Statics, forces in each memeber of a truss

AI Thread Summary
The discussion revolves around solving for the forces in each member of a truss using static equilibrium equations. The initial attempt to analyze joint D revealed three unknowns (ED, BD, CD) but only two equations, leading to confusion about how to proceed. Suggestions include using a free body diagram (FBD) for multiple joints and applying torque calculations to find the necessary forces. It is emphasized that starting from a different joint, such as A, may provide a clearer path to solving the problem. The conversation highlights the importance of considering the entire structure rather than isolating individual joints for accurate analysis.
jonjacson
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Homework Statement



I´ll show the problem with a picture:

izvnkh.jpg


Homework Equations



Sum forces = 0 ;

The Attempt at a Solution



Well I tried to solve the forces equations at the joint D, but I see three unkown quantities ED, BD and CD, and I have two equations, one for the X axis, and one for the Y axis so I cannot solve... or that´s what I thought until I saw that the book gives numerical answers to the problem.

So, Anybody knows how to get three unkown quantities from two equations?

I´m not able to do that, and until 5 minutes ago I thought it was impossible.:confused:
 
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Write out the relevant equations and look carefully at them. See what cancels.
 
Ok, I´ll try it:

I assume ED and BD pointing towards the right, positive sense, so DC is pointing downwards, it´s supposed that the equations will confirm if this is correct or not, so:

x axis--> -2 + ED + BD cos (45) = 0

y axis---> BD sin(45) - DC = 0

Well I could substract the second equation from the first and I would get:

-2 + DC + ED = 0, but I cannot solve yet.

Am I doing something wrong?

THanks for your answer.
 
I made a mistake. I think your equations are correct.
 
Ok.

If it helps I show the answers from the book:

AB=BE=CD= 1 KN T

AE = BD = 1,414 KN C

BC = 2 KN T, DE= 1KN C

T is tension and C compression

Maybe I should start with another joint but I don´t have any data, so it looks like a nonsense.
 
Now that I think about it more, you might have to write a free body diagram for more than 1 node.
 
Try this approach. Use torques. Using point C as a pivot the CCW torque from the 2 kn force is 2X3 = 6. Now what must the force in member AE be in order to counter the torque. Knowing AE you can find DE and the problem is solvable (I think).
 
barryj said:
Try this approach. Use torques. Using point C as a pivot the CCW torque from the 2 kn force is 2X3 = 6. Now what must the force in member AE be in order to counter the torque. Knowing AE you can find DE and the problem is solvable (I think).

Yes you are right.

IF you try it with the method of joints you cannot solve it with only the first joint. I don´t know if using all the joints it´s possible to get the answer.

You have to use the fbd for the whole structure and you start at A.

Thank you very much!
 
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