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Lecture6 (72) (1)

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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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Lecture6 (72) (1)

  1. 1. Engineering Mechanics: Statics Force System Resultants
  2. 2. Chapter Objectives  To discuss the concept of the moment of a force and show how to calculate it in 2-D and 3-D systems.  Definition of the moment of a couple.
  3. 3. Chapter Objectives • To present methods for determining the resultants of non-concurrent force systems in 2D and 3D systems. • Reducing the given system of forces and couple moments into an equivalent force and couple moment at any point.
  4. 4. Chapter Outline Moment of a Force Force Couple Principles of Moments
  5. 5. Moment of a Force Carpenters often use a hammer in this way to pull a stubborn nail. Through what sort of action does the force FH at the handle pull the nail? How can you mathematically model the effect of force FH at point O?
  6. 6. MomentMoment of a force about a point (or an axis) is a measure of the tendency of the force to cause a body to rotate about the point or axis. Moment of a Force 2D System
  7. 7. Case 1: Fx horizontal and acts perpendicular to the handle of the wrench and is located dy from the point O Fx tendstends to turn the pipe about the z axisto turn the pipe about the z axis  The larger the force or the distance dy the greater the Moment of a ForceMoment of a Force 2D System2D System
  8. 8. Moment of a ForceMoment of a Force 2D System2D System The larger the force ‘W’ or the distance ‘D’ the greater the turning effect at point ‘P’.
  9. 9. Note that: Moment axis (z) is perpendicular to shaded plane (x-y). i.e. The remaining third axis: ‘z’ Fx and dy lies on the shaded plane (x-y) Moment axis (z) intersects the plane at point O Moment of a Force 2D System
  10. 10. Case 2: Apply force Fz to the wrench  Pipe does not rotate about z axis  The pipe not actually rotates but Fz creates a tendency for rotation so causing (producing)moment along (Mo)x.  Moment axis (x) is perpendicular to the shaded plane (y-z) Moment of a Force 2D System
  11. 11. Case 3: Apply force Fy to the wrench  NoNo momentmoment is produced about point O  Lack of tendency to rotateLack of tendency to rotate as line of action passesas line of action passes through Othrough O Note that, Fy and dy both lies on the same line (and not forming any Moment of a Force 2D System
  12. 12.  Moment of a force does not always cause rotation  Force F;  tends to rotate the beam clockwise about A with moment MA = dAF  tends to rotate the beam counterclockwise about B with moment MB = dBF Moment of a Force 2D System
  13. 13. In General  Consider the force F and the point O which lies in the shaded plane  The moment MO about point O, or about an axis passing through O and which is perpendicular to the plane, is a vector quantityvector quantity.. Moment of a Force 2D System
  14. 14. Moment MO is a vector having specified magnitudemagnitude and directiondirection. Moment of a Force 2D System
  15. 15. M = F x d M = Magnitude of the moment about point or axis [N.m] F= Magnitude of the force [N] d = perpendicular distance [m] Direction is determined by using the right hand rule Moment of a Force 2D System
  16. 16. PositivePositive direction of the moment :direction of the moment : Anti clockwise Moment of a Force 2D System
  17. 17. PositivePositive momentmoment NegativeNegative momentmoment Moment of a Force 2D System
  18. 18. Magnitude: Use simple multiplication; For magnitude of MO: MO = FF .. dd d = moment armmoment arm or perpendicularperpendicular distancedistance from the axis at point O to its line of action of the force. F = Magnitude of the force  Moment of a Force 2D System Scalar Formulation
  19. 19. Moment of a Force 2D System Scalar Formulation Direction: Direction of MO is specified by using “right hand ruleright hand rule” Fingers of the right hand are curled to follow the sense of rotation when force rotates about point O.
  20. 20. Direction: Thumb points along the moment axis to give the direction and sense of the moment vector Moment vector is upwards and perpendicular to the shaded plane Moment of a Force 2D System Scalar Formulation
  21. 21. Direction MMOO is shown by a vector arrow with a curl tois shown by a vector arrow with a curl to distinguish it from force vectordistinguish it from force vector Fig b.  MO is represented by the counterclockwise curl, which indicates the action of F.  Arrowhead shows the sense of rotation caused by F.  Using the right hand rule, the direction and sense of the moment vector points out of the page. Moment of a Force 2D System Scalar Formulation
  22. 22. FrM  ×= =r  Position vector which runs from the moment reference point to any point on the line of action of the force. In some two dimensional problems and most of the three dimensional problems, it is convenient to use a vector approach for moment calculations. The MOMENT of a force about point A may be represented by the cross productcross product expression. Moment of a ForceMoment of a Force 2D2D System VectorVector FormulatiFormulationon
  23. 23. Without using the right hand rule directly apply the equation through cross product of vectors. MO = d XX F The cross product directly gives the magnitude and the direction. Moment of a ForceMoment of a Force 2D2D System VectorVector FormulatiFormulationon
  24. 24. FrMo  ×= Sarrus’ RuleSarrus’ Rule +k- k )kFr)kFrM xyyxo  (( −= Moment of a ForceMoment of a Force 2D2D System VectorVector FormulatiFormulationon
  25. 25. Example: For each case, determine the moment of the force about point O Moment of a ForceMoment of a Force 2D2D System
  26. 26. Solution  Line of action is extended as a dashed line to establish moment arm d  Tendency to rotate is indicated and the orbit is shown as a colored curl o o (a)M (2m)(100N) 200.000 N.m (CW) (b)M (0.75m)(50N) 37.500 N.m (CW) = = = = Moment of a ForceMoment of a Force 2D2D System
  27. 27. Solution o o o (c) M (4m 2cos30 m)(40N) 229.282 N.m (CW) (d) M (1sin45 m)(60N) 42.426 N.m (CCW) (e) M (4m 1m)(7kN) 21.000 kN.m (CCW) = + = = = = − = o o Moment of a ForceMoment of a Force 2D2D System
  28. 28. Example: Determine the moments of the 800 N force acting on the frame about points A, B, C and D. Moment of a Force 2D System
  29. 29. Solution (Scalar Analysis) Line of action of F passes through C A B C D M = (2.5m)(800N) = 2000 N.m (CW) M = (1.5m)(800N) = 1200 N.m (CW) M = (0m)(800N)= 0 N.m M = (0.5m)(800N) = 400 N.m (CCW) Moment of a ForceMoment of a Force 2D2D System
  30. 30. Moment of a ForceMoment of a Force 2D2D System
  31. 31. Principles of Moments  Principles of MomentsPrinciples of Moments Also known as Varignon’s Theorem This principle states that the moment of a force about a point is equal to the sum of moments of the force’s components about the point.
  32. 32. Principles of Moments “Moment of a force about a point is equal to the sum of the moments of the forces’ components about the point”
  33. 33. Principles of Moments
  34. 34. Principles of Moments Solution Method 1: From trigonometry using triangle BCD, CB = d = 100cos45° = 70.7107mm Thus, MA =dF= (0.07071m) 200N = 14142.136 N.mm (CCW) = 14.142 N.m (CCW) As a Cartesian vector, MA = {14.142 k} N.m
  35. 35. Principles of Moments Solution Method 2:  Resolve 200 N force into x and y components  Principle of Moments MA = ∑dF MA =(200)(200sin45°) – (100)(200cos45°) = 14142.136 N.mm (CCW) = 14.142 N.m (CCW)
  36. 36. Principles of Moments “Moment of a force about a point is equal to the sum of the moments of the forces’ components about the point ( ) m898.275sin3 =°=d ( )( ) mkN489.14 898.25 ⋅−= −= −= FdMO ( )( )[ ] ( )( )[ ] mkN489.14 30cos345sin530sin345cos5 ⋅−= °−°−= −−= xyyxO dFdFM Example:
  37. 37. Example: Determine the moment of the 600 N force with respect to point O in both scalar and vector product approaches. Moment of a ForceMoment of a Force 2D2D System
  38. 38. Example: The force F acts at the end of the angle bracket. Determine the moment of the force about point O. Moment of a ForceMoment of a Force 2D2D System
  39. 39. Solution Method 1: Resolve the given force into components and than apply the moment equation. MO = 400sin30°N(0.2m)-400cos30°N(0.4m) = -98.5641 N.m As a Cartesian vector, MO = {-98. 5641k} N.m Moment of a ForceMoment of a Force 2D2D System
  40. 40. Solution Method 2:  Express as Cartesian vector r = {0.4i – 0.2j} m F = {400sin30°i – 400cos30°j} N = {200.000i – 346.410j}N For moment, { } O i j k M rXF 0.4 0.2 0 200.000 346.410 0 -98.564k N.m = = − − = r r r r rr r Moment of a ForceMoment of a Force 2D2D System
  41. 41. Resultant Moment of System of Coplanar Forces Resultant moment MRo = addition of the moments produced by all the forces algebraically since all moment forces are collinear (for 2D case). MRo = ∑F.d taking counterclockwise (CCW), to be positive.
  42. 42. Resultant moment, MRo = addition of the moments produced by all the forces algebraically, since all moment forces are collinear (for 2D case). MRo =M1 – M2 + M3 =∑dF= d1 F1 – d2 F2 + d3 F3 taking counterclockwise (CCW) to be positive. Resultant Moment of System of Coplanar Forces
  43. 43. Counterclockwise is positive ORM Fd+ = ∑ Resultant Moment of System of Coplanar Forces +
  44. 44. Example: Determine the resultant moment of the four forces acting on the rod about point O. Resultant Moment of System of Coplanar Forces
  45. 45. Solution: (by scalar analysis) Note that always positive moments acts in the +k direction, CCW )CW(m.N923.333 m.N923.333 )m30cos3m4)(N40( )m30sin3)(N20()m0)(N60()m2)(N50(M d.FM Ro Ro = −= +− ++−= ∑=   Resultant Moment of System of Coplanar Forces
  46. 46. Solution: (by vector analysis) (CW)or(-k)N.m333.923 (-k)][263.923(k)][30[0]k)]([100 (-j)]40X)(i)3cos30(4[ (i)]20X(-j)[3sin30(i)]60X[0(-j)]50X(i)[2MRo = +++−= ++ ++= ∑=   dXFMRo Resultant Moment of System of Coplanar Forces
  47. 47. Moment of a CoupleMoment of a Couple
  48. 48. Moment Moment force F about point O can be expressed using cross product MO = r X F where r represents position vector from O to any point lying on the line of action of F.
  49. 49. • Remember M = r (vector) X F (vector) • Find the length ‘r’ vectorially for each force ‘F’ • if not given, find the vectorial representation of ‘F’ also. Moment
  50. 50. In some two dimensional problems and many three dimensional problems, it is convenient to use a vector approach for moment calculations. The MOMENT of a force about point A may be represented by the cross product expression FrM rrr ×= Moment
  51. 51. FrM rrr ×= =r r Position vector which runs from the moment reference point to any point on the line of action of the force MA Due to the principle of transmissibility, can act at any point along its line of action and still create the same moment about point A. =F r FrFrFrM A rrrrrrr ×=×=×= 321 Moment
  52. 52. The Moment Vector The result obtained from r X F doesn’t depend on where the vector r intersects the line of action of F: r = r’ + u r × F = (r’ + u) × F = r’ × F because the cross product of the parallel vectors u and F is zero.
  53. 53. Moment of a CoupleMoment of a CoupleMoment of a CoupleMoment of a Couple CoupleCouple - two parallel forces - same magnitude but opposite direction - separated by perpendicular distance dseparated by perpendicular distance d  Resultant force = 0  Tendency to rotate in specified direction  Couple moment = sum of moments of both couple forces about any arbitrary point
  54. 54. Moment of a CoupleMoment of a Couple A couple is defined as twoA couple is defined as two parallel forces withparallel forces with the samethe same magnitudemagnitude butbut opposite inopposite in directiondirection separated by aseparated by a perpendicular distance d.perpendicular distance d. The moment of a couple is defined asThe moment of a couple is defined as:: MO = F . d (using a scalar analysis; right hand rule for direction), MO = d X F (using vector analysis).
  55. 55. Moment of a CoupleMoment of a Couple The net external effect of a couple is zero since the net force equals zero and the magnitude of the net moment equals F.d Moments due to couples can be added using the same rules as adding any vectors. The moment of a couple is a free vector. It can be moved anywhere on the body and have the same external effect on the body.
  56. 56. O a d F FA B C MO= F (a+d) – F a = F d MMOO==MMAA==MMBB==MMCC Moment of a couple has the same value for allMoment of a couple has the same value for all moment centers.moment centers. Moment of a Couple ‘2D’
  57. 57. M M M M 2D CCW couple 2D CW couple Moment of a Couple ‘2D’
  58. 58. The moment of a couple is a free vector. It can be moved anywhere on the body and have the same external effectexternal effect on the body. == Moment of a Couple ‘2D’
  59. 59. Moment of a Couple ‘2D’ APPLICATIONSAPPLICATIONS
  60. 60. Moment of a Couple ‘2D’ APPLICATIONS (continued)
  61. 61. Moment of a Couple ‘2D’ Scalar Formulation Magnitude of couple moment M = F.d  Direction and sense areDirection and sense are determined by right hand ruledetermined by right hand rule In all cases, M acts perpendicular to plane containing the forces.
  62. 62. Moment of a Couple ‘2D’ Vectorial Formulation M = d X F In all cases, M acts perpendicular to plane containing the forces.
  63. 63. Moment of a Couple ‘2D’ Example: A couple acts on the gear teeth. Replace it by an equivalent couple having a pair of forces that act through points A and B. ==
  64. 64. Moment of a Couple ‘2D’ Solution  Magnitude of couple M = 24 N.m  Direction out of the page since forces tend to rotate CCW  M is a free vector and can be placed anywhere.
  65. 65. Moment of a Couple ‘2D’ Solution  To preserve CCW motion, vertical forces acting through points A and B must be directed as shown  For magnitude of each force, M = F.d 24 N.m = F (0.2m) F = 120.000 N
  66. 66. Moment of a Couple ‘2D’ Example:
  67. 67. Two different couples are equivalent if they produce the same moment, (magnitude as well as direction). Equivalent Couples Moment of a Couple ‘2D’
  68. 68. Two couples are equivalent if they produce the same moment with magnitude and direction. == ====== Moment of a Couple ‘2D’
  69. 69. Example: Moment of a Couple ‘2D’
  70. 70. Due principle of transmissibility, the distance r that can act at any point along the line of action of the Force creates the same moment about point A. MA FrFrFrMA  ×=×=×= 321 Moment of a Couple ‘3D’
  71. 71. • select one of the couple forces to start with, • along the line of action of this selected force, select any point over it, • determine a point also along the line of action of the other force, • get the position vector between those two selected points, • using cross-product obtain the couple moment, since the principle of transmissibility applies M = rXF. Moment of a Couple ‘3D’
  72. 72. Example: For the given force-couple system: 1) Determine the resultant couple moment in Cartesian form? 2) Express the magnitude of the resultant couple moment? 3) Find the direction cosines of this resultant couple moment? Moment of a Couple ‘3D’

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