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BMD of the beam joined at intersection

BMD of the beam joined at intersection

BMD of the beam joined at intersection

(OP)
4 beams are supported by the column at one side, another side is jointed together at the intersection. I have assigned UDL of 20 kN/m on all the beams. 2 beams are longer (3m), 2 beams are shorter (2m) . Surprisingly, the BMD of the longer beams is hogging at the middle part , while the BMD of the shorter beams is sagging at th middle. Does it make sense.

Can someone explain this situation ? Why the BMD of the longer beam is hogging at the middle ? There's no column at the middle , how can there's hogging moment at the middle for the longer beam ?

I think the BMD of the shorter beam make sense and the BMD of the longer beam shall be teh same also (pure sagging in the middle and no sagging at all. ) Correct me if I am wrong ..



RE: BMD of the beam joined at intersection

It looks like a relative stiffness issue. The shorter beam is significantly stiffer, therefore, it provides some support to the other beam. Not like a full vertical restraint, but like a vertical spring. That means the short beam is getting a vertical reaction from the long beam. This would explain both moment diagrams.

RE: BMD of the beam joined at intersection

(OP)

Quote (JoshPlumSE)


So the program automatically determine how's the beam being supported based on the relative stiffness. In real life can I have the long beam support the short beam ? Does it make sense ??

RE: BMD of the beam joined at intersection

You can... if you release the major axis bending within the shorter beam at junction of the longer beam.

RE: BMD of the beam joined at intersection

DCCD -

It's a relative stiffness issue between the two beams. Stiffness is related to EI/L^3 right. So, if one beam is 50% longer, it would have to be 3.4 times stiffer to have the same deflection under a point load. Right?

In order for the long beam to support the short beam you have to have and EI equal to the ratio of lengths to the third power.

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