2 Objectives Students must be able to #1 Course ObjectiveState the conditions of equilibrium, draw free body diagrams (FBDs), analyse and solve problems involving rigid bodies in equilibrium.Chapter ObjectivesAnalyse objects (particles and rigid bodies) in equilibriumsClassify problems in equilibrium into SD and SI categories
3 Equilibrium Definition stationaryAn object is in equilibrium when it is stationary or in steady translationrelative to an “inertial reference frame”.movingwith ConstantvelocityWhether object in stationary(moving in steady translation) or not,depends on “reference frame”.reallyequilibrium?EquilibriumCentrifugalaccelerationNewtonian Mechanics“Inertial Reference Frame”- Earth Frame- Central Universe FrameEarthAt universe
4 3/1 IntroductionA- Equilibrium is the most important subject in statics.In statics, we deal primarily with bodies at rest.(i.e. they are in the state of “equilibrium”).A- More precisely, when a body is in equilibrium, the wrench resultant of all forces acting on it is zero; i.e.These requirements are necessary and sufficient conditions for equilibrium; i.e.If is true body in equilibriumIf body in equilibrium is trueFrom the Newton’s second law of motion, a body that moves with constant velocity, rotates with constant angular velocity; i.e. “zero acceleration”, can also be treated as in a state of equilibrium.
5 Equilibrium in 2D 3/2 Mechanical system isolation Isolate body - All physical bodies are inherently 3D, but many may be treated as 2D; e.g. when all forces are on the same plane.3/2 Mechanical system isolationBefore we apply the equilibrium conditionswe need to know what force or couple are involved.Isolate bodyDraw Free Body Diagram (FBD)FBD is used to isolate body (or bodies / system of bodies) so that force/couple acting on it can be identified.
6 Free Body Diagram (FBD) FBD is the sketch of the body under consideration that is isolated from all other bodies or surroundings.The isolation of body clearly separate cause and effects of loads on the body.A thorough understanding of FBD is most vital for solving problems.
7 Construction of FBD y x y x body in interest 1) Pick body/combination of bodies to be isolated2) Isolating the body. Draw “complete external boundary” of the isolated bodyFree Body Diagram3) Add all forces and moments (including that are applied by the removed surrounding)yx150 cm50 cm4) Indicate a coordinate system5) Indicate necessary dimensionsMost important step is solving problems in mechanics.*** If an FBD is not drawn (when it is needed), you will get no credit ( 0 point ) for the whole problem!!!! ***150 cmyx50 cm
8 Note on drawing FBD y x Free Body Diagram Establish the x, y, z axes in any suitable orientation.Label all the known and unknown force magnitudes and directions on the diagramThe sense of a force having an unknown magnitude can be assumed.yx150 cm50 cmUse different colours in diagramsBody outline blueLoad (force and couple) - redMiscellaneous (dimension, angle, etc.) - black
9 Equilibrium Solving Procedure Formulate problems from physical situations.(Simplify problems by making appropriate assumptions)Draw the free body diagram (FBD) of objects under considerationState the condition of equilibriumSubstitute variables from the FBD into the equilibrium equationsSubstitute the numbers and solve for solutionsDelay substitute numbersUse appropriate significant figuresTechnical judgment and engineering senseTry to predict the answersIs the answer reasonable?
10 Equilibrium Free Body Diagram (FBD) FBD is the sketch of the body under consideration that is isolated from all other bodies or surroundings.The isolation of body clearly separate cause and effects of loads on the body.A thorough understanding of FBD is most vital for solving problems.
11 Equilibrium FBD Construction Select the body to be isolatedDraw boundary of isolated body, excluding supportsIndicate a coordinate system by drawing axesAdd all applied loads (forces and couples) on the isolated body.Add all to support reactions (forces and couples) represent the supports that were removed.Beware of loads or support reactions with specific directions due to physical meaningsAdd dimensions and other information that are required in the equilibrium equation
12 Equilibrium Help on FBD Establish the x, y axes in any suitable orientation.Label all the known and unknown applied load and support reaction magnitudesBeware of loads or support reactions with specific directions due to physical meaningsOtherwise, directions of unknown loads and support reactions can be assumed.
13 Equilibrium On FBD Analyses Objective: To find support reactionsApply the equations of equilibriumLoad components are positive if they are directed along a positive direction, and vice versaIt is possible to assume positive directions for unknown forces and moments. If the solution yields a negative result, the actual load direction is opposite of that shown in the FBD.
14 Contents Equilibrium of Objects Particles (2D & 3D) 2D Rigid Bodies the heart ()of StaticsEquilibrium of ObjectsParticles (2D & 3D)2D Rigid Bodies3D Rigid BodiesRigid BodiesSD and SI ProblemsParticleIdeal particle can not rotate.(no couple acting on it)
15 Particles FBD construction To construct a complete FBD of a particleSelect the particle to be isolatedDraw the particle as a pointIndicate a coordinate systemAdd all active forces/moments (weight, etc.)Add all support reactions (e.g. tension in the tangential direction of a cable, tensile and compressive forces in a compressed and stretched springs)
16 Particles Equilibrium Analyses #2 Equations of EquilibriumApply the equations of equilibriumComponents are positive if they are directed along a positive axis, and negative if they are directed along a negative axis.Assume the directions of unknown forces in the positive x, and y axes. If the solution yields a negative result, this indicates the sense of the force is the reverse of that shown on the FBD.
17 Particles Equilibrium in 3D Particle EquilibriumParticles Equilibrium in 3Dzyx
18 H-Ex3-1#1) The sphere has a mass of 6 kg and is supported as shown H-Ex3-1#1) The sphere has a mass of 6 kg and is supported as shown.Draw a free-body diagram of the sphere, the cord CE, and the knot at C.
19 y x How many unknowns? How many Equations? Action-reaction pair - use same symbol- opposite direction
20 Determine the tension in cables AB and AD for equilibrium of the 250-kg engine shown. FBD of A
22 Example Hibbeler Ex 3-3 #12unknown,2 Eqs(at this stage)If the sack at A has a weight of 20 lb, determine the weight of the sack at B and the force in each cord needed to hold the system in the equilibrium position shown.3unknown,2 Eqs(at this stage)= ?FBD of E
24 Example Hibbeler Ex 3-5 #1A 90-N load is suspended from the hook. The load is supported by two cables and a spring having a stiffness k = 500 N/m. Determine the force in the cables and the stretch of the spring for equilibrium. Cable AD lies in the x-y plane and cable AC lies in the x-z plane.
25 3D particle Equilibrium How many unknowns,how many equations?(no FBD, no score)perfect answer sheet
26 Example Hibbeler Ex 3-7 #1Determine the force developed in each cable used to support the 40-kN (4 tonne) crate shown.3D particleEquilibrium
31 Equilibrium of 2D Rigid Bodies 2D EquilibriumEquilibrium of 2D Rigid BodiesUse similar analyses as the particlesAdditional considerationAction forces in supports/constraintsFree-body diagram (FBD) of 2D rigid bodiesEquilibrium equations (scalar form) for rigid bodiesTwo-force and three-force members
32 Force Reaction (2D)To write an FBD, first, you will need to know what kind of force we will get when eliminating the environment/surrounding.1) Flexible cable, belt, chain, or ropealways away from the bodyTTtangent to the cable
33 Not always this direction only this direction 2) Smooth surfaces- Contact force at contact point normal to the surface/contact planeonly this directionN- always compressive3) Rough surfacesNot alwaysthis directionA rough surface can produce a tangential force (F, friction) as well as a normal force (N)direction of F depend on situations (chapter 6)NFRonly this direction
34 M existence due to its bending resistance 4) Roller supportsN- Roller, rocker, or ball transmits a compressive force normal to the supporting surfaceonly this directionN5) Freely sliding guideThe vector N may be up or down depend on problem. If not known, you may assume any of the two. After further calculation, if N is +, correct sense was assumed. If negative, N goes the other way.Nnot alwaysthis directionNMM existence due to its bending resistance
35 6) Pin connectionPin free to turnPin not free to turnRxRyRxRyMnot alwaysthis directionAs a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction. Similarly, if rotation is prevented, a couple is exerted on the body.7) Built-in or fixed supportAnot alwaysthis directionMFVWeldAA
36 m G W=mg x F F 8) Gravitational attraction Resultant of the gravitational attraction is the weight W = mg and act toward center of the earth passing through the center mass GmGW=mg9) Spring actionFx is positivex is negativexFNormaldistanceFor linear springs , F = kx
37 You may assume either case. EquilibriumYou may assume either case.The sign will indicate its sense of direction later.
38 Equilibrium construction of a FBD 2D EquilibriumEquilibrium construction of a FBDTo construct a complete FBD for a 2D rigid bodiesdetermine which body is to be isolateddraw external boundary of isolated bodyindicate a coordinate system (axes)add all loads (forces and couples, be they applied or support)No FBDs Cannot apply equilibrium conditions NO SCORES
39 V F y x y x V F y x V F Correct? F1 F2 F3 F1 F2 F3 mass m A M P M W=mgmass mAMPMyxmass mPW=mgANNCorrect?BVFPPyxNo internal forceis shown in FBDTmAnotalways “mg”BVFN**Write FBD before allowing any movements of vectors (sliding/free)**
40 3/A Correct the incomplete FBD yxfor all figures belowFAxAy
41 3/B Correct the wrong or incomplete FBD yx3/B Correct the wrong or incomplete FBDfor all figures belowfmgIf rough surfaceFOx
42 3/B Correct the wrong or incomplete FBD yx3/B Correct the wrong or incomplete FBDfor all figures belowDCAxC
43 We will choose problem 1 and 4 as samples EquilibriumSystem Isolation & FBD
54 3/3 Equilibrium Conditions We can find the wrench resultant forcePP“equilibrium”“equilibrium”PEquilibrium conditions (3D) of the body is equivalents to(for some point P )We can prove this!(for all point O)- For co-planer forces only (2D), eq conditions is equivalents toFx= Fy = Mo = (for all point O) (2D)Fx= Fy = Mo = (for some point O) (2D)
55 3/3 Equilibrium Conditions P3/3 Equilibrium ConditionsWe can find the wrench resultant forcePPPif in “equilibrium”OEquilibrium conditions (3D) of the body is equivalents to( moment for some specific point P )- For co-planer forces only (2D), eq conditions is equivalents toFx= Fy = Mo = (for all point O) (2D)We can prove this!Fx= Fy = Mo = (for some point O) (2D)
56 3/3 Equilibrium Conditions We can find the resultant force and couple at any point in the bodyEasy - Why?No forceto cause moment effectOOCan couple exist?Equilibrium conditions (3D) of the body is equivalents to( moment for all point )- For co-planer forces only (2D), eq conditions is equivalents toFx= Fy = Mo = (for all point O) (2D)(trivially easy)We can prove this (later) !Fx= Fy = Mo = (for some point O) (2D)
57 Equilibrium and its Independent Equations (no tendency to initiate translation,no tendency to initiate rotation)AAequilibriumAAany O(trivially easy)(some point: A)(any point: O)Important MeaningA2DAOnly at most 3 independent EquationsIf is true body in equilibriumIf body in equilibrium is trueAFrom the Newton’s second law of motion, a body that moves with constant velocity, rotates with constant angular velocity; i.e. “zero acceleration”, can also be treated as in a state of equilibrium.A
58 Equilibrium Eqn. for 2D Rigid Bodies 2D EquilibriumEquilibrium Eqn. for 2D Rigid BodiesScalar FormParticleRigid BodyThe sum of the moment about any point O is zero.
59 2D Equilibrium Procedure #1 Free-Body DiagramEstablish the x, y axes in any suitable orientation.Draw an outlined shape of the body.Show all the forces and couple moments acting on the body.Label all the loadings and specify their directions relative to the x, y axes. The sense of a force or couple moment having an unknown magnitude but known line of action can be assumed.Indicate the dimensions of the body necessary for computing the moments of forces.
60 2D Equilibrium Procedure #2 Equations of EquilibriumWhen applying the force equilibrium equations,orient the x and y axes along lines that will provide the simplest resolution of the forces into their x and y components.If the solution of the equilibrium equations yields a negative scalar for a force or couple moment magnitude, this indicate that the sense is opposite to that which was assumed on the FBD.
61 2D Equilibrium Procedure #3 Equations of EquilibriumApply the moment equation of equilibrium about a point O that lies at the intersection of the lines of action of two unknown forces.In this way, the moments of these unknowns are zero about O, and a direct solution for the third unknown can be determined.
63 H5.7) Find the tension in the cord at C and the horizontal and vertical component of reactions at pin A. The pulley is frictionless.The body is in equilibrium.Fx = 0 Fy = 0 MAnyPoint = 0T are the sameat both side?+If couple also applies at pully,T are the same in both side?If pulley rotates, T are the same in both side ?with constant angular velocity?
64 Independent Equations y** Whatever 2D-equilibrium conditions you used, at most 3 unknowns (scalar) may be found.xBody in equilibriumHey , why you cannot get anymore, you can take another points to take moment, and get new equation?notindependentequations!Fx= 0 Fy = 0 Mo = 0(for some point O)** However, sometimes you can not get those 3 unknowns in 2D problem.Fx= 0 Fy = 0 Mo = 0(for all point O)
65 The body is in equilibrium. H5.7) Find the tension in the cord at C and the horizontal and vertical component of reactions at pin A. The pulley is frictionless.The body is in equilibrium.Fx = 0 Fy = 0 MAnyPoint = 0NotindependentEquations!+
66 2D EquilibriumExample Hibbeler Ex 5-6 #1Determine the horizontal and vertical components of reaction for the beam loaded as shown. Neglect the weight of the beam in the calculation.
67 2D EquilibriumExample Hibbeler Ex 5-6 #2Find: Ay, Bx, ByFBD of ADB
73 tension is internal force. 3/ kg beam has a C.G. at G. The 80-kg man is exerted a 300-N force on the rope. Calculate the force reaction at A which is a weld pin.3 unknowns, 3 Equations.You can solve it.xy+mg =80(9.81)300 NTmg300 NN=mg?200(9.81)CN300NT= 300N?T=300N80(9.81)80(9.81)200(9.81)4 unknowns, 3 Eq.You can’t solve it?200(9.81)300 NYou can solve it !Use FBD of rodT=300NN is internal force.tension is internal force.C
74 tension is internal force. 3/ kg beam has a C.G. at G. The 80-kg man is exerted a 300-N force on the rope. Calculate the force reaction at A which is a weld pin.80(9.81)80(9.81)Sun’sGravitationalforce200(9.81)tension is internal force.C
77 Two-Force Member In Equilibrium Fx= Fy = Mo = 0Two-Force Member In EquilibriumTo keep the body in equillibrium,the second force must …Body inEquilibrium- be co-linear- direction of opposite side.body inequilibrium- has the same manitudebodies inequilibrium
78 (with proper moment arm) Fx= Fy = Mo = 0Three-Force Member for EquilibriumTo keep the body in equillibrium,the second force must …1)(co-planer)2)ConcurrentOR Parallel (with proper moment arm)concurrentParallel(with proper moment arm)4 forces need to be concurrent or parallelto make body in equilibrium?OM = 0
79 Special Member 2-Force Member #1 2D EquilibriumSpecial Member 2-Force Member #1DefinitionThe object isa two-force memberwhen subjected totwo equivalent forcesacting at different points.
80 Special Member 2-Force Member #2 2D EquilibriumSpecial Member 2-Force Member #2If a two-force member is in equilibrium, the two forces:are equal in magnitude.are opposite in direction.have the same line of action.There are no couple moments.The ability to recognize 2-force members is important in the analyses of structures.
81 Special Member 2-Force Member #3 2D EquilibriumSpecial Member 2-Force Member #3Examplecable with negligible weightlight bar with pin joints and no load
82 Special Member 2-Force Member #4 2D EquilibriumExampleSpecial Member 2-Force Member #4
83 Special Member 3-Force Member #1 2D EquilibriumSpecial Member 3-Force Member #1DefinitionThe object is a three-force member when subjected tothree equivalent forces acting at different points.If a three force member is in equilibrium, the threeforces are:coplanar andeither parallel or concurrent.
84 Special Member 3-Force Member #2 2D EquilibriumSpecial Member 3-Force Member #2Concurrent ForcesParallel Forces
85 Draw the FBD of the foot lever shown. why?Two forcein EquilibriumYou should write the FBD of 2-force (in equilibrium) like this.
86 The lever ABC is pin-supported at A and connected to a short link BD as shown. If the weight of the members is negligible, determine the force of the pin on the lever at A.4 unknowns, 3 eq.Hibbeler Ex 5-13xy+2 unknowns, 3 eq.Two ForceMembers
87 O Solve by using 3-force member Concept Three Forces are concurrent. + HibbelerEx 5-13Oxy+0.5 m=0.7 mAssumeDirections2 forcemembers=0.4 m3-Unknowns, 3-Eqs
88 The lever ABC is pin-supported at A and connected to a short link BD as shown. If the weight of the members is negligible, determine the force of the pin on the lever at A.2 forcemembers0.5 mxy+OAns
89 Any two-force members or three-force members here?
91 3/27 In a procedure to evaluate the strength of the triceps muscle, a person pushed down on a load cell with the palm of his hand as indicated in the figure.If the load-cell reading is 160 N, determine the vertical tensile force F generated by the triceps muscle. The mass of the lower arm is 1.5 kg with mass center at G.
92 y x T Ox 160N Oy 1.5(9.81) 3 Unknown, 3 Equations. You can solve it. Free Body DiagramIf what you wanto know is Oy ?TIf what you wanto know is Ox ?OxOy160N0.150.150.251.5(9.81)3 Unknown,3 Equations.You can solve it.By easier way.Because what we want to know is the force T of triceps.Taking summation of moments at point O will eliminate Ox and Oy which are unknowns out, and the calculation will be easier.+Ans
93 Technique: Finding unknown force only in 1 step Concrete slab with mass of 25,000 kg is pulled by cable of tension P. Determine the tension T in the horizontal cable using only 1 equilibrium equation.PxyGiven: = 60°DW = mgART6 mKeyHere we have 3 unknowns (P,R,T). To use only 1equilibrium eq. To determine T, we need to take moment at the point wherethe other unknown forces (R,P) passes.
96 Alternative Equilibrium equations Recall: In general, you have these three equilibrium equations:(for some point O) Fx= Fy = Mo = (E)- Alternatively, you may use eitherMA = 0MB = 0Fx= 0MA = 0MB = 0Mc = 0line AB not |_ to the x directionA B C not on the same straight lineR=0R=0not |_BBxACA
97 400 N 500 N 80 N-m Calculate the magnitude of the force at pin A. Ax AyFree Body DiagramBy500 Nxy80 N-mDHow many unknowns ?We have 3 unknowns ofAxAyBy
98 ANS 400 N 500 N 80 N-m Calculate the magnitude of the force at pin A. AxAyFree Body DiagramBy500 Nwe can determine Ay from+xy80 N-mwe can determine Ax from+Thus, the resultant of force at pin A:ANS
99 Constraints and Statical determinacy ** need to know when a problem can be solved or what force/couple can be found.Statically DeterminateImproper Constraint- Not enough constraintUnknown 3 ,Independent eq 2Unknown 3 ,Independent eq 32) Statically IndeterminateNot in equilibrium- can’t get all unknownsNo solution: (cant maintain moment)Unknown 4 ,Independent eq 3unknown 3 ,independent 2
100 Independent Equations y** Whatever 2D-equilibrium conditions you used, at most 3 unknowns (scalar) may be found.xBody in equilibriumHey , why you cannot get anymore, you can take another points to take moment, and get new equation?notindependentequations!Fx= 0 Fy = 0 Mo = 0(for some point O)** However, sometimes you can not get those 3 unknowns in 2D problem.Fx= 0 Fy = 0 Mo = 0(for all point O)Independent Equations on categories of 2D problem.Equilibrium1) Co-linearxDepend on x-axis selectionChoose thisdirection as x axisFx= Fy = Mo = 0(Choose point O on the line of action)
101 Depend on x-axis selection 2) Concurrent at a pointFx = 0Fy = 0Mo = 0yxconcurrent atpoint OO3) ParallelFx = 0Fy = 0Mo = 0Depend on x-axis selectionyx4) GeneralFx = 0Fy = 0Mz = 0yx
103 x Note on Solving Problems 1. If we have more unknowns than the number of independent equations, then we have a statically indeterminate situation. We cannot solve these problems using just statics.2. The point (axis) which we check for the moment, affects the simplicity of the solution. Choosing appropriate one is the key for fast problem solving.The order in which we apply equations may affect the simplicity of the solution. For example, solving FX = O first allows us to find the first unknown quickly.x4. If the answer for an unknown comes out as negative number, then the sense (direction) of the unknown force is opposite to that assumed when starting the problem.NIf you get N<0,something wrong!
104 3/48 The small crane is mounted on one side of the bed of a pickup truck. For the position q=45, determine the magnitude of the force supported by the pin at O and the oil pressure P against the 50-mm-diameter piston of the hydraulic cylinder BC.Body in interest:AOBC?AOC?Because We also want to find force P exerted by BC to the object AOC
105 y FBD x W Ox C Oy 0.34 C O E F 0.36 B F 0.11 G D H Ans To find C: + q=45Find P ,OIgnore mass of link CByFBDxCWOxOyCBwhichdirction?2 Force Member// with CBDirection // CBCan you find a ?FTo find C:0.11+0.34qCGqOEDF0.36aHBAns
106 y FDB 0.785 x W 0.34 C F O Ox Oy B A F 0.11 Ans To find reactions at O q=45Find P ,OAFDB0.785xWF0.110.34CFOqOxOyaBTo find reactions at OAns
107 3/46 It is desired that a person be able to begin closing the van hatch from the open position shown with a 40-N vertical force P. As a desired exercise, determine the necessary force in each of the two gas-pressurized struts AB. The mass center of the 40-kg door is 37.5 mm directly below point A. Treat the problem as two-dimensional.Why direction: CB?1125 mmAOxOy550 mm2FOW=mg600 mm175 mmP= 40NBUnknowns: F, Ox, OyEquilibrium eqs:
108 ANS Find F 2F A 2F A O G G O W=(40)(9.81) W=mg B P = 40 B + Oy Ox y x 1125 mm2FyAOy2FA550 mmOGxGOxO600 mmW=(40)(9.81)175 mmW=mgP = 40BB+ANS
109 Assumption: no couple reactions develop at C,D 3/38 The hinge P fits loosely through the frame tube, and the frame tube has a slight clearance between the supports A and B. Determine the reactions on the frame tube at A and B, associated with the weight L of an 80-kg person. Also, calculate the changes in the horizontal reactions at C and D.Assumption: no couple reactions develop at C,DL= 80(9.81)?
110 Find appropriate L to make Bodies in equilibrium.: knownq =: knownx=?
112 3/52 The rubber-tired tractor shown has a mass of 13 3/52 The rubber-tired tractor shown has a mass of 13.5 Mg with the center of mass at G and is used for pusshing or pulling heavy loads. Determine the load P which the tractor can pull at a constant speed of 5 km/h up the 15-percent grade if the driving force exerted by the ground on each of its four wheel is 80% of the normal force under the wheel. Also find the total normal reaction N under the rear pair of wheels at B.Normal reaction at A and B are equal?If not, which one should be larger?