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

Ally Wells answered 2 months ago

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