# Homework Help: Truss analysis, determining internal forces

1. Feb 21, 2013

### pedrok

1. The problem statement, all variables and given/known data

I've been given the above truss and have to analyze it's internal forces as part of the problem, the other part of the problem is to analyze for which lengths members will fail, in what order and for which force F, however this is not the part that I find difficult.

In the actual problem statement it says the following: * Do not analyze segment AB it is assumed strong enough do to the support and AB can only extend or contract along its length.

2. Relevant equations
$\sum F_x = 0$
$\sum F_y = 0$
$\sum M = 0$

3. The attempt at a solution
I know how to work the method of joints, but first off from the statement (marked with * above) I gather that I should introduce a support at either point A or B which gives reaction forces and then also makes it a statically determinate truss structure, then I think the following happens (please correct me if I'm wrong):

part CB : in compression
part AB : in compression
part AD : in tension
part CD : in tension
part BD : in compression
part AC : in tension

Then analysis per joint gives me the follow equations:
A - $\sum F_x = F_{ad} + F_{ac} \cdot cos(45) = 0$

- $\sum F_y = -F_{ab} + F_{ac} \cdot sin(45) = 0$

B - $\sum F_x = Rb_x - F_{bc} - F_{bd} \cdot cos(45) = 0$

- $\sum F_y = Rb_y - F_{ab} + F_{bd} \cdot sin(45) = 0$

C - $\sum F_x = F_{bc} - F_{ac} \cdot cos(45) = 0$

- $\sum F_y = -F - F_{cd} - F_{ac} \cdot sin(45) = 0$

D - $\sum F_x = -F_{ad} + F_{bd} \cdot cos(45) = 0$

- $\sum F_y = F_{cd} - F{bd} \cdot sin(45) = 0$

I now have 8 equations with 8 variables (although F is also given as a variable, I can end up with equation based on F for this purpose) and this should be solvable, but I have no idea how to start, at which point and if the assumption for a support is valid and or logical (or simplifying).

Any help is greatly appreciated!
Peter

2. Feb 21, 2013

### pongo38

Firstly, you need to define the support conditions more precisely as either hinged or as rollers. If they are both hinged, then the structure is externally redundant and cannot be solved with equilibrium equations alone. Secondly, the nature of the joints is not clear. If they are assumed pinned joints, then the structure is once redundant internally and cannot be solved with equilibrium equations alone. If the joints have any rigidity then you have bending moments and shear forces in addition to axial compressions and tensions.