Prestressed beam weight for crane lifting - a practical or theoretical approach? 1

[iain98765]

Hi there,

I'm in a bit of a dilemma. Two arguments for the self weight of a beam.

1. to take a standard 2500kg/m^3 for the beam as (unit weight)which would include for the steel. for a beam with theoretical volume of 20m^3
so 2500*20/1000 = 50tons

or
2. take the cylinder tested density of 2430kg/m^3, the steel strands 1.1tons, the actual steel weight 8.9tons and miscl.of 2 tons for grout, collars etc.
2430*20/1000+1.1+8.9+2 = 60.6tons.

problem
1, in calculation 1, the density is underestimated as the steel is much denser, as is the concrete.
2. in calculation 2, the steel hasn't been taken out of the concrete.

Problem,

For a crane lift it is obvious you want to be as conservative as possible. Do you:
1. take calculation 1 as a blanket calculation and risk underestimating the load, or,
2. factor everything in and give yourself a bit of legroom?

What is the standard, or best approach?

I would really appreciate the thoughts and input on this!

 

[BAretired]

When you have an unusually high volume of steel per unit volume of member, as you appear to have in this example, it is better to use Method 2. but you could deduct the weight of the displaced volume of concrete when determining the total weight of member.

BA

 

[iain98765]

displaced volume? a known volume and the difference between..what exactly? displaced means you know the volume of a bath. although you fill it an what spills out is the displaced volume. so I don't think your getting this.

 

[paddingtongreen]

If it worries you, use method 2 but use a steel weight reduced by the appropriate concrete weight. If, say, you had a square foot of steel along the section, use 490-145= 345# or 2.4#/square inch of section.

Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin

 

[slickdeals]

Where you have rebar, you don't have concrete.

(Area of concrete - Area of rebar) * Length * Density of concrete + Area of steel * Length * Density of Steel

 

[IDS]

Just multiply the weight of steel by (7.85-2.5)/7.85.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

Also I'd check the steel weight. A total steel weight of 10 tonnes is nearly 16% by volume, which sounds way too high, especially for a prestressed beam.

And the grout is filling voids in the cross section, so no need to add that, unless you deducted the ducts when calculating the concrete volume.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[JStephen]

So weigh the trucks and find out.

 

[chicopee]

And once you have figured out the weight to be lifted, the crane lifting capacity at whatever load radius you have, should be twice the value to be lifted. While this is a requirement by the railroad when you work on railroad property, it is a good practice to follow you when you are in other places.

 

[iain98765]

I guess the issue is, we know placed volumes of concrete, and we know the theoretical volume. the placed values are 4m^3 more than the theoretical. (although this is fresh concrete volume and there is some evaporation et.) I guess taking the placed volume, you have the actual volume of concrete.

The beam has a lot of steel in it.

 

[sasa2k]

"... the steel strands 1.1tons, the actual steel weight 8.9tons and miscl.of 2 tons for grout, collars etc.
2430*20/1000+1.1+8.9+2 = 60.6tons."

You likely overestimated re-bar weight. For well design beams rebar to concrete ratio is about 1-1.5%, and for columns around 2.5-3.5%. 4% is a practical limit. In your case you have appx. 5.6% (for longitudinal bars only).

So if we assume 1.5% rebar, 0.015 x 7850 kg/m3 x 20m = 2.36 t, add 30% for ties, overlap etc 3.10 t steel. Thus 2430*20/1000 + 3.1 = 51.7t which is the same for all practical purposes to estimated 50t.

 

[iain98765]

Here is what I have.

The steel weight is known. This is from delivery tickets (fabricators slips)this isn't underestimated, its a heavily reinforced highway bridge beam 8.9Tons.
The tested dry density is known and a little higher than normal OPC concrete 2430kg/m^3.

The volume of the concrete is known from the delivery notes. This is higher than the theoretical by 4m^3.

So if I know the volume of concrete that went in by the concrete tickets(this is the displaced volume), I know the weights of steel etc. surely this is the way to go.

Multiplying the theoretical volume by 2500kg/m^3 is this not underestimating the load?

 

[IDS]

If you have the correct weight of all the elements of the beam then obviously the total weight is the sum of them all. You don't need to deduct anything for the concrete displaced by the steel if you know the actual volume of concrete that went into the beam..

But I'd still suggest checking that the reinforcement weight is not in kN, rather than tonnes (or maybe that the weight supplied is for 10 beams rather than one). Also where has all the extra concrete gone if the actual volume is 20% higher than the theoretical?

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[CANPRO]

"The volume of the concrete is known from the delivery notes" ---> How do you know that the trucks had the exact volume of concrete that was on the ticket? How many trucks of concrete did it take? If they're all of by a small margin it would add up. Plus, how do you know all of the concrete made it into the beam? Did you account for the concrete that was removed for testing? And on that note, how often was the concrete tested and how do you know for sure all of the concrete is at the same density?

"displaced volume? a known volume and the difference between..what exactly? displaced means you know the volume of a bath. although you fill it an what spills out is the displaced volume. so I don't think your getting this." ---> He is definitely getting it. Water in bath = concrete in beam, you = rebar. Rebar displaces concrete. Go take a bath and pretend you're a piece of steel, you'll get it.

I would think the most accurate way to get the concrete volume is to measure the actual finished dimensions of the beam to get the volume and then subtract your known volume of steel.

 

[SlideRuleEra]

iain98765 - You are asking "What is the standard, or best approach?"

IMHO, put aside your detailed assumptions and calculations and go the documentation on the prestressed beam design. In the USA, that is the "Precast / Prestressed Concrete Institute" (PCI). Check their tables on the type beam of interest:

http://www.pci.org/Design_Resources/Transportation_Engineering_Resources/Bridge_Design/

For the various shapes the nominal weight is given in kips / foot.

If you are not in the US, perhaps you have a similar governing body to consult.

Then, spend the time saved making sure that the lift plan makes sense, the lifting equipment is of adequate capacity and in satisfactory condition. Because of possible unintended loading during the lift, this is far more important than a calculation of the beam's precise weight.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[iain98765]

I'm concerned the contractor is underestimating the beam weight. Some of the ancillaries are nearing their max Load. I'm looking for the best approach to get the weight of the beam.

Caneit, thank you for your reply, your approach isn't very practical as it's a 30m beam of irregular shape which tapers. Why work out the volumes when you know what's been paid for? I think the known factor here is what's been delivered, I have known placed volumes of concrete, and known steel weights. I think that's the best approach. Then applying a shock load factor to check the ancillaries. There are 72 beams all complete and we have the placed volumes in each.

Slideruleera- I will check out the site, thanks for the link.

 

[CANPRO]

"Why work out the volumes when you know what's been paid for?" Because you don't always get what you pay for, and as mentioned, you say you have 20% more concrete than expected (how did you determine the exact quantity that was placed?) so something isn't quite right. Just because you order up 8 cubic yards of concrete doesn't mean they're putting in exactly 8 cubic yards of concrete into your beam. Anyway, just a suggestion. And just because the shape is irregular and tapered doesn't make it impossible to calculate the volume.

 

[PEinc]

I'm with CANEIT on this one. Calculate the actual volume of the beam. Use the known density of the concrete to get the weight of a solid concrete beam. Add the weight of placed steel reduced by the weight of an equal volume of concrete.
It sounds to me like you are trying to avoid calculating the volume of the concrete beam. You could have done this calculation in less time than it took you to write and read all of these responses.

http://www.PeirceEngineering.com

 

[SlideRuleEra]

Since you have clarified the beam is an irregular shape my suggestion to look up weights of standard beams won't work. In that case, I'm with CANEIT and PEinc, too. To simply the calculation even more, you do not have to compute the volume of the steel. Specific weight of steel is 7.85, specific weight of concrete is 2.40 (approximately). Do some simple math ratios and all you have to do is add 69.4% of the known Weight of Steel to the calculated weight of the concrete.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[iain98765]

I agree with you guys also. The weight of the steel has to be deducted.

if you work it this way, (without having to measure the beam)
Steel is 7850kg/m^3
our tested dry density is 2430kg/m^3
Ratio of concrete to steel is 3.23.

Theoretical volume is 20m^3
Steel is 8.9Tons
Therefore 8.98*1000/20 gives you 445kg/m^3 of steel
Which is 445/3.23 = 137.77kg/m^3 of concrete to deduct.

137.77*20 = 2755kg or (2.755Tons) has to be deducted

as I used in (2) way at the start of the thread (above) the theoretical volume, If I deduct the steel 60.6-2.755 = 57.85Tons
This I would say is the real beam weight, though for safety sake, I would have left the 2.755 tons in the original calulation.
Still way higher than a blanket calculation using 2500kg/m^3 unit weight......Anyone disagree?