Resolved shear stress(rss) problem

  1. 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
     
  2. jcsd
  3. nam

    nam 2

    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))
     
  4. 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)
     
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