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Okay, first off, I know that there is a very similar problem about statically indeterminate trusses here, but my problem was a bit different and I didn't know how to adapt the answer for that problem to mine.

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So my problem is this: I'm trying to solve for the displacement at node 7 using the energy method, and then I need to solve it using the Finite Element Method. Since the Energy Method seemed the easiest, I went for that first. Through the sum of moments at node 7 and at node 1, I found out that the vertical reaction force at 1 and 3 are the same, and their horizontal reaction forces are equal but in opposite directions, thus cancelling out. 2 doesn't have a horizontal reaction force. Thus I found out that this was statically indeterminate, since now I had $P = 2\cdot(R_{1y}) + (R_{2y})$.

So my next step was to use the energy method from the question I linked to earlier, in order to solve the forces in each member. However, being the idiot that I am, I don't know how to solve the forces in the members to begin with. I don't know which forces point where. I know that if the forces at the ends of a member point away, it's tension. Otherwise, it's compression. But since I know that the reaction forces at 1, 2, and 3 point up, that would mean that the force in the members at 1, 2 and 3 must point down, which means that it is in compression....at least in my head it does. In short, could someone help me get this so that I can go on with my Energy method equation?

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1 Answer 1

First assume each member force is tensile and assign it a positive quantities $F_1$, $F_2$, .. etc.

Next draw the force balance on each node with all the member forces pointing towards the member. Like a spring being pulled (in tension), the force on ends is pointing towards the center of the spring.

For node 2 see example below:

Node_2

With equations

$$ F_4 \cos\theta-F_2 \cos\theta = 0 \\ R_2 - F_3 - F_4 \sin\theta - F_2\sin\theta = 0 $$

If in the end the result is a negative number, then the member is in compression. It all follows quite naturally once you have nice free body diagrams for each node.

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