12 The 3 axises need not interesect at one point Equilibrium in 3DorFx= 0Fy= 0Fz= 0any but onlythree independent(perpendicular) axisBody(bodies)inEquilibriumorany but only one pointMx= 0My= 0Mz= 0ORany but only threeindependent axisThe 3 axises need not interesect at one pointxyzposition (where the axis pass) also matters.Direction of moment axis has a affect on solving problem.yxz
13 Equilibrium in 3D or or Body (bodies) in Equilibrium Fx= 0Fy= 0Fz= 0any but onlythree independent(perpendicular) axisBody(bodies)inEquilibriumorany but only one pointMx= 0My= 0Mz= 0ORany but only threeindependent axisat most 6 unknowns may be found.vector approach may be easierEach 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!
17 Example 3/5The 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 nameUse (at least)3 different colorsNo axisNo 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 AIndependentEq. = 2any point but only one pointA= (2,6,0)B=(0,0,3)G=(1,3,1.5)Vector Cross Product is useful in 3D ProblemAnsAns
20 Hibbeler Ex 5-15The 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-mxyz300 N981 N
21 Selection of moment axis yz200 N-m300 N3 m981 N2 mSelection of moment axisOROR
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: CDMxyzEF3m10mproblemThere are 5 unknowns:2m4muse moment at B:4m(need only 2 components (eq.) from 3)component x , component z
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 directsolution for the tension in cable BC. Assume that the bearing at A is capable of exerting forcecomponents 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 actionScalar 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 onlyFind the tensions, T1 and T2Find total force supported at C and the lateral force R supported at D.
35 this is not a FBD, why?RECxCzCyGet T1 and T2Get R and C
37 Properly Alignment and Moment Support A,B,C:journalbearings“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 with6 independent equations.
38 Properly Alignment and Moment Support “initiallyproperlyaligned”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”“properlyaligned”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.
42 Sample 3/6A 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. ByBxAyAxAzat same pointNo need to be the same pointm(9.81)100 – 173.2(250) = 02D View-Bx(150)-70.7(175)+70.7(250)= 02D FBD2D ViewAy(150)+m(9.81) (325)=0Ax+Bx-70.7=0Az-70.7=0By+Ay-m(9.81) =0convenient when almost forces are othorgonal.
44 Use when almost forces are othorgonal. ByBxAyAxAzat same pointNo need to be the same pointm(9.81)100 – 173.2(250) = 0-Bx(150)-70.7(175)+70.7(250)= 0FBDAy(150)+m(9.81) (325)=0Ax+Bx-70.7=0Az-70.7=0By+Ay-m(9.81) =0Use when almost forces are othorgonal.
45 BearingA,B : Journal bearing 3/83The 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 axiszyBearing A,B :Journal bearingzxyx
48 You will get exact value for some unknowns but not for all of them. Bearing A,B :Journal bearing30-Tsin36.9(0.6)+28(9.81)(0.22)=0T=251NAx=11.74N28(9.81)(0.3)-Tsin36.9(0.5)-Ax(0.6)=0Ax+Bx+Tsin (9.81)=0Bx=112.2NAy(0.6)-Tcos36.9(0.5)=0Ay=167.5NAy+By-Tcos36.9=0By=33.5NBearing A,B :thrust bearingAz+Bz=0Cannot be solvedto get exact value.7 unknowns, 6 equationsYou will get exact value for some unknowns but not for all of them.
49 BearingA,B : thrust bearing 3/83The 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 axiszyBearingA,B :Journal bearingzxyx
51 ignore T // plane x-y 2 force member 2 force member Most forces are parallel with rectangular axiszxy
52 T=251N 30-Tsin36.9(0.6)+28(9.81)(0.22)=0 Ax=11.74N You need Az Bz to answer the problemsThink about its Physics30-Tsin36.9(0.6)+28(9.81)(0.22)=0T=251N28(9.81)(0.3)-Tsin36.9(0.5)-Ax(0.6)=0Ax=11.74NAx+Bx+Tsin (9.81)=0Bx=112.2NAy(0.6)-Tcos36.9(0.5)=0Ay=167.5N300300zxyAy+By-Tcos36.9=0By=33.5N600Az+Bz=0...indeterminate (by math)200200Assume no Az and Bz7 unknowns, 6 equationsAz = 0, Bz = 0200300380220100
54 (some specific problems Statically determinate (SD) problemsFor one obj, usually3 scalar eqs (2D)or 6 scalar eqs (3D)the problem that can be analyzed byusing the equilibrium condition alone.number of unknownsnumber of independentequilibrium equationsmaybe > (3 or 6)(many FBDs)yxzmaybe < ( 3 or 6 )(some specific problemsi.e. point equilibrium)
55 All couples occuring when moving force |_ this direction Categories of Equilibrium (3D)Concurrent at a pointIndependent Equations:yxOzdependantConcurrent with a lineIndependent Equations:xyzAll couples occuring when moving force |_ this direction
56 Categories of Equilibrium (3D) y3) ParallelIndependent Equations:xzAll couples occuring when moving force // the planexyz4) GeneralIndependent Equations:Constraints and statical determinacy: see the book
58 Type of Statics Problem SD/SIType of Statics Problem1) SD (statically determinate)mass mthe problem that can be solved for all unknown with only static equilibrium equations.barUnknown 3 ,Independent eq 32) SI-redundant (redundant statically indeterminate)the problem that can not be solved for all unknown with only static equilibrium equations.Unknown 4 ,Independent eq 33) SI-Improper (improper statically indeterminate)the problem that can not be solved due to improper support.unknown 3 ,independent 2number ofunknownsnumber of independentequilibrium equation
59 Statically Indeterminacy SD/SIStatically IndeterminacyStatically indeterminate (SI) problemsthe problem that can not be analyzed by using the equilibrium condition alone.SI-redundantSI-improperProblem with imporper supportsProblem withredundant supportsNot able to maintain equilibriumNo solution: (cant maintain moment)number ofunknownsnumber of independentequilibrium equationunknown 3 ,independent 2
61 3/91The 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 1.20.8yF solvedxBx solvedzBy solvedAnsCD0.3Ax solvedAz solvedAy 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 unknownsOn x-y plane: take moment at point B to reduce unknownsFBDyzxCxAxBxAx +BxAnsAzBzAz +BzThen sum forces on x-axisWWCxAns
66 Here ends the most important chapter of the subject. EquilibriumHere ends the most important chapter of the subject.
67 Review Quiz #4ReviewClassify problems in equilibrium into SD and SI categoriesHow 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 ConceptsReviewWhen 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 equilibriumsSpecify the condition of equilibriumDescribe objects in equilibrium in physical and idealized worldsChoose appropriate bodies/parts of bodies for the analysesSpecify support reactions from physical worlds and vice versaSpecify loads from physical worlds and vice versaDraw free-body diagramsCalculate the support reactions from condition of equilibriumIdentify 2-force and 3-force members
70 Chapter Objectives Descriptions #2 Classify problems in equilibrium into SD and SI categoriesSpecify types and conditions of SD and SI problemsDescribe physical meanings of SI problemsUse FBDs differentiate between SD and SI problemsObtain the degree of redundancy (when applicable)
71 Review Quiz #1ReviewAnalyse objects (particles and rigid bodies) in equilibriumsWhat 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 #2ReviewAnalyse objects (particles and rigid bodies) in equilibriumsCan 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 #3ReviewAnalyse objects (particles and rigid bodies) in equilibriumsAlthough 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?