pad foundation with two columns example

CSCE LTD.P.O BOX 21030, KAMPALA.

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FOUNDATION ANALYSIS (EN1997-1:2004)

In accordance with EN1997-1:2004 incorporating Corrigendum dated February 2009 and the UK National Annex

incorporating Corrigendum No.1TEDDS calculation version 3.2.02

Pad foundation details

Length of foundation; Lx = 3500 mm

Width of foundation; Ly = 2500 mm

Foundation area; A = Lx Ly = 8.750 m2

Depth of foundation; h = 400 mm

Depth of soil over foundation; hsoil = 600 mm

Level of water; hwater = 0 mm

Density of water; water = 9.8 kN/m3

Density of concrete; conc = 24.5 kN/m3

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2

63.8 kN/m 2

93.4 kN/m 2

84.9 kN/m 2

114.6 kN/m 2

x

y

Column no.1 details

Length of column; lx1 = 300 mm

Width of column; ly1 = 300 mm

position in x-axis; x1 = 750 mm

position in y-axis; y1 = 1750 mm

Column no.2 details

Length of column; lx2 = 300 mm

Width of column; ly2 = 300 mm

position in x-axis; x2 = 2750 mm

position in y-axis; y2 = 750 mm

Soil properties

Density of soil; soil = 18.0 kN/m3

Characteristic cohesion; c'k = 0 kN/m2


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Characteristic effective shear resistance angle; 'k = 30 deg

Characteristic friction angle; k = 20 deg

Foundation loads

Self weight; Fswt = h conc = 9.8 kN/m2

Soil weight; Fsoil = hsoil soil = 10.8 kN/m2

Column no.1 loads

Permanent load in x; FGx1 = 20.0 kN

Permanent load in y; FGy1 = 20.0 kN

Permanent load in z; FGz1 = 200.0 kN

Variable load in x; FQx1 = 10.0 kN

Variable load in y; FQy1 = 10.0 kN

Variable load in z; FQz1 = 100.0 kN

Permanent moment in x; MGx1 = 10.0 kNm

Permanent moment in y; MGy1 = 10.0 kNm

Variable moment in x; MQx1 = 5.0 kNm

Variable moment in y; MQy1 = 5.0 kNm

Column no.2 loads

Permanent load in x; FGx2 = 20.0 kN

Permanent load in y; FGy2 = 20.0 kN

Permanent load in z; FGz2 = 200.0 kN

Variable load in x; FQx2 = 10.0 kN

Variable load in y; FQy2 = 10.0 kN

Variable load in z; FQz2 = 100.0 kN

Permanent moment in x; MGx2 = 10.0 kNm

Permanent moment in y; MGy2 = 10.0 kNm

Variable moment in x; MQx2 = 5.0 kNm

Variable moment in y; MQy2 = 5.0 kNm

Bearing resistance (Section 6.5.2)

Forces on foundation

Force in x-axis; Fdx = FGx1 + FGx2 + FQx1 + FQx2 = 60.0 kN

Force in y-axis; Fdy = FGy1 + FGy2 + FQy1 + FQy2 = 60.0 kN

Force in z-axis; Fdz = A (Fswt + Fsoil) + FGz1 + FGz2 + FQz1 + FQz2 = 780.3 kN

Moments on foundation

Moment in x-axis; Mdx = A (Fswt + Fsoil) Lx / 2 + FGz1 x1 + MGx1 + FGx1 h + FGz2 x2 +

MGx2 + FGx2 h + FQz1 x1 + MQx1 + FQx1 h + FQz2 x2 + MQx2 + FQx2

h = 1419.4 kNm

Moment in y-axis; Mdy = A (Fswt + Fsoil) Ly / 2 + FGz1 y1 + MGy1 + FGy1 h + FGz2 y2 +

MGy2 + FGy2 h + FQz1 y1 + MQy1 + FQy1 h + FQz2 y2 + MQy2 + FQy2

h = 1029.3 kNm

Eccentricity of base reaction

Eccentricity of base reaction in x-axis; ex = Mdx / Fdz - Lx / 2 = 69 mm

Eccentricity of base reaction in y-axis; ey = Mdy / Fdz - Ly / 2 = 69 mm

Pad base pressures

q1 = Fdz (1 - 6 ex / Lx - 6 ey / Ly) / (Lx Ly) = 63.8 kN/m2

q2 = Fdz (1 - 6 ex / Lx + 6 ey / Ly) / (Lx Ly) = 93.4 kN/m2


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q3 = Fdz (1 + 6 ex / Lx - 6 ey / Ly) / (Lx Ly) = 84.9 kN/m2

q4 = Fdz (1 + 6 ex / Lx + 6 ey / Ly) / (Lx Ly) = 114.6 kN/m2

Minimum base pressure; qmin = min(q1, q2, q3, q4) = 63.8 kN/m2

Maximum base pressure; qmax = max(q1, q2, q3, q4) = 114.6 kN/m2

Presumed bearing capacity

Presumed bearing capacity; Pbearing = 200.0 kN/m2

PASS - Presumed bearing capacity exceeds design base pressure

Partial factors on actions - Combination1

Permanent unfavourable action - Table A.3; G = 1.35

Permanent favourable action - Table A.3; Gf = 1.00

Variable unfavourable action - Table A.3; Q = 1.50

Variable favourable action - Table A.3; Qf = 0.00

Partial factors for spread foundations - Combination1

Bearing - Table A.5; R.v = 1.00

Sliding - Table A.5; R.h = 1.00


Force in x-axis; Fdx = G (FGx1 + FGx2) + Q (FQx1 + FQx2) = 84.0 kN

Force in y-axis; Fdy = G (FGy1 + FGy2) + Q (FQy1 + FQy2) = 84.0 kN

Force in z-axis; Fdz = G (A (Fswt + Fsoil) + FGz1 + FGz2) + Q (FQz1 + FQz2) = 1083.3

kN


Moment in x-axis; Mdx = G (A (Fswt + Fsoil) Lx / 2 + FGz1 x1 + FGz2 x2) + G (MGx1 +

MGx2) + Q (FQz1 x1 + FQz2 x2) + Q (MQx1 + MQx2) + (G (FGx1 +

FGx2) + Q (FQx1 + FQx2)) h = 1971.4 kNm

Moment in y-axis; Mdy = G (A (Fswt + Fsoil) Ly / 2 + FGz1 y1 + FGz2 y2) + G (MGy1 +

MGy2) + Q (FQz1 y1 + FQz2 y2) + Q (MQy1 + MQy2) + (G (FGy1 +

FGy2) + Q (FQy1 + FQy2)) h = 1429.8 kNm




Effective area of base

Effective length; L'x = Lx - 2 ex = 3360 mm

Effective width; L'y = Ly - 2 ey = 2360 mm

Effective area; A' = L'x L'y = 7.932 m2

Pad base pressure

Design base pressure; fdz = Fdz / A' = 136.6 kN/m2

Sliding resistance (Section 6.5.3)




Force in z-axis; Fdz = Gf (A (Fswt + Fsoil) + FGz1 + FGz2) + Qf (FQz1 + FQz2) = 580.3

kN


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Sliding resistance verification (Section 6.5.3)

Horizontal force on foundation; H = [Fdx2 + Fdy

2]0.5 = 118.8 kN

Sliding resistance (exp.6.3b); RH.d = Fdz tan(k) / R.h = 211.2 kN

PASS - Foundation is not subject to failure by sliding

Partial factors on actions - Combination2

Permanent unfavourable action - Table A.3; G = 1.00

Permanent favourable action - Table A.3; Gf = 1.00

Variable unfavourable action - Table A.3; Q = 1.30

Variable favourable action - Table A.3; Qf = 0.00

Partial factors for spread foundations - Combination2

Bearing - Table A.5; R.v = 1.00

Sliding - Table A.5; R.h = 1.00




Force in z-axis; Fdz = G (A (Fswt + Fsoil) + FGz1 + FGz2) + Q (FQz1 + FQz2) = 840.3 kN


Moment in x-axis; Mdx = G (A (Fswt + Fsoil) Lx / 2 + FGz1 x1 + FGz2 x2) + G (MGx1 +

MGx2) + Q (FQz1 x1 + FQz2 x2) + Q (MQx1 + MQx2) + (G (FGx1 +

FGx2) + Q (FQx1 + FQx2)) h = 1529.8 kNm

Moment in y-axis; Mdy = G (A (Fswt + Fsoil) Ly / 2 + FGz1 y1 + FGz2 y2) + G (MGy1 +

MGy2) + Q (FQz1 y1 + FQz2 y2) + Q (MQy1 + MQy2) + (G (FGy1 +

FGy2) + Q (FQy1 + FQy2)) h = 1109.7 kNm




Effective area of base

Effective length; L'x = Lx - 2 ex = 3359 mm

Effective width; L'y = Ly - 2 ey = 2359 mm

Effective area; A' = L'x L'y = 7.922 m2

Pad base pressure

Design base pressure; fdz = Fdz / A' = 106.1 kN/m2

Sliding resistance (Section 6.5.3)




Force in z-axis; Fdz = Gf (A (Fswt + Fsoil) + FGz1 + FGz2) + Qf (FQz1 + FQz2) = 580.3

kN

Sliding resistance verification (Section 6.5.3)

Horizontal force on foundation; H = [Fdx2 + Fdy

2]0.5 = 93.3 kN

Sliding resistance (exp.6.3b); RH.d = Fdz tan(k) / R.h = 211.2 kN

PASS - Foundation is not subject to failure by sliding


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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.1TEDDS calculation version 3.2.02

Concrete details (Table 3.1 - Strength and deformation characteristics for concrete)

Concrete strength class; C30/37

Characteristic compressive cylinder strength; fck = 30 N/mm2

Characteristic compressive cube strength; fck,cube = 37 N/mm2

Mean value of compressive cylinder strength; fcm = fck + 8 N/mm2 = 38 N/mm2

Mean value of axial tensile strength; fctm = 0.3 N/mm2 (fck/ 1 N/mm2)2/3 = 2.9 N/mm2

5% fractile of axial tensile strength; fctk,0.05 = 0.7 fctm = 2.0 N/mm2

Secant modulus of elasticity of concrete; Ecm = 22 kN/mm2 [fcm/10 N/mm2]0.3 = 32837 N/mm2

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

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

Design compressive concrete strength (exp.3.15); fcd = cc fck / C = 17.0 N/mm2

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

Des.tens.strength for plain concrete (exp.12.1); fctd,pl = ct,pl fctk,0.05 / C = 1.1 N/mm2

Maximum aggregate size; hagg = 20 mm

Reinforcement details

Characteristic yield strength of reinforcement; fyk = 500 N/mm2

Modulus of elasticity of reinforcement; Es = 210000 N/mm2

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

Design yield strength of reinforcement; fyd = fyk / S = 435 N/mm2

Nominal cover to reinforcement; cnom = 30 mm

Rectangular section in flexure (Section 6.1)

Design bending moment; MEd.x.max = 78.3 kNm

Depth to tension reinforcement; d = h - cnom - y.bot - x.bot / 2 = 352 mm

K = MEd.x.max / (Ly d2 fck) = 0.008

K' = 0.207

K' > K - No compression reinforcement is required

Lever arm; z = min((d / 2) [1 + (1 - 3.53 K)0.5], 0.95 d) = 334 mm

Depth of neutral axis; x = 2.5 (d - z) = 44 mm

Area of tension reinforcement required; Asx.bot.req = MEd.x.max / (fyd z) = 539 mm2

Tension reinforcement provided; 12 No.12 dia.bars bottom (225 c/c)

Area of tension reinforcement provided; Asx.bot.prov = 1357 mm2

Minimum area of reinforcement (exp.9.1N); As.min = max(0.26 fctm / fyk, 0.0013) Ly d = 1325 mm2

Maximum area of reinforcement (cl.9.2.1.1(3)); As.max = 0.04 Ly d = 35200 mm2

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

Crack control (Section 7.3)

Limiting crack width; wmax = 0.3 mm

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

Serviceability bending moment; Msls.x.max = 42.8 kNm

Tensile stress in reinforcement; s = Msls.x.max / (Asx.bot.prov z) = 94.3 N/mm2

Load duration factor; kt = 0.4

Effective depth of concrete in tension; hc.ef = min(2.5 (h - d), (h - x) / 3, h / 2) = 119 mm


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Effective area of concrete in tension; Ac.eff = hc.ef Ly = 296667 mm2

Mean value of concrete tensile strength; fct.eff = fctm = 2.9 N/mm2

Reinforcement ratio; p.eff = Asx.bot.prov / Ac.eff = 0.005

Modular ratio; e = Es / Ecm = 6.395

Bond property coefficient; k1 = 0.8

Strain distribution coefficient; k2 = 0.5

k3 = 3.4

k4 = 0.425

Maximum crack spacing (exp.7.11); sr.max = k3 (cnom + y.bot) + k1 k2 k4 x.bot / p.eff = 589 mm

Maximum crack width (exp.7.8); wk = sr.max max([s – kt (fct.eff / p.eff) (1 + e p.eff)] / Es,

0.6 s / Es) = 0.159 mm

PASS - Maximum crack width is less than limiting crack widthRectangular section in flexure (Section 6.1)

Design bending moment; abs(MEd.x.min) = 71.6 kNm

Depth to tension reinforcement; d = h - cnom - y.top - x.top / 2 = 352 mm

K = abs(MEd.x.min) / (Ly d2 fck) = 0.008

K' = 0.207




Area of tension reinforcement required; Asx.top.req = abs(MEd.x.min) / (fyd z) = 493 mm2

Tension reinforcement provided; 12 No.12 dia.bars top (225 c/c)

Area of tension reinforcement provided; Asx.top.prov = 1357 mm2

Minimum area of reinforcement (exp.9.1N); As.min = max(0.26 fctm / fyk, 0.0013) Ly d = 1325 mm2

Maximum area of reinforcement (cl.9.2.1.1(3)); As.max = 0.04 Ly d = 35200 mm2





Serviceability bending moment; abs(Msls.x.min) = 39.2 kNm

Tensile stress in reinforcement; s = abs(Msls.x.min) / (Asx.top.prov z) = 86.5 N/mm2



Effective area of concrete in tension; Ac.eff = hc.ef Ly = 296667 mm2


Reinforcement ratio; p.eff = Asx.top.prov / Ac.eff = 0.005




k3 = 3.4

k4 = 0.425

Maximum crack spacing (exp.7.11); sr.max = k3 (cnom + y.top) + k1 k2 k4 x.top / p.eff = 589 mm


0.6 s / Es) = 0.145 mm

PASS - Maximum crack width is less than limiting crack widthRectangular section in shear (Section 6.2)

Design shear force; abs(VEd.x.min) = 150.2 kN


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CRd,c = 0.18 / C = 0.120

k = min(1 + (200 mm / d), 2) = 1.754

Longitudinal reinforcement ratio; l = min(Asx.bot.prov / (Ly d), 0.02) = 0.002

vmin = 0.035 N1/2/mm k3/2 fck0.5 = 0.445 N/mm2

Design shear resistance (exp.6.2a & 6.2b); VRd.c = max(CRd.c k (100 N2/mm4 l fck)1/3, vmin) Ly d

VRd.c = 391.8 kN

PASS - Design shear resistance exceeds design shear force

Rectangular section in flexure (Section 6.1)

Design bending moment; MEd.y.max = 95.9 kNm

Depth to tension reinforcement; d = h - cnom - y.bot / 2 = 364 mm

K = MEd.y.max / (Lx d2 fck) = 0.007

K' = 0.207




Area of tension reinforcement required; Asy.bot.req = MEd.y.max / (fyd z) = 638 mm2

Tension reinforcement provided; 18 No.12 dia.bars bottom (225 c/c)

Area of tension reinforcement provided; Asy.bot.prov = 2036 mm2

Minimum area of reinforcement (exp.9.1N); As.min = max(0.26 fctm / fyk, 0.0013) Lx d = 1919 mm2

Maximum area of reinforcement (cl.9.2.1.1(3)); As.max = 0.04 Lx d = 50960 mm2





Serviceability bending moment; Msls.y.max = 52.4 kNm

Tensile stress in reinforcement; s = Msls.y.max / (Asy.bot.prov z) = 74.5 N/mm2



Effective area of concrete in tension; Ac.eff = hc.ef Lx = 315000 mm2


Reinforcement ratio; p.eff = Asy.bot.prov / Ac.eff = 0.006




k3 = 3.4

k4 = 0.425

Maximum crack spacing (exp.7.11); sr.max = k3 cnom + k1 k2 k4 y.bot / p.eff = 418 mm


0.6 s / Es) = 0.089 mm

PASS - Maximum crack width is less than limiting crack widthRectangular section in flexure (Section 6.1)

Design bending moment; abs(MEd.y.min) = 1.2 kNm

Depth to tension reinforcement; d = h - cnom - y.top / 2 = 364 mm

K = abs(MEd.y.min) / (Lx d2 fck) = 0.000

K' = 0.207



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Area of tension reinforcement required; Asy.top.req = abs(MEd.y.min) / (fyd z) = 8 mm2

Tension reinforcement provided; 18 No.12 dia.bars top (225 c/c)

Area of tension reinforcement provided; Asy.top.prov = 2036 mm2

Minimum area of reinforcement (exp.9.1N); As.min = max(0.26 fctm / fyk, 0.0013) Lx d = 1919 mm2

Maximum area of reinforcement (cl.9.2.1.1(3)); As.max = 0.04 Lx d = 50960 mm2





Serviceability bending moment; abs(Msls.y.min) = 0.8 kNm

Tensile stress in reinforcement; s = abs(Msls.y.min) / (Asy.top.prov z) = 1.2 N/mm2



Effective area of concrete in tension; Ac.eff = hc.ef Lx = 315000 mm2


Reinforcement ratio; p.eff = Asy.top.prov / Ac.eff = 0.006




k3 = 3.4

k4 = 0.425

Maximum crack spacing (exp.7.11); sr.max = k3 cnom + k1 k2 k4 y.top / p.eff = 418 mm


0.6 s / Es) = 0.001 mm

PASS - Maximum crack width is less than limiting crack widthRectangular section in shear (Section 6.2)

Design shear force; abs(VEd.y.min) = 33.8 kN

CRd,c = 0.18 / C = 0.120

k = min(1 + (200 mm / d), 2) = 1.741

Longitudinal reinforcement ratio; l = min(Asy.bot.prov / (Lx d), 0.02) = 0.002

vmin = 0.035 N1/2/mm k3/2 fck0.5 = 0.440 N/mm2

Design shear resistance (exp.6.2a & 6.2b); VRd.c = max(CRd.c k (100 N2/mm4 l fck)1/3, vmin) Lx d

VRd.c = 561.2 kN

PASS - Design shear resistance exceeds design shear force

Punching shear (Section 6.4)

Strength reduction factor (exp 6.6N); v = 0.6 [1 - fck / 250 N/mm2] = 0.528

Average depth to reinforcement; d = 358 mm

Maximum punching shear resistance (cl.6.4.5(3)); vRd.max = 0.5 v fcd = 4.488 N/mm2

k = min(1 + (200 mm / d), 2) = 1.747

Longitudinal reinforcement ratio (cl.6.4.4(1)); lx = Asx.bot.prov / (Ly d) = 0.002

ly = Asy.bot.prov / (Lx d) = 0.002

l = min((lx ly), 0.02) = 0.002

vmin = 0.035 N1/2/mm k3/2 fck0.5 = 0.443 N/mm2

Design punching shear resistance (exp.6.47); vRd.c = max(CRd.c k (100 N2/mm4 l fck)1/3, vmin) = 0.443 N/mm2


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Column No.1 - Punching shear perimeter at column face

Punching shear perimeter; u0 = 1200 mm

Area within punching shear perimeter; A0 = 0.090 m2

Maximum punching shear force; VEd.max = 410.2 kN

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

Maximum punching shear stress (exp 6.38); vEd.max = VEd.max / (u0 d) = 1.432 N/mm2

PASS - Maximum punching shear resistance exceeds maximum punching shear stress

Column No.1 - Punching shear perimeter at 2d from column face



Design punching shear force; VEd.2 = 186.4 kN


Design punching shear stress (exp 6.38); vEd.2 = VEd.2 / (u2 d) = 0.227 N/mm2

PASS - Design punching shear resistance exceeds design punching shear stress

Column No.2 - Punching shear perimeter at column face



Maximum punching shear force; VEd.max = 410.2 kN


Maximum punching shear stress (exp 6.38); vEd.max = VEd.max / (u0 d) = 1.432 N/mm2

PASS - Maximum punching shear resistance exceeds maximum punching shear stress

Column No.2 - Punching shear perimeter at 2d from column face



Design punching shear force; VEd.2 = 186.4 kN


Design punching shear stress (exp 6.38); vEd.2 = VEd.2 / (u2 d) = 0.227 N/mm2

PASS - Design punching shear resistance exceeds design punching shear stress

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12 No.12 dia.bars bottom (225 c/c)

12 No.12 dia.bars top (225 c/c)

18 No.12 dia.bars bottom (225 c/c)

18 No.12 dia.bars top (225 c/c)


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pad foundation with two columns example

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