USE OF GRADE 550 AND GRADE 690 REINFORCING BARS IN ACI 318 19

[00:00:00]Uh good evening engineers. Uh today I am going to present a general overview um on the requirements of uh ACI uh 31819 for using highgrades reinforcement.

[00:00:19]uh when we say high grades reinforcement reinforcement in fact I mean grade 50 uh 550 megapascal and 690 megapascal um uh I will discuss this topic from uh the point of view of ACI 31819 uh rather than uh 31825 but uh I believe if you understand the intent of ACI 318 1819 you will understand easily uh the ACI 31825 requirements I believe uh no significant changes between both of them uh okay let's start I I will I will try to uh be fast as kind as possible uh the main objectives we will discuss the types of non prestress steel reinforcing bars development and supplies requirements ments. Uh we will discuss the additional requirements uh in ACI code for uh using high-grade reinforcement for different structural members sim like beams, columns to slabs uh and also some special members. Uh also we will discuss the limits uh in boosted on concrete compressive strength. Also we will discuss uh strength reduction factors uh and uh benefits of high strength steel reinforcing bars quickly. What about ACI 3 35020 for the environmental uh structure? Uh does this code permit to use high-grade enforcement or no? And also I will give some notes about EAP's capability about uh using high-grade reinforcement.

[00:02:12]If you would like to uh read more about this topic, I strong I I strongly recommend uh those technical notes from CSI uh website. Uh and also you can read uh from the main code ACI 318, 19 or 25.

[00:02:30]also uh David Fanila one of the best persons he uh I I I I believe he discussed the same topic also with ICC uh so uh let's start first of all you have to know that in ACI code there are different types of uh nonreress uh deformed bars uh the first type is called A61 15. This is carbon steel. In in fact, this is the most common reinforcement type used in Dubai and in UAE in general. Uh if you if you uh see table 221 uh.3A, this table presents the minimum requirements for ACI uh 40 uh sorry for for A615 reinforcement. So for example, if you would like to use a grade 420 and the type is A615 as per the uh uh code requirement, you have to satisfy minimum tensile strength 550 uh megapascal for this grade and you have also a limit on the uh ratio uh of actual tensile strength to actual yield strength. These are minimum ratios. So you have to satisfy uh those requirements and also uh uh these requirements can be found also in ASM A615.

[00:04:11]Uh also this table is from ASM A615. The same requirements in ACI code but with more additional details. Uh also we have another type uh of reinforcement. This type is called A6 A706.

[00:04:30]A706 mainly is used for special uh systems. This type of reinforcement is very uh good type for uh seismic resistant structures in special or in in in high seismic regions or seismic design category and higher uh seismic design category D and higher. Let's say if you look uh uh if if you see table 20.2.1.3b 2.1.3b here it was a table A here the table is B you will find that for A706 reinforcement we have a additional requirements when compared to A615 because again this type of reinforcement is used for special systems. If you look here, here we have minimum tensile strength and also here we have a ratio of actual tensile strength to actual yield strength.

[00:05:33]Uh these uh first two two requirements are similar to uh what are required for A615. But here we have additional requirements. seem like here you have uh a minimum yield strength uh maximum yield strength and also here we have uh fractal elongation requirement.

[00:06:02]Uh if you look here here the code uh specifies also uh a maximum yield strength. If you look for A615 requirement, you will find that there is no upper limit on the yield strength.

[00:06:19]So this is very important. Uh one of the main reason for that that in if you have a special system you cannot use uh uh you cannot use a reinforcement with um with actual strength or yield strength higher than the value uh specified in the design calculations.

[00:06:42]Uh in in special members we are calculating the shear or induced shear from the probable moment. Probable moment is calculated from the expected strength of the material. So if you increase the strength of the material, you will increase the induced here on the joints or within the joints. That's why uh uh this is very important. If if the code uh need you to use A706, you have to know that we have upper limit on the yield strength. you cannot use any uh high yield strength than the maximum value. So for example, if you would like to use a grade six uh 690, your actual yield strength shall not exceed 81 814 megapascal.

[00:07:34]But this this is not required for A615 reinforcement. Here also we have uniform elongation requirements um uh 9% 7% 6% this is based on the uh bar diameter uh and also for large size bars here we have 6% as a minimum uh elongation again uh these values are minimum Uh also this is uh this table two uh tinsai requirements for A706 from ASM also.

[00:08:17]Okay. Uh uh I believe that ACI 31825 deleted or removed uh these tables uh because these tables already are available in AM now. So uh instead of repeating the same tables in ACI so now ACI refer only to the AM number so you can get the requirements from the ASM instead of repeating the details in the code.

[00:08:48]Also if you uh uh if if the code uh require uh if the code requires to use A706 there is also additional requirement uh for the bars used in the quick resistance structures. uh uh this requirement is related for the radius of deformationations uh for the bars to avoid uh fatigue or inelastic fatigue cracks. So you have to satisfy also that uh also we have other types of non-stress steel bars uh same like ASM A996 ASM A955 this is stainless steel also A1 uh 35 uh this is low carbon reinforcement also uh Okay.

[00:09:48]Uh also here I would like to highlight that uh ASM A615 bar sizes larger than 57 mm are not uh permitted.

[00:10:05]Uh also uh in ACI code we we have different types for blame bars. A615, A706, A955, A 035.

[00:10:17]Blade bars are permitted only for spiral reinforcement used as a transverse reinforcement for columns uh shear and torsion or confining reinforcement for sublices. So you have to know that if you would like to use a blend bars for these uh purposes you have to understand that we have a specific types also other types seem like welded deformed bars heated also deformed bars. So if you would like to use heated deformed bars, you have to know that the uh A970 or or ASM A970 covers the uh requirements for uh for this type of uh bars.

[00:11:04]Okay. And here SEI table 20.2.2.4A.

[00:11:09]This is the most important table uh in SEI in chapter 20.

[00:11:16]Here in this chapter you will find the maximum value uh uh of yield strength uh permitted for design calculations.

[00:11:28]This is very important. The key word here is the calculations.

[00:11:34]So we can understand that we can use higher strength but we cannot assume uh or or we have to or or the code need you to specify upper limit uh of your assumed strength in the calculations.

[00:11:52]For example, I will give you for examp uh one of the uh uh common example uh typically in the uh in the uh market especially in Dubai we are using a grade 500 megapascal reinforcement. So we are using grade 500 megapascal uh when we design our structures. But when we design the sheer reinforcement, we have to assume 4 uh 420 megapascal in the calculation. Also, if for example, if you have a raft and inside this raft, we have uh a reinforcement with 500 megapascal grade and you would like or or or due to side constructions, we have to provide a construction joints. within this construction joints we have to check the sheer friction requirements assuming a field of uh 420 megapascal.

[00:12:49]So there is a different there's a difference between uh specified yield strength in the calculations and the actual yield strength. Uh this is very important you have to understand that.

[00:13:02]Okay. If you look here, if you would like, if you would like to calculate the flexual reinforcement or the reinforcement required for axial strength or the shrinkage and temperature reinforcement, first of all, you have to determine your structure. Do you have special system or or other systems? Uh for example, special systems seem like special moment resisting frames or special shear walls. Other systems seem like ordinary shear walls.

[00:13:33]Okay, if you have a special system and you have special moment resisting frame, the code does not permit you to assume in the calculations uh a value of field strength higher than 550 megapascal. You have above limit.

[00:13:53]Okay. If you would like to use a special shear wall or special structural walls, the code permits you to use 690 megapascal in your calculations. This is very important. Okay. If you if you have for example uh ordinary structural shear walls, you can use uh uh or you can assume a design value of yield strength up to 690 megapascal.

[00:14:23]This is the first point. Second point you have to look here. If you remember just right now we have discussed the type of deformed bars. Why? For this point if you look here uh as per the code uh in in special seismic systems we have to use only A706 bars.

[00:14:46]But for other systems same like ordinary uh shear walls as per the code it's permitted to use A615 A706 A955 A996 A1035 as you can see here because here we have ordinary systems when we say ordinary systems so we can we can expect uh uh low seismic hazard or midsismic hazard.

[00:15:14]So the code permits to use different types of uh deformed bars. Uh but if you have a special seismic system, so you are in in a high seismic region with high seismic design category as per the code, it's not permitted to use A615.

[00:15:33]So you have to understand this point.

[00:15:36]Okay. If you would like to calculate the lateral support uh bars also as per the code uh you can use uh 690 megapascal in special systems but for other systems sim like ordinary systems you have you can assume above value of 550 megapascal.

[00:15:59]So here we have a bar limits on the uh uh calculations also a similar example I will I I I believe that it would be useful for you engineers uh for oneway shear strength as per the code the maximum FC prime using the calculations for oneway shear or two-way shear uh is around 70 megapascal as FC prime or cylinder strength. Uh what does this mean? This mean that you can use FC prime with 80 or 90 megapascal. But for calculations of sheer strength, you can use only FC prime equal to 70 megapascal as up limit on the design values. Uh in the calculations type here also if you look here this table is from 31814.

[00:16:57]This this table is from 31819.

[00:17:01]In this table uh for flexure or or axial strength or shrinkage and temperature uh the maximum permitted uh value is 550 while in ACI code we have 600 in ACI 19 I mean we have 690.

[00:17:19]This is the the first difference. Uh number two in the special systems you can use only 420 megabascal but in ACI 819 one of the major changes that the code uh now permits to use high grade reinforcement in special systems.

[00:17:40]But in in in 14 version the maximum uh uh value or permitted value for calculations uh was 420 megapascal only.

[00:17:55]Also here for here calculations if you have a special systems uh the code permits to use 550 megapascal or 690 megapascal. This is based on if your system is special moment frames or special structural walls. But uh if you would like to uh calculate the stups, ties hoops uh for other other systems same like ordinary structural shear walls you have to use a value 420 megapascal. This value uh we are using frequently in our uh daily design. But for example 550 is not permitted. Not permitted. You have here uh other types of uh or other type of deformed bars. Uh welded deformed bars uh sorry welded deformed wires. Uh I have not used uh uh I have not used this type of deformed bars before. In fact uh here also uh I need to uh highlight that you have always to read the footnote. Footnote is very important. So for example here uh in ordinary systems it's permitted to use a grade reinforcement up to 690 megapascal but you have to read three and four foot node. In fact uh these footnotes are related to intermediate moment resisting frame. So you have a restrictions or uh additional restrictions on using high-grade reinforcement for intermediate moment resisting frames.

[00:19:35]Okay.

[00:19:37]For anchor reinforcement in fact in special system in special seismic systems you can use h a strength or yield strength up to 550 megapascal but as I said before the type of reinforcement shall be a706. Now we can understand again A706 uh I will say has more restrictions in ASM specifications uh more than A615.

[00:20:09]The most uh common type of reinforcement in Dubai is A615.

[00:20:17]Uh again uh in in NCI 318 uh 14 um uh the the the using uh of highgrade uh reinforcement were uh uh prohibited in in in many uh cases but uh ACI 3A 1019 now uh uh permits to use the high-grade reinforcement uh in uh different uh cases in fact.

[00:20:52]Okay. Again for uh grade 550 megapascal used for anchor reinforcement in seismic design category C must conform to ASM E706.

[00:21:08]Uh in fact uh rail of engineers aware for this requirement. This is this is uh required for seismic design category C not only D E or F. So as you can see here the type shall be A706.

[00:21:28]Uh here also uh ACI 20.2.2.5 2.5.

[00:21:35]If you would like to uh calculate the required reinforcement uh uh for uh moment action axial force or combined effect for especially shear walls or structural walls I will say structural walls because shear walls or structural walls can be a sheer walls or a flectoral walls but for this purpose you can use or you have to use only A706 uh And as per ACI code, uh it's permitted to use a grade 420, 550 or 690 megapascal.

[00:22:14]For special moment resisting frames, you cannot use 690 megapascal. The maximum value is 550 megapascal.

[00:22:24]Uh again, A706, A706 here. In fact, ASMA615 grade 550 and 690 are not permitted in special seismic systems and anchor reinforcement in seismic design category CD, E and F. This is very important note but uh the code in fact uh give you an exception if you would like to use uh A615 grid uh 420. So in case if you have a special systems and you would like to use a grid uh uh grade 420 megapascal, it's permitted now by the code as an exception to use A615 type of reinforcement in LEO if uh of A706 grade 550 reinforcement. Uh but you have to satisfy additional requirements. You have to look into the code to see what are the additional requirements uh uh needed to be satisfied to uh uh to use A615 grade 420 megapascal for a special systems.

[00:23:47]Now in fact I would like to discuss the uh development and supplies length requirements.

[00:23:57]uh in SEI 31819 there is a new factor uh added to uh the development length equations uh called uh grade factor. This table 25.4.2.3 uh presents a a simplified equations.

[00:24:20]These simplified equations will uh erh will uh result always in a very conservative values of development length.

[00:24:32]Anyway, if you look here, here we have thigh subg. This is a great factor to take the effect of using high-grade reinforcement.

[00:24:42]But before going into the next step, I would like to uh add a comment here that I don't recommend to use uh these simplified equations. I recommend always to use the detailed equation. This is the detailed equation. I will give you an example from Jim's uh white book. If you go there when he calculate uh the required development length uh from the simplified equation he got a development length almost double the development length required by the detailed equation. Detailed equation always is recommended to be used. In fact in in in in my high-rise structures uh this made a very uh big difference.

[00:25:31]uh um if you would like to use this simplified equation this will be very fast for calculations but this will cost a lot later on that's why I recommend not to use uh uh these u uh simplified equations uh and I would recommend to use the detailed equation in ACI code anyway if you look here here we have reinforcement degrade this is by the way this factor is a new factor in ACI SV 1819 for a grade 420 the value of the factor is 1 for a grade 550 the factor is one uh uh 1.15 as you can see okay as per the code you can use uh high grade reinforcement so 550 in fact if we divide the 550 by 420 you have 30% additional yield strength but again you I have 15% additional require uh requirement uh for the development length. Okay. If you would like to use 690 uh you have to increase the development length by 30%.

[00:26:45]This is very important. Okay. Uh now some engineers will ask what if I have 600 because now in in Dubai uh uh I uh started to see a new type of reinforcement called E600. 600 is something in between between the 550 and 690. In fact the relation here is not proportional. If you look here for example 280 and 420 is one both of them.

[00:27:19]uh so um if if we need to say that it's a proportional so this should be one this should be for example 6 something same like that uh but if if my my personal opinion uh if I need to take a grade in between this is my personal uh opinion in fact this is not the main interpretation of the code uh uh uh I can make interpolation um but I send an inquiry for ACI stuff before and they said to me that in fact the relation is not a proportional and some engineers will take the upper li uh the upper uh factor and some engineers maybe can take uh the interpolation other engineers can take the lower value so in fact it's it's something not a proportional but for me personal opinion I prefer to uh take the interpolation okay so this is the detailed equation of ACI code. Uh here we have um uh a confinement factor. This will take the effect of additional cover and also additional stepups or or confinement bars or transverse reinforcement.

[00:28:33]Uh this equation will give you a very perfect uh values in the vertical members. In fact, uh, in one of my projects, uh, for example, we have T40 T40, uh, um, uh, uh, sorry, T T32, a lot a lot of T32 bars. T32 bars, uh, typically, uh, require more development length. So when you, uh, uh, when you try to use the detailed equation, you will get a reasonable values.

[00:29:06]Okay, I need to check if it's recording. Okay, also uh if you look in the uh or into provision 25.4.2.2, 2 you will find that uh the code need you to provide maybe additional transverse reinforcement along the development length if you would like to use a strength uh or yield a strength higher or equal than 550 mega pascal.

[00:29:44]So as per the code if you if you would like to use high-grade reinforcement he need also additional confinement this additional confinement can be calculated simply uh from this equation by the way this is KTR uh if you look for this uh uh what the the code need you KT K subtr shall not be smaller than.5 diameter of the main bar so this is known this is equal to the half value of the uh diameter of the longitudinal bars. Uh 40 is a fixed value. ATR is the number or the area of the number of uh uh or sorry the the area of the transverse reinforcement enclosed the bars uh being developed.

[00:30:37]Uh so also the number of legs are known and the area of these legs also are known. Uh n is the number of legs. Now we can calculate the spacing. In fact the spacing uh uh we are talking about the spacing along the body of the beam or along the height of the column or along the height of the wall. Uh so this is very important.

[00:31:05]Maximum as of transverse reinforcement shall be satisfied along the LD distance and is this requirement is not to be satisfied along the uh beam length or along the or full height of the column.

[00:31:19]No only along the development length. So the code okay uh the the code says in fact yes you can use high-grade reinforcement and you can you have to increase the required development length by 15% or 30% but again uh you have to uh make sure that the transverse reinforcement along the development length uh is not less than a certain value provide sufficient confinement to prevent splitting failure uh of these bars because if you would like to use high grade reinforcement. In fact, the area of the bars required bars will be less. The strength is more, the stress is more. So if the if the bars are subjected to higher stress, so we have uh uh a risk of splitting failure.

[00:32:10]That's why the code need you to provide a transverse reinforcement. By the way, this requirement is mandatory also. uh if you would like to calculate the standard hook uh development length intension by the way don't use the equation from ACI 19 now you you we have a nice equation in in in ACI 31825 will give you a lesser values significantly than this uh uh this this equation I believe that this equation is uh not correct and uh that's why the ACI um uh uh uh 31825 uh uh returned back the uh the uh the same equation of ACI uh 318 uh from version 14. So uh don't don't use that.

[00:33:08]But anyway, if you'd like to use high-grade reinforcement here, you have to increase the yield strength and this will increase the LDH.

[00:33:18]Again, in ACI code, you cannot consider the full length as a uh as anchorage length. No, you have to satisfy the horizontal length uh and L extension uh uh uh this depth and also the radius uh diameter. So three uh requirements shall be uh satisfied independently.

[00:33:45]Uh again this equation is for LDH in ACI 31819.

[00:33:50]This is equation for LDH in ACI 31825.

[00:33:54]This equation will result uh in a very significant lower value than the equation of ACI 31819. So uh take care.

[00:34:04]As you can see here, this is the commentary of ACI 25. Revisions to 25.4.3.1 introduced in ACI 31819 were found to result in uh appreciably increased value of LDH for a common cases.

[00:34:21]Uh this is the same what I have said just right now. Also, if you'd like to develop your bars using uh a um a heated deformed bars, you have to use a different equation LDT. LD or L subdt uh also if you increase the yield strength, this LD subdt will be increased. also also uh lab supplies lengths you have to to understand that this requirement along the lab supplies what I showed you before here along the development length. So if you look for the both provisions you will feel both of them are identical uh but this is for the lab supplies. So along the lab supplies not the development link. Development link is something different. Uh lab supplies in fact is a function of development length by the way. The code need you to provide a transverse reinforcement of a certain spacing to satisfy a value of specific k subtr as we mentioned just uh before a couple of slides. So uh sometimes you have to increase the transverse reinforcement to satisfy this requirement.

[00:35:41]Also if you would like to calculate the compression lab supplies length as BCI 25.5 you have to know that the code need uh longer compression lab supplies for high grade reinforcement.

[00:35:57]So uh if you look here here we have the difference between the compression lab supplies for a grade 550 and for grade 420. The required compression lab supplies for grade 550 is higher than uh the values required for 420 by uh a 30% more uh sorry by 60% more by 60% more. You have to know that uh uh out of this 160% we or sorry out of the 60% increasing uh the 30% is directly because of the uh ratio of the yield strength. Uh so you have to know that also as you increase the strength of the bars there are many penalties you have uh to satisfy. So as you can see here the code permits you to use high grade reinforcement but you have to satisfy additional penalties on that.

[00:37:03]Now now we will discuss different structural members. As I said before uh I will start with two slabs. If you look here if you would like to find the minimum thickness of two slabs you have first to determine the use of uh yield strings. If you look here for a grade 500 megabascal uh the minimum thickness required by the code is higher than uh uh is higher than the 420 uh mega pascal by around 10%.

[00:37:37]So uh okay you can use higher grade reinforcement but you have to provide more minimum thickness of the uh to a slab. Uh but if you would like to use 690, okay, the code here covers 280, 420, 500, uh 50. Where is the 690?

[00:38:00]Uh I believe the main or or the my interpretation for that that if you would like to use 690, you cannot uh find the minimum thickness of the two-way slabs. uh you have to specify a thickness and calculate the actual deflection because some engineers they are using this table to avoid the uh calculation of long-term deflection. In fact, practically this is not the case.

[00:38:28]We are not using this table at all. But some engineers in some countries they are using this table. So you have to know that for 690 megapascal you cannot use this table and you have to calculate the uh deflection.

[00:38:42]If you also would like to use uh a strength of the steel more than 550 mega pascal the modulus of rupture used to compute the deflection uh uh shall be based on the reduced modulus value. This value is around uh 2/3 of the 100% value. The main value is 62 uh square root of FC prime. But here the code all uh uh permits you to use high grade reinforcement. But you have to calculate the deflection based on reduced value of uh modulus of rupture.

[00:39:24]By the way, practically in all of my projects, I uh I have used this value even though with a grid reinforcement lesser than this value uh in SEI 25. I believe this uh I think this node for the uh modus fracture has been removed in ACI 25.

[00:39:50]uh and again as I said deflection must be calculated for grids higher than 550 megapascal what I mean here here if you have a 690 you cannot use this table and you must calculate the deflection okay also in two slabs we have the in two slabs chapter chapter 8 we have this table this table I think is known for the engineers uh here when you have a connection and you would like to know how much the percentage of the uh uh uh of the load will be transferred as a flexure and how much the percentage of the load will be transferred as a shear.

[00:40:33]So to determine this this percentage is is is important for uh to a shear capacity. By the way if you would like to determine this percentage as you can see here you have to find the strain or tensile strain. The inside strain now is function of yield strain. So uh if you are using u highrade reinforcement you have to uh uh uh you will uh you have to to calculate first the yield strain and then you have to calculate this uh limit. So uh here the code uh connects this the the limit with the grade of reinforcement that you would like to use also. Uh now we I would like to start the beams section. Uh also for the minimum thickness uh or depth of nonrestress beams. As you can see here the code provides a minimum values for the thickness. But here we have an important footnote that uh uh these values are applicable for normal concrete concrete and yield strength uh of 420 megapascal.

[00:41:52]But if you would like to use h a highrade reinforcement please you have to multiply these values by this factor. So as you increase the used u yield strength these values will be higher. So for a grade 550 megapascal the minimum thickness uh may increase by around 18%.

[00:42:18]So here I'm trying to help you to connect how you can understand the additional requirements on using the high-grade reinforcement.

[00:42:34]Okay, here also for minimum flexal reinforcement the maximum value that can be used to compute the minimum flexal reinforcement of the beams is 550 megapascal. So let's assume assume that you would like to use 690 megapascal.

[00:42:51]uh you have when you would like to calculate the minimum flexal reinforcement you have to use uh uh a maximum value of 550 megapascal.

[00:43:04]So uh so if your beams let's say are governed by the minimum value you cannot take the full benefit from 690 megabascal the maximum value can be used at that time is 550 megapascal.

[00:43:25]Okay. Also again uh we have here uh um a limitation on the transverse reinforcement spacing along the development length. Uh so the transverse reinforcement in beams must be provided such that k subtr uh is more than a certain value for a grade 550 and degrade 690 regardless of the bar spacing. This is very important.

[00:43:53]As you can see the same the same provision was in the development length and also in lab supplies and also in beams section here for example this slide uh was presented by ACI uh university here they need they need they need to inform you that okay we have used a high-grade reinforcement uh with a 50% of the reinforcement and we satisfied uh 88% of the nominal moment. Again, let's assume that we have a section or beam section with a great 420 megapascal bars and we have this amount of reinforcement. If we calculate the strength, we will get this value. If we use half of this value of reinforcement, but with high grade reinforcement, we can get this nominal strength. So approximately 50% of reinforcement achieved 88% of the nominal moment. But again we cannot look to the to this example from one uh single angle. We have to look for other penalties in the code. Here you saved reinforcement but what about the additional requirements of transverse reinforcement about the development length? So you have to look to the full pictures. Also you have to be care that if you change the grade of reinforcement tensile strain also will be changed.

[00:45:24]Typically our uh calculations uh are based on the tensile strain equal to 0.005. In fact this assumption is correct for a grade 420 megapascal. But if you increase the yield strength of the bars uh this value will get higher and this will change the design of your section.

[00:45:47]Uh so this is also the same requirement but in the lab supplies of the beams.

[00:45:56]uh they need the code need you to provide a transverse reinforcement with a certain spacing along the lap supplies within the beam if you would like to use a high-grade reinforcement.

[00:46:12]Uh okay. Now we I would like to start the column section. Here I would like to highlight that in in in columns and walls uh it's permitted to use highrade reinforcement up to 690 megapascal. Uh again this depends on your system structural system. But if you look here the maximum axial compressive strength shall be calculated based on the maximum value of 550 megapascal.

[00:46:42]So even even though you would like to use 690 megapascal you have a maximum limit on the calculations of maximum compressive strength. So this also uh you have to keep in mind. So if you look for this example for 80 ksi here we have the same section the same amount of reinforcement but uh uh in first case the the grade of the bars uh is 80 ksi in the second case is 100 ksi. Both of them the maximum compression is the same. Why? Because as per the code uh the maximum uh axial compressive strength shall be calculated on based on 80 megabascal uh sorry 80 ksi or 550 mega pascal. So yes here we increase the strength but the maximum compression uh still equal to the same case of 80ks because of this limitation.

[00:47:46]If you look here for example uh uh we have uh a curve for uh a red curve is for grade 100 and a blue curve is for grade 80. As you can see here at this point uh both of the grades are uh intersecting the same point because of this limit.

[00:48:12]Okay. Redistribution of moments for flexal members in in in continuous flexural members. SI code permits you to make uh uh a redistribution of moments uh due to uh u some rotational capacity in the connections.

[00:48:31]Uh as you can see here as the grade of reinforcement increase the permissible redistribution decrease.

[00:48:40]So at this limit for example uh if you would like to use 550 megabascar reinforcement the maximum redistribution percentage is 10%. But if you'd like to use 420 uh we can go up to uh something in between between 10 and 15. So as you increase the uh uh uh grade of reinforcement the uh allowance of redistribution uh decrease.

[00:49:11]Okay.

[00:49:13]Now for intermediate moment resting frames. In intermediate moment resisting frames we can use only grade 60 or 420 uh or grade 80 550 mega pascal. Now if you look here the spacing of the transverse reinforcement depends on the grade of of bars. So if you would like to use higher grade of bars the code need you to provide smaller spacing to prevent the puckling. So this is also another type uh of penalties from the code also this beams uh this beam this beam section also for enter uh let's say special moment frame um here the minimum size of the column again this is for special moment frames uh uh the minimum size of the column depends also on the grade of the bars.

[00:50:12]If you would like to use a higher grade, let's say grade 80, you have to provide more width for the column. And also the reinforcement here required along the body of the beam also depends on the grade of bars. If uh if you provide uh higher a grade of bars, you have to provide a different uh amount of reinforcement.

[00:50:42]Uh here also spacing at the uh uh plastic hinge region uh spacing also depends on the grade of the bars. If you would like to use higher grade reinforcement the spacing uh will be uh lesser.

[00:51:02]Also if you have a beams with a factory axial force more than this limit uh as per the code the maximum hoop spacing at least or is the least of the following 6 in or uh six diameter the main bar of the smallest grade 600 enclosed longitudinal beam bar or five multiply by the diameter of the bar if you would like to use a grade 100 sorry a grade 80 Again here if you would like to use higher grade you have to use a smaller spacing.

[00:51:37]Also for the columns the dimension limits for the beam column joints also depends on the grade of the reinforcement.

[00:51:46]If you would like to use higher grade reinforcement you have to provide more width.

[00:51:52]Here also the same issue spacing of the transverse reinforcement depends on the grade of the rebars.

[00:52:01]Also for joints uh in special moment frames concrete used in joints with a grade 550 longitudinal reinforcement shall be normal weight concrete. You cannot use lightweight concrete. This also another penalties on using high-grade reinforcement.

[00:52:20]Uh also for special structural shear walls you have to know that the uh maximum vertical spacing of transverse reinforcement also depends on the grade of the primary flexal bars. Uh if you look here if you would like to use 690 megapascal you have to provide uh a transverse reinforcement with lesser spacing.

[00:52:47]uh for 420 the spacing for example is six diameter here four five diameter here four diameter okay also for the diaphragms in seismic design by the way you have to know that in SI code we have a chapter to design the diaphragm chapter 12 I think that's for seismic design category A B or C if you would like to know the additional requirements in case of special systems or the frams and building assigned to seismic design category DE or F you have to check uh chapter 18 section 12 by the way here uh if in the diaphragms or dias in buildings assigned to seismic design category DE or F it's not permitted to use the mechanical supplies for grade 80 and grade 100 ksi I think this uh provision uh uh now changed in ACI I 25 or has been changed in ACI 25 and now ACI 25 permits to use the mechanical sublices at this area but uh with a certain type uh or certain class sub class S by the way now mechanical sublices uh types uh are totally different in SI25 Uh okay. Table 20.2.2.4a permits the maximum design yield strength to be 550 megapascal for portions of a collector for example at the near critical sections. Also here another penalty on you if you would like to use a strength higher than uh 550. So the upper limit here is 550. Again this is for special systems. Also here we have a limits on concrete compressive strength. As you can see for uh a special moment frames uh with a grade reinforcement 420 or 550 the minimum of C prime will be 21 megabascal.

[00:55:00]But if you would like to use 619 as special walls the minimum of C prime is 35 megabascal.

[00:55:08]So here also other requirements on the compressive strength related to the grade of reinforcement strength reduction factor. I'm not sure if it's recording or no.

[00:55:20]Uh okay strength reduction factor. As you know in ASI code if you would like to compute the uh uh design strength of any section you have to know first that this section is attention controlled compression controlled or in the uh transition region. If tension control section the strength reduction factor is 0.9 but if compression control the strength reduction factor can be 65. Anyway, in SEI 2014, if you look here, uh this the tensile strain uh was compared to the the 0.5 I will zoom here.

[00:56:10]0.5 uh and the yield uh tensile strain. In fact, this is in SEI 2014, but in two in 2019, the formula changed uh little bit.

[00:56:24]Uh now the code instead of 0.005 the code need you to calculate the actual strain based on the yield strength. So if you would like to use a high grade reinforcement, this value will change. If you look here, this maybe can be um more clear. If you look here, this is this picture uh uh presents the variation of a strength reduction factor with uh an a tensile strain. This is ACI 2014. Here we have a fixed value. In fact, this fixed value uh was based on yield strength equal to 420 megapascal.

[00:57:06]Now in ACI 2019 as we can use high-grade reinforcement this value shall be calculated based on the actual or based on the on the a grade of reinforcement that you would like to use. So this value will not be the same if you would like to use a grade uh uh 550 uh or 690.

[00:57:29]So 400 if you would like to use four grade 420 the this value will be 0.5 but for grade 550 this value will be 0577 uh 5 for 690 this value will be also higher this again will affect the uh sectional design uh of your member okay here also I need to highlight that uh this curve black curve is for uh is the interaction diagram for this section as per the rules of ACI 14. But here we have a little bit change in the shape of the interaction diagram. If we would like to use 31819 uh so if we have a point here or load combination here this point is safe in ACI 2014 but not safe in ACI 2019. So this uh area uh uh or or or the the the the main reason for for for the uh not matching between the two curves because of the additional requirements on high-grade reinforcement.

[00:58:47]Uh in fact again um the benefits of high-grade reinforcement uh can be sometime to reduce the conjunction uh con uh let's say congestion between uh for the reinforcement uh improve the uh concrete placement uh in fact lower the placement cost as you reduce the number of the bars uh smaller member size more usable space. By the way, the benefits again not guaranteed totally. Uh sometimes if you provide more reinforcement uh with more strength, this will not improve the strength of your section.

[00:59:29]This depend depends in fact uh uh uh on the location or sorry this depends mainly uh on the shape of the interaction diagram. Where where is your govern combination? it's here or here or here or here. So in some locations the reinforcement contribution will be very low. Uh here also uh the main idea that if you would like to take the maximum benefit from using highrade reinforcement try to use the maximum high uh the maximum grade reinforcement with maximum concrete strength. As you can see here as we increase the strength of the concrete and also uh we increase the yield the strength of the bars uh we will get the least size of the section.

[01:00:17]By the way again this is based on axial force only. If we have a moment the equation will change.

[01:00:25]Uh we have discussed a lot of requirements in different seismic design categories. So before using the high-grade reinforcement you have to check the additional requirements in seismic design category A B or C or special requirements for special systems um uh for ACI 35020 this is for the environmental structure in fact this this code mainly is integrated with ACI 31811 is not integrated yet with ACI 31819 that's why if you open this code you will find that the main equations uh formulas and strains are based on grid 420.

[01:01:09]The maximum yield strength permitted for the environmental structures uh is 550 megapascal as per the code uh as you can see here 550 megapascal. uh but the shrinkage and temperature area reinforcement used in the environmental structure uh as BCI 350 must be based on a grade 420 megapascal. This is very important.

[01:01:35]Uh uh here I need you uh to read the manual of EABS uh carefully. EABS can take some additional some requirements for using high-grade reinforcement and a lot of requirements you have to satisfy by yourself. For example, the limits on the FC prime you have to satisfy by yourself. Uh EABS maybe can take the uh some requirements I will show you. For example, the upper limit on the maximum compressive strength is taken by EAPS.

[01:02:10]Uh the strain also is calculated by EAPS based on the grade of reinforcement.

[01:02:14]This is a good point. Uh here also the area of reinforcement for special members also is based on the grade of reinforcement as we have discussed before. Also as you can see here for special moment resisting frames uh the the area of reinforcement is based also of uh the grade of reinforcement. Here I got an email from CSI uh staff. So CSI uh technical support uh that uh I was in fact assuming that the EABS always consider the tensile strain for tension control section is 0005 but they inform me know that the tensile strain is calculated using this equation. This is very nice.

[01:02:58]uh I believe that I have finished this in fact uh uh finally I would like for some people they don't know that I have published a new course the name of this course is the design journey 9 hours uh we I have discussed in fact uh the the real design project from the first step to the last step uh I uh in fact I spend a lot of time to try to make the content easy for people. Um, in fact, I didn't waste the time uh on the or in in the detailed calculations that can be found anywhere in uh rather than concentration uh or concentrating on the uh main concepts that you need to complete your uh real design from A to Zed. Uh thank you engineers. I hope this uh uh topic was useful for you uh and have a nice day.