Chapter Objectives To develop the equations of equilibrium for a rigid body. To introduce the concept of the free-body diagram for a rigid body. To show how to solve rigid-body equilibrium problems using the equations of equilibrium.
In-Class Activities 1. Reading Quiz 2. Applications 3. Conditions for Rigid Equilibrium 4. Free-Body Diagrams 5. Equations of Equilibrium 6. Two and Three-Force Members 7. Free-Body Diagrams 8. Equations of Equilibrium 9. Concept Quiz
READING QUIZ 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
READING QUIZ (cont) 2) Internal forces are shown on the free body diagram of a whole body. a) Always b) Often c) Rarely d) Never
READING QUIZ (cont) 3) Internal forces are not shown on a free-body diagram because the internal forces are. a) Equal to zero b) Equal and opposite and they do not affect the calculations c) Negligibly small d) Not important
READING QUIZ (cont) 4) If a support prevents rotation of a body about an axis, then the support exerts a on the body about that axis. a) Couple moment b) Force c) Both A and B. d) None of the above.
READING QUIZ (cont) 5) When doing a 3-D problem analysis, you have scalar equations of equilibrium. a) 3 b) 4 c) 5 d) 6
APPLICATIONS
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 M 0 R R O
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.
FREE-BODY DIAGRAMS (cont)
FREE-BODY DIAGRAMS (cont)
FREE-BODY DIAGRAMS (cont)
FREE-BODY DIAGRAMS (cont) Internal Forces External and internal forces can act on a rigid body For FBD, internal forces that act between particles contained within the boundary of the FBD are not represented Particles outside this boundary exert external forces on the system
FREE-BODY DIAGRAMS (cont) 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
FREE-BODY DIAGRAMS (cont) Procedure for Drawing a FBD: 1. Draw Outlined Shape Imagine body to be isolated or cut free from its constraints Draw outlined shape 2. Show All Forces and Couple Moments Identify all external forces and couple moments that act on the body 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
EXAMPLE 1 Draw the free-body diagram of the uniform beam. The beam has a mass of 100 kg.
EXAMPLE 1 (cont) Solution Free-Body Diagram: Support at A is a fixed wall Three reactions acting on the beam at A denoted as Ax, Ay, and M A drawn in an arbitrary direction Unknown magnitudes of these vectors Assume sense of these vectors
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 sums of x and y components of all the forces M O represents the sum of the couple moments and moments of the force components
EQUATIONS OF EQUILIBRIUM (cont) 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 x = 0; M A = 0; M B = 0 (Line AB cannot be // to y-axis) or M A = 0; M B = 0; M C = 0 (A, B, and C cannot be on the same line)
STATICALLY INDETERMINATE REACTIONS Beer and Johnston
STATICALLY INDETERMINATE REACTIONS Beer and Johnston
EQUATIONS OF EQUILIBRIUM (cont) 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
EQUATIONS OF EQUILIBRIUM (cont) Please refer to the website for the animation: Equilibrium of a Free Body Procedure for Analysis Equations of Equilibrium Apply M O = 0 about a point O Unknowns moments of force 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
EXAMPLE 2 Determine the horizontal and vertical components of reaction for the beam loaded (roller at A and hinge at B). Neglect the weight of the beam in the calculations.
EXAMPLE 2 (cont) Solution Free Body Diagram 600 N represented by x and y components 200 N force acts on the beam at B
EXAMPLE 2 (cont) Solution Equations of Equilibrium FX 0; 600cos45 N B 0 B 424N M 100N(2m) (600sin 45 A F 319N 600sin 45 B y y B 0; 319N y 405N 0; N)(5m) (600cos 45 N 100N 200N B x y 0 x N)(0.2m) A y (7m) 0
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 Only force magnitude must be determined
TWO- AND THREE-FORCE MEMBERS (cont) Three-Force Members When subjected to three forces, the forces are concurrent or parallel
EXAMPLE 3 The lever ABC is pin-supported at A and connected to a short link BD. If the weight of the members is negligible, determine the force of the pin on the lever at A.
EXAMPLE 3 (cont) Solution Free Body Diagrams BD is a two-force member Lever ABC is a three-force member Equations of Equilibrium tan 1 F F y x 0.7 0.4 0; 0; 60.3 F F A A cos60.3 sin 60.3 F cos 45 F sin 45 400N 0 0 Solving, F A 1.07kN F 1. 32kN
A fixed crane has a mass of 1000 kg and is used to lift a 2400-kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the components of the reactions at A and B. Beer and Johnston
Beer and Johnston
A loading car is at rest on a track forming an angle of 25 with the vertical. The gross weight of the car and its load is 5500 lb, and it is applied at a point 30 in. from the track, halfway between the two axles. The car is held by a cable attached 24 in. from the track. Determine the tension in the cable and the reaction at each pair of wheels. Beer and Johnston
Beer and Johnston
EQUILIBRIUM IN 3-D 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
FREE-BODY DIAGRAMS (cont)
FREE-BODY DIAGRAMS (cont)
FREE-BODY DIAGRAMS (cont)
Couple Moments at the Supports It is a mistake to support a door using a single hinge. The hinge must develop a couple moment M to support the moment of W Two properly aligned hinges resist the moment of W by the couple of F x and -F x Wd= F x d In this case, no couple moments are produced by the hinges
EXAMPLE 4 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.
EXAMPLE 4 (cont) Solution
EXAMPLE 4 (cont) Solution
EXAMPLE 4 (cont) Solution
EQUATIONS OF EQUILIBRIUM Vector Equations of Equilibrium For two conditions for equilibrium of a rigid body in vector form, F = 0 M O = 0 Scalar Equations of Equilibrium If all external forces and couple moments are expressed in Cartesian vector form F = F x i + F y j + F z k = 0 M O = M x i + M y j + M z k = 0
Example: A 20-kg ladder used to reach high shelves in a storeroom is supported by two flanged wheels A and B mounted on a rail and by an unflanged wheel C resting against a rail fixed to the wall. An 80-kg man stands on the ladder and leans to the right. The line of action of the combined weight W of the man and ladder intersects the floor at point D. Determine the reactions at A, B, and C. Beer and Johnston
Beer and Johnston
Example: A uniform pipe cover of radius r=240 mm and mass 30 kg is held in a horizontal position by the cable CD. Assuming that the bearing at B does not exert any axial thrust, determine the tension in the cable and the reactions at A and B. Beer and Johnston
Beer and Johnston
MSE 221 FINAL-FALL 2014 (Only few correct answers!!!) Copyright 2011 Peason Education South Asia Pte Ltd
CONCEPT QUIZ 1) The beam and the cable (with a frictionless pulley at D) support an 80 kg load at C. In a FBD of only the beam, there are how many unknowns? a) 2 forces and 1 couple moment b) 3 forces and 1 couple moment c) 3 forces d) 4 forces
CONCEPT QUIZ (cont) 2) The three scalar equations FX = FY = MO = 0, are equations of equilibrium in two dimensions. a) Incorrect b) The only correct c) The most commonly used d) Not sufficient
CONCEPT QUIZ (cont) 3) A rigid body is subjected to forces. This body can be considered as a member. a) Single-force b) Two-force c) Three-force d) Six-force
CONCEPT QUIZ (cont) 4) For this beam, how many support reactions are there and is the problem statically determinate? a) (2, Yes) b) (2, No) F F F F c) (3, Yes) d) (3, No)
CONCEPT QUIZ (cont) 5) The beam AB is loaded as shown: a) how many support reactions are there on the beam, b) is this problem statically determinate, and c) is the structure stable? a) (4, Yes, No) b) (4, No, Yes) A Fixed support F B c) (5, Yes, No) d) (5, No, Yes)
CONCEPT QUIZ (cont) 6) Which equation of equilibrium allows you to determine F B right away? a) FX = 0 100 lb b) FY = 0 A X A B c) MA = 0 d) Any one of the above. A Y F B
CONCEPT QUIZ (cont) 7) A beam is supported by a pin joint and a roller. How many support reactions are there and is the structure stable for all types of loadings? a) (3, Yes) b) (3, No) c) (4, Yes) d) (4, No)
CONCEPT QUIZ (cont) 8) The rod AB is supported using two cables at B and a ball-and-socket joint at A. How many unknown support reactions exist in this problem? a) 5 force and 1 moment reaction b) 5 force reactions c) 3 force and 3 moment reactions d) 4 force and 2 moment reactions
CONCEPT QUIZ (cont) 9) If an additional couple moment in the vertical direction is applied to rod AB at point C, then what will happen to the rod? a) The rod remains in equilibrium as the cables provide the necessary support reactions. b) The rod remains in equilibrium as the ball-and-socket joint will provide the necessary resistive reactions. c) The rod becomes unstable as the cables cannot support compressive forces. d) The rod becomes unstable since a moment about AB cannot be restricted.
CONCEPT QUIZ (cont) 10) A plate is supported by a ball-and-socket joint at A, a roller joint at B, and a cable at C. How many unknown support reactions are there in this problem? a) 4 forces and 2 moments b) 6 forces c) 5 forces d) 4 forces and 1 moment
CONCEPT QUIZ (cont) 11) What will be the easiest way to determine the force reaction BZ? a) Scalar equation FZ = 0 b) Vector equation MA = 0 c) Scalar equation MZ = 0 d) Scalar equation MY = 0