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## Engineering Mechanics: Statics

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**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 • equations of equilibrium.**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**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**5.1 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**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)**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**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**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**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**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**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**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**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**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**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**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**5.2 Free-Body Diagrams**• Floor beams of this building are welded together and thus form fixed connections • Type 10 connections**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**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**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**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**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**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**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**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**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**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**5.2 Free-Body Diagrams**Example 5.1 Draw the free-body diagram of the uniform beam. The beam has a mass of 100kg.**5.2 Free-Body Diagrams**Solution Free-Body Diagram**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**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.**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,**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**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**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.**5.2 Free-Body Diagrams**Solution • For idealized models, • Free-Body Diagram of pipe A**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**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**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.**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**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**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**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.**5.2 Free-Body Diagrams**Solution**5.2 Free-Body Diagrams**Solution**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