a)There are two types of flow that occur in the given channel at constant discharge, which are described below: Super critical flow: When the flow rate in a channel is higher than the critical value, it is known as supercritical flow. In this type of flow, the water surface is above the critical depth level, and the flow velocity is higher than the velocity of shallow water. Subcritical flow:
If the flow rate in a channel is lower than the critical value, it is known as subcritical flow. In this type of flow, the water surface is below the critical depth level, and the flow velocity is less than the velocity of shallow water.When the specific energy is minimum, the flow condition is known as the critical flow condition.
At this point, the Froude number becomes 1, and the flow rate is equal to the critical flow rate. The channel's specific energy decreases with an increase in flow velocity. When the specific energy reaches its minimum value, it indicates that the flow is critical.b)
Given,Width of rectangular channel (B) = 25 m Slope (S) = 1/1200Depth of flow (y) = 0.99Normal depth (y_n) is given by the equation: y_n = 1.49[(Q²/B³S)^(1/3)]This gives the value of normal depth: y_n = 1.49[(20²/25³(1/1200))^(1/3)] = 0.778 mWe have to find out the distance from the outlet where the depth of flow is equal to 0.99 times the normal depth (y_n).Divide the flow into four equal depth steps, each of size:
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02. Lateral earth pressure
An excavation was made in saturated soft clay (∅=0), with its sides more or less vertical. When the depth of excavation reached 6 m, the sides caved in. What was the approximate value of cohesion of the clay soil? Take unit weight of clay 20 kN/m³.
Lateral earth pressure refers to the horizontal forces exerted by soil or rock onto an object placed within it. It is essential for ensuring the stability of structures built on or in the ground.
an excavation was made in saturated soft clay (∅=0), and its sides were more or less vertical. When the depth of excavation reached 6 m, the sides caved in.
[tex]Pa = Ka * y * h[/tex]
From the given information, the depth of the excavation, h, is 6 m, and the unit weight of clay, y, is 20 kN/m³. As the soil is saturated, the coefficient of earth pressure at rest, Ka, is 0.5.
Therefore,[tex]Pa = 0.5 * 20 * 6 = 60 kN/m²[/tex]
This collapse happened because the lateral earth pressure exceeded the strength of the soil to resist it.
the lateral earth pressure, Pa, is equal to the shear strength,
We can rewrite the formula for cohesion as:
[tex]C = Pa / (tan Ø) = T/ (tan Ø)[/tex]
[tex]C = 60 / (tan 0)[/tex]
the approximate value of cohesion of the clay soil is Undefined because the internal friction angle of the clay soil is zero, and the value of the cohesion cannot be determined.
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A W 533 x 93 simply supported beam with span of 7.8 m carries a uniformly distributed load of 52 kN/m throughout its length. The beam has the following properties: Ix = 0.000556 m² Depth, d 533 mm Web thickness, tw = 10.2 mm Fy 248 MPa. The beam is laterally supported over its entire length. The allowable flexural stress is 0.66Fy, allowable shearing stress is 0.4Fy, and allowable deflection is L/360.
Given data: A W 533 x 93 simply supported beam with span of 7.8 m carries a uniformly distributed load of 52 kN/m throughout its length. The beam has the following properties: Ix = 0.000556 m² Depth, d 533 mm Web thickness, tw = 10.2 mm Fy 248 MP
Calculation of shear force at the right side of the beam (x = L)Shear force at the right side of the beam V = 0 k N Calculation of maximum bending moment Maximum bending moment at the center of the beam (x = L/2)M max= (wL²)/8Mmax= (52 × 7.8²)/8Mmax= 2046.96 k N .m Maximum bending moment at the center of the beam (x = L/2)M max= 2046.96 kN.mStep4: Calculation of maximum shear stress Maximum shear stress occurs at the section where the shear force is maximum. The maximum shear stress is given by,τmax = (VQ)/(Ibt) Where, Q = moment of area of the beam about the neutral axis b = width of the beam t = thickness of the web at the point where shear stress is to be determined. I = Moment of inertia of the beam about the neutral axis. V = Shear force at the section where shear stress is to be determined .
M max= Maximum bending moment at the section where shear stress is to be determined. Q = (bd²)/4Q = [(533 × 10.2³)/4]Q = 361732.4 mm³τmax = (VQ)/(Ibt)τmax = (202.8 × 361732.4)/(0.000556 × 10.2 × 533)τmax = 38.29 M Pa Maximum shear stress = 38.29 MPaStep5: Calculation of deflectionThe maximum deflection is given by,δmax = (5 × w L⁴)/(384 × EI)Where, E = Young's modulus of the Materiali = Moment of inertia of the beam about the neutral axis.δmax = (5 × wL⁴)/(384 × EI)δmax = (5 × 52 × (7.8 × 10⁶)⁴)/(384 × (2.1 × 10⁵) × 0.000556)δmax = 8.54 mm Maximum deflection = 8.54 mm Answer: Therefore, the maximum bending moment is 2046.96 kN.m, the maximum shear stress is 38.29 MPa, and the maximum deflection is 8.54 mm. The beam satisfies the given allowable flexural stress, allowable shearing stress, and allowable deflection.
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Draw a detailed block diagram of an NMR spectrometer and label
all components. Describe the function of each component in
detail.
An NMR spectrometer consists of several components, including sample preparation, RF generator, transmitter, detector, amplifier, data acquisition system, and a computer. These components work together to prepare and load the sample, generate and apply RF pulses, detect the signal, amplify it, convert it to a digital format, and process it using a computer.
A detailed block diagram of an NMR spectrometer illustrates the various components and their functions within the instrument. Each component is labeled to indicate its role in the circuit. The key components of an NMR spectrometer and their functions are as follows:
1. Sample preparation and loading: This section includes sample holders designed to accommodate different sample sizes and orientations. Its purpose is to prepare the sample and load it into the instrument.
2. Radiofrequency generator: The RF generator produces an oscillating voltage at the resonant frequency of the nucleus under study. It supplies the transmitter with the RF pulse required to excite the nuclei and modulates the signal for detection.
3. Transmitter: The transmitter applies the RF pulse to the sample, and its duration and amplitude can be adjusted to control the level of excitation of the nuclei.
4. Detector: The detector senses the magnetic field of the sample and generates a voltage signal proportional to the field strength. It detects the signal emitted by the sample after excitation, which can be further amplified and digitized.
5. Amplifier: The amplifier increases the magnitude of the signal produced by the detector, enhancing its strength for further processing. The gain of the amplifier can be adjusted to control the instrument's sensitivity.
6. Data acquisition: The data acquisition system converts the amplified signal from analog to digital form, making it suitable for processing by a computer. It enables the conversion of the signal into a digital format for subsequent analysis.
7. Computer: The computer governs the instrument's operation, controlling its functions and processing the digitized signal. It manages data acquisition, analysis, and visualization.
In summary, an NMR spectrometer consists of multiple components working together to perform various functions. The sample preparation and loading section prepares and introduces the sample, while the RF generator and transmitter generate and apply the RF pulse. The detector detects the resulting signal, which is then amplified by an amplifier. The data acquisition system converts the amplified signal into a digital format for processing by a computer, which controls the instrument and analyzes the data.
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Given: mechanical spreading of cement at 3.2% by mass; the layer maximum mod.AASHTO density is 3500 kg/m³ and the specified density for stabilised sub-base is 97% of mod.maximum. layer is given as 150mm thick. Calculate the spread rate per square metre (kg/m²) if the portion to be stabilised is 1450 m long the sub-base is 11 m wide (7)
Given data: Mechanical spreading of cement at 3.2% by mass; The layer maximum mod. AASHTO density is 3500 kg/m³The specified density for stabilized sub-base is 97% of mod. maximum. The layer is given as 150 mm thick. The portion to be stabilized is 1450 m long.
The sub-base is 11 m wide. To find: The spread rate per square meter (kg/m²)Calculation: Thickness of layer = 150 mm = 0.15 m Maximum Modified AASHTO density = 3500 kg/m³The specified density for stabilized sub-base = 97% of Maximum Modified AASHTO density= 97/100 × 3500 kg/m³= 3395 kg/m³Length of the portion to be stabilized = 1450 mWidth of sub-base = 11 m.
So, the mass of cement required to achieve the maximum Modified AASHTO density= 0.98 kg/m³ × 3500 kg/m³= 3430 kg/m³Total mass of stabilized soil required to achieve the specified density for stabilized sub-base= Specified density × Maximum Modified AASHTO density= 97/100 × 3500 kg/m³= 3395 kg/m³.
So, the mass of cement required to stabilize 1 m³ of soil= 3395 – 3500= - 105 kg/m³The negative sign indicates that no cement is required for stabilization because the maximum Modified AASHTO density is already achieved. Spreading rate of cement= 0.98 kg/m³ / 0.15 m= 6.53 kg/m²Thickness of stabilized soil layer = 0.15
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Sandstone an unconfined aquifer, the top of the zone of saturation is called the a) Pressure surface. ) Water table () Saturation limit. d) Water level 12. When fine-grained sediments at bottom of shallow water are exposed() air, they dry out to form structures known as Ripple cracks b) Water marks c) Ripple marks d) Mud cracks
Sandstone is an unconfined aquifer. The top of the zone of saturation is called the water table. When fine-grained sediments at the bottom of shallow water are exposed to air, they dry out to form structures known as mud cracks
When the soil is saturated with water, the water table rises. An unconfined aquifer is one in which the water table is free to fluctuate in response to changes in the volume of water stored in the aquifer. Sandstone is an example of an unconfined aquifer, which is porous and permeable, meaning that water can move through it quite easily.
The top of the zone of saturation in an unconfined aquifer is called the water table. Ripple marks are features that form in fine-grained sedimentary rocks like sandstone. When these sediments are deposited under shallow water, they are often subjected to wave action, causing the sediment to ripple.
These ripples are preserved in the rock when it solidifies. Mud cracks are a different type of feature that forms when fine-grained sediments at the bottom of shallow water are exposed to air.
As the sediment dries out, it shrinks and cracks, forming a pattern of interconnected cracks that resemble a dried-up mud puddle.
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A construction activity has budget of $6,000 and is scheduled to be completed in 5 days (8 hours a day), after three days 40% of activity completed and $3,000 of the budget used. 2. What is schedule performance index (SPT)? a. 1.33 b. 0.67 c. 0.33 d. 1.11 3. What is value of schedule variance (SV) in hours? a. +8 hours. b. - 8 hours. c. +16 hours. d. +24 hours. 4. What is value of cost performance index (CPD)? a. 0.8 b. 0.2 c. 0.7 d. 0.3 22. What is value of cost variance (CV)? a. $ 5400 b. $ - 600 c. $ 6000 d. $ - 5400
Given Information: Budget of the construction activity = $6,000Scheduled time to complete the activity = 5 days (8 hours/day)Budget used after 3 days = $3,000% of activity completed after 3 days = 40%1.
To find the Schedule Performance Index (SPI):SPI = % of work completed / % of work scheduled SPI = 40% / (3/5*100%)SPI = 40% / 60%SPI = 0.67Hence, the value of Schedule Performance Index (SPI) is 0.67.2. To find the Schedule Variance (SV) in hours:
As we know, SV = BCWP - BCWS Where, BCWP = Budgeted Cost of Work Performed BCWS = Budgeted Cost of Work Scheduled BCWP = % of work completed * Budget SV = (40% * $6,000) - (3/5 * $6,000)SV = $2,400 - $3,600SV = -$1,200So, the value of Schedule Variance (SV) in hours is -$1,200.
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Safety Management System (SMS) is a safety model that is widely
adopted in aviation industry. Please explain the process in detail
with an example illustrated.
Safety Management System (SMS) is a safety model that is widely adopted in the aviation industry. It is an integrated and proactive approach to safety management that is used to identify and manage safety risks. The process of SMS is detailed below:
1. Establishing Safety Policy: This involves developing a safety policy that outlines the organization's commitment to safety and sets clear safety goals and objectives. The safety policy should also include the roles and responsibilities of all personnel involved in the safety management process.
2. Risk Assessment: This involves identifying and assessing safety risks in the organization's operations. This can be done through hazard identification and risk assessment tools such as checklists, safety audits, and safety assessments.
3. Risk Mitigation: Once the risks have been identified, the next step is to implement controls to mitigate the risks. This can include changes to procedures, equipment, training, and other factors that contribute to safety.
The airline would regularly review and update its safety management system to ensure that it remains effective in managing safety risks. This would involve continuous improvement and a commitment to ongoing safety management in order to ensure the safety of its passengers, employees, and operations.
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Estimating are the fundamental of building construction as it provides all parties on information such as the overall cost of the building, the material needed to construct a building, resources needed to fulfil trades of work and even information on contractual aspect of construction project.
Over the years, measurement have becoming more and more important to construction sector as more and more industry and macro-economic sector started utilizing estimation as part of construction fundamentals. However, since the emergence of the concept "Industrial Revolution 4.0" across the globe, various technologies and software are being developed as an initiative to traditional measurement.
Hence, as the future prospect of our construction sector, you are required to study on the "effect of modernization to estimating" in general.
You are required to discuss on
Traditional estimating
Involvement of technologies in estimating
Pro and cons of the involvement of technologies in estimating
Pick one (1) software or application that is being used in the industry (Eg: Cubicost, Primavera, etc…)
Introduce them in general.
What can they do in terms of estimation.
Why do you choose them as your software.
Conclusion (Please include your findings on the matters discuss above – Technologies is good initiative to estimation, it is not good, etc…)
You shall provide the discussion in the form of report and the following shall be taken into consideration while preparing for your report.
Not more than 15 pages (Minimum 10 Pages)
Include cover page, table of content, reference and necessary formatting
Use ‘Times New Roman’ fonts
Use 14 font size for topic, 12 font size for heading and 11 font size for the body
1.5 line and paragraph spacing
All paragraph should be in justify format
The student has to prepare a slide for presentation in which will cover for all content under the report.
The slides shall be:
Not more than 15 slides
15 minutes’ presentation time
Cover all parts under the report
Use your own creativity as there will be marks for slides Submit your slide along with your report
Estimation is a crucial part of construction work. It helps in providing information about the cost of building a structure, the necessary materials required to construct a building, resources required to carry out various trades, and information on contractual aspects of construction projects.
The importance of measurements in construction has been increasing over the years. As the world witnesses Industry Revolution 4.0, new software and technologies are being developed to replace traditional measurement methods. This report explores the effect of modernization on estimating and the involvement of technology in estimating in general.
The process of estimating involves the preparation of a detailed plan and a detailed estimate. The estimate should include the cost of labor, materials, and other expenses that will be required to complete the construction project. A traditional estimate is usually done by an estimator.
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A thick steel pressure vessel, having an inside diameter of 210 mm and outside diameter of Pressure cylinder 275 mm, is subjected to an internal pressure of 65 MPa and an external pressure of 10 MPa. For this thick steel pressure vessel:
Question 1
a. Using Lamé's Theory, develop expressions for radial and circumferential stresses; [50%]
b. Plot the radial and circumferential stresses against the wall thickness of the pressure vessel and identify the maximum values; [30%]
c. Calculate the longitudinal stress in the pressure vessel wall.
[20%]
A thick steel pressure vessel is subjected to an internal pressure of 65 MPa and an external pressure of 10 MPa. The inside diameter of the pressure vessel is 210 mm and the outside diameter is 275 mm.
Lamé's Theory expression for radial and circumferential stresses:
Radial stress is given by:[tex]σr = (pi*D² /d² - 1)*P/2[/tex]Here, D is the external diameter, d is the internal diameter, and P is the pressure.
[tex]σr = (pi*(275)²/(210)² - 1)*65/2σr = 53.08[/tex]
[tex]σθ = (pi*D² /d² + 1)*P/2σθ = (pi*(275)²/(210)² + 1)*65/2σθ = 122.08 MPa2.[/tex]
For a thick-walled cylinder, the wall thickness, t, is given by:
[tex]t = (D - d) / 2t = (275 - 210) / 2t = 32.5 mm[/tex]
[tex]σr = (pi*D² /d² - 1)*P/2σr = (pi*(275)²/(210)² - 1)*P/2[/tex]
Circumferential Stress,[tex]σθ = (pi*D² /d² + 1)*P/2σθ = (pi*(275)²/(210)² + 1)*P/2[/tex]
For a given internal pressure, the radial and circumferential stresses are directly proportional to the wall thickness of the pressure vessel.
Maximum Radial stress = [tex]110.98[/tex]
MPa Maximum Circumferential stress = 112.1 MPa3.
Longitudinal stress in the pressure vessel wall is given by:
[tex]σL = P*(D/d)σL = 65*(275/210)[/tex]
[tex]σL = 85.7 MPa[/tex]
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In the materials for this Week, we have seen that affirmative action programs in the employment context are harder to get through judicial review than are affirmative action programs in education. Review the assigned cases in this Week and explain both why this difference exists between employment and education affirmative action policies and how it has been applied by the Court.
Affirmative action is the implementation of various policies and practices aimed at ending historical and present-day discrimination and promoting greater diversity within society.
Affirmative action in the employment and education sector is subject to different scrutiny and judicial review, as stated in the cases assigned for this week. Here, we will discuss why employment and education affirmative action policies are different and how it has been applied by the Court. Education and employment are two different domains, so the nature of affirmative action programs may be different.
Affirmative action programs in education aim to ensure that a diverse pool of students attends college, and the selection is based on the student’s merit. On the other hand, employment affirmative action programs strive to make the workplace more diverse, which could lead to the selection of unqualified or less qualified individuals.
v. Bollinger, the Court upheld the University of Michigan Law School's affirmative action program, which did not establish a quota or point system.
v. Pena, the Supreme Court held that the employment affirmative action policy violates the equal protection clause of the Fourteenth Amendment. The court ruled that such policies must withstand strict scrutiny and must serve a compelling governmental interest.
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The following information belongs to a target cost contract:
Target cost $250 M
Target band = $230 M and ends $270 M
Target contractor fee = $10.5 M
Cost saving share = 55/45 by owner
Cost increase share = 65/35 by contractor
Contractor minimum fees =$5 M
The GMC = $320 M
Minimum expected project cost = $ 200 M
?Calculate the contractor maximum fee
a. M 42
b. M 27
c. M 39
d. M 24
Target cost contract is a type of contract which is based on the agreement between the buyer and seller for the cost that would be incurred in performing the project and the amount of profit that would be charged.
The contractor fee is negotiated by the two parties in the target cost contract. The maximum contractor fee can be calculated as follows: Given that, Target cost = 250 M Target band = 230 M and ends 270 MTarget contractor fee = 10.5 M Cost saving share = 55/45 by owner Cost increase share = 65/35 by contractor Contractor minimum fees =5 MGMC = 320 Minimum expected project cost = 200 MWe can find the contractor maximum fee as follows: Contractor maximum fee = (Target cost - (Minimum expected project cost + Target contractor fee)) × Share by contractor + Contractor minimum fees Contractor maximum fee = (250 M - (200 M + 10.5 M)) × (65/100) + 5 M Contractor maximum fee = $15.25 M Therefore, the maximum contractor fee is M 39, option (c).
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define consolidation process. A saturated soil has a compression dex Cc=0.27. Its void ratio at a stress of 125 kN/m2 is 2.04, and its permeability is 3.5X10-8cm/sec. Compute:
The change in void ratio if the stress is increased to 187.5 kN/m² The settlement in (i) if the soil stratum is 5m thick and Time required for 50% consolidation to occur if drainage is one and time factor is 0.196 for 50% consolidation.
The process by which the compression of saturated soils is decreased is known as consolidation. This phenomenon causes a time-dependent reduction in the volume of soil under load, resulting in settlement.
Soil consolidation is a complex process that may be affected by a variety of factors, including soil type, load intensity, and drainage conditions. Consolidation is a fundamental component of foundation design that is used to determine the deformation and settlement of soil beneath structures.
[tex]e₂ = e₁ + Cc log10 (σ₂/σ₁)Where, σ₂ = 187.5 kN/m²[/tex]
[tex]e₂ = 2.04 + 0.27 log10 (187.5/125[/tex]
[tex]e₂ = 2.04 + 0.27 × 0.096910013e₂ = 2.0718[/tex]
Therefore, the change in void ratio if the stress is increased to 187.5 kN/m² is 0.0718.
The equation for calculating settlement is given as, [tex]S = Cc [log10 (σi/σ₀)]²[/tex]Where, σi = initial effective stress = 125 kN/m²σ₀ = final effective stress = 187.5 kN/m²Given, soil stratum thickness, h = 5mCc = 0.27The equation for calculating settlement is,[tex]S = 0.27 [log10 (125/187.5)]² S = 0.27 (-0.1761)² S = 0.0012m[/tex]
[tex]k = 3.5 × 10-8 cm/sec[/tex] = [tex](3.5 × 10-8) × (0.0001) m/seck = 3.5 × 10-12 m²/seck = 3.5 × 10-12 × (86,400) m²/dayk = 0.0003024 m²/day[/tex]
[tex]tv = Cv H²/kt = (0.196 × 5²)/(0.0003024)tv = 3244 days[/tex]
Therefore, the time required for 50% consolidation to occur if drainage is one and time factor is 0.196 for 50% consolidation is 3244 days.
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Check for the adequacy of the mat footing shown. Qallowable is
75 kpa
A mat foundation, also known as a raft foundation, is a sort of footing that covers the entire surface region of a building. It distributes loads evenly throughout the foundation, reduces differential settlements, and provides a level base for the structure.
The volume of the foundation is given by:[tex]V = 3 x 5 x 0.8V = 12 m³[/tex]The soil unit weight is not specified in the problem, so let us take it as 20 kN/m³.[tex]W = 12 x 20W = 240 kN[/tex]
The ultimate bearing capacity can now be calculated using the bearing capacity equation.
[tex]C = 15 kPaqu = 100 kPa[/tex] (Assumed)
[tex]Nq = 36Nγ = 24σ' = 33.75 kPaγB = 20 kN/m³[/tex]
[tex]B = 3 mq = C Nc + YNq + 0.5[/tex]
Bγ B[tex]Nγq = 15 Nc + 0.5 x 3 x 20 x 24q = 15 Nc + 720q = 720 + 15 Nc[/tex]...The soil's unconfined compressive strength is less than the ultimate bearing capacity of the soil, indicating that the soil is safe under the footing
[tex]q = 15 Nc + 720q = 720 + 15 Nc[/tex]
To prevent failure, the bearing capacity must be greater than the total factored load on the foundation.
qallowable = Wu’ / Awhere:A = area of the footingq allowable = 30.23 / (3 x 5)q allowable = 2.015 kPa
The load-bearing capacity is less than the total factored load on the foundation. The mat footing is not sufficient for the given loads.
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(a) m³/s A broad-crested weir is located in a wide prismatic channel with a bed surface slope of So=2×10 with a discharge per unit width of q=1.5 m The bed surface roughness has a Manning's coefficient of n is 0.015. Assuming per m width. that energy losses may be neglected (and no hydraulic jumps do occur):
(i) find the normal and critical flow depths and respective flow energies for this channel flow.
(ii) find the height of the weir, so that the flow over the weir will just obtain the critical flow depth.
(iii) find the flow depths just upstream the weir (approaching), at the weir, and downstream of the weir, for a weir height of 0.2 m.
(iv) find the flow depths just upstream the weir (approaching), at the weir, and downstream of the weir, for a weir height of 0.5 m.
(a) m³/sA broad-crested weir is located in a wide prismatic channel with a bed surface slope of So=2×10 with a discharge per unit width of q=1.5 m. The bed surface roughness has a Manning's coefficient of n is 0.015.
(i) So, the normal depth calculation for a broad-crested weir is given by the following equation:
[tex]qn^2 = (g/2) * (n^2) * Q^2 * S^(3/2[/tex])Where, q = Discharge per unit width n = Manning's coefficient of channel bedS = Bed slope Q = Discharge per unit width g = Acceleration due to gravity[tex]n = 0.015S = 2×10Q = 1.5 m³/s= 0.2306 J/m³[/tex]
(ii) qc = 1.84 * H^(3/2)where qc = Flow over the weir H = Weir height
For a weir height such that the flow over the weir will just obtain the critical flow depth, qc = q, and
[tex]H = hcqc = 1.84 * Hc^(3/2)1.5 = 1.84 * Hc^(3/2)Hc = (1.5 / 1.84)^(2/3)Hc = 0.6007 m[/tex]
The height of the weir for the flow over the weir to just obtain the critical flow depth is Hc = 0.6007 m
(iii) The flow depths just upstream the weir (approaching) is h_1 = h_n.The flow depth at the weir is h_2 = yc.For the downstream of the weir, the flow depth h_3 can be calculated as:[tex]h_3 = sqrt((h_2^2) + ((2/3) * H))h_3 = sqrt((0.2165^2) + ((2/3) * 0.2))h_3 = 0.3012 m[/tex]
(iv) The flow depths just upstream the weir (approaching) is h_1 = h_n.The flow depth at the weir is h_2 = yc.For the downstream of the weir, the flow depth h_3 can be calculated as:
[tex]h_3 = sqrt((h_2^2) + ((2/3) * H))h_3 = sqrt((0.2165^2) + ((2/3) * 0.5))h_3 = 0.2974 m[/tex]
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Design a non-cylinder prestressed concrete pipe of 600 mm internal diameter to withstand a working hydrostatic pressure of 1.05 N/mm 2 , using a 2.5 mm high-tensile wire stressed to 1000 N/mm 2 at transfer. Permissible maximum and minimum stresses in concrete at transfer and service loads are 14 and 0.7 N/mm 2 . The loss ratio is 0.8. Calculate also the test pressure required to produce a tensile stress of 0.7 N/mm 2 in concrete when applied immediately after tensioning and also the winding stress in steel if E s =210kN/mm 2 and E c =35kN/mm 2 .
To design a non-cylinder prestressed concrete pipe with a 600 mm internal diameter to withstand a working hydrostatic pressure of 1.05 N/mm², the ultimate tensile stress developed in the wire is calculated as 1250 N/mm². The stress in the wire at transfer and service loads is determined based on the assumptions of Pp/P = x and Ps/P = y. The design includes calculating the thickness of the pipe and ensuring that the load capacity exceeds the applied load. The winding stress in steel is determined to be 7500 N/mm². The test pressure required to achieve a tensile stress of 0.7 N/mm² in concrete is also calculated. Overall, the design meets the requirements for the given working hydrostatic pressure.
Given:
Internal diameter (D) = 600 mm
Working Hydrostatic pressure (p) = 1.05 N/mm²
Prestressing wire stress (fp) = 1000 N/mm² at transfer
Concrete permissible stress at transfer (fc) = 14 N/mm²
Concrete permissible stress at service load (fcd) = 0.7 N/mm²
Loss ratio (α) = 0.8
Modulus of Elasticity of steel (Es) = 210 kN/mm²
Modulus of Elasticity of concrete (Ec) = 35 kN/mm²
To design a non-cylinder prestressed concrete pipe with a 600 mm internal diameter to withstand a working hydrostatic pressure of 1.05 N/mm², we need to determine the ultimate tensile stress developed in the wire. The formula for ultimate tensile stress is given as:
σpu = fp / α = 1000 / 0.8 = 1250 N/mm²
Thus, the ultimate tensile stress developed in the wire (σpu) is 1250 N/mm².
Next, we calculate the stress in the wire at transfer (σp) using the formula:
σp = σpu × Pp / P = 1250 × Pp / P
Assuming Pp / P = x, we have:
σp = 1250 × x ............. (1)
The initial stress in the wire at transfer is given as 1000 N/mm².
The stress loss in the wire at service is calculated as:
fpe = fp - σpu = 1000 - 1250 = -250 N/mm²
The stress in the wire at service (σps) is given by:
σps = fpe × Ps / P = -250 × Ps / P
Assuming Ps / P = y, we have:
σps = -250 × y ............. (2)
At service load, the stress in the wire should be equal to fcd, so we have:
σps = fcd / -250
y = P / 625000 fcd ............. (3)
We also know that:
Pp + Ps = Pu = P × α
Pp + Ps = P / 0.8
Pp + Ps = 1.25P
Assuming Pp / P = x and Ps / P = y, we have:
x + y = 1.25 ............. (4)
From equations (1) and (2), we get:
1250 × x - 250 × y = σp - σps
= σp + fcd / 625000 × P ............. (5)
To calculate the thickness (t) of the pipe, we assume a value and then use it to calculate the area of the cross-section (A) of the pipe. The load capacity of the pipe should be greater than the load acting on it.
A = π/4 (D² - (D - 2t)²)
The load acting on the pipe is given as:
P = 1.05 × A = 1.05 × π/4 (600² - (600 - 2t)²) = 1647.88t² N
By substituting the value of P in equation (5) and solving for x and y, we can calculate the stresses in the wire at transfer and service loads.
From the given data, we can calculate the winding stress in steel (σw) as:
σw = Es / Ec × σp = 210000 / 35000 × 1250 = 7500 N/mm²
The test pressure required to produce a tens
ile stress of 0.7 N/mm² in concrete when applied immediately after tensioning is given by:
σpe = fp / α
σpe = 1000 / 0.8 = 1250 N/mm²
Using the value of σpe, we can find the stress in concrete as:
σce = σpe / (1 + Es / Ec)
σce = 1250 / (1 + 210000 / 35000) = 35.29 N/mm²
Using the stress in concrete, we can calculate the test pressure (Ptest) required to produce a tensile stress of 0.7 N/mm² in concrete:
Ptest = σce × D / 2t - σw
= 35.29 × 600 / (2 × t) - 7500
= 106 / t - 7500 N/mm²
Therefore, a non-cylinder prestressed concrete pipe with a 600 mm internal diameter can be designed to withstand a working hydrostatic pressure of 1.05 N/mm².
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2.50m x 2.50m square footing supports a 350mm x 400mm reinforced concrete column carrying a factored axial load of 930 kN. Use fc = 28 MPa and fy = 415 MPa. Use 20mm diameter reinforcing steel bars bothways. Clear concrete cover is 70 mm
A. Determine the effective depth of footing for wide-beam action
B. Determine the effective depth of footing based on two way action.
C. Determine the required safe thickness of footing.
a) To find the effective depth of the footing, use the formula for wide-beam action:
be =[tex][bc + {(12AsFy) / (0.85fc(b - d))}] / 2[/tex],where:bc = width of the columnAs = area of steel reinforcement in the footing
Fy = steel yield strengthfc = compressive strength of the concreteb = width of the footingd = effective depth of the footing
Substituting the values given,be =[tex][0.4 + {(12 x 0.031416 x 415) / (0.85 x 28 x (2.5 - 0.07))}] / 2= 0.844 m[/tex]Therefore, the effective depth of the footing for wide-beam action is 0.844 m.
b) To find the effective depth of the footing based on two-way action, use the formula:deff = [tex]0.75{[Asfy + (bt)^2/2]}^(1/2) / (0.85fc)[/tex]bwhere:t = thickness of the footing
Substituting the values given and solving for deff,deff =[tex]0.75{[0.031416 x 415 + (2.5 x 0.35)^2/2]}^(1/2) / (0.85 x 28 x 2.5)= 0.60 m[/tex]
The effective depth of the footing based on two-way action is 0.60 m.c) The required safe thickness of the footing can be calculated as follows:V = factored axial load on the column = 930 kN, or 930,000
[tex]Nq = 1 - V/[4(b + t)fc][/tex ]where:q = non-dimensional bearing capacity factorb = width of the footing d = effective depth of the footingt = thickness of the footing for a footing width of 2.5 m and a value of q = 0.15, the minimum thickness of the footing required for safe bearing capacity is 0.35 m.
the required safe thickness of the footing is 0.35 m.
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3- There are a few important factors in forming a risk/trend management team. One of the following factors shouldn't be on the list, and that is:
risk attitude of team members
knowledge level of team members
proximity of effect of team members
risk appetite of team members
risk blindness of PM
4- There are a few important factors in forming a risk/trend management team. Which composition of people you should work with:
Balance overall
Risk taking in every scenario
Risk blind in every scenario
3. Risk blindness of PM is the factor that should not be on the list while forming a risk/trend management team. Explanation :Risk blindness refers to a person's inability to see the risks that exist in the situation.
This is a particularly problematic condition for risk management, which is focused on identifying and mitigating risks. Therefore, this factor should not be considered in forming a risk/trend management team.Proximity of the effect of team members: The proximity of the effect of team members is important to consider in forming a risk/trend management team. Team members who are close to the areas that are at risk are better positioned to understand and identify potential risks. Risk attitude of team members: It is important to consider the risk attitude of team members in forming a risk/trend management team.
Team members should be willing to take risks and be open to new ideas and solutions .Knowledge level of team members: It is important to consider the knowledge level of team members in forming a risk/trend management team. Team members should have a strong understanding of the risks and trends that exist in the environment.Risk appetite of team members: It is important to consider the risk appetite of team members in forming a risk/trend management team. Team members should be willing to take risks and be comfortable with the level of risk involved.4. To form a risk/trend management team, you should work with a balanced overall composition of people. Explanation:To form an effective risk/trend management team, you need to have a diverse group of people with different backgrounds and skill sets.
You should work with a balanced overall composition of people who have experience in different areas such as finance, operations, technology, and marketing. This will help ensure that you have a well-rounded team that can identify and mitigate risks from different angles. A balanced team composition will help you have a wide range of ideas and solutions to effectively manage risks and trends.
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3.) An electronic store that sales transistors pays their lot space P200k per year. If the transistor can be sold P8 pesos per unit while manufacturing cost P1.25 pesos per unit. Calculate the number of units that must be sold per month to achieve breakeven. 4.) An electrical company manufactures transformer at a cost P6kper transformer. If the maintenance of the equipment cost P100k pesos every six months and the company sells the unit for P75k per unit. The employees' salaries are P20k per month. If there are 10 employees in total. What is the volume of sales that must be made each month to achieve breakeven?
3.) An electronic store that sales transistors pays their lot space P200k per year. If the transistor can be sold P8 pesos per unit while manufacturing cost P1.25 pesos per unit. Calculate the number of units that must be sold per month to achieve breakeven.
Breakeven point is the level at which total cost equals total revenue. Breakeven point in units = fixed costs / contribution margin per unitFixed costs = P200,000 per yearContribution margin per unit = selling price per unit - manufacturing cost per unit = P8 - P1.25 = P6.75Breakeven point in units = P200,000 / P6.75 per unit = 29,630. In order to achieve breakeven, the store must sell 29,630 units per year, or about 2,469 units per month.4.) An electrical company manufactures transformer at a cost P6kper transformer.
Breakeven point in units = fixed costs / contribution margin per unitFixed costs = equipment maintenance cost + salaries = P100,000 per 6 months + 10 x P20,000 per month = P220,000 per 6 months = P36,667 per monthContribution margin per unit = selling price per unit - manufacturing cost per unit = P75,000 - P6,000 = P69,000Breakeven point in units = P36,667 per month / P69,000 per unit = 0.531 units per month. In order to achieve breakeven, the company must sell at least 1 unit per month.
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Prepare a UML class diagram (classes, associations, attributes, and multiplicities) for the specified component of M&M Enterprises. M&M sells its agleclaps to customers through a network of company salespeople. Each type of agleclap is bought from a particular vendor and is given an initial list price. Each salesperson services a separate group of customers and is allowed to offer them various discounts from list to induce sales. Each sale can include one or more types of agieclaps and can be paid for in any one of three ways: (1) immediately in cash, (2) on the 15th of the following month, or (3) over the course of six months. When cash is received, a cashier deposits it into a company bank account. Sales are signaled by invoices; cash receipts by remittance advices. The policies above plus the information elements listed below are relevant to potential users of this data model, and they all should be included in its final structure (attributes with asterisks are hints for some (not all) of the needed relationships). You may assume that cashiers and salespeople both belong to an entity called "employee." -list price -agleclap# -primary vendor for this agleclap type -bank for cash account -customer name -customer# -sale date -actual price of each agleclap -salesperson quarterly sales -fidelity bond rating of cashier -remittance advice# -agleclap description -QOH (quantity-on-hand) -bank account# -a/r amount -invoice# -sale amount -employee name -salesperson comm. rate -cash receipt amount -bank account for receipt -bank-account type -customer salesperson -customer sold to* -inventory quantity sold -receipt date -employee# -employee category -employee's highest degree -cash account balance -number of employees in category -medical plan for this category -monthly pay (salary) of employee -unit cost -amount of cash receipt applied to a certain sale
A UML class diagram is a design visualization that represents classes, interfaces, associations, and objects, among other things. It provides an object-oriented view of a software system from its structure to its behavior.
M&M Enterprises is selling its agleclaps to customers through a network of company salespeople.
Each type of agleclap is bought from a particular vendor and is given an initial list price. Each salesperson services a separate group of customers and is allowed to offer them various discounts from list to induce sales. Each sale can include one or more types of agieclaps and can be paid for in any one of three ways:
(1) immediately in cash,
(2) on the 15th of the following month,
(3) over the course of six months.
When cash is received, a cashier deposits it into a company bank account. Sales are signaled by invoices, cash receipts by remittance advices.
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7. Compute the minimum time headway for cars moving at an average speed of 80 kph. The average length of cars is 4.5 m., and the reaction time is approximately 0.7 seconds.
The minimum time headway refers to the minimum time gap between two cars that are moving in the same direction on the road. It is important to determine the minimum time headway as it helps ensure the safety of vehicles on the road, and it helps to prevent accidents.
In the calculation of the minimum time headway, various factors such as the average speed of cars, average length of cars, and reaction time are taken into account. For instance, given the average speed of cars as 80 kph, the average length of cars as 4.5m, and the reaction time as approximately 0.7 seconds, we can compute the minimum time headway as follows: We can start by converting the average speed of cars to meters per second.
This is the distance that a car would take to come to a complete stop after the driver has applied the brakes. It is given as follows:d1 = vt1 + 1/2 at1^2Where d1 is the stopping distance, v is the initial velocity, t1 is the reaction time, a is the deceleration, and t1^2 is the time taken for the car to come to a stop after the brakes have been applied. In this case, the initial velocity is 22.222 m/s, the reaction time is 0.7 seconds, and the deceleration is 9.81 m/s^2 (which is the acceleration due to gravity).
Finally, we can compute the minimum time headway using the following formula:t2 = d2/v Where t2 is the minimum time headway, d2 is the sum of the stopping distance and the length of the car (which is 4.5 meters), and v is the velocity of the car. Thus, the minimum time headway is given as:t2 = (15.25 + 4.5)/22.222 = 0.77 seconds. Therefore, the minimum time headway for cars moving at an average speed of 80 kph is 0.77 seconds.
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Design a reinforced concrete beam to carry a service live load of 10kN/m and dead load of 6kN/m (including its own weight) Use f 'c = 27MPa, fy = 414MPa, b = d/2. The beam is to be reinforced in tension only. L1 = 7m & L2 = 7m
The design of the reinforced concrete beam to carry a service live load of 10 kN/m and dead load of 6 kN/m (including its own weight) using f'c = 27 MPa, fy = 414 MPa, b = d/2 and reinforced in tension only with L1 = 7m and L2 = 7m is given as follows;
Step 1: Calculation of self-weight of the beam
[tex]A = bd = d²/2As = (fy/0.87) * 0.85A's = 0.85 * AsM_uz = (dl+ll) * L^2 / 8M_u = 1.2M_uzM_u = 1.2(6 + 10) x (7²)/8M_u = 105 kNm/g1c1 = M_u / bd²f_c′ = 27 MPa = 27 N/mm²f_y = 414 MPa = 414 N/mm²From 250 mm deep beam table, taking b/d = 0.5;For d = 250mm;[/tex]
Step 2: Calculation of the Required Area of Steel to resist moment only Calculate d' (effective depth)d' = d - (As/A) * d' = 240 mm M_u = (f_c' * A * d') * g1c1 + As * fy * (d - d') * g1c1 (Equation 1).
By equating Equation 1, we have;(f_c' * A * d') * g1c1 + As * fy * (d - d') * g1c1 = Mu (Equation 2) Substituting d' = 240 mm and A = 31,250 mm² in Equation 2 gives;
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1. What are the major parameters used in water quality assessment? Please identify four parameters and explain why these parameters should be concerned (4 pts)
2. Explain the difference between coagulation and flocculation and name three coagulants. (4 pts)
3. Name all settling types and their respective applications. (4 pts)
4. What filter types and media types are used in filtrations, and name four particle removal mechanisms involved in filtration process. (4 pts)
5. Name three typical disinfection process and explain the potential negative effect of disinfection byproducts. (4 pts)
6. What is the effect of pH on chlorination? (4 pts)
1. The major parameters used in water quality assessment are:Temperature – High temperatures can make the water undesirable for consumption, while low temperatures can provide an ideal breeding ground for pathogens.
Turbidity – Turbidity of water means the cloudiness or haziness of the water, higher turbidity means higher chances of sediments and particles that affect the clarity of the water and it's quality.
2. Coagulation and flocculation are two water treatment processes used to remove suspended solid particles and impurities from water.
3: Sludge blanket settling - removes organic matter and suspended solids from water, and is commonly used in industrial wastewater treatment.
4. There are several types of filter media used in the filtration process, some of them are: Activated carbon, Sand, Anthracite, Crushed glass, Gravel.
5. Three typical disinfection processes are chlorination, ultraviolet (UV) radiation and ozone disinfection. Disinfection byproducts are formed when disinfectants, such as chlorine, react with organic matter in the water.
6. The pH level of water affects the effectiveness of chlorination. If the water is too acidic, chlorine can be consumed quickly and lose its disinfecting power before it can be effective.
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Determine the magnitude of moment for which 3.6 m long tied-column should be designed. The cross- section of the column is 300 300mm. The service dead and live loads are 356kN and 311kN respectively. The dead moment (MD)is zero and the live moment (ML) at top and bottom is 27 kN (single curvature). Assume K-0.85. The concrete is of normal weight with a compressive strength of 21 Mpa and yield stress of steel is 420 Mpa. Using the simpler ACI Equation to compute El.
The column moment for a 3.6 m long tied-column having a cross-sectional area of 300 x 300mm, which has to be designed can be determined by using the simpler ACI equation.
Moments are among the most common types of forces or loads encountered in structures. The force applied is perpendicular to the plane of the object, causing it to rotate about a given axis.
The moment of a force about a point is the measure of the tendency of the force to rotate an object about that point. It is denoted by M and measured in newton-metres (Nm).When the column is subjected to the load, there will be a moment produced as a result of the distance of the force from the center of the column.
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Determine True/False: (1) The CPM method was developed primarily for the construction industry (ii) CPM uses probabilistic value for project duration. (iii) A Dummy Activity may sometimes have non-zero duration (iv) In a CPM arrow network, when two or more arrows begin at the same node, this means that the work must begin at the same time. (v) The Forward & Backward Pass algorithms calculate EST and LST respectively, of all activities. (vi) The Critical Path time is defined as the longest possible path of the network. (vii) Backward Pass algorithm assumes the duration of the project as obtained by the Forward Pass. (viii) On the Activity-on-Arrow network, the nodes represent a point in time. (ix) To draw an Activity-on-arrow network, Activities IPA are needed. (x) Duration of a Project (D.) calculation, using Average Productivity, is usually an optinistic one. (xi) If an activity has an EST equal to its LST then it's on the Critical Path. (xii) An activity A[i,j) is a Critical Activity if E(1) L(). (xiii). A project always has at least one Critical Path. (xiv) Of all the floats, Free Float always has the maximum value. (xv) If all Free Floats along a path are zero, then that's a Critical path (xvi) Cash-flow Analysis using CPM can be used using EST or LST Schedule. (xvii) Resource Leveling using CPM is usually a trial-and-error method (xviii) For a PERT model if each activity has a le-0, then effectively it becomes a CPM model. (ix) The Expected Time for PERT is a weighted average of three times. (xx) In the Monte Carlo (Simulation) Method, Activity Duration are generated by random numbers.
(i) The CPM method was developed primarily for the construction industry. True.(ii) CPM uses probabilistic value for project duration. False.(iii) A Dummy Activity may sometimes have non-zero duration. True.(iv) In a CPM arrow network, when two or more arrows begin at the same node, this means that the work must begin at the same time.
True.(v) The Forward & Backward Pass algorithms calculate EST and LST respectively, of all activities. True.(vi) The Critical Path time is defined as the longest possible path of the network. True.(vii) Backward Pass algorithm assumes the duration of the project as obtained by the Forward Pass.
False.(viii) On the Activity-on-Arrow network, the nodes represent a point in time. False.(ix) To draw an Activity-on-arrow network, Activities IPA are needed. False.(x) Duration of a Project (D.) calculation, using Average Productivity, is usually an optimistic one.
True.(xi) If an activity has an EST equal to its LST, then it's on the Critical Path. True.(xii) An activity A[i,j) is a Critical Activity if E(1) < L().False.(xiii) A project always has at least one Critical Path. True.(xiv) Of all the floats, Free Float always has the maximum value. False.(xv) If all Free Floats along a path are zero, then that's a Critical path. True.(xvi) Cash-flow Analysis using CPM can be used using EST or LST Schedule. False.(xvii) Resource Leveling using CPM is usually a trial-and-error method.
True. The Monte Carlo Simulation Method can be used to generate the activity duration, which uses a random number generator to produce the project completion time frame based on different simulation scenarios.
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h= 300mm b =190mm t = 4mm L/h = 12.5
Question: Now ignore self weight. The beam is loaded by two vertical point loads, each of magnitude P, that act at L/3 and 2L/3 along the beam. What is the maximum value of P so that the stress criteria above are satisfied? Draw the stress and strain distributions (values required) for both the steel and the timber on the critical cross-section for this value of P.
Given that;h = 300 mmb = 190 mmt = 4 mmL/h = 12.5Beam is loaded by two vertical point loads, each of magnitude P, that act at L/3 and 2L/3 along the beam.
The maximum value of P so that the stress criteria above are satisfied will be calculated.To calculate the maximum value of P, we need to calculate the following;Stress in steelStress in timberShear stress in steelShear stress in timberStep-by-step solution is given below.
Let's first draw the FBD of the beam for an easier calculation;[tex]\frac{1}{2}P[/tex] at [tex]\frac{L}{3}[/tex][tex]\frac{1}{2}P[/tex] at [tex]\frac{2L}{3}[/tex]Take a small segment at distance x from the left support and calculate the shear force on that section.
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Discuss and point out the testing requirements to secure the
material properties of the ribbed steel reinforcing bars between
the construction standards CS2:1995 and CS2:2012.
Ribbed steel reinforcing bars are a crucial component of reinforced concrete structures, and their strength and durability are essential for ensuring the safety of structures. The quality of these reinforcing bars must be ensured during their manufacture, transportation, storage, and installation.
1. Chemical Composition Test: The chemical composition of ribbed steel reinforcing bars must be tested to ensure that it meets the requirements specified in the construction standards. The tests are carried out using a spectrometer, and the chemical composition of the steel is determined by analyzing its various components.
2. Tensile Test: Tensile tests are performed to determine the strength and ductility of the ribbed steel reinforcing bars. These tests are performed in accordance with the appropriate ASTM standards.
3. Bend Test: The bend test is performed to determine the ductility and strength of the ribbed steel reinforcing bars. The test is carried out using the appropriate ASTM standards.
4. Rebend Test: The rebend test is performed to determine the ductility and strength of the ribbed steel reinforcing bars. The test is carried out using the appropriate ASTM standards.
These are the various testing requirements necessary for securing the material properties of the ribbed steel reinforcing bars between construction standards CS2:1995 and CS2:2012.
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Ramp Handling Operator (RHO) is an airport ground-handling agent
who provides airside services for airline. Please explain in detail
what services do they provide to airlines.
Ramp Handling Operators (RHOs) provide airside services to airlines. Airside services refer to services offered on the airport apron, taxiway, and runway. They make up the ground handling operations on an airport.
Aircraft marshalling Aircraft marshalling is the act of directing an aircraft to its parking position on the apron. This service is done to ensure the safe arrival and departure of an aircraft . RHOs ensure that the loading and unloading of cargo and baggage is done efficiently, effectively, and within the stipulated time frame.
Cabin cleaning RHOs also offer cabin cleaning services. The cabin cleaning involves cleaning the aircraft's interior, such as the cabin, the lavatories, the galleys, and other compartments. Catering services RHOs also provide catering services to the airlines. They supply meals, snacks, and beverages to the aircraft.
This service involves providing check-in services, verifying travel documents, issuing boarding passes, and directing passengers to their assigned aircraft. They also handle the transfer of passengers from one aircraft to another on connecting flights.
RHOs provide a range of airside services to airlines, and these services are critical in ensuring the safe and efficient handling of aircraft operations at the airport.
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The thickness of the clay layer under a layer of sand is equal to 5 m. The ground surface is subjected to a uniformly distributed load of 40 kPa and resulted to a primary consolidation settlement of 180 mm. Determine the secondary settlement (mm) of the clay layer 5 years after the completion of the primary consolidation settlement. Time of completion of primary settlement is 1.5 yrs. Secondary compression index Cα = 0.02 and void ratio of 0.54.
Given data:The thickness of clay layer (hc) = 5mUniformly distributed load (σ) = 40kPaPrimary consolidation settlement (s) = 180mmTime of completion of primary settlement (tp) = 1.5 yrs Time period after completion of primary settlement (t) = 5 - 1.5 = 3.5 yrs Secondary compression index (Cα) = 0.02Void ratio (e) = 0.54To find:
Secondary settlement (ss)Formula used:Total settlement, s = primary consolidation settlement + secondary settlement ⇒ s = sP + ssThe secondary settlement, ss = Cασlog10(t+tp)log10(tp)The void ratio (e) is given by e = Vv/Vs = (V – Vw)/(Vg – Vw)Where, V = volume of soilVw = volume of waterVg = volume of air and water.
Vs = volume of solidVv = volume of voidsV = Vs + VvSo, Vs = V/(1+e) and Vv = V × e/(1+e)Calculations: As given, total thickness of the soil (h) = hc + hsTo calculate hs, use the formula of Bousiness q. Bousiness q equation is used to calculate the settlement of soil when it is subjected to the loads, and the thickness of the soil layer is very large relative to the footing size and the soil layer is assumed to be homogeneous and isotropic.
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A composite section has an effective flange width of 2100mm and a slab with thickness of 150mm. The steel beam is a W530x101 of A-36 steel with Fy = 250 MPa. Concrete strength fc’ = 20.7 MPa. Modular Ratio = 9.0. Assume full shoring and full composite action.
Determine the modular ratio.
Determine the moment of inertia of the composite beam about the neutral axis.
Determine the resisting moment of the section.
1. The modular aspect ratio is around 5.73. 2. The composite beam's moment of inertia (Icomposite) is equal to the sum of its constituent moments of inertia:Steel and concrete are combined to form composite materials.3. The formula Mresisting = n * (Icomposite / d) may be used to compute the section's resisting moment (Mresisting).
The calculation is as follows:
We must solve the following equation to find the modular ratio:
Es / Ec is the modular ratio (n).
where Ec is the elastic modulus of concrete and Es is the elastic modulus of steel.
You can suppose that steel has an elasticity modulus of 200 GPa (200,000 MPa).
The following equation may be used to get the concrete's elasticity modulus (Ec):
Ec = 4700 √(fc')
where fc' is the concrete's MPa compressive strength.
Since fc' = 20.7 MPa, we can figure out Ec:
Ec = 4700 √(20.7) ≈ 34,877 MPa
We can now determine the modular ratio:
Es / Ec = 200,000 / 34,877 5.73 is the modular ratio.
Therefore, the modular ratio is around 5.73.
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8(a)A coal-fired factory burns 10 tons of coal per hour and discharges the combustion products through a stack with an effective stack height of 100 m. The coal has a sulfur content of 5%, and the wind velocity at the top of the stack is 8.0 m/s. Atmospheric conditions are moderately unstable. Determine the maximum ground-level concentration of SO2 and the distance from the stack at which the maximum occurs. (Please refer to the graphs in appendix).
8(b)(i)A convective kerosene heater is tested in a well-mixed 27 m³ chamber having an air exchange rate of 0.39 hr¹. After 1-hr operation, the NO concentration reached 4.7 ppm. Assume the ambient NO concentration is zero and there is no NO before the test. Treating NO as a conservative pollutant.
Estimate the NO source strength of the heater (mg/hr).
(ii) Estimate the NO concentration (µg/m³) that would be expected in the lab 1 hr after turning off the heater.
(iii)If this heater were to be used in a "tight" 300 m³ home with 0.2 hr¹, what steady-state concentration of NO would you expect to be caused by the heater?
8(a) The ground level concentration of SO2 is given by: Ground level concentration = E × H / (Q × u × (1 + σ × H / d)2 )Given that the factory burns 10 tons of coal per hour, it means that 10000 kg/h is burnt. Thus, the sulfur dioxide produced is 5% of 10000 kg/h = 500 kg/h. 500 kg of SO2/h / 3600s/h = 0.14 kg of SO2/s = 140 g/s. The molar mass of SO2 is 64 g/mol. The SO2 emission rate is:
140 g/s / 64 g/mol = 2.2 mol/s The effective stack height is 100 m The wind velocity at the top of the stack is 8.0 m/s Atmospheric conditions are moderately unstable E = 3.1H = 100 mQ = 2.2 mol/su = 8.0 m/sσ = 0.3d = The graph in Appendix A gives a value of 1.30 for σ/H = 0.003The maximum ground level concentration of SO2 is:
Ground level concentration = E × H / (Q × u × (1 + σ × H / d)2 )= 3.1 × 100 / (2.2 × 8 × (1 + 1.30))2= 0.028 mg/m³The distance from the stack at which the maximum concentration occurs is given by: d = H × sqrt(E / Q × u × sqrt(1 + σ × H / d))Let D = d / H; then solving for D:0.0309D4 + 0.007D3 - 0.004D2 + 0.002D - 0.0008 = 0This is a quartic equation with no easy solution.
We use a spreadsheet program to solve for D = d/H = 1.02. Therefore, d = 102 m.8(b)(i)The source strength of the heater is given by: Source strength = C i V / 1000Where,C i is the initial pollutant concentration (in ppm)V is the volume of the chamber (in m³)C o is the final pollutant concentration (in ppm)t is the duration of the test (in hours).
The ground level concentration of NO is: Ground level concentration = E × H / (Q × u × (1 + σ × H / d)2 )= 3.1 × 100 / (0.0556 × 1.10 × (1 + 0.019 × 100 / d)2)= 0.00103 d - 0.107The maximum concentration will occur when d = 55 m. At this distance, the maximum ground level concentration of NO is:
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