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Truss

Truss is a framework, typically consisting of rafters, posts, and struts, supporting a roof, bridge, or other structure.
a truss is a structure that "consists of two-force members only, where the members are organized so that the assemblage as a whole behaves as a single object"

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Truss

  1. 1. Lecturenotes Prepared by Prem Kumar Soni Asst. Prof. LNCT-S 12/5/2016 1Lecture notes Chapter 5 By Prem Kumar Soni
  2. 2. TRUSSES 12/5/2016 2Lecture notes Chapter 5 By Prem Kumar Soni
  3. 3. CONTENTS : 1. Introduction. 2. Plane truss. 3. Statical determinacy and stability of trusses. 4. Analysis of statically determinate plane truss. 5. Sign conventions used. 6. Method of joints. 7. Identification of zero force members. 8. Examples on method of joints. 9. Method of sections. 10. Examples on method of sections. 11. Exercise problems. 12/5/2016 3Lecture notes Chapter 5 By Prem Kumar Soni
  4. 4. INTRODUCTION : A truss is an articulated (skeletal) structure with hinged or ball and socket joints. It is an assemblage of slender bars fastened together at their ends by smooth pins or ball and socket joints acting as hinges. Plane truss: A truss consisting of members which lie in a plane and are loaded in the same plane is called a plane truss. Ex: Roof truss, bridge truss, etc. Space truss: A truss made up of members not lying in the same plane is referred to as space truss. Ex: Electric power transmission tower, microwave tower, etc. 12/5/2016 4 Lecture notes Chapter 5 By Prem Kumar Soni
  5. 5. APPLICATIONS Plane Truss Space Truss Trussed Bridge Braced Frame 12/5/2016 5Lecture notes Chapter 5 By Prem Kumar Soni
  6. 6. 12/5/2016 6Lecture notes Chapter 5 By Prem Kumar Soni
  7. 7. Truss  In a truss, the joints are of pin type, where end of the members can rotate freely. Moreover, individual truss members should not directly loaded transversely; loads should be applied at the joints of members so that no bending or shear forces will be generated on truss members. Self-weight of the members should be dumped both end for the same reason. 12/5/2016Lecture notes Chapter 5 By Prem Kumar Soni 7
  8. 8. Frame  Unlike truss, the members of frames are connected rigidly at joints, and individual members can carry transverse load, which may generate bending moment, shear forces along with axial forces. 12/5/2016Lecture notes Chapter 5 By Prem Kumar Soni 8
  9. 9.  Truss- In truss, joints are of pin joint type and can freely rotate about the pin. So, they cannot transfer moments.  In practical, truss have no flexibility in joints.  Frame- In frame, joints are joined by welding abd bolting so they are rigid and cannot rotate freely. They can transfer moments as well as axial loads.  Frames are of 2types. Rigid non collapsible and non rigid collapsible. 12/5/2016Lecture notes Chapter 5 By Prem Kumar Soni 9
  10. 10. The basic element of a plane truss is three members (bars, angles,tubes,etc) arranged to form a triangle. To this base triangle, two more members are added to locate a new joint, and the process continued to form the complete truss. The truss built in such a manner is called as ‘Simple Truss’. PLANE TRUSS : 12/5/2016 10Lecture notes Chapter 5 By Prem Kumar Soni
  11. 11. T T Tension Member The free body diagram of a member shows that it is acted upon by two equal and opposite forces. The hinged joint permits members to rotate with respect to each other and hence the members are subjected to purely axial forces. A truss is held in position by the supports and the loads are applied only at joints. 12/5/2016 11 Lecture notes Chapter 5 By Prem Kumar Soni
  12. 12. Compression Member The convention for internal forces, i.e., the action of forces in the members on the joints is shown. TENSION COMPRESSION C C T C 12/5/2016 12Lecture notes Chapter 5 By Prem Kumar Soni
  13. 13. STATICAL DETERMINACY AND STABILITY OF TRUSSES : Total number of unknowns = m + c Let m = number of unknown member forces. c = number of unknown constrained reaction components ( usually three, if statically determinate externally ) j = number of joints Each joint has two equations of equilibrium (H=0 and V=0). Hence number of equilibrium equations available = 2j. 12/5/2016 13Lecture notes Chapter 5 By Prem Kumar Soni
  14. 14. If the number of unknowns is less than the number of available equations, then the truss is unstable. If total number of unknowns is equal to total number of equations available, then the truss is statically determinate. If number of unknowns is more than the number of available equations, then the truss is statically indeterminate. i.e., m+c = 2j i.e., m+c > 2j i.e., m+c < 2j STATICAL DETERMINACY AND STABILITY OF TRUSSES : (contd..) 12/5/2016 14Lecture notes Chapter 5 By Prem Kumar Soni
  15. 15. Degree of indeterminacy, i = (m+c) – 2j Degree of internal indeterminacy. ii = i – ie = m+3 – 2j Degree of external indeterminacy, ie = c – 3 STATICAL DETERMINACY AND STABILITY OF TRUSSES : (contd..) 12/5/2016 15Lecture notes Chapter 5 By Prem Kumar Soni
  16. 16. (iii)When ii > 0, number of members is more than the number required. Such a truss is called statically indeterminate or redundant truss. STATICAL DETERMINACY AND STABILITY OF TRUSSES : (contd..) (ii) When ii < 0, the number of members provided is less than that required for stability. Such a truss is called unstable or deficient truss. (i) When ii = 0, m = 2j – 3 and the structure is statically determinate internally and is stable. Such a truss is called a perfect truss. When we consider the internal indeterminacy, we have the following cases of truss: 12/5/2016 16Lecture notes Chapter 5 By Prem Kumar Soni
  17. 17. STATICAL DETERMINACY AND STABILITY OF TRUSSES : (contd..) When we consider the external indeterminacy, we have the following cases of truss: (i) When c = 3 (=> ie = 0 ), the truss is said to be statically determinate externally. (ii) When c < 3 (=> ie < 0 ), the truss is said to be unstable externally. (iii)When c > 3 (=> ie > 0 ), the truss is said to be statically indeterminate externally. 12/5/2016 17Lecture notes Chapter 5 By Prem Kumar Soni
  18. 18. ANALYSIS OF STATICALLY DETERMINATE PLANE TRUSS : The assumptions made in the analysis are as follows: The members of the truss are connected at the ends by frictionless hinges. The axes of all members lie in a single plane called ‘middle plane of the truss’. All the external forces acting on the truss are applied at the joints only. All the loads are applied in the plane of the truss. 12/5/2016 18Lecture notes Chapter 5 By Prem Kumar Soni
  19. 19. The following methods of analysis will be adopted : A.) Method of joints. B.) Method of sections. METHOD OF JOINTS : A member in a pin - jointed truss has only one internal force resultant i.e., the axial force. Hence the F.B.D of any joint is a concurrent system in equilibrium. SIGN CONVENTIONS USED : Positive sign is used for tension. Negative sign is used for compression. Clockwise moment ( )is taken positive and anti- clockwise moment ( ) is taken as negative. 12/5/2016 19 Lecture notes Chapter 5 By Prem Kumar Soni
  20. 20. METHOD OF JOINTS : (contd..) The procedure for method of joints is as follows :  The support reactions of the truss are first obtained considering the three conditions of equilibrium, applied to the truss as a whole (H=0, V=0 and M=0).  Taking the F.B.D of a joint which has not more than two unknowns (preferably), and applying the equations of equilibrium for a coplanar concurrent force system (H=0 and V=0), the unknowns are evaluated.  The analysis is continued with the next joint with two unknowns (preferably), until the forces in all the members are obtained. 12/5/2016 20Lecture notes Chapter 5 By Prem Kumar Soni
  21. 21. IDENTIFICATION OF ZERO FORCE MEMBERS : 1. When two of the three members meeting at a joint are collinear, and no load is acting at the joint, then the force in the third member is zero. A A F FF F 2. When two members meet at a joint where no load is acting, then the forces in those members are zero. A A A 12/5/2016 21Lecture notes Chapter 5 By Prem Kumar Soni
  22. 22. IDENTIFICATION OF ZERO FORCE MEMBERS : (contd..) 3. When two members meet at a joint where there is a support such that the support reaction is collinear with any one member, then the force in the other member is zero. VA A F=VA 12/5/2016 22Lecture notes Chapter 5 By Prem Kumar Soni
  23. 23. METHOD OF SECTIONS : In this method we can directly determine the force in any member without proceeding to that member by a joint by joint analysis. The method is as follows :  Determine the reactions at the supports of the truss given.  Take an imaginary cutting plane through the truss, dividing it into two parts, such that it passes through members in which forces are required. 12/5/2016 23Lecture notes Chapter 5 By Prem Kumar Soni
  24. 24. METHOD OF SECTIONS : (contd..)  The cutting plane should be taken in such a way that it cuts a maximum of three members in which forces are unknown, preferably.  Now consider any one part to the left or right of the section, and evaluate the unknowns by applying the conditions of equilibrium for the coplanar non-concurrent force system (H=0, V=0 and M=0). 12/5/2016 24Lecture notes Chapter 5 By Prem Kumar Soni
  25. 25. EXERCISE 1: Find the forces in members of the truss shown below. Solution : FAB = 84.9kN(C) FEF = 60kN(T) FBC = 20kN(C) FBE = 56.67kN(C) 20kN20kN 20kN B A C DEF 4m 4m 4m 4m 12/5/2016 25Lecture notes Chapter 5 By Prem Kumar Soni
  26. 26. EXERCISE 2: Find the forces in members of the truss shown below. Solution : FAB = 6.57kN(C) FCE = 6kN(C) FBC = 6.19kN(C) FDE = 3.87kN(T) FAE = 2.71kN(C) FCD = 6.19kN(C) F = 7.15kN(T) E2m 2m 60º 60º 30º 30º 3kN 6kN 3kN A B C D 1kN 12/5/2016 26Lecture notes Chapter 5 By Prem Kumar Soni
  27. 27. EXERCISE 3: Find the forces in members of the truss shown below. Solution : FAB = 7.5kN(T) FBE = 18.75kN(C) FBC = 26.25kN(T) FCE = 43.75kN(C) FAD = 12.5kN(C) FDE = 15kN(T) FBD = 12.5kN(T) BA C D E 1.8m 1.8m 2.4m 10kN 5kN 1.8m 1.8m 12/5/2016 27Lecture notes Chapter 5 By Prem Kumar Soni
  28. 28. EXERCISE 4: Find the forces in members CE, CF and FH of th truss shown below. Solution : FCE = 10kN(T) FCF = 7.07kN(C) FFH =5kN(T) 20kN 3m 3m 3m 3m A B C D E F G H I J 3m 12/5/2016 28Lecture notes Chapter 5 By Prem Kumar Soni
  29. 29. EXERCISE 5: Find the force in member GI of the truss shown below. Solution : FGI = 22.5kN(C) A B C D E F G H I 2m 2m 2m 4m 15kN 15kN 15kN 12/5/2016 29Lecture notes Chapter 5 By Prem Kumar Soni
  30. 30. EXERCISE 6: Find the forces in members CE and CD of the truss shown below. Solution : FCE = 11.92kN(T) FCD = 0A B C D E F G40º 3m 3m3m 10kN 12/5/2016 30Lecture notes Chapter 5 By Prem Kumar Soni
  31. 31. EXERCISE 7: If force in member BC is 30kN(C), find magnitude of load W. Solution : W =10kN A B C DE W W a a a 12/5/2016 31Lecture notes Chapter 5 By Prem Kumar Soni
  32. 32. EXERCISE 8: Find the forces in members AB, BC and CF of the truss shown below. Solution : FAB = 3.6kN(T) 2m2m 2m 2m 2m 2m 4kN A 4kN 4kN B C D E F G 12/5/2016 32Lecture notes Chapter 5 By Prem Kumar Soni
  33. 33. THANK YOU. 12/5/2016 33Lecture notes Chapter 5 By Prem Kumar Soni
  34. 34. 12/5/2016 34Lecture notes Chapter 5 By Prem Kumar Soni

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