Hi, Thank you for this tutorial. this is so far the direct and fastest way to learn Robot structural analysis. I like the way you discuss. Anyways do you have a tutorial on a single storey or two storey building with roof being run by analysis same as this to view calculation... anyways if you have spare time making the tutorial...
I designed a flat sloping roof and the columns were dimensional (60*30 cm) The program gave the shear circumference equal to 5.56 and 6.82 for internal columns although if I calculated the shear circumference manually according to the BS8110, a different value would come out. How does the program calculate the shear circumference? the slab thick is 25 cm cover 2.5 cm i use 16 diameter for top bars and 12 diameter for bottom bars
Maybe you didn't include the self weight of elements when you calculated manually. RSA tends to include the element self-weight automatically when you add the dead load.
@@ARbyEV Sorry sir, maybe you didn't understand my question because of my English is not good I am talking about the length of the shear circumference (u). The program gives a larger circumference length
After solving manually and by using RSA, RSA does not actually use "1.5d". It uses "1.4286d". Also, it doesn't use "0.75d" for the extra width, but "0.6912d" instead. I don't know why, but I'll keep asking until I get the right answer.
Thank you for your wonderful explanation, I am very grateful for what you offer on this channel. Excuse me sir, I have a problem and I want to ask you I design a beam in the required, its dimensions are 60 * 30 cm and its length is 5 meters It is subjected to a positive moment in the middle of 98KN.m and a negative moment at the end of the beam is 100KN.m. The other end of the beam is zero. . It is subjected to a torsion moment in the end of 14KN.m at the end and in the middle of 9KN.m There reinforcement in required was at negative moment 4 bars dia. 16 The lower reinforcement of the positive moment was 5 bars of diameter 16 for links 10 by 12 cm .. When I designed the same beam in the provided reinforcement the program gave a lower reinforcement of 2 bars diameter 16 and an upper reinforcement of 4 bars diameter 16 And gave reinforcement to resist the moment of Torsion 16 bars of diameter 14, and when I changed the diameter to 12, he gave the same number of bars When designed beam and this beam resist moment of Torsion the program always gives 16 bars What is the reason for that? I am using UK code BBS8110 fy=460n/mm2 for main bars fy=250 n/mm2 for links Fcu = 25 N/mm2
1. Can you share with me the concrete cover you use? 2. Are those moments the result of structural analysis calculations? If not, you can specify whether they are live or dead loads. Thanks.
In RSA you can only place design loads and then the software will calculates the internal forces for you. I don't think there is a way of placing internal forces on the structure model. Can you please send the question and the robot file to my email address vkilemo@gmail.com Thanks.
This is very helpful. Good stuff
Hi, Thank you for this tutorial. this is so far the direct and fastest way to learn Robot structural analysis. I like the way you discuss. Anyways do you have a tutorial on a single storey or two storey building with roof being run by analysis same as this to view calculation... anyways if you have spare time making the tutorial...
Kindly do a storey structure using bs8110
I designed a flat sloping roof and the columns were dimensional (60*30 cm) The program gave the shear circumference equal to 5.56 and 6.82
for internal columns although if I calculated the shear circumference manually according to the BS8110, a different value would come out. How does the program calculate the shear circumference?
the slab thick is 25 cm cover 2.5 cm
i use 16 diameter for top bars and 12 diameter for bottom bars
Maybe you didn't include the self weight of elements when you calculated manually. RSA tends to include the element self-weight automatically when you add the dead load.
@@ARbyEV Sorry sir, maybe you didn't understand my question because of my English is not good
I am talking about the length of the shear circumference (u). The program gives a larger circumference length
Let me check and get back to you later.
After solving manually and by using RSA, RSA does not actually use "1.5d". It uses "1.4286d". Also, it doesn't use "0.75d" for the extra width, but "0.6912d" instead. I don't know why, but I'll keep asking until I get the right answer.
@@ARbyEV Thanks so much
Thank you for your wonderful explanation, I am very grateful for what you offer on this channel.
Excuse me sir, I have a problem and I want to ask you
I design a beam in the required, its dimensions are 60 * 30 cm and its length is 5 meters
It is subjected to a positive moment in the middle of 98KN.m and a negative moment at the end of the beam is 100KN.m. The other end of the beam is zero. . It is subjected to a torsion moment in the end of 14KN.m at the end and in the middle of 9KN.m
There reinforcement in required was at negative moment 4 bars dia. 16
The lower reinforcement of the positive moment was 5 bars of diameter 16
for links
10 by 12 cm
..
When I designed the same beam in the provided reinforcement
the program gave a lower reinforcement of 2 bars diameter 16 and an upper reinforcement of 4 bars diameter 16
And gave reinforcement to resist the moment of Torsion 16 bars of diameter 14, and when I changed the diameter to 12, he gave the same number of bars
When designed beam and this beam resist moment of Torsion the program always gives 16 bars
What is the reason for that?
I am using UK code BBS8110
fy=460n/mm2 for main bars
fy=250 n/mm2 for links
Fcu = 25 N/mm2
I'll get back to you later tomorrow.
1. Can you share with me the concrete cover you use?
2. Are those moments the result of structural analysis calculations? If not, you can specify whether they are live or dead loads.
Thanks.
@@ARbyEV the cover is 25mm
the moment that i put in the above comment is for ultimate load
@@ARbyEV yes this moment from calcuations
In RSA you can only place design loads and then the software will calculates the internal forces for you. I don't think there is a way of placing internal forces on the structure model. Can you please send the question and the robot file to my email address vkilemo@gmail.com
Thanks.