---
title: "Well foundation of pier: PCC and steel"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:well-foundation-dia-7-pier"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/well-foundation-dia-7-pier
site: BCT Engineering
---

# Well foundation of pier: PCC and steel

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2081 Bhadra, 4 marks
- Video: [Sinking of Well | Well Foundation |DCRUST](https://www.youtube.com/watch?v=PIE1gTJUDwI) — AJ THE CIVIL
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/well-foundation-dia-7-pier
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

A well foundation for pier of a bridge is to be constructed. It has 7 meters internal diameter and 15 metres depth from river bed level to top of well curb. Well steining of 1 meter thickness is to be constructed up to 7.5 metres depth and remaining of 2 m thickness with M20 RCC. There are two plugs of M20 concrete at top and middle of 1 m thickness. Find out
i) Quantity of PCC for M20 RCC works
ii) Steel reinforcement for M20 RCC work well steining (1.5% of PCC by volume)

## Short answer

M20 concrete in steining is 612.61 m3 (upper 1 m thick 188.50 m3, lower 2 m thick 424.12 m3); adding two 1 m plugs gives 689.58 m3. Steel at 1.5% by volume is 72.13 tonnes.

## Answer

**Meaning of the data.** The well steining is 1.0 m thick for the top 7.5 m of the 15 m depth and 2.0 m thick for the remaining 7.5 m (M20 RCC). Two 1.0 m thick M20 plugs are placed at the top and middle of the well.

#### i) Quantity of concrete in the steining
Annular volume = $\frac{\pi}{4}(D_o^2-D_i^2)\times h$, with $D_i = 7$ m.

- Upper steining ($t=1$ m, $D_o=9$ m): $\frac{\pi}{4}(81-49)\times 7.5 = 188.496$ m³
- Lower steining ($t=2$ m, $D_o=11$ m): $\frac{\pi}{4}(121-49)\times 7.5 = 424.115$ m³

Total M20 concrete in steining = 188.496 + 424.115 = **612.61 m³**.

Plugs (2 nos., 1.0 m thick, 7 m dia): $2\times\frac{\pi}{4}\times 7^2\times 1.0 = 76.97$ m³ of M20 plain concrete.

Total M20 concrete including the plugs = 612.61 + 76.97 = 689.58 m³.

#### ii) Steel reinforcement in steining (1.5 % by volume)
Volume of steel = 0.015 × 612.611 = 9.189 m³.

Weight = 9.189 × 7850 kg/m³ = 72,135 kg = **72.13 tonnes**.

**Answer:** Steining concrete = 612.61 m³ (plus 76.97 m³ in plugs, 689.58 m³ in all); steel = 72.13 t.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For 7 m internal diameter, what are the outer diameters of the 1 m and 2 m steining?
   A. 8 m and 9 m
   B. 9 m and 11 m
   C. 7 m and 9 m
   D. 14 m and 18 m
   Answer: B. Outer dia = 7 + 2t: 9 m for t = 1 m and 11 m for t = 2 m. (Key point: Outer diameters)

2. Which formula gives the steining concrete of each length?
   A. pi/4 x Do^2 x h
   B. pi/4 (Do^2 - Di^2) x h
   C. pi x Di x h
   D. pi/4 x Di^2 x h
   Answer: B. Steining is a ring, so volume = area of annulus x height; total 188.496 + 424.115 = 612.61 m3. (Key point: Annular volume)

3. How is the steel weight computed for 1.5% steel in 612.61 m3?
   A. 0.015 x 612.61 = 9.189 m3, x 7850 kg/m3 = 72.13 t
   B. 0.15 x 612.61 x 7850
   C. 0.015 x 7850 only
   D. 0.015 x 612.61 in tonnes
   Answer: A. Steel volume 9.189 m3 times density 7850 kg/m3 is about 72,135 kg = 72.13 t. (Key point: Steel weight)

4. How are the two 1 m thick plugs treated?
   A. Added to steining as annulus
   B. Computed separately as solid discs of 7 m dia (76.97 m3)
   C. Ignored
   D. Computed with 11 m outer diameter
   Answer: B. Plugs fill the internal 7 m bore: 2 x pi/4 x 7^2 x 1.0 = 76.97 m3. (Key point: Plugs separate)

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