All categories

3 years ago

What is ONE DIFFERENCE between civil structural engineering

Engineering

1 answer:

Rudiy273 years ago

3 0

Answer:

One of the differences is that civil engineering focuses on design elements while structural engineering is more concern on inspecting the materials used for construction. The structural engineers are the one who are supposed to ensure that the materials used for construction can support the design of the structure.

Explanation:

You might be interested in

What do you understand by the term phase angle?<br>

aleksandrvk [35]

Answer:

The angle between the earth and the sun as seen from a planet is called phase angle.

5 0

3 years ago

The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content

MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

$Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf$

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

$\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}$

Therefore, $V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}$

$V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy$

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since $1 yd^{3}= 27 ft^{3}$

The embankment requires water of 6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

$Unit_{weight}=\frac {17451720}{498594}=35.00186 lb$

1 gallon is approximately $8.35 yd^{3}$ hence

$\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}$

That's approximately 4.2 gallons

7 0

3 years ago

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago

A machine raises 20kg of water through a height of 50m in 10secs. What is the power of the machine.

Tomtit [17]

Answer:

hhahhhwghwhwhwhwjwnwjnnnnwnwwnw

Explanation:

jwkwkkwoiwiwiwiwiwowwiwowowiiiiwuuwuwgeevehehsvhsvwhbhhehehwgjjwhwhjwjqwjjuuuwi####!\\\\e

5 0

3 years ago

The pressure gage on a 2.5-m^3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank if the temperature is 28°C

s2008m [1.1K]

Answer:

19063.6051 g

Explanation:

Pressure = Atmospheric pressure + Gauge Pressure

Atmospheric pressure = 97 kPa

Gauge pressure = 500 kPa

Total pressure = 500 + 97 kPa = 597 kPa

Also, P (kPa) = 1/101.325 P(atm)

Pressure = 5.89193 atm

Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)

Temperature = 28 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K

⇒n = 595.76 moles

Molar mass of oxygen gas = 31.9988 g/mol

Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g

7 0

3 years ago