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Burka [1]
3 years ago
9

An equal-tangent vertical curve is to be constructed between grades of -2% (initial) and 1% (final). The PVI is at station 110 0

0 and at elevation 420 ft. Due to a street crossing the roadway, the elevation of the roadway at station 112 00 must be at 424.5 ft. Design the curve, and determine the elevations and stations for PVC and PVI.

Engineering
1 answer:
Mila [183]3 years ago
4 0

Answer:

The curve length (<em>L</em>) will be = 1218 ft

The elevations and stations for PVC and PVI

a. station of PVC = 103 + 91.00

b. station of PVI = 116 + 09.00

c. elevation of PVC = 432.18ft

d. elevation of PVI = 426.09ft

Explanation:

First calculate for the length (<em>L</em>)

To calculate the length, use the formula of "elevation at any point".

where, elevation at any point = 424.5.

and ∴ PVC Elevation = (420 + 0.01L)

Then, calculate for Station of PVC and PVI and elevation of PVC and PVI

You might be interested in
A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

v₂² = 88.88

v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

hence , actual velocity will be equal to 8.39 m/s

6 0
3 years ago
The following are the results of a sieve analysis. U.S. sieve no. Mass of soil retained (g) 4 0 10 18.5 20 53.2 40 90.5 60 81.8
il63 [147K]

Answer:

a.)

US Sieve no.                         % finer (C₅ )

4                                                  100

10                                                95.61

20                                               82.98

40                                               61.50

60                                               42.08

100                                              20.19

200                                              6.3

Pan                                               0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no.             Mass of soil retained (C₂ )

4                                            0

10                                          18.5

20                                         53.2

40                                         90.5

60                                         81.8

100                                        92.2

200                                       58.5

Pan                                        26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂×\frac{100}{w}

∴ we get

US Sieve no.               % retained (C₃ )               Cummulative % retained (C₄)

4                                            0                                           0

10                                          4.39                                      4.39

20                                         12.63                                     17.02

40                                         21.48                                     38.50

60                                         19.42                                     57.92

100                                        21.89                                     79.81

200                                       13.89                                     93.70

Pan                                        6.30                                      100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no.               Cummulative % retained (C₄)          % finer (C₅ )

4                                                     0                                          100

10                                                  4.39                                      95.61

20                                                 17.02                                     82.98

40                                                 38.50                                    61.50

60                                                 57.92                                    42.08

100                                                79.81                                     20.19

200                                                93.70                                   6.3

Pan                                                 100                                        0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = \frac{D60}{D10}

⇒ Cu = \frac{0.4}{0.12} = 3.33

d.)

Coefficient of Graduation = Cc = \frac{D30^{2}}{D10 . D60}

⇒ Cc = \frac{0.22^{2}}{(0.4) . (0.12)} = 1

3 0
2 years ago
Select the correct answer.
Brilliant_brown [7]
The answer is either b or e search it up and you will know the answer by definition
4 0
3 years ago
Read 2 more answers
A spherical container made of steel has 20 ft outer diameter and wal thickness of 1/2 inch. Knowing the internal pressure is 50
anastassius [24]

Answer:

maximum normal stress = 5975 psi

maximum shear stress = 2987.50 psi

Explanation:

Given data

dia = 20 ft

wall thickness = 1/2 inch

internal pressure  = 50 psi

To find out

the maximum normal stress and the maximum shearing stress

Solution

By the Mohr's circle we will find out shear stress

first we calculate inner radius

i.e. r = (diameter/2) - t

r = (20 × 12 in )/2 - ( 1/2 )

r =  120 - 0.5 = 119.5 inch

Now we find out maximum normal stress by given formula

normal stress = ( internal pressure× r ) / 2 t

normal stress = ( 50×119.5 ) / 2 × 0.5

maximum normal stress = 5975 psi

and minimum normal stress is 0, due to very small radius

and maximum shear stress will be

shear stress = ( maximum normal stress - minimum normal stress ) / 2

shear stress = ( 5975- 0 ) / 2

maximum shear stress = 2987.50 psi

5 0
3 years ago
Describe the cartesain coordinate system.
olga nikolaevna [1]

Answer:

Cartesian coordinate system is used to specify any point on a plane.

In two dimensional plane,the two types of axes or coordinates are x and y axis.

Explanation:

Cartesian coordinate system specifies each point and every point on axes.

A Cartesian Coordinate system in two dimension also named as rectangular coordinate system.

The two types of axes or coordinates are x and y axis.

The horizontal axis is x-axis and vertical axis is named as y-axis.

The coordinate system specifies each point as a set of numerical coordinates in a plane which are signed from negative to positive. that is from  (-\infty,\infty).

In three dimensional plane, x, y and z coordinates are used and in two dimensional plane, x and ycoordinates are used to address any point in the interval (-\infty,\infty).

For defining both the coordinates, the two perpendicular directed lines x- axis and x-axis are drawn.

Now, for example (3,4) is a point in which indicates that the value of x is 3 and y is 4.

It is drawn by moving 3 units right from the origin to positive x axis and 4 units upwards from the origin (0,0) to positive y-axis.

5 0
3 years ago
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