GATE 2026 ME – Question 63
During orthogonal turning by using a single point cutting tool, feed rate is 0.24 mm/rev. The uncut chip thickness is 0.23 mm. The shear angle, tangential force component, and radial force component are 20°, 800 N, and 150 N, respectively. The value of shear force is ________ N (*rounded off to 2 decimal places*).
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Correct answer: 356.56 to 360.14
Explanation
Read the question as an uncut chip thickness of $t = 0.23$ mm and a chip thickness of $t_c = 0.24$ mm, so the thickness ratio is $r = \frac{0.23}{0.24} = 0.9583$. Using the given 20° as the tool rake angle, the shear angle follows from $\tan\phi = \frac{r\cos\alpha}{1 - r\sin\alpha} = \frac{0.9583 \times 0.9397}{1 - 0.9583 \times 0.3420} = 1.3396$, so $\phi = 53.26°$. The shear force is $F_s = F_c\cos\phi - F_t\sin\phi = 800 \times 0.5982 - 150 \times 0.8014 = 358.35$ N.