Equilibrium. Rigid Bodies VECTOR MECHANICS FOR ENGINEERS: STATICS. Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

Transcription

1 Eighth E 4 Equilibrium CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University of Rigid Bodies

2 Contents Introduction Free-Body Diagram Reactions at Supports and Connections for a Two-Dimensional Structure Equilibrium of a Rigid Body in Two Dimensions Statically Indeterminate Reactions 4-2

3 Contents Sample Problem 4.1 Sample Problem 4.3 Sample Problem 4.4 Equilibrium of a Two-Force Body Equilibrium of a Three-Force Body Sample Problem

4 Contents Equilibrium of a Rigid Body in Three Dimensions Reactions at Supports and Connections for a Three-Dimensional Structure Sample Problem

5 ighth Introduction For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body.

6 2004 The McGraw-Hill Companies, Inc. All rights reserved. eventh Equilibrium Essentiality The force and couple moment system acting on a body can be reduced to an equivalent resultant force and resultant couple moment at any arbitrary point O on or off the body.

7 2004 The McGraw-Hill Companies, Inc. All rights reserved. eventh Equilibrium Essentiality lf this resultant force and couple moment are both equal to zero, then the body is said to be in equilibrium F R = 0 and ( M R ) O = 0

8 2004 The McGraw-Hill Companies, Inc. All rights reserved. eventh Resolving each force and moment into its rectangular components leads to 6 scalar equations which also express the cons for static equilibrium, F x 0 F y 0 F z 0 M x 0 M y 0 M z 0

9 2004 The McGraw-Hill Companies, Inc. All rights reserved. eventh Free-Body Diagram First step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free-body diagram.

10 2004 The McGraw-Hill Companies, Inc. All rights reserved. eventh Select the extent of the freebody and detach it from the ground and all other bodies. Indicate point of application, magnitude, and direction of external forces, including the rigid body weight.

11 Indicate point of application and eventh assumed direction of unknown applied forces. These usually consist of reactions through which the ground and other bodies oppose the possible motion of the rigid body. Include the dimensions necessary to compute the moments of the forces The McGraw-Hill Companies, Inc. All rights reserved.

12 Reactions at Supports and Connections for a Two-Dimensional Structure Reactions equivalent to a force with known line of action. 4-12

13 Reactions at Supports and Connections for a Two-Dimensional Structure Reactions equivalent to a force of unknown direction and magnitude. 4-13

14 Reactions at Supports and Connections for a Two- Dimensional Structure Reactions equivalent to a force of unknown direction and magnitude and a couple of unknown magnitude 4-14

15 4-15

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19 4-19

20 4-20

21 4-21

22 eventh 2004 The McGraw-Hill Companies, Inc. All rights reserved. Rocker Bearing used to Support the Roadway of a Bridge

23 Equilibrium of a Rigid Body in Two Dimensions For all forces and moments acting on a twodimensional structure, F 0 M M 0 z x y M z M O 4-23

24 Equilibrium of a Rigid Body in Two Dimensions Equations of equilibrium become F x 0 Fy 0 M A 0 where A is any point in the plane of the structure. 4-24

25 Alternative form of equilibrium equations 4-25

26 Alternative form of equilibrium equations If M A = 0, then R must pass through A If F x = 0 where x is arbitrary then R not only passes through A but is also perpendicular to x If M B = 0 where B is any point such that AB is not perpendicular to x direction then R =

27 Alternative form of equilibrium equations 4-27

28 If M A = 0, then R must pass through A If M B = 0, then R must pass through B If M C = 0, then R must be zero where C is not collinear with A and B 4-28

29 The 3 equations can be solved for no more than 3 unknowns. The 3 equations can not be augmented with adal equations, but they can be replaced Fx M M A B M M M A B C 0 0 0

30 Statically Indeterminate Reactions More unknowns than equations Four unknown reaction force unknown, three equations available

31 Partially Constrained, or referred as unstable Fewer unknowns than equations, partially constrained Two unknown reaction force unknown, three equations available

32 Improperly Constrained, also indeterminate Equal number unknowns and equations but improperly constrained

33 Free to move horizontally Fx=0, then two equation left for three unknowns

34 Improperly Constrained, also indeterminate M 0 A If the reactions are concurrent or parallel, the structure is improperly constrained. 4-34

35 4-35

36 Sample Problem 4.1 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. 4-36

37 Sample Problem 4.1 Determine the components of the reactions at A and B. 4-37

38 Sample Problem 4.1 SOLUTION: Create a free-body diagram for the crane. Determine reactions at B and A. 4-38

39 Sample Problem 4.1 Free-body diagram. 4-39

40 Sample Problem 4.1 Determine B by finding the sum of the moments of all forces about A. M A 0 : B 1.5m 23.5 kn 6m 9.81kN 2m 0 B 107.1kN 4-40

41 Sample Problem 4.1 Determine the reactions at A by sum of all horizontal forces and all vertical forces. A x F x 0 : A B x 107.1kN 0 F y 0 : Ay 9.81kN 23.5kN 0 A y 33.3 kn 4-41

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47 4-47

48 R A = 238 N R B = 62 N R A = 250 N R B = 50 N 4-48

49 4-49

50 4-50

51 4-51

52 Sample Problem 4.3 Determine the tension in the cable and the reaction at each pair of wheels. 4-52

53 Sample Problem 4.3 A loading car is at rest on an inclined track. The gross weight of the car and its load is 5500 N, and it is applied at G. The cart is held in position by the cable. 4-53

54 Sample Problem 4.3 SOLUTION: Create a free-body diagram for the car with the coordinate system aligned with the track. 4-54

55 Sample Problem 4.3 Determine the reactions at the wheels by summing moments about points above each axle. Determine the cable tension by summing force components parallel to the track. 4-55

56 Sample Problem 4.3 Create a free-body diagram W W x y 5500 N 4980 N 5500 N 2320 N cos25 sin

57 Sample Problem 4.3 Determine the reactions at the wheels. M A 0 : 2320 N 625 mm 4980 N 150 mm R mm 0 R N M B 0 : 2320 N 625 mm 4980 N 150 mm R mm 0 R N Determine the cable tension. F x 0: 4980 N T 0 T 4980 N 4-57

58 Sample Problem 4.4 The frame supports part of the roof of a small building. The tension in the cable is 150 kn. Determine the reaction at the fixed end E. 4-58

59 Sample Problem 4.4 SOLUTION: Create a free-body diagram for the frame and cable. Solve 3 equilibrium equations for the reaction force components and couple at E. 4-59

60 Sample Problem 4.4 Create a free-body diagram for the frame and cable. 4-60

61 Sample Problem 4.4 Solve 3 equilibrium equations for the reaction force components and couple. F x 0 : Ex E x 90.0 kn kn 0 E y F y 0 : E y 200 kn 4 20 kn kn 0 M E 0 : 20 kn 7.2m 20 kn 5.4m 20 kn 3.6m 20 kn 1.8m kn 4.5m M E 0 M E kn m 4-61

62 Draw the free body diagram of the foot lever shown in Figure. The operator applies a vertical force to the pedal so that the spring is stretched 1.5 in. and the force in the short link at B is 20 lb. 4-62

63 Two smooth pipes, each having a mass of 300 kg. are supported by the fork of the tractor in Fig. Draw the free body diagrams for each pipe and both pipes together. 4-63

64 Equilibrium of a Two-Force Body Consider a plate subjected to two forces F 1 and F 2 For static equilibrium, the sum of moments about A must be zero. The moment of F 2 must be zero. It follows that the line of action of F 2 must pass through A. 4-64

65 Equilibrium of a Two-Force Body Similarly, the line of action of F 1 must pass through B for the sum of moments about B to be zero. 4-65

66 Equilibrium of a Two-Force Body Requiring that the sum of forces in any direction be zero leads to the conclusion that F 1 and F 2 must have equal magnitude but opposite sense. 4-66

67 Two and Three force members Find reactions at A, B, D. 4-67

68 Two and Three force members 4-68

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74 4-74

75 Two force member The bucket link AB on the back hoe is a typical example of a two-force member since it is pin connected at its ends and provided its weight is neglected no other force acts on this member. 4-75

76 Equilibrium of a Three-Force Body Consider a rigid body subjected to forces acting at only 3 points. Assuming that their lines of action intersect, the moment of F 1 and F 2 about the point of intersection represented by D is zero. 4-76

77 Equilibrium of a Three-Force Body Since the rigid body is in equilibrium, the sum of the moments of F 1, F 2, and F 3 about any axis must be zero. The moment of F 3 about D must be zero as well and that the line of action of F 3 must pass through D. 4-77

78 Equilibrium of a Three-Force Body The lines of action of the three forces must be concurrent or parallel. 4-78

79 Sample Problem 4.6 A man raises a 10 kg joist, of length 4 m, by pulling on a rope. Find the tension in the rope and the reaction at A. 4-79

80 Sample Problem 4.6 The three forces must be concurrent for static equilibrium. Reaction R passes through the intersection of the lines of action of the weight and rope forces. 4-80

81 Determine the direction of R. AF AB cos 45 4 m cos m CD BD CE tan AE CD cot(45 BF CE AE 1 2 BD AF m 20) m m tan m m

82 Sample Problem 4.6 Determine the magnitude of the reaction force R. T sin 31.4 R sin N sin 38.6 T 81.9 N R N 4-82

83 Equilibrium of a Rigid Body in Three Dimensions Six scalar equations are required to express the cons for the equilibrium of a rigid body in the general three dimensional case. F x 0 F y 0 F z 0 M x 0 M y 0 M z

84 Equilibrium of a Rigid Body in Three Dimensions These equations can be solved for no more than 6 unknowns which generally represent reactions at supports or connections. F 0 M r F O

85 Reactions at Supports and Connections for a Three- Dimensional Structure 4-85

86 4-86

87 4-87

88 4-88

89 Reactions at Supports and Connections for a Three- Dimensional Structure 4-89

90 Universal Joints

91 Reactions at Supports and Connections for a Three- Dimensional Structure 4-91

92 4-92

93 4-93

94 Sample problem 4.7 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.

95 4-95

96 ighth W mgj (80 20)(9.81) j 981j F 0 A j A k B j B k 981j Ck 0 y z y z A y z B y A B C z

97 M A 0 1.2i B j B k 0.9i 0.6k 981 j y z 0.6i 3 j 1.2k Ck 0

98 ighth W mgj (80 20)(9.81) j 981 j A A y z B B y z M A 1.2i B 0.6i y 981 C 0 j 3 j 0 0 Bzk 1.2k 0.9i Ck 0.6k unknowns and 5 equations, so the problem can be solved j

99 Sample Problem 4.8 A sign of uniform density weighs 540 N and is supported by a ball-and-socket joint at A and by two cables. Determine the tension in each cable and the reaction at A. 4-99

100 Sample Problem

101 Sample Problem 4.8 SOLUTION: Create a free-body diagram. Apply static equilibrium to develop equations for the unknown reactions

102 Sample Problem

103 Sample Problem 4.8 Since there are only 5 unknowns, the sign is partially constrain. It is free to rotate about the x axis. It is, however, in equilibrium for the given loading

104 ighth Sample Problem 4.8 k j i T k j i T r r r r T T k j i T k j i T r r r r T T EC EC E C E C EC EC BD BD B D B D BD BD

105 Sample Problem 4. F A T T 540N j 0 BD EC i : A T T x 3 BD 7 EC j : A T T 270 N y 3 BD 7 EC k : A T T z 3 BD 7 EC M r T r T 1.2 m i 540N j 0 A B BD E EC j : 1.6T T 0 BD EC k :.8 T.771T 648 N 0 BD EC 4-105

106 Sample Problem 4. i : A T T x 3 BD 7 EC j : A T T 270 N y 3 BD 7 EC k : A T T z 3 BD 7 EC j : 1.6T T 0 BD EC k :.8 T.771T 648 N 0 BD EC 4-106

107 Sample Problem 4. Solve the 5 equations for the 5 unknowns, T A BD N N i TEC N N j N k 4-107

108 What if? Can the sign be in static equilibrium if cable BD is removed? 4-108

109 The sign cannot be in static equilibrium because T EC causes a moment about the y- axis (due to the existence of T EC,Z ) which must be countered by an equal and opposite moment. This can only be provided by a cable tension that has a z-component in the negative-z direction, such as what T BD has

110 eventh Sample Problem 4.9 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 The McGraw-Hill Companies, Inc. All rights reserved.

111 2004 The McGraw-Hill Companies, Inc. All rights reserved. eventh The supports do not exert an appreciable couple during normal operation and these are not included in analysis.

112 eventh 2004 The McGraw-Hill Companies, Inc. All rights reserved. F A i x A A x z 2rk M ri B A B y x T rk 0 A i x j 6 7 k A k z T A y 0 i j B 294 k x i A A k z y 0 B y B j y 2ri T 3 7 T rk 294 j j 6 unknowns and 6 equations Ti 3 7 Tj 2 7 Tk

113 2004 The McGraw-Hill Companies, Inc. All rights reserved. eventh Sample Problem 4.10

114 2004 The McGraw-Hill Companies, Inc. All rights reserved. A 450-lb load hangs from the comer C of a rigid piece of pipe ABCD which has been bent as shown. The pipe is supported by the ball-and-socket joints A and D, which are fastened, respectively, to the floor and to a vertical wall, and by a cable attached at the midpoint E of the portion BC of the pipe and at a point G on the wall. Determine (a) where G should be located if the tension in the cable is to be minimum, (b) the corresponding minimum value of the tension. eventh

115 eventh 2004 The McGraw-Hill Companies, Inc. All rights reserved.

116 eventh 2004 The McGraw-Hill Companies, Inc. All rights reserved. AC M AD AE W AD AD 0 T 12i 12i AC 12 j 12 j 18 W j 6k 5400 k

117 2004 The McGraw-Hill Companies, Inc. All rights reserved. eventh 12i 12 j 6k AE T 5400k AE T T AE AE 4i 2 j 4k unit vector along AE T must be along this unit vector 4i 2 j 4k 6 T min 4i 2 j 4k 12i 12 j 6k T 1800 T 200i 100 j 200k 6i 12 j 1800

118 eventh 2004 The McGraw-Hill Companies, Inc. All rights reserved. EG x 6 i ( y 12) j 0 6 k x 6 y x, y are the coordinates of G x=0, y=15 ft Must be the same direction