Hi Guys, I have a 2nd year structural analysis problem which I am hoping someone can help me with.(adsbygoogle = window.adsbygoogle || []).push({});

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

T-shaped, indeterminate frame consisting of two members connected at a node in the middle with the bottom member joined by a fixed pin-joint and the top member connected by a fixed joint on either side, the node in the middle has been modeled as a fixed pin joint as it can still be rotated due to bending of the frame (is this a correct assumption?).

Also for this questions assume E&I are constant all the way throughout the frame so when calculating the distribution factors, E&I just cancel out.

To see the actual problem, please have a look at the attached file which sets out the problem as well as the BMD and reaction forces generated from Space Gass.

2. Relevant equations

FEM at C = WL^2/12 = FEM at B if joint B is locked.

Stiffness factor for BA = Stiffness factor for BC = 4EI/l

Stiffness factor for BD = 3EI/L

3. The attempt at a solution

K(BA) = 4/6 EI ; K(BC) = 2/6 EI ; K(BD) = 4.5/6EI

R(BA) = 0.381 ; R(BC) = 0.19 ; R(BD) = 0.429 ; Sum of R = 1 (So far so good)

If you lock joint B, the UDL will be pushing down on member BC and therefore there will be a clockwise or negative fixed end reaction moment at C and according to our FEM rules, WL^2/12, there should be the same moment in an anti clockwise or positive direction (360kn.m) at B (Moment from B to A). Of course if we do this, there must be an equal and opposite moment on the other side to keep the whole thing in equilibrium, so moment B to A must be -360kn.m..... right?

Then I release the joint B and the the FEM at B disappears but the opposing moment, so I quickly redistribute everything before the frame demolishes itself according to my r-values and carrying over 1/2 the distributed values to their respective ends at A & C but not to D because D is a pin joint.

Okay, so this is all fairly straight forward but, what I cant figure out is why Space Gass is telling me that the moment at BC is -291.43 when I got the same figure but positive 291.43. My reasoning is simple, because the original FEM at BC was 360 and I distributed 0.19% of the opposing moment or (-68.4 kn.m) should the total be a positive when I sum them up?

Infact I don't see how the solution in picture which is a Space Gass printout, can actually work seeing as around B, -154.29 -137.14 - 291.43 = -582.86, and thus the way I see it the moment would cause B to continuously spin clockwise until it broke. Whereas -154.29 -137.14 +291.43 = 0.

Ofcourse, my reaction forces don't match up either because the moments are mixed up.

Please help! I have also included a small table in the picture which shows how I calculated the figures and the carryover.

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

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Hardy Cross Moment Distribution Problem

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