**Problem 31**

One section of the mainline sewer pipe for the city of Annapolis passes through a tunnel 484 ft. long. The cross section of the tunnel is shown in the figure. The sewer pipe is 16 in. in external diameter and is partially embedded, in a mattress of concrete, a cross section of which is shown in the figure. Find the total cost (a) of digging the tunnel at \$5.00 per cubic yard, (b) of pouring of concrete at \$9.00 per cubic yard.

**Solution 31**

**Digging:**

$L = 484 ~ \text{ft.}$

$V_d = A_dL = \frac{145}{12}(484)$

$V_d = 5848.33 ~ \text{ft.}^3 = 5848.33 ~ \text{ft.}^3 \left( \dfrac{1 ~ \text{yd.}}{3 ~ \text{ft.}} \right)^3$

$V_d = 216.60 ~ \text{yd.}^3$

Cost of Digging:

$C_d = 5(216.60) = \$1,083.02$ *answer*

**Concrete:**

$a = 6.65 ~ \text{in.}$

$b = 22 + 2a = 22 + 2(6.183) = 34.37 ~ \text{in.}$

$b = 2.864 ~ \text{ft.}$

$\cos (\frac{1}{2}\theta) = \dfrac{4}{8}$

$\theta = 120^\circ$

$A_1 = \frac{1}{2}r^2(\theta_{rad} - \sin \theta_{deg})$

$A_1 = \dfrac{1}{2}\left(\dfrac{8}{12}\right)^2\left[ 120^\circ \left( \dfrac{\pi}{180^\circ} \right) - \sin 120^\circ \right]$

$A_1 = 0.273 ~ \text{ft.}^2$

$A_c = \frac{1}{2}(3 + b)(\frac{7}{12}) - A_1 = \frac{1}{2}(3 + 2.864)(\frac{7}{12}) - 0.273$

$A_c = 1.437 ~ \text{ft.}^2$

$V_c = A_cL = 1.437(484) = 695.67 ~ \text{ft.}^3$

$V_c = 695.67 ~ \text{ft.}^3 \left( \dfrac{1 ~ \text{yd.}}{3 ~ \text{ft.}} \right)^3$

$V_c = 25.76 ~ \text{yd.}^3$

Cost of concrete pouring:

$C_c = 9.00(25.76) = \$231.89$ *answer*

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