1. A cutting is to be formed as part of a proposed road having a formation width of 7.30 m and side slopes of $$ 1 \mathrm{in} 2 . \mathrm{A} $$ cross-section is to be taken at chainage 2050 m , where the depth of dig to the proposed formation level is 10.89 m and the existing ground level is horizontal at right-angles to the proposed center line. Calculate the plan width and the area of cut required at this cross-section. 2. 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 center line is 1 in 8 and the existing ground levels at chainages $$ 1750 \mathrm{~m}, 1800 \mathrm{~m} $$ and 1850 m are 176.32 $$ \mathrm{m}, 175.18 \mathrm{~m} $$ and 174.87 m , respectively. The reduced level of the formation center 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 .

Question

UpStudy AI · Accepted Answer

To solve the problems, we will break them down step by step.

### Problem 1

**Given:**
- Formation width = 7.30 m
- Side slopes = 1 in 2
- Depth of dig = 10.89 m
- Existing ground level is horizontal

**Calculating Plan Width:**
The plan width of the cutting can be calculated by considering the formation width and the side slopes.

1. **Calculate the horizontal distance from the center line to the toe of the slope:**
   - For a side slope of 1 in 2, for every 2 units of vertical depth, there is 1 unit of horizontal distance.
   - Therefore, for a depth of 10.89 m:
     $$
     \text{Horizontal distance} = \frac{10.89 \, \text{m}}{2} = 5.445 \, \text{m}
     $$

2. **Total plan width:**
   - The total plan width is the formation width plus twice the horizontal distance (once for each side):
   $$
   \text{Plan Width} = \text{Formation Width} + 2 \times \text{Horizontal Distance}
   $$
   $$
   \text{Plan Width} = 7.30 \, \text{m} + 2 \times 5.445 \, \text{m} = 7.30 \, \text{m} + 10.89 \, \text{m} = 18.19 \, \text{m}
   $$

**Calculating Area of Cut:**
The area of the cut can be calculated as a trapezoid:
$$
\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
$$
Where:
- Base_1 = Formation Width = 7.30 m
- Base_2 = Plan Width = 18.19 m
- Height = Depth of Dig = 10.89 m

$$
\text{Area} = \frac{1}{2} \times (7.30 + 18.19) \times 10.89
$$
$$
\text{Area} = \frac{1}{2} \times 25.49 \times 10.89 \approx 138.56 \, \text{m}^2
$$

### Problem 2

**Given:**
- Formation width = 10.00 m
- Side slopes = 1 in 3
- Existing ground levels:
  - Chainage 1750 m: 176.32 m
  - Chainage 1800 m: 175.18 m
  - Chainage 1850 m: 174.87 m
- Reduced level of formation center line at chainage 1750 m = 173.65 m
- Falling gradient = 4% from chainage 1750 m to 1850 m

**Calculating Reduced Levels:**
1. **Calculate the reduced level of the formation center line at chainage 1800 m:**
   - The change in elevation over 50 m (from 1750 m to 1800 m) with a 4% gradient:
   $$
   \text{Change in elevation} = 0.04 \times 50 = 2.00 \, \text{m
   }
   $$
   $$
   \text{Reduced level at 1800 m} = 173.65 - 2.00 = 171.65 \, \text{m}
   $$

2. **Calculate the reduced level of the formation center line at chainage 1850 m:**
   - The change in elevation over another 50 m:
   $$
   \text{Change in elevation} = 0.04 \times 50 = 2.00 \, \text{m}
   $$
   $$
   \text{Reduced level at 1850 m} = 171.65 - 2.00 = 169.65 \, \text{m}
   $$

**Calculating Depth of Cut:**
1. **Depth of cut at chainage 1750 m:**
   $$
   \text{Depth} = \text{Existing Ground Level} - \text{Formation Level} = 176.32 - 173.65 = 2.67 \, \text{m}
   $$

2. **Depth of cut at chainage 1800 m:**
   $$
   \text{Depth} = 175.18 - 171.65 = 3.53 \, \text{m}
   $$

3. **Depth of cut at chainage 1850 m:**
   $$
   \text{Depth} = 174.87 - 169.65 = 5.22 \, \text{m}
   $$

**Calculating Area of Cut:**
The area of cut can be calculated for each section (1750 m to 1800 m and 1800 m to 1850 m) and then summed.

1. **Area from 1750 m to 1800 m:**
   $$
   \text{Average Depth} = \frac{2.67 + 3.53}{2} = 3.10 \, \text{m}
   $$
   $$
   \text{Area} = \text{Formation Width} \times \text{Average Depth} = 10.00 \times 3.10 = 31.00 \, \text{m}^2
   $$

2. **Area from 1800 m to 1850 m:**
   $$
   \text{Average Depth} = \frac{3.53 + 5.22}{2} = 4.375 \, \text{m}
   $$
   $$
   \text{Area} = \text{Formation Width} \times \text{Average Depth} = 10.00 \times 4.375 = 43.75 \, \text{m}^2
   $$

**Total Area of Cut:**
$$
\text{Total Area} = 31.00 + 43.75 = 74.75 \, \text{m}^2
$$

**Calculating Volume of Cut:**
The volume of cut can be calculated by multiplying the area by the length of the sections.

1. **Volume from 1750 m to 1800 m:**
   $$
   \text{Volume} = \text{Area} \times \text{Length} = 31.00 \times 50 = 1550 \, \text{m}^3
   $$

2. **Volume from 1800 m to 1850 m:**
   $$
   \text{Volume} = 43.75 \times 50 = 2187.5 \, \text{m}^3
   $$

**Total Volume of Cut:**
$$
\text{Total Volume} = 1550 + 2187.5 = 3737.5 \, \text{m}^3
$$

### Summary of Results
1. **Problem 1:**
   - Plan Width: 18.19 m
   - Area of Cut: 138.56 m²

2. **Problem 2:**
   - Total Volume of Cut: 3737.5 m³
1. Plan Width: 18.19 m, Area of Cut: 138.56 m² 2. Total Volume of Cut: 3737.5 m³

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Engineering Questions