---
title: "Load bearing building: excavation, brick, plaster"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:building-g-load-bearing-23"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/building-g-load-bearing-23
site: BCT Engineering
---

# Load bearing building: excavation, brick, plaster

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2079 Bhadra, 3+4+3 marks
- Video: [Estimation of One Room - 2023 || Long wall and Short wall method || @CivilConceptandPlans](https://www.youtube.com/watch?v=bqrq47VnFaE) — Civil Concept and Plans
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/building-g-load-bearing-23
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Find the following quantities for the following items of works from the attached building drawing. The building is a load bearing structure with 23 cm wall all around. A non-load bearing wall of thickness 11 cm divides the living room and the bathroom. The half brick thick wall is not connected to the foundation.
a) Earthwork in excavation in foundation
b) First class brick work in cement sand mortar (1:4) in super structure.
c) 12 mm thick cement plaster in 1:4 c/s mortar in inner (opening schedule door: 2 nos 1.2 × 2.1 m, window 3 nos 1 × 1.1 m)
[Building drawing not included in the scanned paper]

## Short answer

For the assumed 5.0 m × 3.0 m building, excavation for 23 cm walls is 15.23 m³, first-class brickwork in 1:4 is 11.05 m³, and 12 mm inner plaster is 54.48 m².

## Answer

No drawing is given, so a layout consistent with the data is assumed: the building is 5.0 m × 3.0 m clear inside (outside 5.46 × 3.46 m), with 230 mm load-bearing walls all around. The 110 mm half-brick partition, placed across the 3.0 m width, divides a living room (3.5 × 3.0 m) from a bathroom (5.0 − 3.5 − 0.11 = 1.39 m, so 1.39 × 3.0 m). The partition is **not connected to the foundation**, so it has no trench.

#### Assumed data
- Centre-line length of 230 mm walls = 2[(5.0 + 0.23) + (3.0 + 0.23)] = 16.92 m.
- Trench 0.90 m wide × 1.00 m deep; superstructure height 3.0 m (plinth to slab soffit).
- Openings: 2 doors 1.2 × 2.1 m (one external, one in the partition), 3 windows 1.0 × 1.1 m (external).

#### a) Earthwork in excavation in foundation
| Item | No. | L (m) | B (m) | H/D (m) | Qty (m³) |
|---|---|---|---|---|---|
| Trench for 230 mm walls only | 1 | 16.920 | 0.900 | 1.000 | 15.228 |
| **Total** | | | | | **15.228** |

#### b) First class brickwork in 1:4 c/s mortar, superstructure
| Item | No. | L (m) | B (m) | H/D (m) | Qty (m³) |
|---|---|---|---|---|---|
| Outer wall 0.23 × 3.0 | 1 | 16.920 | 0.230 | 3.000 | 11.675 |
| Partition 0.11 (3.0 long × 3.0 high) | 1 | 3.000 | 0.110 | 3.000 | 0.990 |
| Deduct external door 1.2×2.1 | -1 | 1.200 | 0.230 | 2.100 | -0.580 |
| Deduct partition door 1.2×2.1 | -1 | 1.200 | 0.110 | 2.100 | -0.277 |
| Deduct windows 1.0×1.1 (3 nos.) | -3 | 1.000 | 0.230 | 1.100 | -0.759 |
| **Total** | | | | | **11.049** |

#### c) 12 mm thick 1:4 inner plaster
Inner faces only (rooms, including both faces of the partition):

| Item | No. | L (m) | B (m) | H/D (m) | Qty (m²) |
|---|---|---|---|---|---|
| Living room 3.5 × 3.0, inside perimeter | 1 | 13.000 | 3.000 | – | 39.000 |
| Bathroom 1.39 × 3.0, inside perimeter | 1 | 8.780 | 3.000 | – | 26.340 |
| Deduct external door (living room face) | -1 | 1.200 | 2.100 | – | -2.520 |
| Deduct partition door (both faces) | -2 | 1.200 | 2.100 | – | -5.040 |
| Deduct windows (3 nos.) | -3 | 1.000 | 1.100 | – | -3.300 |
| **Total** | | | | | **54.480** |

**Answer:** Excavation 15.23 m³; superstructure brickwork 11.05 m³; inner plaster 54.48 m² (assumed layout).

## Check yourself

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

1. Why does the trench cover only the 230 mm walls (16.92 m) and not the 110 mm partition?
   A. The partition is shorter than 3.5 m
   B. The half-brick partition is not connected to the foundation, so it needs no trench
   C. Partition walls are always measured in the brickwork item only, never in excavation
   D. The trench width is too small for a 110 mm wall
   Answer: B. Because the non-load-bearing half-brick wall rests on the floor and is not connected to the foundation, no excavation is required under it. (Key point: Partition not in foundation)

2. How is the centre-line length 16.92 m of the 230 mm walls obtained for a 5.0 m x 3.0 m clear room?
   A. 2[(5.0 + 0.23) + (3.0 + 0.23)]
   B. 2(5.0 + 3.0)
   C. 2[(5.0 + 0.46) + (3.0 + 0.46)]
   D. 2(5.0 + 3.0) + 3.0 for the partition
   Answer: A. Centre-line length per side = clear length + one wall thickness (half on each side), so 2 x (5.23 + 3.23) = 16.92 m. (Key point: Centre-line of 230 mm walls)

3. In the brickwork, what thickness is used when deducting the door in the partition?
   A. 0.23 m, as for all walls
   B. 0.34 m, the sum of both walls
   C. 0.11 m, the partition thickness (deduction 1.2 x 0.11 x 2.1 = 0.277 m3)
   D. No deduction is made because the partition is non-load-bearing
   Answer: C. An opening is deducted using the thickness of the wall it is in; the partition is 110 mm thick, so the door deduction is 0.277 m3. (Key point: Thickness used for partition deductions)

4. Why is the partition door deducted twice (-2 nos.) in the inner plaster calculation?
   A. Because the door is 1.2 m wide, twice the normal width
   B. The partition is plastered on both faces, so the door opening removes plaster area from both sides
   C. Because two doors exist in the building
   D. Because plaster is applied in two coats
   Answer: B. Both faces of the partition lie inside rooms and are plastered, so the opening is deducted once from each face, while the external door is deducted once. (Key point: Plaster on inner faces)

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