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: Understanding the Concept:
This question relates to Schmid's Law, which describes the condition for slip in a single crystal. It's important to clarify the terminology. The Critical Resolved Shear Stress (CRSS) is a material constant; it does not change with orientation. The question is likely intended to ask: "For a given applied tensile stress, when is the resolved shear stress a maximum?" The resolved shear stress (\(\tau_R\)) is the component of the applied stress that acts on the slip system.

Step 2: Key Formula or Approach:
Schmid's Law is given by: \[ \tau_R = \sigma \cos(\phi) \cos(\lambda) \] where:

• \(\tau_R\) is the resolved shear stress.

• \(\sigma\) is the applied tensile stress.

• \(\phi\) is the angle between the tensile axis and the normal to the slip plane.

• \(\lambda\) is the angle between the tensile axis and the slip direction.
The term \(m = \cos(\phi) \cos(\lambda)\) is known as the Schmid factor.

Step 3: Detailed Explanation:
Slip begins when the resolved shear stress (\(\tau_R\)) reaches the critical resolved shear stress (CRSS). To initiate slip with the minimum applied stress (\(\sigma\)), or to get the maximum resolved stress (\(\tau_R\)) for a given applied stress, the Schmid factor \(m = \cos(\phi) \cos(\lambda)\) must be maximized.
We need to find the maximum value of the function \(f(\phi, \lambda) = \cos(\phi) \cos(\lambda)\). In a tensile test, the angles are not independent, but the theoretical maximum value of this product occurs when both cosine terms are as large as possible. For the general case of maximizing the product, the maximum value is achieved when the angles are equal and result in the largest product.
The product \( \cos(x) \cos(y) \) is maximized when \(x = y\). Here, the condition that maximizes the Schmid factor is \( \phi = \lambda = 45^\circ \). At this orientation: \[ m_{max} = \cos(45^\circ) \cos(45^\circ) = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2} = 0.5 \] This is the highest possible value for the Schmid factor.

Step 4: Final Answer:
Assuming the question means "when is the resolved shear stress maximized", the condition is \(\phi = \lambda = 45^\circ\).