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 . 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?


1 Answer 1


Note: a reaction is an external force, such as at A and D. The internal forces in members are not reactions; you could refer to them as "axial loads" in this case.

Remove unknowns one at a time. The easiest are the external reactions at A and D. Take moments about A to get the reaction at D, take moments about D to get the reaction at A. Resolve vertically to make sure they add up!

Next, draw free body diagrams at nodes with the fewest members; these are likely to allow you to eliminate unknowns. In this example, resolving vertically at A will give you the force in member AB, given that you know the reaction at A. Resolving horizontally at A will give you the force in member AF in terms of the force in member AB.

Then you can resolve a free body diagram at B, which will give you that the force in member AF is equal to the force in member FE.

Continue in this manner and you can get the forces in all members.

  • $\begingroup$ do you mean for this type of question , we should always find the reaction (external force ) first before solving the internal forces ? $\endgroup$ Jan 25, 2017 at 13:31
  • 3
    $\begingroup$ @kelvinmacks: Yes. Solving for global reactions is usually done first. $\endgroup$
    – Wasabi
    Jan 25, 2017 at 14:01

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