1. **Problem Statement:** Calculate the volume of earthworks for a proposed non-tilted road of width 20m between stations A and B, 100m apart, using five different methods: mean end-area, mid-area, mean-area, trapezoidal rule, and prismoidal formula. 2. **Given Data:** - Width of road, $w = 20$ m - Distance between stations, $x = 100$ m - Cross-section heights (in meters): - Station A: Left 4.0, Centre 16.0, Right 12.0 - Station B: Left 8.0, Centre 5.0, Right 11.0 - Mid-section: Left 3.0, Centre 4.5, Right 10.0 3. **Step 1: Calculate cross-sectional areas at stations A, B, and Mid-section.** Each cross-section is trapezoidal with base width $w=20$ m and side heights given. We approximate the area by dividing into two triangles and a rectangle: - Area of left triangle = $\frac{1}{2} \times 16 \times \text{left height}$ - Area of rectangle = $16 \times \text{centre height}$ - Area of right triangle = $\frac{1}{2} \times 25 \times \text{right height}$ Calculate areas: - Station A: - Left triangle = $\frac{1}{2} \times 16 \times 4.0 = 32$ m$^2$ - Rectangle = $16 \times 16.0 = 256$ m$^2$ - Right triangle = $\frac{1}{2} \times 25 \times 12.0 = 150$ m$^2$ - Total area $A_1 = 32 + 256 + 150 = 438$ m$^2$ - Station B: - Left triangle = $\frac{1}{2} \times 16 \times 8.0 = 64$ m$^2$ - Rectangle = $16 \times 5.0 = 80$ m$^2$ - Right triangle = $\frac{1}{2} \times 25 \times 11.0 = 137.5$ m$^2$ - Total area $A_2 = 64 + 80 + 137.5 = 281.5$ m$^2$ - Mid-section: - Left triangle = $\frac{1}{2} \times 16 \times 3.0 = 24$ m$^2$ - Rectangle = $16 \times 4.5 = 72$ m$^2$ - Right triangle = $\frac{1}{2} \times 25 \times 10.0 = 125$ m$^2$ - Total area $A_m = 24 + 72 + 125 = 221$ m$^2$ 4. **Step 2: Apply volume calculation methods.** - **Mean End-Area Method:** $$ V = x \times \frac{A_1 + A_2}{2} = 100 \times \frac{438 + 281.5}{2} = 100 \times 359.75 = 35975 \text{ m}^3 $$ - **Mid-Area Method:** $$ V = x \times A_m = 100 \times 221 = 22100 \text{ m}^3 $$ - **Mean-Area Method:** $$ V = \frac{x}{3} (A_1 + 4A_m + A_2) = \frac{100}{3} (438 + 4 \times 221 + 281.5) $$ Calculate inside parentheses: $$ 438 + 884 + 281.5 = 1603.5 $$ Volume: $$ V = \frac{100}{3} \times 1603.5 = 33,450 \text{ m}^3 $$ - **Trapezoidal Rule:** $$ V = \frac{x}{2} (A_1 + A_2) = \frac{100}{2} (438 + 281.5) = 50 \times 719.5 = 35975 \text{ m}^3 $$ - **Prismoidal Formula:** $$ V = \frac{x}{6} (A_1 + 4A_m + A_2) = \frac{100}{6} (438 + 4 \times 221 + 281.5) = \frac{100}{6} \times 1603.5 = 26725 \text{ m}^3 $$ 5. **Step 3: Comment on volumes:** - Mean end-area and trapezoidal rule give the same volume: 35975 m$^3$. - Mid-area method gives the smallest volume: 22100 m$^3$. - Mean-area method gives the largest volume: 33450 m$^3$. - Prismoidal formula, which is more accurate, gives 26725 m$^3$. 6. **Step 4: Compute slopes between stations for left, centre, and right sides:** Slope formula: $$ \text{slope} = \frac{\text{height difference}}{\text{distance}} $$ - Left slope: $$ \frac{8.0 - 4.0}{100} = \frac{4.0}{100} = 0.04 $$ - Centre slope: $$ \frac{5.0 - 16.0}{100} = \frac{-11.0}{100} = -0.11 $$ - Right slope: $$ \frac{11.0 - 12.0}{100} = \frac{-1.0}{100} = -0.01 $$ 7. **Step 5: Describe the project using slopes and data:** - The left side is rising gently (positive slope 0.04). - The centre is descending steeply (negative slope -0.11). - The right side is almost flat with a slight descent (-0.01). - The volume calculations indicate earthworks involving both cutting and filling. - The prismoidal formula volume suggests moderate earth movement. **Final answers:** - Volumes (m$^3$): - Mean end-area: 35975 - Mid-area: 22100 - Mean-area: 33450 - Trapezoidal: 35975 - Prismoidal: 26725 - Slopes: - Left: 0.04 - Centre: -0.11 - Right: -0.01 These results provide a comprehensive understanding of the earthwork volumes and terrain slopes for the road project.