Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Chapter Objectives Develop the equations of equilibrium for a rigid body Concept of the free-body diagram for a rigid body Solve rigid-body equilibrium problems using the equations of equilibrium Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Chapter Outline Conditions for Rigid Equilibrium Free-Body Diagrams Equations of Equilibrium Two and Three-Force Members Free Body Diagrams Constraints and Statical Determinacy Copyright © 2010 Pearson Education South Asia Pte Ltd

5.1 Conditions for Rigid-Body Equilibrium The equilibrium of a body is expressed as Consider summing moments about some other point, such as point A, we require Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams Support Reactions If a 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. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams Internal Forces External and internal forces can act on a rigid body 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 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams Weight and Center of Gravity Each particle has a specified weight 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 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams 3. Identify Each Loading and Give Dimensions Indicate dimensions for calculation of forces Known forces and couple moments should be properly labeled with their magnitudes and directions Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 5.1 Draw the free-body diagram of the uniform beam. The beam has a mass of 100kg. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Free-Body Diagram Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Free-Body Diagram 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 Copyright © 2010 Pearson Education South Asia Pte Ltd

5.3 Equations of Equilibrium Please refer to the Companion CD for the animation: Equilibrium of a Free Body 5.3 Equations of Equilibrium For equilibrium of a rigid body in 2D, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 ∑Fx and ∑Fy represent sums of x and y components of all the forces ∑MO represents the sum of the couple moments and moments of the force components Copyright © 2010 Pearson Education South Asia Pte Ltd

5.3 Equations of Equilibrium Please refer to the Companion CD for the animation: Equilibrium of a Free Body 5.3 Equations of Equilibrium Alternative Sets of Equilibrium Equations For coplanar equilibrium problems, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 2 alternative sets of 3 independent equilibrium equations, ∑Fa = 0; ∑MA = 0; ∑MB = 0 Copyright © 2010 Pearson Education South Asia Pte Ltd

5.3 Equations of Equilibrium Please refer to the Companion CD for the animation: Equilibrium of a Free Body 5.3 Equations of Equilibrium Procedure for Analysis Free-Body Diagram Force or couple moment having an unknown magnitude but known line of action can be assumed Indicate the dimensions of the body necessary for computing the moments of forces Copyright © 2010 Pearson Education South Asia Pte Ltd

5.3 Equations of Equilibrium Please refer to the Companion CD for the animation: Equilibrium of a Free Body 5.3 Equations of Equilibrium Procedure for Analysis Equations of Equilibrium Apply ∑MO = 0 about a point O Unknowns moments of are zero about O and a direct solution the third unknown can be obtained Orient the x and y axes along the lines that will provide the simplest resolution of the forces into their x and y components Negative result scalar is opposite to that was assumed on the FBD Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 5.5 Determine the horizontal and vertical components of reaction for the beam loaded. Neglect the weight of the beam in the calculations. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Free Body Diagrams 600N represented by x and y components 200N force acts on the beam at B Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Equations of Equilibrium Copyright © 2010 Pearson Education South Asia Pte Ltd

5.4 Two- and Three-Force Members Two-Force Members When forces are applied at only two points on a member, the member is called a two-force member Any two-force member to be in equilibrium, the two forces must have the same magnitude, act in opposite directions and have the same line of action, directed along the two joining the two points where these forces act. Copyright © 2010 Pearson Education South Asia Pte Ltd

5.4 Two- and Three-Force Members When subjected to three forces, the forces are concurrent or parallel Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 5.13 The lever ABC is pin-supported at A and connected to a short link BD. If the weight of the members are negligible, determine the force of the pin on the lever at A. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Free Body Diagrams BD is a two-force member Lever ABC is a three-force member Equations of Equilibrium Solving, Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 5.5 Free-Body Diagrams Support Reactions As in the two-dimensional case: A force is developed by a support A couple moment is developed when rotation of the attached member is prevented The force’s orientation is defined by the coordinate angles α, β and γ Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 5.5 Free-Body Diagrams Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 5.5 Free-Body Diagrams Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 5.14 Several examples of objects along with their associated free-body diagrams are shown. In all cases, the x, y and z axes are established and the unknown reaction components are indicated in the positive sense. The weight of the objects is neglected. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Copyright © 2010 Pearson Education South Asia Pte Ltd

5.6 Equations of Equilibrium Vector Equations of Equilibrium For two conditions for equilibrium of a rigid body in vector form, ∑F = 0 ∑MO = 0 Scalar Equations of Equilibrium If all external forces and couple moments are expressed in Cartesian vector form ∑F = ∑Fxi + ∑Fyj + ∑Fzk = 0 ∑MO = ∑Mxi + ∑Myj + ∑Mzk = 0 Copyright © 2010 Pearson Education South Asia Pte Ltd

5.7 Constraints for a Rigid Body Redundant Constraints More support than needed for equilibrium Statically indeterminate: more unknown loadings than equations of equilibrium Copyright © 2010 Pearson Education South Asia Pte Ltd

5.7 Constraints for a Rigid Body Improper Constraints Instability caused by the improper constraining by the supports When all reactive forces are concurrent at this point, the body is improperly constrained Copyright © 2010 Pearson Education South Asia Pte Ltd

5.7 Constraints for a Rigid Body Procedure for Analysis Free Body Diagram Draw an outlined shape of the body Show all the forces and couple moments acting on the body Show all the unknown components having a positive sense Indicate the dimensions of the body necessary for computing the moments of forces Copyright © 2010 Pearson Education South Asia Pte Ltd

5.7 Constraints for a Rigid Body Procedure for Analysis Equations of Equilibrium Apply the six scalar equations of equilibrium or vector equations Any set of non-orthogonal axes may be chosen for this purpose Choose the direction of an axis for moment summation such that it insects the lines of action of as many unknown forces as possible Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 5.15 The homogenous plate has a mass of 100kg and is subjected to a force and couple moment along its edges. If it is supported in the horizontal plane by means of a roller at A, a ball and socket joint at B, and a cord at C, determine the components of reactions at the supports. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Free Body Diagrams Five unknown reactions acting on the plate Each reaction assumed to act in a positive coordinate direction Equations of Equilibrium Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Equations of Equilibrium Components of force at B can be eliminated if x’, y’ and z’ axes are used Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Solving, Az = 790N Bz = -217N TC = 707N The negative sign indicates Bz acts downward The plate is partially constrained as the supports cannot prevent it from turning about the z axis if a force is applied in the x-y plane Copyright © 2010 Pearson Education South Asia Pte Ltd