Manufacturing Processes Ii Practice Questions

This problem involves Electrochemical Machining (ECM) and asks us to find the Material Removal Rate (MRR) for Titanium based on given data for Iron. The fundamental principle governing material removal in ECM is Faraday's laws of electrolysis, which can be expressed by the formula for MRR:
$\text{MRR} = \frac{I \cdot A \cdot \eta}{n \cdot F}$
Here, $I$ is the current, $A$ is the atomic weight, $\eta$ is the current efficiency, $n$ is the valency, and $F$ is Faraday's constant ($96500 \text{ C/mol}$).

We are given the following:
For Iron: $A_{Fe} = 56$, $n_{Fe} = 2$, $I_{Fe} = 1000 \, \text{A}$, $\text{MRR}_{Fe} = 0.26 \, \text{g/s}$.
For Titanium: $A_{Ti} = 48$, $n_{Ti} = 3$, $I_{Ti} = 2000 \, \text{A}$, and $\eta = 0.9$ (for both materials).

Using the formula for Titanium:
$\text{MRR}_{Ti} = \frac{I_{Ti} \cdot A_{Ti} \cdot \eta}{n_{Ti} \cdot F}$
Substituting the values:
$\text{MRR}_{Ti} = \frac{2000 \times 48 \times 0.9}{3 \times 96500}$
Calculating the value:
$\text{MRR}_{Ti} = \frac{86400}{289500} \approx 0.298 \, \text{g/s} \approx 0.3 \, \text{g/s}$
Thus, the expected material removal rate for Titanium is approximately $0.3 \, \text{g/s}$.

In a rolling process, the material is squeezed by the rollers, which exerts a normal force perpendicular to the surface, creating compression stress in the thickness direction. Simultaneously, friction between the rollers and the material pulls it forward, inducing shear stresses parallel to the rolling direction and surface. Thus, the material experiences both compression and shear.

To find the mean undeformed chip thickness, we first calculate the feed per tooth ($f_t$).

Step 1: Calculate Feed per Tooth ($f_t$)

The feed per tooth is given by the formula:
$f_t = \frac{\text{Table Feed Rate (f)}}{\text{Cutter Speed (N)} \times \text{Number of Teeth (z)}}$
Given:

• Table Feed Rate ($f$): 200 mm/min
• Cutter Speed ($N$): 100 rpm
• Number of Teeth ($z$): 4

Plugging in the values:
$f_t = \frac{200 \text{ mm/min}}{100 \text{ rpm} \times 4} = 0.5 \text{ mm/tooth}$

Step 2: Calculate Mean Undeformed Chip Thickness ($t_{mean}$)

For plain milling, the mean undeformed chip thickness is calculated using the formula:
$t_{mean} = f_t \sqrt{\frac{\text{Depth of Cut (d)}}{\text{Cutter Diameter (D)}}}$
Given:

• Feed per Tooth ($f_t$): 0.5 mm/tooth (from Step 1)
• Depth of Cut ($d$): 2 mm
• Cutter Diameter ($D$): 100 mm

Substitute these values into the formula:
$t_{mean} = 0.5 \text{ mm/tooth} \times \sqrt{\frac{2 \text{ mm}}{100 \text{ mm}}}$
$t_{mean} = 0.5 \times \sqrt{0.02}$
$t_{mean} \approx 0.5 \times 0.14142$
$t_{mean} \approx 0.07071 \text{ mm}$

Step 3: Convert to Microns

Since 1 mm = 1000 microns, convert the calculated $t_{mean}$ from millimeters to microns:
$t_{mean} \approx 0.07071 \text{ mm} \times 1000 \text{ microns/mm}$
$t_{mean} \approx 70.71 \text{ microns}$
Rounding this value to the nearest whole number gives 71 microns.

Let's break down the stress types involved in different metal forming processes.

1. Coining: This process uses very high forces to imprint intricate designs. The material is essentially squeezed, experiencing compressive stress ($S$).
2. Wire Drawing: Here, a wire is pulled through a die to reduce its diameter. The pulling action stretches the material, resulting in primary tensile stress ($P$) along the wire's length.
3. Blanking: This is a cutting operation where a punch cuts a piece from a sheet. The material fails due to the cutting action, which is predominantly shear stress ($Q$).
4. Deep Drawing: When forming a hollow part, the material is stretched over a punch (causing tensile stress) and simultaneously compressed in the flange area as it flows into the die ($R$), hence it experiences both tensile and compressive stress.

Therefore, the correct match is 1-S, 2-P, 3-Q, 4-R.

This problem involves applying sequential 3D rotations to a point. First, let's identify the initial coordinates of point C. Since it's a square plate of unit length in the X-Y plane with A at the origin, the initial coordinates of C are $(1, 1, 0)$.

Step 1: Rotation about the Z-axis by $30^\circ$ anti-clockwise.
The rotation matrix for an anti-clockwise rotation by an angle $\theta$ about the Z-axis is:
$R_z(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$
For $\theta = 30^\circ$, this becomes:
$R_z(30^\circ) = \begin{bmatrix} \cos 30^\circ & -\sin 30^\circ & 0 \\ \sin 30^\circ & \cos 30^\circ & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Applying this rotation to the initial coordinates of C $(1, 1, 0)$:
$\begin{bmatrix} x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} - \frac{1}{2} \\ \frac{1}{2} + \frac{\sqrt{3}}{2} \\ 0 \end{bmatrix}$
Substituting the numerical values (\sqrt{3} \approx 1.732): $$ \begin{bmatrix} 0.866 - 0.5 \\ 0.5 + 0.866 \\ 0 \end{bmatrix} = \begin{bmatrix} 0.366 \\ 1.366 \\ 0 \end{bmatrix} $$ So, after the first rotation, the coordinates of C are (0.366, 1.366, 0)$.

Step 2: Rotation about the X-axis by $90^\circ$ anti-clockwise.
The rotation matrix for an anti-clockwise rotation by an angle $\phi$ about the X-axis is:
$R_x(\phi) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \phi & -\sin \phi \\ 0 & \sin \phi & \cos \phi \end{bmatrix}$
For $\phi = 90^\circ$, this becomes:
$R_x(90^\circ) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos 90^\circ & -\sin 90^\circ \\ 0 & \sin 90^\circ & \cos 90^\circ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix}$
Applying this second rotation to the coordinates of C from Step 1 $(0.366, 1.366, 0)$:
$\begin{bmatrix} x'' \\ y'' \\ z'' \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0.366 \\ 1.366 \\ 0 \end{bmatrix} = \begin{bmatrix} 0.366 \\ 0 \\ 1.366 \end{bmatrix}$
Therefore, the final coordinates of C are $(0.366, 0.0, 1.366)$.

For maximum interference in an interchangeable assembly, we consider the largest possible shaft fitting into the smallest possible hole.
Given:
Shaft size: $25.000_{-0.010}^{+0.040}$ mm
Hole size: $25.000_{+0.020}^{+0.030}$ mm

The Maximum Shaft Size is $25.000 + 0.040 = 25.040$ mm.
The Minimum Hole Size is $25.000 + 0.020 = 25.020$ mm.

Maximum Interference is calculated as:
$\text{Maximum Interference} = (\text{Maximum Shaft Size}) - (\text{Minimum Hole Size})$
$\text{Maximum Interference} = 25.040 \text{ mm} - 25.020 \text{ mm} = 0.020 \text{ mm}$
Converting to microns ($1 \text{ mm} = 1000 \text{ microns}$):
$\text{Maximum Interference (microns)} = 0.020 \text{ mm} \times 1000 \text{ microns/mm} = 20 \text{ microns}$

The mould filling time is determined using Torricelli's Law, which gives the velocity of the molten metal, and the principle of continuity for volumetric flow rate.

Initially, for a gate with height $h$ and cross-sectional area $A$, the velocity is $v = \sqrt{2gh}$. The volumetric flow rate $Q_1 = A \cdot v = A \sqrt{2gh}$. The volume of the mould is $V = 100 \times 90 \times 20 \text{ mm}^3$. The initial filling time is $t_1 = \frac{V}{Q_1} = \frac{100 \times 90 \times 20}{A \sqrt{2gh}}$.

In the second scenario, the height is quadrupled to $4h$ and the cross-sectional area is halved to $\frac{A}{2}$. The new velocity becomes $v_2 = \sqrt{2g \times 4h} = 2\sqrt{2gh}$. The new volumetric flow rate is $Q_2 = \frac{A}{2} \times 2\sqrt{2gh} = A \sqrt{2gh}$. The new filling time is $t_2 = \frac{V}{Q_2} = \frac{100 \times 90 \times 20}{A \sqrt{2gh}}$.

Comparing the two filling times:
$\frac{t_2}{t_1} = \frac{V/Q_2}{V/Q_1} = \frac{Q_1}{Q_2} = \frac{A \sqrt{2gh}}{A \sqrt{2gh}} = 1$

To calculate the machining time ($T_m$) for a single pass, we use the formula $T_m = \frac{L}{f N}$, where $L$ is the length of cut, $f$ is the feed rate, and $N$ is the spindle speed. The question unfortunately doesn't provide these necessary parameters. However, assuming this question is from a specific context where these values are implicitly known or given elsewhere, and based on the provided answer key, the correct machining time is found to be 66 seconds.

To find the average cutting velocity in a shaping process, we use the formula that relates the stroke length, number of double strokes per minute, and the quick return ratio.

The average cutting velocity ($V_{avg}$) is given by:
$V_{avg} = L \times N \times (1 + KR)$
Where:
$L = \text{Length of stroke} = 250 \text{ mm} = 0.25 \text{ m}$ (converting to meters for m/min units)
$N = \text{Number of double strokes per minute} = 30 \text{ strokes/min}$
$KR = \text{Quick Return Ratio} = 0.6$

Substituting the values:
$V_{avg} = 0.25 \text{ m} \times 30 \text{ strokes/min} \times (1 + 0.6)$
$V_{avg} = 0.25 \times 30 \times 1.6 \text{ m/min}$
$V_{avg} = 7.5 \times 1.6 \text{ m/min}$
$V_{avg} = 12 \text{ m/min}$

To find the angle of bite ($\alpha$), we first need to calculate the draft ($\Delta t$) and the roller radius ($R$). Given the initial thickness $t_0 = 8$ mm and 10% reduction, the draft is $\Delta t = 0.10 \times 8 \text{ mm} = 0.8 \text{ mm}$. The roller diameter $D = 410$ mm gives a radius $R = D / 2 = 410 \text{ mm} / 2 = 205 \text{ mm}$. Using the formula for the angle of bite, $\alpha \approx \sqrt{\frac{\Delta t}{R}}$, we substitute the values: $\alpha \approx \sqrt{\frac{0.8 \text{ mm}}{205 \text{ mm}}} \approx \sqrt{0.0039024...} \approx 0.06247$ radians. Rounding this, the angle of bite is approximately $0.062$ radians.

In an Electrical Discharge Machining (EDM) process, the cycle time ($T$) for an RC relaxation circuit, considering negligible discharge time, is given by the formula: $T = RC \ln\left(\frac{V_s}{V_d}\right)$.
Given $R = 40 \, \Omega$, $C = 20 \, \mu F = 20 \times 10^{-6} \, F$, $V_s = 220 \, V$, and $V_d = 110 \, V$.
Substitute these values into the formula:
$T = (40 \, \Omega) \times (20 \times 10^{-6} \, F) \times \ln\left(\frac{220 \, V}{110 \, V}\right)$
$T = (800 \times 10^{-6} \, s) \times \ln(2)$
$T \approx (0.8 \times 10^{-3} \, s) \times 0.6931$
$T \approx 0.5545 \times 10^{-3} \, s$
Converting to milliseconds, $T \approx 0.5545 \, ms$.
Rounding to two decimal places, the time for one cycle is approximately 0.55 ms.

In Abrasive Jet Machining (AJM), the Material Removal Rate (MRR) is significantly affected by the Stand-Off Distance (SOD) between the nozzle and the workpiece. Initially, as SOD increases, the abrasive jet spreads into a more effective conical shape, enhancing impact area and efficiency, causing the MRR to increase. After this initial rise, there's a stable phase where MRR is relatively constant or peaks, representing an optimal balance of jet spread and particle energy. Beyond this optimal range, further increasing the SOD causes the jet to diverge excessively, leading to decreased kinetic energy of abrasive particles due to air friction and wider dispersion, which ultimately reduces the MRR. Thus, the MRR first increases, then becomes stable, and finally decreases with increasing SOD.

When a solid cylinder is forged between frictionless flat dies, its volume remains constant. We are given the initial diameter ($d_0 = 100 \text{ mm}$) and height ($h_0 = 50 \text{ mm}$), and the final height ($h_f = 25 \text{ mm}$). Using the principle of constant volume ($V_0 = V_f$), where $V = \pi \left(\frac{d}{2}\right)^2 h$:

$\pi \left(\frac{d_0}{2}\right)^2 h_0 = \pi \left(\frac{d_f}{2}\right)^2 h_f$

This simplifies to:

$d_0^2 h_0 = d_f^2 h_f$

Solving for the final diameter ($d_f$):

$d_f^2 = d_0^2 \frac{h_0}{h_f} = 100^2 \cdot \frac{50}{25} = 10000 \cdot 2 = 20000$

$d_f = \sqrt{20000} \approx 141.42 \text{ mm}$

Now, calculate the percentage change in diameter:

$\text{Percentage change} = \left(\frac{d_f - d_0}{d_0}\right) \times 100\%$

$\text{Percentage change} = \left(\frac{141.42 - 100}{100}\right) \times 100\% = 41.42\%$

To find the milling time, we first determine the total length the cutter travels. The slot is a straight line from (0, 0) to (100, 100). The length ($L$) of this path is calculated using the distance formula:
$L = \sqrt{(100 - 0)^2 + (100 - 0)^2} = \sqrt{100^2 + 100^2} = \sqrt{2 \times 100^2} = 100\sqrt{2} \text{ mm}$
Since the cutter diameter (10 mm) matches the slot width, the cutter's center follows this exact path.
Next, we calculate the milling time in minutes using the given feed rate ($F = 50 \text{ mm/min}$):
$T_{min} = \frac{L}{F} = \frac{100\sqrt{2} \text{ mm}}{50 \text{ mm/min}} = 2\sqrt{2} \text{ minutes}$
Finally, convert this time to seconds by multiplying by 60:
$T_{sec} = 2\sqrt{2} \times 60 \text{ seconds} = 120\sqrt{2} \text{ seconds}$
Approximating the value: $T_{sec} \approx 120 \times 1.4142 \approx 169.7 \text{ seconds}$. This value is closest to 170 seconds.

To find the average power input in an Electrical Discharge Machining (EDM) process using an RC relaxation circuit, we use the formula for average power. This formula is derived from the energy discharged per spark and the frequency of discharge.

The average power, $P_{avg}$, is given by:
$P_{avg} = \frac{V_d^2}{2R \ln\left(\frac{V_s}{V_s - V_d}\right)}$
Given values:
$R = 40 \text{ } \Omega$
$C = 20 \text{ } \mu F = 20 \times 10^{-6} \text{ F}$ (Note: C is not directly used in the final simplified formula but is part of the derivation.)
$V_s = 220 \text{ V}$
$V_d = 110 \text{ V}$

Substitute these values into the formula:
$P_{avg} = \frac{(110 \text{ V})^2}{2 \times (40 \text{ } \Omega) \times \ln\left(\frac{220 \text{ V}}{220 \text{ V} - 110 \text{ V}}\right)}$
$P_{avg} = \frac{12100}{80 \times \ln\left(\frac{220}{110}\right)}$
$P_{avg} = \frac{12100}{80 \times \ln(2)}$
Using $\ln(2) \approx 0.693$:
$P_{avg} = \frac{12100}{80 \times 0.693}$
$P_{avg} = \frac{12100}{55.44}$
$P_{avg} \approx 218.25 \text{ W}$
Converting to kilowatts (kW):
$P_{avg} \approx 0.218 \text{ kW}$

To calculate the current required for electrochemical machining (ECM), we use Faraday's laws of electrolysis. The core relationship is given by the formula relating mass removal rate to current:
$I = \frac{\left(\frac{m}{t}\right) \times n \times F}{M}$
First, convert the desired volumetric metal removal rate to a mass removal rate:
$\text{MRR}_{\text{mass}} = 2 \text{ cc/min} \times 7.8 \text{ g/cc} = 15.6 \text{ g/min}$
Then, convert this to grams per second:
$\text{MRR}_{\text{g/s}} = \frac{15.6 \text{ g/min}}{60 \text{ s/min}} = 0.26 \text{ g/s}$
Now, substitute the values into Faraday's law formula ($M=56 \text{ g/mol}$, $n=2$, $F=96500 \text{ C/mol}$):
$I = \frac{(0.26 \text{ g/s}) \times 2 \times 96500 \text{ C/mol}}{56 \text{ g/mol}} = \frac{50180}{56} \text{ A}$
$I \approx 896.07 \text{ A}$
Thus, a current of $896.07 \text{ A}$ is required to achieve the desired material removal rate.

To find the ultimate shear stress, we first need to determine the shear force and the shear plane area.

1. Calculate Chip Thickness Ratio ($r$): This ratio tells us how much the chip thickens during cutting.
   $r = \frac{\text{Uncut chip thickness}}{\text{Chip thickness}} = \frac{t_1}{t_2} = \frac{0.25 \text{ mm}}{0.75 \text{ mm}} = \frac{1}{3}$

2. Calculate Shear Angle ($\phi$): The shear angle is crucial for understanding the deformation process. We use Merchant's equation, relating $r$ and the rake angle ($\alpha$).
   Given $\alpha = 0^\circ$ and $r = 1/3$:
   $\tan(\phi) = \frac{r \cos(\alpha)}{1 - r \sin(\alpha)} = \frac{(1/3) \cos(0^\circ)}{1 - (1/3) \sin(0^\circ)} = \frac{1/3 \times 1}{1 - 0} = \frac{1}{3}$
   $\phi = \arctan\left(\frac{1}{3}\right) \approx 18.43^\circ$

3. Identify Cutting and Thrust Forces: In orthogonal machining, the given 'Normal force' ($F_n$) is the Cutting Force ($F_c$), and the 'Thrust force' ($F_t$) is the Thrust Force ($F_t'$).
   $F_c = 950$ N
   $F_t' = 475$ N

4. Calculate Shear Force ($F_s$): This is the force acting along the shear plane.
   $F_s = F_c \cos(\phi) - F_t' \sin(\phi)$
   $F_s = 950 \cos(18.43^\circ) - 475 \sin(18.43^\circ)$
   $F_s = 950 \times 0.9487 - 475 \times 0.3162 \approx 901.265 - 150.195 \approx 751.07 \text{ N}$

5. Calculate Shear Plane Area ($A_s$): This is the area over which the shear force acts.
   $A_s = \frac{t_1 \times b}{\sin(\phi)}$
   Given $t_1 = 0.25$ mm and width of cut $b = 2.5$ mm:
   $A_s = \frac{0.25 \text{ mm} \times 2.5 \text{ mm}}{\sin(18.43^\circ)} = \frac{0.625 \text{ mm}^2}{0.3162} \approx 1.9766 \text{ mm}^2$

6. Calculate Ultimate Shear Stress ($\tau$): This is simply the shear force divided by the shear plane area.
   $\tau = \frac{F_s}{A_s} = \frac{751.07 \text{ N}}{1.9766 \text{ mm}^2} \approx 380.0 \text{ N/mm}^2$

Let's break down how to find the shear angle and shear force in this orthogonal machining problem.

First, we calculate the chip thickness ratio ($r$) using the given uncut chip thickness ($t_1 = 0.25 \text{ mm}$) and chip thickness ($t_2 = 0.75 \text{ mm}$):
$r = \frac{t_1}{t_2} = \frac{0.25 \text{ mm}}{0.75 \text{ mm}} = \frac{1}{3}$
Now, we find the shear angle ($\phi$) using the formula that relates it to the chip thickness ratio ($r$) and rake angle ($\alpha = 0^{\circ}$):
$\tan(\phi) = \frac{r \cos(\alpha)}{1 - r \sin(\alpha)} = \frac{(1/3) \cos(0^{\circ})}{1 - (1/3) \sin(0^{\circ})} = \frac{1/3 \times 1}{1 - 1/3 \times 0} = \frac{1}{3}$
$\phi = \arctan\left(\frac{1}{3}\right) \approx 18.435^{\circ}$
Next, we determine the shear force ($F_s$). We identify the given "Normal force" as the Cutting Force ($F_c = 950 \text{ N}$) and the "Thrust force" as the Thrust Force ($F_t = 475 \text{ N}$). Using the formula for shear force:
$F_s = F_c \cos(\phi) - F_t \sin(\phi)$
$F_s = (950 \text{ N}) \cos(18.435^{\circ}) - (475 \text{ N}) \sin(18.435^{\circ})$
$F_s \approx (950 \times 0.9487) - (475 \times 0.3162) \approx 901.27 \text{ N} - 150.20 \text{ N} \approx 751.07 \text{ N}$
Thus, the shear angle is $18.435^{\circ}$ and the shear force is approximately $751.07 \text{ N}$.

Cemented carbide cutting tools are composite materials where super-hard carbide particles, like tungsten carbide (WC), are embedded in a metallic matrix. This metallic binder is crucial for holding the hard grains together, giving the tool toughness and preventing it from breaking. Among the given options, cobalt (Co) is overwhelmingly the most common and effective binding material used for tungsten carbide due to its excellent wetting properties and ability to form a strong matrix. While tungsten itself is the hard phase and nickel can be used in some specialized carbides, cobalt is the standard binder. Graphite is not suitable as a binder for this application. Therefore, cobalt is the correct binding material.