The material is AA 6061 T6

You may do that comparison through the failure criterion. Von Mises criterion is the most conservative and exact for most of material. It compare the distorsion energy (deviatory) of any state of stress with the same for the simple tensile test.

U'_o=sigma_o^2/(6G)          U'=(3/(4G)) Toct^2

G is the shear modulus and Toct is the octaedral shear stress. Finally this gives

(9/2)Toct^2=sigma_o^2

See for example:

https://www.academia.edu/11942882/Mec%C3%A1nica_y_Termodin%C3%A1mica_de_Sistemas_Materiales_Continuos

Sección II.3.5 pp.78-79

Thanks Amit Chimanpure & Andres Granados for your valuable comments. During Experimentation I have found the bending angle of my samples greater than the bending angle of the welded samples(both in face bend and root bend). Form This I can interpret that there has been an increase in the ductility of the welded samples. Thus from optimization point of view when bending angle (ductility) is an output, should I consider it smaller the better and Tensile strength as larger the better? that was actually my question

Dear Barnik, I am sorry to say that the large bending angle corresponds to higher tensile as well ultimate strength because of the upper part of the test piece is subject to the tensile loading compare to the lower part.  Plus the rupture takes place at the tensile side of the test piece. Welding sample mostly  shows weak spats at the welding regions because of the unavoidable stress concentrations there.  Therefore this type of comparison test doesn't mean very much technically but may be in practice YES.  

Try to identify the principal stresses sigma_1, sigma_2 and sigma_3 in your welded sample. Calculate the octaedral stress

Toct=squareroot( (sigma_1-sigma_2)^2+(sigma_2-sigma_3)^2

                                                                                              +(sigma_3-sigma_1)^2 ) / 3

Then make the comparison

Toct=( squareroot(2)/3 ) sigma_o

sigma_o = yield strength in a simple tensile stress test

SOME DEFINITIONS:  a) Ultimate tensile strength corresponds to the peak value of the Enginering stress  S  versus strain plot, where the necking begins. UTS= P/Ao. This may be converted to the scientific stress by multiplying  UTS by  Ao /A = P/A where  A  is the area measured at the neck. That correspons much higher position in the sigma versus strain plot without showing any peaking. b)  Yield strength correspond to Sigma at the transition point where the elastic deformation region is replaced by the plastic region.  This would be sharp transition such as the case for the mild steel (Fe+0.2% C) 

where sigma -strain plot shows the upper yield point.  In the case of  FCC (Aluminum) and HCP(Zinc)  metals no sharp yield points may be observed the yield strength is defined by the intersection of a line with the sigma-strain curve, which as a slope equal to the elastic modulus with   0.2% strain  at zero stress. 

Conclusions:  Therefore with the bending test there is no way to observe the yielding phenomenon but  UTS  may be observed by tracking the necking phenomena closely.

What about plastics? UTS or flexural while bending