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# 6161103 6.6 frames and machines

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• 1. 6.6 Frames and MachinesComposed of pin-connected multi-forcemembers (subjected to more than twoforces)Frames are stationary and are used tosupport the loads while machinescontain moving parts, designated totransmit and alter the effects of forcesApply equations of equilibrium to eachmember to determine the unknownforces
• 2. 6.6 Frames and MachinesFree-Body Diagram Isolate each part by drawing its outlined shape - show all the forces and the couple moments that act on the part - label or identify each known and unknown force and couple moment with reference to the established x, y and z coordinate system
• 3. 6.6 Frames and MachinesFree-Body Diagram - indicate any dimension used for taking moments - equations of equilibrium are easier to apply when the forces are represented in their rectangular coordinates - sense of any unknown force or moment can be assumed
• 4. 6.6 Frames and MachinesFree-Body Diagram Identify all the two force members in the structure and represent their FBD as having two equal but opposite collinear forces acting at their points of application Forces common to any contracting member act with equal magnitudes but opposite sense on the respective members
• 5. 6.6 Frames and MachinesFree-Body Diagram - treat two members as a system of connected members - these forces are internal and are not shown on the FBD - if the FBD of each member is drawn, the forces are external and must be shown on the FBD
• 6. 6.6 Frames and MachinesExample 6.9For the frame, draw the free-body diagram of(a) each member,(b) the pin at B and(c) the two membersconnected together.
• 7. 6.6 Frames and MachinesSolutionPart (a) members BA and BC are not two-force members BC is subjected to 3 forces, the resultant force from pins B and C and the external P AB is subjected to the resultant forces from the pins at A and B and the external moment M
• 8. 6.6 Frames and MachinesSolutionPart (b) Pin at B is subjected to two forces, force of the member BC on the pin and the force of member AB on the pin For equilibrium, these forces and respective components must be equal but opposite
• 9. 6.6 Frames and MachinesSolutionPart (b) But Bx and By shown equal and opposite on members AB ad BC results from the equilibrium analysis of the pin rather from Newton’s third law
• 10. 6.6 Frames and MachinesSolutionPart (c) FBD of both connected members without the supporting pins at A and C Bx and By are not shown since they form equal but opposite collinear pairs of internal forces
• 11. 6.6 Frames and MachinesSolutionPart (c) To be consistent when applying the equilibrium equations, the unknown force components at A and C must act in the same sense Couple moment M can be applied at any point on the frame to determine reactions at A and C
• 12. 6.6 Frames and MachinesExample 6.10A constant tension in the conveyor belt ismaintained by using the device. Draw theFBD of the frame andthe cylinder whichsupports the belt.The suspended blackhas a weight of W.
• 13. 6.6 Frames and MachinesSolution Idealized model of the device Angle θ assumed known Tension in the belt is the same on each side of the cylinder since it is free to turn
• 14. 6.6 Frames and MachinesSolution FBD of the cylinder and the frame Bx and By provide equal but opposite couple moments on the cylinder Half of the pin reactions at A act on each side of the frame since pin connections occur on each side
• 15. 6.6 Frames and MachinesExample 6.11Draw the free-body diagrams of each part ofthe smooth piston and link mechanism usedto crush recycled cans.
• 16. 6.6 Frames and MachinesSolution Member AB is a two force member FBD of the parts
• 17. 6.6 Frames and MachinesSolution Since the pins at B and D connect only two parts together, the forces are equal but opposite on the separate FBD of their connected members Four components of the force act on the piston: Dx and Dy represent the effects of the pin and Nw is the resultant force of the floor and P is the resultant compressive force caused by can C
• 18. 6.6 Frames and MachinesExample 6.12For the frame, draw the free-body diagrams of (a)the entire frame including the pulleys and cords, (b)the frame without the pulleys and cords, and (c)each of the pulley.
• 19. 6.6 Frames and MachinesSolutionPart (a) Consider the entire frame, interactions at the points where the pulleys and cords are connected to the frame become pairs of internal forces which cancel each other and not shown on the FBD
• 20. 6.6 Frames and MachinesSolutionPart (b) and (c) When cords and pulleys are removed, their effect on the frame must be shown
• 21. 6.6 Frames and MachinesExample 6.13Draw the free-bodydiagrams of the bucket andthe vertical boom of the backhoe. The bucket and itscontent has a weight W.Neglect the weight of themembers.
• 22. 6.6 Frames and MachinesSolution Idealized model of the assembly Members AB, BC, BE and HI are two force members
• 23. 6.6 Frames and MachinesSolution FBD of the bucket and boom Pin C subjected to 2 forces, force of the link BC and force of the boom Pin at B subjected to three forces, force by the hydraulic cylinder and the forces caused by the link These forces are related by equation of force equilibrium
• 24. 6.6 Frames and MachinesEquations of Equilibrium Provided the structure is properly supported and contains no more supports and members than necessary to prevent collapse, the unknown forces at the supports and connections can be determined from the equations of equilibrium The selection of the FBD for analysis are completely arbitrary and may represent each of the members of the structure, a portion or its entirety.
• 25. 6.6 Frames and MachinesEquations of Equilibrium Consider the frame in fig (a) Dismembering the frame in fig (b), equations of equilibrium can be used FBD of the entire frame in fig (c)
• 26. 6.6 Frames and MachinesProcedures for AnalysisFBD Draw the FBD of the entire structure, a portion or each of its members Choice is dependent on the most direct solution to the problem When the FBD of a group of members of a structure is drawn, the forces at the connected parts are internal forces and are not shown Forces common to two members which are in contact act with equal magnitude but opposite sense on their respective FBD
• 27. 6.6 Frames and MachinesProcedures for AnalysisFBD Two force members, regardless of their shape, have equal but opposite collinear forces acting at the ends of the member In many cases, the proper sense of the unknown force can be determined by inspection Otherwise, assume the sense of the unknowns A couple moment is a free vector and can act on any point of the FBD
• 28. 6.6 Frames and MachinesProcedures for AnalysisFBD A force is a sliding vector and can act at any point along its line of actionEquations of Equilibrium Count the number of unknowns and compare to the number of equilibrium equations available In 2D, there are 3 equilibrium equations written for each member
• 29. 6.6 Frames and MachinesProcedures for AnalysisEquations of Equilibrium Sum moments about a point that lies at the intersection of the lines of action of as many unknown forces as possible If the solution of a force or couple moment magnitude is found to be negative, it means the sense of the force is the reserve of that shown on the FBD
• 30. 6.6 Frames and MachinesExample 6.14Determine the horizontal and verticalcomponents of the force which the pin Cexerts on member CBof the frame.
• 31. 6.6 Frames and MachinesSolutionMethod 1 Identify member AB as two force member FBD of the members AB and BC
• 32. 6.6 Frames and MachinesSolution ∑ M C = 0; 2000 N (2m) − FAB sin 60o (4m) = 0 FAB = 1154.7 N + → ∑ Fx = 0; 1154.7 cos 60o − C x = 0 C x = 577 N + ↑ ∑ Fy = 0; 1154.7 sin 60o N − 2000 N − C y = 0 C y = 1000 N
• 33. 6.6 Frames and MachinesSolutionMethod 2 Fail to identify member AB as two force member
• 34. 6.6 Frames and MachinesSolutionMember AB ∑MA = 0; Bx (3sin60o m) − By (3cos60o m) = 0 + →∑Fx = 0; Ax − Bx = 0 + ↑ ∑Fy = 0; Ay − By = 0
• 35. 6.6 Frames and MachinesSolutionMember BC∑ M C = 0;2000 N (2m) − B y (4m) = 0+ → ∑ Fx = 0;Bx − C x = 0+ ↑ ∑ Fy = 0;B y − 2000 N + C y = 0B y = 1000 N ; Bx = 577 N ; C x = 577 N ; C y = 1000 N
• 36. 6.6 Frames and MachinesExample 6.15The compound beam is pin connected at B.Determine the reactions at its support.Neglect its weight and thickness.
• 37. 6.6 Frames and MachinesSolution FBD of the entire frame Dismember the beam into two segments since there are 4 unknowns but 3 equations of equilibrium
• 38. 6.6 Frames and MachinesSolutionSegment BC+ → ∑ Fx = 0;Bx = 0∑ M B = 0;− 8kN (1m) + C y (2m) = 0+ ↑ ∑ Fy = 0;B y − 8kN + C y = 0
• 39. 6.6 Frames and MachinesSolutionMember AB+ → ∑ Fx = 0;Ax − (10kN )  + Bx = 0 3   5∑ M A = 0;M A − (10kN ) ( 2m) − B y (4m) = 0 4  5+ ↑ ∑ Fy = 0;Ay − (10kN )  − B y = 0 4  5Ax = 6kN ; Ay = 12kN ; M A = 32kN .m; Bx = 0; B y = 4kN ; C y = 4kN
• 40. 6.6 Frames and MachinesExample 6.16Determine the horizontal and verticalcomponents of the force which the pin at Cexerts on member ABCD ofthe frame.
• 41. 6.6 Frames and MachinesSolution Member BC is a two force member FBD of the entire frame FBD of each member
• 42. 6.6 Frames and MachinesSolutionEntire Frame ∑ M A = 0;−981N (2m) + Dx (2.8m) = 0 Dx = 700.7 N + → ∑ Fx = 0; Ax − 700.7 N = 0 Ax = 700.7 N + ↑ ∑ Fy = 0; Ay − 981N = 0 Ay = 981N
• 43. 6.6 Frames and MachinesSolutionMember CEF ∑ M C = 0;−981N (2m) − ( FB sin 45o )(1.6m) = 0 FB = −1734.2 N + → ∑ Fx = 0;−C x − (−1734.2 cos 45o N ) = 0 C x = 1226 N + ↑ ∑ Fy = 0; C y − ( −1734.2 sin 45o N ) − 981N = 0 C y = −245 N
• 44. 6.6 Frames and MachinesExample 6.17The smooth disk is pinned at D and has a weight of20N. Neglect the weights of others member,determine the horizontal and vertical componentsof the reaction at pins B and D
• 45. 6.6 Frames and MachinesSolution FBD of the entire frame FBD of the members
• 46. 6.6 Frames and MachinesSolutionEntire Frame ∑ M A = 0;−20 N (3cm) + C x (3.5cm) = 0 C x = 17.1N + → ∑ Fx = 0; Ax − 17.1N = 0 Ax = 17.1N + ↑ ∑ Fy = 0; Ay − 20 N = 0 Ay = 20 N
• 47. 6.6 Frames and MachinesSolutionMember AB + → ∑ Fx = 0;17.1N − Bx = 0 Bx = 17.1N ∑ M A = 0;−20 N (6cm) + N D (3cm) = 0 N D = 40 N + ↑ ∑ Fy = 0;20 N − 40 N + B y = 0 B y = 20 N
• 48. 6.6 Frames and MachinesSolutionDisk + → ∑ Fx = 0; Dx = 0 + ↑ ∑ Fy = 0; 40 N − 20 N − D y = 0 D y = 20 N
• 49. 6.6 Frames and MachinesExample 6.18Determine the tension in the cablesand also the force P required tosupport the 600N force using thefrictionless pulley system.
• 50. 6.6 Frames and MachinesSolution FBD of each pulley Continuous cable and frictionless pulley = constant tension P Link connection between pulleys B and C is a two force member
• 51. 6.6 Frames and MachinesSolutionPulley A+ ↑ ∑ Fy = 0;3P − 600 N = 0P = 200 NPulley B+ ↑ ∑ Fy = 0;T − 2 P = 0T = 400 NPulley C+ ↑ ∑ Fy = 0; R − 2 P − T = 0R = 800 N
• 52. 6.6 Frames and MachinesExample 6.19A man having a weight of 750N supportshimself by means of the cable andpulley system. If the seat has aweight of 75N, determine the forcehe must exert on the cable at A andthe force he exerts on the seat.Neglect the weight of the cablesand pulleys.
• 53. 6.6 Frames and MachinesSolutionMethod 1 FBD of the man, seat and pulley C
• 54. 6.6 Frames and MachinesSolutionMan+ ↑ ∑ Fy = 0;TA + N S − 750 N = 0Seat+ ↑ ∑ Fy = 0;TE + N S − 75 N = 0Pulley C+ ↑ ∑ Fy = 0;2TE − TA = 0TA = 550;TE = 275 N ; N E = 200 N
• 55. 6.6 Frames and MachinesSolutionMethod 2 FBD of the man, seat and pulley C as a single system
• 56. 6.6 Frames and MachinesSolution+ ↑ ∑ Fy = 0;3TE − 75 N − 750 N = 0TE = 275 N+ ↑ ∑ Fy = 0;TE + N S − 75 N = 0+ ↑ ∑ Fy = 0;2TE − TA = 0TA = 550; N E = 200 N
• 57. 6.6 Frames and MachinesExample 6.20The hand exerts a force of 35N on the grip of thespring compressor. Determine the force in thespring needed to maintain equilibrium of themechanism.
• 58. 6.6 Frames and MachinesSolution FBD for parts DC and ABG
• 59. 6.6 Frames and MachinesSolutionLever ABG∑ M B = 0; FEA (25mm) − 35 N (100mm) = 0FEA = 140 NPin E+ ↑ ∑ Fy = 0; FEA sin 60o − FEF sin 60o = 0FED = FEF = F+ → ∑ Fx = 0;2 F cos 60o − 140 N = 0F = 140 N
• 60. 6.6 Frames and MachinesSolutionArm DC ∑ M C = 0; − Fs (150mm) + 140 cos 30o (75mm) = 0 Fs = 60.62 N
• 61. 6.6 Frames and MachinesExample 6.21The 100kg block is held in equilibrium bymeans of the pulley and the continuouscable system. If the cable isattached to the pin at B,compute the forces which thispin exerts on each of itsconnecting members
• 62. 6.6 Frames and MachinesSolution FBD of each member of the frame Ad and CB are two force members
• 63. 6.6 Frames and MachinesSolutionPulley B + → ∑ Fx = 0; Bx − 490.5 cos 45o N = 0 Bx = 346.8 N + ↑ ∑ Fy = 0; B y − 490.5 sin 45o N − 490.5 N = 0 B y = 837.3 N
• 64. 6.6 Frames and MachinesSolutionPin E 4 + ↑ ∑ Fy = 0; FCB − 837.3 N − 490.5 N = 0 5 FCB = 1660 N 3 + → ∑ Fx = 0; FAB − (1660 N ) − 346.8 N = 0 5 FAB = 1343 N Two force member BC subjected to bending as caused by FBC Better to make this member straight so that the force would only cause tension in the member