as shown. Draw the free-body diagrams for each pipe and both pipes together and determine the forces acting on the pipes. See the figure on the next slide for details. 38

39 SOLUTION - PROBLEM-1 (2/3) Q1: Is the problem completely defined? Clearly specify all the input data necessary to solve the problem (and nothing more). Q2: What is expected to be calculated? l 39

40 SOLUTION PROBLEM-1 (3/3) 40

41 Contd.. 41

42 Contd.. Q3: Does decomposition of forces along T fix its value? Q4: Why did we decompose the forces along T but not along the vertical direction? Q5: Now why this combined free body diagram will not help us to go further? How going back to the FBD for pipe A will help? Q6: Can we determine F and P now? Which is likely to be higher? 42

43 Free-Body Diagrams Support Reactions It is a mistake to support a door using a single hinge since the hinge must develop a force Cy to support W and a couple moment M to support the moment of W If the door is supported using two aligned hinges W is supported by both hinges and the moment of the door is resisted by the two hinge forces Fx and -Fx 43

44 Example 1 The crate is to be pulled horizontally by 1 m to place it on the cart by tying a knot as shown. What is the force required to complete the job? How does tension T in the string compare with the weight of the crate? List all the required information making sure redundant information is removed. See that it is solved as a static problem! 44

45 Example 1 (contd) Let P be the horizontal force. As θ decreases from 90, P increases. Q1: Why is T larger than weight? Q2: When the crate is pulled horizontally, with top point fixed, what happens to its vertical location? 45

46 Support Reactions and Member Connections Roller Support A y A y A x A y 46

47 Support Reactions and Member Connections Pin Support A A x y A y A A x x A y 47

48 Support Reactions and Member Connections Cantilever Support (Fixed End) y A x M A A x A y 48

49 Look at Diving Board 49

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52 A L 1 L 2 W M h c B 52

53 Free Body Diagram of Diving Board L 1 L 2 M A h W B 53

54 Dynamics 54

55 Dynamics Kinematics Kinetics Kinematics:- The study of the motion of particles and bodies, without reference to the force causing the motion / deals with the way things move. It is the study of the geometry of motion. This analysis involves determination of position, displacement, rotation, speed, velocityetc. t Kinetics:- The study of the motion of particles and bodies, with reference to the forces causing this motion. 55

56 Linear Motion Linear displacement and distance The linear displacement is the length moved in a given direction - it is a vector quantity. The magnitude of the displacement is the distance - a scalar quantity. 56

57 Linear velocity and speed The linear velocity is the rate of change of displacement with time. As displacement is a vector, velocity has to be a vector. The magnitude of the velocity is speed. It is the rate of change of distance with time - hence it is a scalar. If a body moves with uniform velocity then it must move in a fixed direction with constant speed. The average speed of a body is the total distance moved divided by the total time taken. 57

58 Distance - time curve A graph plotted for distance (s) against time (t), might look like this as shown: o 58

59 As speed is the rate of change of distance with time, the slope or gradient of the s-t curve is the speed. Over the linear section OA of the curve the speed is uniform. Between A and Btheslope isdecreasing, hence the body is slowing down. At B the body has stopped (distance is not changing) and remains at rest between B and C. Q1: How does the corresponding velocity versus time curve look like? Q2: If you are given velocity time trace, what additional information is needed to get the s-t curve? 59

60 Linear acceleration The linear acceleration of a body is the rate of change of linear velocity with time. It is a vector. If acceleration is uniform the speed must be increasing by equal amount in equal time intervals. 60

61 Free fall under gravity When a body falls to earth freely (without any other forces involved) it accelerates and this acceleration is called acceleration due to gravity. It is often given the symbol g. Provided that air resistance is negligible all bodies, heavy or light fall at the same acceleration. Although g varies slightly at different points, on the earth s surface g = 9.81 m/s 2 canbeusedinmost calculations. Q1: Can you argue that g has to be same for all bodies? Q2: How does g value change when we reach to higher altitudes? 61

62 See that we have assumed F = mg. It needs a justification also! 62

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64 Terminal Velocity 64

65 Work, Energy and Power Work done by a constant force When the point at which a force acts moves in the direction of force, the force is said to have done work. When the force is constant, the work done is defined as the product of the force and distance moved. Work done = force distance moved in direction of force 65

66 Consider the example in the Figure below. A force F acting at an angle θ moves a body from point A to point B. Figure : Notation for work done by a force F 66

67 The distance moved in the direction of the force is given by Distance in the direction of force = s cos θ. So the work done by the force F is Work done = F s cos θ. If the body moves in the same direction as the force then Work done = Fs. If the angle is 90 then the work done is zero. The SI units for work are joules J (with force, F, innewtons N and distance, s, inmetresm). 67

68 Energy A body which has the capacity to do work is said to possess energy. For example, water in a reservoir is said to possess energy as it could be used to drive a turbine lower down the valley. There are many forms of energy e.g., mechanical, electrical, chemical, heat, nuclear etc. The units are the same as those for work, and the SI unit is Joule J. We mainly consider mechanical energy here which may be of two kinds: potential and kinetic. 68

69 Potential Energy There are different forms of potential energy. Two examples are: i. A pile driver raised ready to fall on to its target possesses gravitational potential energy while ii. A coiled spring which is compressed possesses an internal potential energy. So is a stretched spring. Gravitational potential energy may be described as energy due to position relative to a standard position (normally chosen to be the earth's surface.) The potential energy of a body may be defined as the amount of work it would do if it were to move from the its current position to the standard position. 69

70 Formulae for gravitational potential energy A body is at rest on the earth's surface. It is then raised a vertical distance h above the surface. The work required to do this is the force required times the distance h. Since the force required is its weight, and weight, W=mg,then the work required is mgh. The body now possesses this amount of energy - stored as potential energy -ithasthecapacity to do this amount of work, andwoulddosoifallowedtofalltoearth. Potential energy is thus given by: P.E. = mgh. 70

71 Kinetic energy Kinetic energy may be described as energy due to motion. The kinetic energy of a body may be defined as the amount of work it can do before being brought to rest. For example: when a hammer is used to drive in a nail, work is done on the nail by the hammer and hence the hammer must have possessed energy. 71

72 Kinetic energy and work done When a body with mass m has its speed increased from u to v in a distance s by a constant force F which produces an acceleration a, then we know Multiplying this by m gives an expression for the change in kinetic energy (the difference in kinetic energy at the end and the start) 72

73 Since F= ma, increase in K.E. = F s. But we also know F s = work done. So therelationship hi between kinetici energy can be summed up Work done by forces acting on a body = change of kinetic energy in the body This is sometimes known as the work-energy theorem. 73

74 Conservation of energy The principle i of conservation of energy states that the total energy of an isolated system remains constant. Energy cannot be created or destroyed but may be converted from one form to another. Take the case of a crate on a slope. Initially it is at rest, all its energy is potential energy. As it accelerates, some of it potential energy is converted into kinetic energy and some used to overcome friction. This energy used to overcome friction is not lost but converted into heat. If we consider a body falling freely in air, neglecting air resistance, then mechanical energy is conserved, as potential energy is lost and equal amount of kinetic energy is gained as speed increases. Q1: When the muscles in the human body contract, explain where the work is done. Which is the energy source to do this work? Q2: Heart is pumping blood in the arteries. Assume steady periodic state is reached. Explain how energy balance can be made for the cardiovascular system. 74

75 Power Power is the rate at which work is done, or the rate at which energy is used transferred. The SI unit for power is the watt W. Apowerof1W means that work is being done at the rate of 1 J/s. Larger units for power are the kilowatt kw (1kW = 1,000 W=10 3 W) and the megawatt MW (1 MW = 1,000,000 W =10 6 W). If work is being done by a machine moving at speed v against a constant t force, or resistance, F, then since work done is force times distance, work done per second is Fv, which is the same as power. Power = Fv. 75

76 Review Equilibrium of forces, Moment of Inertia, Moment of a force, Centre of gravity, Couple are discussed. Support reactions are discussed d and calculated. l Free body diagrams are discussed and shown how to draw them and use them. Basic definitions of work, energy and power have been discussed. Some Simple problems have been solved to demonstrate the ideas. 76

77 Thank you 77