Structural analysis of a horizontally curved, non symmetrically loaded (out of plane) pinned beam.

[Vimukthi]

Hi,

I have solved horizontally curved, out of plane loaded fixed beam for BM and torsional moments using Castigliano's method. But can't get proper results for a pinned beam. If anyone can provide a worked example or anything that can help its very much appreciated.

 

[robyengIT]

fixed one end and pinned the other one ?

 

[Vimukthi]

both ends are pinned. or a fixed - pinned example so I can study it.

 

[robyengIT]

Google :
1 - SCI-P281 design of curved steel
2 - NCHRP-12-52 AASHTO-LRFD design example Horizzontally curved steel box girder bridge
3 - FHWA-HIF-16-002- Vol 23 - Dec 2015
4 - FHWA-HIF-16-002- Vol 25 - Dec 2015

PS : both ends pinned = impossible. Possible only with 3 supports

 

[Vimukthi]

ya it seems impossible after applying Castigliano's method. But thanks for your help though! also I have attached a figure (1st figure of the doc) of the problem I'm trying to figure out. For fixed conditions it has three reactions Mb, Tb and Fb and I was trying to eliminate just Mb reaction making it kind of partially pinned. I have used SAP 2000 to model it and it gave results for those boundary conditions. I was trying to get those results manually. BTW I have attached the calculation for the fully fixed condition which works perfectly.

 

[CrabbyT]

Check out AISC Design Guide 33: Curved Member Design. It's available for free if you are a member of the AISC.

Can horizontal members be treated as pinned-pinned, though? Since the center of the member would be in a different plane than the supports, the vertical loads would generate a torsional moment.

 

[Ron]

A curved beam would be unstable in a pinned only condition. You are creating rotation at the support in two directions. At least one end must be fixed, preferably both.

 

[r13]

Bridge usually has more than one bearing point at each end, so it is possible to design for torsion with pin-pin condition.

 

[Vimukthi]

Both ends are restrained for torsion of course. Only pinned for Flexural moment. If you saw the calculation for fixed condition it gives perfect results. Just eliminated Mb by removing Mb components of those equations to make it pinned but getting wrong results.

 

[robyengIT]

for me you should consider both ends restrained for flexural moment and only pinned for torsion

 

[r13]

I don't quite understand. If the formulations were based on fixed end conditions, how could you expect to get correct solutions for pined conditions?

 

[Vimukthi]

I tried eliminating Mb components but it's wrong apparently.

 

[rb1957]

1) surely, reactions should be designed to balance the applied loads ? You can go for the minimum (determinate) set or the maximum (redundant).

2) I thought the "trick" to curved beams was their non-linear stress distribution for moments.

3) I hate saying it, but surely FEA is a good tool to solve this ?

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

 

[r13]

Since this is a single curved girder with two end supports only, so a pin-pin condition is not possible. However, you shall be able to get solutions for pin-fixed condition. The formulation started from support B with an generic moment M[sub]b[/sub], which varies along the curve. If you can identify the integration, and set boundary condition to M[sub]b[/sub] = 0 at the origin, then you shall get the desirable results.

 

[Vimukthi]

@rb1957 I will use Abaqus after figuring it out in more theoretical level.

@retired13 I was imagining something like U socket support at each end. U socket walls will restrain the torsion but still allow it to rotate by flexural moment. So I simply eliminated Mb support reaction from all four equilibrium equations.(1 to 4) Then only have two boundary conditions for Castigliano's method, rotation and vertical translation are zero at support but not the flexural rotation. I have attached what I did to modify the original calculation. All the parts marked with red are eliminated components by making Mb equal to zero.

Thanks everyone for your valuable time though. Appreciate it a lot!

 

[r13]

I don't think you can eliminate M[sub]b[/sub], as moment wouldn't be zero along the span. But due to symmetry, I think you can perform the analysis on one half of the span (say B to C), so M = 0, when Θ = 0, and M is maximum at Φ[sub]o[/sub] = Φ/2.

 

[Vimukthi]

@retired13 Mb is the support reaction. MΘ is the moment along the beam which related Θ by quilibrium equations. And system is not symmetrical, the load can be anywhere on the beam. I'll try only considering the part between support and load.

 

[robyengIT]

ready for translation/explanation (if needed)

 

[r13]

Yes, my mistake.

 

[JoshPlumSE]

This thread really makes me want to read Bow Dowswell's design guide on the design of curved members.

AISC Design Guide 33: Curved Member Design

Now that I'm confined to the house these days / weeks, maybe I'll download it and go over it in detail. My belief (based on how good/practical the AISC design guides usually are, and on my personal interactions with Bo) is that this design guide is probably the best place to start.