Manufacturing Processes Ii Practice Questions

This problem asks us to find the number of pulses needed for an NC machine's stepper motor to achieve a specific linear movement. We'll connect the linear distance to the lead screw's rotation, then to the motor's rotation via the gear ratio, and finally to the number of pulses using the step angle.

Given:

• Linear movement distance: $x$ mm
• Lead screw pitch: $p$ mm/revolution
• Stepper motor step angle: $\alpha$ degrees/pulse
• Gear ratio: $g$ (motor turns per lead screw turn)

First, to achieve a linear movement of $x$ mm, the lead screw must rotate by:
$\text{Lead Screw Revolutions} = \frac{\text{Linear Movement}}{\text{Pitch}} = \frac{x \text{ mm}}{p \text{ mm/revolution}} = \frac{x}{p} \text{ revolutions}$
Next, considering the gear ratio $g$, which is the number of motor turns for each single turn of the lead screw, the motor's total revolutions will be:
$\text{Motor Revolutions} = (\text{Lead Screw Revolutions}) \times (\text{Gear Ratio}) = \frac{x}{p} \times g \text{ revolutions}$
To convert these motor revolutions into angular movement in degrees (since the step angle is in degrees), we multiply by $360$ degrees/revolution:
$\text{Motor Angular Movement (degrees)} = (\text{Motor Revolutions}) \times 360 \text{ degrees/revolution} = \frac{x}{p} \times g \times 360$
Finally, to find the total number of pulses, we divide the total angular movement of the motor by the step angle per pulse:
$\text{Number of Pulses} = \frac{\text{Total Motor Angular Movement (degrees)}}{\text{Step Angle (degrees/pulse)}} = \frac{\frac{x}{p} \times g \times 360}{\alpha}$
$\text{Number of Pulses} = \frac{360gx}{p\alpha}$

This problem asks for the minimum coefficient of friction needed for a steel plate to be rolled without external assistance (unaided rolling). This means friction alone must pull the plate into the rolls.

First, let's break down the given information:

• Initial plate thickness ($T_0$) = $80 \text{ mm}$
• Final plate thickness ($T_f$) = $40 \text{ mm}$
• Number of passes ($n$) = $4$
• Roll diameter ($D$) = $800 \text{ mm}$ (so, Roll Radius $R = D/2 = 400 \text{ mm}$)
• The reduction is equal in each pass.

Step 1: Calculate the reduction per pass.
The total reduction is $T_0 - T_f = 80 \text{ mm} - 40 \text{ mm} = 40 \text{ mm}$.
Since this reduction occurs over 4 equal passes, the reduction per pass ($\Delta T$) is:
$\Delta T = \frac{\text{Total Reduction}}{n} = \frac{40 \text{ mm}}{4} = 10 \text{ mm}$

Step 2: Apply the condition for unaided rolling.
For unaided rolling, the actual reduction per pass ($\Delta T$) must be less than or equal to the maximum possible reduction ($\Delta T_{max}$) that the rolls can achieve by friction alone. The formula relating $\Delta T_{max}$, roll radius ($R$), and coefficient of friction ($\mu$) is:
$\Delta T_{max} = R \cdot \mu^2$
To find the minimum coefficient of friction, we consider the threshold condition where $\Delta T = \Delta T_{max}$:
$\Delta T = R \cdot \mu^2$
Now, substitute the values we know:
$10 \text{ mm} = 400 \text{ mm} \cdot \mu^2$

Step 3: Solve for the coefficient of friction ($\mu$).
Rearrange the equation to find $\mu^2$:
$\mu^2 = \frac{10 \text{ mm}}{400 \text{ mm}} = \frac{1}{40} = 0.025$
Finally, take the square root to find $\mu$:
$\mu = \sqrt{0.025} \approx 0.1581$
The minimum coefficient of friction required is approximately $0.158$.

A mandrel is a tool used in metalworking to shape or support material, especially for hollow items like tubes. Let's analyze the options for where a moving mandrel is typically used:

• Wire Drawing: This process reduces wire diameter. While specialized cases might use mandrels (e.g., hollow wire), a moving mandrel is not a standard or defining feature of general wire drawing.
• Tube Drawing: This process reduces the diameter and/or wall thickness of a tube. Here, a mandrel is often used inside the tube to control the internal diameter and ensure uniform wall thickness. Types include fixed, plug, and floating mandrels. A floating mandrel is a key example of a moving mandrel, as it isn't fixed but moves freely with the tube, optimizing its position due to the forces involved. This significantly aids in achieving precise inner dimensions and smooth finishes.
• Metal Cutting: This involves material removal. Mandrels might hold workpieces, but they aren't directly involved in the cutting action itself, and a moving mandrel is not a typical tool here.
• Forging: This shapes metal using compressive forces. While mandrels can create holes in specific forging operations, a moving mandrel is not a characteristic technique of forging in general.

Therefore, the application where a moving mandrel is a significant and characteristic component is Tube drawing, where it is crucial for maintaining dimensional accuracy and surface quality of the tube's interior during the operation.

Let's break down these common mechanical mechanisms and where you'll find them in machine tools, which is a frequently tested concept in GATE.

P. Quick return $\rightarrow$ 2. Shaping: The quick return mechanism is a defining characteristic of shaping machines. Its purpose is to make the non-cutting return stroke of the ram much faster than the cutting stroke, which saves time and improves machining efficiency.

Q. Apron $\rightarrow$ 1. Lathe: You'll find the apron as part of the carriage on a lathe. This component houses the gears and controls that enable the automatic longitudinal and cross feeds, crucial for operations like turning and threading.

R. Intermittent indexing $\rightarrow$ 4. Milling: Intermittent indexing involves rotating the workpiece by a specific, fixed amount after each cutting pass. This is widely used in milling machines, often with a dividing head, to create evenly spaced features like gear teeth, splines, or holes.

S. Differential mechanism $\rightarrow$ 3. Gear hobbing: The differential mechanism is vital in gear hobbing, especially when cutting helical gears. It ensures the precise synchronization between the workpiece's rotation and the hob's rotation, allowing for the accurate generation of the helix angle.

Based on these associations:

• P matches with 2 (Shaping)
• Q matches with 1 (Lathe)
• R matches with 4 (Milling)
• S matches with 3 (Gear hobbing)

This combination leads to the correct mapping.

The Basic Length Unit (BLU) is the smallest possible increment of movement the machine can achieve. It's determined by the mechanical design. We are given that each angular step moves the table by one BLU. The angular step is determined by the stepper motor's resolution, not the pulse frequency. Therefore, the BLU is constant and does not depend on the pulse generator's frequency. Doubling the frequency of the pulse generator will only double the speed of the motor and thus the speed of the table movement, but the size of each step (BLU) remains unchanged. So, BLU will remain the same.

The tool moves a total distance of $50 \text{ mm}$ (calculated as $\sqrt{30^2 + 40^2}$) at a feed rate of $100 \text{ mm/min}$. We need to find the feed rate component along the X-axis. Since the tool covers $50 \text{ mm}$ in total while moving $30 \text{ mm}$ along the X-axis, the X-axis feed rate is proportionally calculated as:
$\text{Feed rate along X-axis} = \frac{100 \text{ mm/min}}{50 \text{ mm}} \times 30 \text{ mm} = 60 \text{ mm/min}$

In open die forging of a circular disc, the exponential decay of normal stress from the center to the periphery indicates how friction behaves. If the disc were frictionless, the normal stress would be uniform. If there was only sticking friction, the stress distribution would be different from exponential decay. The presence of exponential decay suggests that the material is always sliding relative to the platens, meaning the shear stress at the interface is always less than the shear strength required for sticking. This implies that there is no sticking friction anywhere on the flat face of the disc. Therefore, the correct answer is that there is no sticking friction anywhere on the flat face of the disc.

Reaming is a machining operation that uses a reamer to improve the quality of an existing hole, typically after drilling. Its primary purpose is to enlarge the hole slightly to achieve a precise, final dimension, improve the surface finish, and enhance the accuracy of the hole's diameter and roundness. Unlike drilling, which creates the initial hole, reaming removes a very small amount of material to refine the hole's precision and surface finish. Therefore, reaming is fundamentally a finishing process applied to existing holes to meet tight tolerances and quality standards.

The drilling time ($T_m$) for a through hole is calculated using the formula: $T_m = \frac{L + AP + OR + h}{f \times N}$, where $L$ is the workpiece thickness, $AP$ is approach, $OR$ is overtravel, and $h$ is the necessary approach (here, the cone height).
Given: $L = 50 \text{ mm}$, $h = 5 \text{ mm}$, $AP = 0$, $OR = 0$, cutting speed $V_c = 10 \text{ m/min}$, and feed rate $f = 0.3 \text{ mm/rev}$. Drill diameter $D = 20 \text{ mm}$.
First, calculate the drill's rotational speed ($N$) from the cutting speed:
$V_c = \frac{\pi \times D \times N}{1000}$ (since $V_c$ is in m/min and D in mm)
$N = \frac{1000 \times V_c}{\pi \times D} = \frac{1000 \times 10}{\pi \times 20} = 159.154 \text{ rpm}$.
Now, substitute the values into the drilling time formula (note: the original solution uses $N = \frac{60 \times V_c}{\pi \times D}$ and $V_c$ in m/min, so $D$ must be in meters for unit consistency. In the calculation, $V_c$ is multiplied by 1000, making it mm/min, so $D$ remains in mm):
$N = \frac{60 \times V_c}{\pi \times D} = \frac{60 \times 10 \times 1000}{\pi \times 60 \times 20} = 159.154 \text{ rpm}$.
Therefore, the drilling time is:
$T = \frac{50 + 5}{0.3 \times 159.154} = \frac{55}{47.7462} = 1.152 \text{ minutes}$.
Converting to seconds: $1.152 \times 60 = 69.11 \text{ seconds}$.

To produce metal powders, we need to break down bulk metal into fine particles. Atomization involves disintegrating molten metal using high-velocity fluid or centrifugal force, allowing the droplets to solidify into powder. Machining and grinding reduce solid metals into powder through mechanical processes like milling. Electrolysis is an electrochemical method where metal ions deposit onto a cathode and are then collected as powder. Compaction, on the other hand, is a process where already produced metal powder is pressed under high pressure into a solid shape (green compact). Therefore, compaction consolidates existing powder rather than producing it.

In casting, the solidification time ($T_s$) of molten metal in a riser is critical. Chvorinov's rule describes this, stating that $T_s$ is proportional to the square of the ratio of the riser's volume ($V$) to its surface area ($A$). This relationship is given by:
$T_s \propto \left(\frac{V}{A}\right)^2$
We are given that the riser's volume ($V$) remains constant, and we need to quadruple the solidification time ($T_s$). Let the initial state be subscript 1 and the final state be subscript 2.
$T_{s1} \propto \left(\frac{V}{A_1}\right)^2$
$T_{s2} \propto \left(\frac{V}{A_2}\right)^2$
We are given $T_{s2} = 4 \times T_{s1}$. Taking the ratio of final to initial solidification times:
$\frac{T_{s2}}{T_{s1}} = \frac{k \left(\frac{V}{A_2}\right)^2}{k \left(\frac{V}{A_1}\right)^2} = \left(\frac{A_1}{A_2}\right)^2$
Substituting $T_{s2} = 4 \times T_{s1}$ into the equation:
$4 = \left(\frac{A_1}{A_2}\right)^2$
Taking the square root of both sides gives:
$2 = \frac{A_1}{A_2}$
Rearranging this to solve for $A_2$:
$A_2 = \frac{A_1}{2}$
Therefore, to quadruple the solidification time, the surface area of the riser should be made half.

To size a GO gage for a plated component, we need to ensure it checks the maximum material condition of the part. This means the GO gage must correspond to the largest possible diameter of the pin after plating. For a cylindrical pin, the plating adds thickness to the surface. Since the pin has a circular cross-section, plating increases the diameter by twice the plating thickness (one thickness on each side).

1. Maximum Pin Diameter:
   The pin diameter is $25^{+0.020}_{+0.010}$ mm.
   Maximum Pin Diameter = Nominal Diameter + Upper Tolerance
   Maximum Pin Diameter = $25 + 0.020 = 25.020$ mm.

2. Maximum Plating Thickness:
   The plating thickness is $30^{\pm 2.0}$ micron.
   Maximum Plating Thickness = Nominal Thickness + Upper Tolerance
   Maximum Plating Thickness = $30 + 2.0 = 32$ micron.
   Converting to mm: $32 \times 0.001 = 0.032$ mm.

3. GO Gage Size Calculation:
   For a cylindrical component, the plating adds to the diameter on both sides. Therefore, the total increase in diameter due to plating is $2 \times$ Maximum Plating Thickness.
   GO Gage Size = Maximum Pin Diameter + $(2 \times$ Maximum Plating Thickness$)$
   GO Gage Size = $25.020 \text{ mm} + (2 \times 0.032 \text{ mm})$
   GO Gage Size = $25.020 \text{ mm} + 0.064 \text{ mm}$
   GO Gage Size = $25.084 \text{ mm}$.

The specific cutting energy ($u$) tells us how much energy is needed to remove a tiny amount (unit volume) of material during machining. We find it by dividing the cutting power ($P_c$) by the material removal rate (MRR): $u = \frac{P_c}{MRR}$.

First, let's gather our given parameters:

• Outer Diameter, $D_o = 200 \text{ mm}$
• Inner Diameter, $D_i = 80 \text{ mm}$
• Feed, $f = 0.1 \text{ mm/rev}$
• Depth of Cut, $d = 1 \text{ mm}$
• Cutting Speed, $v = 90 \text{ m/min}$
• Main Cutting Force, $F_c = 200 \text{ N}$

Step 1: Calculate Average Diameter ($D_{avg}$)
For these calculations, we use the average diameter:
$D_{avg} = \frac{D_o + D_i}{2} = \frac{200 \text{ mm} + 80 \text{ mm}}{2} = 140 \text{ mm} = 0.14 \text{ m}$

Step 2: Calculate Cutting Power ($P_c$)
Cutting power is the product of cutting force and cutting speed.
$P_c = F_c \times v = 200 \text{ N} \times 90 \text{ m/min} = 18000 \text{ N·m/min} = 18000 \text{ J/min}$

Step 3: Calculate Material Removal Rate (MRR)
The MRR can be calculated using the formula $MRR = f \times d \times v$. We need to ensure consistent units.
Convert feed and depth of cut to meters: $f = 0.1 \text{ mm/rev} = 0.0001 \text{ m/rev}$ and $d = 1 \text{ mm} = 0.001 \text{ m}$.
$MRR = (0.0001 \text{ m/rev}) \times (0.001 \text{ m}) \times (90 \text{ m/min}) = 0.000009 \text{ m}^3/\text{min}$
Now, convert MRR to $mm^3/min$:
$MRR = 0.000009 \text{ m}^3/\text{min} \times (10^9 \text{ mm}^3 / 1 \text{ m}^3) = 9000 \text{ mm}^3/\text{min}$

Step 4: Calculate Specific Cutting Energy ($u$)
Finally, we use the calculated cutting power and MRR:
$u = \frac{P_c}{MRR} = \frac{18000 \text{ J/min}}{9000 \text{ mm}^3/\text{min}} = 2 \text{ J/mm}^3$
The specific cutting energy is $2 \text{ J/mm}^3$.

The 3-2-1 principle in fixture design aims to constrain a workpiece's six degrees of freedom (three translational and three rotational) using strategically placed locators. The '3' in the 3-2-1 principle refers to the 3 locators placed on the primary datum face. These three locators establish the foundational positioning and restrict movement along the three axes perpendicular to this face (e.g., $X$, $Y$, $Z$ translation). The '2' refers to two locators on the secondary datum face, restricting two rotational movements, and the '1' refers to one locator on the tertiary datum face, restricting the final rotational movement.

This problem asks us to calculate the machining time for a facing operation on a CNC lathe with constant cutting speed.

Given Information:
Outer Diameter ($D_{outer}$) = 200 mm
Inner Diameter ($D_{inner}$) = 80 mm
Feed rate ($f$) = 0.1 mm/rev
Depth of cut ($d$) = 1 mm
Constant cutting speed ($V$) = 90 m/min
Tool approach and over-travel = 0 mm

First, convert all dimensions to consistent units (meters):
Outer Radius ($R$): $R = D_{outer} / 2 = 200 \text{ mm} / 2 = 100 \text{ mm} = 0.1 \text{ m}$
Inner Radius ($r$): $r = D_{inner} / 2 = 80 \text{ mm} / 2 = 40 \text{ mm} = 0.04 \text{ m}$
Feed rate ($f$): $f = 0.1 \text{ mm/rev} = 0.0001 \text{ m/rev}$
Cutting speed ($V$): $V = 90 \text{ m/min}$

For a facing operation with constant cutting speed, the machining time ($T_m$) is given by the integral formula:
$T_m = \frac{\pi (R^2 - r^2)}{f V}$
Note that the depth of cut (1 mm) determines the number of passes ($200 \text{ mm} / 1 \text{ mm} = 200$ passes, if it were radial, but here depth of cut does not affect the time calculation by this formula).

Substitute the values into the formula:
$T_m = \frac{\pi ((0.1 \text{ m})^2 - (0.04 \text{ m})^2)}{(0.0001 \text{ m/rev}) \times (90 \text{ m/min})}$
$T_m = \frac{\pi (0.01 \text{ m}^2 - 0.0016 \text{ m}^2)}{0.009 \text{ m}^2/\text{rev-min}}$
$T_m = \frac{\pi (0.0084)}{0.009} \text{ min}$
$T_m = \pi \times \frac{84}{90} \text{ min} = \pi \times \frac{14}{15} \text{ min}$
$T_m \approx 2.932 \text{ min}$
The calculated machining time is approximately 2.93 minutes.

In orthogonal turning, the main cutting force ($F_c$) acts tangentially to the workpiece. The friction force ($F_{friction}$) acts along the rake face of the cutting tool. The angle between the rake face direction and the direction of $F_c$ is $(90^\circ - \alpha)$, where $\alpha$ is the orthogonal rake angle. Since $F_{friction}$ acts along the rake face, the angle between $F_c$ and $F_{friction}$ is also $(90^\circ - \alpha)$. The problem states that $F_c$ is perpendicular to $F_{friction}$, meaning the angle between them is $90^\circ$.
$90^\circ - \alpha = 90^\circ$
Solving for $\alpha$:
$\alpha = 90^\circ - 90^\circ$
$\alpha = 0^\circ$
The orthogonal rake angle is $0^\circ$. Other parameters like diameter, feed, depth of cut, cutting velocity, and the magnitude of $F_c$ are not needed for this calculation.

We need to calculate the power of the water jet in a Water Jet Machining (WJM) process. We are given the orifice diameter ($d$), pressure ($P$), water density ($\rho$), and coefficient of discharge ($C_d$). Since losses are neglected, $C_d = 1.0$.

Given Data:

• Orifice Diameter, $d = 0.3 \text{ mm} = 0.3 \times 10^{-3} \text{ m}$
• Pressure, $P = 400 \text{ MPa} = 400 \times 10^6 \text{ Pa}$
• Density of water, $\rho = 1000 \text{ kg/m}^3$
• Coefficient of Discharge, $C_d = 1.0$

Step 1: Calculate the Jet Velocity ($v$)
The kinetic energy per unit volume of the jet is equal to the pressure energy. This gives us the velocity of the water jet.
$\frac{1}{2} \rho v^2 = P$
Rearranging for $v$:
$v = \sqrt{\frac{2P}{\rho}}$
Substitute the given values:
$v = \sqrt{\frac{2 \times (400 \times 10^6 \text{ Pa})}{1000 \text{ kg/m}^3}} = \sqrt{800 \times 10^3} \text{ m/s} = \sqrt{800000} \text{ m/s} \approx 894.43 \text{ m/s}$

Step 2: Calculate the Orifice Area ($A$)
The area of the circular orifice is calculated from its diameter.
$A = \frac{\pi d^2}{4}$
Substitute the orifice diameter:
$A = \frac{\pi (0.3 \times 10^{-3} \text{ m})^2}{4} = \frac{\pi \times 0.09 \times 10^{-6}}{4} \text{ m}^2 \approx 7.0686 \times 10^{-8} \text{ m}^2$

Step 3: Calculate the Volumetric Flow Rate ($Q$)
The volumetric flow rate is the product of the orifice area, jet velocity, and the coefficient of discharge.
$Q = C_d \times A \times v$
Substitute the calculated values:
$Q = 1.0 \times (7.0686 \times 10^{-8} \text{ m}^2) \times (894.43 \text{ m/s}) \approx 6.324 \times 10^{-5} \text{ m}^3/\text{s}$

Step 4: Calculate the Jet Power ($P_{jet}$)
The power of the water jet is the rate at which energy is delivered, which can be found by multiplying the pressure by the volumetric flow rate.
$P_{jet} = P \times Q$
Substitute the pressure and flow rate:
$P_{jet} = (400 \times 10^6 \text{ Pa}) \times (6.324 \times 10^{-5} \text{ m}^3/\text{s}) \approx 25296 \text{ W}$

Step 5: Convert Power to Kilowatts (kW)
To express the power in kilowatts, divide by 1000.
$P_{jet} \approx \frac{25296}{1000} \text{ kW} = 25.296 \text{ kW}$
Rounding to one decimal place, the power of the water jet is approximately $25.3 \text{ kW}$.

To determine the Material Removal Rate (MRR) in turning, we first need to ensure all units are consistent. The rotational speed ($N$) is given in revolutions per minute ($rpm$), so we convert it to revolutions per second ($rev/s$):
$N_{rps} = \frac{N_{rpm}}{60 \, s/min} = \frac{160 \, rpm}{60 \, s/min} = \frac{16}{6} \, rev/s = \frac{8}{3} \, rev/s$
The formula for MRR in turning is given by $MRR = \pi \times D \times d_c \times f \times N_{rps}$, where $D$ is the diameter, $d_c$ is the depth of cut, and $f$ is the feed. Substituting the given values:
$MRR = \pi \times (200 \, mm) \times (4 \, mm) \times (0.25 \, mm/rev) \times \left(\frac{8}{3} \, rev/s\right)$
$MRR = \pi \times 200 \times (4 \times 0.25) \times \frac{8}{3} \, mm^3/s$
$MRR = \pi \times 200 \times 1 \times \frac{8}{3} \, mm^3/s = \pi \times \frac{1600}{3} \, mm^3/s$
Calculating the numerical value, $MRR \approx 3.14159 \times \frac{1600}{3} \, mm^3/s \approx 1675.5 \, mm^3/s$.

Here's how to figure out when the carbide tool is better:

First, let's express tool life ($T$) for each tool in terms of cutting speed ($V$) using the given equations:
For the Carbide tool: $VT^{1.6} = 3000 \implies T_{carbide} = \left(\frac{3000}{V}\right)^{\frac{1}{1.6}}$
For the HSS tool: $VT^{0.6} = 200 \implies T_{HSS} = \left(\frac{200}{V}\right)^{\frac{1}{0.6}}$

We want to find the cutting speed ($V$) where the carbide tool gives a longer tool life, meaning $T_{carbide} > T_{HSS}$:
$\left(\frac{3000}{V}\right)^{\frac{1}{1.6}} > \left(\frac{200}{V}\right)^{\frac{1}{0.6}}$
To solve for $V$, we take the natural logarithm of both sides. This allows us to bring the exponents down:
$\frac{1}{1.6} \ln\left(\frac{3000}{V}\right) > \frac{1}{0.6} \ln\left(\frac{200}{V}\right)$
Substitute the decimal values for the exponents ($1/1.6 = 0.625$ and $1/0.6 \approx 1.667$) and expand the logarithms:
$0.625 (\ln(3000) - \ln(V)) > 1.667 (\ln(200) - \ln(V))$
Rearrange the terms to isolate $\ln(V)$:
$0.625 \ln(3000) - 0.625 \ln(V) > 1.667 \ln(200) - 1.667 \ln(V)$
$(1.667 - 0.625) \ln(V) > 1.667 \ln(200) - 0.625 \ln(3000)$
$1.042 \ln(V) > 1.667 \times 5.2983 - 0.625 \times 8.0064$
$1.042 \ln(V) > 8.837 - 5.004$
$1.042 \ln(V) > 3.833$
$\ln(V) > \frac{3.833}{1.042} \approx 3.678$
Finally, to find $V$, we exponentiate both sides:
$V > e^{3.678}$
$V > 39.55 \, \text{m/min}$
Therefore, the carbide tool provides higher tool life when the cutting speed exceeds approximately $39.55$ m/min, which is closest to option B.