Concept:

Minimum eccentricities:

\({e_{\;x\;min}} = {\rm{max}}\left\{ {\begin{array}{*{20}{c}} {\frac{{{\ell _x}}}{{500}} + \frac{D}{{30}}}\\ {20\;mm} \end{array}} \right.\)

\({e_{y\;min}} = {\rm{max}}\left\{ {\begin{array}{*{20}{c}} {\frac{{{\ell _y}}}{{500}} + \frac{b}{{30}}}\\ {20\;mm} \end{array}} \right.\)

ℓ_x, ℓ_y = Unsupported length

Calculation:

ℓ_x = ℓ_y = 3m = 3000 mm

D = 360 mm, b = 240 mm

\({e_{x\;min}} = \max \left\{ {\begin{array}{*{20}{c}} {\frac{{3000}}{{500}} + \frac{{360}}{{30}}}\\ {20\;mm} \end{array}} \right. = 18\;mm\)

= 20 mm

\({e_{y\;min}} = \max \left\{ {\begin{array}{*{20}{c}} {\frac{{3000}}{{500}} + \frac{{240}}{{30}}}\\ {20\;mm} \end{array}} \right. = 14\;mm\)

Concept:

Column load (downward) and load due to soil pressure (upward) will lead to crack 45°.

Thus, the critical section due to one-way shear is at d from the face of the column.

\({\tau _o}\left( {one\;way\;shear} \right) = \frac{{\left( {{q_o} \times \frac{{B - b - 2d}}{2} \times B} \right)}}{{B \times d}}\)

Where

B = width of square footing

b = width of column

d = depth of footing

q_o = load intensity of footing

Calculation:

\(qo\; = \frac{{320 \times 1000}}{{2000 \times 2000}} = 0.08\;{\rm{N}}/{\rm{m}}{{\rm{m}}^2}\)

The critical section for bending moment is the face of the column:-

P_o = Net upward pressure = 1.5 N/mm² 

\(\begin{array}{l} {M_u} = {P_o}B \times \left[ {\frac{B}{2} - \frac{b}{2}} \right] \times \frac{1}{2}\left[ {\frac{B}{2} - \frac{b}{2}} \right]\\ = \frac{{{P_o}B}}{2}{\left[ {\frac{B}{2} - \frac{b}{2}} \right]^2}\\ = \frac{{1.50 \times 4000}}{2}{\left[ {\frac{{4000}}{2} - \frac{{400}}{2}} \right]^2} = 9720{\rm{\;KN}} - {\rm{m}} \end{array}\)

Concept:

The strength reduction coefficient method is used for the design of a long column. It is valid for Working Stress Method only.

Strength Reduction coefficient is given as

\({C_r} = 1.25 - \frac{{{\ell _{eff}}}}{{48\;b}}\)

Where, b = least lateral dimension of core and for the circular column with helical reinforcement it is the diameter of the core.

Calculation:

ℓ_eff = 6 m = 6000 mm, b = 250 mm

Concept:

If \(\frac{{\sigma \begin{array}{*{20}{c}} '\\ {cc} \end{array}}}{{{\sigma_{cc}}}} + \frac{{\sigma\begin{array}{*{20}{c}} '\\ {cbc} \end{array}}}{{{\sigma_{cbc}}}} \le 1\) (Design is done on the basis of uncracked section)

If \(\frac{{\sigma\begin{array}{*{20}{c}} '\\ {cc} \end{array}}}{{{\sigma_{cc}}}} + \frac{{\sigma\begin{array}{*{20}{c}} '\\ {cbc} \end{array}}}{{{\sigma_{cbc}}}} > 1\) (Design is done on the basis of cracked section)

\(\sigma\begin{array}{*{20}{c}} '\\ {cc} \end{array}\)⇒ Developed (Calculated) stress in concrete in direct compression

\(\sigma\begin{array}{*{20}{c}} '\\ {cbc} \end{array}\)⇒ Developed (Calculated) stress in concrete in bending compression

σ_cc ⇒ Permissible stress in concrete in direct compression

σ_cbc ⇒ Permissible stress in concrete in bending compression

Calculations:

For M - 20, concrete

σ_cbc = 7 N/mm²

σ_cc = 5 N/mm²

Now, \(\sigma\begin{array}{*{20}{c}} '\\ {cc} \end{array} = 2.5\;N/m{m^2}\)

\(\sigma\begin{array}{*{20}{c}} '\\ {cbc} \end{array} = 3.6\;N/m{m^2}\)

\(\frac{{2.5}}{5} + \frac{{3.6}}{7} = 1.014 > 1\)

Concept:

Two way/punching shear:

It shall be checked around the column a perimeter half the effective depth of footing slab away from the face of column or pedestal.

\({\tau _{v\;two\;way}} = \frac{{w\left[ {L \times B+\left( {a + d} \right)\left( {b + d} \right)} \right]}}{{2\left[ {\left( {a + d} \right) + \left( {b+d} \right)} \right]d}}\)

Calculation:

The critical section for two-way shear is at a distance of d/2 from the face of the column as shown in figure in dotted lines.

Load = 420 kN, Area of footing = 9 m²

w = \(\frac{{420 \times 1000}}{{9 \times {{10}^6}}}\) = 0.0467 MPa

∴ \({\tau _{v\;two\;way}} = \frac{{0.0467 \times \left( {{{3000}^2} - {{\left( {600 + 300} \right)}^2}} \right)}}{{2 \times \left( {900 + 900} \right) \times 300}}\) = 0.35 Mpa

Important Point:

Footing is designed as per shear criteria. There are two types of shear:

1. One way shear
2. Two way shear

In one way shear the critical section for shear is:

1. At a distance d from the face of the wall or column as the case may be.
2. At a distance d/2 from the face of the wall or column if the footing slab is on piles.

Interaction Diagram:

To draw above interaction diagram, following quantities are required:

1. Grade of concrete

2. Grade of steel

3. Section size (b and D both)

4. Shape of section (Rectangular & circular)

5. \(\frac{{d'}}{D}ratio\)

6. Arrangement of Reinforcement

7. Area of longitudinal steel and ‘%’ of longitudinal steel.

Important Points about Interaction Diagram:

• Point B represents balanced failure of section.
• Portion AB represents compression failure compression failure because concrete attains its permissible value before yielding of steel.
• Portion BC represents tension failure because concrete attains its permissible value after yielding of steel.
• Any load combination failing within interaction curve is safe for section.

Area of footing = 2 × 3 = 6 m²

Section modulus,

\(Z = \frac{{b \times {d^2}}}{6} = \frac{{2 \times {3^2}}}{6} = 3 {m^3}\)

Stress, \(\sigma = \frac{P}{A} \pm \frac{M}{Z}\)

\(95 = \frac{P}{A} + \frac{M}{Z}\) -------------- (i)

\(55 = \frac{P}{A} - \frac{M}{Z}\) -------------- (ii)

From (i) and (ii), we get

\(\begin{array}{l} 40 = 2 \times \frac{M}{Z}\\ M = 2 \times \frac{M}{3} \end{array}\)

Concept:

The load-carrying capacity for a short column loaded axially with minimum eccentricity is given as

P_u = 0.4 f_ck A_c + 0.67 f_y A_sc

Where,

A_c = Area of concrete after reducing steel area, A_sc = Area of steel

Calculation:

f_ck = 25, f_y = 415, b = 300 mm, d = 450 mm

∴ Load carrying capacity,

P = 0.4 × 25 (Total area – Area of steel) + 0.67 × 415 × A_sc

\( = 0.4 \times 25\left( {300 \times 450 - 6 \times \frac{\pi }{4} \times {{16}^2}} \right) + 0.67 \times 415 \times 6 \times \frac{\pi }{4} \times {16^2}\)

= 1673.37 kN

Important Point:

If, \(\frac{{{\ell _{eff\;x}}}}{D}\) or \(\frac{{{\ell _{eff\;y}}}}{b}\) both are less than 12 and

\({e_{x\;min}} = \max \left\{ {\begin{array}{*{20}{c}} {\frac{{{\ell _x}}}{{500}} + \frac{D}{{30}}}\\ {20\;mm} \end{array}} \right.\) < 0.05 D

& \({e_{y\;min}} = \max \left\{ {\begin{array}{*{20}{c}} {\frac{{{\ell _y}}}{{500}} + \frac{b}{{30}}}\\ {20\;mm} \end{array}} \right.\) < 0.05 b

The column is designed as a short axially loaded column with minimum eccentricity.

For purely axial load, i.e. eccentricity = 0

Concept:

Pitch of the helical reinforcement is given by:

\(\frac{{0.36{f_{ck}}}}{{{f_y}}}\left( {\frac{{{A_g}}}{{{A_c}}} - 1} \right) \le \frac{{{V_h}}}{{{V_c}}}\)

Where,

A_g = Gross area of column=  \(\frac{\pi }{4}{\left( {{D_g}} \right)^2}\)

A_c = area of core = \(\frac{\pi }{4}{\left( {{D_g} - 2{d_c}} \right)^2}\)

V_h = Volume of helical reinforcement per pitch

\({{\rm{V}}_{\rm{h}}} = \frac{{1000}}{{\rm{p}}} \times {\rm{\pi }}{{\rm{D}}_{\rm{h}}} \times \frac{{\rm{\pi }}}{4}\phi _{\rm{h}}^2\)

V_c = volume of core = A_c x 1000

d_c = clear cover

D_c = core diameter

As per IS 456:2000, the pitch calculated above should satisfying:

 1) p ≤ 75 mm

2) p ≤ D_c/6

3) p ≥ 75 mm

4) p ≥ 3Φ_h

Calculation:

D_h = D_c - ϕ_h

D_c = D_g – 2xd_c

D_c = 500 – 80 = 420 mm

D_h = 420 – 10 = 410 mm

\({A_c} = \frac{\pi }{4}{\left( {500 - 80} \right)^2} = 138544024\;m{m^2}\)

\({V_c} = 138544024 \times 1000\;m{m^3}\)

\({{\rm{V}}_{\rm{h}}} = \frac{{1000}}{{\rm{p}}} \times {\rm{\pi }} \times 410 \times \frac{{\rm{\pi }}}{4} \times {10^2} = \frac{{144964749.4}}{{\rm{p}}}\)

\(\frac{{0.36 \times 30}}{{415}}\left( {\frac{{196349.54}}{{138544024}} - 1} \right) \le \frac{{144964749.4}}{{138544024 \times 1000 \times {\rm{p}}}}\)

p = 96.36 mm, but it should not be greater than one sixth of core diameter of column

\({\rm{p}} \le \frac{{420}}{6} = 70{\rm{\;mm}}\)

Therefore, the pitch should be 70 mm.