Question

1. how many tons of coarse aggregate will be required to fill a trench 5 ft deep, 3 ft wide, and 300 ft long? the coarse aggregate unit weight equals 93.6 lb per cu ft.

Explanation:

Step1: Calculate the volume of the trench

The volume \(V\) of a rectangular prism (trench) is given by \(V = l\times w\times h\), where \(l = 300\) ft (length), \(w = 3\) ft (width), and \(h=5\) ft (depth).

$$V=300\times3\times5 = 4500\space cu.ft$$

Step2: Calculate the total weight of the aggregate

The unit - weight of the aggregate is \(93.6\space lb/cu.ft\). The total weight \(W\) is \(W=\text{Unit weight}\times V\).

$$W = 93.6\times4500=421200\space lb$$

Step3: Convert pounds to tons

Since \(1\) ton \( = 2000\) lb. Let \(T\) be the weight in tons. Then \(T=\frac{W}{2000}\)

$$T=\frac{421200}{2000}=210.6\space tons$$

Answer:

\(210.6\) tons