Download presentation

Presentation is loading. Please wait.

1
FBD: 3D Force Reaction

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

3
3 unknowns 6 unknowns 6. smooth pin 5 unknown

4
**7. bearing 4 unknown 5 unknown journal-bearing support**

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

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

6
**If glue or friction exist**

If thrust barring If glue or friction exist

7
**3D Equilibrium Supports #2**

8
**3D Equilibrium Supports #3**

9
**3D Equilibrium Supports #4**

10
**3D Equilibrium Supports #5**

Comparison with 2D supports

12
**The 3 axises need not interesect at one point**

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

13
**Equilibrium in 3D or or Body (bodies) in Equilibrium**

Fx= 0 Fy= 0 Fz= 0 any but only three independent (perpendicular) axis Body (bodies) in Equilibrium or any but only one point Mx= 0 My= 0 Mz= 0 OR 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!

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

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
**FBD: frequent mistake Correct FBD System Isolation Don’t forget axis**

Write force name Use (at least) 3 different colors No axis No system isolation (surrounding still exists) 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.

19
**Vector Cross Product is useful in 3D Problem**

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

20
Hibbeler Ex 5-15 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. 200 N-m x y z 300 N 981 N

21
**Selection of moment axis**

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

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
**problem There are 5 unknowns: use moment at B:**

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

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

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

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

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

28
**Orthorgraphic Drawing Reprensentation**

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

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
**Find T only in 1 scalar Equation.**

From any point on AD, to any point on line of action Scalar Equation

34
**3/90 Sign 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, T1 and T2 Find total force supported at C and the lateral force R supported at D.

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

37
**Properly Alignment and Moment Support**

A,B,C: journal bearings “Properly Alignment” 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. 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. 6 unknown reactions with 6 independent equations.

38
**Properly Alignment and Moment Support**

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

39
**Properly Alignment and Moment Support**

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
thrust-bearing “initially properly aligned”

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
**convenient when almost forces are othorgonal.**

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

44
**Use when almost forces are othorgonal.**

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

45
**BearingA,B : Journal bearing**

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
**ignore T // plane x-y 2 force member 2 force member**

Most forces are parallel with rectangular axis z y Bearing A,B : Journal bearing z x y x

47
**T=251N 30-Tsin36.9(0.6)+28(9.81)(0.22)=0 Ax=11.74N**

Ax+Bx+Tsin (9.81)=0 Bx=112.2N Ay(0.6)-Tcos36.9(0.5)=0 Ay=167.5N 300 300 z x y Ay+By-Tcos36.9=0 By=33.5N 600 200 200 200 300 380 220 100

48
**You will get exact value for some unknowns but not for all of them.**

Bearing A,B : Journal bearing 30-Tsin36.9(0.6)+28(9.81)(0.22)=0 T=251N Ax=11.74N 28(9.81)(0.3)-Tsin36.9(0.5)-Ax(0.6)=0 Ax+Bx+Tsin (9.81)=0 Bx=112.2N Ay(0.6)-Tcos36.9(0.5)=0 Ay=167.5N Ay+By-Tcos36.9=0 By=33.5N 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
**BearingA,B : thrust bearing**

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
**ignore T // plane x-y 2 force member 2 force member**

Most forces are parallel with rectangular axis z y BearingA,B : Journal bearing z x y x

51
**ignore T // plane x-y 2 force member 2 force member**

Most forces are parallel with rectangular axis z x y

52
**T=251N 30-Tsin36.9(0.6)+28(9.81)(0.22)=0 Ax=11.74N**

You need Az Bz to answer the problems Think about its Physics 30-Tsin36.9(0.6)+28(9.81)(0.22)=0 T=251N 28(9.81)(0.3)-Tsin36.9(0.5)-Ax(0.6)=0 Ax=11.74N Ax+Bx+Tsin (9.81)=0 Bx=112.2N Ay(0.6)-Tcos36.9(0.5)=0 Ay=167.5N 300 300 z x y Ay+By-Tcos36.9=0 By=33.5N 600 Az+Bz=0 ...indeterminate (by math) 200 200 Assume no Az and Bz 7 unknowns, 6 equations Az = 0, Bz = 0 200 300 380 220 100

54
**(some specific problems**

Statically determinate (SD) problems For one obj, usually 3 scalar eqs (2D) or 6 scalar eqs (3D) 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) y x z maybe < ( 3 or 6 ) (some specific problems i.e. point equilibrium)

55
**All couples occuring when moving force |_ this direction**

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

56
**Categories of Equilibrium (3D)**

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

58
**Type of Statics Problem**

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

59
**Statically Indeterminacy**

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

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
**y x z F solved Bx solved By solved Ans Ax solved Ay solved Az solved**

1.2 0.8 y F solved x Bx solved z By solved Ans C D 0.3 Ax solved Az solved Ay solved

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
**Cx is the key for determining Ax and Bx**

On x-z plane: take moment at line AB to reduce unknowns On x-y plane: take moment at point B to reduce unknowns FBD y z x Cx Ax Bx Ax +Bx Ans Az Bz Az +Bz Then sum forces on x-axis W W Cx Ans

65
Recommended Problem 3/73 3/82 3/92 3/93

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

Equilibrium Here ends the most important chapter of the subject.

67
Review Quiz #4 Review 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?

68
Concepts Review 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.

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
**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
Review Quiz #1 Review 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?

72
Review Quiz #2 Review 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?

73
Review Quiz #3 Review 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?

Similar presentations

OK

ENGINEERING MECHANICS STATICS & DYNAMICS Instructor: Eng. Eman Al.Swaity University of Palestine College of Engineering & Urban Planning Chapter 5: Equilibrium.

ENGINEERING MECHANICS STATICS & DYNAMICS Instructor: Eng. Eman Al.Swaity University of Palestine College of Engineering & Urban Planning Chapter 5: Equilibrium.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on political parties in india Ppt on foreign exchange market mechanism Ppt on george bernard shaw Synthesizing in reading ppt on ipad Ppt on ip address classes hosts Ppt on print media and electronic media Ppt on ram and rom differences Ppt on depth first search vs breadth Ppt on bagasse cogeneration Ppt on ip addressing and subnetting