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ENGINEERING MECHANICS STATICS & DYNAMICS

University of Palestine College of Engineering & Urban Planning. ENGINEERING MECHANICS STATICS & DYNAMICS. Instructor: Eng. Eman Al.Swaity. 1. Lecture 13. Chapter 6: Structural Analysis. FRAMES AND MACHINES (Section 6.6). Today’s Objectives : Students will be able to:

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ENGINEERING MECHANICS STATICS & DYNAMICS

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  1. University of Palestine College of Engineering & Urban Planning ENGINEERING MECHANICS STATICS & DYNAMICS Instructor:Eng. EmanAl.Swaity 1 Lecture 13 Chapter 6: Structural Analysis

  2. FRAMES AND MACHINES (Section 6.6) Today’s Objectives: Students will be able to: a) Draw the free body diagram of a frame or machine and its members. b) Determine the forces acting at the joints and supports of a frame or machine. Lecture 13 Chapter 6: Structural Analysis

  3. APPLICATIONS Frames are commonly used to support various external loads. How is a frame different than a truss? How can you determine the forces at the joints and supports of a frame? Lecture 13 Chapter 6: Structural Analysis

  4. FRAMES AND MACHINES: DEFINITIONS Frames and machines are two common types of structures that have at least one multi-force member. (Recall that trusses have nothing but two-force members). Frames are generally stationary and support external loads. Machines contain moving parts and are designed to alter the effect of forces. Lecture 13 Chapter 6: Structural Analysis

  5. Pin B FAB STEPS FOR ANALYZING A FRAME or MACHINE 1. Draw the FBD of the frame or machine and its members, as necessary. Hints:a) Identify any two-force members, b) Forces on contacting surfaces (usually between a pin and a member) are equal and opposite, and, c) For a joint with more than two members or an external force, it is advisable to draw a FBD of the pin. FAB 2. Develop a strategy to apply the equations of equilibrium to solve for the unknowns. Problems are going to be challenging since there are usually several unknowns. A lot of practice is needed to develop good strategies. Lecture 13 Chapter 6: Structural Analysis

  6. STEPS FOR ANALYZING A FRAME or MACHINE Lecture 13 Chapter 6: Structural Analysis

  7. CONCEPT QUIZ (continued) D 2. The figures show a frame and its FBDs. If an additional force is applied at D, then how will you change the FBD of member BC at B? A) No change, still just one force (FAB) at B.B) Will have two forces, BX and BY, at B.C) Will have two forces and a moment at B.D) Will add one moment at B. Lecture 13 Chapter 6: Structural Analysis

  8. A constant tension in the conveyor belt is maintained by using the device shown in Fig. 6-22a. Draw the free-body diagrams of the frame and the cylinder which supports the belt. The suspended block has a weight of W. Lecture 13 Chapter 6: Structural Analysis

  9. Lecture 13 Chapter 6: Structural Analysis

  10. For the frame shown in Fig. 6-24a, 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 pulleys. Lecture 13 Chapter 6: Structural Analysis

  11. Lecture 13 Chapter 6: Structural Analysis

  12. GROUP PROBLEM SOLVING Given: A frame and loads as shown. Find: The reactions that the pins exert on the frame at A, B and C. Plan: a) Draw a FBD of members AB and BC. b) Apply the equations of equilibrium to each FBD to solve for the six unknowns. Think about a strategy to easily solve for the unknowns. Lecture 13 Chapter 6: Structural Analysis

  13. +  MA = BX (0.4) + BY (0.4) – 1000 (0.2) = 0 +  MC = BX (0.4) + BY (0.6) + 500 (0.2) = 0 BY = 0 and BX = 500 N GROUP EXAMPLE (continued) Equating moments at A and C to zero, we get: Lecture 13 Chapter 6: Structural Analysis

  14. GROUP EXAMPLE (continued) Applying EofE to bar AB: • +  FX = AX– 500 = 0 ; AX = 500 N  +  FY = AY– 1000 + 500 = 0 ; AY = 500 N Consider member BC: • +  FX = 500 – CX = 0 ; CX = 500 N  +  FY = CY– 500 = 0 ; CY = 500 N Lecture 13 Chapter 6: Structural Analysis

  15. Example Lecture 13 Chapter 6: Structural Analysis

  16. Determine the tension in the cables and also the force P required to support the 600-N force using the frictionless pulley system shown in Fig. 6-31a. Solution Free-Body Diagram. A free-body diagram of each pulley including its pin and a portion of the contacting cable is shown in Fig. 6-31b. Since the cable is continuous and the pulleys are frictionless, the cable has a constant tension P acting throughout its length The link connection between pulleys Band C is a two-force member, and therefore it has an unknown tension T acting on it. Equations of Equilibrium. The three unknowns are obtained as Lecture 13 Chapter 6: Structural Analysis

  17. Lecture 13 Chapter 6: Structural Analysis

  18. ATTENTION QUIZ 1. When determining the reactions at joints A, B, and C, what is the minimum number of unknowns for solving this problem? A) 3 B) 4 C) 5 D) 6 2. For the above problem, imagine that you have drawn a FBD of member AB. What will be the easiest way to directly solve for the first unknown? A)  MC = 0 B)  MB = 0 C)  MA = 0 D)  FX = 0 Lecture 13 Chapter 6: Structural Analysis

  19. University of Palestine College of Engineering & Urban Planning ENGINEERING MECHANICS STATICS & DYNAMICS End of the Lecture Let Learning Continue 19 Lecture 13 Chapter 6: Structural Analysis

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