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Pouring of the deck slab of a skew bridge

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AlanLord

Structural
Nov 19, 2014
26
I have to pour the deck slab of a skew bridge soon. After reading a bit of the literature out there on the subject, I’m getting a little nervous about the proper sequence of concrete pour to follow (i.e. perpendicular, parallel to skew, or staggered perpendicular pour).

I have a 260’ single-span 50’ wide bridge on 5 girders 10’deep, and a similar one with the same span, but 37’ wide on 4 girders; both with a skew of 39°. The slab to pour is 9" thick.

The steel installed was a hybrid of no-load fit and steel load fit, and we’re having to live with permanent girder top flange lateral displacements of ½” under steel load, and anticipated concrete pour displacements of ¾” , for a permanent total dead load lateral displacement of 1.5”

I'd appreciate the suggestions of someone who has a lot of experience pouring the decks of skew bridges.

I'm in Montreal, Canada.
 
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Oy... I don't have any experience to help, but just the idea of 1.5" lateral flange displacement makes me twinge.

Best of luck! I'll be curious to hear what the others suggest here.
 
Welcome to the wonderful world of skew decks.
 
I'm surprised the contract plans don't specify the deck pouring sequence. It's typically required on all bridge plans in NJ. Maybe the designer can provide some input.
 
We're in Montreal, Canada - usually we leave that up to the contractor. I'm the designer.
 
The Contractor's available equipment, employees, access to the slab, concrete supplier capability, etc. will be as important considerations as determining an optimum placement sequence. There are usually pros and cons to any reasonable proposal. When the proposal is submitted, consider all factors not just the engineering aspects. For example, the "best" placement sequence is of no value if cold joints occur because the concrete supplier cannot make timely deliveries because of a "forced" schedule.

IMHO, the most important single engineering concern is that on any one placement, ALL of the beams are equally loaded. That would define all placements, regardless of their position, as parallelograms.

[idea]
[r2d2]
 
Yes, you're absolutely right. Thank y'all.
 
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