O slideshow foi denunciado.
Utilizamos seu perfil e dados de atividades no LinkedIn para personalizar e exibir anúncios mais relevantes. Altere suas preferências de anúncios quando desejar.

Lecture7 (37)

628 visualizações

Publicada em

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.

Publicada em: Educação
  • Don't ask yourself if you can afford the "Demolisher" Betting System. Ask yourself if you can afford NOT to! ♣♣♣ http://t.cn/A6zP2wH9
       Responder 
    Tem certeza que deseja  Sim  Não
    Insira sua mensagem aqui
  • Seja a primeira pessoa a gostar disto

Lecture7 (37)

  1. 1. Equivalent Force Systems EQUIVALENT SYSTEMS for SINGLE FORCEEQUIVALENT SYSTEMS for SINGLE FORCE Determining the effect of moving a force. 1. MOVING1. MOVING A FORCEA FORCE ONON ITS LINE OF ACTIONITS LINE OF ACTION 2. MOVING2. MOVING A FORCEA FORCE OFFOFF OF ITS LINE OF ACTIONOF ITS LINE OF ACTION
  2. 2. 1. MOVING A FORCE1. MOVING A FORCE ONON ITS LINE OF ACTIONITS LINE OF ACTION Moving a force from A to O, when both points are on the vectors’ line of action, does not change the external effect. Hence, a force vector is called a sliding vector. (But the internal effect of the force on the body does depend on where the force is applied). the two systems are equivalentthe two systems are equivalent Equivalent Force Systems
  3. 3. 2. MOVING A FORCE2. MOVING A FORCE OFFOFF OF ITS LINE OF ACTIONOF ITS LINE OF ACTION Moving a force from point A to B (as shown above) requires creating an additional couple moment. Since this new couple moment is a “free” vectorfree” vector, it can be applied at any point P on the body. the two systems are equivalentthe two systems are equivalent Use this process repeatedly for systems of forcesUse this process repeatedly for systems of forces Equivalent Force Systems
  4. 4. Moving a force from point A to B requires creating an additional couple moment. ≡ MB= (80+40)(-150 kN) = -18000 kN.mm = -18.000 kN.m Equivalent Force Systems 2. MOVING A FORCE2. MOVING A FORCE OFFOFF OF ITS LINE OF ACTIONOF ITS LINE OF ACTION Example:
  5. 5. 2. MOVING A FORCE2. MOVING A FORCE OFFOFF OF ITS LINE OF ACTIONOF ITS LINE OF ACTION Mo = (200 sin60°)(400 N) = 69282 N.mm = 69.282 N.m Equivalent Force Systems Example: Determine the equivalent system for the 400 N force acting at point O?
  6. 6. Determine equivalent force and couple at A? Example: Equivalent Force Systems
  7. 7. Equivalent Force Systems ‘2D’ Replace by Equivalent System
  8. 8. RESULTANT OF PARALLEL FORCES AT A POINT ‘2D’ AA system of forcessystem of forces andand momentsmoments cancan bebe simplify intosimplify into a single resultanta single resultant forceforce andand momentmoment acting at a specified pointacting at a specified point.. Parallel system of forcesParallel system of forces
  9. 9. If the force system lies in the x-y plane (2-D case), thenIf the force system lies in the x-y plane (2-D case), then the reduced equivalent system can be obtained usingthe reduced equivalent system can be obtained using the following two scalar equations.the following two scalar equations. Parallel system of forcesParallel system of forces ARA yR MM FF ∑ ∑ = = RESULTANT OF PARALLEL FORCES AT A POINT ‘2D’
  10. 10. Parallel system of forcesParallel system of forces ARA yR MM FF ∑ ∑ = = RESULTANT OF PARALLEL FORCES AT A POINT ‘2D’
  11. 11. a a a a O F1 F2 F3 F4 F5 y x Find the equivalent Force ­­­­ couple system at O? F1 = 100 N, F2 = 90 N, F3 = 80 N, F4 = 70 N, F5 = 60 N and a = 0.2 m Parallel system of forcesParallel system of forcesExample: RESULTANT OF PARALLEL FORCES AT A POINT ‘2D’
  12. 12. 4 4 4 333 F1 F2 F3 h1 h2 h2h3 h3 • P Given that: F1 = 20 kN, F2 = 30 kN, F3 = 40 kN and h1 = 2 m h2 = 3 m h3 = 1 m Replace the given system of parallel forces by a force-couple system acting at the point P? Parallel system of forcesParallel system of forces Example: RESULTANT OF PARALLEL FORCES AT A POINT ‘2D’
  13. 13. Further Reduction ofFurther Reduction of ParallelParallel ForceForcess ‘2D’‘2D’ Several parallel forces acting on the stick can be replaced by a single resultant force FR acting at a distance d from the point of grip. The equivalent Force: FR = F1 + F2 + .... + FN To find distance d use: FRd = F1d1 + F2d2 + ..... + FNdN dN FN Parallel system of forcesParallel system of forces
  14. 14. ≡≡ ≡≡ Parallel system of forcesParallel system of forces Further Reduction ofFurther Reduction of ParallelParallel ForceForcess ‘2D’‘2D’
  15. 15. a a a a O F1 F2 F3 F4 F5 y x For the given force system find a representative single force and its location on the x-axis? F1 = 100 N; F2 = 90 N; F3 = 80 N; F4 = 70 N; F5 = 60 N; a = 0.2 m Parallel system of forcesParallel system of forces Example: Further Reduction ofFurther Reduction of ParallelParallel ForceForcess ‘2D’‘2D’
  16. 16. Example: Replace the force system by: 1- a single force – couple resultant at point P, 2- a single force resultant along the x-axis. Parallel system of forcesParallel system of forces Further Reduction ofFurther Reduction of ParallelParallel ForceForcess ‘2D’
  17. 17. Equivalent Force Systems ‘2D’ Replace by Equivalent System
  18. 18. AA system of forcessystem of forces andand momentsmoments cancan bebe simplifsimplifiedied into a single resultantinto a single resultant forceforce andand moment acting at a specified pointmoment acting at a specified point.. General system of forces and moments Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  19. 19. WhenWhen several forces and couple momentsseveral forces and couple moments act on aact on a body, each forcebody, each force should beshould be movemove withwith its associatedits associated couple moment tocouple moment to thatthat common point O.common point O. AAdd all the forces and couple moments togetherdd all the forces and couple moments together and findand find one resultant force-couple moment pone resultant force-couple moment pairair.. General system of forces and moments Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  20. 20. When several forces acting on a given system with couple moments, a single resultant force FR can be replaced to such a distance d along the specified line so that, it will have the same external effect on the given system. i.e. Locate the resultant force FR at such a distance d so that, this resultant force FR will overcome the couple-moment effect also. Determine: FR = F1 + F2 + .... + FN Find distance d: F d = (F d + F d + .... + F d ) + (M +M +...+M ) General system of forces and moments Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  21. 21. M2 M1 ≡≡≡≡ General system of forces and moments Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  22. 22. [1] 20 points Replace the given system of forces and couple by a single force and its location on the line OE? • E 2m 2m Example: General system of forces and moments Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  23. 23. Example: The beam AE is subjected to a system of coplanar forces. Determine the magnitude, direction and the location on the beam of a resultant force which is equivalent to the given system of forces measured from E. i.e. Determine the equivalent force system measured from point E. General system of forces and moments Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  24. 24. C Example: General system of forces and moments Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  25. 25. General system of forces and moments If the force system lies in the x-y plane (2-D case), then the reduced equivalent system can be obtained using the following three scalar equations. Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  26. 26. M2 M1 General system of forces and moments ≡≡ Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  27. 27. Reducing the given system of forces and couple moments into equivalent resultant force and couple moment: 1- add all the forces algebraically, 2- determine the moments of each of these forces with respect to that point 3- add all the existing moments the system. Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  28. 28. For resultant moment calculations if the system contains forces and moment then, MMRoRo = ∑M + ∑(r X F)= ∑M + ∑(r X F) Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  29. 29. Determine equivalent force and couple (moment) at O. Example: General system of forces and moments Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  30. 30. [1] 20 points •E 2m 2m Example: General system of forces and moments Replace the given system of forces and couple by equivalent force- couple system acting at the point E. Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  31. 31. 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. Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  32. 32. Example: Determine the magnitude and directional sense of the resultant moment of the forces about point O and point P. Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  33. 33. Determine equivalent single force and couple at O. Example: Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  34. 34. Example: Replace the forces acting on the brace by an equivalent resultant force and couple moment acting at point A. i.e. Determine equivalent single force and couple (moment) at A. 57.2° Reducing the given system of forces and couple moments into an equivalent SINGLE force and couple moment at ANT POINTat ANT POINT ‘2D’
  35. 35. Example:Example: For the given force – couple system, obtain an equivalent (SINGLE) force along the line OP? Reducing the given system of forces and couple moments into an equivalent SINGLE force ALONG A LINE ‘2D’
  36. 36. Example: Reducing the given system of forces and couple moments into an equivalent SINGLE force ALONG A LINE ‘2D’ Determine equivalent single force along x-axis.
  37. 37. Example: Replace the given forces by: a) an equivalent resultant force and couple moment acting at point H. b) a single force and its location acting along the line A-D-B. 10 mm • 10 mm • • • • • A B C D E H F1 F2 F3 M1 M2 F2 = 42.806 N F1 = 50.215 N F3 = 97.317 N M1 = 2171.575 Nmm M2 = 820.679 Nmm General system of forces and moments Reducing the given system of forces and couple moments into an equivalent SINGLE force ALONG A LINE ‘2D’

×