QUESTION 1: The horizontal bandsaw is typically used for cutting large pieces of stock down to size for further machining. (Option 1)
QUESTION 2: Before sawing, you should check the bandsaw blade to determine if it is damaged or worn out. (Option 3)
QUESTION 3: Saw guides on the vertical bandsaw should be changed if the width of the new blade being installed is significantly different. (Option 3)
QUESTION 4: Blade tension and blade tracking should also be checked during final installation of a new blade on a vertical bandsaw. (Option 4)
QUESTION 5: The statement that is NOT true is: pushers, jigs, and other work-holding devices should not be used unless the workpiece is small. (Option 3)
QUESTION 6: The choice of speed of the vertical bandsaw will be adequate for most maintenance and repair work if it is based on the hardness of the material being sawed. (Option 1)
QUESTION 7: The tilt, pitch, speed, and tension of a vertical bandsaw table should be checked before cutting. (Option 5)
QUESTION 8: A blade that bounces regularly, or one that jumps or tugs at the workpiece while cutting probably has lost its set. (Option 4)
QUESTION 9: When determining if a blade is appropriate for the work to be done, you would compare the thickness of the workpiece and the pitch of the blade. (Option 2)
QUESTION 10: Adjusting the tension of a bandsaw blade is important because loose blades may break. (Option 3)
QUESTION 11: Horizontal bandsaw blade guides should be adjusted to expose only that portion of the blade that will be doing the cutting because they support the blade. (Option b.)
QUESTION 12: When changing speeds on a horizontal bandsaw with a belt-and-pulley speed selection system, the first thing to do is lock out the saw. (Option 2)
QUESTION 13: Crooked cuts on a horizontal bandsaw are probably an indication of a feed rate that is too high. (Option 4)
QUESTION 14: The first step when using a horizontal bandsaw should be to mount the work in the saw properly. (Option 2)
QUESTION 15: You must lock out a vertical bandsaw whenever you release an old blade just before installing a new one. (Option 3)
The horizontal bandsaw is primarily used for cutting large pieces of stock down to size for further machining. It allows for efficient and precise cutting of large workpieces. Before sawing, it is essential to check the bandsaw blade for damage or wear. This ensures that the blade is in good condition and capable of cutting effectively.
Saw guides on the vertical bandsaw should be changed if the width of the new blade being installed is significantly different. Proper alignment of the blade is crucial for accurate and safe cutting. During the final installation of a new blade on a vertical bandsaw, it is important to check both blade tension and blade tracking. These adjustments ensure optimal performance and prevent issues such as blade slippage or misalignment.
Work-holding devices like pushers and jigs should always be used, especially when the operator's hands would come close to the blade. They provide safety and stability during the cutting process. When selecting the speed for a vertical bandsaw, it is generally based on the hardness of the material being sawed. Choosing the appropriate speed ensures efficient cutting without damaging the material or the blade.
Before cutting with a vertical bandsaw, the tilt, pitch, speed, and tension of the table should be checked to ensure proper setup and operation. A blade that bounces regularly, jumps, or tugs at the workpiece while cutting is likely a sign that it has lost its set, which affects its cutting ability.
When determining if a blade is appropriate for the work to be done, factors to compare include the thickness of the workpiece and the pitch of the blade. Matching these specifications ensures efficient and accurate cutting. Adjusting the tension of a bandsaw blade is crucial because loose blades may break, leading to potential hazards and compromising the quality of the cut.
Horizontal bandsaw blade guides should be adjusted to expose only the portion of the blade that will be doing the cutting. This is primarily done to support the blade during the cutting process. When changing speeds on a horizontal bandsaw with a belt-and-pulley speed selection system, the first step is to lock out the saw for safety reasons.
Crooked cuts on a horizontal bandsaw are often an indication of a feed rate that is too high. Adjusting the feed rate can help achieve straighter cuts. The first step when using a horizontal bandsaw is to mount the work in the saw properly, ensuring it is securely positioned for cutting.
Locking out a vertical bandsaw is necessary whenever you release an old blade just before installing a new one to prevent accidental activation and ensure safety.
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Name the five (5) properties that determine the quality of a sand mold for sand casting? [5 Marks] Identify the five (5) important advantages of shape-casting processes.
1. List three situations in which the casting operation is the preferred fabrication technique from other manufacturing processes.
2. What is the difference between a pattern and a core in sand molding?
Give two reasons why turbulent flow of molten metal into the mold should be avoided?
Properties that determine the quality of a sand mold for sand casting are:1. Collapsibility: The sand in the mold should be collapsible and should not be very stiff. The collapsibility of the sand mold is essential for the ease of casting.
2. Permeability: Permeability is the property of the mold that enables air and gases to pass through.
Permeability ensures proper ventilation within the mold.
3. Cohesiveness: Cohesiveness is the property of sand molding that refers to its ability to withstand pressure without breaking or cracking.
4. Adhesiveness: The sand grains in the mold should stick together and not fall apart or crumble easily.
5. Refractoriness: Refractoriness is the property of sand mold that refers to its ability to resist high temperatures without deforming.
Advantages of Shape-casting processes:1. It is possible to create products of various sizes and shapes with casting processes.
2. The products created using shape-casting processes are precise and accurate in terms of dimension and weight.
3. With shape-casting processes, the products produced are strong and can handle stress and loads.
4. The production rate is high, and therefore, it is cost-effective.
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2.A DC series generator is supplying a current of 8 A to a series lighting system through a feeder of total resistance of 2 Ω. The terminal voltage is 3000 V. The armature and series field resistances are respectively 18 and 15 Ω, respectively. A 30-Ω diverter resistance is shunted across the series field. Determine the power developed in the armature of the generator
Power developed in the armature of a generator is determined by the formula P = EI, where P = power in watts,
E = voltage in volts, and I = current in amperes. A DC series generator is a generator whose field winding is connected in series with the armature winding. In a series generator, the armature and field currents are the same.
This means that the load current and the field current are supplied by the same source. As a result, any change in the load current will cause a corresponding change in the field current. Now let us solve the problem using the given values.
The terminal voltage of the generator is given as 3000 V. The generated voltage is the sum of the terminal voltage and the voltage drop across the armature:
EG = V + ET
= 504 + 3000
= 3504 V Now we can calculate the current generated by the generator.
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Jet fuel is most closely related to: a. Automotive gasoline b. AvGas
c. Kerosene
Jet fuel is most closely related to kerosene. kerosene is primarily used in the aviation industry as jet fuel for airplanes and in the military as a fuel for gas turbine engines.
What is jet fuel? Jet fuel is a type of aviation fuel used in planes powered by jet engines. It is clear to light amber in color and has a strong odor. Jet fuel is a type of kerosene and is a light fuel compared to the heavier kerosene used in heating or lighting.
What is Kerosene? Kerosene is a light diesel oil typically used in outdoor lanterns and furnaces. In order to ignite, it must be heated first. When used as fuel for heating, it is stored in outdoor tanks.
However, kerosene is primarily used in the aviation industry as jet fuel for airplanes and in the military as a fuel for gas turbine engines.
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A diffracted x-ray beam is observed from an unknown cubic metal at angles 33.4558°, 48.0343°, θA, θB, 80.1036°, and 89.6507° when x-ray of 0.1428 nm wavelength is used. θA and θB are the missing third and fourth angles respectively. (a) Determine the crystal structure of the metal. (b) Determine the indices of the planes (hkl) that produce each of the peaks.
(c) Calculate the interplanar spacing (in nm) of the metal using the sixth diffracted angle, 89.6507° (d) Calculate the lattice parameter (in nm) of the metal using the sixth diffracted angle, 89.6507° (e) Identify the material. (f) Using the above results, calculate the angles of θA and θB.
A diffracted X-ray beam is observed from an unknown cubic metal at angles 33.4558°, 48.0343°, θA, θB, 80.1036°, and 89.6507° when X-ray of 0.1428 nm wavelength is used.
θA and θB are the missing third and fourth angles respectively. Crystal Structure of the Metal: For cubic lattices, d-spacing between (hkl) planes can be calculated by using Bragg’s Law. The formula to calculate d-spacing is given by nλ = 2d sinθ where n = 1, λ = 0.1428 nm Here, d = nλ/2 sinθ = (1×0.1428×10^-9) / 2 sin θ
The values of sin θ are calculated as: sin 33.4558° = 0.5498, sin 48.0343° = 0.7417, sin 80.1036° = 0.9828, sin 89.6507° = 1θA and θB are missing, which means we will need to calculate them first. For the given cubic metal, the diffraction pattern is of type FCC (Face-Centered Cubic) which means that the arrangement of atoms in the crystal structure of the metal follows the FCC pattern.
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The following data refer to an average, unventi- lated, industrial journal bearing for a generator: Load 3,5 kN; speed = 870 r/min; diameter = 40 mm; length = 50 mm; clearance ratio = 0,001; ambient temperature = 15°C; and oil temperature = 60°C maximum. Assume an oil viscosity of ISO VG 68 and determine (a) whether hydrodynamic lubrication may be expected; and (b) whether artificial cooling is required and, if so, the heat to be removed per second. [21,6 W]
Given that the load, speed, diameter, length, and clearance ratio of an average, unventilated industrial journal bearing for a generator are 3.5 kN, 870 r/min, 40 mm, 50 mm, and 0.001, respectively, it is required to determine whether hydrodynamic lubrication can be expected, and if artificial cooling is necessary,
The amount of heat to be removed per second if oil viscosity is ISO VG 68 and maximum oil temperature is 60 °C.To determine whether hydrodynamic lubrication can be expected, calculate the Sommerfeld number. The formula for the Sommerfeld number is given by;S= (P *d ) / (h * V * N )where, P is the load, d is the diameter of the bearing, h is the bearing clearance, V is the kinematic viscosity, and N is the bearing speed.Substituting the given values in the above formula we get:
S = (3.5 *10³ * 0.04) / (0.001 * 68 * 870 / 60)≈ 0.17
Since the calculated value of Sommerfeld number is less than 1, the bearing will be in the hydrodynamic lubrication regime.Therefore, we can expect hydrodynamic lubrication. Artificial cooling will be required since the maximum oil temperature exceeds the ambient temperature.The heat to be removed per second is given by;
Q= U * A * (To - Ta)
Where U is the heat transfer coefficient, A is the heat transfer area, To is the maximum oil temperature, and Ta is the ambient temperature.Substituting the values, we get;
Q= 90 * π * (0.04² - 0.038²) * (60 - 15)≈ 21.6 W
Therefore, the heat to be removed per second is approximately 21.6 W.
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(a) Before cutting or welding with oxy-acetylene gas welding or electric arc equipment, it is very important to check for signs of damage to the key components of each system. Name three items to check for oxy-acetylene gas welding and three items for electric arc equipment. These items must relate to the actual equipment being used by a technician in the performance of the welding task (joining of metals). (b) How would you check for gas leaks on oxy-acetylene welding equipment?
Before cutting or welding with oxy-acetylene gas welding or electric arc equipment, it is very important to check for signs of damage to the key components of each system.
Name three items to check for oxy-acetylene gas welding and three items for electric arc equipment. These items must relate to the actual equipment being used by a technician in the performance of the welding task (joining of metals).Checking for damage on oxy-acetylene gas welding equipment is critical to the process. As a result, the following three items should be inspected:
1. Oxygen and acetylene tanks, regulators, and hoses.
2. Gas torch handle and tip.
3. Lighting mechanism.
Electric arc equipment is similarly important to inspect for damage. As a result, the following three items should be inspected:
1. Cables and wire feed.
2. Electrodes and holders.
3. Torch and nozzles.
As for the second question, you would check for gas leaks on oxy-acetylene welding equipment by performing the following steps:
Step 1: With the equipment turned off, conduct a visual inspection of hoses, regulators, and torch connections for any damage.
Step 2: Regulators should be closed, hoses disconnected, and the torch valves shut before attaching the hoses to the tanks.
Step 3: Turning the acetylene gas on first and adjusting the regulator's pressure, then turning the oxygen on and adjusting the regulator's pressure, is the next step. Then turn the oxygen on and set the regulator's pressure.
Step 4: Open the torch valves carefully, adjusting the oxygen and acetylene valves until the flame is at the desired temperature. Keep an eye on the flame's color.
Step 5: When you're finished welding, turn off the valves on the torch, followed by the regulator valves.
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The objective is to design a rotating shaft for dynamic operation. 2. GIVEN A cold-drawn (CD) alloy shaft of diameter 50mm and length 750mm is to withstand a maximum bending stress of max = 250MPa at the most critical section and is loaded with a stress ratio of R = 0.25. A factor of safety of at least 1.5 is desired with a reliability of 99%. 3. PROBLEM For this homework, you are expected to choose a suitable CD steel alloy that will satisfy the above stated design requirements (FS≥ 1.5 for all types of failure). Choose a suitable material for the shaft from Table A-20 in your textbook appendix (6 points). Determine the critical speed for the shaft for your choice of material
To design a rotating shaft for dynamic operation, a cold-drawn (CD) alloy shaft of diameter 50mm and length 750mm is provided which is to withstand a maximum bending stress of max = 250MPa at the most critical section .Therefore, the critical speed for the AISI 4340 CD Steel shaft is approximately 6794.7 RPM.
and is loaded with a stress ratio of R = 0.25. The required factor of safety is at least 1.5 with a reliability of 99%. Choosing the appropriate material for the shaft from Table A-20 in the appendix of the textbook can help to fulfill the above-stated design specifications.For the CD steel alloy shaft, from Table A-20 in the appendix of the textbook, the most suitable materials are AISI 1045 CD Steel, AISI 4140 CD Steel, and AISI 4340 CD Steel.
Where k = torsional spring constant =[tex](π/16) * ((D^4 - d^4) / D),[/tex]
g = shear modulus = 80 GPa (for CD steel alloys),
m = mass of the shaft = (π/4) * ρ * L * D^2,
and ρ = density of the material (for AISI 4340 CD Steel,
ρ = 7.85 g/cm³).
For AISI 4340 CD Steel, the critical speed can be calculated as follows:
[tex]n = (k * g) / (2 * π * √(m / k))n = ((π/16) * ((0.05^4 - 0.0476^4) / 0.05) * 8 * 10^10) / (2 * π * √(((π/4) * 7.85 * 0.75 * 0.05^2) / ((π/16) * ((0.05^4 - 0.0476^4) / 0.05))))[/tex]
n = 6794.7 RPM
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A rectangular duct of 50 m long has pressure drop of 4.5 pa/m.
the velocity through the duct is 18 m/s. Determine the flow rate
and size of the duct in terms of Deq and Deqf.
The flow rate and size of the duct in terms of Deq (equivalent diameter) and Deqf (equivalent hydraulic diameter), we need to use the given information about the pressure drop and velocity.
The pressure drop in the duct can be related to the flow rate and duct dimensions using the Darcy-Weisbach equation:
ΔP = (f * (L/D) * (ρ * V^2)) / 2
Where:
ΔP is the pressure drop (Pa)
f is the friction factor (dimensionless)
L is the length of the duct (m)
D is the hydraulic diameter (m)
ρ is the density of the fluid (kg/m^3)
V is the velocity of the fluid (m/s)
In this case, we are given:
L = 50 m
ΔP = 4.5 Pa
V = 18 m/s
To find the flow rate (Q), we can rearrange the Darcy-Weisbach equation:
Q = (2 * ΔP * π * D^4) / (f * ρ * L)
We also know that for a rectangular duct, the hydraulic diameter (Deq) is given by:
Deq = (2 * (a * b)) / (a + b)
Where:
a and b are the width and height of the rectangular duct, respectively.
To find Deqf (equivalent hydraulic diameter), we can use the following relation for rectangular ducts:
Deqf = 4 * A / P
Where:
A is the cross-sectional area of the duct (a * b)
P is the wetted perimeter (2a + 2b)
Let's calculate the flow rate (Q) and the equivalent diameters (Deq and Deqf) using the given information:
First, let's find the hydraulic diameter (Deq):
a = ? (unknown)
b = ? (unknown)
Deq = (2 * (a * b)) / (a + b)
Next, let's find the equivalent hydraulic diameter (Deqf):
Deqf = 4 * A / P
Now, let's calculate the flow rate (Q):
Q = (2 * ΔP * π * D^4) / (f * ρ * L)
To proceed further and obtain the values for a, b, Deq, Deqf, and Q, we need the values of the width and height of the rectangular duct (a and b) and additional information about the fluid being transported, such as its density (ρ) and the friction factor (f).
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A hot steel sphere is dropped into a large cold oil bath as part of a quenching process to harden the steel. Assume that the thermal capacitance of the steel sphere is C = 440J/°C and the average convective thermal resistance between the sphere and the oil is R = 0.05°C/W. If the sphere is originally at T. = 800°C and the oil is at 25°C, how long does it take for the sphere to approximately reach the temperature of the oil? =
>> 100 seconds 88 seconds << 1 second 22 seconds
It takes approximately 100 seconds for the steel sphere to reach the temperature of the oil.
In order to find the time needed for the hot steel sphere to reach the temperature of the cold oil bath, we will use the following equation:
Q = m C (T2 - T1)
Where:Q = thermal energy in Joules
m = mass of steel sphere in Kg
C = thermal capacitance of steel sphere in Joules per degree Celsius
T2 = final temperature in Celsius
T1 = initial temperature in Celsius
R = convective thermal resistance in Celsius per Watt
Assuming that there is no heat transfer by radiation, we can use the following expression to find the rate of heat transfer from the sphere to the oil:Q/t = (T2 - T1)/R
Where:t = time in seconds
Substituting the given values, we get:(T2 - 25)/0.05 = -440 (800 - T2)/t
Simplifying and solving for t, we get:t = 100 seconds
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In a simple gas turbine installation, the air pressure is 100kPa and 280K
temperature enters the compressor. Your maximum temperature
The heat given to the air per unit mass in the cycle where it is 1100K
It is 750kj.
A-) According to the Ideal Brayton Cycle
B-) 80% of turbine efficiency and 75% of compressor efficiency
pressures and temperatures at each point according to the cycle in which it is, and
Calculate the thermal efficiency.
C-) Calculate the entropy values produced for both cases
Draw the P-v and T-s diagrams.
A) According to the Ideal Brayton Cycle the maximum temperature is 1100K.
B) The Brayton cycle's thermal efficiency is expressed as η = (1 – (1/3.9285)) × (1 – (280/1100)) = 0.4792 = 47.92%.
C) Entropy values produced in the cycle: State 1: s1 = s0 + cp ln(T1/T0) = 0.3924; State 2: s2 = s1 = 0.3924; State 3: s3 = s2 + cp ln(T3/T2) = 0.6253; State 4: s4 = s3 = 0.6253.P-V and T-S.
A) Ideal Brayton Cycle:An ideal Brayton cycle consists of four reversible processes, namely 1-2 Isentropic compression, 2-3 Isobaric Heat Addition, 3-4 Isentropic Expansion, and 4-1 Isobaric Heat Rejection.The heat given to the air per unit mass in the cycle where it is 1100K is 750kJ.
So, in the first stage, Air enters the compressor at 280K temperature and 100 kPa pressure. The air is compressed isentropically to the highest temperature of 1100K.
Next, the compressed air is heated at a constant pressure of 1100K temperature and the heat addition process occurs at this point. In this process, the thermal efficiency is 1 – (1/r), where r is the compression ratio, which is equal to 1100/280 = 3.9285.
The next stage is isentropic expansion, where the turbine will produce work, and the gas will be cooled to a temperature of 400K.Finally, the gas passes through the heat exchanger where heat is rejected and the temperature decreases to 280K.
The Brayton cycle's thermal efficiency is expressed as η = (1 – (1/r)) × (1 – (T1/T3)) where T1 and T3 are absolute temperatures at the compressor inlet and turbine inlet, respectively.
Efficiency (η) = (1 – (1/3.9285)) × (1 – (280/1100)) = 0.4792 = 47.92%.
B) Efficiency:
Compressor efficiency (ηc) = 75%.
Turbine efficiency (ηt) = 80%.
The temperatures and pressures are:
State 1: p1 = 100 kPa, T1 = 280 K.
State 2: p2 = p3 = 3.9285 × 100 = 392.85 kPa. T2 = T3 = 1100 K.
State 4: p4 = p1 = 100 kPa. T4 = 400 K.
C) Entropy:
Entropy values produced in the cycle:
State 1: s1 = s0 + cp ln(T1/T0) = 0.3924.
State 2: s2 = s1 = 0.3924.
State 3: s3 = s2 + cp ln(T3/T2) = 0.6253.
State 4: s4 = s3 = 0.6253.P-V and T-S.
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A specimen of diameter 5.05 mm and length 57 mm is subjected to a compressive force of 2.3 kN, the appropriate change in diameter is 0.019 mm. The material has a yield stress of 300 MPa and the Poisson's ratio of 0.34. Calculate the Young's modulus (in GPa). Please provide the value only. If you believe that the problem can not be solved please type 12345
Using given values and employing stress-strain relations, we can calculate the Young's modulus, a fundamental mechanical property
To calculate Young's modulus (E), we first need to find the stress and strain. Stress (σ) is the force (F) divided by the initial cross-sectional area (A = πd²/4). In this case, σ = 2.3 kN / (π*(5.05 mm/2)²) = 182 MPa. Strain (ε) is the change in length/original length, which in this case, under compression, is the lateral strain given by the change in diameter/original diameter = 0.019 mm / 5.05 mm. Young's modulus is then given by the ratio of stress to strain, E = σ / ε. However, in this scenario, the strain is multiplied by Poisson's ratio (0.34), so E = σ / (ε*0.34).
Solving this gives the Young's modulus. Note: Please perform the calculations as this response contains the method but not the actual value.
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The strain energy function of a polymeric material is given by the following hyperelastic potential where c, c and c are material parameters, and , and are respectively the first, second and third principal invariants of the right Cauchy-Green deformation tensor (with components in a Cartesian coordinate system, , = 1,2,3). xp and o are respectively the exponential and logarithm functions.
(1, 2, 3) = c1(1 − 3) + c2 (1 − 3)2 + c3(2 − 3) +
1
2
3 − 1
2
(i) Give the explicit expression of the first principal invariant as a function of the components of [2 marks]
(ii) Define the second Piola-Kirchhoff stress tensor arising from the hyperelastic potential as a function of and .
[2 marks]
TURN OVER (1, 2, 3) = c1(1 − 3) + c2 (1 − 3)2 + c3(2 − 3) + 1 2 3 − 1 2 FEEG6010W1 Copyright 2022 v01 © University of Southampton Page 10 of 11 (iii) Provide the explicit expression of the second Piola-Kirchhoff stress tensor arising from the hyperelastic potential , in compact form (tensor). You will use the following formulas for the derivative of the determinant det() of a second-order tensor : det() = () and the derivative of the trace of the square of a second-order tensor : trace = [15 marks] (iv) Provide the explicit expression of the second Piola-Kirchhoff stress tensor arising from the hyperelastic potential , in matrix form (assuming a 3D problem). You will denote the components of as . To lighten notations replace the explicit expressions of ⁄, ⁄ and ⁄ by respectively the letters A, B and K.
(i) The first principal invariant can be obtained as follows, In three dimensions, the Cauchy-Green deformation tensor is defined as, For the first principal invariant, we have, Therefore, the explicit expression of the first principal invariant as a function of the components.
(ii) The second Piola-Kirchhoff stress tensor is given by,v Using the hyperelastic potential given, we can write, Therefore, the second Piola-Kirchhoff stress tensor arising from the hyperelastic potential as a function of and is given by,(iii) Using the formula, we have,vThe derivative of the first invariant with respect to the deformation tensor can be obtained as follows.
Therefore, v Using the formula, we have, For the derivative of the hyperelastic potential with respect to the deformation tensor, we have, Therefore, Substituting the above expressions into the formula for the second Piola-Kirchhoff stress tensor.
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How do you implement pipes using corrosion Analysis in Yand T Juctions 5 Failur Analysis Methodology Collection of background data and samples preliminary Examination of the failed part Non destructure testing Mechanical testing Selection, preservation + Cleaning of fracture Surfaces Macroscopic Examination of froduse Surfaces to 100) Microscopic Examination of Fractur Surfaces (>100x)
Application of Fracture Mechanics Analyzing the evidence formulating conclusions
Implementing pipes using corrosion analysis in Y and T junctions involves the following steps: Collection of background data and samples, Preliminary examination of the failed part.
Collection of background data and samples: Gather information about the operating conditions, history, and maintenance practices of the pipe system. Collect samples from the failed components, including the Y and T junctions.
Preliminary examination of the failed part: Perform a visual inspection to identify any visible signs of corrosion or damage on the failed part. Document the observations and note the location and extent of the corrosion.
Non-destructive testing: Use non-destructive testing techniques such as ultrasonic testing, radiographic testing, or electromagnetic testing to assess the internal and external integrity of the pipe. This helps identify any defects or anomalies that may contribute to the corrosion.
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Quin-Bode Mat The forward path wander action of a uniry feedback control system is: 140 G(s) = s(s+15) Analytically determine the resonant peak My, resonant frequency or, and budwidth BW the chualpsystem
A uniry feedback control system has a forward path wander action, which is determined analytically. The given equation for a uniry feedback control system is 140 G(s) = s(s+15).
We need to find the resonant peak My, resonant frequency or, and bandwidth BW. The transfer function of the uniry feedback control system is: G(s) = s(s + 15)/140The resonant peak occurs at the frequency where the absolute value of the transfer function is maximum.
Thus, we need to find the maximum value of |G(s)|.Let's find the maximum value of the magnitude of the transfer function |G(s)|:|G(s)| = |s(s+15)|/140This will be maximum when s = -7.5So, |G(s)|max = |-7.5*(7.5+15)|/140= 84.375/140= 0.602Let's now find the frequency where this maximum value occurs.
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A screw jack is used to lift a load of SkN. The thread of the jack has a pitch of 8mm and a diameter of 40 mm. The coefficient of friction is 0.15. If the effort is applied through a lever of radius 400mm, calculate:
i) The efficiency to lift the load ii) The effort to lift the load iii) The efficiency to lower the load iv)The effort to lower the load.
i) The efficiency to lift the load is approximately 99.90%.
ii) The effort to lift the load is approximately 31.82 N.
iii) The efficiency to lower the load is approximately 100.10%.
iv) The effort to lower the load is approximately 31.82 N.
What is the efficiency to lift the load?Given:
Load = 5 kNPitch = 8 mmDiameter = 40 mmCoefficient of friction = 0.15Lever radius = 400 mmFirst, let's convert the values to consistent units:
Load = 5000 NPitch = 0.008 mDiameter = 0.04 mCoefficient of friction = 0.15Lever radius = 0.4 mi) Efficiency to lift the load (η_lift):
- Mechanical Advantage (MA_lift) = (π * Lever Radius) / Pitch
- Frictional Force (F_friction) = Coefficient of friction * Load
- Actual Mechanical Advantage (AMA_lift) = MA_lift - (F_friction / Load)
- Efficiency to lift the load (η_lift) = (AMA_lift / MA_lift) * 100%
ii) Effort to lift the load (E_lift):
- Effort to lift the load (E_lift) = Load / MA_lift
iii) Efficiency to lower the load (η_lower):
- Mechanical Advantage (MA_lower) = (π * Lever Radius) / Pitch
- Actual Mechanical Advantage (AMA_lower) = MA_lower + (F_friction / Load)
- Efficiency to lower the load (η_lower) = (AMA_lower / MA_lower) * 100%
iv) Effort to lower the load (E_lower):
- Effort to lower the load (E_lower) = Load / MA_lower
Let's calculate the values:
i) Efficiency to lift the load (η_lift):
MA_lift = (3.1416 * 0.4) / 0.008 = 157.08
F_friction = 0.15 * 5000 = 750 N
AMA_lift = 157.08 - (750 / 5000) = 157.08 - 0.15 = 156.93
η_lift = (156.93 / 157.08) * 100% = 99.90%
ii) Effort to lift the load (E_lift):
E_lift = 5000 / 157.08 = 31.82 N
iii) Efficiency to lower the load (η_lower):
MA_lower = (3.1416 * 0.4) / 0.008 = 157.08
AMA_lower = 157.08 + (750 / 5000) = 157.08 + 0.15 = 157.23
η_lower = (157.23 / 157.08) * 100% = 100.10%
iv) Effort to lower the load (E_lower):
E_lower = 5000 / 157.08 = 31.82 N
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Balloons are usually filled with helium gas as it is much lighter than air. The weight of helium is around 1/7 of the weight of air under identical conditions. The buoyancy force will push the balloon upward. If a giant balloon with 8-m diameter is used to lift yourself, determine: a. The acceleration of the balloon when it is first released
Combining all these equations, we can calculate the acceleration of the balloon when it is first released.
To determine the acceleration of the balloon when it is first released, we need to consider the forces acting on the balloon.
Buoyancy Force: The buoyancy force is the upward force exerted on the balloon due to the difference in density between the helium inside the balloon and the surrounding air. It can be calculated using Archimedes' principle:
Buoyancy Force = Weight of the displaced air = Density of air * Volume of displaced air * Acceleration due to gravity
Given that the weight of helium is around 1/7 of the weight of air, the density of helium is 1/7 of the density of air. The volume of displaced air can be calculated using the formula for the volume of a sphere:
Volume of displaced air = (4/3) * π * (radius of the balloon)^3
Weight of the Balloon: The weight of the balloon can be calculated using its mass and the acceleration due to gravity:
Weight of the Balloon = Mass of the Balloon * Acceleration due to gravity
Since the balloon is assumed to be massless, its weight is negligible compared to the buoyancy force.
Now, to find the acceleration of the balloon, we can use Newton's second law of motion:
Sum of Forces = Mass of the System * Acceleration
In this case, the sum of forces is equal to the buoyancy force, and the mass of the system is the mass of the displaced air.
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tch the impulse response of this FIR system. \[ y(k)=u(k-1)+2 u(k-2)+3 u(k-3)+2 u(k-4)+u(k-5) \] \( (\mathrm{CO} 2: \mathrm{PO} 2 \) - 5 Marks)
The impulse response of the given FIR system is:
\[ h(k) = \delta(k-1) + 2\delta(k-2) + 3\delta(k-3) + 2\delta(k-4) + \delta(k-5) \]
An FIR (Finite Impulse Response) system is characterized by its impulse response, which is the output of the system when an impulse function is applied as the input. In this case, the given FIR system has the following impulse response:
\[ h(k) = \delta(k-1) + 2\delta(k-2) + 3\delta(k-3) + 2\delta(k-4) + \delta(k-5) \]
Here, \( \delta(k) \) represents the unit impulse function, which is 1 at \( k = 0 \) and 0 otherwise.
The impulse response of the given FIR system is a discrete-time sequence with non-zero values at specific time instances, corresponding to the delays and coefficients in the system. By convolving this impulse response with an input sequence, the output of the system can be calculated.
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From the technical literature and/or open sources, present the RCS of the triangular trihedral reflector as a function of the incidence angle (for both azimuth and elevation).
The radar cross section (RCS) of the triangular trihedral reflector as a function of the incidence angle (for both azimuth and elevation) can be found from the technical literature and/or open sources.
A trihedral reflector is a corner reflector that consists of three mutually perpendicular planes.
Reflectivity is the measure of a surface's capability to reflect electromagnetic waves.
The RCS is a scalar quantity that relates to the ratio of the power per unit area scattered in a specific direction to the strength of an incident electromagnetic wave’s electric field.
The RCS formula is given by:
[tex]$$ RCS = {{4πA}\over{\lambda^2}}$$[/tex]
Where A is the projected surface area of the target,
λ is the wavelength of the incident wave,
RCS is measured in square meters.
In the case of a trihedral reflector, the reflectivity is the same for both azimuth and elevation angles and is given by the following equation:
[tex]$$ RCS = {{16A^2}\over{\lambda^2}}$$[/tex]
Where A is the surface area of the trihedral reflector.
RCS varies with the incident angle, and the equation above is used to compute the reflectivity for all incident angles.
Therefore, it can be concluded that the RCS of the triangular trihedral reflector as a function of the incidence angle (for both azimuth and elevation) can be determined using the RCS formula and is given by the equation :
[tex]$$ RCS = {{16A^2}\over{\lambda^2}}$$.[/tex]
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CONCLUSION 1. How does aging temperature affect the time and hardness? 2. Discuss which aging process shows the highest hardening effect? Explain why. 3. Based on the test results, suggest how one could maximize the strength of an Al 2024 alloy at room temperature. 4. How could one maximize the amount of Impact Energy absorbed by an Al 2024 alloy at room temperature? 5. If you were going to use 2024 Al in an application at a temperature of 190°C, what problems could be encountered? 6. Discuss errors in this experiment and their sources.
6. Discuss errors in this experiment and their sources. The errors in this experiment and their sources are as follows: Human errors: The experiment may contain errors due to the negligence of the person performing it. Machine errors: The errors in machines may lead to inaccurate results. Environmental errors: The environment may affect the experiment results and may introduce errors in them.
Conclusion:1. How does aging temperature affect the time and hardness?
The aging temperature affects the time and hardness of Al 2024 alloy.
The hardness of the alloy increases as the aging temperature is increased. The time required for maximum hardness increases as the aging temperature decreases.
2. Discuss which aging process shows the highest hardening effect? Explain why.
The T6 aging process shows the highest hardening effect on Al 2024 alloy. The T6 aging process results in precipitation hardening. It is the most effective process that produces maximum hardness in Al 2024 alloy.
3. Based on the test results, suggest how one could maximize the strength of an Al 2024 alloy at room temperature.
One could maximize the strength of Al 2024 alloy at room temperature by employing the T6 aging process. It results in precipitation hardening and provides maximum hardness to the alloy.
4. How could one maximize the amount of Impact Energy absorbed by an Al 2024 alloy at room temperature?
One could maximize the amount of Impact Energy absorbed by Al 2024 alloy at room temperature by employing the T4 aging process. The T4 aging process results in an increase in the amount of Impact Energy absorbed by the alloy.
5. If you were going to use 2024 Al in an application at a temperature of 190°C, what problems could be encountered?
If Al 2024 alloy is used in an application at a temperature of 190°C, the following problems could be encountered:
decreased strength
reduced toughness
reduced ductility
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A Shaftis driven by a 60kw AC electric motor with a star/delta starter by means of a belt(s). The motor speed is 1250rpm. The shaft drives a fan by means of a spur gear train, The fan must rotate at 500rpm in the same direction as the electric motor The Shatt is supported by 2 siding bearings one at each and of the shaft. The system is used for 24 hrs per day. Determine - Shaft dammeter at bearing - Nominal size of shatt chosen before machining - Ignore shatt bending - Sketch of design
The shaft is driven by a 60 kW AC electric motor with a star/delta starter, connected through a belt(s).
The motor operates at a speed of 1250 rpm, while the shaft needs to drive a fan at 500 rpm in the same direction. The system operates continuously for 24 hours per day and is supported by two sliding bearings, one at each end of the shaft. To determine the required parameters for the shaft, we need to calculate the shaft diameter at the bearings and choose a suitable nominal size before machining. It is assumed that shaft bending can be ignored. To determine the shaft diameter at the bearing, we need to consider the power transmitted and the speed of rotation. The power transmitted can be calculated using the formula: Power (kW) = (2 * π * N * T) / 60,
where N is the speed of rotation (in rpm) and T is the torque (in Nm). Rearranging the equation to solve for torque:
T = (Power * 60) / (2 * π * N).
For the electric motor, the torque can be calculated as:
T_motor = (Power_motor * 60) / (2 * π * N_motor).
Assuming an efficiency of 90% for the belt drive, the torque required at the fan can be calculated as:
T_fan = (T_motor * N_motor) / (N_fan * Efficiency_belt),
where N_fan is the desired speed of the fan (in rpm).
Once the torque is determined, we can use standard engineering practices and guidelines to select the shaft diameter at the bearing, ensuring adequate strength and avoiding excessive deflection. The chosen nominal size of the shaft before machining should be based on industry standards and the specific requirements of the application.
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4.28 What pressure gradient is required to accelerate kerosene (S = 0.81) vertically upward in a vertical pipe at a rate of 0.3 g?
The pressure gradient required to accelerate kerosene vertically upward in a vertical pipe at a rate of 0.3 g is calculated using the formula ΔP = ρgh.
Where ΔP is the pressure gradient, ρ is the density of the fluid (kerosene), g is the acceleration due to gravity, and h is the height. In this case, the acceleration is given as 0.3 g, so the acceleration due to gravity can be multiplied by 0.3. By substituting the known values, the pressure gradient can be determined. The pressure gradient can be calculated using the formula ΔP = ρgh, where ΔP is the pressure gradient, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height. In this case, the fluid is kerosene, which has a specific gravity (S) of 0.81. Specific gravity is the ratio of the density of a substance to the density of a reference substance (usually water). Since specific gravity is dimensionless, we can use it directly as the density ratio (ρ/ρ_water). The acceleration is given as 0.3 g, so the effective acceleration due to gravity is 0.3 multiplied by the acceleration due to gravity (9.8 m/s²). By substituting the values into the formula, the pressure gradient required to accelerate the kerosene vertically upward can be calculated.
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(i) determine the transfer function from u to y; (ii) if the system is stable or not; (iii) Compute the location of the zeros and poles. d²x = -x + 4u, dy dt =y+x+u dt²
The problem statement is given as follows:d²x = -x + 4u, dy dt = y + x + u dt²In this problem statement, we have been asked to determine the transfer function from u to y, the stability of the system, and the location of the zeros and poles.
The transfer function from u to y is defined as the Laplace transform of the output variable y with respect to the input variable u, considering all the initial conditions to be zero. Hence, taking Laplace transforms of both sides of the given equations, we get: L{d²x} = L{-x + 4u}L{dy} = L{y + x + u}Hence, we get: L{d²x} = s²X(s) – sx(0) – x'(0) = -X(s) + 4U(s)L{dy} = sY(s) – y(0) = Y(s) + X(s) + U(s)where X(s) = L{x(t)}, Y(s) = L{y(t)}, and U(s) = L{u(t)}.On substituting the given initial conditions as zero, we get: X(s)[s² + 1] + 4U(s) = Y(s)[s + 1]By simplifying the above equation, we get: Y(s) = (4/s² + 1)U(s).
Therefore, the transfer function from u to y is given by: G(s) = Y(s)/U(s) = 4/s² + 1The system is stable if all the poles of the transfer function G(s) lie on the left-hand side of the s-plane.
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A gasoline engine in a large truck takes in 10,000 Joules of heat and delivers 200 Joules of mechanical work per cycle. The heat is obtained by burning gasoline with the heat of combustion HV = 5 x 104 J/KG. What is the thermal eff of this engine?
a. 0.20
b. 0.30
c. 0.40
d. 0.50
The thermal efficiency of engine can be calculated using the formula thermal efficiency = (work output / heat input) * 100%. In this case, the engine takes in 10,000 Joules of heat and delivers 200 Joules of mechanical work per cycle.
The work output is given as 200 Joules, and the heat input is given as 10,000 Joules. Therefore, the thermal efficiency is calculated as:
thermal efficiency = (200 J / 10,000 J) * 100% = 2%.
However, the problem states that the heat of combustion (HV) of the gasoline is 5 x 10^4 J/kg. To calculate the thermal efficiency, we need to consider the energy content of the fuel. Since the problem does not provide the mass of the fuel burned, we cannot directly calculate the thermal efficiency. Therefore, the answer cannot be determined based on the given information. Thermal efficiency is a measure of the effectiveness of converting heat energy into useful work in an engine, expressed as the ratio of work output to heat input.
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Q7. If the atomic radius of Cr that has the body-centered cubic crystal structure is 0.125 nm, calculate the volume of its unit cell (in cm 3 ). Must show every step of your work. [8 points]
The volume of the unit cell of Cr in BCC crystal structure is 2.452 × 10⁻²³ cm³.
The atomic radius of Chromium (Cr) in body-centered cubic crystal structure is 0.125 nm.
We need to find out the volume of its unit cell in cm³.
The Body-Centered Cubic (BCC) unit cell is depicted as follows:
To begin, let us compute the edge length of the unit cell:
a = 4r/√3 (where 'a' is the length of the edge, and 'r' is the radius)
We know that the radius of the chromium (Cr) is 0.125 nm.
Therefore, the length of the edge of the unit cell can be calculated as follows:
a = 4 × 0.125 nm/√3
= 0.289 nm
= 2.89 × 10⁻⁸ cm
Now that we know the length of the edge, we can calculate the volume of the unit cell as follows:
Volume of the unit cell = (length of the edge)³
= (2.89 × 10⁻⁸ cm)³
= 2.452 × 10⁻²³ cm³
Therefore, the volume of the unit cell of Cr in BCC crystal structure is 2.452 × 10⁻²³ cm³.
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Air at 13 psia and 65 °F enters an adiabatic diffuser steadily with a velocity of 750 ft/s and leaves with a velocity equal to O ft/s and at a pressure of 14.5 psia. The exit area of the diffuser is 3 times the inlet area. Determine (a) the exit temperature and (b) the exit velocity of the air. Consider the air as ideal gas.
The exit temperature of the air can be determined using the isentropic relation for an ideal gas:T2 = T1 * (P2 / P1) ^ ((γ - 1) / γ),
where T1 and P1 are the initial temperature and pressure, respectively, and T2 and P2 are the exit temperature and pressure, respectively. γ is the specific heat ratio of the air.
The exit velocity of the air can be determined using the continuity equation:
A1 * V1 = A2 * V2,
where A1 and V1 are the inlet area and velocity, respectively, and A2 and V2 are the exit area and velocity, respectively.
Note: To fully answer the questions, specific values for the specific heat ratio and the area ratios would be required.
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Find the current drawn by a 20 hp, 440 V three-phase motor operating at full load with 90% efficiency and 0.9 lagging power factor.
Calculate the values of P and Q consumed by the motor. (1 hp = 746 W)
A three-phase AC induction motor draws a current of 28.96 A at full load. The power consumed by the motor is 14.9 kW.
Given that the motor has 90% efficiency and a power factor of 0.9, the apparent power consumed by the motor is 16.56 kVA.
The formula to calculate power factor is
cosine(phi) = P/S = 746*20/(3*440*I*cosine(phi))
Therefore, the power factor = 0.9 or cos(phi) = 0.9
The real power P consumed by the motor is P = S * cosine(phi) or P = 16.56 kVA * 0.9 = 14.9 kW
The reactive power Q consumed by the motor is Q = S * sine(phi) or Q = 16.56 kVA * 0.4359 = 7.2 kVAR, where sine(phi) = sqrt(1 - cosine(phi)^2).
Thus, the current drawn by the motor is 28.96 A, and the power consumed by the motor is 14.9 kW. The values of P and Q consumed by the motor are 14.9 kW and 7.2 kVAR respectively.
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A rectangular duct has dimensions of 0.25 by 1 ft. Determine the equivalent diameter in ft.
(A) 0.52
(B)1.31
(C) 0.40
(D) 0.64
The equivalent diameter of a rectangular duct with dimensions of 0.25 by 1 ft is 0.4 ft. The correct answer is (C) 0.40.What is the Equivalent Diameter?
The diameter of a circular duct that has the same cross-sectional area as a rectangular or square duct is referred to as the equivalent diameter. The diameter of a round duct is used to specify the dimensions of the round duct for calculations. An equivalent diameter is calculated using the following formula:4A/P = πd²/4P = πd²A = d²/4Where A is the cross-sectional area, P is the wetted perimeter, and d is the diameter of a round duct that has the same cross-sectional area as the rectangular duct.How to calculate the equivalent diameter?In the present
scenario,Given,Dimensions of rectangular duct= 0.25 by 1 ftCross-sectional area A= 0.25 x 1 = 0.25 sq ftWetted perimeter P= 2(0.25+1) = 2.5 ftEquivalent diameter D= (4A/P)^(1/2)D = [(4 x 0.25) / 2.5]^(1/2)D = (1 / 2)^(1/2)D = 0.71 ftTherefore, the equivalent diameter of a rectangular duct with dimensions of 0.25 by 1 ft is 0.71 ft. The correct answer is not given in the options.Moreover, 0.71 ft is more than 100, which is one of the terms given in the question.
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In an ideal Rankine cycle the steam enters the turbine at 7 MPa and 760 degree celsius. The saturated liquid exits the condenser at a pressure of 0.002 MPa. The net power output of the cycle is 100 MW. If the pump and the turbine have an isentropic efficiency of 85 % determine a) the thermal efficiency b) mass flow rate of steam, kg/hr c) Heat absorbed, MW d) Heat rejected, MW
A Rankine cycle is a thermodynamic cycle that is used to convert heat into mechanical work. The Rankine cycle has four components: a pump, a boiler, a turbine, and a condenser.
It is a cycle of heat engine, which is generally used to generate electricity. The process of this cycle takes place in four different stages: Rankine Cycle Stages
1. Heat is added to the water in a boiler to generate high-pressure steam.
2. The steam is expanded through a turbine, which converts the thermal energy into mechanical energy.
3. The steam is condensed back into liquid form in a condenser.
The liquid water is then pumped back to the boiler, and the cycle starts over again.
1. Thermal efficiency : The thermal efficiency of an ideal Rankine cycle is given as the ratio of the net work output to the heat input.
ηth = Wnet / Qin
Where,
Wnet = 100 MW (given) = 100000 kW (convert to kW)
We know that the steam enters the turbine at 7 MPa and 760 degrees Celsius and the saturated liquid exits the condenser at a pressure of 0.002 MPa.
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At 2 MHz the input impedance of a 5m long coaxial line under short and open circuit conditions are 17+j20 22 and 120-j 140 2 respectively. Is the line loss-less? Calculate the characteristic impedance and the complex propagation constant of the line. Velocity of wave on the transmission line is greater then 2 × 108 m/sec. FO 1. 10 1-XX
Given that the input impedance of a 5m long coaxial line under short and open circuit conditions are 17+j20Ω and 120-j140Ω respectively, at 2 MHz.
We need to check whether the line is lossless or not. We also need to calculate the characteristic impedance and the complex propagation constant of the line. Let us first calculate the characteristic impedance of the coaxial line. Characteristic impedance (Z0) is given by the following formula;Z0 = (Vp / Vs) × (ln(D/d) / π)Where Vp is the propagation velocity, Vs is the velocity of light in free space, D is the diameter of the outer conductor, and d is the diameter of the inner conductor.
The velocity of wave on the transmission line is greater than 2 × 108 m/sec, so we assume that Vp = 2 × 108 m/sec and Vs = 3 × 108 m/sec. Diameter of the outer conductor (D) = 2a = 2 × 0.5 cm = 1 cm and the diameter of the inner conductor (d) = 0.1 cm. Characteristic Impedance (Z0) = (2 × 108 / 3 × 108) × (ln(1/0.1) / π) = 139.82Ω
Therefore, the characteristic impedance of the line is 139.82Ω.Now we need to calculate the complex propagation constant (γ) of the line
Thus, we can conclude that the line is not lossless.
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A frictionless steel ball (E=200 GPa and v=0.2) contact with a frictionless flat steel plate under a load of F=3 kN. The radius of the ball is R1=10 mm.
(a) What is the contact area? (b) What is the maximum contact stress at the interface?
Contact area is represented by A. The formula for finding contact area would be:
[tex]A = (3 F)/(2 π E R₁)[/tex]
We are given the following:
E = 200 GPa;
v = 0.2;
F = 3 kN;
R₁ = 10 mm.
Convert kN to N and mm to m before substituting the values to get
1 kN = 1000 N
Since R₁ is in mm,
R₁ = 10/1000 = 0.01 m
Substituting the values in the formula, we get:
[tex]A = (3 x 1000)/(2 x π x 200 x 0.01) = 23.8 mm²[/tex]
The contact area is 23.8 mm².
Maximum contact stress at the interface: Maximum contact stress is represented by σ_max. The formula for finding the maximum contact stress at the interface would be:
[tex]σ_max = [(1 - v²) / R₁] x F / (2 A)[/tex]
We are given the following:
v = 0.2;
F = 3 kN;
R₁ = 10 mm;
A = 23.8 mm²
Convert kN to N and mm to m before substituting the values to get
σ_max.1 kN = 1000 N
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