A roadway is to be designed on a level terrain. The roadway id 500 ft. Five cross-sections have been selected at 0 ft, 125 ft, 250 ft, 375 ft, and 500 ft. the cross sections have areas of 130 ft^2, 140 ft^2, 60 ft^2, 110 ft^2, and 120 ft^2. What is the volume needed along this road assuming a 6% shrinkage?

51112.5 ft^3

Explanation:

Determine the volume needed along the road when we assume a 6% shrinkage

shrinkage factor = 1 - shrinkage  = 1 - 0.06 =  0.94

first we have to calculate the volume between the cross sectional areas (i.e. A1 A5 ) using average end area method

Volume between A1 - A2

= (125 ft - 0 ft) * [(130 ft^2 + 140 ft^2) / 2]

 = 125 ft * 135 ft^2

= 16875 ft^3

Volume between A2 - A3

= (250 ft - 125 ft) * [(140 ft^2 + 60 ft^2) / 2]

= 125 ft * (200 ft^2 / 2)

= 12500 ft^3

Volume between A3 - A4

= (375 ft - 250 ft) * [(60 ft^2 + 110 ft^2) / 2]

= 125 ft * (170 ft^2 / 2)

= 10625 ft^3

Volume between A4 - A5

(500 ft - 375 ft) * [(110 ft^2 + 120 ft^2) / 2]

 = 125 ft * 115 ft^2

= 14375 ft3

Hence the total volume along the 500 ft road

= ∑ volumes between cross sectional areas

=  16875 ft^3 + 12500 ft^3 + 10625 ft^3 + 14375 ft^3 = 54375 ft^3

Finally the volume needed along this road is calculated as

Total volume * shrinkage factor

= 54375 * 0.94  = 51112.5 ft^3

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answer:

rawr

explanation: