What does the prismoidal (Simpson's) rule require of the sections?

An odd number of sections, i.e. an even number of equal intervals

In V = (h/3)[A0 + 4(A1 + A3) + 2A2 + A4], which areas get weight 4?

The odd-numbered intermediate sections

The even-numbered intermediate sections

2075 Asoj · 6 marks

Canal earthwork by prismoidal method

Short answer

All sections are in cutting with bottom width 6 m, side slope 1:1. Prismoidal (Simpson's) rule is applied to five sections at 50 m intervals. Volume of earthwork = 2,650.00 m³.

Find the quantity of earth work of irrigation canal using prismoidal method from the following data:

Distance (m)	0	50	100	150	200
RL of Ground (m)	100.00	101.00	101.00	99.00	100.00
RL of Formation (m)	99.50	99.00	89.50	89.00	88.50

Formation bottom width of canal is 6 meter and side slope 1:1. (RL of formation at 100 m and 150 m as printed.)

Given data and assumptions

Bottom width $b = 6$ m; side slope $1:1$ ; the canal is in cutting at every station. Depth $d$ = GL − formation level.
The printed formation levels at 100 m, 150 m and 200 m (89.50, 89.00, 88.50) are not consistent with the first two (99.50, 99.00), so the pattern of a 0.5 m fall per 50 m is continued: 99.50, 99.00, 98.50, 98.00, 97.50 m.

Sectional areas

$A = (b + s\,d)\,d = (6 + d)\,d$

Distance (m)	GL (m)	Formation RL (m)	Depth $d$ (m)	Area (m²)
0	100.00	99.50	0.50	3.250
50	101.00	99.00	2.00	16.000
100	101.00	98.50	2.50	21.250
150	99.00	98.00	1.00	7.000
200	100.00	97.50	2.50	21.250

Volume by the prismoidal rule

There are 5 sections (4 equal intervals of $h = 50$ m, an even number), so Simpson's rule applies:

\begin{aligned} V &= \frac{h}{3}\left[A_0 + 4(A_1 + A_3) + 2A_2 + A_4\right]\\ &= \frac{50}{3}\left[3.250 + 4(16.000 + 7.000) + 2(21.250) + 21.250\right]\\ &= \frac{50}{3} \times 159.00 = 2650.00\ \text{m}^3 \end{aligned}

Answer: volume of earthwork (prismoidal rule) = 2,650.00 m³. (For comparison, the mean-area method gives 2,825.00 m³.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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