Frame Deflection

[Debra555]

Can someone help me check a solution?

I have got a frame 10 ft high and 10 ft long. I have a UDL on one side @ 400 lb/ft. Icol = 200 in^4, same for Ibeam. One side is pinned, other side a roller.

I calculated a deflection of 0.2 inches?? Does this sound correct using a standard drift equation.
Also If I place a UDL on the beam of 400 lb/ft how will this affect the deflection?

I am grateful if some of the experts in structural online can help.
How would the deflection change if both ends were fixed? What equation can I use?

NOTE: I am looking for a simplified solution and set of equations I might use.
Thank you for helping me with the above problem.

 

[amendale]

Using method of virtual work I got a deflection of 0.45 in. I would suggest using this method, as you're not going to find accurate equations for the various conditions you want.

 

[Debra555]

Sorry, my frame had two pins.......my mistake.....
What do you get with two pins?, and two fixed ends? (as an independent check)

I'm using a formula developed by Rudolf Budeski - Modern Engineering Interchange.
Would you know how to incorporate loads on the Beam into the drift.

Thank you.

 

[amendale]

With two pins I get 0.174 in and with two fixed ends 0.03 in.

 

[Debra555]

Thank you.

So it appears the simplified equation is OK.
Can you , or anyone else suggest how to incorporate a UDL on the beam into the deflection.

Thx again

 

[amendale]

A UDL on the beam would result in a P-delta effect. To incorporate this you can iterate by hand by multiplying the initial deflection by the total gravity load, resulting in a moment. Solve for a new defelection and keep iterating until convergance. Or you can use the amplification factor in the code for P-delta.

 

[Debra555]

Thx

I don't have a copy of the code here with me.

Can you illustrate the calculation using an amplification factor. Assume 400 lbs/ft and one point load at center = 2 kips (as an example).

Can you show this with the deflection with 2 pins?
Thank you so much.

 

[amendale]

From the Canadian code (I think AISC has an identical provision):

U₂ = 1/(1 - (C_f*delta)/(V_f*h))

For your case:

U₂ = 1/(1 - (6kip*0.174in)/(4kip*5ft)) = 1.004

Multiply deflection by U2 to get deflection accounting for P-delta

 

[JAE]

Is this for some kind of school design project?

 

[Debra555]

Thx

Your help was much appreciated.
Any other comments from other would be most welcome as well.


Thx again.

 

[amendale]

JAE,

I think so...but I'm just bored at work.

 

[Debra555]

Hi JAE

No this is not a school project.

 

[Debra555]

JAE,

I'm at work too.....
Can you add anything further to the discussion? Greatly appreciated.

Thx