Ch. 3: Equilibrium. 3.0 Outline Mechanical System Isolation (FBD) 2-D Systems Equilibrium Conditions 3-D Systems Equilibrium Conditions. 3.

Transcription

1 3.0 Outline Mechanical System Isolation (FBD) 2-D Systems Equilibrium Conditions 3-D Systems Equilibrium Conditions 3.0 Outline

2 3.0 Outline When a body is in equilibrium, the resultant on the body is zero. And if the resultant on a body is zero, the body is in equilibrium. So, F = 0 M = 0 is the necessary and sufficient conditions for equilibrium. 3.0 Outline

3 3.1 Mechanical System Isolation (FBD) Free-Body Diagram (FBD) is the most important first step in the mechanics problems. It defines clearly the interested system to be analyzed. It represents all forces which act on the system. The system may be rigid, nonrigid, or their combinations. The system may be in fluid, gaseous, solid, or their combinations. FBD represents the isolated / combination of bodies as a single body. Corresponding indicated forces may be 1. Contact force with other bodies that are removed virtually. 2. Body force such as gravitational or magnetic attraction forces. 3.1 FBD

4 3.1 FBD

5 Remarks 1. Force by flexible cable is always a tension. Weight of the cable may be significant. 2. Smooth surface ideally cannot support the tangential or frictional force. Contact force of the rough surface may not necessarily be normal to the tangential surface. 3. Roller, rocker, smooth guide, or slider ideally eliminate the frictional force. That is the supports cannot provide the resistance to motion in the tangential direction. 4. Pin connection provides support force in any direction normal to the pin axis. If the joint is not free to turn, a resisting couple may also be supported. 3.1 FBD

6 3.1 FBD

7 Remarks 5. The built-in / fixed support of the beam is capable of supporting the axial force, the shear force, and the bending moment. 6. Gravitational force is a kind of distributed non-contact force. The resultant single force is the weight acted through C.M. towards the center of the earth. 7. Remote action force has the same overall effects on a rigid body as direct contact force of equal magnitude and direction. 8. On the FBD, the force exerted on the body to be isolated by the body to be removed is indicated. 9. Sense of the force exerted on the FBD by the removed bodies opposes the movement which would occur if those bodies were removed. 3.1 FBD

8 Remarks 10. If the correct sense cannot be known at first place, the sense of the scalar component is arbitrarily assigned. Upon computation, a negative algebraic sign indicates that the correct sense is opposite to that assigned. 3.1 FBD

9 Construction of FBD 1. Make decision which body or system is to be isolated. That system will usually involve the unknown quantities. 2. Draw complete external boundary of the system to completely isolate it from all other contacting or attracting bodies. 3. All forces that act on the isolated body by the removed contacting and attracting bodies are represented on the isolated body diagram. Forces should be indicated by vector arrows, each with its magnitude, direction, and sense. Consistency of the unknowns must be carried throughout the calculation. 4. Assign the convenient coordinate axes. Only after the FBD is completed should the governing equations be applied. 3.1 FBD

10 3.1 FBD

11 Note 1. Include as much as possible the system in FBD while the unknowns are still being revealed. 2. Internal forces to a rigid assembly of members do not influence the values of the external reactions. And so the external response of the mechanism as a whole would be unchanged. 3. Include the weights of the members on FBD. 4. Try to get the correct sense of unknown vectors by visualizing the motion of the whole system when the supports are pretended to disappear. The correct sense will oppose the motion s direction. 5. Follow the action of force prototypes in determining the forces acted by the removed bodies. 3.1 FBD

12 3.1 FBD

13 A x Ay M O O x O y B x A x A y 3.1 FBD

14 3.1 FBD

15 F F B y A x M A A x 3.1 FBD

16 3.1 FBD

17 3.1 FBD

18 1. T y x mg F N 2. P mg On verge of being rolled over means the normal force N = 0 R N = 0 y x T 3.1 FBD

19 3. T y R x x R y 4. L N y A X mg m O g x A y 3.1 FBD

20 5. O mg T y x F N MO = 0 6. R y A X B X x A y B y 3.1 FBD

21 7. T Ch. 3: Equilibrium y A X x 8. T A y B X mg y A X B y x A y L 3.1 FBD

22 3.2 2-D Equilibrium Conditions A body is in equilibrium if all forces and moments applied to it are in balance. In scalar form, F = 0 F = 0 M = 0 x y O The x-y coordinate system and the moment point O can be chosen arbitrarily. Complete equilibrium in 2-D motion must satisfy all three equations. However, they are independent to each other. That is, equilibrium may only be satisfied in some generalized coordinates. System in equilibrium may stay still or move with constant velocity. In both cases, the acceleration is zero D Eqilibrium Conditions

23 Categories of equilibrium Ch. 3: Equilibrium Some equations are automatically satisfied and so contribute nothing in solving the problems D Eqilibrium Conditions

24 Weights of the members negligible Equilibrium of a body under the action of two force only: The forces must be equal, opposite, and collinear D Eqilibrium Conditions

25 Equilibrium of a body under the action of three force only: The lines of action of the three forces must be concurrent. The only exception is when the three forces are parallel. The system may be reduced to the three-force member by successive addition of the known forces. If all forces are concurrent, then the equilibrium statement calls for the closure of the polygon of forces D Eqilibrium Conditions

26 Alternative Equilibrium Equations Three independent equilibrium conditions: F = 0 M = 0 M = 0 x A B ( AB x-direction) D Eqilibrium Conditions

27 Alternative Equilibrium Equations Three independent equilibrium conditions: M = 0 M = 0 M = 0 A B C A, B, and C are not on the same straight line D Eqilibrium Conditions

28 Constraints and Statical Determinacy The equilibrium equations may not always solve all unknowns in the problem. Simply put, if #unknowns (including geometrical variables) > #equations, then we cannot solve it. This is because the system has more constraints than necessary to maintain the equilibruim. This is call statically indeterminate system. Extra equations, from force-deformation material properties, must also be applied to solve the redundant constraints D Eqilibrium Conditions

29 Constraints and Statical Determinacy P mg Q #unknowns = 2 #equilibrium eqs. = 2 statically determinate A y A x A x Ay B y B x P #unknowns = 4 #equilibrium eqs. = 3 statically indeterminate B y C y B x C x F #unknowns = 6 #equilibrium eqs. = 3 statically indeterminate D Eqilibrium Conditions

30 Adequacy of Constraints Ch. 3: Equilibrium D Eqilibrium Conditions

31 Problem Solution 1. List known unknown quantities, and check the number of unknowns and the number of available independent equations. 2. Determine the isolated system and draw FBD. 3. Assign a convenient set of coordinate systems. Choose suitable moment centers for calculation. 4. Write down the governing equation, e.g. M O = 0, before the calculation. 5. Choose the suitable method in solving the problem: scalar, vector, or geometric approach D Eqilibrium Conditions

32 P. 3/27 In a procedure to evaluate the strength of the triceps muscle, a person pushes 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. State any assumptions D Eqilibrium Conditions

33 P. 3/27 Assumption: contraction force from biceps muscle acts at point O 1. Known: weight of lower hand, pushing force Unknown: triceps force, biceps force 2. FBD: lower hand T C y x MO = 0 -T g = 0 T=1832 N Fy = 0 T-C-1.5g +160=0 C=1977 N 1.5g 160 N D Eqilibrium Conditions

34 P. 3/31 1. Unknown: l,r Known: m, b, T 2. FBD: tensioning system with cut-cable R T equivalent tension forces at the middle pulley F = 2Tcos30 y x F T mg Three-force member with m g, F, and O For equilibrium, three lines of action must be concurrent. MO = 0 F b-mg l = 0 l = ( ) F=0 R= F + mg = 3T + m g 2Tbcos30 mg D Eqilibrium Conditions

35 P. 3/33 The exercise machine consists of a lightweight cart which is mounted on small rollers so that it is free to move along the inclined ramp. Two cables are attached to the cart one for each hand. If the hands are together so that the cables are parallel and if each cable lies essentially in a vertical plane, determine the force P which each hand must exert on its cable in order to maintain an equilibrium position. The mass of the person is 70 kg, the ramp angle is 15, and the angleβis 18. In addition, calculate the force R which the ramp exerts on the cart D Eqilibrium Conditions

36 T P. 3/33 Assumption: negligible rail friction 70g 1. Unknown: P, T, R 2. FBD: exercise machine, pulley T T x 2P R F F x y ' ' = 0 70gsin15-Tcos9 = 0 T = N = 0 R-70gcos15-Tsin9 = 0 R = 691 N x T - 4Pcos9 = 0 P = 45.5 N 2P D Eqilibrium Conditions

37 P. 3/35 A uniform ring of mass m and radius r carries an eccentric mass m o at a radius b and is in an equilibrium position on the incline, which makes an angleαwith the horizontal. If the contacting surfaces are rough enough to prevent slipping, write the expression for the angleθwhich defines the equilibrium position D Eqilibrium Conditions

38 P. 3/35 1. Unknown: F, N, θ 2. FBD: ring+eccentric mass mg m o g F x mogbsinθ MO = 0 Fr - mogbsin θ = 0 F = r r m = ( + ) α θ + α -1 F ' 0 x F - mo m gsin = 0 = sin 1 sin b mo N D Eqilibrium Conditions

39 P. 3/39 The hook wrench or pin spanner is used to turn shafts and collars. If a moment of 80 Nm is required to turn the 200 mm diameter collar about its center O under the action of the applied force P, determine the contact force R on the smooth surface at A. Engagement of the pin at B may be considered to occur at the periphery of the collar D Eqilibrium Conditions

40 P. 3/39 R B 80 Nm y x Three - force member shaft & hook as one system MO = 0 80-P = 0 P = N A N A ( ) MB = 0 NA 0.1sin 60 P cos60 = 0 N = 1047 N D Eqilibrium Conditions

41 P. 3/48 The small crane is mounted on one side of the bed of a pickup truck. For the positionθ=40, 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 D Eqilibrium Conditions

42 P. 3/48 D C O d α 360 α B geometry at BCDO sin cos 40 α = = 340 cos sin 40 d = 360cosα = 200 mm 1 tan D Eqilibrium Conditions

43 P. 3/48 y x 120g C O O x Three - force member 2 2 x y ( ) MO = 0 120g cos 40 C d = 0 C = 5063 N F p = = 2.58 MPa 2 π r Fx = 0 Ox Ccos α = 0 Ox = 2820 N Fy = 0 - Oy 120g + Csin α = 0 Oy = 3030 N O = O + O = 4140 N O y O D Eqilibrium Conditions

44 P. 3/52 The rubber-tired tractor shown has a mass of 13.5 Mg with the C.M. at G and is used for pushing 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 wheels is 80 percent of the normal force under that wheel. Also find the total normal reaction N B under the rear pair of wheels at B D Eqilibrium Conditions

45 P. 3/52 Ch. 3: Equilibrium y x 13500g 0.8N A 0.8NB N A NB 15 ' F = 0 x P - 0.8NA 0.8NB g = F ' = 0 N y A + NB 13500g = MA = 0 NB 1.8 P g g = N = 6.3 kn, N = kn, P = 85.1 kn A B alternative equations: M = 0 M = 0 F = 0 A B x ' D Eqilibrium Conditions

46 P. 3/53 Pulley A delivers a steady torque (moment) of 100 Nm to a pump through its shaft at C. The tension in the lower side of the belt is 600 N. The driving motor B has a mass of 100 kg and rotates clockwise. Determine the magnitude R of the force on the supporting pin at O D Eqilibrium Conditions

47 P. 3/53 mg T 100 Nm by load y 600 N ( ) MC = T = 0 T = N D P x T 100g 600 N O x O y y ( ) MD = 0 Oy g T Tcos Tsin = 0 O = 906 N Fx = 0 Tcos Ox = 0 Ox = N O= O M = 0 F = 0 M = 0 D x O + O = 1.17 kn 2 2 x y Fy = 0 Tsin30-100g - P + Oy = 0 P = 2.8 N spring compressed to resist rotation of the body D Eqilibrium Conditions

48 P. 3/56 When setting the anchor so that it will dig into the sandy bottom, the engine of the 40 Mg cruiser with C.G. at G is run in reverse to produce a horizontal thrust T of 2 kn. If the anchor chain makes an angle of 60 with the horizontal, determine the forward shift b of the center of buoyancy from its position when the boat is floating free. The center of buoyancy is the point through which the resultant of the buoyant force passes D Eqilibrium Conditions

49 P. 3/56 y 40000g x b B x A free floating (no thrust, tension): buoyancy force = weight, acting at C.G. backward motion: new buoyancy force acting at new position to maintain equilibrium Fx = 0 Acos = 0 A = 4 kn Fy = 0 B g - Asin60 = 0 B = N MA = g Bx = 0 x = m b = 8- x = 85.2 mm D Eqilibrium Conditions

50 P. 3/59 A special jig for turning large concrete pipe sections (shown dotted) consists of an 80 Mg sector mounted on a line of rollers at A and a line of rollers at B. One of the rollers at B is a gear which meshes with a ring of gear teeth on the sector so as to turn the sector about its geometric center O. When α= 0, a counterclockwise torque of 2460 Nm must be applied to the gear at B to keep the assembly from rotating. When α = 30, a clockwise torque of 4680 Nm is required to prevent rotation. Locate the mass center G of the jig by calculating r and θ D Eqilibrium Conditions

51 P. 3/59 F Nm 4680 Nm F 2 MB = 0 α = 0 : F = 0, F1 = N α = 30 : F 0.24 = 0, F = N g y N A F N B MO = 0 α = 0 : 80000g rcosθ = 0 α = 30 : g rcos = 0 r = 367 mm, θ = 79.8 ( θ) x D Eqilibrium Conditions

52 3.3 3-D Equilibrium Conditions A body is in equilibrium if all forces and moments applied to it are in balance. In scalar form, F = 0 F = 0 F = 0 x y z M = 0 M = 0 M = 0 O O O x y z The x-y-z coordinate system and the moment point O can be chosen arbitrarily. Complete equilibrium in 3-D motion must satisfy all six equations. However, they are independent to each other. That is, equilibrium may only be satisfied in some generalized coordinates. System in equilibrium may stay still or move with constant velocity. In both cases, the acceleration is zero D Eqilibrium Conditions

53 3.3 3-D Eqilibrium Conditions

54 Categories of equilibrium Ch. 3: Equilibrium Some equations are automatically satisfied and so contribute nothing in solving the problems D Eqilibrium Conditions

55 Constraints and Statical Determinacy The equilibrium equations may not always solve all unknowns in the problem. Simply put, if #unknowns (including geometrical variables) > #equations, then we cannot solve it. This is because the system has more constraints than necessary to maintain the equilibrium. This is call statically indeterminate system. Extra equations, from force-deformation material properties, must also be applied to solve the redundant constraints D Eqilibrium Conditions

56 Adequacy of Constraints Ch. 3: Equilibrium D Eqilibrium Conditions

57 P. 3/67 The light right angle boom which supports the 400 kg cylinder is supported by three cables and a ball-and-socket joint at O attached to the vertical x-y surface. Determine the reactions at O and the cable tensions D Eqilibrium Conditions

58 P. 3/67 Ch. 3: Equilibrium T AC O T BD T BE 400g n = 0.408i j 0.816k AC n = 0.707j 0.707k BD n = k, n = i BE OB OD OE n = 0.6i+ 0.8k n = 0.6i+ 0.8j M OB AC ( ) ( ) ( ) M OD BE ( ) ( ) ( ) ( ) M OE = 0 to find T 0.75i 400gj + 2k TACnAC i nob = 0 TAC = N = 0 to find T 2k TACnAC i+ 2k 400gj + 1.5i TBEnBE i nod = 0 TBE = 654 N = 0 to find T BD ( j) n ( i+ k) F=0 Ox = 1962 N, Oy = 0 N, Oz = 6540 N ( ) ( ) AC AC OE BD 2 TBD BD gj + 2k T n i n = 0 T = N D Eqilibrium Conditions

59 P. 3/68 The 600 kg industrial door is a uniform rectangular panel which rolls along the fixed rail D on its hanger-mounted wheels A and B. The door is maintained in a vertical plane by the floor-mounted guide roller C, which bears against the bottom edge. For the position shown compute the horizontal side thrust on each of the wheels A and B, which must be accounted for in the design of the brackets D Eqilibrium Conditions

60 P. 3/68 Ch. 3: Equilibrium A x B x B z A z 600g N C MAB = 0 600g 0.15 NC 3 = 0 NC = N MA = 0 z NC 0.6 Bx 3 = 0 Bx = N Fx = 0 Ax + Bx NC = 0 Ax = N D Eqilibrium Conditions

61 P. 3/73 The smooth homogeneous sphere rests in the 120 groove and bears against the end plate which is normal to the direction of the groove. Determine the angle θ, measured from the horizontal, for which the reaction on each side of the groove equals the force supported by the end plate D Eqilibrium Conditions

62 P. 3/73 Projection onto two orthogonal planes z mgcosθ y z θ mg x N 1 30 N 2 N r N 1 cos30+n 2 cos30 Fy = 0 N1 = N2 = N Fz = 0 mgcosθ = 2Ncos30 Fx = 0 Nr = mgsinθ if N = N, tanθ = 1/ 2cos30 θ = 30, N = mg/2 r D Eqilibrium Conditions

63 P. 3/74 The mass center of the 30 kg door is in the center of the panel. If the weight of the door is supported entirely by the lower hinge A, calculate the magnitude of the total force supported by the hinge at B D Eqilibrium Conditions

64 P. 3/74 z B x B y A x 30g y A y F = 0 M = 0 30g 0.36 B 1.5 = 0, B = A = 70.6 N y x A x x x F = 0 M = 0 B g 0.9 = 0, B = A = N x y A y y y B= B + B = N 2 2 x y 30g x D Eqilibrium Conditions

65 P. 3/79 One of the three landing pads for the Mars Viking lander is shown in the figure with its approximate dimensions. The mass of the lander is 600 kg. Compute the force in each leg when the lander is resting on a horizontal surface on Mars. Assume equal support by the pads and consult Table D/2 in Appendix D as needed D Eqilibrium Conditions

66 P. 3/79 F DC T CB g=3.73 m/s 2 T CA 200g n = 0.35i k, n = i j k DC M BA ( ) = 0 to find F DC DC CA CA CB DC CA 0.85k+ 0.1i FDC ndc 200g 0.55j i j= 0 FDC = N Fx = 0 and symmetry about x-z plane F n i i 2T = 0 T = T = N D Eqilibrium Conditions

67 P. 3/82 The uniform 15 kg plate is welded to the vertical shaft, which is supported by bearings A and B. Calculate the magnitude of the force supported by bearing B during application of the 120 Nm couple to the shaft. The cable from C to D prevents the plate and shaft from turning, and the weight of the assembly is carried entirely by bearing A D Eqilibrium Conditions

68 P. 3/82 A y 15g A B x y z B x y 15g n = 0.95i 0.316j DC MO = T z + i ndcik = 0, T = N MA = 0 y Bx g Tx 0.68 = 0, Bx = 2265 N MA = 0 x By 0.2 Ty 0.68 = 0, By = 680 N B= B + B = 2635 N 2 2 x y T x D Eqilibrium Conditions

69 P. 3/88 The uniform 900x1200 mm trap door has a mass of 200 kg and is propped open by the light strut AB at the angle θ= atan(4/3). Calculate the compression F B in the strut and the force supported by the hinge D normal to the hinge axis. Assume that the hinges act at the extreme ends of the lower edge D Eqilibrium Conditions

70 P. 3/88 z D z C x T AB D x 200g D y y n = i j k AB 2 2 n y z [ ] MC = 0 x 0.9j TABnAB ii 200g 0.45cos53.13 = 0, TAB = 688 N MC = 0 200g D y z 1.2 = 0, Dz = 981 N ( n ii) MC = 0 z Dy TAB AB 0.9 = 0, Dy = N D = D + D = 992 N D Eqilibrium Conditions

71 P. 3/92 The uniform rectangular panel ABCD has a mass of 40 kg and is hinged at its corners A and B to the fixed vertical surface. A wire from E to D keeps edges BC and AD horizontal. Hinge A can support thrust along the hinge axis AB, whereas hinge B supports force normal to the hinge axis only. Find the tension T in the wire and the magnitude B of the force supported by hinge B D Eqilibrium Conditions

72 P. 3/92 x B y B z 40g T DE z A y n = 0.35i 0.707j+ 0.61k DE ( ) [ ] i MA = g cos30 sin 30 x j k i i i + 1.2j T n i = 0 T = N DE DE DE ( ) MA = g cos 30 sin B y i k i i j z = 0, Bz = N MAE = 0 By = 0 N B = N n y A z A x D Eqilibrium Conditions

73 P. 3/93 Under the action of the 40 Nm torque (couple) applied to the vertical shaft, the restraining cable AC limits the rotation of the arm OA and attached shaft to an angle of 60 measured from the y-axis. The collar D fastened to the shaft prevents downward motion of the shaft in its bearing. Calculate the bending moment M, the compression P, and the shear force V in the shaft at section B. (note: Bending moment, expressed as a vector, is normal to the shaft axis, and shear force is also normal to the shaft axis.) D Eqilibrium Conditions

74 P. 3/93 x y M Bx V x 40 Nm P V y M By T AC section the shaft at B revealing the reaction force and moment n = 0.53i+ 0.38j 0.758k AC B x x y AC [ ] Mz = j TACnAC ik = 0 TAC = N Fz = 0 P + TACnACik = 0 P = N V = V + V = T P = N B y ( ) MB = 0 MB i+ M x B j k+ 0.18j T y ACnAC = M = Nm, M = 20.0 Nm M = M + M = 47.3 Nm b B B x y D Eqilibrium Conditions

75 P. 3/ D Eqilibrium Conditions

76 P. 3/94 z x y FBD of reel only N C N B MO y = P 0.3 = 0, P = 50 N MB = sin x Psin N 0.5 = 0, N = N M B = 0 N y A NC cos sin C Pcos Psin = 0, N = N Fz = 0 NA + NB + 100sin15 + NC Psin = 0, NB = N C A N A D Eqilibrium Conditions

77 P. 3/109 The drum and shaft are welded together and have a mass of 50 kg with mass center at G. The shaft is subjected to a torque (couple) of 120 Nm, and the drum is prevented from rotating by the cord wrapped securely around it and attached to point C. Calculate the magnitudes of the forces supported by bearings A and B D Eqilibrium Conditions

78 P. 3/109 50g z A x A z y T x B x Bz MB = 0 T = 0, T = 800 N y MB = 0 Tcos A z x 0.7 = 0, Ax = N MB = 0 50g 0.3 Tsin A x + z 0.7 = 0, Az = N F = 0 A + B Tcos66.87 = 0, B = N x x x x Fz = 0 z z z A + B 50g Tsin66.87 = 0, B = N T A = A + A = N, B = B + B = N x z x z D Eqilibrium Conditions

79 P. 3/ D Eqilibrium Conditions

80 P. 3/110 double U-joint z TBC O y M O x T AD x O z n = 0.13i 0.91j k, n = 0.48i 0.84j k BC BC MAB = 0 Oz = 0 N AD ( ) i ( ) Mz = 0 1.8i TBCnBC k+ 2.1j TADnAD ik = 0 ( ) i ( ) Mx = 0 2.1j TBCnBC i+ 2.1j TADnAD ii 50g 2.1 = 0 T = 625 N, T = 1024 N AD x y z ( ) My = 0 M + 50gx + 1.5i TBCnBC ij= 0 M = x Fy = 0 Oy + TBCnBCij+ TADnADij= 0, Oy = 1429 N Fx = 0 Ox + TBCnBCii+ TADnAD ii = 0, Ox = 410 N O = O + O + O = 1487 N 50g D Eqilibrium Conditions