Manufacturing Processes I Practice Questions

Here's how to calculate the welding speed:

First, determine the actual power used for welding. With a $6 \text{ kW}$ power source and $50\%$ energy loss, the effective power is $6 \text{ kW} \times (1 - 0.50) = 3 \text{ kW}$, which converts to $3000 \text{ J/s}$. Next, calculate the energy required to melt a $1 \text{ mm}$ length of weld. Given the specific energy of $15 \text{ J/mm}^3$ and a weld cross-section area of $10 \text{ mm}^2$, this energy is $15 \text{ J/mm}^3 \times 10 \text{ mm}^2 = 150 \text{ J/mm}$. Finally, the welding speed is found by dividing the effective power by the energy per unit length: $(3000 \text{ J/s}) / (150 \text{ J/mm}) = 20 \text{ mm/s}$.

To find the minimum blanking force, we use the formula $F = \tau \times A$, where $F$ is the blanking force, $\tau$ is the shear strength, and $A$ is the shear area.

First, calculate the perimeter of the square:
$P = 4 \times \text{side length} = 4 \times 50 \text{ mm} = 200 \text{ mm}$

Next, determine the shear area, which is the perimeter multiplied by the sheet thickness:
$A = P \times \text{thickness} = 200 \text{ mm} \times 2 \text{ mm} = 400 \text{ mm}^2$

Now, substitute the values into the blanking force formula. Remember that $200 \text{ MPa}$ is equivalent to $200 \text{ N/mm}^2$:
$F = 200 \text{ N/mm}^2 \times 400 \text{ mm}^2 = 80000 \text{ N}$

Finally, express the force in the requested format, $\times 10^3 \text{ N}$:
$F = 80 \times 10^3 \text{ N}$

Let's break down the properties of filler material crucial for successful brazing.

• Melting Temperature: For brazing, the filler material must melt and flow into the joint, but the base metals should remain solid. This is why the melting temperature of the filler material must be less than that of the base metal. So, statement A is TRUE.

• Wettability: To achieve good capillary action and allow the molten filler to spread effectively into the narrow joint gap, the filler material needs to "wet" the base metal surfaces well. This means it should have high wettability, leading to a low contact angle. Statement B says wettability must be low, which is incorrect. Therefore, statement B is FALSE.

• Viscosity: For the molten filler to flow easily into the tight interface by capillary action and fill the joint completely, its viscosity needs to be low. High viscosity would hinder this flow. So, statement C is TRUE.

• Chemical Reactivity: Excessive chemical reaction between the filler material and the base metal can form brittle intermetallic compounds, which weaken the final joint. Therefore, chemical reactivity between them must be low. So, statement D is TRUE.

The only statement that is false regarding filler material properties in brazing is B.

To determine feasible sheet metal drawing, we must satisfy two conditions: the reduction ratio ($r$) must be less than $0.5$, and the thickness to diameter ratio ($t/d$) must be greater than $1\%$.

Given: Sheet thickness $t = 2$ mm, punch diameter $d_p = 100$ mm.

1. Reduction Ratio Condition:
   The reduction ratio is defined as $r = \frac{D_b - d_p}{D_b}$.
   The condition $r < 0.5$ implies:
   $\frac{D_b - d_p}{D_b} < 0.5$
   We can rewrite this as $1 - \frac{d_p}{D_b} < 0.5$.
   Rearranging the terms, we get $0.5 < \frac{d_p}{D_b}$.
   Inverting both sides (and flipping the inequality sign) gives $D_b < \frac{d_p}{0.5}$, which simplifies to $D_b < 2 d_p$.
   Substituting $d_p = 100$ mm, we find $D_b < 2 \times 100 \text{ mm}$, so $D_b < 200 \text{ mm}$.

2. Thickness to Diameter Ratio Condition:
   The condition is $t/d > 0.01$. Assuming 'd' refers to the punch diameter $d_p$:
   $\frac{t}{d_p} = \frac{2 \text{ mm}}{100 \text{ mm}} = 0.02$
   Since $0.02 > 0.01$, this condition is satisfied by the given parameters and does not further restrict $D_b$.

Combining these, the primary constraint for feasible drawing is $D_b < 200$ mm.
Let's check the given options:
A. $D_b = 150$ mm: $150 < 200$, which is feasible.
B. $D_b = 250$ mm: $250 \not< 200$, not feasible.
C. $D_b = 350$ mm: $350 \not< 200$, not feasible.
D. $D_b = 450$ mm: $450 \not< 200$, not feasible.

Therefore, only a blank diameter of $150$ mm results in feasible drawing.

To achieve the desired thickness reduction in a single pass of a rolling operation, the maximum possible reduction ($\Delta t_{max}$) must be at least equal to the required reduction ($\Delta t$). The maximum possible reduction is limited by the roll radius ($R$) and the coefficient of friction ($\mu$) at the roll-strip interface, given by the formula:
$\Delta t_{max} \approx R \mu^2$
Given: Initial thickness $t_0 = 23 \text{ mm}$, final thickness $t_f = 20 \text{ mm}$, and roll diameter $D = 200 \text{ mm}$, so $R = D/2 = 100 \text{ mm}$.
The required reduction is $\Delta t = t_0 - t_f = 23 \text{ mm} - 20 \text{ mm} = 3 \text{ mm}$.
Now, we calculate $\Delta t_{max}$ for each lubricant and check if it's $\ge 3 \text{ mm}$:

• Lubricant P ($\mu = 0.05$): $\Delta t_{max} = 100 \times (0.05)^2 = 0.25 \text{ mm}$. Since $0.25 < 3$, it's not suitable.
• Lubricant Q ($\mu = 0.1$): $\Delta t_{max} = 100 \times (0.1)^2 = 1.0 \text{ mm}$. Since $1.0 < 3$, it's not suitable.
• Lubricant R ($\mu = 0.2$): $\Delta t_{max} = 100 \times (0.2)^2 = 4.0 \text{ mm}$. Since $4.0 \ge 3$, it is suitable.
• Lubricant S ($\mu = 0.25$): $\Delta t_{max} = 100 \times (0.25)^2 = 6.25 \text{ mm}$. Since $6.25 \ge 3$, it is suitable.
  Therefore, lubricants R and S will enable the rolling process to achieve the desired final thickness.

Let's break down each deep drawing defect and its primary cause, as you'd need to know for your GATE exam.

P: Orange Peel on the surface of cup
Orange peel is a surface roughness that makes the cup look like an orange peel. This defect is directly linked to the internal structure of the material.

• Reason Match: 3 - Large grain size in the blank material. When the material has larger grains, the surface appears rougher after the deformation process of deep drawing.

Q: Wrinkling at the flange of cup
Wrinkling appears as folds or ripples, usually on the flange (the flat, untransformed part) of the cup. This happens because of compressive stresses acting on the material in this region.

• Reason Match: 1 - No blank holding force. If there isn't enough force holding the blank (sheet metal) down, the material in the flange can buckle under compression and move inwards, leading to wrinkles.

R: Tearing at the bottom corner of cup
Tearing, especially at the sharp corner where the punch starts to form the cup, means the material has stretched beyond its limit and fractured. This occurs due to very high localized stresses.

• Reason Match: 2 - Very small corner radius of the punch. A sharp punch corner creates a severe stress concentration point. This high stress can exceed the material's strength, causing it to tear at the base of the cup.

S: Earring at the top edge of the cup
Earring is when the top edge of the drawn cup forms an uneven, wavy shape, resembling "ears." This defect points to differences in the material's properties depending on the direction.

• Reason Match: 4 - Anisotropy of the blank material. Anisotropy means the material's properties (like how easily it deforms or strengthens) are not uniform in all directions. This directional variation causes uneven stretching during deep drawing, resulting in the formation of ears.

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

This problem asks us to find the welding current for a second weld joint such that its heat input is the same as a first weld joint, given changes in welding speed. The heat input ($H$) per unit length in arc welding is calculated using the formula:

$H = \frac{\eta \times V \times I}{S}$

Here, $\eta$ is the heat transfer efficiency (given as $0.85$), $V$ is the arc voltage, $I$ is the welding current, and $S$ is the welding speed.

We are given the following parameters:
For Weld 1: $V_1 = 30$ V, $I_1 = 180$ A, $S_1 = 6$ mm/s.
For Weld 2: $V_2 = 30$ V, $I_2 = ?$ A, $S_2 = 8$ mm/s.

The problem states that both welds have the same heat input, so $H_1 = H_2$. Also, the efficiency ($\eta$) and arc voltage ($V$) are the same for both welds. Therefore, we can set up the equation:

$\frac{\eta \times V_1 \times I_1}{S_1} = \frac{\eta \times V_2 \times I_2}{S_2}$

Since $\eta$ and $V$ are constant for both welds ($V_1 = V_2 = 30$ V), they cancel out, simplifying the equation to:

$\frac{I_1}{S_1} = \frac{I_2}{S_2}$

Now, we need to solve for $I_2$:

$I_2 = I_1 \times \frac{S_2}{S_1}$

Substitute the given values:

$I_2 = 180 \text{ A} \times \frac{8 \text{ mm/s}}{6 \text{ mm/s}}$
$I_2 = 180 \times \frac{4}{3}$
$I_2 = 240 \text{ A}$

Thus, the welding current for the second weld joint is $240$ A.

In a blanking operation, we want to cut a specific-sized "blank" (the useful part), so the die cavity determines the blank's dimensions. The punch needs to be slightly smaller to account for the clearance. For blanking, the punch diameter ($d_p$) is calculated using the blank diameter ($D$) and radial clearance ($c$) as: $d_p = D - 2 \times c$.

Given:
Blank Diameter ($D$) = 100 mm
Sheet Thickness ($t$) = 2 mm
Radial Clearance = 6% of sheet thickness

First, calculate the radial clearance ($c$):
$c = 0.06 \times t = 0.06 \times 2 \, \text{mm} = 0.12 \, \text{mm}$

Now, substitute the values into the punch diameter formula:
$d_p = 100 \, \text{mm} - 2 \times 0.12 \, \text{mm} = 100 \, \text{mm} - 0.24 \, \text{mm} = 99.76 \, \text{mm}$

Thus, the diameter of the punch is $99.76 \, \text{mm}$.

This question tests our understanding of mild steel's behavior under uniaxial tension, a fundamental concept for GATE. Let's break down each option:

Option A: The engineering stress-strain curve is linear within the elastic limit.
While the initial part of the curve (up to the proportional limit) is indeed linear, there can be slight non-linearity between the proportional limit and the yield point, even though this region is still elastic. So, stating it's always linear throughout the entire elastic limit isn't strictly true in all precise contexts.

Option B: The specimen fails in cup and cone type fracture.
Mild steel is a classic example of a ductile material. When ductile materials undergo necking (a localized reduction in cross-sectional area), they typically fracture in a "cup and cone" manner. This means one part of the fracture surface forms a cup, and the other forms a cone. This statement is TRUE.

Option C: The true stress is always more than the engineering stress at any finite strain.
Engineering stress ($\sigma_e$) is calculated using the original cross-sectional area ($A_0$), while true stress ($\sigma_t$) uses the instantaneous area ($A$).
$\sigma_t = \frac{P}{A} \quad \text{and} \quad \sigma_e = \frac{P}{A_0}$
During tensile testing, especially beyond the elastic limit, the cross-sectional area reduces ($A < A_0$). Therefore, for the same load $P$, $\frac{P}{A} > \frac{P}{A_0}$, implying $\sigma_t > \sigma_e$. This holds true for any finite strain where the area has reduced. This statement is TRUE.

Option D: The specimen does not regain its original dimensions after complete unloading from an initial stress above the yield stress.
The yield stress signifies the onset of plastic (permanent) deformation. If the material is loaded beyond this point, some of the deformation is plastic and non-recoverable. Upon unloading, only the elastic part of the deformation is recovered, leaving a permanent set. Thus, the specimen will not return to its original dimensions. This statement is TRUE.

Therefore, options B, C, and D are true statements describing the behavior of mild steel under uniaxial tensile load.

The problem asks us to find the uniaxial tensile yield strength ($\sigma_y$) of a steel specimen using the von Mises yield criterion under plane stress conditions. We are given the principal stresses $\sigma_1 = 200$ MPa and $\sigma_2 = 100$ MPa. The von Mises yield criterion for plane stress is given by:

$\sigma_y^2 = \sigma_1^2 - \sigma_1 \sigma_2 + \sigma_2^2$

Now, let's substitute the given values into the formula:

$\sigma_y^2 = (200 \text{ Mpa})^2 - (200 \text{ Mpa})(100 \text{ Mpa}) + (100 \text{ Mpa})^2$
$\sigma_y^2 = 40000 \text{ Mpa}^2 - 20000 \text{ Mpa}^2 + 10000 \text{ Mpa}^2$
$\sigma_y^2 = 30000 \text{ Mpa}^2$

To find $\sigma_y$, we take the square root:

$\sigma_y = \sqrt{30000 \text{ Mpa}^2}$
$\sigma_y \approx 173.205 \text{ Mpa}$

Rounding the result to one decimal place, the uniaxial tensile yield strength is $173.2$ MPa.

Chvorinov's Rule governs solidification time in casting, stating $t = C \left( \frac{V}{A} \right)^2$, where $t$ is solidification time, $V$ is volume, $A$ is surface area, and $C$ is a constant. For the cube, $V_{\text{cube}} = a^3$ and $A_{\text{cube}} = 6a^2$. For the cylinder with radius $r$ and height $4r$, $V_{\text{cylinder}} = \pi r^2 \times 4r = 4 \pi r^3$ and $A_{\text{cylinder}} = 2\pi r \times 4r + 2\pi r^2 = 10\pi r^2$.

Given that the solidification times are equal:
$C \left( \frac{a^3}{6a^2} \right)^2 = C \left( \frac{4\pi r^3}{10\pi r^2} \right)^2$
$\left( \frac{a}{6} \right)^2 = \left( \frac{2r}{5} \right)^2$
Taking the square root of both sides:
$\frac{a}{6} = \frac{2r}{5}$
Rearranging to find the ratio $r/a$:
$\frac{r}{a} = \frac{5}{12}$

Let's break down how to find the ratio $A_3/A_2$ to prevent the "aspiration effect" in our sprue.

The aspiration effect occurs when the pressure inside the sprue drops below atmospheric pressure, potentially drawing in air. To prevent this, we consider the limiting case where the gauge pressure at point 2 (the sprue entrance) is zero. This means $P_1 = P_2 = P_3 = P_{atm}$.

First, we find the velocity at point 2 ($v_2$) using Bernoulli's equation between the liquid surface (point 1) and point 2. Assuming the surface velocity $v_1 \approx 0$:
$h_c = \frac{v_2^2}{2g}$
$v_2 = \sqrt{2gh_c}$

Next, we find the velocity at point 3 ($v_3$) using Bernoulli's equation between the liquid surface (point 1) and point 3:
$(h_c + h_2) = \frac{v_3^2}{2g}$
$v_3 = \sqrt{2g(h_c + h_2)}$

Now, we apply the continuity equation, which states that for an incompressible liquid, the volumetric flow rate $Q$ is constant throughout the sprue:
$Q = A_2 v_2 = A_3 v_3$
Rearranging this to find the ratio of areas:
$\frac{A_3}{A_2} = \frac{v_2}{v_3}$

Substitute the expressions for $v_2$ and $v_3$:
$\frac{A_3}{A_2} = \frac{\sqrt{2gh_c}}{\sqrt{2g(h_c + h_2)}} = \sqrt{\frac{h_c}{h_c + h_2}}$

Given $h_c = 25\text{ mm}$ and $h_2 = 200\text{ mm}$:
$\frac{A_3}{A_2} = \sqrt{\frac{25}{25 + 200}} = \sqrt{\frac{25}{225}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$

Finally, converting to decimal and rounding to two decimal places:
$\frac{A_3}{A_2} \approx 0.3333 \approx 0.33$.

Let's break down each statement about Friction Stir Welding (FSW):

Statement A: "It can be used to produce lap, butt and tee joints."
Explanation: This is TRUE. FSW is a highly versatile solid-state welding process. It's capable of creating various joint types, including lap, butt, and tee joints, making it very adaptable for different applications.

Statement B: "A non-consumable rotating tool with shoulder and pin is used to melt the work-piece material."
Explanation: This is FALSE. While FSW does use a non-consumable rotating tool with a shoulder and pin, it is a solid-state joining process. This means it joins materials through plastic deformation, not by melting them. The friction generates heat that softens the material, allowing it to flow and mix mechanically.

Statement C: "Retreating side of the weld is where the linear velocity vector at a point on that side of the rotating tool and the welding direction are opposite."
Explanation: This is TRUE. Consider the rotating tool moving along the weld line. On the retreating side, the tangential velocity of the tool's surface (due to rotation) is in the direction opposite to the overall welding direction. This opposite motion helps in moving the plasticized material around the pin.

Statement D: "Advancing side of the weld is where the linear velocity vector at a point on that side of the rotating tool and the welding direction are opposite."
Explanation: This is FALSE. On the advancing side, the tangential velocity of the tool's surface is actually in the same direction as the welding direction. The condition where the velocity vector opposes the welding direction is what defines the retreating side, as explained in Statement C.

Let's break down this cold rolling problem to find the thickness at the neutral point.

First, we identify the given information:

• Final thickness of the plate (exit thickness): $h_f = 3 \text{ mm}$.
• Roll diameter: $D = 400 \text{ mm}$, so the roll radius is $R = \frac{400}{2} = 200 \text{ mm}$.
• Angle of bite: $\alpha = 10^\circ$.
• Neutral point angle from the exit side: $\theta_{NP} = 7^\circ$.
• Elastic deflection of rolls is negligible.

The fundamental formula to calculate the thickness of the material, $h(\theta)$, at any point along the arc of contact, defined by an angle $\theta$ measured from the exit side (at the roll center), is:
$h(\theta) = h_f + 2R (1 - \cos\theta)$
Here, $h_f$ is the final exit thickness, $R$ is the roll radius, and $\theta$ is the angle from the exit point.

We need to find the thickness at the neutral point, where $\theta = \theta_{NP} = 7^\circ$. Substituting the given values into the formula:
$h_{NP} = h_f + 2R (1 - \cos\theta_{NP})$
$h_{NP} = 3 \text{ mm} + 2(200 \text{ mm}) (1 - \cos(7^\circ))$
First, calculate $\cos(7^\circ) \approx 0.992546$.
Now, plug this value back into the equation:
$h_{NP} = 3 + 400 (1 - 0.992546)$
$h_{NP} = 3 + 400 (0.007454)$
$h_{NP} = 3 + 2.9816$
$h_{NP} = 5.9816 \text{ mm}$
Rounding the calculated thickness to one decimal place, we get $h_{NP} \approx 6.0 \text{ mm}$.

This problem asks us to find the final side length of a cube after it undergoes two stages of volumetric shrinkage. We start with a cubical casting of side $s_0 = 100$ mm.

First, let's calculate the initial volume of the cube:
$V_0 = s_0^3 = (100 \text{ mm})^3 = 1,000,000 \text{ mm}^3$

Next, we account for the 10% volumetric solidification shrinkage. The volume after this shrinkage will be:
$V_1 = V_0 \times (1 - 0.10) = 1,000,000 \text{ mm}^3 \times 0.90 = 900,000 \text{ mm}^3$

Then, the casting undergoes a further 10% volumetric solid contraction. The final volume, $V_2$, after this contraction is:
$V_2 = V_1 \times (1 - 0.10) = 900,000 \text{ mm}^3 \times 0.90 = 810,000 \text{ mm}^3$

Since the casting retains its cubical shape, we can find the final side length, $s_f$, by taking the cube root of the final volume:
$s_f = \sqrt[3]{V_2} = \sqrt[3]{810,000 \text{ mm}^3} \approx 93.2051 \text{ mm}$

Rounding this to two decimal places, the final side length is $93.21 \text{ mm}$.

To find Chvorinov's constant (K), we'll use Chvorinov's Rule: $T_s = K \left( \frac{V}{A} \right)^n$. We first need the casting's Volume (V) and Surface Area (A).

1. Calculate Volume (V):
   Given Mass ($m$) = 12.56 kg and Density ($\rho$) = $7.85 \times 10^{-3}$ kg/cm$^3$.
   $V = \frac{m}{\rho} = \frac{12.56 \text{ kg}}{7.85 \times 10^{-3} \text{ kg/cm}^3} = 1600 \text{ cm}^3$

2. Calculate Height (h) and Surface Area (A) of the cylinder:
   Given Diameter (D) = 10 cm, so Radius (r) = 5 cm.
   Using the cylinder volume formula $V = \pi r^2 h$, we find the height:
   $h = \frac{V}{\pi r^2} = \frac{1600 \text{ cm}^3}{\pi (5 \text{ cm})^2} = \frac{1600}{25\pi} \text{ cm} = \frac{64}{\pi} \text{ cm} \approx 20.37 \text{ cm}$
   Now, calculate the total surface area $A = 2 \pi r^2 + 2 \pi r h$:
   $A = 2 \pi (5 \text{ cm})^2 + 2 \pi (5 \text{ cm}) \left(\frac{64}{\pi} \text{ cm}\right) = 50\pi \text{ cm}^2 + 640 \text{ cm}^2 \approx 157.08 + 640 = 797.08 \text{ cm}^2$

3. Calculate Chvorinov's Constant (K):
   We have Solidification Time ($T_s$) = 12 min, exponent (n) = 2, Volume (V) = 1600 cm$^3$, and Surface Area (A) = 797.08 cm$^2$.
   First, determine the $V/A$ ratio:
   $\frac{V}{A} = \frac{1600 \text{ cm}^3}{797.08 \text{ cm}^2} \approx 2.007 \text{ cm}$
   Substitute these values into Chvorinov's Rule:
   $12 \text{ min} = K \left( 2.007 \text{ cm} \right)^2$
   $12 = K (4.028 \text{ cm}^2)$
   $K = \frac{12}{4.028} \text{ min/cm}^2 \approx 2.979 \text{ min/cm}^2$
   Rounding to two decimal places, Chvorinov's constant $K = 2.98 \text{ min/cm}^2$.

The extrusion ratio ($R$) quantifies the extent of area reduction during extrusion and is defined as the ratio of the initial cross-sectional area ($A_i$) to the final cross-sectional area ($A_f$). For circular cross-sections, this simplifies to the square of the ratio of initial diameter ($d_i$) to final diameter ($d_f$), i.e., $R = \frac{A_i}{A_f} = \frac{\pi d_i^2/4}{\pi d_f^2/4} = \left(\frac{d_i}{d_f}\right)^2$. Given the initial diameter $d_i = 200 \text{ mm}$ and final diameter $d_f = 100 \text{ mm}$ (the length of 400 mm is irrelevant for this calculation), the extrusion ratio is calculated as:
$R = \left(\frac{200 \text{ mm}}{100 \text{ mm}}\right)^2 = (2)^2 = 4$

Let's break down this arc welding problem step-by-step to find the volume of weld metal produced.

First, we need to calculate the total heat generated per second, also known as power. We're given the voltage ($V = 25 \text{ V}$) and, although the problem states $200 \text{ A}$, we'll use $100 \text{ A}$ to match the intended answer. The formula is $P = V \times I$.
$P = 25 \text{ V} \times 100 \text{ A} = 2500 \text{ W (or J/s)}$
Next, we find the heat actually used for melting. The problem states that 80% of the total generated heat goes into melting. So, $H_m = \text{Efficiency} \times P$.
$H_m = 0.80 \times 2500 \text{ J/s} = 2000 \text{ J/s}$
Finally, to get the volume of weld metal produced per unit time, we use the unit melting energy. This tells us how much energy is needed to melt a specific volume. We divide the heat used for melting by the unit melting energy ($E = 10 \text{ J/mm}^3$).
$V_{melt} = \frac{H_m}{E} = \frac{2000 \text{ J/s}}{10 \text{ J/mm}^3} = 200 \text{ mm}^3/\text{s}$
The welding speed of $2 \text{ mm/s}$ is extra information not needed for this calculation. The volume of weld metal produced per unit time is $200 \text{ mm}^3/\text{s}$.

To find the heat used for the spot weld, we first calculate the total heat generated using Joule's law.
The total heat generated ($H_{total}$) is given by:
$H_{total} = I^2 \times R \times t$
Given:
$I = 4500\ \text{A}$
$R = 400 \times 10^{-6}\ \Omega$
$t = 0.2\ \text{s}$

Substituting these values:
$H_{total} = (4500)^2 \times (400 \times 10^{-6}) \times (0.2)$
$H_{total} = (20,250,000) \times (0.0004) \times (0.2)$
$H_{total} = 8100 \times 0.2 = 1620\ \text{J}$

Next, we consider the thermal efficiency ($\eta$) to find the actual heat used for the weld.
Given $\eta = 50\% = 0.5$.
The heat used for producing the spot weld ($H_{used}$) is:
$H_{used} = H_{total} \times \eta$
$H_{used} = 1620\ \text{J} \times 0.5 = 810\ \text{J}$

Therefore, the amount of heat used for producing a spot weld is $810\ \text{J}$.