Manufacturing Processes I Practice Questions

To find the effective power utilized for melting the workpiece, we first calculate the total electrical power input to the welding process. This is given by the product of voltage and current:
$P_{\in} = V \times I = 18 \text{ V} \times 160 \text{ A} = 2880 \text{ W}$
Next, we account for the heat loss during the creation of the arc, which is 40% of the input power:
$P_{arc\_loss} = 0.40 \times P_{\in} = 0.40 \times 2880 \text{ W} = 1152 \text{ W}$
The power available after this arc loss is then:
$P_{available} = P_{\in} - P_{arc\_loss} = 2880 \text{ W} - 1152 \text{ W} = 1728 \text{ W}$
Finally, we subtract the heat loss from the workpiece due to convection and radiation to find the effective power used for melting:
$P_{eff} = P_{available} - P_{workpiece\_loss} = 1728 \text{ W} - 800 \text{ W} = 928 \text{ W}$
Rounding to the nearest integer, the effective power utilized to melt the workpiece is $928 \text{ W}$.

To find the minimum punching force, we need to determine the shear force required to cut the material. This is calculated using the formula $F = \tau \times A$, where $F$ is the force, $\tau$ is the shear strength, and $A$ is the shear area.

First, let's calculate the shear area ($A$). Imagine the punch cutting through the sheet; the area being sheared is the cylindrical surface of the blank.
Diameter ($d$): $10 \, mm$
Thickness ($t$): $2 \, mm$
Circumference ($C$): $\pi \times d = \pi \times 10 \, mm = 10\pi \, mm$
Shear Area ($A$): $C \times t = (10\pi \, mm) \times (2 \, mm) = 20\pi \, mm^2$

Now, we can calculate the punching force ($F$).
Shear Strength ($\tau$): $80 \, MPa = 80 \, N/mm^2$
Force ($F$): $\tau \times A = (80 \, N/mm^2) \times (20\pi \, mm^2) = 1600\pi \, N$

Finally, we convert the force from Newtons to kilonewtons ($kN$).
Force ($F$) in $kN$: $\frac{1600\pi \, N}{1000 \, N/kN} = 1.6\pi \, kN$
Approximate Value: $1.6 \times \pi \approx 1.6 \times 3.14159 \approx 5.0265 \, kN$

Therefore, the minimum punching force required is approximately $5.03 \, kN$.

This problem asks us to find the solidification time for a heavier cube casting, given the solidification time of an original cube casting.

The key concept here is Chvorinov's Rule, which states that solidification time ($T_s$) is proportional to the square of the ratio of the casting's Volume ($V$) to its Surface Area ($A$). Mathematically, this is expressed as:
$T_s \propto \left(\frac{V}{A}\right)^2$
For a cube with side length $L$, the volume is $V = L^3$ and the surface area is $A = 6L^2$. Substituting these into the ratio:
$\frac{V}{A} = \frac{L^3}{6L^2} = \frac{L}{6}$
So, the solidification time for a cube is proportional to the square of its side length:
$T_s \propto \left(\frac{L}{6}\right)^2 \propto L^2$
Alternatively, we can express $L$ in terms of $V$ ($L = V^{1/3}$). Substituting this into the proportionality:
$T_s \propto (V^{1/3})^2 = V^{2/3}$
This means that solidification time is proportional to the volume raised to the power of $2/3$.

Now, let's apply this to the problem:
We are given that the new casting is 8 times heavier than the original. Since the material is the same, weight is directly proportional to volume.
So, if the original casting has volume $V_1$ and the new casting has volume $V_2$:
$W_2 = 8W_1 \Rightarrow V_2 = 8V_1$
Let $T_{s1}$ be the solidification time for the original casting and $T_{s2}$ for the new casting. Using the proportionality $T_s \propto V^{2/3}$:
$\frac{T_{s2}}{T_{s1}} = \left(\frac{V_2}{V_1}\right)^{2/3}$
We are given $T_{s1} = 5$ min and we found $V_2/V_1 = 8$. Substituting these values:
$\frac{T_{s2}}{5 \text{ min}} = \left(8\right)^{2/3}$
First, calculate $8^{2/3}$:
$8^{2/3} = (8^{1/3})^2 = (2)^2 = 4$
Now, substitute this back into the equation:
$\frac{T_{s2}}{5 \text{ min}} = 4$
Solving for $T_{s2}$:
$T_{s2} = 4 \times 5 \text{ min} = 20 \text{ min}$
Therefore, the solidification time for the new, heavier cube casting is 20 minutes.

To find the minimum number of passes in cold rolling, we first determine the maximum possible reduction per pass. The maximum reduction per pass ($\Delta h_{max}$) is given by $\Delta h_{max} \approx \mu^2 R$. Given the coefficient of friction $\mu = 0.1$ and roll radius $R = 300 \text{ mm}$ (from diameter $D = 600 \text{ mm}$), we calculate $\Delta h_{max} \approx (0.1)^2 \times 300 = 0.01 \times 300 = 3 \text{ mm}$. The total thickness reduction required is $\Delta h_{total} = 30 \text{ mm} - 10 \text{ mm} = 20 \text{ mm}$. The minimum number of passes ($n$) is then found by dividing the total reduction by the maximum reduction per pass and rounding up: $n = \left\lceil \frac{\Delta h_{total}}{\Delta h_{max}} \right\rceil = \left\lceil \frac{20}{3} \right\rceil = \left\lceil 6.667 \right\rceil = 7$.

This problem involves calculating the total true strain and final length of a bar after multiple drawing passes.

First, let's find the total true strain.
The true strain for a single pass ($\epsilon_{pass}$) with an area reduction $r = 0.35$ is given by:
$\epsilon_{pass} = \ln\left(\frac{1}{1-r}\right) = \ln\left(\frac{1}{1 - 0.35}\right) = \ln\left(\frac{1}{0.65}\right) \approx 0.43078$
Since there are $n=7$ identical drawing passes, the total true strain ($\epsilon_{total}$) is:
$\epsilon_{total} = n \times \epsilon_{pass} = 7 \times \ln\left(\frac{1}{0.65}\right) \approx 7 \times 0.43078 \approx 3.01546$
Rounding this value, the total true strain is approximately $3.02$.

Next, let's calculate the final length ($L_f$).
During drawing, the volume remains constant, which implies $A_0 L_0 = A_f L_f$. We can relate the final length to the initial length ($L_0 = 100$ mm) and the total true strain using the formula:
$L_f = L_0 \times e^{\epsilon_{total}}$
Substituting the values:
$L_f = 100 \text{ mm} \times e^{3.01546} \approx 100 \times 20.40 \approx 2040 \text{ mm}$
Alternatively, using the area reduction ratio:
$L_f = L_0 \times \left(\frac{1}{1-r}\right)^n = 100 \text{ mm} \times \left(\frac{1}{0.65}\right)^7 \approx 100 \times (1.53846)^7 \approx 100 \times 20.40 \approx 2040 \text{ mm}$
The closest option for the final length is $2043$ mm.

Therefore, the total true strain is $3.02$ and the final length is $2043$ mm.

To find the drawing force, we first calculate the initial and final cross-sectional areas. The initial diameter is $d_0 = 10 \text{ mm}$, so the initial area is $A_0 = \frac{\pi d_0^2}{4} = \frac{\pi (10 \text{ mm})^2}{4} = 25\pi \text{ mm}^2$. With a 35% reduction in area, the final area after the first pass is $A_1 = A_0 \times (1 - 0.35) = 0.65 \times 25\pi \text{ mm}^2 = 16.25\pi \text{ mm}^2$.

Next, we calculate the drawing stress ($\sigma_{draw}$) using the formula $\sigma_{draw} = Y_{avg} \times \ln\left(\frac{A_0}{A_1}\right)$. Since strain hardening is ignored, the average flow stress ($Y_{avg}$) equals the yield strength ($\sigma_y = 200 \text{ MPa}$).
First, find the area ratio: $\frac{A_0}{A_1} = \frac{A_0}{0.65 A_0} = \frac{1}{0.65} \approx 1.5385$.
Then, calculate its natural logarithm: $\ln(1.5385) \approx 0.4308$.
Now, the drawing stress is $\sigma_{draw} = 200 \text{ MPa} \times 0.4308 \approx 86.16 \text{ MPa}$.

Finally, the drawing force ($F$) is calculated as the product of drawing stress and the final cross-sectional area: $F = \sigma_{draw} \times A_1$.
$F = 86.16 \text{ MPa} \times 16.25\pi \text{ mm}^2 \approx 86.16 \text{ N/mm}^2 \times 51.05 \text{ mm}^2 \approx 4398.5 \text{ N}$.
Converting to kilonewtons, $F = \frac{4398.5 \text{ N}}{1000} \approx 4.399 \text{ kN}$.

In resistance seam welding, the electrodes are designed for continuous motion and to apply pressure and current along the joint. A circular disc is the standard electrode form because it rotates along the seam, ensuring consistent contact and heat application for a continuous weld. While a cylinder might offer rolling, it lacks the precise profile of a disc. Flat plates and coils of wire are unsuitable for achieving the continuous, uniform pressure and current distribution required for seam welding.

This problem asks us to compare the solidification times of a sphere and a cylinder with equal volumes in sand casting. The key concept here is Chvorinov's Rule, which states that solidification time ($T_s$) is proportional to the square of the volume-to-surface area ratio ($V/A$):

$T_s = K \left( \frac{V}{A} \right)^2$

Here, $K$ is a constant that depends on the mold material, metal properties, and casting conditions. Since the conditions are identical for both castings, $K$ will cancel out when we take the ratio.

Let's define the dimensions:
For the sphere: radius is $R_s$.
For the cylinder: radius is $R_c$, height is $H_c$. We are given that $H_c = D_c = 2R_c$.

First, we need to express the volumes and surface areas in terms of their respective radii.

1. Volumes:

• Sphere Volume ($V_s$): $V_s = \frac{4}{3}\pi R_s^3$
• Cylinder Volume ($V_c$): $V_c = \pi R_c^2 H_c = \pi R_c^2 (2R_c) = 2\pi R_c^3$

Since the volumes are equal ($V_s = V_c$):
$\frac{4}{3}\pi R_s^3 = 2\pi R_c^3$
$R_s^3 = \frac{3}{2} R_c^3$
Taking the cube root, we find the relationship between the radii:
$R_s = \left(\frac{3}{2}\right)^{1/3} R_c$

2. Surface Areas:

• Sphere Surface Area ($A_s$): $A_s = 4\pi R_s^2$
• Cylinder Surface Area ($A_c$): $A_c = 2\pi R_c^2 + 2\pi R_c H_c = 2\pi R_c^2 + 2\pi R_c (2R_c) = 6\pi R_c^2$

3. Calculate the $V/A$ ratio for each shape:

• For the sphere: $\frac{V_s}{A_s} = \frac{\frac{4}{3}\pi R_s^3}{4\pi R_s^2} = \frac{1}{3} R_s$
• For the cylinder: $\frac{V_c}{A_c} = \frac{2\pi R_c^3}{6\pi R_c^2} = \frac{1}{3} R_c$

4. Determine the ratio of solidification times:
Using Chvorinov's Rule:
$\frac{T_{s,sphere}}{T_{s,cylinder}} = \frac{K \left( \frac{V_s}{A_s} \right)^2}{K \left( \frac{V_c}{A_c} \right)^2} = \frac{\left( \frac{1}{3} R_s \right)^2}{\left( \frac{1}{3} R_c \right)^2} = \frac{R_s^2}{R_c^2}$
Now, substitute the relationship $R_s = (\frac{3}{2})^{1/3} R_c$ into this ratio:
$\frac{T_{s,sphere}}{T_{s,cylinder}} = \frac{\left[ \left(\frac{3}{2}\right)^{1/3} R_c \right]^2}{R_c^2} = \frac{\left(\frac{3}{2}\right)^{2/3} R_c^2}{R_c^2} = \left(\frac{3}{2}\right)^{2/3}$

5. Calculate the numerical value:
$\left(\frac{3}{2}\right)^{2/3} = (1.5)^{2/3} \approx 1.31037$
The ratio of the solidification time of the sphere to that of the cylinder is approximately 1.31.