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Lecture 7

Trusses. Lecture 7. Truss: is a structure composed of slender members ( two-force members ) joined together at their end points to support stationary or moving load. Each member of a truss is usually of uniform cross section along its length.

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Lecture 7

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  1. Trusses Lecture 7 • Truss: is a structure composed of slendermembers (two-force members) joined together at their end points to support stationary or moving load. • Each member of a truss is usually of uniform cross section along its length. • Calculation are usually based on following assumption: • The loads and reactions act only at the joint. • Weight of the individual members can be neglected. • Members are either under tension or compression. • Joints: are usually formed by bolting or welding the members to a common plate, called a gusset plate, or simply passing a large bolt through each member. • Joints are modeled by smooth pin connections.

  2. Analysis of Trusses Lecture 7 Truss Analysis Internal equilibrium External equilibrium To find the force in each member External Equilibrium:to find the reaction forces, follow the below steps: To find the reaction forces Method of joints Method of sections Draw the FBD for the entire truss system. Determine the reactions. Using the equations of (2 D) which states:

  3. Analysis of Trusses Lecture 7 Method of Joints:to find the forces in any member, choose a joint, to which that member is connected, and follow the below steps: Draw the FBD for the entire truss system. Determine the reactions. Using the equations of (2 D) which states: Choose the joint, and draw FBD of a joint with at least one known force and at most two unknown forces. Using the equation of (2 D) which states: The internal forces are determined. Choose another joint.

  4. Analysis of Trusses Lecture 7 Method of section (Internal equilibrium):to find the forces in any member, choose a section, to which that member is appeared as an internal force, and follow the below steps: Draw the FBD for the entire truss system. Determine the reactions. Using the equations of (2 D) which states: Choose the section, and draw FBD of that section, shows how the forces replace the sectioned members. Using the equation of (2 D) which states: The internal forces are determined. Choose another section or joint.

  5. Analysis of Trusses Lecture 7 Analysis of trusses (Zero-force members): Analysis of trusses system is simplified if one can identify those members that support no loads. We call these zero-force members. Examples to follow: If two members form a truss joint and there is no external load or support reaction at that joint then those members are zero-force members. Joints D and A in the following figure are the joints with no external load or support reaction, so: FAF = FAB = FDE = FDC = 0.

  6. Analysis of Trusses Lecture 7 Analysis of trusses (Zero-force members): Examples to follow: If three members form a truss joint and there is no external load or support reaction at that joint and two of those members are collinear then the third member is a zero-force member. In the following figure, AC and AD are zero-force members, because Joints D and A in the following figure are the joints with three members, there is no external load or support reaction, so: FCA = FDA = 0

  7. EXAMPLES of Trusses: Lecture 7 Example 1:Determine the support reactions in the joints of the following truss. Calculate the force in member (BA & BC.) Solution 1. Draw FBD of entire truss and solve for support reactions: • 2. Draw FBD of a joint with at least one known force and at most two unknown forces. We choose joint B. • Assume BC is in compression.

  8. EXAMPLES of Trusses: Lecture 7 Example 2:In the following Bowstring Truss, find the force in member (CF). Solution draw the FBD and find the support reactions which are shown below • Fy = 0 • RE + RA – 5– 3 = 0 • RA = 3.25 kN • MA = 0 • RE * 16 – 5 * 8 – 3 * 12 = 0 • RE = 4.75 kN G F 2 m 6 m O C D E 4 m 4 m X

  9. EXAMPLES of Trusses: Lecture 7 Example 3:In the following truss, find the force in member (EB). Solution Notice that no single cut will provide the answer. Hence, it is best to consider section (a-a and b-b). • MA = 0 • RC * 8 – 1000 * 6 – 1000 * 4 – 3000 * 2 = 0 • RC = 2000 N • Fy = 0 • RA + RC – 1000 – 1000 – 3000 - 1000 = 0 • RA = 4000 N Taking the moment about joint (B), to find (FED), as shown in below figure: • MB = 0 • 1000 * 4 + 3000 * 2 – 4000 * 4 + FED * sin30o * 4 = 0 • FED = 3000 N (compression)

  10. Lecture 7 Continue Example 3: From joint (E) to find (FEB), as shown in below figure: • Fx = 0 • FEF . cos30o– 3000 cos30o = 0 • FEF = 3000 N (compression) • Fy = 0 • FEF . Sin30o + 3000 . sin30o - 1000 - FEB = 0 • FEF = 2000 N (Tension)

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