Question

In: Civil Engineering

A highway embankment of formation width of 15 m and side slopes 1.5:1 is to be...

A highway embankment of formation width of 15 m and side slopes 1.5:1 is to be constructed. The ground level along the centerline is as follows in Table below. The embankment has a rising gradient of 1 in 200, and the formation level at zero chainage is 115.00. Assuming the ground is level across the centerline; compute the volume of earthwork.

Chainage 0 50 100 150 200 250
GL (m) 115.75 114.35 116.80 115.20 118.50 118.25

Solutions

Expert Solution

Solutions Giren dort en fonction eth (B) = 1500 Esde tape is 1.5 iting graderst every son,kice in 200 1 150 -0.25 trx for every so 57 Elis. 25 o 075 119 o in the figure above, ABFON ADFE As explamed previously, 2 of 07 - 1.2 =) 4 1.21. - 2 = 1 4 1 = 13-14 and AOF, FNAFE,E = x, typ so 3, +1.4ax, so 3. : 205 - 2 1 - 2 2.44 - Y = 1.21 3:44 - so -5.1.234)-50 3= so

118.75 116 5 0.5$ US oss 11522 115.2 As explained previously. As explained previously, y = 4.554, =) *, - 4 = 0.42X - - 5. 1x, = 35.21 1&i= am]

RL of ground RL of formation Depth mean. depth volume sd² Bd Area Length so (Botsd?) I cutting Ranking level 115.75 - 0.75 0.375 5.625 0.211 5.836 385 224.7. 38.5 50 114.35 115.25 0.9 0.304 7.054 11.5 81-12 0.45 0.45 6.75 6.75 0.304 7.054 20.5 144.6 70.5 10o 116.8 115.50 -0.65 9.75 0.634 10.384 29.5 286.2 – 1.3 0 135.2 -0.65 1.875 1.875 0.634 10.384 35.2 365.5 0.023 1898 14.8 115.2 ISO 11575 0.55 +0.125 1.875 28.) 159 +0.125 1.875 0.023 1.898 9 17 1 116.00 118.5 -2.5 200 - 1.25 18.75 2.344 21.094 41 864 9. 118.25 116.25 – 20 – 2.25 33.75 7.594 41:344 so 2,067-2 250


Related Solutions

)An embankment of width 15 m and side slopes 2:1 is required to make on a...
)An embankment of width 15 m and side slopes 2:1 is required to make on a ground, which is level in a direction transverse to the centerline. The central heights at 50 m intervals are as follows: 1.70, 2.25, 3.15, 4.50, 2.75, 2.15, and 1.95. Calculate the volume of earthwork according to Trapezoidal formula, and prismoidal formula.
A road of formation width 12 m is to be constructed with side slopes of 1...
A road of formation width 12 m is to be constructed with side slopes of 1 vertical to 2 horizontal in cut and 1 vertical to 3 horizontal in fill. The existing ground surface has a cross fall of 1 vertical to 8 horizontal and it will intersect the formation 1.5 m to the left of the center line and 1.0 m to the right of the center line at two cross sections 20 m apart. Calculate the areas related...
B. A trapezoidal channel with a bottom width of 4 m and side slopes 4H: 1V...
B. A trapezoidal channel with a bottom width of 4 m and side slopes 4H: 1V carries a discharge of 30 m3 /s. The channel has bed slope= 0.1 % and manning roughness coefficient = 0.015. Determine the following: i. the minimum specific energy ii. the normal depth iii. the critical slope iv. state of the flow in the channel
Compute the normal depth (y) in a trapezoidal channel having a bottom-width 10 m, side slopes...
Compute the normal depth (y) in a trapezoidal channel having a bottom-width 10 m, side slopes 1H to 1V and carrying a flow of 30 m3/s. The slope of the channel bottom is So = 0.001 and n = 0.013. A. B. C. D.  
A trapezoidal channel having bottom width 6m ,side slopes 2 horizontal to 1 vertical ,Mannings n=...
A trapezoidal channel having bottom width 6m ,side slopes 2 horizontal to 1 vertical ,Mannings n= 0.025 and bottom slope 0.016 ,carries a discharge of 10 m3/sec . Compute the back water profile created by the dam which backs up the water to a depth of 2.0 m immediately behind the dam ((Direct method if it is possible))
A trapezoidal channel having bottom width 6m ,side slopes 2 horizontal to 1 vertical ,Manning’s n=...
A trapezoidal channel having bottom width 6m ,side slopes 2 horizontal to 1 vertical ,Manning’s n= 0.025 and bottom slope 0.016 ,carries a discharge of 10 m3/sec . Compute the back water profile created by the dam which backs up the water to a depth of (3) immediately behind the dam.                                                                [ 5 Marks ] ( THIS SHOULD BE DONE USING EXCEL SHEET )
Heat Q5 A vertical flat plate (1.5 m) height and (1 m) width ,with uniform temperature...
Heat Q5 A vertical flat plate (1.5 m) height and (1 m) width ,with uniform temperature (120 oC) one face of this plate expressed to the air with the velocity (6 m/s) to the up direction, the second face expressed to the static air ,the temperature of the air with two side (30oC) . Calculate the heat losses from the plate. The properties of the air at (75 oC),[ ρ=0.998 kg/m3 .Cp=1.009 kJ/kg.oC, υ =2.076 *10-5 m 2 /s ,...
In a rectangular cross-section canal with a width of 10 m and a depth of 1.5...
In a rectangular cross-section canal with a width of 10 m and a depth of 1.5 m, uniform and Continuous current monitoring is applied. The channel's Manning coefficient includes 0.015. Base because its slope is 0.0005. Considering this information a) By writing the numerical magnitude of water face slope and energy line slope and its rationale Set. b) Calculate the speed of the current in the channel. c) Calculate the flow of the current in the channel. d ) When...
A wall in a building has a width of 15 m and a height of 3...
A wall in a building has a width of 15 m and a height of 3 m. It is made of several layers — a layer of motionless air with an R value of 0.12 K m^2 / W, a 12.4 mm thick layer of gypsum board with a thermal conductivity of 0.17 W/m K, a layer of brick with an R value of 1.7 m^2/ W, and another layer of motionless air with an R value of 0.12 K...
A trapezoidal channel has size slopes of 3:1 (H:V), a bottom width of 22 feet, a...
A trapezoidal channel has size slopes of 3:1 (H:V), a bottom width of 22 feet, a longitudinal slope of 0.0003, a Manning’s roughness coefficient of 0.023, and carries a flow rate of 340 cfs. a) Calculate the normal depth in this channel. b) Calculate the normal specific energy. c) Calculate the Froude number. State the classification of flow. d) State how the critical depth will compare to the normal depth and why.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT