Normal Stress in members of a stystem

[RustyShackelfor]

Diagram: http://i177.photobucket.com/albums/w222/77whtrocco/NORMALSTRESS.jpg

Given: members AB, CD, and EF have x-sectional area of 25 mm^2. E = 200 GPa. Neglect deformation in member GH.
Find: Normal Stresses in members AB, CD, and EF.

I know that I need to determine the axial forces in these members, but it's been a while since I took a statics class.

Isolating member GH, I have come up with...
Sum of M about G = 0 = 5000(1.6m) - AB(.4) - CD(.8) - EF(1.2)

and I have concluded that the reactions at G, A, C, and E are zero for "x" components (ie. Gx = 0, Ax = 0, Cx = 0, etc)

I should also not that AB is basically equal to reaction Ay, CD = Cy, EF = Ey. I am trying to solve for these values. Thanks.

Now I'm stumped... What else is there? I'm confident that after finding these forces I can solve for the stresses.

 

[RustyShackelfor]

normal stress in a member of a system = axial force / x-sectional area. Need to determine axial forces (AB, CD, EF). Axial forces and the vertical and horizontal displacements can be related using a compatibility equation using an initial bar length (say Lab, Lcd, Lef) and change in bar length (ie "delta L"). So...
delta Lab = (EF*Lab)/(E*A), where E is the modulus of elasticity and A is cross-sectional area... my problem is that there is no specified deformation. I was given the answer, but I'm beginning to wonder if something was left out.

 

[PhanthomJay]

normal stress in a member of a system = axial force / x-sectional area. Need to determine axial forces (AB, CD, EF). Axial forces and the vertical and horizontal displacements can be related using a compatibility equation using an initial bar length (say Lab, Lcd, Lef) and change in bar length (ie "delta L"). So...
delta Lab = (EF*Lab)/(E*A), where E is the modulus of elasticity and A is cross-sectional area... my problem is that there is no specified deformation. I was given the answer, but I'm beginning to wonder if something was left out.

Since the beam is considered rigid, it must rotate counterclockwise about the left hinge to its equilibrium position where the deformation of each vertical member is proportional to its distance from the left end. Use similar triangles to calculate the ratio of each members deflection , and hence each member's force, to each other. This when combimed with your equilibrium equations should give you the solution.