Machining and Cutting Tools Questions and Answers

Answer: A

A bandsaw employs an endless looped blade to cut, and it is suited to irregular and curved shapes. A shaper and a planer use a single point tool in a reciprocating motion for flat surfaces, and a slotter is a vertical shaper for cutting slots and keyways, so none of them uses an endless blade. The key therefore gives the bandsaw.

Answer: A

The new tool life is 4 times the original.

Taylor's equation is \(VT^{n} = C\). With n = 0.5:

\[ \begin{aligned} \frac{T_{2}}{T_{1}} &= \left(\frac{V_{1}}{V_{2}}\right)^{1/n} \\ &= 2^{1/0.5} = 2^{2} = 4 \end{aligned} \]

Halving the speed doubles \(V_{1}/V_{2}\), and the exponent 1/n = 2 squares that ratio. 2 is what a linear relation (n = 1) would give, and 0.5 or 1 would mean slower cutting shortened or did not change tool life.

Answer: C

The shear strain rate is about \(1.0104\times10^{5}\) s^-1.

First the shear angle, from the chip thickness ratio r = 0.4 and rake angle 10°:

\[ \begin{aligned} \tan\phi &= \frac{r\cos\alpha}{1-r\sin\alpha} = \frac{0.3939}{0.9305} \\ &= 0.4233,\ \phi = 22.94^{\circ} \end{aligned} \]

The shear velocity along the shear plane is:

\[ \begin{aligned} V_{s} &= \frac{V\cos\alpha}{\cos(\phi-\alpha)} = \frac{2.5(0.9848)}{\cos 12.94^{\circ}} \\ &= 2.526\ \text{m/s} \end{aligned} \]

Dividing by the shear zone thickness of 25 microns gives the strain rate:

\[ \begin{aligned} \dot{\gamma} &= \frac{2.526}{25\times10^{-6}} \\ &= 1.0104\times10^{5}\ \text{s}^{-1} \end{aligned} \]

Dividing the cutting speed itself by the zone thickness gives exactly \(1.0\times10^{5}\), near but not equal to the keyed value; the velocity that slides across the shear zone is the shear velocity, not the cutting speed. The other three options do not follow from these data.

Answer: A

The correct statement is that Configuration P1 arrests 6 degrees of freedom, while P2 and P4 are over-constrained and P3 is under-constrained.

A cuboid is located by the 3-2-1 principle: three pins under the largest face fix one translation and two rotations, two pins on a side face fix a further translation and rotation, and one pin on the adjacent end face fixes the last translation, six points for six degrees of freedom.

• P1: three pins below and three on the sides, in the 3-2-1 pattern, so exactly 6.
• P2: four pins below and three on the sides, 7 in all; one is redundant, so it is over-constrained.
• P3: three below and only two on the sides, 5 in all; one freedom is left, so it is under-constrained.
• P4: four below and four on the sides, 8 in all; over-constrained.

Each of the other options puts the correct 3-2-1 label on a configuration with too many or too few pins.

Answer: B

The shear strain rate is \(5.20 \times 10^{4}\ \text{s}^{-1}\).

The strain rate is the shear velocity along the shear plane divided by the thickness of the shear zone. From the velocity triangle of orthogonal cutting,

\[ \begin{aligned} V_{s} &= \frac{V\cos\alpha}{\cos(\phi - \alpha)} \\ &= \frac{\cos 5^{\circ}}{\cos 40^{\circ}} \\ &= \frac{0.9962}{0.7660} \\ & = 1.300\ \text{m/s} \end{aligned} \]

\[ \begin{aligned} \dot{\gamma} &= \frac{V_{s}}{\delta} \\ &= \frac{1.300}{25 \times 10^{-6}} \\ &= 5.20 \times 10^{4}\ \text{s}^{-1} \end{aligned} \]

The distractors reuse numbers from the same calculation. 1.84 is the shear strain, \(\cot\phi + \tan(\phi - \alpha) = 1 + 0.839\), a dimensionless quantity, not a rate. 1.30 is the shear velocity in m/s; dividing it by 100 μm instead of 25 μm would give \(1.30 \times 10^{4}\). 0.71 is \(\cos 45^{\circ}\), which does not give the strain rate on its own.