What is the minimum width of a square footing needed to support a total service load (i.e.

Six  challenges facing operations and maintenance of water supply and delivery systems in Africa include;Poor governance and corruptionInsufficient funding for infrastructureInadequate planning.

and technical expertisePoor stakeholder participationWeak policies and regulatory frameworksLack of skilled personnel(b) Seven (7) policy directions for the government in Africa continent towards operations and maintenance of water supply and delivery systems in Africa include;Institutional strengthening and capacity development.

Addressing financial barriers and mobilization of resourcesEstablishment of effective monitoring and evaluation systemsPromoting private sector participationImproving legal and regulatory frameworksDeveloping community involvement and participationStrengthening regional cooperation and collaboration.

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(a) Section class of steel column:    The section class of the steel column can be calculated using the formula: Section modulus, Zx=Ixx/ymaxwhere,Ixx is the moment of inertia about the xx axis and ymax is the maximum distance from the neutral axis to the extreme fibre .Then,    Zx = (Ixx/ymax) = (356 x 368³/12)/(368/2) = 204×10⁴ mm³Effective length factor, Kx = 1.0 ≤ 1.2 (cl. 5.3.3.1 IS 800-2007)

Slenderness ratio, λx = Le/Kx = 5000/1.0 = 5000 mm Radius of gyration,rx = √(Zx/A) = √(204×10⁴/177) = 25.3 mm (A = Cross-sectional area)The buckling parameter is given by,αx = √(P/A)(K/ry)Here, P = Factored axial load = 3500 kN, A = Cross-sectional area = 177×10⁻⁶ m², K = Effective length factor = 1.0, and ry = Radius of gyration = 25.3 mm∴

αx = √(3500×10³/177×10⁻⁶) (1.0/25.3) = 21.3

The equivalent slenderness ratio,λex = λx/√(1+(αx/π)²) = 5000/√(1+(21.3/π)²) = 2234 mm The equivalent slenderness ratio is less than the limiting slenderness ratio (i.e., 300) for all class of sections. Therefore, the column is short and hence plastic design can be done. Section class of the column = Plastic section (Class 1)∴

Section class of the steel column is Plastic section (Class 1).(b) Cross section capacity of the steel column: The design strength of the column for the axial compression, Pn is given by the Euler's formula,Pn = π²ExIxx/(KxLe)²where, Ex is the elastic modulus = 2.1×10⁵ N/mm² for Grade 355 steel∴

Pn = π²×2.1×10⁵×368³/(1.0×5000)² = 7774 kN < Factored axial load (3500 kN)Hence, the cross-section is safe against buckling under axial compression. The design strength of the column for bending about the xx axis, Mn is given by,Mn = Zx x fy/γmwhere, fy is the yield strength of the material, and γm is the partial safety factor for resistance design.

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Given that a one-story steel storage building with a Gable Roof Main wind- force resisting system has the following configuration .Building width (B) = 6 m Building Length (L) = 6 m Height to Apex (H) = 7.5 m Eve Height (h) = 4.5 mEnclosed Structure which has an Exposure Category C with a wind speed of 290 kphGust factor (G) = 0.85Load Combination = 1.2DL +1WLDirection of wind (perpendicular to ridge) = West to EastDL = 1.0 kPa

We are to determine the most critical (consider sign/direction) Leeward Roof Pressure (considering all cases) at the surface of the walls if the wind direction goes from West to East (the Wind direction is perpendicular to the ridge of the structure). Formula used:

Leeward roof pressure = qz – GCp Dynamic Wind pressure (qz) = 0.613kPaCp = Net Pressure Coefficient (Cp)GC = Gust Coefficient Gust Coefficient (GC) = G x (1 + 0.2 Kz)  where Kz is the velocity pressure exposure coefficient.

The leeward roof surface pressure is negative for a gable roof in the windward direction and positive in the leeward direction. Thus, it is the most critical case.1) West Wall: qz = 0.613 kPa Cp = Cpi x GC = - 0.7 x 0.85 = - 0.5955GCp = G x (1 + 0.2 Kz) x Cp = 0.85 x (1 + 0.2 x 1) x (- 0.5955) = - 0.9383Leeward roof pressure = qz – GCp = 0 - (- 0.9383) = 0.9383 kPa (negative)2) East Wall:

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1. Cofferdams: Cofferdams are constructed in an excavation site to keep water out and provide a dry work environment. Design considerations for cofferdams include the structural stability of the cofferdam, the size of the excavation, the type of soil or rock at the excavation site.

2. Earth-Retaining Structures: The primary factors used to determine the type of earth-retaining structure selected and its design are the type of soil or rock at the excavation site, the water table, the load of the structure being retained, the depth of the excavation, the duration of the construction project, and cost considerations.

3. Diaphragm/Slurry Walls: The loss of bentonite can occur as a result of the following: seepage through soil, breaks or cracks in the wall, evaporation of water, and/or the use of inadequate support fluid.

4. Construction Dewatering and Groundwater Control: Factors involved in dewatering include the location of the water source, the depth of the excavation, the soil type, the permeability of the soil, and the size and duration of the excavation.

5. Underground/Tunneling Support: Understanding "CHILE" and "DIANA" ground is essential in tunneling projects. CHILE ground is classified as hard rock, whereas DIANA ground is classified as soft ground. In tunneling, the geological structure of the surrounding rock and soil is critical to understanding the potential impact on the tunneling project.

6. Underpinning: In underpinning, influence lines are essential to determine the load distribution on a structure. Influence lines are the diagrams that show the effect of a load on a structure. They are used to predict the location of stresses in the structure.

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(a)  The design speed for a highway is determined by considering the vehicle and driver characteristics, as well as the road's topography and alignment.. the design speed of the roadway would be 100 km/h, as the highway is being widened, thus making it expressway-like.

(b)

.i. First 10 years traffic forecast = [tex]25,000 vpd × 1.022^10 = 34,536 vpd[/tex]

ii. Traffic in next 5 years (15-10) forecast = [tex]34,536 vpd × 1.011^5 = 37,205vpd[/tex]

DHV in 15 years = [tex]34,536vpd + 37,205vpd = 71,741vpd[/tex]

(c) Number of lanes required to ensure that the level of service (LOS) does not fall below C during the design yearTo determine the number of lanes needed for Level of Service C, we will use the design hour volume (DHV), service flow rate (Q), and the capacity of one lane (pcu/hr).

(d) For Level of Service B, the service flow rate is Q = 2,200 pcu/hr. lane.n = 71,741 vpd / (2,200 pcu/hr.lane × 12,000 pcu/hr) ≈ 3 lanes(e) Capacity of roadway in one direction in vehicles per hou

The appropriate LOS for this project will be D or E as the area lies in an urban region, and we have calculated the number of lanes needed to ensure that the LOS does not fall below C and B.

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Daylighting is the process of using natural light to illuminate the interior spaces of a building, such as a workplace or home. Natural lighting, as compared to artificial lighting, provides a number of benefits that contribute to the well-being of the people who work or live in the area.

Workers who have access to natural light are more content and less likely to experience fatigue or other adverse effects.b. Reduction in electric consumption at peak demand times: When daylight is available, less artificial lighting is required. This can result in significant energy savings, particularly during peak demand periods when energy costs are highest. In addition, the use of less artificial light means less heat generation, which in turn lowers the amount of air conditioning needed.

c. Highly predictable and constant light source: The sun is a reliable and consistent source of light that produces less flicker than artificial light sources. This can be a crucial consideration in environments where the absence of flicker is essential, such as in areas where machinery is being used. In addition, natural light can help to reduce eye strain and other visual impairments that may result from prolonged exposure to artificial light.

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(a) The Young's modulus of the reinforcing steel bar is usually taken to be 29,000 ksi. (b) The term 'grade' on a reinforcing steel bar represents the strength of steel. The strength is determined by the steel's yield strength or the stress at which steel starts deforming or yielding.

(c) The nearly yield strain of Grade-80 steel is about 0.008. Explanation: Reinforcing steel bars are used in reinforced concrete to give the concrete tensile strength to resist cracking and also resist the weight of structures and live loads. These bars are made of high-quality steel and must comply with several standards and specifications like ASTM or ACI.

Reinforcing steel bars are classified by grades according to their tensile strength. The grade is indicated by a number that ranges from 40 to 100 based on the steel's yield strength. The most commonly used grades are 40, 60, and 80, with 80 being the highest.

This means that Grade-80 steel is stronger than Grade-60 and Grade-40.The Young's modulus of steel refers to its elasticity, i.e., its ability to stretch and then return to its original shape. The Young's modulus of reinforcing steel bars is generally considered to be 29,000.

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Net Present Value (NPV) is defind as the present value of cash inflows minus the present value of cash outflows over a given period of time.

It assists in assessing the viability of a project or investment by determining the total value of future cash inflows and outflows in today's dollars. As the given data is as follows:Initial Investment = $100,000Project Life = 16 yearsAnnual Receipts = $25,000Disbursements = $25,000Salvage Value = $45,000Annual.

Discount Rate = 12%To calculate the net present value of the given data, the following formula is used[tex]:NPV = - Initial Investment + ∑ (Cash Flows / (1+Discount Rate) ^ Period Number)where,∑ = Sum ofCash Flows = Annual Receipts - DisbursementsInitial Investment = $100,000Salvage Value = $45,000Annual Receipts.[/tex]

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Civil engineering systems on campus impact the elements of the triple bottom line, which are environment, economy, and society (EST).Civil Engineering System’s Impact on Environment EST relates to the ecosystem in which the civil engineering system operates.

Construction and maintenance activities can have severe environmental impacts, such as soil erosion, water pollution, and deforestation. These can result in habitat destruction, ecosystem imbalance, and loss of biodiversity. As a result, the civil engineering system should be designed and operated in a sustainable manner. For example, minimizing soil erosion by utilizing construction techniques that avoid excavation, preventing soil compaction by reducing the number of vehicles that drive over unpaved areas, and implementing erosion control measures that minimize soil erosion by rainwater or wind. This will have a positive impact on the environment and increase the system's sustainability .Economic Impact of Civil Engineering System The civil engineering system's impact on the economy is another element of the triple bottom line. Construction activities, maintenance, and operation of the system provide jobs and boost the local economy.

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a) Type of marsh creation project A marsh creation project can be of two types: 1. Depositional-based marsh creation2. Terracing-based marsh creation Depositional-based marsh creation involves the placement of sediments onto the surface of the marsh area to be created.

b) Factors controlling the selection of a particular type of marsh creation project There are numerous factors that control the selection of the type of marsh creation project to be used.

c) Step by step procedure for marsh creation project The following are the procedures for creating a marsh: Step 1: Decide on the project's goals Step 2: Select an appropriate location for the project Step 3: Begin the process of marsh creation Step 4: Manage the project's progress Step 5: Monitoring and maintenance Step 6: Documentation of the project The first step in creating a marsh is to establish the project's goals, which should be focused on improving the quality of the local ecosystem or providing some other ecological benefit.

The second step is to choose a site that meets the criteria for a successful marsh creation project. The third step is the actual creation of the marsh, which entails either the deposition or terracing of sediments. The fourth step is the management of the project's progress, including the monitoring of any environmental or economic impacts that may arise.

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Stormwater drainage, slope stability, and sediment control structures are critical elements that must be present in any construction project to guarantee environmental security. These aspects of construction can cause serious concerns if they are not properly managed or ignored.

Here are the most suitable, resilient, effective, and reliable adoption measures to be used in storm water drainage, slope stability, and sediment control structures:

Slope Stabilization Techniques: These techniques are used to maintain the slope's stability and prevent soil erosion. Techniques such as retaining walls, slope stabilization mats, rock bolts, and soil nailing are among them. The soil nailing technique.

Storm-water Drainage: Storm-water drainage systems are critical in managing water flow during heavy rains. The most suitable measures for stormwater drainage management include underground storage systems, bio-retention ponds, and infiltration trenches. The bio-retention ponds provide a natural and ecological method to handle the storm water.

Sediment Control Structures: Sediment control structures are used to contain soil erosion and minimize sediment runoff. Silt fences, sediment ponds, and sediment basins are the most suitable sediment control structures. The sediment ponds are designed to hold storm water and help filter out sediment.

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Given that for a roadway reconstruction project, it was determined that majority of the subgrade soils have PI values ranging 41 to 53 and percent passing No.

40 sieve ranging from 50% to 60%.Since the PI values of the subgrade soils range between 41 to 53, the subgrade soil belongs to group A-2-6 (Clayey soils) according to the AASHTO soil classification system.The percent passing No. 40 sieve for the soil ranges from 50% to 60%. Therefore, the soil is poorly graded or poorly sorted (More than 100) according to the Unified Soil Classification System (USCS).Poorly graded soil is characterized by a wide range of particle sizes and an absence of a well-defined particle size distribution curve.

Such soils are primarily composed of sand-sized particles and contain a limited amount of fines.The soil may have poor drainage characteristics due to the absence of fines to fill the voids between the coarse grains. As a result, it is necessary to test the soil for its moisture content and density before starting the construction process.

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The hydrographs of urban streams are substantially different from forested streams due to human activity and construction. Urbanization and development alter the natural water balance, resulting in changes in streamflow patterns, erosion, sedimentation, and water quality.

Impervious Surface: In urban areas, vast areas of land are paved, which prevents water from penetrating the soil and replenishing groundwater. Instead, the water runs off the surface and into nearby streams, resulting in quick, heavy floods and flash floods.

Land Use Changes: Urbanization results in land-use changes that influence the local climate, such as increased temperatures, less vegetation, and less evapotranspiration.

Stormwater Management: Forested streams provide a variety of ways for water to infiltrate the ground and enter streams, whereas urbanization necessitates the use of impervious surfaces like roads, parking lots, and rooftops. When it rains, water accumulates on these surfaces and is channeled into pipes or stormwater systems.

Stream Channelization: streams are often modified to enhance the flow of water and reduce the risk of flooding. Straightening, widening, and deepening of stream channels are examples of such changes.

These are some of the primary reasons why the hydrographs of urban streams are so different from forested streams.

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a. The main risk factors that a UK-based building contractor is likely to encounter when working on construction projects in an isolated part of the world (e.g. Central America, Asia) include political instability, lack of funding, limited resources, and language barriers .

Political instability: Instability may result from corrupt government officials, political unrest, civil wars, and political leaders changing policies with regards to construction .Lack of funding: Lack of funds can be an issue in isolated locations, where access to banks and financial institutions can be limited. This can cause difficulty in getting loans for construction projects and may require long-term funding options .Limited resources: Construction materials may not be readily available in some areas. In such a case, the contractor may need to consider importing the required materials. Also, reliable and skilled labor may not be easily available in remote areas. Language barriers:

Language barriers between the contractor and the locals can also be a challenge, particularly if the contractors don't understand the local language. This can cause misunderstandings and confusion .b. (i) The Laplace rule weighs each state of nature equally(ii) Hurwicz’s criterion of realism uses an "alpha" value between 0 and 1 to weigh the payoff between the best-case and worst-case scenarios(iii) Maximin rule selects the decision that maximizes the worst outcome of a decision(iv) Minimax rule selects the decision that minimizes the maximum outcome of a decision

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Manufacturing Cost Savings: $22,000Financing Cost Savings: $18,000Total Estimated Savings: $40,000

Given,Cost data for producing a valve:Cost Casting Finishing Inspection Packing TotalDirect materials $ 200 $ 12 $ 0 $ 8 $ 220Direct labor 110 120 20 20 270Variable manufacturing overhead 100 150 20 20 290Allocated fixed overhead 70 80 40 10 200$ 480 $ 362 $ 80 $ 58 $ 980The cost of producing a typical valve is $980 per unit.Annual output (equal in total to 15,000 units).Thus, the total cost of producing 15,000 valves:980 × 15,000 = $14,700,000Reject rate decreased from 5.0% to 3.5%15,000 valves, 5% of which were rejected previously = 750750 valves × $980 = $735,00015,000 valves, 3.5% of which were rejected subsequently = 525525 valves × $980 = $514,500Manufacturing Cost Savings = Cost of reworking/reproducing rejected valves prior - Cost of reworking/reproducing rejected valves now= $735,000 - $514,500 = $220,500 per year or $22,000 per year for the 15,000 units produced. Inventory Holdings:Inventory holdings reduced from $400,000 to $250,000.Annual financing cost associated with inventory holdings = 12%.Thus, the financing cost savings = Annual financing cost associated with inventory holdings × Reduction in inventory holdings= 0.12 × ($400,000 - $250,000) = $18,000Total Estimated Savings: $22,000 + $18,000 = $40,000. Hence, the total estimated savings is $40,000.

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Corporate directors and officers are personally liable for the crimes they commit if the crimes were committed for the corporation's behalf. In some situations, the law holds corporate directors and officers personally accountable for any crime that is committed while they are in charge.

For the crimes committed on behalf of the corporation, directors and officers may be held personally responsible. A corporation can be held liable for violations of state and federal law, but its officers and directors may also be held liable for their own conduct. Officers and directors of corporations can face personal responsibility for actions they took while they were on the job.

Directors and officers must fulfill their legal responsibilities and duties when running a corporation. They must ensure that the corporation operates lawfully and ethically. They must also act in the corporation's best interests while avoiding conflicts of interest. In addition to criminal liability, corporate directors and officers can face civil liability.

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As a civil engineering technician who has been appointed to assist a farmer in estimating the evaporation and infiltration on his vegetation dense land, would be influenced if this plot were developed into an urban neighborhood with roads and less vegetation.

1. The difference between evapotranspiration and transpiration Evapotranspiration is the combined process of transpiration and evaporation of water from the soil, vegetation, and water surfaces in an area.

2. The operation process of the most accurate method you would apply for estimating /calculating evapotranspiration on this plot. The most accurate method to estimate the evapotranspiration rate is the combination of the Penman-Monteith formula and FAO 56 crop coefficient approach,

3. When the potential evaporation value would be negative When the potential evaporation rate (E0) is less than the actual amount of precipitation received by the site in a specific period, the potential evaporation value would be negative

4. How the infiltration and evapotranspiration rate of this place would be influenced if this plot were developed into an urban neighborhood with roads and less vegetation.

the infiltration rate of the area would be reduced, and the runoff rate would be increased. As a result, the evapotranspiration rate will decrease, and the urban heat island impact will increase.

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a) Pressure in the pipe is given as P_1 = 320 kPa and at the outlet is P_2 = 105 kPa.

The difference in pressure at two points in the fluid is given by[tex]∆P = P1 − P2 = 320 − 105 = 215 kPa[/tex]. Density of water, ρ = 1000 kg/m³Length of the pipe, L = 50 mDiameter of the pipe, d = 7 cm = 0.07 mRadius of the pipe,[tex]r = d/2 = 0.07/2 = 0.035 m[/tex]Cross-sectional area of the pipe, [tex]A = πr² = π(0.035)² = 0.00385 m²[/tex]

b) When the pipe instead contains gaseous carbon dioxide in isothermal flow, the pressure would not be the same at this sensor as if the fluid were water. For ideal gases like carbon dioxide, the Bernoulli equation is modified to account for the change in internal energy associated with the flow.

Bernoulli's equation is:

[tex]P_1 + (1/2) ρ V_1² + ρgh_1 + U_1 = P_2 + (1/2) ρ V_2² + ρgh_2 + U_2[/tex]where U is internal energy. If the process is isothermal, internal energy is costant,[tex]U_1 = U_2.[/tex]

P_1 + (1/2) ρ V_1² + ρgh_1 = P_2 + (1/2) ρ V_2² + ρgh_2

This will cause the velocity of the fluid to be much greater for carbon dioxide, which will in turn cause the pressure to be much lower in the carbon dioxide pipe at the same point, as compared to the water pipe.

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Given data, The specific gravity of solids, Gs for a soil = 2.75Moisture content = 20% The formula to find zero air void dry unit weight is: $$ \gamma_ d = {\frac {Gs\ \times\ 100}{100+w}}$$ Where, Gs = Specific gravity of soil100 = Weight of solids in 100 unit weight of soil

w = Water content in soil (percentage) 100 + w = Weight of soil per unit volume (unit weight) So, substituting given values we have, $$\gamma_d = \frac {2.75\ \times\ 100}{100 + 20} = 22.9 \ KN/m^3 $$

Therefore, the zero air void dry unit weight of the soil (KN/m3) at a moisture content of 20 percent is approximately 22.9 KN/m3. Hence option (none of the above) is the correct answer.

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Among the given options, Built-up Roofing System is not used on a steep roof. Steep roofing is used in houses with slanted roofs. It is most commonly found in mountainous areas where there is a risk of heavy snowfall and rainfall.

The purpose of a steep roof is to prevent the accumulation of snow, which could cause roof failure or leaks. To create a steep roof, the roof structure is raised at a 45-degree angle. Materials used for steep roofing must be tough enough to withstand the elements, including heavy rainfall and high wind speeds, without being blown away or eroded. Steep roofing is also known as sloped roofing.

Wood shingles are made from cedar, redwood, or other softwoods. They're popular because of their natural beauty, long lifespan, and energy efficiency. However, they're not fire-resistant, and they need to be kept dry to prevent rotting and warping.

It is created by combining multiple layers of asphalt and felt to create a roof system that is immune to moisture and has excellent insulation properties. It is also durable and long-lasting. it is not suitable for steep roofs because it is heavy and has a low slope. It can also be difficult to install on steep roofs.

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Natural fiber FRC derived from wood pulp has lower maintenance and a wide range of surface finishes. It is made up of cellulose and is processed into fibers that can be used to make clothes and other items.

These fibers have a number of properties that make them useful for a variety of applications. Some of the key benefits of natural fiber FRC include low maintenance, wide range of surface finishes, good thermal stability, low toxicity, good mechanical properties, and environmental sustainability.

These properties make natural fiber FRC an attractive option for a wide range of applications, including clothing, automotive parts, packaging, and construction materials. natural fiber FRC is biodegradable and renewable, making it a sustainable alternative to synthetic materials that are often derived from non-renewable resources.

Natural fiber FRC derived from wood pulp has lower maintenance and a wide range of surface finishes.

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The ejection fraction (EF) is 33%, indicating reduced heart function. The cardiac output is 7.38 L/min.

Ejection fraction (EF) is defined as the proportion of blood ejected by the heart's left ventricle with each contraction. EF may be used to help diagnose and evaluate the severity of heart failure and other heart disorders. The EF is expressed as a percentage, with a higher percentage indicating a stronger heart function.

To calculate EF:

EF = (EDV – ESV) ÷ EDV

EF = (120 – 80) ÷ 120

EF = 40 ÷ 120

EF = 0.33 or 33%

EF is a measurement of how well the heart is pumping blood to the rest of the body. It is considered normal when it is between 50 and 70 percent. An EF of 33% is lower than the normal range, so the answer is no. The patient's EF is not normal. Hence, the answer is: no, EF=33%.

Cardiac output (CO) is defined as the volume of blood the heart pumps per minute. The equation used to calculate cardiac output is:

CO = SV x HR (CO = Stroke volume x Heart rate)

Given, stroke volume = 82 ml and pulse = 90 beats per minute. Therefore,

CO = 82 ml × 90 beats per minute

CO = 7380 ml/min

CO = 7.38 L/min

Therefore, the cardiac output is 7.38 L/min. Hence, the answer is: 7.38 L/min.

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Introduction: A timber beam is used to support loads in a storage facility in Melbourne. It has a span of 3.6m, and it is simply supported. There are two concentrated point loads (P) and a uniformly distributed load of (permanent) g = 1.2 kN/m and imposed floor load (5 months duration) of q = 2.4 kN/m.

The beam is made of 2/240x45 F17 seasoned hardwood sections nailed side by side, and it is laterally restrained along its top edge at each end and at the location of the point loads. We need to calculate the bending moment capacity and the shear force capacity of the beam to resist the combination of permanent and imposed loads and assess whether the beam is adequate in resisting the applied bending moment and shear force.

Calculation of the bending moment capacity of the beam The bending moment capacity of the beam can be calculated using the formula: M = WL²/8where,W = total load on the beam = permanent load + imposed load = (1.2 kN/m + 2.4 kN/m) x 3.6 m= 10.8 kNP = concentrated point loads = 8 kN (permanent load) + 4 kN (imposed load) = 12 kNL = distance between the point loads = 3.6/3 = 1.2

m Now, the bending moment at the point loads can be calculated as:M1 = P x L/4 = 12 kN x 1.2 m/4 = 3.6 kN.mM2 = P x 3L/4 = 12 kN x 3.6 m/4 = 10.8 kN.m The maximum bending moment in the beam occurs at the mid-span and is given by: M max = WL²/8 + M1 = 10.8 kN x 3.6 m²/8 + 3.6 kN. m= 37.44 kN.m.

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Once the interior design project is complete, the deliverables must be accepted properly. Following explains how completed interior design project deliverables shall be accepted.

Acknowledge the designers and any additional workers who assisted in the project. It should also describe what was accomplished and what the final outcome should look like. Explain in detail what was done and if everything meets your needs and specifications. During the review, ask to see samples of the products that were used to complete the design. This is your chance to express any concerns you may have. Finally, after a thorough inspection, once you're satisfied with the final product, you can accept the completed interior design project. To do so, you may have to sign off on the work in order to provide confirmation that the job has been completed to your satisfaction. For instance, in the case of an office space, once the project is finished, you can acknowledge the designers who worked on the project. During the inspection, ask for a demonstration of any furniture items or equipment that were used. You may also want to make certain that everything is in good working order. Finally, once everything has been checked and you're happy with the final product, you can sign off on the work to accept the completed interior design project.

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We need to calculate the application efficiency (%) of a cornfield.

We can calculate it by using the following formula:

Application Efficiency = [tex](Depth of water stored in rootzone / Depth of water applied) × 100[/tex]

Let's substitute the values given in the question and solve it:

Area of cornfield = 5 ha

Discharge = 250 lps

Time of irrigation = 12 hours

Water applied =[tex]Discharge × Time= 250 × 12 × 60 × 60= 1080000 liters[/tex]

Depth of water applied =[tex](Volume of water applied / Area of cornfield)= (1080000 / 50000) m= 21.6 mm[/tex]

Depth of water stored in the rootzone = 160 mm

Now, we can use the formula:

Application Efficiency = [tex](Depth of water stored in rootzone / Depth of water applied) × 100[/tex]

= [tex](160 / 21.6) × 100= 740.74%[/tex]

Therefore, the application efficiency of the cornfield is [tex]740.74%.[/tex]

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1) The diameter of the solid shaft is obtained by using the formula for torsion, T/J = τ/R or T = τJR. Where J = πd⁴/32, R = d/2τ, and τ = 16,000 Nm/[(πd⁴/32)(d/2)] = 32,000/(πd³) MPa. Solving for d gives 64.03 mm.ii) The diameter of the hollow circular shaft is obtained from the expression: τ = T(r² - r₁²)/J where r₁ = r - t and r is the external radius.

Thus, we have r₁ = 0.9r and τ = 16,000 Nm[(0.9r)² - r²]/[(π/32)(r⁴ - (0.9r)⁴)] = 36.61/r MPa. If τ = 65 MPa, then 65 = 36.61/r, r = 55.6 mm and the internal diameter is 50 mm.2) The maximum shear stress in a hollow circular shaft is given by τ = Tr/J. Where T = 12,000 Nm, r = 0.09 m, R = 0.135 m, J = (π/32)(0.135⁴ - 0.09⁴) = 3.57 × 10⁻⁵ m⁴. Thus, τ = (12,000)(0.135)/3.57 × 10⁻⁵ = 454.545 MPa.3) From the formula T/J = τ/R, we can obtain J = πd⁴/32 = (π/32)(d²)².

Substituting the given values, we have 20,000 Nm/(π/32)(d²)² = 50 MPa, d = 160 mm.4) The maximum shear stress can be obtained from the formula τmax = Tr/J where J = πd⁴/32. Thus, we have J = π(0.08)⁴/32 = 1.005 × 10⁻⁶ m⁴. Therefore, τmax = (10 × 10⁶)(0.08/2)/1.005 × 10⁻⁶ = 3.98 × 10⁸ N/m².The angle of twist per unit length can be obtained from the expression θ/L = Tl/GJ.

Thus, θ/L = (10 × 10⁶)(2)/(85 × 10⁹)(π/32)(0.08)⁴ = 0.005 rad/m.5)i) The torque is obtained from T = τJR/(r² - r₁²) where r = 0.06 m and r₁ = 0.045 m. Solving for T gives 47,725 Nm.ii) The maximum shear stress is given by τmax = Tr/J where J = π(r⁴ - r₁⁴)/32. Substituting the given values, we have J = π(0.06⁴ - 0.045⁴)/32 = 2.097 × 10⁻⁶ m⁴.

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(a) The evaporation depth can be calculated using the following formula:

[tex]E= [λCpρ/λ + γ] * [(1/B + Δ/λ) * (Rn/G)][/tex]Where,λ = Latent Heat of Vaporisation = [tex]2.5 x 106 J/kgCp[/tex] = Specific Heat Capacity of Air = 1013 J/kgKρ = Density of Water = 1000 kg/m3γ = Psychometric Constant = 66.2 Pa/KΔ = Slope of Saturation Vapour Pressure Curve = 0.14 kPa/°CRn = Net Radiation = 600 W/m2B = Bowen Ratio = 0.4G = Soil Heat Flux Density = 0 (for open water surface)

Here, E = Evaporation Depth in mm=[tex][2.5x10^6*1013*1000/(2.5x10^6+66.2)] * [(1/0.4+0.14/2.5x10^6) * (600/0)]= 8.66 mm[/tex]Therefore, the expected evaporation depth from a location under the given conditions is 8.66 mm in 24 hours.

(b) When the Bowen ratio is less than 1, the air above the water surface gets warmer because the energy is supplied to the atmosphere from the water surface as latent heat. This is because the energy is needed to transform the water to water vapour and is transferred to the air through a latent heat flux.

As the Bowen ratio decreases, the amount of latent heat flux increases and the amount of sensible heat flux decreases. Hence, the air above the water surface gets warmer, and the water surface temperature decreases.

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The following are the most important TWENTY (20) activities, including their descriptions, dependencies, and durations for a shopping mall construction project:Activity description dependencies.

The Activity on Node diagram for the Shopping mall construction project is depicted in the figure below:The critical path is activities A, C, D, F, G, H, I, J, and K with a total duration of 31 weeks.b)Perform a Forward and Backward passForward Pass: To determine the Early Start (ES) and Early Finish (EF) for each activity,

The Forward Pass technique is used. The duration of the first activity, A, is 4 weeks, and its ES and EF are both 1 week. The remaining ES and EF for all activities are determined by adding the duration of the previous activity to its EF. Thus, the table below depicts the Forward Pass computations for the shopping mall construction project.

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Answer:

Emails and phone calls are both common forms of communication that are used in professional and personal settings. There are several similarities between email and phone calls:

1. Both are asynchronous forms of communication: Unlike instant messaging or face-to-face conversations, both emails and phone calls allow the sender or recipient to respond at their convenience. They don't require immediate attention or an instant response.

2. Both are written forms of communication: While phone calls rely on spoken words, emails are written. As a result, both can be used to convey detailed information and allow the sender to carefully consider their words before sending.

3. Both are forms of direct communication: Emails and phone calls both allow for direct communication between two parties. This can be beneficial for discussing sensitive information or resolving issues quickly.

4. Both can be used for formal and informal communication: Emails and phone calls can be used in both personal and professional contexts. They are both flexible forms of communication that can be adapted to fit different situations.

5. Both require attention to tone and etiquette: Just like with phone calls, emails require attention to tone and proper etiquette. Both forms of communication should be approached professionally and respectfully to ensure effective communication.

In conclusion, while there are differences between emails and phone calls, there are also similarities that make them useful communication tools. Both allow for direct, asynchronous communication and can be adapted to fit different situations.

Explanation:

The drag force on the building is given by[tex]FD2 = (1/2) x ρ x U2 x A x CD[/tex]where,U = 20 m/s (uniform wind speed)A = W x H (area of the building[tex])FD2 = (1/2) x 1.2 x 202 x 30 x 150 x 1.4= 5.04 MN = 5.04 x 103 kN[/tex](kilonewtons) the drag force (FD)2 in kN on this building is 5040 kN.

The power law equation is used to determine the wind speed (V) at a height (Z) where the wind speed at another height (Z0) is known. The power law is given by[tex]V = V0(Z/Z0)n[/tex]where, n = 0.40 for the given profile

[tex](U~ y0.40)V0 = 20 m/s[/tex] at Z0 = 100 m For calculating FD1, the building can be split into several sections, and each section can be considered as a flat plate having the same drag coefficient (CD) and the same wind speed. Using the wind speed and drag coefficient, the drag force can be calculated for each section.

The total drag force on the building is calculated by adding the individual drag forces for each section.

[tex]FD1 = (1/2) x ρ x V2 x A x CD[/tex]where,ρ = 1.2 kg/m3 (density of air)A = W x H (area of the building)

[tex]FD1 = (1/2) x 1.2 x (20 x (150/100)0.4)2 x 30 x 150 x 1.4= 7.38 MN[/tex] (meganeutons)

[tex]7.38 x 103 kN[/tex](kilonewtons) the drag force (FD)1 in kN on this building is 7380 kN.2. Drag force (FD)2 in kN if the incoming wind speed is assumed to be uniform and equal to Uavg for the profile up to the height of the building.

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