How is the sloping footing part computed?

Edge slab plus frustum by the prismoidal formula

2078 Bhadra · 3+5 marks

RCC column with footing: concrete and steel

Short answer

Column RCC 1:2:4: 0.337 m³ (0.3×0.3×3.75 m). Footing RCC: edge slab 0.45 m³ + frustum 0.232 m³ = 0.682 m³. Total 1.02 m³. Steel: main 12 mm (15 nos), distribution 10 mm (15), vertical 16 mm (4) and 12 mm (4), stirrups 6 mm (29 nos @15 cm), total mass 86.3 kg.

From the attached drawing (figure 2) of RCC column, estimate the following items. (i) RCC 1:2:4 in column (ii) Steel reinforcement work excluding formwork [Figure 2: Column footing 1.5 m × 1.5 m in plan, RCC 1:2:4, 20 cm thick at edge (sloping up to column) with 10 cm PCC 1:3:6 below; column 30 cm × 30 cm, 3.50 m above GL and 70 cm below GL; footing reinforcement 12 mm main bars @ 10 cm c/c and 10 mm distribution bars @ 10 cm c/c; column reinforcement 4–16 mm dia and 4–12 mm dia vertical bars; 6 mm dia stirrups @ 15 cm c/c; anchorage/development 45 times bar dia.]

The footing thickness at the column face is not given, so 0.45 m is assumed (0.20 m at the edge, sloping up 0.25 m to the column). The 1.5 m × 1.5 m footing has its bottom (of the RCC) 0.70 m below GL, and the 0.30 × 0.30 m column rises 3.50 m above GL.

i) RCC 1:2:4

Column (above footing top, height = 0.70 − 0.45 + 3.50 = 3.75 m):

0.30 × 0.30 × 3.75 = 0.337 m³

Footing (a slab plus a sloping pyramid frustum):

Edge slab: 1.5 × 1.5 × 0.20 = 0.450 m³
Frustum (prismoidal formula, $h=0.25$ m, $A_1=2.25$ , $A_2=0.09$ ): $\frac{0.25}{3}\left(2.25+0.09+\sqrt{2.25\times0.09}\right)=0.2325$ m³
Footing total = 0.682 m³

Total RCC 1:2:4 (footing + column) = 0.337 + 0.682 = 1.020 m³. (PCC 1:3:6, 10 cm under footing = 1.5 × 1.5 × 0.10 = 0.225 m³.)

ii) Steel reinforcement (formwork excluded)

Cover 50 mm in footing, 25 mm in column; weight per metre = $d^2/162$ kg (d in mm).

Bar	Working	Length (m)	Weight (kg)
12 mm main, 15 bars	15 × (1.5 − 0.10)	21.00	18.67
10 mm distribution, 15 bars	15 × 1.40	21.00	12.96
16 mm vertical, 4 bars	4 × (4.20 − 0.05 + 45 × 0.016)	19.48	30.78
12 mm vertical, 4 bars	4 × (4.20 − 0.05 + 45 × 0.012)	18.76	16.68
6 mm stirrups, 29 nos.	29 × (4 × 0.25 + 2 × 10 × 0.006)	32.48	7.22
Total			86.31

Number of stirrups = (4.20 − 0.10)/0.15 + 1, rounded up = 29. Bars per layer in the footing = 1.40/0.10 + 1 = 15. (Each column bar includes one 45d anchorage (development) allowance; hooks and bend allowances are ignored.)

Answer: RCC 1:2:4 = 1.02 m³ (column 0.337 m³, footing 0.682 m³); steel ≈ 86.3 kg.

✓ Answer checked by Claude Opus 5.5 (high effort)

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RCC column with footing: concrete and steel

i) RCC 1:2:4

ii) Steel reinforcement (formwork excluded)

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