A biaxial eccentrically loaded pad footing occurs when the column transmitting load to the foundation is subjected to compressive axial force and bending moment in the two principal axes. As a result of the biaxial bending, two eccentricities e_{x} and e_{y} of the axial load occur on the pad footing, thereby leading to non-uniform pressure distribution on the foundation. The effects of such non-uniform pressure distribution must be accounted for during the geotechnical and structural design of the pad foundation.

This article will consider the geotechnical and structural design of a biaxial eccentrically loaded pad footing. The footing for a single column may be made square in plan, but where there is a large moment acting about only one axis it may be more economical to have a rectangular base.

When a bending moment *M* and axial force *N* are acting on a pad foundation, the pressures are given by the equation for axial load plus bending. This condition is valid provided there is positive contact between the pad base and the ground along the complete length *D* of the footing so that;

p = N/BD ± My/I

where* I* is the second-moment area of the base about the axis of bending and *y *is the distance from the axis to where the pressure is being calculated.

Substituting for I = BD^{3}/12 and y = D/2, the maximum pressure is;

p_{max} = N/BD + 6M/BD^{2}

and the minimum pressure is;

p_{min} = N/BD – 6M/BD^{2}

For biaxially loaded footings, this pressure must be verified in both directions, and the maximum pressure should not exceed the allowable bearing capacity of the soil. Furthermore, the reinforcement design must also be carried out in both directions.

**Design Example of Biaxial Eccentrically Loaded Pad Footing**

A 1500 x 1500mm pad foundation is subjected to the following loads from a 250 mm x 250 mm square column;

Permanent load axial load; F_{Gz1} = 650.0 kN

Variable laxial load; F_{Qz1} = 135.0 kN

Permanent moment in x; M_{Gx1} = 25.0 kNm

Permanent moment in y; M_{Gy1} = 21.0 kNm

Variable moment in x; M_{Qx1} = 13.0 kNm

Variable moment in y; M_{Qy1} = 11.0 kNm

The design is to be done in accordance with EN1997-1:2004 incorporating Corrigendum dated February 2009 and the UK National Annex incorporating Corrigendum No.1

**Pad foundation details**

Length of foundation; L_{x} = 1500 mm

Width of foundation; L_{y} = 1500 mm

Foundation area; A = L_{x} × L_{y} = 2.250 m^{2}

Depth of foundation (thickness of footing); h = 500 mm

Depth of soil over foundation; h_{soil} = 600 mm

Level of water; h_{water} = 0 mm

Density of water; γ_{water} = 9.8 kN/m^{3}

Density of concrete; γ_{conc} = 25.0 kN/m^{3}

**Column details**

Length of column; l_{x1} = 250 mm

Width of column; l_{y1} = 250 mm

position in x-axis; x_{1} = 750 mm

position in y-axis; y_{1} = 750 mm

Library item: Column details output

**Soil properties**

Density of soil; γ_{soil} = 18.0 kN/m^{3}

Characteristic cohesion; c’_{k} = 15 kN/m^{2}

Characteristic effective shear resistance angle; φ’_{k} = 25 deg

Characteristic friction angle; δ_{k} = 20 deg

**Foundation loads**

Permanent surcharge load; F_{Gsur} = 5.0 kN/m^{2}

Self weight; F_{swt} = h × γ_{conc} = 12.5 kN/m^{2}

Soil weight; F_{soil} = h_{soil} × γ_{soil} = 10.8 kN/m^{2}

**Column loads**

Permanent load in z; F_{Gz1} = 650.0 kN

Variable load in z; F_{Qz1} = 135.0 kN

Permanent moment in x; M_{Gx1} = 25.0 kNm

Permanent moment in y; M_{Gy1} = 21.0 kNm

Variable moment in x; M_{Qx1} = 13.0 kNm

Variable moment in y; M_{Qy1} = 11.0 kNm

**Design Approach 1** **(DA 1)** – **Combination 1**

Partial factors on actions – Combination1

Partial factor set; A1

Permanent unfavourable action – Table A.3; γ_{G} = 1.35

Permanent favourable action – Table A.3; γ_{Gf} = 1.00

Variable unfavourable action – Table A.3; γ_{Q} = 1.50

Variable favourable action – Table A.3; γ_{Qf} = 0.00

**Partial factors for soil parameters – Combination1**

Soil factor set; M1

Angle of shearing resistance – Table A.4; γ_{φ’} = 1.00

Effective cohesion – Table A.4; γ_{c’} = 1.00

Weight density – Table A.4; γ_{g} = 1.00

**Partial factors for spread foundations – Combination1**

Resistance factor set; R1

Bearing – Table A.5; γ_{R.v} = 1.00

Sliding – Table A.5; γ_{R.h} = 1.00

**Bearing Resistance **

**Forces on foundation**

Force in z-axis; F_{dz} = γ_{G} × [A × (F_{swt} + F_{soil} + F_{Gsur}) + F_{Gz1}] + γ_{Q}F_{Qz1} = 1166.0 kN

**Moments on foundation**

Moment in x-axis;

M_{dx} = γ_{G} × (A × (F_{swt} + F_{soil} + F_{Gsur}) × L_{x}/2 + F_{Gz1}x_{1}) + γ_{G}M_{Gx1} + γ_{Q}F_{Qz1}x_{1} + γ_{Q}M_{Qx1} = 927.7 kNm

Moment in y-axis;

M_{dy} = γ_{G} × (A × (F_{swt} + F_{soil} + F_{Gsur}) × L_{y}/2 + F_{Gz1}y_{1}) + γ_{G}M_{Gy1} + γ_{Q}F_{Qz1}y_{1} + γ_{Q}M_{Qy1} = 919.3 kNm

**Eccentricity of base reaction**

Eccentricity of base reaction in x-axis;

e_{x} = M_{dx} / F_{dz} – L_{x} / 2 = 46 mm

Eccentricity of base reaction in y-axis;

e_{y} = M_{dy} / F_{dz} – L_{y} / 2 = 38 mm

**Effective area of base**

Effective length;

L’_{x} = L_{x} – 2 × e_{x} = 1409 mm

Effective width;

L’_{y} = L_{y} – 2 × e_{y} = 1423 mm

Effective area;

A’ = L’_{x} × L’_{y} = 2.005 m^{2}

**Pad base pressure**

Design base pressure; f_{dz} = F_{dz} / A’ = 581.6 kN/m^{2}

Design angle of shearing resistance; φ’_{d} = tan^{-1}(tan(φ’_{k}) / γ_{φ’}) = 25.000 deg

Design effective cohesion; c’_{d} = c’_{k} / γ_{c’} = 15.000 kN/m^{2}

Effective overburden pressure;

q = (h + h_{soil}) × γ_{soil} – h_{water} × γ_{water} = 19.800 kN/m^{2}

Design effective overburden pressure;

q’ = q / γ_{g} = 19.800 kN/m^{2}

**Bearing resistance factors; **

N_{q} = Exp(π × tan(φ’_{d})) × [tan(45 deg + φ’_{d} / 2)]^{2} = 10.662

N_{c} = (N_{q} – 1) × cot(φ’_{d}) = 20.721

N_{γ} = 2 × (N_{q} – 1) × tan(φ’_{d}) = 9.011

**Foundation shape factors; **

s_{q} = 1 + (L’_{x} / L’_{y}) × sin(φ’_{d}) = 1.418

s_{γ} = 1 – 0.3 × (L’_{x} / L’_{y}) = 0.703

s_{c} = (s_{q} × N_{q} – 1) / (N_{q} – 1) = 1.462

**Load inclination factors;**

H = 0.0 kN

m_{y} = [2 + (L’_{y} / L’_{x})] / [1 + (L’_{y} / L’_{x})] = 1.497

m_{x} = [2 + (L’_{x} / L’_{y})] / [1 + (L’_{x} / L’_{y})] = 1.503

m = m_{x} = 1.503

i_{q} = [1 – H / (F_{dz} + A’ × c’_{d} × cot(φ’_{d}))]^{m} = 1.000

i_{γ} = [1 – H / (F_{dz} + A’ × c’_{d} × cot(φ’_{d}))]^{m + 1} = 1.000

i_{c} = i_{q} – (1 – i_{q}) / (N_{c} × tan(φ’_{d})) = 1.000

Ultimate bearing capacity;

n_{f} = c’_{d}N_{c}s_{c}i_{c} + q’N_{q}s_{q}i_{q} + 0.5γ_{soil}L’_{x}N_{γ}s_{γ}i_{γ} = 834.0 kN/m^{2}

PASS – Ultimate bearing capacity exceeds design base pressure

**Design Approach 1 (DA 1) – Combination 2**

**Partial factors on actions – Combination2**

Partial factor set; A2

Permanent unfavourable action – Table A.3; γ_{G} = 1.00

Permanent favourable action – Table A.3; γ_{Gf} = 1.00

Variable unfavourable action – Table A.3; γ_{Q} = 1.30

Variable favourable action – Table A.3; γ_{Qf} = 0.00

**Partial factors for soil parameters – Combination2**

Soil factor set; M2

Angle of shearing resistance – Table A.4; γ_{φ’} = 1.25

Effective cohesion – Table A.4; γ_{c’} = 1.25

Weight density – Table A.4; γ_{g} = 1.00

**Partial factors for spread foundations – Combination2**

Resistance factor set; R1

Bearing – Table A.5; γ_{R.v} = 1.00

Sliding – Table A.5; γ_{R.h} = 1.00

**Bearing resistance (Section 6.5.2)**

**Forces on foundation**

Force in z-axis;

F_{dz} = γ_{G} × (A × (F_{swt} + F_{soil} + F_{Gsur}) + F_{Gz1}) + γ_{Q}F_{Qz1} = 889.2 kN

**Moments on foundation**

Moment in x-axis;

M_{dx} = γ_{G} × (A × (F_{swt} + F_{soil} + F_{Gsur}) × L_{x}/2 + F_{Gz1} × x_{1}) + γ_{G}M_{Gx1} + γ_{Q}F_{Qz1}x_{1} + γ_{Q}M_{Qx1} = 708.8 kNm

Moment in y-axis;

M_{dy} = γ_{G} × (A × (F_{swt} + F_{soil} + F_{Gsur}) × L_{y}/2 + F_{Gz1} × y_{1}) + γ_{G}M_{Gy1} + γ_{Q}F_{Qz1}y_{1} + γ_{Q}M_{Qy1} = 702.2 kNm

**Eccentricity of base reaction**

Eccentricity of base reaction in x-axis;

e_{x} = M_{dx} / F_{dz} – (L_{x} /2) = 47 mm

Eccentricity of base reaction in y-axis;

e_{y} = M_{dy} / F_{dz} – (L_{y}/2) = 40 mm

**Effective area of base**

Effective length;

L’_{x} = L_{x} – 2e_{x} = 1406 mm

Effective area;

A’ = L’_{x} × L’_{y} = 1.997 m^{2}

Effective width;

L’_{y} = L_{y} – 2e_{y} = 1421 mm

**Pad base pressure**

Design base pressure; f_{dz} = F_{dz} / A’ = 445.3 kN/m^{2}

Ultimate bearing capacity under drained conditions (Annex D.4)

Design angle of shearing resistance;

φ’_{d} = tan^{-1}(tan(φ’_{k}) / γ_{f’}) = 20.458 deg

Design effective cohesion;

c’_{d} = c’_{k} / γ_{c’} = 12.000 kN/m^{2}

Effective overburden pressure;

q = (h + h_{soil}) × γ_{soil} – h_{water} × γ_{water} = 19.800 kN/m^{2}

Design effective overburden pressure;

q’ = q/γ_{g} = 19.800 kN/m^{2}

**Bearing resistance factors; **

N_{q} = Exp(π × tan(φ’_{d})) × (tan(45 deg + φ’_{d} / 2))^{2} = 6.698

N_{c} = (N_{q} – 1) × cot(φ’_{d}) = 15.273

N_{γ} = 2 × (N_{q} – 1) × tan(φ’_{d}) = 4.251

**Foundation shape factors; **

s_{q} = 1 + (L’_{x} / L’_{y}) × sin(φ’_{d}) = 1.346

s_{γ} = 1 – 0.3 × (L’_{x} / L’_{y}) = 0.703

s_{c} = (s_{q} × N_{q} – 1) / (N_{q} – 1) = 1.407

**Load inclination factors;**

H = 0.0 kN

m_{y} = [2 + (L’_{y} / L’_{x})] / [1 + (L’_{y} / L’_{x})] = 1.497

m_{x} = [2 + (L’_{x} / L’_{y})] / [1 + (L’_{x} / L’_{y})] = 1.503

m = m_{x} = 1.503

i_{q} = [1 – H / (F_{dz} + A’ × c’_{d} × cot(φ’_{d}))]^{m} = 1.000

i_{γ} = [1 – H / (F_{dz} + A’ × c’_{d} × cot(φ’_{d}))]^{m + 1} = 1.000

i_{c} = i_{q} – (1 – i_{q}) / (N_{c} × tan(φ’_{d})) = 1.000

**Ultimate bearing capacity; **

n_{f} = c’_{d}N_{c}s_{c}i_{c} + q’N_{q}s_{q}i_{q} + 0.5 γ_{soil}L’_{x}N_{γ} s_{g} i_{γ} = 474.1 kN/m^{2}

PASS – Ultimate bearing capacity exceeds design base pressure

**Foundation design (EN1992-1-1:2004)**

In accordance with EN1992-1-1:2004 incorporating Corrigendum dated January 2008 and the UK National Annex incorporating National Amendment No.1

**Concrete details**

Concrete strength class; C25/30

Characteristic compressive cylinder strength; f_{ck} = 25 N/mm^{2}

Characteristic compressive cube strength; f_{ck,cube} = 30 N/mm^{2}

Mean value of compressive cylinder strength;f_{cm} = f_{ck} + 8 N/mm^{2} = 33 N/mm^{2}

Mean value of axial tensile strength; f_{ctm} = 0.3 N/mm^{2} × (f_{ck})^{2/3} = 2.6 N/mm^{2}

5% fractile of axial tensile strength;f_{ctk,0.05} = 0.7 × f_{ctm} = 1.8 N/mm^{2}

Secant modulus of elasticity of concrete; E_{cm} = 22 kN/mm^{2} × [f_{cm}/10]^{0.3} = 31476 N/mm^{2}

Partial factor for concrete (Table 2.1N); γ_{C} = 1.50

Compressive strength coefficient (cl.3.1.6(1)); a_{cc} = 0.85

Design compressive concrete strength (exp.3.15); f_{cd} = a_{cc} × (f_{ck} / γ_{C}) = 14.2 N/mm^{2}

Tens.strength coeff.for plain concrete (cl.12.3.1(1)); a_{ct,pl} = 0.80

Des.tens.strength for plain concrete (exp.12.1); f_{ctd,pl} = a_{ct,pl} × (f_{ctk,0.05} / γ_{C}) = 1.0 N/mm^{2}

Maximum aggregate size; h_{agg} = 20 mm

Ultimate strain – Table 3.1; ε_{cu2} = 0.0035

Shortening strain – Table 3.1;ε_{cu3} = 0.0035

Effective compression zone height factor; λ = 0.80

Effective strength factor; h = 1.00

Bending coefficient k_{1}; K_{1} = 0.40

Bending coefficient k_{2}; K_{2} = 1.00 × (0.6 + 0.0014/ε_{cu2}) =1.00

Bending coefficient k_{3}; K_{3} =0.40

Bending coefficient k_{4}; K_{4} =1.00 × (0.6 + 0.0014/ε_{cu2}) = 1.00

**Reinforcement details**

Characteristic yield strength of reinforcement; f_{yk} = 500 N/mm^{2}

Modulus of elasticity of reinforcement; E_{s} = 210000 N/mm^{2}

Partial factor for reinforcing steel (Table 2.1N); γ_{S} = 1.15

Design yield strength of reinforcement; f_{yd} = f_{yk} / γ_{S} = 435 N/mm^{2}

Nominal cover to reinforcement; c_{nom} = 50 mm

**Rectangular section in flexure** **(x-axis)**

Design bending moment; M_{Ed.x.max} = 160.7 kNm

Depth to tension reinforcement; d = h – c_{nom} – φ_{x.bot} / 2 = 444 mm

K = M_{Ed.x.max} / (L_{y} × d^{2} × f_{ck}) = 0.022

K’ = (2 × h × a_{cc}/γ_{C}) × (1 – λ(d – K_{1})/(2K_{2})) × (λ(d – K_{1})/(2K_{2}))

K’ = 0.207

K’ > K – No compression reinforcement is required

Lever arm; z = min(0.5 + 0.5 × (1 – 2K / (h × a_{cc} /γ_{c}))^{0.5}, 0.95) × d = 422 mm

Depth of neutral axis; x = 2.5(d – z) = 55 mm

Area of tension reinforcement required;

A_{sx.bot.req} = M_{Ed.x.max} / (f_{yd}z) = 876 mm^{2}

Tension reinforcement provided;

10Y12@155 c/c bottom (A_{sx.bot.prov} = 1131 mm^{2})

Minimum area of reinforcement (exp.9.1N);

A_{s.min} = max(0.26 × f_{ctm} / f_{yk}, 0.0013) × L_{y} × d = 888 mm^{2}

Maximum area of reinforcement (cl.9.2.1.1(3));

A_{s.max} = 0.04 × L_{y} × d = 26640 mm^{2}

PASS – Area of reinforcement provided is greater than area of reinforcement required

**Rectangular section in shear** **(x-axis)**

Design shear force;

abs(V_{Ed.x.min}) = 162.8 kN

C_{Rd,c} = 0.18 /γ_{C} = 0.120

k = min(1 + √(200 mm / d), 2) = 1.680

Longitudinal reinforcement ratio;

ρ_{l} = min(A_{sx.bot.prov} / (L_{y} × d), 0.02) = 0.002

v_{min} = 0.035k^{3/2} × f_{ck}^{0.5} = 0.381 N/mm^{2}

Design shear resistance (exp.6.2a & 6.2b);

V_{Rd.c} = max(C_{Rd.c} × k × (100 N^{2}/mm^{4} × ρ_{l} × f_{ck})^{1/3}, v_{min}) × L_{y} × d

V_{Rd.c} = 247 kN

PASS – Design shear resistance exceeds design shear force

**Rectangular section in flexure (y-axis)**

Design bending moment;

M_{Ed.y.max} = 157.5 kNm

Depth to tension reinforcement;

d = h – c_{nom} – f_{x.bot} – φ_{y.bot} / 2 = 432 mm

K = M_{Ed.y.max} / (L_{x} × d^{2} × f_{ck}) = 0.02

K’ = (2h × a_{cc}/γ_{C}) × (1 – λ × (d – K_{1})/(2K_{2})) × (λ × (d – K_{1})/(2K_{2}))

K’ = 0.207

K’ > K – No compression reinforcement is required

Lever arm;

z = min(0.5 + 0.5 × (1 – 2K / (h × a_{cc}/γ_{C}))^{0.5}, 0.95) × d = 410 mm

Depth of neutral axis;

x = 2.5(d – z) = 54 mm

Area of tension reinforcement required;

A_{sy.bot.req} = M_{Ed.y.max} / (f_{yd}z) = 883 mm^{2}

Tension reinforcement provided;

12Y12@125 c/c A_{sy.bot.prov} = 1357 mm^{2}

Minimum area of reinforcement (exp.9.1N);

A_{s.min} = max(0.26f_{ctm} / f_{yk}, 0.0013) × L_{x} × d = 864 mm^{2}

Maximum area of reinforcement (cl.9.2.1.1(3));

A_{s.max} = 0.04 × L_{x} × d = 25920 mm^{2}

PASS – Area of reinforcement provided is greater than the area of reinforcement required

**Crack control **

Limiting crack width; w_{max} = 0.3 mm

Variable load factor (EN1990 – Table A1.1); y_{2} = 0.3

Serviceability bending moment; M_{sls.y.max} = 99 kNm

Tensile stress in reinforcement; s_{s} = M_{sls.y.max} / (A_{sy.bot.prov} × z) = 177.8 N/mm^{2}

Load duration factor; k_{t} = 0.4

Effective depth of concrete in tension;

h_{c.ef} = min(2.5 × (h – d), (h – x) / 3, h/2) = 149 mm

Effective area of concrete in tension;

A_{c.eff} = h_{c.ef} × L_{x} = 223000 mm^{2}

Mean value of concrete tensile strength;

f_{ct.eff} = f_{ctm} = 2.6 N/mm^{2}

Reinforcement ratio;

ρ_{p.eff} = A_{sy.bot.prov} / A_{c.eff} = 0.006

Modular ratio; a_{e} = E_{s} / E_{cm} = 6.672

Bond property coefficient; k_{1} = 0.8

Strain distribution coefficient; k_{2} = 0.5

k_{3} = 3.4

k_{4} = 0.425

Maximum crack spacing (exp.7.11);

s_{r.max} = k_{3} × (c_{nom} + f_{x.bot}) + k_{1}k_{2}k_{4} × φ_{y.bot} / ρ_{p.eff} = **546** mm

Maximum crack width (exp.7.8);

w_{k} = s_{r.max} × max([s_{s} – k_{t} × (f_{ct.eff} / ρ_{p.eff}) × (1 + a_{e} × ρ_{p.eff})] / E_{s}, 0.6 × s_{s} / E_{s}) = **0.277** mm

PASS – Maximum crack width is less than limiting crack width

**Rectangular section in shear (y-axis)**

Design shear force; abs(V_{Ed.y.min}) = 159.1 kN

C_{Rd,c} = 0.18/γ_{C} = 0.120

k = min(1 + √(200 mm / d), 2) = 1.680

Longitudinal reinforcement ratio;

r_{l} = min(A_{sy.bot.prov} / (L_{x} × d), 0.02) = 0.002

v_{min} = 0.035k^{3/2} × f_{ck}^{0.5} = 0.381 N/mm^{2}

Design shear resistance (exp.6.2a & 6.2b);

V_{Rd.c} = max(C_{Rd.c} × k × (100 × r_{l} × f_{ck})^{1/3}, v_{min}) × L_{x} × d

V_{Rd.c} = 247 kN

PASS – Design shear resistance exceeds design shear force

**Punching shear**

Strength reduction factor (exp 6.6N); v = 0.6[1 – f_{ck} / 250] = 0.540

Average depth to reinforcement; d = 438 mm

Maximum punching shear resistance (cl.6.4.5(3)); v_{Rd.max} = 0.5vf_{cd} = 3.825 N/mm^{2}

k = min(1 + √(200 mm / d), 2) = 1.676

Longitudinal reinforcement ratio (cl.6.4.4(1));

r_{lx} = A_{sx.bot.prov} / (L_{y} × d) = 0.002

r_{ly} = A_{sy.bot.prov} / (L_{x} × d) = 0.002

r_{l} = min(√(r_{lx} × r_{ly}), 0.02) = 0.002

C_{Rd,c} = 0.18 / g_{C} =0.120

v_{min} = 0.035 k^{3/2} × f_{ck}^{0.5} = 0.380 N/mm^{2}

Design punching shear resistance (exp.6.47);

v_{Rd.c} = max(C_{Rd.c} k (100r_{l}f_{ck})^{1/3}, v_{min}) = 0.380 N/mm^{2}

Design punching shear resistance at 1d (exp. 6.50);

v_{Rd.c1} = (2d/d)v_{Rd.c} = 0.759 N/mm^{2}

**Punching shear perimeter at column face**

Punching shear perimeter; u_{0} = 1000 mm

Area within punching shear perimeter; A_{0} = 0.063 m^{2}

Maximum punching shear force; V_{Ed.max} = 1046 kN

Punching shear stress factor (fig 6.21N); β = 1.500

Maximum punching shear stress (exp 6.38);

v_{Ed.max} = β V_{Ed.max} / (u_{0} × d) = 3.582 N/mm^{2}

PASS – Maximum punching shear resistance exceeds maximum punching shear stress

**Punching shear perimeter at 1d from column face**

Punching shear perimeter; u_{1} = 3752 mm

Area within punching shear perimeter; A_{1} = 1.103 m^{2}

Design punching shear force; V_{Ed.1} = 480.5 kN

Punching shear stress factor (fig 6.21N); β = 1.500

Design punching shear stress (exp 6.38); v_{Ed.1} = βV_{Ed.1} / (u_{1}d) = 0.439 N/mm^{2}

PASS – Design punching shear resistance exceeds increased design punching shear stress

**Punching shear perimeter at 2d from column face**

Punching shear perimeter; u_{2} = 63 mm

Area within punching shear perimeter; A_{2} = 2.250 m^{2}

Design punching shear force; V_{Ed.2} = 0 kN

Punching shear stress factor (fig 6.21N); β = 1.500

Design punching shear stress (exp 6.38); v_{Ed.2} = βV_{Ed.2} / (u_{2} × d) = 0.001 N/mm^{2}

PASS – Design punching shear resistance exceeds design punching shear stress