Pad Footing Example

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PAD FOOTING ANALYSIS AND DE SIGN (BS8110-1:1997) TEDDS calculation version 2.0.03.00

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Pad footing details Length of pad footing footing;;

L = 2500 mm

Width of pad footing; footing;

B = 1500 m m

Area of pad footing; footing;

A = L × B = 3.750 m 2

Depth of pad footing ;

40 0 mm h = 400

Depth of soil over pad footing; footing ;

h soil = 20 0 mm

Density of concrete; concrete ;

ρ conc = 23.6 kN/m 3

Column details Column base length; length ;

30 0 mm l A = 300

Column base width; width;

b A = 30 0 mm

Column eccentricity in x ;

e PxA = 0 mm

Column eccentricity in y ;

e PyA = 0 mm

Soil details

Design base friction; friction ;

ρ soil = 20.0 kN/m 3 φ’ = 25.0 de g δ = 19.3 deg de g

Allowable Allowable bearing pressure; pressure;

15 0 kN/m 2 P bearing = 150

Density of soil; soil ; Design shear strength; strength ;

Axial loading on column Dead axial load on column ;

P GA = 200.0 kN

Imposed axial load on column; column ;

P QA = 165.0 kN

Wind axial load on column; column ;

0. 0 kN P W A = 0.0

Total axial load load on column; column ;

P A = 365.0 kN

Foundation loads Dead surcharge load; load ;

F Gsur = 0.000 kN/m 2

Imposed surcharge load; load ;

F Qsur = 0.000 kN/m 2

Pad footing self weight; weight ;

F swt = h × ρ conc = 9.440 kN/m 2

Soil self weight; weight ;

F soil = h soil × ρ soil = 4.000 kN/m 2

Total foundation load; load ;

F = A × (F Gsur + F Qsur + F swt + F soil) = 50.4 kN

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Moment on column base Dead moment o n column in x direction;

M GxA = 15.000 kNm

Imposed moment on column in x direction;

M QxA = 10.000 kNm

Wind mo ment on column in x direction;

M Wx A = 0.000 kNm

Total moment on column in x direction;

M xA = 25.000 kNm

Dead moment on column in y direction;

M GyA = 0.000 kNm

Imposed moment on column in y direction;

M QyA = 0.000 kNm

Wind mo ment on column in y direction;

M WyA = 0.000 kNm

Total moment on column in y direction;

M yA = 0.000 kNm

Check stability against overturning in x direction Total overturning mo ment;

M xOT = M xA + H xA × h = 25.000 kN m

Restoring mo ment in x direction Foundation loading;

M xsur = A × (F Gsur + F swt + F soil) × L / 2 = 63.000 kNm

Axial loading on column;

M xaxial = (P GA ) × (L / 2 - e PxA ) = 250.000 kNm

Total restoring mom ent;

M xres = M xsur + M xaxial = 313.000 kNm PASS - Restoring moment is greater than overturning moment in x direction 

Calculate pad base reaction Total base reaction;

T = F + P A = 415.4 kN

Eccentricity of base reaction in x;

e Tx = (P A × e PxA + M xA + H xA × h) / T = 60 mm

Eccentricity of base reaction in y;

e Ty = (P A × ePyA + M yA + H yA × h) / T = 0 mm

Check pad base reaction eccentricity abs(eTx ) / L + abs(e Ty) / B = 0.024 Base reaction acts within middle third of base  Calculate pad base pressures q 1 = T / A - 6 × T × eTx / (L × A) - 6 × T × e Ty / (B × A) = 94.773 kN/m 2 q 2 = T / A - 6 × T × eTx / (L × A) + 6 × T × e Ty / (B × A) = 94.773 kN/m 2 q 3 = T / A + 6 × T × e Tx / (L × A) - 6 × T × e Ty / (B × A) = 126.773 kN/m 2 q 4 = T / A + 6 × T × e Tx / (L × A) + 6 × T × eTy / (B × A) = 126.773 kN/m 2 Minimum base pressure; Maximum base pressure;

q min = min(q 1 , q 2 , q 3 , q 4) = 94.773 kN/m 2 q max = max(q 1 , q 2 , q 3 , q 4) = 126.773 kN/m 2 PASS - Maximum base pressure is less than allowable bearing pressure 

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94.8 kN/m

126.8 kN/m

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94.8 kN/m

126.8 kN/m

Partial safety factors for loads Partial safety factor for dead loads; Partial safety factor for imposed loads; Partial safety factor for wind loads ;

γ fG = 1.40 γ fQ = 1.60 γ fW = 0.00

Ultimate axial loading on column Ultimate axial load on column;

P uA = P GA × γ fG + P QA × γ fQ + P W A × γ fW = 544.0 kN

Ultimate foundation loads Ultimate foundation load;

F u = A × [(F Gsur + F swt + F soil) × γ fG + F Qsur × γ fQ ] = 70.6 kN

Ultimate horizontal loading on co lumn Ultimate horizontal load in x direction;

H xuA = H GxA × γ fG + H QxA × γ fQ + H Wx A × γ fW = 0.0 kN

Ultimate horizontal load in y direction ;

H yuA = H GyA × γ fG + H QyA × γ fQ + H W yA × γ fW = 0.0 kN

Ultimate moment on column Ultimate mo ment on column in x d irection;

M xuA = M GxA × γ fG + M QxA × γ fQ + MW xA × γ fW = 37.000 kNm

Ultimate moment on column in y direction ;

M yuA = M GyA × γ fG + M QyA × γ fQ + M WyA × γ fW = 0.000 kNm

Calculate ultimate pad base reaction Ultimate base reaction;

T u = F u + P uA = 614.6 kN

Eccentricity of ultimate base reaction in x ;

e Txu = (P uA × e PxA + M xuA + H xuA × h) / T u = 60 mm

Eccentricity of ultimate base reaction in y ;

e Tyu = (P uA × ePyA + M yuA + H yuA × h) / T u = 0 mm

Calculate ultimate pad base pressures q 1u = T u /A - 6 × Tu × e Txu /(L × A) - 6 × Tu × eTyu /(B × A) = 140.203 kN/m 2 q 2u = T u /A - 6 × Tu × e Txu /(L × A) + 6 × T u × e Tyu /(B × A) = 140.203 kN/m 2 q 3u = T u /A + 6 × T u × e Txu /(L × A) - 6 × T u × e Tyu /(B × A) = 187.563 kN/m 2 q 4u = T u /A + 6 × T u × e Txu /(L × A) + 6 × T u × e Tyu /(B × A) = 187.563 kN/m 2 Minimum ultimate base pressure;

q minu = min(q 1u , q 2u , q 3u , q 4u ) = 140.203 kN/m 2

Maximum ultimate base p ressure;

q maxu = max(q 1u , q 2u , q 3u , q 4u ) = 187.563 kN/m 2

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Calculate rate of change of ba se pressure in x direction Left hand base reaction;

f uL = (q 1u + q 2u ) × B / 2 = 210.304 kN/m

Right hand base reaction;

f uR = (q 3u + q 4u ) × B / 2 = 281.344 kN/m

Length of base reaction;

L x = L = 2500 mm

Rate of change of base pressure;

C x = (fuR - f uL ) / L x = 28.416 kN/m/m

Calculate pad lengths in x direction Left hand length;

L L = L / 2 + e PxA = 1250 mm

Right hand length;

L R = L / 2 - e PxA = 1250 mm

Calculate ultimate mo ments in x direction M x = f uL × LL 2 /2+ C x× L L3 /6-Fu × L L 2 /(2 × L)+MxuA = 188.500 kN m

Ultimate moment in x direction;

Calculate rate of change of base pressure in y direction Top edge base reaction;

f uT = (q 2u + q 4u ) × L / 2 = 409.707 kN/m

Bottom edge base reaction;

f uB = (q 1u + q 3u ) × L / 2 = 409.707 kN/m

Length of base reaction;

L y = B = 1500 mm

Rate of change of base pressure;

C y = (fuB - f uT) / L y = 0.000 kN/m/m

Calculate pad lengths in y direction Top length;

L T = B / 2 - e PyA = 750 mm

Bottom length;

L B = B / 2 + e PyA = 750 mm

Calculate ultimate moments in y direction M y = f uT × LT 2 /2+ C y× L T3 /6-F u × LT 2 /(2 × B) = 102.000 kNm

Ultimate moment in y direction; Material details Characteristic strength of concrete ;

f cu = 30 N/mm 2

Characteristic strength of reinforcement;

f y = 500 N/mm 2

Characteristic strength of shear reinforcement ;

f yv = 500 N/mm 2

Nominal cover to reinforcement;

c nom = 30 mm

Moment design in x direction Diameter of tension reinforcement;

φxB = 12 mm

Depth of tension reinforcement;

d x = h - c nom - φxB / 2 = 36 4 mm

Design formula for rectangular beams (cl 3.4.4.4) K x = M x / (B × dx 2 × fcu ) = 0.032 K x’ = 0.156 K x  < K x ' compression reinforcemen t is not required 

Lever arm;

z x = d x × min([0.5 + √ (0.25 - K x / 0.9)], 0.95) = 346 mm

Area of tension reinforcement required;

A s_x_req = M x / (0.87 × fy × z x) = 1253 mm 2

Minimum area of tension reinforcement;

A s_x_min = 0.0013 × B × h = 780 mm 2

Tension reinforcement p rovided;

12 No. 12 d ia. bars bottom (125 centres)

Area of tension reinforcement provided;

A s_xB_prov = N xB × π × φxB 2 / 4 = 1357 mm 2

PASS - Tension reinforcement provided exceeds tension reinforcement required 

Moment design in y direction Diameter of tension reinforcement;

φyB = 12 mm

Depth of tension reinforcement;

d y = h - c nom - φxB - φyB / 2 = 352 mm

Design formula for rectangular beams (cl 3.4.4.4) K y = M y / (L × dy2 × fcu ) = 0.011 K y’ = 0.156

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6/27/2012 K y  < K y ' compression reinforcemen t is not required 

Lever arm;

z y = d y × min([0.5 + √ (0.25 - K y / 0.9)], 0.95) = 33 4 mm

Area of tension reinforcement required;

A s_y_req = M y / (0.87 × fy × zy) = 701 mm 2

Minimum area of tension reinforcement;

A s_y_min = 0.0013 × L × h = 1300 mm 2

Tension reinforcement p rovided;

13 No. 12 d ia. bars bottom (200 centres)

Area of tension reinforcement provided;

A s_yB_prov = N yB × π × φyB2 / 4 = 1470 mm 2

PASS - Tension reinforcement provided exceeds tension reinforcement required 

Calculate ultimate shear force at d from right face of column Ultimate pressure for shear;

q su = (q 1u + C x × (L / 2 + e PxA + l A / 2 + d x) / B + q 4u ) / 2 q su = 180.591 kN/m 2

Area loaded for shear;

A s = B × min(3 × (L / 2 - e Tx ), L / 2 - e PxA - l A / 2 - d x) = 1.104 m 2

Ultimate shear force;

V su = A s × (q su - F u / A) = 178.600 kN

Shear stresses at d from right face of column (cl 3.5.5.2) Design shear stress;

v su = V su / (B × d x) = 0.327 N/mm 2

From BS 8110 :Part 1:1997 - Table 3.8 Design concrete shear stress ;

v c = 0.432 N/mm 2

Allowable design shear stress ;

v max = min(0.8N/mm 2 × √ (fcu / 1 N/mm 2), 5 N/mm 2) = 4.382 N/mm 2 PASS - v su  < v c  - No shear reinforcement required 

Calculate ultimate punching shear force at face of column Ultimate pressure for punch ing shear;

q puA = q 1u +[(L/2+e PxA -l A /2)+ (l A)/2]× C x /B- [(B/2+ e PyA -b A /2)+(b A )/2]× C y /L q puA = 163.883 kN/m 2

Average effective depth of reinforcement;

d = (d x + d y) / 2 = 35 8 mm

Area loaded for punching shear at column ;

A pA = (l A) × (b A ) = 0.090 m 2

Length of punching shear perimeter;

u pA = 2 × (l A)+2 × (b A) = 1200 mm

Ultimate shear force at shear perimeter ;

V puA = P uA + (F u / A - q puA ) × ApA = 530.944 kN

Effective shear force at shear perimeter;

V puAeff = V puA × [1+1.5× abs(MxuA )/(VpuA × (b A))] = 715.944 kN

Punching shear stresses at face of column (cl 3.7.7.2) Design shear stress;

v puA = V puAeff / (u pA × d) = 1.667 N/mm 2

Allowable design shear stress ;

v max = min(0.8N/mm 2 × √ (fcu / 1 N/mm 2), 5 N/mm 2) = 4.382 N/mm 2 PASS - Design shea r stress is less than allowable design shear stress 

Calculate ultimate punching shear force at perimeter of 1.5 d from face of column Ultimate pressure for punch ing shear;

q puA1.5d = q 1u +[(L/2+e PxA -l A /2-1.5 × d)+(lA +2 × 1.5 × d)/2]× C x /B-[B/2] × C y /L q puA1.5d = 163.883 kN/m 2

Average effective depth of reinforcement;

d = (d x + d y) / 2 = 35 8 mm

Area loaded for punching shear at column ;

A pA1.5d = ( lA +2 × 1.5 × d) × B = 2.061 m 2

Length of punching shear perimeter;

u pA1.5d = 2 × B = 3000 mm

Ultimate shear force at shear perimeter ;

V puA1.5d = P uA + ( F u / A - q puA1.5d ) × A pA1.5d = 245.018 kN

Effective shear force at shear perimeter;

V puA1.5deff = V puA1.5d × 1.25 = 306.272 kN

Punching shear stresses at perimeter of 1.5 d from face of column (cl 3.7.7.2) Design shear stress;

v puA1.5d = V puA1.5deff / (u pA1.5d × d) = 0.285 N/mm 2

From BS 8110 :Part 1:1997 - Table 3.8 Design concrete shear stress ;

v c = 0.409 N/mm 2

Allowable design shear stress ;

v max = min(0.8N/mm 2 × √ (fcu / 1 N/mm 2), 5 N/mm 2) = 4.382 N/mm 2 PASS - v p uA1.5d  < v c  - No shear reinforcement required 

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13 No. 12 dia. bars btm (200 c /c)

12 No. 12 dia. bars btm (125 c/c) Shear at d from column face Punching shear perimeter at 1.5 × d from column face

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