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Engineering Mechanics: Statics

Engineering Mechanics: Statics. Chapter 5: Equilibrium of a Rigid Body. Chapter Objectives. To develop the equations of equilibrium for a rigid body. To introduce the concept of the free-body diagram for a rigid body. To show how to solve rigid-body equilibrium problems using the

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Engineering Mechanics: Statics

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  1. Engineering Mechanics: Statics Chapter 5: Equilibrium of a Rigid Body

  2. Chapter Objectives • To develop the equations of equilibrium for a rigid body. • To introduce the concept of the free-body diagram for a rigid body. • To show how to solve rigid-body equilibrium problems using the • equations of equilibrium.

  3. Chapter Outline • Conditions for Rigid Equilibrium • Free-Body Diagrams • Equations of Equilibrium • Two and Three-Force Members • Equilibrium in Three Dimensions • Equations of Equilibrium • Constraints for a Rigid Body

  4. 5.1 Conditions for Rigid-Body Equilibrium • Consider rigid body fixed in the x, y and z reference and is either at rest or moves with reference at constant velocity • Two types of forces that act on it, the resultant internal force and the resultant external force • Resultant internal force fi is caused by interactions with adjacent particles

  5. 5.1 Conditions for Rigid-Body Equilibrium • Resultant external force Fi represents the effects of gravitational, electrical, magnetic, or contact forces between the ith particle and adjacent bodies or particles not included within the body • Particle in equilibrium, apply Newton’s first law, Fi + fi = 0

  6. 5.1 Conditions for Rigid-Body Equilibrium • When equation of equilibrium is applied to each of the other particles of the body, similar equations will result • Adding all these equations vectorially, ∑Fi + ∑fi = 0 • Summation of internal forces = 0 since internal forces between particles in the body occur in equal but opposite collinear pairs (Newton’s third law)

  7. 5.1 Conditions for Rigid-Body Equilibrium • Only sum of external forces will remain • Let ∑Fi = ∑F, ∑F = 0 • Consider moment of the forces acting on the ith particle about the arbitrary point O • By the equilibrium equation and distributive law of vector cross product, ri X (Fi + fi) = ri X Fi + ri X fi = 0

  8. 5.1 Conditions for Rigid-Body Equilibrium • Similar equations can be written for other particles of the body • Adding all these equations vectorially, ∑ri X Fi + ∑ri X fi = 0 • Second term = 0 since internal forces occur in equal but opposite collinear pairs • Resultant moment of each pair of forces about point O is zero • Using notation ∑MO = ∑ri X Fi, ∑MO = 0

  9. 5.1 Conditions for Rigid-Body Equilibrium • Equations of Equilibrium for Rigid Body ∑F = 0 ∑MO = 0 • A rigid body will remain in equilibrium provided the sum of all the external forces acting on the body = 0 and sum of moments of the external forces about a point = 0 • For proof of the equation of equilibrium, - Assume body in equilibrium

  10. 5.1 Conditions for Rigid-Body Equilibrium - Force system acting on the body satisfies the equations ∑F = 0 and ∑MO = 0 - Suppose additional force F’ is applied to the body ∑F + F’ = 0 ∑MO + MO’= 0 where MO’is the moment of F’ about O - Since ∑F = 0 and ∑MO = 0, we require F’ = 0 and MO’ - Additional force F’ is not required and equations ∑F = 0 and ∑MO = 0 are sufficient

  11. 5.2 Free-Body Diagrams • FBD is the best method to represent all the known and unknown forces in a system • FBD is a sketch of the outlined shape of the body, which represents it being isolated from its surroundings • Necessary to show all the forces and couple moments that the surroundings exert on the body so that these effects can be accounted for when equations of equilibrium are applied

  12. 5.2 Free-Body Diagrams

  13. 5.2 Free-Body Diagrams

  14. 5.2 Free-Body Diagrams

  15. 5.2 Free-Body Diagrams Support Reactions • If the support prevents the translation of a body in a given direction, then a force is developed on the body in that direction • If rotation is prevented, a couple moment is exerted on the body • Consider the three ways a horizontal member, beam is supported at the end - roller, cylinder - pin - fixed support

  16. 5.2 Free-Body Diagrams Support Reactions Roller or cylinder • Prevent the beam from translating in the vertical direction • Roller can only exerts a force on the beam in the vertical direction

  17. 5.2 Free-Body Diagrams Support Reactions Pin • The pin passes through a hold in the beam and two leaves that are fixed to the ground • Prevents translation of the beam in any direction Φ • The pin exerts a force F on the beam in this direction

  18. 5.2 Free-Body Diagrams Support Reactions Fixed Support • This support prevents both translation and rotation of the beam • A couple and moment must be developed on the beam at its point of connection • Force is usually represented in x and y components

  19. 5.2 Free-Body Diagrams • Cable exerts a force on the bracket • Type 1 connections • Rocker support for this bridge girder allows horizontal movements so that the bridge is free to expand and contract due to temperature • Type 5 connections

  20. 5.2 Free-Body Diagrams • Concrete Girder rest on the ledge that is assumed to act as a smooth contacting surface • Type 6 connections • Utility building is pin supported at the top of the column • Type 8 connections

  21. 5.2 Free-Body Diagrams • Floor beams of this building are welded together and thus form fixed connections • Type 10 connections

  22. 5.2 Free-Body Diagrams External and Internal Forces • A rigid body is a composition of particles, both external and internal forces may act on it • For FBD, internal forces act between particles which are contained within the boundary of the FBD, are not represented • Particles outside this boundary exert external forces on the system and must be shown on FBD • FBD for a system of connected bodies may be used for analysis

  23. 5.2 Free-Body Diagrams Weight and Center of Gravity • When a body is subjected to gravity, each particle has a specified weight • For entire body, consider gravitational forces as a system of parallel forces acting on all particles within the boundary • The system can be represented by a single resultant force, known as weight W of the body • Location of the force application is known as the center of gravity

  24. 5.2 Free-Body Diagrams Weight and Center of Gravity • Center of gravity occurs at the geometric center or centroid for uniform body of homogenous material • For non-homogenous bodies and usual shapes, the center of gravity will be given

  25. 5.2 Free-Body Diagrams Idealized Models • Needed to perform a correct force analysis of any object • Careful selection of supports, material, behavior and dimensions for trusty results • Complex cases may require developing several different models for analysis

  26. 5.2 Free-Body Diagrams Idealized Models • Consider a steel beam used to support the roof joists of a building • For force analysis, reasonable to assume rigid body since small deflections occur when beam is loaded • Bolted connection at A will allow for slight rotation when load is applied => use Pin

  27. 5.2 Free-Body Diagrams Support at B offers no resistance to horizontal movement => use Roller • Building code requirements used to specify the roof loading (calculations of the joist forces) • Large roof loading forces account for extreme loading cases and for dynamic or vibration effects • Weight is neglected when it is small compared to the load the beam supports

  28. 5.2 Free-Body Diagrams Idealized Models • Consider lift boom, supported by pin at A and hydraulic cylinder at BC (treat as weightless link) • Assume rigid material with density known • For design loading P, idealized model is used for force analysis • Average dimensions used to specify the location of the loads and supports

  29. 5.2 Free-Body Diagrams Procedure for Drawing a FBD 1. Draw Outlined Shape • Imagine body to be isolated or cut free from its constraints • Draw outline shape 2. Show All Forces and Couple Moments • Identify all external forces and couple moments that act on the body

  30. 5.2 Free-Body Diagrams Procedure for Drawing a FBD • Usually due to - applied loadings - reactions occurring at the supports or at points of contact with other body - weight of the body • To account for all the effects, trace over the boundary, noting each force and couple moment acting on it 3. Identify Each Loading and Give Dimensions • Indicate dimensions for calculation of forces

  31. 5.2 Free-Body Diagrams Procedure for Drawing a FBD • Known forces and couple moments should be properly labeled with their magnitudes and directions • Letters used to represent the magnitudes and direction angles of unknown forces and couple moments • Establish x, y and coordinate system to identify unknowns

  32. 5.2 Free-Body Diagrams Example 5.1 Draw the free-body diagram of the uniform beam. The beam has a mass of 100kg.

  33. 5.2 Free-Body Diagrams Solution Free-Body Diagram

  34. 5.2 Free-Body Diagrams Solution • Support at A is a fixed wall • Three forces acting on the beam at A denoted as Ax, Ay, Az, drawn in an arbitrary direction • Unknown magnitudes of these vectors • Assume sense of these vectors • For uniform beam, Weight, W = 100(9.81) = 981N acting through beam’s center of gravity, 3m from A

  35. 5.2 Free-Body Diagrams Example 5.2 Draw the free-body diagram of the foot lever. The operator applies a vertical force to the pedal so that the spring is stretched 40mm and the force in the short link at B is 100N.

  36. 5.2 Free-Body Diagrams Solution • Lever loosely bolted to frame at A • Rod at B pinned at its ends and acts as a short link • For idealized model of the lever,

  37. 5.2 Free-Body Diagrams Solution • Free-Body Diagram • Pin support at A exerts components Ax and Ay on the lever, each force with a known line of action but unknown magnitude

  38. 5.2 Free-Body Diagrams Solution • Link at B exerts a force 100N acting in the direction of the link • Spring exerts a horizontal force on the lever Fs = ks = 5N/mm(40mm) = 200N • Operator’s shoe exert vertical force F on the pedal • Compute the moments using the dimensions on the FBD • Compute the sense by the equilibrium equations

  39. 5.2 Free-Body Diagrams Example 5.3 Two smooth pipes, each having a mass of 300kg, are supported by the forks of the tractor. Draw the free-body diagrams for each pipe and both pipes together.

  40. 5.2 Free-Body Diagrams Solution • For idealized models, • Free-Body Diagram of pipe A

  41. 5.2 Free-Body Diagrams Solution • For weight of pipe A, W = 300(9.81) = 2943N • Assume all contacting surfaces are smooth, reactive forces T, F, R act in a direction normal to tangent at their surfaces of contact • Free-Body Diagram at pipe B

  42. 5.2 Free-Body Diagrams Solution *Note: R represent the force of A on B, is equal and opposite to R representing the force of B on A • Contact force R is considered an internal force, not shown on FBD • Free-Body Diagram of both pipes

  43. 5.2 Free-Body Diagrams Example 5.4 Draw the free-body diagram of the unloaded platform that is suspended off the edge of the oil rig. The platform has a mass of 200kg.

  44. 5.2 Free-Body Diagrams Solution • Idealized model considered in 2D because by observation, loading and the dimensions are all symmetrical about a vertical plane passing through the center • Connection at A assumed to be a pin and the cable supports the platform at B

  45. 5.2 Free-Body Diagrams Solution • Direction of the cable and average dimensions of the platform are listed and center of gravity has been determined • Free-Body Diagram

  46. 5.2 Free-Body Diagrams Solution • Platform’s weight = 200(9.81) = 1962N • Force components Ax and Ay along with the cable force T represent the reactions that both pins and cables exert on the platform • Half of the cables magnitudes is developed at A and half developed at B

  47. 5.2 Free-Body Diagrams Example 5.5 The free-body diagram of each object is drawn. Carefully study each solution and identify what each loading represents.

  48. 5.2 Free-Body Diagrams Solution

  49. 5.2 Free-Body Diagrams Solution

  50. 5.3 Equations of Equilibrium • For equilibrium of a rigid body in 2D, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 • ∑Fx and ∑Fy represent the algebraic sums of the x and y components of all the forces acting on the body • ∑MO represents the algebraic sum of the couple moments and moments of the force components about an axis perpendicular to x-y plane and passing through arbitrary point O, which may lie on or off the body

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