# Critical resolved shear stress

CRSS redirects here. For the Pakistan-based think tank, see Center for Research and Security Studies.

Critical resolved shear stress is the component of shear stress, resolved in the direction of slip, necessary to initiate slip in a grain. It is a constant for a given crystal.

Tests have been conducted on single crystals of metals to measure the shear stress required to initiate plastic deformation, or cause atomic planes to slip. Since this is a threshold value, it is referred to as critical; and since it is a component of the applied force or stress, it is said to be resolved; that is, the critical resolved shear stress.

Resolved shear stress is given by τ = σ cos Φ cos λ[1] where σ is the magnitude of the applied tensile stress, Φ is the angle between the normal of the slip plane and the direction of the applied force and λ is the angle between the slip plane direction and the direction of the applied force. whereas, critical resolved shear stress value is given by τ =σ (cosΦ cosλ)max

Resolved shear stress is the shear component of an applied tensile (or compressive) stress resolved along a slip plane that is other than perpendicular or parallel to the stress axis. The critical resolved shear stress is the value of resolved shear stress at which yielding begins; it is a property of the material.

## Resolved Shear Stress in Crystalline Metals

In metals, slip occurs in specific directions on crystallographic planes. Stress projection factors are used to calculate the stresses on these planes. The most common stress projection factor is the Schmid Factor[2] which is the maximum value of the term (cosΦ cosλ) for all possible slip planes. The Schmid Factor is most applicable to face centered cubic (FCC) single crystal metals.[3]

For an FCC system, the Schmid Factor for the $(\overline{1} 1 1)$ plane in the $[1 0 1]$ direction for an applied tensile load in the $[1 2 3]$ direction can be calculated as:

\begin{align} M_{[\text{1 0 1}](\overline{1} 1 1)}=&\frac{(-1*1)+(1*2)+(1*3)}{((-1)^2+1^2+1^2)^{1/2}(1^2+2^2+3^2)^{1/2}}*\frac{(1*1)+(0*2)+(1*3)}{(1^2+0^2+1^2)^{1/2}(1^2+2^2+3^2)^{1/2}}=&\frac{16}{14\sqrt{6}} \end{align}

The values of slip systems to test using the Schmid Factor are from the {1 1 1}<1 -1 0> system. There are 12 different possible pairings of the planes and directions. The maximum Schmid Factor will indicate the operational slip system(s).[4]

For polycrystal metals, the Taylor factor has been shown to be more accurate than the Schmid Factor for polycrystal specimens.[5] Taylor showed [5] that a minimum of five active slip systems are required to accommodate an arbitrary deformation. The five (or more) active systems are those that, combined, provide the greatest amount of work of deformation.

## References

1. ^ Gottstein G., Physical Foundations of Materials Science, Springer, 2004, page 227.
2. ^ Schmid E., Boas W., Plasticity of Crystals with Special Reference to Metals, F.A. Hughes & Co. Ltd., 1935.
3. ^ Hosford W.F., Mechanical Behavior of Materials, 2nd ed., Cambridge University Press, 2010, page 113.
4. ^ Meyers and Chawla. (1999) Mechanical Behaviors of Materials. Prentice Hall, Inc. Page 301.
5. ^ a b Taylor, Sir Geoffrey Ingram. Plastic strain in metals. 1938.