In structural design of timber, span tables are used to determine the size of joist and beams. In this case, the span table required is only for a beam and must include the variables of service class, different variable actions for residential floor, commercial floor, light roof and heavy roof and spacing of beam.
The table must be completed using the design spreadsheet for a beam. The type of engineered timber chosen for this span table is the glulam or glue-laminated timber.Glulam or glue-laminated timber is a type of engineered timber that is made up of several layers of timber that are glued together to create a large and solid beam. It is often used in structures where a long span is required. The table will include the different spans for various types of glulam timber used. The size of the beam is determined by the required span and the load that it will carry.
The spacing of the beam will also be taken into account to ensure that it is strong enough to support the load. A spreadsheet can be used to complete the table to make it easier to calculate the different variables and to ensure accuracy.
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Determine the short-term response of the composite section shown below (based on the adopted reference system). The external applied moment is 350 kNm and the external axial force Ne. is 100 kN (applied with respect to the origin of the reference system). In particular, calculate the strains and stresses induced in the cross-section. Summarise the results by plotting the strain diagram and stress diagram produced by the applied external actions.
Assume: Ec = 32,000 MPa Ess = 200,000 MPa Es = 200,000 MPa
A composite section consists of different materials placed together, to act as a single structural unit with improved structural integrity. Steel-concrete composite sections are used extensively in construction, primarily in buildings, bridges, and tunnels.
The moment of inertia of the steel section about the neutral axis is calculated as follows :Is t = (1/12)(bh3)Is t = (1/12)(300 x 5003 ) Is t = 52.08 x 109 mm4The effective depth of the concrete slab is calculated as follows: d = h steel section + 0.5hconcrete slab d = 500 + 0.5(400)d = 700 mm The cross-sectional area of the composite section is calculated as follows :A = A steel section + A concrete slab A = 500 x 300 + 400 x 700A = 410000 mm2The bending stress in the steel section can be calculated using the following expression:σb = Mc/Istσb = (350 x 106 x 250)/52.08 x 109σb = 1.69 N/mm
2The axial stress in the composite section is calculated using the following expression:σa = Ne/Aσa = 100 x 103/410000σa = 0.244 N/mm2The strain in the steel section is calculated using the following expression:εb = σb/Esεb = 1.69 x 106/200,000εb = 0.0085 mm/mm The strain in the concrete slab is calculated using the following expression:[tex]εc = σc/Ecεc = (σa - σb)/Essεc = (0.244 - 1.69)/200,000εc = -0.00773[/tex] mm/mm The total strain in the composite section is equal to the sum of the strains in the steel section and the concrete slab:εt = εb + εcεt = 0.0085 - 0.00773εt = 0.00077 mm/mm To summarize the results, we can plot the strain diagram and the stress diagram produced by the applied external actions. The strain diagram is shown below :To draw the stress diagram, we can use the following expressions:σb = Mc/Istσa = Ne/Aσc = Essεcσt = σa + σcThe stress diagram is shown below:
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Draw a formation that presents the following terms as used in groundwater hydrology and provide a brief description each:
(i)
Aquiclude
(ii)
Phreatic
Confined aquifer
(iv)
Piezometric surface
(v)
Artesian well
Hint: Combine all in one diagram
Here's a formation that presents the following terms as used in groundwater hydrology:AquicludeAn aquiclude is a geological formation that can absorb and transmit water, although not as rapidly as an aquifer. Aquicludes can be made up of clay, silt, or shale, and they act as a boundary between two aquifers.
Piezometric surface The piezometric surface, also known as the water table, is a plane in the subsurface that represents the pressure head of groundwater and defines the upper boundary of the saturated zone in the subsurface. This surface corresponds to the height of water that can be measured in a well.
Phreatic water refers to groundwater in an unconfined aquifer, and the phreatic surface is the level below which the soil is completely saturated with water. Artesian wellAn artesian well is a type of well that is drilled into an artesian aquifer, which is an aquifer that is confined between two aquicludes. Water flows freely to the surface from an artesian well.
A confined aquifer is an aquifer that is sandwiched between two layers of impermeable rock or clay, preventing water from entering or exiting it through those layers. In such an aquifer, water can only be extracted by a well drilled into it.
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After a re-evaluation of the duration of activity B, it has been provided the following data. Most likely duration = 10 Optimistic Duration = 16 Pessimistic Duration = 25 Using the three-point estimating technique the new duration of activity B is a. 7 b. 15 c. 17 d. 22
Three-point estimating technique: Three-point estimating technique is a method used in project management for calculating the estimates of the project management activity.
The three-point estimating technique is used in estimating the cost and duration of a project. Activity B can be defined as a task that must be completed within a specific time frame and assigned to a specific project member in the project plan.
The duration of Activity B has been reevaluated as follows: Most likely duration = 10Optimistic Duration = 16Pessimistic Duration = 25New duration of Activity B: We can use the three-point estimating technique to determine the new duration of Activity B as follows.
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A dam was designed for a flow rate of 2500 cu-mecs. The dam must be modelled in an hydraulics laboratory to sort out the calibration of such dam (like to find out minor design details) Since it is a retaining structure, it should be modelled as a distorted. Consider H-scale 1:50, V- scale 1:30. What will be the required flows rate in l/s, be at the lab. What are the two components of the EIA and what is the role in planning a dam projects? Discuss NEMA. What is EMP and EA?
The dam was designed for a flow rate of 2500 cu-mecs. But, the dam needs to be modeled in an hydraulics laboratory to sort out the calibration of the dam. Since it is a retaining structure, it should be modeled as a distorted. Consider H-scale 1:50, V- scale 1:30.
The main purpose of EIA is to determine the potential impact of a proposed project on the environment and society. It is a formal study that analyzes and assesses the impact of a proposed project on the environment and social structure.
The organization's main objective is to ensure that environmental considerations are taken into account in the decision-making process. The key functions of NEMA include environmental impact assessment, environmental audit, environmental monitoring, and enforcement of environmental regulations. NEMA also has a mandate to coordinate environmental management activities with other stakeholders.
An Environmental Management Plan (EMP) is a document that outlines how a proposed project will manage its environmental impacts. It is an essential part of the environmental management process, and it provides guidelines for monitoring and managing environmental impacts during the project's construction and operation. The main goal of an EA is to identify potential environmental impacts and to suggest ways to mitigate them.
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How shall completed interior design project deliverables be
accepted? explain with an example.
Once the interior design project is complete, the deliverables must be accepted properly. Following explains how completed interior design project deliverables shall be accepted.
Acknowledge the designers and any additional workers who assisted in the project. It should also describe what was accomplished and what the final outcome should look like. Explain in detail what was done and if everything meets your needs and specifications. During the review, ask to see samples of the products that were used to complete the design. This is your chance to express any concerns you may have. Finally, after a thorough inspection, once you're satisfied with the final product, you can accept the completed interior design project. To do so, you may have to sign off on the work in order to provide confirmation that the job has been completed to your satisfaction. For instance, in the case of an office space, once the project is finished, you can acknowledge the designers who worked on the project. During the inspection, ask for a demonstration of any furniture items or equipment that were used. You may also want to make certain that everything is in good working order. Finally, once everything has been checked and you're happy with the final product, you can sign off on the work to accept the completed interior design project.
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Mr. Jobert commissioned you to estimate the construction of a certain structure. As Engineers, our job includes giving detailed estimate inclusive of labor and materials. If approved, the contract for building the structure will be awarded to you.
However, Mr. Jobert already had a previous consultation with Prudencio, an independent foreman, who offered labor and services at a cheaper price.
How will you convince Mr. Jobert to engage in your services instead of Prudencio's?
Cite applicable provisions of the Code of Ethics for such situation.
As engineers, it is our responsibility to provide a comprehensive estimate that includes labor and materials. The code of ethics for engineers obliges us to uphold professional ethics and practices.
To be able to convince Mr. Jobert to engage in our services instead of Prudencio's, I will take the following steps. First, I will highlight our commitment to providing high-quality services, backed by our wealth of experience.
Furthermore, I will explain to him the numerous benefits of our services, including the high level of professionalism, the use of high-quality materials, and the use of modern technologies.
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1- A shaft is subjected to a torque of 16,000 Nm. If the maximum permissible stress in the material of the shaft is 65 N/mm², find:
1) the diameter of the solid shaft ii) the dimension of the hollow circular shaft if the thickness is 10% of
ii)the internal diameter. Assume it's not thin-walled.
2- A hollow circular shaft is of 180 mm internal diameter and thickness 5 mm. Find the maximum stress in the shaft if the torque is 12,000 Nm.
3- A solid circular shaft is subjected to a torque of 20,000 Nm. If the maximum permissible shear stress is 50 N/mm², find a suitable minimum diameter of the shaft.
4- A solid circular shaft has a diameter of 80 mm. Find the maximum shear stress and the angle of twist in a length of 2m when the shaft is subjected to a torque of 10 kNm. Given: G=85GPa.
5- A hollow circular shaft has an external diameter of 120 mm and the internal diameter is 90 mm. If the stress at a fibre on the inside wall is 36 MPa, due to a torque applied, find:
i) this torque,
ii) the maximum shear stress
iii) the angle of twist per unit length.
Given: G= 85 GPa.
6- A solid circular shaft of 25 cm diameter is to be replaced by a hollow shaft, the ratio of the external to internal diameters being 2 to 1. Find the size of the hollow shaft if the maximum shearing stress is to be the same as for the solid shaft.
1) The diameter of the solid shaft is obtained by using the formula for torsion, T/J = τ/R or T = τJR. Where J = πd⁴/32, R = d/2τ, and τ = 16,000 Nm/[(πd⁴/32)(d/2)] = 32,000/(πd³) MPa. Solving for d gives 64.03 mm.ii) The diameter of the hollow circular shaft is obtained from the expression: τ = T(r² - r₁²)/J where r₁ = r - t and r is the external radius.
Thus, we have r₁ = 0.9r and τ = 16,000 Nm[(0.9r)² - r²]/[(π/32)(r⁴ - (0.9r)⁴)] = 36.61/r MPa. If τ = 65 MPa, then 65 = 36.61/r, r = 55.6 mm and the internal diameter is 50 mm.2) The maximum shear stress in a hollow circular shaft is given by τ = Tr/J. Where T = 12,000 Nm, r = 0.09 m, R = 0.135 m, J = (π/32)(0.135⁴ - 0.09⁴) = 3.57 × 10⁻⁵ m⁴. Thus, τ = (12,000)(0.135)/3.57 × 10⁻⁵ = 454.545 MPa.3) From the formula T/J = τ/R, we can obtain J = πd⁴/32 = (π/32)(d²)².
Substituting the given values, we have 20,000 Nm/(π/32)(d²)² = 50 MPa, d = 160 mm.4) The maximum shear stress can be obtained from the formula τmax = Tr/J where J = πd⁴/32. Thus, we have J = π(0.08)⁴/32 = 1.005 × 10⁻⁶ m⁴. Therefore, τmax = (10 × 10⁶)(0.08/2)/1.005 × 10⁻⁶ = 3.98 × 10⁸ N/m².The angle of twist per unit length can be obtained from the expression θ/L = Tl/GJ.
Thus, θ/L = (10 × 10⁶)(2)/(85 × 10⁹)(π/32)(0.08)⁴ = 0.005 rad/m.5)i) The torque is obtained from T = τJR/(r² - r₁²) where r = 0.06 m and r₁ = 0.045 m. Solving for T gives 47,725 Nm.ii) The maximum shear stress is given by τmax = Tr/J where J = π(r⁴ - r₁⁴)/32. Substituting the given values, we have J = π(0.06⁴ - 0.045⁴)/32 = 2.097 × 10⁻⁶ m⁴.
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The sustainability of water supply systems largely depends on timely and regular maintenance and operation of the systems. However, in most developing countries it has been found that operations and maintenance of water supply systems are in the bad conditions making sustainability of water very challenging. As a well-known water resource engineer in African continent
(a) Discuss six (6) challenges facing operations and maintenance of water supply and delivery systems in Africa
(b) Suggest seven (7) policy directions for government in Africa continent towards operations and maintenance of water supply and delivery systems in Africa
(c) Examine driving factors affecting water quality in a water transmitting and distribution systems in Africa
(d) Develop actions that can be taken to improve water quality or prevent its deterioration in the distribution system
Six challenges facing operations and maintenance of water supply and delivery systems in Africa include;Poor governance and corruptionInsufficient funding for infrastructureInadequate planning.
and technical expertisePoor stakeholder participationWeak policies and regulatory frameworksLack of skilled personnel(b) Seven (7) policy directions for the government in Africa continent towards operations and maintenance of water supply and delivery systems in Africa include;Institutional strengthening and capacity development.
Addressing financial barriers and mobilization of resourcesEstablishment of effective monitoring and evaluation systemsPromoting private sector participationImproving legal and regulatory frameworksDeveloping community involvement and participationStrengthening regional cooperation and collaboration.
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Consider a cylindrical cross-section pipe that is 50 m long with a 7 cm diameter, and which has a starting pressure of 320 kPa and an outlet pressure of 105 kPa. A pressure sensor is located 10 m from the start of the pipe.
(a) Calculate the pressure that this sensor would read if the fluid were water (p = 1000 kg/m3).
(b) If the pipe instead contains gaseous carbon dioxide in isothermal flow, would the pressure be the same at this sensor as if the fluid were water?
a) Pressure in the pipe is given as P_1 = 320 kPa and at the outlet is P_2 = 105 kPa.
The difference in pressure at two points in the fluid is given by[tex]∆P = P1 − P2 = 320 − 105 = 215 kPa[/tex]. Density of water, ρ = 1000 kg/m³Length of the pipe, L = 50 mDiameter of the pipe, d = 7 cm = 0.07 mRadius of the pipe,[tex]r = d/2 = 0.07/2 = 0.035 m[/tex]Cross-sectional area of the pipe, [tex]A = πr² = π(0.035)² = 0.00385 m²[/tex]
b) When the pipe instead contains gaseous carbon dioxide in isothermal flow, the pressure would not be the same at this sensor as if the fluid were water. For ideal gases like carbon dioxide, the Bernoulli equation is modified to account for the change in internal energy associated with the flow.
Bernoulli's equation is:
[tex]P_1 + (1/2) ρ V_1² + ρgh_1 + U_1 = P_2 + (1/2) ρ V_2² + ρgh_2 + U_2[/tex]where U is internal energy. If the process is isothermal, internal energy is costant,[tex]U_1 = U_2.[/tex]
P_1 + (1/2) ρ V_1² + ρgh_1 = P_2 + (1/2) ρ V_2² + ρgh_2
This will cause the velocity of the fluid to be much greater for carbon dioxide, which will in turn cause the pressure to be much lower in the carbon dioxide pipe at the same point, as compared to the water pipe.
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Calculate the ejection fraction (Ef) of a patient with an EDV of 120ml, an ESV of 80ml, and a cardiac output of 4.2L/min. Is the ejection fraction normal? yes, Ef=33% no, EF=33% yes, Ef=66% no, Ef=66% yes, Ef=50%
A patient has a blood pressure of 140/80 with a stroke volume of 82ml and a pulse of 90 beats per minute, what is the cardiac output? 7.38 L/min 7380 L/min 6560 L/min 6.56 L/min 10.98 L/min
The ejection fraction (EF) is 33%, indicating reduced heart function. The cardiac output is 7.38 L/min.
Ejection fraction (EF) is defined as the proportion of blood ejected by the heart's left ventricle with each contraction. EF may be used to help diagnose and evaluate the severity of heart failure and other heart disorders. The EF is expressed as a percentage, with a higher percentage indicating a stronger heart function.
To calculate EF:
EF = (EDV – ESV) ÷ EDV
EF = (120 – 80) ÷ 120
EF = 40 ÷ 120
EF = 0.33 or 33%
EF is a measurement of how well the heart is pumping blood to the rest of the body. It is considered normal when it is between 50 and 70 percent. An EF of 33% is lower than the normal range, so the answer is no. The patient's EF is not normal. Hence, the answer is: no, EF=33%.
Cardiac output (CO) is defined as the volume of blood the heart pumps per minute. The equation used to calculate cardiac output is:
CO = SV x HR (CO = Stroke volume x Heart rate)
Given, stroke volume = 82 ml and pulse = 90 beats per minute. Therefore,
CO = 82 ml × 90 beats per minute
CO = 7380 ml/min
CO = 7.38 L/min
Therefore, the cardiac output is 7.38 L/min. Hence, the answer is: 7.38 L/min.
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how should email be considered similar to a phone call
Answer:
Emails and phone calls are both common forms of communication that are used in professional and personal settings. There are several similarities between email and phone calls:
1. Both are asynchronous forms of communication: Unlike instant messaging or face-to-face conversations, both emails and phone calls allow the sender or recipient to respond at their convenience. They don't require immediate attention or an instant response.
2. Both are written forms of communication: While phone calls rely on spoken words, emails are written. As a result, both can be used to convey detailed information and allow the sender to carefully consider their words before sending.
3. Both are forms of direct communication: Emails and phone calls both allow for direct communication between two parties. This can be beneficial for discussing sensitive information or resolving issues quickly.
4. Both can be used for formal and informal communication: Emails and phone calls can be used in both personal and professional contexts. They are both flexible forms of communication that can be adapted to fit different situations.
5. Both require attention to tone and etiquette: Just like with phone calls, emails require attention to tone and proper etiquette. Both forms of communication should be approached professionally and respectfully to ensure effective communication.
In conclusion, while there are differences between emails and phone calls, there are also similarities that make them useful communication tools. Both allow for direct, asynchronous communication and can be adapted to fit different situations.
Explanation:
What are the principle reasons why the hydrographs of
urban streams are so different from forested streams?
The hydrographs of urban streams are substantially different from forested streams due to human activity and construction. Urbanization and development alter the natural water balance, resulting in changes in streamflow patterns, erosion, sedimentation, and water quality.
Impervious Surface: In urban areas, vast areas of land are paved, which prevents water from penetrating the soil and replenishing groundwater. Instead, the water runs off the surface and into nearby streams, resulting in quick, heavy floods and flash floods.
Land Use Changes: Urbanization results in land-use changes that influence the local climate, such as increased temperatures, less vegetation, and less evapotranspiration.
Stormwater Management: Forested streams provide a variety of ways for water to infiltrate the ground and enter streams, whereas urbanization necessitates the use of impervious surfaces like roads, parking lots, and rooftops. When it rains, water accumulates on these surfaces and is channeled into pipes or stormwater systems.
Stream Channelization: streams are often modified to enhance the flow of water and reduce the risk of flooding. Straightening, widening, and deepening of stream channels are examples of such changes.
These are some of the primary reasons why the hydrographs of urban streams are so different from forested streams.
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impact tools such as chisels with ___________________ heads should be removed from use.
Impact tools such as chisels with damaged or worn heads should be removed from use.
Impact tools like chisels are widely used in various industries for tasks such as cutting, shaping, or removing materials. However, when the heads of these tools become damaged or worn, they can pose significant risks to both the user and the work being performed. Therefore, it is essential to remove such tools from use to ensure safety and maintain efficiency.
When the heads of impact tools like chisels are damaged, they can lead to a loss of control and accuracy during operation. A damaged head may cause the tool to slip, resulting in unintended cuts or injuries to the user. Moreover, worn heads can reduce the effectiveness of the tool, requiring more force to achieve the desired results. This increased force can strain the user's muscles and joints, leading to fatigue and potential long-term injuries.
By removing impact tools with damaged or worn heads from use, the risk of accidents and injuries can be significantly reduced. Regular inspection and maintenance of these tools are crucial to identify any signs of wear or damage early on. If a chisel or any other impact tool is found to have a damaged head, it should be promptly replaced or repaired by a qualified professional.
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Bagasse Ash Stabilization
Answer:
Bagasse ash stabilization refers to the process of treating and utilizing bagasse ash, which is a byproduct generated from the combustion of bagasse, a residue from sugarcane processing. Bagasse ash is rich in silica and possesses pozzolanic properties, making it suitable for various engineering applications.
Explanation:
The stabilization of bagasse ash involves blending it with other materials, such as lime or cement, to enhance its properties and create a stable and durable material. This process helps improve the geotechnical characteristics of bagasse ash, making it suitable for use in construction projects.
Bagasse ash stabilization has several benefits. It helps reduce the environmental impact of bagasse ash by diverting it from landfill disposal. Additionally, it can contribute to the conservation of natural resources by replacing traditional construction materials with a sustainable alternative. The stabilized bagasse ash can be used in applications such as road construction, embankment construction, and soil stabilization.
1) If you were asked to perform an excavation contract for oil pipeline project competitively with limited boring data, what type of contract would you want and why? Please explain briefly.
2) given a single person with a $96000 income, apply the standard deduction and tax rate in 2015, which of the following is closest to the tax this person should pay? Assume single filing status.
3) What is the main driver in petroleum projects, and why?
If I were asked to perform an excavation contract for an oil pipeline project competitively with limited boring data, I would choose a unit price contract. A unit price contract is a construction contract where the contractor bids on a fixed price per unit of work, such as per cubic meter of soil removed or per foot of pipeline installed.
The advantage of this type of contract is that it allows for flexibility in pricing and risk allocation. It is ideal for a situation where there is limited information on the scope of work or soil conditions because the owner is only paying for the actual amount of work performed.
This reduces the risk of unforeseen costs or changes in scope. The unit price contract also promotes efficiency because the contractor has an incentive to complete the work as quickly and efficiently as possible.
Given a single person with a $96000 income, applying the standard deduction and tax rate in 2015, the tax this person should pay would be $18,790. The profitability of petroleum projects is largely dependent on the global market for oil and gas and the balance of supply and demand.
The companies that are successful in petroleum projects are those that can effectively manage costs, mitigate risks, and stay competitive in a dynamic and challenging market.
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7 The beam with a cross-sectional area of 350 mm x 600mm is prestressed by using a bonded (stress relieved tendons) with an fpu =1350MPa. If fc'=30 MPa., determine the safe ultimate moment capacity of the beam. Aps=1230mm2
fpy=1160MPa
14 Compute the safe ultimate moment of a T-beam beam with a flange width 2500mm, flange thickness of 80mm, web thickness of 300mm and an overall depth of 900mm. The beam is reinforced with an unbonded tendon having an area Aps = 1000mm2 and located 80mm from the bottom fiber of the beam, fc' =35MPa. The beam has a simple span of 6m.
The safe ultimate moment capacity of the beam is 1163.025 KN-m. 14 The safe ultimate moment of the T-beam beam is 87.12 KN-m. Safe ultimate moment capacity of the beam.
[tex]Mu = (0.87 x fpu x Aps x (d - a/2)) + 0.35 fc' (b - d/2) x (A - a[/tex])
d = overall depth of beam = 900 mma = 80 mm
Aps = [tex]1000[/tex] mm
2fpu = ultimate strength of prestressed steel = 1350 MPa
fc' = compressive strength of concrete = 35 MPab
width of flange = 2500 mmA =
overall area of the section = (2500 mm x 80 mm) + (300 mm x 820 mm) = 211,600 mm2
[tex](d - a/2), (A - a) and b/2.(d - a/2) = 900 mm - 80 mm/2= 860 mm(A - a)[/tex]
= [tex]211600 mm2 - 80 mm x 1000 mm= 131600 mm2b/2 = 2500 mm/2 = 1250 mm[/tex]Put these values in the equation of ultimate moment;
Mu = [tex](0.87 x 1350 MPa x 1000 mm2 x 860 mm) + (0.35 x 35 MPa x (2500 mm - 900 mm/2) x 131600 mm2)[/tex]
Mu = [tex]87,119,500[/tex] N-mm or 87.12 KN-m
The safe ultimate moment of the T-beam beam is 87.12 KN-m.
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Contract completion date has just lapsed. The Architect (A) did not issue the Certificate of Non- Completion right away. It is likely that the Contractor (C) may need another 10 weeks before substantial completion is possible.
In the last interim payment certification, A has in gross certified 70% of the contract sum of HK$210M with contract retention deducted at a maximum 10%. 2 weeks ago, C has submitted a new interim payment application at a value of total work done at 90% of the original scope of contract and a sum of HK$20,000,000 on contract variation. At the same time, C submitted a notice of substantial completion of the works inviting A for inspection
To C, all the variations were incurred from the Architect's instructed change in -
(a) curtain wall design from glass size width of 800 mm wide to 1000 mm wide, and
(b) extra 6 numbers of piles from the 30 piles originally outlined in the BD approved piling plan as part of the tender drawings.
The contract adopted the HKIA/HKIS/HKICM Standard Form of Building Contract 2006 (Without Quantities) on lump sum basis (the "Contract"). It includes a soil investigation report with was marked for reference.
Today, A (4th week after original contract completion date) issued an Interim Payment Certificate ("IP") with the followings:-
(i) (ii) Liquidated Ascertained Damage retention calculated at 4 weeks of delay; Work done at 80% on original scope; and
(iii) No variation claim.
A did not conducted completion inspection.
Provide your answers for the following question:-
(a) Assuming that you are C who wished to argue, among others, on the IP, prepare your claims with quotes on referenced clauses in the Contract.
(i) monetary claim, and(12%)
(ii) time claim.(12%)
(b)
Provide arguments and the likely defence of A to your proposed claims.(16%)
(a) Assuming that you are C who wished to argue, among others, on the IP, prepare your claims with quotes on referenced clauses in the Contract. (b) Provide arguments and the likely defence of A to your proposed claims.
(i) Monetary claim :C could argue that the certification of the Interim Payment Certificate issued by A is incomplete, inaccurate, or otherwise deficient. He has calculated that his work is worth[tex]HK$20,000,000[/tex], and he could cite clauses 30 and 31 of the Contract to back up his argument.
ii) Time claim :C may claim that he is entitled to an extension of time due to the Architect's changes to the curtain wall design and the piling plan. He may argue that this would take into account the impact of these modifications on his working time.
(i) Monetary claim: A could argue that the Interim Payment Certificate is correct and that the Contractor's work is worth no more than HK$16,800,000. Clause 30 of the Contract states that the quantity surveyor should make a decision on the value of the work done, and A has done so.
(ii) Time claim: A could argue that the modifications to the curtain wall design and piling plan had little impact on the Contractor's working time, and therefore, he is not entitled to an extension. Clause 23.2 of the Contract allows for time extensions only if the modifications cause a significant delay, and A could argue that this was not the case.
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1. A contractor has a project involving the construction of a small manufacturing plant. The contractor has selected a borrow site and has decided to use 27-cubic yard dump trucks to haul the needed fill material. The contractor has decided to use a hydraulic excavator with a shovel attachment to load the trucks. The excavator has expected productivity as 422cy/hr at 100% efficiency, and the truck cycle time excluding loading is 0.48hr. The job site operation efficiency is assumed as 0.8. a) How many dump trucks will be needed to work with the excavator? Select one: a. 8 b. 9 c. 6 d. 7 b)What is the expected fleet production if only 6 trucks are available to the fleet? Select one: a. 225 cy/hr b. 238 cy/hr c. 298 cy/hr d. 281 cy/hr
c) What is the expected fleet production if 11 trucks are available to the fleet?
Select one:a.422 cy/hr
b.516 cy/hr
c.338 cy/hr d.437 cy/hr d) How many dump trucks will be needed to work with the excavator?
Select one:
a.8
b.9
c.6
d.7
The answer is option (c) 6.
a) Number of dump trucks needed to work with excavator The total number of trucks needed to work with excavator can be calculated as; Number of trucks needed =
= [tex](422/27) × 0.48 × 0.8 = 6[/tex]
Therefore, the number of dump trucks needed to work with the excavator is 6.
b) Expected fleet production if only 6 trucks are available to the fleet Given, number of dump trucks available = 6The volume of each truck = 27 cubic yards Expected output = 100%
The job site operation efficiency is assumed as 0.8.Volume of excavator (bucket) = 422 cy/hr
[tex](6 × 27 × 60 / (0.48 × 6)) × 0.8 = 135 × 0.8 = 108 cy/hr[/tex]
Therefore, the expected fleet production if only 6 trucks are available to the fleet is 108 cy/hr.
c) Expected fleet production if 11 trucks are available to the fleet Given, number of dump trucks available = 11The volume of each truck = 27 cubic yards Expected output = 100%
The job site operation efficiency is assumed as 0.8.
Volume of excavator (bucket) = 422 cy/hr
The expected fleet production can be calculated as :- [tex](11 × 27 × 60 / (0.48 × 11)) × 0.8 = 297.5 cy/hr[/tex]
Therefore, the expected fleet production if 11 trucks are available to the fleet is 297.5 cy/hr.
The number of dump trucks needed to work with the excavator is given as 6.
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A steel column in a residential building with 5 m effective height is subjected to factored end loads as listed below. The size of column section is 356 x 368 x 177 kg/m UC (Grade 355).
Given:
Factored axial load = 3500 KN
Factored moment about x-x axis, Mx-x, at top of column = +100 kNm
Factored moment about x-x axis, Mx-x, at bottom of column = -80 kNm
Amplification factor P-A-d effect in x-x axis = 1.1
Factored moment about y-y axis, My-y, at top of column=-60 kNm
Factored moment about y-y axis, My-y, at bottom of column = -40 kNm
Amplification factor P-A-8 effect in y-y axis = 1.1
Factored moment amplified for P-A-8 effect governing M.:
MT1.1 x 100 110 kNm (Note: Positive sign for clockwise moment; negative sign for anti-clockwise
moment)
(a) Determine the section class of the steel column.
(b) Check the cross section capacity of the steel column.
(c) Check the member buckling resistances of the steel column.
(a) Section class of steel column: The section class of the steel column can be calculated using the formula: Section modulus, Zx=Ixx/ymaxwhere,Ixx is the moment of inertia about the xx axis and ymax is the maximum distance from the neutral axis to the extreme fibre .Then, Zx = (Ixx/ymax) = (356 x 368³/12)/(368/2) = 204×10⁴ mm³Effective length factor, Kx = 1.0 ≤ 1.2 (cl. 5.3.3.1 IS 800-2007)
Slenderness ratio, λx = Le/Kx = 5000/1.0 = 5000 mm Radius of gyration,rx = √(Zx/A) = √(204×10⁴/177) = 25.3 mm (A = Cross-sectional area)The buckling parameter is given by,αx = √(P/A)(K/ry)Here, P = Factored axial load = 3500 kN, A = Cross-sectional area = 177×10⁻⁶ m², K = Effective length factor = 1.0, and ry = Radius of gyration = 25.3 mm∴
αx = √(3500×10³/177×10⁻⁶) (1.0/25.3) = 21.3
The equivalent slenderness ratio,λex = λx/√(1+(αx/π)²) = 5000/√(1+(21.3/π)²) = 2234 mm The equivalent slenderness ratio is less than the limiting slenderness ratio (i.e., 300) for all class of sections. Therefore, the column is short and hence plastic design can be done. Section class of the column = Plastic section (Class 1)∴
Section class of the steel column is Plastic section (Class 1).(b) Cross section capacity of the steel column: The design strength of the column for the axial compression, Pn is given by the Euler's formula,Pn = π²ExIxx/(KxLe)²where, Ex is the elastic modulus = 2.1×10⁵ N/mm² for Grade 355 steel∴
Pn = π²×2.1×10⁵×368³/(1.0×5000)² = 7774 kN < Factored axial load (3500 kN)Hence, the cross-section is safe against buckling under axial compression. The design strength of the column for bending about the xx axis, Mn is given by,Mn = Zx x fy/γmwhere, fy is the yield strength of the material, and γm is the partial safety factor for resistance design.
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a) Considering a four-step travel demand modeling process, list the types of data that should be collected and possible data sources associated with each step in the process b) Briefly discuss in what ways transportation (system) influences land use (activity system)?
a) The Three-step travel demand modeling process consists of trip generation, trip distribution, mode choice, and. Each step requires different data and possible data sources:
Trip Generation: This step involves gathering data on the number of trips that originate from or end at specific locations within a geographic area. Data sources for this step include household surveys, census data, and land use surveys.
Trip Distribution: The process of allocating generated trips to their respective destinations is called trip distribution. Data sources for this step include destination surveys, travel surveys, and employment data.
Mode Choice: This step involves predicting the modes of transportation that individuals will use to travel to their destinations. Data sources for this step include mode-specific surveys, household travel surveys, and transit ridership data.
Land use patterns are often influenced by transportation infrastructure, and transportation networks are influenced by land use patterns. Residential areas may be designed with cul-de-sacs, which discourage through-traffic, while commercial areas may be designed with more direct, arterial roads.
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Consider an off-axis coupon under the action of two simultaneous axial stresses ox and oy. Express the safety conditions from "Maximum Stress Theory" as a function of the in-plane strength parameters XT, Xc, YT, Ye, S (ultimate shear stress) and the fiber orientation angle .
In material science, off-axis coupons are used to examine the material properties of composites under tension and shear loads. To express the safety conditions from the maximum stress theory as a function of in-plane strength parameters and fiber orientation angle,
Maximum Principal Stress =[tex]ox/2 + oy/2 +/- sqrt[(ox/2-oy/2)^2+tau^2][/tex]where,
[tex]tau = S/2*sqrt[(2*cos(theta))^2 + (Ye/YT*sin(theta))^2][/tex]Here, theta is the fiber orientation angle, S is the ultimate shear stress, XT and Xc are the tensile and compressive strength of the material in the x-direction, and YT and Ye are the tensile and compressive strength of the material in the y-direction.
If the maximum principal stress is less than or equal to the tensile strength in the x-direction (XT), then the coupon is safe in tension. If the maximum principal stress is less than or equal to the compressive strength in the x-direction (Xc), then the coupon is safe in compression. If the maximum shear stress is less than or equal to the ultimate shear stress (S), then the coupon is safe in shear.
The safety conditions from the maximum stress theory for off-axis coupons under the action of two simultaneous axial stresses ox and oy can be expressed as a function of in-plane strength parameters XT, Xc, YT, Ye, S, and fiber orientation angle.
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Two double headed 16D common nails will be used to attach a Doug Fir-Larch S4S 2x4 to a Doug Fir-Larch S4S 4x4 in side grain under normal temperature conditions for load duration of less than 2
months. Determine the total ASD withdrawal capacity of the connection if assembled under dry conditions but subject to significant wetting and drying in service
The total ASD withdrawal capacity of the connection can be determined if the total number of nails is multiplied by the allowable withdrawal capacity of each nail. The allowable withdrawal capacity of a nail varies with wood species, moisture content, and temperature.
[tex]C = ((N * F * G * M) / (12 * Z * H)) * D[/tex]
where C is the allowable withdrawal capacity of a single nail, N is the number of nails, F is the specific gravity of the wood, G is the adjustment factor for the moisture content and temperature, M is the adjustment factor for the end grain of the wood, Z is the penetration depth of the nail, H is the head diameter of the nail, and D is the diameter of the shank of the nail.
The values of these parameters can be looked up in the appropriate tables.
[tex]C = ((2 * 0.53 * 1.1 * 1.0) / (12 * 1.5 * 0.25)) * 0.162C = 206.1 lb[/tex]
As the connection is subject to significant wetting and drying in service, the moisture content of the wood will vary over time, which will affect the allowable withdrawal capacity of the nails. However, as the load duration is less than 2 months, the effect of moisture content variation can be neglected.
Therefore, the total ASD withdrawal capacity of the connection is [tex]2 * 206.1 = 412.2 lb.[/tex]
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Interpretation of environmental data produced from a well-designed monitoring program can be complex. "One may approach the analysis of the results with pre-conceived outcomes in mind and use or misuse statistics to demonstrate his/her own view point." Give your opinion on this argument using the global warming controversy.
Environmental data produced from a well-designed monitoring program can be complex, but interpreting such data is an important task that helps to ensure the safety and sustainability of the environment. In some cases, however, the interpretation of environmental data can be influenced by preconceived outcomes and statistical misuses.
The global warming controversy is a prime example of how preconceived outcomes can influence the interpretation of environmental data. Many people have different views on the causes and effects of global warming, and this can lead to a biased interpretation of environmental data. For instance, some people believe that global warming is caused by human activities such as burning fossil fuels, while others believe that it is caused by natural factors such as changes in the sun's radiation. In both cases, the interpretation of environmental data is influenced by preconceived outcomes.In addition, statistical misuses can also contribute to a biased interpretation of environmental data.
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1. Cofferdams. Describe design considerations for cofferdams.
2. Earth-Retaining Structures. Describe the principle factors which are used to determine the type of earth retaining structure selected and its design.
3. Diaphragm/Slurry Walls. A frequent accident with diaphragm walls is the loss of bentonite. Describe how the loss could occur and measures to prevent such loss.
4. Construction Dewatering and Groundwater Control. Describe factors involved in dewatering and how these affect the design for a project.
5. Underground/Tunneling Support. Describe "CHILE" and "DIANA" ground and why understanding this is important.
6. Underpinning. Discuss influence lines, why they are important and what to look out for when installing underpinning.
1. Cofferdams: Cofferdams are constructed in an excavation site to keep water out and provide a dry work environment. Design considerations for cofferdams include the structural stability of the cofferdam, the size of the excavation, the type of soil or rock at the excavation site.
2. Earth-Retaining Structures: The primary factors used to determine the type of earth-retaining structure selected and its design are the type of soil or rock at the excavation site, the water table, the load of the structure being retained, the depth of the excavation, the duration of the construction project, and cost considerations.
3. Diaphragm/Slurry Walls: The loss of bentonite can occur as a result of the following: seepage through soil, breaks or cracks in the wall, evaporation of water, and/or the use of inadequate support fluid.
4. Construction Dewatering and Groundwater Control: Factors involved in dewatering include the location of the water source, the depth of the excavation, the soil type, the permeability of the soil, and the size and duration of the excavation.
5. Underground/Tunneling Support: Understanding "CHILE" and "DIANA" ground is essential in tunneling projects. CHILE ground is classified as hard rock, whereas DIANA ground is classified as soft ground. In tunneling, the geological structure of the surrounding rock and soil is critical to understanding the potential impact on the tunneling project.
6. Underpinning: In underpinning, influence lines are essential to determine the load distribution on a structure. Influence lines are the diagrams that show the effect of a load on a structure. They are used to predict the location of stresses in the structure.
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A beam is loaded with the following service loads: Moment due to Dead Load = 312 kN-m Moment due to Live Load = 339 kN-m Section: b-37 cm and d=52 cm. Material properties: f'c-30 MPa and fy=420 MPa Use rhomax = 0.019 for all calculations If required, compression reinforcement centroid is located 70mm from extreme compression face Calculate the the sum of the tension and compression reinforcements (if required) in mm² Use NSCP2015. Consider the displaced area of concrete. Answer in 2 decimal places
The question is about calculating the sum of the tension and compression reinforcements (if required) of a beam with the following specifications: Section: b-37 cm and d=52 cm. Moment due to Dead Load = 312 kN-m. Moment due to Live Load = 339 kN-m. Material properties: f'c-30 MPa and fy=420 MPa. Use rhomax = 0.019 for all calculations.
If required, compression reinforcement centroid is located 70mm from the extreme compression face. The answer is:To calculate the sum of the tension and compression reinforcements, we have to calculate the ultimate moment of resistance, Mu, of the beam first and check whether it is greater than or less than the given factored moment:
Dead Load Moment = 312 kN-m,Live Load Moment = 339 kN-mThe ultimate moment of resistance of the beam, Mu, is calculated as follows:First, we calculate the area of tension steel required: From the given data, rhomax = 0.019 We have to calculate the area of tension steel (As) first:
As = 0.019 × b × d = 0.019 × 37 × 52 = 37.076 mm²As per NSCP Table 7.5.1, the minimum As required for Mu is:Minimum As = 0.45 × fy/f's × b × d= 0.45 × 420/0.85 × 37 × 52 = 163.82 mm²Now, we have to calculate the area of compression steel (Asc) required: From the given data, compression reinforcement centroid is located 70 mm from extreme compression face.
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The helical spring has 10 turns of 20 mm diameter wire. If maximum shearing stress must not exceed 200 MPa and the elongation is 71.125mm. calculate the mean diameter of spring and the spring index(m) if the load is 3498.38N and G-83GPa
Given: Number of turns=10Diameter of wire=20mmMax shearing stress = 200 MP Elongation=71.125 mm Load = 3498.38 NG=83 G Pa To find: Mean diameter of spring and the spring index(m).
Formula Used: Total elongation of spring = (4FL)/(πd^3G)whereF = LoadL = Length of springd = Mean diameter of spring G = Modulus of rigidity of the material Spring index(m) = D/dwhere,D = Mean diameterd = Diameter of wire Calculation: Length of spring = Total length - Compression Length of spring = 20 - (10 x 20)Length of spring = -180mm
Total elongation of spring = 71.125 - (-180)Total elongation of spring = 251.125mm Substituting the given values in the formula[tex]:251.125 = (4 x 3498.38 x -180) / (πd^3 x 83 x 10^9)On solving the above equation, we get:d^3 = (4 x 3498.38 x 180 x 10^3) / (251.125 x π x 83 x 10^9)d^3 = 0.00131223d = 0.108 mm[/tex]Spring index(m) = D/dD = 20 mm Spring index(m) = 20/0.108Spring index(m) = 185.18As we know ,D = (d + D)/2Mean diameter of spring = 2D - d= 2 x 20 - 0.108Mean diameter of spring = 39.892 mm Therefore, the mean diameter of the spring is 39.892 mm and the spring index (m) is 185.18.
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A non vented logger records a pressure of 64.41 kPa while suspended in a piezometer at a depth of 16.85 m below TOC. which is 193.97 m AHD. A nearby barometric logger records 100.2 kPa Assuming p= 1001.7 kg/m³, what is the hydraulic head in the piezometer?
The hydraulic head in the piezometer is 177.12 m. Hydraulic head is defined as the height of the water above a given point of measurement. The hydraulic head is obtained by subtracting the height of the pressure transducer from the elevation of the water table.
Hydraulic Head, denoted as h, is given by the formula:
[tex]h = z + p/γ[/tex]where h = hydraulic head, z = height of pressure transducer above datum, p = pressure, and γ = unit weight of water.
When a non-vented logger records a pressure of 64.41 kPa while suspended in a piezometer at a depth of 16.85 m below TOC and a nearby barometric logger records 100.2 kPa,
we can determine the hydraulic head in the piezometer by using the formula above.
The unit weight of water (γ) is 1001.7 kg/m³.
[tex]h = z + p/γ= 193.97 - 16.85 + (100.2 - 64.41)/(1001.7)[/tex]= [tex]177.12 m.[/tex]
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for credit card processing, stock exchanges, and airline reservations, data availability must be continuous. there are many other examples of mission critical applications. research online to find two other mission critical applications and explain why data availability must be continuous for these applications.
In both cases, continuous data availability is crucial for real-time decision-making, timely response, and maintaining the overall functioning and safety of the systems involved.
1. Emergency Medical Services (EMS):
- Continuous data availability is vital for delivering timely and effective care to patients.
- EMS providers rely on real-time access to critical patient information such as medical history, allergies, medications, and vital signs.
- Interruption in data availability could lead to delays in treatment or jeopardize patient safety.
- Continuous data availability ensures immediate access to accurate patient information, enabling informed decisions and timely interventions.
2. Power Grid Management:
- Power grid management requires continuous data availability for effective monitoring and control.
- Real-time data from sensors, smart meters, and control systems is essential for grid operators.
- Interruptions in data availability can lead to extended power outages and grid instability.
- Continuous data availability enables prompt detection and response to faults or outages, ensuring a stable and reliable power supply.
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A 5-ha cornfield was irrigated for 12 hrs at a discharge of 250 lps. If the average depth of water stored in the rootzone as a result of irrigation was observed to be 160 mm, the application
efficiency(%) is ?
We need to calculate the application efficiency (%) of a cornfield.
We can calculate it by using the following formula:
Application Efficiency = [tex](Depth of water stored in rootzone / Depth of water applied) × 100[/tex]
Let's substitute the values given in the question and solve it:
Area of cornfield = 5 ha
Discharge = 250 lps
Time of irrigation = 12 hours
Water applied =[tex]Discharge × Time= 250 × 12 × 60 × 60= 1080000 liters[/tex]
Depth of water applied =[tex](Volume of water applied / Area of cornfield)= (1080000 / 50000) m= 21.6 mm[/tex]
Depth of water stored in the rootzone = 160 mm
Now, we can use the formula:
Application Efficiency = [tex](Depth of water stored in rootzone / Depth of water applied) × 100[/tex]
= [tex](160 / 21.6) × 100= 740.74%[/tex]
Therefore, the application efficiency of the cornfield is [tex]740.74%.[/tex]
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Soft soil can be identified by various methods before and after conducting site investigation. Plan how to identify soft soil accurately during the desk study, site visit, field test, and laboratory test.
Soft soil can be identified by various methods before and after conducting a site investigation. Before a site investigation, a desk study should be done to determine the geology of the area. In the desk study, geological maps can be used to identify areas where soft soil is likely to occur.
Reports from previous investigations in the area should be studied to identify areas with soft soils. During a site visit, there are several ways of identifying soft soils. if the soil is waterlogged or contains a lot of organic matter, it could also be an indication that the soil is soft.
Before conducting a field test, it is important to visually inspect the soil. Soil with a higher water content is more likely to be soft. After collecting soil samples from the site, laboratory tests can be conducted to determine the physical properties of the soil.
A direct shear test, for instance, is used to determine the shear strength of the soil. In summary, identifying soft soil accurately during the desk study, site visit, field test, and laboratory test is essential to ensure the stability of any structures built on the soil. It is important to carry out a thorough investigation to ensure that all necessary measures are taken to avoid any future problems.
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