Chapter 5: Equilibrium of a Rigid Body Chapter Objectives To develop the equations of equilibrium for a rigid body. To introduce the concept of a free-body diagram for a rigid body. To show how to solve rigid body equilibrium problems using the equations of equilibrium. In this chapter, we will describe the various types of supports that are used with rigid body equilibrium problems. We will then show how free-body diagrams and equilibrium equations are used to determine unknown forces and couples acting on a rigid body. 5.1 Conditions for Rigid-Body Equilibrium When the force and couple are both equal to zero, the external forces form a system equivalent to zero and the rigid body is said to be in equilibrium. Hence two equations of equilibrium for a rigid body can be summarized as follows. F = 0 M = 0 Necessary and sufficient conditions for the equilibrium of a rigid body. We may express the necessary and sufficient conditions for the equilibrium of a rigid body in the following scalar form. Fx = 0 Fy = 0 Fz = 0 Mx = 0 My = 0 Mz = 0 Equilibrium in Two Dimensions First, we will consider the case in which the force system acting on a rigid body is in a single plane. This type of force system is referred to as a two-dimensional or coplanar force system. Forces acting on the body are in the same plane. Couple moments acting on the body are perpendicular to this plane. 5.2 Free-Body Diagrams The best way to account for all of the known and unknown external forces acting on a rigid body is to draw the free-body diagram. 5.1
Support Reactions Reactions are the forces through which the ground and other bodies oppose a possible motion of the free body. Reactions are exerted at points where the free body is supported or connected to other bodies. It is important to understand how to replace certain supports with the appropriate constraining forces. In general, if a support prevents translation, then a force is developed on the body in that direction. Likewise, if rotation is prevented, a couple moment is exerted on the rigid body. Reactions exerted on two-dimensional structure may be divided into three groups, corresponding to three types of supports or connections. 1. Reaction equivalent to a force with a known line of action (1 unknown). This reaction prevents translation of the free body in one direction, but it cannot prevent the body from rotating about the connection. For each of these supports there is only one unknown involved, that is, the magnitude of the reaction. The line of action is known. The most common examples are as follows. a. Roller support A roller support prevents vertical translation. b. Link A link prevents translation along the axis of the link. 5.2
c. Cables (weightless) 2. Reaction equivalent to a force with an unknown line of action (2 unknowns). This reaction prevents translation of the free body in all directions, but it cannot prevent the body from rotating about the connection. The magnitude and direction of the reactive force is unknown. Normally we work with the components, thus fixing the directions but not knowing the magnitude of the components. The most common examples are as follows. a. Pin A pin prevents vertical and horizontal translation. b. Ball and socket joint (2-D) A ball and socket joint prevents vertical and horizontal translation. 3. Reaction equivalent to a force and a couple (3 unknowns). This reaction is caused by fixed supports which oppose any motion of the free body and thus constrain it completely, preventing both translation and rotation. Reactions of this group involve three unknowns, consisting usually of the two components of the force and the moment of the couple. 5.3
The most common example is as follows. a. Fixed end The fixed end prevents vertical and horizontal translation, and prevents rotation. When the sense of an unknown force or couple is not clearly apparent, no attempt should be made to determine the correct direction. Instead, the sense of the force or couple should be arbitrarily assumed. The sign of the answer obtained will indicate whether the assumption is correct or not. Internal Forces If a free-body diagram for the rigid body is drawn, only external forces are shown. Internal forces within the rigid body are not represented on a free-body diagram. Internal forces cancel each other, and as a result, do not create an external effect on the rigid body Weight and the Center of Gravity When the weight of a body must be considered, a force resultant representing the weight is used with its point of application acting through the center of gravity. Idealized Models In order to perform a correct force analysis of any object, it is important to consider a corresponding analytical or idealized model that gives results that approximate as closely as possible the actual situation. Example, consider the underground pump station: Is reinforced concrete slab acting as simply supported, or fixed-fixed to concrete block walls? 5.4
Procedure for Drawing a Free-Body Diagram 1. Draw the isolated body. 2. Add all the applied forces (including the weight of the body). 3. Show the reactive forces ( constraining forces ). On a free body diagram, include the dimensions. The dimensions may be needed to compute moments. 5.5
Example Free Body Diagram Given: The loaded W18 x 40 beam shown. Find: Draw the free body diagram and solve for the reactions. Draw the free body diagram. Steps: 1. Draw the isolated body. 2. Add the applied forces. 3. Show the reactive forces. a. Does the support at A prevent... 1) Horizontal translation? 2) Vertical translation? 3) Rotation? b. Does the support at B prevent... 1) Horizontal translation? 2) Vertical translation? 3) Rotation? Solve for the reactions. Fx = 0 = (4/5) 75 + Ax Ax = - 60 kips MA = 0 = 10 By (3/5) 75 (4) 10 By = 180 By = + 18.0 kips MB = 0 = - 10 Ay + (3/5) 75 (6) 10 Ay = 270 Ay = + 27.0 kips Ax = 60 kips By = 18.0 kips Ay = 27.0 kips 5.6
5.3 Equations of Equilibrium The conditions of equilibrium in two dimensions are as follows. Fx = 0 Fy = 0 Mz = 0 These three equations (called the equations of statics ) allow solution for no more than three unknowns. Alternative Sets of Equilibrium Equations These three equations may be replaced with another set of equations. An alternate set of equilibrium equations may be as follows. Fx = 0 MA = 0 MB = 0 or MA = 0 MB = 0 MC = 0 5.7
Example Given: Beam loaded as shown. Find: Range of values for P for a safe beam: RA and RB 25 kips (i.e. compression in the columns at A and B) Fx = 0 = Ax Ax = 0 kips Let RA = 0 kips: MB = 0 = 6 P - 6 (2) 6 (4) P = 6.0 kips Check RB Fy = 0 = - 6 6 6 + RB RB = 18.0 kips < 25.0 kips OK Let RA = 25 kips: MB = 0 = - 25 (9) + 6 P - 6 (2) 6 (4) 6 P = 225 + 12 + 24 = 261 P = 43.5 kips Check RB Fy = 0 = 25-43.5 6 6 + RB RB = 30.5 kips > 25.0 kips NG Let RB = 25 kips: MA = 0 = - 3 P + 9 (25) 6 (11) 6 (13) 3 P = 225 66 78 = 81 P = 27.0 kips Check RA Fy = 0 = - 27 + 25 6 6 + RA RA = 14.0 kips < 25.0 kips OK Therefore, 6.0 kips P 27.0 kips Answer 5.8
5.4 Two- and Three-Force Members The solution to some equilibrium problems can be simplified if one is able to recognize members that are subjected to only two or three forces. Two-Force Members If a two-force body is in equilibrium, the two forces must have the same magnitude, same line of action, and opposite sense. For the corner plate to be in equilibrium the following equilibrium equations must be satisfied. MA = 0 = F2r d : either F2r = 0 or d = 0, and MB = 0 = F1r d : either F1r = 0 or d = 0 F1r and F2r must be zero; d = 0 is too restrictive (the rigid body would revert to a particle). Characteristics of a two-force member: 1. Coplanar or non-coplanar (any shape). 2. Two forces same magnitude, same line of action, opposite direction. 3. Direction of two forces is collinear with line of action connecting points of application. 4. Points need not be at the end of the member. 5. No couple allowed on member. A truss member is an example of a two-force member. 5.9
Three-Force Members If a three-force body is in equilibrium, the lines of action of the three forces must be either concurrent or parallel. For the rigid body to be in equilibrium the following equilibrium equation must be satisfied. MA = 0 = + F3 d: either F3 = 0 or d = 0 F3 0 (otherwise the system would revert to a two-force system). Thus d = 0. For the rigid body to be in equilibrium, the three forces must be either concurrent (i.e. d = 0), or parallel. Concurrent forces Parallel forces 5.10
Equilibrium in Three Dimensions 5.5 Free-Body Diagrams A simple way of determining the type of reaction corresponding to a given support or connection and the number of unknowns involved is to find which of the six fundamental motions (translations in the x, y, and z directions and rotations about the x, y, and z axes) are allowed and which motions are prevented. Support Reactions Reactions exerted on three-dimensional structures may be divided into three groups, corresponding to three types of supports or connections. 1. Reaction equivalent to a single force and a known line of action (translation prevented in one direction one unknown). Examples: Ball (or roller supports), frictionless surface, cables 2. Reaction equivalent to a single force and an unknown line of action (translation prevented in 3 directions 3 unknowns). Example: Ball and socket joint (3D) 3. Reaction equivalent to a force and a couple with unknown lines of action (translation and rotation prevented 6 unknowns) Example: Fixed support 5.11
5.6 Equations of Equilibrium Vector Equations of Equilibrium The two conditions of equilibrium of a rigid body are as follows. F = 0 M = 0 Necessary and sufficient conditions for the equilibrium of a rigid body. Scalar Equations of Equilibrium Six scalar equations are required to express the conditions for the equilibrium of a rigid body in the general three-dimensional case. Fx = 0 Fy = 0 Fz = 0 Mx = 0 My = 0 Mz = 0 These equations may be solved for no more than six unknowns. 5.7 Constraints and Statical Determinacy A problem is said to be statically determinate when a rigid body is completely constrained by its supports and the reactions can be determined using the equations of statics. Two-Dimensional Analysis In a two-dimensional analysis, only three independent equilibrium equations are available. When there are more than three unknown reactions, not all of the unknown reactions can be determined using the equations of equilibrium. - This type of problem is said to be statically indeterminate. Three-Dimensional Analysis In a three-dimensional analysis, only six independent equilibrium equations are available. When there are more than six unknown reactions, not all of the unknown reactions can be determined using the equations of equilibrium. - This type of problem is said to be statically indeterminate. Redundant Constraints Determinate 5.12
Determinate Determinate Statically indeterminate (with degree of redundancy) SI 1 SI 3 SI 1 Mobile structure (improperly constrained) Improper Constraints Partially constrained The supports provided are not enough to keep the body (structure) from moving. There are fewer unknowns than equations and one of the equilibrium equations will not be satisfied. 5.13
If the reactions involve less than three unknowns (2-D) or less than six unknowns (3-D), there are more equations than there are unknowns, and some of the equations of equilibrium cannot be satisfied under a general loading condition. The rigid body is said to be only partially constrained and should be avoided since an unstable condition may result. Improperly constrained There is a sufficient number of constraints, but these constraints are not properly arranged and the body (structure) is free to move. A rigid body is improperly constrained when the supports, even though they provide a sufficient number of reactions, are arranged in such a way that at least one of the equilibrium equations is not satisfied. - Even with three or more unknowns (2-D), or six or more unknowns (3-D), it is possible that some of the equations of equilibrium will not be satisfied. - This may occur when the supports are such that the reactions are forces that are parallel. The rigid body is said to be improperly constrained and should be avoided since an unstable condition may result. 5.14