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Say we have the following truss, with $1$ kN load pointing upwards at node 5. The goal is the calculate all the reactions and the internal forces within the members using the method of joints.

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I used an online calculator to do this and got the following results:

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I agree with everything except the internal force in member number 4. Why is this in tension? We have a point load of $1$ kN in node 5, when you isolate this load and apply equilibrium equations, you have $\sum F_y = 1 + F_{4} =0$, so force in member 4 is $-1$. Is the software incorrect? Also And if you isolate node 4, the equilibrium equations are not satisfied?

Computed using https://skyciv.com/free-truss-calculator/

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In a truss, the member is in tension when the reaction is pointing away from the node. You have applied 5kN on node 5 in "+y" direction, as direct pulling, there must be an equal amount of force in member 4 but runs away from node 5 in direction of "-y".

Also, look at the diagram below, with the applied load and the reaction both pointing away from the respective node, will this member become longer (due to tension), or shorter (due to compression)?

enter image description here

enter image description here

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  • $\begingroup$ Hi thanks for the answer, yes so the software is incorrect? As you said, there must be an equal amount of force in member 4 but runs away from node 5 in direction of "-y". So that will be compression right? The software indicated tension, and when you apply the equilibrium equation $\sum F_y = 1 + F_{4} =0$, so force in member 4 is $-1$, Positive $1$ as upwards. $\endgroup$
    – CountDOOKU
    Apr 24, 2022 at 14:49
  • $\begingroup$ Alsowhen you look at node 4 the equilibrium equations are not satisfied, the software indicated positive internal forces for both members $\endgroup$
    – CountDOOKU
    Apr 24, 2022 at 14:50
  • $\begingroup$ " ...runs away from node 5 in direction of "-y". So that will be compression right?". No, as all forces (the arrowheads) are running away from the joints, by definition, both the joins and the members are in tension. Your second graph shows the result of member forces, which are subjected to tension and are indicated by the tensile forces at the ends of the members. From the diagram that shows the "joint forces", the sum of the forces about a joint is always zero (the arrowheads on each side are reversed). You shouldn't get confused if sticking to the sign convention. $\endgroup$
    – r13
    Apr 24, 2022 at 16:51
  • $\begingroup$ Can you just explain this part $\sum F_y = 1 + F_{5} =0$ on node 5, this is right is it not? Sum of forces about a joint is always zero. There are only $2$ forces acting on node 5, the externally applied load of 1KN and the internal force in member 4. So $F_5 = -1$, This negative is compression, so arrowhead towards the node. But why is it showing tension as positive? $\endgroup$
    – CountDOOKU
    Apr 25, 2022 at 7:06
  • $\begingroup$ At node 5, the reaction = -1 (pointing DN), the applied load = +1 (pointing up), (-1)+(1) = 0. $\endgroup$
    – r13
    Apr 25, 2022 at 14:03

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