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Lecture 20 ENGR-1100 Introduction to Engineering Analysis.

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1 Lecture 20 ENGR-1100 Introduction to Engineering Analysis

2 Previous Lectures Outline 2D trusses analysis- a) method of joints. b) method of sections.

3 Today’s Lecture Outline Frames

4 Two Important Structures Types 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) 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 - 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 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-bar frame 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  ) = ft  M A = E y (5.648) -500(2)- 400 (4) (6) = 0  F y = A y = 0 A y = lb=779 lb E y = lb  779 lb From a free-body diagram on the complete frame: d

11 From a free-body diagram on member CDE:  M C = 400 (2) (3.464) + E x (6) = 0 E x = lb  583 lb From a free-body diagram on member ABC: x y  M C = 500 (4) + A x (6) (2.184) = 0  F x = C x = 0 A x = lb  617 lb C x = lb = lb

12  F y = C y = 0 C y = lb = 779 lb C = = = lb  788 lb  c = tan -1 =  C  788 lb 81.5  A = = = lb  994 lb  A = tan -1 =  A  994 lb 51.6 

13 E = = = lb  973 lb  E = tan -1 =  E  973 lb 53.2 

14 Class Assignment: Exercise set 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  lb  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:  M B = C cos 45  (2.0) - 40 cos 30  (3.5) = 0 C = lb  85.7 lb  F x = B x + 40 cos 30  cos 45  = 0  F y = B y sin 45  + 40 sin 30  = 0 B x = lb  26.0 lb B y = lb  40.6 lb

16  B = tan -1 = tan -1 = 57.4  B  48.2 lb 57.4  B =  ( B x ) 2 + ( B x ) 2 =  ( 25.98) 2 + ( 40.62) 2 = lb y x B B

17 Example 7-90 Determine all forces acting on member ABE of the frame shown in Fig. P7-90.

18 Solution x y From a free-body diagram on the complete frame:  M A = D (300) (300) = 0 D = N = N   F x = A x = 0 A x = N = N   F y = A y = 0 A y = N = N   A = tan -1 =  A = 212 N 45  A = = = N  212 N

19 From a free-body diagram on member ABE:  M B = E y (100) (100) - 150(100) = 0 E y = 450 N = 450 N  E = = = N  E = tan -1 =  E = 541 N 56.3   M C = E x (100) (200) = 0 E x = 300 N = 300 N  (on ABE) From a free-body diagram on member CEF: 150N CxCx ExEx EyEy CyCy

20  F x = B x = 0 B x = N = N   F y = B y = 0 B y =- 300 N = 300 N  B = = = N  B = tan -1 =  B  335 N 63.4 

21 Class Assignment: Exercise set 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 63.4 o B=424 lb 45 o C=335 lb 26.6 o

22 Class Assignment: Exercise set 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-body diagram for the lever:  M F = 100 (30) - F DE (8) = 0 +   F x = -F BD cos 67  - F CD cos 78  - 375=0   F y = -F BD sin 67  + F CD sin 78  = 0 From a free-body diagram for pin D: F DE = 375 lb F BD = lb F CD = lb

24 From a free-body diagram for the piston at B:  F y = A y sin 67  =0 A y = 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 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. B y = lb = lb  B x = -450 lb = 450 lb  A x = 450 lb = 450 lb  A y = lb = lb  C y = 0C x = lb

26 Solution From a free-body diagram For the complete system: From a free-body diagram for pin F: +   F x = A x + B x = A x = 0 +  M A = - B x (20) (30) = 0 B x = -450 lb = 450 lb  A x = 450 lb = 450 lb  +   F y = T EF -300 sin 45  = 0 T EF =212.1 lb  212 lb (T)

27 From a free-body diagram for bar ACE: +  M C = -A y (10) (10) (10/cos 45  ) = 0 A y = lb = lb  +   F y = A y + C y sin 45  = C y sin 45  = 0 C y = 0 +   F x = C x cos 45  =0 C x = lb

28 From a free-body diagram for the complete system: +   F y = B y + A y = B y = 0 B y = lb = lb  A = = = lb  474 lb  A = tan -1 =  A  474 lb 

29 Example Forces 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: +   F x = C x = 0 C x = 0 +  M D = C y (3) - E(2) = 0 C y = E

31 From a free-body diagram For handle ABC: +  M B = C y (1) - 50 (20) = 0 C y =1000 lb = 1000 lb   F x = B x = 0 B x = 0 C = 1000 lb   F y = B y = 0 B y = lb B = 1000 lb  E = 1.5C y = 1.5 (1000) = 1500 lb Force on the bolt:E = 1500 lb 

32 Class Assignment: Exercise set 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. P Determine all forces acting on member ACE.

33 Solution From a free-body diagram for the cylinder: W = mg = 150 (9.807) = N +   F x = D sin 45  - E sin 45  = 0 +   F y = 2D cos 45  = 0 E  1040 lb 45.0  (on member ACE) From a free-body diagram for the complete frame:  M B = A (2) (1) = 0 A =735.6 N = N  A  736 N  D = E = lb

34 From a free-body diagram for ACE:  M C = T (1) (1) (0.8) = 0 T = N  1568 N T  1568 N   F x = C x + T cos 45  = C x cos 45  = 0 C x =  2300 N   F y = C y sins 45  = 0 C y = 0 C  2300 N 


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