Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

1. Engineering 36 Chp6:Trusses-1 Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. Introduction: MultiPiece Structures • For the equilibrium of structures made of several connectedparts, the internal forces as well the external forces are considered. • In the interaction between connected parts, Newton’s 3rd Law states that the forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense. • The Major Categories of Engineering Structures • Frames: contain at least one multi-force member, i.e., a member acted upon by 3 or more forces • Trusses: formed from two-force members, i.e., straight members with end point connections • Machines: structures containing moving parts designed to transmit and modify forces

3. Tension Compression Definition of a Truss • A truss consists of straight membersconnected at joints. No member is continuous through a joint. • A truss carries ONLY those loads which act in its plane, allowing the truss to be treated as a two-dimensional structure. • Bolted or Welded connections are assumed to be PINNED together. Forces acting at the member ends reduce to a single force and NO couple. Only two-force members are considered  LoACoIncident with Geometry • When forces tend to pull the member apart, it is in tension. When the forces tend to push together the member, it is in compression.

4. Truss Defined • Members of a truss are SLENDER and NOT capable of supporting large LATERAL loads • i.e.; IN-Plane, or 2D, loading only • Members are of NEGLIBLE Weight • Loads MUST be applied at the JOINTS to Ensure AXIAL-ONLY Loads on Members. • Mid-Member Loads Produce BENDING-Loads which Truss Members are NOT Designed to Support Beams Apply RoadBed Load at JOINTS Only

5. Types of Trusses

6. Yet More Trusseshttp://www.caltruss.com/products.htm

7. A rigid truss will not collapse under the application of a load. Simple Trusses • A simple truss is constructed by successively adding two members and one connection to the basic TRIANGULAR truss • In a simple truss, m = 2n− 3 where m is the total number of members and n is the number of joints

8. Method of Joints for Trusses • Dismember the truss and create a freebody diagram for each member and pin. • The two forces exerted on each member are equal, have the same line of action, and opposite sense as the load on the associated Pin • Forces exerted by a member on the pins or joints at its ends are directed along the member and are equal & opposite. • Conditions of equilibrium on the pins provide 2n equations for 2n unknowns. For a simple truss, 2n = m + 3. May solve for m member forces and 3 reaction forces at the supports. • Conditions for equilibrium for the entire truss provide 3 additional equations which are not independent of the pin equations.

9. Forces in opposite members intersecting in two straight lines at a joint are equal (to maintain pin equilibrium). • The forces in two opposite members are equal when a load is aligned with a third member. The third member force is equal to the load (including zero load). If P=0, Then AC Applies NO Force to the pin; AC is then a ZERO-FORCE MEMBER • The forces in two members connected at a joint are equal if the members are aligned and zero otherwise. Zero-Force Members • Recognition of joints under special loading conditions (ZFMs) simplifies truss analysis.

10. Zero Force Members (ZFMs) • TWO-Member Version • When only TWO members form a NON-CoLinear truss joint and NO external load or support reaction is applied to the joint then the members MUST be ZERO-FORCE members. • THREE-Member Version • When THREE members form a truss joint for which two of the members are CoLinear and the third is forms an angle with the first two, then the NON-CoLinear member is a ZERO-FORCE member providedNO external force or support reaction is applied to the joint • NOTE that The CoLinear members carry EQUAL loads.

11. Truss members that canNOT carry load are called Zero Force Members. Examples of Zero Force Members (ZFMs) are the colored members (AB, BC, and DG) in the truss shown at Right Zero Force Members (ZFMs)

12. Again Consider Zero Force Members (ZFMs) • Summing the forces in the y-direction in the AB FBD shows that FABmust be ZERO since it is NOT balanced by another y-force. • With FAB =0 Summing forces in the x-direction shows that FBC must also be zero • FBD for AB & BC

13. Again Consider Zero Force Members (ZFMs) • Summing forces in the y-direction in the DG free-body-diagram, reveals that FDG must be zero since it is not balanced by another y-force • The FBD for DG

14. Space Trusses • An elementary space truss consists of 6 members connected at 4 joints to form a tetrahedron. • A simple space truss is formed and can be extended when 3 new members and 1 joint are added at the same time. • In a simple space truss, m = 3n− 6 where m is the number of members and n is the number of joints. • Conditions of equilibrium for the Ball-and-Socket joints provide 3n equations (no moments). For a simple truss, 3n = m + 6 and the equations can be solved for mmember forces and 6 support reactions. • Equilibrium for the entire truss provides 6 additional equations which are not independent of the joint equations.

15. Example  2D Truss • Solution Plan • Based on a free-body diagram of the entire truss, solve the 3 equilibrium eqns for the reactions at A and D • Joint D is subjected to only two unknown member forces. Determine the member forces from the joint (or pin) equilibrium requirements. • In succession, determine unknown member forces at joints C, B, and A from pin equilibrium requirements. • Use Pin-E equilibrium requirement to check the results • Using the method of joints, determine the force in each member of the truss

16. Black/White Board Work Solve byMethod ofJoints Weight of Members & Truss is Negligible Relative to Applied Load(s)

18. Example  2D Truss • SOLUTION PLAN • Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C. • Joint A is subjected to only two unknown member forces. Determine these from the joint (or pin) equilibrium requirements. • In succession, determine unknown member forces at joints D, B, and E from pin equilibrium requirements. • All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results • Using the method of joints, determine the force in each member of the truss.

19. SOLUTION Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C. Example  2D Truss • So-Far, So-Good

20. Joint A is subjected to only two unknown member forces. Determine these from the pin (or joint) equilibrium requirements. Example  2D Truss • There are now only two unknown member forces at joint D.

21. There are now only two unknown member forces at joint B. Assume both are in tension. Example  2D Truss • There is one unknown member force at joint E. Assume the member is in tension.

22. CHECK All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results. Example  2D Truss • 

23. A Real Truss Joint Idealized 2 in Tension, 3 in Compression

24. WhiteBoard Work Let’s WorkThese NiceProblems

26. Engineering 36 APPENDIXTruss Analysis Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu –

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34. Truss-Member Load Analysis →2049N← ←2898N→ →2049N← →2049N← →776.4N← 0 (ZFM) ←2049N→