Using classical methods to solve statically indeterminate beam.

[StructureMan44]

The propped cantilever in the attachment has a rigid rod supporting the free end, a uniform distributed load and a given deflection at the free end. Assuming the rod does not elongate (infinite elastic modulus) is there a way to determine the moment produced at the wall without utilizing table 3-23 in AISC 325?
I should include that this isn't coursework, is a problem I ran across and was unable to solve.

 

[KootK]

Fun:

1) Treat the beam as a cantilever and calculate the amount of force that would be required at the tip to produce the known deflection. Multiply this force times the length of the beam to get this component of the wall moment (M1).

2) Treat the beam as fixed at one end and simply supported at the other end. Apply the uniform load by itself and work out this component of the moment at the wall (M2).

3) M_total = M1 + M2

If this is a real problem, you've gotta tell us what the the heck it is. Grade beam settlement?

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[JoshPlumSE]

Principal of super-position:

Calculate the tip deflection that you'd get from just the uniform load on a cantilever beam. Draw out the moment diagram for that case.

Calculate the load that would be required at the end of the cantilever to to produce a negative tip deflection of the same magnitude. Draw out the moment diagram for this 2nd case.

Just super-impose those two moment diagrams together.

 

[BAretired]

Calculate the deflection due to the uniform load alone, then calculate the force F in the rod necessary to bring it back to the stipulated deflection. Moment at the fixed end would then be:

M = wL²/2 - FL

BA

 

[CANPRO]

I'm not sure I follow the problem. Does the rigid rod force the deflection and then the UDL is added after? If that is the case I think you would find the force required to produce the initial deflection and the resulting moment at the wall. Then determine the moment at the wall from the UDL on a fixed-pin beam. Add the two resulting moments together. I don't have that table you reference on hand, so not sure if the above is applicable.

 

[JoshPlumSE]

Oops, I didn't realize that you want to maintain that deflection Delta. Therefore, my fist step is the same. But, the tip deflection used in the 2nd step would be the amount of deflection to get you back to the desired Delta.

Concepts still the same though.

 

[BAretired]

It makes no difference which comes first, the UDL or the rod force. The answer is the same.

BA

 

[rb1957]

read up on solving redundant structures.

for the propped cantilever, the easiest (IMHO) is unit force method ...

solve as a cantilever, calculate the tip deflection, DL

apply a unit load at the tip, calculate the unit load deflection, du

Prop reaction is DL/du*unit load.

apply prop reaction and re-solve the fixed end reactions.

another day in paradise, or is paradise one day closer ?

 

[StructureMan44]

I apologize, I believe I wasn't clear in my problem definition. The UDL will cause a deflection at the tip of the cantilever. This deflection is partially restrained by the rigid rod; it only allows a portion of the total deflection.

After the advice from the folks here, my approach was to calculate the deflection at the tip due to the UDL as if the beam is cantilevered. Subtracting the existing deflection from this total deflection we have the deflection restrained by the rod from which we can calculate the force in the rod.

Is the total moment at the wall then just the moment at the wall caused by the UDL acting downwards on the cantilever minus the moment at the wall created by the rigid rod pulling upwards? This seems too simple, especially given the strange shape I expect the beam to take: |--\__

 

[KootK]

Is the total moment at the wall then just the moment at the wall caused by the UDL acting downwards on the cantilever minus the moment at the wall created by the rigid rod pulling upwards?

Yeah, I believe so.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[BAretired]

Simple but true!

BA

 

[rb1957]

since you two guys have replied, I must be lost in translation. why is the UDL causing deflection of the cantilever ? or why comment on it ??

"my approach was to calculate the deflection at the tip due to the UDL as if the beam is cantilevered." ... ok, that's the first step in unit load method.
"Subtracting the existing deflection" ... what existing deflection ?
"... from this total deflection we have the deflection restrained by the rod from which we can calculate the force in the rod." ... that's not how I do unit force method.

another day in paradise, or is paradise one day closer ?

 

[IDS]

Is the total moment at the wall then just the moment at the wall caused by the UDL acting downwards on the cantilever minus the moment at the wall created by the rigid rod pulling upwards? This seems too simple, especially given the strange shape I expect the beam to take: |--\__

As far as the beam knows, it has a moment and vertical reaction at one end, a udl along its length, and a vertical reaction at the other end, so the deflected shape is not very strange. Starting from zero deflection and slope at the left we have a fourth order polynomial curve due to the UDL, minus a cubic polynomial due to the end point load.

It would be quite easy to set up a spreadsheet to calculate moment, curvature, slope and deflection at points along the beam, and confirm that the end deflection is as expected.


Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rb1957]

"Is the total moment at the wall then just the moment at the wall caused by the UDL acting downwards on the cantilever minus the moment at the wall created by the rigid rod pulling upwards?" ... yes, like the other two responders, this is simple superposition.

but now I'm re-reading (for about the 3rd time) your posts, and instead of a propped cantilever you have a cantilever with an elastic support (my bad). the solution is much the same, with a propped cantilever you have zero net deflection at the prop; with an elastic support you have a finite net deflection, x = F/k.

for a rod PL/EA = x

1) calc the deflection of the tip of the cantilever due to the loads; D
2) calc the deflection at the tip due to a unit load, u; d
3) for a rigid prop, the prop reaction, P = D/d*u, ie D-d*P/u = 0
4) but you want D-d*P/u = PL/EA ... P = D/(L/EA+d/u)
5) then adjust the cantilever fixed end reactions to account for reaction P.

another day in paradise, or is paradise one day closer ?