PAT 101 FUNDAMENTAL OF ENGINEERING MECHANICS EQUILIBREQUILIBRIUM OF A RIGID BODY IUM OF A RIGID BODY

Transcription

1 PAT 101 FUNDAMENTAL OF ENGINEERING MECHANICS EQUILIBREQUILIBRIUM OF A RIGID BODY IUM OF A RIGID BODY MARDHIAH FARHANA BINT OMAR Week 5-6

2 EQUILIBRIUM OF A RIGID BODY Conditions for Rigid Equilibrium Free-Body Diagrams Equations of Equilibrium

3 CONDITIONS FOR RIGID-BODY EQUILIBRIUM The equilibrium of a body is expressed as F R = F = 0 (M R ) O = M O = 0 Consider summing moments about some other point, such as point A, we require M A = r F R +(M R ) O = 0

4 Support Reactions FREE BODY DIAGRAMS 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.

5 FREE-BODY DIAGRAM

6 One unk nown. The reaction is a force which acts perpendicular to the surface at the poi nt of con tact. One unknown.the reaction is a force which acts perpendicular to the slot. One un known. The reaction is a force which acts perpendicular to the rod.

7 (8) smooth pin or hinge or Two un know ns. TI1e reactions a re two components of force, or the magnit ude and direction </>of the resu ltan t force. Note that cf> and 8 are not necessarily equal [usually not. unless the rod shown is a link as in (2)]. (9) member fixed connected lo collar on smooth rod (10) fixed support ( )'!- M F or Two unknowns.the reactions are the cou ple momen t and the force which acts perpendicular to the rod. Three unknowns. The reactions are the couple momen t and the two force components. or the couple momen t and the magnitude and direction </> of the resultant force.

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11 Internal Forces FREE BODY DIAGRAMS 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 Ed uc a ti o n

12 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

13 FREE-BODY DIAGRAMS 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.

14 FREE-BODY DIAGRAMS Idealized model Free-body diagram 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 A x, A y, M A. Indicate any necessary dimensions.

15 EXAMPLE 1 Draw the free-body diagram of the uniform beam. The beam has a mass of 100kg.

16 Free-Body Diagram SOLUTION

17 SOLUTION Free-Body Diagram 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

18 EQUATIONS OF EQUILIBRIUM Alternative Sets of Equilibrium Equations For coplanar equilibrium problems, F x = 0; F y = 0; M O = 0 2 alternative sets of 3 independent equilibrium equations, F a = 0; M A = 0; M B = 0

19 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

20 EQUATIONS OF EQUILIBRIUM Procedure for Analysis Equations of Equilibrium Apply M O = 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

21 EXAMPLE 1 Determine the horizontal and vertical components of reaction for the beam loaded. Neglect the weight of the beam in the calculations.

22 SOLUTION + Fx = 0; 600cos 45 N B x = 0 B x = 424N M B = 0; A y = 319N + F y = 0; Equations of Equilibrium 100N(2m) + (600sin 45 N)(5m) (600cos 45 N)(0.2m) A (7m) = 0 319N 600sin 45 N 100N 200N + B B y = 405N y = 0 y

23 EXAMPLE 2

24 EXERCISE 3