Design of an Eccentric Pad Footing

In the design of pad footing, it is desirable that there should be not eccentricity. However, when eccentricity which is brought in by moment occurs due to unbalanced loads on the pad footing, the footing should be designed as an eccentric pad footing. In practice, almost all kinds of pad footings experience certain degrees of eccentricity. Eccentricity of pad footings leads the moments and the moments have the following effects on the pad footing:

Effects of Moment on Footings

1. Dislocate the axis of the load coming on the footing.
2. Increase the earth pressure on the footing on one side and reduce it on the other side.

When software is used for design, these are calculated automatically. However, when manual design process is being used, the following order can be followed to achieve the design.

Design Formulae

1. Eccentricity, e = M/S_k >> M = eS_k (1)

Where,

M = moment on the footing due to eccentricity

S_k = axial load coming on the footing

Due to eccentricity, bearing capacity, q ≠ S_k/A_prov

2. Thus, q = w/A + S_k/A + Mx/I (2)

Where,

I = moment of inertia of the footing

x = distance measured from central axis of footing to any point within the footing

w = self-weight of the footing

For rectangular footing, Area, A = B x L

Where,

B = breadth of footing

L = length of footing

Moment of Inertia, I = LB³/12 = (LB) B²/12 = AB²/12

Substituting Eqn (1) into (2) for rectangular footing

3. q = w/A + S_k/A + (eS_kx)/ (AB²/12) >> w/A + S_k/A + [(eS_kx)/1] x [(12/ AB²)] >> w/A + S_k/A + 12eS_kx/ AB² >> w/A + S_k/A (1 + (12ex/B²))

The maximum and minimum pressures occurs at the edges of the footing where x = ± (B/2)

4. At x = B/2, the maximum pressure, q_max = w/A + S_k/A (1 + (12e (0.5B)/B²)) = w/A + S_k/A (1 + (6e/B))

5. At x = -B/2, the minimum pressure, q_min = w/A + S_k/A (1 – (6e/B))

The formula for q_min and q_max derived above are only applicable where e ≤ B/6

If e ˃ B/6, the formula would not satisfy and this condition serves as control on the footing size. Thus, by trial and error, we would continue to adjust until we have a situation where e ≤ B/6

Secondly, different values of B must be tried until q ≤ q_allowable and at the same time e ≤ B/6

How to Calculate Moment Due to Eccentricity

To determine the moment, first, factor the dead and live loads where applicable

Note, Design load, N = w + S_k >> N/A = w/A + S_k/A

q_max = N/A (1 + (6e/B)) (x = B/2)

q₃ = N/A (1 + (6ea₂/B²)) (x = B/2)

The area of the triangles gives the following magnitude of force (see image below)

F₁ = 1/2 (q₃) j; F₂ = 1/2 (q_max) j

Taking moment about the column face,

Design moment, M = F₁ x j/3 + F₂ x 2j/3

Substitute F₁ and F₂ into the design moment, thus;

Design moment, M = (1/2 x q₃ x j x j/3) + (1/2 x q_max x j x 2j/3) = (q₃ + 2q_max) j²/6

Solved Example

A square pad footing is to be provided to column to support a service, S_k vertical load of 800 kN and a moment of 100 kNm. If the q_allowable = 175 kN/m², design appropriate footing using f_cu = 25N/mm² and f_y = 410 N/mm²

Solution

λ (load factor) = 1.45. The column dimensions are 230 x 230 mm

Eccentricity, e = M/S_k = 100/800 = 0.125

Self-weight of footing, w = 8% of S_k

w = (8/100) x 800 = 64 kN

q_max = 64/A + 800/A (1 + (6 x 0.125/B)) [for a square footing, A = B²]

N.B: q_max = q_allowable = 175 kN/m²; e ≤ B/6

175 = 64/B² + 800/B² (1 + (6 x 0.125/B)) = 64/B² + 800/B² ((B + 0.75)/B))

175 = 64/B² + 800/B² + 600/B³ = (64B + 800B + 600)/B³

175 = (864B + 600)/ B³

175B³ = 864B + 600

175B³ – 864B – 600 = 0

Solving, B = 2.51 or -1.71 or -0.796

Since we cannot have negative B, take B = 2.51. Thus, B = 2.6 m

Check for Bearing Capacity

q_max = 64/(2.6)² + 800/(2.6)² + 600/(2.6)³ >> 9.467 + 118.345 + 34.1337 = 161.946 kN/m²

q_max = 161.946 kN/m² < 175 kN/m². Therefore, B = 2.6 m is Ok

Check for Eccentricity

e ≤ B/6 >> B/6 = 2.6/6 = 0.433

e = 0.125 < 0.433 Ok

Structural design of footing

h = w/24Aprov = 64/ (24 x 2.6 x 2.6) = 0.394 = 394 mm

Try h = 400 mm

Assuming concrete cover (cc) of 50 mm and bar size, θ = 16 mm

Effective depth of bar, d = h – cc – θ/2 = 400 – 50 – (16/2) = 400 – 50 – 8 = 342 mm

Check for Punching Shear at Column Face

ν = N/4a₁d

Where, N = 1.45 x 800 = 1160 kN

ν = (1160 x 10³)/ (4 x 230 x 342) = 3.687 N/mm² < 0.8 √f_cu or 5 N/mm², OK

Determination of Design Moment

j = (B – a)/2 = (2.6 – 0.23)/2 = 1.185 m

q_max = N/A (1 + 6e/B) = 1160/2.6² (1 + (6 x 0.125)/2.6) = 171.598 (1 + 0.28846) >> q_max = 221.097 kN/m²

q₃ = N/A (1 + 6ea₁/B²) = 1160/2.6² (1 + (6 x 0.125 x 0.23)/2.6²) = 171.598 (1 + 0.02552) = 175.977

Design moment, M = (q₃ + 2q_max) j2/6 = (175.977 + (221.097 x 2)) 1.185²/6 =144.64 kNm

K = M/f_cubd² = 144.64 x 10⁶/ (25 x 1000 x 3422) = 0.0495

Lever arm, la = 0.5 + √0.25 – (0.0495/9) = 0.994 ˃ 0.95, use 0.95

Z = lad = 0.95 x 342 = 324.9 mm

As_req = M/0.95f_yZ = 144.64 x 10⁶/ (0.95 x 410 x 324.9) = 1142.96 mm²

As_min = 1.3h = 1.3 x 400 = 520 mm² (As^min is satisfied)

Provide suitable reinforcement that is not less than the required reinforcement

Other Checks

Shear checks are very important on pad footings. Shear check should be carried out at distance of (1 x effective depth, d) from the column face while punching shear check should be carried out distance of (1.5 x effective depth, d) from the column face. Watch out for subsequent post on how to do this.