End Area (ft2) Volume (yd) Volume (yd) Net Volume (yd) Station Mass Shrinkage Total Cut Fill Cut Fill Cut Fill Diagram (12%) Fill Ordinate 0+00 400 3+00 600 152 5556 843 101 945 4611 4611 6+00 500 93 9+00 253 20 The mass ordinate at sta 6+00 is: 9,200 O 6,111 O 4,589 O -4,589

Calculate the end areas for depths of fill from 0 to 20 ft using increments of 2 ft for level sections, a 26-ft-wide level roadbed with side slopes of 1-1/2:1. Part I What is the end area for the depth of 16 ft? Express your answer in feet squared as an integer. End Area = Part J 197| ΑΣΦ | 11 End Area = vec What is the end area for the depth of 18 ft? Express your answer in feet squared as an integer. [Π| ΑΣΦ ↓↑ vec ? ? ft² ft²

Complete the earth work calculation sheet here and plot the resulting mass diagram. Divide ccy by 0.9 to convert to bcy Station 0 + 00 1 + 00 2 +00 3 +00 4 + 00 4 + 50 5 +00 6 +00 6 +50 7 +00 8 +00 9 +00 10+ 00 End- End- area cut (sf) 0 0 0 136 206 256 179 123 98 51 0 0 0 area Volume fill of cut (sf) (bcy) 0 87 162 0 0 3 15 185 225 300 Volume Strip Strip Total Total Adj. of fill cut fill fill Sum (ccy) (bcy) (bcy) fill (ccy) (ccy) (bcy) (bcy) DORMCORES SOO 13 37 56 20 38 16 23 39 22 0 0 0 0 1 5 20 69 73 cut (bcy) Mass ordinate (bcy)

Exercise 1 Using Simpson's Rule, calculate the volume of cut required for a road construction for a national road 200m long, 15m wide with sides slope of 1:2 given the following table. Stake Value/1 X-section | 2 3 5 8 NGL 81.30 82.50 83.80 84.10 87.25 85.80 84.60 83.50 83.00 Formation 80.00 80.20 80.40 80.60 80.80 81.20 81.40 81.60 81.80 200 Level Chainage 100 125 150 175 225 250 275 300 Cut Fill Using the cut/fill you got from the table above fill in the table below. X-Section Depth Overall width Average width X- sectional Multiplier Extended Area Area 1 2 3 4 6. 7 Total Volume = ?