# Hardy Cross Moment Distribution Problem

1. Jul 29, 2010

### iced_maggot

Hi Guys, I have a 2nd year structural analysis problem which I am hoping someone can help me with.

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

#### Attached Files:

• ###### Moment Distribution.jpg
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2. Jul 30, 2010

### iced_maggot

Btw, I'm not sure if the MO here is to actually post in every bit of working out I did. If thats the case please let me know and I can scan them in, its about 4 pages odd though.

3. Jul 30, 2010

### pongo38

Hardy cross uses a clockwise/anticlockwise sign convention, whereas most computer analyses willl use a hogging/ sagging sign convention. That would explain your issue.

4. Jul 30, 2010

### iced_maggot

Does it though? It doesnt matter which sign convention you use, one will still be positive and one will still be negative which will cancel out to leave joint b in static equalibrium... right? if the 291 is negative then shouldn't the other two moments around that joint should be positive so the net effect is zero?

5. Jul 31, 2010

### pongo38

Hardy cross requires an algebraic balance so that the net moment at a joint is zero, even though the actual moment is not. Then, the question is: How is that moment distributed between the members meeting at that joint? Answer: in proportion to their stiffnesses, as determined (in this case) by hardy cross method. I haven't checked the solution but it looks ok. Sketch the approximate deflection diagram and notice that joint B has a rotation. From the popint of view of an observer just to the right of BD, the members are all hogging. This agrees with the computer analysis, and it agrees with the moment distribution. Your problem could be that you think of positive and negative moments as hogging and sagging, but, in Hardy cross, you have to abandon that idea in favour of cloackwise and anticlockwise. This isn't the other interpretation of +and - as sagging/hogging rather than hogging/sagging. It's completely different and gives rise to the question you have raised. Look again at your fem's. They are + and -, as required by HC but you wouldn't put that in a conventional computer program.