To the Creator of this video... please add suitable titles for each lesson instead of just Module number and Lecture number. It is difficult to locate a particular lecture again. Please do it if you can. It would be great help for students. Thank you.
Use phythagoras theorem to find the shift of Cv' from Cv = √y1^2+y2^2 y1= Cx+Tg/2 y2= Cy+ Tg/2 since for equal ISA angles Cx=Cy, so is the result as obtained. Note:: No shift in Cu.
In finding area we have to factored load =1.5*250=375kn in problem it is axial load
In question already the effective length has been given..
Why wasn't rv used as rv in calculating radius of gyration?
How to write the code
as already Leff provided then why we are multiplying it by 0.85 ????
is it because we are using double angle section ???
No.. I think we to take the Leff directly
Can anybody explain me how rv dash has been written by parallel axis theorem
Since it is an equal angle,
Cyy = Cxx
Cvv=sqrt( Cxx^2 + Cyy^2)
Cvv= sqrt(2) * Cxx
= sqrt(2)*Cyy
Multiply sqrt(2) both sides
Sqrt(2) * Cvv = (2)* Cyy
... eq1
Now, { rt(tg^2 +tg^2)} /2
=tg/ rt(2)
Using parallel axes thm.,
I'vv =
2( Ivv + A (Cvv + tg/rt(2)^2)
=
2(Ivv +A(rt(2)* Cvv +tg)^2/2)
=
2(Ivv+A(2Cyy+ tg) ^2 /2)... using eq1
r'vv = rt(I'vv / 2A)
=rt{rvv +(1/2)*(2Cyy+tg)^2}
(1/2) * (2Cyy + tg)^2
=
(4/2)*(Cyy+ tg/2)^2
Great work, thanks for taking the pain to explain.
@@shivamsharanlall672 thanks vro
In calculating Rv . Why is Cy used?? Cv should be used..
Cv will be cy*root2 so after doing square it will be 2cy^2 so it correct
To the Creator of this video... please add suitable titles for each lesson instead of just Module number and Lecture number.
It is difficult to locate a particular lecture again. Please do it if you can. It would be great help for students. Thank you.
Refer NPTEL
How rv dash come
Apply parallel axis theorem between vv axis of individual angle section and vv axis of combined section
Since it is an equal angle,
Cyy = Cxx
Cvv=sqrt( Cxx^2 + Cyy^2)
Cvv= sqrt(2) * Cxx
= sqrt(2)*Cyy
Multiply sqrt(2) both sides
Sqrt(2) * Cvv = (2)* Cyy
... eq1
Now, { rt(tg^2 +tg^2)} /2
=tg/ rt(2)
Using parallel axes thm.,
I'vv =
2( Ivv + A (Cvv + tg/rt(2)^2)
=
2(Ivv +A(rt(2)* Cvv +tg)^2/2)
=
2(Ivv+A(2Cyy+ tg) ^2 /2)... using eq1
r'vv = rt(I'vv / 2A)
=rt{rvv +(1/2)*(2Cyy+tg)^2}
(1/2) * (2Cyy + tg)^2
=
(4/2)*(Cyy+ tg/2)^2
@@shivamsharanlall672 rt...? what is rt here bro after eqn1
@@sonukumarsah4020 root- rt
Square root
same doubt
Use phythagoras theorem to find the shift of Cv' from Cv = √y1^2+y2^2
y1= Cx+Tg/2
y2= Cy+ Tg/2
since for equal ISA angles Cx=Cy, so is the result as obtained.
Note:: No shift in Cu.
rv das how came
Apply parallel axis theorem between vv axis of individual angle section and vv axis of combined section
Since it is an equal angle,
Cyy = Cxx
Cvv=sqrt( Cxx^2 + Cyy^2)
Cvv= sqrt(2) * Cxx
= sqrt(2)*Cyy
Multiply sqrt(2) both sides
Sqrt(2) * Cvv = (2)* Cyy
... eq1
Now, { rt(tg^2 +tg^2)} /2
=tg/ rt(2)
Using parallel axes thm.,
I'vv =
2( Ivv + A (Cvv + tg/rt(2)^2)
=
2(Ivv +A(rt(2)* Cvv +tg)^2/2)
=
2(Ivv+A(2Cyy+ tg) ^2 /2)... using eq1
r'vv = rt(I'vv / 2A)
=rt{rvv +(1/2)*(2Cyy+tg)^2}
(1/2) * (2Cyy + tg)^2
=
(4/2)*(Cyy+ tg/2)^2
Can anybody explain me how rv dash has been written by parallel axis theorem?