ACI Pad Foundation Example

CLIENT CONSTRUCTION COMPANY LIMITED Date 6-Apr-15PROJECT APARTMENT BUILDING By G EngineerLocation 2-BEDROOM APARTMENT MODEL Project: 135-006Sub-Location Pad footing - H1

Reference Calculation Output

References:1 - ACI 318M-05, Building Code requirements for Structural Concrete, 20052 - ASCE 7-10, Minimum Design loads for Buildings & Other Structures, 20103 - Preliminary Geotechnical report, Trintoplan Consultants Limited, October 20134 - STAAD output5 - ASTM A615-04, Standard Specification for Deformed and Plain Carbon-Steel Bars for Concrete Reinforcement

Summary of calculation checks Utilisation ratio (actual vs capacity)Soil bearing capacity 1.00 OK in soil bearing capacity **this condition governs**

Two-way (punching) shear 0.55 OK in two-way (punching) shearOne way (wide beam) shear (x axis) 0 31 OK in one way (wide beam) shear (x axis)One-way (wide beam) shear (x-axis) 0.31 OK in one-way (wide beam) shear (x-axis)One-way (wide beam) shear (z-axis) 0.43 OK in one-way (wide beam) shear (z-axis)

Flexural moment capacity (x-axis) 0.49 OK in flexural rfct (x-axis)Flexural moment capacity (z-axis) 0.49 OK in flexural rfct (z-axis)

Shrinkage rfct (x-axis) 0.64 OK in minimum shrinkage reinforcementShrinkage rfct (z-axis) 0.64 OK in minimum shrinkage reinforcement

Starter bar reinforcement OK for starter bar min. reinforcementStarter bar embedment OK for embedment depth

Starter bar development length OK for development length

FOOTING GEOMTERY & MATERIAL PROPERTIES

Foundation geometrycolumn width (in X-direction), px = 400 mm 400

column breadth (in Z-direction), pz = 400 mm 400column height (in Y-direction), H 25 mm 25

Footing thickness, h 400 mm 400

Note for user and reader: Cells in yellow (or shaded) denote user-input cells. All other cells are calculated via this spreadsheet using the relevant base data, material and guidance from the noted References

Founding depth below GL 1000 mmLength of footing, L (in X-direction) 2150 mm 2150Length of footing, B (in Z-direction) 2150 mm 2150

Footing shape: Square

Calculation of footing & soil surcharge weightRef 4: Tbl C3-2 Concrete density (kN/m3) 23.1 kN/m3

Column weight 0.1 kNPad weight 42.6 kN

Soil weight above pad (assumes =20kN/m3) = 90.5 kNTotal footing & soil eight F 133 3 kN

400

400 400

25

Total footing & soil weight Ff = 133.3 kN

Material properties28-day concrete comp. strength, f'c 25 N/mm2 fcu=25MPa

Ref 1: 7.7.1 Cover to reinforcement 75 mmRef 5 Main reinforcement to be used A615 Gr 60

Reinforcement yield strength, fy = 420 N/mm2Effective depth, d 317 mm

Ref 3 Allowable soil bearing capacity, qa 115.0 kN/m2

2150 2150

LOADING DATA

Serviceability Limit State resultsRef 4 From STAAD output, using Serviceability Limit state analysis:

Fx (kN) Fy (kN) Fz (kN) Mx (kNm) My (kNm) Mz (kNm) NodeMaximum FY occurrence (SLS) 53.5 381.3 79.4 10.2 0.7 4.4 1007

Note; Use Serviceability Limit state values when checking for soil bearing pressures. There is already a Factor of Safety built in to the allowable bearing pressure parameter as outlined in the geotehcnical report.Use Ultimate Limit State results for all other design checks.

Ultimate Limit State resultsRef 4 From STAAD output, using Ultimate Limit state analysis:

Fx (kN) Fy (kN) Fz (kN) Mx (kNm) My (kNm) Mz (kNm) NodeMaximum FY occurrence (ULS) 1.1 700.6 35.6 4.9 0.1 7.6 1007

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CHECK ON SOIL BEARING PRESSURES

Bearing pressure along x-axisPressure distr. under base: qmax,min =

Ref 4 Use Fy (serviceability) = 381.3 kNLoad FV on underside of footing = Fy+Ft = 514.6 kN

FV/LB = 111.3 kN/m2MZ+FXH/BL2 = 2.3 kN/m2

q = 113 6 kN/m2

514.6

4.4

2150

53.5 2BLHFM

LBF XZV

qmax= 113.6 kN/mqmin= 109.0 kN/m2

Bearing pressure along z-axisPressure distr. under base: qmax,min =

Ref 4 Use Fy (serviceability) = 381.3 kNLoad FV on underside of footing = Fy+Ft = 514.6 kN

FV/LB = 111.3 kN/m2MX+FZH/LB2 = 3.4 kN/m2

514.6

0.7 79.4

2150

2LBFzHMx

LBFV

X Z / kN/mqmax= 114.7 kN/m2qmin= 107.9 kN/m2

Use max bearing pressure as pa = 114.7 kN/m2Ref 3 Allowable soil bearing capacity, qa 115.0 kN/m2

Ratio of pa to qa = 0.998 OK in soil bearing capacity

CHECK FOR COLUMN BEARING (transfer of load from column to footing)

OK in soil bearing capacity

700.6

Nominal column or bearing load capacity, PnRef 1: 10.17.1 Pn(0.85f'cA1)*[A2/A1]Ref 1: 9.3.2.4 0.65 [unitless]

A1 = 0.160 m2A2 = (px+4d)*(pz+4d) = 2.782 m2

Ref 1: 10.17.1 A2/A1 (but max. 2.0)= 2.00 [unitless]f'c = 25.0 N/mm2

Pn 4,420 kNRef 4 Fy = 700.6 kN

Ratio of Fy to Pn 0.16 OK in column bearing resistance OK in column bearing resistance

SHEAR CHECKSCheck for Two-way (punching) shear capacity Vn

Ref 1: 9.3.2.3 Where 0.75Ref 1: 11.1.1 VnVc Assuming that no shear reinforcement is used in the footing

Ref 1: 11.12.1.2 Shear perimeter is located at a distance of 1.0d outside of the column/column faceShear perimeter length, bo is given by

bo = 2(px+d)+(2(pz+d)px = 400 mmpz = 400 mm

bearing resistance

pz 400 mmd = 317 mm

bo = 2868 mm

Factored net soil pressure qu = Pu/BLFy = Pu = 700.6 kN

qu = 151.6 kN/m2

Find the total shear force for two-way shear, VU2where VU2 = Pu-qu(px+d)(py+d)

VU2 = 622.7 kN

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Calculate the nominal shear strength V of the footingCalculate the nominal shear strength, VC of the footing(a) 0.17(1+2/)f'cbod

Ref 1: 11.12.2.1 (b) 0.083([sd/bo]+2)f'cbod(c) 0.33f'cbod

Condition (a)Ratio of long side to short side of col, = 1.00 [unitless]

f'c = 25.0 N/mm2bo = 2,868 mmd = 317 mm

V ( ) = 2 318 3 kN

Vc (kN) = min of

2150

Vc-condition (a) = 2,318.3 kNCondition (b)

Column location for determining s = EdgeRef 1: 11.12.2.1 s = 30 [unitless]

f'c = 25.0 N/mm2bo = 2,868 mmd = 317 mm

Vc-condition (b) = 2,005.7 kNCondition (c)

Vc-condition (c) = 1,500.1 kN

21

50

(a) 2,318.3 kN(b) 2,005.7 kN(c) 1,500.1 kN

Use Vc = 1,500.1 kNVc = 1,125.1 kNVU2 = 622.7 kN

Ratio of VU2/Vc = 0.55 OK in two-way (punching) shear

Vc = min of

OK in two-way (punching) shear

Check for one-way (wide beam) shear:Ref 1: 11.12.1.1

One-way shear parallel to x-axisRef 1: 11.3.1.1 Vc = 0.17f'cBd

B = 2150 mmVc = 579.3 kN

Vc = 434.5 kN

Find the projection noted as L'F ti h S

The critical section for one-way (wide beam) shear extends across the width of the footing at a distance d from the face of the loaded area.

50

2150

Footing shape: SquareL' = 558 mm

Shear force on shear plane, Vu1 = BL'qu = 136.3 kNRatio of Vu1/Vc = 0.31

OK in one-way (wide beam) shear (x-axis)

One-way shear parallel to z-axisVc = 0.17f'cLdL = 2150 mm

Vc = 579.3 kN Vc = 434 5 kN

OK in one-way (wide beam) shear (x-axis)

21

5

Vc 434.5 kN

Find the projection noted as L'Footing shape: Square Note: sketch holds for one-way shear in the z-axis as well

L' = 758 mmShear force on shear plane, Vu1 = BL'qu = 187.0 kN

Ratio of Vu1/Vc = 0.43 [unitless] OK in one-way (wide beam) shear (z-axis) OK in one-way (wide beam) shear (z-axis)

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MOMENT AND REINFORCEMENT CHECKS

Bending parallel to x-axisFind the projection noted as Lm

Lm = 875 mmmoment on the bending plane, Mu = quBL2m/2

qu = 151.6 kN/m2 Mu = 124.7 kNm

Coefficient of resistance Ru =Mu / (bd2)

2150

Coefficient of resistance Ru Mu / (bd2)Ref 1: 9.3.2.1 = 0.9

Ru = 0.6415 Required steel ratio, =

0.85f'c / fy = 0.051 2RU/((0.85f'c) = 0.067

= 0.0017 Area of steel required As req = rBd

As req = 1,177 mm2

21

50

875

''

85.021185.0

c

u

y

c

fR

ff

mm

Check for minimum reinforcement required for flexureRef 1: 10.5.1 0.25Bdf'c / fy

1.4Bd/ fy0.25Bdf'c / fy = 2,028 mm2

1.4Bd/ fy = 2,272 mm2 Use As min (flexure) = 2,272 mm2

Check for minimum reinforcement required for shrinkageRef 1: 7.12.2.1 As, min (shrinkage) = 0.0018 (times gross sectional area)

As min (flexure) max of

0.0018Bh = 1,548 mm2Reinforcement selection

Tension Comp.Select bar diameter (mm) 16 10

Select number of bars 12 12Provide bar spacing (to nearest 25mm) 150 150 mm

As prov 2,413 942 mm2As prov (flexure) = 2,413 mm2 OK in minimum flexure reinforcement OK in minimum fle reOK in minimum shrinkage reinforcement OK in minimum shrinkageRatio of As req / As prov = 0.49 [unitless] OK in flexural rfct (x-axis) OK in flexural rfct (x-

a is)

Bending parallel to z-axisFind the projection noted as Lm

Lm = 875 mmmoment on the bending plane, Mu = quLL2m/2

qu = 151.6 kN/m2 Mu = 124.7 kNm

Coefficient of resistance Ru = Mu / (Ld2)Ref 1: 9 3 2 1 = 0 9Ref 1: 9.3.2.1 0.9

Ru = 0.6415 Required steel ratio, =

0.85f'c / fy = 0.051 2RU/((0.85f'c) = 0.067

= 0.0017 Area of steel required As req = rLd

As req = 1,177 mm2

21

50

2150

87

5

''

85.021185.0

c

u

y

c

fR

ff

Check for minimum reinforcement required for flexureRef 1: 10.5.1 0.25Ldf'c / fy

1.4Ld/ fy0.25Ldf'c / fy = 2,028 mm2

1.4Ld/ fy = 2,272 mm2 Use As min (flexure) = 2,272 mm2

As min max of

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Check for minimum reinforcement required for shrinkageRef 1: 7.12.2.1 As, min (shrinkage) = 0.0018 (times gross sectional area)

0.0018Lh = 1,548 mm2Reinforcement selection

Tension Comp.Select bar diameter (mm) 16 10

Select number of bars 12 12Provide bar spacing (to nearest 25mm) 150 150 mm

As prov 2,413 942 mm2As prov (tens) = 2,413 mm2 OK in minimum flexure reinforcement OK in minimum fle reOK in minimum shrinkage reinforcement OK in minimumOK in minimum shrinkage reinforcement OK in minimum shrinkageRatio of As req / As prov = 0.49 [unitless] OK in flexural rfct (z-axis) OK in flexural rfct (z-

a is)

Reinforcement summaryAlong x-axis, bottom mat 12-T16-150B

Along x-axis, top mat 12-T10-150TAlong z-axis, bottom mat 12-T16-150B

Along z-axis, top mat 12-T10-150T

Ref 1: 7.12.2.1 Note on distribution of steel parallel to the axis:Centre band zone, s = 2/(+1), s ( )

= 1.00s = 1.00 No special distribution needed

CHECK ON STARTER BARS

Check on minimum % reinforcement to starter barsdiameter of starter bar, db = 12 mm

Number of starter bars = 8 Nr (note: minimum 4)Cross sectional area of bars = 905 mm2

Ref 1: 15.8.2.1 Minimum As-starter = 0.005Agcolumn cross-section area Ag=(px*pz) = 160,000 mm2

As-starter / Ag = 0.006 OK for starter bar min. reinforcement

Check on starter bar embedment into footingRef 1: 12.3.2 Min of 200 mm

(0.24fy/f'c)db0.043fydb

fy = 420 N/mm2d 12

larger oflength of embedment ldc =

OK for starter bar min. reinforcement

db = 12 mmf'c = 25 N/mm2

(0.24fy/f'c)db = 242 mm0.043fydb = 217 mm

Use ldc as = 242 mm 245 mm (rounded up)Check on ldc versus depth of footing, h, and effective depth, d

d = 317 mmh = 400 mm OK for embedment depth

Check on development length of starter barsCritical sections for the development length (ld) of the starter bars occur at the column/column face

OK for embedment depth

Ref 1: 12.2.2 ld =

For tdepth of freshly cast concrete below ld



SUMMARY OUTPUT

400

25

400

12-T10-150T12-T10-150T

8-T12

245 400

2150 480

400 2150

400

12-T16-150B

fcu=25MPa

12-T16-150B

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Ref 4

Horizontal Vertical Horizontal MomentNode L/C Fx kN Fy kN Fz kN Mx kNm My kNm Mz kNm

SERVICEABILITY LIMIT STATE (1.0DL+1.0LL)

APPENDIXSTAAD ANALYSIS OUTPUT

The user is to carry out the analysis in STAAD and use the post-processing results to obtain the values shown in these tables. Note that two limit state Envelopes are used, Serviceability Limit State and Ultimate Limit State.

Node L/C Fx kN Fy kN Fz kN Mx kNm My kNm Mz kNmMax Fx 1007.00 1 1.0DL+1.0 -53.51 381.35 79.45 10.20 0.65 4.35Min Fx 1007.00 1 1.0DL+1.0 -53.51 381.35 79.45 10.20 0.65 4.35Max Fy 1007.00 1 1.0DL+1.0 -53.51 381.35 79.45 10.20 0.65 4.35Min Fy 1007.00 1 1.0DL+1.0 -53.51 381.35 79.45 10.20 0.65 4.35Max Fz 1007.00 1 1.0DL+1.0 -53.51 381.35 79.45 10.20 0.65 4.35Min Fz 1007.00 1 1.0DL+1.0 -53.51 381.35 79.45 10.20 0.65 4.35Max Mx 1007.00 1 1.0DL+1.0 -53.51 381.35 79.45 10.20 0.65 4.35Min Mx 1007.00 1 1.0DL+1.0 -53.51 381.35 79.45 10.20 0.65 4.35Max My 1007.00 1 1.0DL+1.0 -53.51 381.35 79.45 10.20 0.65 4.35Min My 1007.00 1 1.0DL+1.0 -53.51 381.35 79.45 10.20 0.65 4.35yMax Mz 1007.00 1 1.0DL+1.0 -53.51 381.35 79.45 10.20 0.65 4.35Min Mz 1007.00 1 1.0DL+1.0 -53.51 381.35 79.45 10.20 0.65 4.35

Maximum values -53.51 381.35 79.45 10.20 0.65 4.35Corresponding values at Fy max 1007 53.51 381.35 79.45 10.20 0.65 4.35

Horizontal Vertical Horizontal Moment Node L/C Fx kN Fy kN Fz kN Mx kNm My kNm Mz kNm

Max Fx 1007.00 0.9DL+1.0EQ 1.12 163.02 35.59 4.87 0.11 -7.57Min Fx 1007.00 DL+1.0EQ(+X -96.75 506.75 99.13 13.38 0.91 15.61

ULTIMATE LIMIT STATE (All Load Combs)

Max Fy 1007.00 1.0WL(-Z)+1 -81.34 700.60 296.49 80.00 0.10 6.52Min Fy 1007.00 0.9DL+1.0W -17.95 -18.53 -159.33 -60.82 0.87 2.39Max Fz 1007.00 1.0WL(-Z)+1 -81.34 700.60 296.49 80.00 0.10 6.52Min Fz 1007.00 0.9DL+1.0W -17.95 -18.53 -159.33 -60.82 0.87 2.39Max Mx 1007.00 1.0WL(-Z)+1 -81.34 700.60 296.49 80.00 0.10 6.52Min Mx 1007.00 0.9DL+1.0W -17.95 -18.53 -159.33 -60.82 0.87 2.39Max My 1007.00 1.0WL(+Z)+1 -44.65 179.28 -115.06 -55.73 1.30 4.40Min My 1007.00 0.9DL+1.0W -54.65 502.79 252.22 74.92 -0.33 4.52Max Mz 1007.00 DL+1.0EQ(+X -96.75 506.75 99.13 13.38 0.91 15.61Min Mz 1007.00 0.9DL+1.0EQ 1.12 163.02 35.59 4.87 0.11 -7.57

M i l 1 12 700 60 296 49 80 00 1 30 15 61Maximum values 1.12 700.60 296.49 80.00 1.30 15.61Corresponding values at Fy max 1007 1.12 700.60 35.59 4.87 0.11 7.57

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ACI Pad Foundation Example