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Reinforcement yield strength 1
- Thread starter aladdin76
- Start date
Ron:
I would take your words with reserved cautious. Wish I had spent more time and paid interest in metallurgical studies. But I do agree with you without hesitation that the significance of strength deviation in structural steel and reinforcing steel is not directly comparable, because of the stiffening effect from the concrete for the latter case. You are absolutely right on this.
I would take your words with reserved cautious. Wish I had spent more time and paid interest in metallurgical studies. But I do agree with you without hesitation that the significance of strength deviation in structural steel and reinforcing steel is not directly comparable, because of the stiffening effect from the concrete for the latter case. You are absolutely right on this.
kslee1000...I appreciate your willingness to take my comments at "face value". If you check the respective ASTM requirements for rebar (ASTM A615) and steel shapes (ASTM A36) you will find that there are essentially no chemical restraints placed on rebar; whereas, A36 has specific chemistry requirements to control the metallurgy. Further, rebar is allowed to have less ductility, as measured by elongation, than mild steel shapes. Rebar is required to have a minimum elongation of 12 percent, while A36 steel has to have a minimum elongation of 20 percent.
What this means, at least to me, is that the controlled metallurgy of the mild steel shapes provides much greater predictability of its properties. It is weldable without significant limitation as to process, filler metals, and pre-weld/post-weld thermal conditions (except for very thick sections). Conversely, rebar has limited weldability, specifically because of its lack of controlled chemistry.
Sorry to ramble on....I just wanted to be clear about the source of my comments.
What this means, at least to me, is that the controlled metallurgy of the mild steel shapes provides much greater predictability of its properties. It is weldable without significant limitation as to process, filler metals, and pre-weld/post-weld thermal conditions (except for very thick sections). Conversely, rebar has limited weldability, specifically because of its lack of controlled chemistry.
Sorry to ramble on....I just wanted to be clear about the source of my comments.
This weekend seems not a waste. After all, been able to find two articles discussing properties of reinforcing steels. Please pay attention to Table 5, P.16 of the article from Dubai, Table 5 & Table 7, p.27 & p.28 respecticely, of the article from India. Wish we all gain better understanding after review these articles.
kslee1000...thanks for the article. Only the one from India got attached. You have to do each file separately in separate postings.
It is a good overview of the properties of rebar. In particular the summary notes the things we have discussed here. Note also the non-linearity of the stress-strain curve of cyclic loading on rebar.
It is a good overview of the properties of rebar. In particular the summary notes the things we have discussed here. Note also the non-linearity of the stress-strain curve of cyclic loading on rebar.
To calculate the ultimate moment, I think you can just assume strain compatibility with each bar at its own yield strength. You run into a similar situation when you calculate the ultimate strength of a prestressed beam that also has mild steel.
civilperson
Structural
Ron,
The strains will be identical and the % of elongation will be identical for all stresses less than the lower yield stress.
The strains will be identical and the % of elongation will be identical for all stresses less than the lower yield stress.
Agree that the strain difference might be small, but what we could be concerned about, if there were are larger difference in the yield strength (please note that I've already stated a couple of times that I didn't think the yield difference in this case was significant), is that you would develop a strain incompatibility after the first yielding. That could lead, again if the difference were greater, to a non-ductile failure by quickly overloading the other bar.
If that were not the case, we wouldn't be concerned about mixing Grade 40 and Grade 60 bars in the same section.
If that were not the case, we wouldn't be concerned about mixing Grade 40 and Grade 60 bars in the same section.
civilperson
Structural
No beam in use with service loads achieves yield unless a mistaken design was used or overloaded. The steel is suppose to yield prior to concrete failure for a ductile failure. The strains of the steel will be exactly the same as the concrete surrounding them.
civilperson has it right. The strains in the bars will be the same as the surrounding concrete. Say the strain at a particular level in a beam that has grade 40 and grade 60 bars has reached 0.003. The grade 40 bars would have yielded at 0.0014, and the grade 60 bars would have yielded at 0.002. Because both bars are now in the plastic zone, the stress in each bar when the strain in 0.003 would be 40 ksi and 60 ksi respectively.
If the OP knew the locations and yield stress of each bar, the ultimate moment could be calculated. Alternatively, he or she could calculate the service load stress and decide if the factor of safety is acceptable for the lower yield bars.
If the OP knew the locations and yield stress of each bar, the ultimate moment could be calculated. Alternatively, he or she could calculate the service load stress and decide if the factor of safety is acceptable for the lower yield bars.
I don't think the above statement is correct. Let's go back to basics:
E=29000 ksi, Sy=0.003=l/L
fy = E*Sy = E(l/L)
At fy = 40 kai, the elongation l is:
l = fy*L/E = 40 ksi*L/29000 ksi = 0.00138L
At this stage with increase load, the elongation l is:
l = 0.00138L + (60-40)*L/29000 = 0.00138L + 0.00069L
= 0.00207 L = 60*L/29000 = 0.00207L, check.
Note, after yield at 40 ksi, the lower grade bars will continue to lengthen until the higer yield point is reached, but the stress in the lower grade bars would not change. The final elongation of all bars are identical, however, the stresses in bars and the concrete wrapped around the bars are different, they depend on the yield strengths of the bars.
E=29000 ksi, Sy=0.003=l/L
fy = E*Sy = E(l/L)
At fy = 40 kai, the elongation l is:
l = fy*L/E = 40 ksi*L/29000 ksi = 0.00138L
At this stage with increase load, the elongation l is:
l = 0.00138L + (60-40)*L/29000 = 0.00138L + 0.00069L
= 0.00207 L = 60*L/29000 = 0.00207L, check.
Note, after yield at 40 ksi, the lower grade bars will continue to lengthen until the higer yield point is reached, but the stress in the lower grade bars would not change. The final elongation of all bars are identical, however, the stresses in bars and the concrete wrapped around the bars are different, they depend on the yield strengths of the bars.
graybeach:
Yes, you are right. I also noted it in my response - no stress changes for grade 40 bars after reaching yield. However, it elongates with the grade 60 bars (at the same rate) after that point (40 kai) until the stress 60 ksi is reached.
Yes, you are right. I also noted it in my response - no stress changes for grade 40 bars after reaching yield. However, it elongates with the grade 60 bars (at the same rate) after that point (40 kai) until the stress 60 ksi is reached.
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