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lecture.20.21.Frames.ppt
- 1. Lecture 20 ENGR-1100 Introductionto Engineering Analysis
- 2. Previous Lectures Outline 2Dtrusses analysis- a) method of joints. b) method of sections.
- 3.
- 4. Two Important StructuresTypes • Trusses: Structures composed entirely of two force members. • Frames: Structures containing at least one member acted on by forces at three or more points.
- 5. Trusses Assumptions 1) Trussmembers are connected together at their ends only. 2) Truss members are connected together by frictionless pins. 3) The truss structure is loaded only at the joints. 4) The weight of the member may be neglected.
- 6. Frames vs. Machines •Frames - Rigid structure - Overall equilibrium is sufficient to determine support reaction. • Machines - Not a rigid structure - Overall equilibrium is not sufficient to determine support reaction. additional support reaction is needed for equilibrium
- 7. Beware • Members ofa frame are not necessarily a two force member. • The direction of the force applied by the members on the pins are not necessarily known. F
- 8. Method of solvingframes • Draw a free body diagram for each component • Not all members can be treated as two-force members. • Write the equilibrium equations for each free body diagram. • Solve the equilibrium equations of the system of rigid bodies.
- 9. Example 7-85 A two-barframe is loaded and supported as shown in Fig. F7-85. Determine the reactions at supports A and E and the force exerted on member ABC by the pin at C.
- 10. Solution x y d = 6/tan(70)+ 6/tan(70) = 5.648 ft MA = Ey (5.648) -500(2)- 400 (4)- -300 (6) = 0 Fy = Ay + 779.0 = 0 Ay =- 779.0 lb=779 lb Ey = 779.0 lb 779 lb From a free-body diagram on the complete frame: d
- 11. From a free-bodydiagram on member CDE: MC = 400 (2) + 779.0 (3.464) + Ex (6) = 0 Ex = -583.1 lb 583 lb From a free-body diagram on member ABC: x y MC = 500 (4) + Ax (6) + 779.0 (2.184) = 0 Fx = Cx + 500 - 616.9 = 0 Ax = -616.9 lb 617 lb Cx = 116.9 lb = 116.9 lb
- 12. Fy = Cy- 779.0 = 0 Cy = 779.0 lb = 779 lb C = = = 787.7 lb 788 lb 2 2 x y C C 2 2 116.9 799 c = tan -1 = 81.46 C 788 lb 81.5 779.0 116.9 A = = = 993.7 lb 994 lb 2 2 x y A A 2 2 616.9 799.0 A = tan -1 = -128.38 A 994 lb 51.6 779.0 616.9
- 13. E = == 973.1 lb 973 lb 2 2 x y E E 2 2 583.1 799.0 E= tan -1 = 126.82 E 973 lb 53.2 779.0 583.1
- 14. Class Assignment: Exerciseset P7-83 please submit to TA at the end of the lecture Determine all forces acting on member BCD of the linkage shown in Fig. P7-83.
- 15. C 85.73lb 45 Member AC is a two-force member; Therefore, the line of action of force C is known as shown on the free-body diagram for member BCD: MB = C cos 45 (2.0) - 40 cos 30 (3.5) = 0 C = 85.73 lb 85.7 lb Fx = Bx + 40 cos 30 - 85.73 cos 45 = 0 Fy = By - 85.73 sin 45 + 40 sin 30 = 0 Bx = 25.98 lb 26.0 lb By = 40.62 lb 40.6 lb
- 16. B = tan-1 = tan -1 = 57.4 40.62 25.98 B 48.2 lb 57.4 B = ( Bx)2 + ( Bx)2 = ( 25.98)2 + ( 40.62)2 = 48.22 lb y x B B
- 17. Example 7-90 Determine allforces acting on member ABE of the frame shown in Fig. P7-90.
- 18. Solution x y From a free-bodydiagram on the complete frame: MA = D (300) - 150 (300) = 0 D = 150.0 N = 150.0 N Fx = Ax + 150.0 = 0 Ax =- 150.0 N = 150.0 N Fy = Ay + 150.0 = 0 Ay =- 150.0 N = 150.0 N A = tan -1 = 135.0 A = 212 N 45 150.0 150.0 A = = = 212.1 N 212 N 2 2 x y A A 2 2 150.0 150.0
- 19. From a free-bodydiagram on member ABE: MB = Ey (100) - 300 (100) - 150(100) = 0 Ey = 450 N = 450 N E = = = 540.8 N 2 2 x y E E 2 2 300 450 E = tan -1 = 56.31 E = 541 N 56.3 450 300 MC = Ex (100) - 150 (200) = 0 Ex = 300 N = 300 N (on ABE) From a free-body diagram on member CEF: 150N Cx Ex Ey Cy
- 20. Fx = Bx+ 300 - 150 = 0 Bx =- 150.0 N = 150.0 N Fy = By + 450 - 150 = 0 By =- 300 N = 300 N B = = = 335.4 N 2 2 x y B B 2 2 150 300 B = tan -1 = -116.56 B 335 N 63.4 300 150
- 21. Class Assignment: Exerciseset P7-91 please submit to TA at the end of the lecture Determine all forces acting on member ABCD of the Frame shown in Fig. P7-91. Solution: A=167.7 lb 63.4o B=424 lb 45o C=335 lb 26.6o
- 22. Class Assignment: Exerciseset P7-87 please submit to TA at the end of the lecture A pin-connected system of leaves and bars is used as a toggle for a press as shown in Fig. P7-87. Determine the force F exerted on the can at A when a force P = 100 lb is applied to the lever at G.
- 23. Solution From a free-bodydiagram for the lever: MF = 100 (30) - FDE (8) = 0 + Fx = -FBD cos 67 - FCD cos 78 - 375=0 Fy = -FBD sin 67 + FCD sin 78 = 0 From a free-body diagram for pin D: FDE = 375 lb FBD = -639.5 lb FCD = -601.8 lb
- 24. From a free-bodydiagram for the piston at B: Fy = Ay - 639.5 sin 67 =0 Ay = 588.7 lb = 588.7 lb Force on the can: F 589 lb
- 25. Class Assignment: Exerciseset P7-89 please submit to TA at the end of the lecture A pin-connected system of bars supports a 300 lb load as shown in Fig. P7-87. Determine the reactions at supports A and B and the force exerted by the pin at C on member ACE. By = 150.0 lb = 150.0 lb Bx = -450 lb = 450 lb Ax = 450 lb = 450 lb Ay = 150.0 lb = 150.0 lb Cy = 0 Cx = - 600 lb
- 26. Solution From a free-bodydiagram For the complete system: From a free-body diagram for pin F: + Fx = Ax + Bx = Ax - 450 = 0 + MA = - Bx (20) - 300 (30) = 0 Bx = -450 lb = 450 lb Ax = 450 lb = 450 lb + Fy = TEF -300 sin 45 = 0 TEF =212.1 lb 212 lb (T)
- 27. From a free-bodydiagram for bar ACE: + MC = -Ay (10) + 450 (10) - 212.1 (10/cos 45) = 0 Ay = 150.0 lb = 150.0 lb + Fy = Ay + Cy - 212.1 sin 45 = 150.0 + Cy - 212.1sin 45 = 0 Cy = 0 + Fx = Cx+ 450+212.1 cos 45=0 Cx = - 600 lb
- 28. From a free-bodydiagram for the complete system: + Fy = By + Ay - 300 = By + 150.0 - 300 = 0 By = 150.0 lb = 150.0 lb A = = = 474.3 lb 474 lb 2 2 x y A A 2 2 450.0 150.0 A = tan-1 = 18.434 A 474 lb 18.43 150.0 450.0
- 29. Example 7-101 Forces of50 lb are applied to the handles of the bolt cutter of Fig. P7-101. Determine the force exerted by on the bolt at E and all forces acting on the handle ABC.
- 30. Solution From a free-bodydiagram for member CDE: + Fx = Cx = 0 Cx = 0 + MD = Cy (3) - E(2) = 0 2 3 Cy = E
- 31. From a free-bodydiagram For handle ABC: + MB = Cy (1) - 50 (20) = 0 Cy =1000 lb = 1000 lb Fx = Bx = 0 Bx = 0 C = 1000 lb Fy = By + 50 + 1000 = 0 By = -1050 lb B = 1000 lb E = 1.5Cy = 1.5 (1000) = 1500 lb Force on the bolt: E = 1500 lb
- 32. Class Assignment: Exerciseset P7-110 please submit to TA at the end of the lecture A cylinder with a mass of 150 kg is supported by a two-bar frame as shown in Fig. P7-110. Determine all forces acting on member ACE.
- 33. Solution From a free-bodydiagram for the cylinder: W = mg = 150 (9.807) = 1471.1 N + Fx = D sin 45 - E sin 45 = 0 + Fy = 2D cos 45 - 1471.1 = 0 E 1040 lb 45.0 (on member ACE) From a free-body diagram for the complete frame: MB = A (2) - 1471.1 (1) = 0 A =735.6 N = 735.6 N A 736 N D = E = 1040.2 lb
- 34. From a free-bodydiagram for ACE: MC = T (1) - 735.6 (1) - 1040.2 (0.8) = 0 T = 1567.8 N 1568 N T 1568 N Fx = Cx + T + 1040.2 cos 45 = Cx + 1567.8 + 1040.5 cos 45 = 0 Cx = -2303 2300 N Fy = Cy + 735.6 - 1040.2 sins 45 = 0 Cy = 0 C 2300 N