The value that the resolved shear stress acting along the slip direction on a slip plane must reach for slip to begin. A single crystal yields when the resolved shear stress on some system reaches this value.
Japanese name: 臨界分解せん断応力
Critical resolved shear stress appears in 2 practice questions across 1 topic, including Strengthening mechanisms and dislocations.
When a tensile stress σ is applied to a single crystal, the component of shear stress acting along the slip direction on the slip plane of a given slip system is τ = σ cosφ cosλ. φ is the angle between the slip-plane normal and the tensile axis, λ the angle between the slip direction and the tensile axis, and cosφ cosλ is called the Schmid factor. Slip begins when τ on some system reaches a value τc characteristic of the material, and this τc is the critical resolved shear stress. Schmid and co-workers showed that when tensile tests are made on crystals of different orientations, the tensile stress at yield varies greatly, but converted to resolved shear stress it is nearly constant. This is known as Schmid's law.
The critical resolved shear stress of pure FCC single crystals such as copper and aluminum is very small, about 1 MPa or less, less than a thousandth of the theoretical shear strength. This is because dislocations move a little at a time to carry the shear, and the value is set by the snagging of dislocations on one another and by impurities. Adding solute atoms raises it through solid-solution strengthening, and fine precipitates raise it much more. In BCC metals the resistance of the lattice itself (the Peierls stress) is large, so the critical resolved shear stress depends strongly on temperature and is larger at lower temperatures.
In a polycrystal, several slip systems, including poorly oriented ones, must operate at the same time so that the grains deform without gaps. The yield stress of a polycrystal is therefore the critical resolved shear stress multiplied by the Taylor factor M, which is about 3.06 for randomly oriented FCC metals. In sheet with a texture, in which crystal orientations have been aligned by rolling and similar processes, M varies with direction, so anisotropy appears in the yield stress and the r-value. Strengthening by grain boundaries is added on top of this through the Hall–Petch relationship.
Answer: C. About 2.1 MPa
The resolved shear stress on a slip system is τ = σ cosφ cosλ (Schmid's law), and cosφ cosλ is called the Schmid factor. Here cos60° × cos40° = 0.500 × 0.766 ≈ 0.383, so slip begins at σ = 0.80/0.383 ≈ 2.1 MPa. About 0.31 MPa multiplies the critical resolved shear stress by the Schmid factor instead of dividing, and about 1.6 MPa divides only by cosφ and forgets cosλ. The Schmid factor reaches its maximum of 0.5 when the slip-plane normal and the slip direction both make 45° with the tensile axis, and it is 0 when the slip plane is parallel or perpendicular to the axis, in which case that system does not slip. In a single crystal, the system with the largest factor slips first.
Answer: B. For neighboring grains to deform without opening gaps or overlapping at the grain boundaries, at least five independent slip systems must operate in each grain, including poorly oriented ones
In a polycrystal, if each grain deformed into any shape it liked, gaps and overlaps would open at the grain boundaries. To accommodate an arbitrary change of shape at constant volume, five independent slip systems are needed (the von Mises criterion). As a result, poorly oriented systems must operate along with those of high Schmid factor, and averaging over a randomly oriented FCC polycrystal gives a Taylor factor of about 3.06. If yielding were set only by the most favorably oriented grains, a factor larger than 2 could not be explained. The Taylor factor is set by the distribution of grain orientations and does not depend on grain-boundary area or grain size; the effect of grain size is added separately through the Hall–Petch relationship. The same criterion explains why polycrystals of HCP metals, which lack enough independent slip systems, have poor ductility.
No. What depends on orientation is the tensile stress at which slip begins. In an orientation with a small Schmid factor, a larger tensile stress is needed to reach the same critical resolved shear stress.
If the critical resolved shear stress is the same for every system, the system with the largest Schmid factor slips first. In FCC metals, the direction of the tensile axis determines which of the twelve {111}<110> systems has the largest factor.
The critical resolved shear stress differs greatly between types of slip system. In magnesium the value for basal slip is orders of magnitude smaller than for prismatic or pyramidal slip, so basal slip dominates at room temperature and polycrystals have poor ductility.
About 0.31 MPa multiplies the critical resolved shear stress by the Schmid factor instead of dividing, and about 1.6 MPa divides only by cosφ and forgets cosλ.
A single-crystal rod is pulled in tension. For a slip syste…Schmid factor Taylor factor Hall–Petch relationship Bauschinger effect Dispersion strengthening Orowan mechanism Peierls stress