Question Bank • QB_SWCE_ADV_141 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Surveying and Levelling

Subtopic: Computation of areas and volume

Question Statement:

Earthwork volume for a drainage channel reach is computed from five consecutive parallel cross-sections taken at uniform intervals of $d = 20.0\text{ m}$. The measured cross-sectional areas are: $A_1 = 45.0\text{ m}^2$, $A_2 = 72.0\text{ m}^2$, $A_3 = 95.0\text{ m}^2$, $A_4 = 80.0\text{ m}^2$, and $A_5 = 50.0\text{ m}^2$. Using Simpson's one-third (Prismoidal) rule, the total earthwork volume in cubic meters ($\text{m}^3$) is ________ (round off to nearest integer).

Answer Key:

5953

Step-by-Step Solution & Explanation

**Method 1: Simpson's One-Third Prismoidal Rule** $V = \frac{d}{3} [ (A_1 + A_5) + 4(A_2 + A_4) + 2(A_3) ]$ Given: - $d = 20.0\text{ m}$ - First and last areas: $A_1 + A_5 = 45.0 + 50.0 = 95.0\text{ m}^2$ - Even-indexed areas: $A_2 + A_4 = 72.0 + 80.0 = 152.0\text{ m}^2 \implies 4(152.0) = 608.0\text{ m}^2$ - Odd-indexed internal areas: $A_3 = 95.0\text{ m}^2 \implies 2(95.0) = 190.0\text{ m}^2$ Sum of bracketed terms: $\text{Sum} = 95.0 + 608.0 + 190.0 = 893.0\text{ m}^2$ $V = \frac{20.0}{3} \times 893.0 = \frac{17860.0}{3} = 5953.33\text{ m}^3 \approx 5953\text{ m}^3$ **Method 2: Trapezoidal Volume Comparison** $V_{trap} = 20.0 \times \left[ \frac{45 + 50}{2} + 72 + 95 + 80 \right] = 20.0 \times [ 47.5 + 247.0 ] = 20.0 \times 294.5 = 5890.0\text{ m}^3$ Simpson's rule yields $5953\text{ m}^3$.
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