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CHAPTER 4

CHAPTER 4. Equilibrium of Rigid Body & Friction. Rigid Body. Definition of a rigid body : A rigid body is defined as a body on which the distance between two points never changes whatever be the force applied on it.

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CHAPTER 4

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  1. CHAPTER 4 Equilibrium of Rigid Body & Friction

  2. Rigid Body Definition of a rigid body: • A rigid body is defined as a body on which the distance between two points never changeswhatever be the force applied on it. • Or you may say the body which does not deform under the influence of forces is known as a rigid body. • But, in real life, there would be some force under which the body starts to deform. For example, a bridge does not deform under the weight of a single man but it may deform under the load of a truck or ten trucks. However, the deformation is small.

  3. Condition for rigid body equilibrium • General equation of equilibrium for a rigid body F=0 M=0 • Equilibrium in two dimension Fx=0 Fy=0 M=0 • Equilibrium in three dimension Fx=0 Fy=0 Fz=0 M=0

  4. Free Body Diagram (2D) • Support Reactions • there are various types of reaction that occur at supports and point • As a general rules, • If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction. Likewise , if rotation is prevented , a couple moment is exerted on the body Example : (i) Prevents the beam from translating in the vertical direction Prevents the beam from translating in the vertical direction, the roller can only exert a force on the beam

  5. The pin passes through a hole in the beam and two leaves which are fixed to the ground. The pin can prevent translation of the beam in any direction This support prevent both translation and rotation of the beam, and so to do this a force and couple moment must be developed on the beam at its point of connection.

  6. External and Internal Forces • Internal Forces – not represented on Free Body Diagram • External Forces – do represented on Free Body Diagram • Weight and the center of Gravity • When the body is uniform or made of homogeneous material , the centre of gravity will be located at the body’s geometric centre or centroid • the body nonhomogeneous or has an unusual shape, then the location of its center of gravity will be given W=mg

  7. Idealized Model

  8. 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?

  9. Example 4.1 Draw the free body diagram of the uniform beam shown in figure below. The beam has a mass of 100kg Idealized model 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 typicallyinclude: a) applied loads, b) support reactions, and, c) the weight of the body.

  10. Example 4.4 • The link shown in Figure below is pin-connected at A and rests against a smooth support at B. Compute the horizontal and vertical components of reaction at the pin A.

  11. Determine support reaction for beam • In determining of support reactions, the unknown reactions acted on the beam must be identified. This can be done by referring to Table of support reaction. • By using the Equation of Equilibrium, the support reactions can be solved. For the point load: • the distance is taken from the point of force to moment about point. For distributed load (DL), • the distance is taken from the centroid of DL to the moment about point.

  12. A point load is self-explanatory: a load that is applied at one point • A distributed load/uniform load, on the other hand, is applied along an edge

  13. Practical cases • In buildings, point loads on beams is subjected to loads from other beams or columns and uniformly distributed loads are due to floors, walls and partitions and the weights of the beams themselves Example 1

  14. Constraint for rigid body • Type of constraint • Redundant constraint • When a body has redundant supports, that is more supports than are necessary to hold it in equilibrium, it become statically indeterminate (more unknown loadings on the body than equations of equilibrium).

  15. Improper constraint • Resulting in stability • axis is perpendicular to the plane of the forces and therefore appears as a point, Hence, when all the reactive forces are concurrent at this point, the body is improperly constrained

  16. Equilibrium of rigid body in 3 Dimension Free body diagram of Support Reactions • The magnitude of forces, F = • Force’s orientation defined by the coordinate direction angles ,  and 

  17. Example 4.3 A single bearing or hinge can prevent rotation by providing a resistive couple moment. However, it is usually preferred to use two or more properly aligned bearings or hinges. Thus, in these cases, only force reactions are generated and there are no moment reactions created.

  18. Equation of equilibrium • Vector Equations of Equilibrium , F = 0, vector sum of all the external forces acting on the body Mo = 0, is the sum of couple moments and the moments of all the forces about any point O located either on or off the body • Scalar Equations of Equilibrium If applies the external forces and couple moments in Cartesian Vector. F = Fxi + Fyj + Fzk = 0 Mo = Mxi + Myj + Mzk = 0 • Since the i, j and k components are independent from one another, the above equations are satisfied provided Fx = 0 , Fy = 0 and Fz = 0 Mx = 0 , My = 0 and Mz = 0

  19. Example 4.14 • The homogeneous plate shown in figure below has a mass of 100kg and is subjected to a force and couple moment along it edges. If it is supported in the horizontal plane by means of a roller at A, a ball and socket joint at B and a cord at C, determine the components of reaction at the supports.

  20. Equations of Equilibrium: Fx = 0; Bx = 0 Fy = 0; By = 0 and Fz = 0; Az + Bz + Tc - 300N - 981 N = 0 Moment at axis equation (right hand rules=define direction of moment): Mx = 0; Tc ( 2m) - 981 ( 1m) + Bz (2m) = 0 My = 0; 300 N (1.5m) + 981N ( 1.5m) – Bz (3m) - Az(3m) – 200N.m = 0 The components of force at B can be eliminated if the x’, y’ and z’ axes are used. We obtain Mx’ = 0; 981 N (1m) + 300 N (2m) - Az ( 2m)= 0 My’ = 0; -300 N (1.5m) - 981N (1.5m) – 200 N.m+Tc (3m) = 0  Results Az = 790 N Bz = -217 N, negative sign indicates that Bz acts downward Tc = 707 N

  21. Pulleys system

  22. Example 4.7 • Determine the tension T in the cable of the pulley system shown in Figure The most solution would be as follows: Fy = 0 5T = 1500 N T = 300 N

  23. ∑Fy = 0 T3 = 4T T3 = 1200 N ∑Fy = 0 T2 = 2T T2 = 600 N ∑Fy = 0 T + T4 = 1500 N T = 300 N

  24. Friction

  25. Friction

  26. What is Friction? • Friction is defined as a force of resistance acting on a body which prevents or retards slipping of the body relative to a second body • Friction Law for dry surface • Motion or impending motion of two forces in contact causes a reaction force known friction force. • This friction force is : • Parallel to a flat surface or tangent to a curved surface • Opposite in direction to the motion or impending motion • Dependent on the force pressing the surfaces together • Generally independent of the area of surface of contact • Independent of velocity, except for extreme cases not to be considered in this module • Dependent on the nature of the contacting surface

  27. Friction • How to define friction? • Friction laws for dry surface • On Horizontal plane • On Inclined plane

  28. Friction laws for dry surface on horizontal plane • Condition 1: No applied load (Not moving) • No friction Weight(W) F Normal force(N)

  29. Condition 2 : applied load P pushed the block and the block has impending motion • F acted as a reacting force an not be great enough to balance P and subsequently the block will tend to slip. • In other words P will slowly increases until the block is on verge sliding, F can be to max value (Fmax) but the block still in equilibrium • At this state F is know as static friction force, FS Weight(W) Fs=μsN P F Coefficient of static friction Normal force(N) Notes: if F=P, F< Fs, the equation of FS can not be use to determine the friction since the friction force not achieved the maximum value

  30. Condition 3 : when the block is start moving (motion) • F will reduced fastly because of the momentum influence • At this state F is know as kinetic friction force, Fk • At this condition Fk is less than Fs Weight(W) Fk=μkN P Fk Normal force(N)

  31. Condition 4 : the block is sliding/overturning • When P is increased (P > Fk) until it make the block sliding/overtuning at point B. Moment at B = zero. + ΣMB =P(h)-W(a) Weight(W) P Fk Normal force(N) • Total moment at B is +ve = the block is not sliding/overturning. • Total moment at B is -ve = the block is sliding/overtuning.

  32. W W P P F Fs N N R R • Resultant force • Experimentally Those limiting static frictional force Fs and Fk is directly proportional to the resultant normal force (Fs) and to the magnitude of the resultant normal force (Fk) • Defining the angle of static friction, фs tan Φs =Fs /N =μs N / N = μs • Φs = tan-1μs Φ Φs

  33. W P F N R Φ • Defining the angle of kinetic friction, фk W P Fk N tan Φk =Fk /N = μk N / N = μk • Φk = tan-1μk Φk R

  34. Example 4.18 • The 70N force shown in figure causes impending motion to the right. The block is not moving due to this force. Determine the static coefficient of friction, μs. Given the mass of block is 40kg. Solution; Draw the free body diagram of block. 70N W W = mg = 40(9.81) = 392.4N W = N Ncc Fs 70N

  35. Example 4.19 • The mass of block is 20000 kg is subjected to the applied load as shown in figure. Determine the friction force if μs = 0.5. Solution; Draw the free diagram of block. 80kN Ө=20º 80 sin 20º =27.4kN W=mg =20000(9.81) =196.2kN 80kN Ө=20º 80 cos 20º =75.2kN F N

  36. ΣFy ↑=ΣFy↓ 27.4 + N = 196.2 or can be write as 27.4 + N – 196.2 =0 N = 168.8 kN Fx = 0 , 75.2 – F = 0 75.2 kN= F • The friction force, Fs = μs N = 0.5(168.8) = 84.4kN. • The value of Fs then compare to the F. Show that F < Fs. The block is not moving yet. The equation of Fs can not be used to determine the friction since the friction forces not achieved the maximum value. • The friction force is F = 75.2kN.

  37. Example • Te uniform crate shown in Figure has a mass of 20kg. If a force P=80 N is applied to the crate, determine if it remains in equilibrium. The coefficient of static friction is μs = 0.3

  38. Solution • Free body diagram As shown in Figure, the resultant normal force Nc must act a distance x, from the counteract the tipping effect cause by P. there are three unknowns F, Nc and x, which can be determined strictly from three equations of equilibrium

  39. Friction laws for dry surfaces on an inclined plane • Condition 1 : the block is not moving (no friction) N can be determined using ΣFx = 0, F = W sin θ and ΣFy = 0, N = W cos θ. The block tends to move down on the inclined plane. The friction force acted in the opposite direction. F< Fs

  40. Condition 2 : the block is in the verge of impending motion N can be determined using ΣFx = 0, F = W sin θ and ΣFy = 0, N = W cos θ. F = Fs, F = μs N

  41. Condition 3 : the block is moving N can be determined using ΣFy = 0, N = W cos θ. Fk = μk N Fk = μk N= μk W cos θ

  42. Condition 4 : the block is sliding/overtuning When P is increased (P > Fk) until it make the block sliding/overtuning at point A. + ΣMA = - W sin θ(h) + W cosθ(a)

  43. Example 4.20 • A 100N force acts as shown on a 300N block placed on an inclined plane. The coefficients of friction between the block and plane are μs=0.25 and μk=0.2. determine whether the block is in equilibrium and find the value of the friction force. • Free body diagram

  44. Σfy =0; - 240 + N = 0 N= 240N Fs= μs N = 0.25(240) = 60N Σfx= 0 100 - 180 – F = 0 F = -80N (the friction force is acted opposite direction, thus directed up and to the right) F = 80N ( ) F > Fs , 80N > 60N, the block is moving down due to this condition. Use μk to determine the actual friction force. The block is not in equilibrium condition. The friction force; Fk = μk N = 0.2(240) = 48N

  45. Tips Solving problems in dry friction • The first step is to draw a free body diagram of the body, labeling and directing all forces involved at each surface of contact. • Theresultant, R exerted by a surface on a free body can be resolved into a normal component N and a tangential component F. F known as a friction force. When a body is in contact with a fixed surface, the direction of the friction force, F is opposite to that of the actual or impending motion or applied load of the body. • No motion will occur as long as F does not exceed the maximum value of Fs = μs N where μs is the coefficient of static friction. • Motionwill occur if a value of F larger than Fs is required to maintain equilibrium. As motion takes place, the actual force drops to Fk = μk N where μk is the coefficient of kinetic friction.

  46. Example 4.23 • The static coefficient μs is 0.4 between the point A and rough surfaces meanwhile there are no friction occurred between point B and contact surfaces. Indentify whether the wood is in equilibrium or not. Assumed the weight of wood acts at the centre of A-B. Smooth surface w Rough surface

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