Can anyone tell me how to find the angle between tensile force and slip direction under these conditions:- 1. Tensile force is applied along [010] direction for a single crystal of iron(BCC). 2. rss is directed along (110) plane and in a [-111] direction when tensile stress of 52 MPa is applied. thank you
I am not sure enough but this is my solution: Suppose lambda is your angle: tan(lambda)=a*sqrt(2)/a=sqrt(2) lambda=tan^-1(sqrt(2))
The schmid law allows you to calculate the shear stress as a function of the tensile stress : rss=sigma.cos(phi).cos(lambda) where phi is the angle between the tensile direction and the slip plane normal ([110] in your case) and lambda is the angle between the tensile direction and the slip direction ([-111] in you case). To find cos(phi) and cos(lambda) you can use the scalar product : 1*0+1*1+1*0=sqrt(2)*sqrt(1)*cos(phi) ==> cos(phi)=1/sqrt(2) -1*0+1*1*1*0=sqrt(3)*sqrt(1)*cos(lambda) ==> cos(lambda)=1/sqrt(3) Then: rss=52*1/sqrt(2)*1/sqrt(3)