Three Hinged Circular Arch - Problem No 1

[@its_me0002]

For finding y we can also use Y=(√R^2 - X^2)-R+H

[@shaikshahida5464]

Thanks sir for making this video ,it helps me a lot

[@abhishekmugalakhod8762]

Please make more video on this topic..

[@NagarajuB-pz2fz]

Sir also explain semi circular arch with udl

[@Unknownrider124]

Thnku so much sir

[@naidurongali8732]

Sir.. in case of positive bending moment you didn't consider point load 100kn ..only you are consider horizontal and vertical reactions..please clarify my doubt sir..

[@Stanacademy]

The distance for the point load is zero. So no need to consider that

[@naidurongali8732]

@@Stanacademy ok sir. Then If UDL is given then how will I consider distance x

[@naidurongali8732]

@@Stanacademy please reply sir

[@Stanacademy]

@@naidurongali8732 You have to check which vertical reaction is high. You have to make a section on the half where reaction is high. Then find the distance and moment

[@shaikshahida5464]

We are considering upto the cross section only ,so we neglect the loads at that cross section

[@AnilNuthakki]

Hello Sir I have small doubt Maximum BM negative X^2=29 2x=29 X=29/2=14.5 "But how to calculate" X= 5.38 Please your valuable reply

[@unknownlad_]

I don't think you are very good at maths

[@Egamez-x2h]

X^2=29 X=√29 =5.38