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.

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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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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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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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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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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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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What are the predictable risks that could be mitigated with forward planning of a project? Projects are subject to different risks that can cause delays, affect quality and, consequently, lead to an increase in costs.

As such, it is crucial to plan for risks and manage them throughout the project life cycle. Below are some of the predictable risks that can be mitigated with forward planning of a project:

Financial risks: A project's budget and cash flow are critical components that should be considered during planning. Financial risks that could be mitigated include cost escalation, foreign exchange rates, cash flow shortfalls, among others.

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Buying a house is one of the biggest financial decisions that an individual makes in their lifetime. Buying a house involves a lot of money, time, and effort. Changes are inevitable, and they can occur during the process of buying a house. Scope creep occurs when new and unplanned activities are added to the process of buying a house. Scope creep can lead to an increase in time, money, and effort.

Here are some steps to implement changes in buying a house and avoid scope creep:

1. Develop a clear plan and stick to itA clear plan should be developed at the beginning of the process of buying a house. The plan should include the budget, timeline, and goals. Once the plan is developed, it should be strictly followed. The plan should only be changed when it is absolutely necessary.

2. Clearly define the scope of the projectThe scope of the project should be clearly defined at the beginning of the process of buying a house. The scope should include the activities that are to be performed and the activities that are not to be performed.

3. Monitor the progress of the projectRegular monitoring of the progress of the project is essential. The progress should be compared against the plan. Any deviations should be documented, and corrective action should be taken.

4. Communicate effectivelyEffective communication is critical during the process of buying a house. All stakeholders should be kept informed of any changes. The stakeholders should be provided with regular updates on the progress of the project.

5. Evaluate the impact of changesBefore any changes are implemented, the impact should be evaluated. The impact should be evaluated in terms of time, money, and effort.

6. Manage changes effectivelyChanges should be managed effectively. The changes should be documented, and the impact of the changes should be evaluated. The changes should be communicated to all stakeholders.

7. Establish a change control process A change control process should be established. The change control process should include the steps that are to be taken when changes are requested. The change control process should be strictly followed to avoid scope creep.

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Signe offers Thorne, a health inspector, 5,000 to overlook the violations in her Farm-to-Fork Dining Emporium. Thorne accepts the cash and overlooks the violations. Signe is charged with the crime of bribery.

The crime occurred when Signe actually offered 5,000 to Thorne. The crime of bribery is committed when someone voluntarily offers a payment or incentive to another person with the intent of inducing that person to act in a manner that benefits the briber.

The act of bribery involves both the offer of something of value, such as money or property, and the acceptance of that offer by the person who is being bribed. Signe committed the crime of bribery by offering Thorne money to overlook the violations at her Farm-to-Fork Dining Emporium.

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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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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 correct answer is option a. P 1,230.Expected annual maintenance costs over the [tex]first 5 years = P 2,000Expected  costs in the 15th year = P 10,000Expected maintenance costs.

In the 25th year = P 20,000Present Worth of Annual Cost - PWA(P/A,i%,n)PWA(P/A,12%,5) = PWA(P2,000/4.0374) = P495.94PWA(P/A,12%,50) = PWA(P1,000) = 9.8181PWA(P/A,12%,25) = PWA(P1,000/15.5043) = P64.51Present Worth of Cost - PWC(P/F,i%,n)PWC(P/F,12%,15) = PWC(P10,000/4.4876) = P2,229.68PWC(P/F,12%,25) = PWC(P20,000/16.0151) = P1,248.97Total present.

Worth of costs over 50 years = P4,049.10Equivalent uniform annual maintenance costs over the entire 50-yr period = AW(P/A,i%,n)AW(P/A,12%,50) = P4,049.10/(AWF(P/A,12%,50)) = P4,049.10/3.2888 = P1,230.07 (rounded off to the nearest whole number)Therefore, option a is the correct answer.

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The three ways by which the factor of safety against a circular slip failure can be increased are:Increase the angle of friction between the soil mass and the foundation by:Increasing the surface roughness of the foundation.

A rough surface between the soil mass and the foundation would produce more interlocking between the soil and the foundation, resulting in an increase in the angle of friction, leading to an increase in the factor of safety against the slip failure.Using geotextiles.

These are synthetic fabrics or mats that have high tensile strength and are used to improve the performance of soil structures. They improve the frictional characteristics of the soil and prevent the migration of soil particles, which increases the angle of friction and results in an increase in the factor of safety against the slip failure.

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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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This is frequently used in conjunction with static pile load testing for a more complete picture of pile capacity and performance.

Friction capacity of sand pile =[tex]35 x π/4 x (0.8)2 x (140 - 0) / (2 x 2.5) = 479.67 kN[/tex]

Net ultimate bearing capacity = [tex]479.67 + 45.83 = 525.5 kN[/tex]

Required pile length =[tex]{[(800 + 0)/525.5] + 0} / (2.5 x 0.6 x π/4 x (0.82))= 15.45 m ≈ 15.5 m[/tex]

(ii) Check whether the calculated pile length is sufficient to carry an axial load of 2 MN with a factor of safety of 3.0

Ultimate capacity of pile = Net ultimate bearing capacity + side resistance capacity= [tex]525.5 + (0.6 x π/4 x (0.8)2 x 27.5 x 1000) = 707.9 kN[/tex]

Ultimate capacity of 10 piles = [tex]707.9 kN x 10 = 7079 kN[/tex]

FOS for axial load =[tex]7079 / 2000 x 3 = 10.6 > 3.0,[/tex]

Therefore, the pile length is sufficient.

(iii) Axial load carrying capacity of pile group= (No of piles x Ultimate capacity of single pile x Reduction factor)/FOSEquivalent diameter, Deq =[tex](N x Ap)1/2 = (10 x π/4 x 0.82)1/2 = 3.52 m[/tex]

Reduction factor, [tex]Rf = (1 – 0.3/Deq) x (1 – 0.3/Hf) = 0.763 x 0.8 = 0.61[/tex]

Axial load carrying capacity of pile group = [tex](10 x 707.9 x 0.61) / 2.5= 10960.84 kN = 10.96 MN[/tex]

Therefore, the axial load carrying capacity of the pile group is 10.96 MN.

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Activated sludge is a biological process used to remove organic matter from wastewater. In order to determine the volume of three identical activated-sludge tanks following primary clarification to aerate 38,000 m3 /d with a BOD concentration of 170 mg/l at a BOD loading of 550 g/m3

[tex]Q = AL[/tex], where:Q = Flow rate (m3/d)A = Surface area (m2)L = Liquid depth (m)First, calculate the volumetric BOD loading [tex](lb/ft3•d): 550g/m3•d ÷ 1,000 = 0.550 kg/m3•d0.550 kg/m3•d x 2.2046 = 1.213 lb/m3[/tex]

Next, calculate the total required volume (m3):

[tex]38,000 m3/d ÷ 1.213 lb/m3•d = 31,318 m3[/tex]

Therefore, each of the three identical activated-sludge tanks should have a volume of approximately 10,439 m3

Operating sludge age = [tex](MLSS, kg/m3 x 0.0624 lb/ft3)[/tex]÷ [tex](950 kg/d x 2.2046 lb/kg x 1,440 min/d x 60 min/h ÷ 7.4805 gal/ft3) = 9.75 days[/tex] (rounded to two decimal places).

Therefore, the required volume is 10,439 m3 per activated-sludge tank. The aeration time is unknown, and the organic load in the influent in terms of BOD is 13,613 lb/day.

The MLSS concentration that should be maintained in the aeration tanks is 2,683 mg/L. Lastly, the operating sludge age is 9.75 days.

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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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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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To solve this problem, we need to determine the break-even point. This is the point at which total revenue equals total costs. To calculate the break-even point, we need to find the total cost of producing the generators and then divide that cost by the selling price of the generators.

This will give us the number of generators that need to be sold to break even.Let's begin by calculating the total cost of producing the generators. The total cost is made up of fixed costs and variable costs. Fixed costs are those that do not change with the level of production. Variable costs are those that do change with the level of production.

In this case, we are given that the fixed cost is $150,000. To find the variable cost per unit, we need to use the following formula: Variable cost per unit = (Total cost - Fixed cost) / Number of units produced At 80% of capacity, the number of units produced is:100,000 x 0.8 = 80,000The total cost is made up of the fixed cost and the variable cost:

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The term that should be included in the answer to your question is "Life-cycle cost".Life-cycle cost is computed to take into account first cost, expected lifetime.

Maintenance costs, fuel costs, replacement cost, inflation, and the time value of money (interest). It is a complete estimate of the cost of ownership over the entire life of an asset, including the cost of acquisition, operation, maintenance, and disposal.

It is a comprehensive analysis of the cost of an asset over its lifetime, and it takes into account all the relevant costs of ownership, including the initial purchase price, maintenance costs, and replacement costs over the expected lifespan of the asset.

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The safe ultimate moment of a T-beam beam is 5,930,060.18 N.mm.

Step 1: Calculate the effective depth of the beam Effective depth, [tex]d = h – 0.5tf= 900 – 0.5(80) = 860mm[/tex]

Step 2: Compute the depth of the equivalent rectangular stress block The depth of the equivalent rectangular stress block, a = 0.85d= 0.85 × 860 = 731mm

Step 3: Compute the ultimate compressive strain for the concrete The ultimate compressive strain for the concrete, εcu = 0.0035 + (0.00175/0.45)× (d – a)= 0.0035 + (0.00175/0.45)× (860 – 731)= 0.0042

Step 4: Here we assume da as 0.85d'.d' = d – (tw/2 + a)= 860 – (300/2 + 731) = -191.5mm
[tex]da = 0.85d = 0.85 × 860 = 731mm[/tex]
[tex]Mu = 0.90 × 1.2 × 1860 × 1000 × (860 – 731 – (-191.5))= 3,189,276 N.mm[/tex]
[tex]Mpe = Asfy (d – a – da)/2 = Asfy(0.5d’ + a – da)[/tex]
[tex]415 × 1000 × (0.5(191.5) + 731 – 731) = 80,922 N.mm[/tex]

[tex]As = (Mu)/(0.95×f'c×(d – 0 .5a) + 0.95fy(As)/As)[/tex]
[tex](3,189,276)/(0.95 × 35 × (860 – 0.5 × 731) + 0.95 × 415 × As/1000)[/tex]
[tex]As = 1052.07 mm²[/tex]

Step 5: By substituting the values, we get;
[tex]Mu = 0.95 × 1052.07 × 415 × (860 – 731/2) + 0.95 × 35 × 2500 × 731 × 731/2= 5,930,060.18 N.mm[/tex]
So, the safe ultimate moment of a T-beam beam is [tex]5,930,060.18[/tex] N.mm.

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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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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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To write the string referenced by the variable message to the file output streams using a PrintWriter object referenced by output, you can use the println method.To write the string to the file output streams, you need to call the println method of the PrintWriter object.

The println method writes a string to the output stream and appends a line separator to it. Here's an example statement that accomplishes this:

output.println(message);

This statement calls the println method on the output object and passes the message string as an argument. The println method writes the string followed by a line separator to the output streams associated with the PrintWriter object. If you want to write the string without appending a line separator, you can use the print method instead:

output.print(message);

Both println and print methods allow you to write the string referenced by message to the file output streams using the PrintWriter object output.

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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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In the case of lateral torsional buckling, CLO, or the modification factor, must be included in the design of a beam. The modification factor for lateral torsional buckling of a castellated beam with simply supported edges is expressed as tC, which is equal to 2.38.

Lateral bracing is supplied at the supports (fed ends).When a beam is loaded and subjected to lateral torsional buckling, the lateral torsional buckling coefficient (CLO) becomes significant. The lateral torsional buckling factor is a ratio of the critical moment of buckling to the elastic moment of the beam.

The greater the lateral torsional buckling coefficient (CLO), the greater the critical moment for buckling and the higher the beam's buckling resistance. If the lateral torsional buckling coefficient (CLO) is greater than or equal to 1, the beam is classified as torsionally rigid and doesn't require additional lateral bracing.

Lateral bracing must be given at the supports (fed ends) for asted beam fed at both ends with uniformay dributed load. Therefore, the lateral torsional buckling coefficient (CLO) for asted beam fed at both ends with uniformay dributed load is equal to 2.38. The beam's buckling strength can be increased by using the lateral torsional buckling factor (CLO).

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(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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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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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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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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