Upcoming SlideShare
×

Chapter 2

20,717 views
20,257 views

Published on

1 Comment
11 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• its a wonderfull presentation

Are you sure you want to  Yes  No
Views
Total views
20,717
On SlideShare
0
From Embeds
0
Number of Embeds
7
Actions
Shares
0
546
1
Likes
11
Embeds 0
No embeds

No notes for slide

Chapter 2

1. 1. Chapter 2Fundamental of Statics 2-1
2. 2. Introduction• The objective for the current chapter is to investigate the effects of forces on particles: - replacing multiple forces acting on a particle with a single equivalent or resultant force, - relations between forces acting on a particle that is in a state of equilibrium. 2-2
3. 3. Resultant of Two Forces • force: action of one body on another; characterized by its point of application, magnitude, line of action, and sense. • Experimental evidence shows that the combined effect of two forces may be represented by a single resultant force. • The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs. • Force is a vector quantity. 2-3
4. 4. Vectors• Vector: parameters possessing magnitude and direction which add according to the parallelogram law. Examples: displacements, velocities, accelerations.• Scalar: parameters possessing magnitude but not direction. Examples: mass, volume, temperature• Vector classifications: - Free vectors may be freely moved in space without changing their effect on an analysis. - Sliding vectors may be applied anywhere along their line of action without affecting an analysis.• Equal vectors have the same magnitude and direction.• Negative vector of a given vector has the same magnitude and the opposite direction. 2-4
5. 5. External and Internal Forces• Forces acting on rigid bodies are divided into two groups: - External forces - Internal forces• External forces are shown in a free-body diagram.• If unopposed, each external force can impart a motion of translation or rotation, or both. 2-5
6. 6. Principle of Transmissibility: Equivalent Forces• Principle of Transmissibility - Conditions of equilibrium or motion are not affected by transmitting a force along its line of action. NOTE: F and F’ are equivalent forces.• Moving the point of application of the force F to the rear bumper does not affect the motion or the other forces acting on the truck. 2-6
7. 7. Classification of A Force System 2-7
8. 8. Classification of a Force System 2-8
9. 9. Resultant of Several Forces• When a number of coplanar forces are acting on a rigid body then these forces can be replaced by a single force which has the same effect on the rigid body as that of all the forces acting together, then this single force is known as the resultant of several forces. 2-9
10. 10. Resultant of Several Forces• Definition:• A single force which can replace a number of forces acting one a rigid body, without causing any change in the external effects on the body is known as the resultant force. 2 - 10
11. 11. Resultant of Coplanar Forces• The resultant of coplanar forces may be determined by following two method: 1. Analytical method 2. Graphical method 2 - 11
12. 12. Resultant of Collinear Coplanar Forces• Analytical Method• The resultant is obtained by adding all the forces if they are acting in the same direction.• If any one of the forces is acting in the opposite direction, then resulting is obtained by subtracting that forces. R= F1 + F2 - F3 2 - 12
13. 13. Resultant of Collinear Coplanar Forces• Graphical Method• Some suitable scale is chosen and vectors are drawn to the chosen scale.• These vectors are added/or subtracted to find the resultant. F1 F2 F3 a b c d R= F1 + F2 + F3 2 - 13
14. 14. Resultant of Concurrent Coplanar Forces• Concurrent coplanar forces are those forces which act in the same plane and they intersect or meet at a common point.• Following two cases are consider: i. When two forces act at a point. ii. When more than two forces act at a point. 2 - 14
15. 15. Resultant of Concurrent Coplanar Forces• When two forces act at a point.• (a) Analytical Method: – When two forces act at a point, their resultant is found by the law of parallelogram of forces. • The magnitude of Resultant force R• The direction of Resultant force R with the force P  Q sin α  θ = tan   P + Q cos α  −1    2 - 15
16. 16. Resultant of Concurrent Coplanar Forces• (b) Graphical Methodi. Choose a convenient scale to represent the forces P and Q.ii.From point O, draw a vector OA= P.iii.Now from point O, draw another vector OB= Q and at an angle of α as shown in fig.iv.Complete the parallelogram by drawing lines AC║ to OB and BC ║ to OA.v. Measure the length OC.vi.Resultant R = OC x Chosen Scale• The Direction of resultant is given by angle θ.• Measure the angle θ. 2 - 16
17. 17. Resultant of Concurrent Coplanar Forces• When more than two forces act at a point.• Analytical Method R = (∑ H ) 2 + (∑ V ) 2 tan θ = ∑V ∑H 2 - 17
18. 18. Resultant of Concurrent Coplanar Forces• (b) Graphical Method• The resultant of several forces acting at a point is found graphically with the help of Polygon law of forces.• Polygon law of forces – “if a number of coplanar forces are acting at a point such that they can be represented in magnitude and direction by the sides of a polygon taken in the same order, then their resultant is represented in magnitude and direction by the closing side of the polygon taken in the opposite order.” F2 c F1 F3 d F2 o F4 b F1F3 e F4 R a 2 - 18
19. 19. Resultant of Concurrent Coplanar Forces Vector Addition: The Order Does NOT Matter 2 - 19
20. 20. Coplanar Parallel Forces• A parallel coplanar force system consists of two or more forces whose lines of action are parallel to each other.• Two parallel forces will not intersect at a point.• The line of action of forces are parallel so that for finding the resultant of two parallel forces, the parallelogram cannot be drawn.• The resultant of such forces can be determined by applying the principle of moments. 2 - 20
21. 21. Coplanar Parallel Forces• Moment of Forces• The tendency of a force to produce rotation of body about some axis or point is called the moment of a force. • moment of a force about a point • moment of a force about an axis • moment due to a couple The moment (m) of the force F about O is given by, M=F x d Unit: force x distance =F*L = N-m, kN-m (SI unit) 2 - 21
22. 22. 2 - 22
23. 23. 2 - 23
24. 24. Moment of Forces• The tendency of moment is to rotate about the body in the clockwise direction about O is called clockwise moment.• If the tendency of a moment is to rotate the body in anti clockwise direction, then that moment is known as anti clock wise moment.• Sign conv. : clockwise (-ve), counterclockwise (+ve) 2 - 24
25. 25. Resultant Moment of Forces• The resultant moment of F1, F2, and F3 about O = -F1 x r1 – F2 x r2 + F3 x r3 2 - 25
26. 26. Resultant Moment of Forces 20N 30N D C• For Example:• Find the resultant moment about point A.• Soln:• Forces at point A and B passes through Point 2m A.• So perpendicular distances from A on the line of action of these forces will be zero.• Hence their moments about point A will be zero. A 2m B 10N• Moment of force at C about point A: 20 x 2 =40N(CCW) 40N• Moment of force at D about point A : 30 x 2= 60N(CCW)So resultant moment at point A = 40 + 60 = 100N(CCW) 2 - 26
27. 27. Principle of Moments• The Principle of Moments, also known as Varignons Theorem, states that the moment of any force about any point is equal to the algebraic sum of the moments of the its components of that force about that point. • As with the summation of force combining to get resultant force u uu uu r r r uu r R = F1 + F2 + K + Fn • Similar resultant comes from the addition of moments uuu r u r uu r uu r uu r M 0 = R d R = F1 d1 + F2 d 2 + K + Fn d n 2 - 27
28. 28. Principle of Moments 2 - 28
29. 29. Types of Parallel Forces• Two important types of parallel forces 1. Like parallel forces 2. Unlike parallel forces• Like Parallel forces• Two parallel forces which are acting in the same direction are known as like parallel forces.• The magnitude of a forces may be equal or unequal.• Unlike Parallel forces• Two parallel forces which are acting in the opposite direction are known as like unparallel forces.• The magnitude of a forces may be equal or unequal. 2 - 29
30. 30. Resultant of Two Parallel forces• The resultant of following two parallel forces will be considered: – Two parallel forces are like. – Two parallel forces are unlike and are unequal in magnitude. – Two parallel forces are unlike but equal in magnitude. 2 - 30
31. 31. Resultant of Two Parallel forces• Two parallel forces are like• Suppose that two like but unequal parallel forces act on a body at position A and B as shown in figure.• We have to calculate the resultant force acting on the body and its position.• From condition of static equilibrium; R = F1 + F2 ...(1) 2 - 31
32. 32. Resultant of Two Parallel forces• The position of R can be obtained by using Varignon’s theorem. To use the theorem consider a point O along the line AB, such that• Algebraic sum of moments of F1 and F2 about O = Moment of resultant about O• Now, Moment of F1 about O = F1 × AO (clockwise) Moment of F2 about O = F2 × BO (anti-clockwise) Moment of R about O = R × CO (anti-clockwise) – F1 × AO + F2 × BO = + R × CO …(2) – F1 × AO + F2 × BO = (F1 + F2) × CO F1 (AO + CO) = F2 (BO – CO) F1 × AC = F2 × BC F1 / F2 = BC / AC• Therefore, it can be observed that R acts at a point C which divides the length AB in the ratio inversely proportional to the magnitudes of F1 and F2. 2 - 32
33. 33. Resultant of Two Parallel forces• Two parallel forces are unlike and are unequal in magnitude • Suppose that two unlike and unequal parallel forces act on a body at position A and B as shown in figure. • We have to calculate the resultant force acting on the body and its position. • From condition of static equilibrium, R = F1 – F2 …(1) 2 - 33
34. 34. Resultant of Two Parallel forces• Once again, the position of R can be obtained by using Varigonon’s theorem. Consider a point O along the line AB, such that• Algebraic sum of moments of F1 and F2 about O = Moment of resultant about O• Now, Moment of F1 about O = F1 × AO (clockwise)• Moment F2 about O = F2 × BO (anti-clockwise)• Moment of R about O = R × CO (anti-clockwise) ⇒ – F1 × AO – F2 × BO = – R × CO …(2) ⇒ F1 × AO + F2 × BO = R × CO ⇒ F1 × AO + F2 × BO = (F1 – F2) × CO ⇒ F2 (BO + CO) = F1 (CO – AO) ⇒ F1 / F2 = BC / AC …(3)• Since F1 > F2, BC will be greater than AC. Hence point C will lie outside AB on the same side of F1. Thus, it can be observed that R acts at C which externally divide length AB in the ratio inversely proportional to the magnitude of F1 and F2 . 2 - 34
35. 35. Moment of a Couple• Two parallel forces are unlike but equal in magnitude• Two parallel forces having different lines of action, equal in magnitude, but opposite in sense constitute a couple.• A couple causes rotation about an axis perpendicular to its plane.• The perpendicular distance between the parallel forces is known as arm of the couple. M=F*aUnit: Nm 2 - 35
36. 36. Moment of a Couple• Two couples are equivalent if they cause the same moment: 2 - 36
37. 37. Resolution of a Force into a Force and a Couple• A force, F, acting at point B can be replaced by the force, F, and a moment, MA, acting at point A. F F F F B A B A B A = d F = MA MA = d F 2 - 37
38. 38. Replace a Force-Couple System with Just Forces F F A A F2 d2 MA = F2 C C d2 F2 = MA 2 - 38
39. 39. Reducing a System of Forces to a Resultant Force- Couple System (at a Chosen Point) F1 R r1 r2 F2 A r3 = MA r r F3 R = ∑F r r r MA = ∑ r ×F ( ) 2 - 39
40. 40. Reducing a System of Forces to a Resultant Force-Couple System (at a Chosen Point) 2 - 40
41. 41. Reduce a System of Forces to a Single Resultant Force F1 R R Rr1 r2 F2 BA r3 = MA = MA F3 R R Using method B from prior slide = 2 - 41
42. 42. Reduce a System of Forces to a Single Resultant Force R R Ry B B A A Rx RxMA Ry R R dx dx R = –MA 2 - 42
43. 43. General Case of parallel forces in a plane• R1= Resultant of (F1, F2, F4) and R2= Resultant of (F3, F5)• The resultant R1 and R2 are acting in opposite direction and parallel to each other.• Two important case are possible.• 1. R1 may not be equal to R2. – Then two unequal parallel forces acting in opposite direction. – The resultant R= R1-R2 – The point of application easily found with the help of Varignons Theorem or moments of forces. 2 - 43
44. 44. General Case of parallel forces in a plane• 2. R1 is equal to R2. – Then two equal parallel forces acting in opposite direction. – The resultant R= R1-R2=0 – Now the system may be reduce to a couple or a system is in equilibrium. – The algebraic sum moment of all forces(F1, F2,…, F5) taken about any point. – If ∑ M = 0 then system is in equilibrium . – If ∑ M ≠ 0 the system reduce to a resultant couple. – And the calculated moment gives the moment of that couple. 2 - 44
45. 45. Equivalent System• Two force systems that produce the same external effects on a rigid body are said to be equivalent.• An equivalent system for a given system of coplanar forces, is a combination of a force passing through a given point and a moment about that point.• The force is the resultant of all forces acting on the body.• The moment is the sum of all the moments about that point.• Equivalent system consists of :• (1) a single force R passing through the given point P• (2) a single moment MR 2 - 45
46. 46. Equivalent System• For Examples:• Determined the equivalent system through point O.• These means find:• (1) a single resultant force, R• (2) a single moment through, O 2 - 46
47. 47. Difference between moment and couple Moment Couple• Moment = force x perpendicular • Two equal and opposite forces whose lines of distance M = Fd action are different from a couple• It is produced by a single force not • It is produced by the two equal and opposite passing through Centre of gravity of parallel, non collinear forces. the body.• The force move the body in the • Resultant force of couple is zero. Hence, direction of force and rotate the body. body does not move, but rotate only. It is the resultant force.• To balance the force causing moment, • Couple cannot be balanced by a single force, equal and opposite force is required. it can be balanced by a couple only.• For example, • For example,• To tight the nut by spanner • To rotate the key in lock• To open or close the door • To open or close the wheel valve of water line • To rotate the steering wheel of car. 2 - 47
48. 48. Equilibrium of Rigid Bodies External forces Body start moving or rotating.• If the body does not start moving and also does not start rotating about any point, then body is said to be in equilibrium.• For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body. 2 - 48
49. 49. Equilibrium of Rigid Bodies• Principle of Equilibrium:• ∑F = 0 …………(1)• ∑M = 0 ………....(2)• Eq. (1) is known as the force law of equilibrium and Eq. (2) is known as the moment law of equilibrium.• The forces are generally resolved into horizontal and vertical components. ∑ Fx = 0 ∑ Fy = 0 2 - 49
50. 50. Equilibrium of Rigid Bodies• Equilibrium of non-concurrent forces system:• A non-concurrent forces system will be in equilibrium if the resultant of all forces and moment is zero. ∑ Fx = 0 ∑ Fy = 0 ∑ M = 0• Equilibrium of concurrent forces system:• For the concurrent forces, the line of actions of all forces meet at a point, and hence the moment of those forces about that point will be zero automatically. ∑ Fx = 0 ∑ Fy = 0 2 - 50
51. 51. Equilibrium of Rigid Bodies• Force Law of Equilibrium:• There are three main force Law of Equilibrium: • Two force system • Three force system • Four or more force system 2 - 51
52. 52. Equilibrium of Rigid Bodies(1) Two force system:• According to this principle, if a body is in equilibrium under the action of two forces, then they must be equal, opposite and collinear.• If the two forces acting on a body are equal and opposite but are parallel, as shown in fig., then the body will not be in equilibrium.• Two condition is satisfied:• (1) ∑ Fx = 0 (2) ∑ Fy = 0 as F1 = F2• Third condition is not satisfied:• (3) ∑ M ≠0 MA = -F2 X AB• A body will not be in equilibrium under the action of two equal and opposite parallel forces.• Two equal and opposite parallel forces produce a couple. 2 - 52
53. 53. Equilibrium of Rigid Bodies• (2) Three force system:• According to this principle, if a body is in equilibrium under the action of three forces then the resultant of any two forces must be equal, opposite and collinear with the third force.• Three forces acting on a body either concurrent or parallel• Case (a) When three forces are concurrent• The resultant of F1 and F2 is given by R.• If the force F3 is collinear equal, opposite to the resultant R, then the body will be in equilibrium.• The force F3 which is equal and opposite to resultant R is known as equilibrant.• Hence for three concurrent forces acting on a body when the body is in equilibrium, the resultant of the two forces should be equal and opposite to the third force. 2 - 53
54. 54. Equilibrium of Rigid Bodies• Case (2): When three forces are parallel• If the three parallel forces F1, F2, and F3 are acting in the same direction, then there will be a resultant R= F1 + F2 + F3 and body will not be in equilibrium.• If the three forces are acting in opposite direction and their magnitude is so adjusted that there is no resultant forces and body is in equilibrium.• Apply the three condition of equilibrium:• (1) Σ Fx = 0,(No horizontal forces) (2) Σ Fy = 0, (F1+ F3=F2)• (3) Σ M = 0 about any point. Σ MA= -F2 X AB + F3 X AC• For equilibrium Σ MA should be zero. -F2 X AB + F3 X AC= 0• If the distance AB and AC are such that the above equation is satisfied, then the body will be in equilibrium under the action of three parallel forces. 2 - 54
55. 55. Equilibrium of Rigid Bodies• (3) Four or more force system:• According to this principle, if a body is in equilibrium under the action of four forces then the resultant of any two forces must be equal, opposite and collinear with the resultant of the other two forces. Σ Fx = 0, Σ Fy = 0, Σ M = 0 2 - 55
56. 56. Equilibrium of Rigid Bodies• Two moment equations. • Σ Fy = 0 • Σ MA = 0, • Σ MB = 0• where A and B are any two points in the xy-plane, provided that the line AB is not parallel to the y-axis. 2 - 56
57. 57. Equilibrium of Rigid Bodies• Free Body Diagram of a Body:• The first step in equilibrium analysis is to identify all the forces that act on the body. This is accomplished by means of a free-body diagram.• The free-body diagram (FBD) of a body is a sketch of the body showing all forces that act on it. The term free implies that all supports have been removed and replaced by the forces (reactions) that they exert on the body.• Free-body diagrams are fundamental to all engineering disciplines that are concerned with the effects that forces have on bodies.• The construction of an FBD is the key step that translates a physical problem into a form that can be analyzed mathematically. 2 - 57
58. 58. Equilibrium of Rigid Bodies• Forces that act on a body can be divided into two general categories— Reactive forces (or, simply, reactions) and Applied forces (action)• Reactions are those forces that are exerted on a body by the supports to which it is attached.• Forces acting on a body that are not provided by the supports are called applied forces. 2 - 58
59. 59. Equilibrium of Rigid Bodies• The following is the general procedure for constructing a free-body diagram.1. A sketch of the body is drawn assuming that all supports (surfaces of contact, supporting cables, etc.) have been removed.2. All applied forces are drawn and labeled on the sketch. The weight of the body is considered to be an applied force acting at the center of gravity.3. The support reactions are drawn and labeled on the sketch. If the sense of a reaction is unknown, it should be assumed. The solution will determine the correct sense: A positive result indicates that the assumed sense is correct, whereas a negative result means that the correct sense is opposite to the assumed sense.4. All relevant angles and dimensions are shown on the sketch. 2 - 59
60. 60. Equilibrium of Rigid Bodies• The most difficult step to master in the construction of FBDs is the determination of the support reactions.• Flexible Cable (Negligible Weight).• A flexible cable exerts a pull, or tensile force, in the direction of the cable. With the weight of the cable neglected, the cable forms a straight line. If its direction is known, removal of the cable introduces one unknown in a free- body diagram—the magnitude of the force exerted by the cable. Support Reaction(s) Description Number of of reaction(s) unknowns Tension of unknown one magnitude T in the direction of the cable 2 - 60
61. 61. Equilibrium of Rigid Bodies• Frictionless Surface: Single Point of Contact.• When a body is in contact with a frictionless surface at only one point, the reaction is a force that is perpendicular to the surface, acting at the point of contact.• This reaction is often referred to simply as the normal force.• Therefore, removing such a surface introduces one unknown in a free-body diagram—the magnitude of the normal force. Support Reaction(s) Description Number of of reaction(s) unknowns Force of unknown one magnitude N directed normal to the surface 2 - 61
62. 62. Equilibrium of Rigid Bodies• Roller Support.• A roller support is equivalent to a frictionless surface: It can only exert a force that is perpendicular to the supporting surface.• The magnitude of the force is thus the only unknown introduced in a free- body diagram when the support is removed. Support Reaction(s) Description Number of of reaction(s) unknowns Force of unknown one magnitude N normal to the surface supporting the roller 2 - 62
63. 63. Equilibrium of Rigid Bodies• Surface with Friction: Single Point of Contact.• A friction surface can exert a force that acts at an angle to the surface.• The unknowns may be taken to be the magnitude and direction of the force.• However, it is usually advantageous to represent the unknowns as N and F, the components that are perpendicular and parallel to the surface, respectively.• The component N is called the normal force, and F is known as the friction force. Support Reaction(s) Description Number of of reaction(s) unknowns Force of unknown magnitude N normal Two to the surface and a friction force of unknown magnitude F parallel to the surface 2 - 63
64. 64. Equilibrium of Rigid Bodies• Pin Support.• Neglecting friction, the pin can only exert a force that is normal to the contact surface, shown as R in Fig.(b).• A pin support thus introduces two unknowns: the magnitude of R and the angle α that specifies the direction of R (α is unknown because the point where the pin contacts the surface of the hole is not known). 2 - 64
65. 65. Equilibrium of Rigid Bodies• Built-in (Cantilever) Support.• A built-in support, also known as a cantilever support, prevents all motion of the body at the support. Translation (horizontal or vertical movement) is prevented by a force, and a couple prohibits rotation.• Therefore, a built-in support introduces three unknowns in a free-body diagram: – The magnitude and direction of the reactive force R (these unknowns are commonly chosen to be two components of R, such as Rx and Ry ) – The magnitude C of the reactive couple. Support Reaction(s) Description Number of of reaction(s) unknowns Unknown force R and a couple of Three unknown magnitude C 2 - 65
66. 66. Equilibrium of Rigid Bodies• You should keep the following points in mind when you are drawing free-body diagrams.1. Be neat. Because the equilibrium equations will be derived directly from the free-body diagram, it is essential that the diagram be readable.2. Clearly label all forces, angles, and distances with values (if known) or symbols (if the values are not known).3. Show only forces that are external to the body (this includes support reactions and the weight). Internal forces occur in equal and opposite pairs and thus will not appear on free-body diagrams. 2 - 66
67. 67. Equilibrium of Rigid Bodies• Sample Problem: The mass of the bar is 50 kg. Take g = 9.81 m/s2 2 - 67
68. 68. Equilibrium of Rigid Bodies• Sample Problem:• Neglecting the weights of the members. 2 - 68
69. 69. Equilibrium of Rigid Bodies 2 - 69
70. 70. Equilibrium of Rigid Bodies 2 - 70
71. 71. Equilibrium of Rigid Bodies 2 - 71
72. 72. Equilibrium of Rigid Bodies• Equilibrium analysis of a body• The three steps in the equilibrium analysis of a body are:• Step 1: Draw a free-body diagram (FBD) of the body that shows all of the forces and couples that act on the body.• Step 2: Write the equilibrium equations in terms of the forces and couples that appear on the free-body diagram.• Step 3: Solve the equilibrium equations for the unknowns. 2 - 72
73. 73. Equilibrium of Rigid Bodies• Statically determinate and Statically indeterminate• The force system that holds a body in equilibrium is said to be statically determinate if the number of independent equilibrium equations equals the number of unknowns that appear on its free-body diagram• If the number of unknowns exceeds the number of independent equilibrium equations, the problem is called statically indeterminate.• The solution of statically indeterminate problems requires the use of additional principles.• When the support forces are sufficient to resist translation in both the x and y directions as well as rotational tendencies about any point, the rigid body is said to be completely constrained, otherwise the rigid body is unstable or partially constrained. 2 - 73
74. 74. Equilibrium of Rigid Bodies• Statically indeterminate and Improper Constraints 2 - 74
75. 75. Equilibrium of Rigid Bodies 2 - 75
76. 76. Equilibrium of Rigid Bodies 2 - 76
77. 77. Equilibrium of Rigid Bodies• Statistical determinate and proper Constraints 2 - 77
78. 78. Any question? 2 - 78
79. 79. Thank you 2 - 79