Equilibrium of a Particle

Transcription

1 ME Statics Equilibrium of a Particle Chapter 3

2 Applications For a spool of given weight, what are the forces in cables AB and AC?

3 Applications For a given weight of the lights, what are the forces in the cables? What size of cable must you use?

4 Applications

5 Equilibrium of 2D Particle This is an example of a 2-D or coplanar force system. If the whole assembly is in equilibrium, then particle A is also in equilibrium. To determine the tensions in the cables for a given weight of the engine, we need to learn how to draw a free body diagram and apply equations of equilibrium.

6 The What, Why, How of a Free Body Diagram (FBD) Free Body Diagrams are one of the most important things for you to know how to draw and use. What? - It is a drawing that shows all external forces acting on the particle. Why? - It helps you write the equations of equilibrium used to solve for the unknowns (usually forces or angles).

7 The What, Why, How of a Free Body Diagram (FBD) Why? 1. Imagine the particle to be isolated or cut free from its surroundings. 2. Show all the forces that act on the particle. Active forces: They want to move the particle. Reactive forces: They tend to resist the motion. 3. Identify each force and show all known magnitudes and directions. Show all unknown magnitudes and / or directions as variables. FBD at A Note : Engine mass = 250 Kg

8 Equations of 2D Equilibrium Since particle A is in equilibrium, the net force at A is zero. Written in a scalar form, SFx = 0 and S Fy = 0 These are two scalar equations of equilibrium (EofE). They can be used to solve for up to two unknowns.

9 Write the scalar EofE: + S Fx = TB cos 30º TD = 0 + SFy = TB sin 30º kn = 0 Solving the second equation gives: TB = 4.90 kn From the first equation, we get: TD = 4.25 kn

10 Springs, Cables, and Pulleys Spring Force = spring constant * deformation, or F = k * S With a frictionless pulley, T1 = T2.

11 Example 1 Given: Sack A weighs 20 N and geometry is as shown. Find: Forces in the cables and weight of sack B. Plan: 1. Draw a FBD for Point E. 2. Apply EofE at Point E to solve for the unknowns (TEG & TEC). 3. Repeat this process at C.

12 Solution Example 2 A FBD at E should look like the one to the left. Note the assumed directions for the two cable tensions. The scalar EofE are: + S Fx = TEG sin 30º TEC cos 45º = 0 + S Fy = TEGcos 30º TEC sin 45º 20 N = 0 Solving these two simultaneous equations for the two unknowns yields: TEC = 38.6 N TEG = 54.6 N

13 Solution Example 2 Now move on to ring C. A FBD for C should look like the one to the left. The scalar EofE are: S Fx = cos 45 (4/5) TCD = 0 S Fy = (3/5) TCD sin 45 WB = 0 Solving the first equation and then the second yields TCD = 34.2 N and WB = 47.8 N.

14 Equilibrium of a Rigid Body Applications Equilibrium in 2D Support reactions Free-body diagram

15 Applications A 200 kg platform is suspended off an oil rig. How do we determine the force reactions at the joints and the forces in the cables? How are the idealized model and the free body diagram used to do this? Which diagram above is the idealized model?

16 Applications A steel beam is used to support roof joists. How can we determine the support reactions at A & B? Again, how can we make use of an idealized model and a free body diagram to answer this question?

17 Conditions for Rigid-Body Equilibrium Statics deals primarily with the description of the force conditions necessary and sufficient to maintain the equilibrium of engineering structures. When a body is in equilibrium, the resultant of all forces acting on it is zero. Thus, the resultant force R and the resultant couple M are both zero, and we have the equilibrium equations R F 0 M M 0

18 Conditions for Rigid-Body Equilibrium Equilibrium of a rigid body requires ΣF = 0 and ΣM = 0 or ΣFx=0 ΣFy=0 ΣFz=0 ΣMx=0 ΣMy=0 ΣMz=0 So, in 3-D, we have six equilibrium equations. In 2-D, we have three equilibrium equations.

19 Equilibrium in Two Dimensions Required and necessary condition is ΣFx=0 ΣFy=0 ΣMz=0

20 Forces In these summations, all of the following forces need to be included: external forces: loads applied to the structure by the environment (e.g., weight, service loads, etc.) internal forces: forces in structural members and connections that are generated by straining of material. reactions:* forces that support the structure as a whole. * Reactions are also internal forces, but to draw attention to their importance, we list them separately.

21 Reactions in 2D 1. Flexible cable, belt, chain, or rope Force exerted by a flexible cable is always a tension away from the body in the direction of the cable.

22 Reactions in 2D 2. Smooth surface: Contact force is compressive and is normal to the surface. 3. Rough surface: Rough surfaces are capable of supporting a tangential component F (frictional force) as well as a normal component N of the resultant contact force R.

23 Reactions in 2D 4. Roller support Roller, rocker, or ball support transmits a compressive force normal to the supporting surface

24 Reactions in 2D 5. Freely sliding guide Collar or slider free to move along smooth guides; can support force normal to guide only.

25 Reactions in 2D 6. Pin connection A freely hinged pin connection is capable of supporting a force in any direction in the plane normal to the axis; usually shown as two components Rx and Ry. A pin not free to turn may also support a couple M.

26 Reactions in 2D 7. Built-in or fixed support A built-in or fixed support is capable of supporting an axial force F, a transverse force V (shear force), and a couple M (bending moment) to prevent rotation.

27 8. Gravitational attraction Reactions in 2D

28 Reactions in 2D 9. Spring action Spring force is tensile if spring is stretched and compressive if compressed. For a linearly elastic spring the stiffness k is the force required to deform the spring a unit distance.

29 Reaction forces 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 rotation is prevented, a couple moment is exerted on the body.

30 Free Body Diagram (FBD) Free Body Diagrams are one of the most important things for you to know how to draw and use. What? - It is a drawing that shows all external forces acting on the particle. Why? - It helps you write the equations of equilibrium used to solve for the unknowns (usually forces or angles).

31 The Process of Solving Rigid-Body Equilibrium Problems Step1: For analyzing an actual physical system, first we need to create an idealized model. Step2: Then we need to draw a free-body diagram showing all the external (active and reactive) forces. Step3: Finally, we need to apply the equations of equilibrium to solve for any unknowns.

32 Procedure for Drawing a Free-Body Diagram 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.

33 Procedure for Drawing a Free-Body Diagram 3. Label loads and dimensions: All known forces and couple moments should be labeled with their magnitudes and directions. For the unknown forces and couple moments, use letters like Ax, Ay, MA, etc.. Indicate any necessary dimensions.

34 Procedure for drawing FBDs (Book Version) 1. Decide on the rigid body (or portion of a rigid body) whose equilibrium you want to analyze. 2. Imagine that this body is cut completely free (separated) from the rest of the structure and/or its environment. in 2-D, think of a closed line that completely encircles the body. in 3-D, think of a closed surface that completely surrounds the body. 3. Sketch the body. 4. Sketch all external forces that are applied to the body. 5. Wherever the cut passes through a structural member, sketch the internal forces that occur at that location. 6. Wherever the cut passes through a support, sketch the support reactions that occur at that location.

35 Example

36 Example

37 Example

38 Example Given: The link is pin-connected at A and rests against a smooth support at B. Find: Horizontal and vertical components of reactions at pin A. Plan: 1. Put the x and y axes in the horizontal and vertical directions, respectively. 2. Draw a complete FBD of the boom. 3. Apply the EofE to solve for the unknowns.

39 Solution FBD Reaction N B is perpendicular to the link at B Horizontal and vertical components of reaction are represented at A

40 Example Given: Weight of the boom = 125 N, the center of mass is at G, and the load = 600 N. Find: Support reactions at A and B. Plan: 1. Put the x and y axes in the horizontal and vertical directions, respectively. 2. Determine if there are any two-force members. 3. Draw a complete FBD of the boom. 4. Apply the EofE to solve for the unknowns.

41 Solution

42 Example Draw the FBD for the simply supported I-beam shown. The cross section is a W14x26 shape which has a weight of 26 N/m. z y y A 6.67 m 2000 N B x L = 10 m

43 z y y A 6.67 m 2000 N B x L = 10 m y 260 N 2000 N A x x A y 5 m 6.67 m B y

44 y 260 N 2000 N Now apply equilibrium equations: F x = 0 = A x A x A y 5 m F y = 0 = A y + B y N N 6.67 m M A = 0 = (260 N) 5 m + (2000 N) 6.67 m - B y (10 m) B y =14640 N.m/10 m = 1464 N A y = 796 N x B y Note: we could have used the equations F x, M A and M B to obtain our solution.

45 Examples A C y x B W 2 A C W 1 B W D y x E Draw FBD for ACB Draw FBD for ABCD

46 250 mm Example: Neglecting friction, determine the tension in cable ABD and the reaction at support C. B D F A C A: C x = 80 N C y = 40 N 100 mm 100 mm F=120 N

47 Example: Rod AB is attached to a frictionless collar at A. Neglecting the weight of AB, determine angle. P = 16 N Q = 12 N l = 20 cm a = 5 cm Q A C a l B P A: =??

48 Static Indeterminacy Statically indeterminate structures and mechanisms: 2-D statically indeterminate example: Find the support reactions in terms of P. y x P L /2 L /2

49 Constraints - support conditions For a structure to be in static equilibrium under general loading, the number of support reactions must be equal to or greater than the number of equilibrium equations (three in 2- D and six in 3-D). This is a necessary but not sufficient condition. Redundant supports: When a structure has more supports than is required for static equilibrium, it becomes statically indeterminate. Improper supports: For static equilibrium, the constraints must be sufficient in number and arrangement so that the structure has no rigid body motion capability. A structure that has rigid body motion capability is called a mechanism.

50 example: For the systems shown below, determine the degree of redundancy (i.e., the number of support conditions beyond those required for equilibrium). cable (a) (b) (c) cable (d) (e) (f) # constraints redundancy (a) (b) (c) (d) (e) (f)

51 example: For the systems shown below, determine the degree of redundancy (i.e., the number of support conditions beyond those required for equilibrium). cable (a) (b) (c) cable (d) (e) (f) (a) (b) (c) (d) (e) (f) # constraints 3 2 or or 3 3 redundancy 0-1 or or 0?

52 Two- and Three-Force Members Simplify some equilibrium problems by recognizing members that are subjected to only 2 or 3 forces Two-Force 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 two-force member If we apply the equations of equilibrium to such a member, we can quickly determine that the resultant forces at A and B must be equal in magnitude and act in the opposite directions along the line joining points A and B.

53 Example of Two-Force Members If we neglect the weight, the members can be treated as two-force members. This fact simplifies the equilibrium analysis of some rigid bodies since the directions of the resultant forces at A and B are thus known (along the line joining points A and B).

54 Two- and Three-Force Members Three-Force 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

55 Remarks on pin-connected straight bars, cables and ropes: Consider a pin-connected bar. Observe that the internal force supported by the bar is purely axial. That is, the internal force in the bar is always directed along the axis of the bar. Identical remarks apply to cables and ropes. These are examples of simple, but common, twoforce members.

56 Draw the free-body diagram of member ABC which is supported by a smooth collar at A, roller at B, and short link CD. Explain the significance of each force acting on the diagram.

57 Determine the horizontal and vertical components of reaction at the pin A and the tension developed in cable BC used to support the steel frame.

58 The mass of 700 kg is suspended from a trolley which moves along the crane rail from d = 1.7 m to d = 3.5 m. Determine the force along the pin connected knee strut BC (short link) and the magnitude of force at pin A as a function of positiion d. Plot these results of F BC and F A (vertical axis) versus d (horizantal axis).

59 The jib crane is supported by a pin at C and rod AB. İf the load has a mass of 2 Mg with its center of mass located at G, determine the horizontal and vertical components of reaction at the pin C and the force developed in rod AB.

60 Example

61 Example

62 Example

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64 Example If the breaking strength of cables AC and CE is 750 N and the breaking strength of cables CD and AB is 400 N, what is the maximum mass M that can be safely supported by this system of cables? (The cables are clamped at joints A and C.)