Mechanics of Solids [ ] CIE 101 / 102. First Year B.E. Degree

Transcription

1 Mechanics of Solids [ ] CIE 101 / 102 First Year B.E. Degree

2 Mechanics of Solids PART- I Mechanics of Rigid Bodies PART- II Mechanics of Deformable Bodies

3 COURSE CONTENT IN BRIEF ART I Mechanics of Rigid Bodies 1. Resultant of concurrent and non-concurrent coplanar forces. 2. Equilibrium of concurrent and non-concurrent coplanar forces. 3. Centroid of plane areas 4. Moment of Inertia of plane areas 5. Kinetics: Newton s second law, D Alembert s principle, Work- Energy, and Impulse- Momentum principle. ART II Mechanics of Deformable bodies 6. Simple stresses and strains 7. Statically indeterminate problems and thermal stresses 8. Stresses on inclined planes 9. Stresses due to fluid pressure in thin cylinders

4 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. 5. Machanics of Materials, by E.P.Popov 6. Machanics of Materials, by E J Hearn 7. Strength of materials, by Beer and Johnston 8. Strength of materials, by F L Singer & Andrew Pytel 9. Strength of Materials, by B.S. Basavarajaiah & P. Mahadevappa 10. Strength of Materials, by Ramamruthum 11. Strength of Materials, by S S Bhavikatti

5 PART - I MECHANICS OF RIGID BODIES

6 PART - I Mechanics of Rigid Bodies INTRODUCTION 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.

7 Engineering Mechanics Branches of Mechanics Mechanics of Solids Mechanics of Fluids igid Bodies Deformable Bodies Ideal Fluids Viscous Fluids Compres Fluids tatics Dynamics Strength of Materials Theory of Elasticity Theory of Plasticity inematics Kinetics

8 Concept of Rigid Body : It is defined as a definite amount of matter the parts of which are fixed in position relative to one another under the application of load. 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.

9 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.

10 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: (i) Magnitude (ii) Direction (iii) Point of application (iv) Sense.

11 Force: characteristics of the force 100 kn are : (i) Magnitude = 100 kn (ii) Direction = at an inclination of 30 0 to the x-axis (iii) Point of application = at point A shown (iv) Sense = towards point A A 100 kn 30 0

12 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. 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.

13 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 A B P 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. P The same is true when the force is applied at a point O. O P

14 1. RESULTANT OF COPLANAR FORCES 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. F 2 F 1 = R A A θ F 3 R = F + F + F external effect on particle, A is same

15 Resultant of two forces acting at a point Parallelogram law of forces : 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. B C O P 2 α θ P 1 R A Contd..

16 B C O P 2 α θ P 1 R A In the above figure, P 1 and P 2, 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 = P 12 + P P 1 P 2 Cos α Inclination of the resultant w.r.t. the force P 1 is given by: θ = tan -1 [( P 2 Sin α) / ( P 1 + P 2 Cos α )]

17 Resultant of two forces acting at a point at right angle B C O P 2 α θ P 1 R A If α = 90 0, (two forces acting at a point are at right angle) B C P 2 O θ R P 1 A R = + tanθ = 2 P 1 P P 2 P 1 2 2

18 Triangle law of forces If two forces acting at a point can be represented both in magnitude and direction, by the two sides of a triangle 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.

19 Triangle law of forces et F 1 and F 2 be the two forces acting at a point A and θ is the ncluded angle. A F 1 θ F 2 = R F 2 θ F 1 Arrange the two forces as two sides of a triangle taken in tip to tail order, the third side of the triangle represents both in magnitude and direction the resultant force R. the sense of the resultant force is defined by its tail at the tail of the first force and its tip at the tip of the second force.

20 Triangle law of forces A F 1 θ F 2 = R θ F 1 F 2 R α F 1 β F 2 F1 = sin β F2 = sinα R sin(180 α β ) (180 - α - β) = θ where α and β are the angles made by the resultant force with the force F 1 and F 2 respectively.

21 Component of a force : 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.

22 Rectangular component of a force θ x F F θ x F x F y = F F y θ x F x Consider a force F making an angle θ x with x-axis. Then the resolved part of the force F along x-axis is given by F x = F cos θ x The resolved part of the force F along y-axis is given by F y = F sin θ x

23 Oblique component of a force Let F 1 and F 2 be the oblique components of a force F. The components F 1 and F 2 can be found using the triangle law of forces. F 2 β α N F F 1 M O The resolved part of the force F along OM and ON can obtained by using the equation of a triangle. F α F 1 β F 2 F 1 / Sin β = F 2 / Sin α = F / Sin(180 - α - β)

24 Sign Convention for force components: y y +ve x +ve x 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. Also force components that are directed along +ve y-direction are taken +ve for summation along the y-direction.

25 Classification of force system Force system Concurrent Coplanar Forces Non-concurrent Non-Coplanar Forces Concurrent Non-concurrent Like parallel Unlike parallel Like parallel 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.

26 RESULTANT OF COPLANAR NON CONCURRENT FORCE SYSTEM Coplanar Non-concurrent Force System: This is the force system in which lines of action of individual forces lie in the same plane but act at different points of applications. F1 F2 F1 F2 F3 F5 F4 F3 Fig. 1 Fig. 2

27 1. Parallel Force System Lines of action of individual forces are parallel to each other. 2. Non-Parallel Force System Lines of action of the forces are not parallel to each other.

28 MOMENT OF A FORCE ABOUT AN AXIS The applied force can also tend to rotate the body about an axis in addition to motion. This rotational tendency is known as moment. Definition: Moment is the tendency of a force to make a rigid body to rotate about an axis. This is a vector quantity having both magnitude and direction.

29 MOMENT OF A FORCE ABOUT AN AXIS Moment Axis: This is the axis about which rotational tendency is determined. It is perpendicular to the plane comprising moment arm and line of action of the force (axis 0-0 in the figure) Moment Center: This is the position of axis on coplanar system. (A). Moment Arm: Perpendicular distance from the line of action of the force to moment center. Distance AB = d.

30 Magnitude of moment: It is computed as the product of the of the force and the perpendicular distance from the line of action to the point about which moment is computed. (Moment center). M A = F d = Rotation effect because of the force F, about the point A (about an axis 0-0) Unit kn-m, N-mm etc.

31 Sense of moment: The sense is obtained by Right Hand Thumb rule. If the fingers of the right hand are curled in the direction of rotational tendency of the body, the extended thumb represents the sense of moment vector. For the purpose of additions, the moment direction may be considered by using a suitable sign convention such as +ve for counterclockwise and ve for clockwise rotations or viceversa. M M

32 VARIGNON S THEOREM (PRINCIPLE OF MOMENTS) Statement: The moment of a force about a moment center or axis is equal to the algebraic sum of the moments of its component forces about the same moment center (axis). P P P sinθ θ P cosθ d θ d 1 A d 2 A oment of Force P about the oint A, x d = Algebraic sum of Moments of components of the Force P about the point A, P cosθ x d 1 + P sinθ x d 2

33 VARIGNON S THEOREM (PRINCIPLE OF MOMENTS) Proof (by Scalar Formulation): Let R be the given force. P & Q are component forces of R. O is the moment center. p, r and q are moment arms from O of P, R and Q respectively. Y α, β and γ are the inclinations of P, R and Q respectively w.r.to X axis. A Q q β α γ r R P p O X

34 We have, R y = P y + Q y R Sinβ = P Sinα + Q Sin γ ----(1) From le AOB, p/ao = Sin α From le AOC, r/ao = Sin β From le AOD, q/ao = Sin γ From (1), R (r/ao) = P (p/ao) + Q (q/ao) i.e., R r = P p + Q q R y Q Q y P y α A Y D γ β q C r B p O R P X Moment of resultant R about O = algebraic sum of moments of component forces P & Q about same moment center O.

35 COUPLE Two parallel, non collinear (separated by certain distance) forces that are equal in magnitude and opposite in direction form couple. F The algebraic summation of the two forces forming couple is zero. Hence, couple does not produce any translation and produces only rotation. = d F M = F x d

36 RESOLUTION OF A FORCE INTO A FORCE-COUPLE SYSTEM Replace the force F acting at the point A to the point B F B A Apply two equal and opposite forces of same magnitude & direction as Force F at point B, so that external effect is unchanged F F B A F d

37 Thus, the force F acting at a point such as A in a rigid body can be moved to any other given point B, by adding a couple M. The moment of the couple is equal to moment of the force in its original position about B. F F F B A d F = B A M = F x d Of these three forces, two forces i.e., one at A and the other oppositely directed at B form a couple. Moment of this couple, M = F d. Third force at B is acting in the same direction as that at P.

38 TYPES OF LOADS ON BEAMS 1. Concentrated Loads This is the load acting for very small length of the beam. also known as point load, Total load W is acting at one point ) 2. Uniformly distributed load This is the load acting for a considerable length of the beam with same intensity of w kn/m throughout its spread. L W kn w kn/m Total intensity, W = w L (acts at L/2 from one end of the spread) W = (w x L) kn L/2 L

39 3. Uniformly varying load This load acts for a considerable length of the beam with intensity varying linearly from 0 at one end to w kn/m to the other representing a triangular distribution. Total intensity of load = area of triangular spread of the load W = 1/2 w L. (acts at 2 L/3 from Zero load end) 2/3 L L W = ½ L w L w kn/m 1/3 L

40 EXERCISE PROBLEMS 1. Resultant of force system Q1 A body of negligible weight, subjected to two forces F 1 = 1200N, and F 2 =400N acting along the vertical, and the horizontal respectively, is shown in figure. Find the component of each force parallel, and perpendicular to the plane. F 2 = 400 N Y F 1 = 1200 N X 4 3 Ans : F 1X = -720 N, F 1Y = -960N, F 2X = 320N, F 2Y = -240N

41 EXERCISE PROBLEMS 1. Resultant of force system Q2. Determine the X and Y components of each of the forces shown in the figure. F 2 = 390 N 12 Y 40º 5 30º X F 3 = 400 N F 1 = 300 N (Ans : F 1X = N, F 1Y = -150 N, F 2X = -150N, F 2Y = 360 N, F 3X = N, F 3Y = N )

42 EXERCISE PROBLEMS 1. Resultant of force system Q3. Obtain the resultant of the concurrent coplanar forces shown in the figure 800N 20º 40º 600N 30º 200N (Ans: R = N, θ = 68.43º)

43 EXERCISE PROBLEMS 1. Resultant of force system Q4. A disabled ship is pulled by means of two tug boats as shown in FIG. 4. If the resultant of the two forces T 1 and T 2 exerted by the ropes is a 300 N force acting parallel to the X direction, find : (a) Force exerted by each of the tug boats knowing α = 30º. (b) 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. α T 2 20º R = 300 N X - direction FIG. 4 T 1 ( Ans: a. T 1 = N, T 2 = N b. α = 70º, T 1 = N, T 2(min) = N )

44 EXERCISE PROBLEMS 1. Resultant of force system Q5. An automobile which is disabled is pulled by two ropes as shown in the figure. Find the force P and resultant R, such that R is directed as shown in the figure. 40º 20º P R Q = 5 kn (Ans: P = 9.4 kn, R = kn)

45 EXERCISE PROBLEMS 1. Resultant of force system Q6. A collar, which may slide on a vertical rod, is subjected to three forces as shown in figure. The direction of the force F may be varied Determine the direction of the force F, so that resultant of the three forces is horizontal, knowing that the magnitude of F is equal to (a) 2400 N, (b)1400n 1200 N 60º 800 N θ COLLAR F ROD ( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)

46 EXERCISE PROBLEMS 1. Resultant of force system Q7. Determine the angle α and the magnitude of the force Q such that the resultant of the three forces on the pole is vertically downwards and of magnitude 12 kn. Refer figure 5kN α Q 30º 8kN Fig. 7 (Ans: α = 10.7 º, Q = kn )

47 EXERCISE PROBLEMS 1. Resultant of force system Q8. Determine the resultant of the parallel coplanar force system shown in figure. 600 N 2000 N 60º o 30º 10º 60º 1000 N 400 N (Ans. R=800N towards left, d=627.5mm)

48 EXERCISE PROBLEMS 1. Resultant of force system Q9. Four forces of magnitudes 10N, 20N, 30N and 40N acting respectively along the four sides of a square ABCD as shown in the figure. Determine the magnitude, direction and position of resultant w.r.t. A. 30N D 20N C a A a B 10N 40N (Ans:R=28.28N, θ=45º, x=1.77a)

49 EXERCISE PROBLEMS 1. Resultant of force system Q10. Four parallel forces of magnitudes 100N, 150N, 25N and 200N acting at left end, 0.9m, 2.1m and 2.85m respectively from the left end of a horizontal bar of 2.85m. Determine the magnitude of resultant and also the distance of the resultant from the left end. (Ans: R = 125 N, x = 3.06 m)

50 EXERCISE PROBLEMS 1. Resultant of force system Q11. Reduce the given forces into a single force and a couple at A. 45º 70.7 kn 200 kn 30º 1.5m A 80 N 1m 30º 100 N (Ans:F=320kN, θ=14.48º, M=284.8kNm)

51 Q12. Determine the resultant w.r.t. point A. EXERCISE PROBLEMS 1. Resultant of force system 150 N 150 Nm 1.5m 3m 1.5m A 100 N 500 N (Ans: R = 450 kn, X = 7.5 knm)

52 2. EQUILIBRIUM OF FORCE SYSTEMS

53 2. EQUILIBRIUM OF FORCE SYSTEMS EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS Definition:- If a system of forces acting on a body, keeps the body in a state of rest or in a state of uniform motion along a straight line, then the system of forces is said to be in equilibrium. ALTERNATIVELY, if the resultant of the force system is zero, then, the force system is said to be in equilibrium.

54 EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS Conditions for Equilibrium : coplanar concurrent force system will be in equilibrium if it satisfies the following two conditions: ) Fx = 0; and ii) Fy = 0.e. Algebraic sum of components of all the forces of the system, along two mutually perpendicular directions, is ZERO. Y X

55 Graphical conditions for Equilibrium Triangle Law: If three forces are in equilibrium, then, they form a closed triangle when represented in a Tip to Tail arrangement, as shown in Fig 2.1 F 2 F2 F 3 F 1 Fig 2.1 Polygonal Law: If more than three forces are in equilibrium, then, they form a closed polygon when represented in a Tip to Tail arrangement, as shown in Fig F 3 F 1 F 4 F 3 F 3 F 2 F 4 F 5 F 1 Fig 2.2 F 1 F 5 F2

56 LAMI S THEOREM If a system of three forces is in equilibrium, then, each force of the system is proportional to sine of the angle between the other two forces (and constant of proportionality is the same for all the forces). Thus, with reference to Fig.2.3, we have, Sin α β F Sin F1 2 3 = = F Sin γ F3 β α F1 γ F2 Fig. 2.3 Note: While using Lami s theorem, all the three forces should be either directed away or all directed towards the point of concurrence.

57 EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCE SYSTEM When a body is in equilibrium, it has neither translatory nor rotatory motion in any direction. Thus the resultant force R and the resultant couple M are both zero, and we have the equilibrium equations for two dimensional force system F x = 0; F y = 0 M = 0 These requirements are both necessary and sufficient conditions for equilibrium.

58 SPACE DIAGRAMS & FREE BODY DIAGRAMS Space Diagram (SPD) : The sketch showing the physical conditions of the problem, like, the nature of supports provided; size, shape and location of various bodies; forces applied on the bodies, etc., is known as space diagram. eg, Fig 2.4 is a space diagram 3 m θ Cable 30 P = 2kN Weight of sphere = 0.5 kn, Radius = 1m wall Sphere Fig. 2.4 SPD

59 Free Body Diagram (FBD) : It is an isolated diagram of the body being analyzed (called free body), in which, the body is shown freed from all its supports and contacting bodies/surfaces. Instead of the supports and contacting bodies/surfaces, the reactive forces exerted by them on the free body is shown, along with all other applied forces.

60 5) Forces in Cables (Strings or Chords): These can only be tensile forces. Thus, these forces will be along the cable and directed away from the body. A Few Guidelines for Drawing FBD 1) Tensile Force: It is a force trying to pull or extend the body. It is represented by a vector directed away from the body. 2) Compressive Force: It is force trying to push or contract the body. It is represented by a vector directed towards the body. 3) Reactions at smooth surfaces: The reactions of smooth surfaces, like walls, floors, Inclined planes, etc. will be normal to the surface and pointing towards the body. 4)Forces in Link rods/connecting rods: These forces will be acting along the axis of the rod, either towards or away from the body. (They are either compressive or tensile in nature).

61 Free Body Diagrams of the sphere shown in Fig. 2.4 T = Tension in the cable θ T P = 2kN Rw = Reaction of the wall Rw Sphere 30 W = self weight of the sphere P = external load acting on the sphere W=0.5kN Fig. 2.5 F B D of Sphere Detach the sphere from all contacts and replace that with forces like: Cable contact is replaced by the force tension = T Contact with the smooth wall is replaced by the reaction Rw.

62 Supports: A structure is subjected to external forces and transfers these forces through the supports on to the foundation. Therefore the support reactions and the external forces together keep the structure in equilibrium. There are different types of supports. a) Roller Support b) Hinged or pinned support c) Fixed or built in support

63 Supports Types of Supports (a) Flexible cable,belt,chain, rope Action on body BODY (b) Smooth surfaces BODY T Force exerted by cable is always a tension away from the body in the direction of cable 90 0 F 90 0 F Contact forces are normal to the surfaces

64 Supports A (c) Roller support A Contact force is normal to the surface on which the roller moves. The reaction will always be perpendicular to the plane of the roller. Roller support will offer only one independent reaction component. (Whose direction is known.)

65 Supports (d) pinned Support / hinged support A R h θ A R R v This support does not allow any translatory movement of the rigid body. There will be two independent reaction components at the support. The resultant reaction can be resolved into two mutually perpendicular components. Or it can be shown as resultant reaction inclined at an angle with respect to a reference direction.

66 Supports (e) Fixed or Built-in Support A R AH M A R AV This type of support not only prevents the translatory movement of the rigid body, but also the rotation of the rigid body. Hence there will be 3 independent reaction components of forces. Hence there will be 3 unknown components of forces, two mutually perpendicular reactive force component and a reactive moment as shown in the figure.

67 TYPES OF BEAMS A member which is subjected to predominantly transverse loads and supported in such a way that rigid body motion is prevented is known as beam. It is classified based on the support conditions. A beam generally supported by a hinge or roller at the ends having one span (distance between the support) is called as simply supported beam. A beam which is fixed at one end and free at another end is called as a cantilever beam. A B A B span span (a) Simply supported beam

68 TYPES OF BEAMS A B span R H M A B R v (b) Cantilever beam

69 TYPES OF BEAMS If one end or both ends of the beam project beyond the support it is known as overhanging beam. A B A B (c) Overhanging beam (right overhang)

70 Statically determinate beam Using the equations of equilibrium given below, if all the reaction components can be found out, then the beam is a statically determinate beam the equations of equilibrium Fx = 0; Fy = 0 M = 0

71 FRICTION Friction is defined as the contact resistance exerted by one body upon another body when one body moves or tends to move past another body. This force which opposes the movement or tendency of movement is known as frictional resistance or friction. Friction is due to the resistance offered by minute projections at the contact surfaces. Hence friction is the retarding force, always opposite to the direction of motion. Friction has both advantages & disadvantages. Disadvantages ---- Power loss, wear and tear etc. Advantages ---- Brakes, traction for vehicles etc.

72 FRICTION W P N F (Friction) Hills & Vales Magnified Surface Frictional resistance is dependent on the amount of wedging action between the hills and vales of contact surfaces. The wedging action is dependent on the normal reaction N.

73 FRICTION Frictional resistance has the remarkable property of adjusting itself in magnitude of force producing or tending to produce the motion so that the motion is prevented. When P = 0, F = 0 block under equilibrium When P increases, F also increases proportionately to maintain equilibrium. However there is a limit beyond which the magnitude of this friction cannot increase.

74 FRICTION When the block is on the verge of motion(motion of the block is impending) F attains maximum possible value, which is termed as Limiting Friction. When the applied force is less than the limiting friction, the body remains at rest and such frictional resistance is called the static friction. Further if P is increased, the value of F decreases rapidly and then remains fairly a constant thereafter. However at high speeds it tends to decrease. This frictional resistance experienced by the body while in motion is known as Dynamic friction OR Kinetic Friction.

75 FRICTION Dynamic Friction Sliding friction friction experienced when a body slides over another surface. Rolling friction friction experienced by a body when it rolls over a surface.

76 FRICTION N W φ R F max P F α N F max = µn Where F max = Limiting Friction N= Normal Reaction between the contact surfaces µ =Coefficient of friction µ = F max N Note : Static friction varies from zero to a maximum value. Dynamic friction is fairly a constant.

77 Angle of Friction FRICTION The angle between N & R depends on the value of F. This angle θ, between the resultant R and the normal reaction N is termed as angle of friction. As F increases, θ also increases and will reach to a maximum value of φ when F is F max (limiting friction) i.e. tanφ = (F max )/N = µ N φ W F max R P Angle φ is known as Angle of limiting Friction.

78 FRICTION Angle of limiting friction is defined as the angle between the resultant reaction (of limiting friction and normal reaction) and the normal to the plane on which the motion of the body is impending. Angle of repose When granular material is heaped, there exists a limit for the inclination of the surface. Beyond that angle, the grains start rolling down. This limiting angle upto which the grains repose (sleep) is called the angle of repose of the granular material.

79 FRICTION Significance of Angle of repose: The angle that an inclined plane makes with the horizontal, when the body supported on the plane is on the verge of motion due to its self -weight is equal to the angle of repose. Angle of repose is numerically equal to Angle of limiting friction

80 Laws of dry friction FRICTION 1. The magnitude of limiting friction bears a constant ratio to the normal reaction between the two surfaces. (Experimentally proved) 2. The force of friction is independent of the area of contact between the two surfaces. 3. For low velocities the total amount of friction that can be developed is practically independent of velocity. It is less than the frictional force corresponding to impending motion.

81 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q1. A 10kN roller rests on a smooth horizontal floor and is held by the bar AC as shown in Fig(1). Determine the magnitude and nature of the force in the bar AC and reaction from the floor under the action of the forces applied on the roller. [Ans:FAC=0.058 kn(t),r=14.98 kn] 7kN A 30 0 C kN Fig(1)

82 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q2. A 10 kn weight is suspended from a rope as shown in figure. Determine the magnitude and direction of the least force P required to pull the rope, so that, the weight is shifted horizontally by 0.5m. Also, determine, tension in the rope in its new position. [Ans: P= 2.43 kn, θ = ; T= 9.7kN.] 2m θ P 10kN

83 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q3. Determine the value of P and the nature of the forces in the bars for equilibrium of the system shown in figure. [Ans: P = 3.04 kn, Forces in bars are Compressive.] kN 60 P 45

84 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q4. A cable fixed as shown in Fig. supports three loads. Determine the value of the load W and the inclination of the segment BC. [Ans: W=25kN, θ = ] A 30 B θ 20 Loads are in kn W 60 C D 22.5

85 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q5. Find the reactions at A,B,C and D for the beam loaded as shown in the figure. (Ans.R A =R B =34kN;R C =28.84kN; M C =-140kNm ; θ C = ) 12kN/m 20 kn 12kN/m 4kN/m 4kN/m 30kN A 40kNm B 4 3 C 1m 2m 1m 1m 2m 1m 1m 2m

86 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q6. A uniform bar AB of weight 50N shown in the figure supports a load of 200N at its end. Determine the tension developed in the string and the force supported by the pin at B. (Ans. T=529.12N;R B =807.15N, θ B =64.6 ) A string 60 B 2.5m 200N 2.5m 2.5m

87 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q7. Find the position of the hinged support (x),such that the reactions developed at the supports of the beam are equal.. (Ans.x=2m.) 10kN/m 15kN 18kN/m 2.0m 1.0m m 3.0m x

88 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q8. A right angled bar ABC hinged at A as shown in fig carries two loads W and 2W applied at B &C.Neglecting self weight of the bar find the angle made by AB with vertical (Ans:θ =18.44 ) L m θ A B W 0.5L C 2W

89 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q9. For the block shown in fig., determine the smallest force P required a) to start the block up the plane b) to prevent the block moving down the plane. Take µ = N θ P [Ans.: (a) P min = 59.2N (b) P min = 23.7N (b) θ = 11.3 o ] 25

90 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q10. A block of weight 2000 N is attached to a cord passing over a frictionless pulley and supporting a weight of 800N as shown in fig. If µ between the block and the plane is 0.35, determine the unknown force P for impending motion (a) to the right (b) to the left [Ans.: (a) P = 132.8N (b) P = 1252N] 2000N 30 P 800N

91 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q11. Determine value of angle θ to cause the motion of 500N block to impend down the plane, if µ for all contact surfaces is N 500N θ =? [Ans.: θ = 28.4 ]

92 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q12. A horizontal bar 10m long and of negligible weight rests on rough inclines as shown in fig. If angle of friction is 15 o, how close to B may the 200N force be applied before the motion impends. A 2 m 100N 200N X =? B [Ans.: x = 3.5m]

93 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q13. Determine the vertical force P required to drive the wedge B downwards in the arrangements shown in fig. Angle of friction for all contact surfaces is 12 o.weight of block A= 1600 N. P A B 20 [Ans.: P = N]

94 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q14. Determine the force P which is necessary to start the wedge to raise the block A weighing 1000N. Self weight of the wedge may be ignored. Take angle of friction, φ = 15 o for all contact surfaces. A [Ans.: P = 1192N] P 20 wedge

95 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q15. A ladder of weight 200N, 6m long is supported as shown in fig. If µ between the floor and the ladder is 0.5 & between the wall and the ladder is 0.25 and it supports a vertical load of 1000N, determine a) the least value of α at which the ladder may be placed without slipping b) the reactions at A & B 1000N B [Ans.: (a) α = 56.3o (b) R A = 1193 N, R B = 550N] 5m α

96 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q16. An uniform ladder of weight 250N is placed against a smooth vertical wall with its lower end 5m from the wall. µ between the ladder and the floor is 0.3. Show that the ladder remains in equilibrium in this position. What is the frictional resistance on the ladder at the point of contact between the ladder and the floor? B Smooth wall [Ans.: F A = 52 N] A 12m

97 EXERCISE PROBLEMS 2. EQUILIBRIUM OF FORCE SYSTEMS Q17. A ladder of length 5m weighing 500N is placed at 45 o against a vertical wall. µ between the ladder and the wall is 0.20 & between ladder and ground is If a man weighing 600N ascends the ladder, how high will he be when the ladder just slips. If a boy now stands on the bottom rung of the ladder, what must be his least weight so that the man can go to the top of the ladder. [Ans.: (a) x = 2.92m (b) W boy = 458N]

98 3. CENTROID OF PLANE AREA

99 3. CENTROID Centre of gravity : of a body is the point at which the whole weight of the body may be assumed to be concentrated. A body is having only one center of gravity for all positions of the body. It is represented by CG. or simply G or C. Contd.

100 CENTRE OF GRAVITY Consider a block of uniform thickness and having a uniform mass m. W R It is possible to support (hold) the block in stable position by a rod as shown in the figure provided rod must be positioned exactly at the point of intersection of the diagonals. Or the rod must be supported exactly below where the total weight of the block act. Contd.

101 W CENTRE OF GRAVITY The block can be supported from any position provided the support rod and the line of action of weight are in same line. This indicates that the whole weight of the block act through one point. This point is called as centre of gravity. R Centre of gravity is that point about which the summation of the first moments of the weights of the elements of the body is zero. Contd.

102 W CENTRE OF GRAVITY x X 1 X 2 W 1 W 2 W 3 W 4 To determine mathematically the location of the centre of gravity of any body, we apply the principle of moments to the parallel system of gravitational forces The moment of the resultant gravitational force W, about any axis = the algebraic sum of the moments about the same axis of the gravitational forces dw acting on all infinitesimal Contd. elements of the body.

103 dw 3 dw 4 W CENTRE OF GRAVITY x dw 1 dw 2 X 1 x W = x X 2 dw x W = dw1 x1 + dw2 x2 + dw3 x dw n x n Where W = dw Contd.

104 x CENTRE OF GRAVITY x where = x- coordinate of centre of gravity x = x dw W x Similarly, y and z coordinates of the centre of gravity are y y dw z dw z = = and ----(1) W W

105 x CENTRE OF MASS With the substitution of W= m g and dw = g dm (if g is assumed constant for all particles, then ) x = x dm m, y = y dm m, z = z dm m ----(2) Equation 2 is independent of g and therefore define a unique point in the body which is a function solely of the distribution of mass. This point is called the centre of mass and clearly coincides with the centre of gravity as long as the gravity field is treated as uniform and parallel Contd.

106 When speaking of an actual physical body, we use the term centre of mass. The term centroid is used when the calculation concerns a geometrical shape only. Calculation of centroid falls within three distinct categories, depending on whether we can model the shape of the body involved as a line, an area or a volume. Contd.

107 The centroid C of the Volume segment, x = x dv V y = y dv,, V z = z dv V The centroid C of the line segment, x = x dl L y = y dl L,, z = z dl L

108 The centroid C of the Area segment, AREA: when the density ρ, is constant and the body has a small constant thickness t, the body can be modeled as a surface area. The mass of an element becomes dm = ρ t da. If ρ and t are constant over entire area, the coordinates of the centre of mass also becomes the coordinates of the centroid, C of the surface area and which may be written as x x A da y da y = A =,, z = z da A Contd.

109 Centroid of Simple figures: using method of moment ( First moment of area) Centroid of an area may or may not lie on the area in question. It is a unique point for a given area regardless of the choice of the origin and the orientation of the axes about which we take the moment.

110 The coordinates of the centroid of the surface area about any axis can be calculated by using the equn. (A) x = (a 1 ) x 1 + (a 2 ) x 2 + (a 3 ) x 3 +.+(a n ) x n = First moment of area Moment of Total area A about y- axis = Algebraic Sum of moment of elemental da about the same axis where (A = a 1 + a 2 + a 3 + a a n )

111 AXIS of SYMMETRY: It is an axis w.r.t. which for an elementary area on one side of the axis, there is a corresponding elementary area on the other side of the axis (the first moment of these elementary areas about the axis balance each other) If an area has an axis of symmetry, then the centroid must lie on that axis. If an area has two axes of symmetry, then the centroid must lie at the point of intersection of these axes. Contd.

112 For example: The rectangular shown in the figure has two axis of symmetry, X-X and Y-Y. Therefore intersection of these two axes gives the centroid of the rectangle. X da D/2 Y B/2 B/2 x x D da X da x = da x Moment of areas,da about y-axis cancel each other D/2 B Y da x + da x = 0 Contd.

113 AXIS of SYMMETYRY C must lie on the axis of symmetry C must lie on the axis of symmetry C must lie at the intersection of the axes of symmetry

114 To locate the centroid of simple rectangular area from first principles D B

115 To locate the centroid w.r.t. the base line x-x X D y X Let the distance of centroid from the base line x-x be y Then from the Principle of Moments A y = y da Moment of Total area A about x-axis = Sum of moment of elemental area da about the same axis Contd.

116 Consider a elemental area da at a distance y from the base line (x-x) Let the thickness of the element be dy da x B y y dy x Area of small element = da = B.dy Moment of this elemental area about x-x axis = (area) x (distance) = (B.dy). (y) Contd.

117 Sum of Moment of all such elemental areas comprising the total area = = y da = B dy y D 2 By 2 = = BD 2 2 Then from the Principle of Moments 0 A y = 2 BD 2 y = BD 2A 2, y = 2 BD 2BD, y = D 2 Contd.

118 Similarly we can show Y B x Y x dx A x = x da = x ( D dx ) = 2 Dx 2 B 0 x = B/2

119 To locate the centroid of simple right angle triangular area from first principles H B

120 Contd. To locate the centroid w.r.t. the base line x-x. H Let the distance of centroid from the base line x-x be y x y x Then from the Principle of Moments A y = y da Moment of Total area A about x-axis = Algebraic Sum of moment of elementary area da about the same axis.

121 Consider a small elemental area da at a distance y from the base line (x-x) Let the thickness of the element be dy b Area of small element = da = b.dy x da y y x dy Moment of this small elemental area about x- x axis = (area) x (distance) = (b.dy). (y) Contd.

122 Then from the Principle of Moments A y = y da A y = b dy y ( H y) B da = b dy = dy H b = B ( H y) H A y = B ( H y) H y = H/3 dy y H H/3 X B y X Contd.

123 Similarly we can prove that x = B/3 A x = x da da = h dx = H ( B x) dx B A x = A x = H b dx x ( B x) B dx x Y dx b = H ( B x) B x = B/3 H x x b B/3 Y B Contd.

124 The centroid of simple right angled triangle area from the base B/3 H Centroid B H/3

125 To locate the centroid of Semi Circular Area w.r.t. the diameter AB from first principles Consider a semicircle of radius R, A = area = 2 πd 8 D = 2R Let G be the centroid of the Semicircle, and y is its distance from the diameter AB. Contd.

126 da = r dθ dr Consider a small elemental area da, located at distance y from the diameter AB, Let r =radial distance of area da from centre of the semi circle. da = r dθ dr y = r sinθ Moment of this elemental area about the diameter AB = 2 = r sinθ dr dθ Contd.

127 Moment of all such elemental area da about the diameter AB = π R O O 2 r sinθ dr. dθ = π O r 3 3 R O sinθ. dθ = R 3 3 ( [ ] π cos ) θ O 2 3. R 3 = Contd.

128 R Then from the Principle of Moments A y A y = = y da π R O O 2 r sinθ dr. dθ y Centroid A y y = = 4R 3π 2. R 3 3 x R y x y = 4R 3π Because of symmetry x = 0

129 To locate the centroid of Quarter Circular Area w.r.t. the boundary radial line AB from first principle A y = y da A r θ dr R dθ B y A y = y = π 2 R O O 4R 3π r 2 sinθ dr. dθ Contd.

130 Centroid of Quarter Circular Area w.r.t. the boundary radial lines x = 4R 3π Centroid 4R y = 3π y = 4R 3π

131 To locate the centroid of Circular Sector w.r.t. the y-axis shown from first principle y X c =(2/3)Rcosθ da Consider a triangle of differential area = da = = 2 1 base height 1 = R dθ R 2 Distance of the differential area da, from y-axis = x c = 2 = R cosθ 3 Contd.

132 To locate the centroid of Circular Sector w.r.t. the y-axis shown from first principle Consider a triangle of differential area = da = = 2 1 base height 1 = R dθ R 2 Distance of the differential area da, from y-axis = x c = 2 = R cosθ 3 Contd.

133 Then from the Principle of Moments A α = 1 2 R dθ R = R α 2 α A x = da xc α 2 3 ( 2 ) 2 R α x = R cosθ R dθ α ( 2 R α ) x = R sinα 3 x = 2 R sinα 3 α

134 x = 2 R sinα 3 α For a semicircular area 2α = π, if we use this value in the above formula we get y = 4 R 3 π

135 To locate the centroid of area under the curve x = k y 3 from x = 0 to x = a from first principle

136 Consider a vertical element of area da = y dx at a distance x from the y-axis. da To find x- coordinate, x x X A x = da x A = a y dx 0 At x = a, y = b, i.e. a = k b 3, k = a/b 3 Contd.

137 Substituting y = (x/k) 1/3 and k = a/b 3 = a a xydx ydx x 0 0 dx k x x dx k x x a a = a 2 b x ab = a x 7 4 = Contd.

138 To find y, Coordinate of centroid of the rectangular element is y c = y/2 A y = y c da y a ydx = a y ( ydx) Substituting, y = b( x/a) 1/3 3ab 4 y = 2 5 y b = 3ab 10 2

139 EXERCISE PROBLEMS 3. Centroid of plane area Problem No.1: Locate the centroid of the shaded area shown Ans: x=12.5, y=17.5

140 Ans: x=474mm, y=474mm Problem No.2: Locate the centroid of the shaded area shown EXERCISE PROBLEMS 3. Centroid of plane area mm r=600 D= mm

141 Ans: x=34.4, y=40.3 Problem No.3: EXERCISE PROBLEMS 3. Centroid of plane area Locate the centroid of the shaded area w.r.t. to the axes shown y-axis x-axis r=40

142 Ans: x= -5mm, y=282mm EXERCISE PROBLEMS 3. Centroid of plane area Problem No.4: Locate the centroid of the shaded area w.r.t. to the axes shown y-axis 250 mm mm x-axis

143 Ans:x =38.94, y=31.46 Problem No.5 EXERCISE PROBLEMS 3. Centroid of plane area Locate the centroid of the shaded area w.r.t. to the axes shown y x r=20

144 Ans: x=0.817, y=0.24 Problem No.6 EXERCISE PROBLEMS 3. Centroid of plane area Locate the centroid of the shaded area w.r.t. to the axes shown y m r=0.6 x 1.5

145 Ans: x= , y= Problem No.7 EXERCISE PROBLEMS 3. Centroid of plane area Locate the centroid of the shaded area w.r.t. to the axes shown

146 Problem No.8 Locate the centroid of the shaded area. EXERCISE PROBLEMS 3. Centroid of plane area 20 Ans: x= 0, y= 67.22(about base)

147 Problem No.9 EXERCISE PROBLEMS 3. Centroid of plane area Locate the centroid of the shaded area w.r.t. to the base line. 2 Ans: x=5.9, y= 8.17

148 Problem No.10 EXERCISE PROBLEMS 3. Centroid of plane area Locate the centroid of the shaded area w.r.t. to the axes shown Ans: x=21.11, y= 21.11

149 Ans: x= y= Problem No.11 EXERCISE PROBLEMS 3. Centroid of plane area Locate the centroid of the shaded area w.r.t. to the axes shown

150 Problem No.12 EXERCISE PROBLEMS 3. Centroid of plane area Locate the centroid of the shaded area w.r.t. to the axes shown Y R=25 R=25 80 X Ans: x= y= 4.723

151 4. MOMENT OF INERTIA OF PLANE AREA

152 4. MOMENT OF INERTIA Moment of Inertia( Second moment area) The product of the elemental area and square of the perpendicular distance between the centroid of area and the axis of reference is the Moment of Inertia about the reference axis I ox = da 1 y da 2 y 22 + da 3 y = da y 2 I oy = da 1 x da 2 x da 3 x y x da = da x 2 y o

153 Radius of Gyration A Elemental area A r 1 k r 2 k r 3 k B Radius of gyration is defined as a constant distance of all elemental areas which have been rearranged with out altering the total moment of inertia. B I AB = da k 2 + da k I AB = A k 2 I AB = da k 2 k= I AB /A

154 Polar moment of Inertia (Perpendicular Axes theorem) The moment of inertia of an area about an axis perpendicular to the plane of the area is called Polar Moment of Inertia and it is denoted by symbol I zz or J or I p. The moment of inertia of an area in xy plane w.r.to z. axis is I zz = I p = J = r 2 da = (x 2 + y 2 ) da = x 2 da + y 2 da = I xx +I yy Y x r y z O x

155 Polar moment of Inertia (Perpendicular Axes theorem) Hence polar M.I. for an area w.r.t. an axis perpendicular to its plane of area is equal to the sum of the M.I. about any two mutually perpendicular axes in its plane, passing through the point of intersection of the polar axis and the area.

156 Parallel Axis Theorem da x *G y x d A Moment of inertia of any area about an axis AB is equal to the M.I. about parallel centroidal axis plus the product of the total area and square of the distance between the two axes. B

157 I AB = da (d +y) 2 = da (d 2 + y 2 +2 d y) = da. d 2 + da y d da y = da. d 2 + da y d. y da In the above term (2 d) is constant & y da = 0 I AB = I xx + A.d 2

158 MOMENT OF INERTIA BY DIRECT INTEGRATION Moment of inertia of rectangular area about centroidal horizontal axis by direct integration M.I. about its horizontal centroidal axis : I x x = = = + D / 2 D / 2 + D / 2 ( B dy) D / 2 BD 12 da 3 y 2 y 2 x D D/2 G y dy x B

159 Moment of Inertia of rectangular area about its base- (about the line AB) using Parallel Axis Theorem I AB = I XX + A(d) 2. Where d = D/2, the distance between axes xx and AB =BD 3 /12+(BD)(D/2) 2 x D D/2 G y dy x =BD 3 /12+BD 3 /4 =BD 3 /3 A B B

160 Moment of inertia of Triangular area about the base by direct integration dy (h-y) x A h h/3 x b y B x From similar triangles b/h = x/(h-y) x = b. (h-y)/h

161 dy (h-y) x A h h/3 x b y B x I AB = da.y 2 = (x.dy)y 2 h I xx = (b. (h-y) y 2.dy) /h 0 = b[ h (y 3 /3) y 4 /4 ]/h = bh 3 /12

162 Moment of inertia of Triangular area about the centroidal horizontal axis Using Parallel axis theorem. MI about any line(ab) = MI about cenroidal parallel axis + Ad 2 I AB = I xx + Ad2 I xx = MI about centroidal axis x x I AB = MI about the Base line AB I xx = I AB Ad 2 = bh 3 /12 bh/2. (h/3) 2 x A h x b h/3 dy (h-y) y B x = bh 3 /36 Centroidal horizontal axis

163 I xx = da. y 2 R 2π = (x.dθ.dr) r 2 Sin 2 θ 0 0 R 2π = r 3.dr Sin 2 θ dθ 0 0 Moment of inertia of Circular area about the centroidal horizontal axis R 2π = r 3 dr {(1- Cos2θ)/2} dθ 0 R 0 2π =[r 4 /4] [θ/2 Sin2θ/4] 0 0 = R 4 /4[π -0] = πr 4 /4 I XX = π R 4 /4 = πd 4 /64 x A r θ R dθ y=rsinθ B x

164 Moment of inertia of Semi-circular area about the Base & centroidal horizontal axis I AB = da. y 2 R π = (r.dθ.dr) r 2 Sin 2 θ 0 R 0 π = r 3.dr Sin 2 θ dθ 0 0 y 0 R π = r 3 dr (1- Cos2θ)/2) dθ 0 0 π =[R 4 /4] [θ/2 Sin2θ/4] 0 = R 4 /4[π/2-0] = πr 4 /8 = (πd 4 /32) 4R/3π x A R y 0 x B

165 Moment of inertia of Semi-circular area about the centriodal horizontal axis using parallel axis theorem: I AB = I xx + A(d) 2 I xx = I AB A(d) 2 = π R 4 /8 πr 2 /2. (4R/3π) 2 I xx = 0.11R 4

166 Moment of inertia of Quarter-circular area about the base & centroidal horizontal axis I AB = I CD D y I AB = R π/2 0 0 (r.dθ.dr). r 2 Sin 2 θ R π/2 = r 3.dr Sin 2 θ dθ 0 0 R π/2 = r3 dr (1- Cos2θ)/2) dθ 0 0 π/2 =[R 4 /4] [θ/2 (Sin2 θ)/4] 0 = R 4 (π/16 0) = πr 4 /16 x A C 4R/3π y x 4R/3π B

167 Moment of inertia about Centroidal axis, I xx = I AB -Ad 2 = πr 4 /16 - πr 2. (0. 424R) 2 = 0.055R 4

168 4R/3π l.no Figure b I x0 -x I y0 -y I xx I yy 0 0 Y x 0 x d/2 Y X o d x0 x bd 3 /12 - bd 3 /3 - h x x 0 0 h/3 x x b x 0 O y 0 R x 0 bh 3 /36 - bh 3 /12 - πr 4 /4 πr 4 /4 - - y y 0 0 4R/3π x 0 x y 0 y y 0 x x R 4 πr 4 /8 πr 4 /8 - x 0 4R/3π 0.055R R 4 πr 4 /16 πr 4 /16

169 Q1. Determine the moment of inertia about the centroidal axes. EXERCISE PROBLEMS 30mm 30mm 20 30mm 100mm [Ans: Y = 27.69mm I xx = x 10 6 mm 4 I yy = x 10 6 mm 4 ]

170 EXERCISE PROBLEMS Q2. Determine second moment of area about the centroidal horizontal and vertical axes. 300mm 300mm mm 900mm [Ans: X = 99.7mm from A, Y = 265 mm I xx = x 10 9 mm 4, I yy = x 10 9 mm 4 ]

171 EXERCISE PROBLEMS Q3. Determine M.I. Of the built up section about the horizontal and vertical centroidal axes and the radii of gyration. 200mm mm mm [Ans: I xx = x 10 6 mm 4, I yy = x 10 6 mm 4 r xx = 62.66mm, r yy = 45.63mm]

172 EXERCISE PROBLEMS Q4. Determine the horizontal and vertical centroidal M.I. Of the shaded portion of the figure. X X [Ans: X = 83.1mm I xx = x 10 4 mm 4, I yy = x 10 4 mm 4 ]

173 EXERCISE PROBLEMS Q5. Determine the spacing of the symmetrically placed vertical blocks such that I xx = I yy for the shaded area. 200mm 200mm d 200mm 400mm 600mm [Ans: d/2 = 223.9mm d=447.8mm] 200mm

174 EXERCISE PROBLEMS Q6. Find the horizontal and vertical centroidal moment of inertia of the section shown in Fig. built up with R.S.J. (I-Section) 250 x 250 and two plates 400 x 16 mm each attached one to each. Properties of I section are I xx = x 10 4 mm 4 I yy = x 10 4 mm 4 160mm 2500mm ross sectional area=6971mm 2 160mm 4000mm [Ans: I xx = x 10 7 mm 4, I yy = x 10 7 mm 4 ]

175 EXERCISE PROBLEMS Q7. Find the horizontal and vertical centroidal moment of inertia of built up section shown in Figure. The section consists of 4 symmetrically placed ISA 60 x 60 with two plates 300 x 20 mm 2. roperties of ISA ross sectional area = 4400mm 2 xx = Iyy ;C xx = C yy =18.5mm 200mm 20mm 18.5mm 18.5mm 300mm [Ans: I xx = x 10 7 mm 4, I yy = x 10 7 mm 4 ]

176 EXERCISE PROBLEMS Q8. The R.S. Channel section ISAIC 300 are placed back to back with required to keep them in place. Determine the clear distance d between them so that I xx = I yy for the composite section. Properties of ISMC300 C/S Area = 4564mm 2 Y Lacing I xx = x 10 4 mm 4 I yy = x 10 4 mm 4 C yy = 23.6mm X 23.6mm X 380mm d [Ans: d = 183.1mm] Y

177 90mm [Ans: I xx = x 10 4 mm 4, I yy = x 10 4 mm 4 ] EXERCISE PROBLEMS Q9. Determine horizontal and vertical centroidal M.I. for the section shown in figure. 40mm 160mm 40mm 40mm

178 5. Kinetics of rectilinear motion

179 5. Kinetics of rectilinear motion In this chapter we will be studying the relationship between forces on a body/particle and the accompanying motion Newton s Second law of motion: Newton s first and third law of motion were used extensively in the study of statics (the bodies at rest) whereas Newton s second law of motion is used extensively in the study of the kinetics.

180 Work done by force: Work done by a force is the product of the force and the distance moved by the point of application in the direction of the force. It is a scalar quantity. A Work done = (F cosα ) s F sinθ α F F cosθ s B F sinθ X- component of force F moves through distance a S, S = displacement of force from A to B α F F cosθ Unit: Nm ( Joule )

181 POWER:- It is defined as the time rate of doing work. Power = work done /Time= (force distance) /Time = force velocity Unit: (Nm)/s = [watt] (kn m)/s = [kilo watt] 1 metric H.P= watts.

182 Energy:- It is defined as the capacity to do work. It is a scalar quantity. Unit :- N m (Joule)

183 Momentum:- Quantity of motion possessed by a body is called momentum. It is the product of mass and velocity. It is a vector quantity. Unit:- N s. Impulse of a Force:- It is defined as the product of force and the time over which it acts. It is a vector quantity. Unit:- N s.

184 Newton s second law of motion. If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of the resultant force and its direction is along that of the resultant force. F α a F =Resultant of forces a = Acceleration of the particle. F = ma m= mass of the particle.

185 The constant value obtained for the ratio of the magnitude of the force and acceleration is characteristic of the particle and is denoted by m. Where m is mass of the particle Since m is a +ve scalar, the vectors of force F and acceleration a have the same direction. Units Force in Newtons (N) N = 1 Kgm/s 2 Acceleration in m/s 2

186 Body will move in the direction of resultant R = m a F2 F1 F3 R = Resultant of forces F1,F2 and F 3

187 Using the rectangular coordinate system we have components along axes as, ΣF x = ma x ΣF y = ma y ΣF z = ma z where F x,f y F z and a x, a y,a z are rectangular components of resultant forces and accelerations respectively.

188 Newton s second law may also be expressed by considering a force vector of magnitude ma but of sense opposite to that of the acceleration. This vector is denoted by (ma) rev. The subscript indicates that the sense of acceleration has been reversed and is called the inertia force vector.

189 R = m a Inertia Force Body will move in the direction of resultant R = m a F2 F1 F3 R = Resultant of forces F1,F2 and F 3 F2 F3 F1

190 F 2 F = R = m a F 1 Inertia Force If the inertia force vector is added to the forces acting on the particle we obtain a system of forces whose resultant is zero. F 3 Resultant of forces F 1,F 2, F 3 and Inertia force = 0 F + F + F + ma = The particle may thus be considered to be in equilibrium. (THIS IS DYNAMIC EQUILIBRIUM)

191 It was pointed out by D Alembert (Alembert, Jean le Rond d ( ), French mathematician and philosopher) that problems of kinetics can be solved by using the principles of statics only (the equations of equilibrium) by considering an inertia force in a direction directly opposite to the acceleration in addition to the real forces acting on the system D Alembert s principle states that When different forces act on a system such that it is in motion with an acceleration in a particular direction, the vectorial sum of all the forces acting on the system including the inertia force ( ma taken in the opposite direction to the direction of the acceleration) is zero.

192 The problem under consideration may be solved by using the method developed earlier in statics. The particle is said to be in dynamic equilibrium. If ΣF x = 0 ΣF y = 0 including inertia force vector ΣF z = 0 This principle is known as D Alembert s principle

193 Work-Energy relation for translation From Newton s second law of motion F = m a = (1) Also a = dv/dt =( dv/dt) ( ds/ds) = v dv/ds substituting in (1) F = m v dv/ds ΣF ds = m v dv (2) Let the initial velocity be u and the final velocity after it moves through a distance s be v

194 Integrating both sides, we get F s ds = 0 m v 2 v F s = ( m ) 2 1 F s = m v 2 v u dv v u ( ) 2 2 u Therefore work done by a system of forces acting on a body while causing a displacement is equal to the change in kinetic energy of the body during the displacement.

195 Impulse-momentum relationship F = m a F = m (v - u)/t = (mv mu)/t Force = Rate of change of momentum F t = mv mu Impulse = final momentum Initial momentum The component of the resultant linear impulse along any direction is equal to change in the component of momentum in that direction.

196 EXERCISE PROBLEMS 5. Kinetics Q1. Blocks A and B of mass 10 kg and 30 kg respectively are connected by an inextensible cord passing over a smooth pulley as shown in Fig. Determine the velocity of the system 4 sec. after starting from rest. Assume coefficient of friction =0.3 for all surfaces in contact. B A 60 o 30 o Ans: v=13.6m/s

197 EXERCISE PROBLEMS 5. Kinetics Q2. A tram car weighs 150kN. The tractive resistance being 1% of the weight of car. What power will be required to move the car at uniform speed of 20 kmph (i) Up an incline 1 in 300 (ii) (ii) Down an incline 1 in 250. Take efficiency 75%. Ans: Pull = 2 kn a) Output power=11.12kw

198 EXERCISE PROBLEMS 5. Kinetics Q3. Two masses of 5 kg and 3 kg rest on two smooth inclined plane, each of inclination 30º and are connected by a string passing over a common apex. Find the velocity of 3 kg mass after 2 sec when released from rest. Find the distance it will cover before changing direction of motion, if 5kg mass is cut off after two sec of its release from rest. 5kg 3kg 30º 30º V = 4.45 m/s s = 0.61 m

199 EXERCISE PROBLEMS 5. Kinetics Q4. A locomotive weighing 900 kn pulls a train of 10 coaches each weighing 300 kn at 72 Kmph on a level track against a resistance of 7 N/kN. If the rear 4 coaches get snapped from the train, find the speed of the engine and the remaining coaches after 120 secs. Assume no change in resistance and draw bar pull. Find also distance traveled by detached coaches before coming to rest. 4x300=1200kN 6x300=1800kN 900kN P V = m/s s = 2.9 km

200 EXERCISE PROBLEMS 5. Kinetics Q5. Find the tension in the cord supporting body C in Fig. below. The pulleys are frictionless and of negligible weight. 150 kn A B C 300 kn Assume all blocks moving either downward or upward and accordingly draw FBD 450 kn 0=a A +2a B +c C Ans : T= kn

201 EXERCISE PROBLEMS 5. Kinetics Q6. Two blocks A and B are released from rest on a 30 o inclined plane with horizontal, when they are 20m apart. The coefficient of friction under the upper block is 0.2 and that under lower block is 0.4. compute the time elapsed until the block touch. After they touch and move as a unit what will be the constant forces between them. (Ans : t = 4.85 s, contact force=8.65 N)

202 EXERCISE PROBLEMS 5. Kinetics Q7. An elevator cage of a mine shaft weighing 8kN when empty is lifted or lowered by means of rope. Once a man weighing 600N entered it and lowered at uniform acceleratin such that when a distance of m was covered, the velocity of the cage was 25m/s. Determine the tension in the cable and force exerted by man on the floor of the cage. (Ans: T=7139 N and R=498 N)

203 EXERCISE PROBLEMS 5. Kinetics Q8. A small block starts from rest at point a and slides down the inclined plane. At what distance along the horizontal will it travel before coming to rest. Take µ k =0.3 [Ans :s=6m ] 5m A 3 4 B s C

204 EXERCISE PROBLEMS 5. Kinetics Q9. The system starts from rest in the position shown. How much further will block A move up the incline after block B hits the ground. assume the pulley to be frictionless and massless and µ is 0.2.W A =1000N, W B =2000N. [ Answer s =1.27m] A 4 3 3m B

205 EXERCISE PROBLEMS 5. Kinetics Q10. A 1500Kg automobile travels at a uniform rate of 50kmph to 75kmph. During the entire motion, the automobile is traveling on a level horizontal road and rolling resistance is 2 % of weight of automobile. Find (i) maximum power developed (ii) power required to maintain a constant speed of 75kmph. [ ANSWER: power developed = 6.131KN]

206 EXERCISE PROBLEMS 5. Kinetics Q11. Two bodies A and B weighing 2000N and 5200N are connected as shown in the figure. find the further distance moved by block a after the block B hits Wall. µ=0.2.[ Answer s=1.34m] A 5 12 B 3m

207 600m EXERCISE PROBLEMS 5. Kinetics Q12. A spring is used to stop 60kg package which is sliding on a horizontal surface. the spring has a constant k = 20kN/m and is held by cable such that it is initially compressed at 120mm. knowing that the package has a velocity of 2.5m/s in position shown and maximum additional displacement of spring is 40mm. Determine the coefficient of kinetic friction between package and surface. (Answer µ k =0.2) 2.5 m/s 60kg

208 EXERCISE PROBLEMS 5. Kinetics Q13. The system shown in figure has a rightward velocity of 4m/s, just before force P is applied. Determine the value of P that will give a leftward velocity of 6m/s in a time interval of 20sec. Take µ = 0.2 & assume ideal pulley. [Answer P=645.41N] P 1000N 400N

209 EXERCISE PROBLEMS 5. Kinetics Q14. A locomotive of weight 500kN pulls a train of weight of 2500kN. The tractive resistance, due to friction is 10N/kN. The train can go with a maximum speed of 27kmph on a grade of 1in100. Determine (a) Power of the locomotive. (b) Maximum speed it can attain on a straight level track with the tractive resistance remaining same. [Answer (a) Power= 450kN (b) v=15m/s] Q15. A wagon weighing 400kN starts from rest, runs 30m down a 1% grade & strikes a post. If the rolling resistance of the track is 5N/kN, find the velocity of the wagon when it strikes the post. If the impact is to be cushioned by means of one bumper string, which compresses 1mm per 20 kn weight, determine how much the bumper spring will be compressed. [Answer v=1.716m/s, x=77.5mm]

210 Q16. A train whose weight is 20kN moves at the rate of 60kmph. After brakes are applied, it is brought to rest in 500m. Find the force exerted, assuming it to be uniform a) Use work-energy relation b) Use D Alemberts equation. EXERCISE PROBLEMS 5. Kinetics Ans: F = kn

211 EXERCISE PROBLEMS 5. Kinetics Q17. The blocks A and B having weights 100 N and 300 N start from rest. The horizontal plane and the pulleys are frictionless. Determine the acceleration and the tension in the string. string A Frictionless pulley string Ans: a A =8.403m/s2 a B =4.201 m/s2 Frictionless pulley B T= N

212 EXERCISE PROBLEMS 5. Kinetics Q18. The blocks A and B having weights 100 N and 300 N start from rest. The horizontal plane and the pulleys are frictionless. Determine the velocity of block B after 0.5 seconds and the tension in the string. Use impulsemomentum relation. string A Frictionless pulley string Frictionless pulley B

213 EXERCISE PROBLEMS 5. Kinetics Q19. The blocks A and B having weights 100 N and 300 N start from rest when a load of 100 N is applied on the block A as shown in the figure. The horizontal plane and the pulleys are frictionless. Determine the acceleration and the tension in the string. 100 N A string Frictionless pulley string Frictionless pulley B

214 EXERCISE PROBLEMS 5. Kinetics Q20. Two blocks A and B are connected as shown in the figure. At the instant of their release if the block A, which is on smooth horizontal plane has a left ward velocity of 2 m/s, what would be its velocity 5 seconds after their release. The blocks A and B weigh 100 N and 300 N respectively. string A Frictionless pulley string Frictionless pulley B

215 EXERCISE PROBLEMS 5. Kinetics Q21. An engine weighing 500 kn drags carriages weighing 1500kN up an incline of 1 in 100 against a resistance of 5N/kN starting from rest. It attains a velocity of 36 kmph (10m/s) in 1 km distance with a constant draw bar pull supplied by the engine. What is the power required for the same? What is the tension developed in the link connecting the engine and carriages? 1500N 500N P Ans: Pull=40.19kN, power=401kw, T=30.15kN

216 Q22. what velocity the block A will attain after 2 seconds starting from rest? Take µ = 0.2. W A = 1500N, W B = 2000N. Use impulse-momentum relation. EXERCISE PROBLEMS 5. Kinetics A B 3

217 PART - II Mechanics of Deformable Bodies

218 COURSE CONTENT IN BRIEF 6. Simple stresses and strains 7. Statically indeterminate problems and thermal stresses 8. Stresses on inclined planes 9. Stresses due to fluid pressure in thin cylinders

219 6. Simple stresses and strains The subject strength of materials deals with the relations between externally applied loads and their internal effects on bodies. The bodies are no longer assumed to be rigid and the deformations, however small, are of major interest The subject, strength of materials or mechanics of materials involves analytical methods for determining the strength, stiffness (deformation characteristics), and stability of various load carrying members. Alternatively the subject may be called the mechanics of solids.

220 GENERAL CONCEPTS STRESS No engineering material is perfectly rigid and hence, when a material is subjected to external load, it undergoes deformation. While undergoing deformation, the particles of the material offer a resisting force (internal force). When this resisting force equals applied load the equilibrium condition exists and hence the deformation stops. These internal forces maintain the externally applied forces in equilibrium.

221 STRESS The internal force resisting the deformation per unit area is called as stress or intensity of stress. Stress = internal resisting force / resisting cross sectional area R = A

222 STRESS SI unit for stress N/m 2 also designated as a pascal (Pa) Pa = N/m 2 kilopascal, 1kPa = 1000 N/m 2 megapascal, 1 MPa = N/m 2 = N/(10 6 mm 2 ) = 1N/mm 2 1 MPa = 1 N/mm 2 gigapascal, 1GPa = N/m 2 = MPa = N/mm 2

223 AXIAL LOADING NORMAL STRESS STRESS P P R Consider a uniform bar of cross sectional area A, subjected to a tensile force P. Consider a section AB normal to the direction of force P A B Let R is the total resisting force acting on the cross section AB. R Then for equilibrium condition, R = P P P Then from the definition of stress, normal stress = σ = R/A = P/A Symbol: σ = Normal Stress

224 AXIAL LOADING NORMAL STRESS STRESS Direct or Normal Stress: Intensity of resisting force perpendicular to or normal to the section is called the normal stress. Normal stress may be tensile or compressive Tensile stress: stresses that cause pulling on the surface of the section, (particles of the materials tend to pull apart causing extension in the direction of force) Compressive stress: stresses that cause pushing on the surface of the section, (particles of the materials tend to push together causing shortening in the direction of force)

225 STRESS The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis. The force intensity on that section is defined as the normal stress. σ = lim A 0 F A σ ave = P A

226 STRAIN STRAIN : when a load acts on the material it will undergo deformation. Strain is a measure of deformation produced by the application of external forces. If a bar is subjected to a direct load, and hence a stress, the bar will changes in length. If the bar has an original length L and change in length by an amount δl, the linear strain produced is defined as, δl Change in length Linear strain, ε = = Original length L Strain is a dimensionless quantity.

227 Linear Strain P σ = = stress A δ ε = = normal strain L 2P σ = 2 A δ ε = L = P A P σ = A 2δ ε = = 2L δ L

228 STRESS-STRAIN DIAGRAM In order to compare the strength of various materials it is necessary to carry out some standard form of test to establish their relative properties. One such test is the standard tensile test in which a circular bar of uniform cross section is subjected to a gradually increasing tensile load until failure occurs. Measurement of change in length over a selected gauge length of the bar are recorded throughout the loading operation by means of extensometers. A graph of load verses extension or stress against strain is drawn as shown in figure.

229 STRESS-STRAIN DIAGRAM Proportionality limit Typical tensile test curve for mild steel

230 STRESS-STRAIN DIAGRAM Typical tensile test curve for mild steel showing upper yield point and lower yield point and also the elastic range and plastic range

231 Stress-strain Diagram Limit of Proportionality : From the origin O to a point called proportionality limit the stress strain diagram is a straight line. That is stress is proportional to strain. Hence proportional limit is the maximum stress up to which the stress strain relationship is a straight line and material behaves elastically. From this we deduce the well known relation, first postulated by Robert Hooke, that stress is proportional to strain. Beyond this point, the stress is no longer proportional to strain σ = P P A = Load at proportionality limit Original cross sectional area

232 Stress-strain Diagram Elastic limit: It is the stress beyond which the material will not return to its original shape when unloaded but will retain a permanent deformation called permanent set. For most practical purposes it can often be assumed that points corresponding proportional limit and elastic limit coincide. Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will be thus some permanent deformation when load is removed. σ = E P E A = Load at proportional limit Original cross sectional area

233 Yield point: Stress-strain Diagram It is the point at which there is an appreciable elongation or yielding of the material without any corresponding increase of load. σ = Y P Y A = Ultimate strength: Load at yield point Original cross sectional area It is the stress corresponding to maximum load recorded during the test. It is stress corresponding to maximum ordinate in the stress-strain graph. σ = U P U A = Maximum load taken by the material Original cross sectional area

234 Stress-strain Diagram Rupture strength (Nominal Breaking stress): It is the stress at failure. For most ductile material including structural steel breaking stress is somewhat lower than ultimate strength because the rupture strength is computed by dividing the rupture load (Breaking load) by the original cross sectional area. σ = B P B A = load at breaking (failure) Original cross sectional area rue breaking stress = load at breaking (failure) Actual cross sectional area

235 Stress-strain Diagram After yield point the graph becomes much more shallow and covers a much greater portion of the strain axis than the elastic range. The capacity of a material to allow these large plastic deformations is a measure of ductility of the material Ductile Materials: The capacity of a material to allow large extension i.e. the ability to be drawn out plastically is termed as its ductility. Material with high ductility are termed ductile material. Example: Low carbon steel, mild steel, gold, silver, aluminum

236 Stress-strain Diagram A measure of ductility is obtained by measurements of the percentage elongation or percentage reduction in area, defined as, increase in gauge length (up to fracture) = ercentage elongation original gauge length 100 Percentage reduction in area = Reduction in cross sectional area of necked portion (at fracture) original area 100 Cup and cone fracture for a Ductile Material

237 Brittle Materials : Stress-strain Diagram A brittle material is one which exhibits relatively small extensions before fracture so that plastic region of the tensile test graph is much reduced. Example: steel with higher carbon content, cast iron, concrete, brick Stress-strain diagram for a typical brittle material

238 HOOKE S LAW Hooke s Law For all practical purposes, up to certain limit the relationship between normal stress and linear strain may be said to be linear for all materials stress (σ) α strain (ε) stress (σ) strain (ε) = constant Thomas Young introduced a constant of proportionality that came to be known as Young s modulus. stress (σ) strain (ε) = E = Young s Modulus or Modulus of Elasticity

239 HOOKE S LAW Young s Modulus is defined as the ratio of normal stress to linear strain within the proportionality limit. E = stress (σ) strain (ε) = P A δ L = L PL AδL The value of the Young s modulus is a definite property of a material From the experiments, it is known that strain is always a very small quantity, hence E must be large. For Mild steel, E = 200GPa = MPa = N/mm 2

240 Deformations Under Axial Loading From Hooke s Law: σ σ = E ε ε = = E P AE From the definition of strain: δ ε = L Equating and solving for the deformation, PL δ = AE With variations in loading, crosssection or material properties, P = il δ i A E i i i

241 Consider an element of length, δx at a distance x from A W A d 1 x d 2 Diameter at x, d x B W 2 ( d d ) πd x c/s area at x, = 1 = ( d ) kx = 1 Change in length over a dx 2 length dx is ( d + kx) E 1 Change in length over a length L is d L = d 1 + k x PL Wdx = = AE π 4 = L Wdx 0 π 2 ( ) d1 + kx E 4 π 4

242 Consider an element of length, δx at a distance x from A ( ) + = L E kx d Wdx π () = L E t k dt W π Change in length over a length L is Put d 1 +kx = t, Then k dx = dt L L L kx d Ek W t Ek W t Ek W ) ( = = = + π π π E d d WL d Ed WL = = π π

243 Derive an expression for the total extension of the tapered bar AB of rectangular cross section and uniform thickness, as shown in the figure, when subjected to an axial tensile load,w. W W d d 2 1 A B b L b

244 W W d d 2 1 A B b b x dx Consider an element of length, δx at a distance x from A depth at x, ( d2 d ) 1 + x c/s area at x, ( d + kx)b = 1 d L d + k x = 1 Change in length over a length dx is = PL AE = dx 1 Wdx ( d + kx) b E = 1

245 Change in length over a = L Wdx b E 0 length L is ( d1 + kx) = b P E k ( log d d ) e 2 log e 1 = P b E ( ) ( log d ) 2 log d1 d d 2 L 1

246 Derive an expression for the total extension produced by self weight of a uniform bar, when the bar is suspended vertically. L Diameter d

247 x dx P 1 element dx P 1 = weight of the bar below the section, = volume specific weight = (π d 2 /4) x γ = A x γ Diameter d Extension of the element due to weight of the bar below that, = PL AE dx = P1 dx ( A x ρ ) dx = AE AE

248 Hence the total extension entire bar = 0 L ( A x γ ) dx AE 2 γx = 2E L 0 = 2 γl 2E The above expression can also be written as = 2 γl 2E A A = ( ALγ ) L 2AE = 1 2 PL AE Where, P = (AL) γ = total weight of the bar

249 SHEAR STRESS Consider a block or portion of a material shown in Fig.(a) subjected to a set of equal and opposite forces P. then there is a tendency for one layer of the material to slide over another to produce the form failure as shown in Fig.(b) P Fig. a P Fig. b P R Fig. c R P The resisting force developed by any plane ( or section) of the block will be parallel to the surface as shown in Fig.(c). The resisting forces acting parallel to the surface per unit area is called as shear stress.

250 Shear stress (τ) = Shear strain Shear resistance Area resisting shear This shear stress will always be tangential to the area on which it acts If block ABCD subjected to shearing stress as shown in Fig.(d), then it undergoes deformation. The shape will not remain rectangular, it changes into the form shown in Fig.(e), as AB ' C'D. τ B' τ C C' B C B = P A A τ Fig. d D A τ Fig. e D

251 B A φ B' τ τ C D C' Shear strain is defined as the change in angle between two line element which are originally right angles to one another. Fig. e BB shear strain = = tanφ φ AB The angle of deformation φ is then termed as shear strain The angle of deformation is measured in radians and hence is non-dimensional.

252 SHEAR MODULUS For materials within the proportionality limit the shear strain is proportional to the shear stress. Hence the ratio of shear stress to shear strain is a constant within the proportionality limit. Shear stress (τ) Shear strain (φ) = constant = G = Shear Modulus or Modulus of Rigidity The value of the modulus of rigidity is a definite property of a material For Mild steel, G= 80GPa = 80,000MPa = 80,000N/mm 2

253 example: Shearing Stress Forces P and P are applied transversely to the member AB. Corresponding internal forces act in the plane of section C and are called shearing forces. The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P. The corresponding average shear stress is, τ ave = P A The shear stress distribution cannot be assumed to be uniform.