How Do You Calculate Forces in Truss Members?

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SUMMARY

The discussion focuses on calculating forces in truss members, specifically addressing a scenario with applied forces of 10kN at point F and 25kN at point E. Participants analyze free body diagrams (FBD) to determine the equilibrium conditions and the resultant forces in the truss members. A key point of confusion is the calculation of a 15kN tension force, which is derived from taking moments about point B in triangle ABF. The discussion emphasizes the importance of understanding equilibrium and moment calculations in truss analysis.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Familiarity with free body diagrams (FBD)
  • Knowledge of truss analysis techniques
  • Ability to calculate moments about a point
NEXT STEPS
  • Study the method of joints in truss analysis
  • Learn about calculating reactions at supports in trusses
  • Explore the use of the Pythagorean theorem in truss geometry
  • Investigate software tools for structural analysis, such as SAP2000 or RISA
USEFUL FOR

Engineering students, structural engineers, and anyone involved in analyzing truss structures will benefit from this discussion, particularly those seeking to improve their understanding of force calculations in truss members.

fonseh
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Homework Statement


It's given that the applied force is 10kN at F and 25kN at E . I have trouble of finding the other real force in the member of the trusses .

Homework Equations

The Attempt at a Solution


So , i start by drawing the FBD at F . I have 10kN point downwards , so the reaction is 10kN upwards , so it's in equilibrium . But , i really have no idea how the author get the reaction 15kN ( in tension) pointing to the left and to the right . Can someone explain about it ? ( i have draw the FBD at E inside the second diagram)
 

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fonseh said:

Homework Statement


It's given that the applied force is 10kN at F and 25kN at E . I have trouble of finding the other real force in the member of the trusses .

Homework Equations

The Attempt at a Solution


So , i start by drawing the FBD at F . I have 10kN point downwards , so the reaction is 10kN upwards , so it's in equilibrium . But , i really have no idea how the author get the reaction 15kN ( in tension) pointing to the left and to the right . Can someone explain about it ? ( i have draw the FBD at E inside the second diagram)
Consider the triangle ABF as a rigid body in its own right. Take moments about B.
 
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