Manufacturing Processes Ii Practice Questions

The control resolution represents the smallest detectable movement of the machine table per pulse.
Given the lead screw has 2 threads/cm, its pitch $P = \frac{1 \text{ cm}}{2 \text{ threads}} = \frac{10^4 \text{ microns}}{2 \text{ threads}} = 5 \times 10^3 \text{ microns/thread}$.
Since the lead screw is directly coupled to the servo motor, 1 revolution of the motor corresponds to 1 revolution of the lead screw.
The encoder emits 100 pulses/revolution.
Therefore, for 100 pulses (which is 1 revolution), the movement is $5 \times 10^3$ microns.
Hence, for 1 pulse, the movement (control resolution) is $\frac{5 \times 10^3 \text{ microns}}{100 \text{ pulses}} = 50 \text{ microns}$.

To find the mass flow rate ($\dot{m}$) of the waterjet, we first need to determine the velocity (V) of the water and the cross-sectional area (A) of the nozzle. The mass flow rate is given by $\dot{m} = \rho Q$, where $Q$ is the volumetric flow rate ($Q = AV$) and $\rho$ is the density of water.

The velocity (V) of the waterjet, neglecting losses, can be calculated using Bernoulli's principle, which simplifies to $V = \sqrt{\frac{2P}{\rho}}$.
Given: $P = 500 \text{ MPa} = 500 \times 10^6 \text{ Pa}$, $D = 0.25 \text{ mm} = 2.5 \times 10^{-4} \text{ m}$, and $\rho = 1000 \text{ kg/m}^3$.

Calculation steps:

1. Calculate the velocity (V):
   $V = \sqrt{\frac{2 \times 500 \times 10^6}{1000}} = 1000 \text{ m/s}$

2. Calculate the cross-sectional area (A):
   $A = \frac{\pi}{4} \times D^2 = \frac{\pi}{4} \times (2.5 \times 10^{-4})^2$

3. Calculate the volumetric flow rate (Q):
   $Q = A \times V = \frac{\pi}{4} \times (2.5 \times 10^{-4})^2 \times 1000 = 4.9087 \times 10^{-5} \text{ m}^3/\text{s}$

4. Calculate the mass flow rate ($\dot{m}$) in kg/s:
   $\dot{m} = \rho Q = 1000 \times 4.9087 \times 10^{-5} \text{ kg/s}$

5. Convert the mass flow rate to kg/min:
   $\dot{m} = 1000 \times 4.9087 \times 10^{-5} \times 60 \text{ kg/min} = 2.94 \text{ kg/min}$

To find the new coordinates after translation, we add the translation amounts to the original coordinates. For a translation of $a$ units along the x-axis and $b$ units along the y-axis, a point $(x, y)$ transforms to $(x', y')$ where $x' = x + a$ and $y' = y + b$.

Here, the rectangle is translated 6 units in the positive x-direction ($a = 6$) and 3 units in the positive y-direction ($b = 3$). Applying this rule to each vertex:

• For P(-4, -2): P' = $(-4 + 6, -2 + 3)$ = $(2, 1)$
• For Q(-2, -3): Q' = $(-2 + 6, -3 + 3)$ = $(4, 0)$
• For R(-3, -5): R' = $(-3 + 6, -5 + 3)$ = $(3, -2)$
• For S(-5, -4): S' = $(-5 + 6, -4 + 3)$ = $(1, -1)$

The new coordinates of the vertices are P'(2, 1), Q'(4, 0), R'(3, -2), and S'(1, -1).

To find the actual machining time, we use the formula: ${{T}_{m}} = \frac{{L} + {{A}_{p}}}{{{f} {{N}}}}$, where $T_m$ is machining time, $L$ is the depth of the hole, $A_p$ is the approach length, $f$ is the feed, and $N$ is the spindle speed in RPM.

Given: $L = 100$ mm, $A_p = 2$ mm, $f = 0.1$ mm/rev, and cutting speed $V = 100$ m/min for a diameter $D = 25$ mm.

First, calculate the spindle speed $N$ using the cutting speed formula:
$V = \frac{{\pi}\times {D}\times {N}}{1000}$
$100 = \frac{{\pi} \times {25} \times {N}}{1000}$
$N = 1273.239$ rpm

Now, substitute these values into the machining time formula:
${{T}_{m}} = \frac{{100} + {2}}{{0.1} \times {1273.239}}$
${T}_{m} = 0.80$ minutes.

The heat generated in resistance spot welding is given by the formula:
$H_g = I^2 \times R \times t$
where $H_g$ is the heat generated, $I$ is the current, $R$ is the electrical resistance, and $t$ is the time for which the current is applied.

Given for the first operation: $H_{g1} = 2000 \text{ J}$. We know $H_{g1} = I_1^2 \times R_1 \times t_1 = 2000 \text{ J}$.
For the second operation: $R_2 = R_1$, $I_2 = 1.25 I_1$ (current increased by 25%), and $t_2 = \frac{t_1}{2}$ (time reduced to half).

Now, let's calculate the heat generated in the second operation, $H_{g2}$:
$H_{g2} = I_2^2 \times R_2 \times t_2$
Substitute the values for $I_2$, $R_2$, and $t_2$:
$H_{g2} = (1.25 I_1)^2 \times R_1 \times \frac{t_1}{2}$
$H_{g2} = \frac{(1.25)^2}{2} \times (I_1^2 \times R_1 \times t_1)$
Since $I_1^2 \times R_1 \times t_1 = 2000 \text{ J}$:
$H_{g2} = \frac{(1.25)^2}{2} \times 2000$
$H_{g2} = \frac{1.5625}{2} \times 2000$
$H_{g2} = 0.78125 \times 2000$
$H_{g2} = 1562.5 \text{ J}$

To find the force required for extrusion, we first calculate the total extrusion pressure ($P_e$) and then multiply it by the initial cross-sectional area ($A_0$).

Concept:
Extrusion force = $P_e \times A_0$
The extrusion pressure is given by $p_e=\sigma_m\left[ 0.8+1.2 \ln\left( \frac{A_0}{A_f}\right)\right]$, where $\sigma_m$ is the mean flow stress, $A_0$ is the initial cross-sectional area, and $A_f$ is the final cross-sectional area.

Calculation:
Given: Diameter of cylindrical billet ($d_0$) = 90 mm, $\sigma_m$ = 80 MPa.
Initial cross-sectional area: $A_0 = \frac{\pi}{4} \times {d_0}^2 = \frac{\pi}{4} \times {90}^2 = 6361.73 \text{ mm}^2$.
Final cross-sectional area (for the I-section): $A_f = 50 \times 5 + 5 \times 60 + 50 \times 5 = 800 \text{ mm}^2$.
Area ratio: $\frac{A_0}{A_f} = \frac{6361.73}{800} = 7.9522$.
Now, calculate the extrusion pressure: $P_e = 80 \left[ 0.8 + 1.2 \ln\left(7.9522\right)\right] = 263.05 \text{ MPa}$.
Finally, the extrusion force = $P_e \times A_0 = 263.05 \times 6361.73 = 1673461.08 \text{ N}$.
Converting to kN: Extrusion force = $1673.46 \text{ kN}$.

The question asks for the process used to produce long bars of Fiber Reinforced Plastics (FRP) with a uniform cross-section. This scenario specifically points to Pultrusion.

Pultrusion is a continuous manufacturing process ideal for creating FRP profiles with a constant cross-section. Here's how it works:

1. Fiber Impregnation: Reinforcing fibers (like glass or carbon) are pulled from spools and passed through a liquid resin bath (typically thermosetting resins like epoxy or polyester) to become fully impregnated.
2. Die Shaping and Curing: The saturated fiber bundle is then pulled through a heated die that shapes it into the desired cross-section (e.g., a round bar) and simultaneously cures (hardens) the resin.
3. Pulling and Cutting: A pulling mechanism continuously draws the cured profile through the die, ensuring uniformity. Once the desired length is achieved, a cutting mechanism cuts the continuous bar.

This method is specifically designed for continuous production of long, consistent FRP products.

Let's quickly see why other options are not suitable:

• Extrusion: Primarily used for thermoplastics, not for continuously pulling fiber-reinforced thermoset composites.
• Injection Molding: Best for complex shapes and high volumes of discrete parts, not long, continuous bars.
• Thermoforming: Used for shaping single plastic sheets into parts, not for continuous FRP bar production.

Therefore, Pultrusion is the correct process for producing long FRP bars with a uniform cross-section.

In the spinning process, the total developed length of the initial blank metal remains constant. This means the length of the blank's outer edge will form the entire profile of the spun part.

From the figure, we have:
$\sin 30^\circ = \frac{45}{\text{Hypotenuse}}$
Since $\sin 30^\circ = 0.5$, we get:
$\text{Hypotenuse} = \frac{45}{0.5} = 45 \times 2 = 90 \text{ mm}$
The total initial blank diameter ($D$) is calculated by summing the two slanted lengths (hypotenuses) and the flat bottom section:
$D = (2 \times \text{Hypotenuse}) + 30 \text{ mm}$
$D = (2 \times 90) + 30 = 180 + 30 = 210 \text{ mm}$

This question asks for the difference between the effective diameter (E) and the dimension under the wire (T) for an M 10 x 1.0 mm thread, using the Best-Wire method. We are given the thread's nominal major diameter ($D$) as 10 mm, pitch ($P$) as 1.0 mm, and included thread angle ($\theta$) as 60°.

First, we calculate the best wire diameter ($w$) using the formula:
$w = \frac{P}{2 \cos(\theta/2)}$
Given $P = 1.0$ mm and $\theta/2 = 60^\circ / 2 = 30^\circ$, we substitute these values:
$w = \frac{1.0}{2 \times \cos(30^\circ)} = \frac{1.0}{2 \times \frac{\sqrt{3}}{2}} = \frac{1.0}{\sqrt{3}} \approx 0.57735 \text{ mm}$
The difference between the effective diameter (E) and the dimension under the wire (T) is given by $E - T = \frac{w}{2}$.
$E - T = \frac{0.57735}{2} \approx 0.288675 \text{ mm}$
Rounding to three decimal places, the difference is $0.289$ mm.

To determine the feed force ($F_t$), we first need to calculate the shear force ($F_s$).
The shear force ($F_s$) is given by the formula: $F_s = \frac{t_0 b}{\sin \phi} \times \tau_s$
Given values are: Rake angle ($\alpha$) = $-5^\circ$, friction angle ($\beta$) = $49.2^\circ$, shear angle ($\phi$) = $25.4^\circ$, shear strength ($\tau_s$) = $220 \text{ MPa}$.
The feed ($f$) is the uncut chip thickness ($t_0$) = $0.2 \text{ mm/rev}$, and the depth of cut ($d$) is the width of chip ($b$) = $1 \text{ mm}$.
Substituting these values: $F_s = \frac{0.2 \times 1}{\sin 25.4^\circ} \times 220 = 102.58 \text{ N}$
Now, we can calculate the feed force ($F_t$) using the formula: $F_t = \frac{F_s \sin (\beta - \alpha)}{\cos (\phi + \beta - \alpha)}$
Plugging in the values: $F_t = \frac{102.58 \times \sin (49.2^\circ - (-5^\circ))}{\cos (25.4^\circ + 49.2^\circ - (-5^\circ))} = \frac{102.58 \times \sin (54.2^\circ)}{\cos (79.6^\circ)}$
Therefore, $F_t = 460.88 \text{ N}$.

In Abrasive Jet Machining (AJM), we define the Mixing Ratio (MR) as the ratio of abrasive volume to carrier gas volume, given by $MR = \frac{V_a}{V_g}$. We also need to find the Mass Ratio ($\alpha$), which is the mass of abrasive particles to the total mass of the mixture (abrasive + carrier gas):
$\alpha = \frac{M_a}{M_{a+g}} = \frac{M_a}{M_a + M_g}$
We know that mass ($M$) is density ($\rho$) multiplied by volume ($V$), so $M_a = \rho_a V_a$ and $M_g = \rho_g V_g$. Substituting these into the formula for $\alpha$:
$\alpha = \frac{\rho_a V_a}{\rho_a V_a + \rho_g V_g}$
To simplify calculations, it's often easier to work with the reciprocal $\frac{1}{\alpha}$:
$\frac{1}{\alpha} = \frac{\rho_a V_a + \rho_g V_g}{\rho_a V_a} = 1 + \frac{\rho_g V_g}{\rho_a V_a} = 1 + \left(\frac{\rho_g}{\rho_a}\right) \times \left(\frac{V_g}{V_a}\right)$
Given: $\frac{\rho_a}{\rho_g} = 25$ (so $\frac{\rho_g}{\rho_a} = \frac{1}{25}$) and $\frac{V_a}{V_g} = 0.25$ (so $\frac{V_g}{V_a} = \frac{1}{0.25}$).
Substituting these values:
$\frac{1}{\alpha} = 1 + \left(\frac{1}{25} \times \frac{1}{0.25}\right) = 1 + \left(0.04 \times 4\right) = 1 + 0.16 = 1.16$
Therefore, $\alpha = \frac{1}{1.16} \approx 0.862$. Rounding off to two decimal places, the mass ratio is $0.86$.

This problem asks us to calculate the time taken to drill a through hole. We'll first find the drill's rotational speed, then determine the total depth the drill needs to travel, and finally calculate the drilling time using the feed rate.

First, let's find the rotational speed ($N_{rev/sec}$) of the drill using the cutting speed ($V$) and drill diameter ($D$):
$N_{rev/sec} = \frac{V}{\pi D}$
Given $V = 333.33 \text{ mm/s}$ and $D = 20 \text{ mm}$:
$N_{rev/sec} = \frac{333.33 \text{ mm/s}}{\pi \times 20 \text{ mm}} \approx \frac{333.33}{62.8318} \approx 5.3051 \text{ rev/s}$
Next, we calculate the total length ($L_{total}$) the drill must travel. This includes the plate thickness ($T$) and the extra penetration required for the drill's conical point ($L_{point}$). The drill point's axial length is given by:
$L_{point} = \frac{r}{\tan(\alpha/2)}$
Given drill radius ($r = D/2 = 10 \text{ mm}$) and half point angle ($\alpha/2 = 118\degree/2 = 59\degree$):
$L_{point} = \frac{10 \text{ mm}}{\tan(59\degree)} \approx \frac{10 \text{ mm}}{1.66428} \approx 6.0088 \text{ mm}$
The total drilling length is:
$L_{total} = T + L_{point} = 100 \text{ mm} + 6.0088 \text{ mm} \approx 106.0088 \text{ mm}$
Finally, we calculate the drilling time ($T_{drill}$). First, find the axial feed rate ($F_{axial}$) by multiplying the feed rate ($f$) by the rotational speed:
$F_{axial} = f \times N_{rev/sec} = 0.22 \text{ mm/rev} \times 5.3051 \text{ rev/s} \approx 1.1671 \text{ mm/s}$
Then, the drilling time is:
$T_{drill} = \frac{L_{total}}{F_{axial}}$
$T_{drill} = \frac{106.0088 \text{ mm}}{1.1671 \text{ mm/s}} \approx 90.828 \text{ seconds}$
The calculated drilling time is approximately 90.83 seconds, which is closest to 90 seconds.

Friction welding is a solid-state joining process that uses heat generated by friction and pressure to bond materials without melting them.

Let's evaluate each option:

• A. Heat affected zone is not formed: Incorrect. Significant heat is generated, forming a Heat Affected Zone (HAZ) where material properties change due to thermal cycles.
• B. Flashes are not produced: Generally incorrect. In most friction welding techniques, material is expelled as 'flash' due to pressure.
• C. Dissimilar materials cannot be joined: Incorrect. Friction welding is effective for joining various dissimilar materials (e.g., aluminum to steel).
• D. Melting of the base material(s) is not involved: Correct. The process relies on plastic deformation and material softening below the melting point, making it a solid-state process.

For continuous insulation coating on electrical wires, Extrusion is the most suitable process. In extrusion, molten thermoplastic material (like PVC or polyethylene) is forced through a die while the electrical wire passes through its center. This forms a uniform, continuous insulation layer along the wire, ideal for its continuous length and allowing precise control over thickness. Other processes like Injection molding create discrete parts, Blow molding makes hollow objects, and Deep drawing is a metal forming process, none of which are suitable for continuous wire coating.

This problem asks us to find the minimum rotational speed (in rpm) for a true centrifugal casting process. The core idea is that the centrifugal force must be strong enough to hold the molten metal against the mold wall, which is quantified by the G-factor.

The G-factor ($G$) is the ratio of centrifugal acceleration ($a_c$) to the acceleration due to gravity ($g$):
$G = \frac{a_c}{g}$
The centrifugal acceleration at the outer surface (which determines the minimum speed required) is $a_c = \omega^2 r$, where $\omega$ is the angular velocity and $r$ is the radius. Here, the outer diameter is $D_o = 0.275$ m, so the radius $r = D_o/2 = 0.275/2$ m.
Thus, we have:
$G = \frac{\omega^2 (D_o/2)}{g}$
We need to find the rotational speed $N$ in rpm. The relationship between angular velocity $\omega$ (in rad/s) and rotational speed $N$ (in rpm) is $\omega = \frac{2 \pi N}{60}$.
Substituting $\omega$ into the G-factor equation:
$G = \frac{\left(\frac{2 \pi N}{60}\right)^2 (D_o/2)}{g}$
Rearranging to solve for $N$:
$N = \frac{30}{\pi} \sqrt{\frac{2 G g}{D_o}}$
Given $G = 65$, $g = 9.8$ m/s$^2$, and $D_o = 0.275$ m:
$N = \frac{30}{\pi} \sqrt{\frac{2 \times 65 \times 9.8}{0.275}}$
$N = \frac{30}{\pi} \sqrt{\frac{1274}{0.275}}$
$N = \frac{30}{\pi} \sqrt{4632.727...}$
$N \approx 9.5493 \times 68.064$
$N \approx 649.75 \text{ rpm}$
Comparing this to the options, the closest value is 650 rpm.

The solidification time ($t_s$) in sand casting is described by Chvorinov's Rule: $t_s = C{\left( {\frac{V}{A}} \right)^n}$, where $C$ is the mold constant, $V$ is the volume of the casting, $A$ is the surface area of the casting, and $n$ is the solidification exponent (given as 2).

Given that $n=2$, the formula becomes $t_s = C{\left( {\frac{V}{A}} \right)^2}$.
For a given unit volume of material, $V$ is constant. The mold constant $C$ is also constant.
Therefore, $t_s \propto \left(\frac{1}{A}\right)^2$, which implies $A^2 \propto \frac{1}{t_s}$, or $A \propto \frac{1}{\sqrt{t_s}}$.

We are given that the solidification time is to be doubled, so $t_2 = 2t_1$.
Using the proportionality $A \propto \frac{1}{\sqrt{t_s}}$:
$\frac{A_2}{A_1} = \frac{\sqrt{t_1}}{\sqrt{t_2}} = \frac{\sqrt{t_1}}{\sqrt{2t_1}} = \frac{1}{\sqrt{2}} \approx 0.707$.

The reduction in cast surface area is calculated as:
$\frac{A_2 - A_1}{A_1} = \frac{A_2}{A_1} - 1 = 0.707 - 1 = -0.2928$.
The negative sign indicates a reduction.
Therefore, the percentage reduction in the cast surface area is $0.2928 \times 100 = 29.28\% \approx 29.3\%$.

This problem involves understanding G-codes for NC milling. The trajectory shown is a circular arc. First, we need to identify the direction of the arc from the start point to the end point. Observing the figure (which is implied to be given in the question context), the trajectory of the point is in the clockwise direction. For clockwise circular interpolation, the specific G-code used is G02.

Next, we need to determine the coordinates of the endpoint of this arc and its radius. According to the problem setup, the final point (endpoint) has coordinates X = 120 and Y = 60. The radius of this circular arc is given as R = 60.

In absolute programming, the block for a circular interpolation specifies the G-code, the absolute X and Y coordinates of the endpoint, and the radius. Combining these, the corresponding NC program block will be:

$G02 \text{ X 120.0 Y 60.0 R 60.0}$

For reference:

• G00: Rapid traverse (point-to-point positioning).
• G01: Linear interpolation (straight cut at cutting feed rate).
• G02: Circular Interpolation - Clockwise (CW).
• G03: Circular Interpolation - Counter-Clockwise (CCW).

The melting efficiency ($\eta_m$) in welding is the ratio of the heat required to melt the material ($H_m$) to the total heat supplied ($H_s$).
${\eta _m} = \frac{{{H_m}}}{{{H_s}}}$
The heat supplied per unit volume ($H_s$) can be calculated using the formula:
${H_s} = \frac{{VI}}{{{A} \times v}} \times {\eta _h}\;\;J/m{m^3}$
Given: $H_m = 10 \text{ J/mm}^3$, Arc Voltage $V = 20 \text{ V}$, Current $I = 250 \text{ A}$, Arc efficiency $\eta_h = 0.90$, Travel speed $v = 5 \text{ mm/s}$, and Weld bead area $A = 30 \text{ mm}^2$.
Substitute these values into the formula for $H_s$:
${H_s} = \frac{{20~\times ~250}}{{{30} ~\times ~ 5}} \times {0.90} = 30 \text{ J/mm}^3$
Now, calculate the melting efficiency $\eta_m$:
${\eta _m} = \frac{{{H_m}}}{{{H_s}}}~=~\frac{10}{30} = 0.3333$
Therefore, the melting efficiency is $33.33\%$.