4. A cutting is to be formed on a section of a proposed road of formation width 10.00 m . The transverse slope at right-angles to the proposed centre line is 1 in 8 and the existing ground levels at chainages $$ 1750 \mathrm{~m}, 1800 \mathrm{~m} $$ and 1850 m are $$ 176.32 \mathrm{~m}, 175.18 $$ m and 174.87 m , respectively. The reduced level of the formation centre line at chainage 1750 m is 173.65 m and the formation is to have a falling gradient of $$ 4 \% $$ from chainage 1750 m to chainage 1850 m . The side slopes are to be 1 in 3. Calculate the volume of cut required to form the cutting between chainages 1750 m and 1850 m .

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To calculate the volume of cut required to form the cutting between chainages 1750 m and 1850 m, we will follow these steps:

1. **Determine the formation levels at chainages 1750 m, 1800 m, and 1850 m.**
2. **Calculate the existing ground levels at these chainages.**
3. **Calculate the depth of cut at each chainage.**
4. **Determine the cross-sectional area of the cut.**
5. **Calculate the volume of cut using the average end area method.**

### Step 1: Determine the Formation Levels

The formation level at chainage 1750 m is given as 173.65 m. The formation has a falling gradient of 4% (0.04) from chainage 1750 m to chainage 1850 m.

The change in level over the distance of 100 m (from 1750 m to 1850 m) is:
$$
\text{Change in level} = \text{Gradient} \times \text{Distance} = 0.04 \times 100 \, \text{m} = 4 \, \text{m}
$$

Thus, the formation level at chainage 1850 m is:
$$
\text{Formation level at 1850 m} = 173.65 \, \text{m} - 4 \, \text{m} = 169.65 \, \text{m}
$$

### Step 2: Existing Ground Levels

The existing ground levels are given as:
- At chainage 1750 m: 176.32 m
- At chainage 1800 m: 175.18 m
- At chainage 1850 m: 174.87 m

### Step 3: Calculate Depth of Cut

The depth of cut at each chainage is calculated as:
$$
\text{Depth of cut} = \text{Existing ground level} - \text{Formation level}
$$

- At chainage 1750 m:
$$
\text{Depth of cut} = 176.32 \, \text{m} - 173.65 \, \text{m} = 2.67 \, \text{m}
$$

- At chainage 1800 m:
$$
\text{Depth of cut} = 175.18 \, \text{m} - \text{Formation level at 1800 m}
$$
$$
\text{Formation level at 1800 m} = 173.65 \, \text{m} - (0.04 \times 50) = 173.65 \, \text{m} - 2 \, \text{m} = 171.65 \, \text{m}
$$
$$
\text{Depth of cut} = 175.18 \, \text{m} - 171.65 \, \text{m} = 3.53 \, \text{m}
$$

- At chainage 1850 m:
$$
\text{Depth of cut} = 174.87 \, \text{m} - 169.65 \, \text{m} = 5.22 \, \text{m}
$$

### Step 4: Cross-Sectional Area of the Cut

The cutting has a formation width of 10.00 m and side slopes of 1 in 3. The width of the cut at the top can be calculated using the depth of cut.

The width of the cut at the top (W) can be calculated as:
$$
W = \text{Formation width} + 2 \times \left(\text{Depth of cut} \times \text{Side slope ratio}\right)
$$

For each chainage:
- At chainage 1750 m:
$$
W = 10 + 2 \times (2.67 \times 3) = 10 + 16.02 = 26.02 \, \text{m}
$$
$$
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 10 \times 2.67 = 13.35 \, \text{m}^2
$$

- At chainage 1800 m:
$$
W = 10 + 2 \times (3.53 \times 3) = 10 + 21.18 = 31.18 \, \text{m}
$$
$$
\text{Area} = \frac{1}{2} \times 10 \times 3.53 = 17.65 \, \text{m}^2
$$

- At chainage 1850 m:
$$
W = 10 + 2 \times (5.22 \times 3) = 10 + 31.32 = 41.32 \, \text{m}
$$
$$
\text{Area} = \frac{1}{2} \times 10 \times 5.22 = 26.10 \, \text{m}^2
$$

### Step 5: Calculate the Volume of Cut

Using the average end area method, the volume of cut (V) can be calculated as:
$$
V = \frac{L}{2} \times (A_1 + A_2)
$$
where $$ L $$ is the length between chainages (100 m), $$ A_1 $$ is the area at the first chainage, and $$ A_2 $$ is the area at the second chainage.

Calculating the volume:
$$
V = \frac{100}{2} \times (13.35 + 17.65 + 26.10) = 50 \times 57.10 = 2855 \, \text{m}^3
$$

Thus, the total volume of cut required to form the cutting between chainages 1750 m and 1850 m is approximately **2855 m³**.
The volume of cut required is approximately 2855 m³.

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