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.
Show 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 \(\alpha=\beta=45^{\circ}\).
Step 1: Get the stress from load and area.
The bar has diameter $10\ \text{mm}$, so area $A=\pi(5)^2=78.54\ \text{mm}^2$. Under $F=2200\ \text{N}$,
\[ \sigma=\frac{2200}{78.54}\approx28.01\ \text{MPa} \]
Step 2: Turn the coplanar condition into a single-variable Schmid factor.
Schmid's law gives $\tau=\sigma\cos\alpha\cos\beta$. Because the slip direction, the tensile axis, and the plane normal are coplanar, and the slip direction is perpendicular to the normal, $\beta=90^{\circ}-\alpha$. Substitute this in:
\[ \tau=\sigma\cos\alpha\cos(90^{\circ}-\alpha)=\sigma\cos\alpha\sin\alpha \]
Using the double angle identity $2\sin\alpha\cos\alpha=\sin(2\alpha)$, this is
\[ \tau=\frac{\sigma}{2}\sin(2\alpha) \]
Step 3: Treat $\alpha=\beta$ as the condition that gives the largest possible resolved shear stress.
Since $\sin(2\alpha)$ reaches its largest value of $1$ when $2\alpha=90^{\circ}$, that is when $\alpha=45^{\circ}$, and at that point $\beta=90^{\circ}-45^{\circ}=45^{\circ}=\alpha$. So the condition $\alpha=\beta$ given in the problem is exactly the orientation that maximizes the Schmid factor for this coplanar family of slip systems.
Step 4: Evaluate the Schmid factor at this orientation.
\[ \tau=\frac{\sigma}{2}\sin(90^{\circ})=\frac{\sigma}{2}\times1=\frac{\sigma}{2} \]
So the Schmid factor is exactly $0.5$, the maximum possible value for any slip system under uniaxial tension.
Step 5: Substitute the stress value.
\[ \tau_{CRSS}=\frac{28.01}{2}\approx14.0\ \text{MPa} \]
This matches the direct geometric calculation, confirming the CRSS.
\[ \boxed{\tau_{CRSS}\approx14.0\ \text{MPa}} \]
