GATE 2026 MT – Question 50
A cylindrical copper single crystal of diameter 10 mm is deformed under a uniaxial tensile load of 2200 N. The angle between the slip-plane normal and the tensile axis is $\alpha$, and the angle between the slip direction and the tensile axis is $\beta$. The slip direction lies in the plane defined by the stress axis and the slip-plane normal. If $\alpha=\beta$, the critical resolved shear stress is ________ MPa. (Rounded off to one decimal place.)
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Correct answer: 12.8 to 14.1
Explanation
The slip direction is perpendicular to the slip-plane normal. With the stated coplanar geometry, $\alpha+\beta=90^\circ$. Since they are equal, both are $45^\circ$, giving Schmid factor $\cos\alpha\cos\beta=1/2$.
The tensile stress is $\sigma=2200/[\pi(10)^2/4]=28.0113\ \mathrm{MPa}$. Therefore $\tau_{CRSS}=\sigma/2=14.0056\ \mathrm{MPa}$, or **14.0 MPa**.