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THE SAME STEEL AMOUNT IN TENSION AND COMPRESSION OF CONCRETE BEAMS QUESTION 12
- Thread starter eyemeyar
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A doubly, and equally reinforced beam can, to some degree, be conceived of as a steel beam by ignoring the concrete contribution. That provides a good deal of ductility and easy to quantify deflection control. Those things aside, a beam reinforced in such a way is inefficient and usually unnecessary.
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.
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.
I have designed grade beams with (close to) equal top and bottom reinforcement due to load reversals. For this type of loading, it would be very appropriate to consider this beam as just steel because after a few cycles the concrete will be mostly useless unless it is well confined. Design for Vc = 0. For that condition it was ideal for me.
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Kootk,
Do you want to rethink your comments regarding conceiving as a steel beam and deflection control
- most of the deflection control in a concrete beam comes from the concrete in compression and the uncracked portion in the tension region, not from the reinforcement.
- Concrete shrinkage warping is not negated by equal top and bottom reinforcement as the effect is relative to the distances of the reinforcement from the neutral axis. Seeing the compression face steel is much closer to the neutral axis than the tension face steel (assuming concrete is cracked) then the tension face steel wins and causes significant increase in deflection.
If you try doing the deflection calculations based purely on the reinforcement, you are wasting your time.
All you are getting from the compression reinforcement is increased ductility which should not be needed in a well dimensioned concrete beam and some reduction in deflection, the amount of which is dependant on how heavily the beam is reinforced in the first place.
Do you want to rethink your comments regarding conceiving as a steel beam and deflection control
- most of the deflection control in a concrete beam comes from the concrete in compression and the uncracked portion in the tension region, not from the reinforcement.
- Concrete shrinkage warping is not negated by equal top and bottom reinforcement as the effect is relative to the distances of the reinforcement from the neutral axis. Seeing the compression face steel is much closer to the neutral axis than the tension face steel (assuming concrete is cracked) then the tension face steel wins and causes significant increase in deflection.
If you try doing the deflection calculations based purely on the reinforcement, you are wasting your time.
All you are getting from the compression reinforcement is increased ductility which should not be needed in a well dimensioned concrete beam and some reduction in deflection, the amount of which is dependant on how heavily the beam is reinforced in the first place.
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rapt said:
I'm always happy to rethink my positions when a able colleague suggests that course. So now I've completed that exercise and I stand by my original comments. Consider a case where the symmetrically apportioned top and bottom steel were replaced by a literal wide flange steel beam of an equivalent area encased within the concrete. You're telling me that the stiffness of that composite section would somehow be less than the stiffness of the bare steel section alone? That's a tough pill to swallow.
rapt said:
I'd be wasting my time if my goal were an accurate deflection estimate. If my goal were a lower bound stiffness estimate, I may in fact be saving some of my time as jwilki has intimated.
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.
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Reasons for compression reinforcement as paraphrased from my McGregor book pg 158-9:
1. Reduced long term deflections
2. Ductility
3. Compression zone failure is negated
4. Easier to build
and the elephant in the room that jwilki had to blab about, shhhh.
1. Reduced long term deflections
2. Ductility
3. Compression zone failure is negated
4. Easier to build
and the elephant in the room that jwilki had to blab about, shhhh.
Kootk
Your initial comment contained the following
" be conceived of as a steel beam by ignoring the concrete contribution"
If you ignore the concrete contribution then you are not considering it to be composite, purely steel!
JWilki,
What calculation method are you using to justify the reduction in deflections for shallow beams due to the addition of compression reinforcement?
I find in most cases the "compression face" reinforcement in shallow T beams is normally not very effective as it is fairly close to the neutral axis so only has a low compression stress in it so it does not have much effect on creep and it has very little effect on shrinkage warping deflection due to its small lever arm compared to the tension face reinforcement.
Your initial comment contained the following
" be conceived of as a steel beam by ignoring the concrete contribution"
If you ignore the concrete contribution then you are not considering it to be composite, purely steel!
JWilki,
What calculation method are you using to justify the reduction in deflections for shallow beams due to the addition of compression reinforcement?
I find in most cases the "compression face" reinforcement in shallow T beams is normally not very effective as it is fairly close to the neutral axis so only has a low compression stress in it so it does not have much effect on creep and it has very little effect on shrinkage warping deflection due to its small lever arm compared to the tension face reinforcement.
Working mainly in transport and mining structures, reinforced beams and slabs are almost always reinforced on both faces, and the reinforcement is often equal on both faces. Why? Because:
- The Australian bridge code requires all faces to have at least 500 mm2/m reinforcement
- The minimum reinforcement required to exceed the cracking moment often governs
- Continuous beams over closely spaced supports, such as bridge pier footings and headstocks, often have similar positive and negative moments
- Precast elements often have moment reversals from transport and erection to in service conditions
In a lightly reinforced slab, with typical cover for an external structure, top reinforcement often makes very little difference to deflections, because the steel centroid will be close to the neutral axis under working loads. For example, doing some quick calcs on a 350 mm deep slab, with a moment of 120 kNm (not much over the cracking moment), ratios of curvature without compression steel to curvature with equal compression steel are:
For a lightly reinforced section (5 N20 per metre)
Short term loads: 101%
Including creep: 105%
Including creep and shrinkage: 107%
Total long term curvature/ short term: 138% with compression reo, 147% without
For heavier reinforcement (10 N20 per metre)
Short term loads: 104%
Including creep: 114%
Including creep and shrinkage: 121%
Total long term curvature/ short term: 145% with compression reo, 170% without
For a heavily reinforced section (10 N32 per metre)
Short term loads: 112%
Including creep: 142%
Including creep and shrinkage: 174%
Total long term curvature/ short term: 124% with compression reo, 193% without
So for the heavily reinforced section the compression reinforcement gave a significant reduction in long term deflections, but for the other two cases it would be more effective to put the additional steel in the tension zone, if it wasn't required in compression for other reasons.
As for ignoring the concrete if you have symmetrical reinforcement, in these cases it would increase the final calculated curvature, including creep and shrinkage, by a factor of 3.5 to 5. If you want a conservative upper bound estimate of long term deflections it's just as easy, and much less over conservative, to calculate the short term stiffness, ignoring tension stiffening, and multiply by 2.
Doug Jenkins
Interactive Design Services
- The Australian bridge code requires all faces to have at least 500 mm2/m reinforcement
- The minimum reinforcement required to exceed the cracking moment often governs
- Continuous beams over closely spaced supports, such as bridge pier footings and headstocks, often have similar positive and negative moments
- Precast elements often have moment reversals from transport and erection to in service conditions
In a lightly reinforced slab, with typical cover for an external structure, top reinforcement often makes very little difference to deflections, because the steel centroid will be close to the neutral axis under working loads. For example, doing some quick calcs on a 350 mm deep slab, with a moment of 120 kNm (not much over the cracking moment), ratios of curvature without compression steel to curvature with equal compression steel are:
For a lightly reinforced section (5 N20 per metre)
Short term loads: 101%
Including creep: 105%
Including creep and shrinkage: 107%
Total long term curvature/ short term: 138% with compression reo, 147% without
For heavier reinforcement (10 N20 per metre)
Short term loads: 104%
Including creep: 114%
Including creep and shrinkage: 121%
Total long term curvature/ short term: 145% with compression reo, 170% without
For a heavily reinforced section (10 N32 per metre)
Short term loads: 112%
Including creep: 142%
Including creep and shrinkage: 174%
Total long term curvature/ short term: 124% with compression reo, 193% without
So for the heavily reinforced section the compression reinforcement gave a significant reduction in long term deflections, but for the other two cases it would be more effective to put the additional steel in the tension zone, if it wasn't required in compression for other reasons.
As for ignoring the concrete if you have symmetrical reinforcement, in these cases it would increase the final calculated curvature, including creep and shrinkage, by a factor of 3.5 to 5. If you want a conservative upper bound estimate of long term deflections it's just as easy, and much less over conservative, to calculate the short term stiffness, ignoring tension stiffening, and multiply by 2.
Doug Jenkins
Interactive Design Services
rapt said:
Yes, that was precisely my intent. Shrinkage warping would still work its voodoo on the cross section of course; creep much less so. I contend, as IDS seems to, that the result would still be conservative.
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.
KootK - I don't think anyone disputes that ignoring the concrete is conservative. The point is that it is excessively conservative, for no real benefit, since calculating the fully cracked stiffness is almost as quick, and gives a much better estimate of the true value.
Doug Jenkins
Interactive Design Services
Doug Jenkins
Interactive Design Services
Kootk,
I am not disagreeing about it being conservative. It would give a meaningless number that gives no idea of the real deflection and would normally be very conservative.
You would get a much more meaningful estimate of total long term deflection by calculating the elastic short term deflection and multiplying by 6 for an RC member.
Doug,
Agree with your logic. I have been pointing this out in training courses for years.
- For lightly reinforced members, you get a better deflection result by adding tension reinforcement.
- For heavily reinforced members, you get a better deflection result by adding compression reinforcement as long as the depth is sufficient relative to the cover to the compression face reinforcement for the reinforcement compression stress/strain to be relatively high. T beam action makes this very difficult in shallower beams.
I am not disagreeing about it being conservative. It would give a meaningless number that gives no idea of the real deflection and would normally be very conservative.
You would get a much more meaningful estimate of total long term deflection by calculating the elastic short term deflection and multiplying by 6 for an RC member.
Doug,
Agree with your logic. I have been pointing this out in training courses for years.
- For lightly reinforced members, you get a better deflection result by adding tension reinforcement.
- For heavily reinforced members, you get a better deflection result by adding compression reinforcement as long as the depth is sufficient relative to the cover to the compression face reinforcement for the reinforcement compression stress/strain to be relatively high. T beam action makes this very difficult in shallower beams.
IDS said:
rapt said:
Rapt specifically disputed my assertions and suggested that I rethink my position. So I rethought it and responded.
IDS said:
rapt said:
1) The method that I proposed is a valid, lower bound "idea" of deflection. Obviously.
2) The method that I proposed is very easy to calculate.
3) The method that I proposed has an easy to understand theoretical underpinning.
4) The method that I proposed is employed by many practitioners in my area.
Upper/lower bound estimates are the meat of what structural engineering is. Just because you guys apparently have better ways of doing this doesn't mean that the methods that others are using are meaningless and without benefit.
To my knowledge, neither of the methods that you guys have proposed is used in my area. They sound great and I'd like to know more. Can you elaborate on the methods and explain their theoretical basis? I really would like to start using your methods if they're superior. Unfortunately, I can't just write "6X short term because some interweb guys say so" on my calculation sheets. You know, internal QC and all...
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.
Kootk said:
No-one calculates concrete and steel stresses assuming a cracked section and linear-elastic properties?
Having calculated the stresses it's simple to calculate strains, curvature and flexural stiffness.
Alternatively find the depth of the NA and calculate the transformed second moment of area using standard procedures.
Doug Jenkins
Interactive Design Services
Kootk,
I think Doug meant to end his first sentence with !!!!!! instead of ?
We simply use strain compatibility calculations to determine the strain and stress conditions under several different loadings (short term and permanent loads as a minimum) and concrete properties (short term and long term properties to allow for creep and shrinkage) and include a method to allow for tension stiffening to give curvatures and then integrate the curvatures to get deflections. You cannot use rectangular stress block, you need a proper concrete stress strain curve. There is nothing earth shattering about the methodology. It is just going back to a 1st principles calculation method rather than using gross simplifications that are defined in most design codes. Just that computers make it possible where it was too difficult to do by hand in our younger days.
But it allows you to account for the actual amounts of and positions of the reinforcement on short and long term effects. We have been doing the calculations in RAPT since the mid 1980's.
There is a simplified version of the methodology in BS8110 Part 2 and I think Eurocode. Also Bransons and Gilberts books cover it well as do probably many others.
You can still do lower/upper bound checks based on variations in concrete properties, especially elastic modulus, creep and shrinkage if your requirements are that sensitive.
The 6 times elastic basically comes from 2 approximations
- Ieff = .5 Igross and
- Long term component of deflection = 2 times short term
so total long term = 3 times short term.
Combining the 2 you get 3 * 2 = 6 times elastic. There was a British Concrete Society Paper/Technical Report on Long Term Deflections that came to the same conclusion.
I think Doug meant to end his first sentence with !!!!!! instead of ?
We simply use strain compatibility calculations to determine the strain and stress conditions under several different loadings (short term and permanent loads as a minimum) and concrete properties (short term and long term properties to allow for creep and shrinkage) and include a method to allow for tension stiffening to give curvatures and then integrate the curvatures to get deflections. You cannot use rectangular stress block, you need a proper concrete stress strain curve. There is nothing earth shattering about the methodology. It is just going back to a 1st principles calculation method rather than using gross simplifications that are defined in most design codes. Just that computers make it possible where it was too difficult to do by hand in our younger days.
But it allows you to account for the actual amounts of and positions of the reinforcement on short and long term effects. We have been doing the calculations in RAPT since the mid 1980's.
There is a simplified version of the methodology in BS8110 Part 2 and I think Eurocode. Also Bransons and Gilberts books cover it well as do probably many others.
You can still do lower/upper bound checks based on variations in concrete properties, especially elastic modulus, creep and shrinkage if your requirements are that sensitive.
The 6 times elastic basically comes from 2 approximations
- Ieff = .5 Igross and
- Long term component of deflection = 2 times short term
so total long term = 3 times short term.
Combining the 2 you get 3 * 2 = 6 times elastic. There was a British Concrete Society Paper/Technical Report on Long Term Deflections that came to the same conclusion.
IDS said:
Of course we do. I was asking for some deatail and justification for your 2 x (ST stiffness w/o tension stiffening) shortcut as a reasonable upper bound estimate. Does it show up in print somewhere? Is there a graphical relationship that can be pointed to? Does it have a physical meaning that one might be find intuitive? Is it an extension of old school code recommendations like the 6X algorithm? Or do I just have to be smart enough to infer it from first principles somehow?
rapt said:
Got it, thank you. Certainly, I'm a fan of the the fancier methods when the target is a accurate determination of deflections.
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.
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