Question

A cylindrical single crystal of copper having 10 mm diameter is deformed under a uniaxial tensile load of 2200 N. The angle between the normal to the slip plane and the tensile loading axis is \(\alpha\), whereas the angle between slip direction and the tensile axis is \(\beta\). The slip direction is in the plane defined by the stress axis and the normal to the slip plane.
If \(\alpha=\beta\), the critical resolved shear stress (CRSS), rounded off to one decimal place, is ______ MPa.

Hint:

Use \(\sigma=F/A\), then Schmid's law \(\tau=\sigma\cos\alpha\cos\beta\) with \(\alpha+\beta=90^{\circ}\) from the coplanar condition, and \(\alpha=\beta=45^{\circ}\).

Answer 1

Correct Answer: 12.8

Step 1: Compute the cross-sectional area and the applied normal stress.
The crystal is cylindrical with diameter \(10\ \text{mm}\), so its radius is \(5\ \text{mm}\). The cross-sectional area is
\[ A=\pi r^2=\pi(5)^2=78.54\ \text{mm}^2 \]
The applied tensile load is \(F=2200\ \text{N}\), so the normal (uniaxial) stress on the crystal is
\[ \sigma=\frac{F}{A}=\frac{2200}{78.54}=28.01\ \text{MPa} \]
(force in newtons over area in mm\(^2\) gives stress directly in N/mm\(^2\), which is MPa).

Step 2: Recall Schmid's law.
The resolved shear stress on a slip system is
\[ \tau=\sigma\cos\alpha\cos\beta \]
where \(\alpha\) is the angle between the tensile axis and the normal to the slip plane, and \(\beta\) is the angle between the tensile axis and the slip direction.

Step 3: Use the coplanar condition to link \(\alpha\) and \(\beta\).
The slip direction always lies within the slip plane, so it is perpendicular to the plane's normal. The problem states that the slip direction also lies in the same plane as the tensile axis and the normal to the slip plane. Since all three vectors are coplanar and the slip direction is perpendicular to the normal, the angle the slip direction makes with the tensile axis is simply \(90^{\circ}\) minus the angle the normal makes with the tensile axis:
\[ \alpha+\beta=90^{\circ} \]

Step 4: Apply the given condition \(\alpha=\beta\).
\[ \alpha+\alpha=90^{\circ}\quad\Rightarrow\quad\alpha=\beta=45^{\circ} \]

Step 5: Compute the Schmid factor and the critical resolved shear stress.
\[ \cos\alpha\cos\beta=\cos(45^{\circ})\cos(45^{\circ})=(0.7071)(0.7071)=0.5 \]
Since slip actually begins under this applied load, this resolved shear stress is the critical resolved shear stress (CRSS):
\[ \tau_{CRSS}=\sigma\times0.5=28.01\times0.5 \]
\[ \tau_{CRSS}\approx14.0\ \text{MPa} \]

Final Answer:
The critical resolved shear stress, rounded to one decimal place, is
\[ \boxed{14.0\ \text{MPa}} \]