Estrucura articulada unida a una viga
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The document discusses a structural system analysis involving an isolated truss connected to a beam, detailing methods for calculating reaction forces and axial forces in members. It highlights equilibrium conditions and the use of the method of joints and sections to determine internal forces and stresses. Additionally, specific numerical values and calculations related to the truss design are provided.
Estrucura articulada unida a una viga
- 1. http://www.elrincondelingeniero.com/ The system shown in the figure is made 1. DSI Isolated truss up of a beam connected on its right end to 3 a truss. 3.2. 2 Data: 1 3. 3 4 1 6 0 2 0 Young modulus = 210 GPA Alternatively Determine: a) 0 6 0,04 (diameter) 1 ; 3 2 3 3 2.6 0 DSI of the truss and its possible implications. Thus, structure is entirely linked and the b) Reaction forces in supports A and E. reactions exerted by the supports can be c) Reaction forces in supports F and G. determined using equilibrium conditions. 2. d) Using the method of the joints, the axial forces in all members of the structure indicating if they are tensile or compressive. e) f) 8 kN A 1m C D 2m 1m B 0→5 1m 4.8 2.2.2 D C F G 0→ 2.2 12 8 Solution: 0→ In order to solve the exercise, we will separate both structures and add the 3. Reaction forces in the truss corresponding internal forces as it is shown in the drawings. E 1m 2 kN/m E 1m B VE Normal stress in member EF. A 2 kN/m HE Using the method of the sections, the axial forces in bars FE and FG. 8 kN Reaction forces beam Isolating the system: 0
- 2. http://www.elrincondelingeniero.com/ √ VE HE 5. E Internal forces (using the method of sections) VE HE G F E 0→4 2.4 NEG F 0→ 4 0→ 4. 0 2→ NFG tan 60° Internal forces (using the method of 2 → √ Node F NFE 0 kN 2√3 0 → 2. the nodes) . 2√3 √ 6. 2 kN . cos 30 . sin 60 Calculation of the normal stress in EF. √ √ . 0 ∆ 0→ . cos 60 0 0 0→2 0 0→ NFG . ∆ →∆ 4 .4 3 √3 10 210. 10 . . 0,02 ,