Equilibrium of a Rigid Body. Engineering Mechanics: Statics

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1 Equilibrium of a Rigid Body Engineering Mechanics: Statics

2 Chapter Objectives Revising equations of equilibrium of a rigid body in 2D and 3D for the general case. To introduce the concept of the free-body diagram FBD for a rigid body. Showing how to solve rigid-body equilibrium problems.

3 Moment of a 3D Force Review THREE DIMENSIONAL FORCE SYSTEMS Remember M = r (vector) X F (vector) Find the length r Vectorially for each Force F Also if not given, find the vectorial representation of F Finding the length r Read the coordinates of the two ends of the line Get the difference of the coordinates from the tail of the force coordinate to the point of interest.

4 Moment of a 3D Force Review In determinant form: Sarrus Rule i j k M = r X F = r x r y r z F x F y F z

5 z y x z y x z y x z y x F F F r r r k j i F F F r r r k j i X F r = = + (r y F z r z Fy) i + (r z F x -r x F z ) j + (r x F y r y F x ) k Moment of a 3D Force Review

6 Moment of a Force AT A POINT 3D Example: A man exerts a 485 N pull on the rope which is looped around the branch at point A. Determine the moment about point C of this force that exerts on the branch of the tree and state the magnitude of this moment.

7 Moment of a Force AT A POINT 3D Example: r CA TAB

8 Example: Moment of a Force AT A POINT 3D Three forces act on the rod. Determine the resultant moment they create about the flange at O and determine the coordinate direction angles of the moment axis.

9 Solution Moment of a Force AT A POINT 3D Position vectors are directed from point O to each force r OA = {5j} m and r OB = {4i + 5j - 2k} m For resultant moment about O, M (r X F) r X F + r X F + r X F Ro A 1 B 2 C 3 OA OA OB i j k i j k i j k i 40 j 60k N.m

10 Solution For magnitude Moment of a Force AT A POINT 3D M Ro (30) ( 40) (60) N.m For unit vector defining the direction of moment axis, uλ MRo 30i 40 j 60k M Ro i j k

11 Solution Moment of a Force AT A POINT 3D For the coordinate angles of the moment axis, cos ; cos ; cos ;

Moment of a Force AT A POINT 3D Example (T): a) Determine the moment of the force F about the point P: b) Find the perpendicular distance D acting on the line of

12 Moment of a Force AT A POINT 3D Example (T): a) Determine the moment of the force F about the point P: b) Find the perpendicular distance D acting on the line of action of the given force that causes the same moment. N m m m m The vector from P to the point of application of F is: r = (12 3) i + (6 4) j + ( 5 1) k = 9 i + 2 j 6 k (m)

13 Moment of a Force AT A POINT 3D Example (T): The moment is: i j k M = r F = = 38i 87j+ 28k P (N - m) The magnitude of M P : M P N - m

14 Moment of a Force AT A POINT 3D Example (T): The perpendicular distance D acting on the line of action of the given force that causes the same moment is D = F M P = N - m 9 N = m The direction of M P tells us both the orientation of the plane of the plane containing D and F and the direction of the moment.

15 Moment of a Force AT A POINT 3D Example (T): The cable tension has a magnitude of kn. Calculate the moment of this cable T, which produces about the base O of the construction crane. TAB

16 Moment of a Force AT A POINT 3D EXAMPLE (T): A force is applied to the tool, to open a gas valve. Find the magnitude of the moment of this force about the point A. A B

Moment of a Force AT A POINT 3D A B r AB = {0.25 sin 30 i + 0.25 cos30 j} m = {0.125 i + 0.

17 Moment of a Force AT A POINT 3D A B r AB = {0.25 sin 30 i cos30 j} m = {0.125 i j} m F = {-60 i + 20 j + 15 k} N M z = (r AB F) i j k 0 0 M 1 z = = 1{0.125(20) (-60)} N m = N m

18 Moment of a Force About A LINE 3D Example (T): 3 cables are attached to a bracket. Replace the forces exerted by the cables with: an equivalent resultant force and couple moment acting at point A.

19 Further Reduction of Forces and Couples System

20 Further Reduction of Forces and Couples System

21 Further Reduction of Forces and Couples System

22 Further Reduction of Forces and Couples System

23 Further Reduction of Forces and Couples System

24 Further Reduction of Forces and Couples System

25 Further Reduction of Forces and Couples System Solution: General system of forces and moments x z

26 Equilibrium Conditions Coplanar (2D) General case F 3 F 2 M 2 d 3 o d 1 d 2 F 1 M 1 When an object acted upon by a system of forces & moments is in equilibrium, the following conditions are satisfied: 1. The sum of the forces is zero: F = 0 2. The sum of the moments about any point is zero: M anypo int =0

27 Equilibrium Conditions Coplanar (2D)General case F 3 F 2 M 2 d 3 o d 1 d 2 F 1 M 1 F x = 0, Fy = 0, M o = 0 Necessary Sufficient Body in Equilibrium Fundamental Equations/Conditions for Statics!

28 + 1 unknown 2 unknowns 2 unknowns 3 unknowns + F 0, 0, x = Fy = + M o = 0

29 EQUILIBRIUM EQUATIONS CONDITIONS OF EQUILIBRIUM STUDY THE EQUILIBRIUM OF 3-FORCE SYSTEMS 3 unknowns 5 unknowns 3 unknowns 6 unknowns

30 Equilibrium Conditions 3D General case There are 2 sets of Independent Equilibrium Equations Set 1: Equilibrium of Forces (3 components) Set 2: Equilibrium of Moments (3 components) + i F x = 0, + j F y = 0, F Z = 0 M x = 0, M y = 0, + i + j + k M z = 0

31 The Free-Body Diagram FBD It is the sketch of the body that shows all the applied known and unknown forces and moments acting on this body. This sketch, shows the particle free (isolated) from its surroundings with all the external forces acting on it. Note that; while showing the externally applied forces and moments, their directions should be indicated. If the directions are not given, ASSUME: Force: + ve for all directions (x, y and z) (i.e. Same direction with the defined axis directions) Moments: + ve for i, j, k components.

The Free-Body Diagram FBD To apply equilibrium equations we must account for all known and unknown forces and moments acting on the object by drawing a free-body

32 The Free-Body Diagram FBD To apply equilibrium equations we must account for all known and unknown forces and moments acting on the object by drawing a free-body diagram FBD of the body. Force Types Active Forces : tend to set the body in motion. Reactive Forces : result from constraints or supports and tend to prevent motion

33 Procedure for Analysis Equilibrium Problems 1) Draw Free-Body Diagram FBD Establish the x, y (for 2D) axes and x, y and z (for 3D) axes in any suitable orientation. Label all known and unknown force and moment magnitudes and directions on the FBD. The sense of an unknown force and moment may be assume. 2) Apply equations of equilibrium. Components of the force and moments are positive if directed along a positive axis and negative if directed along a negative axis. 3) If solution yields a negative result the force or the moment is in the opposite sense of that was assumed and showed on the FBD.

34 Supports Forces & couples exerted on an object by its supports are called reactions. E.g. a bridge is held up by the reactions exerted by its supports. support support

35 Support Reactions Reactive Forces General Rule: If a support prevents the translation of a body in a given direction, then a force (REACTION) is developed on the body in that direction. Likewise, if the rotation is prevented, a couple moment (REACTION) is exerted on the body.

36 2 D SUPPORTS

bracket & the object Figure b: side view SYMBOL USED IN

37 Pin Support: Figure a: pin support a bracket to which an object is attached by a smooth pin that passes through the bracket & the object Figure b: side view SYMBOL USED IN NOTEBOOK The arrows indicate the directions of the reactions A x and A y in 2D.

38 Pin Support Details:

40 Roller Support: It can move freely in the direction parallel to the surface on which it rolls, it can t exert a force parallel to the surface but can exert a force normal (perpendicular) to this surface. The arrow indicate the directions of the reaction A SYMBOL USED IN NOTEBOOK

41 Fixed Support The fixed support shows the supported object literally built into a wall (built-in). SYMBOL USED IN NOTEBOOK A fixed support can exert 2 components of force A X A Y and a couple M A

42 Different Support Types in 2D Different

43 Different Support Types in 2D

44 Different Support Types in 2D

45 Different Support Types in 2 D Unknown is directed along the axis of the short link

46 Different Support Types in 2D

47 Different Support Types in 2D

48 Different Support Types Different Support Types in 2D

49 Different Support Types in 2D (a) Pin in a slot (b) Slider in a slot (c) Slider on a shaft

50 Different 2D Support Types in Practice

51 Example: Free Body Diagram FBD 2D For the given supports and loadings of the body draw its FBD?

52 Example: Free Body Diagram FBD 2D

53 Example: Free Body Diagram FBD 2D

54 Example: Free Body Diagram FBD 2D

55 Free Body Diagram FBD 2D 30

56 Free Body Diagram FBD 2D PULLEYS When a cable (cord) wraps over a frictionless pulley, definitely each and every portion of this cables (cords) is under TENSION. To satisfy the static equilibrium, the magnitudes of the tension T 1 = T 2 should be the SAME.

57 Free Body Diagram FBD 2D PULLEYS r =15 cm 7.8 kn 7.8 kn 7.8 kn

58 Example: Free Body Diagram FBD 2D The object in this figure has a fixed support at the left end (point A). A cable passing over a pulley is attached to the object at 2 points. Isolate it from its supports & complete the free-body by showing the reactions at the fixed support & the forces exerted by the cable.

Free Body Diagram FBD 2D Don t forget the couple at the fixed support, Since the tension in the cable is assumed, on both sides of the pulley, the 2 same forces exerted by the cable

59 Free Body Diagram FBD 2D Don t forget the couple at the fixed support, Since the tension in the cable is assumed, on both sides of the pulley, the 2 same forces exerted by the cable having the magnitude T. Once the free-body diagram of an object is obtained, by identifying the loads & reactions acting on it, for equilibrium, the equilibrium equations can be applied.

60 Free Body Diagram FBD 2D Ay Ax

61 Example: Free Body Diagram FBD 2D

62 Example: Free Body Diagram FBD 2D y x y

63 Example: Free Body Diagram FBD 2D

64 Solving rigid-body equilibrium problems 2D CASES

66 F direction of forces = 0 M any point = Only TWO unknowns can be found

67 Example: Find the tensions in cables BG and DH if P = kn and Q = 12.5 kn? G H y x

69 + + F 0, 0, x = Fy = + M o = 0 Only THREE unknowns can be found

70 Example: Find the reactions at the supports A and D ϴ = 57 o Note that the support at A is PIN type and the support at D is ROLLER type

71 Example: Find the reactions at the supports, if P = 2500 N

72 Example: 25 knm

73 Example: 25 knm 30 o

74 Example: 25 knm 0.8 cos 60 m 0.8 sin 60 m 6 sin 30 kn Ay 1.2 sin 60 m 6 cos 30 kn cos 60 m By Bx

75 EXAMPLE: For the given compound beam find the reactions at supports A and B (ignore the size of the supports).

76 EXAMPLE: For the given compound beam find the reactions at supports A and B (ignore the size of the supports). Note: Seperate the given system into 2 parts as system CD and system AB. Find the reactions at Supports C and D. For system AB apply EQUAL MAGNITUDE BUT OPPOSITE DIRECTION of the reaction values obtained for Support C (due to action-reaction).

77 Example: For the given system, determine the reactions at A, B, C and D (ignore the size of the supports)?

78 A SIMPLE PULLEY 100 N 100 N 100 N 100 N 100 N

79 MULTIPLE PULLIES C B A

80 MULTIPLE PULLIES P /2 P 2T=P P /4 T = P 4T=P

81 Example: Calculate the tension T in the cable which supports 50 kn with the given pulley arregement shown. Note that each pulley is free to rotate about its bearing and has negligible weight compared with the carried load. Ignore the weight of ther cable as well. C B A

82 Solution: C = 12.5 kn 25 kn 12.5 kn 12.5 kn B 25 kn A

83 Example: Relationship between pulled length s of the rope and the elevated height h due to different pulley combinations.

84 Example: Neglecting friction and the radius of the pulley, determine: a) The tension in cable ADB b) Reaction at C. D 150 mm A B C 120 N 80 mm 80 mm 200 mm

85 Example: y x T T R mm A B C 120 N R 1 80 mm 80 mm 200 mm FBD of the beam ABC

86 Example (T): The mass of 700 kg is suspended from a trolley which moves along the crane rail d = 3.5 m. Determine the force along short link BC and the magnitude of the reaction force at pin A.

87 Different Support Types in 3D 1 unknown 1 unknown 1 unknown 3 unknowns 4 unknowns

88 6 unknowns Different Support Types in 3D 5 unknowns 5 unknowns 5 unknowns 5 unknowns

89 Different Support Types in 3D

90 All 2D SUPPORTS Some 3D SUPPORTS

91 Free Body Diagram FBD 3D Example: Ball-and-socket joint at A. Smooth journal bearing at B. Roller support at C.

92 Example: Free Body Diagram FBD 3D

93 Example: Free Body Diagram FBD Ball-and-socket joint at A. BD and BE are cables.

94 Free Body Diagram FBD

95 Solving 3D rigid-body equilibrium problems

96 Example: Determine the reactions due to force F = 300 N, at the fixed support O.

97 Example:

98 Solution: FBD

99 Example: Rod AB subjected to 200N force. Determine the reactions at the ball-and-socket joint A and the tension in cables BD and BE. Note that point C is acting at the middle of the rod AB. z y x

Cable BD under tension along y-axis only and cable BE under

100 Solution: At point A, ball and socket reaction means only reaction forces may occur along 3 axes (no moment). Cable BD under tension along y-axis only and cable BE under tension along x-axis only. Note that point C is at the middle of the two points A and B. FBD

101 Example T: Determine the support reaction at C (ball and socket type) if the applied force at B is F = 3.6 kn.

Engineering Mechanics: Statics in SI Units, 12e

Engineering Mechanics: Statics in SI Units, 12e 5 Equilibrium of a Rigid Body Chapter Objectives Develop the equations of equilibrium for a rigid body Concept of the free-body diagram for a rigid body