Hi Dr. Sherif, Thanks very much for your ever enjoyable lectures on designs of reinforced concrete members. The design of Reinforced concrete beam part 2- design example has been valuable to me. I would however like make the following observations; 1. Reinforcement, Clause 3.4.4.4 Use 6T25 Bot. (Asp = 2950) The area is 2945 and not 2950 as indicated. 2. Check Shear vu = lesser (0.8 square root 30, 5) = 4.38. not 4.28 as indicated. 3. Design of Links At d face of support ; Vd = 292.8 - 58.56 x (0.15 + 0.628) = 247.2. Not 248.8 kN as indicated. 4. Extent of shear links = (Vface - Vn)/w = (284 - 152.3)/58.56. Vn is 166.42 and not 152.3kN as indicated. 5. Fig. 3.24 not Fig 3.25 as indicated 6. I did not get the concept of Transverse Steel. It did not come out clearly on how you arrived at T10-300 C/C. Thanks a lot Dr. Hope you will soon lecture on Design of Reinforced Concrete Columns. Regards, E.K.Bett
Hi Dr Sherif your videos are superb.. I m glad that you have mentioned each and everything with references to the code... Could you please also make same series with reference to ACI 318M code.. If possible.. Thank you so much for your efforts
alslam alikom engineer Sharif, thanks for the excellent material and brief explanation, this is the first time I understand the design. I have a question regarding the minute 26:49 the (extent of shear links) we have find that Vn = 166.42 KN and when substituting in the next line we wrote the value is 152.3, I am wondering from where did we get the value of 152.3? or it should be 166.42? thanks again
I had a question, when dividing the total weight F to get the distributed load W (load calculations in first step) why you divided it by 10m not 4.5m? Shouldn’t be divided by 4.5 so it can be distributed in the 10m direction??
No. This load is over the total length of the beam which is 10 m. To change it to distributed load, we divide the concentrated load by the beam length which is 10.
Always we have to give a credit where Credit is due, Bro Dr.sharif Thanks alot your videos are very enlightening. Sincerely i am enjoying . Regarding this video Would like to ask Since our full length of the beam(span-B1) in the plan is 10,000mm inclusive of column width(both sides) which we have a square column of 300mm by 300m, why do we remain with 300mm for bòth end instead of 600mm after this ( 10,000-(2250+5200+2250)=300mm) with reference to the last page?
Nice content, Dr. Thank you, Trying to refresh all this content and I found a good way to do that watching your videos. I have a question, maybe it's very stupid but, is there any criteria to assume properly the As for the links?
in initial proportioning 20 mm bars were assumed as main bars.but finally 25 mm bars were considered. isnt it a problem as it may causes to change the effective depth and so on.
The critical section of shear IA at distance d from the face of the support. This means (d+ column width/2) from the centerline of the column. Column width was 0.3 m and d=0.628. Therefore the distance from center of the column = 0.15+0.628
Very good video I am super grateful for your explanation, I am interested in the design of those columns with that same example of the structure, will you have a video explaining their design?
He'll sir. I have always liked your tutorials. My concern is about Bs 8110 and eurocode 2, which one should we adapt as the latest reference code? Thanks
They are two different codes. This depends on the required code of the project. Eurocode 2 is newer than the BS8110, but the BS8110 is still used in many countries.
Great job sir may God continue to increase your knowledge. Please I have a question: when you design for Linkes in (Svmax= 0.75×628= 471mm >Sv=150) OK Why is it Ok? Since 150mm > Svmax = 471mm. Please I need more enlightenment sir thank you sir Best regard sir.
