Project 2

AIM:

To design the worst-case column and its isolated footing from the given framing plan using IS 456:2000 and SP 16 charts. Check for adequacy under combined axial and moment loading, and ensure safe pressure distribution within the permissible soil bearing capacity.

GIVEN

• Framing plan: 4 bays of 3 m (short span), 3 bays of 8 m (long span)

• Worst-case column: Interior column supporting largest tributary area

• Slab live load: 4 kN/m²

• Slab dead load: 3.75 kN/m²

• Concrete grade: M30 (column), M25 (footing)

• Steel grade: Fe415

• Allowable bearing capacity of soil: 100 kN/m²

• Depth of footing below ground: 2 m

SOLUTION:

Step 01: Load on Worst-Case Column

• Tributary area = 8 m × 6 m = 48 m²

• Total service load = (4 + 3.75) × 48 = 372 kN

• Factored load (1.5 × 372) = 558 kN

• Assume 10% self-weight of footing: Total design load = 558 × 1.1 = ~614 kN

Step 02: Column Design (Using SP 16)

• Size assumed: 300 mm × 500 mm

• Factored axial load (P_u): 486 kN

• Ultimate moment (M_u): 80 kNm

• Non-dimensional parameters:

  • $\frac{P_u}{f_{ck} \cdot b \cdot D} = 0.108$
    

  • $\frac{M_u}{f_{ck} \cdot b \cdot D^2} = 0.0356$
    

The point lies well within the SP 16 Chart 44 curve, Column design is safe.

Step 3: Footing Area and Size

• Allowable pressure = 100 kN/m²

• Required area:
  $A = \frac{614}{100} = 6.14 \, \text{m}^2$A=100614​=6.14m2

Adopt footing size: 2.6 m × 2.4 m = 6.24 m²

• Net pressure:
  $q = \frac{614}{6.24} \approx 98.4 \, \text{kN/m}^2$, Safe

Step 4: Footing Thickness & Reinforcement

(i) Bending Moment at face of column:

• Projection: 1.3 m – 0.15 m = 1.15 m

• Bending moment,
  $M = q \cdot l^2/2 = 98.4 \cdot 1.15^2 / 2 \approx 65.1 \, \text{kNm/m}$
  

Effective depth (d):

M=0.138⋅f_ck⋅b⋅d²

$65.1 \times 10^6 = 0.138 \cdot 25 \cdot 1000 \cdot d^2$
$d = 222 \, \text{mm}$

d=222mmAdopt effective depth = 250 mm, total depth ≈ 300 mm

(ii) Shear Check at d from face of column:

• Shear span = 1.15 – 0.25 = 0.9 m

• Shear force = $q \cdot 0.9 = 88.6 \, \text{kN}$
  

• Nominal shear stress:
  $\tau_v = \frac{88.6 \times 10^3}{1000 \times 250} = 0.354 \, \text{N/mm}^2$
  

• Permissible shear stress (M25, 0.5% steel):
  $\tau_c = 0.36 \, \text{N/mm}^2$
  Safe in shear, no shear reinforcement needed

(iii) Reinforcement (Flexural):

• $M = 65.1 \, \text{kNm/m}$
  

• $M = 0.87 f_y A_{st} z$
  

• $A_{st}=\frac{65.1\times {10}^6}{0.87\cdot 415\cdot 225}$$\approx 650{\text{mm}}^2\mathrm{/}m$

  Provide: 12 mm bars @ 150 mm c/c in both directions

Step 05: Column Interaction Curve Summary:

Refer attached image of graph,

• Factored Axial Load (Pu): 486 kN

• Assumed Ultimate Moment (Mu): 80 kNm

• Column Size: 300 mm × 500 mm

• Concrete Grade: M30

• Steel Grade: Fe415

Non-Dimensional Parameters:

• $\frac{Pu}{f_{ck}\cdot b\cdot D}=0.108$

• $\frac{Mu}{f_{ck} \cdot b \cdot D^2} = 0.0356$
  

Design point lies well within the interaction curve based on SP 16 Chart 44 approximation. Column is safe under combined axial and bending loads.

RESULT:

• Interior column carrying Pu = 486 kN and Mu = 80 kNm is safe using SP 16.

• Footing size of 2.6 m × 2.4 m is adequate for soil capacity of 100 kN/m².

• Footing thickness = 300 mm, effective depth = 250 mm

• Provide 12 mm ⌀ bars @ 150 mm c/c in both directions.

• No shear reinforcement required in slab or footing.

• Beam B1 needs 2-legged 10 mm stirrups @ 300 mm c/c near supports.