---
title: "Building: excavation, PCC, brick, wood work"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:building-f-fig3-four-items"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/building-f-fig3-four-items
site: BCT Engineering
---

# Building: excavation, PCC, brick, wood work

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2080 Baisakh, 3+2+4+3 marks
- Video: [What is T-Type Wall? | Calculate Excavation, PCC, Brick Work & DPC  | Civil Engineering with Tariq](https://www.youtube.com/watch?v=4ES93VyV2EY) — Civil Engineering with Tariq
- 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-f-fig3-four-items
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Prepare detailed estimate of the following items of works from the given building drawing attached fig 3.
a) Earth work in excavation in foundation
b) Plain cement concrete (1:3:6) in foundation.
c) Brickwork in 1:6 C.S. in super structure wall.
d) Wood work for door and window frame.
[Fig 3: plan, overall length 7250 mm (1265 + 1200 + 2300 + 1200 + 1265), width 3730 mm with internal 3500 mm, 230 mm walls, internal clear 3270 mm each room, partition with door P1, door D 1200 × 2100, D1 1000 × 2100, window W 1500 × 1200, W1 1200 × 1200 (as printed '15400×1200'). Section X-X: 600 mm roof projection, 18 mm screed over 2 layers of tarfelt, roof slab, window 3270 wide, sill/lintel levels 2100 and 2400, DPC, plinth with 3 cm cement sand (1:1) pointing over 36 cement concrete (1:2:4), 75 thick brick bats soling, 150 sand filling, foundation PCC (1:3:6) 770 mm wide with 150/250/150 mm layers.]

## Short answer

Centre-line method gives excavation 14.62 m³, PCC (1:3:6) 2.741 m³, superstructure brickwork 11.18 m³, and door/window frame timber 0.197 m³.

## Answer

#### Reading of the drawing
Two rooms of 3.27 m × 3.27 m clear side by side, 230 mm walls (3 × 0.23 + 2 × 3.27 = 7.23 m; 3.27 + 2 × 0.23 = 3.73 m). Centre-line length of outer walls = 2[(7.23 − 0.23) + (3.73 − 0.23)] = 21.00 m. The central partition is 3.27 + 0.23 = 3.50 m between the outer-wall centre lines; at its two T-junctions it is shortened by half the width of the outer-wall item at each end, so it is 3.50 − 0.77 = 2.73 m for the trench and PCC, and 3.50 − 0.23 = 3.27 m for the wall.

Foundation profile assumed from the section: PCC 1:3:6, 0.77 m wide × 0.15 m; brick step 0.69 m × 0.25 m; step 0.46 m × 0.15 m; 230 mm wall, trench depth 0.80 m; plinth 0.45 m above GL. Wall height above plinth (floor to slab soffit) = 2.4 m. Openings assumed: D 1.2 × 2.1 (1 no.), D1 1.0 × 2.1 (1 no., partition), W 1.5 × 1.2 (2 nos.), W1 1.2 × 1.2 (1 no.).

#### a) Earthwork in excavation in foundation
| Item | No. | L (m) | B (m) | H/D (m) | Qty (m³) |
|---|---|---|---|---|---|
| Trench, outer walls | 1 | 21.000 | 0.770 | 0.800 | 12.936 |
| Trench, partition (3.50 − 0.77) | 1 | 2.730 | 0.770 | 0.800 | 1.682 |
| **Total** | | | | | **14.618** |

#### b) PCC (1:3:6) in foundation
| Item | No. | L (m) | B (m) | H/D (m) | Qty (m³) |
|---|---|---|---|---|---|
| PCC 1:3:6, outer walls | 1 | 21.000 | 0.770 | 0.150 | 2.426 |
| PCC 1:3:6, partition | 1 | 2.730 | 0.770 | 0.150 | 0.315 |
| **Total** | | | | | **2.741** |

#### c) Brickwork in 1:6 c.s. in superstructure wall
| Item | No. | L (m) | B (m) | H/D (m) | Qty (m³) |
|---|---|---|---|---|---|
| Wall 0.23 × 2.4 above plinth (21.00 + 3.27) | 1 | 24.270 | 0.230 | 2.400 | 13.397 |
| Deduct D 1.2×2.1 | -1 | 1.200 | 0.230 | 2.100 | -0.580 |
| Deduct D1 1.0×2.1 | -1 | 1.000 | 0.230 | 2.100 | -0.483 |
| Deduct W 1.5×1.2 | -2 | 1.500 | 0.230 | 1.200 | -0.828 |
| Deduct W1 1.2×1.2 | -1 | 1.200 | 0.230 | 1.200 | -0.331 |
| **Total** | | | | | **11.175** |

#### d) Wood work for door and window frames
Frame section assumed 100 × 75 mm (doors have no sill member).

| Frame | No. | L (m) | B (m) | H/D (m) | Qty (m³) |
|---|---|---|---|---|---|
| Frame D (2×2.1 + 1×1.2) | 1 | 5.400 | 0.100 | 0.075 | 0.0405 |
| Frame D1 (2×2.1 + 1×1.0) | 1 | 5.200 | 0.100 | 0.075 | 0.0390 |
| Frame W (2×1.2 + 2×1.5) | 2 | 5.400 | 0.100 | 0.075 | 0.0810 |
| Frame W1 (2×1.2 + 2×1.2) | 1 | 4.800 | 0.100 | 0.075 | 0.0360 |
| **Total** | | | | | **0.1965** |

**Answer:** Excavation 14.62 m³; PCC 2.741 m³; superstructure brickwork 11.18 m³; frame timber 0.197 m³.

## Check yourself

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

1. How is the total centre-line length of walls obtained for the foundation items?
   A. Outer-wall centre-line 21.00 m plus the central partition 3.27 m = 24.27 m
   B. Outer-wall centre-line only, 21.00 m
   C. Overall perimeter 2(7.23 + 3.73) = 21.92 m
   D. Outer-wall centre-line 21.00 m plus twice the partition length
   Answer: A. The centre-line of the outer walls is 21.00 m and the single central partition adds its length of 3.27 m, giving 24.27 m for the trench and PCC. (Key point: Total centre-line length)

2. Which dimensions are multiplied with the centre-line length to get the excavation quantity of 14.95 m3?
   A. Width 0.23 m (wall thickness) and depth 0.80 m
   B. Width 0.77 m (PCC width) and depth 0.80 m
   C. Width 0.69 m (first brick step) and depth 0.45 m
   D. Width 0.77 m and depth 0.15 m (PCC thickness)
   Answer: B. The trench is as wide as the widest foundation layer (PCC, 0.77 m) and as deep as the trench depth of 0.80 m: 24.27 x 0.77 x 0.80 = 14.95 m3. (Key point: Trench dimensions)

3. Why is the superstructure brickwork 11.175 m3 and not the full wall volume of 13.397 m3?
   A. The plinth portion is deducted from the wall
   B. Only the outer walls are counted and the partition is omitted
   C. The volumes of doors and windows (length x 0.23 x height) are deducted from the full wall volume
   D. The mortar volume is deducted from the brickwork
   Answer: C. The wall is first measured solid (24.27 x 0.23 x 2.4) and then each opening is deducted at wall thickness, leaving 11.175 m3. (Key point: Deducting openings in brickwork)

4. Why is the door frame length taken as 2 x 2.1 + 1.2 and not 2 x 2.1 + 2 x 1.2?
   A. The head is hidden inside the lintel
   B. The door frame sill is measured under floor finishes
   C. Door frames are measured only up to the plinth
   D. Door frames have no bottom sill member, only two posts and a head
   Answer: D. A door frame consists of two vertical posts and one head; windows have a sill as well, which is why window frames use 2 x height + 2 x width. (Key point: Door frame length)

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- [Building: excavation, PCC foundation, brick](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Abuilding-q-three-items) — Estimating and Costing, chapter 6
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