Section 5.1 Conditions for Rigid Body Equilibrium Section 5.2 Free-Body Diagrams

Engineering Mechanics: Statics Fifteenth Edition Chapter 5 Equilibrium of a Rigid Body: Section 5.1 Conditions for RigidBody Equilibrium Section 5.2 Free-Body Diagrams Section 5.3 Equations of Equilibrium Section 5.4 Two- and ThreeForce Members Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 5.1: Conditions for Rigid-Body Equilibrium For a rigid body to be in equilibrium, the net force as well as the net moment about any arbitrary point O must be equal to zero. Forces on a particle  F = 0 ( no translation ) and  MO = 0 ( no rotation ) Forces on a rigid body Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 5.2: Free-Body Diagrams (1 of 2) Idealized model Free-body diagram (FBD) 1. Draw an outlined shape. Imagine the body to be isolated or cut “free” from its constraints and draw its outlined shape. 2. Show all the external forces and couple moments. These typically include: a) applied loads, b) support reactions, and, c) the weight of the body. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 5.2: Free-Body Diagrams (2 of 2) 3. Label loads and dimensions on the FBD: All known forces and couple moments should be labeled with their magnitudes and directions. For the unknown forces and couple moments, use letters like Ax, Ay, MA. Indicate any necessary dimensions. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Support Reactions in 2-D A few example sets of diagrams s are shown above. Other support reactions are given in your textbook (Table 5-1). As a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction. Similarly, if rotation is prevented, a couple moment is exerted on the body in the opposite direction. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅰ Given: The operator applies a vertical force to the pedal so that the spring is stretched 1.5 in. and the force in the short link at B is 20 lb. Draw: An idealized model and free-body diagram of the foot pedal. The idealized model The free-body diagram Copyright © 2022 Pearson Education, Inc. All Rights Reserved Problem Solving Ⅰ (1 of 2) Given: Draw: A FBD of the smooth pipe which rests against the opening at the points of contact A, B, and C. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Problem Solving Ⅰ (2 of 2) The idealized model The free body diagram Copyright © 2022 Pearson Education, Inc. All Rights Reserved Engineering Mechanics: Statics Fifteenth Edition Chapter 5 Equilibrium of a Rigid Body Section 5.3 Equations of Equilibrium Section 5.4 Two- and ThreeForce Members Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 5.3: Equations of Equilibrium A body is subjected to a system of forces that lie in the x-y plane. When in equilibrium, the net force and net moment acting on the body are zero (as discussed earlier in Section 5.1). This 2-D condition can be represented by the three scalar equations: F x =0 F y =0 M O =0 where point O is any arbitrary point. Please note that these equations are the ones most commonly used for solving 2-D equilibrium problems. There are two other sets of equilibrium equations that are rarely used. For your reference, they are described in the textbook. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 5.4: Two-Force Members and Three Force-Members The solution to some equilibrium problems can be simplified if we recognize members that are subjected to forces at only two points (e.g., at points A and B in the figure below). If we apply the equations of equilibrium to such a member, we can quickly determine that the resultant forces at A and B must be equal in magnitude and act in the opposite directions along the line joining points A and B. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Steps for Solving 2-D Equilibrium Problems 1. If not given, establish a suitable x - y coordinate system. 2. Draw a free-body diagram (FBD) of the object under analysis. 3. Apply the three equations of equilibrium (E-of-E) to solve for the unknowns. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Important Notes 1. If there are more unknowns than the number of independent equations, then we have a statically indeterminate situation. We cannot solve these problems using just statics. 2. The order in which we apply equations may affect the simplicity of the solution. For example, if we have two unknown vertical forces and one unknown horizontal force, then solving  FX = 0 first allows us to find the horizontal unknown quickly. 3. If the answer for an unknown comes out as negative number, then the sense (direction) of the unknown force is opposite to that assumed when starting the problem. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example (1 of 2) Given: The 4kN load at B of the beam is supported by pins at A and C. Find: The support reactions at A and C. Plan: 1. Put the x and y-axes in the horizontal and vertical directions, respectively. 2. Determine if there are any two-force members. 3. Draw a complete FBD of the boom. 4. Apply the E-of-E to solve for the unknowns. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Examples of Two-Force Members Note: Upon recognizing CD as a two-force member, the number of unknowns at C is reduced from two to one. Now, using E-o-f E, we get, → +  FX = A X + 11.31 cos 45° = 0; A X = – 8.00 kN  +  FY = A Y + 11.31 sin 45° – 4 = 0; A Y = – 4.00 kN Note that the negative signs means that the reactions have the opposite directions to that assumed (as originally shown on FBD). Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving (1 of 3) Given: The beam is supported by the roller at A and a pin at B. Find: The reactions at points A and B on the beam. Plan: a. Establish the x–y axis system. b. Draw a complete FBD of the beam. c. Apply the E-of-E to solve for the unknowns. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving (2 of 3) Note that the distributed load has been reduced to a single force. First, write a moment equation about point B. Why point B? Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving (3 of 3) Recall NA = 3.713 = 3.71 kN Now write the  FX =  FY = 0 equations. → +  FX = 3.713 sin30° – B x = 0  +  FY = 3.713 cos30° – 12 + B y = 0 ↑ ← Solving these two equations, we get Bx = 1.86 kN By = 8.78 kN Copyright © 2022 Pearson Education, Inc. All Rights Reserved Determine internal moment at A and the reaction forces on the bent rod which is supported by a smooth surface at B and by a collar at A, which is fixed to the rod and is free to slide over the fixed inclined rod. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Attention Quiz (1 of 2) 1. Internal forces are not shown on a free-body diagram because the internal forces are_______. (Choose the most appropriate answer.) A. Equal to zero B. Equal and opposite and they do not affect the calculations C. Negligibly small D. Not important Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (1 of 2) 1. If a support prevents translation of a body, then the support exerts a _______ on the body. A. Couple moment B. Force C. Both A and B. D. None of the above Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (2 of 2) 2. Internal forces are _________ shown on the free body diagram of a whole body. A. Always B. Often C. Rarely D. Never Copyright © 2022 Pearson Education, Inc. All Rights Reserved