The document discusses resolved shear stress (RSS), which is the component of an applied stress that acts to cause slip along a slip plane in the slip direction of a crystal. It defines RSS and provides an equation to calculate it based on applied load, cross-sectional area, and angles between the slip plane/direction and the direction of the applied stress. It states that slip begins when RSS reaches the critical resolved shear stress (CRSS), which depends on composition, temperature, and defect density in the crystal.

1. R.S.S

2. Resolved Shear Stress
• All working processes such as rolling,
extrusion, forging etc; (stress system quite
complex) cause plastic deformation by
involving the processes of slip or twinning.
(shear stress)
• These stresses arises in most processes
and test even the applied stress itself is
not a pure shear stress.

4. Illustration:
• A cylindrical single crystal of area A in a conventional
tensile test under a uni axial load P.
• The angle between normal to the slip plane and tensile
axis is φ,
• The angle which the slip direction makes with the tensile
axis is λ.
• The area of slip plane inclined at the angle φ will be A /
cos φ ,
• The component of the axial load acting in the slip plane
in the slip direction i.e shear stress. is P cos λ
• Therefore the resolved shear stress τR
∀ τ R = P cosλ / A/ cos φ = P/ A cos φ cos λ

5. • It can be seen that the resolved shear stress has a
maximum value when the slip plane is inclined at 45o to
the tensile axis i.e
∀ τ R = 1/2 * P /A.
• It becomes smaller for angles either greater than or less
than 45o.
• If φ < 45o ( slip plane nearly perpendicular to the tensile
axis) , the applied stress has a greater tendency to pull
the atoms apart than to slide them. When the slip plane
becomes more nearly parallel to the tensile axis (φ >
45o ) the shear stress is again small because the area of
the slip plane, A/ cos φ, is correspondingly large.

11. • It can be seen that the resolved shear stress has a
maximum value when the slip plane is inclined at 45o to
the tensile axis i.e
∀ τR = 1/2 * P /A.
• It becomes smaller for angles either greater than or less
than 45o.
• If φ > 45o ( slip plane nearly perpendicular to the tensile
axis) , the applied stress has a greater tendency to pull
the atoms apart than to slide them. When the slip plane
becomes more nearly parallel to the tensile axis (φ <
45o ) the shear stress is again small because the area of
the slip plane, A/ cos φ, is correspondingly large.

12. Critical resolved shear stress
• Slip begins when the shearing stress on the slip
plane in the slip direction reaches a threshold
value called critical resolved shear stress.
• This value is the yield stress of the single crystal
in an ordinary stress strain curve. ( In practice it
is very difficult to determine the stress at which
the first slip bands are produced.
• The value of the critical resolved shear stress
depends chiefly on
• composition and ii) temperature.

13. Magnitude of the critical
resolved shear stress
• i) In solid solution, critical resolved shear stress
would be high. (Atoms of solute and solvent size
different)
• ii) Depends on the interaction of its population of
dislocations with each other and with defects such as
vacancies, interstitials, and impurity atoms.
• On the basis of this reasoning, the critical resolved
shear stress should decrease as the density of
defects decreases, provided that the total number of
imperfections is not zero. When the last dislocation is
eliminated, the critical resolved shear stress should
rise abruptly to the high value predicted for the shear
strength of a perfect crystal.