Determine the maximum possible yield strength for a single crystal of Fe pulled in tension

I am trying to solve the problem:

The critical resolved shear stress for iron is 27 MPa. Determine the maximum possible yield strength for a single crystal of Fe pulled in tension.

I think that this is related to the equation $\sigma_{yield} = \dfrac{\tau_{crit}}{\cos{\Phi}\cos{\gamma}}$.

The book says:

The minimum stress necessary to introduce yielding occurs when a single crystal is oriented such that $\Phi=\gamma=45°$; under these conditions, $\sigma_{yield}= 2\tau_{crit}$.

When I tried check my answer, the solution manual said I should use $\sigma_{yield}= 2\tau_{crit}$:

In order to determine the maximum possible yield strength for a single crystal of Fe pulled in tension, we simply employ Equation 7.5 as $\sigma_y=2\tau_{crss}=(2)(27\ \mathrm{MPa})=54\ \mathrm{MPa}$.

I don't get this. Doesn't this calculate the minimum yield strength, not the maximum? The solution manual is considering the possibility of the iron crystal being pulled in tension in a direction that is most favourable and makes one of the slip systems reach the critical value the quickest.

But to know the maximum possible yield strength shouldn't you calculate the situation in which tension is applied in a direction that makes $\cos\Phi\cos\gamma$ as small as possible? So, basically, in a direction in which the resolved shear stress is just a fraction of the applied tensile stress?

The solution manual gets the yield strength when you pull the single iron crystal in such a direction that the resolved shear stress is the largest fraction possible of the applied stress. But if you reorient the single crystal, couldn't it withstand a way larger applied tensile stress and thus have a higher yield strength?

The book should probably have said "minimum possible yield strength" or "maximum possible resolved shear stress." Note that the maximum yield strength of any real single crystal does not depend on one slip system and thus can not be infinite despite what Schmid's Law might imply. The maximum yield strength occurs at a set of angles "furthest" from all of the slip systems and is more difficult to calculate.

There appears to be confusion about what the Schmid's Law actually means. Schmid's Law relates applied tensile stress to resolved shear stress. In other words, a way of transforming an applied tensile stress to a shear stress in a given plane and along a given direction within that plane, based on the angles of the plane and direction with tension.

The yield strength of a single crystal is related, by Schmid's Law, to the critical resolved shear stress necessary to cause slip. The shear stress necessary to cause slip for a given slip system is a material property. The applied tensile stress on a single crystal is related, also by Schmid's Law, to the resolved shear stress on a given slip system, via the angles between slip plane and tension direction, and slip direction and tension direction. Slip plane and slip direction together make a slip system. Note that yield strength (a material property) is related to critical stress (a material property), and applied stress is related to resolved stress. When the applied stress is such that the resolved shear stress equals the critical resolved shear stress, then the applied stress is equal to the yield strength, and the crystal slips.

To determine the yield strength of a single crystal in a given direction, first pick a slip system (plane and direction), then pick the direction of tension. Finally determine what tensile stress is required to cause the actual resolved shear stress to equal the critical resolved shear stress. Schmid's Law allows you to convert tensile stress to actual resolved shear stress. Then replace actual resolved shear stress with critical resolved shear stress, and work backward to determine the yield strength.

To answer your other question, yes it is possible for the yield strength to vary depending on the choice of slip system and tension direction. Single crystals have anisotropic mechanical properties. Consider a (not-real) system with only one slip plane. If that one slip plane is oriented perpendicular or parallel to the direction of tension, slip is impossible because the actual resolved shear stress is 0 in both cases no matter what amount of tension is applied. In a sense, the theoretical tensile yield strength by slip of such a non-physical system is infinite in those directions.

All real crystal systems have slip systems such that slip can occur in every tensile direction, though the tensile strength may vary based on orientation. Which slip system becomes active depends on which system's resolved shear stress exceeds its critical resolved shear stress first, which in turn depends on the relative angles the systems make with the choice of tension direction. It is also important to note many crystalline materials behave isotropically in bulk due to individual grains being oriented randomly, unlike in a single crystal which has a dominant orientation.

• So then the solution the book gives is wrong? It says that for a single iron crystal with a certain critical resolved shear stress (27 MPa, in this case), you can calculate the maximum yield strength with σyield=2τcrit. Yet to me that seems to be the lowest yield strength a single crystal of iron can have. Because that equation is for when the tensile strength makes an angle of 45 degrees to both the normal to the slip plane and 45 degrees to the slip direction. So if you just apply tension in another direction the yield strength of the single crystal will be higher? Apr 13 '16 at 18:16
• The strength for that slip system will be higher. The stength of the crystal overall depends on which slip system is "closest" to the tensile stress, so strength doesn't increase arbitrarily. The book may have meant to write minimum strength. The resolved stress is highest at those angles for a given applied stress. Apr 13 '16 at 19:23
• @starrise, the maximum CRSS of a crystal can be determined by equation 7 of this paper, CRSS(max)= Gamma/b, where Gamma= Stacking Fault Energy (SFE) and b= magnitude of Burgers vector. While it gives CRSS(max)=836 MPa for Nickel single crystal, it is 113 MPa for 316L steel (low SFE alloy, Gamma = 30 mJ/m^2). Is there a way to calculate the maximum possible (theoretical) CRSS of low SFE metals? Aug 9 '18 at 19:54