1. ENGR-1100 Introduction to Engineering Analysis
Lecture 20

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 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 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. 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) = ft
MA = Ey (5.648) -500(2)- 400 (4)-
-300 (6) = 0 x y
Ey = lb  779 lb
Fy = Ay = 0
Ay = lb=779 lb

11. From a free-body diagram on member CDE:
MC = 400 (2) (3.464) + Ex (6) = 0
Ex = lb  583 lb
From a free-body diagram on member ABC: x y
MC = 500 (4) + Ax (6) (2.184) = 0
Ax = lb  617 lb
Fx = Cx = 0
Cx = lb = lb

12. Fy = Cy = 0
Cy = lb = 779 lb
C = = = lb  788 lb
c = tan = 81.46 C  788 lb 81.5
A = = = lb  994 lb
A = tan =  A  994 lb 51.6

13. E = = = lb  973 lb
E= tan =  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. 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 = lb  85.7 lb
C  lb  
Fx = Bx + 40 cos 30 cos 45 = 0
Bx = lb  26.0 lb
Fy = By sin 45 + 40 sin 30 = 0
By = lb  40.6 lb

16. B = ( Bx)2 + ( Bx)2 = ( 25.98)2 + ( 40.62)2 = 48.22 lby x B
40.62
B = tan = tan = 57.4
25.98
B  48.2 lb 

17. Example 7-90
Determine 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) = 0
D = N = N 
Fx = Ax = 0
Ax = N = N 
Fy = Ay = 0
Ay = N = N  x y
A = tan = 135.0 A = 212 N 45
A = = = N  212 N

19. From a free-body diagram on member CEF:
MC = Ex (100) (200) = 0
Ex = 300 N = 300 N  (on ABE)
From a free-body diagram on member CEF:
150N Cx Ex Ey Cy
From a free-body diagram on member ABE:
MB = Ey (100) (100) - 150(100) = 0
Ey = 450 N = 450 N 
E = = = N
E = tan = 56.31 E = 541 N 56.3

20. Fx = Bx = 0
Bx = N = N 
Fy = By = 0
By =- 300 N = 300 N 
B = = = N
B = tan =  B  335 N 

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 o
B=424 lb o
C=335 lb 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. 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) = 0
FDE = 375 lb
From a free-body diagram
for pin D:
+ Fx = -FBD cos 67 - FCD cos 78 - 375=0
 Fy = -FBD sin 67 + FCD sin 78 = 0
FBD = lb
FCD = lb

24. From a free-body diagram
for the piston at B:
Fy = Ay sin 67 =0
Ay = 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. 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 = 0
Cx = lb

26. Solution From a free-body diagram For the complete system:
+  Fx = Ax + Bx = Ax = 0
+ MA = - Bx (20) (30) = 0
Bx = -450 lb = 450 lb 
Ax = 450 lb = 450 lb 
From a free-body diagram
for pin F:
+  Fy = TEF sin 45 = 0
TEF =212.1 lb  212 lb (T)

27. From a free-body diagram
for bar ACE:
+ MC = -Ay (10) (10)
(10/cos 45) = 0
Ay = lb = lb 
+  Fy = Ay + Cy sin 45
= Cy sin 45 = 0
Cy = 0
+  Fx = Cx cos 45=0
Cx = lb

28. From a free-body diagram for the complete system:
+  Fy = By + Ay - 300
= By = 0
By = lb = lb 
A = = = lb  474 lb
A = tan =  A  474 lb 

29. Example 7-101
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: Cx = 0
+  Fx = Cx = 0
Cx = 0
+ MD = Cy (3) - E(2) = 0
Cy = E

31. From a free-body diagram
For handle ABC:
+ MB = Cy (1) - 50 (20) = 0
Cy =1000 lb = 1000 lb 
C = 1000 lb 
Fx = Bx = 0
Bx = 0
Fy = By = 0
By = lb
B = 1000 lb 
E = 1.5Cy = 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
+  Fx = D sin 45 - E sin 45 = 0
+  Fy = 2D cos 45 = 0
D = E = lb
E  lb  (on member ACE)
From a free-body diagram
for the complete frame:
MB = A (2) (1) = 0
A =735.6 N = N 
A  736 N 

34. From a free-body diagram for ACE:
MC = T (1) (1) (0.8) = 0
T = N  1568 N
T  1568 N 
Fx = Cx + T cos 45
= Cx cos 45 = 0
Cx =  2300 N 
Fy = Cy sins 45 = 0
Cy = 0
C  2300 N 