Miami Pedestrian Bridge, Part XIV 78

Hokie,

The top chord and diagonals shrinks/creeps as well. So you have to look at the relative movement/shortening. Also, there is no significant restraint from the joints or top chord. If the bottom chord shrinks relative to the rest of the truss, it will just camber up. The creep and shrinkage does not create a significant shearing force at the end joints.

You can call it a truss or frame (although by far, the stiff load path is the truss). It doesn't mater. You don't have restraint from the top chord to act as enough of a strong back to create significant shears. Even if you had a full web, a girder just typically cambers with PT in the bottom flange and this structure has less internal restraint than a girder.

Even with a vierendeel, the truss will still cambers. The verticals may get high bending curvature if the chords are much stiffer in bending than the verticals and you get differential creep/shrinkage. But in our case, the diagonals are very stiff since they take load axially and the camber is only resisted by bending in the chords which are much less stiff than the truss configuration.

In a true truss, there are zero shear forces due to PT. This structure is much closer to a true truss than a vierendeel.

I know that is your opinion, as stated previously. But I disagree, and believe that axial shortening of the bottom chord played a big role in the cracking, and ultimate failure, of the structure. Hopefully, someone else will comment about my last post in Thread XIII.

Hokie, there’s nothing to resist the axial shortening. It’s a simply supported truss so the beam geometry can change near freely in response to differential shrinkage of the members.

The flexural strength and stiffness of members are no way near enough to resist the shrinkage to that degree.

What do you imagine is resisting the shrinkage force in the deck, sufficient to rupture the node?

Those end diagonals are stiff, and shortening of the deck caused distress in the joint. That's two of you who disagree with my hypothesis. Fair enough, but I would still like to hear from others who have thoughtfully looked at the photos of cracking at those joints.

And further to that, was axial shortening considered in the analysis load cases? I'm not going to try to figure that out, but does someone know?

Hokie,

The stiffness of the diagonals is not relevant. Take them as being absolutely rigid. No mater how rigid they are, they will only bow the top chord (and truss). It is the flexibility of the chords relative to the truss that maters if differential shortening is going to cause shear stress at the joint.

There are many things that Figg did or didn't do that really surprises everyone. So I can't say that they did look at the long term response. I can say that any competent design team would look at both the short and long term behaviour which would include the creep, shinkage, and temperature changes.

The long term response is particularly important in this structure due to the tube stays. If the long term response wants to cambers up, there is big trouble with additional shear at the joints and compression on the diagonals. That is because unlike the truss chords bending stiffness, the stays are really stiff and even with a little camber, will produce large forces. They would produce enough force to keep the truss flat (or in the initial position). This is when the PT can actually stress the joints.

If I find the time, I can model it to get a more affirmative answer. I may approximate the section shapes just to get a feel for the long term response. Once those stays and back span are on, then the analysis gets to be more complex. Without the stays, it is not so bad to analyze by hand (at least with the simply supported case and only the gravity loads).

hokie66 - I'd find it very hard to believe that any shortening of the deck or canopy would have no effect on the diagonals or their connections at the deck and canopy. From the report and info it's rather hard to say exactly which issue was the straw which caused the cracking to start, but what you're pointing out certainly wasn't helping the structure any.

There's been a bunch of comments and theory about the backspan/tower helping support the 11/12 node. I posted way back that the main span needed to support itself without anything else helping for any chance of the whole structure working long term and I still believe that. From the report and structure build info, it's been made clear errors were made so this wasn't the case, but then that was rather obvious by just looking at the pictures of the cracks.

I remember seeing the detail of the faux stays at the top of the canopy, and it looked undeveloped. If these were just for aesthetics, I think they would have needed some type of slip joint.

Will this data help?
From Bridge Factors Attachment 70 FIU Superstructure Longitudinal
FIGG LARSA Calc Pg 5, pdf file page 13

[pre]Sections
Name Section Area Shear Area Shear Area Torsion Inertia Izz Inertia Iyy Plastic
(ft²) in yy (ft²) in zz (ft²) Constant (ft^4) (ft^4) Modulus
(ft^4) (ft³)
Truss_diagonal 3.5000 2.9167 2.9167 1.6798 0.8932 1.1667 0.0000
Truss_diagonal 2 6.1250 5.1042 5.1042 4.2955 1.5632 6.2526
End Diagonal 5.2500 4.3750 4.3750 3.3994 1.3398 3.9375
Truss Top Chord 16.4607 16.4607 16.4607 5.3100 352.4200 2.6925
-16ft wide
Truss Bottom 45.6753 45.6753 45.6753 36.1683 2,995.0267 12.4210
Chord-31ft 8in_wide[/pre]
Concrete Truss_diagonal Rectangle 1.7500 2.0000
Concrete Truss_diagonal 2 Rectangle 1.7500 3.5000
End Truss Diag 2 is 3.5X21" - do not know where that is used

Vance said:

Will this data help?

Click to expand...

Yes it will help. Thank you.

I am not sure I can sort out the joints and member assignments - they are much less evident - self generated, and so forth.

Horizontal shear in a rectangular beam (and more)
If we consider a rectangular solid beam placed horizontally and simply supported with no end restraints, and loaded with a uniform load , we expect it to deflect downward.
It does so because there is compressive stress in the top and therefore the top shortens, and there is tensile stress in the bottom, and therefore the bottom lengthens. Simultaneously a horizontal shear is induced in the web - Vh=1.5V/A.
Does it not seem logical that anything that would create that same curvature in the beam would cause the same internal stress?
What is less obvious is how a vertical load manages to create a horizontal stress in the rectangular beam. Would not a horizontal load like PT forces seem more direct and more obviously create horizontal stresses?
If we think of a pack of paper for the printer, and we want to prevent the sheets from sticking together we flex the sheets to cause them to slip on each other. That slip would represent the horizontal shear from the curvature. If we prestress the sheets on the tight curve side just enough to shorten them and maintain alignment with the sheets above, that shear is negated. But don't we have to transfer some of that PT force to the sheet just below the top sheet and to all the sheets from there to the PT force?
If the object were a truss and not a beam, if it were 18 feet deep and 174 feet long, and if prestress forces in the deck were sufficient to cause maybe an inch of camber in the truss, would it not have lifted off its interior falsework and be spanning 174 feet, and therefore the axial load in diagonal 11 have been the same as when setting on the pylon? Would the load in member 11 been negated if loads were placed on top of the truss to force it to stay straight? Or would the load in 11 be increased?
This brings to question the issue of concrete strength at the time of tensioning and the E value at that time.
I think the drawings allowed tensioning at 6000 psi.

I am slowly working away on the PT model when I can.

The forces in the diagonals due to the initial PT of the deck are 14Kips and 7Kips respectively for #2 and #11. That will go down with creep and shrinkage. Keep in mind that this is only looking at the initial PT of the bottom deck in isolation. I am getting a DL compression in #2 of 1800Kips and 1280 Kips in #11. IN the north end, I get a vertical shear of 110 Kips in the bottom deck, 58 Kips in the canopy and 12 Kips for the weight of #12. So the net vertical component to the diagonal is (938Kips-100-58-12)=770 Kips

The camber up is 0.92" (this excludes the DL which is obviously in the opposite direction.

The initial dead load deflection is 1.3".

The south end reaction is slightly higher than the north end reaction.

The initial deck PT contributed 0.5% to the interface shear at the north end.

Awesome work. Thank you!

Earth,
Thanks. Does your model give an estimation of longitudinal shortening due to drying shrinkage and applied PT?

In my experience, drying shrinkage has been in the order of 5/8 inch in 100 feet in normal weight concrete. I have no experience with Florida aggregates and 8500 psi concrete.
That value is for only PT D1L and D1R? So there are five more sets of strands to be tensioned?
Thanks,

Vance,

The upward camber is short term with all the PT in the deck (no other PT applied yet). I am trying to set the initial temperatures for the PT strands. The temperature will be kept the same through out the analysis. Shrinkage is models as a negative temperature on the concrete. How much high strength concrete shrinks in a Florida environment is a good question. I usually use 0.0003 strain but 0.0002 may be more appropriate. 5/8" 100 feet is about 0.0005 which is realistic but maybe on the high side (once you include thermal changes it may be more realistic).

The model is 2D and stick elements for simplicity. Even at that is takes a while to put in all the PT, determine temperatures etc.

Hokie,

Right now, I have only included the initial PT (full bottom deck) and the initial concrete modulus. I will include for shrinkage and creep later. I will rerun the model with a lower E for concrete to account for creep. The temperature on the concrete will be negative to account for shrinkage. I am getting there but it is a lot of work and I still have to my regular work.

The concrete shortens with the application of the PT.

Hokie,

I am back on the computer. The total shortening of the bottom deck from the initial bottom deck PT is 0.56" but that is taken up mostly by camber of the truss.

I made a slight error. The DL deflection is closer to 1.5" rather than 1.3".

I work more on this later. I have to go back to work.

Thank you for doing that. Certainly no hurry. That PT shortening is about what I would expect, and as Vance reported, the drying shrinkage shortening is usually about twice the PT. So overall, in the range of 1.5" or a bit more, similar to what I said before.