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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 11. Chapter 6: Structural Analysis.

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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 11 Chapter 6: Structural Analysis

  2. To show how to determine the forces in the members of a truss using the method of joints and the method of sections. • To analyze the forces acting on the members of frames and machines composed of pin-connected members. Lecture 11 Chapter 6: Structural Analysis

  3. TRUSSES – METHODS OF JOINTS Today’s Objectives: Students will be able to: a) Define a simple truss. b) Determine the forces in members of a simple truss. c) Identify zero-force members. Lecture 11 Chapter 6: Structural Analysis

  4. TRUSSES • What is truss? • Truss is a structure whose members are connected at their ends to form triangles. • Why triangle? • Triangle is significant because it is stable & cannot collapse as long as a member does not break of deform. • How can I design truss's members? • When designing a truss, it is necessary to find the force in each member of the truss & then select structural members which are adequate to withstand the force • What we aim to do in this course? • In this course we will learn how to find the internal force of each member of the truss. Lecture 11 Chapter 6: Structural Analysis

  5. COMMON TYPES OF TRUSSES Lecture 11 Chapter 6: Structural Analysis

  6. EXAMPLES OF TRUSSES Lecture 11 Chapter 6: Structural Analysis

  7. APPLICATIONS Trusses are commonly used to support a roof. For a given truss geometry and load, how can we determine the forces in the truss members and select their sizes? A more challenging question is that for a given load, how can we design the trusses’ geometry to minimize cost? Lecture 11 Chapter 6: Structural Analysis

  8. APPLICATIONS (continued) Trusses are also used in a variety of structures like cranes and the frames of aircraft or space stations. How can we design a light weight structure that will meet load, safety, and cost specifications? Lecture 11 Chapter 6: Structural Analysis

  9. TRUSSES • Simple Truss • Three bars joined by a pins at their ends are the basic element of a plane truss • Force Analysis • All members to be two-force members • The two forces are applied at the ends of the member & are necessarily equal, opposite and collinear for equilibrium. • The member may be in tension or compression. • In most cases we assume that the weight of the members is neglected as it’s small compare with the force it supports. • It is assumed that all the joints are connected by means of frictionless pins. Lecture 11 Chapter 6: Structural Analysis

  10. FORCE ANALYSIS • There are two common methods for analysis of trusses which are Method of Joints & Method of Sections. • Both methods are based on principles of static equilibrium. • Determinacy • In order to compute all the unknown forces in the truss by using the equilibrium equations the truss has to be statically determined. • To satisfy this condition, the number of unknown has to be equal to the number of equilibrium equations b + r = 2j statically determinate b + r > 2j statically indeterminate b + r < 2j Unstable b is number of truss’s member r is the number of support’s reactions Lecture 11 Chapter 6: Structural Analysis

  11. (a) (b) (c) DETERMINACY • Classify each of the trusses shown as stable, unstable, statically determinate or statically indeterminate. For truss a b = 9 , r = 3 , j = 6 then b + r = 2j or 12 = 12 therefore the truss is statically determinate & stable. For truss b b = 6 , r = 3 , j = 5 then b + r < 2j or 9 < 10 therefore the truss is unstable. For truss a b = 6 , r = 3 , j = 4 then b + r > 2j or 9 > 8 therefore the truss is statically indeterminate & stable. Lecture 11 Chapter 6: Structural Analysis

  12. DEFINING A SIMPLE TRUSS (Section 6.1) A truss is a structure composed of slender members joined together at their end points. If a truss, along with the imposed load, lies in a single plane (as shown at the top right), then it is called a planar truss. A simple truss is a planar truss which begins with a a triangular element and can be expanded by adding two members and a joint. For these trusses, the number of members (M) and the number of joints (J) are related by the equationM = 2 J – 3. Lecture 11 Chapter 6: Structural Analysis

  13. With these two assumptions, the members act as two-force members. They are loaded in either tension or compression. Often compressive members are made thicker to prevent buckling. ANALYSIS and DESIGN ASSUMPTIONS When designing both the member and the joints of a truss, first it is necessary to determine the forces in each truss member. This is called the forceanalysis of a truss. When doing this, two assumptions are made: 1. All loads are applied at the joints. The weight of the truss members is often neglected as the weight is usually small as compared to the forces supported by the members. 2. The members are joined together by smooth pins. This assumption is satisfied in most practical cases where the joints are formed by bolting or welding. Lecture 11 Chapter 6: Structural Analysis

  14. THE METHOD OF JOINTS (Section 6.2) • The Method of Joints • If a truss is in equilibrium, then each of its joints must also be in equilibrium. • The method of joints consists of satisfying the equilibrium conditions & for the forces exerted on the pin at each joint of the truss. Lecture 11 Chapter 6: Structural Analysis

  15. THE METHOD OF JOINTS (Section 6.2) In this method of solving for the forces in truss members, the equilibrium of a joint (pin) is considered. All forces acting at the joint are shown in a FBD. This includes all external forces (including support reactions) as well as the forces acting in the members. Equations of equilibrium ( FX= 0 and  FY = 0) are used to solve for the unknown forces acting at the joints. Lecture 11 Chapter 6: Structural Analysis

  16. THE METHOD OF JOINTS (Section 6.2) Lecture 11 Chapter 6: Structural Analysis

  17. STEPS FOR ANALYSIS 1. If the support reactions are not given, draw a FBD of the entire truss and determine all the support reactions using the equations of equilibrium. 2. Draw the free-body diagram of a joint with one or two unknowns.Assume that all unknown member forces act in tension (pulling the pin) unless you can determine by inspection that the forces are compression loads. 3. Apply the scalar equations of equilibrium,  FX = 0 and  FY = 0, to determine the unknown(s). If the answer is positive, then the assumed direction (tension) is correct, otherwise it is in the opposite direction (compression). 4. Repeat steps 2 and 3 at each joint in succession until all the required forces are determined. Lecture 11 Chapter 6: Structural Analysis

  18. 200 lb B 500 lb 45 º • 5 4 • 3 FBA FBC 525.3 FBD of pin B 45º FCA C CY FBD of pin C EXAMPLE Given: P1 = 200 lb, P2 = 500 lb Find: The forces in each member of the truss. Plan:First analyze pin B and then pin C +   FX = 500 + FBC cos 45° – (3 / 5) FBA = 0 +   FY = – 200 – FBC sin 45° – (4 / 5) FBA = 0 FBA = 214 lb (T) and FBC = – 525.3 lb (C) +   FX = – FCA + 525.3 cos 45° = 0 FCA = 371 (T) Lecture 11 Chapter 6: Structural Analysis

  19. Example 2 Compute the force in each member. The supports reactions THE METHOD OF JOINTS Lecture 11 Chapter 6: Structural Analysis

  20. THE METHOD OF JOINTS Joint A • Example 2 … continued Joint B Lecture 11 Chapter 6: Structural Analysis

  21. THE METHOD OF JOINTS Joint C • Example 2 … continued Joint E Lecture 11 Chapter 6: Structural Analysis

  22. ZERO-FORCE MEMBERS (Section 6.3) If a joint has only two non-colinear members and there is no external load or support reaction at that joint, then those two members are zero-force members. In this example members DE, CD, AF, and AB are zero force members. You can easily prove these results by applying the equations of equilibrium to joints D and A. Zero-force members can be removed (as shown in the figure) when analyzing the truss. Lecture 11 Chapter 6: Structural Analysis

  23. ZERO – FORCE MEMBERS (continued) If three members form a truss joint for which two of the members are collinear and there is no external load or reaction at that joint, then the third non-collinear member is a zero force member. Again, this can easily be proven. One can also remove the zero-force member, as shown, on the left, for analyzing the truss further. Please note that zero-force members are used to increase stability and rigidity of the truss, and to provide support for various different loading conditions. Lecture 11 Chapter 6: Structural Analysis

  24. ZERO-FORCE MEMBERS Using the method of joints, determine all the zero-force members of the Fink roof truss shown in Fig. 6-13a. Assume all joints are pin connected. Lecture 11 Chapter 6: Structural Analysis

  25. ZERO-FORCE MEMBERS Lecture 11 Chapter 6: Structural Analysis

  26. ZERO-FORCE MEMBERS Lecture 11 Chapter 6: Structural Analysis

  27. P A H B C W CONCEPT QUIZ 1. Truss ABC is changed by decreasing its height from H to 0.9 H. Width W and load P are kept the same. Which one of the following statements is true for the revised truss as compared to the original truss? A) Force in all its members have decreased. B) Force in all its members have increased. C) Force in all its members have remained the same. D) None of the above. Lecture 11 Chapter 6: Structural Analysis

  28. CONCEPT QUIZ (continued) F F F • 2. For this truss, determine the number of zero-force members. • A) 0 B) 1 C) 2 • D) 3 E) 4 Lecture 11 Chapter 6: Structural Analysis

  29. GROUP PROBLEM SOLVING Given: P1 = 240 lb and P2 = 100 lb Find: Determine the force in all the truss members (do not forget to mention whether they are in T or C). Plan: a) Check if there are any zero-force members. b) Draw FBDs of pins D and B, and then apply EE at those pins to solve for the unknowns. Solution: Members AB and AC are zero-force members. Lecture 11 Chapter 6: Structural Analysis

  30. Y FDC 240 lb D X 13 5 12 FDB Y FBC 260 lb 13 5 12 X B BX SOLUTION (continued) Analyzing pin D: • + FY = –100 – (5 / 13) FDB= 0 FDB = – 260 lb (C) 100 lb • +  FX = 240 – FDC– (12 / 13) (– 260) = 0 FDC = 480 lb (T) • Analyzing pin B: • +  FY = FBC– (5 / 13) 260 = 0 FBC = 100 lb (T) Lecture 11 Chapter 6: Structural Analysis

  31. FBC B FBD BY ATTENTION QUIZ • 1. Using this FBD, you find that FBC = – 500 N. Member BC must be in __________. • A) tension B) compression • C) Can not be determined Lecture 11 Chapter 6: Structural Analysis

  32. University of Palestine College of Engineering & Urban Planning ENGINEERING MECHANICS STATICS & DYNAMICS End of the Lecture Let Learning Continue 32 Lecture 11 Chapter 6: Structural Analysis

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