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1 FBD: 3D Force Reaction. 2 1 unknown 3 unknown 2 unknown No moment support?

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Presentation on theme: "1 FBD: 3D Force Reaction. 2 1 unknown 3 unknown 2 unknown No moment support?"— Presentation transcript:

1 1 FBD: 3D Force Reaction

2 2 1 unknown 3 unknown 2 unknown No moment support?

3 3 3 unknowns 6 unknowns 6. smooth pin 5 unknown

4 4 For some problem, the couples in both case should be treated as zero to provide statical determinacy thrust-bearing support journal-bearing support 7. bearing 5 unknown 4 unknown

5 5 (for some problem, the couple must be assumed zero to provide statical determinacy) 3 unknowns 6 unknowns

6 6 If thrust barring If glue or friction exist

7 7 3D Equilibrium Supports #2

8 8 3D Equilibrium Supports #3

9 9 3D Equilibrium Supports #4

10 10 3D Equilibrium Supports #5 Comparison with 2D supports

11 11

12 12 Equilibrium in 3D  F x = 0  F y = 0  F z = 0  M x = 0  M y = 0  M z = 0 Body (bodies) in Equilibrium or any but only one point OR any but only three independent (perpendicular) axis any but only three independent axis x y z x y z Direction of moment axis has a affect on solving problem. position (where the axis pass) also matters. The 3 axises need not interesect at one point

13 13 Equilibrium in 3D  F x = 0  F y = 0  F z = 0  M x = 0  M y = 0  M z = 0 Body (bodies) in Equilibrium or any but only one point OR any but only three independent (perpendicular) axis any but only three independent axis at most 6 unknowns may be found. vector approach may be easier - Each of the equation may be applied independently; e.g., an accelerating car on a flat surface may be treated as in equilibrium in the vertical direction. Same for the moment equations. - Not in this class, but be careful about the moment equations, things get very complicated if the body is not spinning in a single plane!

14 14

15 15 A r d  X Y Z r position vector: from any point on line to any point on tline of action of the force. Forces which interest or parallel with axis, do not cause the moment in that axis

16 16

17 17 Example 3/5 The uniform 7m steel shaft has a mass of 200kg and is supported by a ball-and-socket joint at A in the horizontal floor. The ball end B rests against the smooth vertical walls. Find the force exerted by the walls and the floor on the ends of the shaft. this is not FBD.

18 18 FBD: frequent mistake No system isolation (surrounding still exists) Don’t forget axis Write force name Caution: In some slides using in this class, FBD may be drawn wrongly according to the rule introduced, DO NOT IMITATE this style in your homework or examination. Correct FBD Use (at least) 3 different colors No axis System Isolation

19 19 A= (2,6,0) B=(0,0,3) G=(1,3,1.5) any point but only one point Ans Independent Eq. = 2 Vector Cross Product is useful in 3D Problem Use point A

20 20 The homogeneous plate shown has a mass of 100 kg and is subjected to a force and couple moment along its 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. Hibbeler Ex 5-15 x y z 300 N 981 N 200 N-m

21 21 x y z 300 N 981 N 200 N-m OR Selection of moment axis 2 m 3 m

22 22 Boom AB lies in vertical y-z plane, is supported by a ball- and-socket joint at B and by 2 cables at A. Calculate tension in each cable resulting from the 20-kN force acting in the horizontal plane at the midpoint M of the boom. Neglect the weight of the boom.

23 23 A B C D M x y z E F There are 5 unknowns: use moment at B: 3m 2m 4m 10m (need only 2 components (eq.) from 3) component x, component z problem

24 24 A B C D M x y z E F 3m 2m 4m 10m G H

25 25 A B C D M x y z E F 3m 2m 4m 10m G H

26 26 A B C D M x y z E F 3m 2m 4m 10m G H (2D) Top view y x Top view 3m

27 27 A B C D M x y z E F 3m 2m 4m 10m G H

28 28 Orthorgraphic Drawing Reprensentation A B C D M x y z E F 3m 2m 4m 10m G H A B C D M x y z E F 3m 2m 4m 10m P Front view Top view y y x z Front view

29 29

30 30 Technique: Finding unknown in 1 equation The bent rod is supported at A by a journal bearing, at D by a ball-and-socket joint, at B by means of cable BC. Use only one equilibrium equation to obtain a direct solution for the tension in cable BC. Assume that the bearing at A is capable of exerting force components only in the z and y directions (since it is properly aligned on the shaft). Hibbeler Ex 5-19

31 31 From any point on AD, to any point on line of action Find T only in 1 scalar Equation. Scalar Equation

32 32

33 33

34 34 3/90Sign has a mass of 100 kg, center of mass at the center of the sign. - Support at C can be treated as a ball-and- socket joint. - At D, support is provided in y-direction only - Find the tensions, T 1 and T 2 - Find total force supported at C and the lateral force R supported at D.

35 35 E Cz Get T1 and T2 this is not a FBD, why? Get R and C Cy R Cx

36 36

37 37 Properly Alignment and Moment Support A,B,C: journal bearings The 6 force reactions developed by the bearings are sufficient for maintaining the equilibrium since they prevent the shaft from rotating about any of the coordinate axes. 6 unknown reactions with 6 independent equations. “Properly Alignment” If the line of pipe is initially placed without pre-torsion, the 6 couple reactions are not necessary to maintain equilibrium, and then can be considered as not existing.

38 38 Properly Alignment and Moment Support The couple reactions are developed to make the object in equilibrium Even though the line of pipe is initially placed without pre- torsion or properly aligned. “initially properly aligned” The 4 force reactions developed at bearing A and tension T are not sufficient to maintain the equilibrium of the body.

39 39 Properly Alignment and Moment Support “properly aligned” “properly aligned” The 6 force reactions developed by the bearing, hinge and ball are sufficient for maintaining the equilibrium, so the couple reactions will not be developed at the bearing and hinge.

40 40 “initially properly aligned” thrust-bearing

41 41

42 42 Sample 3/6 A 200-N force is applied to the handle of the hoist in the direction shown. The bearing A supports the thrust (force in the direction of the shaft axis), while bearing B supports only radial load (load normal to the shaft axis). Determine the mass m which can be supported and the total radial force exerted on the shaft by each bearing. Each bearing is initially properly align.

43 43 2D View By Bx Ay Ax AzAz at same point m(9.81)100 – 173.2(250) = 0 -Bx(150)-70.7(175)+70.7(250)= 0 Ax+Bx-70.7=0 No need to be the same point By+Ay-m(9.81) =0 Az-70.7=0 Ay(150)+m(9.81) (325)=0 convenient when almost forces are othorgonal. 2D FBD 2D View

44 44 By Bx Ay Ax AzAz at same point m(9.81)100 – 173.2(250) = 0 -Bx(150)-70.7(175)+70.7(250)= 0 Ax+Bx-70.7=0 No need to be the same point By+Ay-m(9.81) =0 Az-70.7=0 Ay(150)+m(9.81) (325)=0 Use when almost forces are othorgonal. FBD

45 45 3/83 The shaft, level and handle are welded together and constitute a single rigid body. Their combined mass is 28 kg with mass center at G. The assembly is mounted in A and B and rotation is prevented by link CD. Determine the forces exerted on the shaft by bearing A and B, while the 30 N-m couple is applied to the handle as shown. BearingA,B : Journal bearing

46 46 z x y x z y T // plane x-y ignore Most forces are parallel with rectangular axis 2 force member Bearing A,B : Journal bearing

47 47 z x y x z y Tsin36.9(0.6)+28(9.81)( 0.22)=0 T=25 1N Ax+Bx+Tsin (9.81)=0 Bx=11 2.2N Ay+By- Tcos36.9=0 By=33. 5N Ay(0.6)- Tcos36.9(0.5)=0 Ay=167. 5N 28(9.81)(0.3)- Tsin36.9(0.5)- Ax(0.6)=0 Ax=11. 74N

48 Tsin36.9(0.6)+28(9.81)( 0.22)=0 T=25 1N Ax+Bx+Tsin (9.81)=0 Bx=11 2.2N Ay+By- Tcos36.9=0 By=33. 5N Ay(0.6)- Tcos36.9(0.5)=0 Ay=167. 5N 28(9.81)(0.3)- Tsin36.9(0.5)- Ax(0.6)=0 Ax=11. 74N Bearing A,B : Journal bearing Bearing A,B : thrust bearing Az+Bz =0 Cannot be solved to get exact value. 7 unknowns, 6 equations You will get exact value for some unknowns but not for all of them.

49 49 3/83 The shaft, level and handle are welded together and constitute a single rigid body. Their combined mass is 28 kg with mass center at G. The assembly is mounted in A and B and rotation is prevented by link CD. Determine the forces exerted on the shaft by bearing A and B, while the 30 N-m couple is applied to the handle as shown. BearingA,B : thrust bearing

50 50 z x y x z y T // plane x-y ignore Most forces are parallel with rectangular axis 2 force member BearingA,B : Journal bearing

51 51 z x y x z y T // plane x-y ignore Most forces are parallel with rectangular axis 2 force member

52 52 You need Az Bz to answer the problemsThink about its Physics z x y x z y Tsin36.9(0.6)+28(9.81)( 0.22)=0 T=25 1N Ax+Bx+Tsin (9.81)=0 Bx=11 2.2N Ay+By- Tcos36.9=0 By=33. 5N Ay(0.6)- Tcos36.9(0.5)=0 Ay=167. 5N Az+Bz= 0...indeterminate (by math ) Az = 0, Bz = 0 7 unknowns, 6 equations 28(9.81)(0.3)- Tsin36.9(0.5)- Ax(0.6)=0 Ax=11. 74N Assume no Az and Bz

53 53

54 54 Statically determinate (SD) problems the problem that can be analyzed by using the equilibrium condition alone. number of unknowns number of independent equilibrium equations maybe > (3 or 6) (many FBDs) maybe < ( 3 or 6 ) For one obj, usually 3 scalar eqs (2D) or 6 scalar eqs (3D) x y z (some specific problems i.e. point equilibrium)

55 55 2)Concurrent with a line Independent Equations: x y z Categories of Equilibrium (3D) O x y z 1)Concurrent at a point Independent Equations: dependant All couples occuring when moving force |_ this direction

56 56 Categories of Equilibrium (3D) x y z 3) Parallel Independent Equations: Constraints and statical determinacy: see the book 4) General Independent Equations: x y z All couples occuring when moving force // the plane

57 57

58 1) SD (statically determinate) Type of Statics Problem SD/SI the problem that can be solved for all unknown with only static equilibrium equations. unknown 3, independent 2 number of unknowns number of independent equilibrium equation 2) SI-redundant (redundant statically indeterminate ) the problem that can not be solved for all unknown with only static equilibrium equations. 3) SI-Improper (improper statically indeterminate) the problem that can not be solved due to improper support. Unknown 3, Independent eq 3 Unknown 4, Independent eq 3 bar mass m

59 59 Statically indeterminate (SI) problems Statically Indeterminacy SD/SI the problem that can not be analyzed by using the equilibrium condition alone. Problem with redundant supports SI-redundant Problem with imporper supports SI-improper - Not able to maintain equilibrium unknown 3, independent 2 No solution: (cant maintain moment) number of unknowns number of independent equilibrium equation

60 60

61 61 3/91 The window is temporarily held open in the position shown. If a=0.8 m and b = 1.2 m and the mass of window is 50 kg with mass center at is geometric center, determine the compressive force F(CD) in the prop and all the component of the forces exerted by the hinges B on the windows. Assume that A is a thrust-bearing hinge but the hinge B is not.

62 C D 0.3 F solved Bx solved By solved Ax solved Az solved Ay solved Ans y z x

63 63 Basic concept: Since all forces are given in orthogonal system, it is easier to solve the problem with orthogonal projections as sample problem 3/6.

64 64 Ans W AzAz BzBz A z +B z W CxCx A x +B x CxCx AxAx BxBx Ans FBD y z y x x z On x-z plane: take moment at line AB to reduce unknowns On x-y plane: take moment at point B to reduce unknowns Then sum forces on x-axis Cx is the key for determining Ax and Bx

65 65 Recommende d Problem 3/73 3/82 3/92 3/93

66 66 Equilibrium Here ends the most important chapter of the subject.

67 67 Review Quiz #4 Classify problems in equilibrium into SD and SI categories –How can I recognize the SD and SI problems? –Why do we need to differentiate between SD and SI problems? –What does it mean when an object has redundant supports? How about the improper supports? –What is the degree of redundancy? Review

68 68 Concepts When a body is in equilibrium, the resultant force and couple about any point O are both zero. Problems can be analysed using free body diagrams (FBDs). Statically determinate (SD) problems can be solved using the equilibrium conditions alone, while the statically indeterminate (SI) problems cannot due to too few or too many supports or constraints. Review

69 69 Chapter Objectives Descriptions #1 Analyse objects (particles and rigid bodies) in equilibriums – Specify the condition of equilibrium – Describe objects in equilibrium in physical and idealized worlds – Choose appropriate bodies/parts of bodies for the analyses – Specify support reactions from physical worlds and vice versa – Specify loads from physical worlds and vice versa – Draw free-body diagrams – Calculate the support reactions from condition of equilibrium – Identify 2-force and 3-force members

70 70 Chapter Objectives Descriptions #2 Classify problems in equilibrium into SD and SI categories – Specify types and conditions of SD and SI problems – Describe physical meanings of SI problems – Use FBDs differentiate between SD and SI problems – Obtain the degree of redundancy (when applicable)

71 71 Review Quiz #1 Analyse objects (particles and rigid bodies) in equilibriums –What are the conditions of equilibrium for each type of bodies? –What are the physical meanings of 2D contact with cable, spring, smooth surface, contact with rough surface, roller, pin, fixed and slider supports? –What are the physical meanings of 3D contact with cable, smooth surface support, roller, ball & socket, journal bearing, thrust bearing, smooth pin, hinge and fixed support? –What are the differences between 2D and 3D FBDs? Review

72 72 Review Quiz #2 Analyse objects (particles and rigid bodies) in equilibriums –Can we analyze objects in equilibrium without FBDs? Why or why not? –What are the steps in the sketching of an FBD? –In drawing the FBDs, how did we choose the directions of support reactions? –Why do we have to delete the physical supports, or isolate objects, before adding the support reactions? –How many independent equilibrium equations can you obtain from 2D and 3D diagrams? What are they? Review

73 73 Review Quiz #3 Analyse objects (particles and rigid bodies) in equilibriums –Although we can choose the point about which the moments are evaluated arbitrarily, do you have some guidelines of choosing for simpler analyses? –Why it is worthwhile to recognize that an object is a two-force member? What about the three-force member? How can I identify 2-force and 3-force members in 2D and 3D problems? Review


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