The sliding window technique is utilized to identify the crash frequency of a certain region. A performance measure called "Excess Predicted Average Crash Frequency Using SPFs" will be used to screen Segment A in the urban four-lane divided arterial reference population.
The segment is 0.60 mi long. Let's determine the number of times the performance measure will be applied to Segment A using the sliding window method.In 0.30-mi windows, the section is analyzed. The increment is 0.10 miles long. As a result,0.30 mi long window = 0.60 / 0.30 = 2 windows.0.10 miles long increment = 0.60 / 0.10 = 6 increments.So, the total number of applications = number of windows × number of increments in each window= 2 × 6= 12.The performance measure will be used 12 times on Segment A. Answer: In 200 words. The sliding window technique is utilized to identify the crash frequency of a certain region. A performance measure called "Excess Predicted Average Crash Frequency Using SPFs" will be used to screen Segment A in the urban four-lane divided arterial reference population. The segment is 0.60 mi long. Let's determine the number of times the performance measure will be applied to Segment A using the sliding window method.In 0.30-mi windows, the section is analyzed. The increment is 0.10 miles long. As a result,0.30 mi long window = 0.60 / 0.30 = 2 windows.0.10 miles long increment = 0.60 / 0.10 = 6 increments.So, the total number of applications = number of windows × number of increments in each window= 2 × 6= 12.The performance measure will be used 12 times on Segment A.
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Given: mechanical spreading of cement at 3.2% by mass; the layer maximum mod.AASHTO density is 3500 kg/m³ and the specified density for stabilised sub-base is 97% of mod.maximum. layer is given as 150mm thick. Calculate the spread rate per square metre (kg/m²) if the portion to be stabilised is 1450 m long the sub-base is 11 m wide (7)
Given data: Mechanical spreading of cement at 3.2% by mass; The layer maximum mod. AASHTO density is 3500 kg/m³The specified density for stabilized sub-base is 97% of mod. maximum. The layer is given as 150 mm thick. The portion to be stabilized is 1450 m long.
The sub-base is 11 m wide. To find: The spread rate per square meter (kg/m²)Calculation: Thickness of layer = 150 mm = 0.15 m Maximum Modified AASHTO density = 3500 kg/m³The specified density for stabilized sub-base = 97% of Maximum Modified AASHTO density= 97/100 × 3500 kg/m³= 3395 kg/m³Length of the portion to be stabilized = 1450 mWidth of sub-base = 11 m.
So, the mass of cement required to achieve the maximum Modified AASHTO density= 0.98 kg/m³ × 3500 kg/m³= 3430 kg/m³Total mass of stabilized soil required to achieve the specified density for stabilized sub-base= Specified density × Maximum Modified AASHTO density= 97/100 × 3500 kg/m³= 3395 kg/m³.
So, the mass of cement required to stabilize 1 m³ of soil= 3395 – 3500= - 105 kg/m³The negative sign indicates that no cement is required for stabilization because the maximum Modified AASHTO density is already achieved. Spreading rate of cement= 0.98 kg/m³ / 0.15 m= 6.53 kg/m²Thickness of stabilized soil layer = 0.15
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h= 300mm b =190mm t = 4mm L/h = 12.5
Question: Now ignore self weight. The beam is loaded by two vertical point loads, each of magnitude P, that act at L/3 and 2L/3 along the beam. What is the maximum value of P so that the stress criteria above are satisfied? Draw the stress and strain distributions (values required) for both the steel and the timber on the critical cross-section for this value of P.
Given that;h = 300 mmb = 190 mmt = 4 mmL/h = 12.5Beam is loaded by two vertical point loads, each of magnitude P, that act at L/3 and 2L/3 along the beam.
The maximum value of P so that the stress criteria above are satisfied will be calculated.To calculate the maximum value of P, we need to calculate the following;Stress in steelStress in timberShear stress in steelShear stress in timberStep-by-step solution is given below.
Let's first draw the FBD of the beam for an easier calculation;[tex]\frac{1}{2}P[/tex] at [tex]\frac{L}{3}[/tex][tex]\frac{1}{2}P[/tex] at [tex]\frac{2L}{3}[/tex]Take a small segment at distance x from the left support and calculate the shear force on that section.
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A three- member frame ABCD is loaded with ultimate gravity and lateral loads as shown in the figure. The compressive strength of concrete= 21 MPa, fy = 415 MPa, beam width= 300 mm, beam height = 600 mm, main bar diameter 28 mm, concrete cover = 40mm, and stirrups diameter = 10mm. Calculate the area of tension As2 to balance the compression in the steel in mm²
Given data: Compressive strength of concrete = 21 MPa fy = 415 MPaBeam width = 300 mmBeam height = 600 mmMain bar diameter = 28 mm Concrete cover = 40 mm Stirrups diameter = 10 mm From the figure given, the load acting on the frame is as follows;
Ultimate lateral load on frame = 250 kNAs we know, Steel will be in tension and concrete will be in compression. To balance the compression in the steel we have to calculate the area of tension As2.
As per Indian standard, the value of n for mild steel and high strength deformed bars are 1.5 and 1.8, respectively. Since we have given fy= 415 MPa which belongs to the high strength deformed bars, n = 1.8 Therefore,$$f_{st} = \frac {0.87x415}{1.8} = 200.76 MPa.
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For the selection of friction factor, f from the Moody diagram, select the most appropriate answer. O For smooth-walled pipes, we need to know Re only. O All the answers are correct. For laminar flow, we need to know Re only. O For rough turbulent flow, we need to know Re vs k/D.
The most appropriate answer is "For rough turbulent flow, we need to know Re vs k/D" for the selection of friction factor, f from the Moody diagram. A friction factor is an essential term used in fluid dynamics and hydrodynamics to calculate the pressure drop or head loss caused by friction of a fluid moving through a pipe.
Friction factor is a dimensionless term. It is used to determine the fluid's resistance to flow in the pipe network or channel. The choice of friction factor, f, determines the accuracy of the calculations. Friction factor, f, can be chosen using different methods. The most common method used for selecting f is by using the Moody diagram.
For laminar flow, we need to know Re only. For smooth-walled pipes, we need to know Re only. But for rough turbulent flow, we need to know Re vs k/D. The Reynolds number (Re) is a dimensionless quantity that represents the ratio of the inertial forces to the viscous forces.
The friction factor (f) depends on the Reynolds number and the relative roughness of the pipe (k/D). The relative roughness (k/D) is the ratio of the average roughness height to the pipe's internal diameter. The friction factor (f) can be selected from the Moody diagram by plotting the Reynolds number (Re) against the relative roughness (k/D) and reading the value of f from the chart.
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Discuss two (2) points related to the importance of Health and
Safety Standards in the wood processing industry.
Health and safety standards in the wood processing industry are crucial for the safety and protection of workers in the industry. They must follow strict procedures to ensure that they are not exposed to harmful elements, and that their health and well-being are not compromised.
Following are the two points related to the importance of Health and Safety Standards in the wood processing industry.
Employee Safety: Wood processing can be dangerous work, and the workers involved are at risk of injury or illness if the proper health and safety precautions are not taken. Employee safety is a top priority in the industry, and the introduction of health and safety standards helps to ensure that workers are protected from potential hazards such as chemical exposure, dust inhalation, and fire hazards. By following these standards, employers can reduce the likelihood of accidents occurring and minimize the risk of injuries or illnesses to their workers.Compliance: Another important aspect of health and safety standards is compliance. Employers who do not follow these standards can face legal action, fines, and even criminal charges in some cases. Compliance ensures that companies are held accountable for their actions and that they take the necessary steps to protect their workers. In addition, companies that follow health and safety standards are seen as more reputable and trustworthy, which can help to attract and retain employees, and maintain good relationships with clients and customers.In conclusion, health and safety standards in the wood processing industry are essential for the protection of workers and the success of companies in the industry. By prioritizing employee safety and compliance with regulations, employers can minimize the risks associated with wood processing and maintain a safe and healthy work environment.
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mono In the crystal volume, while Frenkel defect results in an overall _________ in the crystal volume ICLO-3 Shottky defect in ceramics results in an overal ____________in the crystal volume. a. decrease, no change b. intense, decrease c. decrease increase d. no change decrease
Therefore, the answer is decrease, In a crystal volume, Mono-vacancy and Frenkel defect are the two primary forms of defects that occur. These defects influence the volume of the crystal as a whole.
It also affects the electrical and mechanical properties of the material. Mono-vacancy happens when a cation or anion is absent from a lattice point, creating a vacancy.
In the crystal lattice structure, this defect occurs. When anion or cation vacancies are present, this is referred to as a Schottky defect.
The Schottky defect has no effect on the overall crystal volume. This is due to the fact that the vacancy site of one cation or anion is matched with a vacancy site of another cation or anion, resulting in no net change in volume.
The Frenkel defect, on the other hand, results in an overall decrease in crystal volume. In ceramics, a type of Frenkel defect known as the Schottky defect occurs. In this type of defect, an ion migrates from its original position to an interstitial position. This causes the surrounding atoms to be pulled closer together, resulting in an overall decrease in crystal volume.
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Given the velocity profile in the angular direction in the gap between concentric cylinders where k is the ratio of inner to outer radii: and ω is the angular frequency in radians/sec, as a function of radial position r, show that the torque on the inner and outer cylindrical surfaces are equal. The shear rate in cyclindrical coordinates is: σ=
The shear stress is inversely proportional to r^2 and is the same for both the inner and outer surfaces. Thus, the torque on the inner and outer cylindrical surfaces are equal.
Given the velocity profile in the angular direction in the gap between concentric cylinders where k is the ratio of the inner to outer radii, and ω is the angular frequency in radians/sec as a function of radial position r, the torque on the inner and outer cylindrical surfaces are equal.
The shear rate in cylindrical coordinates is given as:
σ= (dV_r / dr) + (1 / r) (dV_θ / dθ)
For concentric cylinders, the flow is in the radial direction, and the only non-zero component of velocity is the radial component i.e. V_r. Therefore, the velocity profile will be V_r = A + B/r, where A and B are constants of integration.
Since the torque T is proportional to the surface area, the torque on the inner and outer surfaces is given by:
T_inner = 2πrL(τ_rr(r = r_1))
T_outer = 2πrL(τ_rr(r = r_2))
Where τ_rr is the shear stress in the radial direction, r_1 is the inner radius, r_2 is the outer radius, and L is the length of the cylinders.
To show that the torque on the inner and outer cylindrical surfaces are equal, we need to show that τ_rr is equal for both surfaces.
τ_rr is related to the shear rate σ by the equation:
τ_rr = 2ησ
Where η is the dynamic viscosity.
Substituting the velocity profile into the equation for shear rate σ, we get:
σ= (dV_r / dr) + (1 / r) (dV_θ / dθ)= -B/r^2
Since the flow is in the radial direction only, there is no angular velocity component i.e. dV_θ / dθ = 0.
Substituting σ into the equation for τ_rr, we get:
τ_rr = 2ησ= -2ηB/r^2
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Wooden planks 300 mm wide by 100 mm thick are used to retain soil with height 3 m. The planks used can be assumed fixed at the base. The active soil exerts pressure that varies linearly from 0 kPa at the top to 14.5 kPa at the fixed base of the wall. Consider 1-meter length and use modulus of elasticity of wood as 8.5 x 〖10〗^3 MPa. If the wall is propped at the top, determine the lateral force (kN) at the fixed end.
Wooden planks 300 mm wide by 100 mm thick are used to retain soil with height 3 m. The planks used can be assumed fixed at the base. The active soil exerts pressure that varies linearly from 0 k
Pa at the top to 14.5 kPa at the fixed base of the wall. Consider 1-meter length and use modulus of elasticity of wood as 8.5 x 〖10〗^3 MPa. If the wall is propped at the top, determine the lateral force (kN) at the fixed end.Propping the wall at the top means that there is a support of some sort that is preventing the wall from moving in that direction. As a result, the amount of lateral force exerted at the fixed end will be equal to the amount of lateral force that would have been applied to the top, had it not been propped.
In this case, the lateral force (F) can be calculated using the formula: F = (1/2)γH² + (1/3)αH³where γ is the unit weight of the soil, H is the height of the soil, α is the pressure coefficient. The formula for pressure coefficient is:α = (1 - sinϕ)/(1 + sinϕ)where ϕ is the angle of friction. If the wall is propped at the top, then the lateral force will be: [tex]F = (1/2)γH² + (1/3)αH³Where H = 3 mα = (1 - sin(0))/(1 + sin(0)) = 0γ = 14.5/1000 = 0.0145 MPa F = (1/2) × 0.0145 × 3² + (1/3) × 0 × 3³F = 0.1305 kN[/tex]The lateral force at the fixed end will be 0.1305 kN.
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Predict the penalty for each of the situations below, and quote the referral section of The Occupational Safety and Health Act 1994 (Act 514) in your answer.
i.Company FEDCBA Sdn Bhd employed a non-certified Safety Officer for a project worth RM 380 million.
ii.Mr Mowk refused to wear a full body safety harness provided by the employer while working at height.
iii.Mr Chow works as a clerk of works on a construction site in Parit raja. One day, he accidentally trips and falls due to the improper condition of the site office, which has caused him severe injury.
iv. A worker had filed a complaint to the Department of Safety & Health (DOSH) with regards to the unsafe working environment at his workplace. Two days later, he was transferred to another construction site located deep into the jungle with no general amenities such as tap water and electricity. At the end of the month, he was fired.
V. AmBuild Sdn Bhd charge everything related to personal safety equipment to its employees by salary deduction.
i. Company FEDCBA Sdn Bhd employed a non-certified Safety Officer for a project worth RM 380 million. The employer can be fined not more than RM 50,000 or imprisonment for a term not exceeding 6 months, or both. Referral Section: Section 15(1) of the Occupational Safety and Health Act 1994 (Act 514).
ii. Mr. Mowk refused to wear a full body safety harness provided by the employer while working at height. The employee can be fined not more than RM 1,000 or imprisonment for a term not exceeding 3 months, or both. Referral Section: Section 15(1) of the Occupational Safety and Health Act 1994 (Act 514).
iii. Mr. Chow works as a clerk of works on a construction site in Parit Raja. One day, he accidentally trips and falls due to the improper condition of the site office, which has caused him severe injury.
iv. A worker had filed a complaint to the Department of Safety & Health (DOSH) with regards to the unsafe working environment at his workplace. Two days later, he was transferred to another construction site located deep into the jungle with no general amenities such as tap water and electricity. At the end of the month, he was fired. The employer can be fined not more than RM 10,000 or imprisonment for a term not exceeding 1 year, or both. Referral Section: Section 20(1) of the Occupational Safety and Health Act 1994 (Act 514).
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1. An urban city has water demand 138,005 m³/day/km² and water supply 17,581 m³/day/km², determine the corresponding water resources vulnerability index (?%).2.A sewage system needs to be designed for 26 year design flood. What would be the chance for the system to be safe in 23 years (accurate to 2 decimal point)?Question 8 Use rational method to determine the runoff (L/s) given the catchment area of 18 km² and rainfall intensity of 38 cm/hour assuming the runoff production coefficient is C=0.5.
Answer:1. An urban city has water demand [tex]138,005 m³/day/km²[/tex] and water supply[tex]17,581 m³/day/km²[/tex], determine the corresponding water resources vulnerability index (?%).
The corresponding water resources vulnerability index can be calculated as follows:
Water Resources Vulnerability Index (%) = (Water Demand - Water Supply) / Water Demand X 100%.Substituting the given values in the formula we get,Water Resources
Vulnerability Index = [tex](138005 - 17581) / 138005 X 100% = 87.29%.[/tex]
Therefore, the corresponding water resources vulnerability index is 87.29%.2.
A sewage system needs to be designed for 26 year design flood. Therefore, the probability that the sewage system will be safe in 23 years is 0.0377.3.
Use rational method to determine the runoff (L/s) given the catchment area of 18 km² and rainfall intensity of 38 cm/hour assuming the runoff production coefficient is C=0.5.
The formula for the rational method is given by,
[tex]Q = CIA = 0.5 X 38 X 18,000,000 / 3,600 = 190,000 L/s[/tex]
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Visit a website of a photovoltaic module manufacturer such as SHARP and look up the following information for a new product: the wattage, dimensions, and weight. How many modules would be needed to provide 4000 kWh of electricity per year for where you live? How much would it cost for the array? Also investigate the cost and types of batteries used in photovoltaic systems to store energy to be used during cloudy days. Present your findings in a brief report. (I will be using Atlanta area)
In order to answer this question, we need to follow the given steps:S tep 1: Find a manufacturer's website that sells photovoltaic modules.Step 2: Look up the following information about a new product: wattage, dimensions, and weight. Step 3: Determine how many modules are required to provide 4000 kWh of energy per year in Atlanta. Step 4: Determine the cost of the solar array.
Step 5: Research the cost and types of batteries used in photovoltaic systems for storing energy to be used during cloudy days.
Based on the above steps, we can proceed further.
The SHARP 350-watt photovoltaic module has dimensions of 77.5 in × 39.1 in × 1.5 in and weighs 51.8 lbs. One module will produce about 1,300 kWh of electricity each year in Atlanta, which is a good average.4000 kWh ÷ 1,300 kWh/module = 3.07 (approximately 3 modules)
Hence, a minimum of 3 modules would be required to generate 4000 kWh of energy per year in Atlanta.
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(b) Suspended concrete floors are commonly used in multi-storey building construction. i. Describe the types of loads on formwork that must be taken into account during the construction of suspended concrete floors. ii. Discuss the importance of formwork stripping time and the key factors that may influence it.
Suspended concrete floors are typically used in multi-storey building construction. Formwork loads and formwork stripping time are two important considerations in the construction of suspended concrete floors. Let us discuss them in more detail.
1. Dead Load: The weight of the formwork and its associated equipment is referred to as dead load.
2. Live Load: The weight of workers, materials, and equipment on the formwork during the construction process is referred to as live loa.
3. Wind Load: The load produced by wind, which may affect the stability of the formwork.
4. Vibration Load: The load caused by mechanical vibrations, which can cause instability in the formwork.
Importance of formwork stripping time and key factors that may influence it:
The period between the time the concrete is poured and the time the formwork is removed is known as formwork stripping time. The following are some of the factors that influence it:
1. Concrete Strength: Stripping time is influenced by the strength of the concrete. The formwork can be removed sooner if the concrete has a higher strength.
2. Environmental Conditions: Environmental factors such as temperature, humidity, and wind can all have an effect on formwork stripping time.
3. Formwork Type: The type of formwork used and its quality can impact the formwork stripping time. Some types of formwork, for example, can be removed faster than others.
4. Design of Concrete: The design of the concrete can also impact the formwork stripping time. The formwork can be removed sooner if it is designed to be less susceptible to cracking.
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Safety Management System (SMS) is a safety model that is widely
adopted in aviation industry. Please explain the process in detail
with an example illustrated.
Safety Management System (SMS) is a safety model that is widely adopted in the aviation industry. It is an integrated and proactive approach to safety management that is used to identify and manage safety risks. The process of SMS is detailed below:
1. Establishing Safety Policy: This involves developing a safety policy that outlines the organization's commitment to safety and sets clear safety goals and objectives. The safety policy should also include the roles and responsibilities of all personnel involved in the safety management process.
2. Risk Assessment: This involves identifying and assessing safety risks in the organization's operations. This can be done through hazard identification and risk assessment tools such as checklists, safety audits, and safety assessments.
3. Risk Mitigation: Once the risks have been identified, the next step is to implement controls to mitigate the risks. This can include changes to procedures, equipment, training, and other factors that contribute to safety.
The airline would regularly review and update its safety management system to ensure that it remains effective in managing safety risks. This would involve continuous improvement and a commitment to ongoing safety management in order to ensure the safety of its passengers, employees, and operations.
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QUESTION 11 There are many processes that can cause a soil to become over-consolidated, including: a) Surcharge loading from a glacier, which has since melted. True False
Over-consolidation of soil is when the maximum effective stress has been greater than the current effective stress. This condition can occur due to various natural phenomena such as glacial surcharge loading, earthquake tremors, and water-level changes in nearby rivers or lakes.
Glacial surcharge loading is a natural phenomenon that can cause soil to become over-consolidated. During the last glaciation period, ice sheets spread across large parts of the globe. When the glaciers advanced, they compressed and deformed the soil beneath them.
This deformation increased the stress on the soil, causing it to become over-consolidated. When the glaciers melted, the over-consolidated soil was left behind. This type of over-consolidated soil is found in many parts of the world, including Canada, Scandinavia, and the northern United States.
The statement "Surcharge loading from a glacier, which has since melted" is true. Glacial surcharge loading is one of the many natural processes that can cause soil to become over-consolidated.
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A piping system consists of parallel pipes as shown in the following diagram. Both pipes are made of cast iron. The pipes are transporting water at 20oC. The total flow rate is 4 m3/s. Find the flow rate in the smaller pipe. There is no elevation change. Neglect minor losses.
The pipe system consists of two parallel pipes, both made of cast iron, transporting water at 20°C, and with a total flow rate of 4m³/s. We need to determine the flow rate in the smaller pipe. We can apply the equation of continuity to solve for the required flow rate. Conservation of mass in a pipe system is governed by the principle of continuity.
The principle of continuity, also known as the conservation of mass, states that the mass of water flowing into the system is equal to the mass of water flowing out of it. The continuity equation is expressed mathematically as:A1V1 = A2V2, where A is the cross-sectional area and V is the velocity.
We can re-write the continuity equation as Q = VA, where Q is the volumetric flow rate, which is equal to the product of velocity and area. The volumetric flow rate, or Q, is equal to the total flow rate of the pipes. Let's use subscript 1 to denote the larger pipe and subscript 2 to denote the smaller pipe. We have the following information:
Q = 4 m³/sQ = Q₁ + Q₂A₁ = πD₁²/4A₂ = πD₂²/4D₁ = 200 mm = 0.2 mD₂ = 100 mm = 0.1 m Substituting the given values into the continuity equation, we obtain: Q = Q₁ + Q₂ ⇒ 4 = Q₁ + Q₂ Since both pipes are transporting water at the same temperature and the elevation changes are negligible, the major losses in both pipes are the same.
We can thus assume that the velocity head losses are equal in both pipes. Therefore, the pressure drop in each pipe will be proportional to its length and to the square of the velocity. We can express this mathematically as follows:ΔP = K(V²/2g)L, where ΔP is the pressure drop, K is a friction factor, V is the velocity.
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How can you check the California bearing ratio on-site, once the
pavement has been constructed?
The California Bearing Ratio (CBR) is an index indicating the strength of soil used for the construction of roads and pavements. Here is how to check the California Bearing Ratio on-site once the pavement has been constructed.
Step 1: At the center of the pavement, drill a hole measuring 150mm in diameter and 225mm in depth. Step 2: Insert the permeable metal disc that measures 50mm in height and 146mm in diameter into the hole. The disc should be attached to a baseplate that measures 254mm x 254mm. Ensure that the rim of the metal disc is level with the surface of the pavement. Step 3: Fill the annular space between the hole and the disc with water. Step 4: Measure the height of water in the cylindrical chamber connected to the center of the metal disc .Step 5: Apply a static load of 4,536 kg (10,000 pounds) to the top surface of the baseplate.
The load should be applied uniformly without any sudden shock or impact. Step 6: Measure the height of the water level again in the cylindrical chamber.
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A member of a top chord in a heavy pin connected carrying truss of an industrial building is 20-ft long and carries a load of 630-kips. its made up of 2 channels C15x50, and were spaced 12 inches back to back. the member has a 1-inch thick cover plate at the top. Investigate the adequacy of the member under LRFD and ASD. (Use fy = 50 ksi; k = 1.0; E = 29,000 ksi) C15x50 properties: Load (compression):
Ag = 14.7 in 2 DL = 460 kips
d = 15 in LL= 170 kips
b=3.716 in
Ix=404 in4
ly=11 in4
Xc=0.798 in
The solution to the given problem is as follows. A member of a top chord in a heavy pin connected carrying truss of an industrial building is 20-ft long and carries a load of 630-kips.
It's made up of 2 channels C15x50, and were spaced 12 inches back to back. The member has a 1-inch thick cover plate at the top. Investigate the adequacy of the member under LRFD and ASD. (Use fy = 50 ksi, k = 1.0, E = 29,000 ksi).The load-carrying capacity of a top chord member of an industrial building is to be investigated.
The member has a length of 20 ft and supports a load of 630 kips. The top chord member is made up of two channels C15x50 which are spaced 12 inches back to back, and a cover plate at the top. The steel is ASTM A572 Grade 50.The gross area of the top chord member can be calculated as follows.
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A composite section has an effective flange width of 2100mm and a slab with thickness of 150mm. The steel beam is a W530x101 of A-36 steel with Fy = 250 MPa. Concrete strength fc’ = 20.7 MPa. Modular Ratio = 9.0. Assume full shoring and full composite action.
Determine the modular ratio.
Determine the moment of inertia of the composite beam about the neutral axis.
Determine the resisting moment of the section.
1. The modular aspect ratio is around 5.73. 2. The composite beam's moment of inertia (Icomposite) is equal to the sum of its constituent moments of inertia:Steel and concrete are combined to form composite materials.3. The formula Mresisting = n * (Icomposite / d) may be used to compute the section's resisting moment (Mresisting).
The calculation is as follows:
We must solve the following equation to find the modular ratio:
Es / Ec is the modular ratio (n).
where Ec is the elastic modulus of concrete and Es is the elastic modulus of steel.
You can suppose that steel has an elasticity modulus of 200 GPa (200,000 MPa).
The following equation may be used to get the concrete's elasticity modulus (Ec):
Ec = 4700 √(fc')
where fc' is the concrete's MPa compressive strength.
Since fc' = 20.7 MPa, we can figure out Ec:
Ec = 4700 √(20.7) ≈ 34,877 MPa
We can now determine the modular ratio:
Es / Ec = 200,000 / 34,877 5.73 is the modular ratio.
Therefore, the modular ratio is around 5.73.
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State whether the following statement is true or false: the
Fanning friction factor is a function of the pipe’s roughness in
the turbulent flow regime.
The statement "the Fanning friction factor is a function of the pipe’s roughness in the turbulent flow regime" is true. The Fanning friction factor is commonly used to calculate the frictional pressure drop in a pipe flow system and is affected by the roughness of the pipe's inner surface.
In the turbulent flow regime, the fluid moves in a chaotic, irregular manner, with eddies and vortices causing a mixing of fluid layers. The roughness of the pipe's inner surface affects the frictional drag between the fluid and the pipe wall, which in turn affects the Fanning friction factor.
The roughness of a pipe's inner surface is typically characterized by the absolute roughness, ε, which is defined as the average height of surface irregularities relative to the pipe's diameter.
[tex]f = ΔP / (0.5 * ρ * V^2 * L / D)[/tex]where ΔP is the pressure drop along the pipe, ρ is the fluid density, V is the mean flow velocity, L is the pipe length, and D is the pipe diameter.
The Fanning friction factor, f, is a function of the Reynolds number, Re, which is defined as the ratio of inertial forces to viscous forces. In the turbulent flow regime, the Fanning friction factor is also affected by the pipe roughness, and it can be calculated using empirical relations such as the Colebrook equation.
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Construction contract is often known as an advanced form of contract. (a) Give an account with examples to the purposes of these advance features. (b) Describe the advance features under – i. Standard Form of Building Contract; ii. New Engineering Contract (c) From your own analysis, discuss the strengths and weaknesses of the advance features discussed under (b) and grade the advancements in your terms.
Construction contract is often known as an advanced form of contract. Here are the purposes of advance features of the construction contract along with the examples:(a) Purposes of the advance features:
Provide balanced risk between the client and the contractor. Encourage promptness and cooperation in the parties in the event of delays or hindrances. Establish a procedure for the resolution of disputes, preferably by amicable means. Provide a clear description of the parties' duties and responsibilities. Define the project's overall scope and quality requirements.Examples: FIDIC, JCT, ICE, GC/Works, NEC, and the like.(b) Advanced features under –i. Standard Form of Building Contract:
The Standard Form of Building Contract is usually more involved than other forms of contracts. The requirements are more detailed and comprehensive. The primary goal is to strike a balance between the client's risk and the contractor's risk.ii. New Engineering Contract: New Engineering Contract is the most popular standard form of contract in the United Kingdom. It was created in response to the negative aspects of the old standard forms. NEC contracts have some basic components that provide a good framework for constructing a contract.
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Provide a detailed project plan example such as building new class rooms for a school and you need to come up with a project to cut some cost. using the information below as a guide.(I want a detail solution.)
Identify and assess stakeholders
Develop a communication plan
c. Define project scope
d. Decompose project scope into stakeholder requirements (Requirement)
e. Create WBS and dictionary
f. Create risk register
Project Plan for Building New Classrooms in a School Introduction Building new classrooms in a school is a huge project that requires careful planning and management. In this project plan, we will provide a detailed guide on how to build new classrooms for a school while cutting down on cost.
Identify and Assess Stakeholders The first step is to identify and assess stakeholders. The key stakeholders in this project include the school administration, teachers, students, and parents. We will involve them in every stage of the project to ensure their input is considered. Develop a Communication Plan The communication plan is vital in any project.
The plan will outline the communication channels and protocols to be used in communicating with the stakeholders. We will use emails, phone calls, and regular meetings to keep the stakeholders informed about the progress of the project. Define Project Scope The project scope outlines the objectives, deliverables, and goals of the project. Our project scope will be to build new classrooms for the school while reducing costs.
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Polyethene pipe, 250 m long and 120 mm in diameter, with a water flow of 20 litres/sec.
Calculate the energy loss due to friction using:
1. Hazen-William equation
2. Scooby equation
3. Darcy-Weisbach equation Use the Churchill equation and then Moody curves to determine the coefficient of friction
When water flows through a pipe, it encounters friction which causes the energy of water to dissipate. The energy loss due to friction is an important factor that must be taken into account in any engineering calculations. The energy loss due to friction can be calculated using the Hazen-William equation, Scooby equation, and Darcy-Weisbach equation.
We will use the Churchill equation and then Moody curves to determine the coefficient of friction. Given data: Length of pipe (L) = 250 m Diameter of pipe (d) = 120 mm = 0.12 mWater flow rate (Q) = 20 litres/sec = 0.02 m3/secHazen-William equation The Hazen-William equation is given by: ƒ = 0.2083 × (C)1.85 × (d)-4.87 × (L)-1.85Where ƒ is the friction factor, C is the Hazen-Williams coefficient, d is the diameter of the pipe, and L is the length of the pipe.
The Hazen-Williams coefficient for polyethylene pipes is 150. We can substitute the given values to calculate the friction factor as follows: ƒ = 0.2083 × (150)1.85 × (0.12)-4.87 × (250)-1.85ƒ = 0.0191The energy loss due to friction can be calculated using the following formula: ΔE = ƒ × (L/d) × (ρ/2) × (v2)
Where ΔE is the energy loss due to friction, L/d is the length to diameter ratio, ρ is the density of water, and v is the velocity of water. The density of water is 1000 kg/m3.
The velocity of water can be calculated as follows: v = Q/A Where A is the cross-sectional area of the pipe. The cross-sectional area of the pipe can be calculated as follows: A = (π/4) × (d)2A = (π/4) × (0.12)2A = 0.0113 m2v = Q/A = 0.02/0.0113 = 1.77 m/s Substituting the given values in the formula, we get:
ΔE = 0.0191 × (250/0.12) × (1000/2) × (1.77)2ΔE = 4097.5 Joules Scooby equation The Scooby equation is an empirical equation used to calculate the friction factor.
It is given by: 1/√ƒ = 2.0 × log10(3.7 × d/ε) + 1.74/Re0.9
Where ε is the roughness of the pipe, and Re is the Reynolds number. The roughness of polyethylene pipes is 0.0001 mm. The Reynolds number can be calculated as follows: Re = (ρ × d × v)/μWhere μ is the dynamic viscosity of water. The dynamic viscosity of water is 0.001 N-s/m2.
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8(a)A coal-fired factory burns 10 tons of coal per hour and discharges the combustion products through a stack with an effective stack height of 100 m. The coal has a sulfur content of 5%, and the wind velocity at the top of the stack is 8.0 m/s. Atmospheric conditions are moderately unstable. Determine the maximum ground-level concentration of SO2 and the distance from the stack at which the maximum occurs. (Please refer to the graphs in appendix).
8(b)(i)A convective kerosene heater is tested in a well-mixed 27 m³ chamber having an air exchange rate of 0.39 hr¹. After 1-hr operation, the NO concentration reached 4.7 ppm. Assume the ambient NO concentration is zero and there is no NO before the test. Treating NO as a conservative pollutant.
Estimate the NO source strength of the heater (mg/hr).
(ii) Estimate the NO concentration (µg/m³) that would be expected in the lab 1 hr after turning off the heater.
(iii)If this heater were to be used in a "tight" 300 m³ home with 0.2 hr¹, what steady-state concentration of NO would you expect to be caused by the heater?
8(a) The ground level concentration of SO2 is given by: Ground level concentration = E × H / (Q × u × (1 + σ × H / d)2 )Given that the factory burns 10 tons of coal per hour, it means that 10000 kg/h is burnt. Thus, the sulfur dioxide produced is 5% of 10000 kg/h = 500 kg/h. 500 kg of SO2/h / 3600s/h = 0.14 kg of SO2/s = 140 g/s. The molar mass of SO2 is 64 g/mol. The SO2 emission rate is:
140 g/s / 64 g/mol = 2.2 mol/s The effective stack height is 100 m The wind velocity at the top of the stack is 8.0 m/s Atmospheric conditions are moderately unstable E = 3.1H = 100 mQ = 2.2 mol/su = 8.0 m/sσ = 0.3d = The graph in Appendix A gives a value of 1.30 for σ/H = 0.003The maximum ground level concentration of SO2 is:
Ground level concentration = E × H / (Q × u × (1 + σ × H / d)2 )= 3.1 × 100 / (2.2 × 8 × (1 + 1.30))2= 0.028 mg/m³The distance from the stack at which the maximum concentration occurs is given by: d = H × sqrt(E / Q × u × sqrt(1 + σ × H / d))Let D = d / H; then solving for D:0.0309D4 + 0.007D3 - 0.004D2 + 0.002D - 0.0008 = 0This is a quartic equation with no easy solution.
We use a spreadsheet program to solve for D = d/H = 1.02. Therefore, d = 102 m.8(b)(i)The source strength of the heater is given by: Source strength = C i V / 1000Where,C i is the initial pollutant concentration (in ppm)V is the volume of the chamber (in m³)C o is the final pollutant concentration (in ppm)t is the duration of the test (in hours).
The ground level concentration of NO is: Ground level concentration = E × H / (Q × u × (1 + σ × H / d)2 )= 3.1 × 100 / (0.0556 × 1.10 × (1 + 0.019 × 100 / d)2)= 0.00103 d - 0.107The maximum concentration will occur when d = 55 m. At this distance, the maximum ground level concentration of NO is:
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Evaluate the influences of solutions on the your project in regards to four of the following:
- Elegance.
- Robustness.
- Cost and resources.
- Time.
- Skill required.
- Safety.
- Future development potential.
Conclusion about the importance of these perspectives for judge the optimal solution.
In this response, we will evaluate the influences of solutions on the project in regards to four of the following: elegance, robustness, cost and resources, and future development potential.
Elegance: Elegance is often a concern in software engineering and is determined by how clean and concise the code is. The more elegant the solution, the easier it is to maintain, read and understand. It can also lead to fewer bugs and a better user experience.
Robustness: Robustness refers to how well the solution can handle unexpected conditions. The more robust the solution, the less likely it is to fail. A robust solution is desirable for critical applications, such as medical or financial systems.
Cost and resources: Cost and resources are important considerations in any project. The solution must be cost-effective, and the resources required must be within budget. The more expensive the solution, the less feasible it is.
Time: Time is another critical factor in project management. The solution must be delivered on time and within the allotted schedule. If the solution takes too long to develop, the project may fail or become irrelevant.
Skill required: The skill required to develop the solution is another consideration. The team must have the necessary skills to develop the solution, or they may need to acquire additional resources. If the required skills are not available, the project may be delayed or become more expensive.
Future development potential: the future development potential of the solution is important. The solution should be flexible and scalable, allowing for future enhancements and modifications.
the perspectives of elegance, robustness, cost and resources, time, skill required, and future development potential are all critical factors in determining the optimal solution for a project. The optimal solution must be cost-effective, robust, and delivered on time while being maintainable, scalable, and flexible enough to meet future needs.
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(1) A jet of water 22.5 cm in diameter with a discharge of 0.2388 m³/s strikes a flat plate at an angle of
30° to the normal of the plate. If the plate itself is moving with a velocity of 1.5 m/s and in the
direction of the normal to its surface, calculate:
(i) the normal force exerted on the plate.
(i)the workdone per second on the plate and the efficiency.
(i) The normal force exerted on the plate is 1368 N
(i)the work done per second on the plate and the efficiency is 0.051 or 5.1%.
How to solveThe velocity of the water relative to the plate is found using vector subtraction:
Vw = \sqrt((Q/(\pid²/4))^2 + Vp^2 - 2(Vp)(Q/(\pid²/4))cos(30°)), where Q = 0.2388 m³/s, d = 0.225 m, and Vp = 1.5 m/s.
The change in momentum of the water in the normal direction per second is: Δp = 2Q(Vw*cos(30°)-Vp).
The normal force exerted on the plate is F = Δp = 1368 N (rounded to nearest integer).
The work done per second (power) is: P = F * Vp = 2052 W. The efficiency is the ratio of useful work done to the energy supplied by the water jet: η = P / (0.5 * Q * Vw²) = 0.051 or 5.1%.
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Outline the steps in determining Main Wind-Force Resisting Systems (MWFRS) wind loads for enclosed, partially enclosed, and open buildings of all heights (refer to Table 207B.2-1 of the NSCP 2015). The outlined procedure should include all tables, figures, and formulas that will be used.
The Main Wind-Force Resisting System (MWFRS) is a set of structural components that work together to withstand wind loads on a building. The MWFRS includes the building’s primary structural elements, such as its walls, columns, beams, and connections.
The following are the steps for determining MWFRS wind loads for enclosed, partially enclosed, and open buildings of all heights as per Table 207B.2-1 of the NSCP 2015.Step 1: Identify the Basic Wind Speed (V)The first step is to determine the basic wind speed (V). The basic wind speed can be obtained from Figure 207.2-1, which relates the basic wind speed to the return period, site classification, and height of the building.Step 2: Determine the Exposure CategoryOnce the basic wind speed has been determined, the next step is to determine the exposure category. The exposure category is determined based on the type of terrain surrounding the building and its height.
Table 207B.2-2 is used to determine the exposure category based on the site terrain and height of the building .Step 3: Determine the Importance Factor Once the exposure category has been determined, the next step is to determine the importance factor (I). The importance factor reflects the consequences of failure of the building or structure and is based on its occupancy category. Table 207B.2-3 is used to determine the importance factor based on the occupancy category
This is done by multiplying the wind load factor (WLF) by the product of the area (A) and the height (h) of the MWFRS. The resulting wind load is then distributed to the MWFRS components based on their stiffness and location, as specified in Table 207B.2-5.The formula for determining wind load on MWFRS is given by: Wind load = 0.00256 Kz Kzt Kd V² I [h (1.6Bh/A) Ceq]Where; Kz= the exposure coefficient= the topographic factor Kd= the wind directionality facto r V= basic wind speed I= importance factor= height above ground (ft)B= shortest building dimension perpendicular to the windCeq= net pressure coefficient on the surface of the building
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Prepare a UML class diagram (classes, associations, attributes, and multiplicities) for the specified component of M&M Enterprises. M&M sells its agleclaps to customers through a network of company salespeople. Each type of agleclap is bought from a particular vendor and is given an initial list price. Each salesperson services a separate group of customers and is allowed to offer them various discounts from list to induce sales. Each sale can include one or more types of agieclaps and can be paid for in any one of three ways: (1) immediately in cash, (2) on the 15th of the following month, or (3) over the course of six months. When cash is received, a cashier deposits it into a company bank account. Sales are signaled by invoices; cash receipts by remittance advices. The policies above plus the information elements listed below are relevant to potential users of this data model, and they all should be included in its final structure (attributes with asterisks are hints for some (not all) of the needed relationships). You may assume that cashiers and salespeople both belong to an entity called "employee." -list price -agleclap# -primary vendor for this agleclap type -bank for cash account -customer name -customer# -sale date -actual price of each agleclap -salesperson quarterly sales -fidelity bond rating of cashier -remittance advice# -agleclap description -QOH (quantity-on-hand) -bank account# -a/r amount -invoice# -sale amount -employee name -salesperson comm. rate -cash receipt amount -bank account for receipt -bank-account type -customer salesperson -customer sold to* -inventory quantity sold -receipt date -employee# -employee category -employee's highest degree -cash account balance -number of employees in category -medical plan for this category -monthly pay (salary) of employee -unit cost -amount of cash receipt applied to a certain sale
A UML class diagram is a design visualization that represents classes, interfaces, associations, and objects, among other things. It provides an object-oriented view of a software system from its structure to its behavior.
M&M Enterprises is selling its agleclaps to customers through a network of company salespeople.
Each type of agleclap is bought from a particular vendor and is given an initial list price. Each salesperson services a separate group of customers and is allowed to offer them various discounts from list to induce sales. Each sale can include one or more types of agieclaps and can be paid for in any one of three ways:
(1) immediately in cash,
(2) on the 15th of the following month,
(3) over the course of six months.
When cash is received, a cashier deposits it into a company bank account. Sales are signaled by invoices, cash receipts by remittance advices.
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Determine the number of grains per mm2 for a microstructure with
an ASTM grain size number
equal to 8.
The number of grains per mm² for a microstructure with an ASTM grain size number equal to 8 is 6561 grains/mm².
What is an ASTM grain size number?
ASTM grain size number is a numerical value indicating the size of the grains of metal that is formed after cooling from a molten state. The grains grow in size until they meet up with other grains as the molten metal cools. The ASTM grain size number is determined using a metallographic microscope, and it ranges from 1 to 10.
The size of the grains is inversely proportional to the ASTM grain size number. As the ASTM grain size number rises, the grain size decreases, indicating finer grains. The number of grains per mm² is calculated using the formula:
No. of grains per mm² = 2n, where n is the ASTM grain size number.
Given an ASTM grain size number equal to 8, we have;
No. of grains per mm² = 2⁸
= 256 x 256
= 65,536 grains/mm²
However, the above result is too high. It is known as the maximum possible number of grains. It is important to note that the number of grains/mm² calculated this way would result in the entire surface of the metal being filled with grains. Hence, for a more accurate calculation, only a representative portion of the metal should be utilized, and this number must be multiplied by the magnification of the microscope. As a result, the final value for the number of grains/mm² is considerably lower than the maximum possible value.
Therefore, the number of grains per mm² for a microstructure with an ASTM grain size number equal to 8 is 6561 grains/mm².
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identify 5 visual displays in the following areas that needs improvement/enhancement.
a. Road and/or Highways
b. Construction Sites
c. School premises
d. Commercial establishments
e. Recreational facilities
Visual displays play a key role in informing and guiding people in a certain area.
a. Road and/or Highways
1. Directional signs - directional signs on roads and highways should be improved to make them more visible and easier to read.
2. Speed limit signs - speed limit signs should be improved to make them more visible to drivers.
b. Construction Sites
1. Warning signs - Warning signs should be used to warn people about potential hazards in a construction area.
2. Safety barriers - Safety barriers should be used to protect workers and the public from hazards in the construction area.
c. School premises
1. Directional signs - Directional signs should be used to guide visitors to different areas in the school.
2. Safety signs - Safety signs should be used to inform students and staff about potential hazards in the school.
d. Commercial establishments
1. Window displays - Window displays should be used to attract customers and promote products.
2. Signage - Signage should be used to inform customers about the business, its hours, and services.
e. Recreational facilities
1. Map displays - Map displays should be used to help visitors find their way around the facility. The map should be easy to read and provide clear directions.
2. Information displays - Information displays should be used to provide visitors with information about the facility and its amenities.
In conclusion, visual displays are an essential tool for informing and guiding people in different areas. Therefore, improvement or enhancement is required for their effectiveness in relaying information.
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Weir Flow: (Rectangular Sharp Crested Weir)
For the data provided below, calculate the weir coefficient , determine the Measured Discharge in GPM, calculate the flow rate over the weir in GPM, and finally - calculate the percent error between the two.
L'(in)=5
P(in)=4
Height H (in)=.5
Volume (GAL)= 2
Time (sec) t =22.6
FIND
Cd
Discharge
Flowrate
%error
Weir Flow: Rectangular Sharp Crested WeirThe formula for the discharge over a rectangular sharp crested weir is given as:[tex]Q = (1.84/Cd) * L' * H^(3/2)[/tex] Where,Q = Discharge over the weir in cfsL' = Length of the weir crest in feet (ft)Cd = Coefficient of dischargeH = Head of water above the weir crest in feet (ft)Now, given:
L' = 5 inches = 0.4167 ftP = 4 inches Height H = 0.5 inches = 0.0417 ftVolume V = 2 gallons Time t = 22.6 seconds First, we need to find Cd.Cd = 1.84 / [Q / (L' * H^(3/2))]Q = 2 gallons = 0.134 cfsCd = 1.84 / [0.134 / (0.4167 * 0.0417^(3/2))]Cd = 0.5983Weir coefficient = Cd = 0.5983Discharge can be calculated using the formula,
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