4 Two Important Structures Types Trusses: Structures composedentirely of two forcemembers.Frames: Structures containing at least one member acted on by forces at three or more points.
5 Trusses Assumptions1) Truss members 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 Machines - 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 of a 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 solving frames Draw a free body diagram for each componentNot 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 7-85A two-bar frame is loaded and supported as shown in Fig. F Determine the reactions at supports A and E and the force exerted on member ABC by the pin at C.
10 Solution From a free-body diagram on the complete frame: d = 6/tan(70) + 6/tan(70) = ftMA = Ey (5.648) -500(2)- 400 (4)--300 (6) = 0xyEy = lb 779 lbFy = Ay = 0Ay = lb=779 lb
11 From a free-body diagram on member CDE: MC = 400 (2) (3.464) + Ex (6) = 0Ex = lb 583 lbFrom a free-body diagram on member ABC:xyMC = 500 (4) + Ax (6) (2.184) = 0Ax = lb 617 lbFx = Cx = 0Cx = lb = lb
13 E = = = lb 973 lbE= tan = E 973 lb 53.2
14 Class Assignment: Exercise set P7-83 please submit to TA at the end of the lectureDetermine all forces acting on member BCD of the linkage shown in Fig. P7-83.
15 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) = 0C = lb 85.7 lbC lb Fx = Bx + 40 cos 30 cos 45 = 0Bx = lb 26.0 lbFy = By sin 45 + 40 sin 30 = 0By = lb 40.6 lb
16 B = ( Bx)2 + ( Bx)2 = ( 25.98)2 + ( 40.62)2 = 48.22 lb yxB40.62B = tan = tan = 57.425.98B 48.2 lb
17 Example 7-90Determine all forces acting on member ABE of the frame shown in Fig. P7-90.
18 Solution From a free-body diagram on the complete frame: MA = D (300) (300) = 0D = N = N Fx = Ax = 0Ax = N = N Fy = Ay = 0Ay = N = N xyA = tan = 135.0 A = 212 N 45A = = = N 212 N
19 From a free-body diagram on member CEF: MC = Ex (100) (200) = 0Ex = 300 N = 300 N (on ABE)From a free-body diagram on member CEF:150NCxExEyCyFrom a free-body diagram on member ABE:MB = Ey (100) (100) - 150(100) = 0Ey = 450 N = 450 N E = = = NE = tan = 56.31 E = 541 N 56.3
20 Fx = Bx = 0Bx = N = N Fy = By = 0By =- 300 N = 300 N B = = = NB = tan = B 335 N
21 Class Assignment: Exercise set P7-91 please submit to TA at the end of the lectureDetermine all forces acting on member ABCD of the Frame shown in Fig. P7-91.Solution:A=167.7 lb oB=424 lb oC=335 lb o
22 Class Assignment: Exercise set P7-87 please submit to TA at the end of the lectureA pin-connected system of leaves and bars is used as a toggle for a press as shown in Fig. P 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-body diagram for the lever: MF = 100 (30) - FDE (8) = 0FDE = 375 lbFrom a free-body diagramfor pin D:+ Fx = -FBD cos 67 - FCD cos 78 - 375=0 Fy = -FBD sin 67 + FCD sin 78 = 0FBD = lbFCD = lb
24 From a free-body diagram for the piston at B:Fy = Ay sin 67 =0Ay = lb = lb Force on the can:F 589 lb
25 Class Assignment: Exercise set P7-89 please submit to TA at the end of the lectureA pin-connected system of bars supports a 300 lb load as shown in Fig. P Determine the reactions at supports A and B and the force exerted by the pin at C on member ACE.By = lb = lb Bx = -450 lb = 450 lb Ax = 450 lb = 450 lb Ay = lb = lb Cy = 0Cx = lb
27 From a free-body diagram for bar ACE:+ MC = -Ay (10) (10)(10/cos 45) = 0Ay = lb = lb + Fy = Ay + Cy sin 45= Cy sin 45 = 0Cy = 0+ Fx = Cx cos 45=0Cx = lb
28 From a free-body diagram for the complete system: + Fy = By + Ay - 300= By = 0By = lb = lb A = = = lb 474 lbA = tan = A 474 lb
29 Example 7-101Forces of 50 lb are applied to the handles of the bolt cutter of Fig. P Determine the force exerted by on the bolt at E and all forces acting on the handle ABC.
30 Solution From a free-body diagram for member CDE: Cx = 0 + Fx = Cx = 0Cx = 0+ MD = Cy (3) - E(2) = 0Cy = E
31 From a free-body diagram For handle ABC:+ MB = Cy (1) - 50 (20) = 0Cy =1000 lb = 1000 lb C = 1000 lb Fx = Bx = 0Bx = 0Fy = By = 0By = lbB = 1000 lb E = 1.5Cy = 1.5 (1000) = 1500 lbForce on the bolt: E = 1500 lb
32 Class Assignment: Exercise set P7-110 please submit to TA at the end of the lectureA cylinder with a mass of 150 kg is supported by a two-bar frame as shown in Fig. P Determine all forces acting on member ACE.
33 Solution From a free-body diagram for the cylinder: W = mg = 150 (9.807) = N+ Fx = D sin 45 - E sin 45 = 0+ Fy = 2D cos 45 = 0D = E = lbE lb (on member ACE)From a free-body diagramfor the complete frame:MB = A (2) (1) = 0A =735.6 N = N A 736 N
34 From a free-body diagram for ACE: MC = T (1) (1) (0.8) = 0T = N 1568 NT 1568 N Fx = Cx + T cos 45= Cx cos 45 = 0Cx = 2300 N Fy = Cy sins 45 = 0Cy = 0C 2300 N