i need correct answers about this subject

Solution 1:
For open cross-sections consisted of several thin rectangle joint together the
torsion constant is given as;
∑



= k {





Where k = 1 for thin section bent from a sheet





}


And warping constant of the section, Fww is given by the equation;






Let warping function at A and B is wA and wB respectively;
It can be written as;
For cell (1); 1 =


{


}
For cell (2); 1 =


{


}
By solving both equations
wA = 6 mm
2
wB = 4 mm
2
Value of maximum longitudinal stress =





= 0.254 MPa
Solution 2:
Given data:
Side of an equilateral triangle, L = 100 mm
Wall thickness, t = 10 mm
Torque value, M = 1000 kN-mm
This is a closed section so torsional constant value for single cell profile;


∮
Area enclosed by the profile = A
=



=


√











Equation for stress is given as;









For single cell profile the value of Ψ is given as:


∮














When beam of uniform cross-section is twisted by a torque M at each end then it
is uniform torque and warping stress w(s) in this case is zero only shear stress...