Hello Dr am from East East Africa, Ethiopia. I really appreciate ur tutorial. Since as we now there are different adopted code on Rcc Design for instance in our country we are Using EBCS=CES ( which is similar to Euro code) ,so the design parameters and factors are varied. Bytheway when I was in university, I was taught by Indian teachers such as Dr. Vikas anyhow If u have willing to communicate with me please forward ur spread sheat
Doctor thank you very much for all the videos on reinforced concrete design But in this exam I tried to determine the maximum moment by getting the area from shear force diagram and got it to be over 600kNm
hello doctor, thank you for this. I do not understand why we can't in the case you cover for the collapse load, just find the load that will create a full plastic moment at the section of maximum moment. We know the moment diagram, and we know the value of the plastic modulus. So we can find the load that will causes Mp in the section. Max moment at the point load = 0.25*Wc*0.75*L = 0.1875*Wc*L (simple statics). Plastic moment Mp = Fy * Z, therefore collapse load Wc = Fy*Z/(0.1875*L). Is this correct? thanks. I just realized this gives the same result, as 1/01875 = 5.3333
@@okaroh hey, after searching i got how he did it. Cross section area= Pi/4 X (diameter of bar used)^2 Spacing= (cross section area x 1000)/ area required you’ll get ‘180c/c’ but he ended up using ‘150c/c’ Area provided (Asp)= (cross section area X 1000)/ spacing [which is 150 in this case] the answer for Asp will be 523mm^2, that’s how he got it
Salam Dr Sharif, could we have a video on structural design for strip foundation under a brick (9 inch) load bearing walls , I am your brother from Iraq, many thanks.
I dont understand how do they assume the value of -100 in sway, do we need to calculate or not from anywhere. Can we assume it also 500, 600...or only it should be less than or equal to 100
I have a question if you don't mind? In example 4: where did you get that there are 3 independent mechanisms? Nubmer of possibe hinges n = 3 Degree of redundancy r = 4-3 =1 Number of independent mechanisms = n-r = 3-1 = 2 Number of hinges for total collapse = r+1 = 1+1 = 2