Ishik University / Sulaimani Architecture Department. Structure. ARCH 214 Chapter 5 Equilibrium of a Rigid Body


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1 Ishik University / Sulaimani Architecture Department 1 Structure ARCH 214 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 freebody diagram for a rigid body. To show how to solve rigidbody equilibrium problems using the Equations of equilibrium. 1
2 CHAPTER OUTLINE FreeBody Diagrams Equations of Equilibrium Two and ThreeForce Members Constraints for a Rigid Body FBD is the best method to represent all the known and unknown forces in a system. FreeBody Diagrams 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. 4 2
3 FreeBody Diagrams 5 FreeBody Diagrams 6 3
4 FreeBody Diagrams 7 8 4
5 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 9  fixed support 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. 10 5
6 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. 11 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. 12 6
7 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 13 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 14 7
8 Floor beams of this building are welded together and thus form fixed connections Type 10 connections 15 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. 16 8
9 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. 17 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. 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. 18 9
10 Example 5.1 Draw the freebody diagram of the uniform beam. The beam has a mass of 100kg. 19 Solution FreeBody Diagram 20 10
11 Solution Support at A is a fixed wall. Three forces acting on the beam at A denoted as A x, A y, A z, 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. 21 Example 5.2 Draw the freebody diagram of member AB, which is supported by a roller at A and a pin at B. Explain the significance of each force on the diagram
12 Solution 23 Example 5.3 Draw the freebody diagram of the beam which supports the 80 kg load and is supported by the pin at A and a cable which wraps around the pulley at D. Explain the significance of each force on the diagram
13 Solution 25 Equations of Equilibrium For equilibrium of a rigid body in 2D, F x = 0; F y = 0; M O = 0 F x and F y represent the algebraic sums of the x and y components of all the forces acting on the body. M O represents the algebraic sum of the couple moments and moments of the force components about an axis perpendicular to xy plane and passing through arbitrary point O, which may lie on or off the body
14 Alternative Sets of Equilibrium Equations For coplanar equilibrium problems, F x = 0; F y = 0; M O = 0 can be used Two alternative sets of three independent equilibrium equations may also be used. F a = 0; M A = 0; M B = 0 When applying these equations, it is required that a line passing through points A and B is not perpendicular to the a axis. 27 Alternative Sets of Equilibrium Equations Consider FBD of an arbitrarily shaped body. All the forces on FBD may be replaced by an equivalent resultant force F R = F acting at point A and a resultant moment M RA = M A If M A = 0 is satisfied, M RA =
15 Alternative Sets of Equilibrium Equations A second set of alternative equations is M A = 0; M B = 0; M C = 0 Points A, B and C do not lie on the same line. Consider FBD, if M A = 0, M RA = 0 M A = 0 is satisfied if line of action of F R passes through point B. M C = 0 where C does not lie on line AB. F R = 0 and the body is in equilibrium. 29 Example
16 Solution
17 Example 5.5 The cord supports a force of 500N and wraps over the frictionless pulley. Determine the tension in the cord at C and the horizontal and vertical components at pin A. 33 Solution Equations of Equilibrium M A 0; 500N(0.2m) T(0.2m) 0 T 500N Tension remains constant as cord passes over the pulley (true for any angle at which the cord is directed and for any radius of the pulley
18 Solution F A Ax 250N F A A y y x y x 500sin 30 N 0 500N 500cos N 0; 0; N 0 35 Example 5.6 The link is pinconnected at a and rest a smooth support at B. Compute the horizontal and vertical components of reactions at pin A
19 Solution FBD Reaction N B is perpendicular to the link at B. Horizontal and vertical components of reaction are represented at A. 37 Solution Equations of Equilibrium; M 90N. m 60N(1m ) N N F A x B A 0; 200N x 200sin 30 N 0 Ax 100N F A A y y y 233N 0; 0; 60N 200cos 30 B N 0 (0.75m)
20 Example Solution 40 20
21 41 Example
22 Solution
23 Example Solution 46 23
24 47 Two and ThreeForce Members Simplify some equilibrium problems by recognizing members that are subjected top only 2 or 3 forces. TwoForce Members When a member is subject to no couple moments and forces are applied at only two points on a member, the member is called a twoforce member
25 TwoForce Members Hence, only the force magnitude must be determined or stated. Other examples of the twoforce members held in equilibrium are shown in the figures to the right. 49 ThreeForce Members If a member is subjected to only three forces, it is necessary that the forces be either concurrent or parallel for the member to be in equilibrium. To show the concurrency requirement, consider a body with any two of the three forces acting on it, to have line of actions that intersect at point O. To satisfy moment equilibrium about O, the third force must also pass through O, which then makes the force concurrent. If two of the three forces parallel, the point of currency O, is considered at infinity. Third force must parallel to the other two forces to insect at this point
26 Example 5.10 Determine the normal reactions at A and B. 51 Solution 52 26
27 Example 5.11 Determine the tension in the cord and the horizontal and vertical components of reaction at support A of the beam. 53 Solution 54 27
28 Example 5.12 Determine the horizontal and vertical components of reaction at C and the tension in the cable AB for the truss. 55 Solution 56 28
29 Homework 5.1 Determine the horizontal and vertical components of reaction at the pin A and the force in the cable BC. Neglect the thickness of the members. 57 Homework 5.2 Determine the horizontal and vertical components or reaction at the pin A and the reaction on the beam fit C
30 Homework 5.3 Determine the normal reaction at the roller A and horizontal and vertical components at pin B for equilibrium of the member
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