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Lecture 2(57)

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Concept of Particles and Free Body Diagram

Why FBD diagrams are used during the analysis?

It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
It helps to identify the forces and ensures the correct use of equation of equilibrium.

Note:
Reactions on two contacting bodies are equal and opposite on account of Newton's III Law.
The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces.
Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable.

Physical Meaning of Equilibrium and its essence in Structural Application
The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium.

A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium.
The rigid body in equilibrium means the body is stable.
Equilibrium means net force and net moment acting on the body is zero.

Essence in Structural Engineering
To find the unknown parameters such as reaction forces and moments induced by the body.
In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters.
For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.

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Lecture 2(57)

  1. 1. Engineering Mechanics: StaticsStatics Chapter 2: Force VectorsForce Vectors
  2. 2. Objectives To show how to add forces and resolve them into components using the Parallelogram Law. To express force and position in Cartesian vector form and explain how to determine the vector’s magnitude and direction.
  3. 3. Chapter Outline Scalars and Vectors Vector Operations Vector Addition of Forces Addition of System of Coplanar Forces
  4. 4. Chapter Outline Cartesian Vectors Addition and Subtraction of Cartesian Vectors Position Vectors Force Vector Directed along a Line Dot Product
  5. 5. Number Any positive or negative value. Eg: 5, -3.07, π... Scalar A quantity characterized by a positive or negative number and have a unit. Eg: mass (5.1 kg), volume (0.028 m3 )... Scalars and Vectors
  6. 6. Scalars and Vectors Vector A quantity that has both magnitude and direction (i.e. a numbenumber having unitunit and directiondirection) Eg: Position (3.4 km North), force (8.73 N towards right side), ... – Represented by a letter with an arrow over it such as or bold such as A – Magnitude is designated as or simply A – In this course, vector is presented as A and its A  A 
  7. 7. Numbers, Scalars and Vectors
  8. 8. Scalars and Vectors VectorVector Represented graphically as an arrow Length of the arrow = Magnitude of Vector Angle between the reference axis and arrow’s line of action = Direction of Vector Arrowhead =Sense of Vector
  9. 9. Scalars and Vectors Example Magnitude of Vector = 4 units Direction of Vector = 20° measured counterclockwise from the horizontal axis Sense of Vector = Upward and to the right The point O is called tail of the vector and the point P is called the tip or head
  10. 10. Vector Operations Multiplication and Division of a Vector by a Scalar - Product of vector A and scalar a = aA - Magnitude = - If a is positive (+), sense of aA is the same as sense of A - If a is negative (-), sense of aA, it is opposite to the sense of A aA
  11. 11. Vector Operations Multiplication and Division of a Vector by a Scalar - Negative of a vector is found by multiplying the vector by ( -1 ) - Law of multiplication applies Eg: A/a = ( 1/a ) A, if a≠0
  12. 12. Vector Operations Vector Addition Addition of two vectors A and B gives a esultant vector R by the parallelogramparallelogram lawlaw Result R can be found by triangle construction Communicative Eg: R = A + B = B + A
  13. 13. Vector Addition Paralelogram Law Triangular Construction Vector Operations
  14. 14. Vector Operations Vector Addition - Special case: Vectors A and B are collinear (both have the same line of action). Triangular shape does not form.
  15. 15. VectorVector Operations Vector Subtraction Special case of additionaddition Eg: R’ = A – B = A + ( - B ) Rules of Vector Addition Applies
  16. 16. VectorVector Operations Resolution of Vector Any vector can be resolved into two components by the parallelogram law The two components A and B are drawn such that they extend from the tail of R
  17. 17. Resolution of a vector is breaking up a vector into its components. It is the reverse application of the parallelogram law. Vector OperationsVector Operations
  18. 18. Vector Addition of Forces When two or more forces are added, successive applications of the parallelogram law is carried out to find the resultant Eg: Forces F1, F2 and F3 acts at a point O - First, find resultant of F1 + F2 - Resultant, FR = ( F1 + F2 ) + F3
  19. 19. Procedure for Analysis Parallelogram Law - Make a sketch using the parallelogram law - Two components forces add to form the resultant force - Resultant force is shown by the diagonaldiagonal of the parallelogram - The components is the sides of the parallelogram Vector Addition of Forces
  20. 20. Procedure for Analysis Parallelogram Law - To resolve a force into components along two axes directed from the tail of the force - Start at the head, constructing lines parallel to the axes - Label all the known and unknown force magnitudes and angles - Identify the two unknown components Vector Addition of Forces
  21. 21. Procedure for Analysis Trigonometry - Redraw half portion of the parallelogram - MagnitudeMagnitude of the resultant force can be determined by the law of COSINE - DirectionDirection of the resultant force can be determined by the law of SINE Vector Addition of Forces
  22. 22. Magnitude of the two components can be determined by the law of COSINE. Direction of the two components can be determined by the law of SINE. Vector Addition of Forces
  23. 23. Method 1: Parallelogram Law OR Establishing triangles graphical methodMethod 2: Trigonometry se the Cosine or Sine Law (3 forces). Vector Addition of ForcesVector Addition of Forces
  24. 24. Example: The screw eye is subjected to two forces F1 and F2. Determine the magnitude and direction of the resultant force. Vector Addition of Forces
  25. 25. Solution Parallelogram Law Unknown:  magnitude of FR and  angle θ Vector Addition of Forces
  26. 26. Vector Addition of Forces FR= Magnitude of the resultant force (determined by the law of cosine) θ= angle (determined by the law of sine) ϕ = direction of the resultant force = θ +15.0000° Half portion of the parallelogram
  27. 27. Solution Trigonometry Law of Cosine ( ) ( ) ( ) ( ) ( ) N552.212 4226.0300002250010000 115cosN150N1002N150N100F 22 R = −−+= −+=  Vector Addition of Forces
  28. 28. Solution Trigonometry Law of Sines ( )   sin115 N212.552 θsin N150 7613.39 9063.0 N552.212 N150 sin = = = θ θ Vector Addition of Forces
  29. 29. Solution Trigonometry Direction Φ of FR measured from the horizontal   354.761 1539.7613 = +=ϕ Vector Addition of Forces
  30. 30. ∑= FFR F1 F1 Addition of a System of ConcurrentConcurrent Coplanar (2D) Forces Concurrent at a point:
  31. 31. Addition of a System of ConcurrentConcurrent Coplanar (2D) Forces Cartesian Vector Notation - Cartesian unit vectors i and j are used to designate the x and y directions - Unit vectors i and j have dimensionless magnitude of unity ( = 1 )
  32. 32. Cartesian Vector Notation F = Fx i + Fy j F’ = F’x i + F’y (-j) F’ = F’x i – F’y j Addition of a System of Concurrent Coplanar (2D) Forces
  33. 33. Addition of a System of ConcurrentConcurrent Coplanar (2D) Forces Resolve vectors into components using xx and yy axes system. Each component of the vector will have a magnitude and a direction. The directions are based on the x and y axes. The “unit vectors” ii and jj used designate xx and yy axes. F = Fxx ii + Fyy jj xx and yy axes are always perpendicular to each other. Together, they can be directed at any inclination.
  34. 34. Cartesian Vector Notation jFiFF yx  += F  Vector components may be expressed asVector components may be expressed as products of the unit vectors with the scalarproducts of the unit vectors with the scalar magnitudes of the vector componentsmagnitudes of the vector components.. Fx and Fy are referred to as the scalar components of
  35. 35. Addition of a System of ConcurrentConcurrent Coplanar (2D) Forces Resultant of two or moretwo or more forces: Find the components of the forces in the specified axes Add components algebraically Form the resultant
  36. 36. Coplanar Force Resultants To determine resultant of several coplanar forces: - Resolve force into x and y components - Addition of the respective components using scalar algebra - Resultant force is found using the parallelogram law Addition of a System of Concurrent Coplanar (2D) Forces
  37. 37. Coplanar Force Resultants - Positive scalars = sense of direction along the positive coordinate axes - Negative scalars = sense of direction along the negative coordinate axes - Magnitude of FR can be found by Pythagorean Theorem RyRxR FFF 22 += Addition of a System of Concurrent Coplanar (2D) Forces
  38. 38. Coplanar Force Resultants - Direction angle θ (orientation of the force) can be found by trigonometry Rx Ry F F1 tan− =θ Addition of a System of Concurrent Coplanar (2D) Forces
  39. 39. Addition of a System of ConcurrentConcurrent Coplanar (2D) Forces
  40. 40. Example: Add (sum) the given three coplanar forces Addition of a System of ConcurrentConcurrent Coplanar (2D) Forces
  41. 41. Step 1: to resolve each force into its x-yx-y components. Step 3: find the magnitudemagnitude and the angleangle of the rresultant vector. Step 2: to add all the xx components together and add all the yy components together. These two totals become the resultant vector. Cartesian Vector Method (solution by three steps) Addition of a System of ConcurrentConcurrent Coplanar (2D) Forces
  42. 42. Example: Add (sum) the given three coplanar forces Cartesian vector notation FF11 == FF1x1x ii ++ FF1y1y jj FF22 = -= - FF2x2x ii ++ FF2y2y jj FF33 == FF3x3x ii –– FF3y3y jj Addition of a System of ConcurrentConcurrent Coplanar (2D) Forces Step 1
  43. 43. Vector resultant is therefore FFRR == FF11 ++ FF22 ++ FF33 == FF1x1x ii ++ FF1y1y jj -- FF2x2x ii ++ FF2y2y jj ++ FF3x3x ii –– FF3y3yjj = (= (FF1x1x -- FF2x2x ++ FF3x3x))ii + (+ (FF1y1y ++ FF2y2y –– FF3y3y))jj = (= (FFRxRx))ii + (+ (FFRyRy))jj Addition of a System of Concurrent Coplanar (2D) Forces Step 2
  44. 44. ( ) ( )↑+ → + Addition of a System of Concurrent Coplanar (2D) Forces Coplanar Force Resultants FRx = (F1x - F2x + F3x) FRy = (F1y + F2y – F3y) In all cases, FFRxRx == ∑∑FFxx FFRyRy == ∑∑FFyy * Take note of sign conventions
  45. 45. Step 3 Scalar Notation: Rx Ry RyRxR F F FFF 1 22 − = += tanθ Addition of a System of Concurrent Coplanar (2D) Forces The magnitudemagnitude and the angleangle of the rresultant vector
  46. 46. WAYS OF FINDING THE COMPONENTS OF FORCES IN 2D 1- By an ANGLE measured from any axis (Cosine of this angle gives the component of that axis) Addition of a System of Concurrent Coplanar (2D) Forces
  47. 47. WAYS OF FINDING THE COMPONENTS OF FORCES IN 2D 2- By right angle TRIANGLE of known sides Addition of a System of Concurrent Coplanar (2D) Forces
  48. 48. WAYS OF FINDING THE COMPONENTS OF FORCES IN 2D 3- By COORDINATES or known lengths. Addition of a System of Concurrent Coplanar (2D) Forces 12 unit (18, 10) (0, 0) 12 unit 10 cm 18 cm
  49. 49. Example: The end of the boom O is subjected to three concurrent and coplanar forces. Determine the magnitude and orientation of the resultant force. 335˚ Addition of a System of Concurrent Coplanar (2D) Forces
  50. 50. Solution N788.324 N 5 3 200N35cos250F :FF N606.416 N 5 4 200N35sin250N400F :FF Ry yRy Rx xRx =       += = −=       −+−= =   Σ Σ 35˚ 324.788 N 416.606 N Addition of a System of Concurrent Coplanar (2D) Forces
  51. 51. Solution Resultant Force From vector addition, Direction angle θ is ( ) ( ) N250.528 N788.324N606.416F 22 R = +−=   0598.1429402.37180 9402.37 N606.416 N788.324 tan 1 =− −=       − = − θ 324.788 N 416.616 N Addition of a System of Concurrent Coplanar (2D) Forces 142.0598°
  52. 52. Example: Addition of a System of Concurrent Coplanar (2D) Forces
  53. 53. Example: Addition of a System of Concurrent Coplanar (2D) Forces R R

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