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Notes On Basic Engineering Statics

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### Me13achapter1and2 090428030757-phpapp02

1. 1. STATICSCOURSEINTRODUCTION
2. 2. Details of Lecturer Course Lecturer: Dr. E.I. Ekwue Room Number: 216 Main Block, Faculty of Engineering Email: ekwue@eng.uwi.tt , Tel. No. : 662 2002 Extension 3171 Office Hours: 9 a.m. to 12 Noon. (Tue, Wed and Friday)
3. 3. COURSE GOALS This course has two specific goals: (i) To introduce students to basic concepts of force, couples and moments in two and three dimensions. (ii) To develop analytical skills relevant to the areas mentioned in (i) above.
4. 4. COURSE OBJECTIVES Upon successful completion of this course, students should be able to:i) Determine the resultant of coplanar and space force systems. (ii) Determine the centroid and center of mass of plane areas and volumes. (iii) Distinguish between concurrent, coplanar and space force systems (iv) Draw free body diagrams.
5. 5. COURSE OBJECTIVES CONTD. (v) Analyze the reactions and pin forces induces in coplanar and space systems using equilibrium equations and free body diagrams. (vi) Determine friction forces and their influence upon the equilibrium of a system. (vii) Apply sound analytical techniques and logical procedures in the solution of engineering problems.
6. 6. Course Content (i) Introduction, Forces in a plane, Forces in space (ii) Statics of Rigid bodies (iii) Equilibrium of Rigid bodies (2 and 3 dimensions) (iv) Centroids and Centres of gravity (v) Moments of inertia of areas and masses (vi) Analysis of structures (Trusses, Frames and Machines) (vii) Forces in Beams (viii)Friction
7. 7. Teaching Strategies The course will be taught via Lectures and Tutorial Sessions, the tutorial being designed to complement and enhance both the lectures and the students appreciation of the subject. Course work assignments will be reviewed with the students.
8. 8. Course Textbook and Lecture Times Vector Mechanics For Engineers By F.P. Beer and E.R. Johnston (Third Metric Edition), McGraw-Hill. Lectures: Wednesday, 1.00 to 1.50 p.m. Thursday , 10.10 to 11.00 a.m. Tutorials: Monday, 1.00 to 4.00 p.m. [Once in Two Weeks]Attendance at Lectures and Tutorials is Compulsory
9. 9. Tutorial OutlineChapter 2 – STATICS OF PARTICLES2.39*, 41, 42*, 55, 85*, 86, 93*, 95, 99*, 104, 107*, 113Chapter 3 – RIGID BODIES: EQUIVALENT SYSTEM OF FORCES3.1*, 4, 7*, 21, 24*, 38, 37*, 47, 48*, 49, 70*, 71, 94*, 96, 148*, 155Chapter 4 – EQUILIBRIUM OF RIGID BODIES4.4*, 5, 9*, 12, 15*, 20, 21*, 31, 61*, 65, 67*, 93, 115*Chapters 5 and 9 – CENTROIDS AND CENTRES OF GRAVITY, MOMENTS OFINERTIA5.1*, 5, 7*, 21, 41*, 42, 43*, 45, 75*, 77 9.1*, 2, 10*, 13, 31*, 43, 44*Chapter 6 – ANALYSIS OF STRUCTURES6.1*, 2, 6*, 9, 43*, 45, 75*, 87, 88*, 95, 122*, 152, 166*, 169Chapters 7 and 8 – FORCES IN BEAMS AND FRICTION7.30 , 35, 36, 81, 85 8.25, 21, 65* For Chapters 1 to 6 and 9, two groups will do the problems in asterisks; the other twogroups will do the other ones. All the groups will solve all the questions in Chapters 7and 8.
10. 10. Time-Table For Tutorials/Labs MONDAY 1:00 - 4:00 P.M. Week 1,5,9 2,6,10 3,7,11, 4,8,12 Group - ME13A ME16A ME13A K (3,7) L ME13A - ME13A ME16A (4,8) M ME16A ME13A - ME13A (5,9) N ME13A ME16A ME13A - (6,10)
11. 11. Course Assessment (i) One (1) mid-semester test, 1-hour duration counting for 20% of the total course. (ii) One (1) End-of-semester examination, 2 hours duration counting for 80% of the total course marks.
12. 12. ME13A: ENGINEERINGSTATICS CHAPTER ONE: INTRODUCTION
13. 13. 1.1 MECHANICS Body of Knowledge which Deals with the Study and Prediction of the State of Rest or Motion of Particles and Bodies under the action of Forces
14. 14. PARTS OF MECHANICS
15. 15. 1.2 STATICS Statics Deals With the Equilibrium of Bodies, That Is Those That Are Either at Rest or Move With a Constant Velocity. Dynamics Is Concerned With the Accelerated Motion of Bodies and Will Be Dealt in the Next Semester.
16. 16. ME13A: ENGINEERINGSTATICS CHAPTER TWO: STATICS OF PARTICLES
17. 17. 2.1 PARTICLE A particle has a mass but a size that can be neglected. When a body is idealised as a particle, the principles of mechanics reduce to a simplified form, since the geometry of the body will not be concerned in the analysis of the problem.
18. 18. PARTICLE CONTINUED All the forces acting on a body will be assumed to be applied at the same point, that is the forces are assumed concurrent.
19. 19. 2.2 FORCE ON A PARTICLE A Force is a Vector quantity and must have Magnitude, Direction and Point of action. F α P
20. 20. Force on a Particle Contd. Note: Point P is the point of action of force and α and are directions. To notify that F is a vector, it is printed in bold as in the text book. Its magnitude is denoted as |F| or simply F.
21. 21. Force on a Particle Contd. There can be many forces acting on a particle. The resultant of a system of forces on a particle is the single force which has the same effect as the system of forces. The resultant of two forces can be found using the paralleolegram law.
22. 22. 2.2.VECTOR OPERATIONS 2.3.1 EQUAL VECTORS Two vectors are equal if they are equal in magnitude and act in the same direction. pP Q
23. 23. Equal Vectors Contd. Forces equal in Magnitude can act in opposite Directions R S
24. 24. 2.3.2 Vector AdditionUsing the Paralleologram Law, Construct aParm. with two Forces as Parts. Theresultant of the forces is the diagonal. P R Q
25. 25. Vector Addition Contd. Triangle Rule: Draw the first Vector. Join the tail of the Second to the head of the First and then join the head of the third to the tail of the first force to get the resultant force, R R=Q+P P Q
26. 26. Triangle Rule Contd.  Also: Q P R=P+QQ + P = P + Q. This is the cummutative law ofvector addition
27. 27. Polygon Rule Can be used for the addition of more than two vectors. Two vectors are actually summed and added to the third.
28. 28. Polygon Rule contd. S Q P S Q R (P + Q) P R=P+Q+S
29. 29. Polygon Rule Contd. P + Q = (P + Q) ………. Triangle Rule i.e. P + Q + S = (P + Q) + S = R The method of drawing the vectors is immaterial . The following method can be used.
30. 30. Polygon Rule contd. S Q P S Q R (Q + S) P R=P+Q+S
31. 31. Polygon Rule Concluded Q + S = (Q + S) ……. Triangle Rule P + Q + S = P + (Q + S) = R i.e. P + Q + S = (P + Q) + S = P + (Q + S) This is the associative Law of Vector Addition
32. 32. 2.3.3. Vector Subtraction P  P - Q = P + (- Q) Q P P P -Q Q-Q P-QParm. Rule Triangle Rule
33. 33. 2.4 Resolution of Forces It has been shown that the resultant of forces acting at the same point (concurrent forces) can be found. In the same way, a given force, F can be resolved into components. There are two major cases.
34. 34. Resolution of Forces: Case 1 (a)When one of the two components, P is known: The second component Q is obtained using the triangle rule. Join the tip of P to the tip of F. The magnitude and direction of Q are determined graphically or by trignometry. P Q i.e. F = P + Q F
35. 35. Resolution of Forces: Case 2(b) When the line of action of each component is known: The force, F can beresolved into two components having lines of action along lines ‘a’ and ‘b’ using theparalleogram law. From the head of F, extend a line parallel to ‘a’ until it intersects ‘b’.Likewise, a line parallel to ‘b’ is drawn from the head of F to the point of intersection with‘a’. The two components P and Q are then drawn such that they extend from the tail ofF to points of intersection. a Q F P b
36. 36. Example Determine graphically, the magnitude and direction of the resultant of the two forces using (a) Paralleolegram law and (b) the triangle rule. 600 N 900 N 45o o 30
37. 37. SolutionSolution: A parm. with sides equal to 900 N and 600 N is drawn to scale as shown.The magnitude and direction of the resultant can be found by drawing to scale. 600N 900N 45o 600 N R 30o 15o 900 N 45o 30o The triangle rule may also be used. Join the forces in a tip to tail fashion andmeasure the magnitude and direction of the resultant. 600 N R 45o 135o C 900 N B 30o
38. 38. Trignometric Solution Using the cosine law:R2 = 9002 + 6002 - 2 x 900 x 600 cos 1350 600N R 135oR = 1390.6 = 1391 N 30o 900 NUsing the sine law: B R 600 600 sin 135ο = i. e. B = sin −1sin 135ο sin B 1391=17.8οThe angle of the resul tan t =30 +17.8 = 47.8οie. R = 139N 47.8o
39. 39. Example Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 30 kN and Q = 20 kN, determine the magnitude and direction of the resultant force exerted on the bracket. P 25o 50o
40. 40. SolutionSolution: Using Triangle rule: 75o 30 kN 20 kN 105o θ 25o Q RR2 = 302 + 202 - 2 x 30 x 20 cos 1050 - cosine lawR = 40.13 NUsing sine rule:4013 N . 20 − 20 sin 105o = and Sin 1 = =28.8 oSin 105o Sin θ 4013 .Angle R =28.8 o −25o =38 o .i. e R =401 N , . 38 o .
41. 41. 2.5 RECTANGULARCOMPONENTS OF FORCE y Fy = Fy j F j i x Fx = Fx i
42. 42. RECTANGULAR COMPONENTSOF FORCE CONTD. In many problems, it is desirable to resolve force F into two perpendicular components in the x and y directions. Fx and Fy are called rectangular vector components. In two-dimensions, the cartesian unit vectors i and j are used to designate the directions of x and y axes. Fx = Fx i and Fy = Fy j i.e. F = Fx i + Fy j Fx and Fy are scalar components of F
43. 43. RECTANGULAR COMPONENTS OF FORCE CONTD.While the scalars, Fx and Fy may be positive or negative, depending on the sense of Fxand Fy, their absolute values are respectively equal to the magnitudes of the componentforces Fx and Fy,Scalar components of F have magnitudes: Fx = F cos θ and Fy = F sin θF is the magnitude of force F.
44. 44. Example Determine the resultant of the three forces below. y 600 N 800 N 350 N 45o 60o 25o x
45. 45. Solution∑ x = 350 cos 25 o + 800 cos 70 o - 600 cos 60o F = 317.2 + 273.6 - 300 = 290.8 N∑ y = 350 sin 25 o + 800 sin 70o + 600 sin 60o F = 147.9 + 751 + 519.6 = 1419.3 Ni.e. F = 290.8 N i + 1419.3 N j y 800 NResultant, F 600 N 350 N F = 290.82 + .32 = 1419 1449 N 45o 1419.3 60 o 25o θ tan − = 1 = .4 0 78 290.8F = 1449 N 78.4 o
46. 46. Example A hoist trolley is subjected to the three forces shown. Knowing that = 40o , α determine (a) the magnitude of force, P for which the resultant of the three forces is vertical (b) the corresponding magnitude of the resultant. P α α 2000 N 1000 N
47. 47. Solution (a) The resultant being vertical means that thehorizontal component is zero.∑ F x = 1000 sin 40o + P - 2000 cos 40o = 0P = 2000 cos 40o - 1000 sin 40o =1532.1 - 642.8 = 889.3 = 889 kN (b) ∑ Fy = - 2000 sin 40o - 1000 cos 40o = - 1285.6 - 766 = - 2052 N = 2052 N 40o P 40o 2000 N 1000 N
48. 48. 2.6. EQUILIBRIUM OF A PARTICLEA particle is said to be at equilibrium when the resultant of all the forces acting on it iszero. It two forces are involved on a body in equilibrium, then the forces are equal andopposite... 150 N 150 NIf there are three forces, when resolving, the triangle of forces will close, if they are inequilibrium. F2 F1 F2 F3 F1 F3
49. 49. EQUILIBRIUM OF A PARTICLE CONTD.If there are more than three forces, the polygon of forces will be closed if the particle isin equilibrium. F3 F2 F2F3 F1 F4 F1 F4The closed polygon provides a graphical expression of the equilibrium of forces.Mathematically: For equilibrium: R = ∑F = 0i.e. ∑ ( Fx i + Fy j) = 0 or ∑ (Fx) i + ∑ (Fy) j
50. 50. EQUILIBRIUM OF A PARTICLECONCLUDED For equilibrium: ∑ Fx = 0 and ∑ F y = 0. Note: Considering Newton’s first law of motion, equilibrium can mean that the particle is either at rest or moving in a straight line at constant speed.
51. 51. FREE BODY DIAGRAMS: Space diagram represents the sketch of the physical problem. The free body diagram selects the significant particle or points and draws the force system on that particle or point. Steps: 1. Imagine the particle to be isolated or cut free from its surroundings. Draw or sketch its outlined shape.
52. 52. Free Body Diagrams Contd. 2. Indicate on this sketch all the forces that act on the particle. These include active forces - tend to set the particle in motion e.g. from cables and weights and reactive forces caused by constraints or supports that prevent motion.
53. 53. Free Body Diagrams Contd. 3. Label known forces with their magnitudes and directions. use letters to represent magnitudes and directions of unknown forces. Assume direction of force which may be corrected later.
54. 54. Example The crate below has a weight of 50 kg. Draw a free body diagram of the crate, the cord BD and the ring at B. A B ring C 45o D CRATE
55. 55. Solution(a) Crate FD ( force of cord acting on crate) A 50 kg (wt. of crate) B C 45o(b) Cord BD FB (force of ring acting on cord) D CRATE FD (force of crate acting on cord)
56. 56. Solution Contd.(c) Ring FA (Force of cord BA acting along ring) FC (force of cord BC acting on ring) FB (force of cord BD acting on ring)
57. 57. Example
58. 58. Solution Contd. FAC sin 75oFBC = o = 3.73FAC .............(1) cos 75∑ Fy = 0 i.e. FBC sin 75o - FAC cos 75o - 1962 = 0 1962 + 0.26 FAC FBC = = 20312 + 0.27 FAC ......(2) . 0.966From Equations (1) and (2), 3.73 FAC = 2031.2 + 0.27 FAC FAC = 587 NFrom (1), FBC = 3.73 x 587 = 2190 N
59. 59. RECTANGULAR COMPONENTSOF FORCE (REVISITED) y F = Fx + Fy F = |Fx| . i + |Fy| . j Fy = Fy j |F|2 = |Fx|2 + |Fy|2 j | F| = | Fx|2 + | Fy |2 F i x Fx = Fx i
60. 60. 2.8 Forces in Space Rectangular Components j Fy F λ Fx i Fz k
61. 61. Rectangular Components of a Force in Space F = Fx + Fy + Fz F = |Fx| . i + |Fy| . j + |Fz| . k |F|2 = |Fx|2 + |Fy|2 + |Fz|2| F| = | Fx|2 + | Fy|2 + | Fz|2| Fx| = | F | cosθ x | Fy| = | F | cosθ y | Fz| = | F |cosθ zCosθ x , Cosθ y and Cosθ z are called direction cos ines of angles θ x , θ y and θ z
62. 62. Forces in Space Contd.i.e. F = F ( cos θx i + cos θy j + cos θz k) = F λF can therefore be expressed as the product of scalar, Fand the unit vector λ where: λ = cos θx i + cos θy j + cos θz k. λ is a unit vector of magnitude 1 and of the same direction as F. λ is a unit vector along the line of action of F.
63. 63. Forces in Space Contd.Also:λx = cos θx, λy = cos θy and λz = cos θz - Scalar vectorsi.e. magnitudes.λx2 + λy2 + λz2 = 1 = λ2i.e. cos2 θx, + cos2 θy + cos2 θz = 1Note: If components, Fx, Fy, and Fz of a Force, F are known,the magnitude of F, F = Fx2 + Fy2 + Fz2Direction cosines are: cos θx = Fx/F , cos θy = Fy/F and cos2 θz = Fz/F
64. 64. Force Defined by Magnitude and two Points on its Line of Action Contd.Unit vector, λalong the line of action of F = MN/MNMN is the distance, d from M to N. λ= MN/MN = 1/d ( dx i + dy j + dz k )Recall that: F = F λF = F λ= F/d ( dx i + dy j + dz k ) Fd Fd y FdFx = x , Fy = , Fz = z d d dd x = 2 −1 , d y = 2 − 1 , d z =2 −1 x x y y z zd = dx2 + y2 + z2 d d d dy dcosθ = x , cosθ = , cosθ = z x y z
65. 65. 2.8.3 Addition of Concurrent Forces in SpaceThe resultant, R of two or more forces in space is obtained bysumming their rectangular components i.e. R = ∑Fi.e. Rx i + Ry j + Rz k = ∑ ( Fx i + Fy j + Fz k )= (∑ Fx) i + (∑ Fy)j + (∑ Fz )kR x = ∑ Fx, Ry = ∑ Fy , Rz = ∑ FzR = Rx2 + Ry2 + Rz2cos θx = Rx/R cos θy = Ry/R cos θz = Rz/R
66. 66. SolutionSolution:Position vector of BH = 0.6 m i + 1.2 m j - 1.2 m kMagnitude, BH = 0.6 2 + 12 2 + 12 2 . . = 18 m . BH → 1λ BH = = (0.6 m i + 12 m j − 12 m k ) . . | BH | 18 . → BH → 750 NTBH = | TBH |. λ BH = | TBH | = 0.6 m i + 12 m j − 12 m k . . | BH | 18m . →TBH = (250 N ) i→ + (500 N ) j→ − (500 N ) k →Fx = 250 N , Fy = 500 N , Fz = − 500 N
67. 67. 2.9 EQUILIBRIUM OF APARTICLE IN SPACE For equilibrium: ∑Fx = 0, ∑Fy = 0 and ∑Fz = 0. The equations may be used to solve problems dealing with the equilibrium of a particle involving no more than three unknowns.