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Building: PCC, brick, sal wood frame, RCC slab

Short answer

Assuming a 4.0 × 3.5 m clear room with 230 mm walls: PCC in foundation 2.87 m³, brickwork in foundation and plinth 6.22 m³, salwood door/window frames 0.111 m³, and M20 RCC slab including projection 3.26 m³.

Estimate the quantity of the following items of work from the accompanying building drawings: i) PCC (1:3:6) in foundation ii) Brick work in (1:6) cement mortar in foundation and plinth. iii) Salwood work for doors and windows frame iv) PCC M20 for R.C.C. slab [Drawing not included in the scanned paper]

No drawing is given. Assumed: room 4.0 m × 3.5 m clear, 230 mm walls, centre-line length 15.92 m; trench 0.90 m wide × 1.00 m deep; PCC 200 mm; brick steps 0.69 and 0.46 m wide (150 mm each); plinth 0.45 m above GL; door 1.0 × 2.1, two windows 1.2 × 1.2; frames 100 × 75 mm sal; RCC slab 125 mm thick with 0.45 m projection all round (M20).

i) PCC (1:3:6) in foundation

ItemNo.L (m)B (m)H/D (m)Qty (m³)
PCC 1:3:6 under walls 0.90 × 0.20115.9200.9000.2002.866
Total2.866

ii) Brickwork (1:6) in foundation and plinth

Wall height from the last step = 0.95 m.

ItemNo.L (m)B (m)H/D (m)Qty (m³)
Brick footing step 1 (0.69 m wide)115.9200.6900.1501.648
Brick footing step 2 (0.46 m wide)115.9200.4600.1501.098
Wall up to plinth115.9200.2300.9503.479
Total6.225

iii) Sal wood for door and window frames

FrameNo.L (m)B (m)H/D (m)Qty (m³)
Door frame D (2×2.1 + 1.0)15.2000.1000.0750.0390
Window frame W (2×1.2 + 2×1.2)24.8000.1000.0750.0720
Total0.1110

iv) PCC M20 for RCC slab

ItemNo.L (m)B (m)H/D (m)Qty (m³)
RCC slab 125 mm incl. 0.45 m projection: (4.0+0.46+0.90) × (3.5+0.46+0.90)15.3604.8600.1253.256
Total3.256

(Lintels, beams and reinforcement are measured separately.)

Answer: PCC 2.87 m³; brickwork 6.22 m³; sal wood frames 0.111 m³; M20 concrete in slab 3.26 m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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