Engineering College

## Answers

**Answer 1**

The **bearing** between points A and B is approximately 2.25 **radians**, or 129 degrees (rounded to the nearest degree).

To compute the bearing and distance between points A and B, we can use the **Haversine formula**. Given that:

Point A: N 1000.0000 E 1000.0000

Point B: N 928.1000 E 1500.5912

We can first calculate the differences in **longitude** and latitude:

Δlat = latB - latA = 928.1000 - 1000.0000 = -71.9

Δlon = lonB - lonA = 1500.5912 - 1000.0000 = 500.5912

Next, we convert the latitude and longitude values from degrees to radians, using the following formulas:

lat = lat * π/180

lon = lon * π/180

Using this formula, we have:

latA = 1000.0000 * π/180 = 0.01745329252

lonA = 1000.0000 * π/180 = 0.01745329252

latB = 928.1000 * π/180 = 0.01616889051

lonB = 1500.5912 * π/180 = 0.02618553617

We can then calculate the Haversine formula:

a = sin²(Δlat/2) + cos(latA) * cos(latB) * sin²(Δlon/2)

c = 2 * atan2( √a, √(1−a) )

d = R * c

where R is the radius of the Earth (mean **radius** = 6,371km), and d is the distance between the two points.

Substituting the values, we get:

a = sin²(-71.9/2) + cos(0.01745329252) * cos(0.01616889051) * sin²(500.5912/2)

= 0.230616

c = 2 * atan2( √0.230616, √(1−0.230616) )

= 1.956351

d = 6,371 * 1.956351

= 12435.7 km

Therefore, the distance between points A and B is approximately 12435.7 km.

Next, we can calculate the bearing between the two points. This is given by the following formula:

θ = atan2( sin(Δlon) * cos(latB), cos(latA) * sin(latB) − sin(latA) * cos(latB) * cos(Δlon) )

Substituting the values, we get:

θ = atan2( sin(500.5912) * cos(0.01616889051), cos(0.01745329252) * sin(0.01616889051) − sin(0.01745329252) * cos(0.01616889051) * cos(500.5912) )

= 2.248467

Therefore, the bearing between points A and B is approximately 2.25 radians, or 129 **degrees** (rounded to the nearest degree).

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## Related Questions

Two double headed 16D common nails will be used to attach a Doug Fir-Larch S4S 2 x 4 to a Doug Fir- Larch S4S 4 x 4 in side grain under normal temperature conditions for load duration of less than 2 months. Determine the total ASD withdrawal capacity of the connection if assembled under dry conditions but subject to significant wetting and drying in service

### Answers

The total ASD** withdrawal capacity **of the connection cannot be determined without additional information, such as the specific withdrawal capacity values for the nails and the specific **moisture** conditions during wetting and drying in service.

The task is to determine the total ASD (**Allowable Stress Design**) withdrawal capacity of a connection using two double-headed 16D common nails.

The connection is between a **Doug Fir-Larch** S4S 2 x 4 and a Doug Fir-Larch S4S 4 x 4 in side grain. The assembly is under normal temperature conditions and subject to significant wetting and drying in service, with a load duration of less than 2 months.

To determine the total ASD withdrawal capacity of the connection, several factors need to be considered. First, the withdrawal capacity of a single nail must be determined based on the specific wood species and nail size. The **American Wood Council** provides withdrawal capacity tables for different nail sizes and wood species.

Since the connection involves side grain, the withdrawal capacity may be reduced compared to connections in end grain. The withdrawal capacity is also affected by the** moisture content **of the wood. Wetting and drying cycles can lead to changes in wood dimensions, which may affect the withdrawal capacity of the nails.

To calculate the total ASD withdrawal capacity, the withdrawal capacity of each nail is determined and then multiplied by the number of nails in the connection (two in this case). The lower of the two withdrawal capacities is typically used to ensure a** conservative** **design**.

It is recommended to consult the appropriate withdrawal capacity tables provided by recognized standards or consult a **structural engineer **or a specialized professional for accurate and up-to-date information on withdrawal capacities and design considerations for the specific wood species, nail size, and conditions of the connection.

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Which of the following statements is true?

a. Any straight section of any road must be normally sloped always.

b. Roads with Lower normal cross slopes improve road drainage.

c. The normal cross slope of the road is essential since it improves road drainage and thus enhance tire friction.

d. Roads with higher normal cross slopes indicate that the materials used in road construction have better proge vilian team

### Answers

The argument that the typical cross slope of the road is necessary because it promotes road **drainage** and hence increases tire **friction** is accurate. Option (C) is hence the appropriate response.

Kinetic energy, or work, is converted into thermal energy, or heat, through friction between two surfaces that are in contact and moving in respect to one another.

The utilization of **friction **produced by rubbing pieces of wood together to ignite a fire is an illustration of this trait's propensity for spectacular outcomes. Kinetic energy is changed into thermal energy every time there is frictional motion, for as when a **drainage **viscous fluid is churned.

Numerous frictional phenomena have a substantial side effect known as wear that can impair performance or cause component damage. Friction is a topic in the science of tribology.

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Using which type(s) of the following triaxial tests, both total and effective stress shear strength parameters will be found. (find different values for total and effective stress)?

Both UU and CD

CU

UU

CD

### Answers

Both UU (Unconsolidated Undrained) and CD (**Consolidated** Drained) triaxial tests can be used to determine both total and effective stress shear strength parameters.

In the UU test, the specimen is not **allowed **to drain during shearing, so the shear strength parameters obtained represent the undrained conditions. This gives the total stress shear strength.In the CD test, the specimen is allowed to drain during **shearing**, simulating drained conditions. The shear strength parameters obtained in this test represent the effective stress shear strength, taking into account the pore water pressure changes.Therefore, both UU and CD triaxial tests can provide separate values for total **stress** shear strength (UU) and effective stress shear strength (CD), allowing the analysis of soil behavior under different stress conditions.On the other hand, the CU (Consolidated Undrained) test only provides the undrained (total stress) shear strength parameters, while the UU test only provides the undrained shear strength parameters.

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Which among the features can be represented by multiple vector types? What conditions might lead you to choose one vector type over another?

a. Barangay boundaries

b. Telephone poles

c. Buildings

d. Stream networks

e. Mountain peaks

f. Soil types

g. Hurricane tracks

This is an informative geospatial property that describes the connectivity, area definition, and contiguity of interrelated points, lines, and polygons.

a. Vector data

b. Georelational data model

c. Raster data

d. Topology

e. Object-based data model

This is a common conceptualization of spatial data wherein it utilizes points, lines and polygons to represent the spatial features in a map.

a. Raster data

b. Object-based data model

c. Georeleational data model

d. Topology

e. Vector data

### Answers

Among the given features, **barangay boundaries**, buildings, stream networks, and hurricane tracks can** **be represented by multiple vector types. The choice of vector type depends on factors such as precision, storage requirements, and the need for **topological relationships.**

Barangay boundaries, buildings, **stream networks**, and hurricane tracks can be represented using different vector types depending on the specific requirements and characteristics of the data. For example, barangay boundaries and buildings can be represented using both line and polygon vectors.** Line vectors** can be used to represent the precise boundaries of these features, while polygon vectors can be used to represent their area and shape.

Stream networks can also be represented using different vector types. They can be represented as lines to capture the flow and connectivity of the streams. Alternatively, they can be represented as a series of points along the stream network, which is useful for capturing the topology and connectivity of the network.

Hurricane tracks, on the other hand, can be represented using both point and line vectors. Point vectors can be used to represent the locations of specific points along the track, such as the position of the hurricane at different times. Line vectors can be used to represent the trajectory of the hurricane, showing the path it takes over a** specific period.**

The choice of vector type depends on various factors. Precision and accuracy requirements play a role in selecting the appropriate vector type. For example, if a high level of detail is needed, polygon vectors may be preferred over line vectors for representing barangay boundaries or buildings. **Storage requirements** also come into play, as certain vector types may require more storage space. Additionally, the need to capture and analyze topological relationships among features, such as connectivity and adjacency, may influence the choice of vector type.

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Part A and B are entered into a contract for a building construction in a town. Unfortunately, the town is hit by an earth quake and result in impossible to implement the contract. Analysis if they are beaching the contract or can be described by other contractual terms. You are required to provide descriptions on your provided contractual terms.

### Answers

In this scenario, if an earthquake** occurs** and makes it impossible to implement the **contract** between Part A and Part B for building construction in the town.

It may not necessarily be considered a breach of contract. Instead, it could be described as a case of "**force majeure**" or an "act of God." These terms refer to unforeseeable circ*mstances beyond the control of the parties involved that prevent the **performance **of the contract.

Force majeure typically releases the **parties** from their contractual obligations, as the event was unforeseeable and unavoidable. It is essential to review the specific **contractual **provisions and applicable laws to **determine** the rights and obligations of both parties in this situation.

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D Question 9 6 pts Two wells are located at points A and B respectively. Point C is the middle point between A and B. When water is pumped out from Well A only, the drawdown at Cis 7 in. if water is p

### Answers

The** drawdown **at point C is 7 inches when water is** pumped out** from Well A only.

Drawdown refers to the decrease in the water level in a well due to pumping. In this scenario, when water is pumped out from Well A only, the drawdown at point C is measured to be 7 inches. This means that the water level at point C is 7 inches below its **original level **when Well A is actively pumping water.

The drawdown at a specific point depends on various factors, including the rate of pumping, the** geology of the aquifer,** and the distance between the well and the point of measurement. In this case, since point C is the midpoint between Well A and Well B, it is expected to experience a drawdown when water is pumped from Well A.

The drawdown at point C being 7 inches indicates that Well A is effectively lowering the water level in the vicinity. The rate of pumping from Well A and the** hydrogeological properties **of the aquifer determine the magnitude of drawdown. It is important to monitor and manage drawdown levels to ensure sustainable water supply and minimize potential impacts on** neighboring wells **or water sources.

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In a manufacturer's test, a valve resulted in an average head loss of 1.21 m when the flow speed was controlled at 2.00 m/s. What is most approximately the minor loss coefficient of this valve? O 1.32 O 7.88 O 5.94 O 3.61

### Answers

To calculate the minor loss coefficient (K) for the valve, we can use the relationship between head loss (hL), **velocity **(V), and the minor loss **coefficient**:

hL = K * (V^2 / (2g))

Given data:

**Average **head loss (hL) = 1.21 m

Flow speed (V) = 2.00 m/s

**Acceleration **due to gravity (g) ≈ 9.81 m/s²

**Rearranging **the equation, we can solve for the **minor **loss coefficient (K):

K = (2g * hL) / V^2

Now, substituting the given values:

K = (2 * 9.81 * 1.21) / 2^2

K ≈ 7.88

Therefore, the most approximate minor loss coefficient for this valve is 7.88 (option O 7.88).

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A weir is to be installed in a river that is initially flowing at a depth of 3.5 ft and velocity of 2.5 ft/s. The width of channel is 250 ft. The weir installation requires that the water level upstream must not rise more than 1.9 ft. Find the required height of the weir for a weir length of 210 ft.

### Answers

The required height of the **weir **for a weir length of 210 ft is approximately 1.6 ft.

To determine the required **height** of the weir, we need to consider the flow conditions and the desired water level upstream. In this case, the water level upstream should not rise more than 1.9 ft.

Using the energy equation for flow over a weir, we can calculate the required height of the weir. The equation is:

h = (H + Z) - (h₁ + z₁)

Given:

Initial water depth (h₁) = 3.5 ft

Initial **velocity **(V) = 2.5 ft/s

Weir length (L) = 210 ft

Desired maximum rise in water level (H) = 1.9 **ftInitial** elevation of the river surface (z₁) = 0 ft (assuming same level as the weir crest)By substituting the given values into the equation and solving for h, we find that the required height of the weir is approximately 1.6 ft. This height will ensure that the water level upstream does not rise more than the specified limit.

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Why does the cross-sectional shape and orientation (Major x-x or Minor y-y axis) of a beam cross-section have such a major influence over its structural performance? In providing your response, include a simple and practical example to quantify the impact of shape and orientation (use a different example than those the module!). Expectation:100-150words.

### Answers

The** cross-sectional** shape and orientation of a beam have a significant influence on its structural performance due to their impact on key structural properties, such as moment of inertia and section modulus.

The moment of **inertia **determines a beam's resistance to bending and deflection. A larger moment of inertia implies greater stiffness and resistance to bending, while a smaller moment of inertia results in increased bending and **deflection**. The shape of the cross-section directly affects the distribution of material away from the neutral axis, which determines the moment of inertia. For example, a hollow circular section with a larger outer **diameter **and thinner walls will have a higher moment of inertia compared to a solid circular section with the same outer diameter.

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Need help asap. PLEASE SHOW THE CORRECT FIGURE AND DETAILED

SOLUTION. Answer on the book: 0.64

During a test on a 245 cm long rectangular suppressed weir which was 100 cm high, the head was maintained constant at 30 cm. In 38 sec, 28.80 m³ of water were collected. Find the correction factor C'

### Answers

The **correction** factor C' 0.484x[tex]C^{0.5}[/tex]

Where, C is the discharge **coefficient.**

We know that the formula for **discharge** Q using,

Q = (C/2)L[tex](2gH)^{0.5}[/tex]

where,

C is the discharge coefficient,

L is the length of the weir,

g is the **acceleration **due to gravity,

And H is the **head **over the weir.

Put the given values, we get:

⇒ Q = (C/2)245(2x9.81x0.3[tex])^{0.5}[/tex]

⇒ Q = 115.04[tex]C^{0.5}[/tex]

Find the** volume** of water collected using the formula,

⇒ V = At

where A is the **area **of the collection tank, and t is the time taken to collect the water.

Put the values, we get:

⇒ V = 28.80/1000 = 0.0288 m³

⇒ t = 38 seconds

Since we know that the area of the collection tank is much larger than the area of the weir, we can assume that the** water level** in the collection tank remained constant at 30 cm, so the volume collected is equal to the area of the **collection** tank times the change in water level,

⇒ V = A(H2 - H1)

where H1 is the** initial** water level and H2 is the final water level.

Put the given values, we get,

⇒ 0.0288 = A(0.3 - 0)

⇒ A = 0.096 m²

Finally, we can find the correction** factor** C' using the formula,

⇒ C' = Q/(AL)

Plugging in the values we calculated, we get,

⇒ C' = (115.04x[tex]C^{0.5}[/tex])/(0.096x245)

Hence, **Simplifying **we get,

⇒ C' = 0.484x[tex]C^{0.5}[/tex]

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Compute the rectangular survey description for Dodger Stadium. Keep in mind that the tools you have

may not be accurate enough to be definitive. In principle, you can find the description somewhere, but

you must calculate it for this assignment.

Rubric: 2 pts: submission, 1 pt: References, 2: Input data, 4: Compute required distances, 2: Compute

Range/Tier, 1: Compute Section, 1: Compute location within section (bonus).

### Answers

The **rectangular** **survey** description for Dodger Stadium can be computed using the tools used in rectangular **surveying**.

Rectangular surveying is a land surveying method that divides land into square-shaped areas or townships. It makes use of the township and range system, which is a **grid**-**like** **system** that is used to identify locations and boundaries of land.The first step in computing the rectangular survey description for Dodger Stadium is to gather input data, including the latitude and longitude of the stadium, and the meridian and baseline that are used to define the township and range lines. The meridian and baseline are typically set by the government or a surveying authority.Next, the required distances must be calculated. This includes the distance from the baseline to the stadium, and the distance from the meridian to the **stadium**. These distances are used to determine the township and range lines that intersect at the location of the stadium. The township and range lines are numbered based on their distance from the baseline and meridian, respectively.Once the township and range lines are determined, the stadium can be located within a specific section. Sections are 1-mile square areas that are numbered within each **township** and range. The section number is determined by counting the number of sections between the township line and the stadium, and the number of sections between the range line and the stadium.Finally, the location of the stadium within the section can be determined by dividing the section into smaller portions. This is typically done using a system of fractions, where the section is divided into halves, quarters, and so on. The location of the stadium is then described in terms of the **fraction** that it is located within. For example, if the stadium is located in the southeast quarter of the section, it would be described as being located in section 27, township 1 north, range 1 west, southeast quarter.

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What is the recommended value of effective length if the column is effectively held in position and fixed against rotation in both ends? Select the correct response: A.0.7 B.0.5 C.08 D.0.65.

### Answers

The recommended value of **effective length** for a column effectively held in position and fixed against **rotation **in both ends is 0.65. Therefore, **option D **is correct.

The **effective length **of a column refers to the distance between the points of **rotation **or points of zero moment in a column that is effectively held in position and fixed against rotation at both ends.

It is a critical parameter in structural engineering as it determines the column's buckling behavior and overall stability. The recommended value of **effective length**, in this case, is 0.65.

This value is based on extensive research and analysis to ensure that the column can resist buckling and maintain structural integrity under various loading conditions. Choosing the appropriate **effective length **helps engineers design columns that can effectively support the intended loads and maintain structural safety.

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Find the number of solar panels required for A class room in HCT. The daily energy consumption is given below 1. Ten Computers 5 x10 -2 KW , Using 8 hours .2. Two AC 2x10 -4 MW, using 6 hours 30 minutes.3. Six LED lights 20 Watt, using 7 hour 15 minutes.

### Answers

As per the details given, 40 **solar** **panels** would be required for the classroom in HCT, considering the given daily **energy** **consumption** of computers, AC units, and LED lights.

The total daily **energy** usage must be calculated in order to determine the number of solar panels needed for a classroom in **HCT**, and the solar panels' requisite capacity must then be established.

**Computers**:

Energy consumption of 10 computers = 10 * 0.05 kW * 8 hours

= 4 kWh

**AC**:

Energy consumption of 2 AC units = 2 * 0.0002 MW * 6.5 hours

= 0.0026 MWh = 2.6 kWh

**LED** **lights**:

Energy consumption of 6 LED lights = 6 * 20 W * 7.25 hours

= 870 Wh = 0.87 kWh

Total daily **energy** **consumption** = 4 kWh + 2.6 kWh + 0.87 kWh = 7.47 kWh

Capacity of solar panels needed = 7.47 kWh / (0.15 * 5 hours) = 9.96 kW

Number of panels = Capacity of solar panels needed / Capacity of each panel

Number of panels = 9.96 kW / 0.25 kW = 39.84

Thus, approximately 40 solar panels would be required for the **classroom** in HCT.

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Using the information from the previous problem, find the shrinkage properties of the backfilled soil. Given, during the shrinkage test of the backfilled soil the soil patty had a moist mass of 44gm with initial volume 20.11ml. While the volume of the dry soil patty was 19.2 ml. Find the following,

a. mass of the dry soil patty

b. shrinkage limit of the soil

C. shrinkage ratio of the soil

d. specific gravity of the soil and explain if the specific gravity matches with the initially excavated soil.

e. volumetric shrinkage

f. linear shrinkage.

### Answers

This** soil** is to be excavated and transported to a **construction site** for use in a compacted Volume fill is 8680.73 m³

The specific gravity of solids = 2.75

**situ moisture** content of a soil is 15% with saturation level equal to 70% and void ratio of 0.9.

Dry weight = 103.5

Yd = 16.5/(1 + 18/100) = 14.34KN/m³

**Volume** of soil to be excavated to produce 7651m³

V = Yd compacted / Yd insite × compacted fill volume

V = 16.27/14.34 × 7651 = 8680.73 m³

No of trucks = V × Yinsity / 1.78 = 8680 × 16.5/178 = 826trucks.

Using the **human body,** such as by pinching or measuring an object's volume using the size of your hand, is the most traditional method of doing so. The human body, however, is incredibly unpredictable due to its differences.

A more accurate technique to measure** volume **is to utilize fairly constant and long-lasting natural objects, like gourds, sheep or pig stomachs, and bladders. Small volumes are typically measured using standardized human-made containers today since **metallurgy** and glass manufacture have advanced through time.

This approach is frequently used to measure small volumes of fluids or granular **materials.** To create a nearly flat surface, the container is shook or leveled off for granular items. The most exact approach to measure is not with this technique.

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A solid shaft of diameter d and a tube of outer diameter D = 60

mm and thickness of 15 mm are to carry the same torque at the same

maximum shear stress. What is the required d?

### Answers

The required **diameter** (d) for the** solid shaft** is approximately 48.97 mm.

To determine the required diameter (d) of a solid shaft that can carry the same torque and have the same maximum **shear stress** as a tube with an outer diameter (D) of 60 mm and a thickness of 15 mm, we need to consider the torque and shear stress equations for solid and hollow circular sections.

For a solid circular shaft, the **torque** (T) can be calculated using the equation:

T = (π/16) * τ_max * d^3

where τ_max is the maximum** shear stress.**

For a hollow circular tube, the torque (T) can be calculated using the equation:

T = (π/16) * τ_max * (D^4 - d^4) / D

In this case, we want to find the diameter (d) of the solid shaft that can carry the same torque and have the same maximum shear stress as the given hollow tube with an outer diameter (D) of 60 mm and a thickness of 15 mm.

By equating the torque equations for the solid shaft and the hollow tube, we have:

(π/16) * τ_max * d^3 = (π/16) * τ_max * (D^4 - d^4) / D

We can simplify the equation by canceling out common terms:

d^3 = (D^4 - d^4) / D

Next, we can rearrange the equation:

D * d^3 = D^4 - d^4

Substituting the values for D and rearranging further, we get:

D * d^3 + d^4 = D^4

d^3 = D^4 - (D - 2t)^4

d^3 = (60 mm)^4 - (60 mm - 2 * 15 mm)^4

d^3 = 129,600,000 mm^4 - 20,736,000 mm^4

d^3 = 108,864,000 mm^4

Take the cube root of both sides to solve for d:

d = ∛(108,864,000 mm^4)

d ≈ 48.97 mm

Now, we can solve this equation for the required diameter (d) of the solid shaft. By **iterating** through different values of d and comparing the left and right sides of the equation, we can find the value of d that satisfies the equation and gives the required diameter for the solid shaft.

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The cross section of roadways should be level

True or False

### Answers

False. The **cross section of roadways **should not be completely** level**.

The statement that the cross section of roadways should be level is false. In fact, roadways are designed with a specific cross-sectional profile that **includes slopes** and gradients to ensure proper drainage and safety.

The cross section of a roadway typically consists of several components, including the** pavement surface**, shoulders, and ditches or curbs. These components are designed to facilitate the flow of water off the road surface and prevent water from pooling or accumulating, which can lead to hazardous conditions such as** hydroplaning. **

Therefore, roadways are deliberately sloped to allow water to drain effectively. Additionally, roadways may also have crowned surfaces, where the center of the road is slightly higher than the edges, to further aid in water runoff and provide** improved visibility** for drivers. So, the notion that roadways should be completely level is incorrect.

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Determine the friction factor for water flowing at a rate of 1 cfs in a cast iron pipe 2 in. in diameter at 80°F. Problem# 2: What

### Answers

The **friction factor** for water flowing at a rate of 1 cfs in a cast iron pipe 2 in. in diameter at 80° F is 0.00104.

The friction factor for fluids flowing in pipes is determined using the Darcy-Weisbach equation. The Darcy-Weisbach equation is expressed as:

f = 64/Re × [1.325log(Re)+(-2.45)]

Where f is the friction factor, Re is the Reynold’s number, and all parameters are dimensionless.

In order to calculate the Reynold’s number, the following equation is used:

Re = ([tex]D_h[/tex]×ρ×V)/μ

Where [tex]D_h[/tex] is the **hydraulic diameter** of the pipe ([tex]D_h[/tex] = 4×A/P, where A is the area and P is the perimeter of the pipe), ρ is the density of the fluid, V is the average velocity of the fluid, and μ is the dynamic viscosity of the fluid.

Since we are given the following conditions:

- Water flowing at a rate of 1 cfs in a cast iron pipe 2 in. in diameter at 80°F

We can calculate the Reynold’s number as follows:

Calculate the hydraulic diameter:

[tex]D_h[/tex] = 4×A/P

A = (π/4) × (2 in.)² = 3.14 in.²

P = 2π×2 in = 12.56 in.

[tex]D_h[/tex] = 4 × 3.14 / 12.56 = 0.8145 in

Calculate the fluid density:

- Water at 80°F = 62.368 lb/ft³

Calculate the fluid velocity:

- V = 1 cfs = 448.8 gpm

Calculate the **dynamic viscosity** of the fluid:

- μ = 0.01007 lb-sec/ft²

Re = (0.8145 in × 62.368 lb/ft³ × 448.8 gpm) / 0.01007 lb-sec/ft²

= 176083.9

Now that the Reynold’s number has been calculated, we can use it to solve for the friction factor:

f = 64/Re × [1.325log(Re)-2.45]

= 0.00104

Therefore, the **friction factor** for water flowing at a rate of 1 cfs in a cast iron pipe 2 in. in diameter at 80° F is 0.00104.

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The position of a particle in millimeters is given by s=84-19t+t^2 where t is in seconds. Plot the s-t and v-t relationships for the first 12 seconds. Determine the net displacement As during that interval and the total distance D traveled. By inspection of the s-t relationship. what conclusion can you reach regarding the acceleration?

### Answers

The net **displacement** (Δs) during the interval from t = 0 to t = 12 seconds is -216 mm.

The total distance (D) traveled during the interval from t = 0 to t = 12 seconds is 198 mm.

By inspecting the s-t **relationship**, we conclude that the particle is experiencing uniform acceleration with a magnitude of 2 mm/s².

Given: s = 84 - 19t + t²

To find the **velocity**, we take the derivative of the position equation with respect to time (t):

v = ds/dt = d/dt(84 - 19t + t^2)

v = -19 + 2t

To find the **acceleration**, we take the derivative of the velocity equation with respect to time (t):

a = dv/dt = d/dt(-19 + 2t)

a = 2

Now we have the equations for velocity (v) and acceleration (a).

To plot the s-t and v-t **relationships** for the first 12 seconds, we can calculate the values for each second from t = 0 to t = 12:

For the s-t relationship:

We substitute the values of t into the position equation (s = 84 - 19t + t²) to calculate the corresponding values of s.

For the v-t relationship:

We **substitute** the values of t into the velocity equation (v = -19 + 2t) to calculate the corresponding values of v.

Let's calculate and plot the s-t and v-t relationships:

t (s) s (mm)v (mm/s)

0 84 -19

1 66 -17

2 48 -15

3 30 -13

4 12 -11

.--- ---- ---

12 -132 5

The equation s-t graph is s=-18t+84

The equation v-t graph is v=2t-19

Net **displacement** (Δs) can be calculated by finding the difference between the initial position (s_initial) and the final position (s_final):

As = s_final - s_initial

As = -132 - 84

As = -216 mm

Total distance (D) can be calculated by summing the **absolute values** of the individual displacements during the interval:

D = |s_1 - s_0| + |s_2 - s_1| + |s_3 - s_2| + ... + |s_final - s_n-1|

D = |-18| + |-18| + |-18| + ... + |-18|

D = 18 + 18 + 18 + ... + 18 (11 terms)

D = 198 mm

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The slenderness ratio of a column determines if it is short, intermediate or long column. Select the correct response: I don't know True O False None enough data

### Answers

It is true that the **slenderness** **ratio** of a column determines if it is short, intermediate or long **column**.

A critical factor that affects a column's **stability** and buckling behaviour is its slenderness ratio. It is described as the ratio of the column's effective length to its smallest **gyration** **radius**.

The column's **terminal** **circ*mstances** and supporting structure are taken into account while determining the effective length.

The column's **resistance** to buckling about its weakest axis is represented by its least gyrating radius.

Thus, the answer is **true**.

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Derive the equations of motion for an airplane in descending gliding

flight (T=0) in a vertical plane. First, draw a free body diagram

showing an aircraft in gliding flight and all the coordinate systems,

angles, and forces. Here, assume that the velocity vector is at an

angle φ below the horizon and that the aircraft is at a positive angle

of attack α. Show that these equations have one mathematical

degree of freedom and are the same as those obtained from Eqs.

(2.24) with T = 0 and γ = −φ

### Answers

In descending gliding flight, an **airplane **experiences several forces and moments. The derived equation is ΣFx = W cos(γ + α) - L = 0.

To derive the equations of motion, let's start by drawing a free body diagram.

The diagram includes the following elements:

The aircraft, represented by a body with the longitudinal and vertical axes.

A **coordinate system** with an x-axis (horizontal) and a y-axis (vertical) that are fixed with respect to the Earth.

The velocity vector, which makes an angle φ (phi) below the horizon.

The weight force acting vertically downward.

The lift force perpendicular to the velocity vector.

The drag force opposite to the velocity vector.

The thrust force, assumed to be zero for gliding flight.

Now, let's consider the forces acting on the aircraft. The weight force can be decomposed into components: Wx in the x-direction and Wy in the y-direction.

The lift force can be decomposed into components: Lx in the x-direction and Ly in the y-direction.

The** drag force **can be decomposed into components: Dx in the x-direction and Dy in the y-direction.

In the vertical plane, the equations of motion are given by:

ΣFy = Wy + Ly - Dy - W = 0, where W is the weight of the aircraft.

ΣFx = Wx + Lx - Dx = 0.

We can rewrite these equations using trigonometric relationships:

ΣFy = W sin(γ + α) - D - W = 0, where γ is the glide path angle (equal to -φ in this case).

ΣFx = W cos(γ + α) - L = 0.

Since the aircraft is in **gliding fligh**t, the thrust force T is assumed to be zero.

These equations of motion have only one degree of freedom because the aircraft's motion is constrained to the vertical plane.

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In a simple circular curve, the coordinates of the curve's starting point (P.C) are (10.30) meters, the coordinates of the end point (P.T) being (80, 100) meters and the angle of deviation (60) Calculate the radius of the curve.

### Answers

In a simple **circular curve**, the coordinates of the curve's starting point (P.C) are (10.30) meters, the coordinates of the end point (P.T) being (80, 100) meters and the **angle of deviation** (60), the radius of the curve is approximately 98.99 meters.

Given: Coordinates of PC: (x1, y1) = (10, 30) meters **Coordinates **of PT: (x2, y2) = (80, 100) meters Angle of deviation: θ = 60 degrees

We can start by calculating the **chord length** (C) between the PC and PT using the distance formula:

C = √((x2 - x1[tex])^2[/tex] + (y2 - y1[tex])^2[/tex])

C = √((80 - 10[tex])^2[/tex] + (100 - 30[tex])^2[/tex])

= √(7[tex]0^2[/tex] + 7[tex]0^2[/tex])

= √(2 × 7[tex]0^2[/tex])

= √(2 × 4900)

= √(9800)

= 98.99 meters (approximately)

Next, one can calculate the radius (R) of the curve using the following formula:

R = C / (2 × sin(θ/2))

R = 98.99 / (2 × sin(60/2))

= 98.99 / (2 × sin(30))

= 98.99 / (2 × 0.5)

= 98.99 / 1

= 98.99 meters (approximately)

Therefore, the radius of the curve is approximately 98.99 meters.

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An existing sag vertical curve on highway joins a +2.5% grade with a -5.1% grade. If the posted speed limit is 40 mph. Determine the length of the sag curve based on a. Stopping sight distance b. Comfort c. Appearance

### Answers

a) The **length **of the **sag **curve based on the **stopping **sight distance is 175.38 ft.

b) The **sag **curve is **not **required for this particular combination of grades and speed limit.

c) The **length **of the sag curve based on **appearance **is 104,330.71 feet.

a) Stopping Sight Distance:

the **minimum stopping **sight distance using the formula

SSD = V² / (2 g f),

First, let's calculate the stopping sight distance for the ascending grade:

a1 = 0.025 (rate of change of grade)

V = 40 mph (speed limit)

SSD1 = = (40² x 1.2) / (30 x (0.025 + 32.2))

≈ 118.84 ft

and, a2 = -0.051 (**rate **of **change **of grade)

SSD2 = (40² x 1.2) / (30 x (-0.051 + 32.2))

≈ 56.54 ft

The **total length **of the sag curve is the sum of the stopping sight distances for both grades:

= 118.84 ft + 56.54 ft

≈ 175.38 ft

b) Comfort:

Design **speed **(V) = 40 mph

Ascending grade (A) = +2.5%

Descending grade (B) = -5.1%

Converting the percentages to decimal form:

A = 2.5% = 0.025

B = -5.1% = -0.051

**Substituting **the values into the formula:

L = (40²/2540) x ((0.025 + (-0.051)) / 0.025)

L = 0.6299 x (-1.04)

L ≈ -0.655 feet

Since the resulting length is negative, it indicates that a **sag **curve is **not **required for this particular combination of grades and speed limit.

c) Appearance:

Design **speed **(V) = 40 mph

Rate of change of grade (R) = 0.01 per foot

**Difference **in grade (∆G₁) = 2.5% + 5.1% = 7.6%

Difference in grade (∆G₂) = 5.1% + 2.5% = 7.6%

**Converting **the **percentages **to decimal form:

∆G₁ = 0.076

∆G₂ = 0.076

Substituting the values into the formula:

L = (40²/ (2540x 0.01)) x ((0.076 + 0.076) / (0.076 x 0.076))

L = (1600/0.254) x 26.354

L ≈ 104,330.71 feet

Therefore, the **length **of the sag curve based on **appearance **is 104,330.71 feet.

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Determine the maximum steel area for a tension-controlled singly reinforced condition in positive bending with the given material strengths. "A rectangular beam 250 mm wide. 500 mm deep is reinforced at the bottom with 4-20-mm-diameter bars and at the top with 2-16-mm bars. Concrete cover to bar centroid at the top is 80 mm and at the bottom is 70 mm. Use concrete strength f. = 29.91 MPa and steel yield strength fy = 345 MPa for 20-mm bars and fy = 275 MPa for 16-mm bars."

### Answers

A **rectangular** **beam** 250 mm wide. 500 mm deep is reinforced at the bottom with 4-20-mm-diameter bars and at the top with 2-16-mm bars, the maximum steel area for the given **configuration** is approximately 402.12 mm².

The smaller of the two values represents the maximum **steel** **area**.

The effective **depth** (d') of the beam:

d' = d - c_bottom

d' = 500 mm - 70 mm = 430 mm

Number of 20-mm bars (n_bottom): 4

Diameter of 20-mm bars (diameter_bottom): 20 mm

Steel area for one 20-mm bar (A_bottom): (π/4) * (diameter_bottom²)

A_bottom = (π/4) * (20 mm)^2 ≈ 314.16 mm²

A_total_bottom = 4 * 314.16 mm²

= 1256.64 mm²

The **steel** **area** for the top reinforcement:

Number of 16-mm bars (n_top): 2

Diameter of 16-mm bars (diameter_top): 16 mm

Steel area for one 16-mm bar (A_top): (π/4) * (diameter_top²)

A_top = (π/4) * (16 mm)² ≈ 201.06 mm²

Finally comparing the **steel** **areas **and determine the maximum steel area:

Maximum steel area (A_max) = min(A_total_bottom, A_total_top)

A_max = min(1256.64, 402.12) ≈ 402.12 mm²

Thus, the maximum steel** area** for the given configuration is approximately 402.12 mm².

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Just type the number, do not write any words. Do not round your answer. Answer to two decimal places. You have a masonry structure that you are asked to estimate cost and quantities. The structure has

### Answers

Based on the number of hours in the work week, the **cost** of the laborers and masons, worth of the** materials** and the profit, the estimated cost of the job can be written as a mathematical statement of A. 3 x $15 x 40 + 4 x $25 x 40 + $3,268 + $700 = $

The total** estimated cost** is $9,768

The total estimated cost of the** job **can be found by the formula:

= (Number of laborers needed x Cost of laborers x Number of hours in work week) + (Number of masons x cost of masons x number of hours in work week) + materials worth + **Required profit**

The total estimated cost is therefore:

= (3 x $15 x 40) + (4 x $25 x 40) + $3,268 + $700

= 1,800 + 4,000 + 3,268 + 700

= $9,768

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WHAT ARE THE TYPICAL RESULTS FOR A GRADIENT RATIO

TEST)

### Answers

The typical results for a** gradient ratio test **may include the determination of the **hydraulic conductivity **or **permeability** of a soil sample.

**Gradient ratio** **test** results rely on test parameters and goals. Geotechnical engineers use it to measure** soil shear strength** and **stability**. Test findings may include:

**Failure envelope:** The soil specimen's shear stress and normal stress can be plotted using the test's data points. The failure envelope aids soil shear strength analysis and stress prediction.

**Soil classification: **The gradient ratio test can be used to categorise soil by shear strength using the Unified Soil Classification System (USCS) or other comparable classification systems. This classification aids engineering and construction.

**Stability analysis:** Results can be analysed to determine slope, retaining wall, and other geotechnical structural safety. Stability analyses like limit equilibrium or numerical modelling use test results.

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A hollow shaft with an outside diameter of 130 mm and an inside diameter of 95 mm is 3 m long. A flywheel with a weight of 15kN is shrunk on the shaft 1 m from the one bearing. The allowable tensile stress for the shaft is 135 MPa.

Calculate the following:

The maximum torque that can be transferred by the shaft

The power that can be transmitted at 500 r/min by the shaft if there is an overload of 12 per cent on the shaft

The shear stress in the shaft for the above conditions

### Answers

The **maximum torque** that can be transferred by the shaft is 146.9 MPa.

Maximum Torque:

The maximum torque that can be transferred by the shaft is calculated using the following formula:

T = 16πT×([tex]D^4_{out}[/tex] - [tex]D^4_{in}[/tex])/(64×L)

Where:

T = Tensile Stress in MPa

[tex]D_{out[/tex] = Outside diameter of the shaft in mm

[tex]D_{in[/tex] = Inside diameter of the shaft in mm

L = Length of the shaft in m

Substituting the given values,

T = 16π×135×(130⁴ - 95⁴)/(64×3) = 146,929 N.m

Power transmitted:

The power that can be transmitted at 500 r/min by the shaft if there is an overload of 12% on the shaft is calculated using the following formula:

P = (T×ω)/(9550×(1+0.12))

Where:

P = Power transmitted in kW

T = Maximum torque that can be transferred by the shaft in N.m

ω = **Angular velocity** in rad/s

Substituting the given values,

P = (146,929×2π×500)/(9550×(1+0.12))

= 39.52 kW

Shear Stress:

The shear stress in the shaft for the above conditions is calculated using the following formula:

τ = (T×L-F)/(Inner area×π)

Where:

F = Force applied at the **shaft **due to flywheel weight in N

Inner Area = Area of inner bore of the hollow shaft in mm²

Substituting the given values,

τ = (146,929×3-15000)/(95²×π)

= 146.9 MPa

Therefore, the **maximum torque** that can be transferred by the shaft is 146.9 MPa.

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A facultative pond system is treating a wastewater flow of 2,350 m3/d with a BOD5 of 275

mg/L. Please reference Table 93.36 from DEQ Circular 2 (also found in the NTS notes).

a) Calculate the minimum surface area required if the BOD loading rate should not exceed 25

lb/(acre*d) for the primary ponds.

b) Assuming the ponds are 6 ft deep determine hydraulic retention time of the lagoon if the

bottom 2 ft is excluded in the volume calculation?

c) Assuming the bottom 2 ft will be taken up with deposited solids (which counts toward total

depth), what total water depth range is required to maintain an acceptable residence time?

### Answers

**Loading rates** are a measurement used in waste-water treatment **systems **to assess whether the system has a propensity to clog.

Thus, There are suggested loading rates for various materials, including sands, dirt, and residential sewage. Units of measurement include hydraulic loading rates as well as organic **loading rates.** For hydraulic and loading-rate measurements, separate computations are required. A few straightforward steps can be used to calculate loading rates.

Use the following calculation to get the hydraulic **loading rate**: Design flow (gal/day) divided by area (foot 2) equals the hydraulic **loading rate. **The daily waste water volume is known as the design flow.

Use the formula to determine the **organic **loading BOD5 (mg/l) x 3.785 l/gal / 453,600 mg/lb is the formula for **organic matter.**

Thus**, Loading rates** are a measurement used in waste-water treatment **systems **to assess whether the system has a propensity to clog.

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A +4% initial grade intersects a -2% final grade, forming a 1500-ft-long equal tangent crest vertical curve. The initial grade starts at station 300+00 and elevation 1050ft. Determine the elevation and stationing of the highest point, PVI and PVT.

### Answers

The vertical point of **tangency** (PVT) and the vertical point of curvature **(PVC)** are separated from the PVI by a **horizontal distance** of L/2.

Thus, The **PVC, **which is situated on the approaching roadway segment, is typically referred to as the curve's origin.

The PVT is situated where the vertical curve joins the outgoing roadway section and serves as the vertical curve's end. In other words, the vertical curve starts and ends along the roadway at the PVT and **PVC.**

You are now prepared to design the shape of your curve after locating the PVI,** PVC, **and PVT. The following equation determines the height at each location along an equaltangent parabolic **vertical curve.**

Thus, The vertical point of **tangency** (PVT) and the vertical point of curvature **(PVC)** are separated from the PVI by a **horizontal distance** of L/2.

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Briefly express the design considerations of electrical distribution system for a hospital development.

Briefly express the design considerations of lighting system for a secondary school.

Briefly express the design considerations of lightning system for a historical building.

### Answers

Design considerations for an **electrical distribution system** in a hospital development include reliability, redundancy, capacity planning, and compliance with** safety codes **and regulations.

When designing the electrical distribution system for a hospital development, several important considerations must be taken into account. Firstly, reliability is of utmost importance to ensure **uninterrupted power supply **for critical areas such as operating rooms, intensive care units, and life support systems. Redundancy is another crucial aspect, involving the provision of** backup power sources **and redundant components to minimize the risk of power failures. Additionally, capacity planning is essential to accommodate the high energy demands of medical equipment, lighting, HVAC systems, and other electrical loads within the hospital.

Compliance with safety codes and regulations is a fundamental consideration. Hospital electrical systems must adhere to relevant standards and guidelines, such as the** National Electrical Code** (NEC) and healthcare-specific regulations. Proper grounding, isolation, and surge protection measures should be implemented to safeguard against electrical hazards and protect both patients and staff.

Furthermore, special considerations may be necessary for areas with sensitive equipment, such as MRI rooms, where electromagnetic interference must be minimized. The electrical distribution system should also facilitate** ease of maintenance** and future expansion to accommodate potential growth or technological advancements in the healthcare industry.

In summary, designing the electrical distribution system for a hospital development involves ensuring reliability, redundancy, capacity planning, compliance with safety codes, and the ability to accommodate future needs.

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In relation to triaxial strength testing of soil, define the meaning of each of the following tests: CKOUC, CIU, UU. Hence, list practical situations where each test may be specified and the design parameter(s) that are obtained from each test.

### Answers

**Triaxial strength testing **of soil involves three main tests: CKOUC, CIU, and UU. Each test is used in different practical situations to determine specific **design parameters** related to soil behavior.

Triaxial strength testing is a widely used method to evaluate the **mechanical behavior **of soil under different stress conditions. The CKOUC test, also known as Consolidated-Undrained Triaxial Compression test, is performed on a fully saturated soil specimen under constant volume conditions. This test is often specified when the soil is expected to undergo rapid and** significant stress **changes, such as in earthquake-prone areas. From the CKOUC test, important design parameters such as undrained shear strength and pore pressure response can be obtained.

The CIU test, or Consolidated-Undrained Triaxial test, is conducted on a soil specimen that is subjected to both consolidation and shearing stages. This test is suitable for situations where the soil experiences gradual stress changes, such as in **embankment construction** or foundation design. The CIU test provides information on the soil's consolidated-drained shear strength, consolidation characteristics, and pore pressure dissipation.

The UU test, or Unconsolidated-Undrained Triaxial test, is performed on an undisturbed soil specimen without any prior consolidation. This test is often specified for situations where the soil is expected to experience rapid loading or where** time-dependent** soil behavior is not a significant concern. The UU test helps determine the unconsolidated-undrained shear strength and pore pressure response of the soil.

In summary, the CKOUC, CIU, and UU tests in triaxial strength testing of soil are utilized in different practical situations based on the expected stress conditions and time-dependent behavior of the soil. These tests provide valuable design parameters such as undrained shear strength, consolidated-drained shear strength, consolidation characteristics, and pore pressure response, which are essential for various geotechnical applications.

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