Engineering Mechanics Statics

Transcription

1 Mechanical Systems Engineering _ 2016 Engineering Mechanics Statics 7. Equilibrium of a Rigid Body Dr. Rami Zakaria

2 Conditions for Rigid-Body Equilibrium Forces on a particle Forces on a rigid body The difference between the forces on a particle and the forces on a rigid-body is that the forces may cause rotation of the body (because of the moments created by the forces). We say that a rigid body is in equilibrium when the net force and the net moment about any arbitrary point O is equal to zero. F = 0 (no translation) and M O = 0 (no rotation) Note: We will discuss how to find the centre of gravity in another chapter (the location where the weight force is applied).

3 Steps of Solving Rigid Body Equilibrium Problems 1. Create an idealized model (above right). 2. Draw a free-body diagram (FBD) showing all the external (active and reactive) forces. 3. Apply the equations of equilibrium (E-O-E) to solve for any unknowns. Note.1: In this lecture we will only discuss 2D force systems. Note2: Internal forces always cancel each others and they don t create external effect. So in a free body diagram we only draw the external forces.

4 Support Reaction Prevents translation in one direction Prevents translation in two directions Prevent translation in two directions and also prevents rotation A few example sets of diagrams are shown above. See the other support reactions 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 a support prevents rotation, then a couple moment is exerted on the body in the opposite direction.

5 How to Draw a Free Body Diagram? Idealized model Free-body diagram (FBD) 1. Draw an outlined shape. (imagine the body is isolated) 2. Show all the external forces and couple moments. These usually include: a) applied loads/forces b) support reactions, and c) the weight of the body. 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, etc.. Indicate any necessary dimensions.

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7 W is the weight (effect of gravity on the roll). N A, N B are the surface reactions on the roll.

8 Q. If a support only prevents translation of a body, then the support exerts a on the body. A) Couple moment B) Force C) Both force and couple moment. D) None of the above N A is the reaction of the surface. B (B X, B Y ) is the pin force

9 Q. Internal forces are shown on the free body diagram of a whole body. A) Always B) Often C) Rarely (Sometimes) D) Never W is the weight A x, A y are the pin reaction FBC is the hydrologic cylinder reation

10 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. 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 other sets of equations that are rarely used (we will not discuss them). Steps of Solving Free Body Equilibrium problems 1. If not given, establish a suitable x - y coordinate system. 2. Draw a free body diagram (FBD) of the object. 3. Apply the three equations of equilibrium (E-of-E) to solve for the unknowns.

11 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 F X = 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. Q. Which equation of equilibrium allows you to calculate F B right away? A) F X = 0 B) F Y = 0 C) M A = 0 D) Any one of the above. A X A B A Y F B 100 lb

12 EXAMPLE 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. Establish the x- y axes. 2. Draw a complete FBD of the boom. 3. Apply the E-of-E to solve for the unknowns.

13 Solution: FBD of the beam: A Y 1.5 m 1.5 m 4 kn A X A 45 C B F C Using E-o-f E, we get: M A = (F C sin 45 ) (1.5) (4) (3) = 0 F c = kn + F X = A X cos 45 = 0; + F Y = A Y sin 45 4 = 0; A X = 8.00 kn A Y = 4.00 kn Note that the negative signs means that the reactions have the opposite direction to that shown on FBD.

14 Plan: GROUP PROBLEM SOLVING a) Establish the x y axes. Given: The jib crane is supported by a pin at C and rod AB. The load has a mass of 2000 kg with its center of mass located at G. Assume x = 5 m. Find: Support reactions at B and C. b) Draw a complete FBD of the jib crane beam. c) Apply the E-of-E to solve for the unknowns.

15 Solution:

16 Summary: A rigid body is in equilibrium when the net force and the net moment about any arbitrary point O is equal to zero. F = 0 (no translation) and M O = 0 (no rotation) F x = 0 and F y = 0 - We also learned that before we solve any rigid body problem we need to draw the Free Body Diagram (FBD). - We should put all external forces on the FBD. Let s look at some examples

17 Note: The pin (B) prevents the movement in two directions, while the rocker (A) prevents the movement in one direction only.

18 FBD

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22 N B cos30 N B sin30

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