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Frames and Machines: Structural Analysis

Learn to draw free-body diagrams, determine joint forces, and solve for unknowns in frames and machines. Understand the differences and applications. Master equilibrium equations for analysis.

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Frames and Machines: Structural Analysis

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  1. Chapter 5 Structural Analysis FAUZIAH MAT Applied Mechanics Division School of Mechatronic Engineering Universiti Malaysia Perlis (UniMAP) fauziah@unimap.edu.my

  2. 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. FRAMES AND MACHINES • In-Class Activities: • Check Homework, if any • Reading Quiz • Applications • Analysis of a Frame/Machine • Concept Quiz • Group Problem Solving • Attention Quiz

  3. Frames are commonly used to support various external loads. APPLICATIONS How is a frame different than a truss? To be able to design a frame, you need to determine the forces at the joints and supports.

  4. “Machines,” like those above, are used in a variety of applications. How are they different from trusses and frames? APPLICATIONS (continued) How can you determine the loads at the joints and supports? These forces and moments are required when designing the machine’s members.

  5. FRAMES AND MACHINES: DEFINITIONS Frame Machine 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.

  6. STEPS FOR ANALYZING A FRAME OR MACHINE 1. Draw a FBD of the frame or machine and its members, as necessary. Hints: a) Identify any two-force members, b) Note that 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

  7. STEPS FOR ANALYZING A FRAME OR MACHINE 2. Develop a strategy to apply the equations of equilibrium to solve for the unknowns. Look for ways to form single equations and single unknowns. Problems are going to be challenging since there are usually several unknowns. A lot of practice is needed to develop good strategies and ease of solving these problems. FAB

  8. Given: The frame supports an external load and moment as shown. Find: The horizontal and vertical components of the pin reactions at C and the magnitude of reaction at B. Plan: EXAMPLE a) Draw FBDs of the frame member BC. Why pick this part of the frame? b) Apply the equations of equilibrium and solve for the unknowns at C and B.

  9. +  MC = FAB sin45° (1) – FAB cos45° (3) + 800 N m + 400 (2) = 0 FAB = 1131 N EXAMPLE (continued) . 800 N m 400 N CX CY 1 m 1 m 2 m FBD of member BC (Note AB is a 2-force member!) B 45° FAB Please note that member AB is a two-force member. Equations of Equilibrium: Start with  MC since it yields one unknown. .

  10. +  FX = – CX+ 1131 sin 45° = 0 CX = 800 N  +  FY = – CY+ 1131 cos 45° – 400 = 0 CY = 400 N EXAMPLE(continued) . 800 N m 400 N CX CY 1 m 1 m 2 m FBD of member BC B 45° FAB

  11. GROUP PROBLEM SOLVING Given:A frame supports a 50-kN load as shown. Find: The reactions exerted by the pins on the frame members at B and C. Plan: a) Draw a FBD of member BC and another one for AC. b) Apply the equations of equilibrium to each FBD to solve for the four unknowns. Think about a strategy to easily solve for the unknowns.

  12. +  MA = – CY (8) + CX (6) + 50 (3.5) = 0 (1) Applying E-of-E to member AC: GROUP PROBLEM SOLVING (continued) FBDs of members BC and AC CY CX 50 kN 3.5 m 6 m 8 m AX AY • +  FX = CX– AX = 0 • +  FY = 50 – AY– CY= 0

  13. +  MB = – 50 (2) – 50 (3.5) + CY (8) = 0 ; CY = 34.38 = 34.4 kN From Eq (1), CX can be determined; CX = 16.67 = 16.7 kN Applying E-of-E to member BC: GROUP PROBLEM SOLVING (continued) FBDs of members BC and AC CY CX 50 kN 3.5 m 6 m 8 m AX AY • +  FX = 16.67 + 50 – BX = 0 ; BX = 66.7 kN • +  FY = BY – 50 + 34.38= 0 ; BY = 15.6 kN

  14. End of the Lecture Let Learning Continue

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