Question

The CRSS for a single crystal is maximum when

• A. \(\phi = \lambda = 45^\circ\)
• B. \(\phi = 45^\circ, \lambda = 60^\circ\)
• C. \(\phi = 30^\circ, \lambda = 45^\circ\)
• D. \(\phi = \lambda = 60^\circ\)

Hint:

Remember that CRSS is a material constant. The variable part is the orientation, captured by the Schmid factor \( \cos(\phi) \cos(\lambda) \). This factor is always maximized at 45 degrees, giving a maximum value of 0.5.

Answer 1

The Correct Option is A

Step 1: Write down Schmid's law.
The resolved shear stress on a slip system is $\tau_R = \sigma \cos\phi \cos\lambda$, where $\phi$ is the angle between the loading axis and the slip plane normal, and $\lambda$ is the angle between the loading axis and the slip direction.
Step 2: Treat the Schmid factor as a simple optimisation problem.
We want to know which combination of $\phi$ and $\lambda$ makes the product $\cos\phi \cos\lambda$ as large as possible, since that gives the largest resolved shear stress, and therefore the easiest activation of slip, for a given applied stress.
Step 3: Solve and check the numbers.
Treating the two cosine terms symmetrically, the product is maximised when the two angles are equal at $\phi = \lambda = 45^\circ$, giving $\cos 45^\circ \cos 45^\circ = \dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}} = 0.5$, which is the well known maximum Schmid factor of one half. Any other combination, such as $30^\circ$ and $45^\circ$, or $60^\circ$ and $60^\circ$, gives a smaller product.
\[ \boxed{\phi = \lambda = 45^\circ} \]