Resolved shear stress(rss) problem

[custer]

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

 

[nam]

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))

 

[CharlesMareau]

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)