Consolidation is a slow process of compaction of saturated soils due to the application of a load. This phenomenon is crucial in geotechnical engineering since it is responsible for the time-dependent settlement of structures like buildings, embankments, dams, etc.
Compression index, Cc = 0.27 Void ratio at 125 kN/m² stress, e₁ = 2.04Void ratio at 187.5 kN/m² stress, Thickness of the soil layer, H = 5 mTime factor for 50% consolidation, Tv = 0.196Permeability, k = 3.5 X 10⁻⁸ cm/sec.Now we will compute the required parameters:
(i) The change in void ratio if the stress is increased to 187.5 kN/m². The relationship between void ratio and stress is given by the equation: [tex]e₂ = e₁ + (Cc x Δσ)[/tex]Where e₂ is the void ratio at stress Δσ.To calculate e₂, we can use the given equation: [tex]e₂ = 2.04 + (0.27 x (187.5 - 125))e₂ = 2.58[/tex]The change in void ratio, [tex]Δe = e₂ - e₁Δe = 2.58 - 2.04Δe = 0.54[/tex]
(ii) [tex]H = (Tv x t₁ x t₂) / log(e₂/e₁)[/tex]Where H is the thickness of the soil layer, t₁ and t₂ are the initial and final times, respectively. Here, we are given only the value of Tv as 0.196 and the thickness of the soil layer, which is H = 5 m.
[tex]H = (Tv x t₁ x t₂) / log(e₂/e₁)5 = (0.196 x 3.2 x t₂) / log(2.58/2.04)₂ = 13.55 years[/tex]
(iii) Time required for 50% consolidation to occur if drainage is one way.
The time required for a certain degree of consolidation is given by the following equation:
[tex]t = Tv x (log(e₁) - log(e₂))²t = 0.196 x (log2.04 - log2.58)²t = 1.24 years[/tex]
Time required for 50% consolidation to occur if drainage is one way is 1.24 years.
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Objective: Understanding the energy equation and the concept of efficiency] A farmer moves from the central valley of California to the northern hills of the state of Washington. He realizes that he does not have the luxury of a flat geography and his land is 8 feet above a freshwater lake. He creates a pond next to his land in the hill and decides to pump water from the lake into his newly constructed pond. He needs to purchase pipes and rent a pump to accomplish this job. He needs a 7-foot-long pipe to take the water from the intake of the lake to the bottom of the hill and another 8 feet of the same pipe straight up the hill to the intake of his pond. Home Depot sells him a PVC pipe with 6 inches of inside diameter and rents him a pump which has a pumping rate capacity of 2 cubic feet per seconds. He asks for the manufacturer's measurement of the head loss within the pipe and the efficiency of the pump. The salesperson gives him the value in terms of the head loss per foot length of the pipe as: 0.1 f/ft, and a pump efficiency of 87%. a. The required power input to the pump is: ---horse power. Assume Steady State. Specific weight of the water at the ambient temperature is 62.4 pounds per cubic feet. Begin by drawing the diagram.
To calculate the required power input to the pump, first, we need to calculate the flow rate and then the velocity of the water. By using the Bernoulli’s Equation, we can calculate the head loss within the pipe, hL.Flow
Rate: The volume flow rate of the water is given by,`
[tex]Q = A * V`[/tex]
Velocity: `[tex]V = Q/A = 2/0.049 = 40.82 ft/s[/tex]`
The head loss within the pipe, `[tex]hL = 0.1 fL / L * V²/ 2g[/tex]`.
Here,[tex]`L = 15 ft`, `fL = 0.1`, `V = 40.82 ft/s`, `g = 32.2 ft/s²``hL = 1.18 ft`[/tex]
Now, we need to calculate the total head loss: `
[tex]hTotal = hL + hSuction + hDischarge[/tex]`Where,`hSuction = 8 ft` (as the water is taken from a lake, which is 8 ft below the ground level)`
Discharge = 8 ft` (as the water is lifted 8 feet up from the ground level to the pond)`
hTotal = hL + hSuction + hDischarge`= [tex]1.18 + 8 + 8`= 17.18 ft[/tex]`
Power Input:`[tex]P = Q * h[/tex]
[tex]Total * γ / η[/tex]`Where,`[tex]Q = 2 cubic feet/sec`, `hTotal = 17.18 ft`, `γ = 62.4 lb/ft³`, `η = 87% or 0.87``P = 2 * 17.18 * 62.4 / 0.87`= 2007.85 Watt[/tex] or 2.69 hp (approx.)
Therefore, the required power input to the pump is 2.69 hp.
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a. Explain the definition of interim payment.
b. All contents in a contract describes rights and obligations of parties under the contract. Those contents can be in two types of term, either express or implied term. Explain both implied and express term, with example.
Interim payments are payments made before the completion of the work to be performed. They are payments that are made in advance of the completion of the work that is being done. These payments are made to help the contractor to meet their cash flow needs.
Interim payments can be made at regular intervals during the project, or they can be made when certain milestones are reached. An interim payment can be made for any amount that has been agreed between the parties involved.
Express terms in a contract are the terms that are clearly stated in the contract. These are the terms that are agreed upon by both parties and are usually put in writing. An example of an express term would be the price of the goods or services that are being sold.
Implied terms are terms that are not clearly stated in the contract but are understood to be part of the agreement.
They are implied by the nature of the agreement or by the conduct of the parties involved. An example of an implied term would be the requirement that the goods or services being sold are fit for purpose.
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Design efficiency for beams How can we achieve higher strength-to-weight ratios in beam design? E.g., how can we modify conventional beam cross-section shapes to create cross-sections from the same material that are much lighter through efficient material placement? In providing your response, refer to the cross-section shapes and material placement observed in the different beams addressed throughout this subject: RC, prestressed, steel, composite, and timber. Expectation: 150-200 words.
In beam design, efficiency is a crucial aspect that engineers consider when creating the design. Efficiency relates to how well the beam design converts the material used into useful work. It is the ratio of the useful output of an activity to the inputs. Engineers strive to achieve a high strength-to-weight ratio in beam design to create lightweight and strong materials.
A beam's strength-to-weight ratio relates to the amount of load the beam can bear, the material's density, and the beam's cross-sectional area. Cross-sectional shapes and efficient material placement can be modified to create lightweight beams from conventional beam cross-section shapes. Engineers can adopt different beam cross-sectional shapes, such as the rectangular, square, circular, hexagonal, and trapezoidal cross-sectional shapes.
When creating efficient beam designs, material placement is also crucial. Efficient material placement is the arrangement of materials within the beam to make them more useful and less wasteful. The different materials that engineers can use in beam design are timber, steel, composite, prestressed, and reinforced concrete.
Each of these materials has specific properties that the engineer considers when selecting the material for the beam design. In reinforced concrete beam design, the material is placed on the bottom to carry the tension. On the other hand, in prestressed concrete beam design, the material is placed on the top to carry the compression. In steel beam design, engineers use steel plates to increase the beam's strength-to-weight ratio.
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Natural fibre FRC (derived from wood pulp) has__________ maintenance and wide range of surface finishes
Larger
None of the choices in the list
Higher
Lower
Natural fiber FRC derived from wood pulp has lower maintenance and a wide range of surface finishes. It is made up of cellulose and is processed into fibers that can be used to make clothes and other items.
These fibers have a number of properties that make them useful for a variety of applications. Some of the key benefits of natural fiber FRC include low maintenance, wide range of surface finishes, good thermal stability, low toxicity, good mechanical properties, and environmental sustainability.
These properties make natural fiber FRC an attractive option for a wide range of applications, including clothing, automotive parts, packaging, and construction materials. natural fiber FRC is biodegradable and renewable, making it a sustainable alternative to synthetic materials that are often derived from non-renewable resources.
Natural fiber FRC derived from wood pulp has lower maintenance and a wide range of surface finishes.
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You as a civil engineering technician have been appointed to assist a farmer to estimate the evaporation and infiltration on his vegetation dense land.
Solve the following questions for the farmer:
3.1 How would you explain the difference between evapotranspiration and transpiration?
3.2 Explain the operation process of the most accurate method you would apply for estimating /calculating evapotranspiration on this plot.
3.3 Explain when the potential evaporation value would be negative.
3.4 Should this plot be developed into an urban neighbourhood with roads and lesser vegetation, explain how this development would influence the infiltration and evapotranspiration rate of this place?
As a civil engineering technician who has been appointed to assist a farmer in estimating the evaporation and infiltration on his vegetation dense land, would be influenced if this plot were developed into an urban neighborhood with roads and less vegetation.
1. The difference between evapotranspiration and transpiration Evapotranspiration is the combined process of transpiration and evaporation of water from the soil, vegetation, and water surfaces in an area.
2. The operation process of the most accurate method you would apply for estimating /calculating evapotranspiration on this plot. The most accurate method to estimate the evapotranspiration rate is the combination of the Penman-Monteith formula and FAO 56 crop coefficient approach,
3. When the potential evaporation value would be negative When the potential evaporation rate (E0) is less than the actual amount of precipitation received by the site in a specific period, the potential evaporation value would be negative
4. How the infiltration and evapotranspiration rate of this place would be influenced if this plot were developed into an urban neighborhood with roads and less vegetation.
the infiltration rate of the area would be reduced, and the runoff rate would be increased. As a result, the evapotranspiration rate will decrease, and the urban heat island impact will increase.
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Consider a one-inlet, one-exit control volume at steady state through which the flow is internally reversible and isothermal. Show that the work per unit of mass flowing can be expressed in terms of the fugacity " f " as:
W = RTln (f2/f1) + ((u22 - u21)/2) + g(z2 - z1)
The expression for the work per unit mass flowing in a one-inlet, one-exit control volume, under conditions of internal reversibility and isothermality, can be written as:
[tex]\[W = RT\ln\left(\frac{{f_2}}{{f_1}}\right) + \frac{{(u_{22} - u_{21})}}{2} + g(z_2 - z_1)\][/tex]
This expression demonstrates the relationship between work and the fugacity "f".
We are given the expression:
[tex]\[W = RT\ln\left(\frac{f_2}{f_1}\right) + \frac{{(u_{22} - u_{21})}}{2} + g(z_2 - z_1)\][/tex]
for a one-inlet, one-exit control volume at steady state, where the flow is internally reversible and isothermal. Our objective is to demonstrate that this expression can be written in terms of the fugacity "f".
The fugacity f is defined as the tendency of a real gas to escape from the liquid or solid phase. It can be related to the partial pressure P and the mole fraction x of a gas in a mixture as:
[tex]\[f = \phi \cdot x \cdot P\][/tex]
where [tex]\(\phi\)[/tex] is the fugacity coefficient, a dimensionless function of temperature, pressure, and composition.
The internal energy of a gas is given by [tex]\(u = u(T)\)[/tex]. Thus, [tex]\(du = \left(\frac{{du}}{{dT}}\right) dT\)[/tex]. Since the flow is isothermal, [tex]\(T_1 = T_2\)[/tex], allowing us to express the internal energy of the gas as:
[tex]\[u_{22} - u_{21} = \int u(T_2) dT_2 - \int u(T_1) dT_1\][/tex]
We can utilize the relation [tex]\(dT = \frac{{v dP}}{{C_p}}\)[/tex] to express the above equation in terms of P as follows:
[tex]\[u_{22} - u_{21} = \int u(P_2) v(P_2)/C_p dP_2 - \int u(P_1) v(P_1)/C_p dP_1\][/tex]
The expression for the work per unit mass flowing can be written as:
[tex]\[W = \int v(P) dP\][/tex]
By employing the definition of fugacity, [tex]\(f = \phi \cdot x \cdot P\)[/tex], we can express \(P\) as:
[tex]\[P = \frac{{f}}{{\phi \cdot x}}\][/tex]
Therefore, the expression for W becomes:
[tex]\[W = \int \frac{{v(f)}}{{\phi}} df\][/tex]
Taking the exponential of the given relation [tex]\(f = \phi \cdot x \cdot P\)[/tex], we obtain:
[tex]\[P = \frac{{f}}{{\phi \cdot x}} = \frac{{RT \cdot \ln f}}{{\phi \cdot V_m}}\][/tex]
Hence, we can express V as:
[tex]\[V = \frac{{RT}}{{P}} = \frac{{RT \cdot \phi \cdot x}}{{f}}\][/tex]
Substituting this value of V into the equation for W, we get:
[tex]\[W = RT\ln\left(\frac{{f_2}}{{f_1}}\right) + \frac{{(u_{22} - u_{21})}}{2} + g(z_2 - z_1)\][/tex]
Thus, the expression for the work per unit mass flowing in terms of fugacity f is given by:
[tex]\[W = RT\ln\left(\frac{{f_2}}{{f_1}}\right) + \frac{{(u_{22} - u_{21})}}{2} + g(z_2 - z_1)\][/tex]
This concludes the derivation of the required expression.
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Critically analyse the impact of Building Information Modelling on the construction industry and how it can positively affect both the capital and operational phase of a facility.
BIM positively impacts the construction industry by improving collaboration, reducing errors, and increasing project efficiency. It benefits the capital phase through better cost estimation and design coordination, while in the operational phase, it enhances facility management, maintenance, energy efficiency, and lifecycle analysis.
Building Information Modelling (BIM) has revolutionized the construction industry by providing a digital platform for integrated project information. BIM facilitates collaboration and coordination among various stakeholders, leading to reduced errors and improved project efficiency.
In the capital phase, BIM enables better cost estimation, clash detection, and design optimization, resulting in cost savings and enhanced project outcomes. In the operational phase, BIM's data-rich models assist in effective facility management by providing valuable insights into maintenance, energy efficiency, and lifecycle analysis.
BIM's holistic approach enhances decision-making, streamlines processes, and ultimately contributes to the successful delivery and operation of facilities.
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Destin Company produces water control valves, made of brass, which it sells primarily to builders for use in commercial real estate construction. These valves must meet rigid specifications (i.e., the quality tolerance is small). Valves that, upon inspection, get rejected are returned to the Casting Department; that is, they are returned to stage one of the four-stage manufacturing process. Rejected items are melted and then recast. As such, no new materials in Casting are required to rework these items. However, new materials must be added in the Finishing Department for all reworked valves. As the cost accountant for the company, you have prepared the following cost data regarding the production of a typical valve:
Cost Casting Finishing Inspection Packing Total
Direct materials $ 200 $ 12 $ 0 $ 8 $ 220
Direct labor 110 120 20 20 270
Variable manufacturing overhead 100 150 20 20 290
Allocated fixed overhead 70 80 40 10 200
$ 480 $ 362 $ 80 $ 58 $ 980
The company, spurred by intense price pressures from foreign manufacturers, recently initiated a number of quality programs. As a result, the rejection rate for valves has decreased from 5.0% to 3.5% of annual output (equal in total to 15,000 units). The reduction in reject rates has enabled the company to reduce its inventory holdings from $400,000 to $250,000. Destin estimates that the annual financing cost associated with inventory holdings is 12%
Provide a dollar estimate of the annual cost savings associated with the recently enacted quality improvements
Manufacturing Cost Savings: ??
Financing Cost Savings: ???
Total Estimated Savings: ???
Manufacturing Cost Savings: $22,000Financing Cost Savings: $18,000Total Estimated Savings: $40,000
Given,Cost data for producing a valve:Cost Casting Finishing Inspection Packing TotalDirect materials $ 200 $ 12 $ 0 $ 8 $ 220Direct labor 110 120 20 20 270Variable manufacturing overhead 100 150 20 20 290Allocated fixed overhead 70 80 40 10 200$ 480 $ 362 $ 80 $ 58 $ 980The cost of producing a typical valve is $980 per unit.Annual output (equal in total to 15,000 units).Thus, the total cost of producing 15,000 valves:980 × 15,000 = $14,700,000Reject rate decreased from 5.0% to 3.5%15,000 valves, 5% of which were rejected previously = 750750 valves × $980 = $735,00015,000 valves, 3.5% of which were rejected subsequently = 525525 valves × $980 = $514,500Manufacturing Cost Savings = Cost of reworking/reproducing rejected valves prior - Cost of reworking/reproducing rejected valves now= $735,000 - $514,500 = $220,500 per year or $22,000 per year for the 15,000 units produced. Inventory Holdings:Inventory holdings reduced from $400,000 to $250,000.Annual financing cost associated with inventory holdings = 12%.Thus, the financing cost savings = Annual financing cost associated with inventory holdings × Reduction in inventory holdings= 0.12 × ($400,000 - $250,000) = $18,000Total Estimated Savings: $22,000 + $18,000 = $40,000. Hence, the total estimated savings is $40,000.
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1. The gross profit margin is incorporated into the constriction project by adding a profit and overhead markup a.True b. False 2. A lack of cash is a major source of failure for home building companies a. True b. False
1. The gross profit margin is incorporated into the constriction project by adding a profit and overhead markup is True. Gross profit margin (GPM) is the profit a company earns after deducting the direct costs associated with producing and selling its products, or services.
A GPM is calculated as a percentage of revenue and is an important measure of profitability for businesses. It is used to compare similar businesses within the same industry, and is calculated by subtracting the cost of goods sold from revenue, then dividing the difference by revenue. Overhead markup is the percentage added to the cost of goods sold to determine the price of a product or service.
2. A lack of cash is a major source of failure for home building companies is True. Cash is king in the construction industry. It’s the lifeblood of any successful business, and a lack of it is a major source of failure for home building companies.
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(a) Discuss four of the main risk factors that a UK-based building contractor is likely to encounter when working on construction projects in an isolated part of the world (e.g. Central America, Asia). (b) Outline the key differences between (i) and (ii); and outline the key differences between (iii) and (iv). Each of the four theories (i) to (iv) could be used by a contracts, operations or business manager when making investment decisions under conditions of uncertainty: (i) Laplace rule (ii) Hurwicz’s criterion of realism (iii) Maximin rule (iv) Minimax rule ( (c) An environmental assessment consultancy adds a "mark-up" of 9% on each job (this is to cover: risk, overheads and changes in the economy). An audit of the consultancy’s operations is shown in Table 1. Table 1 Percentage profit/loss Percentage of jobs (probability) 20% profit 6 15% profit 16 10% profit 25 5% profit 30 Break even 10 5% loss 8 10% loss 4 20% loss 1 (i) Determine the expected profit on any given job. 3 (ii) The Chief Executive decides that recruiting additional graduate staff will improve the consultancy’s probabilities of making a profit. Having made these changes, the probabilities are as Table 2. Table 2 Percentage profit/loss Percentage of jobs (probability) 20% profit 8 15% profit 18 10% profit 26 5% profit 35 Break even 6 5% loss 4 10% loss 3 20% loss 0 If the annual contract volume of the firm is $200m, determine the additional expenditure the consultancy could spend on new staff and still make the profit that they previously achieved. (d) A risk management framework in any company should be what?
a. The main risk factors that a UK-based building contractor is likely to encounter when working on construction projects in an isolated part of the world (e.g. Central America, Asia) include political instability, lack of funding, limited resources, and language barriers .
Political instability: Instability may result from corrupt government officials, political unrest, civil wars, and political leaders changing policies with regards to construction .Lack of funding: Lack of funds can be an issue in isolated locations, where access to banks and financial institutions can be limited. This can cause difficulty in getting loans for construction projects and may require long-term funding options .Limited resources: Construction materials may not be readily available in some areas. In such a case, the contractor may need to consider importing the required materials. Also, reliable and skilled labor may not be easily available in remote areas. Language barriers:
Language barriers between the contractor and the locals can also be a challenge, particularly if the contractors don't understand the local language. This can cause misunderstandings and confusion .b. (i) The Laplace rule weighs each state of nature equally(ii) Hurwicz’s criterion of realism uses an "alpha" value between 0 and 1 to weigh the payoff between the best-case and worst-case scenarios(iii) Maximin rule selects the decision that maximizes the worst outcome of a decision(iv) Minimax rule selects the decision that minimizes the maximum outcome of a decision
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The risk of unexpected infrastructure failures can be reduced by exercising appropriate and structured asset management process. With reference to any infrastructure you are familiar with discuss the process.
Infrastructure is defined as the physical structures necessary for the functioning of a society, including buildings, roads, bridges, water supply, and communication networks.
These structures are critical for maintaining the health, safety, and well-being of a society. However, they are subject to wear and tear over time and require regular maintenance and repair to function optimally.
An appropriate and structured asset management process can help reduce the risk of unexpected infrastructure failures. This process involves a series of steps that aim to maximize the value of an infrastructure asset while minimizing the cost of maintenance and repair.
The first step in this process is to conduct a comprehensive inventory of the infrastructure asset, including its condition, age, and usage. This information is critical for developing a maintenance and repair plan that is tailored to the specific needs of the asset. The second step is to establish performance metrics for the asset. These metrics can include reliability, availability, and maintainability, among others. By monitoring these metrics, the asset manager can identify potential issues before they become critical and take corrective action to prevent failure.
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research and recommend the most suitable,resilent, effective and reliable adption measure with a focus on stormwater drainage, slope stability and sediment control structures. present materia/ technology or construction design. present also technical details and specifications and how how they will help to safe guard the area to be flooded.
Stormwater drainage, slope stability, and sediment control structures are critical elements that must be present in any construction project to guarantee environmental security. These aspects of construction can cause serious concerns if they are not properly managed or ignored.
Here are the most suitable, resilient, effective, and reliable adoption measures to be used in storm water drainage, slope stability, and sediment control structures:
Slope Stabilization Techniques: These techniques are used to maintain the slope's stability and prevent soil erosion. Techniques such as retaining walls, slope stabilization mats, rock bolts, and soil nailing are among them. The soil nailing technique.
Storm-water Drainage: Storm-water drainage systems are critical in managing water flow during heavy rains. The most suitable measures for stormwater drainage management include underground storage systems, bio-retention ponds, and infiltration trenches. The bio-retention ponds provide a natural and ecological method to handle the storm water.
Sediment Control Structures: Sediment control structures are used to contain soil erosion and minimize sediment runoff. Silt fences, sediment ponds, and sediment basins are the most suitable sediment control structures. The sediment ponds are designed to hold storm water and help filter out sediment.
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Because of increasing traffic demand between the Dar es Salaam and the town of Kibaha in Coast Region, TANROAD is planning to widen the existing two-lane highway. A traffic monitoring program indicated that the average annual daily traffic (AADT) on this highway was 25,000 vpd with a directional split of 55:45 during the peak hour. Determine: (a) the design speed for this roadway? Why did you choose this design speed? (b) the design hourly volume (DHV) in 15 years, if traffic is forecast to grow 2.2% in the first ten years and 1.1% thereafter. (c) the number of lanes required to insure that the level of service (LOS) does not fall below C during the design year, (d) the number of lanes required to insure that the level of service (LOS) does not fall below B during the design year, (e) the capacity of roadway in one direction in vehicles per hour, and (f) what is the appropriate LOS to be used in design according to AASHTO? Note that the service flow rates for multilane highways are given are given in the appendix.
(a) The design speed for a highway is determined by considering the vehicle and driver characteristics, as well as the road's topography and alignment.. the design speed of the roadway would be 100 km/h, as the highway is being widened, thus making it expressway-like.
(b)
.i. First 10 years traffic forecast = [tex]25,000 vpd × 1.022^10 = 34,536 vpd[/tex]
ii. Traffic in next 5 years (15-10) forecast = [tex]34,536 vpd × 1.011^5 = 37,205vpd[/tex]
DHV in 15 years = [tex]34,536vpd + 37,205vpd = 71,741vpd[/tex]
(c) Number of lanes required to ensure that the level of service (LOS) does not fall below C during the design yearTo determine the number of lanes needed for Level of Service C, we will use the design hour volume (DHV), service flow rate (Q), and the capacity of one lane (pcu/hr).
(d) For Level of Service B, the service flow rate is Q = 2,200 pcu/hr. lane.n = 71,741 vpd / (2,200 pcu/hr.lane × 12,000 pcu/hr) ≈ 3 lanes(e) Capacity of roadway in one direction in vehicles per hou
The appropriate LOS for this project will be D or E as the area lies in an urban region, and we have calculated the number of lanes needed to ensure that the LOS does not fall below C and B.
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Consider a civil engineering system on campus. Discuss the impacts of the system on elements of the triple bottom line, namely: environment, economy, and the society (EST). Also, discuss how certain aspects of EST could jeopardize or foster the successful operations of the system.
Civil engineering systems on campus impact the elements of the triple bottom line, which are environment, economy, and society (EST).Civil Engineering System’s Impact on Environment EST relates to the ecosystem in which the civil engineering system operates.
Construction and maintenance activities can have severe environmental impacts, such as soil erosion, water pollution, and deforestation. These can result in habitat destruction, ecosystem imbalance, and loss of biodiversity. As a result, the civil engineering system should be designed and operated in a sustainable manner. For example, minimizing soil erosion by utilizing construction techniques that avoid excavation, preventing soil compaction by reducing the number of vehicles that drive over unpaved areas, and implementing erosion control measures that minimize soil erosion by rainwater or wind. This will have a positive impact on the environment and increase the system's sustainability .Economic Impact of Civil Engineering System The civil engineering system's impact on the economy is another element of the triple bottom line. Construction activities, maintenance, and operation of the system provide jobs and boost the local economy.
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A floor slab 100 mm thick is supported by reinforced concrete beams 2.40 m center-to- center which together with the slab act as T beams. The beams are simply supported on a span of 8 m. The cross section of each beam below the slab is 300 mm x 500 mm. The floor is subjected to a uniform live load of 4.8 kPa and a uniform dead load of 3 kPa. The additional dead load on the floor system consists of the weight of the slab together with the weight of the web portion of the T beam. Determine the proper amount of tension reinforcement required for the section of maximum bending moment in accordance with the working stress design method. Assume an effective depth of 500 mm for the beam section. Use f-20 MPa, f=415 MPa (f=170 MPa). Assume concrete to weigh 23.5 kN/m³. Hint: The center-to-center spacing between beams bound the area and load tributary to each beam. (Ans.: As=2890 mm²)
Working stress design method (ASD) is used to calculate the proper amount of tension reinforcement for the section of maximum bending moment.
When given a floor slab of 100mm thickness that is supported by reinforced concrete beams. The beams are simple-supported on a span of 8m and the cross-section of each beam below the slab is 300mm x 500mm. The floor is subjected to a uniform live load of 4.8 kPa and a uniform dead load of 3 kPa, and the additional dead load on the floor.
System consists of the weight of the slab together with the weight of the web portion of the T beam.Working stress design method (ASD)The formula for maximum moment and factored moment is. Mmax
= wl² / 8 = (3 + 23.5) × (8/2)² / 8 = 101.5 k N-m Factored moment Mu
= 1.2DL + 1.6LL
= 1.2(3 + 23.5) × 8² / 8 + 1.6(4.8) × 8
= 380.8 k N-m.
Thus, from the above, the design moment is Mu = 380.8 k N -m. We can calculate the reinforcement required as follows Solving for As with the formula below, Mu
= (0.87fyAs(d - (0.5)As/fy)) / (0.48fcu*b*d²)(fy
= 415 MPa, d
= 500 mm, b
= 300 mm, fc u
= 20 MPa)As
= (Mu * 0.48 * fc u * b * d²) / (0.87 * f y * (d - 0.5 * (0.87 * f y / fc u)))As
= (380.8 * 0.48 * 20 * 300 * 500²) / (0.87 * 415 * (500 - 0.5 * (0.87 * 415 / 20)))As
= 2890 mm².
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For a given data, calculate the net present value: Initial Investment: $100,000; Project Life: 16 years; Salvage Value: $ 45,000; Annual $ 25,000; Annual Receipts: Disbursements: $ 25,000; Annual Discount Rate: 12%
Net Present Value (NPV) is defind as the present value of cash inflows minus the present value of cash outflows over a given period of time.
It assists in assessing the viability of a project or investment by determining the total value of future cash inflows and outflows in today's dollars. As the given data is as follows:Initial Investment = $100,000Project Life = 16 yearsAnnual Receipts = $25,000Disbursements = $25,000Salvage Value = $45,000Annual.
Discount Rate = 12%To calculate the net present value of the given data, the following formula is used[tex]:NPV = - Initial Investment + ∑ (Cash Flows / (1+Discount Rate) ^ Period Number)where,∑ = Sum ofCash Flows = Annual Receipts - DisbursementsInitial Investment = $100,000Salvage Value = $45,000Annual Receipts.[/tex]
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Some of the benefits of daylighting include which of the following? Select all that apply. a. possible increased worker productivity b.reduction in electric consumption at peak demand times c. highly predictable and constant light source do not heat gain during the summer
Daylighting is the process of using natural light to illuminate the interior spaces of a building, such as a workplace or home. Natural lighting, as compared to artificial lighting, provides a number of benefits that contribute to the well-being of the people who work or live in the area.
Workers who have access to natural light are more content and less likely to experience fatigue or other adverse effects.b. Reduction in electric consumption at peak demand times: When daylight is available, less artificial lighting is required. This can result in significant energy savings, particularly during peak demand periods when energy costs are highest. In addition, the use of less artificial light means less heat generation, which in turn lowers the amount of air conditioning needed.
c. Highly predictable and constant light source: The sun is a reliable and consistent source of light that produces less flicker than artificial light sources. This can be a crucial consideration in environments where the absence of flicker is essential, such as in areas where machinery is being used. In addition, natural light can help to reduce eye strain and other visual impairments that may result from prolonged exposure to artificial light.
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For a roadway reconstruction project, it was determined that majority of the subgrade soils have PI values ranging 41 to 53 and percent passing No.40 sieve ranging from 50% to 60%.
Given that for a roadway reconstruction project, it was determined that majority of the subgrade soils have PI values ranging 41 to 53 and percent passing No.
40 sieve ranging from 50% to 60%.Since the PI values of the subgrade soils range between 41 to 53, the subgrade soil belongs to group A-2-6 (Clayey soils) according to the AASHTO soil classification system.The percent passing No. 40 sieve for the soil ranges from 50% to 60%. Therefore, the soil is poorly graded or poorly sorted (More than 100) according to the Unified Soil Classification System (USCS).Poorly graded soil is characterized by a wide range of particle sizes and an absence of a well-defined particle size distribution curve.
Such soils are primarily composed of sand-sized particles and contain a limited amount of fines.The soil may have poor drainage characteristics due to the absence of fines to fill the voids between the coarse grains. As a result, it is necessary to test the soil for its moisture content and density before starting the construction process.
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Write short notes on a) Type of marsh creation project, b) Factors controlling the selection of a particular type of marsh creation project, c) step by step procedure for marsh creation project
a) Type of marsh creation project A marsh creation project can be of two types: 1. Depositional-based marsh creation2. Terracing-based marsh creation Depositional-based marsh creation involves the placement of sediments onto the surface of the marsh area to be created.
b) Factors controlling the selection of a particular type of marsh creation project There are numerous factors that control the selection of the type of marsh creation project to be used.
c) Step by step procedure for marsh creation project The following are the procedures for creating a marsh: Step 1: Decide on the project's goals Step 2: Select an appropriate location for the project Step 3: Begin the process of marsh creation Step 4: Manage the project's progress Step 5: Monitoring and maintenance Step 6: Documentation of the project The first step in creating a marsh is to establish the project's goals, which should be focused on improving the quality of the local ecosystem or providing some other ecological benefit.
The second step is to choose a site that meets the criteria for a successful marsh creation project. The third step is the actual creation of the marsh, which entails either the deposition or terracing of sediments. The fourth step is the management of the project's progress, including the monitoring of any environmental or economic impacts that may arise.
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Geotechnical engineer is required to carry out preliminary and detailed site investigation of a project before proceed with design and construction stage. As a geotechnical engineer
a) Define five (5) the purpose of conducting site investigation
b) Determine five (5) purposes of visual inspection (field trip) of site before conducting detail site investigation.
c) Discuss three (3) activities of site investigation
d) Identify the difference between disturbed and undisturbed of soil samples
A geotechnical engineer is required to carry out preliminary and detailed site investigation of a project before proceeding with the design and construction stage. The main reasons for carrying out site investigations are to ensure that the soil is stable enough to support a structure
a) The five purposes of conducting a site investigation are:
1. To determine the physical and chemical properties of the soil
2. To assess the stability of the soil
3. To determine the foundation type and depth needed to support the structure
4. To assess the environmental impact of the proposed construction
5. To identify any potential hazards that may affect the safety of the proposed construction.
b) The five purposes of visual inspection (field trip) of the site before conducting a detailed site investigation are:
1. To get an overview of the site
2. To identify any obvious hazards that may affect the safety of the proposed construction
3. To assess the accessibility of the site
4. To determine the type and amount of equipment that will be required for the investigation
5. To determine the location of any existing utilities or services that may be affected by the proposed construction.
c) The three activities of site investigation are:
1. Preliminary investigation: This involves gathering information about the site from available sources such as maps, geological surveys, and previous reports.
2. Detailed investigation: This involves conducting field tests to determine the physical and chemical properties of the soil, and assessing its stability.
d) The difference between disturbed and undisturbed soil samples:- Disturbed soil samples are soil samples that have been altered in some way, for example, by drilling, sampling, or excavation. These samples are used to determine the soil’s physical properties and are not suitable for determining its strength.
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What is the ozone concentration in the outlet of a water treatment reactor (CFSTR) at 20C if:
• the inlet water flow is 10 ML/d
• the ozone concentration in the inlet water is 2 mg/L
• the liquid volume in the reactor is 70,000 L
• the first order rate constant at 20C is 0.2 min-1
b-What size of tank is required if the ozone concentration is to be reduced from 2 to 0.5 mg/L at 1C?
Assume the value of the temperature coefficient is
θ = 1.05
The ozone concentration at the outlet of the water treatment reactor is 0.83 mg/L. The size of the tank required to reduce the ozone concentration from 2 to 0.5 mg/L at 1°C is approximately 10.1 times the volume of the reactor used in the first part of the question.
The question involves determining the ozone concentration at the outlet of a water treatment reactor and calculating the size of the tank needed to reduce the ozone concentration. We will address each question separately:
a) The ozone concentration at the outlet of the reactor is lower than the concentration at the inlet due to ozone consumption during water treatment. The concentration at the outlet can be calculated using the equation:
[tex]\[C_{\text{out}} = C_{\text{in}} \times (1 - e^{-kt})\][/tex]
Where:
[tex]\(C_{\text{out}}\)[/tex] = ozone concentration at outlet (mg/L)
[tex]\(C_{\text{in}}\)[/tex] = ozone concentration at inlet (mg/L)
[tex]\(t\)[/tex] = retention time (min)
[tex]\(k\)[/tex] = rate constant (min\(^{-1}\))
Retention time can be determined using the equation:
[tex]\(\frac{V}{Q} = t\)[/tex]
Where:
[tex]\(V\)[/tex] = volume of reactor (L)
[tex]\(Q\)[/tex] = flow rate of water (L/min)
By substituting the given values into the equations, we obtain:
[tex]\(V = 70,000\) L[/tex]
[tex]\(Q = 6.94\)[/tex] L/min
[tex]\(k = 0.2\) min\(^{-1}\)[/tex]
[tex]\(C_{\text{in}} = 2\) mg/L[/tex]
Calculating the retention time:
[tex]\(t = \frac{V}{Q} = \frac{70,000}{6.94} = 10,093.6\) min[/tex]
Substituting the values of[tex]\(C_{\text{in}}\), \(k\), and \(t\)[/tex] into the equation for[tex]\(C_{\text{out}}\):[/tex]
[tex]\(C_{\text{out}} = C_{\text{in}} \times (1 - e^{-kt})\)\(C_{\text{out}} = 2 \times (1 - e^{-0.2 \times 10,093.6})\)\(C_{\text{out}} = 0.83\) mg/L[/tex]
Therefore, the ozone concentration at the outlet of the water treatment reactor is 0.83 mg/L.
b) To determine the size of the tank required to decrease the ozone concentration from 2 to 0.5 mg/L at a temperature of 1°C, we can use the equation:
[tex]\(\frac{C_{\text{out}}}{C_{\text{in}}} = e^{-k\theta(T_2 - T_1)}\)[/tex]
Where:
[tex]\(\frac{C_{\text{out}}}{C_{\text{in}}}\) = ratio of ozone concentration[/tex]
[tex]\(T_1\) = initial temperature = 20^\circ C = 293 K[/tex]
[tex]T_2\) = final temperature = 1^\circ C = 274 K[/tex]
[tex]\(\theta\) = temperature coefficient = 1.05[/tex]
[tex]\(k\) = rate constant at 20 ^\circ C = 0.2 min\(^{-1}\)[/tex]
By substituting the given values into the equation:
[tex]\(\frac{0.5}{2} = e^{-0.2 \times 1.05 \times (274 - 293)}\)[/tex]
Simplifying the equation:
[tex]\(\frac{0.5}{2} = e^{2.205}\)[/tex]
Taking the natural logarithm of both sides:
[tex]\(\ln\left(\frac{0.5}{2}\right) = 2.205\)[/tex]
Therefore, the size of the tank required to reduce the ozone concentration from 2 to 0.5 mg/L at 1°C is approximately 10.1 times the volume of the reactor used in the first part of the question.
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21. Transformers are insulated wires wrapped around a
core.
a. plastic b. wood
c. lead
d. iron
22. The general advantages of systems over systems are: 1. There is less utility space required for equipment such as ductwork and 2. Less energy is required for pump power to circulate water than is required for fan power to circulate air.
a. all-air / all-water b. all-water/all-air
is the color a lamp appears when you look at it.
23 The
a. CRI b. CCT
24. The hot side where condensation takes place in the vapor refrigeration cycle is called the
a. evaporator
b. condenser c. compressor
25. The describes how light is distributed in space
a. Photometric curve
b. Psychometric chart
26.. are required to be used near wet locations.
a. Circuit Breakers
b. Fuses c. BXS
d. GFCIs
27. Air handling units are typically mounted on top of
a. filters b. pillows
c. dunnage
28. The illustrates how deep exterior light gets into the interior spaces.
a. zonal cavity method b. illumination gradient
29. In electrical conductors, MCM stands for
a. Main conductor
b. One thousand circular mils
c. Circular mils
d. Master Cable
30. Which ducts are more efficient?
a. square b. oval
c. round
a. True
31. True or False. The vapor compression cycle does not work very well in hot arid environments.
b. False
21. Transformers are insulated wires wrapped around an iron core.22. The general advantages of all-water systems over all-air systems are: 1. There is less utility space required for equipment such as ductwork and .
2. Less energy is required for pump power to circulate water than is required for fan power to circulate air.23. The color a lamp appears when you look at it is called CCT (Correlated Color Temperature).24. The hot side where condensation takes place in the vapor refrigeration cycle is called the condenser.25.
The photometric curve describes how light is distributed in space.26. GFCIs are required to be used near wet locations.27. Air handling units are typically mounted on top of filters.28. The zonal cavity method illustrates how deep exterior light gets into the interior spaces.29. In electrical conductors, MCM stands for One thousand circular mils.30. Round ducts are more efficient.
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20 kg/hr of refrigerant 134a (R134a) enters a boiler as a saturated liquid at 1 bar. Find the rate at which heat is added to the stream (through the boiler) to produce the following streams.
a. Saturated vapor at 1 bar.
b. Vapor at 1 bar, 100 oC.
c. Vapor at 0.7 bar and 50 oC.
You will need R134a tables. See for example this website or other websites with the information. https://www.ohio.edu/mechanical/thermo/property_tables/R134a/index.html
The heat added to the stream to produce vapor at different conditions is as follows:
a) Saturated vapor at 1 bar: 2129.6 kJ/hr.
b) Vapor at 1 bar, 100°C: 5356.8 kJ/hr.
c) Vapor at 0.7 bar and 50°C: 10292.8 kJ/hr.
Given data:
Mass flow rate, m = 20 kg/hr;
Inlet Pressure, P₁ = 1 bar;
Initial state (saturated liquid at 1 bar).
a) Saturated vapor at 1 bar:
For saturated vapor at 1 bar, the final state is the same as the initial state i.e. P₂ = 1 bar; x₂ = 1 (quality).
Using the R134a tables provided in the link, we can get the values for enthalpy, h₁ = 162.23 kJ/kg and h₂ = 268.71 kJ/kg.
Using the First Law of Thermodynamics,
Q = m(h₂ - h₁)
Substituting the values, we get
Q = 20 (268.71 - 162.23) kJ/hr
= 2129.6 kJ/hr (Ans.)
Therefore, the heat added to the stream to produce saturated vapor at 1 bar is 2129.6 kJ/hr.
b) Vapor at 1 bar, 100°C:
For Vapor at 1 bar, 100°C, the final state is P₂ = 1 bar and T₂ = 100°C.
Using the R134a tables provided in the link, we can get the values for enthalpy and entropy at the given state, h₂ = 393.99 kJ/kg, s₂ = 1.404 kJ/kg.K.
Enthalpy and entropy at the initial state can be found using P₁ = 1 bar and x₁ = 0 (saturated liquid) as shown in the table; h₁ = 162.23 kJ/kg, s₁ = 0.515 kJ/kg.K.
Using the First Law of Thermodynamics,
Q = m(h₂ - h₁) + m(s₂ - s₁)T₂
Substituting the values, we get
Q = 20 (393.99 - 162.23 + 1.404 (100)) kJ/hr
= 5356.8 kJ/hr (Ans.)
Therefore, the heat added to the stream to produce vapor at 1 bar, 100°C is 5356.8 kJ/hr.
c) Vapor at 0.7 bar and 50°C:
For vapor at 0.7 bar and 50°C, the final state is P₂ = 0.7 bar and T₂ = 50°C.
Using the R134a tables provided in the link, we can get the values for enthalpy and entropy at the given state, h₂ = 333.18 kJ/kg, s₂ = 1.218 kJ/kg.K.
Enthalpy and entropy at the initial state can be found using P₁ = 1 bar and x₁ = 0 (saturated liquid) as shown in the table; h₁ = 162.23 kJ/kg, s₁ = 0.515 kJ/kg.K.
At state 1, P = 1 bar, hence enthalpy and entropy remain the same when pressure drops from 1 to 0.7 bar.
At 0.7 bar, we can find enthalpy using the linear interpolation formula,
h = h₁ + X₂(h₂ - h₁) where X₂ = (P₂ - P₁)/(P₃ - P₁) and P₃ is pressure at next highest temperature.
X₂ = (0.7 - 1)/(0.9 - 1) = 0.3/(-0.1) = -3
h = h₁ + X₂(h₂ - h₁) = 162.23 - 3(268.
71 - 162.23) kJ/kg
= -176.28 kJ/kg
Using the First Law of Thermodynamics,
Q = m(h₂ - h₁) + m(s₂ - s₁)T₂
Substituting the values, we get
Q = 20 (333.18 + 176.28 + 1.218 (50)) kJ/hr
= 10292.8 kJ/hr (Ans.)
Therefore, the heat added to the stream to produce vapor at 0.7 bar and 50°C is 10292.8 kJ/hr.
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In roofing applications ballast can perform all of the following functions except: a. contribute to air circulation under the roof membrane b. protect the roof membrane from physical wear (foot traffic) C. hold the roof membrane against wind uplift d. protect the roof membrane against UV radiation
In roofing applications, ballast can perform all of the following functions except protecting the roof membrane against UV radiation. Ballast is a material that is added to a flat roof to protect the membrane underneath from damage caused by foot traffic and wind uplift.
It can also contribute to air circulation under the roof membrane, which helps to regulate the temperature of the roof.Ballast is an effective method for protecting flat roofs from damage caused by UV radiation. It is important to note that ballast is typically used on roofs that are not exposed to direct sunlight.
This is because ballast can absorb heat and cause the roof membrane to overheat, which can lead to premature failure of the membrane. Ballast is typically composed of materials such as gravel, crushed stone, or sand. The amount of ballast required depends on the size and slope of the roof, as well as the type of ballast being used.
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A simply supported timber beam 3.6 m in span in a storage facility in Melbourne carries two concentrated point loads (P) at one third and two third span points of the beam and a uniformly distributed load of (permanent) g=1.2 kN/m and imposed floor load (5 months duration) of q= 2.4 kN/m. Each point load (P) consists of 8 KN permanent load and 4 kN imposed (floor live load) of 5 months duration. The beam is made up by nailing 2/240x45 F17 seasoned hardwood sections nailed side by side, thus making a beam of 240X90 mm in cross section. It is laterally restrained along its top edge at each end and at the location of the point loads (ie at one third and two third span points). Calculate the bending moment capacity and the shear force capacity of the beam to resist the combination of permanent and imposed loads and assess whether the beam is adequate in resisting the applied bending moment and shear force. Assume that there are no other beams running parallel to this beam and that the beam is a primary element. Ignore the self-weight of the beam.
Introduction: A timber beam is used to support loads in a storage facility in Melbourne. It has a span of 3.6m, and it is simply supported. There are two concentrated point loads (P) and a uniformly distributed load of (permanent) g = 1.2 kN/m and imposed floor load (5 months duration) of q = 2.4 kN/m.
The beam is made of 2/240x45 F17 seasoned hardwood sections nailed side by side, and it is laterally restrained along its top edge at each end and at the location of the point loads. We need to calculate the bending moment capacity and the shear force capacity of the beam to resist the combination of permanent and imposed loads and assess whether the beam is adequate in resisting the applied bending moment and shear force.
Calculation of the bending moment capacity of the beam The bending moment capacity of the beam can be calculated using the formula: M = WL²/8where,W = total load on the beam = permanent load + imposed load = (1.2 kN/m + 2.4 kN/m) x 3.6 m= 10.8 kNP = concentrated point loads = 8 kN (permanent load) + 4 kN (imposed load) = 12 kNL = distance between the point loads = 3.6/3 = 1.2
m Now, the bending moment at the point loads can be calculated as:M1 = P x L/4 = 12 kN x 1.2 m/4 = 3.6 kN.mM2 = P x 3L/4 = 12 kN x 3.6 m/4 = 10.8 kN.m The maximum bending moment in the beam occurs at the mid-span and is given by: M max = WL²/8 + M1 = 10.8 kN x 3.6 m²/8 + 3.6 kN. m= 37.44 kN.m.
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All Loads Given Are Service (Unfactored) Fy = 50 Ksi E = 29,000 Ksi Pinned Support Conditions Full Lateral Bracing For Beams Moments Applied About Strong Axis Beam Cb =1.0 Recommended Design Values For Effective Length "K" Factors Note That "Capacity" And "Strength" Have The Same Meaning Where Applicable, Select The Nearest Answer That Applies
all loads given are service (unfactored)
fy = 50 ksi
E = 29,000 ksi
pinned support conditions
full lateral bracing for beams
moments applied about strong axis
beam Cb =1.0
recommended design values for effective length "k" factors
note that "capacity" and "strength" have the same meaning.The shear strength Rn of one Group B (A490) bolt in double shear, thread condition X (threads excluded from shear plane), 1-inch diameter is:
49.5 kips
98.9 kips
80.1 kips
490.0 kips
The shear strength Rn of one Group B (A490) bolt in double shear, thread condition X (threads excluded from shear plane), 1-inch diameter is 98.9 kips (Option B).Given that:• fy = 50 ksi• E = 29,000 ksi• Pinned support conditions Full lateral bracing for beams
Moments applied about the strong axis• Beam Cb =1.0• Recommended design values for effective length "k" factors. From the above-mentioned information, we can calculate the Rn of one Group B (A490) bolt in double shear. Thread condition X (threads excluded from shear plane), 1-inch diameter.
We know that the formula for the shear strength of one bolt in double shear is:[tex]Rn = 2.5 x F_u x A_s[/tex]Where F_u is the tensile strength and A_s is the shear area of the bolt.For Group B (A490) bolt, tensile strength F_u = 150 ksi. Shear area A_s = 0.758 in²Putting all values into the above formula, we get;[tex]Rn = 2.5 x F_u x A_s= 2.5 x 150 ksi x 0.758 in²= 284.625 kips[/tex]
Now, the given bolt is in thread condition X (threads excluded from shear plane)
Therefore, the shear strength of one Group B (A490) bolt in double shear, thread condition X (threads excluded from shear plane), 1-inch diameter = [tex]284.625 / 2= 142.3125 kips[/tex]
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Shopping mall construction project
1)Identify the most important TWENTY (20) activities including their (descriptions, dependencies, and duration) for the above construction project. 2)By using the activity list in Q2, complete the following tasks: a)Draw the Activity on Node diagram and find the critical path
b)Perform a Forward and Backward pass.
c)Replace the duration of the first eight activities and draw the network diagram.
The following are the most important TWENTY (20) activities, including their descriptions, dependencies, and durations for a shopping mall construction project:Activity description dependencies.
The Activity on Node diagram for the Shopping mall construction project is depicted in the figure below:The critical path is activities A, C, D, F, G, H, I, J, and K with a total duration of 31 weeks.b)Perform a Forward and Backward passForward Pass: To determine the Early Start (ES) and Early Finish (EF) for each activity,
The Forward Pass technique is used. The duration of the first activity, A, is 4 weeks, and its ES and EF are both 1 week. The remaining ES and EF for all activities are determined by adding the duration of the previous activity to its EF. Thus, the table below depicts the Forward Pass computations for the shopping mall construction project.
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Problem S. For a typical reinforcing steel bar permitted for use by AC1318-14. (a) What is generally the Young's modulus of the rebar? (b) What does the 'grade' on a rebar represent? (c) What is nearly the yield strain of a Grade-80 steel?
(a) The Young's modulus of the reinforcing steel bar is usually taken to be 29,000 ksi. (b) The term 'grade' on a reinforcing steel bar represents the strength of steel. The strength is determined by the steel's yield strength or the stress at which steel starts deforming or yielding.
(c) The nearly yield strain of Grade-80 steel is about 0.008. Explanation: Reinforcing steel bars are used in reinforced concrete to give the concrete tensile strength to resist cracking and also resist the weight of structures and live loads. These bars are made of high-quality steel and must comply with several standards and specifications like ASTM or ACI.
Reinforcing steel bars are classified by grades according to their tensile strength. The grade is indicated by a number that ranges from 40 to 100 based on the steel's yield strength. The most commonly used grades are 40, 60, and 80, with 80 being the highest.
This means that Grade-80 steel is stronger than Grade-60 and Grade-40.The Young's modulus of steel refers to its elasticity, i.e., its ability to stretch and then return to its original shape. The Young's modulus of reinforcing steel bars is generally considered to be 29,000.
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Determine the design strength of the cantilever beam (properties given below) and the safe uniform service live load if the service dead load is 320 kN-m. Use f 'c = 22 MPa, fy = 415
MPa.
L = 4m
b = 400 mm
d' = 70 mm
d = 620 mm
As 10-28 mm bars
A's 3-25 mm bars
Cantilever beam is a beam supported on one end and free on the other end. There are different types of cantilever beams including single fixed end, single free end, and the double-ended cantilever beam.
We are to use f'c = 22 MPa, fy = 415 MPa. We are also given that L = 4m, b = 400 mm, d' = 70 mm, d = 620 mm, As 10-28 mm bars and A's 3-25 mm bars.
Service dead load is 320 kN-m
Determine the ultimate moment of resistance of the beam Ultimate moment resistance of the beam is given by;
[tex]Mu = 0.138f'cbd^2(1-0.416as/fy)[/tex](where as is the area of tensile reinforcement)Substituting values;
[tex]Mu = 0.138(22)x(0.4)x(0.62)^2(1-0.416((10xpi x (0.028)^2)/4)/(415))Mu = 15.53 kN-m2.[/tex]
Determine the design moment of resistanceDesign moment resistance (Md) is given by;Md = 0.9Mu
Substituting;[tex]Md = 0.9 x 15.53Md = 13.98 kN-m3. D[/tex]
[tex]M = 1.2 (Dead Load Moment) + 1.6 (Live Load Moment)[/tex]Where;Dead Load Moment = Service dead load x (L/2) = 320/2 x 4 = 640 kN-mLive Load Moment = wL^2/10 where w is the uniform load in kN/m and L is the span in meters.
Substituting;[tex]Live Load Moment = w(4)^2/10 = 0.4w kN-m[/tex]
Factored Moment = [tex]1.2(640) + 1.6(0.4w) = 768 + 0.64w[/tex]
Design moment capacity (Md) =[tex]13.98 kN-m Equating;Md >= M13.98 >= 768 + 0.64w0.64w <= -754w <= 1178 kN/m[/tex]
Therefore the safe uniform service live load is 1178 kN/m.
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A one-story steel storage building with a Gable Roof Main wind- force resisting system have the following configuration. Building width=6m, Building Length = 6m, Height to Apex = 7.5m, Eve Height 4.5m. And is an enclosed structure which have an Exposure Category C with a wind speed of 290 kph. Use a Gust factor of 0.85.
Considering the Load Combination = 1.2DL +1WL, Determine the most critical (consider sign/direction) Leeward Roof Pressure (considering all cases) at the surface of the walls if the wind direction goes from West to East (the Wind direction is perpendicular to the ridge of the structure).
DL=1.0kPa
Choices
-1.32
0.23
-0.55
+2.3
+1.32
-0.23
-2.3
-1.0
+2.098
+0.55
Given that a one-story steel storage building with a Gable Roof Main wind- force resisting system has the following configuration .Building width (B) = 6 m Building Length (L) = 6 m Height to Apex (H) = 7.5 m Eve Height (h) = 4.5 mEnclosed Structure which has an Exposure Category C with a wind speed of 290 kphGust factor (G) = 0.85Load Combination = 1.2DL +1WLDirection of wind (perpendicular to ridge) = West to EastDL = 1.0 kPa
We are to determine the most critical (consider sign/direction) Leeward Roof Pressure (considering all cases) at the surface of the walls if the wind direction goes from West to East (the Wind direction is perpendicular to the ridge of the structure). Formula used:
Leeward roof pressure = qz – GCp Dynamic Wind pressure (qz) = 0.613kPaCp = Net Pressure Coefficient (Cp)GC = Gust Coefficient Gust Coefficient (GC) = G x (1 + 0.2 Kz) where Kz is the velocity pressure exposure coefficient.
The leeward roof surface pressure is negative for a gable roof in the windward direction and positive in the leeward direction. Thus, it is the most critical case.1) West Wall: qz = 0.613 kPa Cp = Cpi x GC = - 0.7 x 0.85 = - 0.5955GCp = G x (1 + 0.2 Kz) x Cp = 0.85 x (1 + 0.2 x 1) x (- 0.5955) = - 0.9383Leeward roof pressure = qz – GCp = 0 - (- 0.9383) = 0.9383 kPa (negative)2) East Wall:
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